Percentage - Aptitude MCQ Questions and Solutions with Explanations

We know that
⇒ Total surface area of a cube
⇒ 6a²
⇒ If each side is doubled i.e. 2a
⇒ Total surface area of a cube
⇒ 6 × (2a)²
⇒ 24a²
Increase in surface area = 24a² - 6a² = 18a²
∴ % Surface area of cube will increase :
$$\frac{{18{a^2}}}{{6{a^2}}} \times 100 $$
= 300%

Failed students in Mathematics = 35%
Failed students in English = 25%
Student failed in both subject Mathematics and english = 10%
Student only fail in Mathematics = 35 - 10 = 25%
Student only fail in English = 25 - 10 = 15%
Percentage of passed students in both subjects :
= 100 - [ student fail in Mathematics + student fail in english + student fail in both subject]
= [100 - (25 + 15 + 10)]
= 50%

4% = $$\frac{1}{25}$$
\[\begin{gathered} {\text{Initial }}\,\,\,\,\,\,{\text{Final}} \hfill \\ \,\,\,25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,26 \hfill \\ \,\,\,25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,26 \hfill \\ \overline {\,\,\,\,\,625\,\,\,\,\,\,\,\,\,\,\,676\,\,\,} \hfill \\ \end{gathered} \]
According to the question,
625 units = 50000
1 unit     = $$\frac{50000}{625}$$ = 80
676 units = 80 × 676 = 54080
Hence, population after two years = 54080

Note :In such type of questions use allegation method to save your valuable time.

Required number of males
= $$\frac{8000}{(1 + 1)}$$  × 1
= 4000

Number of females :
= 25000 × $$\frac{1}{5}$$
= 5000
Number of males :
= 25000 - 5000
= 20000
Number of educated females :
= 5000 × $$\frac{60}{100}$$
= 3000
Number of educated males
= 20000 × $$\frac{95}{100}$$
= 19000
Total educated population = 22000
Percentage of educated population
= $$\frac{22000}{25000}$$  × 100
= 88%

The production of cycles rose to 48400 from 40000 in 2 years
⇒ Present production = 40000
⇒ After two years = 48000
⇒ Time = 2 years
⇒ Rate of increment = ?
According to the question,
Production after 2 years = Present production $${\left( {1 + \frac{{\text{R}}}{{100}}} \right)^t}$$
$$\eqalign{ & \Rightarrow 48400 = 40000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{484}}{{400}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{22}}{{20}} \cr & \Rightarrow \frac{{\text{R}}}{{100}} = \frac{1}{{10}} \cr & \Rightarrow {\text{R}} = 10\% \cr} $$
⇔ Rate of increment = 10%

Total marks
= (300 + 200) × $$\frac{46}{100}$$
= 230
Marks obtained by the students in science
= 300 × $$\frac{32}{100}$$
= 96
Required marks in language paper
= (230 - 96)
= 134
Required percentage
= $$\frac{134}{200}$$ × 100
= 67%
Hence, required percentage = 67%

Let the initial population = x
According to the question,
$$\eqalign{ & x \times \frac{{104}}{{100}} \times \frac{{104}}{{100}} = 67600 \cr & x = \frac{{67600 \times 100 \times 100}}{{104 \times 104}} \cr & \,\,\,\,\,\, = 62500 \cr} $$
Hence required population = 62500

Let total number of student apply in college 100x
40% student alloted in group A , i.e. 40x
Remaining Student 100x - 40x = 60x
Now 75% of remaining student got alloted in Group B
i.e. 75% of 60x = 45x
Remain student is 60x - 45x = 15x (which alloted in Group C)
According to question
15x = 12
x = $$ \frac{12}{15} $$
Total Number of Student applied in college = 100x
= 100 x $$ \frac{12}{15} $$
= 80

Required apples :
= $$\frac{420}{(100 - 40)}$$  × 100
= $$\frac{420}{60}$$ × 100
= 700