CSAT Solved Papers/ 2023/Q60

2023 CSAT — Q60

Quant Data sufficiency 2.5 marks Medium

For five children with ages $a < b < c < d < e$ ; any two successive ages differ by $2$ years.

Question: What is the age of the youngest child?

Statement-1: The age of the eldest is $3$ times the youngest.

Statement-2: The average age of the children is $8$ years.

Which one of the following is correct in respect of the above Question and the Statements?

A The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
B The Question can be answered by using either Statement alone Answer
C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
D The Question cannot be answered even by using both the Statements together

Worked rationale

The five ages form an arithmetic progression with common difference $2$ : $a, a+2, a+4, a+6, a+8$ , so finding $a$ (the youngest) is the goal.

Statement-1 alone (eldest $= 3\times$ youngest): $a + 8 = 3a \Rightarrow 2a = 8 \Rightarrow a = 4$ . A single value — sufficient.

Statement-2 alone (average $= 8$ ): the mean of an AP is its middle term, $a + 4 = 8 \Rightarrow a = 4$ . A single value — sufficient.

Each statement alone pins $a = 4$ .

Answer: (b) The Question can be answered by using either Statement alone.

Why the other options miss

A

judged the second statement short when it isn’t: solves one statement but wrongly judges the other under-determined (e.g. doesn’t see the AP average equals the middle term).
C

thought both were needed when either alone suffices: thinks both are needed, missing that each single equation already fixes $a$ .
D

ignored the fixed spacing: fails to use the “successive differ by $2$ ” structure, treating the ages as five free unknowns.

Specialist insight

The fixed spacing ( $+2$ ) reduces five unknowns to one ( $a$ ); after that, any single linear fact about the set determines $a$ . St-1 gives $a+8 = 3a$ ; St-2 uses the AP shortcut “mean $=$ middle term” $a+4 = 8$ . Both independently yield $a = 4$ , so the answer is (b) “either alone.” The DS lesson: once a structure collapses the unknowns to one, one equation suffices — beware of reflexively answering (c).

The trap, in one line

Ages are $a, a{+}2,\dots,a{+}8$ ; St-1 ( $a{+}8{=}3a$ ) and St-2 (mean $=a{+}4{=}8$ ) each give $a=4$ alone $=$ (b).

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