Practice MCQ Questions and Answer on Number System
31.
The remainder obtained when any prime number greater than 6 is divided by 6 must be :
(A) either 1 or 2
(B) either 1 or 3
(C) either 1 or 5
(D) either 3 or 5
Solution:
Let the required prime number be p. Let p when divided by 6 give n as quotient and r as remainder. Then p = 6n + r, where 0 $$ \leqslant $$ r 6 Now, r = 0, r = 2, r = 3 and r = 4 do not give p as prime. ∴ r $$ \ne $$ 0, r $$ \ne $$ 2, r $$ \ne $$ 3, and r $$ \ne $$ 4 Hence, r = 1 or r = 5
32.
A boy added all natural numbers from 1 to 10, however he added one number twice due to which the sum becomes 58. What is the number which he added twice?
(A) 3
(B) 4
(C) 7
(D) 8
Solution:
Sum of natural number from 1 to 10 $$\eqalign{ & = \frac{{\left( n \right)\left( {n + 1} \right)}}{2}\,\,\,\,\,\,\,\left( {{\text{where }}n = 10} \right) \cr & = \frac{{10 \times 11}}{2} \cr & = 55 \cr} $$ Clearly, we can see the number which is added twice = 58 - 55 = 3
33.
Let x be the product of two numbers 3, 659, 893, 456, 789, 325, 678 and 342, 973, 489, 379, 256. The number of digits in x is :
(A) 32
(B) 34
(C) 35
(D) 36
Solution:
Sum of digits in the two numbers = 19 + 15 = 34 So, the product will have 33 or 34 digits Since 36 × 34 = 1224 (i.e., product has 2 + 2 = 4 digits) So, the number of digits in x is 34
34.
476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's places are respectively:
(A) 7 and 4
(B) 7 and 5
(C) 8 and 5
(D) None of these
Solution:
Let the number be 476ab0 476ab0 is divisible by 3 ⇒ 4 + 7 + 6 + a + b + 0 is divisible by 3 ⇒ 17 + a + b is divisible by 3 - - - - - - - (i) 476ab0 is divisible by 11 [(4 + 6 + b) - (7 + a + 0)] is 0 or divisible by 11 ⇒ [3 + (b - a)] is 0 or divisible by 11 - - - - - - - -(ii) Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation (i) and Equation (ii). if a = 6 and b = 2, 17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation(i).Hence this is not the answer if a = 8 and b = 2, 17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- Meet equation(i) [3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11 --- Does not meet equation(ii).Hence this is not the answer if a = 6 and b = 5, 17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation (i) .Hence this is not the answer if a = 8 and b = 5, 17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- Meet equation 1 [3 + (b - a)] = [3 + (5 - 8)] = 0 ---Meet equation 2 Since these values satisfies both equation 1 and equation 2, this is the answer
35.
In a certain series, each number except the first and second is obtained by adding the previous two numbers. If the first no is 2 and sixth no is 26, then the seventh number is:
If the number 8764x5 is divisible by 9, then find the least possible value of x where x is a two-digit number.
(A) 15
(B) 06
(C) 14
(D) 18
Solution:
If 8764x5, is divisible by 9 then, sum of digit 8 + 7 + 6 + 4 + x + 5 = 0/9/18/36/45 30 + x = 45 x = 15
37.
The number of times 99 is subtracted from 1111 so that the remainder is less then 99 is :
(A) 10
(B) 11
(C) 12
(D) 13
Solution:
On dividing 1111 by 99, the quotient is 11 and the remainder is 22. Hence, the required number is 11
38.
The least among the fractions $$\frac{{15}}{{16}}$$ , $$\frac{{19}}{{20}}$$ , $$\frac{{24}}{{25}}$$ , $$\frac{{34}}{{35}}$$ is :
(A) $$\frac{{34}}{{35}}$$
(B) $$\frac{{15}}{{16}}$$
(C) $$\frac{{19}}{{20}}$$
(D) $$\frac{{24}}{{25}}$$
Solution:
If difference between numerator and denominator is same in all fractions and numerator is smaller than denominator in all fractions, then smaller will be smaller and larger will larger. So, $$\frac{{15}}{{16}}$$ is the smallest value among all. So this is the answer
39.
For the integer n, if n3 is odd, then which of the following statements are true ?
I. n is odd
II. n2 is odd
III. n2 is even
(A) I only
(B) II only
(C) I and II only
(D) I and III only
Solution:
n3 is odd ⇒ n is odd and n2 is odd ∴ I and II are true.
40.
If m = - 4, n = - 2, then the value of m3 - 3m2 + 3m + 3n + 3n2 + n3 is :