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Quant Aptitude: Percentage Solved Questions

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Practicing Aptitude solved problems from section Percentage and Fraction

Practicing fully solved Percentage and fraction problems from Aptitude section will be useful for exams like IIFT, MAT,FMS, SAT etc. Online practice of percent and fraction, aptitude questions with detailed explanation will help you in building your concepts well in percentage and fraction section. Multiple choice aptitude problems on percents and fractions with solutions are provided for online practice. Fully solved questions on compound interest and discount margin are also available in this aptitude section.

Why practice online aptitude question on Percent and Fraction?

Only theory of profit loss and compound interest will not be sufficient for clearing MBA exams, online practice of aptitude problems is required. In this section solved questions of aptitude on profit loss and percentage fractions are provided.

Full explanation of aptitude problems on percentages

A complete description and explanation on percentages and fractions problems is given in this section aptitude. These online questions with detailed answer to each problem will be useful for exams like CAT, XAT,MAT etc. Tips and tricks provided while explain the concepts on aptitude will be useful for solving questions in this category of percentage and fraction questions. Analysis of practiced aptitude questions is very important at the end of your practice session.


Aptitude Questions of moderate difficulty on the topic of "Percentages or Fractions" are presented on this page. ALL the problems are fully solved and ABSOLUTELY FREE for everyone. To get a good score in MAT, XAT, GRE, GMAT, NMIMS, IRMA etc. MBA or similar competition practicing these questions will prove to be very helpful.

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1.

In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?

A.

300,000

B.

168,000

C.

36,000

D.

24,000

View Ans

 

Answer - (C)

Solution:

Let the percentage of the total votes secured by Party D be x%
Then the percentage of total votes secured by Party $R = (x – 12)%$
As there are only two parties contesting in the election, the sum total of the votes secured
by the two parties should total up to 100%
i.e.,$ x + x – 12 = 100$
$2x – 12 = 100$
or $2x = 112$ or $x = 56%.$

If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.

44% of the total votes = 132,000 
i.e., $(44/100)×T = 132,000$
⇒ $T = ({132,000×100}/44) = 300,000$ votes.
 
The margin by which Party R lost the election = 12% of the total votes 
= 12% of 300,000 = 36,000

2.

A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?

A.

17

B.

23

C.

77

D.

None of these

View Ans

 

Answer - (B)

Solution:

Let the number of apples be 100.
On the first day he sells 60% apples i.e.,60 apples. Remaining apples =40.
He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has $40-6 = 34$ apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in total he throws$ 6 + 17 =$ 23 apples.

3.

If the cost price of 20 articles is equal to the selling price of 16 articles, What is the percentage of profit or loss that the merchant makes?

A.

20% Profit

B.

25% Loss

C.

25% Profit

D.

33.33% Loss

View Ans

 

Answer - (C)

Solution:

Let Cost price of 1 article be Re.1.
Therefore, Cost price of 20 articles = Rs. 20.
Selling price of 16 articles = Rs. 20
Therefore, Selling price of 20 articles = $(20/16) × 20 = 25$
$\text"Profit = Selling price - Cost price"$
$= 25 - 20 = 5$
$\text"Percentage of profit" = \text"Profit" / \text"Cost price" × 100$
= $(5 / 20) × 100 =$ 25% Profit

4.

When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?

A.

1.5 kgs

B.

1.7 kgs

C.

3.33 kgs

D.

None of these

View Ans

 

Answer - (B)

Solution:

Flower-nectar contains 50% of non-water part. 
In honey this non-water part constitutes 85% (100-15).
Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg 
Therefore amount of flower-nectar needed $= ({0.85/0.5} × 1)kg = $1.7 kgs

5.

30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?

A.

15%

B.

20%

C.

80%

D.

70%

View Ans

 

Answer - (C)

Solution:

20% of the men are above the age of 50 years. 20% of these men play football. Therefore, 20% of 20% or 4% of the total men are football players above the age of 50 years. 

20% of the men are football players. Therefore, 16% of the men are football players below the age of 50 years.

Therefore, the % of men who are football players and below the age of $50 = (16/20)×100 =$ 80%