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Delhi University MCA Previous Year Questions (PYQs)
Delhi University MCA Time And Distance PYQ
Delhi University MCA PYQ
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ
In a race of 1 Kilometer, athlete A beats athlete B by 50 meters, in another race of 500 meters, athlete B beats athlete C by 50 meters. By how many meters (Approximately) can athlete A beat athlete C in a race of 400 meters?
(a) 50 meters
(b) 58 meters
(c) 62 meters
(d) 72 meters
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ
Solution
$\dfrac{v_A}{v_B}=\dfrac{1000}{950}=\dfrac{20}{19}$, $\dfrac{v_B}{v_C}=\dfrac{500}{450}=\dfrac{10}{9}$
$\Rightarrow \dfrac{v_A}{v_C}=\dfrac{200}{171}$
When A runs $400$ m, C runs $400\times\dfrac{171}{200}=342$ m.
Difference $=400-342=58$ m.
Answer: 58 meters ✅
Delhi University MCA PYQ
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ
An
athlete has to cover a distance of 6 Kms in 90 Minutes. He covers two-third of
the distance in two-thirds of the total time. To cover the remaining distance
in the remaining time, his speed should be____ Km/Hr. ?
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ
Solution
Total distance = 6 km, Total time = 90 min = 1.5 hr
Distance covered = $\frac{2}{3}\times6=4$ km, Time taken = $\frac{2}{3}\times90=60$ min = 1 hr
Speed in first part = $\dfrac{4}{1}=4$ km/hr
Remaining distance = 2 km, Remaining time = 0.5 hr
Required speed = $\dfrac{2}{0.5}=4$ km/hr
Answer: $\boxed{4\text{ km/hr}}$ ✅
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