TRUE OR FALSE — FRACTIONS AND PERCENTAGES
Set 1: Classifying Fractions as Proper or Improper
PREAMBLE
The following fraction is a proper fraction.
1. $\dfrac{3}{5}$
ANSWER: True
2. $\dfrac{9}{4}$
ANSWER: False
3. $\dfrac{6}{6}$
ANSWER: False
SOLUTION
A proper fraction must have a numerator strictly smaller than its denominator. $\dfrac{3}{5}$ qualifies since $3<5$. $\dfrac{9}{4}$ does not, since $9>4$, making it an improper fraction. $\dfrac{6}{6}$ equals 1, and since its numerator is not strictly less than its denominator, it is also classed as improper, not proper.
Set 2: Recognising Equivalent Fractions
PREAMBLE
The following fraction is equivalent to $\dfrac{2}{3}$.
1. $\dfrac{4}{6}$
ANSWER: True
2. $\dfrac{6}{8}$
ANSWER: False
3. $\dfrac{10}{15}$
ANSWER: True
SOLUTION
Two fractions are equivalent if they reduce to the same value in lowest terms. $\dfrac{4}{6}$ simplifies to $\dfrac{2}{3}$, and $\dfrac{10}{15}$ also simplifies to $\dfrac{2}{3}$, so both are equivalent. $\dfrac{6}{8}$ simplifies to $\dfrac{3}{4}$ instead, so it is not equivalent to $\dfrac{2}{3}$.
Set 3: Calculating Percentage Change
PREAMBLE
The stated percentage change for the given price movement is correct.
1. A price rises from $40 to $50; the percentage increase is stated as $25\%$.
ANSWER: True
2. A price falls from $90 to $72; the percentage decrease is stated as $20\%$.
ANSWER: True
3. A price rises from $60 to $75; the percentage increase is stated as $20\%$.
ANSWER: False
SOLUTION
Percentage change is always the change divided by the ORIGINAL value, multiplied by 100%. For item 1, the increase of $10 over an original $40 gives $\frac{10}{40}\times100\%=25\%$, so the statement is correct. For item 2, the decrease of $18 over an original $90 gives $\frac{18}{90}\times100\%=20\%$, also correct. For item 3, the increase of $15 over an original $60 gives $\frac{15}{60}\times100\%=25\%$, not $20\%$, so the statement is false.
Set 4: Applying the Simple Interest Formula
PREAMBLE
Using $I=\dfrac{PRT}{100}$, the interest calculated below is correct.
1. $P=\$800$, $R=5\%$, $T=3$ years; interest stated as $\$120$.
ANSWER: True
2. $P=\$1500$, $R=4\%$, $T=2$ years; interest stated as $\$100$.
ANSWER: False
3. $P=\$600$, $R=8\%$, $T=5$ years; interest stated as $\$240$.
ANSWER: True
SOLUTION
Item 1: $I=\dfrac{800\times5\times3}{100}=\$120$, correct. Item 2: $I=\dfrac{1500\times4\times2}{100}=\$120$, not $\$100$, so the statement is false. Item 3: $I=\dfrac{600\times8\times5}{100}=\$240$, correct.
Set 5: Applying the Compound Interest Formula
PREAMBLE
Using $A=P\left(1+\dfrac{R}{100}\right)^T$, the total amount after the stated period is correct.
1. $P=\$1000$, $R=10\%$, $T=2$ years; amount stated as $\$1210$.
ANSWER: True
2. $P=\$2000$, $R=5\%$, $T=2$ years; amount stated as $\$2200$.
ANSWER: False
3. $P=\$300$, $R=10\%$, $T=3$ years; amount stated as $\$399.30$.
ANSWER: True
SOLUTION
Item 1: $1000(1.1)^2=\$1210$, correct. Item 2: $2000(1.05)^2=\$2205$, not $\$2200$, so the statement is false. Item 3: $300(1.1)^3=\$399.30$, correct.
Set 6: Determining Profit or Loss
PREAMBLE
For the given cost price and selling price, the transaction described results in a profit.
1. Cost price $\$40$, selling price $\$55$.
ANSWER: True
2. Cost price $\$90$, selling price $\$90$.
ANSWER: False
3. Cost price $\$120$, selling price $\$95$.
ANSWER: False
SOLUTION
A profit occurs only when the selling price exceeds the cost price. Item 1 has a selling price above cost price, so it is a profit. Item 2 has equal cost and selling price, which is neither a profit nor a loss, so calling it a profit is false. Item 3 has a selling price below cost price, which is a loss rather than a profit, so it is also false.
Set 7: Evaluating Fraction Operations
PREAMBLE
1. $\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{5}{12}$
ANSWER: True
2. $\dfrac{2}{3}\times\dfrac{3}{4}=\dfrac{3}{4}$
ANSWER: False
3. $\dfrac{2}{3}\div\dfrac{1}{4}=\dfrac{8}{3}$
ANSWER: True
SOLUTION
Item 1: using a common denominator of 12, $\dfrac{3}{12}+\dfrac{2}{12}=\dfrac{5}{12}$, correct. Item 2: multiplying numerators and denominators gives $\dfrac{2\times3}{3\times4}=\dfrac{6}{12}=\dfrac{1}{2}$, not $\dfrac{3}{4}$, so this is false. Item 3: using keep-change-flip, $\dfrac{2}{3}\times\dfrac{4}{1}=\dfrac{8}{3}$, correct.