Problems on Time and Distance - Solved Examples (Set 1)

Therefore,

time needed for riding one way = time needed for waking one way - $2$ hours

Given that time taken in walking one way and riding back $=5$ hours $45$ min

Hence, the time he would take to walk both ways

$=5$ hours $45$ min + $2$ hours

$=7$ hours $45$ min

Solution 2Let the distance be $x$ km. Then,

Time taken to walk $x$ km + Time taken to ride $x$ km

$=5$ hour $45$ min $=5\dfrac{45}{60}$ hour

$=5\dfrac{3}{4}$ hour $=\dfrac{23}{4}$ hour $\cdots(1)$

Time taken to ride $2x$ km

$=5$ hour $45$ min $-2$ hour

$=3$ hour $45$ min $=3\dfrac{45}{60}$ hour

$=3\dfrac{3}{4}$ hour $=\dfrac{15}{4}$ hour $\cdots(2)$

Solving $(1)$ and $(2)$

$(1)×2 \implies $

Time taken to walk $2x$ km + Time taken to ride $2x$ km $=\dfrac{23}{2}$ hour $\cdots(3)$

$(3)-(2)\implies$

Time taken to walk $2x$ km

$=\left(\dfrac{23}{2}-\dfrac{15}{4}\right)=\left(\dfrac{46}{4}-\dfrac{15}{4}\right)$

$=\dfrac{31}{4}=7\dfrac{3}{4}$ hours $=7$ hours $45$ minutes

Time $=5$ minutes $=\dfrac{1}{12}$ hour

$\text{Speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{0.6}{\left(\dfrac{1}{12}\right)}$ $=7.2\text{ km/hr}$

Solution 2Distance $=600$ metre

Time $=5$ minutes $=5×60$ seconds $=300$ seconds

$\text{Speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{600}{300}=2\text{ m/s}\\=2×\dfrac{18}{5}\text{ km/hr}=\dfrac{36}{5}\text{ km/hr}$ $=7.2\text{ km/hr}$

Speed of the bus including stoppages $=45$ kmph

Loss in speed when including stoppages $=54-45=9\text{ kmph}$

$\Rightarrow$ In $1$ hour, bus covers $9$ km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover $9$ km

$=\dfrac{\text{distance}}{\text{speed}}= \dfrac{9}{54}\text{ hour}=\dfrac{1}{6}\text{ hour }$ $=\dfrac{60}{6}\text{ min}=10\text{ min}$

Average Speed $=\dfrac{2×21×24}{21+24}=22.4\text{ km/hr}$

Total distance $=22.4×10=224\text{ km}$

Solution 2distance = speed × time

Let time taken to travel the first half $=x$ hr

Then, time taken to travel the second half $=(10-x)$ hr

Distance covered in the first half $=21x$

Distance covered in the second half $=24(10-x)$

But distance covered in the first half = Distance covered in the second half

$\Rightarrow 21x=24(10-x)\\\Rightarrow 21x=240-24x\\\Rightarrow 45x=240\\\Rightarrow 9x=48\\\Rightarrow 3x=16\\\Rightarrow x=\dfrac{16}{3}$Hence, distance covered in the first half

$=21x=21×\dfrac{16}{3}=7×16=112\text{ km}$

Total distance $=2×112=224\text{ km}$

$=1$ hr $+\dfrac{40}{60}$ hr $+\dfrac{48}{3600}$ hr

$=1+\dfrac{2}{3}+\dfrac{1}{75}=\dfrac{126}{75}\text{hr}$

distance $=42$ km

$\text{speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{42}{\left(\dfrac{126}{75}\right)}$ $=\dfrac{42×75}{126}=25\text{ km/hr}$

$\Rightarrow \dfrac{5}{7}$ of the actual speed $=25$

$\Rightarrow$ Actual speed $=25×\dfrac{7}{5}=35\text{ km/hr}$

1. 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. $11$ hrs | B. $8$ hrs $45$ min |

C. $7$ hrs $45$ min | D. $9$ hrs $20$ min |

| Discuss |

answer with explanation

Answer: Option C

Explanation:

Solution 1Given that time taken for riding both ways will be $2$ hours lesser than the time needed for waking one way and riding back.Therefore,

time needed for riding one way = time needed for waking one way - $2$ hours

Given that time taken in walking one way and riding back $=5$ hours $45$ min

Hence, the time he would take to walk both ways

$=5$ hours $45$ min + $2$ hours

$=7$ hours $45$ min

Solution 2Let the distance be $x$ km. Then,

Time taken to walk $x$ km + Time taken to ride $x$ km

$=5$ hour $45$ min $=5\dfrac{45}{60}$ hour

$=5\dfrac{3}{4}$ hour $=\dfrac{23}{4}$ hour $\cdots(1)$

Time taken to ride $2x$ km

$=5$ hour $45$ min $-2$ hour

$=3$ hour $45$ min $=3\dfrac{45}{60}$ hour

$=3\dfrac{3}{4}$ hour $=\dfrac{15}{4}$ hour $\cdots(2)$

Solving $(1)$ and $(2)$

$(1)×2 \implies $

Time taken to walk $2x$ km + Time taken to ride $2x$ km $=\dfrac{23}{2}$ hour $\cdots(3)$

$(3)-(2)\implies$

Time taken to walk $2x$ km

$=\left(\dfrac{23}{2}-\dfrac{15}{4}\right)=\left(\dfrac{46}{4}-\dfrac{15}{4}\right)$

$=\dfrac{31}{4}=7\dfrac{3}{4}$ hours $=7$ hours $45$ minutes

2. A person crosses a $600$ metre long street in $5$ minutes. What is his speed in km per hour? | |

A. $8.2$ | B. $4.2$ |

C. $6.1$ | D. $7.2$ |

| Discuss |

answer with explanation

Answer: Option D

Explanation:

Solution 1Distance $=600$ metre $=0.6$ kmTime $=5$ minutes $=\dfrac{1}{12}$ hour

$\text{Speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{0.6}{\left(\dfrac{1}{12}\right)}$ $=7.2\text{ km/hr}$

Solution 2Distance $=600$ metre

Time $=5$ minutes $=5×60$ seconds $=300$ seconds

$\text{Speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{600}{300}=2\text{ m/s}\\=2×\dfrac{18}{5}\text{ km/hr}=\dfrac{36}{5}\text{ km/hr}$ $=7.2\text{ km/hr}$

3. Excluding stoppages, the speed of a bus is $54$ kmph and including stoppages, it is $45$ kmph. For how many minutes does the bus stop per hour? | |

A. $12$ | B. $11$ |

C. $10$ | D. $9$ |

| Discuss |

answer with explanation

Answer: Option C

Explanation:

Speed of the bus excluding stoppages $=54$ kmphSpeed of the bus including stoppages $=45$ kmph

Loss in speed when including stoppages $=54-45=9\text{ kmph}$

$\Rightarrow$ In $1$ hour, bus covers $9$ km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover $9$ km

$=\dfrac{\text{distance}}{\text{speed}}= \dfrac{9}{54}\text{ hour}=\dfrac{1}{6}\text{ hour }$ $=\dfrac{60}{6}\text{ min}=10\text{ min}$

4. A man complete a journey in $10$ hours. He travels first half of the journey at the rate of $21$ km/hr and second half at the rate of $24$ km/hr. Find the total journey in km. | |

A. $121$ km | B. $242$ km |

C. $224$ km | D. $112$ km |

| Discuss |

answer with explanation

Answer: Option C

Explanation:

Solution 1reference: formula 4Average Speed $=\dfrac{2×21×24}{21+24}=22.4\text{ km/hr}$

Total distance $=22.4×10=224\text{ km}$

Solution 2distance = speed × time

Let time taken to travel the first half $=x$ hr

Then, time taken to travel the second half $=(10-x)$ hr

Distance covered in the first half $=21x$

Distance covered in the second half $=24(10-x)$

But distance covered in the first half = Distance covered in the second half

$\Rightarrow 21x=24(10-x)\\\Rightarrow 21x=240-24x\\\Rightarrow 45x=240\\\Rightarrow 9x=48\\\Rightarrow 3x=16\\\Rightarrow x=\dfrac{16}{3}$Hence, distance covered in the first half

$=21x=21×\dfrac{16}{3}=7×16=112\text{ km}$

Total distance $=2×112=224\text{ km}$

5. A car traveling with $5/7$ of its actual speed covers $42$ km in $1$ hr $40$ min $48$ sec. What is the actual speed of the car? | |

A. $30$ km/hr | B. $35$ km/hr |

C. $25$ km/hr | D. $40$ km/hr |

| Discuss |

answer with explanation

Answer: Option B

Explanation:

time $=1$ hr $40$ min $48$ sec$=1$ hr $+\dfrac{40}{60}$ hr $+\dfrac{48}{3600}$ hr

$=1+\dfrac{2}{3}+\dfrac{1}{75}=\dfrac{126}{75}\text{hr}$

distance $=42$ km

$\text{speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{42}{\left(\dfrac{126}{75}\right)}$ $=\dfrac{42×75}{126}=25\text{ km/hr}$

$\Rightarrow \dfrac{5}{7}$ of the actual speed $=25$

$\Rightarrow$ Actual speed $=25×\dfrac{7}{5}=35\text{ km/hr}$

Niveda

2015-07-08 16:37:16

The speed of a bus increases by 4 km after every 2 hours. If the distance travelling in the first 2 hour was 70 km. what was the total distance travelled in 24 hours?

Megha

2015-07-15 05:45:39

Since the speed increases every 2 hrs, in 24 hrs the speed increases 12 times.

Arithmetic Progression

Sum of n terms=n/2*(2a+(n-1)d)

=12/2*(2*70+(11*4))

=1104

Arithmetic Progression

Sum of n terms=n/2*(2a+(n-1)d)

=12/2*(2*70+(11*4))

=1104

Vishal Jaiswal

2015-04-09 06:35:46

Give the HCF of two no.is 16.find the LCM if their product is 19712.

Dev

2015-04-09 18:40:46

Product of two numbers = Product of their HCF and LCM

19712 = 16 * LCM

LCM = 19712/16 = 1232

19712 = 16 * LCM

LCM = 19712/16 = 1232

varthini

2015-04-06 14:21:29

Walking at a rate of 4 km an hour, a man covers a distance in 3 hours. Running at a speed of 8 km, he will cover the same
distance in

a)5 hrs 45 min

b) 9 hrs 60 min

c)11 hrs 90 min

d) 1 hr 30 min

a)5 hrs 45 min

b) 9 hrs 60 min

c)11 hrs 90 min

d) 1 hr 30 min

Shubham gupta

2015-11-07 08:35:59

Total distance covers= 4×3=12km

At running @8km/hr the time taken

= 12/8 = 1hr 30 min

At running @8km/hr the time taken

= 12/8 = 1hr 30 min

Dev

2015-04-08 19:54:36

speed = 4 km/hr

time = 3hrs

distance = speed * time = 12 km

New speed = 8 km

required time = distance/speed = 12/8 = 3/2 = 1.5 hour = 1 hr 30 minute

or more easily

4 km/hr : 3 hrs

8 km/hr : 1.5 hrs (as speed is inversely proportional to time when distance is constant)

time = 3hrs

distance = speed * time = 12 km

New speed = 8 km

required time = distance/speed = 12/8 = 3/2 = 1.5 hour = 1 hr 30 minute

or more easily

4 km/hr : 3 hrs

8 km/hr : 1.5 hrs (as speed is inversely proportional to time when distance is constant)

junaid

2015-04-04 15:08:21

if a cyclist had gone 3km/hr faster , he would have taken 1 hour and 20 min less to ride 80 kms. what time did he take

Dev

2015-04-08 19:21:18

Let his normal speed = x km/hr

distance = 80 km

Normal time he takes = 80/x hr

If he had gone 3km/hr faster, time taken = 80/(x+3) hr

Savings of time = 1 hr 20 min = 1

80/x - 80/(x+3) = 4/3

1/x - 1/(x+3) = 1/60

60(x+3) - 60x = x(x+3)

180 = x^2 + 3x

x^2 + 3x - 180 = 0

(x+15)(x-12)=0

x = 12 (taking +ve value)

Required time = distance/speed = 80/12 = 20/3 hrs

distance = 80 km

Normal time he takes = 80/x hr

If he had gone 3km/hr faster, time taken = 80/(x+3) hr

Savings of time = 1 hr 20 min = 1

^{1}/_{3}hr = 4/3 hr80/x - 80/(x+3) = 4/3

1/x - 1/(x+3) = 1/60

60(x+3) - 60x = x(x+3)

180 = x^2 + 3x

x^2 + 3x - 180 = 0

(x+15)(x-12)=0

x = 12 (taking +ve value)

Required time = distance/speed = 80/12 = 20/3 hrs

Kaustubh kurlekar

2015-04-02 07:39:57

A man walking with a speed of 4 km/hr reach the office 5 min late. If he walk with 5 km/hr he will reach 2.5 min earlier. What is the distance between his home and office ? (question frm postal exam march 2015)

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