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­Is this the same as (1.1)^9:(1.1)^12 = 3:4 = 0.75?
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­Is this the same as (1.1)^9 :(1.1)^12 = 3:4 = 0.75?

Yes. This is if you assume the power of the 1-cylinder engine is 1. However, notice that here also you are basically calculating the same as in the above: (1.1)^9/(1.1)^12 = 1/(1.1)^3.
 
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Hi Bunuel,

Can you please help identify where did I go wrong in my approach.
Power of 9 Cylinder = 1
Power of 12 Cylinder= 1*(110/100)^3
9 Cylinder Power/12 Cylinder Power= (10/11)^3= (0.9)^3=0.729
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Nikhil_Rai
Hi Bunuel,

Can you please help identify where did I go wrong in my approach.
Power of 9 Cylinder = 1
Power of 12 Cylinder= 1*(110/100)^3
9 Cylinder Power/12 Cylinder Power= (10/11)^3= (0.9)^3=0.729
10/11 is a little more that 0.9, it's actually 0.(90). So, 0.(90)^3 is more than 0.729.
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Hi Bunuel! I got the same answer but I did it this way, can you let me know if I'm wrong and why?

1 cylinder 10%, 9 cylinders be 90%, 12 cylinders be 120%. Therefore: 90/120= 3/4= 0.75
Bunuel
If adding one cylinder to an engine increases its power by 10%, which of the following is closest to the ratio of the power of a 9-cylinder engine to a 12-cylinder engine?

A. 0.69
B. 0.71
C. 0.72
D. 0.75
E. 0.78
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Hi Bunuel! I got the same answer but I did it this way, can you let me know if I'm wrong and why?

1 cylinder 10%, 9 cylinders be 90%, 12 cylinders be 120%. Therefore: 90/120= 3/4= 0.75
Bunuel
If adding one cylinder to an engine increases its power by 10%, which of the following is closest to the ratio of the power of a 9-cylinder engine to a 12-cylinder engine?

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

No, this is not correct. When the problem says that adding one cylinder increases the engine’s power by 10%, it means that if the power of an engine with x cylinders is p, the power of an engine with x + 1 cylinders becomes p * 1.1. So, you multiply incrementally by 1.1 for each added cylinder rather than just adding fixed increments of 10%.
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Thank you Bunuel!
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sssdd1234
Hi Bunuel! I got the same answer but I did it this way, can you let me know if I'm wrong and why?

1 cylinder 10%, 9 cylinders be 90%, 12 cylinders be 120%. Therefore: 90/120= 3/4= 0.75
Bunuel
If adding one cylinder to an engine increases its power by 10%, which of the following is closest to the ratio of the power of a 9-cylinder engine to a 12-cylinder engine?

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

No, this is not correct. When the problem says that adding one cylinder increases the engine’s power by 10%, it means that if the power of an engine with x cylinders is p, the power of an engine with x + 1 cylinders becomes p * 1.1. So, you multiply incrementally by 1.1 for each added cylinder rather than just adding fixed increments of 10%.
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It is interesting that the answer worked because you made a consistent mistake - you applied the same logic on both sides, so in some way you made a mistake but in another, you still answered it correctly 😮
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Bunuel, if we are taking the power of 1 cylinder as 1, then we are adding 8 cylinders and for the other one adding 11 cylinders. Is solving it as (1.1)^8 / (1.1)^11 an incorrect way of thinking? Please help, thanks.
Bunuel
nsbcak
­Is this the same as (1.1)^9 :(1.1)^12 = 3:4 = 0.75?

Yes. This is if you assume the power of the 1-cylinder engine is 1. However, notice that here also you are basically calculating the same as in the above: (1.1)^9/(1.1)^12 = 1/(1.1)^3.
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ragnarok13
Bunuel, if we are taking the power of 1 cylinder as 1, then we are adding 8 cylinders and for the other one adding 11 cylinders. Is solving it as (1.1)^8 / (1.1)^11 an incorrect way of thinking? Please help, thanks.
Bunuel
nsbcak
­Is this the same as (1.1)^9 :(1.1)^12 = 3:4 = 0.75?

Yes. This is if you assume the power of the 1-cylinder engine is 1. However, notice that here also you are basically calculating the same as in the above: (1.1)^9/(1.1)^12 = 1/(1.1)^3.

That's correct and also gives the correct answer (1.1^8 /1.1^11 = 0.75...).
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Bunuel Thank you :)
ragnarok13
Bunuel, if we are taking the power of 1 cylinder as 1, then we are adding 8 cylinders and for the other one adding 11 cylinders. Is solving it as (1.1)^8 / (1.1)^11 an incorrect way of thinking? Please help, thanks.
Bunuel
nsbcak
­Is this the same as (1.1)^9 :(1.1)^12 = 3:4 = 0.75?

Yes. This is if you assume the power of the 1-cylinder engine is 1. However, notice that here also you are basically calculating the same as in the above: (1.1)^9/(1.1)^12 = 1/(1.1)^3.
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I like the solution - it’s helpful.
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I solved, if x+1 cylinder increases 1.1y power of the engine so 1 cylinder have 1.1y/x+1 power

Where x is the no. of initial cylinder and y is the initial power of engine.

so, 9 cylinder will have 1.1y*9/x+1 and 12 cylinder will have 1.1y*12/x+1

Upon taking the ratio, it will become 9/12 =3/4=0.75

Bunuel Is this way of approaching question is right? or it is just the coincidence which may lead to error in approaching other similar type of questions?
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abhinavvisen
I solved, if x+1 cylinder increases 1.1y power of the engine so 1 cylinder have 1.1y/x+1 power

Where x is the no. of initial cylinder and y is the initial power of engine.

so, 9 cylinder will have 1.1y*9/x+1 and 12 cylinder will have 1.1y*12/x+1

Upon taking the ratio, it will become 9/12 =3/4=0.75

Bunuel Is this way of approaching question is right? or it is just the coincidence which may lead to error in approaching other similar type of questions?
Your approach happens to give the same numerical result, but it’s based on an incorrect assumption, so it’s a coincidence.

The problem states that each additional cylinder increases power by 10%, meaning the increase is multiplicative, not linear. So you multiply by 1.1 for every added cylinder, not add a fixed fraction of total power.

In other words, power grows as:

1-cylinder power * (1.1)^(number of added cylinders).

That’s why the correct ratio is (1)/(1.1^3) = 0.75.

Your method treats the relationship as directly proportional (linear), which works only here by coincidence but would fail for any case involving repeated percentage increases.

Please study the discussion above for more.
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I like the solution - it’s helpful.
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