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).
Clearly total persons are increased in 28/35 :: 52/65 = 4:5.
As time is inversely proportional to men, so total time will decrease in the ratio 5:4.
Hence, 22.5 x 4/5 = 18 days.
Rate of leakage = 8.33% per hour
Net efficiency = 50 - (16.66 + 8.33)= 25%
Time required = 100/25 = 4 hours
(16M + 12W) x 20 = 18W x 40
=> 2M = 3W
Then,
Convert all men into women
12M + 27W = 27 + 12 x 3/2W = 45W
Let number of days required be 'D'
=> 18 x 40 = 45 x D
=> D = 16 days.
(20 x 18) men can complete the work in in one day.
one man's one day work = 1/360
(18 x 15) women can complete the work in 1 day
1 woman's one day work = 1/270
So, required ratio = = 4:3
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.
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.
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.
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.
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.
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