percentages

Question

"In the previous government , Party Q was in the opposition. Now Increasing the seats by 33.33% Q is the ruling party and thus party q enjoys twice the majority than that of party P in the previous government. If there were only two parties P and Q and the and the fix no. of seats be 500 in the parliament of Hum-tum, then the no. of seats of the Q in the new government is ?

Answer 1

Solving this problem involves a lot of substitutions....at least the way I did it. 

 

 

Here's what we have

Q₁ + P₁ = 500

Q₂ + P₂ = 500

Q₂ = 1.3333Q₁

Q₂-P₂ = 2(P₁-Q₁) and P₂ = Q₂ - 2(P₁-Q₁)

 

Now

Q₁+P₁=Q₂+P₂  since they both equal 500

 

Now the substitution begins

Q₁+P₁ = 1.3333Q₁ + Q₂-2(P₁-Q₁)

Q₁+P₁ = 1.3333Q₁ + 1.3333Q₁-2(P₁-Q₁)

 

Since all my subscripts are now 1, I'm going to drop the subscript since formatting so many is a bit of a pain.

Q+P = 1.3333Q + 1.3333Q -2(P-Q)

Q+P = 2.6666Q - 2P + 2Q

Q+P = 4.6666Q - 2P

 

Bring the P terms to the left and the Q terms to the right

 

3P = 3.6666Q

P = 1.2222Q

 

Back to the original equation

P + Q = 500

1.2222Q + Q = 500

2.2222Q = 500

Q = 225  This is Q₁, the old Q

P = 500-225 = 275   This is P₁, the old P

 

Q₂ = 1.3333Q1 

Q₂ = 1.3333(225) = 300  <=====  There are 300 Q seats in the new parliament

P₂ = 500-300 = 200

 

Note that P₁-Q₁ = 275-225 = 50

and

Q₂-P₂ = 300-200 = 100

 

and 100 = 2(50)