Question : The population of country A decreased by p% and the population of country B decreased by q% from the year 2020 to 2021. Here 'p' is greater than ‘q’. Let 'x' be the ratio of the population of country A to the population of country B in the given year. What is the percentage decrease in 'x' from 2020 to 2021?
Option 1: $\frac{100(p-q)}{(100-q)}$
Option 2: $\frac{100(p-q)}{(100+P)}$
Option 3: $\frac{100(p-q)}{(100-p)}$
Option 4: $\frac{100(p-q)}{(100+q)}$
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Correct Answer: $\frac{100(p-q)}{(100-q)}$
Solution : Let the initial population of Country A and B be $J$ and $K$ respectively. Hence, $x = \frac{J}{K}$ Let the ratio of the population of country A to the population of country B for the next year be $x_{1}$. According to the question, $(J \times (1 - p\%)) : (K \times (1 - q\%)) = x_{1}$. ⇒ $\frac{J}{K}\times \frac{100−p}{100−q}=x_{1}$. Now, required % decrease = $\frac{x-x_{1}}{x}\times 100$% = $\frac{\frac{J}{K}-\frac{J}{K}\times \frac{100−p}{100−q}}{\frac{J}{K}}×100$% = $ \frac{p-q}{100-q}\times 100$% = $ \frac{100(p-q)}{100-q}$% ∴ The percentage decrease in 'x' from 2020 to 2021 is $ \frac{100(p-q)}{100-q}$%. Hence, the correct answer is $\frac{100(p-q)}{100-q}$.
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Question : Successive discounts of $p$% and $q$% on a catalogue price of an article is equivalent to a single discount of:
Option 1: $\left(p -q + \frac{pq}{100} \right )$%
Option 2: $\left(p - q - \frac{pq}{100} \right )$%
Option 3: $\left(p + q - \frac{pq}{100} \right )$%
Option 4: $\left(p + q + \frac{pq}{100} \right )$%
Question : From 1980 to 1990, the population of the country increased by 20%. From 1990 to 2000, the population of the country increased by 20%. Over 2000-2010, the population of the country increased by 20%. Then, the overall increased population (in percentage) of the country from 1980 to 2010 was:
Option 1: 72.2%
Option 2: 60%
Option 3: 72.8%
Option 4: 62.8%
Question : If A borrowed Rs. $P$ at $x$% and B borrowed Rs. $Q (> P)$ at $y$% per annum in simple interest at the same time, then the amount of their debts will be equal after:
Option 1: $100(\frac{Q-P}{Px-Qy})$
Option 2: $100(\frac{Px-Qy}{Q-P})$
Option 3: $100(\frac{Px-Qy}{P-Q})$
Option 4: $100(\frac{P-Q}{Px-Qy})$
Question : A mask manufacturing company manufactured an ‘$x$’ number of masks in 2018. It increased its manufacturing capacity by 30% in 2019 and further increased its manufacturing by 15% in 2020. In 2021, due to the machinery breakdown, its manufacturing declined by 40%. What is the value of ‘X’ if it manufactured 179400 masks in 2021?
Option 1: 180000
Option 2: 230000
Option 3: 200000
Option 4: 210000
Question : A water tap fills a tub in '$p$' hours and a sink at the bottom empties it in '$q$' hours. If $p < q$, both tap and sink are open, and the tank is filled in '$r$' hours, then:
Option 1: $\frac{1}{r}$ = $\frac{1}{p}+\frac{1}{q}$
Option 2: $\frac{1}{r}$ = $\frac{1}{p}-\frac{1}{q}$
Option 3: $r = p + q$
Option 4: $r = p - q$
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