Test: Percentage- 1 - CLAT MCQ

 Number of runs made by running:-

110- (3*4+8*6) 

= 110- 60 

= 50

Therefore, required percentage :-

[( 50/ 110)  * 100] %

=>

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%

Option ( c) 341 is correct. 

Explanation:- 

{Let the initial radius of circle = 10 cm

Then, it's area = π r^ 2 

=  π (10) ^2 

=  100 π cm 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 π cm square. 

Increase in area=  ( 441 π - 100 π ) cm square. 
=> 341 cm square. }

Number of valid votes
= 80% of 7500
= 6000
∴ Valid votes polled by other candidate
= 45% of 6000
=  45100×600045100×6000 = 2700

Let n(A) = no. of students pass in english
    n(B) = no. of students pass in math
    n(A U 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(A U 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(A U B)

⇒ 40 = x - 90x/100

 ⇒ 40 =  x/10

⇒  x = 400
total no. of students = 400 

• 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.

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

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 

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%.

The correct option is B.
Let maximum marks be x.
Then
36% of x = (113+85). So x = (198*100/ 36) = 550

The correct option is B.
Present value - 4000
Depreciation - 10% of present value annually
After one year value of machine will be =  4000 - 400(10% of 4000) = 3600 
After 2nd year value of machine will be = 3600 - 360(10% of 3600) = 3240

Let  y be the population of the town initially
y x 105/100 x 105/100 = 15435
y = (15435 x 100 x 100) / (105 x 105)
On solving we get,
y = 14000

using the formula A + B + AB/100
where A = - 50(as the wages were decreased) , B = 50
-50 + 50 - (50 x 50/100)
-2500/100 =  - 25 = 25% decrease

• Let the number be y
• according to the question.
• y + y x (37.5/100) = 33
• y x 137.5/100 = 33
• y = 3300/137.5 = 24