# AMCAT Quants Questions with Solutions Previous Year

Set 1

1. The cost price of 10 articles is equal to the selling price of 9 articles. find the profit percent.
a. 101/9 %           b. 100/9 %           c. 102/9 %            d. 103/9 %

Ans: 100/9 %

Let Cost Price be x and selling price be y
Then given that cost price of 10 articles is equal to the selling price of 9 articles
That means 10x=9y
Y= 10x/9
Profit percent = (( selling price – cost price )/cost price ) * 100
= 100/9 %
2. The ratio of radii of two right circular cylinders is 6:7 and their heights are in the ratio 5:9. The ratio of their respective curved surface areas is
a. 14:15                 b. 17:19                c. 23:29                 d. 10:21

Ans: 10 : 21

Curved surface area of a cylinder = 2 * Pi * r * h
Ratio = (6/7) * (5/9) = 10:21

3. In how many ways can the 7 letters A,B,C,D,E,F and G be arranged so that C and E never together.
a. 5040                  b. 6480                  c. 3600                  d. 1440

Ans: 3600

C and E never together = Total arrangements – C and E together
Total arrangements are 7!
C and E together = pack  c and e into one unit + 5 other alphabets = 6! 2! ( 2! Is two arrange c and e internally)
C and E never together = Total arrangements – C and E together = 7! – 6! 2! = 3600

4. How many numbers are there in all from 4000 to 4999 (both 4000 and 4999 included) having at least one of their digits repeated?
a. 356                    b. 216                    c. 496                     d. 504

Ans:  496

Atleast one of their digits repeated = Total numbers – None of the digits repeated
Total numbers from 4000 to 4999 = 1000
None of the digits repeated =  _ _ _ _
There are total 4 places
1st place is filled with 4 only. So only one choice
2nd place is filled with any 9 digits except 4 as we have used 4 in 1st place. So 9 choices
Similarly 3rd place is filled with any 8 digits. So we have 8 choices
4th place is filled with any 7 digits. So we have 7 choices.
So total arrangements = 1 * 9 * 8 * 7 = 504
Ans= 1000 – 504 = 496

5. if 1/2x +1/4x+1/8x=14  Then the value of x is:
a. 8         b. 12      c. 4         d. 16

Ans: x = 16

6. Which of the following expressions will always be true?

Ans:  D

Verify from options

Ans(C)

8. Find the value of  h[f(1,2,3), g(2,1,-2), h(1,-1,-1)].
a. 0.5                     b. none                                c. 1                         d. 0

Ans(D)

9. A trapezium with an area of 5100 cm2 has the perpendicular distance between the two parallel sides of 60m . if one of the parallel sides be 40m. find the length of the other side.
a. 130 m               b. 110 m               c. 120 m                d. 145 m

Ans:  130 m

Area of a trapezium = (1/2) (a+b) h

10. Find the simple interest on Rs. 306.25 from March 3rd to July 27th( In the same year ) at 3.75 percent.
a. Rs. 4.57            b.  Rs. 4.59           c. Rs. 4.53            d. Rs 4.58

Ans: 4.59

from March 3rd to July 27th( In the same year) = 146 days
(306.25 * 146 * 3.75 )/ ( 365 * 100) = 4.59

11. Dhruv and Naksh drive at the speeds of 36 Kmph and 54 kmph respectively. If Naksh takes 3 hours lesser than what Dhruv takes for the same distance. Then distance is :
a. 324 km   b. 524 km       c. 320 km             d. 420 km

Ans:  324 km

Let dhruv takes t hours then naksh takes t-3 hours
Because distance is same in both cases
So 36 * t = 54 (t-3)
t=9
ans: 36 * 9 = 324 km

12. The radius of wheel of axis’s car is 50 cm. What is the distance that the car would cover in 14 revolutions?
a. 11 m                  b. 22 m                 c. 33 m                  d. 44 m

Ans: 44 m

Distance covered in one revolution is equal to wheel surface area = 2 * Pi * r
Distance covered in 14 revolutions = 14 ( 2 * (22/7) * 50) = 44000 cm= 44 m
13.  P can do a piece of work in 5 days of 8 hours each and Q can do in 4 days of 6 hours each. How long will they take do it working  5 hours a day?
a. 2 days               b. 3 days              c. 4 days               d. 5 days

Ans:  3 days

P can do in 5* 8 hours= 40 hours
Q can do in = 24 hours
Working together in one hour = (1/40) + (1/24) = 1/15
Total work can be finished in 15 hours
They 5 hours a day so total number of days = 15/5 = 3 days

14. Libra had three diamond weighing equal. One of the diamond fell and broke into 4 equal pieces weighing 20gm each. what was the total weight of three diamonds.
a. 200 gm             b. 280 gm             c. 320 gm             d. 240 gm

Ans: 20 * 4 * 3 = 240 gm

16. if the antecedent and consequent of a ratio are increased by 5 and 6 respectively then the ratio is 5:6. find the original ratio.     a. 5:6                     b. 1:2                     c. 2:3                      d. 3:4
Ans: let’s say original ratio is x:y
(x+5)/(y+6) = 5/6
Then x/y = 5/6

17. Rohit and Rahul start from the same point and move away from each other at right angle. After 4 hours they are 80 km apart. if the speed of Rohit is 4 kmph more than Rahul. what is the speed of Rohit?
a. 16 kmph          b. 20 kmph          c. 12 kmph          d. none

Ans:  x is the speed of rahul then (x+4) will be rohit speed

802 = (4x)2 + ((x+4)4)2
X=12
Rohit speed = 12 + 4 = 16kmph

18. Abhimanyu and supreet can together finish a work in 50 days. They worked together for 35 days and then supreet left. After another 21 days, Abhimanyu finished the remaining  work. In how many days Abhimanyu alone can finish the work?
a. 70 days            b. 75 days            c. 80 days             d. 60 days

Ans: 35 days worked together +  21 days abhimayu worked = finished the work

35(1/50) + 21(x) = 1
X=70 days
19. if two fair dice are thrown simultaneously. then what is the probability that sum of the numbers appearing on the top faces of the dice is less than 4?  a. 6/14                  b. none                                c. 1/12                   d. 3/18
Ans:  possible cases are (1,1) (1,2) and (2,1) = 3

3/36 = 1/12

20.

21. 3 individuals john wright, greg chappell and gary kristen are in the race for the appointment of new coach of team india. The probabilities of their appointment are 0.5, 0.3 and 0.2 respectively. If john wright is appointed then probability of ganguly appointed as a captain will be 0.7 and corresponding probability if greg chappell or gary kristen is appointed are 0.6 and 0.5 respectively. find the overall probability that ganguly will appointed as a captain.
a. 0.63                   b. 0.35                   c. 0.18                   d. 0.89
Ans: 0.63

22. A man spends Rs 660 on tables and chairs. the price of each table is Rs. 150 and the price of each chair is Rs. 20. If he buys maximum number of tables, what is the ratio of chairs to tables purchased?
a. 2: 5                    b. 3:5                     c. 2:3                      d. 3:4
4 tables + 3 chairs =660
Chairs to tables ratio is 3:4

23. two packets are available for sale.
packet a: peanuts 100 gms for Rs 48 only
packet b: peanuts 150 gms for Rs 72 only
a. both have the same value       b. packet b          c. data insufficient           d. packet a
Ans: a. both have the same value

Packet-a : 1 gm cost = 48/100
Packet-b : 1 gm cost = 72/150

24.find the surface area of a piece of metal which is in the form of a parallelogram whose base is 10 cm and height is 6.4 cm
a. 64 cm2              b. 65 cm2              c. 45 cm2              d. 56 cm2
Ans:

25. Sridevi is younger than Rajeev by 4 years. if their ages are in the ratio of 7:9. how old is Sridevi?
Ans: if Sridevi is x then Rajeev will be (x+4)
x/(x+4) = 7/9
x=14

26. A sum of Rs. 900 amounts to Rs. 950 in 3 years at simple interest. If the interest rate is increased by 4%, it would amount to how much?

27. two trains for Palwal leave Kanpur at 10a.m and 10:30 am and travel at the speeds of 60 kmph and 75 kmph respectively.  After how many kilometres from Kanpur will the two trains be together?
Ans: 150 km

28.  (x + 1/x) = 6 the value of ( x5  + 1/x5 ) = ?
Ans: 6726

29. In how many ways can 44 people be divided into 22 couples?
Ans:  Short cut how many ways n people be divided into n/2 couples
(n!)/{(2!)n/2 (n/2)!}  so ans is b. (44!)/{(2!)22 (22!)}

30. Find the remainder when  ( x3 + 4 x2 + 6 x – 2 ) is divided  (x+5)
Ans: -57

31. a solid cylinder has total surface area of 462 cm2 . If total surface area of the cylinder is thrice of its curved surface area. then the volume of the cylinder is:
a. 539 cm3            b. 545 cm3           c. 531 cm3            d. 562 cm3

Ans:539

32.

In which year was there lowest wheat import?
a. 1973                  b. 1974                  c. 1975                  d. 1982
Ans: a

33. What is the ratio of number of years which have imports above the average imports to those which have imports below the average imports?
a. 5:3                     b. 2: 6                    c. 3: 8                     d. none
Ans: d

34. The increase in imports in 1982 was what percent of the imports in 1981?
a. 25%                   b. 5%                     c. 125%                 d. 80%
Ans: a

35. The section of a solid right circular cone by a plane containing vertex and perpendicular to base is an equilateral triangle of side 10 cm. find the volume of the cone?
a. 221.73 cm3  b.223.73 cm3    c.228.73 cm3    d.226.61 cm3

36. A sum of Rs 468.75 was lent out at simple interest and at the end of 1 year and 8 months, the total amount of Rs 500 is recieved. find the rate of interest.
a. 2%     b. 4%     c. 1%      d. 3%
Ans: 4%

37.Consider the following two curves in the X-Y plane
y=(x3+x2+5)
y=(x2+x+5)
Which of the following statements is true for -2<= x <=2?
a. The two curves do not intersect.      b. The two curves intersect thrice.
c. The two curves intersect twice.        d. The two curves intersect once.
Ans: b

38.Give a model for maximising the profit in a company or minimising the loss in a conflict with optimisation techniques.where quantity f(x) is referred to as the object function while the vector ‘x’consists of decision variables.
A. None of the mentioned options.      B.  x* =arg min f(x)     C. x* =arg max f(x)      D.  x* =an-1+an arg min f(x)n

39. A positive integer is selected at random and is divided by 7,what is the probability that the remainder is 1?
A. 3/7    B. 4/7   C. 1/7    D. 2/7
Ans: 1/7

40. A mixture of 40 litres of salt and water contains 70%of salt.how much water must be added to decrease the salt percentage to 40%?
A. 40 litres  B. 30 litres  C. 20 litres  D.2 litres
Ans: x=30
Ques. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
1. 34
2. 77
3. 12
4. 45

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number =
693×11/99
693×11/99
= 77

Ques. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A.    276
B.    299
C.    322
D.    345

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322
Ques. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively
1. 125
2. 127
3. 129
4. 131

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 – 10) and (2045 -13)
= HCF (1651, 2032) = 127

Ques. Find the largest number which divides 62,132,237 to leave the same reminder
1. 30
2. 32
3. 35
4. 45

Explanation:

Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35

Ques. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A.    4
B.    5
C.    6
D.    8

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A.    101
B.    107
C.    111
D.    185

Explanation:

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

Greater number = 111.

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A.    40
B.    80
C.    120
D.    200

Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A.    3
B.    13
C.    23
D.    33

Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A.    26 minutes and 18 seconds
B.    42 minutes and 36 seconds
C.    45 minutes
D.    46 minutes and 12 seconds

Explanation:

L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Ques.     Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.
1. 15
2. 16
3. 30
4. 31

Explanation:

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.

Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31

Ques. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is
1. 10:28 am
2. 10:30 am
3. 10:31 am
4. None of above

Explanation:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.
Sometimes questions on red lights blinking comes in exam, which can be solved in the same way
Ques. Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively
1. 19
2. 17
3. 13
4. 9

Explanation:
Answer will be HCF of (400-9, 435-10, 541-14)
HCF of (391, 425, 527) = 17

There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
1. 80
2. 82
3. 85
4. 87

Explanation:

As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
So first number is : 551/29 = 19
Third number = 1073/29 = 37
So sum of these numbers is = (19 + 29 + 37) = 85
Ques. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A.    123
B.    127
C.    235
D.    305

Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A.    55/601
B.    601/55
C.    11/120
D.    120/11
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum =    1/a    +    1/b    =    a + b/ab    =    55/600    =    11/120

Ques  : Choose the correct answer.
What is the value of 4^-2 ?
Option 1 : 1/4
Option 2 : 1/16
Option 3 : -16
Option 4 : None of these

Correct Op: 2

Ques  : Choose the correct answer.
What is the value of (0.081)1/4?
Option 1 : 0.3
Option 2 : 0.03
Option 3 : 0.003
Option 4 : None of these

Correct Op: 1

Ques. A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?
3 ⅕ min
3 ⅖ min
3 ⅗ min
3 ⅘ min

Do not be confused, Take this question same as that of work done question’s. Like work done by 1st puncture in 1 minute and by second in 1 minute.
Lets Solve it:

1 minute work done by both the punctures =
(1/9+1/6)=(5/18)

So both punctures will make the type flat in
(18/5)mins=3 ⅗ mins

Correct Op: 3

Ques. A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will B alone finish the work.
27 days
54 days
56 days
68 days

Correct Op: 2

Explanation:

As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18

To get days in which B will finish the work, lets calculate work done by B in 1 day =
=(1/18∗1/3)=1/54

[Please note we multiplied by 1/3 as per B share and total of ratio is 1/3]

So B will finish the work in 54 days
Ques. To complete a work A and B takes 8 days, B and C takes 12 days, A,B and C takes 6 days. How much time A and C will take
24 days
16 days
12 days
8 days

Explanation:

A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6

We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)

So A 1 day work =
1/6−1/12=1/12

Similarly C 1 day work =
1/6−1/8=(4−3)/24=1/24

So A and C 1 day work =

1/12+1/24=3/24=1/8
So A and C can together do this work in 8 days

Ques. A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in
5 days
6 days
7.5 days
8.5 days

Explanation:

B’s 5 days work =
1/10∗5=½
Remaining work =1−1/2=½
A can finish work =15∗1/2=7.5days

A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
35 ½
36 ½
37 ½
38 ½

Explanation:

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 (1/2) days
Ques. 4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it ?
30 days
40 days
50 days
60 days

Explanation:

Let 1 man’s 1 day work = x
and 1 woman’s 1 days work = y.
Then, 4x + 6y = 1/8
and 3x+7y = 1/10
solving, we get y = 1/400 [means work done by a woman in 1 day]

10 women 1 day work = 10/400 = 1/40

10 women will finish the work in 40 days

Ques, 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
6 days
7 days
8 days
9 days

Explanation:

1 woman’s 1 day’s work = 1/70
1 Child’s 1 day’s work = 1/140
5 Women and 10 children 1 day work =

(5/70 +10/140)=1/7

So 5 women and 10 children will finish the work in 7 days.

Ques. 5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio
1:2
1:3
2:1
2:3

Explanation:

Let 1 man 1 day work = x
1 boy 1 day work = y

then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1

Rahul and Sham together can complete a task in 35 days, but Rahul alone can complete same work in 60 days. Calculate in how many days Sham can complete this work ?
84 days
82 days
76 days
68 days

Explanation:

As Rahul and Sham together can finish work in 35 days.
1 days work of Rahul and Sham is 1/35
Rahul can alone complete this work in 60 days,
So, Rahul one day work is 1/60
Clearly, Sham one day work will be = (Rahul and Sham one day work) – (Rahul one day work)
=1/35−1/60=1/84
Hence Sham will complete the given work in 84 days.
Ques. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

A.    4 days
B.    5 days
C.    6 days
D.    7 days

Explanation:

Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.

Then, 6x + 8y =    1/10   and 26x + 48y =    1/2 .

Solving these two equations, we get : x =    1/100    and y =    1/200    .
(15 men + 20 boy)’s 1 day’s work =       15/100    +    20/200       =    1/4    .
15 men and 20 boys can do the work in 4 days.

Ques. A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?

A.    40
B.    50
C.    54
D.    60

Explanation:

(A + B)’s 20 day’s work =    (    1/30  x 20    )    =    2/3   .
Remaining work =    (    1  –    2/3    )    =    1/3   .
Now,    1/3   work is done by A in 20 days.

Therefore, the whole work will be done by A in (20 x 3) = 60 days.

Ques. A person incurs a loss of 5% be selling a watch for Rs. 1140. At what price should the watch be sold to earn 5% profit.
Rs.1200
Rs.1230
Rs.1260
Rs.1290

Option C

Explanation:

Let the new S.P. be x, then.
(100 – loss%):(1st S.P.) = (100 + gain%):(2nd S.P.)

=>(95/1140=105/x)=>x=1260

Ques. A book was sold for Rs 27.50 with a profit of 10%. If it were sold for Rs. 25.75, then would have been percentage of profit and loss ?
2% Profit
3% Profit
2% Loss
3% Loss

Explanation:

S.P.=(((100+gain%)/100)∗C.P)
So, C.P. = ((100/110)∗25.75)
When S.P. = 25.75 then Profit=25.75−25=Re.0.75
Profit%=(0.75/25)∗100=3%

Ques. If the cost price is 25% of selling price. Then what is the profit percent.
150%
200%
300%
350%

Explanation:
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = 75/25 * 100 = 300%

Ques. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x
13
14
15
16

Explanation:

Let the Cost Price of one article = Rs. 1
CP of x articles = Rs. x
CP of 20 articles = 20
Selling price of x articles = 20

Profit = 25% [Given]
⇒((SP−CP)/CP) = 25/100=1/4⇒(20−x)/x=¼
⇒80−4x=x
⇒5x=80
⇒x=80/5=16

Ques. A man buys an item at Rs. 1200 and sells it at the loss of 20 percent. Then what is the selling price of that item
Rs. 660
Rs. 760
Rs. 860
Rs. 960

Explanation:

Here always remember, when ever x% loss,
it means S.P. = (100 – x)% of C.P
when ever x% profit,
it means S.P. = (100 + x)% of C.P

So here will be (100 – x)% of C.P.
= 80% of 1200
= 80/100 * 1200
= 960

Ques. Sahil purchased a machine at Rs 10000, then got it repaired at Rs 5000, then gave its transportation charges Rs 1000. Then he sold it with 50% of profit. At what price he actually sold it.
Rs. 22000
Rs. 24000
Rs. 26000
Rs. 28000

Explanation:

Question seems a bit tricky, but it is very simple.
Just calculate all Cost price, then get 150% of CP.

C.P. = 10000 + 5000 + 1000 = 16000

150% of 16000 = 150/100 * 16000 = 24000

Ques. A plot is sold for Rs. 18,700 with a loss of 15%. At what price it should be sold to get profit of 15%.
Rs 25300
Rs 22300
Rs 24300
Rs 21300

Explanation:

This type of question can be easily and quickly solved as following:

Let at Rs x it can earn 15% pr0fit
85:18700 = 115:x [as, loss = 100 -15, Profit = 100 +15]

x = (18700*115)/85
= Rs.25300

Ques. 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.
130%
140%
150%
160%

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 %

Ques. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25% then determine the value of x.
14
15
16
17

Explanation:

Let the cost price 1 article = Re 1
Cost price of x articles = x
S.P of x articles = 20

Gain = 20 -x

=>25=(((20−x)/x)∗100)
=>2000−100x=25x
=>x=16

In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit
70%
80%
90%
None of above

Explanation:

Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295

Required percentage = (295/420) * 100
= 70%(approx)

A man bought an article and sold it at a gain of 5 %. If he had bought it at 5% less and sold it for Re 1 less, he would have made a profit of 10%. The C.P. of the article was
Rs 100
Rs 150
Rs 200
Rs 250

Explanation:

Let original Cost price is x
Its Selling price = 105/100 * x = 21x/20
New Cost price = 95/100 * x = 19x/20
New Selling price = 110/100 * 19x/20 = 209x/200
[(21x/20) – (209x/200)] = 1

=> x = 200

A fruit seller sells mangoes at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%
Rs 8.81
Rs 9.81
Rs 10.81
Rs 11.81

Explanation:

85 : 9 = 105 : x
x= (9×105/85)
= Rs 11.81

Ques. A shopkeeper sold an article for Rs 2564.36. Approximately what was his profit percent if the cost price of the article was Rs 2400
4%
5%
6%
7%

Explanation:

Gain % = (164.36*100/2400) = 6.84 % = 7% approx

Ques. A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount
1000
1500
2000
2500

Explanation:
SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 – 240 = 2000

Qu.In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%
8
9
10
11

Explanation:
Clue:
Firstly we need to calculate the SI with prinical 800,Time 3 years and Rate 9/2%, it will be Rs. 108
Then we can get the Time as
Time = (100*108)/(150*8) = 9

Ques. A financier claims to be lending money at simple interest, But he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes.
10.25%
10%
9.25%
9%

Explanation:

Let the sum is 100.

As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:

SI for first Six Months = (100*10*1)/(100*2) = Rs. 5

Important: now sum will become 100+5 = 105

SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25

So amount at the end of year will be (100+5+5.25)
= 110.25

Effective rate = 110.25 – 100 = 10.25

Ques. A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
9%
10%
11%
12%

Explanation:

We can get SI of 3 years = 12005 – 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 12005 – 3675 = 6125

So Rate = (100*3675)/(6125*5) = 12%

Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit.
Rs. 8510
Rs. 8620
Rs. 8730
Rs. 8820

Explanation:

=>(8000×(1+5100)^2)
=>8000×21/20×21/20
=>8820

Ques. A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.
Rs 662
Rs 662.01
Rs 662.02
Rs 662.03

Explanation:

=[200(21/20×21/20×21/20)+200(21/20×21/20)+200(21/20)]
=662.02

Ques. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually
Rs 312
Rs 412
Rs 512
Rs 612

Explanation:

Amount=P(1+R/100)^n
C.I. = Amount – P
The present worth of Rs.169 due in 2 years at 4% per annum compound interest is
Rs 155.25
Rs 156.25
Rs 157.25
Rs 158.25

Explanation:

In this type of question we apply formula
Amount=P/(1+R/100)^n
Amount=169(1+4/100)^2
Amount=169∗25∗25/26∗26
Amount=156.25

Ques. At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years
3%
4%
5%
6%

Explanation:

Let Rate will be R%

1200(1+R/100)^2=134832/100
(1+R/100)^2=134832/120000
(1+R/100)^2=11236/10000
(1+R100)=106100
=>R=6%
Ques. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is
4 years
5 years
6 years
7 years

Explanation:

As per question we need something like following

P(1+R/100)^n > 2P
(1+20/100)^n > 2
(6/5)^n > 2
6/5×6/5×6/5×6/5 > 2

In what time will Rs.1000 become Rs.1331 at 10% per annum compounded annually
2 Years
3 Years
4 Years
5 Years

Explanation:

Principal = Rs.1000;
Amount    = Rs.1331;
Rate    = Rs.10%p.a.

Let the time be n years then,

1000(1+10/100)^n=1331
(11/10)^n=1331/1000
(1110)^3=1331/1000

If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, what will be the compound interest on same values
Rs.51.75
Rs 51.50
Rs 51.25
Rs 51

Explanation:

S.I.=P∗R∗T/100
P=(50∗100)/5∗2=500
Amount=500(1+5/100)^2
500(21/20∗21/20)=551.25
C.I.=551.25−500
=51.25

Ques. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum
Rs 600
Rs 625
Rs 650
Rs 675

Explanation:

Let the Sum be P
S.I. = P∗4∗2/100=2P/25
C.I. = P(1+4/100)^2−P
=(676P/625)−P
=51P/625
As, C.I. – S.I = 1
=>51P/625−2P/25=1
=>(51P−50P)625=1
P=625
Ques. A sum of money invested at compound interest to Rs. 800 in 3 years and to Rs 840 in 4 years. The rate on interest per annum is.
4%
5%
6%
7%

Explanation:
S.I. on Rs 800 for 1 year = 40
Rate = (100*40)/(800*1) = 5%

Ques. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. Find the speed at which the train must run to reduce the time of journey to 40 minutes.
50 km/hr
60 km/hr
65 km/hr
70 km/hr

Explanation:

We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.

Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km

New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph

Ques. Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What time will they take to be 8.5km apart, if they walk in the same direction?
15 hours
16 hours
17 hours
18 hours

Explanation:

In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph

Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs

Ques. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour ?
8 minutes
10 mintues
12 minutes
14 minutes

Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = (9/54) hour
= (1/6)*60 minutes
= 10 minutes

Ques. 2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds ?
11:9
13:9
17:9
21:9

Explanation:

We know total distance is 200 Km
If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km]
So ratio of their speed = 110:90 = 11:9

Ques. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
16 km
14 km
12 km
10 km

Explanation:

Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 – x) hr

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

So distance traveled on foot = 4(4) = 16 km

A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.
70.11 km/hr
71.11 km/hr
72.11 km/hr
73.11 km/hr

Explanation:

We know Time = Distance/speed

So total time taken =
(160/64+160/80)=92hours
Time taken for 320 Km
= 320∗2/9
=71.11km/hr
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
9 km/hour
10 km/hour
11 km/hour
12 km/hour

Explanation:

We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,

Let the distance travelled by x km
Time = Distance/Speed

x/10−x/15=2
[because, 2 pm – 12 noon = 2 hours]

3x−2x=60
x=60
Time=Distance/Speed
Time@10km/hr=60/10=6hours
So 2 P.M. – 6 = 8 A.M
Robert starts at 8 A.M.

He have to reach at 1 P.M. i.e, in 5 hours

So, Speed = 60/5 = 12 km/hr

A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
44 km/hour
46 km/hour
48 km/hour
50 km/hour

Explanation:

Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr

Average speed =
2xy/(x+y)
=2∗40∗60/(40+60)
=4800/100
=48Km/hr

A man in a train notices that he can count 41 telephone posts in one minute. If they are known to be 50 metres apart, then at what speed is the train travelling?
60 km/hr
100 km/hr
110 km/hr
120 km/hr

Explanation:

Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour

A person travels equal distances with speed of 3 km/hr, 4 km/hr and 5 km/hr and takes a total of 47 minutes. Find the total distane
3 km
4 km
6 km
9 km

Explanation:

Let the distance be 3x km,
then,
x/3+x/4+x/5=47/60
47x/60=47/60
x=1

So total distance = 3*1 = 3 Km

Ques. A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 7:30 am. They shall first cross each other at ?
7:15 am
7:30 am
7: 42 am
7:50 am

Explanation:

Relative speed between two = 6-1 = 5 round per hour

They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.

So 7:30 + 12 mins = 7:42

The ratio between the speeds of two trains is 7: 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is ?
83.5 km/hr
84.5 km/hr
86.5 km/hr
87.5 km/hr

Explanation:

Let the speeds of two trains be 7X and 8X km/hr.

8X=400/4
=>X=12.5Km/hr

So speed of first train is 12.5*7 = 87.5 km/hr
Ques. A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream.
0,5
5,5
15,5
10,5

Explanation:

If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)

=> Rate in still water = 1/2(20+10) = 15 kmph
=> Rate of current = 1/2(20-10) = 5 kmph
Ques. In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still water
6
7
8
9

Explanation:

We know we can calculate it by 1/2(a+b)

=> 1/2(11+5) = 1/2(16) = 8 km/hr

Ques. If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream is
5 km/hr
4 km/hr
2 km/hr
1 km/hr

Explanation:

Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 – 5)kmph = 1 kmph

Ques. A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is
4 kmph
5 kmph
6 kmph
7 kmph

Explanation:

Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph

If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of the boat in still water is
12 kmph
13 kmph
14 kmph
15 kmph

Explanation:

Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 3 kmph.
Let speed in sttil water is x km/hr
Then, speed upstream = (x —3) km/hr.
x-3 = 10 or x = 13 kmph

Ques. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat in still water and stream is
3:1
1:3
2:4
4:2

Explanation:

Let speed downstream = x kmph
Then Speed upstream = 2x kmph

So ratio will be,
(2x+x)/2 : (2x-x)/2
=> 3x/2 : x/2 => 3:1

A man’s speed with the current is 20 kmph and speed of the current is 3 kmph. The Man’s speed against the current will be
11 kmph
12 kmph
14 kmph
17 kmph

Explanation:

If you solved this question yourself, then trust me you have a all very clear with the basics of this chapter.

If not then lets solve this together.
Speed with current is 20,
speed of the man + It is speed of the current
Speed in still water = 20 – 3 = 17

Now speed against the current will be
speed of the man – speed of the current
= 17 – 3 = 14 kmph

Ques. A man can row at 5 kmph in still water. If the velocity of the current is 1 kmph and it takes him 1 hour to row to a place and come back. how far is that place.
.4 km
1.4 km
2.4 km
3.4 km

Explanation:

Let the distance is x km
Rate downstream = 5 + 1 = 6 kmph
Rate upstream = 5 – 1 = 4 kmph
then
x/6 + x/4 = 1 [because distance/speed = time]
=> 2x + 3x = 12
=> x = 12/5 = 2.4 km

Ques. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is
1.6 km
2 km
3.6 km
4 km

Explanation:

Speed downstreams    =(15 + 3)kmph
= 18 kmph.
Distance travelled    = (18 x 12/60)km
= 3.6km

Ques.Sahil can row 3 km against the stream in 20 minutes and he can return in 18 minutes. What is rate of current ?
1/2 km/hr
1/3 km/hr
2 km/hr
4 km/hr

Explanation:

Speed Upstream=
3/(20/60)=9km/hrSpeed
Downstream=3(18/60)=10km/hr
Rate of current will be (10−9)/2
=1/2km/hr
Ques. In how many words can be formed by using all letters of the word BHOPAL
420
520
620
720

Explanation:

Required number
=6!=6∗5∗4∗3∗2∗1=720

Ques. In how many way the letter of the word “APPLE” can be arranged
20
40
60
80

Explanation:

Friends the main point to note in this question is letter “P” is written twice in the word.
Easy way to solve this type of permutation question is as,

So word APPLE contains 1A, 2P, 1L and 1E
Required number =
=5!/(1!∗2!∗1!∗1!)
=(5∗4∗3∗2!)2!
=60

In how many way the letter of the word “RUMOUR” can be arranged
2520
480
360
180

Explanation:

In above word, there are 2 “R” and 2 “U”
So Required number will be

=6!/(2!∗2!)
=6∗5∗4∗3∗2∗1/4
=180

In how many ways can the letters of the word “CORPORATION” be arranged so that vowels always come together.
5760
50400
2880
None of above

Explanation:

Vowels in the word “CORPORATION” are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)

This has 7 lettes, where R is twice so value = 7!/2!
= 2520

Vowel O is 3 times, so vowels can be arranged = 5!/3!

= 20

Total number of words = 2520 * 20 = 50400

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there
109
128
138
209

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have

(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)

This combination question can be solved as

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done
456
556
656
756

Explanation:

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)

A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw
64
128
132
222

Explanation:

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.

Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]

=1+[3×6]+[3×((6×5)/(2×1))]=1+18+45=64

Ques. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
12
24
48
168

Explanation:

This question seems to be a bit typical, isn’t, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways

Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24

Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?
1/3
1/6
1/2
1/8

Explanation:

Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]

Favoured cases are = [TTH, THT, HTT, TTT] = 4

So required probability = 4/8 = ½

Ques. In a throw of dice what is the probability of getting number greater than 5
1/2
1/3
1/5
1/6

Explanation:

Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]

So probability = ⅙

Ques. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?
3/4
1/4
7/4
1/2

Explanation:

Total number of cases = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So Probability = 27/36 = ¾

Ques. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?
2/3
8/21
3/7
9/22

Explanation:

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.

Ques. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is
1/13
2/13
1/26
1/52

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26

Ques. A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident
30%
35%
40%
45%

Explanation:

Let A = Event that A speaks the truth
B = Event that B speaks the truth

Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5

P(A-lie) = 1-3/4 = 1/4
P(B-lie) = 1-4/5 = 1/5

Now
A and B contradict each other =
[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
= (3/5*1/5) + (1/4*4/5) = 7/20
= (7/20 * 100) % = 35%
Ques. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings
2/121
2/221
1/221
1/13

Explanation:

Total cases =52C2=52∗51/2∗1=1326
Total King cases =4C2=4∗3/2∗1=6
Probability ==6/1326=1/221

Ques. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random. What is the probability that they are not of same colour.
52/55
3/55
41/44
3/44

Ques. Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white
1
2/3
1/3
4/3

Explanation:

Total cases = 10 + 20 = 30
Favourable cases = 20

So probability = 20/30 = ⅔

There is a pack of 52 cards and Rohan draws two cards together, what is the probability that one is spade and one is heart ?
11/102
13/102
11/104
11/102