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

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Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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New post 27 Jun 2018, 21:55
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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. \(\frac{1}{2}\frac{(a+b)}{(x+y)}\)

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Re: Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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New post 27 Jun 2018, 23:03
1

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: B

Note: 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  [#permalink]

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New post 28 Jun 2018, 03:17
2
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  [#permalink]

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New post 28 Jun 2018, 11:02
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  [#permalink]

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New post 28 Jun 2018, 20:57
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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New post 28 Jun 2018, 23:37
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  [#permalink]

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New post 29 Jun 2018, 00:38
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  [#permalink]

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New post 04 Aug 2018, 07:05
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  [#permalink]

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New post 04 Aug 2018, 07:13
1
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 = 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 :)



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 [#permalink] 04 Aug 2018, 07:13
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