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Percentage - Aptitude MCQ Questions and Solutions with Explanations

Percentage MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Percentage with easy and logical explanations. Percentage MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-10 section-5

Each section contains 100 MCQs on Percentage .
91
The population of a town is 4.2 × 106. If the population increases by 75 per 1000 per annum, then what will be the population after 2 years ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Percentage annual increase :
$$\eqalign{ & = \left( {\frac{{75}}{{1000}} \times 100} \right)\% \cr & = 7\frac{1}{2}\% \cr & = \frac{{15}}{2}\% \cr} $$
Population after 2 years :
$$\eqalign{ & = 4200000{\left( {1 + \frac{{15}}{{2 \times 100}}} \right)^2} \cr & = \left( {4200000 \times \frac{{43}}{{40}} \times \frac{{43}}{{40}}} \right) \cr & = 4853625 \cr} $$
92
A's marks in Biology are 20 less than 25% of the total marks obtained by him in Biology, Maths and Drawing. If his marks in Drawings be 50, what are his marks in Maths ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let B + M + D = $$x$$
Then,
B = 25 of $$x$$ - 20
$$\eqalign{ & = \left( {\frac{{25x}}{{100}} - 20} \right) \cr & = \left( {\frac{x}{4} - 20} \right) \cr} $$
And D = 50
∴ $$\frac{x}{4}$$ - 20 + M + 50 = $$x$$
or M = $$\left( {\frac{{3x}}{{4}} - 30} \right)$$
So, marks in Maths cannot be determined.
93
If x = 63% of y, then y2 is approximately what percent of x2 ?
Discuss
Answer & Solution
Answer: Option C
Solution:
x = 63% of y
⇒ x = $$\frac{63y}{100}$$
⇒ y = $$\frac{100x}{63}$$
∴ Required percentage :
$$\eqalign{ & = \left( {\frac{{{y^2}}}{{{x^2}}} \times 100} \right)\% \cr & = \left[ {{{\left( {\frac{{100x}}{{63}}} \right)}^2} \times \frac{1}{{{x^2}}} \times 100} \right]\% \cr & = \left( {\frac{{10000}}{{3969}} \times 100} \right)\% \cr & = 251.96\% \approx 250\% \cr} $$
94
A number is first decreased by 10% and then increased by 10%. The number so obtained is 50 less than the original number. The original number is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the original number be $$x$$
Final number obtained :
= 110% of (90% of $$x$$)
= $$\left( {\frac{{110}}{{100}} \times \frac{{90}}{{100}} \times x} \right)$$
= $$\frac{{99x}}{{100}}$$
$$\eqalign{ & \therefore x - \frac{{99x}}{{100}} = 50 \cr & \Rightarrow \frac{x}{{100}} = 50 \cr & \Rightarrow x = 50 \times 100 \cr & \Rightarrow x = 5000 \cr} $$
95
A city has a population of 300000 out of which 180000 are males. 50% of the population is literate. If 70% of the males are literate, then the percentage of female who are literate is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Number of males = 180000
Number of females :
= (300000 - 180000)
= 120000
Number of literates :
= 50% of 300000
= 150000
Number of literate males :
= 70% of 180000
= 126000
Number of literate females :
= (150000 - 126000)
= 24000
∴ Required percentage :
= $$\left( {\frac{{24000}}{{120000}} \times 100} \right)\% $$
= 20%
96
The price of an articles was increased by r%. Later the new price was decreased by r%. If the latest price was Rs. 1, then the original price was :
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the original price be Rs. $$x$$
∴ (100 - r)% of (100 + r)% of $$x$$ = 1
$$\eqalign{ & \Rightarrow \frac{{\left( {100 - r} \right)}}{{100}} \times \frac{{\left( {100 + r} \right)}}{{100}} \times x \cr & \Rightarrow x = \frac{{100 \times 100}}{{\left( {100 - r} \right)\left( {100 + r} \right)}} \cr & \Rightarrow x = \frac{{10000}}{{\left( {10000 - {r^2}} \right)}} \cr} $$
97
In September 2009, the sales of a product were $$\frac{2}{3}$$rd of the that in July 2009. In November 2009, the sales of the product were higher by 5% as compared to September 2009. How much is the percentage of increase on sales in November 2009 with respect to the base figure in July 2009 ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the total sales in July 2009 be Rs. $$x$$
Then, sales in September 2009 = Rs. $$\frac{2x}{3}$$
Sales in November 2009 :
= 105% of Rs. $$\frac{2x}{3}$$
= Rs. $$\left( {\frac{{105}}{{100}} \times \frac{{2x}}{3}} \right)$$
= Rs. $$\frac{7x}{10}$$
Decrease in sales :
= Rs. $$\left( {x - \frac{{7x}}{{10}}} \right)$$
. = Rs. $$\frac{3x}{10}$$
∴ Decrease % :
= $$\left( {\frac{{3x}}{{10}} \times \frac{1}{x} \times 100} \right)\% $$
= 30%
98
Two numbers are respectively 25% and 60% more than a third number. The ratio of the two numbers is:
Discuss
Answer & Solution
Answer: Option C
Solution:
\[\begin{array}{*{20}{c}} {\text{I}}&{}&{{\text{II}}}&{{\text{III}}} \\ {125}&:&{160}&{100} \\ {25}&:&{32}&{} \end{array}\]
99
The income of A is 80% of B's income and the expenditure of A is 60% of B's expenditure. If the income of A is equal to 90% of B's expenditure, then by what percentage are the saving of A more than B's savings?
Discuss
Answer & Solution
Answer: Option A
Solution:
Income of A is 80% of B
$$\eqalign{ & A = B \times \frac{{80}}{{100}} \cr & \frac{A}{B} = \frac{{4x}}{{5x}} \cr} $$
Expenditure of A is 60% of expenditure of B
$$\eqalign{ & A = B \times \frac{{60}}{{100}} \cr & \frac{A}{B} = \frac{{3y}}{{5y}} \cr} $$
Income of A is 90% of income of B
$$\eqalign{ & 4x = \frac{{90}}{{100}} \times 5y \cr & 8x = 9y \cr & \frac{x}{y} = \frac{9}{8} \cr} $$
Saving ratio of A and B
= (4x - 3y) : (5x - 5y)   [∴ 8x = 9y]
= (4 × 9 - 3 × 8) : (5 × 9 - 5 × 8)
= (36 - 24) : (45 - 40)
= 12 : 5
More = 12 - 5 = 7
More% $$ = \frac{7}{5} \times 100 = 140\% $$

Alternate solution
\[\begin{array}{*{20}{c}} {}&{{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}} \\ {{\text{Income}} \to }&{4 \times 5 \times 9 = 180\,\,\,\,\,\,\,\,\,\,5 \times 5 \times 9 = 225} \\ {{\text{Expenditure}} \to }&{3 \times 4 \times 10 = 120\,\,\,\,\,\,\,\,\,\,5 \times 4 \times 10 = 200} \\ {{\text{Saving}} \to }&{\overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,60\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,25\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} } } \end{array}\]
Ratio of A saving to B saving = 60 : 25 = 12 : 5
More = 12 - 5 = 7
More% $$ = \frac{7}{5} \times 100 = 140\% $$
100
What is to be added to 15% of 180 so that the sum is equal to 20% of 360?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let be the number is }}x \cr & 180 \times \frac{{15}}{{100}} + x = \frac{{360 \times 20}}{{100}} \cr & 27 + x = 72 \cr & x = 72 - 27 \cr & x = 45 \cr} $$