CSAT Solved Papers/ 2024/Q40

2024 CSAT — Q40

Quant Logical & quantitative reasoning 2.5 marks Easy

What is the angle between the minute hand and hour hand when the clock shows $4{:}25$ hours?

A $12{\cdot}5^\circ$
B $15^\circ$
C $17{\cdot}5^\circ$ Answer
D $20^\circ$

Worked rationale

Measure both hands from $12$ o’clock. The minute hand moves $6^\circ$ per minute; the hour hand moves $0.5^\circ$ per minute (it does not sit on the $4$ ).

\text{minute hand at } 25 \text{ min} = 25 \times 6 = 150^\circ.

\text{hour hand at } 4{:}25 = 4 \times 30 + 25 \times 0.5 = 120 + 12.5 = 132.5^\circ.

Angle between $= 150 - 132.5 = 17.5^\circ$ .

Answer: (c) $17{\cdot}5^\circ$ .

Why the other options miss

A

solved the wrong question: reports the hour hand’s advance past the 4 ( $12.5^\circ$ ) instead of the gap between the two hands.
B

wrong formula: parks the hour hand exactly on $4$ ( $120^\circ$ ) and uses $\tfrac{1}{2}\times(\text{something})$ , ignoring the $0.5^\circ/\text{min}$ creep.
D

wrong formula: uses the standard formula $|30H - 5.5M|$ but mis-evaluates $|120 - 137.5|$ , or rounds the hour-hand drift to a full degree.

Specialist insight

The decisive subtlety is that the hour hand drifts: by $4{:}25$ it has moved $25\times0.5 = 12.5^\circ$ past the $4$ , sitting at $132.5^\circ$ , not $120^\circ$ . Forgetting that drift produces the trap (b). The one-line formula $|30H - 5.5M| = |30\cdot4 - 5.5\cdot25| = |120 - 137.5| = 17.5^\circ$ encodes both hand speeds at once — but only if you apply it cleanly. Always credit the hour hand its half-degree per minute.

The trap, in one line

The hour hand is at $132.5^\circ$ (it has crept $12.5^\circ$ past the $4$ ), not $120^\circ$ — so the gap is $150-132.5=17.5^\circ$ .

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