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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?

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??

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 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

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

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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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.

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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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)

+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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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 so answer is D

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 so answer is D

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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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.