Ratio : Help

Hi, can someone please help.

Q : Ratio of the number of toy A to the number of toy B in a large storage box was 9 : 2. This ratio became 16 : 3 after Lucy dropped 10 toy A into the box. Find the initial number of toy A in the box.

Not sure what is the answer, I've tested and approached it this way but stuck along the way.

A : B change to A : B
9 : 2 16 : 3

This would mean ratio changes in A = 7, B = 1
Ratio change of 7 = 10 toy A, that would mean a change of 1 = 10/7
Hence original A should be 9 x 10/7 = seriously i'm lost..as the answers are A) 72, B) 45, C) 63 and D) 54

Ratio of the number of toy A to the number of toy B in a large storage box was 9 : 2. This ratio became 16 : 3 after Lucy dropped 10 toy A into the box. Find the initial number of toy A in the box.

Click to expand...

First equation: A/B = 9/2
Rearrange this equation to get B = (2/9)A

Second equation: (A + 10)/B = 16/3
Rearrange this equation to get B = (3A + 30)/16

By substitution/transitive property, set equations equal to each other:

(2/9)A = (3A + 30)/16

Solve for A.

Thanks I got it for the earlier question.
I've got another question and also got stuck at a certain portion. Please help.

Q 2 : The respective numbers of green beans in Tin K and Tin N were in the ratio nof 25 : 17. 39 green beans were later transferred from Tin N to Tin K. the ratio of the number of green beans in Tin K to the number of green beans in Tin N became 3 : 1. FInd the initial number of green beans in Tin K.

The approach I have taken.
K / N = 25 / 17 K / N-39 = 3 / 1
N = (17/25)K ...equation 1 3 (N-39) = K.....equation 2

I then substitute equation 1 into 2 and but my answer is not right. The answer given A) 200, B) 175 C) 125 D) 150

Can you point out where i went wrong ?

Hello, ctmckoay!

Equation 2 is incorrect.

Q 2: The respective numbers of green beans in Tin K and Tin N were in the ratio of 25 : 17.
39 green beans were later transferred from Tin N to Tin K.
The ratio of the number of green beans in Tin K to the number of green beans in Tin N became 3 : 1.
FInd the initial number of green beans in Tin K.

. . \(\displaystyle (A)\;200 \qquad (B)\;175 \qquad (C)\;125 \qquad (D)\;150\)

Click to expand...

Tin K has \(\displaystyle K\) green beans.
Tin N has \(\displaystyle N\) green beans.

. . \(\displaystyle \text{The ratio is: }\:\frac{K}{N} \,=\,\frac{25}{17} \quad\Rightarrow\quad N \<=\<\tfrac{17}{25}K \;\;{\bf[1]}\)


39 beans are moved from Tin N to Tin K.
Tin K has \(\displaystyle K + 39\) green beans.
Tin N has \(\displaystyle N - 39\) green beans.

. . \(\displaystyle \text{The ratio is: }\:\frac{K+39}{N-39} \:=\:\frac{3}{1} \quad\Rightarrow\quad K \:=\:3N - 156\;\;{\bf[2]}\)


\(\displaystyle \text{Substitute [1] into [2]: }\:K \:=\:3\left(\tfrac{17}{25}K\right) - 156 \quad\Rightarrow\quad K \:=\:\tfrac{51}{25}K - 156\)


\(\displaystyle \text{Multiply by 25: }\;25K \:=\:51K - 3900 \quad\Rightarrow\quad 26K \:=\:3900\)


\(\displaystyle \text{Therefore: }\;K \:=\:150\;\;\hdots\;\;\text{answer (D)}\)