Percentage - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & n\left( A \right) = \left( {\frac{{75}}{{100}} \times 600} \right) = 450 \cr & n\left( B \right) = \left( {\frac{{45}}{{100}} \times 600} \right) = 270 \cr & \therefore n\left( {A \cap B} \right) \cr} $$
$$ = n\left( A \right) + n\left( B \right) - n\left( {A \cup B} \right)$$     $$n\left( {A \cup B} \right)$$   $$ -\, 600$$
$$\eqalign{ & = \left( {450 + 270 - 600} \right) \cr & = 120 \cr} $$

Let original income = Rs. 100
Then, saving = Rs. 20 and expenditure = Rs. 80
New income :
$$=$$ Rs. $$116\frac{2}{3}$$
$$=$$ Rs. $$\frac{350}{3}$$
New expenditure :
$$=$$ $$137\frac{1}{2}$$ % of Rs. 80
$$=$$ Rs. $$\left( {\frac{{275}}{2} \times \frac{1}{{100}} \times 80} \right)$$
$$=$$ Rs. 110
New saving :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{350}}{3} - 110} \right) \cr & = {\text{Rs}}{\text{.}}\frac{{20}}{3} \cr} $$
∴ Required percentage
$$\eqalign{ & = \left( {\frac{{20}}{3} \times \frac{3}{{350}} \times 100} \right)\% \cr & = \frac{{40}}{7}\% \cr & = 5\frac{5}{7}\% \cr} $$

Let the total number of candidates be x
Number of capable candidates who got admitted to college :
= 80% of 40% of x
$$\eqalign{ & = \left( {\frac{{80}}{{100}} \times \frac{{40}}{{100}} \times x} \right) \cr & = \frac{{8x}}{{25}} \cr} $$
Number of incapable college students :
= 25% of (100 - 40)% of x
$$\eqalign{ & = \left( {\frac{{25}}{{100}} \times \frac{{60}}{{100}} \times x} \right) \cr & = \frac{{3x}}{{20}} \cr} $$
Total number of candidates who got admitted to college :
$$=$$ $$\frac{{8x}}{{25}}$$ + $$\frac{{3x}}{{20}}$$
$$=$$ $$\frac{{47x}}{{100}}$$
∴ Required percentage :
$$\eqalign{ & = \left( {\frac{{8x}}{{25}} \times \frac{{100}}{{47x}} \times 100} \right)\% \cr & = \left( {\frac{{3200}}{{47}}} \right)\% \cr & = 68.09\% \approx 68\% \cr} $$

Required percentage :
$$\eqalign{ & = \left( {\frac{{18}}{{7200}} \times 100} \right)\% \cr & = \frac{1}{4}\% \cr & = 0.25 \% \cr} $$

Let x% of 450 + 46% of 285 = 257.1
$$ \Rightarrow \left( {\frac{x}{{100}} \times 450} \right) + \left( {\frac{{46}}{{100}} \times 285} \right)$$     $$ = 257.1$$
$$\eqalign{ & \Rightarrow \frac{{9x}}{2} = 257.1 - 131.1 \cr & \Rightarrow \frac{{9x}}{2} = 126 \cr & \Rightarrow x = \frac{{126 \times 2}}{9} \cr & \Rightarrow x = 28 \cr} $$

Error = (45° 27' - 45° ) = 27'
Accurate measure = 45° = (45 × 60)' = 2700'
∴ Percentage error :
$$\eqalign{ & = \left( {\frac{{27}}{{2700}} \times 100} \right)\% \cr & = 1\% \cr} $$

Let the capacity of the tank be 100 litres.
Then,
Initially :
A type petrol = 100 litres
After first operation :
A type petrol = $$\left( {\frac{{100}}{{2}}} \right)$$ = 50 litres ;
B type petrol = 50 litres
After second operation :
A type petrol = $$\left( {\frac{{50}}{2} + 50} \right)$$  = 75 litres ;
B type petrol = $$\left( {\frac{{50}}{2}} \right)$$ = 25 litres
After third operation :
A type petrol = $$\left( {\frac{{75}}{2}} \right)$$ = 37.5 litres
B type petrol = $$\left( {\frac{{25}}{2} + 50} \right)$$  = 62.5 litres
∴ Required percentage = 37.5%

Let the capacity of the tank be 100 litres.
Then,
Initially :
A type petrol = 100 litres
After first operation :
A type petrol = $$\left( {\frac{{100}}{{2}}} \right)$$ = 50 litres ;
B type petrol = 50 litres
After second operation :
A type petrol = $$\left( {\frac{{50}}{2} + 50} \right)$$  = 75 litres
B type petrol = $$\left( {\frac{{50}}{2}} \right)$$ = 25 litres
After third operation :
A type petrol = $$\left( {\frac{{75}}{2}} \right)$$ = 37.5 litres
B type petrol = $$\left( {\frac{{25}}{2} + 50} \right)$$  = 62.5 litres
∴ Required percentage = 37.5%

15% of 578 + 22.5% of 644
$$ = \left( {\frac{{15}}{{100}} \times 578} \right) + $$   $$\left( {\frac{{225}}{{10}} \times \frac{1}{{100}} \times 644} \right)$$
$$\eqalign{ & = 86.7 + 144.9 \cr & = 231.6 \cr} $$

105.27% of 1200.11 + 11.80% of 2360.85 = 21.99% of (?) + 1420.99
⇒ 105% of 1200 + 12% of 2360 = 22% of (?) + 1421
⇒ $$\frac{{105 \times 1200}}{{100}} + \frac{{12 \times 2360}}{{100}} = $$     $$\frac{{22 \times (?)}}{{100}}$$   $$ + \,1421$$
$$ \Rightarrow\frac{{22 \times (?)}}{{100}} = 1260 + 283.30$$       $$ - \,1421$$
$$\eqalign{ & \Rightarrow \frac{{22 \times (?)}}{{100}} = 1543 - 1421 \cr & \Rightarrow \frac{{22 \times (?)}}{{100}} = 122 \cr & \therefore (?) = \frac{{122 \times 100}}{{22}} \cr & \Rightarrow (?) = \frac{{122 \times 50}}{{11}} \cr & \Rightarrow (?) = 554.5\,\, (approximate\,\, 550) \cr} $$