Percentage And Its Applications Test-1

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Percentage And Its Applications Test-1

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Question 1
1 / -0.25

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A
45%

B

C

D
55%

Solution

Number of runs made by running:-

110- (3*4+8*6)

= 110- 60

= 50

Therefore, required percentage :-

[( 50/ 110) * 100] %

=>
Question 2
1 / -0.25

If two numbers are respectively 20% and 50% more than the 3rd number then what % is the 1st number of the 2nd one?

A
10

B
20

C
40

D
80

Solution

Two numbers are respectively 20% and 50% more than the 3rd number. What percent is the first of the second? Consider 3rd number be 100%.
Then percent of 1st number to 2nd number =(120%/150%)×100=80%
Question 3
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If the radius of circle is increased by 110% its area is increased by

A
110

B
301

C
341

D
501

Solution

Option ( c) 341 is correct.
Explanation:-
{Let the initial radius of circle = 10 cm
Then, it 's area = πr^ 2
= π(10) ^2
= 100 πcm square.
Given, (radius is increased by 110%)
New radius = ( 100+110) % of 10 cm
=>210% of 10 cm
{ 210 /100 * 10} cm
=>21 cm
Therefore, area of circle of radius (21) cm
=>π(21) ^2
=> 441 πcm square.
Increase in area= ( 441 π- 100 π) cm square.
=>341 cm square. }
Question 4
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In 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:

A
2700

B
2900

C
3000

D
3100

Solution

Number of valid votes
= 80% of 7500
= 6000
∴Valid votes polled by other candidate
= 45% of 6000
= 45100 ×600045100 ×6000 = 2700
Question 5
1 / -0.25

In an exam 80% of the students passed in Eng, 85% in Math and 75% in both. if 40 students failed in both subjects, the total number of students is?

A
800

B
400

C
900

D
750

Solution

Let n(A) = no. of students pass in english
    n(B) = no. of students pass in math
    n(AU B) = no. of students pass in either math or english
    n(A ∩B) = no.of students pass in both math and english
let x are no. of students
⇒n(AU B) = n(A) + n(B) - n(A ∩B)
               = 80x/100 + 85x/100 - 75x/100
               = 90x/100
no. of students fail in both either math or english = x - n(AU B)
⇒40 = x - 90x/100
⇒40 =  x/10
⇒  x = 400
total no. of students = 400
Question 6
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Mrs. Ruhani travels with 80% of her usual speed and thus gets late by 40 minutes. What will be the usual time taken by her in covering the same distances?

A
160 minutes

B
120 minutes

C
180 minutes

D
200 minutes
Solution
Let T be the usual time taken by Mrs. Ruhani to cover the distance.
Let S be her usual speed .
Since the distance is the same, we can say:
Distance= S*T
Now, she is traveling at 80% of her usual speed, so her speed becomes: 0.8S
Using this reduced speed, the time taken for the same distance is:
So, at this reduced speed, she takes 1.25 times her usual time .
According to the problem, she gets 40 minutes late . This means the additional time she takes is:
1.25T- T=0.25T
Since this extra time is 40 minutes:
0.25T=40
Now solving for T=
Answer: The usual time taken by Mrs. Ruhani is indeed 160 minutes .
So, the correct answer is: (a) 160 minutes .
Question 7
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The price of rice is increased by 25% and as a result, a person can have 4kg less rice for Rs.400. What is the increased price of rice per kg?

A
Rs. 22.5 kg

B
Rs. 20 kg

C
Rs. 25 kg

D
Rs. 30 kg

Solution
Question 8
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In an examination, Summit got 30% marks and failed by 15 marks. In the same examination, Rohit got 40% marks, which is 35 marks more than the pass marks. Then what is the total marks in the examination?

A
500

B
250

C
600

D
450

Solution

The correct option is A.
Let the total marks be x.
Sumit got = 30% of x
= 0.3x
Passing marks = 0.3x+15
Rohit got = 40% of x
= 0.4x
Passing marks = 0.4x - 35
Equating both passing marks , we get
0.3x+15 = 0.4x - 35
50 = 0.1x
x = 500
Question 9
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After spending 40% in machinery, 25% in building, 15% in raw material and 5% in furniture, Harilal had a balance of Rs. 1305. The money with him was

A
Rs. 6500

B
Rs. 7225

C
Rs. 8700

D
Rs. 1390

Solution

Let the total money harilal had = y
Percentage of money harilal has spent already = 40% + 25% + 15% + 5% = 85% of the money
Money left = 100% - 85% = 15%
Now, 15/100 x y = 1305
therefore, y = (1305 x 100 / 15)
= 8700
Question 10
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In Mathematics exam, a student scored 30% marks in the first paper, out of the total of 180. How much should he score in second paper out of a total of 150, if he is to get an overall average of at least 50%

A
74%

B
76%

C
70%

D
80%

Solution

Firstly, marks he scored in first paper out of 180 will be
180 x 30 ÷100 = 54 marks
So he scored 54 marks out of 180 in the first paper.
Now total marks of the examination is 330 i.e 150+180.
50% of 330 is 165.
So in second paper he has to score 165 –54=111 marks to get 50% marks over all.
Now the ans is asked in terms of percent so,
111 ÷150 x 100 = 74%
So he has to get 74 % in the second paper to get a overall percent of 50%.