1 women ? 1/2 man
10 men + 8 women ? 10 men + 4 men = 14 men
4 men + 6 women ? 4 men + 3 men = 7 men
More men, Less days (Indirect proportion) 7 : 14 : : 5 : N
? N = (14 x 5)/7 = 10 days
So, ratio of times taken = 1 : 2.
B's 1 day's work = | 1 | . |
12 |
∴ A's 1 day's work = | 1 | ; (2 times of B's work) |
6 |
(A + B)'s 1 day's work = | ❨ | 1 | + | 1 | ❩ | = | 3 | = | 1 | . |
6 | 12 | 12 | 4 |
So, A and B together can finish the work in 4 days.
Angle traced by hour hand in 5 hrs 10 min. i.e., | 31 | hrs = | ❨ | 360 | x | 31 | ❩ | ° | = 155°. |
6 | 12 | 6 |
Given that, a = 6 cm, b = 4 cm and c = 5 cm
Required perimeter = a + b + c
= 6 + 4 + 5 cm
= 15 cm
7 | 4 | m |
7 |
A : C = 200 : 182.
C | = | ❨ | C | x | A | ❩ | = | ❨ | 182 | x | 200 | ❩ | = 182 : 169. |
B | A | B | 200 | 169 |
When C covers 182 m, B covers 169 m.
When C covers 350 m, B covers | ❨ | 169 | x 350 | ❩m | = 325 m. |
182 |
Therefore, C beats B by (350 - 325) m = 25 m.
Quantity of A in mixture left = | ❨ | 7x - | 7 | x 9 | ❩ | litres = | ❨ | 7x - | 21 | ❩ litres. |
12 | 4 |
Quantity of B in mixture left = | ❨ | 5x - | 5 | x 9 | ❩ | litres = | ❨ | 5x - | 15 | ❩ litres. |
12 | 4 |
∴ |
|
= | 7 | |||||
|
9 |
⟹ | 28x - 21 | = | 7 |
20x + 21 | 9 |
⟹ 252x - 189 = 140x + 147
⟹ 112x = 336
⟹ x = 3.
So, the can contained 21 litres of A.
Amount of milk left after 3 operations = | [ | 40 | ❨ | 1 - | 4 | ❩ | 3 | ] litres |
40 |
= | ❨ | 40 x | 9 | x | 9 | x | 9 | ❩ | = 29.16 litres. |
10 | 10 | 10 |
The number of ways of selecting 3 men and 3 women out of 6 men and 5 women = 6C3 x 5C3
= 6!/(3! x 3!) + 5/(3! x 2!)
= 20 + 10 = 30
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 |
Sum of the weight of A, B, C and D = 67 x 4 = 268 kg
and average weight of A, B , C , D and E = 67 - 2 = 65kg
? Sum of the weight of A, B, C, D and E = 65 x 5 = 325 kg
? Weight of E = 325 - 268 = 57 kg
? Weight of F = 57 + 4 = 61 kg
Now, average weight of F, B, C, D and E = 64 kg
? Sum of the weight of F, B, C, D and E = 64 x 5 = 320 Kg
? Sum of the weights of B, C and D = 320 - 57 - 61 = 202 kg
? Weight of A = 268 - 202 = 66 kg
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