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Test: Percentages - 2 - UGC NET MCQ


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10 Questions MCQ Test Mathematical Reasoning and Aptitude for UGC NET - Test: Percentages - 2

Test: Percentages - 2 for UGC NET 2024 is part of Mathematical Reasoning and Aptitude for UGC NET preparation. The Test: Percentages - 2 questions and answers have been prepared according to the UGC NET exam syllabus.The Test: Percentages - 2 MCQs are made for UGC NET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Percentages - 2 below.
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Test: Percentages - 2 - Question 1

 In a school, 40 % of the students play football and 50 % play cricket. If 18 % of the students play neither football nor cricket, the percentage of students playing both is

Detailed Solution for Test: Percentages - 2 - Question 1

Let A = set of students who play football and 

B = set of students play cricket. 

Then n(A) = 40, n (B) = 50 and 

n(A U B) = (100 - 18) = 82 

n(A U B) = n(A) + n(B) “ n(A ∩ B) 

n(A∩B) = n(A) + n(B) “ n(AUB) = (40 + 50 -82) = 8 

Percentage of the students who play both = 8%

Test: Percentages - 2 - Question 2

The price of petrol is increased by 25 %. By how much percent a car owner should reduce his consumption of petrol. So that expenditure on petrol would not be increased?

Detailed Solution for Test: Percentages - 2 - Question 2

Reduction % in consumption = {r/(100+ r) Ã -100}%
= (25/125 Ã - 100)%
= 20%

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Test: Percentages - 2 - Question 3

If the numerator of a fraction is increased by 140 % and the denominator is increased by 150 % the resultant fraction is 4/15 what is the original fraction?

Detailed Solution for Test: Percentages - 2 - Question 3

Let the original fraction be x/y. 
Then new fraction

Therefore 24/25(x/y) = 4/15 
⇒ x/y = (4/15 Ã - 25/24) = 5/18 
Therefore Original Fraction = 5/18

Test: Percentages - 2 - Question 4

Fresh grapes contain 80 % water dry grapes contain 10 % water. If the weight of dry grapes is 250 kg. What was its total weight when it was fresh?

Detailed Solution for Test: Percentages - 2 - Question 4

Let the weight of fresh grapes be x kg 
Quantity of water in it = (80/100 Ã - x)kg = 4x/5 kg 
Quantity of pulp in it = (x “ 4x/5)kg = x/5 kg 
Quantity of water in 250 kg dry grapes 
= (10/100 Ã- 250)kg = 25kg 
Quantity of pulp in it = (250 - 25)kg = 225 kg 
Therefore, x/5 = 225 
⇒ x = 1125

Test: Percentages - 2 - Question 5

In two successive years, 100 and 75 students of a school appeared at the final examination. Respectively 75 % and 60 % of them passed. The average rate of pass is:

Detailed Solution for Test: Percentages - 2 - Question 5

Total candidates = (100 + 75) = 175 
Total passed = (75/100 Ã- 100) + (60/100 Ã- 75) 
= (75 + 45) = 120 
Therefore Pass % = (120/175 Ã -100)% 
= 480/7 % = 68 4/7 %

Test: Percentages - 2 - Question 6

The sum of two numbers is 2490. If 6.5 % of one number is equal to 8.5 % of the other then the numbers are:

Detailed Solution for Test: Percentages - 2 - Question 6

Let the number be x and (2490 - x). 
Then 6.5 % of x = 8.5 % of (2490 - x) 
⇒ 6.5/100 Ã - x = 8.5/100 Ã - (2490 - x) 
⇒ 65x/1000 = 85(2490 - x)/1000 
⇒ 65x = (85 Ã - 2490) “ 85x 
⇒ 150x =(85 Ã - 2490) 
⇒ x = 211650/150 = 1411 
Hence the numbers are 1411 and (2490 - 1411) = 1079

Test: Percentages - 2 - Question 7

8 % of the voters in an election did not cast their votes in the election, there were only two candidates. The winner by obtaining 48 % of the total votes defeated his contestant by 1100 votes. The total number of voters in the election was.

Detailed Solution for Test: Percentages - 2 - Question 7

Let the total number of voters be x 
Votes cast = 92 % of x = (92/100 Ã - x) = 23x/25
Votes in favour of winning candidate = 48/100 Ã - x 
= 12x/25
Votes polled by defeated candidate = (23x/25 “ 12x/25) 
= 11x/25 12x/25 “ 11x/25 = 1100 
⇒ 12x “ 11x = 27500 
⇒ x = 27500

Test: Percentages - 2 - Question 8

If the price of the eraser is reduced by 25% a person buy 2 more erasers for a rupee. How many erasers available for a rupee?

Detailed Solution for Test: Percentages - 2 - Question 8

Let n erasers be available for a rupee 
Reduced Price = (75/100 Ã - 1) = Re ¾ 
3/4 rupee fetch n erasers = 1 Rupee will fetch (n à - 4/3) erasers 
Therefore, 4n/3 = n +2 ⇒ 4n = 3n +6 ⇒ n =6

Test: Percentages - 2 - Question 9

In measuring the sides of a rectangle errors of 5 % and 3 % in excess are made. The error percent in the calculates area is

Detailed Solution for Test: Percentages - 2 - Question 9

Let length = x units and breadth = Y units 
Then actual area = xy sq.units 
Length shown = (105/100 Ã - x)units = 21x/20 units; 
Breadth shown = (103/100 Ã - Y) 
Calculated area = (21x/20 Ã - 103y/100)sq.units 
Error = 2163xy/2000 sq.units 
Error = (2163xy/2000 - xy) 
Error = 163xy/2000 
Error % = (163xy/2000 à - 1/xy à - 100)% 
= 163/20 % = 8.15 %

Test: Percentages - 2 - Question 10

Fresh fruit contains 68 % water and dry fruit contains 20 % water. How much dry fruit can be obtained from 100 kg of fresh fruits?

Detailed Solution for Test: Percentages - 2 - Question 10

Quantity of water in 100kg of fresh fruits =(68/100 × 100)kg 
Quantity of pulp in it = (100 - 68)kg = 32 kg 
Let the dry fruit be x kg 
Water in it = (20/100 Ã - x)kg = x/5 kg 
Quantity of pulp in it = (x “ x/5)kg = 4x/5 kg 
Therefore, 4x/5 = 32 ⇒ x = 160/4 = 40 kg

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