S.P. of 1 litre of mixture = Re.1, Gain = | 50 | %. |
3 |
∴ C.P. of 1 litre of mixture = | ❨ | 100 x | 3 | x 1 | ❩ | = | 6 |
350 | 7 |
By the rule of alligation, we have:
C.P. of 1 litre of water C.P. of 1 litre of milk | |||||
0 | Mean Price
|
Re. 1 | |||
|
|
∴ Ratio of water and milk = | 1 | : | 6 | = 1 : 6. |
7 | 7 |
1 |
3 |
1 |
4 |
1 |
5 |
1 |
7 |
1 |
5 |
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 |
1 |
3 |
2 |
3 |
2 |
5 |
3 |
5 |
2 |
3 |
Strength of first jar Strength of 2nd jar | ||
40% | Mean Strength 26% |
19% |
7 | 14 |
So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
∴ Required quantity replaced = | 2 |
3 |
C.P. of 1 kg of the mixture = Rs. | ❨ | 100 | x 68.20 | ❩ | = Rs. 62. |
110 |
By the rule of alligation, we have:
Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. | ||
Rs. 60 | Mean Price Rs. 62 |
Rs. 65 |
3 | 2 |
∴ Required ratio = 3 : 2.
∴ C.P. of 1 kg of mixture = Rs. | ❨ | 100 | x 9.24 | ❩ | = Rs. 8.40 |
110 |
By the rule of allilation, we have:
C.P. of 1 kg sugar of 1st kind Cost of 1 kg sugar of 2nd kind | ||
Rs. 9 | Mean Price Rs. 8.40 |
Rs. 7 |
1.40 | 0.60 |
∴ Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
⟹ x = | ❨ | 7 x 27 | ❩ | = 63 kg. |
3 |
Cost of 1 kg of 1st kind Cost of 1 kg of 2nd kind | ||
720 p | Mean Price 630 p |
570 p |
60 | 90 |
∴ Required ratio = 60 : 90 = 2 : 3.
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.
Then, quantity of wine left in cask after 4 operations = | [ | x | ❨ | 1 - | 8 | ❩ | 4 | ] litres. |
x |
∴ | ❨ | x(1 - (8/x))4 | ❩ | = | 16 |
x | 81 |
⟹ | ❨ | 1 - | 8 | ❩ | 4 | = | ❨ | 2 | ❩ | 4 |
x | 3 |
⟹ | ❨ | x - 8 | ❩ | = | 2 |
x | 3 |
⟹ 3x - 24 = 2x
⟹ x = 24.
B runs | 45 | m in 6 sec. |
2 |
∴ B covers 300 m in | ❨ | 6 x | 2 | x 300 | ❩sec | = 80 sec. |
45 |
∴ B covers 200 m in | ❨ | 7 | x 200 | ❩ | = 40 sec. |
35 |
B's time over the course = 40 sec.
∴ A's time over the course (40 - 7) sec = 33 sec.
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.
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