If (3/2)X = (5/7)Y = (6/5)Z, then what is X : Y : Z?

 

Let us say x boys and x girls joined the group.

(64 + x)/(40 + x) = 4/3

192 + 3x = 160 + 4x => x = 32

Number of members in the group = 64 + x + 40 + x

= 104 + 2x = 168.

504/M = 384/800

(504 x 800) / 384 = M

M = 1050

Total student= 39, boys : girls = 2:1, hence no of boys= 2/3(39)= 26
num of girls= 1/3(39) = 13 , since radhika ranked 15 among 39 student from top
and ranked 8 among girls from bottom , this means 7 girls are below radhika, rest 13-9= 5 girls are above her, now from 14 toper 5 are girls hence 14-5=9 boys.
Hence num of boys below radhika are 26-9=17

The wages of labourers in a factory increases in the ratio 22:25 and there was a reduction in the number of labourers in the ratio 15:11. Find the original wage bill if the present bill is Rs 5000 ?
Ratio of increase of wages = 22:25
Ratio of decrease of labourers = 15:11
Compound ratio of wages of labourers = 22 x 15 : 25 x 11 = 330:275
Final bill = Rs. 5000
For 275 ratio wages = Rs. 5000
For 1 ratio wages = 5000/275
For 330 ratio wages = 5000/275 x 330 = Rs. 6000

To get the solution that contains 1 part of milk and two parts of water,
they must be mixed in the ratio as
7x+6x/5y+11y = 1/2
26x = 16y
x/y = 16/26
x/y = 8/13

Here the number of cogs is inveresly proportional to number of revolutions

=> More cogs less revolution
=> 6 : 14 :: x : 21
=> (21x6)/14 = 9

a:b = 60:40
a:c = 60:45
c/a x a/b = 45/60 x 60/40 = 45/40 = 90/80
So C gives B => 10 points.

Ratio C1:C2 = 3:5 and C2:C3 = 7:11
So C1:C2:C3 = 21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then
Given that 21x + 35x + 55x = 333
=> 111x = 333 or x=3
So strength of the class with highest number of students =55x = 55x3 = 165.