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SSC :: SSC CHSL Quantitative Aptitude

SSC CHSL » Quantitative Aptitude » General Questions - Discussion

1 .


A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is

[ A ]    4.4 km/h
[ B ]    4.25 km/h right
[ C ]    4.25 km/h right
[ D ]    5.15 km/h
Answer : Option B
Explanation :
Speed of A = 5 km/h = 5 $\times$ $5 \over 18$ = $25 \over 18$ m/s

Time taken by A to cover 100 m = $Distance \over Speed$ = $100 \over {25 \over 18}$ = 72 sec

It is given that, A gives B a start of 8 m and still beats him by 8 seconds

B takes (72 + 8) = 80 sec to cover (100 - 8) = 92m

Speed of B = $Distance \over Time$ = $92 \over 80$ m/s

= $92 \over 80$ $\times$ $18 \over 5$ = $23 \times 18 \over 100 $ = $414 \over 100 $ = 4.14 km/h
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