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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-9 section-6

Each section contains 100 MCQs on Number System .
81
Each member of a club contributes as much rupees and as much paise as the number of members of the club. If the total contribution is Rs. 2525, then the number of members of the club is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the total member be x.
x member contribute is Rs. x and each member also contributes x paise.
So total rupees is
x member Rs. x = Rs. x2
x member Rs. x paise $$ = \frac{{x.x}}{{100}}$$
Total rupees is
$$\eqalign{ & {x^2} + \frac{{{x^2}}}{{100}} = 2525 \cr & 101{x^2} = 2525 \times 100 \cr & 101{x^2} = 101 \times 25 \times 100 \cr & \boxed{x = 50} \cr} $$
82
If X and Y are the two digits of the number 347XY such that the number is completely divisible by 80, then what is the value of X + Y?
Discuss
Answer & Solution
Answer: Option A
Solution:
Given: 347XY where X, Y two digit number is completely divisible by 80.
$$\eqalign{ & \frac{{347XY}}{{80}} = \frac{{347XY}}{{8 \times 10}} \cr & \frac{{7 + X + Y}}{8} \cr} $$
∴ Y must be zero
So, go through option A
Put X = 2, then number 3472 is divisible by 8.
83
When 12, 16, 18, 20 and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousand's place in x?
Discuss
Answer & Solution
Answer: Option D
Solution:
LCM of 12, 16, 18, 20, 25 = 3600
For remainder 4 = 3600k + 4
If divisible by 7 $$ = \frac{{7 \times 514k + 2k + 4}}{7}$$
$$\frac{{2k + 4}}{7},$$  If k = 5 divisible
x = 3600 × 5 + 4
= 18000 + 4
$$ = \mathop {\mathop {18004}\limits_ \downarrow }\limits_{\boxed8} $$
84
The value of (1018)2 - 1019 × 1017 + 1015 × 1012 - 1016 × 1011 is:
Discuss
Answer & Solution
Answer: Option D
Solution:
10182 - 1019 × 1017 + 1015 × 1012 - 1016 × 1011
= 10182 - (1018 + 1) × (1018 - 1) + (1016 - 1) × (1011 + 1) - 1016 × 1011
= 10182 - 10182 + 1 + 1016 × 1011 + 1016 - 1011 - 1 - 1016 × 1011
= 5
85
If the sum of digits of a two digit number is 10 and if the digits of two digits number is reversed, then the number is decreased by 36. Which of the following is correct regarding the number?
I. The difference of the digits is 4.
II. The value of number can be 84.
III. Number is always a composite number.
Discuss
Answer & Solution
Answer: Option D
No explanation is given for this question. Let's Discuss on Board
86
When (7777 + 77) is divided by 78, the remainder is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & \frac{{{{77}^{77}} + 77}}{{78}} \cr & = {\left( { - 1} \right)^{77}} + \left( { - 1} \right) \cr & = - 1 - 1 \cr & = - 2 \cr & {\text{Remainder}} = 78 - 2 = 76 \cr} $$
87
What is least possible number when it is divided by 13 leaves 8 and when divided by 7 leaves remainder 6?
Discuss
Answer & Solution
Answer: Option B
Solution:
In this type of questions go through option, we take option B.
$$\frac{{34}}{{13}}$$ remainder is 8.
$$\frac{{34}}{7}$$ remainder is 6.
It satisfy the condition so answer is option B.
88
If the arithmetic mean of 3a and 4b is greater than 50, and a is twice b, then the smallest possible integer value of a is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{3a + 4b}}{2} > 50 \cr & 3a + 4b > 100 \cr & a = 2b\,\,\left( {{\text{Given}}} \right) \cr & 6b + 4b > 100 \cr & 10b > 100 \cr & b > 10 \cr & {\text{If }}b = 10.1 \cr & \therefore a = 2 \times 10.1 \cr & a = 20.2 \cr & \therefore a = 21\left( {{\text{Approx}}} \right) \cr} $$
89
If x is the remainder when 361284 is divided by 5 and y is the remainder when 496 is divided by 6, then what the value of (2x - y)?
Discuss
Answer & Solution
Answer: Option C
Solution:
x is the remainder when 361284 is divided by 5
So, $$\frac{{{3^{61284}}}}{5} = \frac{{{3^{4 \times 15321}}}}{5} = \frac{{{3^4}}}{5} = \frac{{81}}{5}$$
Remainder = 1
x = 1
y is the remainder when 496 is divided by 6
$$\frac{{{4^{96}}}}{6} = \frac{{{4^{4 \times 24}}}}{6} = \frac{{{4^4}}}{6} = \frac{{256}}{6}$$
Remainder = 4
y = 4
Now, (2x - y)
= 2 - 4
= -2

Alternate solution:
As we know,
Unit digit of 361284
⇒ 34
⇒ 1
So we can say if we divide 361284 by 5 then we get remainder 1
x = 1
As we know,
4 ÷ 6, then remainder 4
42 ÷ 6, then remainder 4
43 ÷ 6, then remainder 4
4n ÷ 6, then remainder 4
When n = natural number
So we can say 496 is divided by 6, then the remainder is 4
y = 4
Now, (2x - y)
⇒ 2 × 1 - 4
⇒ 2 - 4
⇒ -2
90
If (px3 - 8x2 - qx + 1), the expression is completely divisible by the expression (3x2 - 4x + 1). Then what will be the value of p and q respectively?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & 3{x^2} - 4x + 1 = 0 \cr & 3{x^2} - 3x - x + 1 = 0 \cr & \left( {3x - 1} \right)\left( {x - 1} \right) = 0 \cr & 3x - 1 = 0,\,\,x - 1 = 0 \cr & \Rightarrow x = 1,\,\frac{1}{3} \cr & {\text{Put }}x = 1{\text{ in }}p{x^3} - 8{x^2} - qx + 1 \cr & \Rightarrow p - 8 - q + 1 = 0 \cr & \Rightarrow p - q = 7 \cr & {\text{Go through the option C}} \cr & p = \frac{{33}}{4},\,q = \frac{5}{4} \cr & \Rightarrow \frac{{33 - 5}}{4} = 7 \cr & \Rightarrow 7 = 7 \cr} $$