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very simple problem - hard to crack !!!

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by desiguy » Tue Jul 15, 2008 11:55 pm
In a formulae a=b(square)c, if b is multiplied by 2 and c is multiplied by 3, what whoudl a be multiplied by ?

4
6
9
12
18

Ans : 12

I know it’s a very very dummbo questin. But my answer is 6. Not sure how its 12 ?
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Source: — Problem Solving |

by sudhir3127 » Wed Jul 16, 2008 12:19 am
can i know how u got 6. i got 12 ..

a= b^2c

((2)^2)* 3 = 12
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by desiguy » Wed Jul 16, 2008 12:28 am
Thanks...but u assume a and b to be 1 each. I assume the numbers to be different
:
a=2
b=3

(2^2)*3=12.................1
(4*2)*(3*3) = 72..............2

Thus multiply 12 by 6 to get 72 ??
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by sudhir3127 » Wed Jul 16, 2008 12:32 am
you cant assume a value for "a" simply becoz its a function of b^2*c ..hence the value of 'a' will always depends on what b and c are...

hope thast helps
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by desiguy » Wed Jul 16, 2008 12:34 am
sorryyyyy

replace a & b in my previous answer to b&c. Ie i assume b=2 and c=3. hence the answer of 12.
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by sudhir3127 » Wed Jul 16, 2008 12:36 am
on a second thought even in ur example.. u shud take (4^2)* 3*3= 144 and not 72..

so 12*12 = 144 I think u missed whole square and nultiplied by 2 instead.

even then the answer is 12
Last edited by sudhir3127 on Wed Jul 16, 2008 12:37 am, edited 1 time in total.
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desiguy wrote:In a formulae a=b(square)c, if b is multiplied by 2 and c is multiplied by 3, what whoudl a be multiplied by ?

4
6
9
12
18

Ans : 12

I know it’s a very very dummbo questin. But my answer is 6. Not sure how its 12 ?
the easiest way to answer this question is to plug in your own numbers and watch what happens. the setup of the answer choices (i.e., all different positive factors) guarantees that any choice of numbers will uniquely yield the correct answer, unless you're unfortunate enough to have selected 0 as one of the initial numbers.

the easiest choice is: b(initial) = c(initial) = 1. then a(initial) = (1^2)(1) = 1.
then you double b and triple c, which means that b(new) = 2 and c(new) = 3. therefore, a(new) = (2^2)(3) = 12.
a has gone from 1 to 12, so it's been multiplied by a factor of 12.
you should find that the same happens regardless of your initial choice of numbers. if it doesn't, then you have miscalculated something.

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if you want to prove this formally, do this:
at the start, the formula is (b^2)(c).
after doubling b and tripling c, you have ((2b)^2)(3c) = (4b^2)(3c) = 12(b^2)(c).
the original quantity has been multiplied by 12.

easier to pick numbers, though - no need to deal with messy things like distributing of exponents
Ron has been teaching various standardized tests for 20 years.

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by desiguy » Wed Jul 16, 2008 12:48 am
Thank Ron and Sudhir...

Got it!! thanks. Instead of doubling the b and then squaring it, I doubled the squared b !!

S.
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