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Number System - 3 - Bank Exams Free MCQ Test with solutions


MCQ Practice Test & Solutions: MCQ: Number System - 3 (15 Questions)

You can prepare effectively for Bank Exams with this dedicated MCQ Practice Test (available with solutions) on the important topic of "MCQ: Number System - 3". These 15 questions have been designed by the experts with the latest curriculum of Bank Exams 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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MCQ: Number System - 3 - Question 1
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When a certain number is multiplied by 13, the product consists entirely of fives. The smallest such number is ?

Detailed Solution: Question 1

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
∴ Req. smallest number =42735.

MCQ: Number System - 3 - Question 2
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Every prime number of the form 3k + 1 can be represented in the form 6m + 1 (where k, m are integers), When

Detailed Solution: Question 2

The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

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MCQ: Number System - 3 - Question 3
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If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, then the resulting fraction is equal to the 1/2. If how ever the numerator is increased by 1, then denominator is decreased by 2, then the resulting fraction is equal to 3/5. Find the original fraction?

Detailed Solution: Question 3

Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
⇒ 2x + 4 = y+1
⇒ 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
⇒ 5x +5 = 3y - 6
∴ 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
⇒ 2 x 2 -y = -3
⇒ -y = -3 - 4
∴ y = 7
∴ Original fraction = x/y =2/7

MCQ: Number System - 3 - Question 4
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Out of three given numbers, the first number is twice the second and thrice the third. If the average of these three numbers is 154, then what is the difference between the first and the third numbers?

Detailed Solution: Question 4

Let the third numbers = p
Then, the first number = 3p
and second number = 3p/2
According to question,
( p + 3p + 3p/2 )/3 = 154
⇒ ( 2p + 6p + 3p )/6 = 154
∴ p = 154 x 6/11 = 84
∴ Required difference = 3p - p = 2p
= 2 x 84 = 168

MCQ: Number System - 3 - Question 5
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On children's Day, sweets were to be equally distributed amongst 300 children. But on that particular day 50 children remained absent; hence each child got one extra sweet. How many sweet were distributed?

Detailed Solution: Question 5

  • Let each student get 'y' sweets
  • As there are 300 students, total sweets = 300 x y
  • Now, 50 students are absent
  • Present students = 300 - 50 = 250
  • each students is getting 1 extra sweet because of the absentees
  • total sweets = 250 x (y +1)
  • Now, 300y = 250 (y + 1)
  • 300y = 250 y + 250
  • 300y - 250y = 250
  • 50 x = 250
  • x = 5
  • number of sweets distributed = 250(5 + 1) = 250 x 6 = 1500
  • Hence, option C is correct. 

MCQ: Number System - 3 - Question 6
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There are two examination halls P and Q. If 10 students shifted P to Q, then the number of students will be equal in both the examination halls. If 20 students shifted from Q to P, then the students of P would be doubled to the students of Q. The number of students would be in P and Q respectively are ?

Detailed Solution: Question 6

Let number of students in examination halls P and Q is x and y, respectively
Then as per the first condition,
x - 10 = y + 10
⇒ x - y = 20 ..............(i)
As per the second condition,
⇒ x + 20 = 2( y - 20)
⇒ x - 2y = - 60 ...............(ii)
On subtraction Eq. (ii) from Eq. (i), we get
- y + 2y = 20 + 60
⇒ y = 80
Putting the value of y Eq. (i) we get,
x - 80= 20 ⇒ x = 100
Hence, number of students in examination halls P and Q is 100 and 80, respectively.

MCQ: Number System - 3 - Question 7
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There are 200 question in a 3 hour examination. Among 200 question, 50 are Mathematics, 100 are from GK and 50 are from Sciences. Ram spent twice as much time on each Mathematics question for each other question. How many minute did he spend on Mathematics questions?

Detailed Solution: Question 7

Let Ram spends p minute on each Mathematics question.
According to the question.
50 x p + 100 x p/2 + 50 x p/2 = 3 x 60
p(50+ 50 + 25) = 180
⇒ p = 180/125
∴ Required time = 50 x 180/125
∴ Required time = 2 x 180/5
∴ Required time = 2 x 36 = 72 min

MCQ: Number System - 3 - Question 8
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In an examination paper of five questions 5% of the candidates answered all of them and 5% answered none. Of the rest 25% candidates answered only one question and 20% answered 4 question. If 396 candidates answered either 2 question or 3 question. The number of candidates that appeared for the examination was

Detailed Solution: Question 8

Let the total number of candidates be x.

5% answered all 5 questions, 5% answered none, 25% answered only 1 question, and 20% answered 4 questions.

Total of these percentages = 5% + 5% + 25% + 20% = 55%.

Therefore, the remaining percentage who answered either 2 or 3 questions = 100% − 55% = 45%.

45% of x = 396

(45/100)x = 396

0.45x = 396

x = 396 / 0.45

x = 880

The number of candidates who appeared for the examination was 880.

MCQ: Number System - 3 - Question 9
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In a two digit positive number the unit digit is equal to the square of ten's place digit. The difference between the original number and the number formed by interchanging the digit is 54. What of 40% of the original number?

Detailed Solution: Question 9

Let ten's place digit be a and unit's place digit be a2
Original number = 10 x a + 1 x a2 = 10a + a2
The number formed by interchanging the digits,
New number = 10 x a 2 + 1 x a = 10a 2 + a
According to the question
(10a2 + a) - ( 10a + a2) = 54
⇒ 10a2 + a - 10a - a 2 = 54
⇒ 9a2 - 9a = 54
⇒ 9( a2 - a) = 54
⇒ ( a2 - a) = 54/9
⇒ ( a2 - a) = 6
⇒ a2 - a - 6 = 0
⇒ a2 - 3a + 2a - 6 = 0
⇒ a (a - 3) + 2 (a - 3) = 0
∴ (a - 3) (a + 2) = 0
∴ a = 3, - 2
∴ Ten,s digit = a = 3
Unit's digit = a2 = 32 = 9
Original number = 39
∴ Required number = 39 x 40/100 = 15.6

MCQ: Number System - 3 - Question 10
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When 1/7 of a number is subtracted from the number itself, it gives the same value as the sum of all the angels of a triangle. What is the number ?

Detailed Solution: Question 10

Let the number be P.
According to the question,
P - P/7 = 180
⇒ 6P/7 = 180
∴ P = 180 x 7/6 = 210

MCQ: Number System - 3 - Question 11
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The product of two consecutive odd number is 6723. What is the greater number ?

Detailed Solution: Question 11

Let two consecutive odd number be (A + 1) and (A + 3).
According to the question.
(A + 1) (A + 3) = 6723
⇒ A2 + 3A + A + 3 = 6723
⇒ A2 + 4A + 3 - 6723 = 0
⇒ A2 + 4A - 6720 = 0
⇒ A2 + 84A - 80A - 6720 = 0
⇒ A(A + 84) - 80 (A + 84) = 0
⇒ (A - 80) (A + 84) = 0
∴ A = 80, (A = ≠ - 84)
Hence, the greater number = 80 + 3 = 83

MCQ: Number System - 3 - Question 12
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The sum of the digits of a two-digit number is 14 and the difference between the two digits of the number is 2. What is the product of the two digits of the two-digit number ?

Detailed Solution: Question 12

Let be the ten's digit be P and unit's digit be Q.
The two-digit number = 10P + Q
(where, P > Q)
According to the question,
         P + Q = 14 ......(i)
and   P - Q = 2 .......(ii)
solving Eqs. (i) and (ii), we get
P = 8 and Q = 6
∴ Required product = 8 x 6 = 48

MCQ: Number System - 3 - Question 13
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Find the greatest among the following given numbers.
(3)1/3, (2)1/2, 1, (6)1/6

Detailed Solution: Question 13

LCM of 3, 2 and 6 = 6
Now we can write as below
∴ (3)1/3 = (32)1/6 = (9)1/6
21/2 = (23)1/6 = (8)1/6
1 = (1)1/6
(6)1/6 = (6)1/6

Clearly we can see that power 1/6 is same for all the base numbers. So the number which has greatest base with be greatest among these all number.
∴ (9)1/6 = (3)1/3 is the greatest number.

MCQ: Number System - 3 - Question 14
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The sum of all natural numbers between 100 and 200, which are multiples of 3 is :

Detailed Solution: Question 14

  1. Find the first multiple of 3 ≥ 100:
    – 100 ÷ 3 ≈ 33.33 ⇒ next integer is 34 ⇒ first term = 34×3 = 102

  2. Find the last multiple of 3 ≤ 200:
    – 200 ÷ 3 ≈ 66.66 ⇒ integer part is 66 ⇒ last term = 66×3 = 198

  3. Number of terms n = (last term – first term)÷3 + 1
    = (198 – 102)÷3 + 1
    = 96÷3 + 1
    = 32 + 1
    = 33

  4. Sum of an AP = n × (first term + last term) ÷ 2
    = 33 × (102 + 198) ÷ 2
    = 33 × 300 ÷ 2
    = 33 × 150
    = 4950

MCQ: Number System - 3 - Question 15
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In a two–digit number, the digit at the unit’s place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is

Detailed Solution: Question 15

Let us assume Ten’s digit = P
Unit’s digit = 2P – 1
∴  Original number = 10P + (2P – 1) = 12P – 1
New number = 10 (2P – 1) + P
New number = 20P – 10 + P = 21P – 10
According to given question,
∴  (21P – 10) – (12P – 1) = 12P – 1 – 20
⇒  9P – 9 = 12P – 21
⇒  3P = 12
⇒ P = 4
⇒  Original number = 12P – 1 = 12 × 4 – 1 = 47

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