Decimal Fractions Questions and Answers

Answer: A

The value is 0.212121.

Line the decimals up by place value:

\[ \begin{aligned} &0.002100 \\ +\ &0.210000 \\ +\ &0.000021 \\ =\ &0.212121 \end{aligned} \]

The other options come from misaligning the digits, for example treating 0.21 as 0.021.

Answer: D

Line the decimal points up and add each column.

(i)

\[ 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025 = 6891.59775 \]

(ii)

\[ 5.064 + 3.98 + 0.7036 + 7.6 + 0.3 + 2 = 19.6476 \]

The first sum has five decimal places because its longest term does, and the second has four for the same reason. Checking the number of decimal places is often enough to eliminate wrong options without adding at all.

Answer: D

The unknown is subtracted, so move it across and move the right-hand side back:

\[ 3889 + 12.952 - x = 3854.002 \]

\[ x = 3889 + 12.952 - 3854.002 \]

\[ 3889.000 + 12.952 = 3901.952 \]

\[ 3901.952 - 3854.002 = 47.950 \]

Written without the trailing zero this is \( 47.95 \). The distractors sit within a few hundredths of it, so the decimal columns must be lined up carefully.

Answer: A

Put everything over a common denominator of 24 and compare:

\[ \frac{1}{3} = \frac{8}{24}, \qquad \frac{7}{8} = \frac{21}{24} \]

So the fraction must lie strictly between \( \tfrac{8}{24} \) and \( \tfrac{21}{24} \).

• \( \tfrac{17}{24} \) - between them ✓
• \( \tfrac{23}{24} \) - above \( \tfrac{21}{24} \)
• \( \tfrac{11}{12} = \tfrac{22}{24} \) - also above
• \( \tfrac{1}{4} = \tfrac{6}{24} \) - below \( \tfrac{8}{24} \)

As decimals the answer is 0.708, comfortably between 0.333 and 0.875.

Answer: C

Write each decimal over the power of ten its last digit sits on, then cancel.

(i) \[ 0.75 = \frac{75}{100} = \frac{3}{4} \]

(ii) \[ 3.004 = \frac{3004}{1000} = \frac{751}{250} \]

(iii) \[ 0.0056 = \frac{56}{10000} = \frac{7}{1250} \]

The denominator is always \( 10^{n} \) for \( n \) decimal places, and the cancelling divides by 25, 4 and 8 respectively.

Answer: D

The numerator is a difference of two squares, so it factorises and the denominator cancels - no squaring is needed at all.

\[ \frac{a^{2}-b^{2}}{a-b} = \frac{(a+b)(a-b)}{a-b} = a+b \]

With \( a = 3.8 \) and \( b = 1.2 \):

\[ \frac{3.8^{2}-1.2^{2}}{3.8-1.2} = 3.8 + 1.2 = 5 \]

Checking the long way: \( 14.44 - 1.44 = 13 \) and \( 13 \div 2.6 = 5 \).

Answer: C

\(\dfrac 23=0.66\\ \dfrac 34=0.75\\ \dfrac 45=0.8\\ \dfrac 56=0.833\)

So biggest is \(\dfrac 56\) and smallest is \(\dfrac 23\)
Their difference is \(\dfrac 56 - \dfrac 23 = \dfrac 16\)

Answer: C

The expression is set up to be the square of a sum. Comparing with

\[ (a+b)^{2} = a^{2} + 2ab + b^{2} \]

the outer terms give \( a = 11.98 \) and \( b = 0.02 \), so the middle term must be

\[ 2ab = 2(11.98)(0.02) = 0.4792 \]

The middle term as written is \( 11.98x \), so

\[ 11.98x = 0.4792 \quad\Longrightarrow\quad x = 2(0.02) = 0.04 \]

The square is then \( (11.98 + 0.02)^{2} = 12^{2} = 144 \). Notice the shortcut: \( x \) is simply twice the smaller number, so the decimals never need multiplying out.

Answer: A

The expression is set up to be the square of a sum. Comparing with

\[ (a+b)^{2} = a^{2} + 2ab + b^{2} \]

the outer terms give \( a = 11.98 \) and \( b = 0.02 \), so the middle term must be

\[ 2ab = 2(11.98)(0.02) = 0.4792 \]

The middle term as written is \( 11.98x \), so

\[ 11.98x = 0.4792 \quad\Longrightarrow\quad x = 2(0.02) = 0.04 \]

The square is then \( (11.98 + 0.02)^{2} = 12^{2} = 144 \). Notice the shortcut: \( x \) is simply twice the smaller number, so the decimals never need multiplying out.

Answer: A

The unknown is being subtracted, so it equals everything on the left minus the result on the right.

\[ 4211.01 + 22.261 - x = 2645.759 \]

\[ x = 4211.01 + 22.261 - 2645.759 \]

\[ 4211.010 + 22.261 = 4233.271 \]

\[ 4233.271 - 2645.759 = 1587.512 \]

The three near-identical distractors differ only in the last three digits, so the subtraction has to be carried out to full precision rather than estimated.