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

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Math Expert
Joined: 02 Sep 2009
Posts: 48044
Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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27 Jun 2018, 21:55
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35% (medium)

Question Stats:

76% (01:38) correct 24% (01:52) wrong based on 140 sessions

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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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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.

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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28 Jun 2018, 03:17
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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.

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

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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?
Math Expert
Joined: 02 Sep 2009
Posts: 48044
Re: Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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28 Jun 2018, 20:57
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  [#permalink]

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28 Jun 2018, 23:37
Bunuel 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
Intern
Joined: 14 Jun 2018
Posts: 36
Re: Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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29 Jun 2018, 00:38
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
Director
Joined: 09 Mar 2016
Posts: 769
Integers a and b are consecutive multiples of 6. Integers x and y are  [#permalink]

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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.

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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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.

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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