TIPS FOR SOLVING QUESTIONS RELATED TO TIME AND WORK:
1. The total work is taken as 1.
2. If A can do a piece of work in 'n' days, ,then A's 1 day's work is equal to = 1/n
3. If A's 1 day's work = 1/n, then A can complete the work in 'n' days.
4. If A is thrice as good a workman as B, then
Ratio of work done by A and B = 3 : 1.
Ratio of time taken by A and B to finish the work = 1 : 3.
5.If ‘M1’ persons can do ‘W1’ work in ‘D1’ days and ‘M2’ persons can do W2 work in D2, days then we have a very general formula in the relationship of M1 D1 W2 = M2 D2 W1. The above relationship can be taken as a very basic and all-in-one formula. We also derive
A) More men less days and conversely more days less men.
B) More men more work and conversely more work more men.
C) More days more work and conversely more work more days.
6.If we include the working hours (say T1 and T2) for the two groups then the relationship is
6.If we include the working hours (say T1 and T2) for the two groups then the relationship is
M1 D1 T1 W2 = M2 D2 T2 Q1
Again, if the efficiency (Say E1 and E2) of the persons in two groups is different then the relationship is
Again, if the efficiency (Say E1 and E2) of the persons in two groups is different then the relationship is
M1 D1 T1 E1 W2 = M2 D2 T2 E2 W1
Now, we should go ahead starting with simpler to difficult and more difficult questions.
Questions on Above Formulaes
1.A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
A. Rs. 375
B. Rs. 400
C. Rs. 600
D.Rs. 800
2. Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm
B. 2 pm
C. 1:00 pm
D. 11 am
3. A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
A. 20 days
B. 22 1/2days
C. 25 days
D. 30 days
4. P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work?
A. 9
B. 11
C. 13
D. 15
5. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
A. 4 days
B. 5 days
C. 6 days
D. 7 days
6.40 men can cut 60 trees in 8 hours. If 8 men leave the job how many trees wiil be cut in 12 hours?
A. 74
B. 82
C. 72
D. 73
7.A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
A. 8 hours
B. 10 hours
C. 12 hours
D. 24 hours
8. P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last?
A. 5 days
B. 10 days
C. 14 days
D. 22 days
9. If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12
B. 14
C. 16
D. 18
10. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days?
A. Rs.1104
B. Rs.1000
C. Rs.934
D. Rs.1210
ANSWERS AND SOLUTION :
1(B)Explanation:
C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24
A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4 : 3 : 1.
C's share (for 3 days) = Rs. (3 x 1/24 x 3200) = Rs. 400.
2(C)Explanation :
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm.
3(B)Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (3/2 x 60) = 90 days.
So, A takes 30 days to do the work.
A's 1 day's work = 1/30
B's 1 day's work = 1/90
(A + B)'s 1 day's work = (1/30 1/90) = 4/90 = 2/45
So A and B together can do the work in 45/2 = 22 1/2 days.
4(C)Explanation :
Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q × (130/100) = 1/23
=> q = 100/(23×130) = 10/(23×13)
Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13
=> P and Q together can do the work in 13 days
5(A)Explanation:
Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2,.
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)'s 1 day's work = (15/100 + 20/200) = 1/4
So, 15 men and 20 boys can do the work in 4 days.
6(C) Explanation
By basic formula:
M1 = 40, D1 = 8, W1 = 60(cutting of trees is taken as work)
M2 = 40 - 8 = 32, D = 12, W2 = ?
Putting the values in the formula
M1 D1 W2 = M2 D2 W1
We have , 40 x 8 x W2 = 32 x 12 x 60
or, W2 = (32 x 12 x 60)/(40 x 8) = 72
7(C)Explanation:
A's 1 hour's work = 1 /4 ;
(B + C)'s 1 hour's work = 1/3 ;
(A + C)'s 1 hour's work = 1/2 .
(A + B + C)'s 1 hour's work = (1/4 + 1/3) = 7/12
B's 1 hour's work = (7/12 - 1/2) = 1/12 .
So, B alone will take 12 hours to do the work.
8(B)Explanation :
Work done by P in 1 day = 1/20
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10
9(C)Explanation :
Wages of 1 woman for 1 day = (21600)/(40 x 30)
Wages of 1 man for 1 day = (21600×2)/(40×30)
Wages of 1 man for 25 days = (21600×2×25)/(40×30) = 1080000/1200=900
Number of men = 14400/900 =16
10(A)Explanation :
Assume that cost of keeping a cow for 1 day = c ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 ---(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56×23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104
ANSWERS AND SOLUTION :
1(B)Explanation:
C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24
A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4 : 3 : 1.
C's share (for 3 days) = Rs. (3 x 1/24 x 3200) = Rs. 400.
2(C)Explanation :
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm.
3(B)Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (3/2 x 60) = 90 days.
So, A takes 30 days to do the work.
A's 1 day's work = 1/30
B's 1 day's work = 1/90
(A + B)'s 1 day's work = (1/30 1/90) = 4/90 = 2/45
So A and B together can do the work in 45/2 = 22 1/2 days.
4(C)Explanation :
Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q × (130/100) = 1/23
=> q = 100/(23×130) = 10/(23×13)
Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13
=> P and Q together can do the work in 13 days
5(A)Explanation:
Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2,.
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)'s 1 day's work = (15/100 + 20/200) = 1/4
So, 15 men and 20 boys can do the work in 4 days.
6(C) Explanation
By basic formula:
M1 = 40, D1 = 8, W1 = 60(cutting of trees is taken as work)
M2 = 40 - 8 = 32, D = 12, W2 = ?
Putting the values in the formula
M1 D1 W2 = M2 D2 W1
We have , 40 x 8 x W2 = 32 x 12 x 60
or, W2 = (32 x 12 x 60)/(40 x 8) = 72
7(C)Explanation:
A's 1 hour's work = 1 /4 ;
(B + C)'s 1 hour's work = 1/3 ;
(A + C)'s 1 hour's work = 1/2 .
(A + B + C)'s 1 hour's work = (1/4 + 1/3) = 7/12
B's 1 hour's work = (7/12 - 1/2) = 1/12 .
So, B alone will take 12 hours to do the work.
8(B)Explanation :
Work done by P in 1 day = 1/20
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10
9(C)Explanation :
Wages of 1 woman for 1 day = (21600)/(40 x 30)
Wages of 1 man for 1 day = (21600×2)/(40×30)
Wages of 1 man for 25 days = (21600×2×25)/(40×30) = 1080000/1200=900
Number of men = 14400/900 =16
10(A)Explanation :
Assume that cost of keeping a cow for 1 day = c ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 ---(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56×23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104
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