Percentage - Aptitude MCQ Questions and Solutions with Explanations

Let the original taxed amount be Rs. x and new taxed amount be Rs. y
Let the original revenue be Rs. 100
Then, 4% of x = 100 or
x = $$\left( {\frac{{100 \times 100}}{{4}}} \right)$$
x = Rs. 2500
New revenue = 110% of Rs. 100 = Rs. 110
Then, 5% of y = 110 or
y = $$\left( {\frac{{110 \times 100}}{{5}}} \right)$$
y = Rs. 2200
Decrease in taxed amount :
= Rs. (2500 - 2200)
= Rs. 300
∴ Decrease % :
$$=$$$${\text{ }}\left( {\frac{{300}}{{2500}} \times 100} \right)$$
$$= 12$$ %

Population at the end of 2 years
= 189000 $$\left( {1 - \frac{8}{{100}}} \right)$$ $$\left( {1 + \frac{5}{{100}}} \right)$$
= $$\left( {189000 \times \frac{{23}}{{25}} \times \frac{{21}}{{20}}} \right)$$
= 182574

Let each vessel contain 100 litres of 40% alcohol
Suppose Sachine added x litres of pure alcohol.
Then,
$$\eqalign{ & \Leftrightarrow \frac{{40 + x}}{{100 + x}} = \frac{{50}}{{100}} \cr & \Leftrightarrow \frac{{40 + x}}{{100 + x}} = \frac{1}{2} \cr & \Leftrightarrow 80 + 2x = 100 + x \cr & \Leftrightarrow x = 20 \cr} $$
Suppose Vivek replaced y litres.
Then, alcohol in y litres = 40% of y = $$\frac{2y}{5}$$
$$\eqalign{ & \therefore \frac{{40 - \frac{{2y}}{5} + y}}{{100}} = \frac{{50}}{{100}} \cr & \Rightarrow \frac{{40 - \frac{{2y}}{5} + y}}{{100}} = \frac{1}{2} \cr & \Rightarrow 80 + \frac{{6y}}{{25}} = 100 \cr & \Rightarrow y = \frac{{20 \times 5}}{6} \cr & \Rightarrow y = \frac{{50}}{3} \cr} $$
Required percentage :
$$\eqalign{ & = \left[ {\frac{{\left( {20 - \frac{{50}}{3}} \right)}}{{\left( {\frac{{50}}{3}} \right)}} \times 100} \right]\% \cr & = \left( {\frac{{10}}{3} \times \frac{3}{{50}} \times 100} \right)\% \cr & = 20\% \cr} $$

Actual cost price of an article :
= Rs. (5844 + 156)
= Rs. 6000
Selling price of that article = Rs. 5700
Loss = Rs. (6000 - 5700) = Rs. 300
Loss percent :
= $$\frac{300}{6000}$$ × 100
= 5%

Rebate = 6% of Rs. 6650
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {\frac{6}{{100}} \times 6650} \right) \cr & = {\text{Rs}}{\text{. 399}} \cr} $$
Sales tax = 10 % of (6650 - 399)
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {\frac{{10}}{{100}} \times 6251} \right) \cr & = {\text{Rs}}{\text{. 625}}{\text{.10}} \cr} $$
∴ Final amount = Rs. (6251 + 625.10)
                         = Rs. 6876.10

Let the required number of chicken be x
Then, 65% of x = 47775
$$\eqalign{ & \Rightarrow \frac{{65}}{{100}}x = 47775 \cr & \Rightarrow x = \frac{{47775 \times 100}}{{65}} \cr & \Rightarrow x = 73500 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Rightarrow x = 125 = 37\frac{1}{2}\% {\text{ of }}x \cr & \Rightarrow x - 125 = \frac{{75}}{2} \times \frac{1}{{100}} \times x \cr & \Rightarrow x - 125 = \frac{{3x}}{8} \cr & \Rightarrow x - \frac{{3x}}{8} = 125 \cr & \Rightarrow \frac{{5x}}{8} = 125 \cr & \Rightarrow x = \left( {\frac{{125 \times 8}}{5}} \right) \cr & \Rightarrow x = 200 \cr & \therefore 25\% {\text{ of 200}} \cr & = \left( {\frac{{25}}{{100}} \times 200} \right) \cr & = 50 \cr} $$

Let the maximum marks be x.
Then,
$$\eqalign{ & \left( {40 - 4} \right)\% {\text{ of }}x = 261 \cr & \Rightarrow 36\% {\text{ of }}x = 261 \cr & \Rightarrow \frac{{36}}{{100}}x = 261 \cr & \Rightarrow x = \left( {\frac{{261 \times 100}}{{36}}} \right) \cr & \Rightarrow x = 725 \cr} $$

$$\frac{x}{100}$$ × y = $$\frac{4}{5}$$ × 80
⇒ xy = 64 × 100
⇒ xy = 6400

Total number of fruits = (14 + 23) = 37
Let x oranges be removed.
Then,
70% of (37 - x) = 14
⇒ 7 (37 - x) = 140
⇒ 37 - x = 20
⇒ x = 17