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CSAT · 140 questions

Number system

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Number system questions per year: 2016: 3, 2017: 7, 2018: 5, 2019: 13, 2020: 19, 2021: 10, 2022: 13, 2023: 21, 2024: 15, 2025: 22, 2026: 12 Asked in 11 of 11 years · most in 2025 (22)
Timed · 30 minutes a set Practise Number system as timed tests 7 sets of about 20 questions with UPSC's CSAT marking. Start practising

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” Sub-topics in the detailed syllabus →

CSAT 2020 · Q7

Medium

गुणनफल

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

के अंत में कितने शून्य होंगे?

Read this question in English

How many zeroes are there at the end of the following product?

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

(a) 10 · (b) 12 · (c) 14 · (d) 15

उत्तर और व्याख्या · explanation in English

Answer: (a) 10

Each trailing zero needs one pair of 2 and 5. Writing the product as 5¹² × 12! shows 14 fives but only 10 twos, so the twos run out first and there are 10 zeroes.

  1. The terms after 1 are 5 × 1, 5 × 2, …, 5 × 12, so the product = 5¹² × (1 × 2 × … × 12) = 5¹² × 12!.
  2. Fives: 12 from 5¹², plus 2 in 12! (from 5 and 10) = 14.
  3. Twos: only from 12! — 6 + 3 + 1 = 10 (multiples of 2, 4 and 8 up to 12).
  4. Zeroes = min(10, 14) = 10.
  5. Check: 12! = 479001600 has two 5s and ten 2s, as used.

Remember · Trailing zeroes = the smaller of the counts of 2s and 5s. Do not assume 5s are always scarcer.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q8

Easy

माना कि XYZ तीन अंकों की एक संख्या है, जहाँ (X + Y + Z), 3 का गुणांक नहीं है। तब (XYZ + YZX + ZXY) विभाज्य नहीं है

Read this question in English

Let XYZ be a three-digit number, where (X + Y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by

(a) 3 · (b) 9 · (c) 37 · (d) (X + Y + Z)

उत्तर और व्याख्या · explanation in English

Answer: (b) 9 से

Adding the three rotations puts each digit once in the hundreds, tens and units places, so the sum is 111(X + Y + Z) = 3 × 37 × (X + Y + Z). It is divisible by 3, 37 and (X + Y + Z), but by 9 only if (X + Y + Z) were a multiple of 3 — which it is not.

  1. XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) = 111(X + Y + Z).
  2. 111 = 3 × 37, so the sum = 3 × 37 × (X + Y + Z): divisible by 3, 37 and (X + Y + Z).
  3. For 9, a second factor 3 must come from (X + Y + Z), which is not a multiple of 3.
  4. So the sum is not divisible by 9.
  5. Check: 124 + 241 + 412 = 777 = 111 × 7; 777 ÷ 9 is not whole.

Remember · Cyclic rotations of a 3-digit number always add up to 111 × (digit sum) = 3 × 37 × (digit sum).

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q9

X-Ray · Core #4 Easy

माना कि p, q, r और s इस प्रकार की प्राकृतिक संख्याएँ हैं कि

p − 2016 = q + 2017 = r − 2018 = s + 2019

है। निम्न में से कौन-सी सबसे बड़ी प्राकृतिक संख्या है?

Read this question in English

Let p, q, r and s be natural numbers such that

p − 2016 = q + 2017 = r − 2018 = s + 2019

Which one of the following is the largest natural number?

(a) p · (b) q · (c) r · (d) s

उत्तर और व्याख्या · explanation in English

Answer: (c) r

Call the common value k. Then each number is k plus or minus a constant, and r = k + 2018 carries the largest addition.

  1. Let p − 2016 = q + 2017 = r − 2018 = s + 2019 = k.
  2. Then p = k + 2016, q = k − 2017, r = k + 2018, s = k − 2019.
  3. The largest is r = k + 2018.

Remember · Set equal expressions to one variable k and compare what is added to k in each.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q10

X-Ray · Core #4 Easy

पाँच अंकों की कितनी अभाज्य संख्याएँ अंकों 1, 2, 3, 4 और 5 के द्वारा, बिना अंकों की पुनरावृत्ति किए हुए, बनाई जा सकती हैं?

Read this question in English

How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?

(a) Zero · (b) One · (c) Nine · (d) Ten

उत्तर और व्याख्या · explanation in English

Answer: (a) शून्य

Every such number uses the digits 1 to 5 once, so its digit sum is 15. A digit sum of 15 makes every one of these numbers divisible by 3, so none can be prime.

  1. Digit sum = 1 + 2 + 3 + 4 + 5 = 15, whatever the order.
  2. 15 is divisible by 3, so every such five-digit number is divisible by 3.
  3. A number greater than 3 that is divisible by 3 is not prime, so the count is zero.

Remember · Before listing permutations for primes, check the digit sum: if it is a multiple of 3, no arrangement is prime.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q11

Easy

योग

⊗ + 1⊗ + 5⊗ + ⊗⊗ + ⊗1 = 1⊗⊗

में प्रतीक ⊗ किस अंक को दर्शाता है?

Read this question in English

In the sum

⊗ + 1⊗ + 5⊗ + ⊗⊗ + ⊗1 = 1⊗⊗

for which digit does the symbol ⊗ stand?

(a) 2 · (b) 3 · (c) 4 · (d) 5

उत्तर और व्याख्या · explanation in English

Answer: (b) 3

Write the symbol as a digit x and expand each number by place value. The sum becomes a simple linear equation, 24x + 61 = 100 + 11x, which gives x = 3.

  1. Let ⊗ = x. Then ⊗ = x, 1⊗ = 10 + x, 5⊗ = 50 + x, ⊗⊗ = 11x, ⊗1 = 10x + 1, and 1⊗⊗ = 100 + 11x.
  2. Left side = x + (10 + x) + (50 + x) + 11x + (10x + 1) = 24x + 61.
  3. 24x + 61 = 100 + 11x, so 13x = 39 and x = 3.
  4. Check: 3 + 13 + 53 + 33 + 31 = 133 = 1⊗⊗ with ⊗ = 3.

Remember · In symbol-for-digit sums, expand every number by place value and solve the resulting equation; then check it back.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q12

Medium

यदि आपके पास दो सीधी 7·5 फुट और 3·25 फुट की छड़ें हैं, तो आप कम-से-कम कितनी लम्बाई नाप सकते हैं?

Read this question in English

If you have two straight sticks of length 7.5 feet and 3.25 feet, what is the minimum length can you measure?

(a) 0.05 foot · (b) 0.25 foot · (c) 1 foot · (d) 3.25 feet

उत्तर और व्याख्या · explanation in English

Answer: (b) 0·25 फुट

Laying the sticks end to end, forwards or backwards, measures any whole-number combination of 7·5 and 3·25 feet. The smallest positive length such combinations can give is their HCF, 0·25 foot.

  1. Convert to hundredths of a foot: 7·5 = 750/100 and 3·25 = 325/100.
  2. HCF(750, 325) = 25 (750 = 2 × 3 × 5³ and 325 = 5² × 13), so HCF of the lengths = 0·25 foot.
  3. Any length made by laying the sticks forwards and backwards is a multiple of 0·25 foot, and 0·25 itself can be made.
  4. Check: 7 × 3·25 = 22·75 and 3 × 7·5 = 22·5; 22·75 − 22·5 = 0·25 foot.

Remember · The smallest length measurable with two rods (by adding and subtracting whole lengths) is the HCF of their lengths.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q13

Easy

अनुक्रम 14, 18, 20, 24, 30, 32, ... की प्रत्येक संख्या में एक सरल गणितीय संक्रिया द्वारा अभाज्य संख्याओं का एक अनुक्रम प्राप्त किया जाता है। निम्न में से कौन-सी अनुक्रम की अगली संख्या है?

Read this question in English

A simple mathematical operation in each number of the sequence 14, 18, 20, 24, 30, 32, ... results in a sequence with respect to prime numbers. Which one of the following is the next number in the sequence?

(a) 34 · (b) 36 · (c) 38 · (d) 40

उत्तर और व्याख्या · explanation in English

Answer: (c) 38

Subtracting 1 from each term gives 13, 17, 19, 23, 29, 31 — consecutive prime numbers. The next prime is 37, so the next term is 37 + 1 = 38.

  1. Subtract 1: 14 → 13, 18 → 17, 20 → 19, 24 → 23, 30 → 29, 32 → 31.
  2. 13, 17, 19, 23, 29, 31 are consecutive primes.
  3. The next prime after 31 is 37 (33, 35 are composite).
  4. Next term = 37 + 1 = 38.

Remember · When a series 'relates to primes', try adding or subtracting 1 or 2 to each term and look for consecutive primes.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q14

X-Ray · Core #4 Easy

एक पुस्तिका, जिसके पन्ने सामान्य रूप से हैं, पहले पन्ने की संख्या 1 से प्रारम्भ करते हुए अंकित हैं। इस पुस्तिका से एक पन्ना फाड़ लिया जाता है। बचे हुए पन्नों पर अंकित पृष्ठ संख्याओं का योग 195 है। फटे हुए पन्ने पर निम्न में से कौन-सी संख्याएँ हैं?

Read this question in English

One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers?

(a) 5, 6 · (b) 7, 8 · (c) 9, 10 · (d) 11, 12

उत्तर और व्याख्या · explanation in English

Answer: (b) 7, 8

The full booklet's page total must be just above 195. With 20 pages the total is 210, so the torn leaf carries numbers adding to 15 — pages 7 and 8.

  1. Sum of pages 1 to n is n(n + 1)/2. For n = 19 it is 190 (too small); for n = 20 it is 210.
  2. Missing sum = 210 − 195 = 15.
  3. A leaf carries two consecutive numbers, odd then even: 7 + 8 = 15.
  4. Check: n = 21 gives 231 − 195 = 36, which cannot be an odd + next-even pair (such pairs add to an odd number).

Remember · Missing leaf: find the first 1 + 2 + … + n above the given total; the gap is the two page numbers' sum.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q16

X-Ray · Core #4 Hard

माना कि A3BC और DE2F चार अंकों की संख्याएँ हैं, जहाँ प्रत्येक वर्ण 3 से बड़े भिन्न अंक को दर्शाता है। यदि संख्याओं का योग 15902 है, तो A और D के मानों के बीच अंतर क्या है?

Read this question in English

Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D?

(a) 1 · (b) 2 · (c) 3 · (d) 4

उत्तर और व्याख्या · explanation in English

Answer: (c) 3

Working column by column from the units, the carries force B = 7 and E = 5, and C and F must be 4 and 8. That leaves 6 and 9 for A and D, whose difference is 3.

  1. Units: C + F ends in 2; both are at least 4, so C + F = 12, carry 1.
  2. Tens: B + 2 + 1 ends in 0, so B + 3 = 10 and B = 7, carry 1.
  3. Hundreds: 3 + E + 1 ends in 9, so E = 5, carry 0.
  4. Thousands: A + D = 15.
  5. Digits left from 4–9 after B = 7, E = 5: {4, 6, 8, 9}. C + F = 12 needs {4, 8}; so A and D are 6 and 9 (6 + 9 = 15).
  6. Difference = 9 − 6 = 3.
  7. Check: 6374 + 9528 = 15902.

Remember · In letter-addition puzzles, start from the units column, track every carry, and use 'different digits' to fix the pairs.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q37

X-Ray · Core #4 Medium

1 और 100 के मध्य कितनी ऐसी पूर्ण संख्याएँ हैं जिनमें एक अंक 4 है लेकिन वे 4 से विभाज्य नहीं हैं?

Read this question in English

How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?

(a) 5 · (b) 11 · (c) 12 · (d) 13

उत्तर और व्याख्या · explanation in English

Answer: (c) 12

There are 19 numbers below 100 containing the digit 4: 4, 14, 24, 34, 40 to 49, and 54, 64, 74, 84, 94. Seven of them are multiples of 4, which leaves 12.

  1. Units digit 4: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94 = 10 numbers.
  2. Tens digit 4 (other than 44): 40, 41, 42, 43, 45, 46, 47, 48, 49 = 9 numbers. Total = 19.
  3. Multiples of 4 among them: 4, 24, 40, 44, 48, 64, 84 = 7.
  4. Required count = 19 − 7 = 12.

Remember · Count the whole set first, then subtract the ones to exclude — fewer chances to miss a number.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q50

Easy

निम्न में कौन-सी सबसे बड़ी संख्या है?

Read this question in English

What is the largest number among the following?

(a) (1/2)⁻⁶ · (b) (1/4)⁻³ · (c) (1/3)⁻⁴ · (d) (1/6)⁻²

उत्तर और व्याख्या · explanation in English

Answer: (c) (1/3)⁻⁴

A negative power of a fraction flips it: (1/a)⁻ⁿ = aⁿ. The four values are 2⁶ = 64, 4³ = 64, 3⁴ = 81 and 6² = 36, so (1/3)⁻⁴ is the largest.

  1. (1/2)⁻⁶ = 2⁶ = 64.
  2. (1/4)⁻³ = 4³ = 64.
  3. (1/3)⁻⁴ = 3⁴ = 81.
  4. (1/6)⁻² = 6² = 36.
  5. The largest is 81, i.e. (1/3)⁻⁴.

Remember · (1/a) raised to −n equals aⁿ; rewrite every option this way before comparing.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q51

Medium

अधिकतम लम्बाई x क्या होगी, जिससे कि 3½ m और 8¾ m, x के पूर्णांक गुणज हों?

Read this question in English

What is the greatest length x such that 3½ m and 8¾ m are integral multiples of x?

(a) 1½ m · (b) 1⅓ m · (c) 1¼ m · (d) 1¾ m

उत्तर और व्याख्या · explanation in English

Answer: (d) 1¾ m

The greatest length that divides both is their HCF. In quarters of a metre the lengths are 14/4 and 35/4, and HCF(14, 35) = 7, so x = 7/4 = 1¾ m.

  1. 3½ = 14/4 m and 8¾ = 35/4 m.
  2. HCF(14, 35) = 7, so the greatest common length = 7/4 m = 1¾ m.
  3. Check: 3½ ÷ 1¾ = 2 and 8¾ ÷ 1¾ = 5, both whole numbers.

Remember · HCF of fractions: put them over a common denominator, take the HCF of the numerators, keep the denominator.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q53

Easy

आवर्त दशमलव निरूपण 1·272727... किसके समान है?

Read this question in English

The recurring decimal representation 1.272727... is equivalent to

(a) 13/11 · (b) 14/11 · (c) 127/99 · (d) 137/99

उत्तर और व्याख्या · explanation in English

Answer: (b) 14/11

The repeating block 27 has two digits, so 0·2727... = 27/99 = 3/11. Adding the whole part gives 1 + 3/11 = 14/11.

  1. Let x = 1·2727...; then 100x = 127·2727...
  2. 100x − x = 126, so x = 126/99 = 14/11.
  3. Check: 14 ÷ 11 = 1·2727...

Remember · A pure recurring decimal 0·(ab) repeating = ab/99; handle the whole-number part separately.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q54

Easy

चार अंकों की सबसे छोटी संख्या कौन-सी है जिसे 3, 4, 5 और 6 से विभाजित करने पर प्रत्येक बार 2 शेष बचता है?

Read this question in English

What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

(a) 1012 · (b) 1022 · (c) 1122 · (d) 1222

उत्तर और व्याख्या · explanation in English

Answer: (b) 1022

A number leaving remainder 2 with each divisor is 2 more than a common multiple of 3, 4, 5 and 6. Their LCM is 60, the first four-digit multiple of 60 is 1020, and so the answer is 1022.

  1. LCM(3, 4, 5, 6) = 60.
  2. Smallest four-digit multiple of 60: 60 × 17 = 1020 (60 × 16 = 960 has three digits).
  3. Required number = 1020 + 2 = 1022.
  4. Check: 1022 = 3 × 340 + 2 = 4 × 255 + 2 = 5 × 204 + 2 = 6 × 170 + 2.

Remember · Same remainder r for every divisor: the number = k × LCM + r. Find the smallest k that gives the required size.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q56

Easy

जब 51 × 27 × 35 × 62 × 75 को 100 से विभाजित किया जाता है, तो शेषफल क्या है?

Read this question in English

What is the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100?

(a) 50 · (b) 25 · (c) 5 · (d) 1

उत्तर और व्याख्या · explanation in English

Answer: (a) 50

Only the last two digits matter. 62 × 75 = 4650 ends in 50, and multiplying a number ending in 50 by any odd number again ends in 50. The other factors 51, 27 and 35 are all odd, so the remainder is 50.

  1. 62 × 75 = 4650, which ends in 50.
  2. 50 × (odd number) always ends in 50, since 50 × odd = 100k + 50.
  3. 51 × 27 × 35 = 48195 is odd, so the whole product ends in 50.
  4. Remainder = 50.

Remember · Remainder on division by 100 = last two digits; pair factors that quickly give a round ending like 50 or 00.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q58

X-Ray · Core #4 Medium

n के किस मान के लिए संख्या (10ⁿ + 1) में अंकों का योग 2 है?

Read this question in English

For what value of n, the sum of digits in the number (10ⁿ + 1) is 2?

(a) For n = 0 only · (b) For any whole number n · (c) For any positive integer n only · (d) For any real number n

उत्तर और व्याख्या · explanation in English

Answer: (b) किसी पूर्ण संख्या n के लिए

For any positive integer n, 10ⁿ + 1 is 1 followed by zeros and a final 1, so its digit sum is 2. For n = 0 it is 1 + 1 = 2, whose digit sum is also 2. So it holds for every whole number 0, 1, 2, …, but not for fractional n, where 10ⁿ + 1 is not a whole number.

  1. n = 1, 2, 3 …: 11, 101, 1001 … — digit sum 2.
  2. n = 0: 10⁰ + 1 = 2 — digit sum 2.
  3. n = 1/2: 10ⁿ + 1 = √10 + 1 ≈ 4·16, not a whole number, so 'any real number' fails.
  4. So the statement holds for any whole number n.

Remember · Always test the edge value n = 0 before choosing 'positive integers only'.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q75

X-Ray · Core #4 Medium

प्राकृतिक संख्याओं के ऐसे कितने युग्म हैं, जिनके वर्गों का अंतर 63 है?

Read this question in English

How many pairs of natural numbers are there such that the difference of whose squares is 63?

(a) 3 · (b) 4 · (c) 5 · (d) 2

उत्तर और व्याख्या · explanation in English

Answer: (a) 3

x² − y² = (x − y)(x + y) = 63, so each way of writing 63 as a product of two factors gives one pair. 63 = 1 × 63 = 3 × 21 = 7 × 9, giving (32, 31), (12, 9) and (8, 1).

  1. x² − y² = (x − y)(x + y) = 63, with x − y < x + y.
  2. Factor pairs of 63: (1, 63), (3, 21), (7, 9) — all odd, so x and y come out whole.
  3. (1, 63): x = 32, y = 31. (3, 21): x = 12, y = 9. (7, 9): x = 8, y = 1.
  4. 3 pairs.
  5. Check: 32² − 31² = 63, 144 − 81 = 63, 64 − 1 = 63.

Remember · Difference of squares: factor the number as (x − y)(x + y) with both factors of the same parity; each factor pair gives one solution.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q76

Medium

यदि भिन्नों 2/3, 3/4, 4/5 और 5/6 के अंश और हर दोनों में 5 जोड़ दिया जाता है, तो निम्नलिखित में से किस एक के मान में न्यूनतम परिवर्तन होगा?

Read this question in English

Which one of the following will have minimum change in its value if 5 is added to both numerator and the denominator of the fractions 2/3, 3/4, 4/5 and 5/6?

(a) 2/3 · (b) 3/4 · (c) 4/5 · (d) 5/6

उत्तर और व्याख्या · explanation in English

Answer: (d) 5/6

Each fraction has the form n/(n + 1). Adding 5 to both parts changes it by 5/[(n + 1)(n + 6)], which shrinks as n grows. So 5/6, the largest, changes least (by 5/66).

  1. 2/3 → 7/8: change 7/8 − 2/3 = 5/24 ≈ 0·208.
  2. 3/4 → 8/9: change 5/36 ≈ 0·139.
  3. 4/5 → 9/10: change 5/50 = 0·1.
  4. 5/6 → 10/11: change 5/66 ≈ 0·076 — the smallest.

Remember · Adding the same number to top and bottom pushes a proper fraction towards 1; fractions already closer to 1 move less.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q77

Easy

एक अंक n > 3 भाज्य है 3 से लेकिन 6 से भाज्य नहीं है। निम्नलिखित में से कौन-सा एक 4 से भाज्य है?

Read this question in English

A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?

(a) 2n · (b) 3n · (c) 2n + 4 · (d) 3n + 1

उत्तर और व्याख्या · explanation in English

Answer: (d) 3n+1

The digits greater than 3 that are divisible by 3 are 6 and 9; 6 is divisible by 6, so n = 9. Then 3n + 1 = 28 is the only option divisible by 4.

  1. Digits above 3 divisible by 3: 6 and 9. 6 is divisible by 6, so n = 9.
  2. 2n = 18, 3n = 27, 2n + 4 = 22 — none divisible by 4.
  3. 3n + 1 = 28 = 4 × 7 — divisible by 4.

Remember · When a condition pins a single digit, find that digit first and then test every option numerically.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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1 of its 5 sub-topics is Core

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