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Simple Math problem

by AndyB » Sun Oct 09, 2011 3:38 am
Hello Everyone,

Anybody!!! Please help me in solving the below problem.

5) The relationship between x and y is given as:

x 0 1 2 3 4 5 6
y 0 10 40 90 160 250 360


If x=2.5, what will be the value of y??

a)45
b)62.5
c)65
d)67.5

I am not able to understand how to derive a relationship between x and y...and then use x=2.5 and then get the result of y.
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by Anurag@Gurome » Sun Oct 09, 2011 3:49 am
AndyB wrote:The relationship between x and y is given as:

x 0 1 2 3 4 5 6
y 0 10 40 90 160 250 360


If x=2.5, what will be the value of y?
y is nothing but 10 multiplied with the square of x.

Hence, for x = 2.5, y = 10*(2.5)^2 = 62.5

The correct answer is B.
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by parul9 » Sun Oct 09, 2011 4:45 am
B!
the relation is y = 10x^2
And then substitution of x = 2.5 gives y = 62.5
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by Abhishek009 » Sun Oct 09, 2011 5:59 am
AndyB wrote:Hello Everyone,

Anybody!!! Please help me in solving the below problem.

5) The relationship between x and y is given as:

x 0 1 2 3 4 5 6
y 0 10 40 90 160 250 360


If x=2.5, what will be the value of y??

a)45
b)62.5
c)65
d)67.5

I am not able to understand how to derive a relationship between x and y...and then use x=2.5 and then get the result of y.

0 = 0 * 10

10 = 1^2 * 10

40 = 2^2 * 10

90 = 3^2 * 10

160 = 4^2 * 10

250 = 5^2 * 10

360 = 6^2 * 10



So actually y = ( x^2 ) *10


hence when x = 2.5


Y = 2.5^2 * 10

y =>6.25
Abhishek
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by GmatMathPro » Sun Oct 09, 2011 7:02 am
AndyB wrote: 5) The relationship between x and y is given as:

x 0 1 2 3 4 5 6
y 0 10 40 90 160 250 360


If x=2.5, what will be the value of y?
The best thing here is just to be so familiar with the sequence of the first few perfect squares, that the relationship just jumps out at you. But what if it doesn't?

If you absolutely cannot see what the formula should be just by looking at it, you can use the following:

Write down the differences between successive y-values, and then write down the differences between those first differences:

y= 0 10 40 90 160 250 360
differences= 10,30,50,70,90,110.
2nd diff= 20,20,20,20,20,20

There is a rule that says if the second differences of a sequence are constant(in this case 20), then the sequence can be modeled by a quadratic equation(that is y=ax^2+bx+c). This would tell you that the formula definitely involves x^2, and hopefully you would be able to see the relationship from there. If you still can't, proceed as follows:

We know the formula is of the form y=ax^2+bx+c and that y=0 when x=0, so plug in these values:
0=a*0^2+b*0+c
0=c

we know c is zero, so it must be that y=ax^2+bx. Now plug in another set of values, x=1, y=10

10=a+b

and another set of values: x=2 y=40

40=4a+2b

Solving this system of equations gives a=10, b=0 so the model is y=10x^2.

But again, this should be a last resort because of the time involved.
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by AndyB » Mon Oct 17, 2011 2:22 am
Thanks everybody guys...For your time..
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by GmatKiss » Mon Oct 17, 2011 3:49 am
IMO: B, wish to get such questions in the real CAT :D
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