18. Box A and B coiatailaed some 18d and blue ribbons. In Box A. the number 2
of red tti>ons.was-3 of the runbef of blle rl>bons. In box B. the ratio of the number of red rl>bona to the·runbet of the blue ri>bol• waa 8 : 7. The runber· of ribbons In Box B was twice the runber of ribbons fn BoxA
(a) What was 1he ratio of the runber of red ri>bol,a in Box A to the runbef-of '8d ri>bol• in Box B? Give your answer In the simplest form.
(b) The runber of blue ribbons in Box B was 50 more than the number of blue ribbons In Box A. How many ribbons are there In Box B?
To solve this problem, let's first assign variables to the unknown quantities.
Let:
- x be the number of red ribbons in Box A
- y be the number of blue ribbons in Box A
- 2x be the number of ribbons in Box A
- 8z be the number of red ribbons in Box B
- 7z be the number of blue ribbons in Box B
- 2(2x) be the number of ribbons in Box B
Let's solve part (a) first.
Given that the number of red ribbons in Box A was 2 less than 3 times the number of blue ribbons, we can set up the equation as follows:
x = 3y - 2
Given that the ratio of the number of red ribbons to the number of blue ribbons in Box B was 8:7, we can set up the equation as follows:
8z / 7z
To find the ratio of the number of red ribbons in Box A to the number of red ribbons in Box B, we need to divide the equation for Box A by the equation for Box B:
(x) / (8z)
Now, let's simplify this expression:
x = 3y - 2 ------(1)
(x) / (8z / 7z)
To divide by a fraction, we can multiply by its reciprocal. Therefore:
(x * 7z) / (8z)
Now, we can substitute the value of x from equation (1) into our expression:
(3y - 2) * 7z / (8z)
Simplifying further:
(21yz - 14z) / (8z)
The ratio of the number of red ribbons in Box A to the number of red ribbons in Box B is:
(21yz - 14z) / (8z) (simplified form)
Now, let's move on to part (b).
Given that the number of blue ribbons in Box B was 50 more than the number of blue ribbons in Box A, we can set up the equation as follows:
7z = y + 50
Let's solve this equation to find the value of y:
7z - 50 = y
Since the number of blue ribbons in Box B is equal to 2 times the number of blue ribbons in Box A, we can set up the equation as follows:
7z = 2y
Now, we can substitute the value of y from the previous equation into this equation:
7z = 2(7z - 50)
Now, let's solve for z:
7z = 14z - 100
100 = 14z - 7z
100 = 7z
z = 14
Substituting this value of z back into the equation for y:
y = 7z - 50
y = 7(14) - 50
y = 98 - 50
y = 48
Now, we need to find the number of ribbons in Box B, which is equal to twice the number of ribbons in Box A:
2(2x) = 4x
Substituting the value of x from equation (1):
2(3y - 2) = 4(3y - 2)
6y - 4 = 12y - 8
8y = 4
y = 0.5
However, y cannot be a fraction in this case, so our previous calculations must be incorrect.
Please double-check the information provided and confirm if there are any mistakes or missing details in the problem statement.