As A1 speaks always after A2, they can speak only in 1st to 9th places and
A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place
A2 can speak in 9 places the remaining
A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e
when A2 speaks in the first place, the number of ways they can speak is 9.8!.
When A2 speaks in second place, the number of ways they can speak is 8.8!.
When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!
Therefore,Total Number of ways they can speak = (9+8+7+6+5+4+3+2+1) 8! = = 10!/2
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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