Question : A number is first increased by 16% and then increased by 20%. The number so obtained has now decreased by 40%. The net decrease percentage in the original number is:
Option 1: $16 \frac{12}{25}$%
Option 2: $15 \frac{9}{25}$%
Option 3: $11 \frac{13}{25}$%
Option 4: $13 \frac{7}{25}$%
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Correct Answer: $16 \frac{12}{25}$%
Solution : Let the number be $100x$. First increase = 16% Second increase = 20% Third decrease = 40% Resultant number = $100x × \frac{100+16}{100} × \frac{100+20}{100} × \frac{100-40}{100}$ = $\frac{2088}{25}x$ Percentage decrease = $\frac{100x - \frac{2088}{25}x}{100x}×100$ = $\frac{412}{25}$ = $16\frac{12}{25}$% Hence, the correct answer is $16\frac{12}{25}$%.
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Question : If the numerator of a fraction is increased by 140% and the denominator of the fraction is decreased by 20%, the resultant fraction is $\frac{12}{7}$. Find the original fraction.
Option 1: $\frac{4}{9}$
Option 2: $\frac{4}{7}$
Option 3: $\frac{7}{6}$
Option 4: $\frac{8}{9}$
Question : A number is first decreased by 20%. The decreased number is then increased by 20%. The resulting number is less than the original number by 20. Then the original number is:
Option 1: 200
Option 2: 400
Option 3: 500
Option 4: 600
Question : If a number is increased by 25% and the resulting number is decreased by 25%, then the percentage increase or decrease finally is:
Option 1: no change
Option 2: decreased by $6\tfrac{1}{4}$%
Option 3: increased by $6\tfrac{1}{4}$%
Option 4: increased by $6$%
Question : If $\left(4y-\frac{4}{y}\right)=11$, find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $7 \frac{9}{16}$
Option 2: $5 \frac{9}{16}$
Option 3: $9 \frac{11}{16}$
Option 4: $9 \frac{9}{16}$
Question : If the price of a product is first increased by 15% and then decreased by 20%, then what is the percentage change in the price?
Option 1: 5% decrease
Option 2: 8% increase
Option 3: 5% increase
Option 4: 8% decrease
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