This set of Aptitude Questions and Answers (MCQs) focuses on “Number System – Set 3”.

1. What is the 999^{th} digit of (56700)^{499} from right side?

a) 3

b) 7

c) 0

d) 9

View Answer

Explanation: (56700)

^{499}has 998 zeroes from right and 999

^{th}term depends on 7

^{499}.

The last digit of 7

^{499}= 7

^{124*4+3}is 3.

Therefore, the 999

^{th}digit of (56700)

^{499}from right side is 3.

2. A person starts multiplying consecutive positive integers from 120. How many numbers should he multiply before he will have a result of 4 zeroes?

a) 6

b) 10

c) 11

d) 5

View Answer

Explanation: We know that product of 5 and 2 yields zero.

There are as many numbers of 2’s compared to 5’s. Therefore, the number of zeroes depends on number of 5’s.

125 = 5*5*5, which yields 3 zeroes and there is already a zero in 120.

Therefore, 120*121*122*123*124*125 yields 4 zeroes, i.e., he should multiply 6 numbers from 120.

3. Find the sum of first 50 terms of the series.

\(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}+\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)……

a) 1/2

b) 3/8

c) 5/8

d) 3/4

View Answer

Explanation: The terms are repeating after every 6 terms.

Every six terms cancel each other and result in zero.

Sum of 48 terms is also 0.

Therefore, sum of 50 terms = 0 + ½ + ¼ = ¾.

4. Find the possible integral values of x in x^{2} + |x – 2| = 2.

a) 1

b) -1

c) -1 or 0

d) 1 or 0

View Answer

Explanation: x

^{2}+ |x – 2| = 2

x

^{2}+ x – 2 = 2 or x

^{2}– x + 2 = 2

x

^{2}+ x – 4 = 0 or x

^{2}– x = 0

The roots are complex. or the integral roots of x are 0 or 1.

5. If x = y + 1, where y is the product of four consecutive natural numbers, then which of the following is/are true?

I. x is odd.

II. x is a perfect square.

III. x is not a multiple of 3.

a) I & II

b) II & III

c) I & III

d) I, II & III

View Answer

Explanation: Consider y = 2*3*4*5 = 120 and x = 120+1 = 121, which is odd, perfect square and not a multiple of 3.

Consider y = 3*4*5*6 = 360 and x = 360+1 = 361, which is odd, perfect square and not a multiple of 3.

Hence, all the three statements are correct.

6. How many pairs of natural numbers are there whose difference of their squares is 100?

a) 1

b) 3

c) 2

d) 4

View Answer

Explanation: Let the pair of natural number by x & y.

x

^{2}– y

^{2}= (x+y)(x-y) = 100.

The factor pairs are, (1,100), (2,50), (4,25), (5,20), (10,10).

The only possible pair is from factor pair (2,50) i.e., x=26 and y=24.

Therefore, there is only 1 of natural number whose difference of their squares is 100.

7. Find the sum two numbers, whose product is 200 and difference is minimum.

a) 33

b) 35

c) 27

d) 30

View Answer

Explanation: x & y be the two numbers.

Given that x*y = 200 and x-y = minimum.

The factor pairs are (1,200), (2,100), (4,50), (5,40), (8,25) and (10,20).

The only pair which satisfy both the condition is 10,20.

Hence, their sum is 30.

8. The mean of 1, 2, 2^{2}, 2^{3}, ……, 2^{63} lies between which of the following number?

a) 2^{57} and 2^{58}

b) 2^{59} and 2^{58}

c) 2^{57} and 2^{56}

d) 2^{56} and 2^{55}

View Answer

Explanation: The series is in GP with a=1, r=2 and n=64.

Sum = 2

^{63}– 1

Mean = \(\frac{2^{63}–1}{64}\)=2

^{57}–\(\frac{1}{2^6}\)

Therefore, the mean lies between 2

^{57}and 2

^{56}.

9. For which of the following conditions is \(\sqrt{p}*\sqrt{q}=\sqrt{pq}\) true?

a) p<0 and q>0

b) p<0 and q<0

c) p>0 and q>0

d) p>0 and q<0

View Answer

Explanation: Square root of negative real numbers is imaginary. Hence both the numbers must be positive real numbers. Therefore, for \(\sqrt{p}*\sqrt{q}=\sqrt{pq}\) to be true p>0 and q>0.

10. Find the number x, such that sum of the number with its square is least.

a) 0.5

b) -0.5

c) -1.5

d) -0.75

View Answer

Explanation: 0.5 + 0.5

^{2}= 0.75.

-0.5 + (-0.5)

^{2}= -0.25.

-1.5 + (-1.5)

^{2}= 0.75.

-0.75 + (-0.75)

^{2}= -0.1875.

Therefore, the least value is -0.1875 i.e., for x=-0.75.

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