B.Tech/B.E. | Igniting the minds | 2002 |
3 Years
I have taught unofficially for a span of 1 year for class 8 to 10 mathematics while working. I have handled quantitative aptitude unofficially for a span of 1 year exhaustively.
Home Tuition Only
400.00
Class 9 - 10 | Mathematics, Physics, Chemistry, Algebra, All Boards, All Medium | INR 400.00 /hour |
Class 11 - 12 | Mathematics, Physics, Chemistry, All Boards, All Medium | INR 350.00 /hour |
Class 6 - 8 | Mathematics, Physics, English, Chemistry, CBSE | INR 400.00 /hour |
Competitive Exams | Bank Clerical, IBPS, SBI Exam, TNPSC | INR 800.00 /hour |
Number of students good in mathematics is 72% of 25. That's (72/100)*25=18. Hence number of students not good in mathematics is 25-18=7
Let variables P,M,C denote the marks obtained in subjects Physics, Mathematics,Chemistry respectively. Then, P+M+C-120=C P+M=120 Average =120/2 =60
Number of students learning mathematics and phscics = 30............ (1) Number of students learning mathematics, physics,chemistry =18.......(2) Hence number of students learning mathematics and physics only = (1)-(2)=12....... (3) Similarly, number of students learning chemistry = 28-18=10.......... (4) Hence number of students learning mathematics alone = 100-(10+12+18)=60
students studying maths alone= students studying maths -(students studying maths and physics + students studying maths and chemistry -students studying all 3 subjects) M= 120-(40+50-x)................................... (1) Similarly, P=90-(30+40-x).........................................(2) and C=70-(50+30-x)......................................... (3) In the above equations, M,P, C denote students studying maths, physics, chemistry alone respectively and x denotes students studying all subjects. Hence 200= M+P+C+x+20+(40-x)+(50-x)+(30-x) Solving we get x=20
teachers teaching only mathematics = teachers teaching maths - teachers teaching physics and mathematics =12-4=8 20-12=8=teachers teaching only physics and teachers teaching physics =12
Their next immediate ringing at the same moment will occur in the lcm of their ringing frequency. LCM =120 seconds. Hence theiron next immediate ringing together will occur at 11.27 am
Let the unit digit be y and let the tenth's digit be x. As unit digit is 5 more than the tenth's digit, x+5=y.....................(1) Exact value of the number =10x+y............ (2) Upon reversing the digits, Value becomes 10y+x.................. (3) After reversing, the new number is 4 less than twice the original number. Hence, 2 (10x+y)=10y+x+4 2 (11x+5)=11x+54 (substituting the value of your from equation (1)) Solving, we get x=4,y=9 Hence the original number is 49
Let the first number be x. As the second number is 3 more than the first, it is (x+3). As sum of their squares is 369, x^2 +(x+3)^2 =369 2x^2 +6x -360=0 Solving by prime factorization, x=12 and x+3=15. Hence sum of the 2 numbers =27
Let the number of horses be x. Let the number of ducks be y. As the sum total of their legs =120, 4x+2y = 120..................................... (1) As their total heads is 50, x+y=50................................................ (2) Multiplying equation (2) throughout and by elimination by Equating the coefficients, Equation (1)-Equation (2) gives x=10 Hence number of horses =10
Let the 2 numbers be x and y. As product of the numbers is 93, x*y=93........................ (1) Also, their difference is 28 x-y=28........................... (2). Substituting the value of y from equation (1), x-(93/x)=28 x^2-93=28x x^2-93-28x=0 Solving by prime factorization, we get, x=31. So, y=3. Difference of their reciprocally is, (1÷31)-(1÷3)=(-28/93)
Let the quantity of first fuel be x. It is akin to the concept of average speed. Here it is average cost. Average cost =22.5= (21x+(24*30))/(x+30) Solving, we get x =30 litres.
As there is no net loss or gain in the transaction, average cost price per kg equals average selling price per kg. Let x be the selling price per kg of 2nd variety of sugar. Then, average C.P=AVERAGE S.P=28=((32*60)+(x*60))/(60+48) 3024=1920+48x Solving, we get x=23. Hence the price of the new type of sugar=Rs23/kg
As there is 25% gain in the transaction, C.P of the product is given as, (450-x)/x=25/100 where x denotes C.P (Cost Price of the product ). Hence average C.P =360 Let z denote the quantity of first type of fluor and y denote the cost/kg of another type of fluor. Average C.P=360= (620z+130y)/(z+130) As the above equation has 2 unknowns,it has infinite solutions. one of the solution is got by assuming cost per kg of second fluor. Let it be 200. Now 360 (z+130)=620z+(130×200) here we get z=80 Hence by assuming second fluor cost/kg to be 200, we get the quantity of first fluor=80 kg
In both the containers, initially 7 litres of milk and 3 litres of water are added.Further, 4 litres of mixture in container 2 is added to container 1 and it gets filled. Hence capacity of container 1 is 14 litres. Hence capacity of container 2 is 140 litres (10 times first). Let x be the quantity of milk added to container 2.subsequently, 134-x becomes the quantity of water added. From the existing 6 litres in container 2, milk=(9/5)and water=(21/5) so, ((9/5)+x)/((21/5)+(134-x))=2/3 solving, we get x=54.2 litres water added =134-54.2=79.8 litres Hence ratio of milk to water added =54.2:79.8 =0.68:1
In the total 60 litres, acid and water are in the ratio 2:1.Hence 2x+x=60 hence x=20 so, acid and water are 40 and 20 litres respectively. 40/(20+y)=1/2 where y denotes quantity of water to be added to make the ratio of acid: water 1:2 hence y=60 litres.
By selling the mixture at Rs350 per quintal, the merchant gets a profit of 25% Hence, (350-x)/x=25/100 where x is the average cost price per quintal of the mixture . hence average C.P =Rs 280 let the quantity of first quality tea mixed be y quintal. average C.P =280=(290y+(250×150))/(y+150) Solving, we get y=450 quintal Hence 450 quintal of first quality has to be added.
80% of 50 litres amountso to 40 litres and hence in a 50 litre quantity, wine=40 litres and water=10 litres. Let the amount of water added to the mixture to make water 50% be x. so, (10+x)÷(50+x)=50÷100 solving, we get x=30 litres. Hence 30 litres of water need to be added.
By selling the rice mixture at Rs16/kg,the profit % attained is 15. Hence average C.P /kg of rice mixture is given by, (16-x)/x=15/100 where x denotes average C.P /kg of rice mixture. Solving, C.P=Rs 13.91 Let the quantity of rice costing Rs 12.85 to be mixed with another rice costing Rs14.8 be y. so, 13.91=(12.85y+(14.8*25))/(y+25) Solving, we get y=21 kg Hence 21 kg of rice worth Rs12.85/kg has to be mixed.
Speed of the bicycle is 125/5=25m/s speed of the man=(4/5)×Speed of bicycle =20m/s For 20 metres, the man takes 1 second. Therefore for 40 metres, the man will take 2 seconds.
Upstream speed= Speed in still water-Speed of current = 5-1= 4 km/hr Downstream speed =Speed in still water+speed of current=5+1=6 km/hr. Let the distance covered be 'd' . As both in upstream and downstream, distance covered is sameternal and the total time taken is 1 hour, (d/6)+(d/4)=1 Solving, we get d= 2.4 km Hence distance =2.4 km
Let the speed of boat in still water be x km/hr As distance upstream and downstream are both same, (x+3)*1=(x-3)*1.5 Solving, we get, x=15 Speed of boat in still water = 15 km/hr
Speed in the upstream = (d/3)=U............(1) Speed in the downstream=(d/2)=D.........(2) where 'd' denotes the distance covered,'U' denotes upstream speed and 'D' denotes downstream speed. Speed of boat in still water=( D+U)/2=5d/12 Speed of stream = (D-U)/2=d/12 Hence ratio of speed of boat in still water to speed of stream =5:1
As shika runs slower than vikash, let us say shika has to start x metres ahead of vikash so as to finish the race in dead heat. Distance covered by shika is such that she completes the race in same 4 minutes. For 4 minutes and 10 seconds (25/6) minutes, shika travels 25m For 4 minutes, she would have travelled (4×2)/(6/25)km=1.92km x=(2-1.92)=0.08km=80m Hence vikash must give shika a start of 80 metres.
As both the boats move towards each other, One is moving with the tide and the other,against it. As speed of boat in still water is 20k/hr, t*(20+x+20-x)=50 Where x denotes speed of the stream and t,the time taken by both boats from starting pount to meeting point. t=5/4 hours=75 minutes. Hence they will meet after 75 minutes.
Let the time taken in the first half be denoted by t1. In order to compensate the idle time of 2 hours, the train's speed is increased by 40 km/hr .Let the time taken in the second half be t2. t2=t1-2....................... (1) If S1 denotes the speed in first half, (S1*t1)+((S1+40)*t2)=1000 (S1*t1)=500................... (2) ((S1+40)*t2)=500............ (3) Equating (2) and (3), we get, S1*t1=(S1+40)*t2 S1*t1=(S1+40)*(t1-2) (from equation (1)).... (4) Replacing S1 with 500/t1 obtained from equation (2) in equation (4) and solving the quadratic equation, we get, t1=6.385 hours t1-2 = 4.385 hours S1/(S1+40)=(t1-2)/t1=4.385/6.385 =0.6867:1 Hence ratio of speeds is 0.6867:1
Length of the train = relative speed of train and person*time taken to pass the person completely. Hence, d= (S-2)*(9/3600)...... (1) d= (S-4)*(10/3600)....... (2) where d and S denotes the length and speed of the train respectively. EquationS (1) and (2) are applicable for persons travelling at 2km/hr and 4km/hr respectively in the direction of train. Equating (1) and (2) we get S=22km/hr Hence length of train (22-2)*(9/3600)km =50metres
The given data is sufficient. As Lucy'so mother is 43 years 5 yearso ago,her mother 'so current age is 48years. As Lucy is half her mother's age , she is (1/2)*48=24 years. As Kate is (1/3) of her mother's age , she is (1/3)*48=16. Hence difference in ages of Kate and Lucy is (24-16)=8 years and hence Kate is 8 years younger to lucy
1. Let the initial speed and initial time taken by ram be denoted by 'S' m/s and 't' seconds respectively. If he travels at 25% greater speed, he reaches 20 seconds earlier. Hence (1.25S)*(t-20)=(t*S) Cancelling S and solving for 't', we get t=100 seconds. 2.As per 2nd condition, (t/2) +50=the Solving, we get, t=100 seconds. 3.As per condition 3, (S/4)*(t+300)=(t*S) Solving, we get, t= 100 seconds. Hence we observe that in all cases , time taken by Ram is 100 seconds.
Let the initial number of apples possessed by John and James be x and y respectively . By giving 7 apples to James, John will have only half the apples as that of James. So, 2*(x-7)=y+7........................... (1) If James gives one apple to John, both will have equal number of apples. So, x+1=y-1............................... (2) Solving (1) and (2), we get x=23,y=25. Therefore John had initially 23 apples and James had initially 25 apples.
As heath is neither taller nor shorter than Ricky, both are of equal height. Hence heath is 9 inches taller than stacy (as given that Ricky is 9 inches taller than stacy).