# Question

Hi,

I really appreciate someone can help for this question. Thank you.

Ricky had 260 more red marbles than blue marbles at first. After selling 1/7 of the blue marbles and 5/9 of the red marbles, he had an equal number of number of blue marbles and red marbles left. He then bought another 30 blue marbles. What was the ratio of the number of red marbles to the number of blue marbles Ricky had in the end? Express your answer in its simplest form.

Start off with drawing two equal rectangles to illustrate equal number of red and blue marbles left. Cut the model for the red rectangles into 6 equal parts and the draw in an extra unit for the red marbles. Cut the model for the blue model into 6 equal parts as well. In each of the 6 units, write 2u and draw another 15units for the blue model.  Red marbles should have 27 units at first while blue marbles should have 14units at first . Hence, 13 units will be 260. 1 unit will be 20. In the end, there will be 260 blue and 240 red marbles. ratio of blue to red will be 13:12

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Ricky had 260 more red marbles than blue marbles at first. After selling 1/7 of the blue marbles and 5/9 of the red marbles, he had an equal number of number of blue marbles and red marbles left. He then bought another 30 blue marbles. What was the ratio of the number of red marbles to the number of blue marbles Ricky had in the end? Express your answer in its simplest form.

9 units R – 7 units B = 260

4 Units Red – 6 Units B = 0

5 Units R – 1 Unit B = 260

35 units R -7 units B = 1820

26 units R= 1560

R = 540

B = 280

B at end = 280*6/7+30=270

ans = 240:270 = 8:9

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This is a Fractional Transfer question.

LCM    1/7 = 9/63

5/9 = 35/63

We shall start with the LCM unit. All number shall only be written in fraction and not decimal.

 Red Blue Before 63u + 260 63u After 28u + 1040/9 54u

63u x 6/7  = 54u

63u x 4/9 = 28u

260 x 4/9 = 1040/9

28u + 1040/9 = 54u

54u – 28u = 1040/9

26u × 9 = 1040

234u = 1040

1u = 40/9

54u = 240  (red)

240 + 30 = 270 (blue)

Red : Blue = 240 : 270 = 8 : 9

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Note to MusicStar: To draw Ace Starling model of 2u and 3u, there is some pre-planning involved looking at LCM, similar to what I try to show below.

[There is no need to draw the table 3 times. Below is to illustrate my thought process.]

Sell                 Left                     Total

Blue       1/7                   6/7                      7/7

Red         5/9                 4/9                      9/9

Note       (Equal number left. make same units)

Sell                 Left                     Total

Blue       #2. 2u             #1. 12u                #4. 14u

Red        #3. 15u           #1. 12u                 #5. 27u

Note                 (#1. make same 12u=LCM of 4 and 6)

Compare Total: original Red is 260 more than original Blue

so Difference of 13u = 260 –> u=20

Sell                 Left                     Total

Blue                             240+30=270

Red                              240

Ratio of red:blue = 24:27 = 8:9#

Just a note that it is useful to recognise that this is a common MCQ-type question:

E.g. 6/7 of blue marbles is equal to 4/9 of red marbles:

6/7 of B = 4/9 of R

key is to make numerator the same….    12/14 of B = 12/27 of R

So total B is 14u and total R is 27u. Then answer to what the question wants.

The problem sum works like a smoke-screen with additional info to deduce 6/7 of B=4/9 of R.

SAHMom,

Good analogy !!!

The example you shown applies the same concept as this problem sum.

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Alternatively, I would solve the question as follows :

red marbles :
7u + 260 – (5/9) x (7u + 260)

blue marbles :
7u – 1u = 6u

6u = 7u + 260 – (5/9) x (7u + 260)
54u = 63u + 2340 – 35u – 1300
1u ——- (2340 – 1300)/(54 – 63 + 35) = 1040/ 26 = 40

red marbles :
7 x 40 + 260 – (5/9) x (7 x 40 + 260) = 540 – 300 = 240

blue marbles :
6 x 40 + 30 = 270

240 : 270
8 : 9

Ans : 8 : 9.

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Can explain how you get 2u & 3u?

B “left-over” is 6 block ( unshaded ).

R “left-over” is 4 block ( unshaded ).

The question says B and R marbles have equal number of “left-over”.

In order to make the left-over “equal”, have to find the common multiples of 6 & 4 ( which is 12 ).

So, objective is to subdivided the B and R “left-over” both each into 12 units.

Therefore, each block in B needs to subdivided into 2 units and each block in R needs to subdivided into 3 units.

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