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: C

As printed, the smallest number is 0.33.

Leading zeros do not change the value of a whole number: 007 is 7 and 002 is 2. So the four numbers are 7, 0.80, 0.33 and 2, and the only ones below 1 are the two decimals. Of these, 0.33 is less than 0.80.

The trap is reading 007 and 002 as decimals; neither has a decimal point, so both are larger than 0.80.

Answer: C

The order is d<c<b<a.

Work out each value:

\[ \begin{aligned} a &= \sqrt{0.25} = 0.5 \\ b &= 0.25 \\ c &= (0.25)^{2} = 0.0625 \\ d &= 0.05 \end{aligned} \]

So \(0.05 \lt 0.0625 \lt 0.25 \lt 0.5\), that is d < c < b < a. The usual slip is to think squaring makes a number bigger; for numbers between 0 and 1 it makes them smaller, and taking the square root makes them bigger.

Comparing c and d is the only close call: 0.0625 against 0.05.

Answer: C

.023 lies between .03 and .02.

Compare in thousandths: .023 is 23 thousandths, .02 is 20 thousandths and .03 is 30 thousandths, so

\[ 0.02 < 0.023 < 0.03 \]

.2 and .3 are ten times too large. The range -.02 to -.03 holds only negative numbers, so it cannot contain .023. The range -.03 to .02 stops at .02, and .023 is just above it. Writing every number with the same count of decimal places (.020, .023, .030) makes such comparisons easy.

Answer: C

Half per cent as a decimal is 0.005.

'Per cent' means 'out of 100', so half per cent is half divided by 100:

\[ \begin{aligned} \tfrac{1}{2}\% &= \frac{0.5}{100} \\ &= 0.005 \end{aligned} \]

0.05 is 5 per cent, 0.02 is 2 per cent and 0.2 is 20 per cent. The usual slip is to write 0.5% as 0.05, forgetting to divide by 100 after halving.

A quick rule: to turn any percentage into a decimal, move the decimal point two places to the left, so 0.5% becomes 0.005.

Answer: A

The answer is 0.008.

Multiply both numbers by 10 to make the divisor a whole number:

\[ \begin{aligned} 0.012 \div 1.5 &= 0.12 \div 15 \\ &= 0.008 \end{aligned} \]

Check: \(1.5 \times 0.008 = 0.012\). 0.08 and 0.8 put the decimal point one and two places too far right, which happens if the shift is made in only one of the two numbers.

An alternative is to write both as fractions: 12/1000 divided by 15/10 is 12/1000 times 10/15, which is 120/15000, or 8/1000.

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: B

A vulgar fraction is a fraction of two whole numbers. Write the decimal over the power of ten matching its number of decimal places, then cancel.

\[ 0.75 = \frac{75}{100} = \frac{75 \div 25}{100 \div 25} = \frac{3}{4} \]

Two decimal places give a denominator of \( 10^{2} = 100 \), and the highest common factor of 75 and 100 is 25.

The distractors are easy to rule out by dividing: \( \tfrac{3}{5} = 0.6 \), \( \tfrac{3}{2} = 1.5 \) and \( \tfrac{3}{7} \approx 0.4286 \).

Answer: B

Again the numerator is a difference of two squares, so the whole expression collapses to the sum of the two numbers.

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

\[ \frac{(2.39)^{2}-(1.61)^{2}}{2.39-1.61} = 2.39 + 1.61 = 4 \]

The awkward decimals are deliberate: they punish anyone who squares first (\( 5.7121 - 2.5921 = 3.12 \), then \( 3.12 \div 0.78 = 4 \)) and reward spotting the identity.

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.