Mr Lim put some red pens and blue pens into a box.The ratio of the number of red pens to the number of blue pens was 3:4.

He added 20 red pens into the box and the ratio became 2:1.
a. were there more red pens or blue pens in the box at the end?
b. how many red pens were there in the end?
c. how many blue pens were there?

User Icon for Steve Steve answered
12 years ago

well, since the ratio r:b = 2:1 I'd say there were more r's...

(r+20)/b = 2/1
r/b = 3/4
so, b = 4/3 r
(r+20)/(4/3 r) = 2
r+20 = 8/3 r
5/3 r = 20
r = 12

so, there were 32 reds at the end
there were 4/3*12 = 16 blue pens. Note that 16 = 32/2 as well.

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8 years ago

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5 years ago

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3 years ago

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3 years ago

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2 years ago

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User Icon for Explain Bot Explain Bot answered
11 months ago

To solve this problem, we can use a system of equations. Let's denote the initial number of red pens as R and the initial number of blue pens as B.

According to the given information, the ratio of red pens to blue pens was 3:4, which can be written as:

R/B = 3/4 --(1)

After Mr. Lim added 20 red pens, the new ratio became 2:1:

(R + 20)/B = 2/1 --(2)

Let's solve these equations to find the answers to the questions.

a. To determine whether there were more red pens or blue pens in the box at the end, we need to compare the values of R and B.
We can solve the equations (1) and (2) simultaneously to find the values of R and B.

Multiply equation (2) by 4, so that the denominators of both fractions become equal:

4(R + 20)/B = 8/1

Simplifying equation (2):

4(R + 20) = 8B
4R + 80 = 8B
4R = 8B - 80

Now substitute the expression for 4R from equation (1) into the right-hand side of the above equation:

(3/4)B = 8B - 80
3B = 32B - 320

Move all terms containing B to one side:

32B - 3B = 320
29B = 320

Divide both sides by 29:

B = 320/29 ≈ 11.03

This means that there were approximately 11 blue pens at the end.

Now we can substitute the value of B into equation (1) to find the value of R:

R/B = 3/4
R/(11.03) = 3/4

Multiply both sides by 11.03:

R = (3/4)(11.03)
R ≈ 8.27

This means there were approximately 8 red pens at the end.

Therefore, there were more blue pens (approximately 11) than red pens (approximately 8) in the box at the end.

b. The number of red pens at the end is approximately 8.

c. The number of blue pens at the end is approximately 11.