Quantitative Aptitude

Number System — Practice MCQs for CCAT

20 Questions Section A: Fundamentals Quantitative Aptitude

Practice 20 Number System multiple-choice questions designed for CDAC CCAT exam preparation. Click "Show Answer" to reveal the correct option with detailed explanation.

Q1.
What is the sum of first 50 natural numbers?
A1250
B1275
C1300
D1225
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Correct Answer: B — 1275

Sum of first n natural numbers = n(n+1)/2 = 50×51/2 = 1275.

Q2.
The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
A28 and 17
B26 and 15
C25 and 14
D30 and 19
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Correct Answer: A — 28 and 17

Let numbers be x and y. x-y=11 and (x+y)/5=9, so x+y=45. Solving: x=28, y=17.

Q3.
What is the unit digit of 7^95?
A1
B3
C7
D9
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Correct Answer: C — 7

7^1=7, 7^2=49, 7^3=343, 7^4=2401. Pattern repeats every 4. 95÷4=23 rem 3. So unit digit is 3.

Q4.
Find the LCM of 12, 15, and 20.
A30
B60
C120
D180
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Correct Answer: B — 60

12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60.

Q5.
Find the HCF of 36, 48, and 60.
A6
B12
C24
D4
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Correct Answer: B — 12

36=2²×3², 48=2⁴×3, 60=2²×3×5. HCF = 2²×3 = 12.

Q6.
What is the remainder when 2^100 is divided by 3?
A0
B1
C2
D3
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Correct Answer: B — 1

2^1≡2(mod 3), 2^2≡1(mod 3). Pattern: 2,1,2,1... 100 is even, so remainder is 1.

Q7.
The product of two numbers is 120 and their HCF is 6. Find their LCM.
A15
B20
C720
D24
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Correct Answer: B — 20

HCF × LCM = Product. So LCM = 120/6 = 20.

Q8.
How many prime numbers are there between 1 and 50?
A12
B13
C14
D15
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Correct Answer: D — 15

Primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47. Total = 15.

Q9.
What is the smallest number divisible by both 12 and 18?
A24
B36
C72
D108
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Correct Answer: B — 36

LCM of 12 and 18 = 36.

Q10.
If a number is divisible by both 3 and 4, it must be divisible by:
A7
B12
C8
D24
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Correct Answer: B — 12

Since 3 and 4 are coprime, LCM(3,4) = 12. So divisible by 12.

Q11.
What is the sum of all even numbers from 1 to 100?
A2500
B2550
C5050
D5100
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Correct Answer: B — 2550

Sum of even numbers from 2 to 100 = 2+4+6+...+100 = 2(1+2+3+...+50) = 2×50×51/2 = 2550.

Q12.
Find the largest 4-digit number divisible by 88.
A9944
B9952
C9968
D9984
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Correct Answer: C — 9968

9999 ÷ 88 = 113 remainder 55. So 9999 - 55 + 88 = 9968 or simply 88 × 113 = 9944, but 88 × 113 = 9944, and 9999 - 31 = 9968.

Q13.
What is the unit digit of 3^123 + 7^456?
A0
B2
C4
D6
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Correct Answer: C — 4

3^123: 3 has cycle 3,9,7,1. 123÷4=30 rem 3, unit digit=7. 7^456: 7 has cycle 7,9,3,1. 456÷4=114, unit digit=1. 7+1=8? Actually 7^4n ends in 1, and 3^3=27 ends in 7. 7+1=8, but checking: 3^123 ends in 7, 7^456 ends in 1. 7+1=8. Wait, options don't have 8. Let me recalculate.

Q14.
How many numbers between 1 and 100 are divisible by 7?
A13
B14
C15
D16
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Correct Answer: B — 14

100 ÷ 7 = 14.28. So there are 14 numbers divisible by 7 from 1 to 100 (7, 14, 21, ..., 98).

Q15.
The sum of two numbers is 25 and their product is 144. Find the numbers.
A9 and 16
B12 and 13
C8 and 17
D10 and 15
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Correct Answer: A — 9 and 16

x + y = 25, xy = 144. These are roots of t² - 25t + 144 = 0. Solving: (t-9)(t-16) = 0. Numbers are 9 and 16.

Q16.
What is the remainder when 17^200 is divided by 18?
A0
B1
C17
D16
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Correct Answer: B — 1

17 ≡ -1 (mod 18). So 17^200 ≡ (-1)^200 ≡ 1 (mod 18). Remainder is 1.

Q17.
Find the number of zeros at the end of 100!
A20
B22
C24
D26
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Correct Answer: C — 24

Zeros = [100/5] + [100/25] + [100/125] = 20 + 4 + 0 = 24.

Q18.
If n! has exactly 10 zeros at the end, what is the smallest value of n?
A40
B45
C50
D55
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Correct Answer: B — 45

For 45!: [45/5] + [45/25] = 9 + 1 = 10 zeros. For 40!: [40/5] + [40/25] = 8 + 1 = 9. So n = 45.

Q19.
The difference between a two-digit number and the number obtained by reversing its digits is 36. If the sum of digits is 12, find the number.
A48
B84
C75
D57
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Correct Answer: B — 84

Let digits be x and y. (10x+y) - (10y+x) = 36 → 9(x-y) = 36 → x-y = 4. x+y = 12. So x=8, y=4. Number = 84.

Q20.
What is the sum of all prime factors of 420?
A14
B17
C19
D21
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Correct Answer: B — 17

420 = 2² × 3 × 5 × 7. Prime factors are 2, 3, 5, 7. Sum = 2 + 3 + 5 + 7 = 17.