Given, SI2 = 60, SI1 = 30, T1 = 4 yr, T2 = 8 yr
According to the question,
[(1500 x R x 8)/100] - [(1500 x R x 4)/100] = 60 - 30
? (6000 x R)/100 = 30
? R = 30/60 = 1/2 = 0.5%
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 |
6 |
5 |
12 |
1 |
2 |
7 |
9 |
5 |
12 |
Let E = Event that the sum is a prime number.
Then E | = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } |
∴ n(E) = 15.
∴ P(E) = | n(E) | = | 15 | = | 5 | . |
n(S) | 36 | 12 |
3 |
20 |
29 |
34 |
47 |
100 |
13 |
102 |
13 |
102 |
Then, n(S) = 52C2 = | (52 x 51) | = 1326. |
(2 x 1) |
Let E = event of getting 1 spade and 1 heart.
∴ n(E) | = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 |
= (13C1 x 13C1) | |
= (13 x 13) | |
= 169. |
∴ P(E) = | n(E) | = | 169 | = | 13 | . |
n(S) | 1326 | 102 |
Average speed of train = 240/12 = 20 km/h
Average speed of bus = (3/4) x 20 = 3 x 5 = 15 km/h
Required distance = 15 x 7 = 15 km/h
A person completes 5/8 part of the project in 10 days.
A person completes the project in 10/5/8 days.
A person completes the project in 10 x 8 /5 days.
A person completes the project in 2 x 8 days.
A person completes the project in 16 days.
1 |
15 |
25 |
57 |
35 |
256 |
1 |
221 |
1 |
221 |
Then, n(S) = 52C2 = | (52 x 51) | = 1326. |
(2 x 1) |
Let E = event of getting 2 kings out of 4.
∴ n(E) = 4C2 = | (4 x 3) | = 6. |
(2 x 1) |
∴ P(E) = | n(E) | = | 6 | = | 1 | . |
n(S) | 1326 | 221 |
[725 x 725 x 725 + 371 x 371 x 371] / [725 x 725 - 725 x 371 + 371 x 371]
= 725 + 371 = 1096
[? (a3 + b3) / (a2 - ab + b2) = (a + b) (a2 - ab + b2)/(a2 - ab + b2) = a + b ]
4 mat-weavers in 4 days can weave 4 mats.
1 mat-weavers in 4 days can weave 4 / 4 mats.
1 mat-weavers in 1 days can weave 4 / 4 x 4 mats.
8 mat-weavers in 1 days can weave 4 x 8 / 4 x 4 mats.
8 mat-weavers in 8 days can weave 4 x 8 x 8 / 4 x 4 mats.
8 mat-weavers in 8 days can weave 4 x 2 x 2 mats.
8 mat-weavers in 8 days can weave 16 mats.
Largest 4-digit number = 9999 88) 9999 (113 88 ---- 119 88 ---- 319 264 --- 55 --- Required number = (9999 - 55) = 9944.
Then, | x | = 15 ⟹ y = | x | . |
y | 15 |
∴ | x + 100 | = | x |
25 | 15 |
⟹ 15(x + 100) = 25x
⟹ 15x + 1500 = 25x
⟹ 1500 = 10x
⟹ x = 150 m.
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