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Bunuel
If the function f is defined for all x by f(x) = ax^2 + bx − 43, where a and b are constants, what is the value of f(3)?

(1) f(4) = 41
(2) 3a + b = 17

It is one of those trick questions in which if you are not careful, you could fall for the trap and never know it. The impressions that "two unknowns need two equations to solve" and "a quadratic gives two solutions" are often exploited by the test makers. So be very wary about them.

Statement 1 is straight forward here:
f(x) = ax^2 + bx - 43
f(4) = 41 = 16a + 4b - 43
4a + b = 21
a and b could take different values so you cannot find f(3)

Statement 2 gives 3a + b = 17
You need to find f(3) = 9a + 3b - 43
Note that 9a + 3b = 3(3a + b) so you actually have the value of the desired expression.
So though statement 2 gives you an equation, you have enough information to solve for f(3).

It is good to write down what you need in such questions.
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1) f(4) =41 f(3) =?

16a+4b-43=41
4a+b=21 .. NS

2) 3a+b =17

f(3) =9a+3b-43
= 3(3a+b)-43
=3*17-43
= 8 ... S

B
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I have a little doubt here
statement 1 gives us an equation 16a+4b-43=41 nd from the question stem we have the equation 9a+3b-43=0
why cant be solve the two equation with two unknown to get the resultant value?
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AmritaSarkar89
I have a little doubt here
statement 1 gives us an equation 16a+4b-43=41 nd from the question stem we have the equation 9a+3b-43=0
why cant be solve the two equation with two unknown to get the resultant value?

How did you get the red part? We don't know the value of f(3) = 9a + 3b - 43, that's what we are asked to find out.
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At first this Q looks like having 2 variables which necessitates 2 equations for finding a solution. This might make us believe that C could be the answer.

St 1 :
f(x) = ax^2 + bx - 43
f(4) = 41 = 16a + 4b - 43
4a + b - 21 = 0 ..... a and b can have any values.
NSUF

St 2: 3a + b = 17
f(3) = 9a + 3b - 43
= 3(3a + b) - 43
= 3( 17 ) - 43
SUF

Hence B
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Bunuel
AmritaSarkar89
I have a little doubt here
statement 1 gives us an equation 16a+4b-43=41 nd from the question stem we have the equation 9a+3b-43=0
why cant be solve the two equation with two unknown to get the resultant value?

How did you get the red part? We don't know the value of f(3) = 9a + 3b - 43, that's what we are asked to find out.


if we put the value of x=3 then don't we get the given equation. Maybe I am totally wrong here, can you please explain?
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AmritaSarkar89
Bunuel
AmritaSarkar89
I have a little doubt here
statement 1 gives us an equation 16a+4b-43=41 nd from the question stem we have the equation 9a+3b-43=0
why cant be solve the two equation with two unknown to get the resultant value?

How did you get the red part? We don't know the value of f(3) = 9a + 3b - 43, that's what we are asked to find out.


if we put the value of x=3 then don't we get the given equation. Maybe I am totally wrong here, can you please explain?

The question asks to find the value of f(3), which is 9a + 3b - 43. You are equating it to 0, which is wrong.
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Bunuel
If the function f is defined for all x by f(x) = ax^2 + bx − 43, where a and b are constants, what is the value of f(3)?

(1) f(4) = 41
(2) 3a + b = 17

Given
=> f(x)=ax^2+bx-43
=> f(x)=x(ax+b)-43 ----(1)

Therefore
=> f(3)=3(3a+b)-43

So if we can find the value of a & b or (3a+b) we can find f(3)

Statement 1 says f(4)=41
Now f(4)=4(4a+b)-43 from (1)
=> 4(4a+b)-43=41
=> 4(4a+b)=84
=> 4a+b=21
We cannot find the value of a & b or (3a+b)
Therefore NOT sufficient

Statement 2 says (3a+b)=17 =>SUFFICIENT

Therefore 'B"

Thanks
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Bunuel
If the function f is defined for all x by f(x) = ax^2 + bx − 43, where a and b are constants, what is the value of f(3)?

(1) f(4) = 41
(2) 3a + b = 17

f(3) = \(a3^2 + 3b - 43\)
= \(9a + 3b - 43\)
= \(3 (3a + b) - 43\)

Looking at each statement, we see that only Statement (2) will be able to satisfy the equation. Since we have the Value of 3a + b given.

Hence, Negate A, D, C, E

Answer B
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