Ratio P6 2009 SA1 P2 Pei Hwa Q17

Gibson had 50-dollar, 20-dollar and 10-dollar tickets for a school funfair in the ratio 1 : 5 : 4. The total value of the funfair tickets was $3420.
a) Find the number of each type of tickets he had. [2]
b) After selling 1/4 of the 10-dollar tickets and a few 20-dollar tickets, he found that 2/11 of the remaining tickets were 50-dollar tickets. How many 20-dollar tickets did he sell? [3]


Gibson had 50-dollar, 20-dollar and 10-dollar tickets for a school funfair in the ratio 1 : 5 : 4.
You may assume the number of 50-dollar, 20-dollar and 10-dollar tickets to be 1, 5 and 4 respectively.

The total value of the funfair tickets was $3420.
Convert the number of tickets to dollar value. Then equate to $3420 to solve for the unknown unit (which is the number of tickets per unit)

50 x 1 + 20 x 5 + 10 x 4 = 190

190 u --> 3420

1 u -- > 3420 ÷ 190 = 18

5 u -- > 5 x 18 = 90

4 u -- > 4 x 18 = 72

a) He had 18, 90 and 72  50-dollar, 20-dollar and 10-dollar tickets respectively.

After selling 1/4 of the 10-dollar tickets and a few 20-dollar tickets, he found that 2/11 of the remaining tickets were 50-dollar tickets.
2/11 -- 2 units of 50-dollar tickets, balance 9 units of 10-dollar and 20-dollar tickets.
So convert 1 : 5 : 4 to 2 : 10 : 8 to get 2 units of 50-dollars.
Now, the number of 10-dollar tickets is 8 units, so 1/4 of it is 2 units, leaving 6 units.
Since number of 10-dollar tickets is 6 units, the number of 20-dollar tickets is 9 - 6 = 3 units. So the number of 20-dollar ticket sold is 10 - 3 = 7 units.
Tabulate this out in a table form.

$50     $20    $10

-----------------------
   1   :   5    :   4

= 2   :  10   :   8   

           - ?     - 2       

----------------------

   2   :   3    :   6

----------------------

10 – 3 = 7

10 u -- > 90

1 u -- > 90 ÷ 10 = 9

7 u -- > 7 x 9 = 63  20-dollar tickets

b) He sold 63  20-dollar tickets.