Question

SAT Question

A box contains only red pens and white pens. If one red pen is removed at random the probability that it is red is 1/5. After putting another 600 red pens in the box, the probability of selecting a red pen is 4/5. How many pens were originally in the box?

Can someone explain the steps?

(Answer is 200)

Answer 1

iam uncertain about the method
but i know why the ans is correct

Answer 2

Let \(r\) and \(w\) be the number of red and white pens, respectively.

In any given draw, the probability of getting a red pen would be \(\dfrac{r}{r+w}\).

Say the number of pens in the first draw is \(P_1\), and the second draw has \(P_2\) pens.
\[\frac{r}{P_1}=\frac{1}{5}\]
\[\frac{r}{P_2}=\frac{4}{5}~\text{ where }P_2=P_1+600\]
Basically, you have a system of two equations with two unknowns. You should be able to solve for \(r\) and \(P_1\), I think.

Answer 3

Hmm, apparently that system has negative solutions... weird

Answer 4

Ah, got it, it should be
\[\begin{cases}
\dfrac{r}{P_1}=\dfrac{1}{5}\\\\
\dfrac{r+600}{P_1+600}=\dfrac{4}{5}
\end{cases}\]

Answer 5

Do I just cross multiply your equations?

Answer 6

Well I would first establish a relation between \(r\) and \(P_1\), then substitute into the second equation.

Answer 7

Cross-multiplication will happen when you solve for the substituted variable.

Answer 8

Let 'r' be the number of red pins and 'w' be the number of white pens at the start.
\(\Large \frac{r}{r+w} = \frac 15;~~~~ \frac{r+600}{r+600+w} = \frac 45 \)
Eliminate 'w' and solve for 'r'

From (1) :
5r = r + w
w = 4r

From (2):
5r + 3000 = 4r + 2400 + 4w
r + 600 = 4w
put w = 4r
r + 600 = 16r
15r = 600
r = 40
w = 4r = 160

r + w = 40 + 160 = 200

(Assumption: The first pen that was drawn was put back).

Answer 9

^what @aum said :)

Answer 10

Ah I got confused around

r + 600 = 4w
put w = 4r

This part. Thanks for your help guys.

Answer 11

yw

Answer 12

Wish I could give two medals.

Answer 13

ikr

Answer 14

letz do 1 one thing u give to @aum i will giv to @SithsAndGiggles

Answer 15

thanks both of u for sharing ur knowledge

Answer 16

you are welcome.

Answer 17

Everyone's welcome