Number System MCQ Question

Number System: Quantitative Aptitude MCQ Question with Details Solution

by

Number System MCQ Question

The number system or the numeral system is the system of naming or representing numbers. There are various types of number systems in Mathematics like integer, prime, binary, decimal, etc. In this chapter Examportal Discussed All type Number System MCQ Question with details Explanation.

Q1. Find the remainder of $\frac{9^{99}}{8}$

(a) 1

(b) 2

(c) 3

(d) 5

Answer: (a)

Explanation: $\frac{9^{99}}{8}=\frac{\left( 8+1 \right) ^{99}}{8}\Rightarrow \mathrm{Remainder} =1$

Rules: If  $\frac{\left( \mathrm{a}+1 \right) ^n}{a}$ then remainder will always be 1, whether n is even or odd.

Q2. Minimum difference between x and y such that 1x 71y 61 is exactly divisible by 11 is,

(a) 2

(b) 3

(c) 1

(d) 0

Answer: (a)

Explanation: As 1x 71y 61 is exactly divisible by 11. (1 + 7 + y + 1) – (x + 1 + 6) = 0 or multiple of 11 for minimum difference

9 + y – 7 – x = 0

⇒ x – y = 2

Q3. The four integers next lower than 81, and the four next higher than 81, are written down and added together, this sum is divisible by,

(a) 7

(b) 9

(c) 11

(d) 13

Answer: (b)

Four integers next lower than 81 is 80, 79, 78, 77 four integers next higher than 81 is 82, 83, 84, 85

Sum = (80 + 82) + (79 + 83) + (78 + 84) + (77+ 85) = 2 ×81 + 2 ×81 + 2 ×81 + 2 ×81 = 8 × 81

Sum is divisible by 9 as 81 is divisible by 9

 

Q4. How many numbers, between 1 and 300 are divisible by 3 and 5 together?

(a) 16

(b) 18

(c) 20

(d) 100

Answer: (c)

Explanation: LCM of 3 and 5 = 15

Number divisible by 15 are 15, 30, 45 …..300.

Let total numbers are n

300 = 15 + (n – 1) × 15

⇒ 300 = 15 + 15n –15

⇒ n = 20

Q5. What is the remainder when 1! + 2! + 3! + …… + 100! Is divided by 7 ?

(a) 0

(b) 5

(c) 6

(d) 3

Answer: (b)

Explanation:  7! + 8! + 9! + 10! + ……. + 100 =  7.6! + 8.7.6! + 9.8.7.6! + ……. + 100! is completely divisible by 7 as each of the terms contain at least one 7 in it.

Now, 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 +  24 + 120 +  720 = 873

which leaves a remainder of 5 when divided by 7.

Q6. How many numbers, lying between 1 and 500, are divisible by 13?

(a) 40

(b) 38

(c) 41

(d) 46

Answer: (b)

Explanation:  Number divisible by 13, 26, 39, …. 494

Let n be the total numbers

494 = 13 + (n – 1) × 13

⇒ 13n = 494

⇒  n = 38

Q7. Two different numbers when divided by the same divisor, left remainder 11 and 21 respectively, and when their sum was divided by the same divisor, remainder was 4. What is the divisor?

(a) 36

(b) 28

(c) 14

(d) 9

Answer: (b)

Explanation:  Divisor = [Sum of remainders] – [ Remainder when sum is divided]

= 11 + 21 – 4 = 28

Q8. If ‘n’ is a natural number then the greatest integer less than that or equal to $\left( 2+\sqrt{3} \right) ^{\mathrm{n}}$  is

(a) odd

(b) even

(c) even when ‘n’ is even and odd when ‘n’ is odd

(d) even when ‘n’ is odd and odd when n is even

Answer: (a)

Explanation:  putting n = 1, we get  $\left( 2+1.\sqrt{3} \right) $  = whose integral part is 3

putting n= 2, we get $\left( 2+\sqrt{3} \right) ^2=4+3+4\sqrt{3}$  ; whose integral part is 11 which is again an odd number

Now, through the options it can be judged that the greatest integer must always be an odd number.

Q9. A number when divided by a divisor, left remainder 23. When twice of the number was divided by the same divisor, remainder was 11. Find the divisor.

(a) 12

(b) 34

(c) 35

(d) data inadequate

Answer: (d)

Explanation:  Let number be N.

Then, N = Divisor × Q1 + 23

2N = Divisor × Q2 + 11,

where Q1 and Q2 are quotients respectively.

Here, we have two equations and 3 variables. There equations cannot be solved.

Pages: 1 2
Pages ( 1 of 2 ): 1 2Next »

Leave a Comment

error: Content is protected !!