Question
A 220-litre mixture contains petrol and water, only in
ratio 6:5, respectively. How much water should be added in the mixture so that the quantity of water becomes 50% of quantity of resultant mixture?Solution
Quantity of water in the initial mixture = 220 × (5/11) = 100 litres Quantity of petrol in the initial mixture = 220 × (6/11) = 120 litres Since, we’re not adding petrol, 120 litres of petrol will represent 50% of the new mixture. So, total quantity of resultant mixture = 120 ÷ 0.5 = 240 litres Quantity of water to be added = 240 – 220 = 20 litres
Find the simplified value of the given expression:
7 of 9 ÷ 3 × 5² + √81 – 14- Calculate the value of the following expression:
{(481 + 426) 2 - 4 × 481 × 426} = ?
Find the value of the expression:
18 + 12 – 4 × [22 + 6 – 2 × (38 – 23)]What value should come in the place of (?) in the following questions?
162 – 12 * 20 = 6 * ? - 3√8000
654.056 + 28.9015 × 44.851 – 43.129 = ?
150% of 850 ÷ 25 – 25 = ?% of (39312 ÷ 1512)
58% of 3300 - ?% of 2900 = 1740
What will come in the place of question mark (?) in the given expression?
(√441 + √729) X 8 = ? + 518 ÷ 2
What will come in place of the question mark (?) in the following expression?
?% of 600 + 15 × 28 = 30% of 1800
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