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18 Dec 2023, 22:38
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:09) correct
36%
(01:46)
wrong
based on 132
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On Saturday, Heena travelled from New York to Chicago in a vehicle with circular wheels of 25 inches in diameter. While returning on Sunday along the same route, she used another vehicle with circular wheels of 30 inches in diameter. If the vehicles travelled no other distance except for this round trip, what is the approximate percentage change in total number of revolutions made by the wheels from Saturday to Sunday?
(A) Decrease by 20%
(B) Decrease by 16.67%
(C) Increase by 16.67%
(D) Increase by 20%
(E) Decrease by 18%
(A) Decrease by 20%
(B) Decrease by 16.67%
(C) Increase by 16.67%
(D) Increase by 20%
(E) Decrease by 18%
19 Dec 2023, 05:36
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Distance travelled by a wheel = 2πr*n, where n = # of revolutions per unit
No. of revolutions on Saturday = n
No. of revolutions on Sunday = N
2πr*n = 2πR*N
25π*n = 30π*N
n=\(\frac{6N}{5}\)
% change = \(\frac{final-initial}{initial}\)*100
% change = \(\frac{N-\frac{6N}{5}}{\frac{6N}{5}}\)*100
% change = \(\frac{-1}{6}*100\) = -16.67% (B)
No. of revolutions on Saturday = n
No. of revolutions on Sunday = N
2πr*n = 2πR*N
25π*n = 30π*N
n=\(\frac{6N}{5}\)
% change = \(\frac{final-initial}{initial}\)*100
% change = \(\frac{N-\frac{6N}{5}}{\frac{6N}{5}}\)*100
% change = \(\frac{-1}{6}*100\) = -16.67% (B)
19 Dec 2023, 08:26
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Gprabhumir
Distance is fixed, but rotations will varry...
On 1st trip no of rotations will be 30 & on second trip no of rotations will be 25
So, net decrease will be 5/30*100 = 16.67%, Answer must be (B)
