Question
In a certain pond, there are 60 more crabs than fish.
Unfortunately, 20% of the crabs have perished. Additionally, the pond has welcomed 80 new fish. As a result, the current ratio of crabs to fish is 3:4. Determine the original quantities of crabs and fish in the pond.Solution
Let the initial number of crabs in the pond be ‘x’ Therefore, initial number of fish in the pond = (x – 60) According to the question, 0.8x/(x – 60 + 80) = 3/4 Or, 3.2x – 3x = 60 Or, x = 60/0.2 = 300 Therefore, initial number of fish and crabs in the pond = (x + x – 60) = 540
5983.987 + 59832.999 – 598.873 = ?
24.98% of 1682 × (18.2659 ÷ 9.04965)(–4) = ?
√81.02 + 11.836 of 24.98 = ?2 + 20.01
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.)...
[√ (121.23) ÷ √ (12100.04)] × √ 80.95 = 3/10 + ? ÷ 4
...63.95 – 21.12 – 24.89 + 6.04 = 3.98 × ? + 3.88
7480 ÷ 16.98 – 34.32 ÷ (4.99/99.9) = ?
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.)...
(9/10 of 3999.79) - √2499.83 + (17.81% of 1199.81) = ?
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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