Primary 6a and 6b collected from a charity

Question

The ratio of the amount of money Primary 6A collected to the amount of money Primary 6B collected from a charity sale was 4:5. A pupil from Primary 6A donated $6 more than the ratio because 7:8. How much did Primary 6B collect from the charity sale.

Initial Ratio (before)

6A: 6B
4: 5

Initial Ratio (after)

6A:6B

7:8

*then I times a common number of the 6A, which is 28

28: 35 (before)
28:32 (after)

*ten I subtracted 35 - 32 to get 3 units. And we know the difference is $6.

=> 3units = 6 => 1 unit = 2

So I got the answer of $64 because 32 (6b after) *2 = 64.

So is the way I did it, correct?

Is there someone who knows a way to create an algebraic expression for this because I couldn't find a good way to do it?

Answer 1

$$\frac{a}{b} = \frac45 \implies 5a=4b $$

$$\frac{a+6}{b} = \frac78 \implies 8a+48=7b $$

Multiply the second equation by $4$, $$32a+4(48) = 7(4b)=35a$$

$$4(48)=3a$$

$$a=4(16)=64$$

Hence $b=\frac{5a}{4}=\frac{5(4)(16)}{4}=5(16)=80$

Remark about your attempt:

The class $B$ donation amount doesn't change so it should be held fixed rather than class $A$ donation amount.

Answer 2

The way you did it is wrong. You need the second number in the ratios (corresponding to 6B) and not the first number in the ratios (corresponding to 6A) to be the same because there is no change in the amount 6B donated.

The first ratio is $32:40$ and the second ratio is $35:40$. Then we have $6$ dollars corresponding to $35-32=3$ units, so 6B donated $\frac{40}{3}\times 6=80$ dollars.

Answer 3

$\frac{7}{8} - \frac{4}{5} = \frac{6}{B}$

$\frac{35 - 32}{40} = \frac{6}{B}$

$\frac{3}{40} = \frac{6}{B}$

$B = \$80$

($A = \$70$)