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CSAT ┬╖ 132 questions

Arithmetic: percentage, ratio, averages, time & work

Every UPSC CSAT question on this topic, 2016тАУ2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Arithmetic: percentage, ratio, averages, time & work questions per year: 2016: 16, 2017: 11, 2018: 8, 2019: 15, 2020: 15, 2021: 16, 2022: 13, 2023: 2, 2024: 11, 2025: 9, 2026: 16 Asked in 11 of 11 years ┬╖ most in 2026 (16)
Timed ┬╖ 30 minutes a set Practise Arithmetic: percentage, ratio, averages, time & work as timed tests 7 sets of about 20 questions with UPSC's CSAT marking. Start practising

UPSC syllabus: тАЬBasic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. тАФ Class X level);тАЭ Sub-topics in the detailed syllabus тЖТ

тВ╣1,840 рдХреЗ рдПрдХ рдмрд┐рд▓ рдХрд╛ тВ╣50, тВ╣20 рдФрд░ тВ╣10 рдореВрд▓реНрдпрд╡рд░реНрдЧ рдХреЗ рдиреЛрдЯреЛрдВ рдореЗрдВ рднреБрдЧрддрд╛рди рдХрд┐рдпрд╛ рдЬрд╛рддрд╛ рд╣реИред рдХреБрд▓ рдорд┐рд▓рд╛рдХрд░ 50 рдиреЛрдЯ рдХрд╛рдо рдореЗрдВ рдЖрдПред рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдХрдердиреЛрдВ рдкрд░ рд╡рд┐рдЪрд╛рд░ рдХреАрдЬрд┐рдП :

  1. 1.тВ╣50 рд╡рд╛рд▓реЗ 25 рдиреЛрдЯ рдХрд╛рдо рдореЗрдВ рдЖрдП рдФрд░ рд╢реЗрд╖ рднреБрдЧрддрд╛рди тВ╣20 рд╡рд╛рд▓реЗ рдФрд░ тВ╣10 рд╡рд╛рд▓реЗ рдиреЛрдЯреЛрдВ рдореЗрдВ рдХрд┐рдпрд╛ рдЧрдпрд╛ред
  2. 2.тВ╣20 рд╡рд╛рд▓реЗ 35 рдиреЛрдЯ рдХрд╛рдо рдореЗрдВ рдЖрдП рдФрд░ рд╢реЗрд╖ рднреБрдЧрддрд╛рди тВ╣50 рд╡рд╛рд▓реЗ рдФрд░ тВ╣10 рд╡рд╛рд▓реЗ рдиреЛрдЯреЛрдВ рдореЗрдВ рдХрд┐рдпрд╛ рдЧрдпрд╛ред
  3. 3.тВ╣10 рд╡рд╛рд▓реЗ 20 рдиреЛрдЯ рдХрд╛рдо рдореЗрдВ рдЖрдП рдФрд░ рд╢реЗрд╖ рднреБрдЧрддрд╛рди тВ╣50 рд╡рд╛рд▓реЗ рдФрд░ тВ╣20 рд╡рд╛рд▓реЗ рдиреЛрдЯреЛрдВ рдореЗрдВ рдХрд┐рдпрд╛ рдЧрдпрд╛ред

рдЙрдкрд░реНрдпреБрдХреНрдд рдХрдердиреЛрдВ рдореЗрдВ рд╕реЗ рдХреМрди-рд╕реЗ рд╕рд╣реА рдирд╣реАрдВ рд╣реИрдВ?

Read this question in English

A bill for тВ╣1,840 is paid in the denominations of тВ╣50, тВ╣20 and тВ╣10 notes. 50 notes in all are used. Consider the following statements:

1. 25 notes of тВ╣50 are used and the remaining are in the denominations of тВ╣20 and тВ╣10.

2. 35 notes of тВ╣20 are used and the remaining are in the denominations of тВ╣50 and тВ╣10.

3. 20 notes of тВ╣10 are used and the remaining are in the denominations of тВ╣50 and тВ╣20.

Which of the above statements are not correct?

(a) 1 and 2 only ┬╖ (b) 2 and 3 only ┬╖ (c) 1 and 3 only ┬╖ (d) 1, 2 and 3

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (d) 1, 2 рдФрд░ 3

Let a, b, c be the numbers of тВ╣50, тВ╣20 and тВ╣10 notes. The two totals reduce to 4a + b = 134, and none of the three statements gives whole, non-negative numbers of notes, so all three are incorrect.

  1. Let the numbers of тВ╣50, тВ╣20 and тВ╣10 notes be a, b and c. Then a + b + c = 50 and 50a + 20b + 10c = 1840, that is 5a + 2b + c = 184.
  2. Subtract the first equation from the second: 4a + b = 134.
  3. Statement 1: a = 25 gives b = 134 тИТ 100 = 34, and then c = 50 тИТ 25 тИТ 34 = тИТ9. Impossible.
  4. Statement 2: b = 35 gives 4a = 99, which is not a whole number. Impossible.
  5. Statement 3: c = 20 gives a + b = 30; with 4a + b = 134 this means 3a = 104, not a whole number. Impossible.
  6. Check: a valid payment does exist, e.g. a = 30, b = 14, c = 6 тАФ that is 50 notes and 1500 + 280 + 60 = тВ╣1,840 тАФ so the bill itself is fine; only the three statements fail.
  • тЬЧ 1. a = 25 forces c = тИТ9, a negative number of тВ╣10 notes.
  • тЬЧ 2. b = 35 forces 4a = 99, so a is not a whole number.
  • тЬЧ 3. c = 20 forces 3a = 104, so a is not a whole number.

Remember ┬╖ With a count total and a value total, subtract to get one clean equation, then test each statement for whole, non-negative answers.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

X рдФрд░ Y, 300 m рд▓рдореНрдмреЗ рд╡реГрддреНрддреАрдп рдорд╛рд░реНрдЧ рдореЗрдВ рджреМрдбрд╝рддреЗ рд╣реБрдП 3 km рдХреА рджреМрдбрд╝ рд▓рдЧрд╛рддреЗ рд╣реИрдВред рдЙрдирдХреА рдЪрд╛рд▓ 3 : 2 рдХреЗ рдЕрдиреБрдкрд╛рдд рдореЗрдВ рд╣реИред рдЕрдЧрд░ рдЙрдиреНрд╣реЛрдВрдиреЗ рдПрдХ-рд╕рд╛рде рдПрдХ рд╣реА рджрд┐рд╢рд╛ рдореЗрдВ рджреМрдбрд╝ рд╢реБрд░реВ рдХреА рд╣реИ, рддреЛ рдХрд┐рддрдиреА рдмрд╛рд░ рдкрд╣рд▓рд╛ рд╡реНрдпрдХреНрддрд┐, рджреВрд╕рд░реЗ рд╡реНрдпрдХреНрддрд┐ рдХреЗ рдкрд╛рд╕ рд╕реЗ рдЧреБрдЬрд░реЗрдЧрд╛ (рджреМрдбрд╝ рд╢реБрд░реВ рдХрд░рдиреЗ рдХреА рд╕реНрдерд┐рддрд┐ рдХреЛ рдкрд╛рд╕ рд╕реЗ рдЧреБрдЬрд░рдиреЗ рдореЗрдВ рдирд╣реАрдВ рдЧрд┐рдирд╛ рдЧрдпрд╛ рд╣реИ)?

Read this question in English

X and Y run a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3:2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?

(a) 2 ┬╖ (b) 3 ┬╖ (c) 4 ┬╖ (d) 5

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (b) 3

By the time X finishes 3 km, Y has run 2 km, so X has gained 1000 m тАФ three full laps of 300 m and a little more. Each full lap gained is one pass, so X passes Y 3 times.

  1. The race ends when X, the faster runner, completes 3 km = 3000 m.
  2. Speeds are 3 : 2, so in that time Y runs 2/3 ├Ч 3000 = 2000 m.
  3. X gains 3000 тИТ 2000 = 1000 m on Y.
  4. X passes Y each time the lead reaches a whole lap: at 300 m, 600 m and 900 m. 1000 ├╖ 300 = 3.33, so there are 3 passes.

Remember ┬╖ On a circular track, passes = lead gained by the faster runner ├╖ track length, counting whole laps only.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

рдПрдХ рд╡рд╕реНрддреБ рдХреА рдХреАрдордд рдореЗрдВ 25% рд╡реГрджреНрдзрд┐ рдХреА рдЧрдИред рддрддреНрдкрд╢реНрдЪрд╛рддреН рдХреАрдордд рдХреЛ 20% рдШрдЯрд╛ рджрд┐рдпрд╛ рдЧрдпрд╛ рдФрд░ рдлрд┐рд░ 10% рдмрдврд╝рд╛ рджрд┐рдпрд╛ рдЧрдпрд╛ред рдХреАрдордд рдореЗрдВ рдкрд░рд┐рдгрд╛рдореА рд╡реГрджреНрдзрд┐ рдХреНрдпрд╛ рд╣реИ?

Read this question in English

The increase in the price of a certain item was 25%. Then the price was decreased by 20% and then again increased by 10%. What is the resultant increase in the price?

(a) 5% ┬╖ (b) 10% ┬╖ (c) 12.5% ┬╖ (d) 15%

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (b) 10%

A 25% rise followed by a 20% fall brings the price exactly back to where it started (1.25 ├Ч 0.8 = 1). Only the last 10% rise remains.

  1. Let the price be тВ╣100. After a 25% increase it is тВ╣125.
  2. A 20% decrease: 125 ├Ч 0.8 = тВ╣100.
  3. A 10% increase: 100 ├Ч 1.1 = тВ╣110.
  4. The resultant increase is тВ╣10 on тВ╣100, i.e. 10%.

Remember ┬╖ Successive percentage changes multiply: +25% then тИТ20% cancel, because 5/4 ├Ч 4/5 = 1.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

1┬╖01 km рд▓рдореНрдмреА рд╕рдбрд╝рдХ рдХреЗ рдПрдХ рдХрд┐рдирд╛рд░реЗ рдПрдХ-рджреВрд╕рд░реЗ рд╕реЗ рд╕рдорд╛рди рджреВрд░реА рдкрд░ 101 рдкреМрдзреЗ рд░реЛрдкреЗ рдЧрдП рд╣реИрдВред 5 рдХреНрд░рдорд╛рдЧрдд рдкреМрдзреЛрдВ рдХреЗ рдмреАрдЪ рдХреБрд▓ рдХрд┐рддрдиреА рджреВрд░реА рд╣реИ?

Read this question in English

On one side of a 1.01 km long road, 101 plants are planted at equal distance from each other. What is the total distance between 5 consecutive plants?

(a) 40 m ┬╖ (b) 40.4 m ┬╖ (c) 50 m ┬╖ (d) 50.5 m

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (b) 40┬╖4 m

With plants at both ends of the road, 101 plants make 100 equal gaps along 1010 m, so each gap is 10.1 m. Five consecutive plants span 4 gaps: 4 ├Ч 10.1 = 40.4 m.

  1. 101 plants in a line from one end of the road to the other leave 101 тИТ 1 = 100 gaps.
  2. Road length = 1.01 km = 1010 m, so each gap = 1010 ├╖ 100 = 10.1 m.
  3. 5 consecutive plants have 4 gaps between them: 4 ├Ч 10.1 = 40.4 m.
  4. Trap: dividing 1010 by 101 (counting plants instead of gaps) gives 10 m and the wrong answer 40 m.

Remember ┬╖ n objects in a line make n тИТ 1 gaps; k consecutive objects span k тИТ 1 gaps.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

A рдХреЗ рдкрд╛рд╕ рдХреБрдЫ рд╕рд┐рдХреНрдХреЗ рд╣реИрдВред рд╡рд╣ рдЙрдирдореЗрдВ рд╕реЗ рдЖрдзреЗ рд╕рд┐рдХреНрдХреЛрдВ рдореЗрдВ 2 рдФрд░ рд╕рд┐рдХреНрдХреЗ рдорд┐рд▓рд╛рдХрд░ B рдХреЛ рджреЗрддрд╛ рд╣реИред B рдЙрдирдореЗрдВ рд╕реЗ рдЖрдзреЗ рд╕рд┐рдХреНрдХреЛрдВ рдореЗрдВ 2 рдФрд░ рд╕рд┐рдХреНрдХреЗ рдорд┐рд▓рд╛рдХрд░ C рдХреЛ рджреЗрддрд╛ рд╣реИред C рдЙрдирдореЗрдВ рд╕реЗ рдЖрдзреЗ рд╕рд┐рдХреНрдХреЛрдВ рдореЗрдВ 2 рдФрд░ рд╕рд┐рдХреНрдХреЗ рдорд┐рд▓рд╛рдХрд░ D рдХреЛ рджреЗрддрд╛ рд╣реИред D рдХреЗ рдкрд╛рд╕ рдЕрдм рдЬрд┐рддрдиреЗ рд╕рд┐рдХреНрдХреЗ рд╣реИрдВ, рд╡рд╣ рджреЛ рдЕрдВрдХреЛрдВ рдХреА рд▓рдШреБрддрдо рд╕рдВрдЦреНрдпрд╛ рд╣реИред рд╢реБрд░реВ рдореЗрдВ A рдХреЗ рдкрд╛рд╕ рдХрд┐рддрдиреЗ рд╕рд┐рдХреНрдХреЗ рд╣реИрдВ?

Read this question in English

A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?

(a) 76 ┬╖ (b) 68 ┬╖ (c) 60 ┬╖ (d) 52

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (d) 52

Work backwards from DтАЩs 10 coins. Each giver handed over half his coins plus 2, so he had 2 ├Ч (coins given тИТ 2): C had 16, B had 28 and A had 52.

  1. D now has 10 coins (the smallest two-digit number), all received from C.
  2. C gave half his coins + 2 = 10, so half = 8 and C had 16 тАФ all received from B.
  3. B gave half + 2 = 16, so half = 14 and B had 28 тАФ all received from A.
  4. A gave half + 2 = 28, so half = 26 and A had 52.
  5. Check: A (52) gives 26 + 2 = 28 to B; B gives 14 + 2 = 16 to C; C gives 8 + 2 = 10 to D.

Remember ┬╖ For тАШgives half and k moreтАЩ chains, reverse each step: original = 2 ├Ч (amount given тИТ k).

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

x рд╕рдкреНрддрд╛рд╣, x рджрд┐рди, x рдШрдгреНрдЯреЗ, x рдорд┐рдирдЯ рдФрд░ x рд╕реЗрдХрдгреНрдб рдореЗрдВ рдХреБрд▓ рдХрд┐рддрдиреЗ рд╕реЗрдХрдгреНрдб рд╣реИрдВ?

Read this question in English

How many seconds in total are there in x weeks, x days, x hours, x minutes and x seconds?

(a) 11580x ┬╖ (b) 11581x ┬╖ (c) 694860x ┬╖ (d) 694861x

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (d) 694861x

Convert each unit to seconds: a week is 604800 s, a day 86400 s, an hour 3600 s, a minute 60 s. Adding these and the lone second gives 694861 seconds for every x.

  1. 1 week = 7 ├Ч 24 ├Ч 3600 = 604800 s; 1 day = 24 ├Ч 3600 = 86400 s; 1 hour = 3600 s; 1 minute = 60 s.
  2. Total = (604800 + 86400 + 3600 + 60 + 1)x = 694861x seconds.
  3. Check: the x seconds make the coefficient end in 1, and one week alone is 604800x, so only 694861x fits.

Remember ┬╖ Use the last digit as a filter: the extra тАШx secondsтАЩ makes the total end in 1.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

рдкрд╛рдБрдЪ рдорд┐рддреНрд░ P, Q, X, Y рдФрд░ Z рдиреЗ рдХреБрдЫ рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВред рд╕рдВрдЧрдд рд╕реВрдЪрдирд╛рдПрдБ рдиреАрдЪреЗ рджреА рдЧрдИ рд╣реИрдВ :

  1. 1.X рдиреЗ рдЬрд┐рддрдиреА рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВ рдЙрд╕рд╕реЗ 8 рдЕрдзрд┐рдХ рдиреЛрдЯрдмреБрдХ Z рдиреЗ рдЦрд░реАрджреАрдВред
  2. 2.P рдФрд░ Q рдиреЗ рдорд┐рд▓рдХрд░ 21 рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВред
  3. 3.P рдиреЗ рдЬрд┐рддрдиреА рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВ рдЙрд╕рд╕реЗ 5 рдХрдо рдиреЛрдЯрдмреБрдХ Q рдиреЗ рдЦрд░реАрджреАрдВред
  4. 4.X рдФрд░ Y рдиреЗ рдорд┐рд▓рдХрд░ 28 рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВред
  5. 5.X рдиреЗ рдЬрд┐рддрдиреА рдиреЛрдЯрдмреБрдХ рдЦрд░реАрджреАрдВ рдЙрд╕рд╕реЗ 5 рдЕрдзрд┐рдХ рдиреЛрдЯрдмреБрдХ P рдиреЗ рдЦрд░реАрджреАрдВред

рдпрджрд┐ рдкреНрд░рддреНрдпреЗрдХ рдиреЛрдЯрдмреБрдХ рдХреА рдХреАрдордд тВ╣40 рд╣реИ, рддреЛ рд╕рднреА рдиреЛрдЯрдмреБрдХ рдХреА рдХреБрд▓ рд▓рд╛рдЧрдд рдХрд┐рддрдиреА рд╣реИ?

Read this question in English

Five friends P, Q, X, Y and Z purchased some notebooks. The relevant information is given below:

1. Z purchased 8 notebooks more than X did.

2. P and Q together purchased 21 notebooks.

3. Q purchased 5 notebooks less than P did.

4. X and Y together purchased 28 notebooks.

5. P purchased 5 notebooks more than X did.

If each notebook is priced тВ╣40, then what is the total cost of all the notebooks?

(a) тВ╣2,600 ┬╖ (b) тВ╣2,400 ┬╖ (c) тВ╣2,360 ┬╖ (d) тВ╣2,320

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (a) тВ╣2,600

Statements 2 and 3 give P = 13 and Q = 8; then X = 8, Z = 16 and Y = 20. The five bought 65 notebooks, which cost 65 ├Ч тВ╣40 = тВ╣2,600.

  1. From 2 and 3: P + Q = 21 and P тИТ Q = 5, so P = 13 and Q = 8.
  2. From 5: X = 13 тИТ 5 = 8. From 1: Z = 8 + 8 = 16. From 4: Y = 28 тИТ 8 = 20.
  3. Total notebooks = (P + Q) + (X + Y) + Z = 21 + 28 + 16 = 65.
  4. Cost = 65 ├Ч тВ╣40 = тВ╣2,600.

Remember ┬╖ Start with the pair of equations that closes on its own (sum and difference), then chain outwards.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

рдХреЛрдИ рд╡реНрдпрдХреНрддрд┐ рдШрд░ рд╕реЗ 14:30 рдмрдЬреЗ рдирд┐рдХрд▓рд╛ рдФрд░ рдпрд╛рддреНрд░рд╛ рдХрд░ рдЧрд╛рдБрд╡ рдкрд╣реБрдБрдЪрд╛, рддрдм рдЧрд╛рдБрд╡ рдХреА рдШрдбрд╝реА рдиреЗ 15:15 рдмрдЬреЗ рдХрд╛ рд╕рдордп рджрд┐рдЦрд╛рдпрд╛ред рд╡рд╣рд╛рдБ 25 рдорд┐рдирдЯ рд░реБрдХрдХрд░, рд╡рд╣ рдкрд╣рд▓реЗ рд╡рд╛рд▓реЗ рдорд╛рд░реНрдЧ рдХреА рдЕрдкреЗрдХреНрд╖рд╛ 1┬╖25 рдЧреБрдирд╛ рд▓рдореНрдмреЗ рдорд╛рд░реНрдЧ рд╕реЗ, рджреЛрдЧреБрдиреА рдЪрд╛рд▓ рд╕реЗ рдпрд╛рддреНрд░рд╛ рдХрд░ 16:00 рдмрдЬреЗ рдЕрдкрдиреЗ рдШрд░ рдкрд╣реБрдБрдЪрд╛ред рдШрд░ рдХреА рдШрдбрд╝реА рдХреА рддреБрд▓рдирд╛ рдореЗрдВ рдЧрд╛рдБрд╡ рдХреА рдШрдбрд╝реА

Read this question in English

A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1.25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is

(a) 10 minutes slow ┬╖ (b) 5 minutes slow ┬╖ (c) 10 minutes fast ┬╖ (d) 5 minutes fast

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (d) 5 рдорд┐рдирдЯ рддреЗрдЬ рд╣реИ

Of the 90 minutes he was away by the home clock, 65 were spent driving. The return took 0.625 of the outward time, so the outward drive took 40 minutes and he reached the village at 15:10 home time тАФ the village clock, showing 15:15, is 5 minutes fast.

  1. By the home clock he was away from 14:30 to 16:00 = 90 minutes; 25 minutes were spent in the village, so driving took 65 minutes.
  2. Let the outward drive take t minutes. The return route is 1.25 times as long at twice the speed, so it takes 1.25 ├╖ 2 ├Ч t = 0.625t.
  3. t + 0.625t = 65, so 1.625t = 65 and t = 40 minutes.
  4. He reached the village at 14:30 + 40 min = 15:10 by the home clock, when the village clock showed 15:15.
  5. So the village clock is 5 minutes fast.
  6. Check: he leaves the village at 15:35 (home time); the return takes 0.625 ├Ч 40 = 25 minutes, reaching home at 16:00.

Remember ┬╖ Use only one clock for elapsed time; compare the other clockтАЩs reading with the true time at that moment.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

24 рдкреБрд░реБрд╖ рдФрд░ 12 рдорд╣рд┐рд▓рд╛рдПрдБ рдПрдХ рдХрд╛рдо рдХреЛ 30 рджрд┐рдиреЛрдВ рдореЗрдВ рдкреВрд░рд╛ рдХрд░ рд╕рдХрддреЗ рд╣реИрдВред 12 рдкреБрд░реБрд╖ рдФрд░ 24 рдорд╣рд┐рд▓рд╛рдПрдБ рдЙрд╕реА рдХрд╛рдо рдХреЛ рдХрд┐рддрдиреЗ рджрд┐рдиреЛрдВ рдореЗрдВ рдкреВрд░рд╛ рдХрд░реЗрдВрдЧреЗ?

Read this question in English

24 men and 12 women can do a piece of work in 30 days. In how many days can 12 men and 24 women do the same piece of work?

(a) 30 days ┬╖ (b) More than 30 days ┬╖ (c) Less than 30 days or more than 30 days ┬╖ (d) Data is inadequate to draw any conclusion

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (d) рдХреЛрдИ рдирд┐рд╖реНрдХрд░реНрд╖ рдирд┐рдХрд╛рд▓рдиреЗ рдХреЗ рд▓рд┐рдП рдЖрдБрдХрдбрд╝реЗ рдЕрдкрд░реНрдпрд╛рдкреНрдд рд╣реИрдВ

The answer depends on how a manтАЩs work rate compares with a womanтАЩs, and that is not given. Equal rates give exactly 30 days, faster men give more than 30 and faster women fewer тАФ so the data are inadequate.

  1. Let a man do m units and a woman w units of work a day. Work = 30 ├Ч (24m + 12w).
  2. Days for 12 men and 24 women = 30(24m + 12w)/(12m + 24w) = 30(2m + w)/(m + 2w).
  3. If m = w this is 30 days; if m > w it is more than 30; if m < w it is less than 30.
  4. The relation between m and w is not given, so no conclusion can be drawn. Option (c) fails because it leaves out the possibility of exactly 30 days.

Remember ┬╖ In time-and-work items with two kinds of workers, you need their relative efficiency before you can swap one for the other.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

рдЬрдм рдПрдХ рд╕рдВрдЦреНрдпрд╛ x рдХрд╛ 70% рдПрдХ рдЕрдиреНрдп рд╕рдВрдЦреНрдпрд╛ y рдореЗрдВ рдЬреЛрдбрд╝рд╛ рдЬрд╛рддрд╛ рд╣реИ, рддреЛ рдЙрдирдХрд╛ рдпреЛрдЧ y рдХреЗ рдорд╛рди рдХрд╛ 165% рд╣реЛ рдЬрд╛рддрд╛ рд╣реИред рдЬрдм рд╕рдВрдЦреНрдпрд╛ x рдХрд╛ 60% рдПрдХ рдЕрдиреНрдп рд╕рдВрдЦреНрдпрд╛ z рдореЗрдВ рдЬреЛрдбрд╝рд╛ рдЬрд╛рддрд╛ рд╣реИ, рддреЛ рдЙрдирдХрд╛ рдпреЛрдЧ z рдХреЗ рдорд╛рди рдХрд╛ 165% рд╣реЛ рдЬрд╛рддрд╛ рд╣реИред рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдореЗрдВ рд╕реЗ рдХреМрди-рд╕рд╛ рдПрдХ рд╕рд╣реА рд╣реИ?

Read this question in English

When 70% of a number x is added to another number y, the sum becomes 165% of the value of y. When 60% of the number x is added to another number z, then the sum becomes 165% of the value of z. Which one of the following is correct?

(a) z < x < y ┬╖ (b) x < y < z ┬╖ (c) y < x < z ┬╖ (d) z < y < x

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (a) z < x < y

The first condition gives 0.7x = 0.65y, so y = 14x/13, just above x. The second gives 0.6x = 0.65z, so z = 12x/13, just below x. Hence z < x < y.

  1. 0.7x + y = 1.65y, so 0.7x = 0.65y and y = 70x/65 = 14x/13.
  2. 0.6x + z = 1.65z, so 0.6x = 0.65z and z = 60x/65 = 12x/13.
  3. For positive x, 12x/13 < x < 14x/13, so z < x < y.
  4. Check with x = 13: y = 14 and z = 12; 9.1 + 14 = 23.1 = 1.65 ├Ч 14 and 7.8 + 12 = 19.8 = 1.65 ├Ч 12.

Remember ┬╖ Turn each percentage condition into тАШpart of x = part of yтАЩ and compare the resulting multiples of x.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

рджреЛ рдЙрдореНрдореАрджрд╡рд╛рд░реЛрдВ, X рдФрд░ Y, рдиреЗ рдПрдХ рдирд┐рд░реНрд╡рд╛рдЪрди рдореЗрдВ рднрд╛рдЧ рд▓рд┐рдпрд╛ред 80% рд▓реЛрдЧреЛрдВ рдиреЗ рдорддрджрд╛рди рдХрд┐рдпрд╛ рдФрд░ рдХреЛрдИ рдЕрд╡реИрдз рдордд рдирд╣реАрдВ рд╣реБрдЖред NOTA (рдЙрдкрд░реНрдпреБрдХреНрдд рдореЗрдВ рд╕реЗ рдХреЛрдИ рдирд╣реАрдВ) рдХрд╛ рд╡рд┐рдХрд▓реНрдк рдирд╣реАрдВ рдерд╛ред рдорддрджрд╛рди рдореЗрдВ рдкрдбрд╝реЗ рдХреБрд▓ рдорддреЛрдВ рдХрд╛ 56% рдордд X рдХреЛ рдорд┐рд▓рд╛ рдФрд░ рд╡рд╣ 1440 рдорддреЛрдВ рд╕реЗ рдЬреАрдд рдЧрдпрд╛ред рдорддрджрд╛рддрд╛ рд╕реВрдЪреА рдореЗрдВ рдорддрджрд╛рддрд╛рдУрдВ рдХреА рдХреБрд▓ рдХрд┐рддрдиреА рд╕рдВрдЦреНрдпрд╛ рд╣реИ?

Read this question in English

Two candidates X and Y contested an election. 80% of voters cast their vote and there were no invalid votes. There was no NOTA (None of the above) option. X got 56% of the votes cast and won by 1440 votes. What is the total number of voters in the voters list?

(a) 15000 ┬╖ (b) 12000 ┬╖ (c) 9600 ┬╖ (d) 5000

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (a) 15000

X got 56% and Y 44% of the votes cast, so the winning margin is 12% of the votes cast: 1440 votes. That makes 12000 votes cast, which is 80% of the list тАФ so the list has 15000 voters.

  1. With two candidates and no invalid votes, Y got 100% тИТ 56% = 44% of the votes cast.
  2. Margin = 56% тИТ 44% = 12% of the votes cast = 1440, so votes cast = 1440 ├╖ 0.12 = 12000.
  3. Votes cast are 80% of the voters list, so total voters = 12000 ├╖ 0.8 = 15000.
  4. Check: 12000 is 80% of 15000; X gets 6720, Y gets 5280, and 6720 тИТ 5280 = 1440.

Remember ┬╖ In two-candidate elections, the margin is (winner% тИТ loser%) of the votes cast; convert back to the full list last.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

X рдФрд░ Y рджреЛ рдкрд╛рддреНрд░ рд╣реИрдВред X рдореЗрдВ 100 ml рджреВрдз рд╣реИ рдФрд░ Y рдореЗрдВ 100 ml рдкрд╛рдиреА рд╣реИред X рдореЗрдВ рд╕реЗ 20 ml рджреВрдз рдирд┐рдХрд╛рд▓рдХрд░ Y рдореЗрдВ рдбрд╛рд▓рд╛ рдЬрд╛рддрд╛ рд╣реИред рдЗрдиреНрд╣реЗрдВ рдЕрдЪреНрдЫреА рддрд░рд╣ рдорд┐рд▓рд╛рдХрд░, Y рдХрд╛ 20 ml рдорд┐рд╢реНрд░рдг рдирд┐рдХрд╛рд▓рдХрд░ рд╡рд╛рдкрд╕ X рдореЗрдВ рдорд┐рд▓рд╛рдпрд╛ рдЬрд╛рддрд╛ рд╣реИред рдпрджрд┐ X рдореЗрдВ рджреВрдз рдХрд╛ рдЕрдиреБрдкрд╛рдд m рд╕реЗ рдирд┐рд░реНрджрд┐рд╖реНрдЯ рд╣реЛрддрд╛ рд╣реИ рдФрд░ Y рдореЗрдВ рдЬрд▓ рдХрд╛ рдЕрдиреБрдкрд╛рдд n рд╕реЗ рдирд┐рд░реНрджрд┐рд╖реНрдЯ рд╣реЛрддрд╛ рд╣реИ, рддреЛ рдирд┐рдореНрдирд▓рд┐рдЦрд┐рдд рдореЗрдВ рд╕реЗ рдХреМрди-рд╕рд╛ рдПрдХ рд╕рд╣реА рд╣реИ?

Read this question in English

There are two containers X and Y. X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of milk from X is transferred to Y. After mixing well, 20 ml of the mixture in Y is transferred back to X. If m denotes the proportion of milk in X and n denotes the proportion of water in Y, then which one of the following is correct?

(a) m = n ┬╖ (b) m > n ┬╖ (c) m < n ┬╖ (d) Cannot be determined due to insufficient data

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (a) m = n

After the two transfers each container again holds 100 ml. Whatever water went into X must be matched by the same amount of milk left in Y, so XтАЩs milk share equals YтАЩs water share: m = n = 5/6.

  1. After the first transfer: X has 80 ml milk; Y has 100 ml water + 20 ml milk = 120 ml.
  2. 20 ml of YтАЩs mixture carries 20 ├Ч 20/120 = 10/3 ml milk and 20 ├Ч 100/120 = 50/3 ml water back to X.
  3. X: milk = 80 + 10/3 = 250/3 ml out of 100 ml, so m = 5/6.
  4. Y: water = 100 тИТ 50/3 = 250/3 ml out of 100 ml, so n = 5/6.
  5. So m = n.

Remember ┬╖ Back-and-forth transfers of equal volume: the milk in the water jar always equals the water in the milk jar.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

A, B, C рдХрд╛ рдФрд╕рдд рднрд╛рд░ 40 kg рд╣реИ; B, D, E рдХрд╛ рдФрд╕рдд рднрд╛рд░ 42 kg рд╣реИ рдФрд░ F рдХрд╛ рднрд╛рд░ B рдХреЗ рднрд╛рд░ рдХреЗ рдмрд░рд╛рдмрд░ рд╣реИред A, B, C, D, E рдФрд░ F рдХрд╛ рдФрд╕рдд рднрд╛рд░ рдХреНрдпрд╛ рд╣реИ?

Read this question in English

The average weight of A, B, C is 40 kg, the average weight of B, D, E is 42 kg and the weight of F is equal to that of B. What is the average weight of A, B, C, D, E and F?

(a) 40.5 kg ┬╖ (b) 40.8 kg ┬╖ (c) 41 kg ┬╖ (d) Cannot be determined as data is inadequate

рдЙрддреНрддрд░ рдФрд░ рд╡реНрдпрд╛рдЦреНрдпрд╛ ┬╖ explanation in English

Answer: (c) 41 kg

A + B + C = 120 and B + D + E = 126. Since F weighs the same as B, the six weights total 120 + 126 = 246 kg, and the average is 41 kg even though B itself is unknown.

  1. A + B + C = 3 ├Ч 40 = 120 kg and B + D + E = 3 ├Ч 42 = 126 kg.
  2. Since F = B, A + B + C + D + E + F = (A + B + C) + (F + D + E) = 120 + (B + D + E) = 120 + 126 = 246 kg.
  3. Average = 246 ├╖ 6 = 41 kg.

Remember ┬╖ Before answering тАШcannot be determinedтАЩ, check whether the unknown cancels тАФ here F = B fills the gap exactly.

Question and answer: UPSC's official GS Paper II (2022, Series A) тАФ paper тЖЧ ┬╖ answer key тЖЧ. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ┬╖

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