**Ques 1. B as a % of A is equal to A as a % of (A+B). Find B as a % of A.**

**Solution:**

→B/A = A/ (A+B)→Assume, B = Ax (B is some percent of A. So consider that some percent as ‘x’.)

→Ax/A = A/(A+Ax)

→X=1/(1+x)

→X^2+ X -1 = 0

→Solving the above equation, we get 0.62(consider only positive terms),

**So the answer is 62%.**

**Solution:**

→Reduction = (% increased - % additional)/ (100 + % increased)

→Reduction in litres = (25-15)/ (100+25) =

**8 litres**

#### Solution:

→42% - 12% = 30% (Actual pass mark)→30% of x – 20% of x = 10

→10% of x = 10

→X = (10*100)/10 =

**100 Maximum mark**

#### Solution:

→Difference of % of marks obtained = difference of the mark obtained→40% of x - 30% of x = 15 – (-10)

→10% of x = 25

→X = 25*100/10 = 250 (maximum marks)

→Passing marks = 30% of 250 + 10 =

**85%**

#### Solution:

→Assume the population be 100.→58% of 100 = 58 votes got by winning party.

→42% of 100 = 42 votes got by loser.

→Difference between them = 58% – 42% = 16% (majority)

→Now in the given question, majority = 960 = 16%

→If 16% = 960 then 100% is 6000.

→So the total number of votes is

**6000.**

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