Speed Time And Distance - Study Mode

[#251] A starts 3 Minutes after B for a place 4.5 km distant. B, on reaching his destination, immediately returns and after walking a km meets A. If A can walk 1 Km in 18 minutes,then what is B's speed?
Correct Answer

(D) 5 km/h

Explanation

Solution: P__________3.5Km__________M__1km____Q
Speed of A = 1 km in 18 min = $$frac{{1000}}{{18}}$$xa0 = 55.55 m/min. When A travels 3.5 km. B already has been traveled (4.5 + 1) = 5.5 km. A will take time to travel 3.5 km = $$frac{{3500}}{{55.55}}$$ xa0= 63 min. So, B will take 66 min to travel 5.5 km (As B has started 3 min before of A) Thus, speed of B = $$frac{{5500}}{{66}}$$xa0 = 83.33 m/min = 5 km/h

[#252] A car travels 50% faster than a bike. Both start at the same time from A to B. The car reaches 25 minutes earlier than the bike. If the distance from A to B is 100 km, find the speed of the bike.
Correct Answer

(C) 80 kmph

Explanation

Solution: P __________100km__________Q Let car takes time T hours to reach destination. So, Bike will take $$left( {{ ext{T}} + frac{{25}}{{60}}}
ight)$$
Let speed of the bike = S kmph Speed of Car = S + 50% of S = $$frac{{3{ ext{S}}}}{2}$$ kmph For the both the case distance is constant. And when distance remain constant then time is inversely proportional to speed (As ST + D) So, $$eqalign{
& frac{{ {frac{{3S}}{2}} }}{{left( S
ight)}} = frac{{ {T + {frac{5}{{12}}} } }}{T} cr
& 3T = 2T + frac{{10}}{{12}} cr
& T = frac{{10}}{{12}}{ ext{hours}} cr
& { ext{Speed}},{ ext{of}},{ ext{the}},{ ext{car}} cr
& frac{{3S}}{2} = frac{{100}}{{ {frac{{10}}{{12}}} }} cr
& frac{{3S}}{2} = 120 cr
& S = 80,kmph cr
& { ext{Speed}},{ ext{of}},{ ext{the}},{ ext{bike}} = 80,kmph cr} $$

[#253] A and B start moving from places X to Y and Y to X, respectively, at the same into on the same day. After crossing each other, A and B take $$5frac{4}{9}$$xa0hours and 9 hours, respectively, to each their respective destinations. If the speed of A is 33 km/h, then the speed (in km/h) of B is:
Correct Answer

(C) $$25frac{2}{3}$$

Explanation

Solution: $$eqalign{
& frac{{{S_1}}}{{{S_2}}} = sqrt {frac{{{t_2}}}{{{t_1}}}} cr
& frac{{33}}{{{S_2}}} = sqrt {frac{{9 imes 9}}{{49}}} cr
& frac{{33}}{{{S_2}}} = frac{9}{7} cr
& {S_2} = 25frac{2}{3}{ ext{ km/h}} cr} $$

[#254] A train of length 287 m, running at 80 km/h, crosses another train moving in the opposite direction at 37 km/h in 18 second. What is the length of the other train?
Correct Answer

(B) 298 m

Explanation

Solution: $$eqalign{
& frac{{left( {287 + x}
ight)}}{{left( {80 + 37}
ight) imes frac{5}{{18}}}} = 18 cr
& left( {287 + x}
ight) = 18left( {80 + 37}
ight) imes frac{5}{{18}} cr
& 287 + x = 117 imes 5 cr
& 287 + x = 585 cr
& x = 585 - 287 cr
& x = 298{ ext{ m}} cr} $$

[#255] A train x running at 74 km/h crosses another train y running at 52 km/h in the opposite direction in 12 seconds. If the length of y is two-thirds that of x, then what is the length of y (in m)?
Correct Answer

(C) 168

Explanation

Solution: According to question, $$eqalign{
& frac{{x + y}}{{left( {74 + 52}
ight)frac{{{ ext{km}}}}{{{ ext{hr}}}}}} = 12 cr
& x + y = 12 imes 126 imes frac{5}{{18}} cr
& x + y = 420{ ext{ m }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}{ ext{. }}left( { ext{i}}
ight) cr} $$ y = 3a + 2a = 5a x = 3a So, from equation (i) 5a = 420 a = 84 Hence length of train y = 2 × 84 = 168