Percentage - Aptitude MCQ Questions and Solutions with Explanations

\[\begin{array}{*{20}{c}} {D = }&{12\% }& \to &{2100} \\ {}&{1\% }& \to &{175} \\ {}&{100\% }& \to &{17500} \end{array}\]

According to the question,

1 unit → 2
3 units → 2 × 3 = 6 ereaser
4 units → 2 × 4 = 8 ereaser

$$\eqalign{ & \frac{{11}}{5}A = \frac{{22}}{{100}}B \cr & \frac{A}{B} = \frac{1}{{10}} \cr & B = \frac{{25}}{{1000}}C \cr & \frac{B}{C} = \frac{1}{{40}} \cr} $$
\[\begin{array}{*{20}{c}} {A\,\,\,:\,\,\,B\,\,\,:\,\,\,C} \\ {\,1\,\,\,\,\,\,\,\,\,\,10\,\,\,\,\,\,\,\,\,\,10} \\ {\,1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,40} \\ {\overline {\,\,1\,\,:\,\,40\,\,:\,\,400\,} } \end{array}\]
$$\eqalign{ & 400 \to 5500 \cr & 400 \to \frac{{55}}{4} \cr & A \to \frac{{55}}{4} \cr & B \to \frac{{550}}{4} \cr & {\text{According to the question,}} \cr & \left( {\frac{{80}}{{100}} \times \frac{{55}}{4}} \right) + \left( {\frac{{40}}{{100}} \times \frac{{550}}{4}} \right) \cr & = 11 + 55 \cr & = 66 \cr} $$

$$\eqalign{ & {\text{Base}} - 40\% = \frac{2}{5} \cr & {\text{Area}} - 60\% = \frac{3}{5} \cr & {\text{Area}} \to {\text{5:8}} \cr & {\text{Base}} \to {\text{5:7}} \cr & {\text{Height}} \to \frac{5}{5}:\frac{8}{7} \cr & \underbrace {7\,\,:\,\,8}_1 \cr & \frac{1}{7} \times 100 = 14.29\% \cr} $$

Number of students of all sections = 25, 30, 40, 45, 60
$$\eqalign{ & 25 \times \frac{1}{5} = 5 \cr & 30 \times \frac{3}{{10}} = 9 \cr & 40 \times \frac{7}{{20}} = 14 \cr & 45 \times \frac{2}{5} = 18 \cr & 60 \times \frac{{100}}{{100}} = 60 \cr} $$
106 → Passed student
Total student = 200
$${\text{Pass}}\% = \frac{{106}}{{200}} \times 100 = 53\% $$

Let the number of Crate = x
Spoiled Fruit = x
$$\eqalign{ & {\text{Now, }}\frac{{x \times 60}}{{100}} = 48 \cr & x = 80 \cr} $$
The number of Fruits = 80 × 25 = 2000

Given:
New selling price = Rs. 29 per kg
Old selling price = Rs. 25 per kg
Expenditure = constant
Formula used:
(i) Expenditure = Price × Quantity
(ii) Percentage difference = $$\frac{{{\text{New price}} - {\text{Old price}}}}{{{\text{Old price}}}} \times 100$$
Calculations:
Let expenditure, initial selling price and initial quantity be C, P₁ and Q₁ respectively.
C = P₁ × Q₁
⇒ C = 25 × Q₁
⇒ Q₁ = $$\frac{{\text{C}}}{{25}}$$
Let final selling price and final quantity be P₂ and Q₂ respectively.
C = P₂ × Q₂
⇒ C = 29 × Q₂
⇒ Q₂ = $$\frac{{\text{C}}}{{29}}$$

Percentage difference $$ = \frac{{{{\text{Q}}_2} - {{\text{Q}}_1}}}{{{{\text{Q}}_1}}} \times 100$$
$$\eqalign{ & \Rightarrow \left[ {\frac{{\frac{{\text{C}}}{{29}} - \frac{{\text{C}}}{{25}}}}{{\frac{{\text{C}}}{{25}}}}} \right] \times 100 \cr & \Rightarrow \left[ {\frac{{\frac{{25{\text{C}} - 29{\text{C}}}}{{25 \times 29}}}}{{\frac{{\text{C}}}{{25}}}}} \right] \times 100 \cr & \Rightarrow \frac{{ - \left( {\frac{{4{\text{C}}}}{{725}}} \right)}}{{\frac{{\text{C}}}{{25}}}} \times 100 \cr & \Rightarrow \frac{{ - \left( {4{\text{C}} \times {\text{25}}} \right)}}{{725{\text{C}}}} \times 100 \cr & \Rightarrow - 13.79\% \approx - 14\% \cr} $$
∴ By 14% should a family reduce its consumption, so as to keep the expenditure the same as before.

Alternate solution
\[\begin{array}{*{20}{c}} {}&{{\text{Price}}}&{{\text{Consumption}}} \\ {{\text{Old}}}&{25}&{{\mathbf{29}}} \\ {{\text{New}}}&{29}&{{\mathbf{25}}} \\ {{\text{Expenditure}}}&{25 \times 29}&{29 \times 25} \end{array}\]
Thus, family should reduce their consumption by 4 kg
% Reduction $$ = \frac{4}{{29}} \times 100 = 13.79 \approx 14\% $$
∴ The family should reduce the consumption by 14%.

Let number of boys = B and number of girls = G
According to the question,
B × 80% - G × 85% = 70
⇒ 80B - 85G = 7000 . . . . . . (i)
Again,
B × 20% + G × 15% = 90
⇒ 20B + 15G = 9000
⇒ 80B + 60G = 36000 . . . . . . (ii)
From equation (i) - (ii)
145G = 29000
⇒ G = 200
20B + 15 × 200 = 9000
⇒ 20B = 6000
⇒ B = 300
∴ Total students = 300 + 200 = 500

$$\eqalign{ & y \times \frac{1}{4} \times \frac{{30}}{{100}} \times 2.5 = x \times \frac{1}{4} \times \frac{1}{2} \cr & \frac{x}{y} = \frac{3}{2} \cr & x:y = \underbrace {3\,:\,2}_{ + 1} \cr & \% = \frac{1}{2} \times 100 = 50\% \cr} $$

$$\eqalign{ & {\text{Let total numbe of oranges}} = x \cr & \left\{ {\left( {\frac{{x \times 55}}{{100}} - 1} \right) \times \frac{{80}}{{100}} - 2} \right\} \times \frac{{10}}{{100}} = 5 \cr & \left( {\frac{{x \times 55}}{{100}} - 1} \right) \times \frac{{80}}{{100}} = 50 + 2 \cr & \left( {\frac{{x \times 55}}{{100}} - 1} \right) = \frac{{52 \times 10}}{8} \cr & \frac{{x \times 55}}{{100}} - 1 = 65 \cr & \frac{{x \times 55}}{{100}} = 66 \cr & x = \frac{{66 \times 100}}{{55}} \cr & x = 120 \cr} $$