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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-8 section-4

Each section contains 100 MCQs on Number System .
71
A number is divisible by 11 if the difference between the sums of the digit in odd even places respectively is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Clearly, (D) is true.
72
6 × 3 (3 - 1) is equal to :
Discuss
Answer & Solution
Answer: Option C
Solution:
= 6 × 3 (3 - 1)
= 6 × 3(2)
= 6 × 6
= 36
73
325325 is a six-digit number. It is divisible by :
Discuss
Answer & Solution
Answer: Option D
Solution:
(325 - 325) = 0.
Which is divisible by 7
So, the given number is divisible by 7
(5 + 3 + 2) - (2 + 5 + 3) = 0
So, the given number is divisible by 11
And,
$$\frac{{325325}}{{13}} = 25025$$
So, 325325 is divisible by 13
74
If a and b are two numbers such that ab = 0, then -
Discuss
Answer & Solution
Answer: Option B
Solution:
ab = 0
⇒ a = 0 or b = 0 or both are zero
75
The number of zeros at the end of 60! is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Clearly, highest power of 2 is much higher as compared to that of 5 in 60!,
So, Required number of zeros
= Highest power of 5
= $$ \left[ {\frac{{60}}{5}} \right] + \left[ {\frac{{60}}{{{5^2}}}} \right]$$
= 12 + 2
= 14
76
In a division problem, the divisor is 7 times of quotient and 5 times of remainder. If the dividend is 6 times of remainder, then the quotient is equal to :
Discuss
Answer & Solution
Answer: Option B
Solution:
Divisor :
= 7 × quotient
= 5 × remainder and dividend
= 6 × remainder
Number System mcq solution image
Let remainder be x
Then, divisor = 5x and dividend = 6x
On dividing 6x by 5x, we get 1 as quotient and x as remainder.
∴ Quotient = 1
77
What should be the maximum value of q in the following equation?
5P9 - 7Q2 + 9R6 = 823
Discuss
Answer & Solution
Answer: Option C
Solution:
⇒ 5P9 - 7Q2 + 9R6 = 823
⇒ (500 + 10P + 9) - (700 + 10Q + 2) + (900 + 10R + 6) = 823
⇒ (500 + 900 - 700) + 10 (P + R - Q) + (9 + 6 - 2) = 823
⇒ 700 + 10 (P + R - Q) = 810
⇒ 700 + 10 (P + R - Q) = 700 + 110
⇒ 10 (P + R - Q) = 110
⇒ P + R - Q = 11
⇒ Q = (P + R - 11)
To get maximum value of Q we take P = 9 and R = 9
This gives Q = (9 + 9 - 11) = 7
Hence, the maximum value of Q is 7
78
Let n be a natural number such that $$\frac{1}{2}$$ + $$\frac{1}{3}$$ + $$\frac{1}{7}$$ + $$\frac{1}{n}$$   is also a natural number. Which of the following statements is not true ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & = \left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{7}} \right) + \frac{1}{n} \cr & = \frac{{\left( {21 + 14 + 6} \right)}}{{42}} + \frac{1}{n} \cr & = \left( {\frac{1}{{42}} + \frac{1}{n}} \right) \cr} $$
This sum is a natural number when n = 42
So, each one of the statements that 2 divides ; 3 divides n and 7 divides n is true.
Hence, n > 84 is false
79
If 37 X 3 is a four-digit natural number divisible by 7, then the place marked as X must have the value :
Discuss
Answer & Solution
Answer: Option A
Solution:
37 × 3 is divisible by 7
⇒ (7 × 3 - 3) is either 0 or divisible by 7
⇒ 7 × 0 is divisible by 7
⇒ X = 0 or X = 7
80
If m and n are positive integers, then the digit in the unit's place of 5n + 6m is always :
Discuss
Answer & Solution
Answer: Option A
Solution:
In 5n we have 5 as unit digit and in 6m we have 6 as unit digit.
∴ 5 + 6 = 11(i.e unit digit always be 1)