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# Ratio problem watch

1. The ratio of black counters to red counters in a box is 3:4. After adding eight counters to the box the ratio is 3:5. How many counters were in the box at the beginning?
Could somebody give me a way to get this started please? Sunday morning brain is not working!!
2. the only thing I can think I that one ratio thing is worth 8 counters. therefore, if there are 3:4 to start with then in total there would be 7(because 3+4) x8 counters = 56. but the qn doesn't specify the colors of the counters added so I can't say for sure
3. (Original post by Stupidanon)
the only thing I can think I that one ratio thing is worth 8 counters. therefore, if there are 3:4 to start with then in total there would be 7(because 3+4) x8 counters = 56. but the qn doesn't specify the colors of the counters added so I can't say for sure
I was thinking that because the "3" doesn't change and the other just increases by 1, then we find the LCM of 4, 5 and 8 (the eight because the 8 gets added on).

This is 40, which then means that the ration would become 24:40 (3 x 8, 5 x 8) and then take off the 8 that was added on it becomes 24:32 which then reduces to 3:4.I am not confident that this is not purely coincidental though.
4. I got 56? My working is related to what you said kinda.

So the initial number of counters has to be divisible by 7 (X=7K), and number of counters plus eight has to be divisible by 8 (X+8=8C). Therefore (7k+8)/8 = a whole number. Therefore (7k/8)+1 = a whole number. K is therefore some multiple of 8, which if we try for K=8 works. (you get 24:32 and 24:40)

and then X=7K, so X = 56
5. Also to eliminate other values of k, initial black counters (A) is (3X/7), and black counters + 8 counters is (3/8)*(X+8) (B). As you can't take counters away, B must be bigger or equal to A. (B>=A). Then if you do that inequality, you get 3>=(3X/56), or 3>=(21K/56), and therefore only K=8 works

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Updated: October 15, 2017
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