Question

Two boxes contain pictures of hockey and basketball players.
In one box, the ratio of hockey players to basketball players is 4:3.
In the other box, the ratio is 3:2.
The boxes contain the same number of pictures.
a) What could the total number of pictures be?
b) Which box contains more pictures of hockey players?

Answer 1

box1:
Hockey cards --- 4x
basketball cards --- 3x

box2
hockey cards --- 3y
basketball cards -- 2y
7x = 5y
y = 7x/5
so x must be a multiple of 5 in order for y to be a whole number
x -- y
5 7
10 14
15 21
....

notice that as x increases by 5 , the y increases by 7
and the slope of y = 7x/5 is 7/5 ?? , mmmhhhh

Answer 2

can't it be 7, 10 too?

Answer 3

The minimum number in one box would be 4 + 3 = 7, and in the other it would be 3 + 2 = 5

Now, find the least common multiple of those two, which is 7 ∙ 5 = 35

So in each box, there are 35 pictures.


H = hockey players

B = basketball players


In one box, the ratio of hockey players to basketball players is 4:3 mean:

H / B = 4 / 3

Multiply both sides by B

H = 4 B / 3

H + B = 35

4 B / 3 + B = 35

4 B / 3 + 3 B / 3 = 35

7 B / 3 = 35

Multiply both sides by 3

7 B = 105

Divide both sides by 7

B = 105 / 7

B = 15

H = 35 - 15

H = 20

In the first box, there would then be 20 hockey players and 15 basketball players.


In the other box, the ratio is 3:2 mean:

H / B = 3 / 2

Multiply both sides by B

H = 3 B / 2

H + B = 35

3 B / 2 + B = 35

3 B / 2 + 2 B / 2 = 35

5 B / 2 = 35

Multiply both sides by 2

5 B = 70

Divide both sides by 5

B = 70 / 5

B = 14

H = 35 - 14

H = 21

In the other box, there would then be 21 hockey players and 14 basketball players.

Answer 4

You asked: "What could the total number of pictures be?"
Bosnian found the smallest such total, but there there are an infinite number of solutions

e.g. take one of my x and y pairs
x = 15, y = 21
so in Box1, there would be 60 hockey cards and 45 basketball cards
notice 60:45 = 4:3
the total would be 105 cards in box1

in box2 there would be 63 hockey cards and 42 basketball cards
for a total of 63+42 = 105 , the same as box1

You said: "can't it be 7, 10 too?"
The answer is no
First of all what are the 7 and 10 supposed to be??
7:10 ≠ 4:3 nor 3:2