CSAT Solved Papers/ 2023/Q36

2023 CSAT — Q36

Quant Statement validity 2.5 marks Hard

A box contains $14$ black balls, $20$ blue balls, $26$ green balls, $28$ yellow balls, $38$ red balls and $54$ white balls. Consider the following statements:

The smallest number $n$ such that any $n$ balls drawn from the box randomly must contain one full group of at least one colour is $175$ .
The smallest number $m$ such that any $m$ balls drawn from the box randomly must contain at least one ball of each colour is $167$ .

Which of the above statements is/are correct?

A 1 only
B 2 only
C Both 1 and 2 Answer
D Neither 1 nor 2

Worked rationale

Both are worst-case (pigeonhole) guarantees. Total balls $= 14+20+26+28+38+54 = 180$ .

Statement 1 — guarantee a full group (every ball of some one colour). The unluckiest draw avoids a complete colour by leaving at least one ball of every colour behind: that means leaving $6$ balls (one per colour) undrawn, so up to $180 - 6 = 174$ balls can be drawn with no complete colour. The next ball ( $175$ th) forces some colour to be complete.

n = 174 + 1 = 175. \quad\text{Statement 1 is true.}

Statement 2 — guarantee at least one of each colour. The unluckiest draw misses a colour entirely; to draw the most while still missing one, omit the smallest colour (black, $14$ ) and take all the rest: $180 - 14 = 166$ balls can still lack a colour. The next ball ( $167$ th) forces all six colours present.

m = 166 + 1 = 167. \quad\text{Statement 2 is true.}

Answer: (c) Both 1 and 2.

Why the other options miss

A

for Statement 2, omits a largest colour ( $54$ ) or mis-identifies which colour to skip, getting $m \neq 167$ and rejecting it.
B

miscounts the undrawn balls: for Statement 1, leaves the wrong number of balls undrawn (e.g. subtracts $1$ total instead of $1$ per colour), missing $n = 175$ .
D

confuses the two guarantees (full-group vs one-of-each), swapping the " $-6$ " and “ $-$ smallest-colour” logic.

Specialist insight

The two statements use opposite worst cases. “Force a complete colour” leaves one of every colour behind ( $180 - 6 = 174$ , so $n = 175$ ). “Force one of each colour” skips the smallest colour entirely ( $180 - 14 = 166$ , so $m = 167$ ). Naming which extremal configuration defeats the guarantee — then adding $1$ — is the entire method. Swapping the two extremes is the engineered trap, and both ledger numbers check out, so the answer is (c).

The trap, in one line

Full-group: leave $1$ of each ( $180-6=174 \Rightarrow n=175$ ); one-of-each: skip smallest colour ( $180-14=166 \Rightarrow m=167$ ); both true $=$ (c).

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