Aptitude

# Aptitude Questions and Answers

1

Find the product of

(8.77 xx 8.77xx 8.77+ 4.23xx 4.23 xx 4.23)/(8.77 xx 8.77 -8.77 xx 4.23+ 4.23 xx 4.23)   ?

a

11

b

4.54

c

8.77

d

13

= ((8.77)^3 + (4.23)^3) / ((8.77)^2 - (8.77 xx 4.23)+ (4.23)^2)

= (a^3 + b^3)/ (a^2 - ab+ b^2)

= (a+b)

= 8.77+4.23

= 13

2

What is the unit digit in 2^30?

a

3

b

4

c

1

d

2

Cyclicity of 2 = 4
D = d * Q + R
30 = 4 * 7 + 2
30/4 = Remainder 2, 22 = 4

Notes:
Cyclicity of 2, 3, 7, 8 => 4
Cyclicity of 4, 9 => 2
Cyclicity of 1, 5, 6 => 1

3

Find the remainder when 6799 is divided by 7.

a

4

b

6

c

1

d

2

Remainder of ((67^99)/7)==R==>((63+4)^99)/7

63 is divisible by 7 for any power, so required remainder will depend on the power of 4.

Require remainder:
(4^99)/7==R==>(4^(96+3))/7
4^3/7==> 64/7==> (63+1)/7==R==>1

4

The remainder when 1010+ 10100+101000+ . . . . . . + 101000000000 is divided by 7 is

a

0

b

1

c

2

d

5

Number of terms in the series = 10.
(We can get it easily by pointing the number of zeros in power of terms.
In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on.)

1010/7, Written as,
(7+3)(4*2+2)/7
The remainder will depend on,
32/7
So, remainder will be 2.
10100/7, Can be written as,
3100/7.
Or, 325*4/7
Or, 34/7.
Remainder = 1
101000/7, Remainder = 2
1010000/7, Remainder = 1

So, we get alternate 2 and 1 as remainder, five times each.
So, required remainder is given by

(2+1+2+1+2+1+2+1+2+1)/7
= 15/7

Remainder: 1

5

In a two-digit number, the digit in the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged, difference between the newly formed number and the original number is less than the original number by 1. What is the original number ?

a

35

b

36

c

37

d

39

Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

6

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

a

4

b

5

c

6

d

8

N = H.C.F of (4655-1305), (6905-4655) and (6905-1305) = H.C.F of 3360, 2240 and 5600

N= 1120, hence sum of digits in N= (1+1+2+0)= 4