Question
A lab technician has a 500-litre solution containing a mix
of salt and water in a 9:11 ratio. If 40% of this solution is removed and replaced with an equal amount of water, what is the new ratio of salt to water in the final mixture?Solution
ATQ;
Quantity of salt in the solution = (9/20) × 500 = 225 litres
Quantity of water in the solution = (500 – 225) = 275 litres
Quantity of solution taken out = (0.4 × 500) = 200 litres
New ratio = {225 – (225 × 0.4)} : {275 – (275 × 0.4) + 200}
= 135 : 335
= 27:67
2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 19Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 70...
6000 3002 1503 ? 378.75 191.375 97.6875
...If  204    196       223   x  284
Then, what is the average of the numbers of the above series?
...8   24    12    ?   18     54
3 ? 7 16 71 346
...104   106   110   113   ?   126
12, 18, 28, 42, 52, ?
18Â Â Â Â Â Â Â Â Â Â Â Â 29 Â Â Â Â Â Â Â Â Â Â Â Â Â Â 51 Â Â Â Â Â Â Â Â 84 Â Â Â Â Â Â Â Â 128 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 182
5, 8, 17, ?, 37, 48
(32.03 + 111.98) ÷ 18.211 = 89.9 – 20.23% of ?
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