Let us say x boys and x girls joined the group.(64 + x) / (40 + x) = 4/3192 + 3x = 160 + 4x => x = 32Number of members in the group = 64 + x + 40 + x= 104 + 2x = 168.

Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively. New ration = [9x + 1/4(9x)]/[10x - 1/4(10x)]=> [9x(1 + 1/4)]/[10x(1 - 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.

Let the present ages of Giri and Hari be 5x and 8x years respectively.(5x + 12) / (8x + 12) = 11/1470x + 168 = 88x + 132 => x= 2Difference in their ages will be the same at all times.This difference = Difference of their present ages= 8x - 5x = 3x = 6 years.

Let the shares of Amar, Bhavan and Chetan be Rs. 3x, Rs. 6x and Rs. 5x respectively.
3x + 6x + 5x = 5600 => 14x = 5600 => x = 400.Required ratio = 3x - 400 : 6x - 400 : 5x - 400= 3x - x : 6x - x : 5x - x= 2x : 5x : 4x => 2 : 5 : 4

Let the numbers be 5x, 6x, 7x.Largest number = 7x.Smallest number = 5x.Third number = 6x.7x + 5x = 6x + 486x = 48 => third number is 48.

Let the present ages of Anil and his son be 7x and 3x years respectively.Let the present age of the son's  mother be m years.(m + 6) / (3x + 6) = 2/1m + 6 = 6x + 12 => m = 6x + 6Ratio of the present ages of Anil and the son's mother = 7x / (6x + 6).This cannot be found uniquely.

Let the numbers be a, b and c.Given that a, b and c are in the ratio 3 : 2 : 4. let, a = 3x, b = 2x and c = 4xGiven, (a+b) + (c - b) = 21   = > a + b  + c - b = 21      = >  a + c = 21= > 3x + 4x = 21               = >7x = 21  = > x = 3a , b , c are 3x, 2x, 4x.a, b, c are 9 , 6 , 12.

Let the income and the expenditure of the person be Rs. 5x and Rs. 4x respectively.Income, 5x = 18000  => x = 3600Savings = Income - expenditure = 5x - 4x = xSo, savings = Rs. 3600.

Let the two numbers be 5x and 6x.
Let the numbers added to both so that their ratio becomes 7 : 8 be k.(5x + k) / (6x + k) = 7/8=> 40x + 8k = 42x + 7k => k = 2x.6x - 5x = 10 => x = 10k = 2x = 20.