Executive Assessment Quantitative Practice Questions

The correct answer is: C)

To satisfy the conditions of the problem, \(x\) must be a factor of 72 less than 10, which are 1, 2, 3, 4, 6, 8, and 9.
$$\begin{align} 1: \frac{72}{1}-1 &= 71\\
2: \frac{72}{2}-2 &= 34\\
3: \frac{72}{3}-3 &= 21\\
4: \frac{72}{4}-4 &= 14\\
6: \frac{72}{6}-6 &= 6\\
8: \frac{72}{8}-8 &= 1\\
9: \frac{72}{9}-9 &= -1\\
\end{align}$$
Therefore, 6 integer values less than 10 for \(x\) produce positive integer values for \(y\).

The correct answer is: E)

The first seven positive multiples of 3 are 3, 6, 9, 12, 15, 18, and 21.

The range is the difference between the largest and smallest numbers in a set. In this case, the largest number is 21 and the smallest number is 3, so the range is:

$$\text{Range}=21-3=18$$

To find the average of these numbers, we can add them up and divide by the number of numbers:

$$\text{Average}=\frac{3+6+9+12+15+18+21}{7}=12$$

Therefore,

$$\begin{align*}\text{A} + \text{a} &= \text{range} + \text{average} \\&= 18 + 12\\& = 30\end{align*}$$

Hence, the value of A + a is 30.

The correct answer is: C)

If X is the total amount of games the team played, then \(.75(112) + .25(X-112) = .6X\).

$$ \begin{align*}
84 + .25X – 28 & = .6X \\
56 & = .35X \\
X &= 160.
\end{align*} $$

The correct answer is: A)

Note that evaluating the discriminant of \(b^2 – 4ac\) against 0 will be sufficient to determine the number of solutions.

If the discriminant is greater than 0, then there are two solutions for the equation. If it is equal to 0, there is a single solution, and if it is less than 0, there are no real solutions for the equation.

The discriminant of this equation is:
$$\begin{align*} (-b)^2-4(\dfrac{1}{2})(b^2) &=b^2-2b^2\\
&=-b^2 < 0\\ \end{align*}$$

Since the discriminant is negative, the equation has no real solutions.

The correct answer is: C)

It is useful to note that the condition [y] = 4 is equivalent to \(4 < y \leq 5\)

(1) A value of y = 2.5 satisfies this statement, and yields [y] = 2.
A value of y = 4.5 also satisfies this statement, and yields [y] = 4; NOT sufficient.

(2) This statement can be rewritten as y > 4, which allows both by y = 4.5 (yielding [y] = 4) and y = 10.5 (yielding [y] = 10); NOT sufficient.
(Together) The two inequalities yield \(4 < y \leq 5\), which yields [y] = 4 for any possible value of y; SUFFICIENT.

The correct answer is C; the statements together, but not alone, are sufficient.

The correct answer is: D)

Find the area of the blue rug by multiplying its length and width: \( 24 × 8 = 192\).

Now, use the area formula to find the length of the red rug as \(192 = 6 × length\).

192 divided by 6 is 32, and 32 is 8 feet longer than 24.

The correct answer is: E)

When every dimension is multiplied by 2, the volume increases by a factor of 2 × 2 × 2 = 8, regardless of the exact dimensions.

Therefore, 20 × 8 = 160 square feet.

The correct answer is: B)

Point P will be equidistant from (2, 4) and (6, 4) if and only if P lies on the perpendicular bisector of the line segment with endpoints (2, 4) and (6, 4).

Therefore, point P will be equidistant from (2,4) and (6,4) if and only if it lies on the vertical line \(x = \frac {(6+2)}{2} = 4\). Therefore, determine if the x-coordinate of P is 4.

(1) Given that y = 5, then both x = 4 and x \(\neq\) 4 are possible; NOT sufficient.

(2) Because x = 4, point P must be equidistant from (2,4) and (6,4); SUFFICIENT.

The correct answer is B; statement 2 alone is sufficient.

The correct answer is: C)

The mass ratio of Concrete X is \(\frac{\text{gravel mixture}}{\text{gravel mixture}+\text{cement}} = \frac{7}{1+7} = \frac{7}{8}\)

So, if x is the number of grams of gravel mixture in 112 grams of concrete X, it follows that \(\frac{x}{112} = \frac{7}{8}\)

Solve for x. \(x = \frac{7}{8} * \frac{112}{1} = \frac{7 \times 7 \times 2 \times 2 \times 2 \times 2}{2 \times 2 \times 2} = 7 \times 7 \times 2 = 98\) grams of gravel in 112 grams of Concrete X.