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Two numbers are less than a third number by 37% and 30%, respectively. By what percentage is the second number more than the first? |
$\frac{100}{9}$ $\frac{10}{9}$ 10 $\frac{100}{3}$ |
$\frac{100}{9}$ |
Let these three numbers are :- A , B & C Let C = 100 ATQ, A = 100 - 37 = 63 & B = 100 - 30 = 70 Required percentage = \(\frac{70 - 63}{63}\) × 100 = \(\frac{100}{9}\) % |
