2014 CSAT, all questions
UPSC Prelims 2014 · CSAT
Question 68: Basic Numeracy
Location of B is north of A and location of C is east of A. The distances AB and AC are 5 km and 12 km respectively. The shortest distance (in km) between the locations B and C is
a60
b13
c17
d7
AnswerUPSC’s answer is (b).
B is north of A and C is east of A, so AB and AC are at right angles and ABC is a right triangle with the right angle at A. By Pythagoras, BC squared = 5 squared + 12 squared = 25 + 144 = 169, so BC = 13 km. The 5-12-13 triple is worth remembering for such questions.
Why each option is right or wrong
- (a) 60 is 5 x 12, the product, not a distance.
- (b) Right. The square root of 25 + 144 is 13.
- (c) 17 is 5 + 12, the distance along both legs, not the direct distance.
- (d) 7 is the difference of the legs, not the hypotenuse.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2014, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.