percentage concept
Concept  Practice set 1  Practice set 2
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
No. of Questions 
10 
Time 
10 min 
Medium 
english 
Marks 
10 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionnMark review to see the question later 
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 easy 0%
 medium 0%
 tough 0%

your performance is poor!!! 🙁 🙁
Attemt more quiz to enhance your score, good luck

your performance is below average!!!!! 🙁
Attempt more quiz to enhance your score, good luck

your performance is average!! good going 🙂
Attempt more quiz to enhance your score, good luck

nice!! your score is above average 🙂 🙂
Attemt more quiz to enhance your score, good luck

your are scoring good :)) 😀
Attemp more quiz to score 100%, good luck

wow you are scoring very good :)) 😀
attempt more quiz to score 100%, good luck

voila!!!! you made it.. you are scoring 100%:)) 😀
attempt more quiz to keep the same, good luck
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
1 pointsIn an acoustics class, 120 students are male and 100 students are female. 25% of the male students and 20% of the female students are engineering students.
20% of the male engineering students and 25% of the female engineering students passed the final exam.
What percentage of engineering students passed the exam?
Correct
Solution:
Option(D) is correctThere are 100 female students in the class, and 20% of them are Engineering students.
Now, 20% of 100 equals (20/100)×100=20.
Hence, the number of female engineering students in the class is 20.Now, 25% of the female engineering students passed the final exam:
2525.
Hence, the number of female engineering students who passed is 5.There are 120 male students in the class. And 25% of them are engineering students.
Now, 25% of 120 equals(25/100)×120=(1/4)×120=30.
Hence, the number of male engineering students is 30.Now, 20% of the male engineering students passed the final exam:
2020.
Hence, the number of male engineering students who passed is 6.Hence, the total number of engineering students who passed is:the total number of engineering students who passed is:
(Female Engineering students who passed)+(Female Engineering students who passed)+
(Male Engineering students who passed)(Male Engineering students who passed)=5+6=11=5+6=11
The total number of engineering students in the class is:The total number of engineering students in the class is:
(Number of female engineering students)+(Number of female engineering students)+
(Number of male engineering students)=(Number of male engineering students)==30+20=50=30+20=50
Hence, the percentage of engineering students who passed is
(Total number of engineering students who passed/Total number of engineering students)×100
=(11/50)×100
= 22%Incorrect
Solution:
Option(D) is correctThere are 100 female students in the class, and 20% of them are Engineering students.
Now, 20% of 100 equals (20/100)×100=20.
Hence, the number of female engineering students in the class is 20.Now, 25% of the female engineering students passed the final exam:
2525.
Hence, the number of female engineering students who passed is 5.There are 120 male students in the class. And 25% of them are engineering students.
Now, 25% of 120 equals(25/100)×120=(1/4)×120=30.
Hence, the number of male engineering students is 30.Now, 20% of the male engineering students passed the final exam:
2020.
Hence, the number of male engineering students who passed is 6.Hence, the total number of engineering students who passed is:the total number of engineering students who passed is:
(Female Engineering students who passed)+(Female Engineering students who passed)+
(Male Engineering students who passed)(Male Engineering students who passed)=5+6=11=5+6=11
The total number of engineering students in the class is:The total number of engineering students in the class is:
(Number of female engineering students)+(Number of female engineering students)+
(Number of male engineering students)=(Number of male engineering students)==30+20=50=30+20=50
Hence, the percentage of engineering students who passed is
(Total number of engineering students who passed/Total number of engineering students)×100
=(11/50)×100
= 22% 
Question 2 of 10
2. Question
1 pointsIf b equals 10% of a and c equals 20% of b, then which one of the following equals 30% of
$c$?
Correct
Solution:
b=10% of a=(10/100)×a=0.1
$a$c=20% of b=(20/100)×
$b$=0.2b=0.2×0.1a
Now, 30% of c=(30/100)×
$c$=0.3c=(0.3)(0.2)(0.1a)
=0.006a=0.6%a
Incorrect
Solution:
b=10% of a=(10/100)×a=0.1
$a$c=20% of b=(20/100)×
$b$=0.2b=0.2×0.1a
Now, 30% of c=(30/100)×
$c$=0.3c=(0.3)(0.2)(0.1a)
=0.006a=0.6%a

Question 3 of 10
3. Question
1 pointsIn an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid.
If the total number of votes was 7500, the number of valid votes that the other candidate got, was
Correct
Solution:
Number of valid votes = 80% of 7500 = 6000.Valid votes polled by other candidate = 45% of 6000
=(45/100)×6000
$=$2700.
Incorrect
Solution:
Number of valid votes = 80% of 7500 = 6000.Valid votes polled by other candidate = 45% of 6000
=(45/100)×6000
$=$2700.

Question 4 of 10
4. Question
1 pointsIf 50% of x equals the sum of y and 20, then what is the value of x–2y?
Correct
Solution:
50% of
$x$equals the sum of
$y$and 20. Expressing this as an equation yields:
(50/100)×x=y+20
x/2=y+20
x=2y+40
x–2y= 40
Incorrect
Solution:
50% of
$x$equals the sum of
$y$and 20. Expressing this as an equation yields:
(50/100)×x=y+20
x/2=y+20
x=2y+40
x–2y= 40

Question 5 of 10
5. Question
1 pointsThe price of an item changed from Rs. 120 to Rs. 100. Then later the price decreased again from Rs. 100 to Rs. 80.
Which can be said about the two decreases in percentage term?
Correct
^{ }
Solution:
First decrease in percent,
partwhole=(120−100)120partwhole=(120−100)120
=0.17=17%
Second decrease in percent,
part/whole=(100−80)/100
=0.20=20%
The second decrease is larger in percent term.
The parts were the same in both cases, but the whole was smaller in the second decrease.
^{ }
The price of an item changed from Rs. 120 to Rs. 100. Then later the price decreased again from Rs. 100 to Rs. 80.
Incorrect
^{ }
Solution:
First decrease in percent,
partwhole=(120−100)120partwhole=(120−100)120
=0.17=17%
Second decrease in percent,
part/whole=(100−80)/100
=0.20=20%
The second decrease is larger in percent term.
The parts were the same in both cases, but the whole was smaller in the second decrease.
^{ }
The price of an item changed from Rs. 120 to Rs. 100. Then later the price decreased again from Rs. 100 to Rs. 80.

Question 6 of 10
6. Question
1 pointsIn an election contested by two parties, Party D
$D$secured 12% of the total votes more than Party R
$R$.
If party R
$R$got 132,000 votes, by how many votes did it lose the election?
Correct
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
$=132,000$i.e., (44/100)×T=132,000
⇒ T=(132,000/44×100)=300,000 votes.
The margin by which Party
$R$lost the election = 12% of the total votes
= 12% of 300,000 = 36,000
Incorrect

Question 7 of 10
7. Question
1 pointsA 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?
Correct
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=3440−6=34 apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in total he throws 6+17=6+17= 23 apples.
Incorrect
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=3440−6=34 apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in total he throws 6+17=6+17= 23 apples.

Question 8 of 10
8. Question
1 pointsA. 1.5 kgs B. 1.7 kgs C. 3.33 kgs D. None of these Correct
Solution:
Flowernectar contains 50%of nonwater part.In honey this nonwater part constitutes 85%(100−15)
Therefore 0.5X Amount of flowernectar = 0.85X Amount of honey = 0.85×1 kg
Therefore amount of flowernectar needed
=(0.85/0.5)×1
=1.7=1.7 kg
Incorrect
Solution:
Flowernectar contains 50%of nonwater part.In honey this nonwater part constitutes 85%(100−15)
Therefore 0.5X Amount of flowernectar = 0.85X Amount of honey = 0.85×1 kg
Therefore amount of flowernectar needed
=(0.85/0.5)×1
=1.7=1.7 kg

Question 9 of 10
9. Question
1 points30%
$30\mathrm{\%}$of the men are more than 25 years old and 80%
$80\mathrm{\%}$of the men are less than or equal to 50 years old. 20%
$20\mathrm{\%}$of all men play football. If 20%
$20\mathrm{\%}$of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
Correct
Incorrect

Question 10 of 10
10. Question
1 pointsA shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by x
$x$during the year. A famine hits his village in the next year and many of his sheep die. The sheep population decreases by y
$y$during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheep. Which of the following is correct?
Correct
Solution:
Let us assume the value of x to be 10%
Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10%of 1 million = 1.1 million
In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.
i.e., 0.1 million sheep have died in 2001.In terms of the percentage of the number of sheep alive at the beginning of 2001,
it will be (0.1/1.1)×100%=9.09%(0.1/1.1)×100%=9.09%.
From the above illustration it is clear that x>y
Incorrect
Solution:
Let us assume the value of x to be 10%
Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10%of 1 million = 1.1 million
In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.
i.e., 0.1 million sheep have died in 2001.In terms of the percentage of the number of sheep alive at the beginning of 2001,
it will be (0.1/1.1)×100%=9.09%(0.1/1.1)×100%=9.09%.
From the above illustration it is clear that x>y