Percentage - Aptitude MCQ Questions and Solutions with Explanations

Population after 2 years :
$$\eqalign{ & = 8500 \times \left( {1 + \frac{{20}}{{100}}} \right)\left( {1 + \frac{{25}}{{100}}} \right) \cr & = \left( {850 \times \frac{6}{5} \times \frac{5}{4}} \right) \cr & = 12750 \cr} $$

$$\eqalign{ & n\left( A \right) = 90 \cr & n\left( B \right) = 15 \cr & n\left( {A \cup B} \right) = 100 \cr & {\text{So, }} \cr & {\text{n}}\left( {A \cap B} \right) \cr & = n\left( A \right) + n\left( B \right) - n\left( {A \cup B} \right) \cr & = 90 + 15 - 100 \cr & = 5 \cr} $$
∴ Percentage of people who own both = 5%

Original length of AB = 5 cm
Original length of AC = 3 cm
Original length of CB = (5 - 3) cm = 2 cm
New length AC :
= 106% of 3 cm
= $$\left( {\frac{{106}}{{100}} \times 3} \right)$$  cm
= 3.18 cm
New length of CB :
= (5 - 3.18) cm
= 1.82 cm
Decrease in length of CB :
= (2 - 1.82) cm
= 0.18 cm
∴ Decrease % :
= $$\left( {\frac{{0.18}}{2} \times 100} \right)\% $$
= 9%

$$\eqalign{ & n\left( A \right) = 34 \cr & n\left( B \right) = 42 \cr & n\left( {A \cap B} \right) = 20 \cr & {\text{So, }} \cr & {\text{n}}\left( {A \cup B} \right) \cr & = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & = 34 + 42 - 20 \cr & = 56 \cr} $$
∴ Percentage failed in either or both the subjects = 56%

$$\eqalign{ & x\% {\text{ of }}a = y\% {\text{ of }}a \cr & \Rightarrow \frac{x}{{100}}a = \frac{y}{{100}}b \cr & \Rightarrow b = \left( {\frac{x}{y}} \right)a \cr & \therefore z\% {\text{ of }}b: \cr & = \left( {z\% {\text{ of }}\frac{x}{y}} \right)a \cr & = \left( {\frac{{xz}}{{y \times 100}}} \right)a \cr & = \left( {\frac{{xz}}{y}} \right)\% {\text{ of }}a \cr} $$

Let the original price of sugar be Rs. $$x$$ per kg
Then, reduced price :
$$\eqalign{ & 97\frac{1}{2}\% {\text{ of Rs}}{\text{. }}x \cr & = {\text{Rs}}{\text{. }}\left( {\frac{{195}}{2} \times \frac{1}{{100}} \times x} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{39x}}{{40}} \cr & \therefore \frac{{1260}}{{\left( {\frac{{39x}}{{40}}} \right)}} - \frac{{1260}}{x} = 9 \cr & \Rightarrow \frac{{16800}}{{13x}} - \frac{{1260}}{x} = 9 \cr & \Rightarrow 13x = \frac{{420}}{9} \cr & \Rightarrow x = \frac{{140}}{{39}} \cr} $$
Increased price :
$$\eqalign{ & 112\frac{1}{2}\% {\text{ of Rs}}{\text{. }}\frac{{140}}{{39}} \cr & = {\text{Rs}}{\text{. }}\left( {\frac{{225}}{2} \times \frac{1}{{100}} \times \frac{{140}}{{39}}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{105}}{{26}} \cr} $$
∴ Quiantity of sugar bought for Rs. 1260 :
$$\eqalign{ & = \left( {1260 \times \frac{{26}}{{105}}} \right){\text{ kg}} \cr & = {\text{312 kg}} \cr} $$

Let the original price of each ticket be Rs. 100
Then, original price of 5 tickets = Rs. 500
Sale price of 5 tickets = Rs. 300
Amount saved :
= Rs. (500 - 300)
= Rs. 200
∴ Required percentage :
= $$\left( {\frac{{200}}{{500}} \times 100} \right)$$  %
= 40%

Let original consumption = 100 kg and new consumption = $$x$$ kg
So,
⇔ 100 × 16 = $$x$$ × 20
⇔ $$x$$ = 80
∴ Reduction in consumption = 20%

$$\eqalign{ & 50\% {\text{ of }}\left( {x - y} \right) = 30\% {\text{ of }}\left( {x + y} \right) \cr & \Rightarrow 5\left( {x - y} \right) = 3\left( {x + y} \right) \cr & \Rightarrow 5x - 5y = 3x + 3y \cr & \Rightarrow 2x = 8y \cr & \Rightarrow y = \frac{x}{4} \cr} $$
∴ Required percentage :
$$\eqalign{ & = \left( {\frac{y}{x} \times 100} \right)\% \cr & = \left( {\frac{x}{4} \times \frac{1}{x} \times 100} \right)\% \cr & = 25\% \cr} $$

Let the number of students in the class be 100 and let the required average be $$x$$
Then,
⇒ (10 × 95) + (20 × 90) + (70 × $$x$$) = (100 × 80)
⇒ 70$$x$$ = 8000 - (950 + 1800)
⇒ 70$$x$$ = 5250
⇒ $$x$$ = 75