Question : A person goes from P to Q at a speed of 20 km/hr. Then he goes from Q to R at a speed of $q$ km/hr. Finally, the person goes from R to S at a speed of $r$ km/hr. The distances from P to Q, Q to R and R to S are equal. If the average speed from P to R is $\frac{280}{11}$km/hr, and the average speed from Q to S is $\frac{112}{3}$ km/hr, then what is the value of $r$?
Option 1: 40
Option 2: 37.5
Option 3: 42.5
Option 4: 45
Correct Answer: 40
Solution :
The distances from P to Q, Q to R and R to S are equal.
Let the distance from P to Q = distance from Q to R = distance from R to S = $y$
Speed of the person from P to Q = 20 km/hr
Speed of the person from Q to R = $q$ km/hr
Average speed from P to R = $\frac{280}{11}$ km/hr
$\frac{y +y}{\frac{y}{20} + \frac{y}{q}} = \frac{280}{11}$
$⇒\frac{2y}{\frac{qy+20y}{20q}} = \frac{280}{11}$
$⇒\frac{2y × 20q}{y(q+20)} = \frac{280}{11}$
$⇒\frac{q}{q+20} = \frac{70}{11}$
$⇒7q + 140 = 11q$
$⇒4q = 140$
$\therefore q = 35$ km/hr
Speed of the person from Q to R = $q$ = 35 km/hr
Speed of the person from R to S = $r$ km/hr
Average speed from P to R = $\frac{112}{3}$ km/hr
So, $\frac{y +y}{\frac{y}{35} + \frac{y}{r}} = \frac{112}{3}$
$⇒\frac{2y}{\frac{ry+35y}{35r}} =\frac{112}{3}$
$⇒\frac{2y × 35r}{y(r+35)} = \frac{112}{3}$
$⇒\frac{5r}{r+35} = \frac{8}{3}$
$⇒15r = 8r + 280$
$⇒15r-8r = 280$
$⇒7r = 280$
$⇒r = 40$ km/hr
Hence, the correct answer is 40.
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