SAT Fractions and Decimals

To solve problems involving fractions and decimals on the SAT, you must know how to

Examples

A chemistry lab lasts one hour. What fraction of the lab is completed $45$ minutes after it begins?

(A) $\ \ \frac{1}{45}$

(B) $\ \ \frac{1}{4}$

(C) $\ \ \frac{3}{4}$

(D) $\ \ 1$

(E) $\ \ \frac{4}{3}$

Correct Answer: C

Tip: Pay attention to units.
One hour has $60$ minutes. The fraction of the lab completed after $45$ minutes is

$\begin{aligned} \frac{45 \ \mbox{elapsed minutes}}{60 \ \mbox{minutes total}} =\frac{45}{60} =\frac{3 \cdot 15}{4 \cdot 15} =\frac{3}{4}. \end{aligned}$

Incorrect Choices:

(A)
Tip: Pay attention to units.
If you forget to convert one hour to $60$ minutes, you will get this wrong answer.
Let's say that $\frac{1}{45}$ of the lab is completed. Then $\frac{1}{45} \times 60 \ \mbox{minutes} = 1.\overline{3} \ \mbox{minutes}$ have passed. But the problem states that $45$ minutes have passed. This is a contradiction and therefore choice (A) is wrong.

(B)
Tip: Pay attention to units.
Tip: Read the entire question carefully.
If you answer the question "What fraction of the lab has yet to elapse after $45$ minutes?", you will get this wrong answer.

One hour has $60$ minutes. If $\frac{1}{4}$ of the lab is completed, then $\frac{1}{4} \times 60 \ \mbox{minutes} = 15 \ \mbox{minutes}$ have passed. But the problem states that $45$ minutes have passed. This is a contradiction and therefore choice (B) is wrong.

(D)
Tip: Eliminate obviously wrong answers.
When dealing with fractions, $1$ represents a whole unit. In this problem, the whole unit is the duration of the chemistry lab: one hour. According to this answer choice then, $60$ minutes have elapsed and the lab is over. But we are told that $45$ minutes have passed. This is a contradiction and therefore choice (D) is wrong.

(E)
Tip: Eliminate obviously wrong answers.
$\frac{4}{3} > 1,$ which means that if $\frac{4}{3}$ of the lab was completed, then the lab lasted longer than one hour.
One hour has $60$ minutes. So $\frac{4}{3} \times 60 \ \mbox{minutes} = 80 \ \mbox{minutes}$ have passed. But the problem states that $45$ minutes have passed. This is a contradiction and therefore choice (E) is wrong.

Which of the following lists the fractions $\frac{1}{7}, \frac{2}{5}, \frac{4}{20}, 1\frac{7}{9}$ , and $\frac{1}{6}$ in order from greatest to least?

(A) $\ \ \frac{1}{7} > \frac{1}{6} > \frac{4}{20} > \frac{2}{5} > 1\frac{7}{9}$

(B) $\ \ \frac{1}{6} > \frac{1}{7} > \frac{2}{5} > \frac{4}{20} > 1\frac{7}{9}$

(C) $\ \ 1\frac{7}{9} > \frac{2}{5} > \frac{4}{20} > \frac{1}{6} > \frac{1}{7}$

(D) $\ \ 1\frac{7}{9} > \frac{2}{5} > \frac{4}{20} > \frac{1}{7} > \frac{1}{6}$

(E) $\ \ \frac{4}{20} > 1\frac{7}{9} > \frac{2}{5} > \frac{1}{6} > \frac{1}{7}$

Correct Answer: C

Solution 1:

Tip: Eliminate obviously wrong answers.
(A) There are two obvious reasons to eliminate this choice:

If $a>b>0$ , then $\frac{1}{a}<\frac{1}{b}.$
$7>6>0 \implies \frac{1}{7}<\frac{1}{6}$ . But in this list, $\frac{1}{7} > \frac{1}{6}$ , which is false.

If $0<a < b$ , then $0 < \frac{a}{b} < 1.$
$\frac{1}{7}, \frac{2}{5}, \frac{4}{20}$ , and $\frac{1}{6}$ are all smaller than $1$ .
$1\frac{7}{9}=1+\frac{7}{9}>1$ . Therefore, of the given fractions it is the greatest and should appear first in the ordered list, not last.

(B) Of the given fractions $1\frac{7}{9}$ is the greatest and should appear first in the ordered list, not last. Eliminate this choice.

(D) In this list, $\frac{1}{7} > \frac{1}{6}$ , which is false. Eliminate this choice.

(E) Of the given fractions $1\frac{7}{9}$ is the greatest and should appear first in the ordered list, not second. Eliminate this choice.

The only choice we did not eliminate is (C).

Solution 2:

Tip: Use a calculator.
An easy way to solve this problem is using a calculator. We convert each fraction into a decimal:

$\frac{1}{7} = 0.142857$

$\frac{2}{5} = 0.4$

$\frac{4}{20} = 0.2$

$1\frac{7}{9}\ = 1+\frac{7}{9}=1 +0.\overline{7}=1.\overline{7}$

$\frac{1}{6}\ = 0.1\overline{6}$ .

Then we order them from greatest to least: $1.\overline{7} > 0.4 > 0.2 > 0.1\overline{6} > 0.142857$ , and therefore $1\frac{7}{9}>\frac{2}{5}>\frac{4}{20}>\frac{1}{6}>\frac{1}{7}$ .

Solution 3:

This lengthy solution involves finding the common denominator.
The lowest common multiple of $7, 5, 20, 9$ , and $6$ is $1260$ . So each fraction becomes as follows:

$\frac{1}{7} = \frac{1 \times 180}{7 \times 180} = \frac{180}{1260}$

$\frac{2}{5} = \frac{2 \times 252}{5 \times 252} = \frac{504}{1260}$

$\frac{4}{20} = \frac{4 \times 63}{20 \times 63} = \frac{252}{1260}$

$1\frac{7}{9}\ = 1+\frac{7}{9}=\frac{1\times9+7}{9}=\frac{16}{9}=\frac{16 \times 140}{9 \times 140} = \frac{2240}{1260}$

$\frac{1}{6}\ = \frac{1 \times 210}{6 \times 210} = \frac{210}{1260}$ .

Now that all fractions have the same denominator, we can compare their numerators and create an ordered list, from greatest to least:

$\frac{2240}{1260} > \frac{504}{1260} > \frac{252}{1260} > \frac{210}{1260} > \frac{180}{1260}\implies1\frac{7}{9}>\frac{2}{5}>\frac{4}{20}>\frac{1}{6}>\frac{1}{7}.$

If $ABCD$ is a positive four-digit integer with digits $A, B, C,$ and $D$ , what is the decimal equivalent of $ABCD \times 10^{-3}?$

(A) $\ \ 0.ABCD$
(B) $\ \ A.BCD$
(C) $\ \ AB.CD$
(D) $\ \ ABC.D$
(E) $\ \ A,BCD,000$

Correct Answer: B

Solution 1:

Tip: Replace variables with numbers.
Tip: Use a calculator.
Let's choose any positive four-digit number, say $ABCD=1234$ . Use your calculator to find $1234 \times 10^{-3} = 1.234$ . Only answer choice (B) matches this form.

Solution 2:

$ABCD$ has an implied decimal point after its last digit $D$ . Multiplying $ABCD$ by $10^{-3}$ will move that decimal point to the left $3$ digits, as shown:

Solution 3:

Begin by rewriting the integer $ABCD$ in its place-value notation:

$\begin{array}{lcl} ABCD &=& A \times 10^{3} + B \times 10^{2} + C \times 10^{1} + D \times 10^{0}\\ ABCD \times 10^{-3} &=& \big(A \times 10^{3} + B \times 10^{2} + C \times 10^{1} + D \times 10^{0}\big) \times 10^{-3}\\ &=& A \times 10^{3} \times 10^{-3} + B \times 10^{2} \times 10^{-3} + C \times 10^{1} \times 10^{-3} + D \times 10^{0} \times 10^{-3}\\ &=& A \times 10^{0} + B \times 10^{-1} + C \times 10^{-2} + D \times 10^{-3}\\ &=& A + 0.B + 0.0C + 0.00D\\ &=& A.BCD. \end{array}$

Incorrect Choices:

(A)
$ABCD$ has an implied decimal point after its last digit $D$ . Multiplying $ABCD$ by $10^{-4}$ instead of by $10^{-3}$ will move that decimal point to the left $4$ digits, yielding this wrong answer:

(C)
$ABCD$ has an implied decimal point after its last digit $D$ . Multiplying $ABCD$ by $10^{-2}$ instead of by $10^{-3}$ will move that decimal point to the left $2$ digits, yielding this wrong answer:

(D)
$ABCD$ has an implied decimal point after its last digit $D$ . Multiplying $ABCD$ by $10^{-1}$ instead of by $10^{-3}$ will move that decimal point to the left $1$ digit, yielding this wrong answer:

(E)
Tip: Be careful with signs!
$ABCD$ has an implied decimal point after its last digit $D$ . Multiplying $ABCD$ by $10^{3}$ instead of by $10^{-3}$ will move that decimal point to the right $3$ digits, yielding this wrong answer:

Review

If you thought that these examples were difficult and that you needed to review the material, these links will help:

SAT Tips for Fractions and Decimals

The following are some useful tips:

Eliminate obviously wrong answers.
Use a calculator.
Pay attention to units.
Replace variables with numbers.
Review SAT General Tips.

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