If the ratio of the number of students learning Chess in College C to College…
2024
If the ratio of the number of students learning Chess in College C to College E (the fifth college) is 7 : 9, the number of students learning Squash in College E is 20% more than that of College B, and the number of students learning Carrom in College E is 30% less than that of College D, then what is the total number of students in College E?
Answer: B. 2004 — Concept — splitting a total when the difference is known. If two groups have a known combined strength S and a known difference D, and one of them is stated…
- A.
2200
- B.
2004
- C.
1964
- D.
2444
Show answer & explanation
Correct answer: B
Concept — splitting a total when the difference is known. If two groups have a known combined strength S and a known difference D, and one of them is stated to be the larger, then the larger group = (S + D) ÷ 2 and the smaller group = (S − D) ÷ 2.
Concept — recovering a whole from a percentage share. If one category holds p% of a college, the remaining categories together hold (100 − p)% of it. So a college’s strength = (its known part) ÷ (100 − p)%, and the percentage category itself is the strength minus that part.
Application — the four figures College E is defined against.
Chess in College C: combined Chess and Squash strength is 1260 with a difference of 140, and Chess is the larger group, so Chess = (1260 + 140) ÷ 2 = 700.
Squash in College B: combined strength is 1170 with a difference of 30, and Squash is the smaller group, so Squash = (1170 − 30) ÷ 2 = 570.
Carrom in College D: its 1800 Chess-and-Squash students fill the 75% left by a 25% Carrom share, so College D’s strength = 1800 ÷ 0.75 = 2400 and its Carrom group = 2400 − 1800 = 600.
Chess in College E: the 7 : 9 ratio links the two Chess groups, so Chess in College E = 700 × 9 ÷ 7 = 900.
Squash in College E: 20% more than College B’s Squash group, so 570 × 1.20 = 684.
Carrom in College E: 30% less than College D’s Carrom group, so 600 × 0.70 = 420.
Every student learns exactly one game, so College E’s strength = 900 + 684 + 420 = 2004.
Cross-check — the three College E groups against their anchors.
Game in College E | Number of students | How it is fixed |
|---|---|---|
Chess | 900 | 900 : 700 reduces to 9 : 7, matching the given ratio |
Squash | 684 | 684 − 570 = 114, which is 20% of 570 |
Carrom | 420 | 600 − 420 = 180, which is 30% of 600 |
Total | 2004 | 900 + 684 + 420 |
Contrast — three readings of the table that change the total.
The 20% condition is tied to College B’s Squash group of 570, not to its Chess group of 600; using 600 raises College E’s strength to 2040.
College D’s Carrom group is 25% of its full strength of 2400, that is 600; reading it as 25% of the 1800 Chess-and-Squash figure gives 450 and lowers College E’s strength to 1899.
The 7 : 9 ratio links the Chess groups only; scaling the 1260 Chess-and-Squash total instead gives 1620 and a College E strength of 2724.