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: $13 \frac{7}{25}$%
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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 $16\frac{12}{25}$%.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
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.
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:
Question : If a number is increased by 25% and the resulting number is decreased by 25%, then the percentage increase or decrease finally is:
Question : If $\left(4y-\frac{4}{y}\right)=11$, find the value of $\left(y^2+\frac{1}{y^2}\right)$.
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?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile