Contents
Chapter 3 Time and Work and Time and Distance
3.1 Time and Work
3.1.1 Important Facts and Formulae
1. If A can do a piece of work in n days, then A’s 1 day’s work \(=\dfrac {1}{n}\)
2. If A’s 1 day’s work \(=\dfrac {1}{n}\), then A can finish the work in n days.
3. A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1.
Ratio of times taken by A and B to finish a work = 1 : 3.
3.1.2 Solved Examples
1. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job? (IGNOU, 2003)
Solution :
A’s 1 hour’s work = \(\dfrac {1}{8}\)
B’s 1 hour’s work = \(\dfrac {1}{10}\)
(A + B)’s 1 hour’s work = \(\dfrac {1}{8}+\dfrac {1}{10}=\dfrac {9}{40}\)
Both A and B will finish the work in \(\dfrac {40}{9}\) days.
2. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work? (Bank P.O. 2003)
Solution :
(A + B)’s 1 day’s work = \(\dfrac {1}{4}\). A’s 1 day’s work = \(\dfrac {1}{12}\)
B’s 1 day’s work =\(\dfrac {1}{4}-\dfrac {1}{12}=\dfrac {1}{6}\)
Hence, B alone can complete the work in 6 days.
3. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together 8 hours a day?
Solution :
A can complete the work in \((7\times \ 9)\) = 63 hours.
B can complete the work in \((6\times \ 7)\) = 42 hours.
A’s 1 hour’s work = \(\dfrac {1}{63}\) and B’s 1 hour’s work =\(\dfrac {1}{42}\)
(A + B)’s 1 hour’s work \(=\dfrac {1}{63}+\dfrac {1}{42}=\dfrac {5}{126}\)
Both will finish the work in \(\dfrac {126}{5}\) hrs.
Number of days. of \(\dfrac {42}{5}\) hrs each \(=\dfrac {126\times \ 5}{5\times \ 42}=3\ \)days
4. A and B can do a piece of work in 18 days; Band C can do it in 24 days A and C can do it in 36 days. In how many days will A, Band C finish it together and separately?
Solution :
(A + B)’s 1 day’s work \(=\dfrac {1}{18}\)
(B + C)’s 1 day’s work \(=\dfrac {1}{24}\)
and (A + C)’s 1 day’s work \(=\dfrac {1}{36}\)
Adding, we get: 2 (A + B + C)’s 1 day’s work \(=\dfrac {1}{18}+\dfrac {1}{24}+\dfrac {1}{36}\ \ =\dfrac {9}{72}=\dfrac {1}{8}\)
(A +B + C)’s 1 day’s work \(=\dfrac {1}{16}\ \)
Thus, A, Band C together can finish the work in 16 days.
Now, A’s 1 day’s work = [(A + B + C)’s 1 day’s work] - [(B + C)’s 1 day work: \(\ =\dfrac {1}{16}--\dfrac {1}{24}=\dfrac {1}{48}\)
A alone can finish the work in 48 days.
Similarly, B’s 1 day’s work \(= \dfrac {1}{16} - \dfrac {1}{36} =\dfrac {5}{144}\)
B alone can finish the work in \(\dfrac {144}{5}=28\dfrac {4}{5}\) days
And C’s 1 day work \(=\dfrac {1}{16}-\dfrac {1}{18}=\dfrac {1}{144}\)
Hence C alone can finish the work in 144 days.
5. A is twice as good a workman as B and together they finish a piece in 18 days. In how many days will A alone finish the work?
Solution :
(A’s 1 day’s work):(B’s 1 day’s work) = 2 : 1.
(A + B)’s 1 day’s work = \(\dfrac {1}{18}\)
Divide \(\dfrac {1}{18}\) in the ratio 2 : 1.
:. A’s 1 day’s work \(=\left (\dfrac {1}{18}\times \dfrac {2}{3}\right )=\dfrac {1}{27}\)
Hence, A alone can finish the work in 27 days.
6. A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?
Solution :
Ratio of times taken by A and B = 160 : 100 = 8 : 5.
Suppose B alone takes x days to do the job.
Then, \(8\ :\ 5\ ::\ 12\ :\ x\Rightarrow \ 8x\ =\ 5\times \ 12\ \Rightarrow \ x\ =\ 7\dfrac {1}{2}\) days.
7. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?
Solution :
Work done by A in 10 days \(=\left (\dfrac {1}{80}\times 10\right )=\dfrac {1}{8}\ \)
Remaining work \(= (1- \dfrac {1}{8} )\ =\dfrac {7}{8}\ \)
Now, \(\dfrac {7}{8}\) work is done by B in 42 days.
Whole work will be done by B in \(\left (42\times \dfrac {8}{7}\right )\ =\ 48\) days.
A’s 1 day’s work \(=\dfrac {1}{80}\) and B’s 1 day’s work \(=\dfrac {1}{48}\)
(A+B)’s 1 day’s work \(=\ \left (\dfrac {1}{80}+\dfrac {1}{48}\right )=\dfrac {8}{240}=\dfrac {1}{30}\)
Hence, both will finish the work in 30 days.
8. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of each.
Solution :
C’s 1 day’s work\(\ =\dfrac {1}{3}-\left (\dfrac {1}{6}+\dfrac {1}{8}\right )=24\)
A : B : C = Ratio of their 1 day’s work \(=\dfrac {1}{6}:\dfrac {1}{8}:\dfrac {1}{24}=\ 4\ :\ 3\ :\ 1.\)
A’s share = Rs.\(\ \left (600\times \dfrac {4}{8}\right )\) = Rs.300,
B’s share = Rs. \(\left (600\times \dfrac {3}{8}\right )\) = Rs. 225.
C’s share = Rs. \(\left [600\ -\ \left (300\ +\ 225\right )\right ]\) = Rs. 75.
9. A and B working separately can do a piece of work in 9 and 12 days respectively, If they work for a day alternately, A beginning, in how many days, the work will be completed?
Solution :
(A + B)’s 2 days’ work \(=\left (\dfrac {1}{9}+\dfrac {1}{12}\right )=\dfrac {7}{36}\)
Work done in 5 pairs of days \(=\left (5\times \dfrac {7}{36}\right )=\dfrac {35}{36}\)
Remaining work \(=\left (1-\dfrac {35}{36}\right )=\dfrac {1}{36}\)
On 11th day, it is A’s turn. \(\dfrac {1}{9}\) work is done by him in 1 day.
\(\dfrac {1}{36}\) work is done by him in \(\left (9\times \dfrac {1}{36}\right )=\dfrac {1}{4}\ \ \)day
Total time taken \(=\ \left (10\ +\dfrac {1}{4}\right )\ \) days\(\ =\ 10\dfrac {1}{4}\ \)days.
10. 45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?
Solution :
\((45\times \ 16)\) men can complete the work in 1 day.
1 man’s 1 day’s work\(\ =\dfrac {1}{720}\)
45 men’s 6 days’ work \(=\left (\dfrac {1}{16}\times 6\right )=\dfrac {3}{8}\)
Remaining work\(\ =\left (1-\dfrac {3}{8}\right )=\dfrac {5}{8}\)
75 men’s 1 day’s work \(=\dfrac {75}{720}=\dfrac {5}{48}\ \)
Now, \(\dfrac {5}{48}\) work is done by them in 1 day.
\(\dfrac {5}{48}\) work is done by them in \(\left (\dfrac {48}{5}+\dfrac {5}{8}\right )\) days.
11. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?
Solution :
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, \(2x+3y\ =\dfrac {1}{10}\ \) and \(3x+2y\ =\dfrac {1}{8}\ \)
Solving, we get: \(\ x\ =\dfrac {7}{200}\) and \(y\ =\dfrac {1}{100}\)
(2 men + 1 boy)’s 1 day’s work \(=\left (2\times \dfrac {7}{200}+1\times \dfrac {1}{100}\right )=\dfrac {16}{200}=\dfrac {2}{25}\ \)
So, 2 men and 1 boy together can finish the work in \(\dfrac {25}{2}=12\dfrac {1}{2}\) days.
3.2 Time and Distance
3.2.1 Important Facts and Formulae
1. Speed \(=\left (\dfrac {\text { Distance }}{\text { Time }}\right )\); Time \(=\left (\dfrac {\text { Distance }}{\text { Speed }}\right )\); Distance \(=(\) Speed \(\times \) Time \()\)
2. \(x \mathrm {~km} / \mathrm {hr}=\left (x \times \dfrac {5}{18}\right ) \mathrm {m} / \mathrm {sec}\)
3. \(x \mathrm {~m} / \mathrm {sec}=\left (x \times \dfrac {18}{5}\right ) \mathrm {km} / \mathrm {hr}\)
4. If the ratio of the speeds of \(A\) and \(B\) is \(a: b\), then the ratio of the times taken by them to cover the same distance is \(\dfrac {1}{a}:\dfrac {1}{b} \) or \(b:a\).
5. Suppose a man covers a certain distance at \(x \mathrm {~km} / \mathrm {hr}\) and an equal distance at \(y \mathrm {~km} / \mathrm {hr}\). Then, the average speed during the whole journey is \(\left (\dfrac {2 x y}{x+y}\right ) \mathrm {km} / \mathrm {hr}\).
3.2.2 Solved Examples
1. How many minutes does Aditya take to cover a distance of \(400 \mathrm {~m}\), if he runs at a speed of \(20 \mathrm {~km} / \mathrm {hr}\).
Solution :
Aditya’s speed \(=20 \mathrm {~km} / \mathrm {hr}=\left (20 \times \dfrac {5}{18}\right ) \mathrm {m} / \mathrm {sec}=\dfrac {50}{9} \mathrm {~m} / \mathrm {sec}\).
\(\therefore \quad \) Time taken to cover \(400 \mathrm {~m}=\left (400 \times \dfrac {9}{50}\right ) \mathrm {sec}=72 \mathrm {sec}=1 \dfrac {12}{60} \mathrm {~min}=1 \dfrac {1}{5} \mathrm {mig}\).
2. A cyclist covers a distance of \(750 \mathrm {~m}\) in \(2 \mathrm {~min} 30 \mathrm {sec}\). What is the speed in \(\mathrm {km} / \mathrm {hr}\) of the cyclist?
Solution :
Speed \(=\left (\dfrac {760}{150}\right ) \mathrm {m} / \mathrm {sec}=5 \mathrm {~m} / \mathrm {sec}=\left (5 \times \dfrac {18}{5}\right ) \mathrm {km} / \mathrm {hr}=18 \mathrm {~km} / \mathrm {hr}\).
3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.
Solution :
Let the distance covered in 1 leap of the dog be \(x\) and that covered in 1 leap of the hare be \(y\).
Then, \(3 x=4 y \Rightarrow x=\dfrac {4}{3} y \Rightarrow 4 x=\dfrac {16}{3} y\).
\(\therefore \) Ratio of speeds of dog and hare \(=\) Ratio of distances covered by them in the same tine
\[=4 x: 5 y=\dfrac {16}{3} y: 5 y=\dfrac {16}{3}: 5=16: 15\]
4. While covering a distance of \(24 \mathrm {~km}\), a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was \(\dfrac {5}{7}\) of the remaining distance. What was his speed in metres per second?
Solution :
Let the speed be \(x \mathrm {~km} / \mathrm {hr}\).
Then, distance covered in \(1 \mathrm {hr} .40\) min. i.e, \(1 \dfrac {2}{3} \mathrm {hrs}=\dfrac {5 x}{3} \mathrm {~km}\).
Remaining distance \(=\left (24-\dfrac {5 x}{3}\right ) \mathrm {km}\).
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \dfrac {5 x}{3}=\dfrac {5}{7}\left (24-\dfrac {6 x}{3}\right ) & \Leftrightarrow \dfrac {6 x}{3}=\dfrac {5}{7}\left (\dfrac {72-5 x}{3}\right ) \Leftrightarrow 7 x=72-5 x \\ & \Leftrightarrow 12 x=72 \Leftrightarrow x=6 \end{align*} Hence, speed \(=6 \mathrm {~km} / \mathrm {hr}=\left (6 \times \dfrac {5}{18}\right ) \mathrm {m} / \mathrm {sec}=\dfrac {5}{3} \mathrm {~m} / \mathrm {kec}=1 \dfrac {2}{3} \mathrm {~m} / \mathrm {sec}\).
5. Peter can cover a certain distance in 1 hr 24 min by covering tow third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.
Solution :
Let the distance be \(x\) km. Then
\[ \dfrac {\dfrac {2}{3}x}{4} + \dfrac {\dfrac {1}{3}x}{5} = \dfrac {7}{5} \Rightarrow \dfrac {x}{6} + \dfrac {x}{15} = \dfrac {7}{5} = \Rightarrow 7x = 42 \Rightarrow x =6 .\]
Total distance \(=6 \mathrm {~km}\).
6. A man travelled from the village to the post-office at the rite of \(25 \mathrm {kmph}\) and walked back at the rate of \(4 \mathrm {kmph}\). If the whole journey took 5 hours \(48\) minutes find the distance of the post-office from the village.
Solution :
Average speed \(=\left (\dfrac {2 x y}{x+y}\right ) \mathrm {km} / \mathrm {hr}=\left (\dfrac {2 \times 25 \times 4}{25+4}\right ) \mathrm {km} / \mathrm {hr}=\dfrac {200}{29} \mathrm {~km} / \mathrm {hr}\).
Distance travelled in 5 hours 48 minutes i.e, \(5 \dfrac {4}{5} \mathrm {hr}=\left (\dfrac {200}{29} \times \dfrac {29}{5}\right ) \mathrm {km}=40 \mathrm {~km}\).
Distance of the post-office from the village \(=\left (\dfrac {40}{2}\right )=20 \mathrm {~km}\).
7. An aeroplane flies along the four sides of a square at the speeds of 200,400 , 600 and \(800 \mathrm {~km} / \mathrm {hr}\). Find the average speed of the plane around the field.
Solution :
Let each side of the square be \(x \mathrm {~km}\) and let the average speed of the plane around the field be \(y \mathrm {~km} / \mathrm {hr}\). Then,
\[ \dfrac {x}{200}+\dfrac {x}{400}+\dfrac {x}{600}+\dfrac {x}{800}=\dfrac {4 x}{y} \Rightarrow \dfrac {25 x}{2400}=\dfrac {4 x}{y} \Rightarrow y=\left (\dfrac {2400 \times 4}{25}\right )=384\]
\(\therefore \quad \) Average speed \(=384 \mathrm {~km} / \mathrm {hr}\).
8. Walking at \(\dfrac {5}{6}\) of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.
Solution :
New speed \(=\dfrac {5}{6}\) of the usual speed
\(\therefore \) New time taken \(=\dfrac {6}{5}\) of the usual time
So, \(\quad \left (\dfrac {6}{5}\right .\) of the usual time \()-\) (usual time) \(=10 \mathrm {~min}\).
\(\Rightarrow \quad \dfrac {1}{5}\) of the usual time \(=10 \mathrm {~min} \Rightarrow \) usual time \(=50 \mathrm {~min}\),
9. If a man walks at the rate of \(5 \mathrm {kmph}\), he misses a train by 7 minutes. However, if to walks at the rate of \(6 \mathrm {kmph}\), he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
Solution :
Let the required distance be \(x \mathrm {~km}\). Difference in the times taken at two speeds \(=12 \mathrm {~min}=\dfrac {1}{5} \mathrm {hr}\).
\(\dfrac {x}{5}-\dfrac {x}{6}=\dfrac {1}{5} \Rightarrow 6 x-5 x=6 \Rightarrow x=6\).
Hence, the required distance is \(6 \mathrm {~km}\).
10. \(A\) and \(B\) are two stations \(390 \mathrm {~km}\) apart. A train starts from \(A\) at 10, at and travels towards \(B\) at \(65 \mathrm {kmph}\). Another train towards \(A\) at \(35 \mathrm {kmph}\). At what time do they met ?
Solution :
Suppose they meet \(x\) hours after \(10 \mathrm {a} \mathrm {m}\). Then,
(Distance moved by first in \(x\) hrs) + [Distance moved by second in \((x-1)\) hours ] = 390.
\(65 x+35(x-1)=390 \Rightarrow 100 x=425 \Rightarrow x=4 \dfrac {1}{4}\).
So, they meet 4 hours. \(15 \mathrm {~min}\) after 10 a.m. i.e., at \(2.15\) p.m.
11. A goods train leaves a station at a certain time and at a fixed speed, After 6 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.
Solution :
Let the speed of the goods train be \(x \mathrm {kmph}\).
Distance covered by goods train in 10 hours = Distance covered by express train in 4 hours
\(\therefore \quad 10 x=4 \times 90\) or \(x=36\).
So, speed of goods train \(=36 \mathrm {kmph}\).
12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be \(8 \mathrm {~km} / \mathrm {hr}\) and that of the policeman \(10 \mathrm {~km} / \mathrm {hr}\), how far the thief will have run before he is overtaken?
Solution :
Relative speed of the policeman \(=(10-8) \mathrm {km} / \mathrm {hr}=2 \mathrm {~km} / \mathrm {hr}\).
Time taken by policeman to cover \(100 \mathrm {~m}=\left (\dfrac {100}{1000} \times \dfrac {1}{2}\right ) \mathrm {hr}=\dfrac {1}{20} \mathrm {hr}\).
In \(\dfrac {1}{20} \mathrm {hrs}\), the thief covers a distance of \(\left (8 \times \dfrac {1}{20}\right ) \mathrm {km}=\dfrac {2}{5} \mathrm {~km}=400 \mathrm {~m}\).
13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?
Solution :
Let the distance be \(x \mathrm {~km}\). Then,
(Time taken to walk \(x \mathrm {~km}\) ) \(+\) (Time taken to ride \(x \mathrm {~km}\) ) \(=37 \mathrm {~min}\).
\(\Rightarrow \) (Time taken to walk \(2 x \mathrm {~km}\) ) \(+\) (Time taken to ride \(2 x \mathrm {~km}\) ) \(=74 \mathrm {~min}\).
But, time taken to walk \(2 x \mathrm {~km}=55 \mathrm {~min}\).
\(\therefore \) Time taken to ride \(2 x \mathrm {~km}=(74-55) \mathrm {min}=19 \mathrm {~min}\).
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