InfyTQ Percentage Quiz 1

Number of runs made by running = 110 - (3 x 4 + 8 x 6)

= 110 - (60)

= 50.

Required percentage = (50/110)*100 = 45 ( 5/11) %

Lets suppose marks gained by one student= X

And marks gained by second student= X+9

Therefore total sum of marks obtained by both students= 2X+9

now according to question

(X+9/ 2X+9)×100=56

100X +900=112X +504

12X=396

X=33, i.e. the marks gained by first student

And X+9 =42 ,i.e. marks gained by second student

let the no. of apple be x

x-x(40/100)=420

x-40x/100=420

or, 60x/100=420

x=42000/60

x= 700

x% of y=x/100 × y

=y/100 × x

=y% of x

∴A=B

20% of a = b
=> (20/100)a = b
therefore, b% of 20 = (b/100)*20
= (20a/100)*(1/100)*20
= 4a/100 = 4% of a.

Let the number of students be x. Then,

Number of students above 8 years of age =
(100 - 20)% of x = 80% of x.

Therefore, 80% of x = 48 + 2/3 of 48
=> 80/100 x = 80
=> x = 100.

Explanation:

5% of A + 4% of B= 2/3(6% of A + 8%of B)

5A / 100 + 4B/100= 2/3 (6A/100+ 8B/100)

1A/20 + 1B / 25= 1A / 25+ 4B/75

(1/20- 1/25) A= (4/75- 1/25)B

1/100 A= 1/75 B

A/B= 100/75

4/3

Required ratio= 4:3

 

 

(60/100) * 50 – (40/100) * 30
30 - 12 = 18

Marks obtained by A = 360 marks

marks obtained by C=(360/1250)×100

= 288 marks

marks obtained by D=(288/80)×100

= 360 marks

Required % marks obtained by D

=(360/500)×100=72%