2018 CSAT, all questions

UPSC Prelims 2018 · CSAT

Question 5: Basic Numeracy

The figure drawn below gives the velocity graphs of two vehicles A and B. The straight line OKP represents the velocity of vehicle A at any instant, whereas the horizontal straight line CKD represents the velocity of vehicle B at any instant. In the figure, D is the point where perpendicular from P meets the horizontal line CKD such that PD = 1/2 LD :

What is the ratio between the distances covered by vehicles A and B in the time interval OL?

a velocity-time graph: vehicle A is a straight line OKP rising from the origin O; vehicle B is a horizontal line CKD; K is where they cross; P is on line A above D; the perpendicular from P meets the time axis at L and the horizontal line at D
a1 : 2
b2 : 3
c3 : 4
d1 : 1

AnswerUPSC’s answer is (c).

On a velocity-time graph, distance covered is the area under the line. Vehicle B moves at constant velocity LD, so its distance in time OL is the rectangle OL × LD. Vehicle A's line runs from the origin to P, so its distance is the triangle OLP, area ½ × OL × LP. Since PD is half of LD, LP = LD + ½LD = 1.5 LD. So A covers ½ × OL × 1.5 LD = 0.75 × OL × LD. Ratio A : B = 0.75 : 1 = 3 : 4.

Why each option is right or wrong
  • (a) 1 : 2 would need A's final velocity to equal B's velocity.
  • (b) 2 : 3 does not follow; A's triangle is 0.75 of B's rectangle.
  • (c) Correct: triangle area 0.75 × OL × LD against rectangle OL × LD.
  • (d) 1 : 1 would need A's final velocity to be twice B's.

Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.

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Source: UPSC, Civil Services (Preliminary) Examination 2018, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.