Solution:
When n is divided by 4: n = 4p + 1
When n is divided by 6: n = 6q + 5
Since both expressions are equal
4p + 1 = 6q + 5
⇒ 4p - 6q = 4
⇒ 2p - 3q = 2
⇒ - 2p + 3q = - 2
Therefore, 3q - 2p = - 2
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Solution:
When n is divided by 4: n = 4p + 1
When n is divided by 6: n = 6q + 5
Since both expressions are equal
4p + 1 = 6q + 5
⇒ 4p - 6q = 4
⇒ 2p - 3q = 2
⇒ - 2p + 3q = - 2
Therefore, 3q - 2p = - 2
Solution:
Let,
First multiple: 3x
Second multiple: 3(x + 1) = 3x + 3
Third multiple: 3(x + 2) = 3x + 6
ATQ,
3x + (3x + 3) + (3x + 6) = 396
⇒ 9x + 9 = 396
⇒ 9x = 387
⇒ x = 387/9
∴ x = 43
∴ The largest number = 3x + 6 = 3 × 43 + 6 = 135
Solution:
Let,
the number is 'x'
then ,
(1/3) × (1/4) × x = 15
⇒ x/12 = 15
⇒ x = 180
Now,
(3/10) × x = (3/10) × 180 = 18 × 3 = 54.
∴ three-tenths of that number is 54.
Solution:
We know that,
= a × a + 2ab + b × b
Now, (11.98 × 11.98 + 2 × 11.98 × 0.02 + 0.02 × 0.02) =
We can say that, (11.98 × 11.98 + 11.98 × Q + 0.02 × 0.02) will be a perfect square
if Q = 2 × 0.02 = 0.04
Solution:
Given,
7 people need to be served 1 cake
So, 98 people need to be served = 98/7 cakes
= 14 cakes
Solution:
x - 2 = 12 - 2 = 10
x/2 = 12/2 = 6
2/x = 2/12 = 1/6 ≈ 0.17
2 - x = 2 - 12 = - 10
The least value among the calculated options is - 10 or 2 - x