SIMPLE INTEREST QUIZ 001
 A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
(a) Rs. 650
(b) Rs. 690
(c) Rs. 698
(d) Rs. 700
(e) None of these
Answer Option :3
Solution »
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs. (39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698
 What will be the ratio of simple interest earned by a certain amount at the same rate of interest for 6 years and that for 9 years?
(a) 3:2
(b) 2:3
(c) 1:3
(d) Data inadequate
(e) None of these
Answer Option :2
Solution »
Let the principal be P and rate of interest be R%.
Required ratio =
(( P x R x 6 )/100)
= (6PR/9PR) = 6/9= 2 : 3.
 What annual instalment will discharge a debt of Rs. 1,092 due in 3 years at 12% simple interest?
(a) Rs. 325
(b) Rs. 545
(c) Rs. 560
(d) Rs. 540
(e) None of these
Answer Option :1
Solution »
Let each instalment be Rs. x.
1st year = [x + (x * 12 * 2)/100]
2nd year = [ x + (x *12 * 1)/100]
3rd year = x
Then, [x + (x * 12 * 2)/100] + [ x + (x *12 * 1)/100] + x =1092
3x + ( 24x/100 ) + ( 12x/100 ) = 1092
336x =109200
x = 325
Each instalment = Rs. 325
 Reena took a loan of Rs. 1,200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
(a) 3.6
(b) 6
(c) 18
(d) cannot be determined
(e) None of these
Answer Option :2
Solution »
Let rate = R% and time = R years.
Then, (1200 * R * R)\100 = 432
= R<sup>2</sup> = 36
= R = 6
 Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3,508, what was the amount invested in Scheme B?
(a) Rs. 6,400
(b) Rs. 6,500
(c) Rs. 7,200
(d) Rs. 7,500
(e) None of these
Answer Option :1
Solution »
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13,900 – x).
Then,
=> 28x – 22x = 350800 – (13900 x 22)
=> 6x = 45000
=> x = 7500.
So, sum invested in Scheme B = Rs. (13,900 – 7,500) = Rs. 6,400.
 A father left a will of Rs. 35 lakhs between his two daughters aged 8.5 and 16 such that they may get equal amounts when each of them reach the age of 21 years. The original amount of Rs. 35 lakhs has been instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at the time of the will?
(a) Rs. 17.5 lakhs
(b) Rs. 21 lakhs
(c) Rs. 20 lakhs
(d) Rs. 15 lakhs
(e) Rs. 25 lakhs
Answer Option :2
Solution »
Let Rs. x be the amount that the elder daughter got at the time of the will. Therefore, the younger daughter got (3,500,000 – x).
 How long will it take for a sum of money to grow from Rs. 1,250 to 10, 000, if it is invested at 12.5% p.a. simple interest?
(a) 56
(b) 45
(c) 68
(d) 78
(e) 43
Answer Option :1
Solution »
Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the number of years for which it is invested, r is the rate of interest per annum
In this case, Rs. 1,250 has become Rs. 10,000.
Therefore, the interest earned = 10,000 – 1,250 = 8,750.
8750 = [(1250*n*12.5)/100]
=> n = 700 / 12.5 = 56 years.
 If the simple interest on a certain sum of money is 4/25 of the sum and the rate per cent equals the number years, then the rate of interest per annum is:
(a) 4%
(b) 5%
(c) 8%
(d) 10%
(e) None of these
Answer Option :1
Solution »
Let the principal be Rs x.
Then the SI =4/25x
Rate of interest = Time
r= (100*4/25x)/ x*r
r<sup>2</sup> =400/25
r=20/5=4%
 A financier lend money at simple interest, but he includes the interest every six months for calculating the principal. If he is changing an interest of 10%, the effective rate of interest becomes?
(a) 10%
(b) 11.5%
(c) 10.25%
(d) 12%
(e) None of these
Answer Option :3
Solution »
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 * 1/2)/100] = Rs. 5
Next 6 months 10% of 5 is Rs. 2 is added.
S.I. for last 6 months = Rs. [(102 * 10 *1/2)/100] = Rs. 5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
R = (110.25 – 100) = 10.25%
 Ragav purchases a coat for Rs. 2,400 cash or for Rs. 1,000 cash down payments and two monthly instalments of Rs. 800 each. Find the rate of interest.
(a) 80%
(b) 100%
(c) 110%
(d) 120%
(e) None
Answer Option :4
Solution »
Amount as a principal for 2 month = 2400 – 1000 = 1400
At the rate of r% per annum after 2 months,
D 1400 will amount to Rs 1400 + (1400*r*2/100*12)
Total amount for 2 instalments at the end of second month
D 800+(800+(800*r*1/100*12))
Then 1400 + 2800*r/1200=1600 + 800*r/1200
R=120%
 Two persons P and Q borrowed Rs. 40,000/ and Rs. 60,000/ respectively from R at different rates of simple interest. The interest payable by P at the end of the first four years and that payable by Q at the end of the first three years is the same. If the total interest payable by P and Q for one year is Rs. 8,400/, then at what rate did Q borrow the money from R?
(a) 8
(b) 10
(c) 12
(d) 9
(e) None
Answer Option :1
Solution »
40000*4*R<sub>1</sub>/100=60000*3*R<sub>2</sub>/100
R<sub>1</sub>=9/8R<sub>2</sub>
1yr interest
40000*1*r<sub>1</sub>/100+60000*1*R<sub>2</sub>/100=8400
4R<sub>2</sub>+6R<sub>2</sub>=84
Then, substitute 4(9/8R<sub>2</sub>)+6R<sub>2</sub>=84==>
R<sub>2</sub>=8
 The rate of interest on a sum of money is 4% per annum for the first 2 years, 6% per annum for the period next 4 years, 8% per annum for the period beyond 6 years. If the simple interest accrued by the sum for a total period of 9 years is Rs. 1680, what is the sum?
(a) Rs. 3,000
(b) Rs. 5,000
(c) Rs. 4,700
(d) Rs. 5,500
(e) Rs. 7,580
Answer Option :1
Solution »
SI at the rate of 4% for 2 years ,
=( P* 4*2)/100 = 8P/100
SI at the rate of 6% for 4 years ,
(P*6*4)/100 = 24P/100
SI for the next 3 years
SI = (P*8*3)/100 = 24P/100
Total SI = 8P/100 + 24P/100 + 24P/100
=>P = (1680*100 )/56 = 3000
 The rates of simple interest in two banks A and B are in the ratio 5 : 4. A person wants to deposit his total savings in two banks in such a way that he receives equal halfyearly interest from both. He should deposit the savings in banks A and B in the ratio
(a) 2:5
(b) 4 :5
(c) 5:2
(d) 5:4
(e) None of these
Answer Option :2
Solution »
 The rate of interest on a sum of money is 4% per annum for the first 2 years, 6% per annum for the next 4 years and 8% per annum for the period beyond 6 years. If the simple interest accrued by the sum for a total period of 9 years is Rs. 1,120, then the sum is
(a) Rs. 2,400
(b) Rs. 2,200
(c) Rs. 2,000
(d) None of these.
Answer Option :3
Solution »
 A person invested some amount at the rate of 12% simple interest and a certain amount at the rate of 10% simple interest. He received yearly interest of Rs. 130. But if he had interchanged the amounts invested, he would have received Rs. 4 more as interest. How much did he invest at 12% simple interest?
(a) Rs. 700
(b) Rs. 800
(c) Rs. 500
(d) Rs. 400
(e) None of these
Answer Option :3
Solution »
 Published in SIMPLE INTEREST, Uncategorized
TIME AND DISTANCE QUIZ 001
 Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
(a) Insufficient data
(b) 3 : 1
(c) 1 : 3
(d) 3 : 2
(e) None of these
Explanation:
Let the speed of the trains be x and y respectively
Length of train1 = 27x
Length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x + y) = 23
=> (27x + 17 y) = 23(x + y)
=> 4x = 6y
=> x/y = 6/4 = 3/2
 A train starts from Delhi at 6.00 a.m. and reaches Meerut at 10 a.m. The other train starts from Meerut at 8 a.m. and reaches Delhi at 11.30 a.m. If the distance between Delhi and Meerut is 200 km, then at what time did the two trains meet each other?
(a) 8.56 a.m
(b) 8.46 a.m
(c) 7.56 a.m
(d) 8.30 a.m
(e) None of these
Explanation:
The Speed of the train starting from Delhi = 200/4 = 50 km/h
The Speed of train starting from Meerut = 200/3.5 = 400/7 km/h
Suppose the two trains meet x hours after 6.00 am
Then x X 50 + (x – 2) x 400/7 = 200
or, 350x + 400x – 800 = 1400
or, 750x = 2200
or, x = 2200/750 = 2h 56 min
Hence, the required time = 8.56 am
 A train does a journey without stopping in 8 hours. If it had traveled 5 km an hour faster, it would have done the journey in 6 hours 40 min. What is its slower speed?
(a) 35 kmph
(b) 25 kmph
(c) 40kmph
(d) 30 kmph
(e) None of these
Explanation:
Let its slower speed = V km per hour
Here distance is same in both the cases
Using the formula = V1 x t1 = V2 x t2
or, V x 8 = (V + 5) x 20/3
or, 24V = (V + 5) x 20
V= 25 km/h
Thus, Slower speed of train is 25 km/h.
 A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
(a) 81 km/hr
(b) 88 km/hr
(c) 62 km/hr
(d) 46 km/hr
(e) None of these
Explanation:
Let x is the length of the train in meter and y is its speed in kmph
x/8.4 = (y – 4.5)(10/36) —(1)
x/8.5 = (y – 5.4)(10/36) —(2)
Dividing 1 by 2
8.5/8.4 = (y – 4.5)/ (y – 5.4)
=> 8.4y – 8.4 × 4.5 = 8.5y – 8.5 × 5.4
0.1y = 8.5 × 5.4 – 8.4 × 4.5
=> .1y = 45.9 – 37.8 = 8.1
=> y = 81 km/hr
 A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
(a) 36
(b) 38
(c) 40
(d) 42
(e) None of these
Explanation:
Let the distance be x km, the speed in which he moved = v kmph
Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes
=> (x/v) – (x/(v+3)) = 40/60
⇒2v(v+3)=9x…………….(Equation1)
Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
=> (x/ (v2)) – (x/v) = 40/60
⇒v(v−2)=3x…………….(Equation2)
Eq 1/ Eq 2 = 2(v+3)/ (v – 2) = 3
⇒2v+6=3v−6⇒v=12
Substituting this value of v in Equation 1⇒2×12×15=9x
Hence distance = 40 km
 The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours?
(a) 422km
(b) 552km
(c) 502km
(d) 492km
(e) None of these
Explanation:
Given that distance travelled in 1st hour = 35 km
and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
Total Distance Travelled = [35 + 37 + 39 + … (12 terms)]
This is an Arithmetic Progression(AP) with first term, a=35, number of terms, n = 12 and common difference, d=2.
Hence, [35+37+39+… (12 terms)]=S12=12/2[2×35+(12−1)2]=6[70+22]=6×92=552
Hence the total distance travelled = 552 km
 Two trains A and B, 100 m long are moving on parallel tracks at speeds of 20 m/s and 30 m/s respectively. They are travelling in opposite direction. The driver of train A sees the driver of train B when he is closest to high. He throws a ball at a speed of 2 m/s which hits the tail of train B. What is the distance between the two trains?
(a) 0 m
(b) 10 m
(c) 4 m
(d) 8 m
(e) 5 m
Explanation:
Since the trains are travelling in opposite direction velocity for the driver of the faster train = 50 m/s
Distance travelled = length of the train = 100 m
Time taken by the ball from one train to the other = 100/50 = 2 seconds
Ball in thrown at 2 m/s, distance between the two trains = 2 × 2 = 4 m.
 Two trains, A and B, start from stations X and Y towards each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach Y and X respectively after they meet if train A is moving at 45 km/hr., then the speed of the train B is
(a) 60 km/hr
(b) 64.8 km/hr
(c) 54 km/hr
(d) 37.5 km/hr
(e) None of these
Explanation:
 An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 (2/3) hours, it most travel at a speed of:
(a) 300 kmph
(b) 360 kmph
(c) 600 kmph
(d) 720 kmph
(e) None of these
Explanation:
Distance = 240 X 5= 1200 km.
 A train is travelling at 48 kmph. It crosses another train having half of its length, travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
(a) 500 m
(b) 360 m
(c) 480 m
(d) 400 m
(e) 450 m
Directions (11): The ratio of time taken by Hunny and Bunny to swim a certain distance downstream in a river is 3 : 4 respectively. The time taken by Bunny to cover a certain distance upstream is 50% more than the time taken by him to cover the same distance downstream.
 What is the ratio of speed of Hunny to that of Bunny?
(a) 7 : 5
(b) 7 : 9
(c) 2 : 5
(d) 6 : 7
(e) None of these
Explanation:
Let, speed of Hunny be ‘a ’
Speed of Bunny be ‘b ’
And speed of stream be ‘r ’,
 Both of them hired a boat that runs with a speed equal to the sum of their individual speeds. If Hunny can cover a straight path of length 14 km in 60 minutes, then find the time taken by both of them to travel a distance of 48 km to and fro by the hired boat?
(a) 5 (4/143) hr.
(b) 2 (4/143) hr.
(c) 3 (4/143) hr.
(d) 4 (4/143) hr.
(e) None of these
Explanation:
 Two ship travelling at 30 km/hr and 90 km/hr head directly towards each other, they are 120 km apart at the starting time. How far apart are they (in Km.) at one minute before they collide.
(a) 1 km
(b) 2 km
(c) 3 km
(d) 4 km
(e) None of the above
Explanation:
If the final one minute before collision, the two ship
 A, B and C start from the same place and travels in same direction at speeds of 20, 30 and 40 km/hr respectively. B starts 3hour after A. If B and C overtake A at the same instant. How many hours after A did C start.
(b) 3.25
(c) 4.5
(d) 5.5
(e) None of the above
Explanation:
Speed of A, B and C are 30 km/hr, 40 km/hr and 60 km/hr respectively.
B Start when A already travelled for 3 h and covered
= 3 × 20 = 60 km
It means when B overtake A, A has travelled for 9 hr and B for 6 hr.
It is given that B and C overtake A at a same instance, It means when C overtakes A, both of them will have covered the same distance.
Let C takes hours to cover the same as covered by A in 9 hr.
9*20= t*40
t= 4.5
C started after (9 – 4.5 = 4.5), when A started.
 There is from point A and finish at the same point after touching points B, C and D. You, then drive 20 km interior towards the tree from point A and from there, reach somewhere in between B and C on
the ring road. How much distance do you have to travel from the tree to reach the point between B and C on the ring road?
(a) 80 km
(b) 15 m
(c) 20 km
(d) 40 m
Explanation:
 Published in TIME AND DISTANCE
TIME AND WORK QUIZ 001
 A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
(a) 20 days
(b) 22 1/2 days
(c) 25 days
(d) 30 days
(e) None of these
Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 – 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes = 3/2 x 60 = 90 days.
So, A takes 30 days to do the work.
A’s 1 day’s work = 1/30
B’s 1 day’s work = 1/90
A’s and B’s one day work = 1/30 + 1/90 = 2/45.
Therefore, A and B together can do the work in = 45/2 = 22(1/2) days.
 A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work was?
(a) 12
(b) 11
(c) 10
(d) 9
(e) None of these
Explanation:
Let the total units of work to be done be 360.
The units of work done by A in a single day = 8
Similarly, the units of work done by B in a single day = 9.
A and B’s one day work = 17 units
A and B worked together for some days = 17X ( Assume)
B’s work alone for 23 days = 23 x 9 = 207
So, the work done by A and B together = (360 – 207) = 153 units
Therefore, 17X = 153
=> 9 units
Therefore, the number of days after which A left the work was 9 days.
 A can do a piece of work in 10 days, B in 15 days. They work for 5 days. The rest of work finished by C in 2 days. If they get Rs 1500 for the whole work, the daily wages of B and C are?
(a) 275
(b) 250
(c) 225
(d) 300
(e) None of these
Explanation:
Part of work done by A = 5/10 = 1/2
Part of work done by B = 1/3
Part of work done by C = (1 (1/2 + 1/3)) = 1/6
A’s share : B’s share : C’s share = 1/2 : 1/3 : 1/6 = 3 : 2 : 1.
A’s share = (3/6) x 1500 = 750
B’s share = (2/6) x 1500 = 500
C’s share = (1/6) x 1500 = 250
A’s daily wages = 750/5 = 150/
B’s daily wages = 500/5 = 100/
C’s daily wages = 250/2 = 125/
Daily wages of B & C = 100 + 125 = 225/
 A is twice as good a workman as B and is therefore able to finish a piece of work in 30 days less than B. In how many days they can complete the whole work; working together?
(a) 20 days
(b) 22.5 days
(c) 25 days
(d) 35 days
(e) 40 days
Explanation:
Ratio of times taken by A and B = 1 : 2.
The time difference is (2 – 1) 1 day while B take 2 days and A takes 1 day.
If difference of time is 1 day, B takes 2 days.
If difference of time is 30 days, B takes 2 x 30 = 60 days.
So, A takes 30 days to do the work.
A’s 1 day’s work = 1/30
B’s 1 day’s work = 1/60
(A + B)’s 1 day’s work = 1/30 + 1/60 = 1/20
A and B together can do the work in 20 days.
 Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?
(a) 4:3
(b) 5:6
(c) 7:9
(d) 8:9
(e) 2:3
Explanation:
(20 x 16) women can complete the work in 1 day.
1 woman’s 1 day’s work = 1/320 .
(16 x 15) men can complete the work in 1 day.
1 man’s 1 day’s work = 1/240
So, required ratio
= 1/240 : 1/320
= 240 : 320
= 1/3 : 1/4
= 4 : 3 (cross multiplied)
 A and B can do a job together in 7 days. A is 13/4 times as efficient as B. The same job can be done by A alone in:
(a) 9 (1/3)
(b) 2 (5/6)
(c) 11
(d) 5 (2/8)
(e) 4 (6/8)
Explanation:
(A’s 1 day’s work) : (B’s 1 day’s work) = 7 /4: 1 = 7 : 4.
Let A’s and B’s 1 day’s work be 7x and 4x respectively.
Then, 7x + 4x = 1 => 11x = 1/7 => x =1/77
Therefore A’s 1 day’s work = ( 1/77 x 7 ) = 1/11 .
 A can do a piece of work in 8 days which B can destroy in 3 days. A has worked for 6 days, during the last 2 days of which B has been destroying. How many days must A now work alone to complete the work?
(a) 7 days
(b) 7(1/3) days
(c) 7(2/3) days
(d) 8 days
(e) None of these
Explanation:
In 6 days part of the work done by
A = 6/8 = ¾ during 2 days, part of the work destroyed by B = 2/3
 A can complete a piece of work in 10 days, B in 15 days and C in 20 days. A and C worked together for two days and then A was replaced by B. In how many days, altogether, was the work completed?
(a) 12
(b) 10
(c) 6
(d) 8
Explanation:
 A can do a piece of work in 20 days. He works at it for 5 days and then B finishes it in 10 more days. In how many days will A and B together finish the work?
(a) 8 days
(b) 10 days
(c) 12 days
(d) 6 days
(e) 16 days
Explanation:
 A and B can do a piece of work in 30 days while B and C do the same work in 24 days and C and A in 20 days. They work for 10 days after that B and C left. How many days more A takes to finish the work?
(a) 15 days
(b) 18 days
(c) 14 days
(d) 12 days
(e) None of these
Explanation:
 A group of 30 men, working 4 hours a day can do a piece of work in 10 days. Find the number of days in which another group of 45 men working 8 hrs a day can do twice the work. Assume that 2 men of the first group do as much work in 2 hours as 4 men of the second group do in 1 hr.
(a) 6(1/3) days
(b) 6(2/3) days
(c) 5(5/6) days
(d) 3(1/6) days
(e) None of these
Explanation:
Let the first group is man can do x unit in one hour and second group’s man can do y unit in one hour
2 × x × 2 = 4 × y × 1
x = y
2 × 30 × 4 × 10 = 45 × 8 × t
t = 6 (2/3) days
 P can complete a work in 12 days working 8 hours a day. Q can complete the same work in 8 days working 10 hours a day. If both P and Q work together. working 8 hours a day. in how many days can they complete the work?
(a) 5 (5/11)
(b) 5 (6/11)
(c) 6 (8/11)
(d) 6 (6/11)
(e) None of these
Explanation:
 If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn for 30 days are ₹
(a) 12
(b) 14
(c) 16
(d) 18
(e) 20
Explanation:
 A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of ₹ 250 and ₹ 300 per day respectively. In addition, a male operator gets ₹ 15 per call he answers and female operator gets ₹ 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 and maximum 12 number of the females?
(a) 15
(b) 14
(c) 12
(d) 10
(e) None of these
Explanation:
Let us form both equations first:
40m + 50f = 1000
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850m + 8000 f = A
When M and F are the number of Males and Females and A is the amount paid by the service provider.
Then the possible values of F are 8, 9, 10, 11, 12
If F = 8, then, M = 15
If F = 9, 10, 11 then M will not be an integer while F = 12 then M will be 10.
By putting F = 8 and M = 15, A = 18800. When F = 12 and M = 10, then A = 18100.
Hence the number of males will be 10.
 Published in TIME AND WORK
RATIO AND PROPORTION QUIZ 001
 Out of the total 390 students studying in a college of Arts and Science, boys and girls are in the ratio of 7 : 6 respectively and the number of students studying Arts and Science are in the ratio of 3 : 7 respectively. The boys and girls studying Arts are in the ratio of 4 : 5 respectively. How many boys are studying Science?
(a) 52
(b) 65
(c) 115
(d) 158
(e) None of these
Explanation:
Given, boys and girls are in the ratio of 7 : 6,
So, Number of boys = 7/13 x 390 = 210
Number of girls = 390 – 210 = 180
Students studying Arts and Science are in the ratio of 3 : 7,
Number of students studying Arts = 3/10 x 390 = 117
Number of students studying Science = 390 – 117 = 273
Also boys and girls studying Arts are in the ratio of 4 : 5,
Number of boys studying arts = 4/9 x 117 = 52
Number of boys studying science = 210 – 52 = 158
 From a number of mangoes, a man sells half the number of existing mangoes plus 1 to the first customer, then sells 1/3rd of the remaining number of mangoes plus 1 to the second customer, then 1/4th of the remaining number of mangoes plus 1 to the third customer and 1/5th of the remaining number of mangoes plus 1 to the fourth customer. He then finds that he does not have any mangoes left. How many mangoes did he have originally?
(a) 12
(b) 14
(c) 15
(d) 13
(e) None of these
Explanation:
Let the No. of mangoes that the man had originally = X
No. of mangoes sold balance
1st customer = (X/2) + 1 (X – 2)/2
2nd customer = (X – 2)/6 + 1 (X – 5)/3
3rd customer = (X – 5)/12 + 1 (X – 9)/4
4th customer = (X – 9)/20 + 1 0
(X – 9)/20 + 1= (X – 9)/4 => X = 14
3. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
(a) 2:3:4
(b) 6:7:8
(c) 6:8:9
(d) 7:8:9
(e) None of these
Explanation:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
(140 * 5x)/100 , (150 * 7x)/100 and ( 175 * 8x )/100
7x, (21/2)x and 14x.
The required ratio = 7x : (21/2)x : 14x
14x : 21x : 28x
2 : 3 : 4.
 4. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s new salary?
(a) 17000
(b) 20000
(c) 25000
(d) 38000
(e) None of these
Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, 2x + 4000 = 40
3x + 4000 57
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
5. The maximum marks per paper in 3 subjects in Mathematics , Physics and Chemistry are set in the ratio 1 : 2 : 3 respectively. Giri obtained 40% in Mathematics, 60% in Physics and 35% in Chemistry papers. What is the overall percentage of mark that he secured?
(a) 44
(b) 32
(c) 50
(d) 60
(e) None of these
Explanation:
40*1/100 : 60*2/100 : 35*3/100 = 0.4:1.2:1.05
Overall % =100* [0.4+1.2+1.05]/1+2+3 = 265/6 = 44.16 = 44%
 Between two stations, the first, second and third class fares are in the ratio 9 : 7 : 2. The number of passengers travelling in a day are in the ratio 5 : 3 : 2, respectively, in the above classes. If the sale of tickets generated revenue of Rs. 98,000 that day and if 200 passengers travelled by third class, what was the fare for a first class ticket?
(a) Rs. 84
(b) Rs. 92
(c) Rs. 106
(d) Rs. 126
(e) Rs. 116
Explanation:
Ratio of number of passengers is 5 : 3 : 2
If 200 passengers travelled by third class,
500 must have travelled by first class
Sum of ratio of amount collected
= (5 × 9 + 3 × 7 + 2 × 2) = 45 + 21 + 4 = 70
Amount collected from 1st class
= (45/70) x 98000 = Rs. 63,000
Fare for the first class
= Rs. (36000/500) = Rs. 126
 If 378 coins consist of Rs. 1, 50 paise and 25 paise coins, whose values are in the ratio of 13 : 11 : 7, the number of 50 paise coins will be
(a) 132
(b) 128
(c) 136
(d) 133
(e) None of these
Explanation:
 A certain product C is made of two ingredients A and B in the proportion of 2: 5. The price of A is three times that of B. The overall cost of C is Rs. 5.20 per kg including labour charges of 80 paisa per kg. Find the cost of B per kg?
(a) Rs. 8.40
(b) Rs. 4.20
(c) Rs. 4.80
(d) Rs. 2.80
(e) None of these
Explanation:
 A factory employs skilled workers, unskilled workers and clerks in the proportion 8:5:1 and the wage of a skilled worker, an unskilled worker and a clerk are in the ratio 5:2:3. When 20 unskilled workers are employed, the total daily wages of all amount to Rs. 3180. Find the daily wages paid to each category of employees.
(a) 2100, 800,280
(b) 2400, 480, 300
(c) 2400, 600, 180
(d) 2200, 560, 420
(e) None of these
Explanation:
 The cost of a bat increased by 10 per cent and the cost of a ball increased by 18 per cent. Before the price rise, the ratio of the cost of the bat to the cost of the ball was 9:2. If the cost of 12 bats and 54 balls before the price rise was Rs. C, what is their cost (in Rs.) now?
(a) 1.12 C
(b) 1.13 C
(c) 1.14 C
(d) 1.15 C
(e) None of these
Explanation:
 From a number of mangoes, a man sells half the number of existing mangoes plus 1 to the first customer, then sells onethird of the remaining number of mangoes plus 1 to the second customer, then onefourth of the remaining number of mangoes plus 1 to the third customer and onefifth of the remaining number of mangoes plus 1 to the fourth customer. He then finds that he does not have any mango left. How many mangoes did he have originally?
(a) 12
(b) 14
(c) 15
(d) 13
(e) None of these
Explanation:
 The incomes of A, B and C are in the ratio 7:9:12 and their spendings are in the ratio 8:9:15. If A saves one fourth of his income, then the savings of A, B and C are in the ratio of
(a) 69:56:48
(b) 47:74:99
(c) 37:72:49
(d) 56:99:69
(e) None of these
Explanation:
 On Republic Day, sweets were to be equally distributed among 450 children. But on that particular day, 150 children remained absent. Thus, each child got 3 sweets extra. How many sweets did each child get?
(a) 6
(b) 12
(c) 9
(d) Cannot be determined
(e) None of these
Explanation:
 Mr. Pandit owned 950 gold coins all of which he distributed amongst his three daughters Lalita, Amita and Neela. Lalita gave 25 gold coins to her husband, Amita donated 15 gold coins and Neeta made jewellery out of 30 gold coins. The new respective ratio of the coins left with them was 20:73:83. How many gold coins did Amita receive from Mr. Pandit?
(a) 380
(b) 415
(c) 400
(d) 350
(e) None of these
Explanation:
 The number of candidates writing three different entrance exams is in the ratio 4:5:6. There is a proposal to increase these numbers of candidates by 40%, 60% and 85% respectively. What will be the ratio of increased numbers?
(a) 14:15:16
(b) 12:15:19
(c)13:19:21
(d) 11:14:17
(e) None of these
Explanation:
Given ratio of number of candidates is 4:5:6
Let the number of candidates for 3 exams be 4k, 5k and 6k respectively.
After increasing, number of candidates become (140% of 4k), (160% of 5k) & (185% of 6k)
That is, (140x4k)/100, (160x5k)/100 and (185x6k)/100
= 56k/10, 80k/10 and 111k/10
Now, the required new ratio = 56k/100 : 80k/10 : 111k/10
= 56 : 80 : 111
Hence the answer is option d.
 Published in RATIO AND PROPORTION
PERCENTAGE QUIZ 001
 A number is increased by 20% and then decreased by 20%, the final value of the number is?
(a) does not change
(b) decreases by 2%
(c) increases by 4%
(d) decreases by 4%
(e) None of these
Explanation:
Here, x = 20 and y = – 20
Therefore, the net % change in value
= ( x + y + xy/100)%
= [20 – 20 + {20 x ( 20)}/100]% or – 4%
Since the sign is negative, there is a decrease in value by 4%.
 If the income of Mohan is 150% higher than Mahesh, then by what percent the income of Mahesh is less than Mohan?
(a) 40%
(b) 50%
(c) 60%
(d) 45%
(e) None of these
Explanation:
(c); Required percentage = (150/250) x 100% = 60%
 A man’s wages were decreased by 50%. Again, the reduced wages were increased by 50%. He has a loss of?
(a) 35%
(b) 25%
(c) 20%
(d) 15%
(e) None of these
Explanation:
Here, x = – 50 and y = 50
Therefore, the net % change in value
= ( x + y + xy/100) %
= [ 50 + 50 + ( 50) x (50)/100]% or – 25%
Since the sign is negative, there is loss of 25%
 By selling 45 lemons for Rs 40, a man loses 20 %. How many should he sell for Rs 24 to gain 20 % in the transaction?
(a) 16
(b) 18
(c) 20
(d) 22
(e) None of these
Explanation:
 In an election between two candidates, 75 % of the voters cast their votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes.
(a) 13000
(b) 17000
(c) 19000
(d) 16800
(e) 12000
Explanation:
Let the total number of votes enrolled are x.
Number of votes cast = 75% of x
Valid votes = 98% of 75% of x
Now, as 9261 is the 75% of valid casted votes so,
75% of 98% of 75% of x = 9261
=>(75×98×75×x)/(100×100×100)=9261=>x=16800
 Rahul’s Mathematics test had 75 problems, 10 arithmetic, 30 algebra, 35 geometry problems. Although he answered 70% of arithmetic, 40% of arithmetic and 60% of geometry problems correctly, still he got less than 60% problems right. How many more questions he would have to answer more to get passed
(a) 5
(b) 6
(c) 7
(d) 8
(e) 9
Explanation:
Number of questions attempted correctly = (70% of 10 + 40% of 30 + 60% of 35)
= 7 + 12 + 21 = 40.
Questions to be answered correctly for 60% = 60% of total questions
= 60 % of 75 = 45.
He would have to answer 45 – 40 = 5
 Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calculate average marks of the remaining students of the class.
(a) 20
(b) 30
(c) 75
(d) 60
(e) 80
Explanation:
Let’s assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)
=> 70x = 8000 – (950 + 1800) = 5250
=> x = 75.
 40% of Ram’s marks is equal to 20% of Rahim’s marks which percent is equal to 30% of Robert’s marks. If Robert’s marks is 80, then find the average marks of Ram and Rahim?
(a) 90
(b) 100
(c) 120
(d) 130
(e) 140
Explanation:
Given, 40% of Ram’s marks = 20% of Rahim’s marks = 30% of Robert’s marks.
Given, marks of Robert = 80
30% of 80 = 30/100 * 8 = 24
Given, 40% of Ram’s marks = 24.
=> Ram’s marks = (24 * 100)/40 = 60
Also, 20% of Rahim’s marks = 24
=> Rahim’s marks = (24 * 100)/20 = 120
Average marks of Ram and Rahim = (60 + 120)/2 = 90.
 Anil spends 40% of his income on rent, 30% of the remaining on medicines and 20% of the remaining on education. If he saves Rs. 840 every month, then find his monthly salary?
(a) 1800
(b) 2500
(c) 2700
(d) 2300
(e) 3000
Explanation:
Let’s Anil’s salary be Rs. 100.
Money spent on Rent = 40% of 100 = Rs. 40.
Money spent on medical grounds = 30% of (100 – 40) = 3/10 * 60 = Rs. 18.
Money spent on education = 20% of (60 – 18) = 1/5 * 42 = Rs. 8.40
Anil saves 100 – (40 + 18 + 8.40) i.e., Rs. 33.60
for 33.6 —> 100 ; 840 —> ?
Required salary = (840/33.6) * 100 = Rs. 2500
 A vendor fixed Selling Price of a product at 40% above the cost price. He sells half the stock at this price, onequarter of that stock at a discount of 25% on the original selling price and rest at a discount of 30% on the original selling price. Find the gain percentage all together?
(a) 20.75%
(b) 0.1575
(c) 0.1475
(d) 0.1675
(e) none of these
Explanation:
Let CP = 100;
marked price=140;
Revenue = [(1/2)*140 + (1/4)*0.75*140 + (1/4)*0.7*140] = 70 + 26.25 + 24.5 = 120.75;
% profit = 20.75/100 = 20.75%
 From the salary of Rahul, 20% is deducted as house rent, 10% of the rest he spends on children’s education and 10% of the balance, he spends on clothes. After this expenditure he is left with Rs 1,377. His salary is?
(a) Rs 2,125
(b) Rs 2,040
(c) Rs 2,100
(d) Rs 2,200
(e) None of these
Explanation:
Suppose that his salary = Rs 100
House rent = Rs 20, balance = Rs 80
Expenditure on education = Rs (10 x 80)/100 = Rs 8
Balance = Rs 72
Expenditure on clothes Rs (10 x 72)/100 = Rs 7.2
Balance now = Rs 64.8
If balance is Rs 64.8, salary = Rs 100
If balance is Rs 1,377, salary
= Rs (100 x 1377) / 64.8
= Rs 2,125
 In a class, 60% of the students are boys and in an examination, 80% of the girls scored more than 40 marks (Maximum Marks:150). If 60% of the total students scored more than 40 marks in the same exam, what is the fraction of the boys who scored 40 marks or less.
(a) 8/15
(b) 2/5
(c) 4/5
(d) 1/5
(e) none of these
Explanation:
Assume Total no of students = 100
60% of the students are boys. so Boys=60,Girls=40
No. of girls who scored more than 40 marks = 80% of girls = 80% of 40 = 32.
No. of students who scored more than 40 marks = 60% of Total Students = 60
Therefore No. of boys who scored more than 40 marks = 6032=28
No. of boys who scored less= Total boys – Boys(scored more) = 6028=32
Fraction=(scored less)/Total boys = 32/60 =8/15
 A man purchases some oranges at the rate of 25 for rupee 100 and same amount of oranges at the rate of 50 for rupees 100. He mixes them and sell at the rate of 25 for rupees 75. Find the profit/loss percentage.
(a) 4% loss
(b) 4% profit
(c) 5% loss
(d) No profit no loss
(e) None of these
Explanation:
Let man buys X unit of oranges at the rate of 25 for rupee 100 and X unit of oranges at the rate of 50 for rupee 100.
CP = (100/25)* X + (100/50)*X = 6X
SP = (75/25)*2X = 6X
So, no profit no loss
 A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition.
(a) 80%
(b) 70.08%
(c) 87.8%
(d) 90%
(e) None of these
Explanation:
Total number of fruits shopkeeper bought = 600 + 400 = 1000
Number of rotten oranges = 15% of 600
= 15/100 × 600
= 9000/100
= 90
Number of rotten bananas = 8% of 400
= 8/100 × 400
= 3200/100
= 32
Therefore, total number of rotten fruits = 90 + 32 = 122
Therefore Number of fruits in good condition = 1000 – 122 = 878
Therefore Percentage of fruits in good condition = (878/1000 × 100)%
= (87800/1000)%
= 87.8%
15. A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is reduced onto sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closest to:
(a) 20%
(b) 5%
(c) 30%
(d) 35%
(e) None of these
Explanation:
No. of Characters in one line = 65
No. of characters in one sheet = No. of lines × No. of characters per line = 55 × 65
Total number of characters = No. of sheets × No. of characters in one sheet = 20 × 55 × 65 = 71500
If the report is retyped –
New sheets have 65 lines, with 70 characters per line
No. of characters in one sheet = 65 × 70
Number of pages required,
Hence, 16 pages will be required if report is retyped.
Hence, reduction of (20 – 16) = 4 pages
% reduction is = (4/20) x 100 = 20%
 Published in PERCENTAGE
PROFIT AND LOSS QUIZ 001
 A shopkeeper allows 10% discount on goods when he sells without credit. Cost price of his goods is 80% of his selling price. If he sells his goods by cash then his profit percent is
(a) 25
(b) 35
(c) 68
(d) 85
(e) 15
Explanation:
S.p = 100 C.P = 80 Profit = 10080 = 20 profit percent =20/80×100=25%
 ’Aman’ sold an article to ’Bijoy’ at a profit of 20%. ’Bijoy’ sold the same article to ’Chitra’ at a loss of 25% and ’Chitra’ sold the same article to ’Dev’ at a profit of 40%. If ’Dev’ paid Rs 252 for the article, then find how much did ’Aman’ pay for it?
(a) Rs 175
(b) Rs 200
(c) Rs 180
(d) Rs 210
(e) None of these
Explanation:
Let the article costs ‘X’ to Aman
Cost price of Bijoy = 1.2X
Cost price of Chitra = 0.75(1.2X) = 0.9X
Cost price of Dev = 1.4(0.9X) = 1.26X = 252
Amount paid by Aman for the article = Rs 200
 Cost price of 12 apples is equal to the selling price of 9 apples and the discount on 10 apples is equal to the profit on 5 apples. What is the percentage point difference between the profit percentage and discount percentage?
(a) 1.5 %
(b) 1.75 %
(c) 1.85 %
(d) 2 %
(e) None of these
Explanation:
Cost price of 12 apples is equal to the selling price of 9 apples,
Let the C.P. of one apple = Re. 1
C.P. of 9 apples = Rs. 9
S.P. of 9 apples = Rs. 12
Profit % of 9 apples = 3/9 x 100 = 33.33 %
Profit % of 1apple = 33.33/9 = 3.703 %
Profit % of 5 apples = 3.703 x 5 = 18.51 %
Given, the discount on 10 apples is equal to the profit on 5 apples,
Discount on 10 apples = 18.51 %
Discount on 1 apple = 1.851 %
Therefore, Profit % of 1 apple – Discount on 1 apple = 3.703 – 1.851 = 1.85 %
 A man gains 20% by selling an article for a certain price. If he sells it at double the price, the percentage of profit will be.
(a) 1.3
(b) 1.4
(c) 1.5
(d) 1.6
(e) None of these
Explanation:
Let the C.P. = x,
Then S.P. = (120/100)x = 6x/5
New S.P. = 2(6x/5) = 12x/5
Profit = 12x/5 – x = 7x/5
Profit% = (Profit/C.P.) * 100
=> (7x/5) * (1/x) * 100 = 140 %
 A sells a set of books to B for Rs. 300 at a profit of 25%. B sells it to C at a loss of 10%.
price paid by A?
 ii) What was the price paid by C to B?
(a) 240, 260
(b) 250, 270
(c) 250, 260
(d) 240, 270
(e) None of these
Explanation:
i. Price paid by A (cost price paid by A) = [100/(100+25)]*300 = Rs. 240.
ii. Price paid by C (Selling price by B to C) = [(10010)/100]*300 = Rs. 270.
 The sale price of an article including the sales tax is Rs. 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 12%, then the cost price of the article is:
(a) 500
(b) 560
(c) 340
(d) 780
(e) 800
Explanation:
110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500
 The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
(a) Rs 2000
(b) RS 5000
(c) Rs 5400
(d) Rs 6000
(e) Rs 7000
Explanation:
Let C.P. be Rs. x.
Then, (1920 – x)/x * 100 = (x – 1280)/x * 100
1920 – x = x – 1280
2x = 3200 => x = 1600
Required S.P. = 125 % of Rs. 1600 = 125/100 * 1600 = Rs. 2000.
 Virat and Dhoni wants to make 25% profit on selling good. Virat calculating it on cost price while Dhoni on the selling price, the difference in the profits earned by both being Rs. 100 and selling price being the same in both the cases. Find out the selling price of both goods?
(a) Rs.1500
(b) Rs.1600
(c) Rs.1200
(d) Rs.2000
(e) None of these
Explanation:
Profit = 25% = 1/4
Let CP of Virat’s article = 4x & Profit = x; SP = 4x + x = 5x
Let CP of Dhoni’s article = 4y = 5x —(1)[Selling price of Virat’s article = Cost Price of Dhoni’s article] x – y = 100—(2)
y = 500
Selling price = 5x = 4y = 2000
 The price of the book is marked 20% above the C.P. If the marked price of the book is Rs. 180, then what is the cost of the paper used in a single copy of the book?
(a) Rs. 36
(b) Rs. 37.50
(c) Rs. 42
(d) Rs. 44.25
(e) None of the above
Explanation:
 A trader sells two bullocks for Rs. 8,400 each, neither losing nor gaining in total. If he sold one of the bullocks at a gain of 20%, the other is sold at a loss of
(a) 20%
(b) 18(2/9)%
(c) 14(2/7)%
(d) 21%
(e) None of these
Explanation:
S.P. of two bullock = 8400 + 8400 = 16800 Rs.
 Gopal goes from place A to B to buy an article costing 15% less at B. although he spends Rs. 150 on travelling, still he gains Rs. 150 compared to buying it at A. His profit percent is:
(a) 4.5
(b) 6
(c) 7.5
(d) 8
(e) None of these
Explanation:
Let CP = 2000
SP = 1700
so, after spend price = 1850
 A man bought a mobile and a laptop for Rs. 78000. He sold the mobile at a gain of 25% and the laptop at a loss of 15%, thereby gaining 5% on the whole. Find the cost price of mobile.
(a) Rs. 39000
(b) Rs. 34000
(c) Rs. 30000
(d) Rs. 38000
(e) Rs. 32000
Explanation:
 The percentage profit earned when a watch is sold for Rs. 546 is double the percentage profit earned when the same watch is sold for Rs. 483. If the marked price of the watch is 40% above the cost price, then what is the marked price of the watch?
(a) Rs. 588
(b) Rs 608
(c) Rs. 616
(d) Rs. 596
(e) Rs. 586
Explanation:
 The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is
(a) 25
(b) 18
(c) 16
(d) 15
(e) None of these
Explanation:
 A watch dealer incurs an expense of Rs.150 for producing every watch. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each watch for Rs. 100.If he produces 1500 watches, what is the number of watches that he must sell during the season in order to breakeven, given that he is able to sell all the watches produced?
(a) 580
(b) 620
(c) 650
(d) 700
(e) None of these
Explanation:
Total cost to produced 1500 watches = (1500 × 150 + 30000) = Rs. 2,55,000
Let he sells x watches during the season, therefore number of watches sold after the season =
(1500 – x)250 × x + (1500 – x) × 100 = 150x + 150000
Now, breakeven is achieved if production cost is equal to the selling price.
150x + 150000 = 2,55,000
x = 700
 Published in PROFIT AND LOSS
PIPES AND CISTERNS QUIZ 001
 Two pipes A and B can fill a tank in 4 and 5 hours respectively. If they are used alternately for one hour each, the time taken to fill the tank is?
(a) 4 hrs 24 min
(b) 5 hrs
(c) 6 hrs
(d) 7 hrs
(e) 5hrs 50 min
Explanation:
A and B combine work 2 hour work=1/4+1/5=9/20
4hour work =9/20+9/20=9/10
remaining work=19/10=1/10
now it’s A’s turn so =1/10*4=2/5
to convert it into minute
5/2 = 60/X
cross multiplication x=60*2/5= 24 min
So Answer is 4hr and 24min
 A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 A.M. When will the cistern be empty?
(a) 3 A.m.
(b) 3.30 A.M.
(c) 4 A.M.
(d) 5 P.M.
(e) 5.30 P.M.
Explanation:
In 1 hour A can fill the cistern in = 1/4hr.
In 1 hour B can fill the cistern in = 1/5hr.
In 1 hour C can empty the cistern in =1/2hr.
Now consider in X hrs cistern will be empty after opening C then C is working X hrs
B is working (X+1) hrs
A is working (X+2) hrs
So,
(X+2)/4 + (X+1)/5 – x/2=0
X=14hrs
So, Answer is 5 p.m
 A tank is filled in eight hours by three pipes A, B and C. Pipe A is twice as fast as pipe B, and B is twice as fast as C. How much time will pipe B alone take to fill the tank?
(a) 20
(b) 45
(c) 30
(d) 28
(e) 29
Explanation:
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours
 One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill tank in 36 min., then the slower pipe alone will be able to fill the tank in?
(a) 81 min
(b) 144 min
(c) 168 min
(d) 167 min
(e) 187 min
Explanation:
Let the slower pipe alone fill the tank in x min.
Then, faster pipe will fill it in x/3 min.
1/x + 3/x = 1/36
4/x = 1/36 => x = 144 min.
 A large tanker can be filled by two pipes A and B in 60 and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
(a) 30 min
(b) 40 min
(c) 60 min
(d) 50 min
(e) 70 min
Explanation:
Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes.
Then, x/2(1/24 + 1/40) = 1
x/2 * 1/15 = 1 => x = 30 min.
 A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank?
(a) 144 min
(b) 150 min
(c) 160 min
(d) 170 min
(e) 185 min
Explanation:
Let the slower pipe alone fill the tank in x minutes then faster will fill in x/3 minutes.
Part filled by slower pipe in 1 minute = 1/x
Part filled by faster pipe in 1 minute = 3/x
Part filled by both in 1 minute = 1/x+3/x=1/36
=>4/x=1/36
x =36∗4=144 min
 Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately is 5 hrs, 10 hrs and 15 hrs, respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at threefourths of their normal rates for the 1st hrs after which they supplied water at the normal rate. Pipe C supplied water at twothirds of its normal rate for first 2 hrs, after which it supplied at its normal rate. In how much time, tank would be filled?
(a) 1.05 hrs
(b) 2.05 hrs
(c) 3.05 hrs
(d) 4.05 hrs
(e) 2.55 hrs
Explanation:
(c); The part of the tank filled by A and B in first two hrs
 Two pipes A and B can fill a water tank in 20 and 24 minutes respectively and a third pipe C can empty at the rate of 3 gallons per minute. If A, B and C opened together fill the tank in 15 minutes, the capacity (in gallons) of the tank is:
(a) 180
(b) 150
(c) 120
(d) 60
(e) None of these
Explanation:
let the capacity of the tank be x gallons.
Quantity of water filled in the tank in 1 minute when all the
pipes A, B and C are opened simultaneously
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 There are two leakages in the bottom of a cistern. Both together empty the cistern in 12 hours. If the first leakage alone empties it in 30 hours, then in how many hours will the second leakage alone empty it?
(a) 20 hours
(b) 25 hours
(c) 30 hours
(d)15 hours
(e) 12 hours
Explanation:
For better understanding of the topic, please refer
to our videos and other relevant notes. Please visit
our website for more information.
 Pipe A, B and C are kept open and fill a tank in ‘t’ minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipes A and B are kept open continuously, the tank would be filled completely in ‘t’ minutes. How long will C alone take to fill the tank?
(a) 18
(b) 36
(c) 27
(d) 24
(e) 20
Explanation:
A is kept open for all t minutes and fills onethird the tank. Or, A should be able to
fill the entire tank in ‘3t’ minutes.
A and B together can fill the tank completely in t minutes.
A takes 60 minutes to fill the entire tank, B takes 30 minutes to fill the entire tank.
A is kept open for all 20 minutes. B is kept open for 10 minutes.
C, which is kept open for = (102) = 8 minutes also fills one third of the tank.
Or, c alone can fill the tank in = (8*3) = 24 minutes.
 Two taps can fill a tank in 20 mins and 30 mins respectively. There is an outlet tap at exactly half level of that rectangular tank which can pump out 50 litres of water per minute. If the outlet tap is open, then it takes 24 mins to fill an empty tank. What is the volume of the tank?
(a) 1200 litres
(b) 1500 litres
(c) 1800 litres
(d) 2400 litres
(e) None of these
Explanation:
12. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hrs faster than the first pipe and 4 hrs slower than the third pipe. The time required by the first pipe is
(a) 6 hrs
(b) 10 hrs
(c) 15 hrs
(d) 30 hrs
(e) None of these
Explanation:
13. A bathtub can be filled by a cold water pipe in 20 mins and by a hot water pipe in 30 mins. A person leaves the bathroom after turning on both pipes simultaneously and returns at the moment when the bathtub should be full. Finding however, that the waste pipe has been open, he now closes it. In 3 mins more the bathtub is full. In what time would the waste pipe empty it?
(a) 38 mins
(c) 43 mins
(c) 45 mins
(d) 48 min
(e) None of these
Explanation:
14. Pavan builds an overhead tank in his house, which has three taps attached to it. While the first tap can fill the tank in 12 hrs, the second one takes one and a half times more than the first one to fill it completely. A third tap is attached to the tank which empties it in 36 hrs. Now, one day, in order to fill the tank. Pavan opens the first tap and after two hrs opens the second tap as well. However, at the end of the sixth hour, he realizes that the third tap has been kept open right from the beginning and promptly closes it. What will be the total time required to fill the tank?
(a) 8 hrs 48 mins
(c) 9 hrs 36 mins
(b) 9 hrs 12 mins
(d) 8 hrs 30 mins
(e) None of these
Explanation:
 Published in PIPES AND CISTERNS
AVERAGE QUIZ 001
 The average marks of Rahul in 10 papers is 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest?
(a) 55
(b) 60
(c) 62
(d) Cannot be determined
(e) None of these
Explanation:
Total marks of 10 papers = 80 x 10 = 800
Total marks of 8 papers = 81 x 8 = 648
Total marks of two papers = (800 – 648) = 152
If highest total is 92, then the lowest total is
(152 – 92) = 60.
 The average daily wages of A, B and C is Rs 120. If B earns Rs 40 more than C per day and, A earns double of what C earns per day, the wages of A per day are?
(a) Rs 80
(b) Rs 120
(c) Rs 160
(d) Rs 100
(e) None of these
Explanation:
Let daily wages of C = X
Then, daily wages of A = 2X
and daily wages of B = X + 40
Hence, average daily wages of A, B and C
= (X + 2X + X + 40) / 3 = (4X + 40)/3
Thus, (4X + 40)/3 =120
=>4X + 40 = 360
=>4X = 320
X = 80.
Therefore, Wages of A per day = 2 x 80 = Rs 160.
3. The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?
(a) 4
(b) 8
(c) 12
(d) 16
(e) None of these
Explanation:
Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.
Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.
or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) – 57 = 8
4. Biswajit was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x ?
(a) 6
(b) 7
(c) 8
(d) 9
(e) None of these
Explanation:
Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12.
or 137 + x = 144 or x = 144 – 137 = 7.
5. The average salary of 90 employees in an organization is Rs.14,500 per month. If the no. of executives is twice the no. of clerks, then find the average salary of clerks?
(a) 11,500
(b) 12000
(c) 13200
(d) can’t be determined
(e) None of these
Explanation:
90 => 2:1=> 60:30
Total Salary = 60*salary of executive +30 * salary of clerk
90*14500 = 60*x+30*y
x, y not given so we can’t determine the answer.

 The average salary of 120 employees in a bank is Rs.15,000 per month. If the no. of Assistants is thrice the no of POs and the average salary of Assistants is 1/3rd of the average salary of Pos, then find the average salary of POs ?
(a) 18,000
(b) 25,000
(c) 36,000
(d) 30,000
(e) None of these
Explanation:
120 = 1:3 =30:90 = PO : Assistant
120*15000 = 30*x+ 90 *x/3
18,00,000 = 90x+90x/ 3
54,00,000*3 = 180x
x = 30000
 The average value of the properties of Akhil, Munna and Anil is Rs.130cr. The Property of Akhil is Rs. 20cr greater than the property value of Munna and Anil’s property value is Rs. 50cr greater than the Akhil’s property value. The value of Anil’s property is
(a) Rs. 120cr
(b) Rs. 170cr
(c) Rs. 100cr
(d) Rs. 150 cr
(e) None of these
Explanation:
Property value of Munna x
130*3 = x+x+20+x+20+50
390 = 3x+90
3x=300
X=100
Anil = 100+20+50 = 170
 The average monthly expenditure of a family was Rs. 2,200 during the first 3 months; Rs. 2,250 during the next 4 months and Rs. 3,120 during the last 5 months of a year. If the total savings during the year were Rs. 1,260, then the average monthly income was:
(a) Rs. 2,605
(b) Rs. 2,805
(c) Rs. 2,705
(d) Rs. 2,905
(e) Rs. 2,995
Explanation:
Total expenditure for the year
= [2200 × 3 + 2250 × 4 + 3120 × 5]
= 6600 + 9000 + 15600 = Rs. 31200
Total saving = Rs. 1260
Total income = expenses + savings = 31200 + 1260 = Rs. 32460
Average income = 32460/12 = Rs. 2705
 The average marks of 100 students were found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct average corresponding to the correct score.
(a) 38.7
(b) 39
(c) 39.7
(d) 41
(e) None of these
Explanation:
Average of 100 students = 40
Total = 40 × 100 = 4000
Error = 83 – 53 = 30 (high)
So correct average = 3970/100 = 39.7
 In a oneday cricket match, Agarkar, Sehwag, Sachin, Dravid and Ganguly scored an average of 39 runs. Dravid scored 7 more than Ganguly. Ganguly scored 9 fewer than Agarkar. Sehwag scored as many as Dravid and Ganguly combined; and Sehwag and Sachin scored 110 runs between them. How many runs did Sachin score?
(a) 47
(6)51
(c) 53
(d) 49
(e) None of the above
Explanation:
 Having scored 98 runs in the 19th innings, a cricketer increases his average score by 4. What will be his average score after the 19th innings?
(a) 28
(b) 26
(c) 24
(d) 22
(e) None of these
Explanation:
 A Mathematics teacher tabulated the marks secured by 35 students of 8th class. The average of their marks was 72. If the marks secured by Reema were written as 36 instead of 86, then find the correct average marks up to two decimal places.
(a) 73.41
(b) 74.3
(c) 72.43
(d) 73.43
(e) None of these
Explanation:
 The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are
(a) 2 and 9
(b) 3 and 8
(c) 4 and 7
(d) 5 and 6
(e) None of these
Explanation:
 If p, q, r be three positive numbers such that p > q > r. When the smallest number is added to the difference of the other two numbers, then the average of the resultant number and the original numbers except the smallest number is 21 more than the average of all the three original numbers. The value of (p q) is
(a) 7
(b) 14
(c) 63
(d) 42
(e) None of these
Explanation:
 The average age of A and B is 20 years. If C were to replace A, the average would be 19 and if C were to replace B, the average would be 21. The ages of A, B and C are (in years)
(a) 22, 17,16
(b) 22, 18, 20
(c) 30, 18, 15
(d) 23, 17 ,15
Explanation:
A + B = 2 x 20 ——(1)
C + B = 2 x 19 ———(2)
A + C = 2 x 21 ——–(3)
Now,
By adding equation (1), (2) and (3) we get,
A + B + C = 60 ———(4)
Subtracting equation (4) to equation (1) we get,
C = 20
Therefore,
B = 18
A = 22
 Published in AVERAGE
AGE PROBLEMS QUIZ 001
 Ten years ago, Kumar was thrice as old as Sailesh was but 10 years hence, he will be only twice as old. Find Kumar’s present age.
(a) 60 years
(b) 80 years
(c) 70 years
(d) 76 years
(e) None of these
Explanation:
Let Kumar’s present age be x years and Sailesh’s present age be y years.
Then, according to the first condition,
x – 10 = 3(y – 10) or, x – 3y = – 20 ..(1)
Now. Kumar’s age after 10 years = (x + 10) years
Sailesh’s age after 10 years = (y + 10)
(x + 10) = 2 (y + 10) or, x – 2y = 10 ..(2)
Solving (1) and (2), we get x = 70 and y = 30
Kumar’s age = 70 years and Sailesh’s age = 30 years..
 Five years hence, father’s age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages.
(a) 40 years and 10 years
(b) 42 years and 12 years
(c) 35 years and 5 years
(d) 41 years and 11 years
(e) None of these
Explanation:
Let the present age of father be x years and the present age of son be y years.
five year’s hence
x + 5 = 3(y + 5)
=>x – 3y 10 = 0 —–(1)
Five year’s ago
x – 5 = 7(y – 5)
=>x – 7y + 30 = 0 —–(2)
Solving Equations (1) and (2), we get
x = 40 and y = 10
Hence, present age of father is 40 years and present age of son is 10 years.
 A person’s present age is twofifth of the age of his mother. After 8 years, he will be onehalf of the age of his mother. How old is the mother at present?
(a) 32
(b) 36
(c) 38
(d) 40
(e) None of these
Explanation:
Let the mother’s age be x
Present age of the person = 2/5x
(2/5)x+8 = 0.5x +4
on solving we get x = 40
 The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:
(a) 5:2
(b) 7:3
(c) 9:2
(d) 13:4
(e) None of these
Explanation:
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
3x + 20 = 2x + 40
x = 20.
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
 Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
(a) 16
(b) 18
(c) 28
(d) 24.5
(e) None of these
Explanation:
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
x – 7 /x = 7/ 9
9x – 63 = 7x
2x = 63
x = 31.5
Hence, Sachin’s age =(x – 7) = 24.5 years.
 The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
(a) 8, 20, 28
(b) 16, 28, 36
(c) 20, 35, 45
(d) 21, 30, 42
(e) None of these
Explanation:
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56
20x = 80
x = 4.
Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
 Suresh has three daughters namely Ramya, Anita and Kiran. Ramya is the eldest daughter of Mr. Suresh while Kiran is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44. What is the age of Ramya after two years?
(a) 15 years
(b) 13 years
(c) 18 years
(d) 17 years
(e) 16 years
Explanation:
Square numbers – a, b, c
(a + 5) + (b + 5)+ (c + 5) = 44
a + b + c = 44 – 15 = 29
Possible values of a, b, c = 4, 9, 16 [Out of 1, 4, 9, 16, 25]
Ramya’s present age = 16; after two years = 18
 Ajay got married 6 years ago. His present age is 5/4 times his age at the time of his marriage. Ajay’s brother was 5 years younger to him at the time of his marriage. What is the present age of Ajay’s brother?
(a) 22 years
(b) 11 years
(c) 25 years
(d) 19 years
(e) 15 years
Explanation:
Present age of Ajay = x ; Present age of Ajay’s sister = y
x = (x6)(5/4)
x = 30
present age of Ajay’s brother = 30 – 5 = 25
 The ratio between the present ages of A and B is 6:7. If B is 4 years older than A. What will be the ratio of the ages of A and B after 4 years?
(a) 8:7
(b) 7:8
(c) 1:8
(d) Data inadequate
(e) None of these
Explanation:
Let A’s age and B’s age be 6X years and 7X years respectively
Then, 7X – 6X = 4
X = 4
Required ratio = (6X + 4) : (7X + 4) = (6*4+4) : (7*4 +4) =28 : 32 = 7 : 8
 Ram’s present age is three times his son’s present age and twofifth of his father’s present age. The average of the present ages of all of them is 46 years. What is the difference between the Ram’s son’s present age and Ram’s father’s present age?
(a) 68 years
(b) 88 years
(c) 58 years
(d) can not determined
(e) None of these
Explanation:
 The ratio of the ages of Anil and his son at present is 7 : 3. Six year hence, the ratio of the ages of the Anil’s wife and the son will be 2 : 1. Find the ratio of the present ages of Anil and his wife.
(a) 6 : 5
(b) 4 : 3
(c) 5 : 4
(d) Cannot be determined.
(e) None of these
Explanation:
Let the present age of Anil and his son be 7x years and 3x years respectively.
Let the present age of the son’s mother be m years.
(m + 6) / (3x + 6) = 2 / 1
m + 6 = 6x + 12
m = 6x + 6
The ratio of the present ages of Anil and his wife =
7x / (6x + 6)
This cannot be found uniquely.
 Five years ago, Mr Sunil was thrice as old as his son and ten years hence he will be twice as old as his son. Mr. Sunil’s present age (in years) is
(a) 35
(b) 45
(c) 50
(d) 55
(e) None of these
Explanation:
Let Mr. Sunil’s age (in years) = X and his son’s age = Y
Then, X – 5 = 3(Y – 5)
i.e., X – 3Y + 10 = 0 —– ’(1)
and, X + 10 = 2(Y + 10)
i.e. X – 2Y – 10 = 0 —– (2)
Solving (1) and (2), we get Y = 20
Substituting Y in eq (1), we get X = 50
Hence, the present age of the father is 50.
 If the present age of A be is twice the present age of B. 18 years ago age of C was half the age of B six years hence. If the average of the present ages of A, B and C is 42 years. Find the age of A, 9 years hence?
(a) 69 years
(b) 70 years
(c) 73 years
(d) 77 years
(e) None of these
Explanation:
 Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eightmember joint family is nearest to
(a) 21 years
(b) 25 years
(c) 24 years
(d) 23 years
(e) None of these
Explanation:
 A boy was asked of his age by his friend. The boy said, “The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age.” If the friend’s age is 14, then the age of the boy is
(a) 28 years
(b) 21 years
(c) 14 years
(d) 25 years
(e) None of these
Explanation:
 Published in AGE PROBLEMS