Practice MCQ Questions and Answer on Number System

651.

A number consists of two digits such that the digit in the ten's place is less by 2 than the digit in the unit place. Three times the number added to $\frac{6}{7}$ times the number obtained by reversing digits equals 108. The sum of digits in the number is -

(A) 8
(B) 9
(C) 6
(D) 7

Solution: $\begin{aligned} Let the unit digit \\ Ten digit = x - 2 \\ ∴ Number : \\ = 10 (x - 2) + x \\ = 10 x - 20 + x \\ = 11 x - 20 \end{aligned}$ New number obtained after reversing the digits $\begin{aligned} = 10 x + x - 2 \\ = 11 x - 2 \\ According to the question, \\ 3 (11 x - 20) + \frac{6}{7} (11 x - 2) = 108 \\ 7 (11 x - 20) + 2 (11 x - 2) = 36 \times 7 \\ 77 x - 140 + 22 x - 4 = 252 \\ 99 x = 252 + 144 \\ x = \frac{396}{99} = 4 \\ Number : \\ = 11 x - 20 \\ = 11 \times 4 - 20 \\ = 24 \\ Sum of digit = 2 + 4 = 6 \end{aligned}$

652.

The product of two numbers is 120 and the sum of their squares is 289. The sum of the two number is -

(A) 23
(B) 7
(C) 13
(D) 169

Solution: Let the two number are a and b According to question, $\begin{aligned} a^{2} + b^{2} = 289 \\ a b = 120 \\ {(a + b)}^{2} = a^{2} + b^{2} + 2 a b \\ {(a + b)}^{2} = {(289)}^{2} + (2 \times 120) \\ {(a + b)}^{2} = 529 \\ (a + b) = 23 \end{aligned}$

653.

n being any odd number greater than 1, n⁶⁵ - n is always divisible by :

(A) 5
(B) 13
(C) 24
(D) None of these

Solution: $\begin{aligned} \Leftrightarrow n^{65} - n \\ = n (n^{64} - 1) \\ = n (n^{32} - 1) (n^{32} + 1) \\ = n (n^{16} - 1) (n^{16} + 1) (n^{32} + 1) \end{aligned}$ $= n (n^{8} - 1) (n^{8} + 1) (n^{16} + 1)$ $(n^{32} + 1)$ $= n (n^{4} - 1) (n^{4} + 1) (n^{8} + 1)$ $(n^{16} + 1)$ $(n^{32} + 1)$ $= n (n^{2} - 1) (n^{2} + 1) (n^{4} + 1)$ $(n^{8} + 1)$ $(n^{16} + 1)$ $(n^{32} + 1)$ $= (n - 1) n (n + 1) (n^{2} + 1) (n^{4} + 1)$ $(n^{8} + 1)$ $(n^{16} + 1)$ $(n^{64} + 1)$ $(n^{32} + 1)$ Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd. Thus, the given number is definitely a multiple of 24

654.

Which one of the following is a prime number ?

(A) 161
(B) 221
(C) 373
(D) 437

655.

If a + b + c = 0, (a + b) (b + c) (c + a) equals

(A) ab (a + b)
(B) (a + b + c)2
(C) - abc
(D) a2 + b2 + c2

656.

When a certain positive integer P is divided by another positive integer, the remainder is $r_{1}$ . When a second positive integer Q is divided by the same integer, the remainder is $r_{2}$ and when (P + Q) is divided by the same divisor, the remainder is $r_{3}$ . Then the divisor may be :

(A) $r_{1}$ $r_{2}$ $r_{3}$
(B) $r_{1}$ + $r_{2}$ - $r_{3}$
(C) $r_{1}$ - $r_{2}$ + $r_{3}$
(D) $r_{1}$ + $r_{2}$ - $r_{3}$

Solution: Let P = x + $r_{1}$ and Q = y + $r_{2}$ , where each of x and y are divisible by the common divisor. Then, P + Q = (x + $r_{1}$ ) + (y + $r_{2}$ ) = (x + y) + ( $r_{1}$ + $r_{2}$ ) (P + Q) leaves remainder $r_{3}$ when divided by the common divisor. ⇒ [(x + y) + ( $r_{1}$ + $r_{2}$ ) - $r_{3}$ ] is divisible by the common divisor. Since (x + y) is divisible by the common divisor, so divisor = $r_{1}$ + $r_{2}$ - $r_{3}$

657.

Between two distinct rational numbers a and b, there exists another rational number which is :

(A) $\frac{a}{2}$
(B) $\frac{b}{2}$
(C) $\frac{a b}{2}$
(D) $\frac{a + b}{2}$

Solution: If a and b are two rational numbers, then $\frac{a + b}{2}$ is a rational number lying between a and b.

658.

$\frac{2}{3}$ of three-fourth of a number is :

(A) $\frac{1}{2}$ of the number
(B) $\frac{1}{3}$ of the number
(C) $\frac{8}{9}$ of the number
(D) $\frac{11}{12}$ of the number

Solution: $\begin{aligned} = \frac{2}{3} \times \frac{3}{4} \times n (n is a number) \\ = \frac{1}{2} n \end{aligned}$

659.

The numbers 2, 4, 6, 8 ..... 98, 100 are multiplied together. The number of zeros at the end of the product must be :

(A) 10
(B) 11
(C) 12
(D) 13

Solution: N = 2 × 4 × 6 × 8 × ..... × 98 × 100 = 250 (1 × 2 × 3 × ..... × 49 × 50) = 250 × 50! Clearly, the highest power of 2 in N is much higher than that of 5 ∴ Number of zeros in N = Highest power of 5 in N = $[\frac{50}{5}] + [\frac{50}{5^{2}}]$ = 10 + 2 = 12

660.

7 is added to a certain number; the sum is multiplied by 5 ; the product is divided by 9 and 3 is subtracted from the quotient. Thus, if the remainder left is 12, what was the original number ?

(A) 20
(B) 30
(C) 40
(D) 60

Solution: Let the original number be x. Then, $\frac{(x + 7) \times 5}{9}$ - 3 = 12 ⇒ $\frac{5 x + 35}{9}$ = 15 ⇒ 5x + 35 = 135 ⇒ 5x = 100 ⇒ x = 20

«
1
2
...
59
60
61
62
63
64
65
66
67
68
Next ›
Last »

Indian Polity Mcq IAS GK Question India GK World GK GK questions GK Quiz Biology GK Science GK Physics GK Chemistry GK General Awareness Computer GK Political GK Indian Political GK Economics GK History GK Sports GK Geography GK Teaching Aptitude GK CTET/TET GK B Ed Entrance GK Banking GK SSC GK UPSC GK Indian Army GK Hindi Grammar GK Hindi Grammar MCQ Reasoning questions Bihar GK Rajasthan GK Jharkhand GK MP GK UP GK Chhattisgarh GK Haryana GK HP GK Maharashtra GK Electronics GK Agriculture GK States/Capitals GK Books Name & Author Arthimetic Verbal Reasoning Non Verbal Reasoning English

Practice MCQ Questions and Answer on Number System

General Knowledge