Model Aptitude questions and answers for placement - Set 1

1)   The unit digit in the product (624 * 708 * 913 * 463) is:

a. 2
b. 5
c. 6
d. 8
Answer  Explanation 

ANSWER: 8

Explanation:
Unit digit in the given product = Unit Digit in (4*8*3*3) = 8


2)   The ratio of two numbers is 4 : 5 and their H.C.F is 4. Find their L.C.M.

a. 96
b. 80
c. 73
d. 48
Answer  Explanation 

ANSWER: 80

Explanation:
Let the numbers be 4x and 5x.

H.C.F = x. So, x=4.
So, the numbers are 16 and 20
L.C.M of 16 and 20 = 80.


3)   3639 + 11.95 - x = 3054. Find the value of x.

a. 407.09
b. 479.75
c. 523.93
d. 596.95
Answer  Explanation 

ANSWER: 596.95

Explanation:
Let 3639 + 11.95 – x = 3054
Then, x = (3639 + 11.95) – 3054

= 3650.95 – 3054
= 596.95


4)   If 2x +3y = 30 and (x+y)/y = 11/8, then find the value of 5y + 6x

a. 72
b. 58
c. 64
d. 29
Answer  Explanation 

ANSWER: 58

Explanation:
The given equations are :

2x + 3y = 30 --------- (i)
and, (x+y)/y = 11/8
8x + 8y = 11y
8x – 3y =0 ----(ii)

Adding (i) & (ii), we get : 10 x = 30 or x = 3.

Put x = 3 in (i), we get : y = 8
Therefore, 5y + 6x = (5 * 8 + 6 * 3) = 40 + 18 = 58.


5)   Given that √12 = 3.464 and √120 = 10.95, find the value of √1.2 + √1200 + √0.012.

a. 32.164
b. 35.844
c. 36.164
d. 37.304
Answer  Explanation 

ANSWER: 35.844

Explanation:
Given exp. = √1.2 +√1200 +√0.0120 = √120/100 +√12*100 + √120/10000
= (√120)/10 + √12 * 10 + (√120)/100 = 10.95/10 + 3.464 * 10 + 10.95/100
= 1.095 + 34.64 + 0.1095
= 35.8445


6)   The average weight of three boys P, Q and R is 54 kg, while the average weight of three boys Q, S and T is 60 kg. What is the average weight of P, Q, R, S and T?

a. 66.4 kg
b. 63.2 kg
c. 58.8 kg
d. Data Inadequate
e. None of these
Answer  Explanation 

ANSWER: Data Inadequate

Explanation:
Total weight of (P + Q + R) = {54*3} kg = 162 kg
Total weight of(Q + S + T) = (60 *3) kg = 180 kg
Adding both, we get : P + 2Q + S + R + T = (162 + 180) kg = 342 kg
So, to find the average weight of P, Q, R, S & T, we ought to know Q's weight, which is not given.
The data is inadequate.


7)   Two-third of a positive number and 16/216 of its reciprocal are equal. Find the positive number.

a. 9/25
b. 14/4
c. 4 /12
d. 144/25
Answer  Explanation 

ANSWER: 4 /12

Explanation:
Let the positive number be x.
Then, 2/3 x = 16/216 * 1/x
x2 = 16/216 * 3/2
= 16/144
x = √16/144 =4 /12.


8)   The ratio between the present ages of A and B is 3:5 respectively. If the difference between B's present age and A's age after 4 years is 2 , what is the total of A's and B's present ages?

a. 24 years
b. 32 year
c. 48 years
d. cannot be determined
e. None of these
Answer  Explanation 

ANSWER: 24 years

Explanation:
Let the present ages of A and B be 3x years and 5x years respectively.
5x – (3x + 4) = 2
2x = 6
x = 3.
Therefore,
Required sum = 3x + 5x = 8x = 24 years


9)   (1000)6/1015 = ?

a. 10
b. 100
c. 1000
d. 10000
Answer  Explanation 

ANSWER: 1000

Explanation:
(1000)6/1015
= (1000)6 / 1015
= (103)6/1015
= 10(3*6)/1015
= 1018/1015
= (10)(18 – 15)
= 103 = 1000


10)   Evaluate :

28% of 400 + 45 % of 250


a. 220.3
b. 224.5
c. 190.3
d. 150
Answer  Explanation 

ANSWER: 224.5

Explanation:
28% of 400 + 45 % of 250
= (28/100 *400 + 45/100 * 250)
= (112 + 112.5)
= 224.5