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

\[\begin{align} \frac{45 \ \mbox{elapsed minutes}}{60 \ \mbox{minutes total}} =\frac{45}{60} =\frac{3 \cdot 15}{4 \cdot 15} =\frac{3}{4}. \end{align}\]

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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