In a class the number of boys and girls is in the ratio of 4 : 5. If 10 more boys join the class, the ratio of numbers of boys and girls become 6 : 5. How many girls there in the class?

Let number of boys = 4N and number of girls = 5N
Accounting to the question
4N / (5N - 100) = 6/7
? 28N = 6 ( 5N - 100 )
? 28N = 30N - 600
? 2N = 600
? N = 600/2 = 300
? Number of boys = 4N = 4 x 300 = 1200

Given that
Sanjay + Suresh = 120 ...(i)
Sanjay - Suresh = 30 ...(ii)
By adding Eqs (I) and (ii) we get
2 x sanjay = 150
? sanjay = 150/2 = 75 kg
? Weight of Suresh = 75 - 35 = 45 kg
? Suresh : Sanjay = 45 : 75 = 3 : 5 = 3/5 = 0.6

Given that, x/2y = 6/7
? x/y = 12/7
By componenod dividendo,
(x - y) / (x + y) = (12 - 7) / (12 + 7) = 5/19
? (x - y) / (x + y) + 14 /19 = 5/19 + 14/19 = 19/19 = 1

(a + b) / (a - b) = 5/3
? 3a + 3b = 5a - 5b
? 2a = 8b
? a = 4b,
? a/b = 4/1

Now, ( a² + b² ) / (a² - b² ) = [a² / (b²+1)] / [a² / (b² -1)]
= [( a/b )² + 1] / [( a/b )² - 1] = [( 4/1 )² + 1] / [( 4/1 )² - 1]
= [16 + 1] / [16 - 1] = 17/15
? ( a² + b² ) : ( a² - b² ) = 17 : 15

Given that
A : B = 3 : 4 = ( 3 x 2 ) : (4 x 2 ) = 6 : 8
B : C = 8 : 9
? A : B : C = 6 : 8 : 9
As consequent of the first ratio is equal to the antecedent of second ratio.

Total number of student = 8670
Total number of boys = 4545
? Total number of girls = 8670 - 4545 = 4125
? Required ratio = 4545 : 4125 = 303 : 275

Speed of truck 640/10 = 64 km/h
Speed of car = 640/8 = 80 km/h
So, the required ratio = 64/80 = 4/5 = 4 : 5

Case I Speed = Distance/ Time
Case II Speed = (Distance/2) / (2 x Time) = Distance / (4 x Time)
? Required ratio = 1/4 = 1:4

The ratio among A, B, C and D = 1/3 : 1 / 4 : 1/5 : 1/6
On rearranging the ratio = 60/3 : 60/4: 60/5 : 60/6 = 20 : 15 : 12 : 10
So minimum number of pens can be when the common ration is 1.
So minimum number of pen = 20 + 15 + 12 + 10 = 57

If three parts be P, Q and R,
then, P/2 = Q/3 = R/5
? P : Q : R = 2 : 3 : 5
? P = (2/10) x 990 = 198
Q = (3/10) x 990 = 297 and
R = (5/10) x 990 = 495