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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-3 section-6

Each section contains 100 MCQs on Number System .
21
If (a2 - b2) ÷ (a + b) = 25, then (a - b) = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
We know (a2 - b2) = (a - b) (a + b)
So, (a2 - b2) ÷ (a + b) = 25
⇒ $$\frac{(a - b) (a + b)}{a + b}$$   = 25
⇒ (a - b) = 25
22
The sum of the digits of a natural number (10n - 1) is 4707, where n is a natural number. The value of n is :
Discuss
Answer & Solution
Answer: Option B
Solution:
10n has (n + 1) digits, Then, 9 will appear n times in (10n - 1).
So, sum of digits in (10n - 1) = 9n
∴ 9n = 4707
⇒ n = $$\frac{4707}{9}$$  = 523
23
The sum of digits of a two-digit number is 12 and the difference between the two-digits of the two-digit number is 6. What is the-digit number ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the two - digit number be 10a + b where a > b
According to the question,
a + b = 12 ..... (i)
a - b = 6 ..... (ii)
On adding equation (i) and (ii)
2a = 18
⇒ a = 9
From equation (i),
9 + b = 12
⇒ b = 12 - 9 = 3
∴ Number is 10a + b
= 9 × 10 + 3
= 93
∴ When a < b
Then required number is = 39
24
7386038 is divisible by :
Discuss
Answer & Solution
Answer: Option D
Solution:
Given number is 7386038
Sum of its digits = 35, which is not divisible by any of 3 and 9
So, the given number is not divisible by any of 3 and 9
Also, 38 is not divisible by 4
So, the given number is not divisible by 4
Also, (8 + 0 + 8 + 7) - (3 + 6 + 3)
= (23 - 12)
= 11
So, the given number is divisible by 11
25
The number of terms between 11 and 200 which are divisible by 7 but not by 3 are :
Discuss
Answer & Solution
Answer: Option A
Solution:
Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ..... , 189, 196
Tm = 196
⇒ 14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 26
⇒ m = 27
Multiples of 7 and 3 both, i.e., that of 21 are 21, 42, 63, ....., 189
Tn = 189
⇒ 21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 8
⇒ n = 9
∴ Required number of terms = (27 - 9) = 18
26
Unit's digit in (784)126 + (784)127 is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Unit digit of 42 is 6
So, unit digit in (42)63 is 6
∴ Unit digit of (784)126 = Unit digit in 4126, which is 6
Unit digit in 4127 = Unit digit in (4126 × 4) = Unit digit in (6 × 4), which is 4
∴ Unit digit in (784)127 is 4
Hence, unit digit of [(784)126 + (784)127] = Unit digit of (6 + 4) = Unit digit of 10, which is 0



27
If n is any positive integer, 34n - 43n is always divisible by :
Discuss
Answer & Solution
Answer: Option C
Solution:
Putting n = 1, we get (34n - 43n)
= (34 - 43)
= (81 - 64)
= 17
Which is divisible by 17
28
When 10025 - 25 is written in decimal notation, the sum of its digits is :
Discuss
Answer & Solution
Answer: Option A
Solution:
10025 - 25
= (102)25 - 25
= 1050 - 25
$$ = \underbrace {\,1000\,.....00\,}_{50\,{\text{zeros}}} - 25$$     $$ = \underbrace {\,9999\,.....\,9975\,}_{48\,{\text{times}}}$$
∴ Sum of digits
= (48 × 9) + 7 + 5
= 432 + 7 + 5
= 444
29
Which is the greatest 5-digit number exactly divisible by 279 ?
Discuss
Answer & Solution
Answer: Option C
Solution:
The greatest 5-digit number = 99999
On dividing 99999 by 279, we get 117 as remainder
∴ Required number
= (99999 - 117)
= 99882
30
If p,q,r are all real numbers then (p - q)3 + (q - r)3 + (r - p)3 is equal to :
Discuss
Answer & Solution
Answer: Option B
Solution:
Let a = p - q, b = q - r, c = r - p
∴ a + b + c = p - q + q - r + r - p
⇒ a + b + c = 0
∴ a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 = 3(p - q) (q - r) (r - p)