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M09-12

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M09-12  [#permalink]

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New post 16 Sep 2014, 00:39
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (01:29) correct 29% (01:19) wrong based on 97 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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Re M09-12  [#permalink]

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New post 16 Sep 2014, 00:39
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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 \(\frac{X}{Y} = \frac{1}{1\frac{1}{3}} = \frac{3}{4} = 0.75\)


Answer: D
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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

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Re: M09-12  [#permalink]

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New post 24 Sep 2015, 07:47
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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
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Re: M09-12  [#permalink]

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New post 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 \(\frac{X}{Y} = \frac{1}{1\frac{1}{3}} = \frac{3}{4} = 0.75\)


Answer: D


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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Re: M09-12  [#permalink]

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New post 12 Sep 2016, 14:21
anupamadw wrote:

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.
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Re: M09-12  [#permalink]

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New post 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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Re: M09-12  [#permalink]

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New post 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.?
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Re: M09-12  [#permalink]

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New post 30 Aug 2018, 23:32
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Re: M09-12 &nbs [#permalink] 30 Aug 2018, 23:32
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