Quiz-1

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.

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

Full marks =500
Then
A got 10℅less than b
A got 360
So
360*10/100
36
So
B got 360+36=396
Then
B got 25℅ more than c
So
B=396
396*25/100
99
So
C got 396-99=297
Then
C got 20℅ less than d
So100
got 297
So
297*20/100
59
D got 297+59=356
Then
D got 356
356/500*100
71.2 %

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.

SOLUTION:

Given,
(5% of a) + (4% of b) = (2/3)×[(6% of a) + (8% of b)

=> (5a/100) + (4b/100) = (2/3)×[(6a/100) + (8b/100)]

=> (5a/100) + (4b/100) = (4a/100) + (16b/300)

=> (5a/100) - (4a/100) = (16b/300) - (4b/100)

=> a[(5/100) - (4/100)] = b[(16/300) - (4/100)]

=> a(1/100) = b(4/300)

=> a = b(4/3)

=> (a/b) = (4/3)

⇒ a : b = 4 : 3

Thus, the required ratio is 4 : 3.

 

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%ofy= x/100×y

           = y/100×x

           = y%of x

∴A=B

Let their marks be (x + 9) and x.

Then, x + 9 = 56 (x + 9 + x)
100
25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

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

= 110 - (60)

= 50.

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