CSAT Solved Papers/ 2021/Q77

2021 CSAT — Q77

Quant Statement validity 2.5 marks Hard

$X$ said to $Y$ , “At the time of your birth I was twice as old as you are at present.” If the present age of $X$ is $42$ years, then consider the following statements:

$8$ years ago, the age of $X$ was five times the age of $Y$ .
After $14$ years, the age of $X$ would be two times the age of $Y$ .

Which of the above statements is/are correct?

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

Worked rationale

Let $Y$ ‘s present age be $y$ ; $X$ is $42$ . At $Y$ ‘s birth ( $y$ years ago), $X$ was $42 - y$ . The quoted condition “I was twice as old as you are at present” gives

42 - y = 2y \;\Rightarrow\; 42 = 3y \;\Rightarrow\; y = 14.

So $X = 42$ , $Y = 14$ now.

Statement 1 ( $8$ years ago): $X = 34$ , $Y = 6$ . Is $34 = 5\times 6 = 30$ ? No. Incorrect.

Statement 2 ( $14$ years hence): $X = 56$ , $Y = 28$ . Is $56 = 2\times 28 = 56$ ? Yes. Correct.

Answer: (b) 2 only.

Why the other options miss

A

an arithmetic slip: mis-solves the birth condition (e.g. $y = 21$ ) so the ” $8$ years ago” multiple appears to work.
C

solved the wrong question: checks only Statement 2 and assumes Statement 1 also holds without arithmetic.
D

an arithmetic slip: errs in the ” $14$ years hence” check ( $56$ vs $2\times 28$ ), rejecting the true Statement 2.

Specialist insight

The crux is decoding “at the time of your birth I was twice your present age” into $42 - y = 2y$ , fixing $y = 14$ . After that, each statement is a one-line plug-in at a shifted time — but you must shift both ages by the same amount. Statement 1 fails ( $34 \ne 30$ ); Statement 2 holds ( $56 = 2\times 28$ ). The trap is setting up the birth equation wrong; once $y = 14$ is locked, the rest is mechanical verification.

The trap, in one line

$42 - y = 2y \Rightarrow y = 14$ ; then $34 \ne 5\cdot 6$ (S1 ✗) but $56 = 2\cdot 28$ (S2 ✓) $\Rightarrow$ (b).

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