How to subtract - SSAT Upper Level Quantitative

Card 1 of 48

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

4593-1296=

Tap to reveal answer

Answer

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.

1296+3297=4593

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

9div 2=4 remainder of 1

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

Now subtract:

4.8-4.5=0.3

← Didn't Know|Knew It →

Question

Subtract $\left ( 2a-3b+7c\right )$ from $\left ( 5a-4b-3c\right )$

Tap to reveal answer

Answer

You set up the expression with $\left ( 5a-4b-3c\right )$ on the top and $\left ( 2a-3b+7c\right )$ on the bottom. Because this is subtraction, remember to distribute the negative sign to the expression on the bottom, and then add, so

you get .

← Didn't Know|Knew It →

Question

and are prime integers. and .

How many possible values of are there?

Tap to reveal answer

Answer

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).

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

Define an operation $\Upsilon$ as follows:

For all real numbers ,

$a \Upsilon b = $a^{5}$ - $b^{4}$$ .

Evaluate: $\frac{1}{2}$; \Upsilon; $\frac{1}{2}$ .

Tap to reveal answer

Answer

$a \Upsilon b = $a^{5}$ - $b^{4}$$

$\frac{1}{2}$; \Upsilon; $\frac{1}{2}$ =\left ( $\frac{1}{2}$ \right $)^{5}$ - \left ( \frac {1}{2} \right $)^{4}$

$= $\frac{1}{32}$ - $\frac{1}{16}$$

$= $\frac{1}{32}$ - $\frac{2}{32}$$

$= -$\frac{1}{32}$$

← Didn't Know|Knew It →

Question

Define a function as follows:

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

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

Tap to reveal answer

Answer

$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$

← Didn't Know|Knew It →

Question

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 )$ .

Tap to reveal answer

Answer

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

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

← Didn't Know|Knew It →

Question

Define a function as follows:

$g(x)=\left | $x^{5}$$-x^{2}\right |$

Evaluate .

Tap to reveal answer

Answer

$g(x)=\left | $x^{5}$$-x^{2}\right |$

$g(2)=\left | $2^{5}$$-2^{2}\right | = \left | 32-4\right | = \left | 28\right | = 28$

$g(-2)=\left |( $-2)^{5}$-\left (-2 \right $)^{2}\right | = \left | -32-4\right | = \left | -36\right | = 36$

← Didn't Know|Knew It →

Question

Define an operation $\bigtriangleup$ on the real numbers as follows:

For all real numbers :

$a \bigtriangleup b = \sqrt[3]{a} - \sqrt[4]{b}$ .

Evaluate $(-1) \bigtriangleup (-1)$ .

Tap to reveal answer

Answer

$a \bigtriangleup b = \sqrt[3]{a} - \sqrt[4]{b}$

$(-1) \bigtriangleup (-1) = \sqrt[3]{-1} - \sqrt[4]{-1}$

However, $\sqrt[4]{-1}$ is undefined in the real numbers; subsequently, so is $(-1) \bigtriangleup (-1)$ .

← Didn't Know|Knew It →

Question

Define a function as follows:

$f(x) = 10 - \sqrt[3]{x} - \sqrt[5]{x}$

Evaluate .

Tap to reveal answer

Answer

$f(x) = 10 - \sqrt[3]{x} - \sqrt[5]{x}$

$f(-1) = 10 - \sqrt[3]{-1} - \sqrt[5]{-1} = 10 - (-1)- (-1)= 10+1+1 = 12$

← Didn't Know|Knew It →

Question

Define a function on the real numbers as follows:

$f(x) = 100 -\left ( $\sqrt{x}$ - \sqrt[4]{x} \right )$

Evaluate .

Tap to reveal answer

Answer

$f(x) = 100 -\left ( $\sqrt{x}$ - \sqrt[4]{x} \right )$

$f(-1) = 100 -\left ( $\sqrt{-1}$ - \sqrt[4]{-1} \right )$

Since even-numbered roots of negative numbers are not defined for real-valued functions, the expression is undefined.

← Didn't Know|Knew It →

Question

4593-1296=

Tap to reveal answer

Answer

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.

1296+3297=4593

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

9div 2=4 remainder of 1

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

Now subtract:

4.8-4.5=0.3

← Didn't Know|Knew It →

Question

Subtract $\left ( 2a-3b+7c\right )$ from $\left ( 5a-4b-3c\right )$

Tap to reveal answer

Answer

You set up the expression with $\left ( 5a-4b-3c\right )$ on the top and $\left ( 2a-3b+7c\right )$ on the bottom. Because this is subtraction, remember to distribute the negative sign to the expression on the bottom, and then add, so

you get .

← Didn't Know|Knew It →

Question

and are prime integers. and .

How many possible values of are there?

Tap to reveal answer

Answer

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).

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

Define an operation $\Upsilon$ as follows:

For all real numbers ,

$a \Upsilon b = $a^{5}$ - $b^{4}$$ .

Evaluate: $\frac{1}{2}$; \Upsilon; $\frac{1}{2}$ .

Tap to reveal answer

Answer

$a \Upsilon b = $a^{5}$ - $b^{4}$$

$\frac{1}{2}$; \Upsilon; $\frac{1}{2}$ =\left ( $\frac{1}{2}$ \right $)^{5}$ - \left ( \frac {1}{2} \right $)^{4}$

$= $\frac{1}{32}$ - $\frac{1}{16}$$

$= $\frac{1}{32}$ - $\frac{2}{32}$$

$= -$\frac{1}{32}$$

← Didn't Know|Knew It →

Question

Define a function as follows:

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

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

Tap to reveal answer

Answer

$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$

← Didn't Know|Knew It →

Question

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 )$ .

Tap to reveal answer

Answer

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

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

← Didn't Know|Knew It →

0

Knew It