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Number System - Aptitude MCQ Questions and Solutions with Explanations

Number System MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Number System with easy and logical explanations. Number System MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-2 section-5

Each section contains 100 MCQs on Number System .
11
If a = 7, b = 5, c = 3, then the value of a2 + b2 + c2 - ab - bc - ca is :
Discuss
Answer & Solution
Answer: Option D
Solution:
= (a2 + b2 + c2 - ab - bc - ca)
= (a + b + c)2 - 3 (ab + bc +ca)
= (7 + 5 + 3)2 - 3 (35 + 15 + 21)
= (15)2 - 3 × 71
= (225 - 213)
= 12
12
While writing all the numbers from 700 to 1000, how many numbers occur in which the first digits greater than the second digit, and the second digit is greater than the third digit ?
Discuss
Answer & Solution
Answer: Option D
Solution:
When the second digit is 1, third digit can be 0, i.e., there is one such number.
When the second digit is 2, third digit can be 0 or 1, i.e., there are 2 such numbers.
When the second digit is 3, third digit can be 0, 1 or 2, i.e., there are 3 such numbers, and so on.
When the first digit is 7, second digit can be 1, 2, 3, 4, 5 or 6
So, there are (1 + 2 + 3 + 4 + 5 + 6) = 21 such numbers between 700 and 799
When the first digit is 8, second digit can be 1, 2, 3, 4, 5, 6 or 7
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7) = 28 such numbers between 800 and 899
When the digit is 9, second digit can be 1, 2, 3, 4, 5, 6, 7 or 8
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 such numbers between 900 and 999
Hence, the required number
= (21 + 28 + 36)
= 85
13
98th term of the infinite series 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, .... is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Given series is 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, .....
∴ 96th term is 4
So, $${T_{97}}$$ = 1 and $${T_{98}}$$ = 2
Hence, the 98th term is 2
14
If x and y are two digits of the number 653xy such that the number is divisible by 80, then x + y is equal to :
Discuss
Answer & Solution
Answer: Option D
Solution:
Since, 653xy is divisible by 80, we must have y = 0
Now, 653x0 must be divisible by both 5 and 16
Clearly, it is divisible by 5 for all values of x
Now, the number 53x0 must be divisible by 16
The least value of x is clearly 6
So, x + y
= 6 + 0
= 6
15
P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P + Q = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
We may have (64 and 1), (32 and 2), (16 and 4) and (8 and 8)
In any case, the sum is not 35
16
How many numbers less than 1000 are multiple of both 10 and 13 ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Required numbers are multiples of (10 × 13), i.e., 130
These numbers are 130, 260, 390, 520, 650, 780 and 910
They are 7 in number.
17
If (1012 + 25)2 - (1012 - 25)2 = 10n, then the value of n is :
Discuss
Answer & Solution
Answer: Option C
Solution:
(1012 + 25)2 - (1012 - 25)2
= 4 × 1012 × 25
[∵ (a + b)2 - (a - b)2 = 4ab]
= 1012 × 100
= 1012 × 102
= 1014
Hence, n = 14
18
Which of the following numbers is divisible by 3, 7, 9 and 11 ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Clearly, 639 is not divisible by 7
Consider 2079
Sum of its digits
= (2 + 0 + 7 + 9)
= 18
So, it is divisible by both 3 and 9
Also, (79 - 2) = 77, which is divisible by 7
So, 2079 is divisible by 7
Also, (9 + 0) - (7 + 2) = 0
So, 2079 is divisible by 11
Hence, 2079 is divisible by each one of 3, 7, 9 and 11
19
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 :
Discuss
Answer & Solution
Answer: Option D
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 = $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$
20
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
Discuss
Answer & Solution
Answer: Option C
Solution:
n3 is odd ⇒ n is odd and n2 is odd
∴ I and II are true.