1 | A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km? A. 36 B. 38 C. 40 D. 42 Answer : C. 40 Explanation: Let the distance be x km , Time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes \(\begin{align} &\Rightarrow \dfrac{x}{v} - \dfrac{x}{v + 3} = \dfrac{40}{60}\\ &\Rightarrow x \left[\dfrac{1}{v} - \dfrac{1}{v + 3}\right] = \dfrac{2}{3}\\ &\Rightarrow x \left[\dfrac{v + 3 - v}{v\left(v + 3\right)}\right] = \dfrac{2}{3}\\ &\Rightarrow 2v(v+3) = 9x\text{................(Equation1)}\\\\\\\\ \end{align}\) Time taken when moving 2 kmph slower - Time taken when moving at normal speed = 40 minutes \(\begin{align} &\Rightarrow \dfrac{x}{v-2} - \dfrac{x}{v} = \dfrac{40}{60}\\ &\Rightarrow x \left[\dfrac{1}{v-2} - \dfrac{1}{v}\right] = \dfrac{2}{3}\\ &\Rightarrow x \left[\dfrac{v - v + 2}{v\left(v - 2\right)}\right] = \dfrac{2}{3}\\ &\Rightarrow x \left[\dfrac{2}{v\left(v - 2\right)}\right] = \dfrac{2}{3}\\ &\Rightarrow x \left[\dfrac{1}{v\left(v - 2\right)}\right] = \dfrac{1}{3}\\ &\Rightarrow v(v-2) = 3x\text{................(Equation2)}\\\\\\\\ &\dfrac{\text{Equation1}}{\text{Equation2}}\Rightarrow \dfrac{2(v+3)}{(v-2)}= 3\\ &\Rightarrow 2v + 6 = 3v -6 \\ &\Rightarrow v = 12\\\\\\\\ &\text{Substituting this value of v in Equation 1}\Rightarrow 2 \times 12 \times 15 = 9x\\ &=> x = \dfrac{2 \times 12 \times 15}{9} = \dfrac{2 \times 4 \times 15}{3} = 2 \times 4 \times 5 = 40\\ &\text{Hence distance = 40 km}\end{align}\) |
2 | Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart? A. 17 Hrs B. 14 Hrs C. 12 Hrs D. 19 Hrs Answer : A. 17 Hrs Explanation: Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction) distance = 8.5 km \(\text{time = }\dfrac{\text{distance}}{\text{speed}} = \dfrac{8.5}{.5} = 17\text{ hr}\) |
3 | A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour? A. 70.24 km/hr B. 74. 24 km/hr C. 71.11 km/hr D. 72.21 km/hr Answer : C. 71.11 km/hr Explanation: \(\boxed{\text{If a car covers a certain distance at x kmph and an equal distance at y kmph,}\\ \text{the average speed of the whole journey = }\dfrac{2xy}{x+y}\text{ kmph.}}\\ \begin{align} &\text{By using the same formula, we can find out the average speed quickly}\\\ &\text{average speed = }\dfrac{2 \times 64 \times 80}{64 + 80} = \dfrac{2 \times 64 \times 80}{144} = \dfrac{2 \times 32 \times 40}{36}\\\\ &= \dfrac{2 \times 32 \times 10}{9} = \dfrac{64 \times 10}{9} = 71.11\text{ kmph} \end{align}\) |
4 | Walking \({6 \over7}^{th}\) of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance? A. 1 hr 42 minuts B. 1 hr C. 2 hr D. 1 hr 12 minuts Answer : D. 1 hr 12 minuts Explanation: New speed = \(6\over7\) of usual speed Speed and time are inversely proportional. Hence new time = \(7\over6\) of usual time Hence, \(7\over6\) of usual time - usual time = 12 minutes => \(1\over6\) of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes |
5 | A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is A. 7 hrs 45 min B. 11 hrs C. 9 hts 20 min D. 8 hrs 45 min Answer : A. 7 hrs 45 min Explanation: \(\begin{align} &\text{Let the distance be x km. Then}\\\\ &\text{(Time taken to walk x km) + (Time taken to ride x km) = 5 hour 45 min }\\ &= 5\dfrac{45}{60}\text{ hour = }5\dfrac{3}{4}\text{ hour = }\dfrac{23}{4}\text{ hour .......(Equation1)}\\\\\\\\ &\text{(Time taken to ride 2x km) = 5 hour 45 min - 2 = 3 hour 45 min}\\ &= 3\dfrac{45}{60}\text{ hour = }3\dfrac{3}{4}\text{ hour = }\dfrac{15}{4}\text{ hour .......(Equation2)}\\\\\\\\ &\text{Solving equations 1 and 2}\\ &\text{(Equation 1 )}\times 2 \Rightarrow \text{(Time taken to walk 2x km)+ (Time taken to ride 2x km) = }\\ &\dfrac{23}{2}\text{ hour .......(Equation3)}\\\\\\\\ &\text{(Equation 3 - Equation 2)}\Rightarrow\text{ Time taken to walk 2x km} = \dfrac{23}{2} - \dfrac{15}{4} \\ &= \dfrac{46}{4} - \dfrac{15}{4} = \dfrac{31}{4}\text{ hours = }7\dfrac{3}{4}\text{ hours = }7\text{ hours 45 minutes }\\ \end{align}\) |
6 | A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, what should be his speed in km/hr? A. 8 km/hr B. 10 km/hr C. 12 km/hr D. 14 km/hr Answer : C. 12 km/hr Explanation: The person needs to cover 6 km in 45 minutes Given that he covers one-half of the distance in two-thirds of the total time => He covers half of 6 km in two-thirds of 45 minutes => He covers 3 km in 30 minutes Hence, now he need to cover the remaining 3 km in the remaining 15 minutes Distance = 3 km Time = 15 minutes = 1/4 hour \(\text{Required Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{3}{\left(\dfrac{1}{4}\right)} = 12\text{ km/hr}\) |
7 | A truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes. What is the ratio of their speed? A. 2 : 1 B. 1 : 2 C. 4 : 3 D. 3 : 4 Answer : D. 3 : 4 Explanation: \(\begin{align}&\text{Speed of the truck = }\dfrac{\text{Distance}}{\text{Time}} = \dfrac{550}{1} = 550\text{ meters/minute}\\\\ &\text{Speed of the train = }\dfrac{\text{Distance}}{\text{Time}} = \dfrac{33}{45}\text{ km/minute = }\dfrac{33000}{45}\text{ meters/minute}\\\\ &\dfrac{\text{Speed of the truck}}{\text{Speed of the train}} = \dfrac{550}{\left(\dfrac{33000}{45}\right)} = \dfrac{550 \times 45}{33000} = \dfrac{55 \times 45}{3300}\\\\ &= \dfrac{11\times 45}{660} = \dfrac{11\times 9}{132} = \dfrac{9}{12} = \dfrac{3}{4}\end{align}\) Hence, Speed of the truck : Speed of the train = 3 : 4 |
8 | A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in metres)? A. 1250 B. 1280 C. 1320 D. 1340 Answer : A. 1250 Explanation: \(\begin{align}&\text{Speed = 5 km/hr}\\\\ &\text{Time = 15 minutes = }\dfrac{1}{4}\text{ hour}\\\\ &\text{Length of the bridge = Distance Travelled by the man = Speed } \times \text{ Time = }5 \times \dfrac{1}{4}\text{ km}\\\\ &= 5 \times \dfrac{1}{4} \times 1000 \text{ metre} = 1250\text{ metre}\end{align}\) |