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At a certain stage of a soccer tournament, the score ratio of teams A,
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Updated on: 07 Oct 2017, 02:30
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At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B? A. 8 B. 10 C. 20 D. 40 E. 80 I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled: a:b = 2* 3:4 = 3:2
halved a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this.
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Originally posted by manukumar on 01 Dec 2012, 06:28.
Last edited by Bunuel on 07 Oct 2017, 02:30, edited 1 time in total.
Renamed the topic and edited the question.



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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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01 Dec 2012, 06:44
manukumar wrote: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8 2)10 3)20 4)40 5)80
I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled: a:b = 2* 3:4 = 3:2
halved a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this. A to B = 3 : 4 So, on doubling we get 6 : 4 A to C = 3 : 5 So, on halving we get 1.5 : 5 or 3 : 10 or 6 : 20 So final ratio = 6 : 4 : 20. If 20x = 40 4x = 8 Hence, answer is A
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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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17 Feb 2017, 11:26
Ratio of A:B is originally 3:4. Then this ratio is doubled and we have \(\frac{3}{2}\) * 2 = \(\frac{6}{4}\) Ratio of A:C is originally 3:5. Then this ratio is halved so we have \(\frac{3}{5}\) * \(\frac{1}{2}\) = \(\frac{3}{10}\).
In order to compare the two ratio we must express the common term (in this case A) in the same quantity. Thus we multiply the second ratio (\(\frac{3}{10}\)) for 2 and we get \(\frac{6}{20}\)
So now the new score ratio of team A, B and C is 6:4:20. If we divide the score of team C by the ratio for team C we get the multiplier (in this case is 2). Finally by applying the multiplier to B we get 8.
Hope it is clear.



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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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17 Feb 2017, 21:24
Answer: [A] A : B : C Initially: 3 : 4 : 5 6 : 4 (ratio A:B is doubled ) 3 : __ : 10 (ratio is A: C is halved) 6 : __ : 20 6 : 4 : 20 Finally, 12 : 8 : 40 (since C has to become 40) Therefore, B = 8
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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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12 Mar 2017, 10:49
I like making it in a table:
A:B:C 3:4:5 then A:B doubled 6:4:? then A:C halved 3:?:10 let's make a common ration for all 3: 6:4:20 and since C earned 40 points: 12:8:40 B is 8 Answer is A (8)



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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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15 Mar 2017, 16:52
manukumar wrote: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8 2)10 3)20 4)40 5)80 We are given that the original ratio of A : B : C = 3x : 4x : 5x. After the ratio of A to B doubles and the ratio of A to C is halved, we have: A/B = 6x/4x AND A/C = 3x/10x Since we need term A to be the same in both ratios, we can multiply our second ratio by 2/2 and we have: A/C = 6x/20x Now the ratio of A : B : C = 6x : 4x : 20x Since team C had a final of 40 points, we have: 20x = 40 x = 2 So, the final score of team B was 4 x 2 = 8. Answer: A
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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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07 Oct 2017, 02:28
Hello Moderators, I got this question as a part of daily practice questions from my workbook. Do you think it would be a good Idea to hide the highlighted portion of the question? The highlighted text is giving away a possible approach that can be taken to solve the question. Hiding this will help users when they attempt it using Timer. Thanking you in advance! manukumar wrote: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8 2)10 3)20 4)40 5)80
I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled: a:b = 2* 3:4 = 3:2
halved a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this.
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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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07 Oct 2017, 02:31
susheelh wrote: Hello Moderators, I got this question as a part of daily practice questions from my workbook. Do you think it would be a good Idea to hide the highlighted portion of the question? The highlighted text is giving away a possible approach that can be taken to solve the question. Hiding this will help users when they attempt it using Timer. Thanking you in advance! manukumar wrote: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8 2)10 3)20 4)40 5)80
I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled: a:b = 2* 3:4 = 3:2
halved a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this. _______________ Edited. Thank you.
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Re: At a certain stage of a soccer tournament, the score ratio of teams A,
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07 Oct 2017, 03:21
manukumar wrote: At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B? A. 8 B. 10 C. 20 D. 40 E. 80 I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled: a:b = 2* 3:4 = 3:2
halved a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this. Initially A:B:C = 3:4:5 Finally, A: B = 6:4 = 3:2 and, A:C = 3: 10 So, A: B: C = 3: 2: 10 Also final score of C = 40 So , A: B: C = 12 : 8 : 40 So, B = 8 Answer A
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