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# M09#12

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Senior Manager
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28 May 2010, 08:54
1
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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

OA
[Reveal] Spoiler:
D

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OE
[Reveal] Spoiler:
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:
$$Y = 1.1^3X = 1.21*1.1*X = 1.331X ~= 1\frac{1}{3}X$$
The required ratio is $$\frac{X}{Y} = 1/1\frac{1}{3} = \frac{3}{4} = 0.75$$

My approach - Could someone validate if this approach is right??
[Reveal] Spoiler:
Let $$P$$ be the power of the engine.
$$1$$ cylinder results in a $$10%$$ increase in power.
Therefore $$0.1P = 1$$ Cylinder.
Power of a $$9$$ cylinder engine $$= 9 * 0.1P = 0.9P$$
Power of a $$12$$ cylinder engine $$= 12 * 0.1P = 1.2P$$
Ratio of power of $$9$$ cylinder engine to $$12$$ cylinder engine$$= \frac{0.9P}{1.2P} = \frac{3}{4} = 0.75$$

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Last edited by sidhu4u on 28 May 2010, 22:33, edited 2 times in total.
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28 May 2010, 11:03
hope thats a good way.... Can you post the OE? I got bogged down by trying to compound the 10%everytime. Is that not the right way to do this?

(P+0.1P)0.1P .... and so on?
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28 May 2010, 22:00
1
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(1*1*1)/1.1*1.1*1.1 ~ 0.75

Does this help?
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08 Jun 2010, 03:18
I think its D.

No. of Engines 9 10 11 12
Power(goes up by 10%) 10 11 12.1 13.31

10/13.31 = 0.75
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08 Jun 2010, 05:13
1
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I solved it using the pricipal of compound intrest
A= P(1+r)^n
Assuming principal P is 1 i.e 1st year
r= rate of intrest
n= no. of years
1) for 9 years
A1=1(1+ .10)^8
2) for 12 years
A2=1(1+ .10)^12

hence ratio is A1/A2 = 1/1.10^3 ~ 1/1.33 = 100/133 = .75
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08 Jun 2010, 07:36
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

I agree with everyone who chose D

A good question to pick numbers:

Let the power of 9-cylinder engine be 1,
Increases from 9 - 12 cylinder has 3 steps of power increment
Therefore, the ratio = 1/ (110/100*110/100*110/100) =
100*100*100/110*110*110 =
1000/1331=0.75
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08 Jun 2010, 09:51
1
KUDOS
This is a geometric sequence.

the nth term in a geometric sequence is a*( r^(n-1) )

the ratio of 9th to 11th is ar^8/ar^11 = 1/r*r*r = 1/1.331 = 0.75
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10 Jun 2010, 08:04
2
KUDOS
since power increase by 10% => common ratio = 1.1
let the power of a 9-engine = p (1st term of a GP series)
Power of a 12(4th term) engine = p(1.1)^3
9- engine / 12-engine = p/(p)1.1^3 = 1/1.331
0.751
The closest is D (0.75)
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15 Jun 2010, 07:41
D too.

by choosing the power of 9 cylinders is 1.
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13 Aug 2010, 02:30
Its C.

The ratio is 1: 1.1 to the power 3. Close to 0.75!
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19 Aug 2010, 07:16
sidhu4u wrote:

My approach - Could someone validate if this approach is right??
[Reveal] Spoiler:
Let $$P$$ be the power of the engine.
$$1$$ cylinder results in a $$10%$$ increase in power.
Therefore $$0.1P = 1$$ Cylinder.
Power of a $$9$$ cylinder engine $$= 9 * 0.1P = 0.9P$$
Power of a $$12$$ cylinder engine $$= 12 * 0.1P = 1.2P$$
Ratio of power of $$9$$ cylinder engine to $$12$$ cylinder engine$$= \frac{0.9P}{1.2P} = \frac{3}{4} = 0.75$$

I think you got lucky in this case. You should have to do it exponentially.

I took a long approach, assuming the power of a 9 cylinder engine is 100:

10 cylinder power: 110
11: 121
12: 133

100/133 was not something I could compute quickly, so I picked .71 first because it was the easiest calculation, then went to .75 and found that it was correct.
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18 Dec 2010, 01:49
1
KUDOS
I think the simplest way is: an increase in 10% simply means a1.1 increase so
9*1.1=9.9
12*1.1=13.2
Ratio of the two is 9.9/13.2=0.75
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11 Jun 2011, 15:03
I used the compound formaule approach as the power compounds....

so....y=x(1+10/100)^3=x(11/10)^3
The question is what is x/y ?
or what is (10/11)^3...

I agree that the ANS. is (D) if you actually solve putting the values...

But what I did was
[(11-1)/11]^3=(1-.09)^3=(.91)^3
Now I approx. this to (.9)^3 giving .729
So I chose (C) that was wrong

Do these types of Questions where you have to actually solve the values rather than apply logic really come in GMAT ?
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11 Jun 2011, 22:07
My approach was,

We are asked to find the ration of power with 9 cylinder to that of 12 cylinder engine.
Given: addition of 1 cylinder will increase the power by 10%. Say if power of engine is 1, then adding 1 cylinder to it makes the power as 1.1

$$=P(9 Cylinders)/ P(12 Cylinders)$$
$$=P(9 C)/ (P(9 C) and P(3 C))$$
$$= 1/ P(3 C)$$
$$= 1/ (1.1* 1.1 * 1.1) = 1/ 1.33 = 1/ 1.3 (approx)$$
$$= .75 (approx)$$ Simplistic, calculation will give nearest answer (100/13 is 7.6)

Ans: D
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12 Jun 2011, 01:16
+1 for D..
Using the approx technique..1/13 is 0.77 and 1/14 is 0.71..
Now we have 1/13.33.. it should be more towards 0.75 rather than 0.72...
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12 Jun 2011, 13:34
since ratio of power for 9 and 12 cylinders is asked we dont have to really calculate the p9 and p12 as we know p/c = p9/c9 = p12/c12
since we know c9=9 cylinders and c12=12 cylinders
so after substituting the value p9/p12=9/12=0.75
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13 Jun 2011, 01:51
I think the easiest way to solve this problem is,

9*10% = 0.9
12*10%= 1.2

9+1.1 = 9.9
12+1.2=13.2

Ratio of 9.9/13.2 = 0.75.

Please let me know if this is correct.

Thanks
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13 Jun 2011, 08:12
sameershintrein wrote:
since ratio of power for 9 and 12 cylinders is asked we dont have to really calculate the p9 and p12 as we know p/c = p9/c9 = p12/c12
since we know c9=9 cylinders and c12=12 cylinders
so after substituting the value p9/p12=9/12=0.75

By this logic, p9/p13 would be 0.69 (9/13)..and p9/p15 would be 0.6 (9/15)

Whereas adding 10% to each increase in cylinder gives 0.68(1/1.46) in case of 13 cylinders and 0.56 (1/1.77) in case of 15 cylinders..

You can see the respective difference between values computed through the two methods is increasing...

The method that you used fitted well because of the values involved but it doesnt take into account the 10% increase..

Please let me know if I'm missing something here..
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14 Jun 2011, 18:35
use 1 for 9 cyl
9= 1
10= 1.1 (10% more=.1)
11= 1.1* 1.1(10% of 1.1 not 1) = 1.21
12 = 1.21 * 1.1 = 1.33

9cly/12cly = (1) / (4/3)<--converted to closest fraction = (1) * (3/4) = 9/12 = 3/4 or .75 so D
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14 Jun 2012, 05:24
Say 9C = 100

Then 12C = 100 * (1.1)^3

So the Ratio = 100 / 100 (1.1)^3 = (10/11)^3
=> Ratio = (0.909)^3
Using approximation: 0.9 * 0.9 * 0.9 = 0.81 * 0.9 = 0.729
So, the ratio must be greater than 0.729 ; 0.75 is the likely value. D is a good choice of answer.
Re: M09#12   [#permalink] 14 Jun 2012, 05:24

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