The total weight $1\text{ kg}$ is shared equally by $3$ guavas, so each guava weighs $1\div3=\frac{1}{3}\text{ kg}$.
$\frac{1}{3}\text{ kg}$.
The total weight $1\text{ kg}$ is divided equally into $4$ packets, so each packet weighs $\frac{1}{4}\text{ kg}$.
$\frac{1}{4}\text{ kg}$.
$3$ glasses shared equally by $4$ friends gives $3\div4=\frac{3}{4}$ glass each.
$\frac{3}{4}$ glass.
$\frac{1}{2}+\frac{1}{4}=\frac{2}{4}+\frac{1}{4}=\frac{3}{4}$.
$\frac{3}{4}\text{ kg}$.
Convert the words to fractions and compare: quarter $=\frac{1}{4}$, half $=\frac{1}{2}$, three quarters $=\frac{3}{4}$, one and a quarter $=1\frac{1}{4}$, one and a half $=1\frac{1}{2}$, two and a half $=2\frac{1}{2}$.
$\frac{1}{4},\frac{1}{2},\frac{3}{4},1\frac{1}{4},1\frac{1}{2},2\frac{1}{2}$.
Each piece is $\frac{1}{6}$ of the same whole chikki. The shapes may look different, but equal fractional parts of the same whole have equal area.
Yes.
Continue the pattern by adding one more half each time.
$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=6$ times $\frac{1}{2}$, and adding one more $\frac{1}{2}$ gives $7$ times $\frac{1}{2}$.
Repeat the fractional unit $\frac{1}{4}$ and count how many quarters have been added.
$\frac{1}{4}$ is one quarter; $\frac{1}{4}+\frac{1}{4}$ is two quarters; $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$ is three quarters; $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$ is four quarters.
Fold the strip into $3$ equal parts to make thirds. Folding each third into $2$ equal parts divides the whole into $6$ equal parts, making sixths.
Yes.
Write one $\frac{1}{4}$ for each quarter counted. Five quarters make $\frac{5}{4}$; nine quarters make $\frac{9}{4}=2\frac{1}{4}$.
a. $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{5}{4}$; b. nine quarters $=\frac{9}{4}=2\frac{1}{4}$.
Divide the unit interval into $10$ equal parts. Then $\frac{1}{10}$ is one part, $\frac{3}{10}$ is three parts, and $\frac{4}{5}=\frac{8}{10}$ is eight parts.
Mark lengths $\frac{1}{10}$, $\frac{3}{10}$, and $\frac{4}{5}$ from $0$ on the number line.
Choose any five fractions, divide the number line into suitable equal parts, and mark each fraction at its distance from $0$.
Answers may vary.
Between any two fractions we can find another fraction, for example by taking their average. Therefore there are infinitely many fractions between $0$ and $1$.
Infinitely many fractions.
$\frac{7}{2}=3+\frac{1}{2}$, so it contains $3$ complete units.
$3$ whole units.
$\frac{4}{3}=1+\frac{1}{3}$ and $\frac{7}{3}=2+\frac{1}{3}$.
$\frac{4}{3}$ has $1$ whole unit; $\frac{7}{3}$ has $2$ whole units.
$\frac{8}{3}=2+\frac{2}{3}$, $\frac{11}{5}=2+\frac{1}{5}$, and $\frac{9}{4}=2+\frac{1}{4}$.
a. $2$; b. $2$; c. $2$.
Fractions greater than $1$ that are whole numbers, such as $\frac{8}{4}=2$, are usually written as whole numbers rather than mixed numbers with a fractional part.
No.
Divide the numerator by the denominator. The quotient is the whole part and the remainder over the denominator is the fractional part.
a. $4\frac{1}{2}$; b. $1\frac{4}{5}$; c. $1\frac{2}{19}$; d. $5\frac{2}{9}$; e. $1\frac{1}{11}$; f. $3\frac{1}{6}$.
For each mixed number $a\frac{b}{c}$, compute $\frac{ac+b}{c}$.
a. $\frac{13}{4}$; b. $\frac{23}{3}$; c. $\frac{85}{9}$; d. $\frac{19}{6}$; e. $\frac{25}{11}$; f. $\frac{39}{10}$.
$\frac{3}{6}$ simplifies to $\frac{1}{2}$, so the lengths are equal.
Yes.
$\frac{4}{6}$ simplifies by dividing numerator and denominator by $2$, giving $\frac{2}{3}$.
Yes.
$\frac{1}{2}=\frac{3}{6}$, so three pieces of length $\frac{1}{6}$ make $\frac{1}{2}$.
$3$ pieces.
$\frac{1}{3}=\frac{2}{6}$, so two pieces of length $\frac{1}{6}$ make $\frac{1}{3}$.
$2$ pieces.
Each fraction simplifies to $\frac{1}{2}$: $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$.
Yes.
$\frac{2}{6}$ simplifies to $\frac{1}{3}$, and multiplying numerator and denominator of $\frac{1}{3}$ by $3$ gives $\frac{3}{9}$.
$\frac{1}{3}$ and $\frac{3}{9}$.
Simplify $\frac{4}{6}$ to $\frac{2}{3}$, then multiply numerator and denominator by the same number to form more equivalent fractions.
$\frac{4}{6}=\frac{2}{3}=\frac{6}{9}=\frac{8}{12}=\frac{10}{15}=\cdots$.
Division fact: $3\div4=\frac{3}{4}$. Addition fact: $3=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$. Multiplication fact: $3=4\times\frac{3}{4}$.
Each child gets $\frac{3}{4}$ roti.
Division fact: $2\div4=\frac{2}{4}=\frac{1}{2}$. Addition fact: $2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$. Multiplication fact: $2=4\times\frac{1}{2}$.
Each child gets $\frac{1}{2}$ roti.
$2$ cakes divided equally among $5$ children gives each child $2\div5=\frac{2}{5}$ cake.
$\frac{2}{5}$ cake.
a. $\frac{5}{4}=\frac{10}{8}$. b. $\frac{4}{3}=\frac{12}{9}$. c. Multiplying numerator and denominator of $\frac{7}{5}$ by $2$ gives $\frac{14}{10}$.
a. $10$; b. $9$; c. one choice is $14$ rotis among $10$ children.
With the same denominator, the fraction with the larger numerator represents more equal parts. Hence $\frac{1}{5}<\frac{2}{5}$ and $\frac{3}{7}<\frac{4}{7}$. Also, $\frac{1}{2}=\frac{4}{8}<\frac{5}{8}$.
Each child's share becomes larger when more units are shared among the same number of children.
For the first pair, compare $\frac{3}{4}$ and $\frac{7}{10}$: $\frac{3}{4}=\frac{30}{40}$ and $\frac{7}{10}=\frac{28}{40}$, so Group 1 gets more. For the second pair, $\frac{5}{7}>\frac{4}{7}$ directly.
1. Group 1; 2. Group 2. The second pair is easier to compare because the denominators are the same.
Use a common denominator for each pair by multiplying numerator and denominator of each fraction by a suitable number.
a. $\frac{35}{10}$ and $\frac{6}{10}$; b. $\frac{16}{6}$ and $\frac{5}{6}$; c. $\frac{15}{20}$ and $\frac{12}{20}$; d. $\frac{30}{35}$ and $\frac{56}{35}$; e. $\frac{9}{4}$ and $\frac{10}{4}$; f. $\frac{9}{90}$ and $\frac{20}{90}$; g. $\frac{32}{12}$ and $\frac{33}{12}$; h. $\frac{39}{18}$ and $\frac{2}{18}$.
Divide numerator and denominator by their common factors: $17/51=1/3$, $64/144=4/9$, $126/147=6/7$, and $525/112=75/16$.
a. $\frac{1}{3}$; b. $\frac{4}{9}$; c. $\frac{6}{7}$; d. $\frac{75}{16}$.
Use common denominators: a. $16/6>15/6$; b. $28/63>27/63$; c. $49/70>45/70$; d. same denominator; e. $9/4<10/4$.
a. $\frac{8}{3}>\frac{5}{2}$; b. $\frac{4}{9}>\frac{3}{7}$; c. $\frac{7}{10}>\frac{9}{14}$; d. $\frac{12}{5}>\frac{8}{5}$; e. $\frac{9}{4}<\frac{5}{2}$.
Compare by using common denominators. For a, use $30$: $12/30<21/30<22/30$. For b, use $24$: $14/24<19/24<20/24$.
a. $\frac{2}{5}<\frac{7}{10}<\frac{11}{15}$; b. $\frac{7}{12}<\frac{19}{24}<\frac{5}{6}$.
Use common denominators. For a, denominator $32$ gives $104/32,50/32,28/32,17/32$. For b, denominator $60$ gives $144/60,75/60,45/60,35/60$.
a. $\frac{13}{4}>\frac{25}{16}>\frac{7}{8}>\frac{17}{32}$; b. $\frac{12}{5}>\frac{5}{4}>\frac{3}{4}>\frac{7}{12}$.
Fractions with the same denominator are added by adding numerators: $4+6=10$, denominator $7$ remains the same.
Yes, $\frac{4}{7}+\frac{6}{7}=\frac{10}{7}=1\frac{3}{7}$.
Convert each group to equivalent fractions with a common denominator, add the numerators, and simplify where appropriate.
a. $\frac{13}{7}$; b. $\frac{13}{12}$; c. $\frac{3}{2}$; d. $\frac{20}{21}$; e. $\frac{77}{60}$; f. $\frac{22}{15}$; g. $\frac{22}{15}$; h. $\frac{49}{40}$; i. $\frac{23}{4}$; j. $\frac{62}{21}$; k. $\frac{77}{60}$; l. $\frac{199}{105}$; m. $\frac{83}{12}$.
$\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{17}{12}=1\frac{5}{12}$.
$1\frac{5}{12}$ litres.
$\frac{2}{5}+\frac{3}{4}=\frac{8}{20}+\frac{15}{20}=\frac{23}{20}=1\frac{3}{20}\text{ m}$. Since this is more than $1\text{ m}$, it is sufficient.
$1\frac{3}{20}\text{ m}$; yes, it is sufficient.
Subtract the numerators because the denominators are the same: $\frac{5-3}{8}=\frac{2}{8}=\frac{1}{4}$.
$\frac{2}{8}=\frac{1}{4}$.
$\frac{7-5}{9}=\frac{2}{9}$.
$\frac{2}{9}$.
$\frac{10-1}{27}=\frac{9}{27}=\frac{1}{3}$.
$\frac{9}{27}=\frac{1}{3}$.
Use common denominators and subtract: a. $5/15=1/3$; b. $6/15-4/15=2/15$; c. $15/18-8/18=7/18$; d. $4/6-3/6=1/6$.
a. $\frac{1}{3}$; b. $\frac{2}{15}$; c. $\frac{7}{18}$; d. $\frac{1}{6}$.
a. $\frac{10}{3}-\frac{13}{4}=\frac{40}{12}-\frac{39}{12}=\frac{1}{12}$. b. $\frac{23}{3}-\frac{18}{5}=\frac{115}{15}-\frac{54}{15}=\frac{61}{15}$. c. $\frac{45}{7}-\frac{29}{7}=\frac{16}{7}$.
a. $\frac{1}{12}$; b. $\frac{61}{15}$; c. $\frac{16}{7}$.
a. $\frac{7}{10}-\frac{1}{2}=\frac{7}{10}-\frac{5}{10}=\frac{2}{10}=\frac{1}{5}$. b. $\frac{10}{3}=\frac{40}{12}$ and $\frac{13}{4}=\frac{39}{12}$, so Namit takes less time by $\frac{1}{12}$ minute.
a. $\frac{1}{5}\text{ km}$; b. Namit takes less time by $\frac{1}{12}$ minute.