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A = P { 1 - r%}^n where A is the depreciated value after n years at the depreciation rate of r% , So , A = 20,00,000 { 1 - 20/100}^3 = 20,00,000 {4/5}^3 = 20,00,000 x 4/5x4/5x4/5 = 10,24,000 Depreciated value after 3 years will be Rs 10,24,000
The ball rebounces only 80% of orginal height i.e. it is going 20% less height, so by using formula A = P { 1 - r%}^n where A is final height after third bounce, P = original height achived, r% = rate of loss in height & n = number of bounces A = 20 {1 - 20/100}^3 = 20 x 4/5x4/5x4/5 = 10.24 m so after third bounce it will achieve a height of 10.24 m
CI=P[{1+R/100}^2-1]=9650[{1+8.3/100}^2-1]=1668.378=1668.38
Simple interest (SI)=(time×rate×principle)/100 Thus principle=(1240×100)/3×6 =6880
Using the formula for compound interest, CI = P { (1+r%)^n - 1 } where CCI = compond interest, P = princcipal, r% = rate, n = time in years. Putting the values 210 = P { (1 + 4%)^2 - 1 } = P { 1.0816 - 1 } = P x 0.0816 or P = 210/0.0816 = 2573.53 so principal = Rs 2573.53 Now SI = {PXrxt}/100 = { 2573.53 x 8 x 2 }/100 = 411.76 so simple interest will be Rs 411.76
A =P(1+%r)^n. Hence 1296=1089(1+%r )^2 i.e.sq.rt(1296/1089)= 36/33=12/11=1+%r, hence r= 100/11= 9.09%.
Amount/Principal =1323/1200 =(1+%r)^2. Hence sq.root(1323/1200)=1.05,which makes interest rate as 5%.
principal P (1+.05)^2 =1323 amount .Hence P= Rs.1200.
Let length and breadth of the rectangle be l and b. Given: l = 1.5b ------ (1) Perimeter of a rectangle P = 2(l+b) Using equation (1), P = 2(1.5b + b) = 2(2.5 b) P = 5b ------------ (2) Rs.60 - 1m can be fenced Rs.12,000 - (1/60)*12000 = 200meters So, perimeter of the rectangle is 200 meters ------------ (3) From (2), P = 5b Hence, 5b = 200 b = 40 m l = 1.5b = 1.5 * 40 = 60 m Length = 60m Breadth=40m
C.P= 1800 Sales tax= 8% He has to pay, 1800+1800*8/100= 1944Rs But he wants to pay 1800Rs(Including sales Tax) Let new C.P =x x + x*8/100=1800 X=1666.66 (New C.P after reduction) Reduction Percentage= (1800-1666.66)/1800*100= 7.40%
CP = (108/100)*(98/100)*(97/100)*(94/100)*Rs.3,500 = Rs. 3,377.67.
Let the distance be 4km. Time taken to cover 1/4th distance i.e 1/4*4km i.e 1km=1/10 hr. Time taken to cover next 1/4th distance i.e 1/4*4km i.e 1km=1/5 hr. Time taken to cover last 1/2 distance i.e 1/2*4km i.e 2km=2/10 hr. Total time taken=1/10+1/5+2/10=5/10=1/2hr speed=distance/time=4km/(1/2hr)=8km/hr
After allowing 10% discount price of vase = 1500-(10/100 *1500) = 1350 Now, the shopkeeper earns a profit of 25% on selling it at 1350 Cost price = 1350 * (100/125) = 1080
Let the distance be 40km. Time taken to cover 1/4th i.e 10km=10km/10kmph=1hour
Time taken to cover next 1/4th i.e 10km=10km/5kmph=2hour. Time taken to cover remaing 1/2 i.e 20km = 20km/10kmph=2hour.
Average speed =total distance/total time=40/(1+2+2)=40/5 = 8 kmph.
Amount=P* (1+.05)^2 =1323 amount .Hence P= Rs.1200.
let assume "x" meter be the trains length
When it pass an electric pole it only cover its own length, whereas when it cross a platform it has to cover distance equal to sum of its length and platform's length.
Then assuming it is moving in same speed we get
x/18 = (x+100)/30
or, 30x = 18x + 1800
or 12x = 1800
so x = 150
So the train's length is 150 meter.
As trains are moving in opposite direction
So combined speed of the two trains = (50+40) km/hr = 90 km/hr = 25 m/s
Let length of train B=x metre
Total length of two trains = (400+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 .
Simple interest for 3 years=225. So interest for one year=225/3=75
When it is compounded, Interest for first year+Interest for second year=153
75+Interest for second year=153. Interest for second year=153-75=78
Difference interest for first and second year=78-75=3
This Rs3 extra interest is of Rs75(Interset of first year) for one year.
Rate of interest=3*100/75=4%
Now using the simple interest of one year
Principal=75*100/4=1875
Ans--Principal=Rs1875
Let, the distance be x km.
Time taken to reach office if speed is 40kmph=x/40 hr
Time taken to reach office if speed is 50kmph=x/50 hr
Since he reaches 10 min late if speed is 40kmph and 5 min early if speed is 50kmph. The difference in time is 15min
15 min=15/60 hr =1/4 hr.
Now x/40-x/50=1/4
=> x/200 = 1/4 or x=200/4 =50km.
Ans. The required distance is 50 km .
Present age of Pravash=10+4=14 years
Present age of Rajat=3*Pravash age=3*14=42 years
Present age of Basu=42-5=37 years
Basu age after 6 years=37+6=43 years.
After arranging letters of word TIGER in ascending order alphabetically, it will become EGIRT. So, no letter will remain in the same position.
Here it is given that P is mother of X
Also Q is husband of P .
Hence X is the daughter of Q or say Q is the father of X.
Here it is given that Riton is the mother of Hyder.
Prashanta is the husband of Riton. And Hyder is the wife of Parash. So Hyder is the daughter of Riton and Prashanta
So Prashanta is the father in law of Parash.
Position of Nilesh from right end=17
Position of sharmila from right end=14
Position of Rohit from right end=19
Position of Rohit from left end=20
Total number of children=20+19-1(Position of Rohit from both end-1)
Total children=38
The position will be as follows.
Left end Rohit Nilesh Sharmila Right end
20(L)19(R) 18(R) 17(R) 16(R) 15(R) 14(R)
Since we have to divide the boys and girls into equal section of either boys or girls, We have to first find the HCF of 408 and 312.
HCF of 408 and 312=24
Number of sections of boys=408/24=17
Number of section of girls=312/24=13
Total number of section=17+13=30
For finding the distance covered by all three men before they 'IN STEP' again , we will have to find the LCM of their steps.
LCM of 68,51,85=1020 cm.
Ans--1020 cm
Here the length of wire is sold in 1m=100cm
Lenghth of wire required=85 cm
To find the minimum length of wire that has to be purchased, we have to find the LCM of 100cm and 85 cm.
LCM of 100 and 85=1700 cm
Ans--1700 cm=17 m
To find when the bells will simultaneously again we have to find the LCM
LCM of 6,8,12,18=72seconds
So the bells will again ring simultaneously after 72 seconds
6 minutes=6*60=360 seconds
Number of times bells will ring in 6 minutes i.e 360 second=360/72=5 times
But the bells rang simultaneously for first time at 12 o'clock
So total number of times=5+1=6 times.
To find the least number of coconuts we will have to find the LCM
LCM of 2,3,5=30
Ans--30 coconuts.
To find the time when both will again be at starting point we have to find the LCM
LCM of 10 minutes and 9 minutes=90 minutes
Ans--Both cyclist will be at starting point after 90 minutes.
To find the capacity we have to find the HCF of 27,33 and 45
HCF of 27,33 and 45=3
Ans--3 Litre
Total number of chocolates Priyanka had in the begining=30
Number of chocolates collected in 5 weeks=5*5=25
Total number of chocolates after 5 weeks=30+25=55
Weight of diferent varities of tea are 132kg,156kg,84kg.
To find the size of box we have to find the HCF of 132,156 and 84
HCF of 132,156 and 84=12kg
Ans--12kg
To find the time when all the three traffic lights will change simultaneously we will have to find the LCM of 48,72 and 108
LCM of 48,72 and 108=432 seconds=432/60 minutes=7minutes and 12 seconds
So the traffic lights will change simultaneously at 7:07:12 am
ANS--A
To find the number of books in each stack we have to find the HCF of 420 and 130
HCF of 420 and 130=10
Ans--10
C.P of 1000 items. = Rs. 1000*3 = Rs. 3000
S,P, of 1000 items = Rs. (700*4.5 + 300*2.5) = Rs. 3900
Profit = SP - CP = Rs. (3900 - 3000) = Rs. 900
Hence, Avg. Profit per item : Rs. 900/1000=.Re. 0.90 = 90 paise
Let, the original price and consumption be P1 and C1 respectively .
So, original expenditure = P1*C1
New price P2=1.25*P1
Let new consumption be C2
Since expenditure remains the same so P2*C2=P1*C1
or 1.25*P1*C2=P1*C1
C2=P1*C1/1.25P1 i.e C2=4/5*C1
% change in consumption=(1-4/5)*100=1/5*100=20%
Ans--20%
From the given condition we can find
Total coins they have = 3*100 -50 = 250 (as 100 is average coins one has)
Now 50+a+b = 250 => a+b=200.----(i)
a - 15 = 1/3*(b + 15) . => 3a - b = 60 ... (ii)
Adding (i) and (ii), 4a = 260 . => a = 65 .
b = 200-a=200-65=135
So a:b= 65:135 = 13:27.
Ans--13:27
Let first and second numbers be a and b respectively.
Now b- 30%a = 3/4b
=> b-3/4b = 30/100a
=>b/4 = 3/10a
=>a/b = 5/6
Total distance covered by boat=1km
Total time taken=15 minutes=1/4 hour
Speed of boat=distance/time=1/(1/4)=4kmph
Let the speed of river Vr
Speed of boat=speed of boat in still water-speed of river
4=5-Vr
Vr=5-4=1kmph
Velocity of river=1kmph
Ans--B
Total Outcomes = 25C3
Favourable Outcomes = 10C1*15C2
Probability = Favourable/Total = 10C1*15C2/25C3 = 10*7*15/23*4*25 = 21/46
ANS--C
Favouralbe case = 10C1
Total cases=14C1
Probability=Favouralbe/Total = 10C1/14C1 = 10/14 = 5/7
ANS--D
Total cases, n(S) = 52C2 = (52∗51) / (2∗1) =1326
One King and One queen case, n(E) = 4C1 * 4C1 = 4 ∗ 4 = 16
Probability = n(E) / n(S) = 16 / 1326 = 8 / 663
ANS--D
p ( getting an even number ) = 3 / 6 = 1 / 2
q ( not getting an even number ) = 3 / 6 = 1 / 2
n ( number of times die is thrown ) = 100
Variance = n * p * q = ( 1 / 2 ) * ( 1 / 2 ) * 100 = 25
Therefore, the variance of the number of successes = 25
Let the speed of boat in still water is x.
Speed upstream=(x-3) kmph
Speed downstream=(x+3) kmph
Since distance is constant
(x-3)*2=(x+3)*1.5
on solving we get x=21 kmph
ANS--C
ANS--B
We know that sum of all angles of a triangle=180
So, 94+41+3rd angle=111+x+3rd angle=180
Since 3rd angle is common so 94+41=111+x
Hence x=24
ANS--D
Angle AMC=180-(56+78)=46
Angle DMB=46
Angle MBD=180-(62+46)=72
ANS--B
Let angle B=x
Then Angle A=2x
2x+x=90
3x=90 i.e x=90/3=30
Angle B=30
Angle A=2*30=60
Ans. D
As both the trains are moving in same direction the relative speed = 30-20=10m/s
Total distance = (200 + 300) m = 500 m .
Hence, time taken = 500/10 s = 50 s .
ANS--A
Let the daughter's share is x
sons share is 2*x=2x
mother's share is 4times of son's share, then mothers share is 4*2x=8x
fatrher's share is 3times than of mother's share, thus father's share is 3*8x=24x
thus the ratio of father's, mother's, son's, daughter's shares are 24x:8x:2x:x=24:8:2:1
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. C
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--A
CP for the manufacturer for 1kg wheat = 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--B
Cost Price = 12500
Marked up = 12500*10/100 = 1250
Toal Marked up price = 12500+1250 = 13750
Discount = 13750*5/100 = 687.50
Sales Price after discount = 13750-687.50 = 13062.50
Profit = sales Price - Cost Price
=> 13062.50-12500 = 562.50
% Profit = 562.50/12500*100 = 4.5%
ANS--C
Average speed = total distance/total time
Total time = 6h + 4h = 10h
distance 1 = 30 X 6 = 180 miles
distance 2 = 45 X 4 = 180 miles
Total distance = distance 1 + distance 2
Total distance = 180 + 180 = 360 miles
Average speed = total distance/total time
Average speed = 360/10= 36 miles per hour
ANS--D
Part of cistern filled in 1 min = (1/16 + 1/12 - 1/8) = 1/48.
Hence, total time to fill cistern=48 mins.
ANS--C
Length of bigger rectangle=30 cm
Breadth of bigger rectangle=15 cm
Area of bigger rectangle=L*B=30*15=450 sq cm
Length of smaller unshaded rectangle=30-8=22cm
Breadth of smaller unshaded rectangle=15-4=11 cm
Area of smaller unshaded rectangle=22*11=242 sq cm
Area of shaded portion =Area of bigger rectangle - Area of smaller rectangle
Area of shaded portion=450-242=208 sq cm
ANS--D
First let us extend the cut portion to form arectangle.
Area of that rectangle=20*15=300 sq m
The cut portion will be a triangle having base=20-8=12 m
and its height=15-10=5m
Area of this triangle=12*5/2=30 sq m
Area of given figure=300-30=270 sq m
Ans. C .
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--B
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 litres
ANS--B
Age of new friend = [(24*5) - {21*4)+(4*4)}] = (120 - 100) yrs = 20 yrs.
In given problem RMO make a right angle triangle
Where Distance between RM = 15km and MO = 9 km
Then (OR)2 =(RM)2 - (MO)2
i.e. (OR)2 = (15*15) - (9*9)
=> (OR)2 = 225 - 81
=> (OR)2 = 144
=> OR = 12km
T's paternal grandfather is P
T's paternal grandmother is R
P and R are husband and wife
Perimeter=sum of the sides Let X be a variable. The given ratio is 13:12:5 Therefore the sum of the sides=13x+12x+5x Perimeter=30x Given perimeter=450 30x=450 x=15m The sides are 12*15,13*15,5*15 It is a right angle triangle since 13square equal to sum of the square of 12 and 5. Area =1/2*b*h 1/2*12*15*5*15=6750sq.m
ANS--A
Total outcomes = 2×2×2×2= 16
Getting at least one tail means one or more tails. P(at least 1 tail) = 1- P(no tail) = 1- 1/16 = 15/16
ANS--C
Total number of balls=2+3+4=9
Number of ways of selecting 3 balls=9C3
Since we have to draw 3 balls and they should be of same colours. So we can take out black or white balls.
This can be done in 3C3 and 4C3 ways
probability=(3C3+4C3)/9C3=5/84
ANS--C
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).
ANS--A
Mr.A has x children while Mrs.B has (x+1) children. After they got married, totally they have 10 children. But it is not given how many children do they have after they got married. Let they be 'y'. Then, we get, x+(x+1)+y=10 i.e. 2x+y=9. Here, if y can take only odd values, otherwise, x will not be a positive integer for any even value of y. Then there are following cases: 1) for y=1, x=4 and x+1=5, 2) for y=3, x=3 and x+1=4 3) for y=5, x=2 and x+1=3 4) for y=7, x=1 and x+1=2 5) for y=9, x=0 and x+1=1. As the children of same parents cannot quarrel, only y can quarrel with x and (x+1) and x can quarrel with (x+1). Then from above five cases, we have to consider the case which will give maximum value. It is the the 2nd case. That is, the product xy+y(x+1)+x(x+1) will be maximum for y=3, x=3 and (x+1)=4, that is 33. Hence 33 is the answer.
ANS--A
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%
ANS--D
When sister was born father was of 28 years and when brother was born sister was of 4 years, so when the brother was born father was of 28 + 4 = 32 years of age.
Since, the average formula is:
(sum of student age) / 22 = 21
sum of student age = 22 x 21
Now the average increase by one when a teacher age included
(sum of student age + teacher's age) / (22 + 1) = 21 + 1
(sum of student age + teacher's age) = 22 x 23
teacher's age = 22 x 23 - sum of student age = 22 x 23 - 22 x 21 = 22 (23 - 21) = 22 x 22 = 44 years.
So, the age of teacher is 44 years.
ANS--B
If each wager is for half the money remaining, each win lead to multiplying the amount by 1.5 times and each loss multiplying the amount by 0.5 times.
the person won three times and lost three times. So we need to multiply the initial amount by 1.5 thrice and the loss by 0.5 times thrice in any order.
So the final amount will be Rs. 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5) = Rs 27 (ina ll caes)
The loss will be 64-27 =37
ANS--A
Let me explain this question with an easy example. If 100 is increased by 20% it will become 120. Now if 120 is decreased by 20% u will get 120-24 = 96 a lesser value. So, to get same value of 100 the % increase of decrease can not be same. There decrease in % will always be less than increase in % Now, according to question, if x = y the value will not come as 1 million, it will be less. Therefore answer will be x>y
ANS--C
relative speed = (68-8)km/s= 60km/s= 60*5/18 m/s= 50/3 m/s
total distance= 150m (Train's length)
time taken to pass a man = 150*3/50 = 9s
ANS--D
Rs. 1 get double in 4 Years i.e. become Rs. 2
In next 4 years get doubled and become Rs. 4
In next 4 years get doubled and become Rs. 8
Hence in 4+4+4 = 12 years amount will become 8 times.
ANS--D
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--C
Time taken for walking one side = 45/2 mins = 22.5 mins.
So, time taken for riding one side = (33-22.5) mins = 10.5 mins.
Hence, time taken for riding both ways = 10.5*2 mins = 21 mins
ANS--A
Let the length of one side be 2 cm
Then surface area=6*2*2=24 sq cm
After 50% increase the side becomes 3 cm
Then surface area=6*3*3=54
Percentage increase=(54-24)/24*100=30/24*100=125%
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--B
Profit=SP-CP=60-50=10
Gain %=Profit/CP*100=10/50*100=20%
ANS-A
Let speed of train= 1/x and speed of car = 1/y Time = distance / speed Therefore, 120x + 480y = 8....... equation 1 200x + 400y = 25/3... equation 2 Solving 1/x = 60 and 1/y = 80 Therefore ratio 60:80 = 3:4
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-C
Speed of the train = (142/6)*(18/5) = 84.9 km/h (approx)
ANS--A
red marbles=24
probability to pick red =3/7
total marbles= X= 24x7/3=56
blue marbles= 56-24=32
Difference in no. of red and blue=8
ANS--D
total marbles 15
White=5
Probability choosing white=5/15
addition of white=X
5+X/15+X=3/5
X=10
ANS--A
Total number of coins in piggy bank = 10 + 20 +10 = 40.
A be event of number of Rs 5 coin = 10 P(A) = (nuber of out comes favorable to A) / (Total number of possible out comes) = 10/40 = 1/4
ANS--A
Weight of 11 people=11*48=528
Weight of 12 people=12*46=552
Weight of 12th man =552-528=24kg
Area of larger square=5*5=25
Area of smaller square=4*4=16
Area of shaded portion=25-16=9
Probability of point in shaded portion =9/25
ANS--A
We know that the time period is proportional to the roo of length of a pendulam, 44% increase in legth means, it is now 144%, i.e, 1.44, root of that is 1.2, so, it os 20% increase.
ANS-A
Total Mass of Solution is 800 g Ratio of Solute is 40℅ Hence, Mass of Solute is (40/100)x800 = 320 g Now, 100 g of the solute is precipitated. Hence, the remaining mass of Solution is 700 g And, the remaining mass of Solute is 220 g Now the Percentage of Solute in the Solution is (220/700)x100℅ = 31.4%
ANS--B
Apply the formula- v^2 = u^2 + 2gR (where u = 0)
V^2=0+2gR
V^2=2gR
V=Sqrt 2gR
ANS-C
AngleDOB = [180 - (132 + 141)] = 93
ANS--D
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
4:1
ANS--A
Total surface area of cylinder=2*3.14*R*H+2*3.14*7*7
or 224*3.14=2*3.14*R*H+2*3.14*7*7
Dividing both sides by 2*3.14
112=7*H+49
or 7*H=112-49
or H=63/7=9cm
ANS--D
Case 1 Probability of picking up a white marble = 5/15=1/3
Case-2 x white marbles were added.
Probability of selecting white marble = (5+x)/(15+x)=3/5 or 25+5x= 45+3x or x=10
ANS--B
Red=X
Green = Y
Yellow = Z
X= 2Y= 2x 3Z=6Z
Y=3Z
Probability of getting Green Marble: Y/X+Y+Z
3Z/6Z+3Z+Z=3/10
ANS--D
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.
No of students taking neither tea nor coffee=600-(150+225-100)=600-275=325
Number of boxes damaged=700/35=20
Let total number of boxes be x
Then 2*x/100=20
x=20*100/2=1000 boxes
From the first statement T>Q
From the second statement T>Q>P>S
Since R is remaining so R will be shortest
T>Q>P>S>R is the order of heights.
ANS--R is the shortest.
ANS--A
Increase in total age of 7 women=7*5=35
Age of women=35+25=60 years
ANS--A
A's 1 day work=1/12
B's 1 day work=1/18
Total work done by both A and B in one day=1/12+1/18=5/36
C's 1 day work=1/4-5/36=1/9
Time taken to complete the work by c alone=9 days
ANS--A
Portion of tank filled in 1 hour by pipe A=1/8
Portion of tank filled in 1 hour by pipe B=1/6
Portion of tank filled in 2 hours when A and B are opened alternately=1/8+1/6=7/24
Portion filled in 6(2*3) hours=7/24*3=21/24
Portion of tank empty=1-21/24=3/24=1/8
To fill this portion pipe A will take 1 hour
Total time taken=6+1=7 hours
ANS--A
Ratio given 4 years ago 19 : 6
Father age = 19x
Son age =6x
ii)Difference 19x - 6x =26
13x =26
x =2
Therefore Their ages 4 years back in the ratio of 38 : 12
4 years hence (Present age) 38+4:12+4 = 42:16 = 21:8
ANS--D
Smita's position from left after interchange=14
From right this is 17th
Total number of girls=17+14-1=30
Sn = n/2 [2a+(n-1)d] to find the sum of n terms of an AP Where a= First term d = difference
S20 = 20/2 [2a+ (20-1)d] = 10[2a+19d] = 20a+190d. ------(1)
S10 = 10/2 [2a+ (10-1)d] = 5[2a+9d] = 10a+45d. ------(2)
S30 = 30/2 [2a+ (30-1)d] = 15[2a+29d] = 30a+435d. -----(3)
Eqn (1) - (2) gives S20 - S10 = (20a+190d) - (10a+45d) = 10a+145d 3(S20-S10) = 3(10a+145d) = 30a+435d = S30
(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--D
(1+ r/100) = 23958/21780 [taking the 3rd year only]
r/100= 1/10
r=10
Hence, rate of interest is 10%
ANS--C
Let the length of train=x
(x+162)/(x+120) = 18/15 = 6/5 .
So, x = 5*162 - 6*120 = 90 .
ANS--B
nitial water = 20% of 25 kg = 5 kg.
Let, water added be x kg.
ATP, 5 + x = 33.33% of (25 + x) . => x = 5 .
R has more number of pages than P but less number of pages than Q. Q>R>P
V has more number of pages than Q i.e V>Q
but it does not have maximum number of pages and since R has 51 pages & T has 76 pages, the equations can be written as T>V>Q>R>P
Since exactly 2 books have less number of pages than P and let those books be x & y, so P>x>y i.e T>V>Q>R>P>x>y
since Q & S does not have min pages, y cannot be S. So that leaves x as S and y as U. The eqn becomes T>V>Q>R>P>S>U
So the num of books with pages less than Q are 4
ANS--A
Daughter has the least share. Therefore the ratios of their shares are (3×4×2×1):(4×2×1):(2×1):1 = 24:8:2:1
Since tangents have equal length we have XP = XQ and AO = AR and BQ = BR
XB+BR = XB + BQ = XQ and similarly XA+AR = XP and we already have XQ = XP
Let d be the distance between tower and chimney. Then tan 60 = h/d where h is height of tower.
Also tan 30 = (h-40)/d
Dividing 3 = h/(h-40) Or 3h-120 = h
h = 60 m.
Height is less than 60 hence meets pollution norms.
Use section formula for the 4 points as ratio 1:4, 2:3, 3:2, 4:1.
First point = (2, 6) and second point = (3, 5), III point = (4, 4) IV point = (5, 3)
If h is the height of the light house, and d the distance between the first ship and light house, then
tan 30 = h/d and tan 45 = h/(100-d)
Divide to get
1/sq rt 3 = (100-d)/d
d = 100 rt 3 - rt 3 d
= 173.2 - 1.732 d
d = 173.2/2.732 = 63.4
h = 100-d = 36.6 m
S7 = 7/2 (2a+6d) = 63 and S14 = 7(2a+13d)
S14 = S7+161 = 224
2a+13d =32
2a+6d = 18 From S7
7d =14 and d =2. a = 9-3d=3
28th term= 3+27(2) = 57
the possibilities of sum 10 (considering both dice) is three-(6,4),(4,6),(5,5)
so, number of envent is -3
total number of events for 2 dice is-6*6=36
probability is= number of event of 10/ total number of event occour= 3/36=1/12
Area of rectangle = curved SA of cylinder
i.e. 880 = 2(3.14) (r)(40)
r = 7/2 = 3.5
ANS--C
We have cube of edge is 15 so radius of sphere half of the edge i.e 7.5 cm Then volume of sphere is= 4/3×pi×r^3 =4/3×22/7×7.5×7.5×7.5 = 1767.86 cm^3
ANS--B
The reqd. volume of water = [70*50*20 - 50*(5^3)] cm^3 = (70000 - 6250) cm^3 = 63750 cm^3
ANS--C
We have cube of edge is 15 cm so radius of sphere is half of edge i.e 7.5 cm Now we calculate volume = 4/3×pi×r^3 = 4/3×22/7×7.5×7.5×7.5 = 1767.86 cm^3
ANS--B
We have (30 + x)(20 + x) = 5*(30*20) . => x = 30 or -80. -ve value to be ignored.
Hence, x = 30 m .
ANS--D
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 .
ANS--C
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 .
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--D
Reqd area = [35*25 - {2*pi/4*13^2)] = [875 - 265.57) cm^2 = 609.43 cm^2
Let nC(r-1) = 45, nCr = 120 and nC(r+1) = 210 => nCr / nC(r-1) = 120/45=8/3 and nC(r+1)/nCr = 210/120=7/4 nCr / nC(r-1)=[ ni /(r)i (n-r)i ]/[ni /(r-1)i (n-r+1)i] = (n-r+1)/r = 8/3 and (n-r)/(r+1) = 7/4 => 4n -11r =7… eq-1 and 3n - 11r =- 3… eq-2 Subtracting 2nd eqn. from the 1st, n = 10 and r = 3
To reach the path represented by the equation 6x - 7y+ 8=0 in the shortest possible time, the family has to travel through the path which is perpendicular to the path 6x - 7y + 8=0 and passes through the intersection of the paths whose equations are 2x - 3y - 4 =0 and 3 x + 4y -5 =0 To get the point of intersection of the two lines: 2x - 3y - 4 =0 and 3 x + 4y -5 =0, we need to solve the equation which will give the the point of intersection,O (x1,y1) 2x1 - 3y1 - 4 =0 ....line1 3x1 + 4y1 -5 =0......line 2 x1=31/17 and y1=(-2/17) The slope of the given line(line-3) is 6x - 7 y + 8=0 , m1=6/7 The slope of the required line(shortest path) will be m2= -7/6 ,ince the shortest path will be perpendicular to line 3 and product of slopes of perpendicular lines are -1 The equation of the required line is : (y - y1)/(x - x1)=- 7/6 ⇒ [y-(-2/17)]/(x -31/7)= (-7/6) ⇒ (17y+2)/(17x-31)=(-7/6) ⇒ 119x +102y= 205 Hence the equation of the path they should follow is 119x +102y- 205=0
Let the one end of the diameter be (x,y). The center (2,-5) is the midpoint of the diameter, whose other end is (-2,-3) By using the midpoint formula x+(-2)/2=2 and y+(-3)/2=-5 => x=6 and y = -7 Coordinates of the other end of the diameter is (6,-7)
ANS--B
Let the perpendicular hit the base at O. As we have A and B on either side of, we need to calculate, AB=OA+OB. Using trignometry we have Tan 45= 120/ OA, i.e. 1=120/OA, i.e.OA=120. Similarly, we have Tan 30=120/OB ,i.e=OB=120(1.732). Thus, the total length= AB=OA+OB= 120*2.732=327.84.
ANS--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
ANS--D
Total volime of two cones = volume of cylinder
=> (1/3)*pi*h*a^2 + (1/3)*pi*h*b^2 = pi * h* r^2
=> r = sqrt((a^2 + b^2)/3)
ANS--C
(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--D
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--B
Volume of cylinder= pir^2h 22/7×7.5×7.5×75 13258.92 cm^3 Given that 1cm^3 is equal to 10 g or 100 kg so mass of the solid is 13258.92/100 132.589 or 132.59
ANS--A
Surface Area of a Sphere = 4πr^2
Curved Surface Area of Cylinder = 2πrh
Thus, ratio of Surface Area of a Sphere : Curved Surface Area of Cylinder = 4πr2 : 2πrh
or 2r:h
or d=h
Ratio 1:1
ANS--C
The average speed is 8 miles per hour.
The total time taken to complete tw course are 20 +25 = 45 minutes.
! course = 3 miles. So 2 course = 2 X 3 = 6 miles.
45 minutes taken to complete miles = 6 miles
1 minute = 6 / 45
60 minutes = (6 * 60) / 45 = 8 miles
ANS--A
Let us assume that the quantity of alcohol as 4x and the quantity of water as 3x.
then he said that 5 liters of water is added
4x/(3x+5)=4/5
20x=12x+20
8x=20
x=2.5
alcohol=4x
4*2.5=10 litres
ANS--C
The investment ratio of A&B is 3:2,
then their profit ratio must be 3:2
but out of 100% profit, 5% was given to charity, the remaining 95% be shared in the ratio 3:2
say profit as X
then A share 95/ 100 X * 3/5 = 855 => X = 1500
ANS--A
Length and bredth of rectangle is l cm and b cm
(l2 + b2)^(1/2) = 41.
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
(l + b) = 9.
Perimeter = 2(l + b) = 18 cm.
ANS--B
The fruit content in both the fresh fruit and dry fruit is the same.
Given, fresh fruit has 68% water.so remaining 32% is fruit content. weight of fresh fruits is 100kg
Dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.
fruit % in freshfruit = fruit% in dryfruit
(32/100) * 100 = (80/100 )* y
we get, y = 40 kg
ANS--D
Total Cost => Set up Cost 2800 + cost of Paper 7 ink 1600+ printing cost 3200 => 7600
Total Sales => 1500*5 =>7500
Let, money obtained from advertising = x
then Profit = sales - cost
=> 25% of 7500 = (7500+x) -7600
=> 1875 = -100 + x
=> x = 1975
ANS--B
Let, B1 : B2 : B3 = 3x : 4x : 5x Again, B1 : B2 : B3 = 5y : 4y : 3y Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets. Hence, 5x = 3y and, 3x : 4x : 5x → 9y /5 : 12y /5 : 15y /5 = 9y : 12y : 15y And, 5y : 4y : 3y → 25y : 20y : 15y Therefore, Increment in first basket = 16. Increment in second basket = 8. Thus, Required ratio = 16 /8 = 2 : 1
ANS--C
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--A
Temp. on Wednesday = [40*3 - (41*3 - 42)] deg = 39 deg
ANS--A
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--A
Sima:Ayesa= 21000:17500= 6:5 Total profit= Rs. 26400 So, Sima gets= 6×26400/(6+5)= 6×26400/11=6×2400=14400
ANS--D
For only women selected means we chose 2 from 5 so 5C2 and total is 12C2
so probability= 5C2/12C2=5/33
ANS--B
Let Bikash runs = x meters
Then, x*25/100 = 7
=> 25x =700
=> x = 28 meters
The length of race = 28+7 => 35 meters
ANS--D
Relative speed = 68 -x Distance traversed= 0.1km Time interval =1/36 hrs 1/36 = 0.1 /(68-x) 5/18 = 1/(68-x) X=32
Let,point C lie closer to point A and point B closer to D. Given that, AC=BD. Now,AB= AC+ BC and CD= BD+ BC. BC being common and AC= BD, we have AB = CD. Had the case been reverse,that's B been closer to A and C being closer to D, AB= AC- BC and CD=BD- BC. Then again, AB= CD
Rohit is standing facing sun in the evening, i.e. he is facing west
he starts walking towards right i.e. walking towards north
Then he take left turn i.e. walking towards west
Then he take right turn i.e. walking towards north
And then he take right turn ie. walking towards east.
so, now he is walking towards east.
ANS--A
4 years ago the the ratio between the ages is 19:6. Ao the difference=19-6=13 The ratio 13:26=1:2 Therefore the ages four years back will be 19×2 and 6×2. That is 38 and 12. So the present ages are 38 + 4 and 12+4. That is 42 and 16. Therefore the ratio between the present ages are 42:16. That is 21:8
Simple Intereest =(P x R x T)/100
240 = (P x 5 x 6)/100
Hence P =Rs. 800
ANS--C
the total no. of employees at present is 5000
the no. increases @2% every year
so for the first year the no. of employees would be 5000+(2%*5000)=5100
during the second year the no. of employees would be 5100+(2%*5100)=5202
so total no. of employees after 2 years =5202
Water content = 25 (20%) = 5 kg.
If x kg is added then percent = (5+x)/(25+x) = 1/3
i.e. 15+3x = 25+x
x = 5 kg.
It "crosses" the platform. Means it travels the distance equal to platform pluse its own length. so, x = 280x3 + 280 = 1120m t =50 v = x/t = 1120/50 m/s to convert it into kmph multiply by 18/5 (remember this) v = 112/5x 18/5 = 896/25 = 36kmph approx
A, B, C together can finish the work in 4 days
Hence the work is done in one day by A, B, C together is = 1/4
It can get by adding one day work of all the individuals i.e. 1/4=(1/12+1/18+1/C)
1/4=5/36+1/C
1/4 - 5/36=1/C
1/C=1/9
Therefore C can finish the work alone in 9 days.
As per given data, A can fill 1/8 tank in 1 hr and B can fill 1/6 in 1 hr. If A is opened first 1/8 + 1/6 + 1/8 + 1/6 + 1/8+1/6+1/8+1/6=1 in this fashion the tank will get full in 7 hrs.
ANS--D
{(x -16)/x}*100 = x.
=> By solving, x = 80 or 20
so x=20
ANS--C
Let Cost Price = Rs. 100
Marked Price = 100+(100*30/100) = Rs. 130
Discount = 130*20/100 => Rs. 26
Sales Price = 130-26 => Rs.104
Profit = 104-100 => Rs. 4
% Profit => 4/100*100 => 4%
ANS--A
Sara can finish the work in 12 Days. In I day she does 1/12 parts of the work
Bula can finish the work in 18 Days. In I day she does 1/18 parts of the work
Kamal can finish the work in d Days. In I day she does 1/d parts of the work
They together can complete in 4 days. In I day they can do 1/4 parts of the work
Hence 1/12+1/18+1/d=1/4 so 1/d= 1/4-1/12-1/18= 1/9
Hence Kamal can complete the work in 9 days
ANS--A
P+C+M = C+120
=> P+M = C+120-C
=> P+M = 120
Average = 120/2 => 60
Let Total Boxes = 100
Damaged Boxes = 100*2/100 => 2
Loss per Damaged Box = 2*35 => Rs. 70
When loss is 70 there are total no of boxes =100
When loss is 700 there are total no of boxes =100 / 70*700 => 1000
If total population is 500, then 300 males and 200 females
Of these, 60 males and 50 females are graduates.
Hence 110/500 i.e. 22% are graudates.
Non graduates are 78%
P is the mother-in law of Z
Since K and P are immediate neighbours.
Total number of people=38+13=51
A can do work in 4 days. One day work of A is 1/4.
B can complete same work in 8 days.( as A is twice efficient as B) One day work of B is 1/8.
C can do same work in 16 days. (as B is twice efficient as C) One day work of C is 1/16.
If all of them work together then in one day they can complete (1/4+1/8+1/16= 7/16) of the work .
So they will take 16/7 days for them to complete work when all work together.
Relative velocity = v1+v2 = 60+v2 kmph
Distance = l1+l2 = 180+270 m = 0.45 km.
0.45 = 10.8(60+v2)/3600
60+v2 = 150
v2= 90kmph
Speed of second train=90 kmph
let the two numbers be x and y.
y-0.3x=3y/4
y-3y/4=0.3x
y/4=0.3x
dividing both sides by y we get
1/4=0.3(x/y)
so x/y=10/12=5/6
so the rato=5:6
ANS--B
The number of visitors in January was 960.
The average number of visitors January-July is 870
∴ Total number of visitors January-July is 870 X 7= 6,090.
∴ Total number of visitors February-July =6,090-960= 5,130
Actual number of visitors in August was 827
∴ Total number of visitors February -August =5,130+827= 5,957
∴ Average number of visitors February-August= 5,957/7=851
We multiply the capital by number of months remained in the business.
For Sreekumar, 30000(6) +21000(6)= 306000
For Pete 30000(6)+18000(6) = 288000
For Ahmed 30000(6)+35000(6) = 390000
Hence ratio of profit share = 306:288:390
Ahmed share = 390(88560)/984
=35100
ANS--B
Side of the square = 2(16+24)/4 = 20 cm.
Circumference of semi circle with radius 10 cm = 3.14(10) = 31.4 cm=31 cm
ANS--D
In 1 hour P will fill up 1/12th of the tank and in 1 hour Q will fill up 1/18th of the tank.
In 2 days (1/12 + 1/18)= 5/36 parts of the tank will be filled.
In 14 days 5/36*7=35/36 part will be filled
The remaining portion is 1/36 which can be filled by P in 12/36 days i.e 0.33
Total days=14.33 days
ANS--C
Let the MRP be 100
After 20% discount S.P=80 and profit%=20
C.P=S.P/1.2=80/1.2
When 10% discount is given, S.P=90
profit=SP-CP=90-80/1.2=108-80/1.2=28/1.2
Profit%=Profit/cp*100
=(28/1.2)/(80/1.2)*100=28/80*100=35%
ANS--D
Let AB= x and AC=y
Tan30 = x/2a => 1/√3= x/2a => x=2a/√3
Tan60=y/a => √3 =y/a => y= a√3
AB/BC = x/(y-x) => (2a/√3)/(√3a-2a/√3) = 2:1
ANS--D
Ratio between sides = square root of ratio between areas
= 3:1
ANS--A
l^2 + b^ = d^2, l^2 + b^2 = 41 and lb = 20.
(l+b)^2 = l^2+ b^2 + 2lb = 41 + 2x20 = 81, l+b = 9,
P = 2(l+b) = 2x9 =18
Here,
Distance=1.56 x10^16m
Speed of light =3 x10^8m/s
Time =?
speed= dist/time
time=dist/speed
=1.56 x10^-16 /3 x10^8
=0.52 x10*-24sec
f length =l, then time for 162+l metres is 18 seconds and time for 120+l is 15 sec
Speed is the same. Hence 162+l/18 = 120+l/15
Cross multiply and simplify : 810+5l = 720+6l
l = 90 metres.
ANS--B
Let the speed be 2v and 3v and distance be 3d and 7d
Time for first car=3d/2v
Time for second car=7d/3v
Ratio of time taken=3d/2v:7d/3v=9:14
4% of CP = 1080-1026 = 54 so CP= 1350
SP for 4 % profit = 1350+ 54 = 1404
P is the mother-in-Law of Z
U sits immediate right of the Q.., from left to right: P,T,R,W,U,Q,V,S..., as Q is from India his rigth is left as per answer.
T,W,Q,S are from India, and remaining people from Japan.
Distance between T and Z =[(Distance between T and U) +(Distance between U and Z) ] units
=(7.5+8)units=15.5 units
1 unit=2 km then 15.5 units=15.5*2=31 km
F is paternal grandmother to Z
Sonal is at 5th to the left of Atul who is at 12th. So the posititon of Sonal is 7th from left and 15th from right. So, there are 21 persons in the row.
ANS--B
The probability is 1/36.
Any number can appear on first dice in 6 ways.
The same has to appear on second and third which is possible in one way only.
So required probability = 1*1/6*1/6 = 1/36.
ANS--A
2j+b+s = 47 -(i)
j+3b+s= 54 - (ii)
3j+b+3s = 98 - (iii)
(i) + (ii)-(iii) gives
3b-s=3 - (iv).
2*(ii)-(i) gives
5b+s=61-(v)
adding (iv) and (v)
8b =64=> b=8.
Age of ben is 8 years
ANS--B
A completes 1/24 and B completes 1/30 part of work in a day.IF A & B alone have worked for 10 days then part of work done by them would be = 10*(1/24 + 1/30) = 10* (9/120) = 3/4 of the work. Now remaining work has been completed by C.
Hence out of total Rs. 2000/- C will get money according to the ratio 25% of Rs.2000 = Rs 500/-
C will get Rs. 500/-
ANS--D
Father in law.
ANS--B
The wheel moves at 140 rev/min. In 1 hour it is 140 X 60 rev.
In I rev the wheel will travel 2πr cm = 2X 22/7X 35 cm = 2X 22/7X 35 X10-5 kms
In 1 hour it will travel 140 X 60 X2X 22/7X 35 X10-5 = 18,480 X 10-3 Kms = 18.48 Kms
ANS--A
Let the price of 1gm rice be Re.1
Therefore, price of 1000gm i.e, 1kg is Rs.1000 (this is the C.P of the manufacturer)
S.P of the manufacturer= 114/100 of Rs.1000 =Rs.1140
Now for calculating the net profit %, it should be kept in mind that the manufacturer is giving 900gm and not 1000gm.
So, C.P of the manufacturer should be taken as Rs.900 ( price of 1gm is Re.1)
Net profit % = (1140-900)/900 * 100
= 26.67%
a) Speed of cycle/speed of scooter=15/30=1/2=1:2
b) 5m/10km=5/10000=1:2000
c)50paise/Rs5=50/500=1:10
ANS--D
Let the distance =x
Speed upstream=5-1=4km/h
speed downstream=5+1=6km/h
Time taken upstream=x/4
Time taken downstream=x/6
Total time taken=x/4+x/6
Now x/4+x/6=75/60
(6x+4x)/24=5/4
10x=5/4*24
10x=30
x=3 km
ANS--C
Since it is a Right angle triangle
Area of triangle=1/2*16*12=96 sq m
Area of plot grazed=x/360*22/7*7*7+y/360*22/7*7*7+z/360*22/7*7*7
22/7*7*7(x+y+z)/360
Since x+y+z=180
So Area=22/7*7*7*180/360=77 sq m
Ratio=77:96
Perimeter of the square = 44
Or, 4*a = 44
Therefore, a = 11 (where a is the lenth of a side of square)
Hence, area of square is a2 = 11*11 = 121 cm2
According to question,
Circumference of circle = 44
Or, = 44
Therefore, r = 44*7/44 = 7 cm
Radius(r) of the circle is 7 cm
Thus, area of the circle = = 22*7*7/7 = 154 cm2
Therefore the difference between the area of square and circle is 154 – 121 = 33 cm2
ANS--B
According to question,
Principle(P)=500
Compound Amount(C.A)=583.20
Time(T)=2 years
Now, applying the formula of compound amound, we have,
C.A = P(1+R/100)2
Or, 583.20 =500(1+R/100)2
Or, 583.20/500 = (1+R/100)2
Or, (1.1664)1/2 = 1+R/100
Or, 1.08 – 1 = R/100
Or, 0.08 = R/100
Now, cross multiplying,
R = 8 (where R = Rate)
Now, from question,
Rate has been increased by 2%
Previous rate as we found(R) = 8%
And , the given time(T) is 1 year
Therefore, the new rate of interest (Rn) = 8% + 2% = 10%
So new Compound Amount with new Rate,
C.A = P(1+R/100)T
= 500(1+10/100)1
= 500(110/100)1
= 500(1.1)1
= 500*1.1
= 550
Therefore, the new Compound Amount after 1 year with 10% Rate will be Rs.550
ANS--B
Let the sum borrowed by A be x and by B be y
Interest paid by A = IA= x*4*5/100 and Interest paid by B = IB= y*5*4/100
Since interest paid by A is double that of B's; IA=2(IB) => x/5 = 2y/5 => x = 2y
x + y =30,000 => 2y + y = 30,000 => y = 10,000. Therefore, x = 20,000
Sum borrowed by A is Rs. 20,000
ANS--B
Let, both the trains start at t=0 from their respective stations.
Given that they proceed towards each other.
Let T1 be the train starting from station A and T2 be the train from station B.
T1 has speed 16 kmph, and T2 has speed 21kmph.
Let us suppose that they meet each other after ‘x’ hours, also suppose that T2 has travelled 60 km more than T1.
Now,
Distance travelled by T1 = 16*x
Distance travelled by T2 = 21*x
So, according to the given condition,
21x – 16x = 60
Or, 5x = 60
Or, x = 60/5
Or, x = 12
Now,
Distance travelled by T1 = 16*12 =192
Distance travelled by T2 = 21*12 = 252
Therefore,
The distance between station A and station B = Distance travelled by T1 + Distance travelled by T2
= 192+252
= 444 Km
ANS--B
Due to stoppage the bus travels (45-36) = 9 kms less.
Time taken for traveling 9 kms = 9/45*60= 12 mins
ANS--B
Marked Price of the laptop(M.P) = $1050, Discount(D) = 10%, Profit(P) = 25%
Now, Price after 10% discount = 1050 – 10*1050/100 = 1050 – 105 = 945 (dollar), So selling price (S.P) = $945
After giving discount profit is 25%
Now, P = (S.P – C.P)*100/C.P (C.P = cost price)
Cross multiplying, Or, 25 * C.P = 100*S.P –100* C.P Or, 125*C.P = 100*945 Or, C.P = 94500/125 = 756
Cost price of the laptop was $756
ANS--A
Let Reena's age be x years and her daughter's be y years. x=8y
Eight years from now, (x+8)/(y+8)= 10/3 => 3x+24 =10y+80 => 24y+24= 10y+80 => y=4. Therefore, x = 8*4=32
Reena's present age is 32 years
ANS--C (Water)
ANS--D (1)
ANS--B([email protected])
ANS--C
ANS--B
1_2_3_4_5_6_7_8_9_10_11_12_13_14_15_16_17_18
Before, 5th fm right → B P → 6th from left
After P B
Bimal is fifth from left and Prakash 6th from right. When they interchange Bimal is thirteenth from left. i.e. 13th from right = 6th from left => there are 13+6-1 =18 positions. Hence after interchange, Prakash is 14th from the right.
ANS--A
ANS--C
ANS--A
ANS--A
ANS--C
ANS--D
ANS--A
ANS--B
Total dist. = 4*2*(22/7)*35 m =880 m
Time = 15 mins
Speed = (880/15)*60/1000 k/hr = 3.52 km/hr
ANS--B
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--B
ANS--B
ANS--E
ANS--B
ANS--A
ANS--B
ANS--D
ANS--A
ANS--A
ANS--A
ANS--A
ANS--C
ANS--A
ANS--B
ANS--C
ANS--C
ANS--A
ANS--E
ANS--E
ANS--A
ANS--D
ANS--C
ANS--D
ANS--C
ANS--C
ANS--A
ANS--A
ANS--B
ANS--D
ANS--B
ANS--A
ANS--D
ANS--D
LET B=X
A=2/9X
C=3/4*2/9=X/6
A+B+C=2/9X+X+X/6=625
X=450
A'share=2/9*450=100
B'sahre=x=450
C'share=450/6=75
ANS--B
100*90/100*80/100=72%
Hence discount=100-72=28%
ANS--A
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--D
(25249.54 - 23820.32)/23820.32*100 = 6
ANS--C
45/100×66+28/100×450 =297+126 =423
ANS--C
(45000*9%) + (45000+9%)*9%
ANS--D
There are 4 significant digits
ANS--A
First applying, v = u + a.t,
Here v=0, u=16 and t=4 sec
we get a = -4 m/s^2
Again applying v^2 = u^2 - 2.a.s , we get, s = 32 m
ANS--D
Information given is not adequate to provide answer so
ANS--D
Let N1, N2, N3 be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3 and remainders R1, R2, R3… respectively.
Now, Remainder of (N1X N2XN3)/D => Remainder of (R1XR2XR3)/D.
Therefore, Remainder of (14X 21X1423X1425)/12 => (2X9X7X5)/12 => (18 X35)/12 => (6X9)/12 => 6
ANS--A
ANS--B
ANS--B
n(T)=35
n(C)=25
n(T and C)=35+25-50=10
n(T) only=35-10=25
ANS--C
2^(73) – 2^(72) – 2^(71)
= 2^(71) x {2^(2) – 2^(1) – 1}
=2^(71) x (4 – 2 – 1)
=2^(71) x (1)
=2^(71)
ANS--C
[ - ]
1) Total no of cards= 52 No of Aces = 4 P(ace) = 4/52 = 1/13. 2) P(not red) = 1 - P (red) Total no of cards =52 No of red cards = 26 P(not red)= 1 - (26/52) = 1 - (1/2) = 1/2
Please mention the time taken to return to original position. It is missing from the question.
ANS--B
Total value of share = 200 x 15 = Rs 3000
Total dividend received = Rs 300
So rate of dividend = (300/3000) x 100 = 10%
ANS--C
Each shares (of Rs 20/-) gives 8% dividend = 8/100*20 = Rs 8/5
The man can earn Rs 8/5 from 1 shares
؞ He can earn Rs 1,000 from 1/8/5*1000 = 5000/8 = 625 shares.
Now each Rs. 20 shares is selling at Rs 40. ؞ To buy 635 shares he need to invest Rs 40 * 625 = Rs 25,000
ANS--C
By using distance formula, Distance between two possible vertices = √{(x2-x1)^2 + (y2-y1)^2} = √{(14-8)^2 + (3-3)^2} = √ 6^2 = 6 cm
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
Let the height of the tower be x m and the separation distance between the cliff and the ower be y m
So tan 45 = 1 = 80/y. This gives y = 80 m
Also tan 30 = 0.577 = (80-x)/y = (80-x)/80 . This gives x = 33.84 metre
Shopkeeper buys at 20% discount or at Rs 4000. ST paid by him = 5% on Rs 4000 =Rs 200
Selling price Rs 5000. Sales tax paid by customer = 7% pn 5000 = Rs 350
So Value Added Tax = 350 -200 = Rs 150
Area of triangle COA = 60 = (1/2) x 5 x AC. This gives AC = 24 cm
By proportionality, BD/AC = OD/OC
So BD = (OD/OC) x AC = (8/5) x 24 This gives BD = 38.4 cm
Area of triangle BOD = (1/2) x 8 x 38.4 = 153.6 sqare cm
ANS--B
Volume of the original hollow cylinder = 22/7 x 5 x (10 x 10 - 8 x 8)= 22/7 x 5 x 36 cubic cm.....(1)
Volume of one cone = 1/3 x 22/7 x 3 x 3 x 5 = 22/7 x 5 x 3 cubic cm.......(2)
Dividing (1) by (2) we get the number of cones = 36/3 = 12
ANS--D
Let the number of children be x
So [(6800/x) - (6800(/(x +9)] = 153
This gives x(x+9) = 400
Or soving the quadratic equatuib x = 16
ANS--B
The compound interest in the first two successive years is Rs. 2000 and Rs. 2500 respectively, So, extra Rs. 500 interest is paid on the interest on first year, i.e., on Rs. 2000 for 1 year, therefore, rate of interest = 500 * 100 / (2000* 1) = 25%
ANS--B
Simple interest for 2 years will be 40000 x 2 x 15/100 = 12000
Amount due after two years at 15 % per annum = Rs 52000
Compund interest for the first six months for six-monthly compounding for the third year = 52000 x 0.05 = 2600
Amount after 6 months - 52000 + 2600 = 54600
Interest for the second six months = 54600 x 0.05 = 2730
So amount due after third year = 54600 + 2730 = 57330
Radius of circle ‘p’ = 15
So Radius of circle Q = 15+ 7 = 22
Similarly Radius of circle R = 15 + 7 + 7 = 29
In Same way find areas of all circles individually as
Area of circle P = 22/7*15*15 = 706.5
Similarly area of circle Q And R are resp. 1519 and 2640.74 (with help of 3.14*r*r)
So area of ring Q = Area of circle Q — area of circle P = 1519– 706.5 = 812.5 cm
Similarly area lf ring R = area of circle R —area of circle Q = 2640.74–1519=1121.74 centimetre
ANS--C
Let d be the distance between the building and the tower.
We get tan 60 = 30/d = 1.732. This gives d = 30/1.732 = 1.73
Let H be the height of the tower above the height of the building
Now we also get tan 45 = 1 = H/17.32 = 1 This gives the dimension H = 17.32 m
So total height of tower = 30 + 17.32 = 47.32 metre
ANS--C
To find th eheight of building, consider one triangle with 30 and other with 60 degree with top of building.
As per one triangle with 30 point, tan 30 = h/(50+y)
As per second triangle with 60 degrees , tan 60 = h/y
now substituting value of y and then solving above equation , the h comes out to be 43.3 metre
ANS--B
Market price = 2500
10% discount on the market price = 2500 * 10/100 = 250
Reduced market Price = 2500 - 250 = 2250
5% of Cash discount = 2250 * 5/100 = 112.50
New price for payment through cash = 2250 - 112.50 = 2137.50
10% Sales Tax on the price for cash payment = 2137.50 *10/100 = 213.75
FINAL amount to be paid by customer by CASH = 2137.50 + 213.75 = Rs. 2351.25
ANS--D
ANS--B
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 metre
ANS--B
an theta = 450/778.5 = 0.578 = 1/root3 (approx)
Hence, theta = 30 degree
ANS--C
By simple division we get Price in Rs per oz for W = 0.035, for X it is 0.03. for Y it is 0.28 and for Z it is 0.03 respectively.
So the least expensive brand is Y
ANS--B
N is the LCM of first 100 natural numbers.
Now 101 is a prime number, so 101 and N do not have any common factors. Hence their LCM = 101XN
102= 2X51, since 2<100 and 51<100, 102 divides N
103 is a prime number, no common factors between 103 and N
104= 52X2; since 2<100 and 52<100, 105 divides N
105 = 21X5; since 5<100 and 21<100, 105 divides N
Hence, LCM of first 105 natural number is NX101X103
ANS--E
Let the desired weight of the alloy containing 40 % gold be W
So (0.40 W + 0.20 x 10) / (W + 100 ) = 0.32
This gives W = 150 gms
ANS--B
numbers 4569 4571 4576 4586 4603 4629 4666
Difference 2 5 10 17 26 37
Difference 3 5 7 9 11
ANS--E
ANS--B
North is vertically upwards 10 KM, Coming vertically downwards 6 KM . So 4 KM to North from starting point
Then East is to the right 3 KM
Therefore he is SQRT(3^2 + 4^2) or 5 KM
ANS--A
A ______30___________C____10_____B
Distance AC=30 Kms, AB=40 kms Let speed of boat P be 'x' speed of boat Q be 'y' and speed of stream be 'z'
Relative speed of P upstream = x-y and speed of Q upstream z-y. Relative speed of P downstream = x+y and speed of Q upstream z+y. Now P travels from A to B = 40 kms upstream in 10 hours and B to A downstream in 5 hours.
time =distance/speed Therfore, x-y= 40/10 =4..(i) and x+y = 40/5=8 ...(ii) solving (i) and (ii) we get x= 6 and y=2
Speed of boat P = 6 kms/hr and speed of stream = 2 kms/hr and Speed of boat Q = z-2= 30/5=6 => z= 8 kms/hr
Now, P starts from A and Q starts from C and Pand Q will meet at D after t hours.
t = 30/(Speed of P upstream+speed of Q downstream) = 30/(4+10) = 15/7
In 15/7 hours P will travel 15/7*4= 60/7 kms upstream from A., = 40-60/7 =220/7 kms from B.
Therefore P and Q will meet at 220/7= 31.4 kms form B
ANS--D
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--C
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--C
Let the time taken by the nth tap to fill up the tank be tn. The efficiency of the nth tap is equal to the sum of the efficiencies of all previous taps. i.e. the time taken by the 3rd tap= sum of the time taken by 1st and 2nd tap.
t3= t2+t1.
t4= t3+t2+t1= (t2+t1) + (t2+t1) = 2(t2+t1)
t5= t4+t3+t2+t1= 2(t2+t1)+(t2+t1) + (t2+t1) = 4(t2+t1)= 2^2(t2+t1).
؞tn= 2^(n-3)(t2+t1). the 8th tap takes 48 hours to fill the tank. ؞ t8= 2^(8-3)(t2+t1) ⇒ 2^5(t2+t1)= 48
⇒ 2^5(t2+t1) = 48
When tap number 1to9 are working together to fill the tank it will be equal to the time taken by 10th tap
2^(10-3)(t2+t1) = 2^7(t2+t1) = 2^2*t^5(t2+t1) = 4*48 = 192
ANS--C
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--C
Speed of A = 36 km/hr. Therefore, Distance traveled by A in 2 mins = 36/60*2= 12/10kms = 1200 meters
Speed of B = 15 m/s. Therefore, Distance traveled by B in 2 mins = 15*120= 1800 meters
Distanc ebetween them after 2 mins = 1800-1200 = 600 meters
ANS--A
If average age is increased by 2 years, total age is increased by 16 years
Total age of the two women = (35+45)+16 = 96 years Average of the two women is 96/2 = 48 years
ANS--E
In the initial solution let the volume of A be 3x and that of B be 2x. When 10 litres of solution is taken out, 6 liters of A and 4 liters of B is taken out. Then 10 liters of B is added.
Therefore, (3x-6)/(2x-4+10)=1/2 => 3x-6 = x + 3 => 2x=9 => x= 9/2 => 5x = 45/2 = 22.5
Volume of initial solution = 22.5 liters
ANS--C
William share: John's share = 7:9 Total amount deposited $ 72,000. William's share = $7/16*72,000= $31,500
John's share = $40,500. William withdrew $ 7,500 and John withdrew $ x.
(31,500-7,500)/(40,500- x) = 4/5 => 24,000/(40,500-x) = 4/5 => x = 40,500-30,000 = $10,500
ANS--C
40/100×620+35/100*4500+X=324×324/18 4×62+35×45+X=18×324 248+1575+X=5832 X=5832-1823=4009
ANS--A
ANS--A
ANS--B
ANS--C
Depositing Rs 4000 in a recurring deposit each month for two years, which is equal to 24 months. So, n=24, PMT = 4000 and FV = 106000.
Now for Annuity FV = C*/i * ((1+i)^n - 1) here i is monthly interest rate, Calculating i=0.0085 or 0.85% per month, annualizing this returns = 10% pa
ANS--C
Maturity value of a recurring deposit is given by M =
R [ (1+i)n – 1] ——————– 1- (1+i) -1/3 Where, M = Maturity value R = Monthly Installment r = Rate of Interest (i) / 400 (Annual interest rate converted into quarterly rate) n = Number of Quarters (Assuming the compounding is done quarterly)
By putting the values given we get M= Rs21,244.00
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 square cm
ANS--C
et 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 metre
Let d be the distance of the building from the foot of the hill
We can see from the sketch tan 30 = 50/d = 0.5773
That gives d = 86.61 m
Let H be the height of the hill
Again from the sketch tan 60 = H/86.61 = 1.7320
That determines the height of the hill = 1.732 x 86.61 = 150 metre
Area of the vertical walls = 2 x 40 x 7 = 560 square metre
Area of the dome = 22/7 x 5 x 40 = 628 .5 square metre
Cost of plastering = 5 x (560+628.5) = Rs 5942.50
ANS--A
A = P(1+r/n)^nt where A=amount, P= principal, r= rate of interest in decimal, n= number of times interest compounded per year, t= time in years
A = 36000(1+0.16/2)^2*3/2 = 36000 (1.08)^3 = 36000*1.25972 = 45,349.63
Therefore, interest = Rs. A - P = 45,349.63 - 36000= Rs. 9,349.63
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] centi metre
ANS--C
Joining the two centers divide the cords into two equal parts, with 12/2 = 6.5cm each.
Applying pythagorus theorem in each circle and triangle sqrt(10^2 - 6^2) + sqrt(15^2 - 6^2)
After Calculating these all You will get 21.75 cm
ANS--A
We use the formula s = ut -1/2 a t^2
So s = 15 x 60 - 0.5 x 0.3 x 60 x 60 = 360 m
So distance from signal would be 400 - 360 =40 metre
The difference between the third generation computers and the fourth generation computers are as follows:
Third Generation
Fourth Generation:
You can just do copy then paste by the office ClipBoard
#Name error occurs when excel didn't recognise a text in Formula
Total thickness of books = 5x20= 100mm Total thickness of papers = 5x0.016 = 0.08 Thickness of stack = 100+0.08 = 100.08 mm Thus 10.008 centimetre
from above we get m+3 = 5
therfore m=2
Distance covered in 25 min= 14 km Distance covered in 1 min= 14/25 km Distance covered in 60 min or 1 hr= 0.56×60 km Distance covered in 5 hr. = 33.6×5 km = 168 km
i shadow = 3.2/5.6 x 10.5 = 6 metre
ii length = 5.6/3.2 x 5 = 8.75 m = 8m 75 centimetre
18 km = 1 cm
72km = 72/18 = 4 cm
Given that: 2kg sugar = 9*10^6 crystals
By Unitary Method 1 kg will have = 9*10^6 /2 = 45*10^5 crystals
1.) 5 kg sugar = 5*45*10^5 = 225*10^5 crystals
2.) 1.2 kg sugar = 1.2* 45*10^5 = 54*10^5 crystals
Let x be the length of model Then, 9/x=12/28 9×28/12= x x=21 centimetre
840 bottles in 6 hr. Then In 1 hr. 840/6=140 bottles then In 5hr. 140*5=700 bottles
Actual length of the bacteria is 5/50,000= 0.0001 cm (ii) Length after 20,000 enlargement 0.0001×2= 2 cm
11 as the number ending in 1 must have a digit having 1 in ones place
The volume of each cuboid = 5 x 5 x 2 (lbh) = 50 cu cm
volume of cube = l x l x l = 10 x 10 x 10 = 1000 cu cm
no. Of cubes required = 100/50 = 20
Value after one year =42000 - (8/100)×42000 =42000-3360 =38640
In 2001 54000*(100/105)*(100/105)=48980 In 2005 54000*(105/100)*(105/100)=59535
For 2 year
A=P (1+r/100)^n
8000* 105/100 * 105/100
=8820
For 3 yr
A=P (1+r/100)^n
8000* 105/100 * 105/100 * 105/100
=9261
interest paid during 3rd year=
amount for 3yr - amount for 2 yr
9261-8820
=441
C.P of VCR = C.P of TV = 8,000. 4% loss on VCR. S.P. of VCR =96% of Rs. 8,000 = Rs. 7,680.
8% gain on TV. S.P. of TV =108% of Rs. 8,000 = Rs. 8640.
Total S.P. = 16,320 Total C.P. 8,000 + 8,000 = 16,000
Profit/Loss % = (S.P. - C.P.)/C.P/ X 100 = 2%
cost = 15500, repair cost = 450
total cost after repairing = 15950
profit on cost price= 15%
selling price = (15950x15/100)+15950 = 18342.50
Selling price of 80 items =2400+(16/100)×2400 =2400+384=2784Rs hence selling price of one item =2784/80=Rs 34.80
percentage of 169 out of 845 is 169/845 X 100=20%
% decrease on monday = (100-20)%=80%
100-(60+30) =100-90= 10℅ like other games,no of people like cricket = (60/100)×5000000=3000000(thirty lakh), no of people like football = (30/100)×5000000=1500000(fifteen lakh), no of people like other game =(10/100)×5000000=500000(fife lakh
Let they played Xmatch , (40/100 )×X=10 or X =(10×100)/40 = 25
answer is 72% of 25 = 18 are good in mathematics Therefore 25-18 = 7 are not good in mathematics
a) 15:30 = 1:2
(b) 1 km = 1000m. So 10 km = 10000 m. Ratio is 5:10000 = 1:2000
(c) 1 Re = 100 paise. So 5 Rs = 500 paise. Ratio is 50 : 500 = 1:100
20+30=50
The sequence is the squqre of odd numbers. Hence the answer is 11^2=121
A compiler is computer software that transforms computer code written in one programming language into another programming language.
Here A/B=3/7 and B/C =2/5
So A:B:C=6:14:35
No. of A officers=6/55*1375=150
ANS--A (Resolution)
Sum of ages of 25 students=25*12=300
Sum of ages of 25 students+ one teacher=26*13=338
Age of teacher=338-300=38
Let CP=100, Then Marked Price=140
Disount=10%, Hence SP=140-14=126
So profit=126-26=26
When profit=26 then CP=100
When profit=325 the CP =325*100/26=Rs1250
Actual CP=RS1250
The given series is an AP
Here First term a=-15, Common diff d=9
Sum of 30 terms=30/2(2*(-15)+29*9)
=15(-30+261)
=15*231
=3465
Let the age of son=x
Then age of father=4x
After 10 years
Son's age=x+10
Fathers age=4x+10
so 2(x+10)=4x+10
Solving we get x=5 years
Son's age=5 years
Focal length=10mm=10/1000 m=0.01m
Power =1/f =1/0.01=100D
Each medium has different refractive index. when light travels from one medium to another it bends due to difference in refractive index.
The amount of energy required to move an electron from valance band to conduction band is called Forbidden enegy gap.
The amount of enegy required to move an electron from valance band to conduction band is called forbidden energy gap
The three quantities with no dimension are
1.solid angle
2.coefficient of friction
3.refractive index
On analysis we found that next term can be found by multiplying the previous number by 2 and adding 11 to it
19=4*2+11
49=19*2+11
Next term=49*2+11=109 (Answer)
Area of parallelogram=Base*Altitude
Area=4.2*5 Cm^2=21 cm^2
ANS--B
X^2+1=0
If we square any number then it will be a positive number and when 1 is added to it the result cannot be zero
ANS--C
ANS--B
ANS--A
ANS--D
ANS--C(300)
ANS--D
Since intelligent student cannot be defined
ANS--B
n(A)=3
n(B)=3
n(A*B)=n(A)*n(B)=3*3=9
Let CP=100
Then Marked price=100*140/100=140
Selling price=140*80/100=112
Profit=112-100=12
Profit %=12/100*100=12%
Ratio of investments=12000:15000:18000 i.e 4:5:6
Share of profits will be in the ratio of investments
Ratio of profits=4:5:6
A' share =4/(4+5+6)*Total investment=3000
Total investment=3000*15/4= Rs11250
Total cost of AC=12500+800+300=13600
Selling price=13600*115/100=15640
ANS--c 2^n
Let Principal=P
Then Time=10 years
SImple Interest=2P-P=P
SI=P*R*T/100
P=P*R*10/100
R=10%
Total travel time=1hr 30 min=90 min
Time when it stopped=12*2=24 min
Total time when bus was moving=90-24=66 min
distance covered =40/60*66=44 Km
ANS
41472/8=5184
51884/9=576
576/8=72
72/9=8
8/8=1
The next term is 1
The portion of field reaped in 10 days by 20 workers=1/3
No. of workers after 10 days=20+20=40
20 workers can reap the remaining fiels in 20 days
40 workers will reap in 10 days
Total no of days=10+10=20 days
ANS--62 is wrong
ANS--207
963-927=36
927-855=72=36*2
855-747=108=36*3
747-603=144=36*4
603-423=180=36*5
427-36*6=207(ANS)
ANS--63
The sequence is of the form 2^n-1
2^6-1=63
ANS--24
The series is in form n^2*6
2^2*6=24
The terms are 15pq+15+9q+25p
=(15pq+9q)+(25p+15)
=3q(5p+3)+5(5p+3)
=(3q+5)(5p+3) ANS
Perimeter=2*22/7*4.9/2=15.4cm
Total lenth used=25*15.4=385 cm
Principal=10000
Interest for 1st year=10000*0.04=400
Principal for second year=10000+400=10400
Interest for second year=10400*0.05=520
Principal for 3rd year=10400+520=10920
Interest for 3rd year=10920*0.06=655.20
Total interest for all 3 years=655.20+520+400=1575.20.
Let P be the principal
When it doubles in 5 years then SI=P
SI=P*R*T/100
P=P*R*5/100
R=20%
When it becomes 7 times the SI=6P
6P=P*20*T/100
T=30 years
Let Cp of 100Litres=Rs100
Then SP of 100 Litres=110
When 20 litres of water is added then amount of milk becomes 120 Litres
SP of 120 litre=120*1.1=132
Increase in profit=132-110=22%
Let the total distance=2X
Then half distance=X
X/21+X/24=10
8X+7X/168=10
15X/168=10
X=168*10/15
X=112
Total distance=2X=112*2=224 KM
Let speed of boat in still water=X
and speed of stream=Y
Upstream speed of boat=X-Y
Downstream speed of boat=X+Y
(X+Y)*2=4(X-Y)
X+Y=2(X-Y)
3Y=X
Putting X=3Y
(X+Y)*2=16
(3Y+Y)*2=16
8Y=16
Y=16/8=2
X=3Y=3*2=6KM/hr
Speed of boat in still water=6KM/hr
Time=2 yr 73 days=2.2 yr
Amount=30000(1+0.1)^2.2=36998.58
Interest=36998.58-30000=6998.58
Let P be the principal and R be the rate of interest
P(1+R/100)^3=8000
Squaring we get
P^2(1+R/100)^6=8000*8000
P(1+R/100)^6=8000*8000/P---(1)
Now P(1+R/100)^6=12167---(2)
Putting the value from (1) into (2)
8000*8000/P=12167
P=8000*8000/12167=5260
P=2400
Let R be rate of interest and T time period
SI=PRT/100
864=2400*T*T/100
T^2=864/24=36
T=6 years
Let he buys 3 litres of milk.
CP of 3 litres=6*3=18
The he adds 1/3 water so total milk becomes 3+1=4 litres
SP of 4 litres=4*7.2=28.8
Gain%=(28.8-18)/18*100
=10.8/18*100=60%
Let cost of 1cm=Rs1
Then CP=Rs33
SP=RS36
Profit %=3/33*100=100/11=9.09%
Ratio of capitals=25000:40000=5:8
Profit ratio=5:8
SP of 36 oranges=Rs1
CP of 36 oranges=36*100/99=1.01
For 8% gain
SP=1.01*108/100=1.09
Let cost of one Pen=X and cost of 1 pencil=Y
So 2X+5Y=240--(1)
4X+2Y=336--(2)
Dividing (2) by 2
2X+Y=118---(3)
Subtracting (3) from (1)
4Y=122
Y=122/4=30.5
Putting Y=30.5 in (3) and solving we get X=43.75
Cost of 1pen=Rs43.75
Cost of 1 pencil=Rs30.5
Let the numbers be 7x and 11x
Product of numbers=HCF*LCM
7x*11x=13*1001
x^2=13*1001/77
x^2=16
x=13
Larger number=11*13=143
LCM of 12,25,45,60=900
The smallest 5 digit number is 10000
So 10000/900=11.11
So smallest number=900*12=10800
The product of two prime numbers is always equal to its LCM
Hence LCM=493
Let X be 4th proportional
So 100*X=50*40
X=2000/100=20
Total amount Julie has=Rs250
Amount spent on sweater=Rs75
amount left=15
Amount spent on jacket=250-(75+15) =250-90=Rs160
ANS--B--Lactobacillus
CNG stands for Compressed Natural Gas
ANS--A-- Resolution
ANS--C--Superscript
The two main products of photosynthesis are Oxygen and Glucose
Polar star is the brightest star located in north and it remains stationary with respect to earth
The utensils are made of metals and hence it is coated balck to absorb maximum heat.
Metals have free electrons which makes it good conductor of heat and electricity
ANS--D--Zinc
Sound waves move in a medium by producing vibration.
Electrical power P=V*I
Switch is an electrical component of circuit which makes or breaks a circuit
We should wear light colured clothes in summer because light coloured clothes reflects most of light and we feel less heat.
The motion of an object which repeats itself in fixed interval of time is called periodic motion. The motion of a pendulum in clock and movement of earth around sun and on its own axis are examples of periodic motion.
The thickness of five rupee coin is 1.9 mm
The resistor opposes the flow of current, Hence it reduces the current if resistance is increased.
I=V/R
Current is inversely proportonal to resistance.
An inductor oppese sudden change of current in a circuit. It stores magnetic energy
capacitors are used to store electric energy.
Plasma is the fourth state of matter.
Relative density has no unit because it is the ratio of same units which gets cancelled.
Power=Work/Time
Trivial, Unimportant
First you should have good knowledge of grammar and literature. Thoroughly study the chapter that you are going to teach. Make some interesting stories in your teaching class. Create some humour in your teaching.
Function of mumcipal corporation are as follows.
1) providing water and electricity to residents
2)Providing roads and sewage facility
3) Registration of birth, death marriages etc
4)Providing education for kids
First war of indian independence
Raja ram mohan roy was the founder of bramha sabha and he played an important role in abolition of sati pratha.
Doctrine of lapse was a rule started by lord dalhousie according to which if a king died without a natural heir then the british will take over the control of that state.
Bronze is an alloy of copper , zinc and tin
Arya samaj was founded in 1875 by dayanand saraswati. He was a sanskrit scholar.The DAV group of schools are run by this samaj.
Bell metal is an alloy of copper and tin
SAARC was formed for collaboration among south asian countries.To grow trade and tourism among these countries.
Let the distance be X
Time taken uphill =X/10
Time taken downhill=X/36
Total time taken=X/10+X/36=23X/180
Average Speed=2X/(23X/180)=15.65KM/HR
n(A*B)=n(A)*n(B)=3*3=9
ANS--B
Womesh chandra bonerjee
The vice president of India
The king Ashoka
British naval force
Vijayalaya
Lord Mountbatten
ANS--A (0.1)
F=1/Power=1/0.5=2metre
Sun is the main source of solar energy
modified drum technology
Balochistan
gujrat
1907 at Jamshedpur
Wheat is Rabi crop
Hirakunad dam
Bajra
Limestone, clay,shale etc
kerala
karnataka
odisha
Assam
Mangrove
Bokaro and durgapur
puttalam
Up, punjab, Haryana, MP
Solar, Wind energy, Tidal energy, Geothermal
F=1/power=1/0.2=5m
Design of ship, submarine, boats
water and carbon dioxide
sabaceous gland
Hepatic artery
solid waste, liquid waste, organic waste
stratosphere
water erosion,wind erosion,soil erosion, glacier erosion,wave erosion
Reduce, Recover, Reuse,Recycle
GOODS
29 is wrong
As other numbers are powers of the number
Line
No of ways of picking 3 balls=15C3=455
No of ways to pick 2 green balls=6C2=15
No of way to pick one blue ball=5C1=5
Probability of picking 2 green and 1 blue ball=15*6/455=15/91
Spokes
One newton is the force required to accelerate one kilogram mass at the rate of one metre per second squared in the direction of force.
LASER stands for Light amplification by stimulated emission of radiation.
20
Answer-- The boy swam ACROSS the river and REACHED the base of the river.
I am sorry FOR the mistake.
He TOLD her that she would pass.
You NEED NOT worry. I'm a very careful driver.
The PLACE chosen for construction of the building is in the heart of the city.
AS pollution control measures are expensive, many industries hesitate to adopt them.
Although there is HEAVY gunfire,there is no stiff resistance to the army.
ANS --C-EDRIRL
OPEC stands for Organization of petroleum exporting countries
Aluminium is more reactive than iron but it does not corrode easily because it forms a layer of aluminium oxide which prevents it from further reaction with air.On the other hand iron keeps on reacting with air and keeps on corroding.
Since
Magnetite as it has high iron content
Lionel Robbins
Agriculture is the main stay of Indian Economy
Third five year plan
Napolean was defeated by Two armies of seventh coalition-- A British led allied army under the command of Duke wellington and a Prussian army under the command of Field Marshal Blucher
Khan abdul gaffar khan was the leader of khilafat movement
Ural mountains separate Asia and Europe.
Chauri chaura is situated in Gorakhpur district of UP
There are 18 puranas
Indra is the main GOD in Rig veda
Portugal
Iceland
Russia
Champaran is situated in Bihar and it has been divided into two districts east champaran and west champaran.
A--Voter
Ghiyas-uddin-Tughlaq was the founder of Tughlaq dynasty
Qutub-uddin Aibak was the founder of slave dynasty.
Bahlol Lodhi was the founder of Lodhi dynasty.
Babar was the founder of Mughal empire.
Vijayalaya chola was the founder of Chola empire.
Simuka was the founder of Satavahana dynasty.
Kajula Kadaphis was the founder of Kushan empire.
Simhavishnu was the founder of Pallavas dynasty.
Pulakesin was the founder of Chalukyas dynasty.
Danti Durga was the founder of Rashtrakutas.
Jalal-uddin Khilji was the founder of Khilji dynasty.
AYE means YES
No of bottles filled in one hour=1120/7=160
No of bottles in 6 hours=160*6=960 bottles
Sources of fresh water are lakes, rivers,glaciers
Fourth proportional=50*40/100=20
We know that K.E=1/2 mv^2
When speed is halved, then KE=1/4th of Original KE
When speed is tripled, KE=9 times of original KE.
Change in KE=9-1/4=35/4
Mother Teresa
Mother Teresa
Margaret Thatcher
Gertrude Caroline Ederle of USA in 1926
Kailashanatha temple in Ellora caves
Krem Liat Prah in Meghalaya
Sonepur in Bihar
Sri Sanmukhanand hall in Mumbai
ITC Grand Chola Sheraton in Chennai
Highest airport is in Leh , Jammu and Kashmir
St. Thomas Syro-Malabar catholic church in kerala
Shri Darbar Sahib in amritsar
Deepest river valley in india are Bhagirathi and Alaknanda.
Air is a mixture
George Boole
Howrah Bridge in Kolkata
Indira Gandhi Canal in Rajasthan
Bharat Ratna
Birla Planetarium in Kolkata
Statue of unity in Gujarat
majuli in Assam
Jama Masjid in Delhi
Buland Darwaza
Delhi to Kolkata via Patna
Siachen glacier
Dorothy arzner
ANS--D
Except option D all the letter pairs are equal distance from beginning and from last.
The first earth satellite named sputnik was launched by Soviet Union(USSR) on 4th october 1957.
Nacl is the salt and it is neither acidic nor basic.
Power of lens =1/f(metres)
f = 1/(-2)=-50cm
Negative sign indicates that lens is concave. Focal length = (-50cm)
We get Silk from silkworms.
Sum of ratio=3+2+5=10
Total money=1260
C's share=(5/10)*1260=630
23+x=345
or, x=345-23
or, x=322
The electrons can reside in valence band or conduction band. The electron can be moved from valence band to conduction band when enegry is supplied to then.So this band gap is called forbidden energy gap.
Let the income be x
He gives 20% and 30% to his sons so remaining income is 50% of income.
50% of x=x/2
He gives 10% of x/2 to the beggar
So he is left with 90% of x/2 which equals 10080
(90/100)*x/2=10080
x=(10080*20)/9
x=22400
So his income is Rs22400
No of elements=3
No of proper subsets=2^3-2=8-2=6
Hence option C is correct answer
Officials from Nazi Germany,Fascist Itlay and Imperial Japan were members of Rome -Berlin and Tokyo Axis.
Incisors, Premolars and Canines form the milk teeth of 3-4 year child.
The dimension of given cuboid is 5cm,2 cm and 3 cm.
For making the cube each side should be of same length.
LCM of 5,2,3=30
Hence the cube should have each side of 30 cm.
Volume of cube=30cm*30cm*30cm=27000cm^3
Volume of one cuboid=5cm*2cm*3cm=30cm^3
Number of cuboids requred =27000/30=900
Hence 900 cuboids are required.
The force of gravitation between two bodies does not depend on the sum of the masses of two bodies.
In young's double slit experiment, bright fringes are of (A) equal widths and unequal intensities
ANS--OPTION B Lactobacillus bacteria converts milk into curd.
ANS B
The two conductors will repel each other as the current flowing in two wires are in opposite direction.
The most fertile plain in the world is INDO--GANGETIC Plain as it is made up of alluvial soil deposited by rivers.
The battle of waterloo sealed the fate of France in 1815.
1200 has two significant digits.
Average age of 10 persons is 60
So the sum of ages of 10 persons =10*60=600
Sum of ages of 8 persons =488
Sum of ages of remaining 2 persons=600-488=112
Average age of remaining 2 persons=112/2=56 years
E-mail stands for Electronic mail. It is sent via internet. Some email service providers are Gmail,Yahoo etc.
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