SSAT Upper Level Quantitative (Math)

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SSAT Upper Level: Quantitative › SSAT Upper Level Quantitative (Math)

Questions 1 - 10

Define an operation $\forall$ as follows:

For all real numbers :

$a \forall b = a^{4} + \frac{1}{b^{4}}$ .

Evaluate $\frac{1}{2} \forall (-2 )$ .

$\frac{1}{8}$

$16\frac{1}{16}$

$-15\frac{15}{16}$

The correct answer is not among the other responses.

Explanation

$a \forall b = a^{4} + \frac{1}{b^{4}}$

$\frac{1}{2} \forall (-2 ) = \left (\frac{1}{2} \right )^{4} + \frac{1}{(-2)^{4}}= \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$

$\dpi{100} 4593-1296=$

$\dpi{100} 3297$

$\dpi{100} 3397$

$\dpi{100} 3497$

$\dpi{100} 497$

Explanation

Subtract the two numbers, being careful not to forget to remove the borrowed number.

You can also add the solution back to 1296 to check your work, as addition is easier than subtraction.

$\dpi{100} 1296+3297=4593$

$\dpi{100} 4.8-\frac{9}{2}$

$\dpi{100} 0.3$

$\dpi{100} -5.2$

$\dpi{100} \frac{5}{2}$

$\dpi{100} 2$

Explanation

First convert $\dpi{100} \frac{9}{2}$ into a decimal.

$\dpi{100} 9\div 2=4\ remainder\ of\ 1$

So we are left with $\dpi{100} 4\frac{1}{2}$ , which is 4.5 in decimal form.

Now subtract:

$\dpi{100} 4.8-4.5=0.3$

and are prime integers. and .

How many possible values of are there?

Five

Six

Four

Seven

Eight

Explanation

The prime integers between 65 and 75 are 67, 71, and 73, so assumes one of those values; the prime integers between 45 and 55 are 47 and 53, so assumes one of those values. Therefore, one of the following holds true:

There are five possible values for (20 appears twice here).

and are prime integers. and . What is the greatest possible value of ?

Explanation

The greatest possible value of is the least possible value of subtracted from the greatest possible value of . The least prime between 55 and 65 is 59, and the greatest prime between 85 and 95 is 89, so and give the greatest possible value of , which is equal to

Define a function as follows:

$g(x)=\left | x^{2} - \frac{1}{x^{2}}\right |$

Evaluate $g(2) - g \left ( \frac{1}{2} \right )$ .

$\frac{1}{2}$

$-\frac{1}{2}$

$7\frac{1}{2}$

$-7\frac{1}{2}$

Explanation

$g(x)=\left | x^{2} - \frac{1}{x^{2}}\right |$

$g(2)=\left | 2^{2} - \frac{1}{2^{2}}\right | = \left | 4- \frac{1}{4}\right | = \left | 3 \frac{3}{4}\right |= 3 \frac{3}{4}$

$g\left ( \frac{1}{2} \right )=\left | \left ( \frac{1}{2} \right )^{2} - \frac{1}{\left ( \frac{1}{2} \right )^{2}}\right | = \left | \frac{1}{4}- \frac{1}{ \frac{1}{4}} \right | = \left | 3 \frac{3}{4}\right | = \left | \frac{1}{4}- 4 \right | = \left | - 3 \frac{3}{4} \right | = 3 \frac{3}{4}$

$g(2) - g \left ( \frac{1}{2} \right ) = 3 \frac{3}{4} - 3 \frac{3}{4} = 0$

Function 2

What equation is graphed in the above figure?

$y = \left \lfloor x\right \rfloor - 4$

$y = \left \lceil x\right \rceil - 4$

$y = \left \lfloor x\right \rfloor - 3$

$y = \left \lceil x\right \rceil - 3$

$y = \left \lfloor x\right \rfloor +3$

Explanation

The greatest integer function, or floor function, $y = \left \lfloor x\right \rfloor$ , pairs each value of with the greatest integer less than or equal to . Its graph is below.

Floor function

The given graph is the above graph shifted downward four units. The graph of any function shifted downward four units is , so the given graph corresponds to equation $y = \left \lfloor x\right \rfloor - 4$ .