Let the Efficiency of pavan be E(P)
Let the Efficiency of sravan be E(S)
Here Work W = LCM(25,20) = 100
Now, E(P+2S) = 100/25 = 4 ....(1)
E(2P+S) = 100/20 = 5 ....(2)
Hence, from (1) & (2) we get
E(S) = 1
=> Number of days Savan alone work to complete the work = 100/1 = 100 days.
1/3 ---- 8
1 -------?
Hari can do total work in = 24 days
As satya is 60% efficient as Hari, then
Satya = 1/24 x 60/100 = 1/40
=> Satya can do total work in 40 days
1 ----- 40
2/3 ---- ? => 26 2/3 days.
One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)
One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)
Divide both sides by 2 in (I) and then multiply both sides by 5
Now we get, 15b + 20g = 1/4.
Therefore, 15 boys and 20 girls can do the same work in 4 days.
Number of pages typed by Adam in 1 hour = 36/6 = 6
Number of pages typed by Smith in 1 hour = 40/5 = 8
Number of pages typed by both in 1 hour = (6 + 8) = 14
Time taken by both to type 110 pages = (120 * 1/14) =
= 8 hrs 34 min.
50 men can build a tank in 40 days
Assume 1 man does 1 unit of work in 1 day
Then the total work is 50×40 = 2000 units
50 men work in the first 10 days and completes 50×10 = 500 units of work
45 men work in the next 10 days and completes 45×10 = 450 units of work
40 men work in the next 10 days and completes 40×10 = 400 units of work
35 men work in the next 10 days and completes 35×10 = 350 units of work
So far 500 + 450 + 400 + 350 = 1700 units of work is completed and
Remaining work is 2000 - 1700 = 300 units
30 men work in the next 10 days. In each day, they does 30 units of work.
Therefore, additional days required = 300/30 =10
Thus, total 10+10+10+10+10 = 50 days required.
After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days
-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).
Ratio of efficiencies of Priya and Sai is
Sai : Priya = 160 : 100 = 8 : 5
Given Priya completes the work in 16 days
Let number of days Sai completes the work be 'd'
=> 5×16 = 8×d
d = 10 days.
Therefore, number of days Sai completes the work is 10 days.
Efficiency of kaushalya = 5%
Efficiency of kaikeyi = 4%
Thus, in 10 days working together they will complete only 90% of the work.
[(5+4)*10] =90
Hence, the remaining work will surely done by sumitra, which is 10%.
Thus, sumitra will get 10% of Rs. 700, which is Rs.70
Given,
P can fill in 12 hrs
Q can fill in 15 hrs
R can fill in 20 hrs
=> Volume of tank = LCM of 12, 15, 20 = 60 lit
=> P alone can fill the tank in 60/12 = 5 hrs
=> Q alone can fill the tank in 60/15 = 4 hrs
=> R alone can fill the tank in 60/20 = 3 hrs
Tank can be filled in the way that
(P+Q) + (P+R) + (P+Q) + (P+R) + ....
=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit
=> In 6 hrs = 17 x 6/2 = 51 lit
=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit
=> So, total tank will be filled in 7 hrs.
Work done by P and Q in the first two hours, working alternately
= First hour P + Second hour Q
work is completed in 2 hours
Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours
Thus, work will be completed at 3pm.
Amount of work K can do in 1 day = 1/16
Amount of work L can do in 1 day = 1/12
Amount of work K, L and M can together do in 1 day = 1/4
Amount of work M can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 ? 1/12 = 5/48
=> Hence M can do the job on 48/5 days = 9 (3/5) days
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