Whole work is done by A in | ❨ | 20 x | 5 | ❩ | = 25 days. |
4 |
Now, | ❨ | 1 - | 4 | ❩ | i.e., | 1 | work is done by A and B in 3 days. |
5 | 5 |
Whole work will be done by A and B in (3 x 5) = 15 days.
A's 1 day's work = | 1 | , (A + B)'s 1 day's work = | 1 | . |
25 | 15 |
∴ B's 1 day's work = | ❨ | 1 | - | 1 | ❩ | = | 4 | = | 2 | . |
15 | 25 | 150 | 75 |
So, B alone would do the work in | 75 | = 37 | 1 | days. |
2 | 2 |
Part filled by 1st pipe in 1 min = 1/20
Part filled by 2nd pipe in 1 min = 1/24
Part filled by all the pipes in 1 min = 1/15
Work done by the waste pipe 1 min
= 1/15 - (1/20 + 1/24 )
= 1/15 - 11/120
= (8 - 11)/120 = ( - 3/120 ) = ( -1/40 )
[-ve sign indicates emptying]
Now, volume of 1/40 part = 6 gallon
? Volume of whole tank = 40 x 6 = 240 gallon
Part of the tank filled by the taps A, B and C in 3 min = 1/20 + 1/30 - 1/15 = (3 + 2 - 4)/60 = 1/60
? Time taken to fill [ 1 - ( 1/20 + 1/30 )] or 55/60th part of the tank = 3 x 55 = 165 min
Remaining part of the tank = 1 - 55/60 = 5/60 = 1/12
Tap A fill 1/0 part in 1 min, then
Remaining part = 1/12 - 1/20 = (5 -3)/60 = 2/60 = 1/30
i.e, 1/30th part is filled by B in 1 min
Hence, required time to fill the whole tank = (165 + 1 +1 ) min = 167 min
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120
? Part filled by C in 1 min = 1/10 - 7/120 = (12 - 7)/120 = 5/120 = 1/24
? Tap C will fill the cistern in 24 min.
Investment of Ramesh = 10% of 10000 = ? 1000
After 3 yr = 2500 - 1000 = ? 1500
We know.
SI = (P x R x T)/100
? 1500 = (1000 x R x 3)/100
? R = (1500 x 100) / (1000 x 3) = 50%
? 4 times of 50% = 200%
Let. principal = P
Given, SI = 55, Time T = 9 months = 9/12 yr,
Rate R = 32/3% = 11/3 %
? SI = (P x R x T) / 100
? P = (100 x SI) / (R x T)
? (55 x 100) / [(11 x 9) / (3 x 12)] = 2000
? Principal (P) = ? 2000
Required probability = P(A). P(B) + P (A ) . P(B)
= 5/7 . 3/10 + 2/7 . 7/10
= 29/70
Given, M1 = 40, W1 = 200, T1 = 8, D1 = 12,
M2 = 30, W2 = 300, T2 = 6 and D2 = ?
Then, using formula
M1T1D1W2 = M2T2D2W1
? 40 x 8 x 12 x 300 = 30 x 6 x D2 x 200
? D2 = (40 x 8 x 12 x 300)/(30 x 6 x 200) = 32 days.
Three day work or Ram, Rahim and Robert = 1/7 + 1/8 + 1/6
= (24 + 21 + 28)/168 = 73/168
Their six days work = (2 x 73)/168 = 146/168 = 73/84
Now, its Ram's turn to work on seventh day.
? 73/84 + 1/7 = 73 + 12/84 = 85/84 > 1
So, Ram will finish the work before the end of day.
Hence, Ram is working for the last day.
12 men =18 women
? 1 man = 18/12 women
? 8 men = 18/12 x 8 = 12 women
Given m1 = 18 M2 = 12 +16 = 28,
D1 , D2 = ? and W1 = W2 = 1
According to the formula
M1D1W2 = M2D2W1
? 18 x14 x1= 28 x D2 x1
? D2 = (18 x 14)/28 = 9 days
Given. T1 = 2 yr and T2 = 4 yr,
P1 = 600, P2 = 150.
According to the question,
SI1 + SI2 = 80
[(600 x R x 2)/100] + [(150 x R x 4)/100] = 80
? 120R + 60R = 800
? 180R = 800
? R = 800/180 = 80/18 = 40/9
= 44/9%
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.