We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways | = (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
| ||||||||||||
| ||||||||||||
= (525 + 210 + 21) | ||||||||||||
= 756. |
Let the number be x.
Then, | 1 | of | 1 | of x = 15 x = 15 x 12 = 180. |
3 | 4 |
So, required number = | 3 | x 180 | = 54. | ||
10 |
Required decimal = | 1 | = | 1 | = .00027 |
60 x 60 | 3600 |
(.000216)1/3 | = | 216 | 1/3 | ||
106 |
= | 6 x 6 x 6 | 1/3 | ||
102 x 102 x 102 |
= | 6 |
102 |
= | 6 |
100 |
= 0.06
Let the third number be x.
Then, first number = 120% of x = | 120x | = | 6x |
100 | 5 |
Second number = 150% of x = | 150x | = | 3x |
100 | 2 |
Ratio of first two numbers = | 6x | : | 3x | = 12x : 15x = 4 : 5. | ||
5 | 2 |
Part filled by (A + B + C) in 3 minutes = 3 | 1 | + | 1 | + | 1 | = | 3 x | 11 | = | 11 | . | ||||
30 | 20 | 10 | 60 | 20 |
Part filled by C in 3 minutes = | 3 | . |
10 |
Required ratio = | 3 | x | 20 | = | 6 | . | ||
10 | 11 | 11 |
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = | 3 - | 3x | + x | litres | ||
8 |
Quantity of syrup in new mixture = | 5 - | 5x | litres | ||
8 |
3 - | 3x | + x | = | 5 - | 5x | |||||
8 | 8 |
5x + 24 = 40 - 5x
10x = 16
x = | 8 | . |
5 |
So, part of the mixture replaced = | 8 | x | 1 | = | 1 | . | ||
5 | 8 | 5 |
A : B = 100 : 90.
A : C = 100 : 72.
B : C = | B | x | A | = | 90 | x | 100 | = | 90 | . |
A | C | 100 | 72 | 72 |
When B runs 90 m, C runs 72 m.
When B runs 100 m, C runs | 72 | x 100 | m | = 80 m. | |
90 |
B can give C 20 m.
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = | n(E) | = | 9 | . |
n(S) | 20 |
T.D. |
| |||||||||||
| ||||||||||||
| ||||||||||||
= Rs. 400. |