Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time & Distance
Total Questions : 422
| Page 41 of 43 pages
Answer: Option C. -> 108 km/hr
$$\eqalign{
& {\text{Speed of train :}} \cr
& = \frac{{450 + 150}}{{20}} \cr
& = 30{\text{ m/s}} \cr
& {\text{Speed (in km/hr):}} \cr
& = 30 \times \frac{{18}}{5} \cr
& = 108{\text{ km/hr}} \cr} $$
$$\eqalign{
& {\text{Speed of train :}} \cr
& = \frac{{450 + 150}}{{20}} \cr
& = 30{\text{ m/s}} \cr
& {\text{Speed (in km/hr):}} \cr
& = 30 \times \frac{{18}}{5} \cr
& = 108{\text{ km/hr}} \cr} $$
Answer: Option B. -> $$\frac{7}{4}$$ km
Difference between his reaching time :
= (14 - 10) hrs
= 4 hrs
= 4 hrs → 6m + 6m
(late + before)
= 4 hrs → 12 minutes
= 1 unit = $$\frac{12}{4 × 60}$$ km
($$\because $$ 1 m = 60 seconds)
1 unit = $$\frac{1}{20}$$ km
Then, 35 units :
= 35 × $$\frac{1}{20}$$ km
= $$\frac{7}{4}$$ km
Then the distance between his house and school is = $$\frac{7}{4}$$ km
Difference between his reaching time :
= (14 - 10) hrs
= 4 hrs
= 4 hrs → 6m + 6m
(late + before)
= 4 hrs → 12 minutes
= 1 unit = $$\frac{12}{4 × 60}$$ km
($$\because $$ 1 m = 60 seconds)
1 unit = $$\frac{1}{20}$$ km
Then, 35 units :
= 35 × $$\frac{1}{20}$$ km
= $$\frac{7}{4}$$ km
Then the distance between his house and school is = $$\frac{7}{4}$$ km
Answer: Option B. -> 36 km/hr
Total distance travelled :
$$\eqalign{
& = \left( {30 \times \frac{{12}}{{60}} + 45 \times \frac{8}{{60{\text{ }}}}} \right){\text{ km}} \cr
& = {\text{12 km}} \cr} $$
Total time taken :
= (12 + 8) min
= 20 min
= $$\frac{1}{3}$$ hr
∴ Average speed :
= (12 × 3) km/hr
= 36 km/hr
Total distance travelled :
$$\eqalign{
& = \left( {30 \times \frac{{12}}{{60}} + 45 \times \frac{8}{{60{\text{ }}}}} \right){\text{ km}} \cr
& = {\text{12 km}} \cr} $$
Total time taken :
= (12 + 8) min
= 20 min
= $$\frac{1}{3}$$ hr
∴ Average speed :
= (12 × 3) km/hr
= 36 km/hr
Answer: Option B. -> 4.30 pm
Speed of Ashok = 39 km/ph
Speed of Rahul = 47 km/ph
Distance between place A and place B = 637 km
Distance covered by Ashok (from 8 am to 10 am) in 2 hours = 2 × 39 = 78 km
∴ Remaining distance = 637 - 78 = 559
Relative speed = 39 + 47 = 86 km/ph
∴ Time taken to travel 559 km = $$\frac{559}{86}$$ = 6.5 hours
So, they meet at = (10 am + 6.5 hrs) = 4.30 pm
Speed of Ashok = 39 km/ph
Speed of Rahul = 47 km/ph
Distance between place A and place B = 637 km
Distance covered by Ashok (from 8 am to 10 am) in 2 hours = 2 × 39 = 78 km
∴ Remaining distance = 637 - 78 = 559
Relative speed = 39 + 47 = 86 km/ph
∴ Time taken to travel 559 km = $$\frac{559}{86}$$ = 6.5 hours
So, they meet at = (10 am + 6.5 hrs) = 4.30 pm
Answer: Option A. -> 75 km
Relative speed = (200 + 200) km/hr = 400 km/hr
Distance covered in 1 hr 30 min
$$\eqalign{
& = \left( {400 \times \frac{1}{2} + 200 \times \frac{1}{2} + 100 \times \frac{1}{2}} \right){\text{km}} \cr
& = \left( {200 + 100 + 50} \right){\text{km}} \cr
& = 350{\text{ km}} \cr} $$
[$$\because $$ Speed reduces by half every half an hour]
Hence, distance between the trains :
= (425 - 350) km
= 75 km
Relative speed = (200 + 200) km/hr = 400 km/hr
Distance covered in 1 hr 30 min
$$\eqalign{
& = \left( {400 \times \frac{1}{2} + 200 \times \frac{1}{2} + 100 \times \frac{1}{2}} \right){\text{km}} \cr
& = \left( {200 + 100 + 50} \right){\text{km}} \cr
& = 350{\text{ km}} \cr} $$
[$$\because $$ Speed reduces by half every half an hour]
Hence, distance between the trains :
= (425 - 350) km
= 75 km
Question 406. A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities ?
Answer: Option A. -> 420 km
Let distance = x km and usual rate y kmph
Then,
$$\eqalign{
& \Rightarrow \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr
& \Rightarrow y\left( {y + 10} \right) = 10x.....(i) \cr} $$
And,
$$\eqalign{
& \Rightarrow \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr
& \Rightarrow y\left( {y + 20} \right) = \frac{{80x}}{7}.....(ii) \cr} $$
On dividing (i) by (ii) we get :
y = 60
Substituting y = 60 in (i), we get: x = 420 km
Let distance = x km and usual rate y kmph
Then,
$$\eqalign{
& \Rightarrow \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr
& \Rightarrow y\left( {y + 10} \right) = 10x.....(i) \cr} $$
And,
$$\eqalign{
& \Rightarrow \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr
& \Rightarrow y\left( {y + 20} \right) = \frac{{80x}}{7}.....(ii) \cr} $$
On dividing (i) by (ii) we get :
y = 60
Substituting y = 60 in (i), we get: x = 420 km
Question 407. Amit travelled black to home in a car, after visiting his friend in a distant village. When he stated at his friend's house the car had exactly 18 litres of petrol in it. He travelled along at a steady 40 kilometres per hour and managed a 10 kilometres per litre of petrol. As the car was old, the fuel tank lost fuel at the rate of half a litre per hour. Amit was lucky as his car stopped just in front of his home because it had rum out of fuel and he only just made it. How far was it from his friend's home to Amit's home ?
Answer: Option D. -> None of these
Quantity of petrol consumed in 1 hour:
$$\eqalign{
& = \left( {\frac{{40}}{{10}} + \frac{1}{2}} \right){\text{ litres}} \cr
& = 4\frac{1}{2}{\text{ litres}} \cr} $$
Time for which the fuel lasted :
$$\eqalign{
& = \left[ {\frac{{18}}{{\left( {4\frac{1}{2}} \right)}}} \right]{\text{ hrs}} \cr
& = \left( {18 \times \frac{2}{9}} \right){\text{ hrs}} \cr
& = 4{\text{ hrs}} \cr} $$
∴ Required distance = (40 × 4) km = 160 km
Quantity of petrol consumed in 1 hour:
$$\eqalign{
& = \left( {\frac{{40}}{{10}} + \frac{1}{2}} \right){\text{ litres}} \cr
& = 4\frac{1}{2}{\text{ litres}} \cr} $$
Time for which the fuel lasted :
$$\eqalign{
& = \left[ {\frac{{18}}{{\left( {4\frac{1}{2}} \right)}}} \right]{\text{ hrs}} \cr
& = \left( {18 \times \frac{2}{9}} \right){\text{ hrs}} \cr
& = 4{\text{ hrs}} \cr} $$
∴ Required distance = (40 × 4) km = 160 km
Answer: Option A. -> $${2\frac{1}{2}}$$ hours
Let the original planned duration of the flight be x hours
Then,
$$\eqalign{
& \Leftrightarrow \frac{{6000}}{x} - \frac{{600}}{{\left( {x + \frac{1}{2}} \right)}} = 400 \cr
& \Leftrightarrow \frac{{6000}}{x} - \frac{{12000}}{{\left( {2x + 1} \right)}} = 400 \cr
& \Leftrightarrow \frac{{15}}{x} - \frac{{30}}{{\left( {2x + 1} \right)}} = 1 \cr
& \Leftrightarrow 2{x^2} + x - 15 = 0 \cr
& \Leftrightarrow \left( {x + 3} \right)\left( {2x - 5} \right) = 0 \cr
& \Leftrightarrow x = \frac{5}{2} \cr
& \Leftrightarrow x = 2\frac{1}{2} \cr} $$
Let the original planned duration of the flight be x hours
Then,
$$\eqalign{
& \Leftrightarrow \frac{{6000}}{x} - \frac{{600}}{{\left( {x + \frac{1}{2}} \right)}} = 400 \cr
& \Leftrightarrow \frac{{6000}}{x} - \frac{{12000}}{{\left( {2x + 1} \right)}} = 400 \cr
& \Leftrightarrow \frac{{15}}{x} - \frac{{30}}{{\left( {2x + 1} \right)}} = 1 \cr
& \Leftrightarrow 2{x^2} + x - 15 = 0 \cr
& \Leftrightarrow \left( {x + 3} \right)\left( {2x - 5} \right) = 0 \cr
& \Leftrightarrow x = \frac{5}{2} \cr
& \Leftrightarrow x = 2\frac{1}{2} \cr} $$
Answer: Option B. -> 1 hr 12 min
New speed = $$\frac{6}{7}$$ of usual speed
New time = $$\frac{7}{6}$$ of usual time
$$\therefore \left( {\frac{7}{6}{\text{ of usual time}}} \right) - $$ $$\left( {{\text{usual time}}} \right)$$ $$ = \frac{1}{5}{\text{ hr}}$$
$$\eqalign{
& \Rightarrow \frac{1}{6}{\text{ of usual time = }}\frac{1}{5}{\text{ hr}} \cr
& \Rightarrow {\text{usual time = }}\frac{6}{5}{\text{ hr}} \cr
& \Rightarrow {\text{usual time = 1 hr 12 min}} \cr} $$
New speed = $$\frac{6}{7}$$ of usual speed
New time = $$\frac{7}{6}$$ of usual time
$$\therefore \left( {\frac{7}{6}{\text{ of usual time}}} \right) - $$ $$\left( {{\text{usual time}}} \right)$$ $$ = \frac{1}{5}{\text{ hr}}$$
$$\eqalign{
& \Rightarrow \frac{1}{6}{\text{ of usual time = }}\frac{1}{5}{\text{ hr}} \cr
& \Rightarrow {\text{usual time = }}\frac{6}{5}{\text{ hr}} \cr
& \Rightarrow {\text{usual time = 1 hr 12 min}} \cr} $$
Answer: Option B. -> 130 km
Time taken = 3 hrs 15 min = $$3\frac{1}{4}$$ hrs = $$\frac{13}{4}$$ hrs
∴ Required distance :
$$\eqalign{
& = \left( {40 \times \frac{{13}}{4}} \right){\text{ km}} \cr
& = 130{\text{ km}} \cr} $$
Time taken = 3 hrs 15 min = $$3\frac{1}{4}$$ hrs = $$\frac{13}{4}$$ hrs
∴ Required distance :
$$\eqalign{
& = \left( {40 \times \frac{{13}}{4}} \right){\text{ km}} \cr
& = 130{\text{ km}} \cr} $$
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