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For which value of x will y=ax^2+20x+b have a minimum in the

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by Max@Math Revolution » Thu Aug 08, 2019 11:52 pm

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[GMAT math practice question]

For which value of x will y=ax^2+20x+b have a minimum in the xy-plane?

1) b=10
2) a=2
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Source: — Data Sufficiency |
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by Max@Math Revolution » Sun Aug 11, 2019 11:46 pm
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the numbers of variables and equations.

We can modify the original condition and question as follows:

If a > 0, the function will have a minimum at x = (-20)/(2a) = (-10)/a.
If a < 0, the function has no minimum. So, to answer the question, we need to find the value of a.
Thus, condition 2) is sufficient.

Note: condition 1) cannot be sufficient as it provides no information about the value of a.

Therefore, the answer is B.
Answer: B
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by deloitte247 » Thu Aug 15, 2019 1:39 pm
$$Question:\ For\ which\ value\ of\ x\ will\ y=ax^2+20x+b\ have\ a\ \min imum\ in\ the\ xy-plane?$$
The minimum value or maximum value of a quadratic function in the form
$$ax^2+bx+c\ occurs\ at\ x=-\frac{b}{2a}$$
If a>0, then minimum value will occur and parabola will open upwards and if a<0, then maximum value will occur and parabola will open downwards.
$$For\ y=ax^2+20x+b$$
$$Minimum\ value\ will\ be\ at\ x=-\frac{20}{2a}where\ a>0$$
$$Statement\ 1:\ b=10$$
The value of a was not provided, so, we cannot find the value of x in x=-20/2a. Hence, statement 1 is NOT SUFFICIENT.
$$Statement\ 2:\ a=2$$
$$Value\ of\ a\ is\ >0,\ so\ for\ x=\frac{20}{2a}$$
$$x=\frac{20}{2a}=\frac{20}{2\cdot2}=\frac{20}{4}=5$$
$$Statement\ 2\ alone\ is\ SUFFICIENT$$

Option B is the correct answer.
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