Question
A mixture contains 70% acid and rest water. A man
withdrew some mixture and replaced it with same quantity of kerosene, such that the quantity of acid remaining in the mixture now is 56%. If the man wants to bring down the quantity of acid to less than 36%, then how many more times should he repeat the same process?Solution
Let the percentage of mixture withdrawn first time be ‘p%’. ATQ; 70 × {1 − (p/100)} = 56 Or,
1 − (p/100) = 0.8 Or,
(p/100) = 0.2 So,
p = 20 So, 20% mixture is drawn every time. Now,
Quantity of acid remaining after withdrawing once more = 56 × 0.8 = 44.8 And,
Quantity of acid remaining after withdrawing once more = 44.8 × 0.8 = 35.84 So, after withdrawing two more times, the quantity of acid reaches less than 36%.
20.05% of 150.05 – 12.15% of 99.99 × 2.02 = ?
(?)2 + 8.113 = 28.92 – 73.03
11.89 × 2.10 × 4.98 × 4.03 ÷ 7.98 of 15.03 = ?
(?)2 + 4.113 = 23.92 – 28.03
25.11% of 199.99 + √143.97 ÷ 6.02 = ?
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
`(601/49)xx(399/81)` ÷ `(29/201)` = ?
(10.013 – 12.04) = ? + 7.98% of 4999.98
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value....
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