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19 Sep 2024, 01:30
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Dinesh and Rahim start their journey from P and Q respectively at the same time. They meet at a point R, reverse their directions and interchange their speeds. Dinesh took t hours more to reach back P than Rahim took to reach back Q. By what percentage is the initial speed of Dinesh greater than the speed of Rahim if it is known that the ratio of the time taken by Rahim to reach R to the time t is 3:8?
A. 500
B. 350
C. 300
D. 200
E. 150
A. 500
B. 350
C. 300
D. 200
E. 150
19 Sep 2024, 04:54
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Bunuel
Assuming speeds as D and R Respectively for Dinesh and Rahim
D ----> (P) ---------------------------(R)---------------------------- (Q) <---- R
Given that the ratio of the time taken by Rahim to reach R to the time t is 3:8
means the Ratio of time taken by them to meet at R (assume m) to the t, means \(\frac{m}{t} = \frac{3}{8}\)
Meeting time m we can calculate with respect to their distance traveled till they reach R,
lets assume Dinesh traveled P distance then m = P/D
P = m*D
and, similarly Rahim traveled Q distance then m = Q/R
Q = m*R
After meeting at point they reverse their directions and interchange their speeds. Dinesh speed is now R and Rahim speed is D
(P) -------------------------- R <------- (R)------> D ----------------------------- (Q)
and, Dinesh took t hours more to reach back P than Rahim took to reach back Q.
lets assume Rahim took x time to reach Q Therefore dinesh took x+t time to reach P
now with new speed of Dinesh, the same distance P = R*(x+t)
P = mD = R(x+t)
Similarly, for rahim Q = mR = Dx
now from here get the value for x,
\(x=\frac{mR}{D}\)
Substitute this in " mD = R(x+t) "\(mD = R(\frac{mR}{D} + t)\)
plus we know the ratio of \(\frac{m}{t} = \frac{3}{8}\)So, taking values for m and t as 3k and 8k respectively and putting them in equation, we will get equation in terms of D and R
\(3D^2 = 3R^2 + 8DR\)
now after solving this you will get value of D = 3R
which means initial speed of Dinesh greater than 200% the speed of Rahim.
Answer is D.
30 Sep 2025, 04:54
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