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
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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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Option 2: 8.25 km/hr
Option 3: 7.14 km/hr
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Option 2: 38 km/hr
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Option 1: 30
Option 2: 32
Option 3: 31.25
Option 4: 27.5
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Option 2: $\frac{630}{11}$ km/hr
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