Question
A 152-litre mixture contains petrol and water, only in
ratio 3:2, respectively. How much water should be added in the mixture so that the quantity of water becomes 48% of quantity of resultant mixture?Solution
Quantity of water in the initial mixture = 152 × (2/5) = 60.8 litres Quantity of petrol in the initial mixture = 152 × (3/5) = 91.2 litres Since, we’re not adding petrol, 91.2 litres of petrol will represent 48% of the new mixture. So, total quantity of resultant mixture = 91.2 ÷ 0.48 = 190 litres Quantity of water to be added = 190 – 152 = 38 litres
(22 × 52 ) + 4 × 6 = ? - √324
What should come in place of (?) question mark in the given expression.
 (25% of 320) + (3/8 of 400) − 30 = ?
(5832)1/3  × 10.11 × 11.97 ÷ 16.32 = ? + 45.022
82% of 400 + √(?) = 130% of 600 - 85% of 400
If (x + 1/x) = 5, then value of x3 + 1/x3 is:
Simplify: (1 ÷ 0.08)
What should come in place of (?) question mark in the given expression.
{ (144 ÷ 12) × 5 } − (18 ÷ 3) = ?
Simplify the following expressions and choose the correct option.
(3/4 of 256) + (2/5 of 150) - (72 ÷ 7)
464 + 181 +? = (154 × 25) - (15) 2 Â
15% of 1800 + 22 = ?Â
