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Integers a and b are consecutive multiples of 6. Integers x and y are
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27 Jun 2018, 20:55
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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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27 Jun 2018, 22:03
Solution Given:• Integers a and b are consecutive multiples of 6 • Integers x and y are consecutive multiples of 8 To find:• In terms of a, b, x, and y, the ratio of the average of a and b and the average of x and y Approach and Working:• The average of a and b = \(\frac{a+b}{2}\) • The average of x and y = \(\frac{x+y}{2}\) Therefore, the ratio of averages =\(\frac{a+b}{2}\) / \(\frac{x+y}{2}\) = \(\frac{a+b}{x+y}\) Hence, the correct answer is option B. Answer: BNote: In an alternate way, we can also get the ratio by assuming values. For example, we can assume a =6, b = 12, and x = 8, y = 16. Putting this values in the option, we can check which one satisfies the answer.
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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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28 Jun 2018, 02:17
Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) a and b are consecutive multiples of 6. a = 6*1 = 6 b = 6*2 = 12 a+b = 18 a+b/2 = 18/2 = 9 x and y are consecutive integer of 8. x = 8*1 = 8 y = 8*2 = 16 x+y = 24 x+y/ 2 = 24/2 =12 we are asked to find out the ratio of the average of a and b to x and y. 9 : 12 =3:4 Scan the answer choices by plugging in the value assumed above. Only option B matches. The best answer is B.



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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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28 Jun 2018, 10:02
Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) I am really puzzled ...'Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8' ..this portion is totally irrelevent 'In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y' Ans for this is always ..a+b/x+y how OA is D?



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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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28 Jun 2018, 19:57
janadipesh wrote: Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) I am really puzzled ...'Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8' ..this portion is totally irrelevent 'In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y' Ans for this is always ..a+b/x+y how OA is D? The OA is B. Edited. Thank you.
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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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28 Jun 2018, 22:37
Bunuel wrote: janadipesh wrote: Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) I am really puzzled ...'Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8' ..this portion is totally irrelevent 'In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y' Ans for this is always ..a+b/x+y how OA is D? The OA is B. Edited. Thank you. Sir still i have confusion how this '...multiple of 6 and multiple of 8 ' significant to the question ....pls help



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Re: Integers a and b are consecutive multiples of 6. Integers x and y are
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28 Jun 2018, 23:38
janadipesh wrote: Sir still i have confusion how this '...multiple of 6 and multiple of 8 ' significant to the question ....pls help irrelevant data to confuse people



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Integers a and b are consecutive multiples of 6. Integers x and y are
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04 Aug 2018, 06:05
selim wrote: Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) a and b are consecutive multiples of 6. a = 6*1 = 6 b = 6*2 = 12 a+b = 18 a+b/2 = 18/2 = 9 x and y are consecutive integer of 8. x = 8*1 = 8 y = 8*2 = 16 x+y = 24 x+y/ 2 = 24/2 =12 we are asked to find out the ratio of the average of a and b to x and y. 9 : 12 =3:4 Scan the answer choices by plugging in the value assumed above. Only option B matches. The best answer is B. Hi selim i solved just like you and got 3:4, but i stiill dont see how to plug in these values (3;4 ) into B option can you pls explain. thanks



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Integers a and b are consecutive multiples of 6. Integers x and y are
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04 Aug 2018, 06:13
dave13 wrote: selim wrote: Bunuel wrote: Integers a and b are consecutive multiples of 6. Integers x and y are consecutive multiples of 8. In terms of a, b, x, and y, what is the ratio of the average of a and b and the average of x and y?
A. \(\frac{(x+y)}{(a+b)}\)
B. \(\frac{(a+b)}{(x+y)}\)
C. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{4})}\)
D. \(\frac{(a+\frac{b}{4})}{(y\frac{x}{2})}\)
E. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\) a and b are consecutive multiples of 6. a = 6*1 = 6 b = 6*2 = 12 a+b = 18a+b/2 = 18/2 = 9 x and y are consecutive integer of 8. x = 8*1 = 8 y = 8*2 = 16 x+y = 24x+y/ 2 = 24/2 =12 we are asked to find out the ratio of the average of a and b to x and y. 9 : 12 =3:4 Scan the answer choices by plugging in the value assumed above. Only option B matches. The best answer is B. Hi selim i solved just like you and got 3:4, but i stiill dont see how to plug in these values (3;4 ) into B option can you pls explain. thanks In the description , u get the value of a+b = 18 and x +y = 24. Option B: \(\frac{a+b}{x + y}\) = 18 / 24 = 3:4 Hope it's clear. Ratio matches.




Integers a and b are consecutive multiples of 6. Integers x and y are &nbs
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