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A compact article on a vital part of Motion in Physics for all boards.
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Immensely helpful write-up in Maths for students reading in classes 6, 7, 8 of all boards.
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Immensely helpful list for all secondary level students and aspirants for competitive exams
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Compact Mensuration formulae guide for all Secondary level students
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A complete guide for school level as well as competitive level students
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Briefing of the conceptual studies on Whole Numbers catering to all students of ICSC and CBSE for the pre-secondary level.
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Essential geometrical chapter covering all required facts & formulae for both CBSE & ICSC students of Class 8.
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The formulae will immensely help all students of Class 8, 9, 10 of both CBSE and ICSC to solve the problems on Algebraic Identities
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Proofs and Justifications by Method of Contradiction - catering ICSC Class 9 students and CBSE class 10 students.
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These notes have been prepared in a consolidated way. Smart and easily understandable tips have been provided in the same. Students of class VIII, as well as, aspirants for competitive exams will surely be benefited.
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Catering for both Physics and Chemistry
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Thermal decomposition in Chemistry
DownloadAns: Required velocity v' = v.h / H
Ans: Any number divided by 1 gives that number only. Any number divided by .1 gives 10 times the number. Any number divided by .01 gives 100 times the number. And so on. Thus we see, as the denominator decreases, quotient increases. If the denominator now on gradual decreasing becomes zero, the quotient will be too large to be measured. It has crossed all limits.Then we can call it undefined or infinity.
Ans: Ultra-Violet rays come alongwith the beam of sunrays that travel all the way from the sun to the earth. Hence, Sun is the source of U-V Rays and the depletion of Ozone layers in the upper strata of atmosphere help them pass through.
Well, the real answer is still obscure to many geologists also. One school of thoughts say ice from the asteroids and metors might have transformed to water owing to the huge collision with the earth. Others say that hot paarticles coming out due to the vulcanic eruption in Sun might have cooled with time coming in contact to the earth surface to form water.
Except Carbon (C), all other non-metals are below Hydrogen {H} in the Activity Series. So they cannot displace Hydrogen from the dilute acids in any chemical reaction.
On combustion of dry wood, carbon dioxide (CO2) is formed. But there is always rovision of incomplete combustion which results in the formation of carbon monoxide (CO) which is more pollutant than CO2. So dry wood is not preferred as domestic fuel these days.
Air contains Oxygen. It may react with (oxidise) the acids present in the fruits and fermentise the same. Nitrogen is unreactive. It is used as the preservant in cryogenic process. It helps to keep the fruits fresh. So Nitrogen is always preferred over air in plastic bags for dried fruits.
sin 30 = 1/2. So, 3400/x = sin 30 =1/2 (where x = dist. travelled by the aircraft in 10s) Or, x = 6800 m Thus, speed of the aircraft = 6800/10 m/s = 680 m/s .
The capital of Cameroon is Yaoundé .
William Shakespeare wrote Macbeth .
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Charge of a body indicates its nature in the perspective of current conduction. Hence it's a chemical propery. On the other hand, mass indicates the matter content of the body. So, it's a physical property. A particle with huge mass may carry zero charge, but the reverse is never possible.
To convert galvanometer into voltmeter, a very high resistance is connected in series with it in the circuit. To convert galvanometer into ammeter, a very low resistance called 'shunt' is connected in parallel with it in the circuit. A potentiometer in the ac circuits majorly where speed is required to vary. As we know, variation of resistance of the pot causes the variation of voltage in the circuit, hence the frequency and hence the speed of any electro-mechanical appliance.
Lightning conductors are arranged at the top of the building to pevent the building from being electrocuted by conducting the huge current, owing to lightning, through the earthing strip or plate straight down into the earth.
Ordinary rubber will easily melt due to the tremendous amount of heat generated during the rolling of tyres at the time of take-oof and landing. So a special type of rubber of low temperature coefficient is used for the purpose.
A magnetic coil galvanometer works as an actuator, by producing a rotary deflection of its pointer, in response to the electric current flowing through a coil in a constant magnetic field. The ponter deflections perhaps indicate the direction of current flow. If there is no current flow or no magnetic field, there will be no pointer deflection.
The set-up can be at 0 deg Celsius. But, the water requires to be heated upto 100 deg Celsius, to make its boiling start.
4xsin(30° - 10°)/sin(2x10°) = 4 Ans.
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Instrumentation is an essential part for integration of expertise of Fluid Mechanics & Piping, Electrical, Electronics and Control Engineering complying with the site conditions and requirements.
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Cylinder :: Surface area of top & bottom- 2πr^2 Lateral/ Curved surface area- 2πrh Total surface area- 2πr(r+h)
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General Organic acid coined as Carboxylic acid and is formulated as R-COOH
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4. none of these Reason- infinite numbers
1. It gives us information about the number of elements in set A as well as the number of elements in set B. Reason- Only the cardinal numbers of the two sets are taken into consideration
4. {1, 8} Reason- The elements of set A mismatching with that of set B
Options 1, 2 and 3 Reason- None of the facts is ever possible
2. {7, 31, 9, 15} Reason- 7 = 2x3 + 1, 31 = 2x15 + 1 ... so on
3. Co(NO3)2 4. CrCl3 will be coloured Reason- The cations Co+2 and Cr+3 are pink and green coloured respectively
Hissing sound is produced alongwith bubbles with the evolution of of Carbon-dioxide gas and this phenomenon is called 'effervescence'
Evaporation takes place at all temperatures, but only from the surface of the liquid.
The actual average molar mass is given as 12.017 g/mole, the standard being taken as 12.
4. Electrostatic and gravitational forces Explanation- Electrostatic force acts against the flow of charge, while Gravitational force always pulls downward.
3. Force As per Newton's 2nd Law of Motion: F = m.a , ie. F = m.(v-u)/t = (mv - mu)/t = Δp/t . Hence, force - rate of change of momentum .
Here, v being 0, u² = 2as So, 50x50 = 2a.50. Or, a = 25 m/s² Now, m = 20 gm = 0.02 kg Hence, F = m.a = 0.02x25 N = 0.5 N (option 3)
2. 1.20 x 10^7 Dyne
{14x + 15(100-x)}/100 = 14.0067 Thus, 1500 - x = 1400.67 Or, x = 1500 - 1400.67 = 99.33 and, 100-x = 0.67 Hence, the respective percentage abundances of 14x and 15x isotopes are 99.33% and 0.67%.
Note first: Na2CO3 does not decompose on heating, but NaHCO3 does. 168 g NaHCO3 suffers a loss of 62 g. (steam + C02) So, 0.12 g is lost from 3.25 g (app.) of NaHCO3. Thus, in 2 g mixture, there is (20 - 3.25)g = 16.75 g of Na2CO3 ie. 83.75% Na2CO3.
2. 6.022 x 10^23 atoms of carbon = 12 g of carbon - is correct. Reason- By Avogadro's Law
Ans- 1.344x10^18 molecules Exp- 6,023x10^23x0.05/22400 = the ans.
25x = Rs. 10,000 So, 20x = 4/5 of Rs. 10,000 = Rs. 8,000 is the cost on shopping
T= D/S = (200+250)/(10+20) = 450/30 s = 15 s
Ans- 4.84x10^16 molecules Desc- 6.023x10^23x18/22400 = the ans.
From given: Total Distance D = 200+250= 450m, S1 = 36km/h or 10m/s, S2 = 72km/h or 20 m/s Thus, Total Relative Speed S = S1+S2 = 30m/s Now, Total Time T = D/S = Total Distance /Total Relative Speed = 450/30 s = 15 seconds.
Ans: 32% Expl: 40% of 20 l = 8 l alchol Now, after adding 5 l water, new mixture = 25 l. So, % of alcohol in new mixture = 8/100x25 = 32% .
Ans: KE will get doubled. Expl: New KE = 1/2.(1/2m).(2v)^2 = mv^2 = Double of Original KE
Here, Work done = P.E. = M.g/2.l
CP of customer = MP - D + ST = MP - D%/100.MP + ST%/100.(MP - D) So here, CP = Rs.(12500 - 1250 + 900) = Rs. 12,150
Ans-1. Rs. 5000
Avg. speed = Total Distance / Total Time = 2x/ (x/25 + x/35) = 29.167 km/h (approx)
v = 54 km/h = 15 m/s, m = 1000 kg, t = 5 s So, KE = 1/2. 1000.15^2 J = 112500 J Thus, Power = KE/T = 112500/5 W = 22500 W = 22.5 kW
Ans- Rs. 3382.49 Expl: P = 42152x100x100x100x100x100/(104.5x104.5x104.5x104.5x104.5) = the ans.
Ans- Rs. 2125 @ 5%, Rs. 375 @ 7%
Ans: Rs. 28,800 Expl: A = Rs. 20,000x(1+20/100)^2
Ans- 2,07,648 Expl: (105x96x103/100x100x100).2,00,000 = the ans.
24 u
18th from bottom = 13th from top So, 1 student between Gurpreet and Jasleen
0.69 g of Aluminium, 0.31 g of Magnesium
Let, MP = x By the prob, 15x/100 = Rs. 160. So, MP = x = Rs. 160*100/15 = Rs. 1066.67 (app.) and SP = Rs (1066.67 - 160) = Rs. 906.67
SP of Rahul = (100/120)*SP of Manish = (120/100)*SP of Kalra . So, SP of Manish = Rs. 120*(120/100) = Rs. 144 and SP of Kalra = Rs. 120*(100/120) = Rs. 100 . Thus, Gain of customer = Rs 44 and Reqd. Gain % = 44/144*100% = 30.56% (if he buys from Kalra instead of Manish)
Take for example, Nitrogen and Hydrogen react to form Ammonia, It is given by the equation: N2 + 3H2 = 2NH3. Here, 1 vol. of Nitrogen combines with 3 vols. of Hydrogen to give 2 vols. of Ammonia. Thus, the ratio of the reactants and product gas can be expressed as 1:3:2, which is a simple ratio of whole numbers, as explained in Gay-Lussac's Law of Gases.
Molar mass of NaHCO3 is 84 and that of CO2 is 44. So, from 6.3 g NaHCO3, 44*6.3/84 = 3.3 g of CO2 is released in the reaction.
Incl. 14% ST, the price = 114/100*Rs.40,000 = Rs. 45,600
Let, the price be reduced to the extent Rs. x. So, the reduced price is Rs(1800 - x). Thus, 108/100*(1800 - x) = 1800. Or, x = Rs. 14400/108 = Rs. 133.33. Hence, the reduction percent is 7.4% (approx)
Speed = Distance/Time = 624/6.5 km/h = 96 km/h
Speed = 90 kmph. So, Distance covered in 1 hr or 60 mins = 90 km. Hence, Distance covered in 10 mins = (90/60)x10 km = 15 km
80% energy lost. So, 20% energy left. So, it will rise to a height of 20%of 100 m = 20 m.
Given- KE1:KE2 = 4:1, p1:p2=1:1. Or, p1^2:p2^2=1:1. So, m1:m2 = 1:4 [Since, KE varies inversely as m, by eqn KE = p^2/(2m)]
1st discounted price = 90/100*Rs. 2500 = Rs. 2250 After 5% discount, the price = 95/100*Rs. 2250 = Rs. 2137.50. With 10% ST, the final SP = Rs. (2137.50 + 213.75) = Rs. 2351.25.
Ultimate CP = (108/100)*(98/100)*(97/100)*(94/100)*Rs.3,500 = Rs. 3,377.67.
In 5 days, 25 men can complete 5/14 part of work, 20 men can complete 4/14 part of work, 15 men can complete 3/14 part of work, 10 men can complete 2/14 of work. Thus, (5+4+3+2)/14 = whole work can be completed in 5x4 days = 20 days, if after every 5 days, 5 men are called off.
The average of 2 sets is (44+48)/2 = 46. So, the sum of scores of all four = 46x4 = 184.
By compound ratio rule, the ratio of the simple interests = (4*8):(9*10) = 32:90 = 16:45 .
PE = m.g.H. Or, H = PE/(mg) = 9800/(500x9.8) = 2m.
PE = mgH = 2x10x20 J = 400 J. Reason- PE of a body is determined by virtue of its position.
Here, A = 4P. Or, SI = A - P = 3P. Thus, 3P = P*10*T/100. So, T = 30 years.
Let, normal price be Rs. x. So, with 10% ST, the price = Rs. 110/100*x = Rs. 11/10*x. BTP, 11x/10 = 1507. Thus, x = 1370. So, for 6% ST, the customer pays less by 4% of Rs. 1370 = Rs. 54.80.
Since, ST is 10%, 110/100*Rs. x = Rs. 850 (where x = SP of the article). So, x = Rs. 772.72.
Total cost = Rs. (2500 + 20/100*2500) = Rs. 3000. Actual SP excluding ST = 100/105*Rs. 3475 = Rs. 3309.52. Hence, Profit of shopkeeper = Rs. (3309.52 - 3000) = Rs. 309.52 = 10.32% (approx)
Let the SP be x. ATP, 112/100*x = Rs. 4096. So, when ST reduced to 8%, 108/100*x = Rs. 3949.71. Ans.
12% of Rs. 90 = Rs. 10.80 So, after 3 years, the difference between CI and SI = Rs. {90 + (90 + 10.80)} = Rs. 190.80 Reason- Rs. 90 will the fixed part of increment every year and the other part is the interest on the increment for the previous year.
ATP, X + Y = Rs. 15,500. So, 15X + 15Y = Rs. 23,2500. --- (i) Again, 15X/100 + 12Y/100 = Rs. 4,140/2 = Rs. 2070. So, 15X + 12Y = Rs. 20,7000. --- (ii) Subtracting (ii) from (i) we get, 3Y = Rs. 25,500. Y = Rs. 25,500/3 = Rs. 8,500. Hence, the amount invested by the person in scheme Y is Rs. 8,500. Ans.
Speed of bike = 288/8 = 36 km/h. So, Speed of car = 2*(36) = 72 km/h. and Speed of bus = (3/4)*72 = 54 km/h. Thus, distance covered by bus in 5 hours = 54*5 = 270km. Hence, distance left uncovered = (288 - 270)km = 18km.
Let, the marks in 1st mid-term exam be x. So, the marks in the next 4 exams are x+20, x+40. x+60 and x+80 respectively. Hence, the difference between maximum and minimum marks is (x+80)-x = 80. Ans.
R for half-yearly = 5% , No. of terms in 1 year = 2 So, A = Rs. 35,000 * (1+ 5/100)^2 = Rs. 38,587.50.
Here, Time period T = 6 months = 1/2 year. Difference between two CI = Rs. (236.25 - 225) = Rs. 11.25. Let, the rate of interest p.a. = R So, 11.25 = (225*R*1/2)/100. Or, R = 10. Hence, the required annual rate of interest is 10%.
Seasonal Discount =(10/100)*12500 = Rs. 1250. So, discounted price = Rs. (12500-1250) = Rs. 11250. Thus, the amount to be paid by the customer = Rs. 12150. So, Sales Tax = Rs.(12150-11250) = Rs. 900. Hence, ST% = (900/11250)*100 = 8%.
Savings after 1st year = Rs.10000*(1+5/100) = Rs.10500. Now he invests further Rs.10000. So, principal at beginning of second year= Rs.(10500+10000) = Rs.20500. Hence, amount at the end of second year= Rs.20500*(1+5/100) = Rs.21525.
Let, total number of employees be 100x. So, 20% of 60% of 100x = 12x, and 15% of 40% of 100x = 6x. ATP- 12x + 6x = 81. Or, 18x = 81. So, 100x = (81/18)*100 = 450. Hence, no. of employees not getting bonus = 450-81 = 369.
Volume of H2 at 27degC or 300K = 22.4 L. So, vol. of H2 at 273K = 22.4*273/300 L = 20.384 L. Now, 24 g of Mg reacts with 2 mols of HCl to give 22.4 L of H2 at 273K and 1atm. Thus, 20.384 L of H2 is given by 24*20.384/22.4 g = 21.84 g of Mg. 10 g of Mg is present in 12.5 g of the sample wire. Hence, 21.84 g of Mg is present in 12.5*21.84/10 g = 27.3 g of the sample wire. Ans. The reqd. mass of the sample of the wire is 27.3 g.
4. solenoid
29 days. Reason- Since, it becomes full right on 30th day.
42-24=18 runs was actually counted in excess by mistake. Now, 18/25 = 0.72 is in excess in the average score for 25 innings. Thus, the correct average score = 52.36 - 0.72 = 51.64.
Net Distance in North direction = (10 - 6) km = 4 km. Distance in East direction = 3 km. Hence, actual distance covered in reference to his starting point = sq.rt(4^2 + 3^2) = 5 km.
M:L = 5:4 = 15:12. L:K = 3:2 = 12:8. So, M:L:K = 15:12:8. Hence, K will get (8/(15+12+8))*Rs.1400 = Rs.320.
Greater the force, more the displacement. Now, for 200 N force, displacement = 2 m. Hence, for 286 N force, displacement = (2/200)*286 N = 2.86 m. [Not necessary, he will do the same amt. of work repeatedly]
Let, each gave Rs. x. ATP, (15 - x ) ^2 = (12 - x ) * (19 - x ). Or, 225 - 30 x + x^2 = 228 - 31 x + x^2. Or, 31x - 30 x = 228 - 225; x = 3. Hence, each friend gave Rs. 3.
Let, the number added to both numerator and denominator be x. ATP, (16 + x)/ (27 + x) = 4/5 . Or, 80 + 5x = 108 + 4x . Or, x = 28. Hence, the reqd, no, is 28.
The weight of the red ball pen is (5/6)*180 g = 150 g.
Let, x part of 1st mixture and y part of 2nd mixture are combined. ATP, (4x + 5y)/ (5x + y) = 5:4 . Or, 16x + 20y = 25x + 5y. Or, 9x = 15y. Thus, x:y = 15:9. Hence, two mixtures should be combined in the ratio 15:9.
Let, the incomes of two employess be 3x and 5x, and their expenditures be 5y and 9y respectively. ATP. 3x - 5y = 5x - 9y. Or, 2x = 4y. Or, x = 2y. Again, 3x - 5y = Rs. 300 (given). So, 3*2y - 5y = Rs. 300. Or, y = Rs. 300. Or, x = 2* Rs. 300 = Rs. 600. Or, 3x = Rs. 1800. Hence, the monthly income of the first employee is Rs. 1800.
In 21 kg alloy, mass of copper = 5/(5+2)*21 kg = 15 kg and that of zinc = 2/(5+2)*21 = 6 kg. Now, after adding 4 kg of copper, the new ratio of zinc and copper = 6:(15+4) = 6:19. Ans.
Refractive index of the material of prism = sin i/ sin r= sin 45⁰/sin 30⁰ = (1/√2)/(1/2) = √2 = 1.414 .
100 g water contains 0.025 g of Chlorine. So, 1000000 g water contains ( 0.025* 1000000) / 100 = 250 gm of Chlorine. Hence, concentration of Chlorine in ppm = 250 . Ans.
1st case: v = a.t, Or, t = v/a. v^2 = 2.a.s. Or, s = (v^2)/(2a). [all symbols are conventional] 2nd case: nv is new velocity, a' is new decceleration. ATP, s = nv.t - (a'*t^2)/2 . Or, (v^2)/(2a) = nv*v/a - {a'*(v/a)^2}/2 . Or, {a'*(v/a)^2}/2 = (v^2/a)*(n - 1/2) . Or, a' = 2a.n*(n - 1/2) = a.n.(2n -1) . Ans.
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ATP: x*(x+2) = 4*(x+x+2) - 1. Or, x^2 + 2x = 8x + 8 - 1. Or, x^2 - 6x - 7 = 0 . Or, (x+1)(x-7) = 0. So, x = 7 should be taken. Thus, x=2 = 9. Hence, the larger number is 9.
Let, the numbers of 1 Re. coin be 7x and that of 2 Rs. coin be 4x. So, their values are Rs. 7x*1 = Rs. 7x and Rs. 4x*2 = Rs. 8x. ATP. 7x + 8x = 60. Or, 15x = 60. Or, x = 4. Hence, total number of coins = (7x+4x) = 11x = 11*4 = 44 . Ans.
Applying, v - u + a.t, we get velocity of the car after 5 secs = 0 + 5*5 =25 m/s. Hence, momentum = mass*velocity = 1000*25 kg.m/s = 25000 kg.m/s . Ans.
Since, body comes to rest, final velocity = 0. So, initial velocity = acceleration*time = 2*10 m/s = 20 m/s. Ans.
Let, height of the tower be x metre. So, tan 45 = (x-40)/(diff. between the feet of building & tower). Thus, diff. between their feet = (x-40) metre [since, tan 45 =1]. Again, tan 60 = x/(x-40) . Or, rt3*(x-40) = x . Or, x(rt3 - 1) = 40*rt3 . Or, x*0.732 = 40*1.732 . Thus, x = 94.64 (approx) . Hence, height of the tower is 94.64 m Ans.
When, angle of elevation is 45, the length of shadow = height of tower = 100 m [since, cot45 = 1]. When, angle of elevation is 30, the length of shadow = root3*height of tower = 100*root3 m [since, cot45 = 1] = 100*1.732 m = 173.2 m. Hence, difference between the lengths of the shadows = (173.2 - 100)m = 73.2 m. Ans.
Height of the ladder is the hypotenuse and the distance of the foot of ladder from wall is the base. We know, hyp./base = sec 60 = 2. Or, Height of the ladder = 2*2 m = 4 m [since, base = 2 m (given)]. Hence, the ladder reaches to a height of 4 m on the wall. Ans.
Ans. 72 g (approx). Clue: Weight of acid radical is fixed. Here, weight of Magnesium is (2.34-1.8) g = 0.54 g is more than that of Hydrogen, both combining with the same weight of acid radical. In general, 24 g of Mg can displace 2 g of Hydrogen. So, 0.6 g (app.) of Mg reacts with 1.8 g of acid. Hence, 24 g of Mg can combine with 72 g of the acid.
1st 160 m covered in 160/64 sec = 2.5 sec. 2nd 160 m covered in 160/80 sec = 2 sec. So, total distance (160+160) m = 320 m covered in total time (2.5+2) sec = 4.5 sec. Hence, average speed = 320/4.5 = 71.11 m/sec . Ans.
Simply applying, v = u - at --> since, v = 0, u = a.t = 2*10 = 20 m/s. Hence, initial velocity is 20 m/s . Ans.
Since, v = 1/2*u, 1/v = 2/u. Now, 1/v - 1/u = 1/f means -- 2/u -1/u = 1/f. Or. 1/u = 1/f. Or, u = f. u = 30 cm. Hence, focal length f = 30 cm . Ans.
First, weight of chilli in the 14 g mixture is 2/(2+5) = (2/7)*14 g = 4 g. After adding 2 g, the new weight of chilli = 4 g. The weight of salt is (14-4) g = 10 g. Hence, new ratio of chilli and salt = 6:10 = 3:5 . Ans.
Sum of ages of 4 friends 4 yrs ago = 21*4 = 84 yrs. So, sum of their present ages = (84+16)yrs = 100 yrs. With new friend joining, the sum of their ages = 24*5 = 120 yrs. Hence, age of new friend = 120 - 100 = 20 yrs. Ans.
Ans: Rs. 1000. Let the sum be Rs, x. ATP, (x*5*4)/100 = 200. Or, x = 200*100/20 = 1000.
x-y = 19, x+y = 33. Adding, 2x = 52. Or, Speed of boat in still water = 26 km/h . Ans.
30% of 240 marks = 72 marks. 40% of 400 marks = 160 marks. So, in language papers out of 160, marks obtained = 160 - 72 = 88 . Ans.
Let, the length be 7x m and breadth be 4x m. ATP, (7x+8)*(4x+8) = 416 m^2. By solving, x = 2.3 (approx). So, breadth 4x = 2.3*4 m = 9.2 m . Ans.
Let, the population 6 yrs ago be x. Here, n = 2, since each term = 3 yrs. So, x*(1 - 10/100)^2 = 240000. Or, x = 240000*100/81 = 296296. Again, 6 yrs later, the population = 240000*(1 - 10/100)^2 = 194400. Ans.
A and B together do the work in 54 min. So, A alone does it in (54+12 )min = 66 min. Part of work B completes in 1 min is [1/(A+B)) - (1/A)] = (1/54) - (1/66) = 1/297 . Therefore, percentage of work done by B in 1 min is (1/297)*100 = 0.3367 % . Ans.
Let, the reqd. amount of sugar be x kg. ATP, (x+75)*5.5 = x*5.75 + 75*4.5 . Or, 0.25x = 75*(5.5-4.5) . Or, x = 75/0.25 = 300 . Hence, 300 kg of sugar must be added. Ans.
30% of 240 marks = 72 marks. 40% of 400 marks = 160 marks. Thus, in language papers out of 160, marks obtained = 160 - 72 = 88 . Ans
Let, speed of boat in still water be x km/h and speed of water be y km/h . x-y = 19, x+y = 33. Adding, 2x = 52. Or, Speed of boat in still water = 26 km/h . Ans.
Average speed = Total Distance/ Total Time = [(30*6) + (45*4)]/(6+4) miles/hr = (180 + 180)/10 miles/hr = 360/10 miles/hr = 36 miles/hr . Ans.
Let, CP of 1 kg rice be Rs. x. ATP, the SP of 0.9 kg = Rs. 114x/100. Or, SP of 1 kg rice = Rs. 126.67*x/100 . Hence, net profit % = 26.67% . Ans.
Ans. 4:1 . Expl: Distance half and Time double. Speed = Dist/Time = (1/2)/2 = 1/4 of original.
ATP: 2*(b + 8 + b) = 64 . So, 4b + 16 = 64. Or, b = 12 and b + 8 = 20. Hence, the length is 20 cm and breadth is 12 cm. Ans.
(8x+10)/(3x+10) = 2/1. Or, 8x+10 = 6x+20. Or, 2x = 10. So, x = 5. Hence, age of father = 8x = 40 years. Ans.
At point Z, Amrita is facing South. Point Z is exactly 8 kms left to the point A . Ans.
Maturity Amount A = Rs.4000*(1 + 5/100)^2 = Rs. 4410 . Ans.
ATP, (33 - 20)% of total marks = 13% of total marks = (24 + 15) = 39. So, total marks = (39/13)*100 = 300. Hence, Robin scored (33/100)*300 = 99 marks . Ans.
Let, the actual CP be Rs. x . ATP, (112x/100)*(88x/100) = 6776 . Or, x = 6875 . Hence, it costed Rs. 6875 to K . Ans.
Total marks of 15 students = 15*x. So, total marks of 16 students = 15*x + 112. ATP, 15x + 112 = 16(x + 2) . Or, x = 80 (by solving). Hence, average score of 16 students = x + 2 = 82 . Ans.
In 1 hour, (1/6 - 1/8) = 1/24 pat of the wall is destroyed. So, full wall is destroyed in 24 hours. Hence, the half-built wall will be destroyed in 12 hours . Ans.
In 5 days, Shruti completes 5/60 = 1/12 of the work, Bharti completes 5/30 = 1/6 of the work, Atul completes 5/20 = 1/4 of the work. So, (1/12 + 1/6 + 1/4) = 6/12 = 1/2 of work is completed. Now, Bharti and Atul both can complete (1/30 + 1/20) = 1/12 of the work in 1 day. Hence, they can complete the rest or 1/2 of the work in the next (1/2)/(1/12) = 6 days . Ans.
Total CP of 50 litres mixture of chemicals = Rs.[(10*10) + (40*13)] = Rs.620. So, total SP to make 10% profit = Rs. (11/10)*620 = Rs.682 . Ans.
Probability of drawing a red ball = 3/(4+5+3) = 3/12 = 1/4 . Probability of drawing a yellow ball = 4/(4+5+3) = 4/12 = 1/3 . Ans.
They will meet after every [LCM of 8, 15, 12]minutes = 120 minutes or 2 hours . Ans.
7x - 2x = 5x = 30 marks. So, marks scored by Batas ie. 3x = 30*3/5 = 18 . Ans.
Z:X = 6:2 . Or, Z = 3X. Again, (Z+5)/(X+5) = 7/3 . Or, 3*(3X+5) = 7*(X+5) . By solving, X = 10. Now, X:Y = 2:5 . Hence, present age of Y = (5/2)*10 = 25 years . Ans.
In x litres, Boric acid = 4x/5. ATP, 4x/5 = 3/5*(x+20) . Or, x/5 = 12 . So, x = 5*12 = 60 . Ans.
Since, Density = Mass/(Length^3), error % in given case = (3/100)/ (2/100)^3 = 37.5% . Ans.
Let,, speed of the boat = x km/h and speed of the current = y km/h . ATP, x + y = 56 ...(i) and x - y = 28 ...(ii) . Subtracting (ii) from (i), 2y = 28. Or, y = 14 . Hence, speed of the current is 14 km/h . Ans.
Let,, speed of the boat = x km/h and speed of the current = y km/h . ATP, x + y = 8 ...(i) and x - y = 4 ...(ii) . Subtracting (ii) from (i), 2y = 4. Or, y = 2 . Hence, speed of the current is 2 km/h . Ans.
Amount to be return after 4 years = P + SI = Rs. (102200 + (102200*12*4)/100) = Rs. (102200 + 49056) = Rs. 151256 . Ans.
The price of the shirt = 55/100*Rs.2000 = Rs. 1100 . Ans .
Let,, speed of the boat = x km/h and speed of the current = y km/h . ATP, x + y = 8 ...(i) and x - y = 4 ...(ii) . Subtracting (ii) from (i), 2y = 4. Or, y = 2 . Hence, speed of the current is 2 km/h . Ans.
V = Rs.80000*[1-7/100]^5 = Rs. 55655.06 (approx). Expl: Depriciation Rule
SP = (93/100)*(86/100)*Rs.350 = Rs.279.93 . Ans.
Here, final velocity v = 0 . So, initial velocity = acceleration*time . Or, u = a.t = 2*10 m/s = 20 m/s . Ans.
Speed of train = 72 km/h = 20 m/s . Speed of bird = 18 km/h = 5 m/s. So, Total collective speed = 25 m/s (since they are in opp. dir.) . Hence, time taken to cross 175 m = 175/25 s = 7 s . Ans.
By Pythagoras Theorem, the last part = rt.(4^2 + 3^2) km = 5 km. So, total distance covered = (4+3+5) km = 12 km. Ans.
Just 4 km east to his starting point
3/7 of total sweets = 36 . So, total no. of sweets = 36*7/3 = 84. Ans.
The larger number = 23*14 = 322 . Ans.
Sum of ages of 5 members in a family at present = 27*5 + 25 years = 160 years. Sum of ages of 6 members in a family at present = 31*6 years = 186 years. Hence, the age of the new member = (186 - 160) years = 26 years. Ans.
Ans. Nt/N0=(1/2)^t/T1/2=(1/2)^1/2=1/√2
(where Nt is the number of undisintegrated atoms after time t and N0 is the initial number of atoms)
Normally, Frequency of sound = Velocity of sound/Wavelength of sound . So, number of beats heard per second by Piyal n = v/(u/V) = V.v/u Ans.
Water level will rise to a volume of 'm/k cc' [taking mass of stone = m gms] as the stone sinks in it, by Archimedes' Principle.
Since, x is displacement, its dimension is L. Hence :- The dimension of a is L^1.T^(-1). The dimension of b is L^1.T^(-2). The dimension of c is L^1.T^(-3).
We know, T ∝ √ l . Now, when l increased by 44%, it becomes 144/100*l . So, T' ∝ √ (144/100*l) or 12/10*√ l . Thus, T' = 12/10*T. So, the percentage increase in time period = [(12/10*T - T)/T] *100% = 20% . Ans.
Rate of interest = [(23958 - 21780)/21780]*100% = 10% . Ans.
ATP, (PR6/100)/(PR4/100) = 1.5 . But, 6/4 = 1.5 already. Hence, the conditions will satisfy for all values of P and R. There is no unique value of the rate of interest p.a.
ATP, age of Mrs. Asha = [25 + (7*5)] years = (25 + 35) years = 60 years. Ans.
Ans. 1/36 .
Exp- Total combinations = 6^3 = 216. Equal combinations = 6 (1,1,1 ; 2,2,2 ; ... 6,6,6)
Hence, required probability = 6/216 = 1/36.
Definition: The number of magnetic field lines passing normally through a surface is called Magnetic Flux.
SI unit: Weber (Wb)
Dimensional formula: L^{2}M^(1)T^{-2}I^{-1}
KE = 1/2.*m.v^2 = (mv)^2/(2m) = p^2/(2m) .
So, KE varies as square of the momentum.
On 20% increase of p, KE varies as (1.2p)^2 or 1.44p^2.
Hence, increase in KE is 44%. [since, mass is const.]
Since, it's rolling friction and two KE are there, it's a solid sphere.
Work done = Force × Displacement = N×10 =10N J
Displacement = (3 - 1) m = 2m. Hence, Work done = (5*2 - 4) N = 6 J . Ans.
Distance = (Power / Force) * time [where, Power and Force are constants]
Max. height attained by the particle = H = u^2/(2g) = 5^2/(2*10) m = 1.25 m.
So, work done by force of gravity = m.g.H = 0.1*10*1.25 J = 1.25 J . Ans.
Power delivered = Work done per unit time = (1/2.m*v^2)/T . Ans.
Let, the length of the train be x m.
ATP, (x+100)/x = 30/18 . => 12x = 1800 . => x = 150 .
Hence, the length of the train is 150 m . Ans.
Combined speed of the two trains = (50+40) km/hr = 90 km/hr = 25 m/s .
Combined length of two trains = (400+x) m. [taking length of train B = x m]
Time taken = 30s .
So, (400+x) = 25*30 = 750 .
Thus, x = 750 - 400 = 350 .
Ans. The length of train B is 350 m .
Rate of water flow = 6000 kg/min = 100 kg/s .
So, Power given to the turbine = 100*100*10 W = 0.1 MW .
Principal investment = Rs. 1875 Ans.
Let, the distance be x km.
ATP, x/40 - x/50 = 15 min = 1/4 hr . => x/200 = 1/4 (by simplifying) . => x = 50 .
Ans. The required distance is 50 km .
After arranging letters of word TIGER in ascending order alphabetically, it will become EGIRT. So, no letter will remain in the same position.
Actually, Rohit is 2 place left of Nilesh. One child is between them.
So, total number of children = 17 + 20 + 1 = 38 . Ans,
Actually, Rohit is 2 place left of Nilesh. One child is between them.
So, total number of children in the row = 17 + 20 + 1 = 38 . Ans,
Let, the common ratio for income be x and for savings be y . Again, income - savings = expenditure.
For Asha 5x-4y= 400 ... (i) For Robin 4x-3y=400 ... (ii)
=> x=400 . (on solving two eqns) .
Hence, monthly income of Asha and Robin are 5x and 4x i.e. Rs. 2000 and Rs. 1600 respectively
Let, the two numbers be a and b.
ATP, a - (3/10)b= (3/4)a . => (1/4)a = (3/10)b . => a/b = 6/5 .
Ans. The ratio of the first and second numbers is 5:6 .
C.P. = Rs. 1000*3 = Rs. 3000
S,P, = Rs. (700*4.5 + 300*2.5) = Rs. 3900
Profit = SP - CP = Rs. (3900 - 3000) = Rs. 900
Hence, Avg. Profit is : Rs. 900/1000=.Re. 0.90 = 90 paise . Ans.
Let, the original price and consumption be x and y respectively . So, original expenditure = x*y.
ATP, (125/100)*x.*y' = x*y . => y' = (4/5)*y . So, 1/5 of y consumption is reduced.
Hence, the reduction is 20% . Ans.
Less in SP = Rs. (1080 - 1026) = Rs. 54. Rs. 54 is 4% of the CP. So, CP = Rs. 1350 .
Hence, The SP to make 4% profit = Rs. (1350 + 54) = 1404.
Biren works twice as fast as Satish. Hence, together they will need 1/(1/9+1/18) days = 6 days . Ans.
Let, the speed of 2nd train be x m/s.
ATP, (270+180)/(60*5/18 + x) = 10.8 . => (on solving) x = 25 .
Hence, required speed = 25 m/s or 90 km/hr . Ans.
a - 15 = 1/3*(b + 15) . => 3a - b = 60 ... (i)
50 + a + b + 50 = 300 . => a + b = 200 ... (ii)
Adding (i) and (ii), 4a = 260 . => a = 65 . => b = 135 .
Since, T = D/S , Ratio of the time taken by them- T1/T2 = (D1/S1)/(D2/S2) = (D1/D2)*(S2/S1) = 3/7*3/2 = 9/14 . Ans .
HCF of 88 and 105 = 17 . Ans.
Always welcome.
HCF of 132, 156, 84 = 12 kg. Ans.
Ans. Zero . Since, it's uniform veelocity, acceleration = 0, and hence the force.
SP = (80/100)*(130/100)*CP = (52/50)*CP
So, Profit = (2/50)*CP = (1/25)*CP = 4% (option C)
ATP, {(x -16)/x}*100 = x.
=> By solving, x = 80 or 20.
So, option D
E = 5000*(1 + 2/100)^2 = 5202 . So, Ans is option C
P = 100*I/(R*T) = Rs. 100*240/(5*6) = Rs. 800 . So, Ans is option C .
Ans, 10/(10+4) = 10/14 = 5/7 (option D)
Non-graduate males = (4/5)*(3/5) = 12/25 of total population.
Non-graduate females = (3/4)*(2/5) = 3/10 of total population
So, total non-graduates = (12/25 + 3/10) = 78/100 = 78% of total population. Ans. Option D
No. of damaged boxes = 700/35 = 20.
So, 2% of total boxes = 20.
Thus, 100% boxes = (20/2)*100 = 1000.
Hence, number of boxes shipped is 1000 . Ans.
Actual work or PE = 2*10*10 J = 200 J .
So, work done against friction = (300 - 200)J = 100 J . Ans. (option B)
Bouyant force = (5 - 2)N = 3N (option C)
Since, the outer volume of the solid sphere is less than that of the hollow sphere, its moment of inertia is also less.
Taking, speed of boat x km/h, frame the equation : (x-3)*2 = (x+3)*3/2 . => x = 21 .
Hence, speed of boat is 21 km/h (option C) . Ans.
Total speed of train and boy = (60+6)km/h = 18.33 m/s
Distance covered = 120 m
Hence, time taken = 120/18.33 s = 6.54 s Ans. (option A)
Speed of the train = (142/6)*(18/5) = 84.9 km/h (approx) . Ans. ( (option C)
Gain on Rs. 50 = Rs. (60-50) = Rs. 10.
So, gain of Rs. 100 = Rs. 20.
Hence, gain percentage = 20% . Ans (option B)
ATP, present sum of ages of Subho and Sudip = (32*2 + 3*2) years = 70 years.
Again, present sum of ages of Subho, Sudip and Subir = 30*3 years = 90 years.
Hence, age of Subir = (90-70) years = 20 years. Ans.
After 50% increase, each edge of the cube becomes 150/100 of the original.
Thus, the surface area of the cube becomes (150/100)^2 = 225/100 of the original.
Hence, increase in perecentage of surface area = (225-100) = 125% . Ans.
Time taken to walk one side = 45/2 mins = 22.5 mins.
So, time taken to ride one side = (33-22.5) mins = 10.5 mins.
Hence, time taken to ride both ways = 10.5*2 mins = 21 mins . Ans. (option C)
The man completes 40/60 = 2/3 part of work in 40 hrs.
So, his son completes (1 - 2/3) = 1/3 part of work in 40 hrs.
Hence, the son can alone complete the job in 40*3 = 120 hrs. Ans. (option D)
Here, A1 = 2P, A2 = 8P.
ATP, 2P = P*(1 + R/100)^4 . => (1 + R/100)^4 = 2 . => (1 + R/100) = 2^(1/4) .
Using this, 8P = P*[2^(1/4)]^t . [where, t = reqd. no. of years]
Or, [2^(1/4)]^t = 2^3 . => t/4 = 3 . => t = 12 .
Hence, reqd. time is 12 years . Ans. (option D)
Temp. on Wednesday = [40*3 - (41*3 - 42)]0C = 39 deg C . Ans. (option A)
Net increase in oranges in the two baskets are 16x/3 and 8x/3 respectively .
So, the reqd, ratio is 2:1 . Ans. (option B)
Taking original number as (10x + y), we get, (10y + x) - (10x + y) = (10x + y) - 1. => 19x - 8y = 1 .... (i)
Again, y = 2x + 1 ... (ii)
Solving the eqns, x = 3 and y = 7 .
Hence, the original no. is 37 . Ans. (option C)
ATP, (l^2 + b^2) = 41 ... (i) and l.b = 20 ... (ii) .
So, (l + b)^2 = 81 . => (l + b) = 9 . => 2(l+b) = 18 .
Hence, perimeter of the rectangle = 18 cm . Ans. (option A)
ATP, 4x/(3x + 5) = 4/5 . => Solving, x = 5/2 . Or, 4x = 10 .
Hence, quantity of alcohol is 10 lites . Ans. (option A)
Let, time taken by Ashish be x hrs.
So, his speed = 40/x km/h and that of Tushar = 40/(x+2) km/h .
ATP, 40/[2*(40/x+2)] = x - 1 . Or, x + 2 = 2x - 2 . Or, x = 4 .
Hence, Tushar's speed = 40/(x+2) = 40/6 km/h = 6.67 km/h . Ans.
CI for the 4th yr = CI after 4 yrs - CI after 3 yrs = Amt. after 4 yrs - Amt. after 3 yrs = P*[(1.2)^4 - (1.2)^3] = 0.3456P .
SI for the 4th yr = SI for the 1st yr = 0.2P .
So, diff. betn CI and SI for 4th yr = 0.3456P - 0.2P = 0.1456P .
ATP, 0.1456P = Rs. 7280. => P = Rs. 50,000 . Ans. (option C)
Ans. Option C (9 cm) .
Ratio of area = Suare of ratio of sides . => 9:25
So, area of smaller square = [9/(9+25) of 304] sq. cm = 81 sq. cm.
Hence, side length of smaller square = rt.81 = 9 cm .
Ans. Option A (26.67%) .
Let, CP for the manufacturer for 1kg wheat be Rs. x.
ATP, his SP for 0.9 kg wheat = Rs. (114/100)*x .
So, SP of 1 kg wheat = Rs. (1.14/0.9)*x = Rs.1.2667x .
Hence, profit% = [(SP - CP)/CP]*100 = 26.67% .
Ans. Option B (4.5%)
SP = Rs. (95/100)*(110/100)*12500 = Rs. 13062.50 .
So, profit = Rs.(13062.50 - 12500) = Rs.562.50 .
Hence, profit% = (562.5/12500)*100 = 4.5% .
Ans. Option C (36 miles/hr) .
Total distance = [(30*6) + (45*4)] km = 360 km .
Total time = (6 + 4) hrs = 10 hrs
Hence, average speed of the train = 360/10 miles/hr = 36 miles/hr .
Ans. Option D (48 minutes)
Part of cistern filled in 1 min = (1/16 + 1/12 - 1/8) = 1/48.
Hence, total cistern filled in 48 mins.
Ans. Option B (48 litres)
Let, total volume be x ltrs. It remains same throughout. But, change in liquor is (18 - 15)% = 3% = 3x/100
ATP, 3x/100 = (18/100)*8 . => x = 48 .
Ans. Option B (20 yrs) .
Age of new friend = [(24*5) - {21*4)+(4*4)}] = (120 - 100) yrs = 20 yrs.
Ans. Option C (Rs. 64.10)
Diff, between CI and SI = Dif. between the Amounts = Rs.1000[(1 + 10/100)^4 - (1 + 10*4/100)] = Rs.1000[1.4641 - 1.4] = Rs. 64.10 .
Ans. Option D (50 sec)
Total distance = (200 + 300) m = 500 m .
Comparitive speed = (30 - 20) m/s = 10 m/s .
Hence, time taken = 500/10 s = 50 s .
Ans, Option A (31.4%)
Original amt. of solute in 800 g solution = 40/100*800 = 320 g.
After 100 ppt, 220 g remains in 700 g of solution.
Hence, new solute percentage is (220/700)*100 = 31.4%
Taking, no. of cold drinks bottles = x, no. of water bottles becomes 2/3*x .
ATP, x + 30 = 4*(2/3*x) . => 5x = 90 . => x = 18 .
So, 1/3*x = 6 .
Hence, The reqd. no. of cold drink bottles is 6 . Ans.
Ans. Option C (mgR)
Ans. Option C (3)
ATP, (p - q)/(4 - 0) = 3/4 . => (p - q) = 3
Ans, Option D - C6H10O4
In the compound, Weight of Carbon = 72 g (approx), weight of Hydrogen = 10 g (approx), weight of Oxygen = 64 g (approx) .
Respective atomic masses of C, H and O are 12, 1 and 16.
So, atoms of C, H and O combined are 72/12 = 6; 10/1 = 10 and 64/16 = 4 respectively.
Hence, the formula.
Since, total current in the circuit is constant, we get 2.25R = 0.75R + 1.5*30 .
Or, 1.5R = 1.5*30 .
Hence, R = 30 ohm Ans. (option B)
Ans. Option A (x = 3y)
Join OR. Now, OR = OP (radii of same circle) . So, OR = OQ in Tri. ORQ and OR = OS in Tri. ORS . Relate between the respective angles of the 2 Triangles to get the answer.
Ans. Option D (14 cm)
Draw the figure with given data and join the two centres A and B. Let, that line AB bisect PQ at O. Take, two rt. ang Tri. AOP and BOP. to find out AO and BO by Pyth. Theo. PO= 12 cm, AP = 13 cm and BO = 15 cm. Thus, AO = 5 cm, BO = 9 cm. Hence, AB = (5 + 9) = 14 cm.
Ans. Option C (√3R)
Circumradius = 2/3 rd of height = (2/3)*(√3/2) of each side = R .
Hence, each side = √3R.
Ans. BD = DC = (a/√3)
Join AD. AD = 2r = 2*(a/√3) . Join OB and OC. OBDC is a rhombus. OB = OC = r = (a/√3). Hence, the answer.
Ans. Option A (5/4)
Real depth = 40 cm. Apparent depth = (40 - 8) cm = 32 cm.
Hence, Refractive Index = 40/32 = 5/4 .
Ans. Option C (12 V)
Equivalent resistance of the circuit = [1 + 2 + (12 || 6 || 4)] ohm = [3 + 2] ohm = 5 ohm .
So, current through the circuit = 30/5 A = 6 A.
Hence, voltage drop across 2 ohm resistance = 6*2 V = 12 V.
Ans. Option C
Sine inverse of the ratio of the two RI - 4/5:5/3 or 4/5 . Hence, the ans.
Ans. The arrows will not collide. The time of flight for either side for each arrow is 10 s. So, when the 2nd arrow is fired, the 1st arrow is midway in the upward direction, when the 2nd arrow is fired. 2nd arrow is midway in the upward direction, when 1st arrow attains the max. height. Again, when 2nd arrow attains the max. height when 1st arrow is midway in the downward direction, and so on.
Ans, Option D - C6H10O4
In the compound, Weight of Carbon = 72 g (approx), weight of Hydrogen = 10 g (approx), weight of Oxygen = 64 g (approx) . [By percentage calculations]
Respective atomic masses of C, H and O are 12, 1 and 16.
So, atoms of C, H and O combined are 72/12 = 6; 10/1 = 10 and 64/16 = 4 respectively.
Hence, the formula.
Since, T varies to the root of l, when l is increased by 44% or becomes 144/100 of the original, new T is 12/10 or 1.2 times of the original. Hence, net increase in T is 20%.
Ans. Option D
Molecular mass of Na2SO4 = 142. Molecular mass of H2O = 18.
Let, there be x molecules of water.
ATP, 18x = 55.9*(142 + 18x) . => x = 10 (approx).
Hence, the formula
Ans. Option A (10%)
[(2000 - 1800)/2000]*100 = 10% .
Ans. Option A (15/16)
Only one combination of 4 heads (H, H, H, H) to be excluded. Other 15 combinations have 1 or more than 1 tails.
Ans. Option C (Mg3N2)
Valencies of Mg and N are 2 & 3 respectively, which are exchanged during combination.
Ans. Option C (48 Ohm)
Let, galvanometer's resistance be x Ohm.
ATP, x/(x||12) = 50/10 . => (12+x)/12 = 5 . => x = 48 .
Ans. 93 degree
<DOB = [180 - (132 + 141)] = 93 degree
Ans. Option D (4:1)
Triangles DCQ and XCD are similar. [Since, <C is common, <X = 90 degree for both triangles]
DQ:DC = 2:1. We know, ratio of areas = Square of the ratio of the sides. Hence, the answer.
Ans. Option B
Apply the formula- v^2 = u^2 + 2gR (where u = 0) . Hence, get the answer.
Ans. Option B (17.94 cm)
Volume of the hollow cylinder = Volume of the cone
pi*(81^2 - 80^2)*200 = (1/3)*pi*r^2*300 . => r^2 = 322 .
Hence, r = root.322 = 17.94 cm (approx)
Ans. Option C (47.32 m)
Distance between the feet of tower and building = 30/root3 m = 10*root3 m = 17.32 m .
So, the height extra of the tower than the building = 17.32 m .
Hence, total height of the tower = (30 + 17.32) m = 47.32 m .
Ans. Option D
Volume of cylindrical part = pi*3^2*5 = 45pi m^3
Volume of conical part = (1/3)*pi*3^2*2 = 6pi m^3
Hence, total volume = (45 + 6)pi m^3 = 51pi m^3 = 160.29 m^3 (approx)
Slant height of the cone l = root(3.5^2 + 12^2) cm = 12.5 cm .
Hence, total surface area of the figure = curved SA of the conical part + curved SA of the hemispherical part = pi*3.5*12.5 + 2pi*3.5^2 = pi*(43.75 + 7) cm^3 = 159.5 cm^3 . Ans.
Ans. Option D
Let, radius of the cylinder be r. Now, put the values in a proper equations relating the two cones and a cylinder.
Ans.Option C (13)
Let, no. of cones made be n.
So, n*(1/3)*pi*3^2*3 = pi*(7.5/2)^2*10. => n = 13.61
13 being the max. whole number of cones to considered.
ATP, (4/3)*pi*R^3 = 8*(4/3)*pi*r^3 . => R^3:r^3 = 8:1 .
So, R:r = 2:1 and 4*pi*R^2:4*pi*r^2 = 4:1 .
Hence, the ratio of their radii and surfaceareas are 2:1 and 4:1 respectively.
Ans. Option C (12:5).
ATP, (pi*r*l + pi*r^2):(pi*r^2) = 18:13 . => r:l = 5:13 . => r^2:(r^2 + h^2) = 25:169 . => h^2:r^2 = 144:25 .
Hence, h:r = 12:5 .
Ans. Option D .
ATP, Total length of the wire = 66*3 cm = 198 cm .
So, circumference of the circle = 198 cm. So, radius of the circle = 198/(2*pi) cm = 31.5 cm
Hence, area of the circle = pi*(31.5)^2 cm^2 = 3118.5 cm^2 .
Ans. Option A (9 cm)
ATP, 2*pi*7*(7 + h) = pi*224 . => 7 + h = 16 . => h = 9 cm .
ATP, (22/7)*(R^2 - 14^2) = 770 . => R^2 = 245 + 196 = 441 . => So, r = root.441 = 21 cm
Hence, radius of the outer circle is 21 cm . Ans.
Ans. Option C (8:1)
Ratio of radii of large circle to small circle = 3:1 . So, ratio of their areas = 3^2:1^2 = 9:1 .
Hence, unshaded area:shaded area = (9 - 1):1 = 8:1 . Ans.
BD = root.(AB^2 + AD^2) = root.(64 + 225) = root.289 = 17 cm.
So, radius of the circle or OD = 1/2 of BD = 8.5 cm .
Hence, area of the circle = 3.14*(8.5)^2 cm^2 = 226.865 cm^2 . Ans.
Ans. Option D (201.14 cm^2)
Apply the formula pi*(R^2 - r^2)
Ans. Option D (Rs. 4224)
Area of the field = 4928/0.5 cm^2 = 9856 m^2 .
So, radius of the field = root(9856/pi) cm = 56 m.
Thus, circumference = 2*pi*56 = 352 m.
Hence, cost = Rs. 352*12 = Rs. 4224 .
The image is: i) real, ii) inverted, iii) formed on the same size of the object, iv) magnified by 1.51 times of the size of the object . Furthermore, it's a concave mirror.
Ans. Option D
Reqd area = [35*25 - {2*pi/4*13^2)] = [875 - 265.57) cm^2 = 609.43 cm^2 .
Ans. Option C
Area of triangle ABD = 1/2*12*6 = 35 cm^2 .... (i)
Area of semi-circle CBDO = pi/2*6^2 cm^2. .... (ii)
Area of smallest circle = pi*3^2 cm^2 .... (iii)
Hence, area of the shaded region = (i) + (ii) + (iii) = 64.29 cm^2 (approx)
Students taking either of or both tea and coffee = 150 + 225 - 100 = 275 .
Hence, students taking neither tea, nor coffee = 600 - 275 = 325 . Ans.
Ans. Option B (30 m)
ATP, (30 + x)(20 + x) = 5*(30*20) . => x = 30 or -80. -ve value to be ignored.
Hence, x = 30 m .
Ans. Option C
The edge of the cube = Diameter of the sphere . Now use the formula .
Ans. Option B
The reqd. volume of water = [70*50*20 - 50*(5^3)] cm^3 = (70000 - 6250) cm^3 = 63750 cm^3 . Ans.
Ans. Option B (12)
Let, no of cones be n . So, n = [pi*(10^2 - 8^2)*5]/[(1/3)*pi*3^2*5] = 36/3 = 12 . Ans.
Ans. Option B (50.72 m)
Apply, cot 45 - cot 60 = Difference the houses/Height of the tower
Or, 1 - 1/root3 = D/120 . => D = 50.72 m (approx) . Ans.
Ans. Option A (60 years) .
Increase in sum of the ages = 7*5 = 35 years .
So, Age of Shilpa = Increased age + Age of the woman left = (35 + 25) years = 60 years. Ans.
Ans. Option A (9 days)
In 4 days, A & B together do 4*(1/12 + 1/18) = 5/9 of the work .
So, in 4 days C does (1 - 5/9) = 4/9 of the work .
So, C does 1/9 of work in 1 day.
Hence, C alone completes the work in 9 days .
Ans. Option C (90 m)
ATP, (x+162)/(x+120) = 18/15 = 6/5 .
So, x = 5*162 - 6*120 = 90 .
Ans. 13:27
ATP, (50 + a + b) + 50 = 3*100 . => a + b = 200 ... (i)
Again, a - 15 = 1/3*(b + 15) . => 3a - b = 60 ... (ii)
Adding, (i) and (ii), 4a = 260 . => a = 65 . So, b = 135 .
Hence, a:b = 65:135 = 13:27 . Ans.
Ans. Option D (10%)
(1+ r/100) = 23958/21780 [taking the 3rd year only]
r/100= 1/10
r=10
Hence, rate of interest is 10% . Ans.
Ans. Option B (5 kg)
Initial water = 20% of 25 kg = 5 kg.
Let, water added be x kg.
ATP, 5 + x = 33.33% of (25 + x) . => x = 5 .
Ans. Option D (25 m)
The car comes to rest in 15/0.3 s = 50 s.
In 50 s, dist. travelled by the car is 15/2*50 m = 375 m.
Hence, dist. left from the signal = (400 - 375) m = 25 m.
Ans. Option B (4 m/s)
Avg velocity = Total dist./Total time = 240/60 m/s = 4 m/s .
2/3 of 6 cm = 4 cm is the reqd. height raised. Ans .
Ans. 4/5
RI = (40 - 8)/40 = 32/40 = 4/5
When we view a virtual image through our eye, it becomes the object. Again the real image of that is formed by our eye lens on the retina.
Ans. Option D (2.016) .
Mol. mass of CaCO3 (limestone) = 100 g.
It liberates 22.4 L of CO2.
So, 10 g of 90% CaCO3 can liberate (10/100)*(90/100)*22.4 L = 2.016 L of CO2 .
Ans. Option C (6.25 s)
To avoid accident, max. speed of the express train should be 5 m/s.
So, applying v = u + (-a).t, 5 = 30 - 4.t
Hence, reqd. time t = 25/4 m/s = 6.25 m/s
Ans. Option D (3 Ohm)
Here, 2||x = 6/5 . Or, 2x/(2 + x) = 6/5 . By solving, x = 3 .
Ans. Option C (9:14)
Ratio of time = Ratio of distance:Ratio of speed = (3:7):(2:3) = 9:14 . Ans.
Ans. 16/7 days
In 1 day, they complete (1/4 + 1/8 + 1/16) = 7/16 of the work. Hence, the ans.
More the angle of refraction, more the velocity of light and vice versa. So, here velocity of light is minimum in medium with angle of refraction 15 deg.
Ans. Option C (Rs. 30,600)
Mr. Sen's monthly income = Rs. (77,520/12) = Rs. 64,600.
Let, Sunita's monthly income be 100x and Nirmala's be 90x.
ATP, 190x = Rs. 64600 .
=> 90x = Rs. (64600*90/190) = Rs. 30600 .
Ans. Option D (216 cm^2)
Applying the formula P = 2(l+b) with b = 12 cm, we get l =18 cm.
Hence, area of rectangle = l*b = 18*12 cm^2 = 216 cm^2 .
Ans. Option C (1,461)
Total cost = Rs. (45*25 + 28*12) = Rs. 1461 .
Ans. Option A (30 days)
1 woman = 1/3 man and 1 boy = 1/5 man .
So, 1 man + 1 woman + 1 boy = (1 + 1/3 + 1/5) men = 23/15 men .
Since, 1 man takes 46 days to do the work, 23/15 men will take 46/23*15 days = 30 days to do the work .
Ans. Option D (154 cm)
Average height = (146 + 154 + 164 + 148 + 158)/5 cm = 154 cm.
Ans. Option C (12 yr)
Let John's age be x yrs.
ATP, (7/2x):(x + 3) = 14:5
By solving, x = 12 .
Ans. Option D
CP of 9 kg mixture = Rs. (60*7 + 105*2) = Rs. 630
SP of 9 kg mixture = Rs. 100 *9 = Rs. 900 .
Hence, profit percentage = [(900 - 630)/630]*100 = (42+6/7)% .
Ans. Option C (12:48:35 pm)
124 km covered by train A in 2 hrs.
So, train A and train B need to cover (827 - 124) km = 703 km with combined speed (62 + 59) km/h = 121 km/h .
Thus, required time period = 703/121 h = 5.81 hrs (approx)
Hence, 7 am + 5.81 hrs = 12:48:35 pm .
Ans. Option A (79.4 J)
Work done = 0.15*(60*9.8)*0.9 J = 79.4 J (approx.)
Ans. Option B (4:1)
Voltage will not change. Equivalent resistance will be 1/4th the original. Hence, Power being inversely proportional to Resistance (P = V^2/R), P2 will be 4 times of P1.
Ans. Option (Rs. 1.5 lakh)
Let, the broken parts be x, 2x, 3x and 4x. So, total parts = 10x.
ATP, (10x)^2 - [(x^2 + (2x)^2 + (3x)^2 + (4x)^2] = Rs. 105,000 . => 70x^2 = Rs. 105000 .
Hence, original price of diamond = 100x^2 = Rs. 150,000 .
Ans. Option A (4.5 km/h and 0.5 km/h)
Let, the speed of boat and water current be x km/h and y km/h respectively.
ATP, x + y = 22.5/4.5 = 5 ... (i) and x - y = 28/7 = 4 .... (ii)
Solving (i) and (ii), x = 4.5 and y = 0.5
Ans. Option E (15 m)
From eqn- 2(22 + 18)*h = 1200, we get, h = 15 .
Ans. Option A (HTHEY)
G= G + 1 = H, R = R +2 = T, E = E + 3 = H, A = A + 4 = E, T = T + 5 = Y .
Hence, GREAT = HTHEY
Ans. Option C (6)
5x + 6x + 8x = 19x = 95 . => x = 5 . => 8x = 40 = 36 + 4 = 6*6 + 4
Ans. Option B (Rs. 540.90)
SP of 1 table = Rs. 1.2*7212/16 = Rs. 540.90 . Ans.
Let, time taken by A is t and that by B is thus (t - 1) .
ATP, 60*(t - 1) = 50*t. => By solving, t = 6 .
Ans. The reqd time is 6 hrs.
P1 for covex lens = 100/40 D = 2.5 D .
P2 for concave lens = -100/25 D = - 4 D .
So, combined P = Pi + P2 = [2.5 + (- 4 D)] = - 1.5 D . Ans.
Ans. Option D (-cos A)
It's in 2nd quadrant. Must follow the all, sin, tan, cos rule .
Ans. Option B (0.5)
sin 3P = cos (P + 14) = sin [90 - (P + 14)] .
By solving, P = 19 .
Hence, sin (P + 11) = sin 30 = 0.5 Ans.
Ans. Option B (0)
Since, sin (50 - θ) = cos [90 - (50 - θ)] = cos (40 + θ)
Ans. Option C (4/3)
cos (90 - A) = sin A = 4/5 .
So, cos A = 3/5 .
Hence, tan A = (sin A)/(cos A) = 4/3 . Ans.
Ans. Option D (4:1)
Given: R + r = 21 cm ... (i) , 2*(22/7)*(R - r) = 44 cm . => R - r = 7 cm ... (ii)
From (i) & (ii), R = 14 cm, r - 7 cm .
So R:r = 14:7 = 2:1
Hence, pi*R^2:pi*r^2 = 4:1 Ans .
Ans. Option D (28 cm)
ATP, 2*(22/7)*r*25000 cm = 44*100000 cm
Hence, r = 28 .
Ans. Option B (14 cm)
ATP, (22/7)*(r^2)/4 - (r^2)/2 = 56 sq cm
By solving, r^2 = 7*28 sq cm .
Hence, r = 14 cm . Ans.
Ans. Option D (6%)
(25249.54 - 23820.32)/23820.32*100 = 6 (approx)
Ans. Option D (4)
Since, cos 60 = 1/2, cos^2 60 = 1/4. Here, x*(1/4) = 1. So, x = 4 . Ans.
Ans. Option C (4,1)
By solving we get, x^2 - 5x + 4 = 0 (for consideration of real roots only). [ x^2 - 5x + 10 = 0 to be ignored]
Again solving we get, x = 4 or 1 .
Ans. Option D (sin^2 θ)
The given exp. = cosθ.sinθ/(cosθ/sinθ) = sin^2 θ .
Ans. Option A (Rs. 425.25)
SI for 1 yr = Rs. 420 . => P = Rs. 8400 . => CI for 6 months = Rs 210 .
Thus, P for next 6 months = Rs. (8400 + 210) = Rs. 8610 .
Now, CI for next 6 months = Rs. 215.25 .
Hence, total CI for 1 year = Rs. (210 + 215.25) = Rs. 425.25 . Ans.
Ans. Option B (Green)
Ans. Option B (a text)
Ans. Option D (Rs. 1352)
CI for 1st yr = SI for 1 yr = Rs 1300/2 = Rs. 650 .
CI for 2nd yr only = Rs. [650(1 + 8/100)^1] = Rs. 702 .
Hence, total CI for 2 yrs = Rs. (650 + 702) = Rs. 1352 . Ans.
Ans. Option C (Rs. 600)
Let, SI for 1yr be Rs. x .
ATP, x(1 + 10/100)^1 = 420 - x [i.e. the CI for 2nd yr only]
By solving, x = 200
Hence, reqd. SI = Rs. 400*3/2 = Rs. 600 . Ans.
Ans. Option C (40)
Let, no. of original employees be x and time taken by them be t . Later, y people join them.
ATP, 1st case: (x + 20).2/3*t = x.t => Solving, x = 20
2nd case: (y + 20).1/3*t = 20.t => Solving, y = 40
Hence, reqd no. of employees is 40 . Ans.
Ans. Option B (3.52 km/hr)
Total dist. = 4*2*(22/7)*35 m =880 m .
Time = 15 mins.
So, speed = (880/15)*60/1000 k/hr = 3.52 km/hr . Ans.
Ans. Option B (3.520 km/hr)
Total dist. = 4*2*(22/7)*35 m =880 m .
Time = 15 mins.
Hence, speed = (880/15)*60/1000 k/hr = 3.52 km/hr . Ans.
Ans. Option C (Water)
Ans. Option E (vacant)
Ans. Option B (diffuse)
Ans. Option E (Modern)
Ans. Option B (E)
Ans. Option B (One)
Ans. Option A (R eq < 1) [= 3/19]
Ans. Option A (0.5)
Applying eqn. v^2 = u^2 - 2*a*s , and putting v = 0, u =20 m/s, s = 40 m, we get a = 5 m/s^2 .
Hence, co-efficient of dynamic friction = a/g = 5/10 = 0.5 . Ans.
Pick up time will be 3 times of the retarding time.
Using eqn: s = 1/2*a*t^2 twice, => s = 1/2.1.(3t)^2 and 1215 - s = 1/2.3*t^2,
we get, t^2 + 9t^2 = 1215 .
By solving, t = 33.08 s (approx) Ans.
Ans. Option C (100 degree)
After joining OS,
<OQS = 90 - 70 = 20 degree . So, <OSQ = 20 . Thus, <SOQ = 180 - (20 + 20) = 140 degree
Similarly, <SOR = 120 degree . So, Reflex <QOR = 260 degree .
Hence, <QOR = 360 - 260 = 100 degree .
Ans. Option A (50 and 10)
Taking triangle CAB, <ABC = 40 (given) and <CAB = 90 (known) .
So, <CBA = [180 - (40 + 90)] = 50 degree.
Now joining OC, <OCA = 50 degree . Hence, <PCA = (90 - 50) = 40 degree.
Hence, in triangle PCA, <CPA = <CAB - <PCA = 50 - 40 = 10 degree .
Ans. Option D (105)
From the given data, <DCE = 15 and <CDE = 75 [Since, < CED = 90]
Now, < EFC = 180 - 75 = 105 [opposite <s of a cyclic quad. are supplementary]
Area of the shaded region = [(pi/2)*20^2 + 2*(pi/2)*10^2] cm^2 = 300*pi cm^2 . Ans.
Perimeter of the shaded region = [4*pi*10 + pi*20 + 20] cm = [60pi + 20] cm . Ans.
QR = root(PR^2 - PQ^2) = 5 cm.
So, area of triangle PQR =(1/2)*5*12 = 30 sq. cm.
Triangle PQR is similar to Triangle STP [Since, <P is common and <Q = <T = 90]
Now, ST/PQ = 4/12 = 1/3 .
Hence, Triangle STP is (1/3)^2 of Triangle PQR = 30/9 sq cm = 3.33 sq cm .
Ans. Option B (30 degree)
tan theta = 450/778.5 = 0.578 = 1/root3 (approx)
Hence, theta = 30 degree .
Taking the respective distances x and y,
Here, cot 30 = x/25 and cot 45 = y/25
Or, x = 25*root3 and y = 25
Hence, total width of the river = x + y = 25(root3 + 1) m = 25*2.732 m = 68.3 m . Ans.
Let, the reqd. height be h m.
Here, h/2 = tan 60 = root3 .
Hence, h = 2*root3 m = 2*1.732 m = 3.464 m . Ans.
Ans. Option C (47.32 m)
Let the dist. betn. the feet of the tower and the building be d m.
ATP, 30/d = tan 60 = root3 . => d = 30/(root3) m = 10*root3 m = 17.32 m .
Since, angle of elevation is 45, the part of height of the tower excess to the height of the building = d = 17.32 m.
Hence, the total height of the tower = (30 + 17.32) m = 47.32 m .
Ans. Option D (27)
Taking radius of cylinder as R and framing the equation 2*pi*(r^2 + 8*r) = 528 .
By solving, R = 6 cm [pi = 22/7]
Again let, the no. spheres be n .
ATP, n.(4/3).*pi*2^3 = pi*6^2*8 .
By solving, n = 27
Triangles COA and DOB are similar by A-A axiom [since, < COA = vert. opp. <BOD, < ACO = <BDO = 90]
Since, ratio of similar sides OC and OD is 5:8, the ratio of areas of triangle DOB: triangle COA = 64:25.
Hence, area of triangle DOB = (64/25)*60 = 153.6 sq. cm. Ans.
Ans. Option A (32 m)
First applying, v = u - a.t, we get a = 4 m/s^2
Again applying v^2 = u^2 - 2.a.s , we get, s = 32 m
Ans. Option B (Babble)
Ans. Option D (Scalar)
Path distance implies normal Distance
Ans. Option B (the total length of the path traversed by an object)
Actually, the potential energy of a body is higher at a greater height. It can be explained by the fact that, we require to apply larger force to lift a body to more height. So, greater the height, greater the force applied, greater the work done. Hence, the energy possessed by a body at greater height is also larger.
When theta is 0 degree, work done is maximum and when theta is 90 degree, work done is minimum (zero). This is because, cos 0 = 1 and cos 90 = 0.
Yes. Classical example of this case is any nuclear (radioactive) reaction. That's why 'the combined law of conservation of matter and energy' has been formulated.
Ans. Option A (the motion has taken place)
Ans. Option A (is increasing in equal intervals of time)
Ans. Option B (Vector)
Ans. Option B (the total length of the path traversed by an object)
Ans. Option D (Ways to describe motion without going into the causes of motion)
On strong heating, first the Boric acid (H3BO3) breaks up into Metaboric acid (HBO2) and Water (H2O). On further heating, the Metaboric acid breaks up into Boric oxide (B2O3) and Water.
Ans. Option D (Mg)
Ans. Option B (Ca and P)
Ans. Option B (Same velocity) {at that particular coinciding point]
Ans. Option B (v-t)
Ans. Option D (120 m/s^2) [ 6^2/0.3 = 120 ]
Ans. Option D (30 cm)
6^2/120 m = 0.3 m = 30 cm
Ans. Option A (True)
Ans. Option D (Kinetic Friction)
Ans. Option C (Centrifugal force)
Ans. Option B (at the equator)
Ans. Option A
Ans. Option C
Ans. Option C [ coulomb (C) ]
Ans. Option B (a voltmeter)
Ans. Option D ( V = W/Q )
Ans. Option B (parallel)
Ans. Option C
Ans. Option A (potential difference)
Ans Option D (a resistor)
Ans. Option B (detect direction of current)
Ans. Option A (the heating effect of the current)
Ans. Option A (normal)
Ans. Option B (in a straight line)
Ans. Option B
Ans. Option C (plane)
Ans. Option A (The fical point)
Ans. Option D (air)
Ans. Option C (always be virtual)
Ans. Option C (1.3 m)
Applyinf the Mirror formula : 1/f = 1/v + 1/u = 1/4 + 1/2 = 3/4 .
Hence, f = 4/3 m = 1.3 m (approx)
Solution. LCM of 12 minutes and 18 minutes is 36 minutes.
Hence, time after they will again meet at the starting point is 36 minutes. Ans.
We know, LCM of 2 numbers = Product of the 2 numbers/ HCF of the numbers .
So here, LCM = 306*657/9 = 22338 . Ans.
Here, all we need to do is to find the HCF of (33 - 3) = 30, (55 - 5) = 50 and (64 - 4) = 60 .
HCF of 30, 50, 60 = 10 .
Hence, 10 is the required greatest number . Ans.
Ans. Radius = 24 cm [no options given though]
1/3*pi*r^2*7 = 1232 . => pi*r^2 = 1232*3/7 = 176*3 .
pi*r*l = 550 . So, r^2/(r.*l) = 24/25
Putting, l^2 = r^2 + h^2 , we get, r^2 = 576 .
Hence, r = 24 cm . Ans.
Ans. Avogadro's number [6.023 x 10^23]
6 (Ans.)
Ans. Option C
ND Bhatt's book is highly recommended. RK Dhawan's book is also good.
Ans. Option C (25 days)
Let, B alone completes the work in x days.
ATP, 10*[(1/x) + (1/x + 1/50)] = 1 ... => 20/x = 4/5 ... => x = 25
Hence, B alone completes the work in 25 days.
Ans. Option B (7)
Total numbers = 12 . Their mean = 12 . So, their sum = 12*12 = 144
Sum of other 11 numbers = 137
Hence, x = 144 - 137 = 7
Ans. Option B (18, 25)
Ans. Option C (25:21)
--- 5*5:7*3 = 25:21
Ans. Option B (Rs. 2660)
Investment ratio = 65*6:84*5:100*3 = 13:14:10
After A's 5% profit, out of 95/100*Rs.7400 = Rs. 7030, B' share of profit = (14/37)*Rs.7030 = Rs. 2660 . Ans.
Solution: (A+B) - (B+C) = 12 .
So, A - C = 12.
Hence, C is 12 years younger than A . Ans.
Ans. Ratio of the present ages of father and son is 7:3
Solution: Let, 10 years ago their ages be x yrs and y yrs .
ATP, x = 3y and (x + 20)/(y + 20) = 2 . Or, (3y+20) = 2*(y + 20) .
Solving, y = 20. Thus, x = 60 .
Hence, present ages ie. x + 10 = 70 and y + 10 = 30 .
Ans. Difference between ages of her parents is 6 yrs.
Soln: Father is 38 + 4 = 42 yrs when her brother is born.
Hence, father is older than mother by 42 - 36 = 6 yrs
Ans. 36 years.
Let, 1 year ago Sooraj and Vimal were 6x years and 7x years respectively. So, 5 yrs hence, they were 6x + 5 yrs and 7x + 5 yrs.
ATP, (6x + 5)/(7x + 5) = 7/8 . => Solving, x = 5
So, 7x + 1= 36
Ans. Data inadequate
Reason: Age of A = Avg. of ages of B & C = 50/2 = 25.
We know, B > A . But, their age diff. can't be determined.
Ans. Value of each resistance is R.
[In parallel circuits, if n no. of resistors of identical resistance R are connected in parallel, the equivalent resistance is R/n .]
Let, the breadth be x.
So, length = 115/100*x .
ATP, (115/100*x)*x = 460 sq. m. Or, 115/100*x^2 = 460 sq. m => x^2 = 400 sq. m.
Hence, x = 20 m .
Ans. The breadth of the plot is 20 m.
Ans. Area of the rectangle is 2520 sq. m.
L - B = 23 m ....(i) , L + B = 206/2 m = 103 m ... (ii)
So, L = 63 m, B = 40 m .
Hence, LxB = 63*40 sq. m. = 2520 sq. m. Ans.
Ans. Increase in the area of the rectangle is 44%.
After 20% increase in Length and Breadth, (120/100*L)*(120/100*B) = 144/100*LB .
Hence, increase in area = 144/100*LB - LB = 44/100*LB = 44% Ans.
Ans. Area of the rectangle is 27 sq. feet.
Length of longer side = root(7.5^2 - 4.5^2) feet = 6 feet .
Hence, area = 6*4.5 sq. feet = 27 sq. feet . Ans.
Ans. Length of the rectangle is 18 cm .
ATP, 2(l + b) = 5b .. => l = 3b/2 .
Again, (3b/2)*b = 216 sq. c. => b = 12 cm .
Hence, l = 3b/2 = 18 cm . Ans.
Ans. Decrease in area is 28%.
After 20% decrease in L, 10% decrease in B, new area = (80/100*L)*(90/100*B) = 72/100*LB.= 72%LB
So, decrease = (100 - 72) % = 28% . Ans.
Ans. Percentage of excess error in calculating area = 4.04% .
For an error of 2% in excess while measuring side 'a', the erroneous area = (102/100*a)^2 = 10404/10000*a^2 .
Thus, excess error = (10404/10000 - 1)*a^2 = 404/100*a^2 = 4.04% . Ans.
Ans. The width of the road is 3 m .
Let, width be x .
ATP, 60*x + 40*x - x^2 = 60*40 - 2109 .
=> x^2 - 100x + 291 = 0
=> (x - 97)*(x - 3) = 0 .
So,either, x = 97 or, x = 3 . But, the width of the road can't exceed the park dimensions.
Hence, x = 3 ...
Ans. Resistance is increased by 10.25% .
When stretched, length is increased by 5%. But, volume of the wire remains samne. So, area of the wire decreases proportionally at the same time.
Thus, l/A now becomes [(105/100*l]/[(100/105)*A] = 11025/10000*l/A .
Hence, increase % in resistance = (11025 - 10000)/10000 *100 % = 10.26 %% . Ans.
Ans. Ratio of heat dissipation in the two wires is 4:1 .
More the cross-sectional area, less the resistance, hence less the heat dissipation.
So, heat ratio will be inverse to the CSA. Hence, 4:1 . Ans.
Ans. Area of the field is 37500 sq. m.
Let, length be L. So, breadth = 60/100*L.
ATP, 2*(L + 60/100*L) = 800 m . => Solving, L = 250 m . and 60/100*L = 150 m .
Hence, area = 25*150 sq. m = 3750 sq. m . Ans.
Ans. Option C (greater than 1)
Area of the square is more than that of the rhombus, for having greater height.
Ans. The total cost of paving = Rs. 5.5*3.75*800 = Rs. 16500
Ans. Option C (88)
One side = 20 feet . Thus, other side = 680/20 feet = 34 feet.
Hence, the length to be fenced = (34 + 20 + 34) feet = 88 feet. Ans.
Ans. Area of the parking space = 126 sq. feet .
One side = 9 feet . So, other side = (37 - 9)/2 = 14 feet .
Hence, area = 14*9 sq. feet = 126 sq. feet . Ans.
Ans. Option B (30%)
Taking each side of square plot as 'a' and value of root2 as 1.4, length of the diagonal = 1.4 a.
So, length saved = 2a - 1.4a =0.6a.
Hence, % of length saved = 0.6a/2a *100 = 30% . Ans.
Ans. Option A (176)
HCF of 544 and 374 = 34 .
So, no. of max square tiles = 544*374/(34^2) = 176 . Ans.
Ans. Length of the plot is 60 m.
ATP, B = L - 20 and, 2*(L + L - 20)*26.5 = 5300 .
=> 2L - 20 = 100 . => L = 60
Speed = 12 km/hr = 200 m/min . , Time = 8 mins .
So, distance travelled = perimeter of the park = 200*8 m = 1600 m .
ATP, 2(3x + 2x) = 1600 m . => x = 160 m .
Thus, length = 3x = 480 m and breadth = 2x = 320 m.
Hence, area of the rectangular park = 480*320 sq. m = 153600 sq. m. (Ans)
Ans. Man's speed against the current is 10 km/hr .
[ 15 - (2*2.5)] = 10
Ans. The avg. speed of the whole journey is 37.5 km/hr .
[ Total dist. = 300 km, Total time = (3 + 5) hrs = 8 hrs ]
Ans. The avg. for last 4 matches is 34.25
[ (389 - 252)/4 ] = 34.25
Ans. Option B (174)
x + y = [ 50*40 - 48*38 ] = 176
x - y = 172 => Solving, x = 174
Ans. The age of the teacher is 51 years .
[ (37*15) - (36*14) = 555 - 504 = 51 ]
Ans. The value of x is 7
[ 12*12 - 137 = 7 ]
Ans. The reqd. ratio between the speed of the boat and that of water current is 8:3
Taking, spped of boat as x and speed of current as y,
x + y = d/240
x - y = d/528
(x+y)/(x - y) = 528/240
x/y = 768/288 = 8/3
Ans. The average marks of all the students is 54.69 (app.)
[ {(55*50) + (60*55) + (45*60)}/(55 + 60 + 45) = 54.6875 ]
Ans. The weight of B is 31 kg .
[ (40*2) + (43*2) - (45*3) = 31 ]
Ans. Option C (110)
[ (x*100) = (x - 10)*110 . => Solving, x = 110 ]
Ans. The remaining food will last for 42 days more .
2nd case: 125 men, 35/45 = 7/9 th part of food
So, no. of days = (45*150*7)/(125*9) = 42 . Ans.
Ans. The height of the building is 12.5 m .
By similarity, x/28.75 = 17.5/40.25 . => x = 12.5
Ans. The price of 49 dozen apples is Rs. 2499 .
(1517.25/357)*588 = 2499
Ans. The food will last for 50 days.
[ Reqd no. of days = 40*250/200 = 50 ]
Ans. Option B (2.5 kW)
[ m = V*d = 0.01*1000 kg = 10 kg ; Power of motor = (mgh + 0.5*m.*v^2)/t = (10*10*20 + 0.5*10*100)/1 W = 2500 W = 2.5 kW ]
Ans. Option B (12)
3 pumps work for 2*8 hrs = 16 hrs .
So, by Rule of 3, 4 pumps will work for 16*3/4 = 12 hrs . Ans.
Ans. Option D (195)
[ 25/0.128 = 195.3125 = 195 (approx.) ]
Ans. Option B
Ans. 60 degree .
[ 18/(6root3 = root3 = tan 60 ]
Ans. Speed of the car = 24.root3 km/hr
[ Total dist = 50(root 3 + 1/root 3) m = 200/root3 m ; Total time = 10 sec
So, speed = 200/root3/10 m/sec = 20/root3 m/sec = (20/root3)*18/5 km/hr = 24.root3 km/hr ]
Ans. Resistance is the property of the conducting material by which current flow through it is opposed. It is caused by the non-free electrons of it.
The factors affecting Resistance are : 1) Nature of Material (directly) 2) Lengt of the conductor (directly) 3) Temperature (directly) 4) Cross-sectional area (inversely) .
Ans. Option C (6.02)
[In between 5 and 7]
Ans. Option B (3)
Let, each part of dist. be x km
So, x*(1/3 + 1/4 + 1/5) = 47/60 => Solving, x = 1 . Thus, 3x = 3 .
Hence, total distance is 3 km . Ans.
Ans. Option C (increase by 8%)
Original area of rectangle = l.b
New area = (120/100)l*(90/100)b = (108/100)*l.b
Hence, increase in area = (108/100 -1).l.b = 8/100*lb = 8% . Ans.
Ans. Option B (700)
[ 2800/4 mins = 700 mins ]
Ans. Option A (160)
[ Rs. {250 - (75 +15)} = Rs. 160 ]
Ans. Percentage of users having camera phones is 7%
[ 35/500*100 = 7% ]
Ans. Option C (5xy)
Clearly, 5x > x^2 , for the given value range.
Now, since y is negative, we will get the most negative ie. least value for 5xy . x^2*y is less negative and xy^2 is positive.
Ans. Option A (30)
2 mins = 120 secs . So green light will flash at 120/5 = 24 secs interval.
1 min = 60 secs. So red light will flash at 60/3 = 20 secs interval.
LCM of 24 and 20 = 120 .
So, both light will flash together after every 120 secs or 2 mins.
Hence, in each hour, or 60 mins, they will flash together 60/2 = 30 times . Ans.
Ans. Option D (1590)
Let, other multiplier be x.
BTP, (53 - 35)*x = 540 , => 18x = 540 , => x = 30 .
Hence, 53*30 = 1590 . Ans.
Ans. 72 cm^2
[ 1i2*(12^2) sq. cm = 144/2 sq. cm. = 72 sq. cm ]
150
The series is the form n^3 + n^2
Ans. The reading on Ammeter is 0.5 A .
Equivalent resistance of the circuit is 12 ohm. Voltage is 6 V (given).
So, current = 6/12 A = 0.5 A . Ans.
Ans. Option B (noise)
Ans. Option D (2)
Ans. Option B (digital signal)
Ans. Option D (0.3 m)
Taking it as light wave
Ans. Option C (wavelength)
More the wavelength, less the frequency and vice versa. Frequency is always reciprocal of time period.
Ans. Option E (15 kHz)
Ans. Option C (constant)
[ Frequency is independent of amplitude ]
Ans. Option C (modulation)
Ans. Option D (signal)
Ans. Option A (bit)
Ans. Option C (highest value)
[ most significant value (MSB) ]
Ans. Option B (same as Earth)
[ stationary w.r.t. Earth ]
Ans. Option B (1001)
2^3 + 2^0 = 8 + 1= 9
Ans. The required probability is 1/3 .
No. of blue balls = 7 . Total no. of balls = (8+7+6) = 21 .
Hence, probability = 7/21 = 1/3 . Ans.
Ans. The required gain percentage is 5.45% .
Total CP = Rs.(4700+800)= Rs.5500 , SP= Rs. 5800 . So, Gain= Rs.(5800-5500) = Rs.300 .
Hence, Gain percentage = 300*100/5500 = 5.45%
Ans. The required number of notes is 90 .
Let the notes of each denomination be 'x' .
So, total amount = x *Rs.(1+5+10) = Rs. 16x = Rs. 480
Thus, x = 30 . Hence, total notes of 3 denominations = 30*3 = 90 . Ans.
Ans. Rs. 400 is paid to C .
Part of work done by C in 1 day = {1/3 - (1/6 +1/8)} = 1/24 .
So, their work ratio = 1/6:1/8:1/24 = 4:3:1 .
Since, payment ratio = work ratio, payment of C is 1/(4+3+1) of Rs. 3200 = 1/8*Rs. 3200 = Rs. 400 . Ans.
Ans. Jan 1. 2010 was Friday .
For every next normal year, 1 day is increased and for every next leap-year, 2 days are increased.
So, total 5 days are increased in 4 yrs, 2008 being a leap-year.
Hence, Sunday + 5 = Friday . Ans.
Ans, Option A (Rs. 480)
[ Average of 524 and 436 is (524 + 436)/2 = 480 ]
Ans. Option E (None of these)
[ Actual answer is 32 mins ]
B alone can fill half-tank in 20 mins. A & B can together fill half-tank in (1/2)/{(1/60) + (1/40)} mins = 12 mins .
Hence, total time = (20 + 12) mins = 32 mins . Ans.
Ans. Option A (Sister)
Bother's father = Father, Grandfather's only son = Father . So, Father is common for both.
Hence, they are siblings .
ATP, Maximum Velocity v = α*t1 or v = β*t2
=> t1 = v/α and t2 = v/β
=> t = t1 + t2 = v(1/α + 1/β) = v*(α+β)/αβ
=> v = αβt/(α+β) . Ans.
Total dist = (850 + 150)m = 1000 m = 1 km
Speed = 45 km/h
Hence, time = 1/45 h = 60/45 mins = 4/3 mins = 1 min 20 secs . Ans.
Ans. The required height = u^2/2g = 28.2*28.2/(2*9.8) m = 40.57 m = 40.6 m (approx)
Applying, H = 1/2*g*t^2 ,
8.52 = 1/2*9.8*t^2
=> t^2 = 2*8.52/9.8 = 1.74
=> t = rt.1.74 s = 1.32 s (approx)
Ans. Required time is 1.32 secs
Max. height attained by the ball above the building = u^2/(2g) = 10^2/(2*10) m = 100/20 m = 5 m .
So, total height while falling to the ground H = (40 + 5) m = 45 m
Thus, fall time t2 = rt.(2H/g) = rt.(90/10) = rt.9 = 3 s
But, rise time t1 = u/g = 10/10 s = 1 s.
Hence, total time to hit the ground t = t1 + t2 = (1 + 3) s = 4 s . Ans.
Ans. Option C (acceleration)
[ acceleration due to gravity (g) remains same for a place ]
Ans. Option C (Frequency)
[ f = 1/T ]
Ans. Option B (associative)
[ A+(B+C) = (A+B)+C ]
Ans. Option D (Decrease)
Ans. Option D (to supply the sidewise (centripetal) acceleration required to make the direction change)
[ prevent from falling ]
Soln:. Total time of flight T = 2u/g = 10 s (given)
So, 2u = 10*10 m/s . => u = 50 m/s .
Hence, max. height attained H = u^2/2g = 2500/20 m = 125 m . Ans.
Soln: Max height attained H = u^2/2g = 100/(2*10) m = 5 m . Ans. i)
Time elapsed to attain H is t = u/g = 10/10 s = 1 s . Ans. ii)
Soln: The work done by the Gravitational Force = Gravitional PE of the body = mgH
Hence, W = 0.002*10*1000 J = 2 J . Ans.
Ans. Option B (acceleration)
[ Acceleration due to gravity (g) is constant ]
Ans. Option B (current)
Ans. When he is suddenly appying the brakes, his body still tends to remain in the state of motion, while the car has come to rest. Hence, he will fall forward and hit his head against the steering wheel, if he is not wearing the seat-belt.
Ans. Power = Work/Time = Potential Energy/Time
=> P = 100*10*10/10 W = 1000 W = 1 kW
Hence, required power of the crane is 1 kW .
Ans. Here, no work is done against gravity, since gravity always acts vertically downwards while the displacement is horizontal.
Ans. The reqd. no. of atoms in 39.4 g gold = 39.4*6.023*10^23/197 = 1.204*10^23 ..
Soln: Relative acceleration = (10 -9) m.s^-2 = 1 m.s^-2 .
Hence, reading on the weighing scale = 50*1 N or 5 kgf . Ans.
Soln: The body having greater mass will have greater momentum. So, it will move a distance before coming to rest, compared to the body with less mass when same retarding force is applied to both of them, as we know force can be defined as the rae of change of momentum.
Soln: Taking, C/5 = (F - 32)/9 and putting C = F,
9C = 5C - 160 . => 4C = -160 . => C= -40
Hence, at 40 degree temperature, both Celsius and Fahrenheit scale will show the same numerical value . Ans.
Ans. The ivory ball will rise to a greater height after striking the floor, since ivory is more elastic than wet-clay .
Let, the 3rd number be 100x.
BTP, 1st number = 50x and 2nd number = 46x .
So, 2nd number is 4x less than 1st number
Hence, 2nd number is 4x/50x*100 = 8% less than the 1st number . Ans.
Ans. c) Oil
{ the rest are dairy products ]
Ans. 50 boys
[ Let, no. of boys be x . So, no. of girls = 70x/100 = 7x/10 . BTP, 17x/100 = 85 . => x = 50 ]
Ans. Option D (white)
[ white is composition of seven diff. colurs, while others are not ]
Ans. First number is 89.6% of the Second number .
[ Taking 3rd number as 100x, 1st no. = 112x and 2nd no. = 125x. So, 1st no./2nd no. = 112/125 = 0.896 ]
Ans. C gets Rs. 250
[ 5/(3+2+5)*500 = 250 ]
Let, their incomes be 9x and 7x, and expenditures be 4y and 3y respectively.
BTP, 9x - 4y = 7x - 3y = 200
=> 27x - 12y = 600 ... (i) , 28x - 12y = 800 ... (ii) .
=> x = 200 , => 9x + 7x = 16x = 16*200 = 3200
Ans. The sum of their daily incomes is Rs. 3200 .
Ans. Option C (NABARD)
Ans. A : D = (8/15)*(5/8)*(4/5) = 4 : 15
Ans. Option D (2000 J)
Initially, PE = mgH = 40*5*10 J = 2000 J . Again total ME is always constant.
So, at 2 m also, total ME = same value as PE = 2000 J .
Ans, Q/R = 7/11 .
Let, the respective proportionals be x and y .
Given: Q/x = R/y and x/y = 7/11
Hence, by alternendo, Q/R = x/y = 7/11 . (Ans.)
Ans. The first mixture and the second mixture should be mixed in the ratio 2:1 .
Let, the qty. of first mixture be x and qty. of second mixture be y.
BTP, (2x + 3y)/(y + 7y) = 7/9 . => Solving, 11x = 22y . => x/y = 2/1
Ans. The diference of the reciprocals is (-28)/93 .
[ 1/x - 1/y = (y - x)/xy = -28/93 ]
Soln: a = 1, n = 2000, d = 2 .
So, l = a + (n - 1)*d = 1 + 1999*2 = 3999 .
Hence, the reqd. no. is 3999 . Ans.
Soln: (a - b) = 3 ... (i) (given) => (a - b)^2 = 9
Again, a^2 + b^2 = 369 => (a - b)^2 + 2.ab = 369 . So, 2.ab = 360
Thus, (a + b)^2 = a^2 + b^2 + 2.ab = 369 + 360 = 729
=> (a + b) = rt.729 = 27
Hence, the reqd. sum of the numbers is 27 . Ans.
Ans. Option D (Cursor)
Ans. Option A (Left-aligned)
10y + x = 2(10x + y) - 4 . => 8y = 19x - 4 ... i)
y = x + 5 ... ii)
=> 8x + 40 = 19x - 4 => 11x = 44 . => x = 4 . So, y = 9 .
Hence, the original no. = 10x + y = 49 . Ans.
Product of the 2 nos. = HCF * LCM = 13*1001 = 13013 .
Let, larger no. be x . So, smaller no. is 7/11x .
BTP, (7/11)x^2 = 13013 . => x^2 = 20449 .
So, x = rt.20449 = 143 .
Hence, the larger no. is 143 Ans.
Ans. Option B (Editing)
Ans. Option D (All of the Above)
Ans. The relation between °C and °F is : C/5 = (F - 32)/9
Ans. Sound can hardly be heard in space, because sound requires any medium to propagate and in space, it is almost vacuum.
Ans. Sound travels fastest in the solid mediums.
Ans. More optically denser the medium, more slower the light will pass through it.
Ans. NH4Cl is neither a base, nor an acid. It's a salt. So it can be treated as neutral. But, NH4+ is a weak basic radical, and Cl- is a strong acidic radical. So, pH of it is <7 indicating bit acidic nature.
Ans. The usual taste of bases is bitter.
Ans. K2CO3 is actually a neutral salt. But, K+ is strong basic, while CO3 -2 is weak acidic So, pH of K2CO3 is > 7 indicating its bit basic nature.
Ans. CH3COOH is a very weak acid.
Ans. Red litmus paper turns blue in a base.
Ans. Noise pollution may lead to permanent deafness, cardiac arrest for old and weak people, chaos in the environment and disturbance in the telemetric signals.
Ans. As per Boyle's law, at a fixed temperature, pressure and volume of a gas are inversely proportional. Again, volume and density of a substance are also inversely proportional. So, in normal conditions, pressure and density are directly proportional. Hence, when the pressure increases, density also increases.
Ans. The gases which strictly obey the Universal Gas Law Equation - PV = n.RT are called 'Ideal Gases'. According to this, ideal gas should have no volume hypothetically at Absolute Zero temperature. But, this is not possible in reality. So, the real gases are different from the ideal gases, though they tend to follow many ideal gas characteristics.
Ans. Yes. Biogas ia a type of biofuel.
Ans. Acceleration actually means the rate of change of velocity. So, it can be given as : a = (v - u)/t ,
(where v = final velocity, u = initial velocity, t = time taken)
Ans. Depending on the direction, a vector can either be positive or negative.
Ans. The SI unit for Acceleration is m/(s^2) .
Ans. Picking up speeds gradually from rest by a train, or a car, or a man are a few examples of acceleration.
Ans. Carbon dioxide (CO2)
Ans. Resource conservation is the process of properly conserving our natural resources for our next generations, before they get exhausted through limiting wastages of energy consumptions, recycling and renewing energies and using non-conventional sources of energies.
Solid Carbon-dioxide (CO2) is called 'dry ice'.
At around -78 degree Celsius, CO2 gas directly freezes to its solid form by the process of deposition.
Dry ice resembles normal ice in appearence, and is also used as a freezing agent. It is named so.
Ans. Option B (Stack)
Ans. Option B (Stack)
Ans. Option D (England)
Ans. Number of people who speak at least one of these two languages is (160 + 130 - 120) = 170 .
There are many formulae to calculate power in electricicity :- 1) Power = Voltage *Current, 2) Power = (Voltage)^2/Resistance, 3) Power = (Current)^2*Resistance .
To calculate power in general : Power = Work/Time or Energy/Time .
There are many formulae to determine power output :- 1) Power = Voltage *Current, 2) Power = (Voltage)^2/Resistance, 3) Power = (Current)^2*Resistance . The SI unit of Power is Watt (W). The larger units are kW, MW, etc.
By conventional notations, (5/2)*((2 + 14d) = 4*(5/2)*(2 + 4d),
Or, 2 + 14d = 8 + 16d , => 2d = -6 , => d = -3
So, 20th term = 1 + 19*(-3) = -56 . Ans.
Ans. A switch fuctions as the 'make/break' operator or simply as the 'on/off' device of an electric circuit.
BTP, in 1 day C can do (1/6 - 1/8) = 1/24 part of the work and A can do (1/6 - 1/12) = 1/12 of the work.
So, in 1 day A and C together can do (1/12+ 1/24) = 1/8 of the work.
Hence, A and B together can complete the work in 8 days. Ans.
Let, the distance be x km.
So, the up time is x/5 hrs and down time is x hrs, while the total distance travelled by the man is 2x km.
Hence, average speed = 2x/(x/5 + x) km/hr = 5/3 km/hr = 1.67 km/hr . Ans.
Let, the present ages of A and B be 7x kg and 2x kg respectively.
BTP, (7x - 2)/2x = 7/6 .
=> x = 12/28 (by solving)
So, 7x = (12/28)*7 = 3
Hence, present weight of A is 3 kg . Ans.
COM = Component Object Model
Valency is the combining capacity of an element or radical.
It is so named, as it actually depends on the number of valence electrons.
Eg. Valency of Sodium is 1, since it can donate 1 electron. Valency of Chlorine is also 1, since it can gain 1 electron.
Optical microscope or in some cases, magnifying glass can be used as an instrument to magnify objects.
Usually, when non-metals combine to form compounds, they do so by the sharing of their valence electrons, and it is called co-valency.
When a body moves in the same direction of the force applied on it, positive displacement is said to be happened.
The benefits of bio-diesel are:
1) It is very much eco-friendly. 2) It can be easily reproduced. 3) The set-up doesn't require much expenses or large land.
In 1 hour or 60 mins, the train covers 90 km.
Hence, in 10 mins it will cover (90/60)*10 km =15 km . Ans.
Bell metal is an alloy of 78% Copper and 22% Tin .
It is used to prepare bells and some musical instruments.
Bronze is an alloy of 88% Copper and 12 % Tin .
It is used to make medals, plates, bangles and several decorative items.
The two major types of friction are : 1) Static and 2) Dynamic or Kinetic. Dynamic friction can be further divided into two types : i) Sliding and ii) Rolling .
Had there been no friction, we could have fallen anywhere, anytime while just moving. Friction actually prevents us from slipping. So, it's a necessary evil.
Solution: Let, the dist. be x km.
BTP, x/5 + x/4 = 9/2 [applying, dist./speed = time]
Or, 9x/20 = 9/2 . => x = 10.
Hence, the reqd. distannce is 10 km. (Ans.)
Ans. The area of the trapezium = 1/2*(3 + 6)*8 sq. m = 36 sq. m .
Solution:
BTP, 5M + 2B = 4(1M + 1B)
Or, 5M + 2B = 4M + 4B . => M = 2B . => M:B = 2:1 .
Hence, the reqd. ratio of efficiency of a man and a boy is 2:1 . (Ans.)
Ans. Option C (NABARD)
Ans. Power Output = Voltage * Current..cos(theta)
Soln: 2(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2)
=> 2(ab + bc + ca) = 0^2 - 20
=> 2(ab + bc + ca) = 0 - 20 = -20
=> ab + bc + ca = -20/2 = -10 Ans.
Soln: The area of the trapezium = 1/2*(3+6)*8 sq.m. = 36 sq.m.
Soln: C/5 = (F - 32)/9
Acceleration is given as the change of velocity w.r.t. time.
The formula is given as : a = (v - u)/t
[ where, a = acceleration, u = initial time, v = final velocity, t = time taken ]
Charles Babbage
By Avagadro's law, 1 Nitrogen mole or 28 g of Nitrogen contains 6.023 x 10^23 no. of molecules.
Nitrogen, being a diatomic element, the no. of atoms = 2 x 6.023 x 10^23 atoms.
So, 1.4 g of Nitrogen contains [(2 x 6.023 x 10^23)/28]x1.4 = 6.023 x 10^22 no. of atoms. (Ans.)
Yes. The data is quite sufficient.
Initially, the time is 9 o'clock. After 6 hours it's 3 o'clock. After further 3 hours, it's 6 o'clock.
Ans. The reqd. next number of the sequence is 20.
24 - 2 = 22, 22 + 3 = 25. So, here we see, initially 2 is decreased and then 3 is increased. Again, 25 - 4 = 21, 21 + 5 = 26. Here we find, 4 is decreased and then 5 is increased. Hence, for the next sequence, 6 must be decreased. It'll become 26 - 6 = 20.
Ans. The wrong number in the sequence is 29.
The other nos. are 1, 4, 256 and 3125. 1 can be written as 1^1, 4 can be written as 2^2, 256 can be written as 4^4, while 3125 can be written a 5^5. So, all the numbers can be represented in the form n^n except 29 .
When we shoot, the bullet goes out of the gun at a very high velocity by the impact of the 'action force'. Now as per Newton's 3rd Law of Motion, the gun in turn recoils making the shooter fall backwards by the impact of the 'reaction force'.
Since, Focal length of the lens = 1/(Power of the lens), here f = 1/0.2 = 5 m.
Hence, none of the 4 options is correct.
Ans: River Jamuna .
(Not Yamuna at all)
Ans. RF stands for Representative Fraction .
1. Visakhapatnam
2. Mumbai
Ans: Since, red colour is scattered the least for having maximum wavelength, it is easily visible from a long distance. Hence, red colour light is used at the traffic signals.
Ans. Option D - Pop (mistaken as Pod)
1. Displacement is the least distance between the initial position and final position of the body.
2. Displacement can be positive, zero or negative.
3. Since displacement possesses both magnitude and direction, it is a vector quantity.
4. Displacement can be either eual to or less than than the actual distance.
Lakshadweep (and Minicoy)
Ans. The molecular weight of CuSO45H2O or the weight of its 6.023x10^23 molecules = 249.6 g
So, the weight of 1x10^22 molecules of CuSO4.5H2O = 249.6/6.023/10 g = 4.14 g (approx)
Ans. The ratio of the atomic masses of Nitrogen and Oxygen is 14:16 = 7:8.
In the given compound, the ratio of the masses of N and O is 28:80 = 7:20 .
Let, x paets of N combine y parts of O in the compound.
So, 7x:8y = 7:20 . => x:y = 2:5 .
Hence, the formula of the compound will be N2O5. (Option C) .
Ans. 6.023x10^23 atoms of Hydrogen weigh 1.008 g.
So, 1 atom of Hydrogen weighs 1.008/(6.023x10^23) g = 0.167x10^(-23) g = 1.67x10^(-27) kg .
Ans. New Delhi
Solution: Let, the speed of boat in still water be x km/hr and speed of streambe y km/hr.
We know, Speed = Distance/Time.
ATP, x + y = 16/2 =8 ... i) x - y = 16/4 = 4 ... ii)
Adding, eqn i) & ii), we get, 2x = 12.
So, x = 12/2 = 6 .
Ans. The speed of the boat in still water is 6 km/hr
Solution: The sum of ages of 25 students = 25x12 years = 300 years.
The sum of ages of 25 students and the teacher = 26x13 years = 338 years.
Hence, the age of the teacher = (338 - 300) years = 38 years. (Ans.)
Solution: P1 = Rs. 30,000. R = 10% p.a., T1 = 2 yrs.
So, A after 2 yrs = Rs. 30000(1 + 10/100)^2 = Rs. 36,300.
Thus, CI after 2 yrs = Rs. (36,300 - 30,000) = Rs. 6,300 .
Now, P2 = Rs. 36,300 , R = 10% p.a. , T2 = 73 days = 1/5 yr.
So, SI after 1/5 yr = Rs. {36300x10x(1/5)}/100 = Rs. 726 .
Hence, total CI after 2 yrs, 73 days is Rs. (6300 + 726) = Rs. 7026 . (Ans.)
Solution : It's an A.P. series.
Here, 1st term a = -15, C.d. d = 9, no. of terms n = 30 .
So, the sum is (30/2)[2x(-15) + (30 - 1)x9] = 15x231 = 3465 . Ans.
Ans The rate of doing work is known as Power.
Mathematically, Power = Work/ Time .
Solution: -2x^2 + 7x - 6 = 0
-(2x^2 - 7x + 6) = 0
2x^2 - 4x - 3x + 6 = 0
2x(x - 2) - 3(x - 2) = 0
(x - 2)(2x - 3) = 0
either, x = 2 or, x = 3/2 Ans.
Solution: 10 mm = 10/1000 m = 1/100 m
We know, P = 1/f .
So, here P = 1/(1/100) D = 100 D Ans.
Solution: MassxPressure/Density = (Mass/Density)xPressure = Volume x(Force/Area)
(AreaxLength) x (Force/Area) = ForcexLength.
Hence, it may be a quantity of Moment of torque or Work done or Energy .
Solution: 15pq + 15 + 9q + 25p
= 15pq + 9q + 25p + 15
= 3q(5p + 3) + 5(5p + 3)
= (5p + 3)(3q + 5) Ans .
Solution:
Let, CP of 1 orange be Re. x
1st case: SP of 36 oranges is Re. 1.
So, SP of 1 orange = Re. 1/36.
Since, Loss = 1 %. SP = 99 % of CP = Re. (99/100)*x.
ATP, (99/100)*x = 1/36
2nd case: Since Gain = 8%, SP = 108 % of CP = Re. (108/100)*x.
Now, (108/100)*x = (1/36)*(100/99)*(108/100) = 1/33 (by simplifying)
Thus, SP of 1 orange = Re. 1/33 .
Hence, he must sell 33 orange at Re. 1 to gain 8% . Ans.
NB: None of the above answers is correct.
Ans. Nizam used to be the monarch of the Hyderabad state, comprising present Telengana province, and many parts of Andhra Pradesh and Karnataka, and the Marathawada region of Maharashtra region. So, the post was very important in terms of administraion of the Deccan territory.
Ans. Fathima Beevi
Gravitational force will become 4 times of the original, when distance is halved. This is because, the force is inversely proportional to the square of the distance.
Multiples:
100 kg = 1 Quintal,
1000 kg = 10 Quintals = 1 Ton = 1 Mega gram
Submultiples :
1 gm = 1/1000 kg
1 mg = 1/1000000 kg
Basically, the protons are very uch stable subject to any chemical changes. It's all about the transfering and sharing of the electrons of the atoms in any chemical reaction. Electrons keep on revolving along the orbits, while the Protons sit pretty static inside the nucleus of the atom along with the neutrons. Only a highly radioactive reaction may lead to the unstability of the protons.
1) The main difference between chloroplast and chromoplast is that chloroplast is the green coloured pigment in plants, while chromoplast is a colourful pigment whose colour may be yellow or orange or even red.
2) Chloroplasts are responsible for undergoing photosynthesis, while chromoplasts synthesize and store pigments.
3) Chloroplasts are usually present in cells of leaves, while chromoplasts are present in fruits, flowers, root and ageing leaves of the plants.
Ans. His speed with the stream will be (6+2) km/hr = 8 km/hr .
Data missing in the problem.
Jerry Ross of USA was the first ever astronaut to achieve this feat.
Ans. Air is a mixture of gases.
Ans. Aluminium oxide or Alumina is formed by the combination reaction between Aluminium and Oxygen with evolution of soe heat energy. Alumina is a completely new entity resulted out of the reaction. It's a compound having its unique physical and chemical properties and no similarities with the Aluminium element. So, the formation of Alumina on Aluminium is a chemical change.
Ans. George Boole first developed Boolean algebra. Actually, the word 'Boolean' is derived from its name itself.
Ans. Multiples - 1 Ton = 1000 kg, 1 Quintal = 100 kg, etc
Sub-Multiples - 1 gram = 1/1000 kg , 1 mg = 1/1000000 kg, etc.
Since we know, KE = 1/2.m.v^2 , so when speed (v) is halved, the KE of the body becomes 1/4 th and while speed is tripled, the KE becomes 9 times .
Arti Pradhan, from Mumbai, India at the age of 15+ on 6th Aug, 1987 was the 1st youngest Asian woman to cross the English channel.
Ans. Total distance = 60 km.
Total time = 70 mins = 7/6 hrs
So, average velocity = 60/(7/6) km/hr = = 51.43 km/hr (approx.)
Ans. Since, Physics exists in our every walk of life with its pros n cons, it's called the exact science.
1323 = 3x3x3x7x7/
So, there is short of a '7' to form a trio.
Hence, 1323 must be multiplied by 7, so that the product is a perfect cube.
Total cards in the pack = 52.
Total black cards = 26.
Total no. of Jacks = 4. But out of them 2 are black. So, effective no. of Jacks to be considered = 4 - 2 = 2.
Hence, probability of getting a black card or a Jack = (26 +2)/52 = 28/52 = 7/13 . (Ans.)
Part of men in the commitee = 5/11 and that of women = 6/11.
New part of men = 5/11 + 20% of 5/11 = 1.2*5/11= 6/11 .
New part of women = 6/11 +10% of 6/11 = 1.1*6/11 = 6/10 .
So, new ratio of men and women = (6/11)/(6/10) = 10:11 . (Ans.)
Ans :- Option A - Atom
Ans :- Option C - Electrons
Ans :- Option A - Oxidation
Mg(CO3)2 is more insoluble in water than Mg(OH)2. Actually the carbonate salts of almost all Grp II elements exhibit this property.
Silicon can be turned to red hot since it is a poor conductor of heat.
So, to me the question is incorrect. It might be Iron instead of Silicon.
When steam is passed over red hot Iron, tri-ferric tetra-oxide is formed along with the liberation of Hydrogen.
Reaction: 3Fe + 4H2O = Fe3O4 + H2
Copper (II) oxide and Carbon monoxide (CO) react together to liberate Carbon-dioxide gas and Copper deposits.
Reaction: CuO + CO = Cu + CO2
It's a redox reaction, where CuO acts as the oxidant while CO acts as a reductant.
The amount of force required to accelerate an object of mass 1 kg by 1 m/ second square is called 1 newton (N).
The weight of a body is given as the product of its mass and earth's acceleration due to gravity. [ W = m.g ]
Simply, it is the amount of gravitational force by which a body is attracted towards the centre of the earth.
Unit of G in SI is m³.kg^(-1).s^(-2)
Black colourisa very good absorber of heat. So, in summer black clothes absorb all the heat from the sun directly and reflect out nothing which make us feel even more hot if we wear them. Thus, in summer white or light-coloured clothes are srictly recommended to wear instead of black clothes.
Since, Power P = -2 D, Focal length f = 100/(-2) = - 50 cm.
The minus (-) signifies that the lens should be Concave .
Chlorophyll is found in the thylakoid sacs of chloroplasts of the leaves of a plant. It is that pigment which is responsible for the green colour of te leaves.
We find, Zinc is placed higher than Copper in the Electro-chemical Series, since Zinc is more reactive than Copper. Now, the reason is that Zinc loses its valence electrons more easily than Copper to take part in a reaction.
The chief components of Bronze are Copper (around 88%) and Tin (around 12%). It may too contain Manganese, or Nickel, or Aluminium, or Zinc, or Silicon in traces.
NaCl (Sodium chloride) being a normal salt is always neutral in its pure state, and neither acidic nor alkaline.
Fog is not an example for sol, rather it's an aerosol .
Let, the number be x.
By the prob, x*(25/100)x = x + 2x .
Or, (x^2)/4 = 3x . Or, x = 12 .
Hence, right answer is Option 1 : 12
LCM of 2, 3 and 5 = 30.
So, each edge of cube has to be 30 cm.
So, its volume = 27000 cc .
Hence, no. of cuboids reqd. to make such a cube = 27000/ (2*3*5) = 900 . (Ans.)
Let, the no. be x.
25% of it = x/4 .
200% more of it = x + 2x = 3x.
BTP, x*x/4 = 3x
Thus, x = 12 .
So, The reqd. no. is 12 --- (Option 1)
The most 2 popular sub-multiples of the unit kilogram are as following :
gram (1 g = 1/1000 kg) and milligram (1 mg = 1/1000000 kg)
An amalgam always must comprise Mercury alongwith any metal(s).
Pickles and jams are acidic in nature. If they are kept in metal containers of Aluminium or Tin, they will slowly corrode the metal surface due to their toxicity. Moreover, Aluminium or Tin will react with the acid(s) contained in the pickles and jam to form salts, which may be harmful for us when we consume the same. Hence, pickles and jams should not be stored in the metal containers.
The chief ore of Aluminium is Bauxite. The other ores are Cryolite and Diaspore.
Ans. Option D - 500 .
Explanation: The compound should contain at least 1 oxygen atom. Now, Atomic mass of Oxygen = 16.
So here, 3.2% of the compound = 16.
Thus, 100% of the compound = (16/3.2)*100 = 500 .
1 N force may be defined as the force that is required to change the velocity of a body of mass 1 kg by 1 m/s in a time period of 1 second.
194 gm caffine contains 28.9% of Nitrogen = 56 gm (approx.) of Nitrogen.
Now, Atomic mass of Nitrogen = 14 gm.
So, in 56 gm of Nitrogen, there are 56/14 = 4 Nitrogen atoms.
Hence, in 1 molecule of caffine, there are 4 Nitrogen atoms. Ans. - Option C
Load is one type of specific Force that always acts downwards.
1439/16 = 89.9375
89.9375x14.99 = 1348.1631
root228 = 15.0997
So, 1348.16 + 15.1 = 1363.26 = 1365 (approx) - Option B
Dopant to convert an intrinsic semi-conductor to a) an n-type semi-conductor is Phosphorus ; b) an p-type semi-conductor is Aluminium .
23 + x = 345
So, x = 345 -23 = 322 . Ans.
Let, the no. of books be x.
So, each book costs Rs. 60/x.
BTP, (x + 5)(60/x - 1) = 60 .
=> 60 - x + 300/x - 5 = 60 => x^2 + 5x - 300 = 0 .
Solving the eqn, we get, x = -20 or x = 15 .
Since, no. of books can't be -ve, the only possible value is 15 .
LCM of 2, 3, 5 = 30.
So, each side of cube = 30 cm. So, its volume = 30x30x30 cm^3.
Thus, no. of cuboids reqd. = 30x30x30/(2x3x5) = 900 .
Any liquid that evaporates at room temperature when kept open may be called a volatile liquid. Eg. Alcohol, Petrol
Water freezes to ice. The density of ice is less than that of water. We know, density is inversely proportional to volume. So, on freezing the volume of water does not decrease at all, rather increases .
Area of a regular octagon of side a = 2a^2(root2 + 1)
The number of proper subsets = 2^3 - 2 = 8 - 2 = 6 .
Sums of ratios = 2 + 3 + 5 = 10 . Total amount = Rs. 1260 .
So, C receives = 5/10 of Rs. 1260 = Rs. 630 .
The avg. ages of 10 persons is 60 yrs.
So, sum of ages of 10 persons = 10x60 yrs = 600 yrs.
Now, Sum of ages of 8 persons is 488 yrs.
Thus, sum of ages of those 2 persons = (600 - 488) yrs = 112 yrs.
Hence, the avg. of their ages = 112/2 yrs = 56 yrs. (Ans.)
CP = 4/5*Rs.53500 = 80/100*Rs.53500.
SP = 106/100*Rs.53500.
Gain % = (SP - CP)/CP*100% = 26/80*100 % = 32.5% Ans. (Option C)
MP = 140% of CP.
D = 20% of MP
So, SP = 80% of MP = (80/100)*(140/100)*CP = 112/100*CP = 112% of CP.
Hence, Profit % = (112 - 100)% = 12% . (Ans.)
Nitrogen and Hydrogen are the two non-metals. In water, they form the compound Ammonium hydroxide (NH4OH) which is an alkali.
We know, SI = PxRXT/100 .
But here, Si = P and T = 10 yrs.
So, 10R/100 = 1.
Hence, R = 100/10 % = 10% p.a.
Ans. Rate of interest = 10% p.a.
Area of a square = 1/2 * (diagonal)^2
So, here area = 1/2*2.5*2.5 m^2 = 3.125 m^2 . Ans.
Using the formula, (a-b)^2 = (a+b)^2 - 4ab, here we get,
(tanA-cotA)^2 = 2^2 - 4x1 = 4 - 4 = 0.
So, (tanA - cot A) = 0.
Now, tan^2 A - cot^2 A = (tanA + cotA).(tanA - cotA) = 2x0 = 0 .
Ans. 0
When Aluminium (Al) is exposed to air, Aluminium oxide or Alumina (Al2O3) is formed. Silver white Aluminium turns greyish in appearance.
Chemical reaction : 4 Al + 3 O2 = 2 Al2O3
Time period (T) is given as the reciprocal of frequency (f). So here, T = 1/10 s = 0.1 s . Ans.
Velocity is defined as the rate of displacement. Mathematical expression : v = s/t .
Power = Voltage*Curent
But, Voltage = Current*Resistance Or, Current = Voltage/Resistance
So, Power can be written as Voltage^Voltage/Resistance = (Voltage)^2/Resistance [P = V^2/R ]
Again, Power can be given as Current*Resistance*Current = (Current)^2*Resistance [ P = I^2.R ]
Fluorine (F) is the only halogen that always shows oxidation state -1 .
a) Loudness of the sound depends on its amplitude. Loudness varies as square of the amplitude.
b) Pitch of the sound depends on its frequency. Higher the frequency, higher is the pitch and lower the frequency, lower is the pitch..
c) Quality or timbre of the sound depends on its waveform.
Gujarat
Ans. m = 5
Solution: 5^m*5^(-2) = 5^3
So, 5^m = 5^(3+2)
=> 5^m = 5^5 .
=> m = 5
Since, x is in 'm', t is in 's',
at is in 'm' . => a is in 'm/s' .
Again, bt^2 is in 'm' . => b is in 'm/(s^2)' .
So, (a/b) will be in (m/s)/[m/(s^2)] = s (same as Time) .
Hence, unit of a/b is sec or s and dimension of a/b is [T] .
Here, the angle of incidence i = (90 - 30)° = 60°. So, the angle of refraction r = (60 - 15)° = 45° .
Now, the refractive indexof the medium = sin i / sin r = sin 60°/ sin 45° = (√3/2)/(1/√2) = √6 = 2.45 (approx.) [Ans.]
Lesser the no. of animals, more will be the no. of days of food for them.
Clearly, it's the case of Inverse Proportion.
So, the reqd. no. of days = (30x4)/20 = 6 . (Ans.).
The 4 characteristics of Distance are :
1) It's the total path length covered by an object.
2) It's a Scalar quantity, since it has magnitude only.
3) It's always positive and can never be negative or 0.
4) It's dimensional formula is L^1.M^0.T^0 .
The 2 common applications of Archimedes Principle are :
1) Determination of purity of any substance, like Gold, on the basis of its actual Density.
2) Building Ships on the basis of making Average Density of the ship less than the density of water.
The ionic bond or the electrovalent bond is the strongest chemical bond formed by the transfer of valence electrons. Sodium chloride (NaCl) is a classical example of the compound formed by ionic bond with Na+ and Cl- ions. Here, we see, a highly electropositive metal reacts with a highly electonegative non-metal. Hence, the bond is so strong.
The 23° 27' N latitude better known as Tropic of Cancer passes through India.
The valency of Carbonate radical is -2 .
The amount of heat energy required to completely convert 1 kg of a liquid to its vapour (gaseous) form without any rise in temperature is called the (latent) Heat of Evaporation.
We can measure the distance travelled by a car with the help of an instrument called Odometer .
The power of lens is defined as the reciprocal of its focal length in metres, Mathematically, D = 1/f, (where D is the power in diopters and f is the focal length in metres) .
The unit of resistivity or specific resistance is ohm-meter, which is written as Ω⋅m .
The negative of any real number is actually its additive inverse .
For a positive real number, its negative will be just the same magnitude with the minus ( - ) sign. Similarly, for a negative real number, its negative will be just the same magnitude with the plus ( + ) sign. So in these cases, always the sign just flips keeping the magnitude or value of the number exactly the same.
Option B, Hinduism
Valency of Zinc (Zn) is 2 and that of Chlorine (Cl) is 1 .
So, after their combination, by criss-cross method, the formula of Zinc chloride is ZnCl₂ .
The Sun is the actual source from which the Earth receives ultraviolet (UV) radiations.More the Ozone layer depletion due to CFC, more the UV radiations received by the Earth.
Classical examples of having both physical and chemical changes at the same time are : i) Burning of candle, ii) Boiling of milk.
i) Burning of candle - The molten wax can be recovered after cooling. So it's a physical change. Carbon-dioxide and water vapours get liberated while burning. So it's a chemical change.
ii) Boiling of milk - No new substance is formed actually. So it's a physical change. The milk loses some of its water content while boiling. So it's a chemical change.
Ans. Option B) to decrease the pressure on the ground
Reason: Caterpillar tracks make greater surface area of contact with the ground. We know, greater the area of contact, lower the pressure exerted.
The branch of science that deals with the appearance, structure, composition, nature and uses of all the substances of matter, as well as, their properties and conditions for reactions among them is called Chemistry.
The Battle of Waterloo
The statement means 19 gms of CuSO4 completely dissolves in 100 gms of water at 15 ⁰C temperature.
Gas diffuses more than solid. Gas molecules are very loosely bonded. So they possess high kinetic energy and can move freely. Thus we can smell a good perfume or flavour while cooking of food from a distance. On the other hand, solid molecules hardly exhibit any movement being closely packed.
Upon frosting, water is transformed to ice. It's just a change of state of the same substance - from liquid to solid. So it's only a physical change.
Biosphere - since here we have to choose between the two options.
Actually it should be Oxygen (gas)
CP of the article = SP + Loss = Rs. (300 + 60) = Rs. 360 .
So to make 15% profit, reqd. SP = Rs. [(100 + 15)/100]x360 = Rs. 115x360/100 = Rs. 414 .
Let the sum of money be Rs. P .
ATQ, P(1 + 10/100)^3 = 6655 . => P = 6655x10x10x10/(11x11x11) = 5000 .
Hence, the reqd. sum of money is Rs. 5000 .
Option B : Specific Gravity
Reason : It is the ratio of 2 same quantity - Density. Actually, it's also called Relative Density.
Two multiples of kilogram are ton and quintal. Although there are many more.
1 ton = 1000 kg . 1 quintal = 100 kg .
Given: 1/f1 = 2d , 1/f2 = -4d .
So, combined power = 1/f = [2 + (-4)] d = -2 d (dioptre).
Thus, combined focal length = 1/(-2) = - 0.5 m = - 50 cm or 50 cm on the same side of the lens as the object.
Instantaneous acceleration can be expressed as the double derivative of displacement .
Yes. Apache Web Server is absolutely FREE !
Taking the respective times for individual taps as t1 hrs & t2 hrs and the time taken by two together as T hrs,
T = 1/[(1/t1) + (1/t2)] . <---- Formula
Hence, here the reqd. time = 1/[1/10 + 1/15] = 1/[5/30] = 30/5 = 6 hrs (Option b).
IISCO stands for The Indian Iron and Steel COmpany .
Siachen glacier of Ladakh (UT) is the highest battlefield of India.(above 6000 metres from the sea-level).
The atoms of the all the elements belonging to Group 16 (VIA) of the periodic table viz. Oxygen (O), Sulphur (S), Selenium (Se), etc contain 6 electrons in their outermost shells.
1 mole of H20 = Avogadro's number i.e. 6.023x10^23 number of molecules = 18 gms in mass .
Power of lens P = -40 D
So, Focal length f = 100/40 cm = 2.5 cm on the same of lens as the object .
Since, power is negative, it's obviously a concave lens .
Thermosetting Plastic, or simply thermoset, is a rigid polymeric material that is resistant to higher temperatures than ordinary thermoplastics. It can never be remould as it is so hardened by curing from a soft solid or viscous liquid prepolymer or resin resulting extensive cross-linking between its polymer chains to produce an infusible and insoluble polymer network. Examples of thermosetting platics are melamine, bakelite, vulcanized rubber, epoxy resin, etc.
We know, Focal length = 1/Power
So here, Focal length = 1/(-2.00) = - 0.5 m = - 50 cm .
The '-' sign indicates that it's a concave lens .
Density of an object depends on its mass and the space or volume occupied by it. So mathematically, Density is expressed as D = M/V. or D = M^1.L^(-3) .
Here, mass (m) in two cases is supposed to be equal, taking the same object in consideration.
So, f1 = m.a1 and f2 = m.a2 (from Newton's 2nd law)
Now, (m.a1)xa2 = (m.a2)xa1
Hence, f1.a2 = f2.a1 . (Proved)
The set of all odd integers may be defined in the form {Odd Z = 2n - 1, n ∈ Z] .
Translatory motion is the type of motion in which all the particles of an object move through the same distance in the same time period. 3 Examples of the translatory motion are : 1) Rectiliniear - a man moving along a straight path, 2) Curvilinear - a car moving along a circular track, 3) Random - moving of flies or mosquitoes.
The main mechanism of the battery is that it converts the chemical energy to electrical energy. Now, the chemical energy is stored in it either in alkaline or acidic form, acting as electrolyte. When the battery terminals are connected to any unit to be powered, positive charges are generated from the battery chemical.
Applying Pythagoras theorem, QR= root(PR^2 - PQ^2) = root(13^2 - 12^2) cm = root(25) cm = 5 cm .
Hence, area of triangle PQR = (1/2).PQ.QR = (1/2).12.5 cm^2 = 30 cm^2 . (Ans.)
Required time = (675/5)*1890 hrs = 14 hrs .
x^2 - (13/24).x - 1/12
= (1/24).(24.x^2 - 13x - 2)
= (1/24).(24.x^2 + 3x - 16x - 2)
= (1/24).{3x(8x + 1) - 2(8x + 1)}
= (1/24).(3x - 2).(8x + 1) Ans.
=> Option (A)
√(2.108304) = √(2108304/1000000) = 1452/1000 = 1.452
Ans. Square root of 2.108304 = 1.452
(2+2)/(2x2 - 2) = 4/(4 - 2) = 4/2 = 2
Some main multiples of gram (g) are decagram (dag = 10 g), hectogram (hg = 100 g), kilogram (kg = 1000 g) .
Let, the sum of money be Rs. 100x.
Since, the rate of interest is 10%, after 1st yr the sum is compounded to Rs. 110x .
After 2nd yr, the sum becomes Rs. 121x and after 3rd yr, the final amount becomes Rs. 133.1x .
ATP, 133.1 x = 6655 . So, x = 6655/133.1 = 50 .
Thus, 100x = 5000 .
Hence, the required sum of money is Rs. 5000 . Ans.
Two multiple units of kg are quintal (= 100 kg) and metric ton (=1000 kg) .
The region of the shadow of partial light or darkness is caled the Penumbra .
An Inductor stores energy in the form of a magnetic field.
Main function of a leaf is to conduct photosynthesis.
Taking North-facing as positive direction, first that Prerna faced south and moved right should be negative. So, actually she went 10 m North and (60 - 20) m = 40 m East. So, ultimately her position from the starting point is root(10^2 + 40^2) = root(1700) = 10.root17 m due North-east.
The principle of moments for a lever states that at equilibrium of the lever about its fulcrum, the sum of all anti-clockwise moments must be equal to the sum of all clockwise moments.
Hooke's Law of Elasticity states that within elastic limit, stress acting on a body is directly proportional to the strain produced.
Overall, the ionic bond or the electrovalent bond is the strongest bond.
Force is an external agent that changes or tends to change the state of rest or of uniform motion of a body or its shape and size. Mathematically, Force is given as the product of Mass of the body and its Acceleration => F = m.a .
Solution:
The given exp. can be written as : 3^x.(1 - 1/9) = 72 . => 3^x.8/9 = 72 => 3^x = 81 = 3^4
Hence, x = 4 . (Ans.)
The physical process by which a substance transforms directly from its solid state to its gaseous state without passing through a liquid state is called Sublimation.
An everyday substance that exhibits sublimation is napthalene.
The process of inserting an element in a stack is called Push .
The loudness of sound depends upon (directly varies as) the amplitude of the sound wave, while the pitch of sound depends upon (directly varies as) the frequency of the sound wave. The loudness actually determines the intensity (high or low) of the sound, whereas the pitch determines how much shrill is there in the sound wave.
The full form of LIFO is " Last In First Out "
The flying of kite in air is an example of Random or Irregular motion.
Quite a few non-metals react with hydrogen forming covalent bonded compounds. Some well-known examples are : Nitrogen and Hydrogen react to form Ammonia (NH3), Carbon and Hydrogen react to form Methane (CH4), Chlorine and Hydrogen react to form Hydrogen chloride (HCl),Sulphur reacts with Hydrogen to form Hydrogen sulphide (H2S), Phosphorus reacts with Hydrogen to form Phosphine (PH3), and Oxygen forms Water (H2O) reacting with Hydrogen.
Following the other 2 patterns : -
? = 4x5 + 1 - 9 = 12.
Ans. 12
Given, s = 10 m, t = 2 sec, u = 0, m = 2 kg
Now, s = ut + 1/2.a.t^2
So, 10 = 1/2.a.4 = 2a
So, a = 10/2 = 5 m/(s^2)
Again, F = m.a
Hence, F = 2x5 N = 10 N ... (Ans)
Since, high temperature is required for the reaction, it's an Endothermic reaction. It involves the combination of Nitrogen and Hydrogen, so it's a Combination reaction too. Again, it's a complete gaseous reaction, where all the reactants and products are gases at NTP.
The balanced equation of the reaction is : N2 (g) + 3H2 (g) = 2NH3 (g) .
Zero (0) group was in practice in old days' Periodic Table. In modern periodic table, it's depicted as Group 18. It's actually the group of all inert gases or noble gases viz. Helium (He), Neon (Ne), Argon (Ar), etc. They hardly react with any other element of the table, since they are chemically inert having valency zero (0).
344 m/s actually signifies the velocity of sound in air. Sound can travel appoximately 344 m in 1 sec in air.
When gas bubbles (or fizz) are found with a hissing sound, the phenomenon is called 'Effervescence'. It is observed when Carbon dioxide (CO2) gas evolves during a chemical reaction.
Both Ammonium (NH4) and Nitrate (NO3) ions have the same valency i.e. 1. So, by criss-cross method, the molecular formula of Ammonium nitrate is NH4NO3 .
Let, their present ages be x yrs & 2x yrs respectively.
So, the sum of their ages = 3x yrs.
18 yrs ago, their ages were (x-18) yrs & (2x - 18) yrs respectively.
ATP, 2x - 18 = 3.(x - 18)
=> 3x - 2x = 54 - 18 => x = 36
Thus, 3x = 3x18 = 108.
Hence, the sum of the ages of the man and his son is 108 years. (Ans.)
We know, W = m.g
So here, the weight of the student on earth = 78x9.8 N = 764.4 N (Ans.)
The seat of memory and intelligence in human brain is the cerebrum .
Electrostatic potential or simply Electric potential is the amount of work done to move a unit charge from a reference point to a specific point against an electric field. Its SI unit is volt (V) .
Ans. Option A - BILE
Since, the formula of its chloride is MCl2, valency of M = 2. So, formula of its oxide is MO.
Since, weight of oxygen is 20% in MO, weight of M in MO = (100 - 20)% = 80%,.
Thus atomic mass of M is 4 times more than that of oxygen = 4x16 u = 64 u - Option (d) Ans.
We know, 35.5 g of Chlorine combine with 23 g of Sodium to form 58.5 g of Sodium chloride (NaCl) .
So, 710 g of Chlorine will combine with 23x710/35.5 = 460 g of Sodium to form (710+460) g = 1170 g of NaCl .
Thus, 1170 g of Sodium chloride contains 710 g of chloride atoms. Ans.
After 1st replacement, in 60 L mixture, milk = 54 L, water = 6 L.
So, 6 L of 1st mixture contains 5.4 L of milk and 0.5 L of water.
So after 2nd replacement, milk = (54 - 5.4) L = 48.6 L , water = (6 - 0.6 + 6) L = 11.4 L .
Hence, the final ratio of milk and water in the resulting mixture = 48.6:11.4 = 81:19 . (Ans)
By the formula, a = v^2/r , the angular acceleration of the object = (4^2)/0.25 = 64 m/(s^2) . (Ans.)
Since, A = 2P, SI = A - P = 2P - P = P.
Now, T = 5 yrs, R = ?
ATP, P = P.R.5/100 . => R = 100/5 % = 20 % .p.a.
So, the reqd. rate of interest per annum is 20 % . (Ans.)
Cell membrane is made up of lipids and proteins .
Non-linear data structure
Camphor
The main good sides of the Friction are : i) It helps us in walking on ithe ground properly without slipping, ii) It helps us keep our palms warm during winter on rubbing, iii) It enables the car come to rest slowly by applying brakes.
Kanha Kisli, or popularly known as Kanha Tiger Reserve is in the central part of India in the state of Madhya Pradesh encompassing an area of about 940 km^2 in the two districts Mandla and Balaghat.
All other gates can be designed by using NAND gate and NOR gate in different combinations. So, we can get any desired output with NAND or NOR gate, even if we don't have the respective gate. That's why, NAND and NOR gates are regarded as Universal gates.
Paramoecium is a unicellular organism .
Since, 5.4 cm = 405 units,
1 cm = 405/5.4 units = 75 units.
Ans. k = 75 uniis
In 25 mins, a truck travels 14 km.
So, in 5 hrs = 300 mins, the truck travels 300x14/25 km = 168 km. (Ans.)
Ampere's swimming rule states that, "if a man swims along the current-carrying wire, such that his face is always towards the magnetic needle through which the current is entering his feet and emerging from his head, then the north pole of the magnetic needle will always be deflected towards his left hand."
This rule is actually used to determine the direction of the magnetic field around a current carrying conductor.
The displacement varies as square of time for a freely falling body. Here, the 'g' remaining same, the ratio of the displacements of the body after 1, 2 and 3 seconds will be 1^2:2^2:3^2 = 1:4:9 . So, option A - 1:4:9 is correct
Hardness is the physical property of a material that enables it to resist plastic deformation, penetration, indentation or scratching. The SI unit of hardness is N/mm².
Mercury is not tilted on its axis.
One-fourth of the perimeter of a square gives the length of its each side.
Since the man is running in opposite direction to the train, the relative speed = (120 + 10) km/hr = 130 km/hr.
So, to cover the distance of 150 m or 0.15m, the man takes 0.15/130 = 0.001154 hr = 0,001154x360 secs = 4.1544 secs.
Hence, the reqd. time taken is 4.1544 secs .
Let. the 1st number be x.
So the 2nd number = x/2 and 3rd number = x/3.
BTP, x + x/2 + x/3 = 121*3 . => 11x/6 = 121*3 . => x = 22 .
Now, 1st number - 3rd number = x - x/3 = 2/3*x = 2/3*22 = 44/3 (Ans.)
60% of the wall is unpainted. So, 40% is painted. Thus, Umar actually painted (40 - 25)% = 15% of the wall.
BTP, 15% = 30 sq. m . => 25% = (30/15)*25 sq. m = 50 sq. m
Hence, Anil painted 50 sq. m of the wall . Ans.
Option B : External
Ans. Option a) f(x).g(x)
Reason: f(x).g(x) is the only polynomial given, which already contains the factor (x-2).
White light is composed of 7 different colurs and when it enters a glass prism it splits into those seven constituents by the phenomenon called 'dispersion'. This is due to the fact, these seven colours having different wavelengths scatter differently by refraction. Red (R) has longest wavelength and bends to the top by least refraction, while violet (V) has shortest wavelength and refracts the most right to the bottom thus forming the colour spectrum VIBGYOR in the bottom-to-top order.
Ans. Option E) 25 m/s
Explanation :Let. the distance covered by 2nd train in 20 secs be x m.
In 20 secs, the 1st train coversits own length + the 2nd train's length + the 2nd train's covered distance + the distance ahead
ATP, 120 + 80 + x + 100 = 40*20 . => x + 300 = 800 . => x = 500
Hence, speed of the 2nd train = 500/20 m/s = 25 m/s .
Correct option - owns
1) I am going to marry an heiress.
2) The report was accepted by the unanimous vote.
Sentence : Coming events cast their shadows before.
Finite verb : cast
ATP, SI = P in 3 yrs.
So, P = (P.R.3)/100 . => R = 100/3% = 33.33% .
Ans. ROI = 33.33% p.a.
The chromosome acts a bearer of heredity units and thus coined as heredity vehicle.
ATP, (-4)^2 - (-4) - 2(k+1) = 0
=> 16 + 4 = 2(k+1) => 2(k+1) = 20 => k+1 = 10
=> k = 9 ... Ans.
ZnO is amphoteric in nature since it reacts with both alkali and acid producing salt and water.
i) ZnO + 2NaOH = Na2ZnO2 (Sodium zincate) + H2O ... (where NaOH is an alkali)
ii) ZnO + 2HCl = ZnCl2 (Zinc chloride) + H2O ... (where HCl is an acid)
Al being higher in the reactivity series than Zn reacts more with the corrosive components and hence Al gets corroded more easily than Zn.
An electrovalent bond or ionic bond is formed by the transfer of valence electron(s) usually between the atoms of a metal and a non-metal, while a covalent bond is formed by the sharing of valence electrons usually between the atoms of two non-metals.
Ahmed usually STAYS at the apartment, but this time he IS STAYING at the city centre.
Biogas is a type of biofuel naturally produced from the decomposition of organic matter, like food scraps and animal waste, in the absence of oxygen. It is a renewable source of energy comprising of mostly Methane (about 75%) and carbon dioxide too.
A) Dhaka
A cell functions as the main power supply in an electric circuit. It is a device, which converts chemical energy into electrical energy.and when switched on provides the same to all the circuit components.
Option C : root5 + 3 is an irrational number. Root5 itself is irrational. When added to a rational number 3, the result is irrational. The other options A, B nd D are all rational numbers.
The last ten 4-digit numbers are 9990 to 9999.
Alcohol is a liquid that can diffuse rapidly in water.
The SI unit of Surface Tension is Newton/metre (Option C)
[Since, Surface tension is given as the force per unit length]
Power of lens = 2 .
So, Focal length f = 100/2 cm = 50 cm.
Since, f is positive, the lens is Convex.
Since frequency is given as the reciprocal of time period, here frequency = 1/0.002 Hz = 1000/2 Hz = 500 Hz . Ans,
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