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M09-12

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Math Expert
Joined: 02 Sep 2009
Posts: 58464

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16 Sep 2014, 00:39
00:00

Difficulty:

35% (medium)

Question Stats:

74% (02:12) correct 26% (02:11) wrong based on 70 sessions

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If the power of an engine grows by 10% when the number of its cylinders is increased by one, which of the following is closest to the ratio of the power of a 9-cylinder engine to that of a 12-cylinder engine?

A. 0.69
B. 0.71
C. 0.72
D. 0.75
E. 0.78

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Math Expert
Joined: 02 Sep 2009
Posts: 58464

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16 Sep 2014, 00:39
1
4
Official Solution:

If the power of an engine grows by 10% when the number of its cylinders is increased by one, which of the following is closest to the ratio of the power of a 9-cylinder engine to that of a 12-cylinder engine?

A. 0.69
B. 0.71
C. 0.72
D. 0.75
E. 0.78

Let $$X$$ denote the power of the 9-cylinder engine and $$Y$$ the power of the 12-cylinder engine. It follows from the stem that:
$$Y = 1.1^3 X = 1.21*1.1*X = 1.331X \approx 1\frac{1}{3}X$$

The required ratio is $$\frac{X}{Y} = \frac{1}{1\frac{1}{3}} = \frac{3}{4} = 0.75$$

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Joined: 01 Oct 2014
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24 Sep 2015, 07:47
1
Bunuel , I solved this question correct however I am not sure about my approach I took the assumption that 9 cylinders will increase the capacity of cylinder by 90% and 12 cylinders will increase by 120% so 90/120=3/4=0.75 is this approach is correct
Manager
Joined: 31 Jul 2014
Posts: 125
GMAT 1: 630 Q48 V29

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05 Oct 2015, 08:42
Bunuel wrote:
Official Solution:

If the power of an engine grows by 10% when the number of its cylinders is increased by one, which of the following is closest to the ratio of the power of a 9-cylinder engine to that of a 12-cylinder engine?

A. 0.69
B. 0.71
C. 0.72
D. 0.75
E. 0.78

Let $$X$$ denote the power of the 9-cylinder engine and $$Y$$ the power of the 12-cylinder engine. It follows from the stem that:
$$Y = 1.1^3 X = 1.21*1.1*X = 1.331X \approx 1\frac{1}{3}X$$

The required ratio is $$\frac{X}{Y} = \frac{1}{1\frac{1}{3}} = \frac{3}{4} = 0.75$$

it is tough to think about such fractions under time pressure.
i was trying to do (10/11)^3 = approx (0.9)^3 but it does not give correct answer
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Joined: 18 Jun 2015
Posts: 38

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12 Sep 2016, 14:21

it is tough to think about such fractions under time pressure.
i was trying to do (10/11)^3 = approx (0.9)^3 but it does not give correct answer

I tried 1/1.331 = 9.09/12.1 which approx the answer.
I agree it is a bit tricky to get the 1.331X≈1 1/3 in the real test pressure.
Current Student
Joined: 28 Aug 2016
Posts: 87
Concentration: Strategy, General Management

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25 Sep 2016, 23:52
I set power of 9 cyl engine = 100

10 cyl = 110
11 cyl = 121
12 cyl = 133.1

then 100/133.1 = approx. .75

Therefore D.
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Joined: 16 Jun 2018
Posts: 8

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30 Aug 2018, 23:24
Bunuel

Can we solve it like this?

P(1.1)^8/P(1.1)^11 =0.75

Is this a correct way of doing it.?
Math Expert
Joined: 02 Sep 2009
Posts: 58464

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30 Aug 2018, 23:32
psych77 wrote:
Bunuel

Can we solve it like this?

P(1.1)^8/P(1.1)^11 =0.75

Is this a correct way of doing it.?

_______________
Yes, this is correct.
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Joined: 02 Oct 2019
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04 Oct 2019, 06:03
I didn't get how it is 1.1^3 . can some one explain it in a detailed way
Re: M09-12   [#permalink] 04 Oct 2019, 06:03
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