Basic Addition, Subtraction, Multiplication and Division - SSAT Upper Level Quantitative

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Question

$\dpi{100} .24+2.5+\frac{1}{4}=$

Answer

First, convert $\dpi{100} \frac{1}{4}$ into a decimal. We are left with

$\dpi{100} .24+2.5+.25$

Now add the three numbers, being careful to line the decimal points up exactly. From right to left we will add:

4 and 5 = 9

5, 2, and 2 = 9

2 is alone = 2

This leaves us with 2.99 for the final answer.

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Question

$Solve: 7+(3-9)^{2}+36\div 4$

Answer

Use order of operations, so you would first complete the operation in the parentheses, which would give you .

Then you do exponenets, so $\left ( -6\right)^{2} = 36$ .

Then you do division which is $36\div4 = 9$ .

Then you do addition from left to right, so .

Last, you add to , which is .

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Question

What is the sum of the two percentages in decimal form?

7 percent and 135 percent

Answer

You add the percentages and you get 142 percent. To convert a percent to a decimal, you divide by 100, so the answer is 1.42.

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Question

Which of the following angles is supplementary to an angle that is 45 degrees?

Answer

Supplementary angles add up to 180 degrees. Given that , the correct answer is 135.

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Question

$3+8\div4-(2\times 5)^{^{2}}$

Answer

Do the expression in the parantheses and you get . Then you do $10^{2}$ and you get . Then you do $8\div4$ and you get . Then you add .

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Question

If Susie bought a pair of shoes for , a shirt for , and a pair of pants for . How much did she spend in total?

Answer

You add all three values and you get .

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Question

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

Answer

To find the greatest possible value of , add the greatest possible value of to the greatest possible value of . This happens when is equal to the greatest prime between 35 and 45, which is 43, and when is equal to the greatest prime between 85 and 95, which is 89. Add the two numbers:

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Question

and are prime integers: and . How many possible values of are there?

Answer

can be equal to any of the prime numbers between 45 and 55, of which there are two - 47 or 53.

can be equal to any of the prime numbers between 15 and 25, of which there are three - 17, 19, and 23.

Their sum can be any of the following:

This makes five possible sums.

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Question

In a math contest, the team from Lincoln High comprised five students - Melanie, Ned, Olivia, Phyllis, and Quinn. Melanie outscored Ned by 12 points; Ned outscored Quinn by 19 points; Phyllis outscored Ned by 45 points and Olivia by 20 points. Who came in third among the five?

Answer

Since this reasoning is valid regardless of the actual scores, suppose Quinn scored 100. Then Ned scored 19 more than Quinn, or 119, and Melanie scored 12 more than Ned, or 131. Phyllis scored 45 more than Ned, or 164, and Olivia scored 20 points fewer than Phyllis, or 144. This makes the order, from greatest to least:

Phyllis, Olivia, Melanie, Ned, Quinn

Melanie scored third among the five.

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Question

Simplify:

Answer

To simplify this expression, regroup and combine like-terms.

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Question

Divide:

$18\textrm{ lbs }2\textrm{ oz } \div 5$

Answer

We can write 18 pounds, 2 ounces as just ounces as follows:

$18\textrm{ lbs }2\textrm{ oz } = (18 \times 16 + 2) \textrm{ oz } = 290 \textrm{ oz }$

Divide:

$290 \div 5 = 58$

Convert 58 ounces to pounds and ounces as follows:

$58 \div 16 = 3 \textrm{ R }10$

That is, 3 pounds 10 ounces.

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Question

Simplify by rationalizing the denominator:

$\frac{16}{4-\sqrt{6}}$

Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $4+\sqrt{6}$ :

$\frac{16}{4-\sqrt{6}}= \frac{16(4+\sqrt{6})}{(4-\sqrt{6}) (4+\sqrt{6})}$

$= \frac{16(4+\sqrt{6})}{4^{2}-(\sqrt{6})^{2} }$

$= \frac{16(4+\sqrt{6})}{16-6}$

$= \frac{16(4+\sqrt{6})}{10}$

$= \frac{8(4+\sqrt{6})}{5}$

$= \frac{ 32+8\sqrt{6}}{5}$

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Question

Simplify by rationalizing the denominator:

$\frac{\sqrt{6}}{6-\sqrt{3}}$

Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $6+\sqrt{3}$ :

$\frac{\sqrt{6}}{6-\sqrt{3}}$

$= \frac{\sqrt{6}(6+\sqrt{3})}{(6-\sqrt{3})(6+\sqrt{3})}$

$= \frac{6\cdot \sqrt{6}+\sqrt{3} \cdot \sqrt{6}}{6^{2}-(\sqrt{3})^{2}}$

$= \frac{6 \sqrt{6}+ \sqrt{18}}{36-3 }$

$= \frac{6 \sqrt{6}+ \sqrt{9} \cdot \sqrt{2}}{3 3 }$

$= \frac{6 \sqrt{6}+3\sqrt{2}}{3 3 }$

$= \frac{2 \sqrt{6}+\sqrt{2}}{11 }$

$= \frac{\sqrt{2}+ 2 \sqrt{6}}{11 }$

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Question

Simplify by rationalizing the denominator:

$\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{6}}$

Answer

Multiply both numerator and denominator by $\sqrt{6}$ :

$\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{6}}$

$= \frac{(3\sqrt{2}+2\sqrt{3})\cdot \sqrt{6}}{\sqrt{6}\cdot \sqrt{6}}$

$= \frac{3\sqrt{2}\cdot \sqrt{6}+2\sqrt{3}\cdot \sqrt{6}}{6}$

$= \frac{3\sqrt{12} +2\sqrt{18} }{6}$

$= \frac{3\cdot \sqrt{4} \cdot \sqrt{3}+2\cdot \sqrt{9}\cdot \sqrt{2} }{6}$

$= \frac{3\cdot 2 \cdot \sqrt{3}+2\cdot3 \cdot \sqrt{2} }{6}$

$= \frac{6 \cdot \sqrt{3}+6 \cdot \sqrt{2} }{6}$

$= \sqrt{3}+ \sqrt{2}$

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Question

Simplify by rationalizing the denominator:

$\frac{12}{\sqrt{14}-\sqrt{5}}$

Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{14}+\sqrt{5}$ :

$\frac{12}{\sqrt{14}-\sqrt{5}}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{(\sqrt{14}-\sqrt{5})(\sqrt{14}+\sqrt{5})}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{(\sqrt{14})^{2}-(\sqrt{5})^{2} }$

$= \frac{12(\sqrt{14}+\sqrt{5})}{14-5}$

$= \frac{12(\sqrt{14}+\sqrt{5})}{9}$

$= \frac{4(\sqrt{14}+\sqrt{5})}{3}$

$= \frac{4 \sqrt{14}+4\sqrt{5}}{3}$

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Question

Simplify by rationalizing the denominator:

$\frac{6}{\sqrt{5}+\sqrt{7}}$

Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}-\sqrt{7}$ :

$\frac{6}{\sqrt{5}+\sqrt{7}}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{(\sqrt{5}+\sqrt{7})(\sqrt{5}-\sqrt{7})}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{(\sqrt{5})^{2}-(\sqrt{7})^{2}}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{(\sqrt{5})^{2}-(\sqrt{7})^{2}}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{5-7}$

$=\frac{6(\sqrt{5}-\sqrt{7})}{-2}$

$= -3(\sqrt{5}-\sqrt{7} )$

$= 3\sqrt{7}-3\sqrt{5}$

This is not among the given choices.

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Question

Simplify by rationalizing the denominator:

$\frac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}$

Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{5}+\sqrt{3}$

$\frac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}$

$=\frac{\sqrt{15}( \sqrt{5}+\sqrt{3})}{\left (\sqrt{5}-\sqrt{3} \right )( \sqrt{5}+\sqrt{3})}$

$=\frac{ \sqrt{15} \cdot \sqrt{5}+ \sqrt{15} \cdot \sqrt{3}}{ (\sqrt{5})^{2}-(\sqrt{3} )^{2} }$

$=\frac{ \sqrt{15 \cdot 5}+ \sqrt{15 \cdot 3}}{5-3 }$

$=\frac{ \sqrt{75}+ \sqrt{45}}{2 }$

$=\frac{ \sqrt{25} \cdot \sqrt {3}+ \sqrt{9} \cdot \sqrt{5}}{2 }$

$=\frac{ 5\sqrt {3}+3 \sqrt{5}}{2 }$

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Question

Divide: $\frac{\sqrt{64}}{\sqrt{8}}$

Answer

The values are of the same power. Divide the terms.

$\frac{\sqrt{64}}{\sqrt{8}}= \sqrt{\frac{64}{8}}= \sqrt{8} = \sqrt{4}\times \sqrt{2}= 2\sqrt2$

The correct answer is: $2\sqrt2$

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Question

$\dpi{100} \frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{9}}=$

Answer

First multiply the fraction in the numerator.

$\dpi{100} \frac{1}{2}\times \frac{1}{3}=\frac{1\times 1}{2\times 3}=\frac{1}{6}$

Now we have $\dpi{100} \frac{\frac{1}{6}}{\frac{1}{9}}$

Never divide fractions. We multiply the numerator by the reciprocal of the denominator.

$\dpi{100} \frac{\frac{1}{6}}{\frac{1}{9}}=\frac{1}{6}\div \frac{1}{9}=\frac{1}{6}\times \frac{9}{1}=\frac{1\times 9}{6\times 1}=\frac{9}{6}=\frac{3}{2}$

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Question

Which of these expressions is the greatest?

Answer

The easiest way to see that all four are equivalent is to convert each to a decimal product, noting that twenty percent and one-fifth are equal to , and twenty-five percent and one fourth are equal to .

One fourth of 0.2: $\frac{1}{4}0.2=\frac{1}{4}\frac{2}{10}=\frac{2}{40}=\frac{1}{20}$

One fifth of 0.25: $\frac{1}{5}0.25=\frac{1}{5}\frac{25}{100}=\frac{25}{500}=\frac{5}{100}=\frac{1}{20}$

Twenty-five percent of one fifth: $\frac{1}{5}=\frac{25}{100}\frac{1}{5}=\frac{25}{500}=\frac{1}{20}$

Twenty percent of one fourth: $\frac{1}{4}=\frac{20}{100}\frac{1}{4}=\frac{20}{400}=\frac{1}{20}$

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