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Number System & Basic Numeracy — UPSC CSAT Paper II Quantitative Aptitude previous year questions
20 free solved previous year questions on Number System & Basic Numeracy, papers from 2019 to 2026, drawn from UPSC CSAT, UPSC CDS, covering questions of the kind UPSC asks in CSAT Paper II. Every answer key has been checked again against published sources and solved independently, and each explanation is written fresh.
Read all 20 questions with answers
- #1UPSC CSAT 2026X, Y and Z jump forward 4′, 6′ and 5′ respectively. At 8 AM, they all land on mark 199′. How many times will they all land on the same mark (need not be at the same moment) between mark 195′ and 1000′, if all of them cross mark 1000′ by 9 AM?
- A11
- B12
- C13
- D14
Answer: (D) 14Explanation
They all stand on 199′ at the start.
They land together again after every LCM(4, 6, 5) = 60 feet.
Common marks are 199 + 60k.
We need 199 + 60k ≤ 1000, so 60k ≤ 801 and k ≤ 13.35.
So k = 0, 1, 2, ..., 13. That is 14 marks, all above 195′.
The 9 AM condition only says they all travel beyond 1000′, so it adds nothing.
Answer: 14, option (d).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #2UPSC CSAT 2026If 10ᵐ × 1000 × n = 75²⁵ × 25³² × 32⁷⁵, where n is not divisible by 10, then the value of m is
- A101
- B111
- C121
- D131
Answer: (B) 111Explanation
Break each number into primes.
75²⁵ = 3²⁵ × 5⁵⁰.
25³² = 5⁶⁴.
32⁷⁵ = 2³⁷⁵.
Product = 2³⁷⁵ × 3²⁵ × 5¹¹⁴.
A factor of 10 needs one 2 and one 5. Only 114 pairs are possible, because there are 114 fives.
So the power of 10 is 10¹¹⁴. Since n has no factor 10, 10ᵐ × 1000 = 10ᵐ⁺³ = 10¹¹⁴.
m + 3 = 114, so m = 111, option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #3UPSC CSAT 2026What is the minimum number of times one needs to measure to get 298 litres of water from a tank, if the measuring cylinders have capacities 1 litre, 6 litres, 25 litres and 100 litres?
- A4
- B5
- C9
- D13
Answer: (B) 5Explanation
Each fill of a cylinder counts as one measurement. Excess water may be poured back into the tank, also one measurement.
298 = 300 – 2.
Fill the 100 L cylinder 3 times: 3 measurements, 300 L.
Pour back 2 L using the 1 L cylinder twice: 2 measurements.
Total = 5 measurements.
No way of 4 or fewer works. With at most 4 fills or pour-backs of 1, 6, 25 or 100 litres, 298 cannot be reached. (Using additions only, the best is 12.)
Answer: 5, option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #4UPSC CSAT 2026Three variables x, y and z take values 2, 3, 4 or 5 such that their values are always distinct. If M and N denote the largest possible value and the smallest possible value, respectively, for the expression { (x × y) + z }; then M – N is
- A11
- B12
- C13
- D14
Answer: (C) 13Explanation
We pick three different numbers from 2, 3, 4, 5 for x, y, z. The expression is x × y + z.
Largest value: multiply the two biggest, add the next: 5 × 4 + 3 = 23. (Check 5 × 3 + 4 = 19, which is smaller.)
Smallest value: multiply the two smallest, add the next smallest left over: 2 × 3 + 4 = 10. (Check 2 × 4 + 3 = 11, bigger.)
M = 23 and N = 10.
M – N = 13, option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #5UPSC CSAT 2026The digit in the unit place of the number 6¹²⁹ × 7³⁰⁷ is
- A2
- B4
- C8
- D6
Answer: (C) 8Explanation
Any power of 6 ends in 6. So 6¹²⁹ ends in 6.
Powers of 7 end in 7, 9, 3, 1 and repeat every 4.
307 ÷ 4 leaves remainder 3, so 7³⁰⁷ ends in the third digit, 3.
Unit digit of the product = unit digit of 6 × 3 = 18, which is 8.
Answer: 8, option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #6UPSC CSAT 2026How many three-digit numbers can be expressed as an integral power of 2?
- A1
- B2
- C3
- D4
Answer: (C) 3Explanation
Powers of 2 are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, ...
Three-digit numbers lie from 100 to 999.
2⁶ = 64 is too small. 2¹⁰ = 1024 is too large.
The ones in range are 2⁷ = 128, 2⁸ = 256 and 2⁹ = 512.
Count = 3.
Answer: option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #7UPSC CSAT 2026How many times does 5 appear in all two-digit positive integers?
- A18
- B19
- C20
- D21
Answer: (B) 19Explanation
Count the digit 5 by place.
Tens place: the numbers 50 to 59 have a 5 in the tens place. That is 10 fives.
Units place: 15, 25, 35, 45, 55, 65, 75, 85, 95. That is 9 fives.
The number 55 has two fives and is counted once in each list, which is right.
Total = 10 + 9 = 19.
Answer: option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #8UPSC CSAT 2026If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct?I. The difference of the numbers is an even number.II. One of the numbers is a perfect square.Select the answer using the code given below.
- AI only
- BII only
- CBoth I and II
- DNeither I nor II
Answer: (C) Both I and IIExplanation
Let the two numbers be a and b. HCF × LCM = a × b.
Given a × b = a³ (cube of one number). So b = a².
The numbers are a and a², with a greater than 1 (they are distinct).
I: a² – a = a(a – 1). One of a and a – 1 is even, so the difference is even. True.
II: a² is a perfect square. True.
Both statements are correct. Answer: option (c).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #9UPSC CSAT 2026If x and y are two digits and the number 4x5y790 is divisible by 11, then what is the remainder, if x+y is divided by 11?
- A1
- B3
- C5
- D7
Answer: (D) 7Explanation
Test for 11: (sum of digits at odd places) – (sum at even places) must be 0 or a multiple of 11.
Digits of 4x5y790 from the left: 4, x, 5, y, 7, 9, 0.
Odd places: 4 + 5 + 7 + 0 = 16. Even places: x + y + 9.
Difference = 16 – 9 – (x + y) = 7 – (x + y).
This must be a multiple of 11, so x + y = 7 or 18 (as x + y is at most 18).
7 ÷ 11 leaves 7. 18 ÷ 11 leaves 7.
Remainder = 7, option (d).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #10UPSC CSAT 2026A is a 2-digit number with different digits. B is also a 2-digit number and is obtained by reversing the digits of A. If A – B is a multiple of 27, where A > B, how many such different A’s are possible?
- A6
- B9
- C12
- D18
Answer: (B) 9Explanation
Let A = 10a + b with a > b. Then B = 10b + a.
A – B = 9(a – b).
For this to be a multiple of 27, (a – b) must be a multiple of 3.
B is a 2-digit number, so b cannot be 0.
a – b = 3: (4,1), (5,2), (6,3), (7,4), (8,5), (9,6). That is 6 values.
a – b = 6: (7,1), (8,2), (9,3). That is 3 values.
a – b = 9 needs b = 0, which is not allowed.
Total = 6 + 3 = 9, option (b).Source: UPSC CSAT 2026 Paper II (Set A); answer key checked by independent computation and against UPSC's provisional key
- #11UPSC CSAT 2024Consider the following statements in respect of the sum S=x+y+z, where x, y and z are distinct prime numbers each less than 10:1. The unit digit of S can be 0.2. The unit digit of S can be 9.3. The unit digit of S can be 5.Which of the statements given above are correct?
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- D1, 2 and 3
Answer: (C) 1 and 3 onlyExplanation
Primes below 10 are 2, 3, 5, 7; taking three at a time gives sums 10, 12, 14, 15.
Unit digits possible: 0, 2, 4, 5.
So statements 1 and 3 hold; 2 does not. Answer: option (c).Source: UPSC CSAT 2024 question; answer key checked by independent computation
- #12UPSC CSAT 2024Let p, q, r and s be distinct positive integers. Let p, q be odd and r, s be even. Consider the following statements:1. (p - r)^2 (qs) is even.2. (q - s) q^2 s is even.3. (q + r)^2 (p + s) is odd.Which of the statements given above are correct?
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- D1, 2 and 3
Answer: (D) 1, 2 and 3Explanation
Statements 1 and 2 contain the even factor s, so each product is even.
In 3, q + r is odd and p + s is odd, so the product of odd numbers is odd.
All three are correct. Answer: option (d).Source: UPSC CSAT 2024 question; answer key checked by independent computation
- #13UPSC CSAT 2023Consider the following in respect of prime number p and composite number c.1. (p+c)/(p-c) can be even2. 2p + c can be odd3. pc can be odd.Which of the statements given above are correct?
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- D1, 2 and 3
Answer: (D) 1, 2 and 3Explanation
Each statement only says 'can be', so one example is enough.
1: p = 2, c = 6 gives 8/(-4) = -2, which is even.
2: p = 3, c = 15 gives 21, which is odd.
3: p = 3, c = 15 gives 45, which is odd.
All three hold.
Answer: option (d).Source: UPSC CSAT 2023 question; answer key checked by independent computation
- #14UPSC CSAT 2022Let A, B and C represent distinct non-zero digits. Suppose x is the sum of all possible 3-digit numbers formed by A, B and C without repetition.Consider the following statements:1. The 4-digit least value of x is 1332.2. The 3-digit greatest value of x is 888.Which of the above statements is/are correct?
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (A) 1 onlyExplanation
Each digit appears twice in each place, so x = 222(A+B+C).
The smallest sum of distinct non-zero digits is 6, so the least x is 222 x 6 = 1332, a 4-digit number. Statement 1 is true.
A 3-digit x would need A+B+C <= 4, which is impossible; 888 needs sum 4. Statement 2 is false.
Answer: option (a).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #15UPSC CSAT 2022Consider the following statements in respect of two natural numbers p and q such that p is a prime number and q is a composite number:1. p x q can be an odd number.2. q/p can be a prime number.3. p + q can be a prime number.Which of the above statements are correct?
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- D1, 2 and 3
Answer: (D) 1, 2 and 3Explanation
1: p = 3, q = 9 gives 27, odd.
2: p = 5, q = 10 gives q/p = 2, a prime.
3: p = 5, q = 6 gives 11, a prime.
All three are possible.
Answer: option (d).Source: UPSC CSAT 2022 question; answer key checked by independent computation
- #16UPSC CSAT 2021Consider the following statements:1. The sum of 5 consecutive integers can be 100.2. The product of three consecutive natural numbers can be equal to their sum.Which of the above statements is/are correct?
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (C) Both 1 and 2Explanation
Five consecutive integers n..n+4 sum to 5n + 10 = 100, giving n = 18: 18+19+20+21+22 = 100. Statement 1 is true.
For 1, 2, 3: product 6 = sum 6. Statement 2 is true.
Answer: option (c).Source: UPSC CSAT 2021 question; answer key checked by independent computation
- #17UPSC CDS 2023Consider the following statements in respect of all factors of 360:1. The number of factors is 24.2. The sum of all factors is 1170.Which of the above statements is/are correct?
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (C) Both 1 and 2Explanation
360 = 2^3 x 3^2 x 5.
Number of factors = (3+1)(2+1)(1+1) = 24, so statement 1 is true.
Sum of factors = (1+2+4+8)(1+3+9)(1+5) = 15 x 13 x 6 = 1170, so statement 2 is true.
Answer: option (c).Source: UPSC CDS 2023 question; answer key checked by independent computation
- #18UPSC CDS 2022Consider the following statements:1. The sum of the cubes of three consecutive natural numbers is divisible by 9.2. Every even power of every odd number (> 1) when divided by 8 gives 1 as remainder.Which of the above statements is/are correct?
- A1 only
- B2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Answer: (C) Both 1 and 2Explanation
For statement 1: the sum expands to 3a^3 + 9a^2 + 15a + 9 = 9(a^2 + 2a + 1) + 3a(a - 1)(a + 1), and a(a-1)(a+1) is a multiple of 6, so the total is a multiple of 9.
For statement 2: an odd number squared is 1 mod 8, so any even power is also 1 mod 8.
Both statements hold.
Answer: option (c).Source: UPSC CDS 2022 question; answer key checked by independent computation
- #19UPSC CDS 2020If A_n = P_n + 1, where P_n is the product of the first n prime numbers, then consider the following statements:1. A_n is always a composite number.2. A_n + 2 is always an odd number.3. A_n + 1 is always an even number.Which of the above statements is/are correct?
- A1 only
- B2 only
- C3 only
- D2 and 3 only
Answer: (D) 2 and 3 onlyExplanation
Statement 1 fails for n = 1: P = 2 and A = 3, which is prime.
P_n contains 2, so P_n is even and A_n = P_n + 1 is odd. Hence A_n + 2 is odd (true) and A_n + 1 is even (true).
Answer: option (d).Source: UPSC CDS 2020 question; answer key checked by independent computation
- #20UPSC CDS 2019Consider the following statements:1. Unit digit in 17^174 is 7.2. Difference of the squares of any two odd numbers is always divisible by 8.3. Adding 1 to the product of two consecutive odd numbers makes it a perfect square.Which of the above statements are correct?
- A1, 2 and 3
- B1 and 2 only
- C2 and 3 only
- D1 and 3 only
Answer: (C) 2 and 3 onlyExplanation
Statement 1: unit digit of 7^n cycles 7, 9, 3, 1; 174 leaves remainder 2 on division by 4, so the digit is 9, not 7. False.
Statement 2: (2a+1)^2 - (2b+1)^2 = 4(a-b)(a+b+1), and (a-b)(a+b+1) is always even, so it is a multiple of 8. True.
Statement 3: (2n+1)(2n+3) + 1 = (2n+2)^2. True.
Answer: option (c).Source: UPSC CDS 2019 question; answer key checked by independent computation
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