Question 13 Marks
Find the sum:
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
AnswerWe have,
L.C.M. of 3, 4 and 6 = (2 × 2 × 3) = 12
$\begin{array}{c|c}3&3,4,6\\\hline2&1,4,2\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
Therefore,
$3\frac{1}{3}+4\frac{1}{4}+6\frac{1}{6}$
$=\frac{10}{3}+\frac{17}{4}+\frac{37}{6}$
$=\frac{(40+51+74)}{12}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
$\Big(\frac{12}{4}=3,3\times17=51\Big)$
and $\Big(\frac{12}{6}=2,2\times37=74\Big)$
$=\frac{165}{12}$
$=\frac{55}{4}$
$=13\frac{3}{4}$
View full question & answer→Question 23 Marks
Convert $\frac{1}{4},\frac{5}{8},\frac{7}{12}$ and $\frac{13}{24}$ into like fractions.
AnswerThe given fractions are $\frac{1}{4},\frac{5}{8},\frac{7}{12}$ and $\frac{13}{24}$L.C.M. of 4, 8, 12 and 24 = (4 × 2 × 3) = 24
So, we convert the given fractions into equivalent fractions with 24 as the denominator. (But, one of the fractions already has 24 as its denominator. So, there is no need to convert it into an equivalent fraction.) Thus, we have: $\frac{1}{4}=\frac{1\times6}{4\times6}=\frac{6}{24}$, $\frac{5}{8}=\frac{5\times3}{8\times3}=\frac{15}{24}$, $\frac{7}{12}=\frac{7\times2}{12\times2}=\frac{14}{24}$ Hence, the required like fractions are $\frac{6}{24},\frac{15}{24},\frac{14}{24}$ and $\frac{13}{24}$.
View full question & answer→Question 33 Marks
Find the sum:
$3\frac{2}{3}+1\frac{5}{6}+2$
AnswerWe have,
L.C.M. of 3 and 6 = (2 × 3) = 6
$\begin{array}{c|c}3&3,6\\\hline2&1,2\\\hline&1,1\end{array}$
Therefore,
$3\frac{2}{3}+1\frac{5}{6}+2$
$=\frac{11}{3}+\frac{11}{6}+\frac{2}{1}$
$=\frac{(22+11+12)}{6}$
$\Big(\frac{6}{3}=2,2\times11=22\Big)$
$\Big(\frac{6}{6}=1,1\times11=11\Big)$
and $\Big(\frac{6}{1}=6,6\times2=12\Big)$
$=\frac{45}{6}$
$=\frac{15}{2}$
$=7\frac{1}{2}$
View full question & answer→Question 43 Marks
Lalita read 30 pages of a book containing 100 pages while Sarita read $\frac{2}{5}$ of the book. Who read more?
AnswerLalita read 30 pages of a book having 100 pages
Sarita read $\frac{2}{5}$ of the same book
$\frac{2}{5}$ of 100 pages = $\frac{2}{5}\times100$
$=\frac{200}{5}=40$
= 40 pages
Hence, Sarita read more pages than Lalita as 40 is greater than 30.
View full question & answer→Question 53 Marks
In a school 20 students out of 25 passed in VI A, while 24 out of 30 passed in VI B. Which section gave better result?
AnswerFraction of students who passed in VI A $=\frac{20}{25}$
$=\frac{20\div5}{25\div5}=\frac{4}{5}$
Fraction of students who passed in VI B $=\frac{24}{30}$
$=\frac{24\div6}{30\div6}=\frac{4}{5}$
In both the sections, the fraction of students who passed is the same, so both the sections have the same result.
View full question & answer→Question 63 Marks
Reduce the following fractions into its simplest form:
$\frac{150}{60}$
AnswerHere, numerator = 150 and denominator = 60 Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150 Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30 H.C.F. of 150 and 60 is 30.$\therefore\frac{150}{60}=\frac{150\div30}{60\div30}=\frac{5}{2}$
Hence, the simplest form of $\frac{150}{60}$ is $\frac{5}{2}$.
View full question & answer→Question 73 Marks
Show that the following fractions is in the simplest form:
$\frac{8}{11}$
AnswerHere, numerator = 8 and denominator = 11
Factors of 8 are 1, 2, 4 and 8
Factors of 11 are 1 and 11
Common factor of 8 and 11 is 1
Thus, H.C.F. of 8 and 11 is 1
Hence, $\frac{8}{11}$ is the simplest form.
View full question & answer→Question 83 Marks
Subtract the sum of $3\frac{5}{9}$ and $3\frac{1}{3}$ from the sum of $5\frac{5}{6}$ and $4\frac{1}{9}$.
Answer$\Big(5\frac{5}{6}+4\frac{1}{9}\Big)-\Big(3\frac{5}{9}+3\frac{1}{3}\Big)$$\Big(\frac{35}{6}+\frac{37}{9}\Big)-\Big(\frac{32}{9}+\frac{10}{3}\Big)$
$\begin{array}{c|c}2&6,9,3\\\hline3&3,9,3\\\hline3&1,3,1\\\hline&1,1,1\end{array}$
L.C.M. of 3, 6 and 9 = (2 × 2 × 3) = 18
$=\frac{[105+74]-[64+60]}{18}$
$\Big(\frac{18}{6}=3,3\times35=105\Big)$
and $\Big(\frac{18}{9}=2,2\times37=74\Big)$
$\Big(\frac{18}{9}=2,2\times32=64\Big)$
and $\Big(\frac{18}{3}=6,6\times10=60\Big)$
$=\frac{[179]-[124]}{18}$
$=\frac{55}{18}$
$=3\frac{1}{18}$
View full question & answer→Question 93 Marks
Find the sum:
$3\frac{1}{8}+1\frac{5}{12}$
AnswerL.C.M. of 8 and 12 = (2 × 2 × 2 × 3) = 34
$\begin{array}{c|c}2&8,12\\\hline2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore,
$3\frac{1}{8}+1\frac{5}{12}$
$=\frac{25}{8}+\frac{17}{12}$
$=\frac{(75+34)}{24}$
$\Big(\frac{24}{8}=3,3\times25=75\Big)$
and $\Big(\frac{24}{12}=2,2\times17=34\Big)$
$=\frac{109}{24}$
$=4\frac{13}{24}$
View full question & answer→Question 103 Marks
The weight of an empty gas cylinder is $16\frac{4}{5}\text{kg}$ and it contains $14\frac{2}{3}\text{kg}$ of gas. What is the weight of the cylinder filled with gas?
AnswerWeight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder Thus, we have: (L.C.M .of 5 and 3 = (5 × 3) = 15$\Big(16\frac{4}{5}+14\frac{2}{3}\Big)\text{kg}$
$=\Big(\frac{84}{5}+\frac{44}{3}\Big)\text{kg}$
$=\frac{(252+220)}{15}\text{kg}$
$=\frac{472}{15}\text{kg}$
$=31\frac{7}{15}\text{kg}$
Hence, the weight of the cylinder filled with gas is $31\frac{7}{15}\text{kg}$.
View full question & answer→Question 113 Marks
Simplify:$\frac{5}{8}+\frac{3}{4}-\frac{7}{12}$
AnswerWe have:$\frac{5}{8}+\frac{3}{4}-\frac{7}{12}$
$\begin{array}{c|c}2&4,8,12\\\hline2&2,4,6\\\hline2&1,2,3\\\hline3&1,1,3\\\hline&1,1,1\end{array}$
L.C.M. of 4, 8 and 12 = (2 × 2 × 2 × 3) = 24 $=\frac{(15+18-14)}{24}$ $\Big(\frac{24}{8}=3,3\times5=15\Big)$ $\Big(\frac{24}{4}=6,6\times3=18\Big)$ and $\Big(\frac{24}{12}=2,2\times7=14\Big)$$=\frac{(33-14)}{24}$
$=\frac{19}{24}$
View full question & answer→Question 123 Marks
Mrs Soni bought $7\frac{1}{2}$ litres of milk. Out of this milk, $5\frac{3}{4}$ litres was consumed. How much milk is left with her?
AnswerAmount of milk left with Mrs. Soni = Total amount of milk bought by her - Amount of milk consumed
$\therefore$ Amount of milk left with Mrs. Soni $=7\frac{1}{2}-5\frac{3}{4}$
$=\frac{15}{2}-\frac{23}{4}$
L.C.M. of 2 and 4 = (2 × 2) = 4
$=\frac{30-23}{4}$
$\Big(\frac{4}{2}=2,2\times15=30\Big)$
and $\Big(\frac{4}{4}=1,1\times23=23\Big)$
$=\frac{7}{4}$
$=1\frac{3}{4}\ \text{litres}$
Therefore, Milk left with Mrs. Soni $=1\frac{3}{4}\ \text{litres}$
View full question & answer→Question 133 Marks
What should be added to $9\frac{2}{3}$ to get 19?
AnswerLet x be added to $9\frac{2}{3}$ to get 19 Therefore, $9\frac{2}{3}+\text{x}=19$ Thus, we have: $\text{x}=19-9\frac{2}{3}$ $=\frac{19}{1}-\frac{29}{3}$ L.C.M. of 1 and 3 is 3$=\frac{(57-29)}{3}$
$\Big(\frac{3}{1}=3,3\times19=57\Big)$
and $\Big(\frac{3}{3}=1,1\times29=29\Big)$
$=\frac{28}{3}$
$=9\frac{1}{3}$
View full question & answer→Question 143 Marks
Show that the following fractions is in the simplest form:
$\frac{9}{14}$
AnswerHere, numerator = 9 and denominator = 14
Factors of 9 are 1, 3 and 9
Factors of 14 are 1, 2, 7 and 14
Common factor of 9 and 14 is 1
Thus, H.C.F. of 9 and 14 is 1
Hence, $\frac{9}{14}$ is the simplest form.
View full question & answer→Question 153 Marks
Sohini bought $4\frac{1}{2}\text{m}$ of cloth for her kurta and $2\frac{2}{3}\text{m}$ of cloth for her pyjamas. Ho much cloth did she purchase in all?
AnswerTotal cloth purchased by Sohini = Cloth for kurta + Cloth for pyjamas
Thus, we have:
$\Big(4\frac{1}{2}+2\frac{2}{3}\Big)\ \text{m}$
$=\Big(\frac{9}{2}+\frac{8}{3}\Big)\ \text{m}$
(L.C.M. of 2 and 3 = (2 × 3) = 6)
$=\Big(\frac{(27+16)}{6}\Big)\ \text{m}$
$\Big(\frac{16}{2}=3,3\times9=27\Big)$
and $\Big(\frac{6}{3}=2,2\times8=16\Big)$
$=\frac{43}{6}\ \text{m}$
$=7\frac{1}{6}\ \text{m}$
$\therefore$ Total length of cloth purchased $=7\frac{1}{6}\ \text{m}$
View full question & answer→Question 163 Marks
Simplify:$5\frac{3}{4}-4\frac{5}{12}+3\frac{1}{6}$
AnswerWe have:$5\frac{3}{4}-4\frac{5}{12}+3\frac{1}{6}$
$\frac{23}{4}-\frac{53}{12}+\frac{19}{6}$
$\begin{array}{c|c}2&4,12,6\\\hline2&2,6,3\\\hline3&1,2,3\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
L.C.M. of 4, 12 and 6 = (2 × 2 × 3) = 12 $=\frac{(69-53+38)}{12}$ $\Big(\frac{12}{4}=3,3\times23=69\Big)$ $\Big(\frac{12}{12}=1,1\times53=53\Big)$ and $\Big(\frac{12}{6}=2,2\times19=38\Big)$$=\frac{(107-53)}{12}$
$=\frac{54}{12}$
$=\frac{9}{2}$
$=4\frac{1}{2}$
View full question & answer→Question 173 Marks
Neelam has 25 pencils. She gives $\frac{4}{5}$ of them to Meena. How many pencils does Meena get? How many pencils are left with Neelam?
AnswerNeelam gives $\frac{4}{5}$ of 25 pencils to Meena
$\Big(\frac{4}{5}\times25\Big)$ = 20 Pencils
Thus, Meena gets 20 pencils.
Therefore, Number of pencils left with Neelam = 25 - 20 = 5 pencils
Thus, 5 pencils are left with Neelam
View full question & answer→Question 183 Marks
Rohit bought a pencil for Rs. $3\frac{2}{5}$ and an eraser for Rs. $2\frac{7}{10}$. What is the total cost of both the articles?
AnswerTotal cost of both articles = Cost of pencil + Cost of eraser
Thus, we have:
$\text{Rs.}\ 3\frac{2}{5}+\text{Rs.}\ 2\frac{7}{10}$
$=\frac{17}{5}+\frac{27}{10}$
$=\frac{(34+27)}{10}$
(L.C.M. of 5 and 10 = (5 × 2) = 10)
$=\frac{61}{10}$
$=\text{Rs.}\ 6\frac{1}{10}$
Hence, the total cost of both the articles is $\text{Rs.}\ 6\frac{1}{10}$.
View full question & answer→Question 193 Marks
Show that the following fractions is in the simplest form:
$\frac{8}{15}$
AnswerHere, numerator = 8 and denominator = 15
Factors of 8 are 1, 2, 4 and 8
Factors of 15 are 1, 3, 5 and 15
Common factor of 8 and 15 is 1
Thus, H.C.F. of 8 and 15 is 1
Hence, $\frac{8}{15}$ is the simplest form.
View full question & answer→Question 203 Marks
Reduce the following fractions into its simplest form:
$\frac{48}{60}$
AnswerHere, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84
Factors of 98 are 1, 2, 7, 14, 49 and 98
Common factors of 84 and 98 are 1, 2, 7 and 14
H.C.F. of 84 and 98 is 14
$\therefore\frac{48}{60}=\frac{48\div12}{60\div12}=\frac{4}{5}$
Hence, the simplest form of $\frac{48}{60}$ is $\frac{4}{5}$.
View full question & answer→Question 213 Marks
What should be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$?
AnswerLet x be added to $6\frac{7}{15}$ to get $8\frac{1}{5}$ Therefore, $6\frac{7}{15}+\text{x}=8\frac{1}{5}$ Thus, we have: $\text{x}=8\frac{1}{5}-6\frac{7}{15}$ $=\frac{41}{5}-\frac{97}{15}$ L.C.M. of 5 and 15 = (5 × 3) = 15$=\frac{(123-97)}{15}$
$\Big(\frac{15}{5}=3,3\times41=123\Big)$
and $\Big(\frac{15}{15}=1,1\times97=97\Big)$
$=\frac{26}{15}$
$=1\frac{11}{15}$
View full question & answer→Question 223 Marks
Simplify:$3+1\frac{1}{5}+\frac{2}{3}-\frac{7}{15}$
AnswerWe have:$3+1\frac{1}{5}+\frac{2}{3}-\frac{7}{15}$
$=\frac{3}{1}+\frac{6}{5}+\frac{2}{3}-\frac{7}{15}$
$\begin{array}{c|c}5&5,3,15\\\hline3&1,3,3\ \\\hline&1,1,1\ \end{array}$
L.C.M. of 5, 3 and 15 = (5 × 3) = 15 $=\frac{(45+18+10-7)}{15}$ $\Big(\frac{15}{1}=15,15\times3=45\Big)$ $\Big(\frac{15}{5}=3,3\times6=18\Big)$ $\Big(\frac{15}{3}=5,5\times2=10\Big)$ and $\Big(\frac{15}{15}=1,1\times7=7\Big)$$=\frac{(73-7)}{15}$
$=\frac{66}{15}$
$=\frac{22}{5}$
$=4\frac{2}{5}$
View full question & answer→Question 233 Marks
Show that the following fractions is in the simplest form:
$\frac{21}{10}$
AnswerHere, numerator = 21 and denominator = 10
Factors of 21 are 1, 3, 7 and 21
Factors of 10 are 1, 2, 5 and 10
Common factor of 21 and 10 is 1
Thus, H.C.F. of 21 and 10 is 1.
Hence, $\frac{21}{10}$ is the simplest form.
View full question & answer→Question 243 Marks
Rafiq exercised for $\frac{2}{3}$ hour, while Rohit exercise for $\frac{3}{4}$ hour. Who exercised for a longer time?
AnswerTo know who exercised for a longer time, we have to compare $\frac{2}{3}$ hour with $\frac{3}{4}$ hour
On cross multiplying:
4 × 2 = 8 and 3 × 3 = 9
Clearly, 8 < 9
$\therefore\frac{2}{3}$ hour < $\frac{3}{4}$ hour
Hence, Rohit exercised for a longer time.
View full question & answer→Question 253 Marks
Convert $\frac{3}{5},\frac{7}{10},\frac{8}{15}$ and $\frac{11}{30}$ into like fractions.
AnswerThe given fractions are $\frac{3}{5},\frac{7}{10},\frac{8}{15}$ and $\frac{11}{30}$
$\begin{array}{c|c}5&5,10,15,30\\\hline2&1,2,3,6\ \ \ \ \ \ \\\hline3&1,1,3,3\ \ \ \ \ \\\hline&1,1,1,1\ \ \ \ \ \end{array}$
L.C.M. of 5, 10, 15 and 30 = (5 × 2 × 3) = 30
So, we convert the given fractions into equivalent fractions with 30 as the denominator.
(But, one of the fractions already has 30 as its denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:
$\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$,
$\frac{7}{10}=\frac{7\times3}{10\times3}=\frac{21}{30}$,
$\frac{8}{15}=\frac{8\times2}{15\times2}=\frac{16}{30}$
Hence, the required like fractions are $\frac{18}{30},\frac{21}{30},\frac{16}{30}$ and $\frac{11}{30}$.
View full question & answer→Question 263 Marks
Find the sum:
$2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12}$
AnswerWe have,
L.C.M. of 3, 4, 6 and 12 = (2 × 2 × 3) = 12
$\begin{array}{c|c}2&3,4,6,12\\\hline2&3,2,3,6\ \ \\\hline3&3,1,3,3\ \ \\\hline&1,1,1,1\ \ \end{array}$
Therefore,
$2\frac{1}{3}+1\frac{1}{4}+2\frac{5}{6}+3\frac{7}{12}$
$=\frac{7}{3}+\frac{5}{4}+\frac{17}{6}+\frac{43}{12}$
$=\frac{(28+15+34+43)}{12}$
$\Big(\frac{12}{3}=4,4\times7=28\Big)$
$\Big(\frac{12}{4}=3,3\times5=15\Big)$
$\Big(\frac{12}{6}=2,2\times17=34\Big)$
and $\Big(\frac{12}{12}=1,1\times43=43\Big)$
$=\frac{120}{12}$
$=10$
View full question & answer→Question 273 Marks
Simplify:$2+5\frac{7}{10}-3\frac{14}{15}$
AnswerWe have:$2+5\frac{7}{10}-3\frac{14}{15}$
$=\frac{2}{1}+\frac{57}{10}-\frac{59}{15}$
$\begin{array}{c|c}5&4,10,15\\\hline2&1,2,3\ \ \\\hline3&1,1,3\ \ \\\hline&1,1,1\ \ \end{array}$
L.C.M. of 10 and 15 = (2 × 5 × 3) = 30 $=\frac{(69+171-118)}{30}$ $\Big(\frac{30}{1}=30,30\times2=60\Big)$ $\Big(\frac{30}{10}=3,3\times57=171\Big)$ and $\Big(\frac{30}{15}=2,2\times59=118\Big)$$=\frac{(231-118)}{30}$
$=\frac{113}{30}$
$=3\frac{23}{30}$
View full question & answer→Question 283 Marks
Reduce the following fractions into its simplest form:
$\frac{84}{98}$
AnswerHere, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84
Factors of 98 are 1, 2, 7, 14, 49 and 98
Common factors of 84 and 98 are 1, 2, 7 and 14
H.C.F. of 84 and 98 is 14
$\therefore\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7}$
Hence, the simplest form of $\frac{84}{98}$ is $\frac{6}{7}$.
View full question & answer→Question 293 Marks
Simplify:
$\frac{5}{6}-\frac{4}{9}+\frac{2}{3}$
AnswerWe have, $\frac{5}{6}-\frac{4}{9}+\frac{2}{3}$ $\begin{array}{c|c}3&3,6,9\\\hline3&1,2,3\\\hline2&1,2,1\\\hline&1,1,1\end{array}$ L.C.M. of 3, 6 and 9 = (2 × 3 × 3) = 18 $=\frac{(15+8+12)}{18}$ $\Big(\frac{18}{6}=3,3\times5=15\Big)$ $\Big(\frac{18}{9}=2,2\times4=8\Big)$ and $\Big(\frac{18}{3}=6,6\times2=12\Big)$$=\frac{(27-8)}{18}$
$=\frac{19}{18}$
$=1\frac{1}{18}$
View full question & answer→Question 303 Marks
Find the difference:
$2\frac{7}{9}-1\frac{8}{15}$
Answer$\begin{array}{c|c}3&9,15\\\hline3&3,5\ \ \\\hline5&1,5\ \ \\\hline&1,1\ \ \end{array}$L.C.M of 9 and 15 = (3 × 3 × 5) = 45
$2\frac{7}{9}-1\frac{8}{15}$
$=\frac{25}{9}-\frac{23}{15}$
$=\frac{(125-69)}{45}$
$=\frac{56}{45}$
$=1\frac{11}{45}$
$\Big(\frac{45}{9}=5,5\times25=125\Big)$
and $\Big(\frac{45}{15}=3,3\times23=69\Big)$
View full question & answer→Question 313 Marks
Simplify:$8\frac{5}{6}-3\frac{3}{8}+2\frac{7}{12}$
AnswerWe have:$8\frac{5}{6}-3\frac{3}{8}+2\frac{7}{12}$
$=\frac{53}{6}-\frac{27}{8}+\frac{31}{12}$
$\begin{array}{c|c}2&6,8,12\\\hline2&3,4,6\ \\\hline3&3,2,3\ \\\hline2&1,2,1\ \\\hline&1,1,1\ \end{array}$
L.C.M. of 6, 8 and 12 = (2 × 2 × 2 × 3) = 24 $=\frac{(212-81+62)}{24}$ $\Big(\frac{24}{6}=4,4\times53=212\Big)$ $\Big(\frac{24}{8}=23,3\times7=81\Big)$ and $\Big(\frac{24}{12}=2,2\times31=62\Big)$$=\frac{(274-81)}{24}$
$=\frac{193}{24}$
$=8\frac{1}{24}$
View full question & answer→Question 323 Marks
Show that the following fractions is in the simplest form:
$\frac{25}{36}$
AnswerHere, numerator = 25 and denominator = 36
Factors of 25 are 1, 5 and 25
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36
Common factor of 25 and 36 is 1
Thus, H.C.F. of 25 and 36 is 1
Hence, $\frac{25}{36}$ is the simplest form.
View full question & answer→Question 333 Marks
Find the sum:
$2+\frac{3}{4}+1\frac{5}{8}+3\frac{7}{16}$
AnswerWe have,
L.C.M. of 4, 8 and 16 = (2 × 2 × 2 × 2) = 16
$\begin{array}{c|c}2&4,8,16\\\hline2&2,4,8\\\hline2&1,2,4\\\hline2&1,1,2\\\hline&1,1,1\end{array}$
Therefore,
$2+\frac{3}{4}+1\frac{5}{8}+3\frac{7}{16}$
$=\frac{2}{1}+\frac{3}{4}+\frac{13}{8}+\frac{55}{16}$
$=\frac{(32+12+26+55)}{16}$
$\Big(\frac{16}{1}=16,6\times2=32\Big)$
$\Big(\frac{16}{4}=4,4\times3=12\Big)$
$\Big(\frac{16}{8}=2,2\times13=26\Big)$
and $\Big(\frac{16}{16}=1,1\times55=55\Big)$
$=\frac{125}{16}$
$=7\frac{13}{16}$
View full question & answer→Question 343 Marks
Compare the fractions given below:
$\frac{7}{8},\frac{9}{10}$
AnswerL.C.M. of 8 and 10 = (2 × 5 × 2 × 2) = 40
Now, we convert $\frac{7}{8}$ and $\frac{9}{10}$ into equivalent fractions having 40 as the denominator.
$\therefore\frac{7}{8}=\frac{7\times5}{8\times5}=\frac{35}{40}$
and $\frac{9}{10}=\frac{9\times4}{10\times4}=\frac{36}{40}$
Clearly, $\frac{35}{40}<\frac{36}{40}$
$\therefore\frac{7}{8}<\frac{9}{10}$
View full question & answer→Question 353 Marks
Simplify:$8-3\frac{1}{2}-2\frac{1}{4}$
AnswerWe have:$8-3\frac{1}{2}-2\frac{1}{4}$
$=\frac{8}{1}-\frac{7}{2}-\frac{9}{4}$
$\begin{array}{c|c}2&1,2,4\\\hline2&1,1,2\\\hline&1,1,1\end{array}$
L.C.M. of 1, 2 and 4 = (2 × 2) = 4 $=\frac{(32-14-9)}{4}$ $\Big(\frac{4}{1}=4,4\times8=32\Big)$ $\Big(\frac{4}{2}=2,2\times7=14\Big)$ and $\Big(\frac{4}{4}=1,1\times9=9\Big)$$=\frac{(32-23)}{34}$
$=\frac{9}{4}$
$=2\frac{1}{4}$
View full question & answer→Question 363 Marks
In one day, a rickshaw puller earned $\text{Rs.}\ 137\frac{1}{2}$. Out of this money, he spent $\text{Rs.}\ 56\frac{3}{4}$ on food. How much money is left with him?
AnswerMoney left with the rickshaw puller = Money earned by him in a day - Money spent by him on food
$=\text{Rs.}\ \Big(137\frac{1}{2}-56\frac{3}{4}\Big)$
L.C.M. of 2 and 4 = (2 × 2) = 4
$=\text{Rs.}\ \Big(\frac{275}{2}-\frac{227}{4}\Big)$
$\Big(\frac{4}{2}=2,2\times275=550\Big)$
and $\Big(\frac{4}{4}=1,1\times227=227\Big)$
$=\text{Rs.}\ \Big(\frac{550-227}{4}\Big)$
$=\text{Rs.}\ \Big(\frac{323}{4}\Big)$
$=\text{Rs.}\ 80\frac{3}{4}$
Hence, $\text{Rs.}\ 80\frac{3}{4}$ is left with the rickshaw puller.
View full question & answer→Question 373 Marks
Find the sum:
$\frac{2}{3}+3\frac{1}{6}+4\frac{2}{9}+2\frac{5}{18}$
AnswerWe have,
L.C.M. of 3, 6, 9 and 18 = (3 × 3 × 2) = 18
$\begin{array}{c|c}3&3,6,9,18\\\hline3&1,2,3,6\ \ \\\hline2&1,2,1,2\ \ \\\hline&1,1,1,1\ \end{array}$
Therefore,
$\frac{2}{3}+3\frac{1}{6}+4\frac{2}{9}+2\frac{5}{18}$
$=\frac{2}{3}+\frac{19}{6}+\frac{38}{9}+\frac{41}{18}$
$=\frac{(12+57+76+41)}{18}$
$\Big(\frac{18}{3}=6,6\times2=12\Big)$
$\Big(\frac{18}{6}=3,3\times19=57\Big)$
$\Big(\frac{18}{9}=2,2\times38=76\Big)$
and $\Big(\frac{18}{18}=1,1\times41=41\Big)$
$=\frac{186}{18}$
$=\frac{31}{3}$
$=10\frac{1}{3}$
View full question & answer→Question 383 Marks
Find the difference:
$\frac{5}{6}-\frac{4}{9}$
Answer$\begin{array}{c|c}3&6,9\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array}$L.C.M of 6 and 9 = (3 × 2 × 3) = 18
Now, we have:
$\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
$\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$
Therefore,
$\frac{5}{6}-\frac{4}{9}$
$=\frac{15}{18}-\frac{8}{18}$
$=\frac{(15-8)}{18}$
$=\frac{7}{18}$
View full question & answer→Question 393 Marks
Convert the fractions $\frac{1}{2},\frac{ 2}{3}, \frac{4}{9}$ and $\frac{5}{6}$ into like fractions.
AnswerThe given fractions are $\frac{1}{2},\frac{ 2}{3}, \frac{4}{9},\frac{5}{6}$ L.C.M. of 2, 3, 9 and 6 = (2 ⨯ 3 ⨯ 3) = 18 Now, we have: $\frac{1}{2}=\frac{1\times9}{2\times9}=\frac{9}{18}$ $\frac{2}{3}=\frac{2\times6}{3\times6}=\frac{12}{18}$ $\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$ $\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$ Hence, $\frac{9}{18},\frac{12}{18}, \frac{8}{18}$ and $\frac{15}{18}$ are like fractions.
View full question & answer→Question 403 Marks
Find the sum:
$3+1\frac{4}{15}+1\frac{3}{20}$
AnswerWe have,
L.C.M. of 15 and 20 = (2 × 2 × 3 × 5) = 60
$\begin{array}{c|c}5&15,20\\\hline3&3,4\ \ \ \ \\\hline2&1,4\ \ \ \ \\\hline2&1,2\ \ \ \ \ \\\hline&1,1\ \ \ \ \ \end{array} $
Therefore,
$3+1\frac{4}{15}+1\frac{3}{20}$
$=\frac{3}{1}+\frac{19}{15}+\frac{23}{20}$
$=\frac{(180+76+69)}{60}$
$\Big(\frac{60}{1}=60,60\times3=180\Big)$
$\Big(\frac{60}{15}=4,4\times19=76\Big)$
and $\Big(\frac{60}{20}=3,3\times23=69\Big)$
$=\frac{325}{60}$
$=\frac{65}{12}$
$=5\frac{5}{12}$
View full question & answer→Question 413 Marks
Simplify:$6\frac{1}{6}-5\frac{1}{5}+3\frac{1}{3}$
AnswerWe have:$6\frac{1}{6}-5\frac{1}{5}+3\frac{1}{3}$
$=\frac{37}{6}-\frac{26}{5}+\frac{10}{3}$
$\begin{array}{c|c}2&6,5,3\\\hline3&3,5,3\\\hline5&1,5,1\\\hline&1,1,1\end{array}$
L.C.M. of 6, 5 and 3 = (2 × 5 × 3) = 30 $=\frac{(185-156+100)}{30}$ $\Big(\frac{30}{6}=5,5\times37=185\Big)$ $\Big(\frac{30}{5}=6,6\times26=156\Big)$ and $\Big(\frac{30}{3}=10,10\times10=100\Big)$$=\frac{(285-156)}{30}$
$=\frac{129}{30}$
$=\frac{43}{10}$
$=4\frac{3}{10}$
View full question & answer→Question 423 Marks
Find the sum:
$2\frac{7}{10}+3\frac{8}{15}$
AnswerWe have,
L.C.M. of 10 and 15 = (2 × 3 × 5) = 30
$\begin{array}{c|c}5&10,15\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore,
$2\frac{7}{10}+3\frac{8}{15}$
$=\frac{27}{10}+\frac{53}{15}$
$=\frac{(81+106)}{30}$
$\Big(\frac{30}{10}=3,3\times27=81\Big)$
and $\Big(\frac{30}{15}=2,2\times53=106\Big)$
$=\frac{187}{30}$
$=6\frac{7}{30}$
View full question & answer→Question 433 Marks
Reduce $\frac{84}{98}$ to the simplest form.
AnswerThe factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
The factors of 98 are 1, 2, 7, 14, 49, 98
The common factors of 84 and 98 are 1, 2, 7, 14
The H.C.F. of 84 and 98 is 14
Dividing both the numerator and the denominator by the H.C.F
$\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7}$
View full question & answer→Question 443 Marks
Compare the fractions given below:
$\frac{4}{5},\frac{7}{10}$
AnswerL.C.M. of 5 and 10 = (5 × 2) = 10
Now, we convert $\frac{4}{5}$ into equivalent fractions having 10 as the denominator.
$\therefore\frac{4}{5}=\frac{4\times2}{5\times2}=\frac{8}{10}$
Clearly, $\frac{8}{10}<\frac{7}{10}$
$\therefore\frac{4}{5}<\frac{7}{10}$
View full question & answer→Question 453 Marks
Simplify:$2+\frac{11}{15}-\frac{5}{9}$
AnswerWe have:$2+\frac{11}{15}-\frac{5}{9}$
$\begin{array}{c|c}3&1,15,9\\\hline3&1,5,3\ \\\hline5&1,5,1\ \\\hline&1,1,1\ \end{array}$
L.C.M. of 15 and 9 = (3 × 3 × 5) = 45 $=\frac{(90+33-25)}{45}$ $\Big(\frac{45}{1}=45,45\times2=90\Big)$ $\Big(\frac{45}{15}=3,3\times11=33\Big)$ and $\Big(\frac{45}{9}=5,5\times5=25\Big)$$=\frac{(98+8)}{45}$
$=\frac{98}{45}$
$=2\frac{8}{45}$
View full question & answer→Question 463 Marks
Compare the fractions given below:
$\frac{4}{9},\frac{5}{6}$
AnswerL.C.M. of 9 and 6 = (3 × 3 × 2) = 18
Now, we convert $\frac{4}{9}$ and $\frac{5}{6}$ into equivalent fractions having 18 as the denominator.
$\therefore\frac{4}{9}=\frac{4\times2}{9\times2}=\frac{8}{18}$
and $\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
Clearly, $\frac{8}{18}<\frac{15}{18}$
$\therefore\frac{4}{9}<\frac{5}{6}$
View full question & answer→Question 473 Marks
Of $\frac{3}{4}$ and $\frac{5}{7}$, which is greater and by how much?
AnswerLet us compare $\frac{3}{4}$ and $\frac{5}{7}$
By cross multiplying:
3 ⨯ 7 = 21 and 4 ⨯ 5 = 20
Clearly, 21 > 20
$\therefore\frac{3}{4}>\frac{5}{7}$
Their difference:
$\frac{3}{4}-\frac{5}{7}$
L.C.M. of 4 and 7 = (2 × 2 × 7) = 28
$(28\div4=7,7\times3=21)$
and $(28\div7=4,4\times5=20)$
$=\frac{21-20}{28}$
$=\frac{1}{28}$
Hence, $\frac{3}{4}$ is greater than $\frac{5}{7}$ by $\frac{1}{28}$.
View full question & answer→Question 483 Marks
Find the equivalent fraction of $\frac{3}{5}$ having denominator 30.
AnswerLet $\frac{3}{5}=\frac{}{30}$ 30 = 5 ⨯ 6 So, we have to multiply the numerator by 6 to get the equivalent fraction having denominator 30$\therefore\frac{3}{5}=\frac{3\times6}{5\times6}=\frac{18}{30}$
Thus, $\frac{18}{30}$ is the equivalent fraction of $\frac{3}{5}$.
View full question & answer→Question 493 Marks
Find the difference:
$\frac{5}{8}-\frac{7}{12}$
Answer$\begin{array}{c|c}2&8,12\\\hline2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array}$L.C.M of 2 and 8 = (2 × 2 × 2 × 3) = 24
Now, we have:
$\frac{5}{8}=\frac{5\times3}{8\times3}=\frac{15}{24}$
$\frac{7}{12}=\frac{7\times2}{12\times2}=\frac{14}{24}$
Therefore,
$\frac{5}{8}-\frac{7}{12}$
$=\frac{15}{24}-\frac{14}{24}$
$=\frac{(15-14)}{4}$
$=\frac{1}{24}$
View full question & answer→Question 503 Marks
While coming back home from his school, Kishan covered $4\frac{3}{4}\text{km}$ by rickshaw and $1\frac{1}{2}\text{km}$ on foot. What is the distance of his house from the school?
AnswerDistance from Kishan's house to school = Distance covered by him by rickshaw + Distance covered by him on foot Thus, we have:$\Big(4\frac{3}{4}+1\frac{1}{2}\Big)\text{km}$
$=\frac{19}{4}+\frac{3}{2}\text{km}$
$=\frac{(19+6)}{4}\text{km}$
$=\frac{25}{4}\text{km}$
$=6\frac{1}{4}\text{km}$
(L.C.M .of 2 and 4 = (2 × 2) = 4 $\begin{array}{c|c}2&2,4\\\hline2&1,2\\\hline&1,1\end{array}$ Hence, the distance from Kishan's house to school is $6\frac{1}{4}\text{km}$.
View full question & answer→Question 513 Marks
Of $\frac{3}{4}$ and $\frac{5}{7}$, which is greater and by how much?
AnswerLet us compare $\frac{3}{4}$ and $\frac{5}{7}$ 3 × 7 = 21 and 4 × 5 = 20 Clearly, 21 > 20 Therefore, $\frac{3}{4}>\frac{5}{7}$ Required difference:$=\frac{3}{4}-\frac{5}{7}$
L.C.M. of 4 and 7 = (2 × 2 × 7) = 28$=\frac{21-20}{28}$
$\Big(\frac{28}{4}=7,7\times3=21\Big)$
and $\Big(\frac{28}{7}=4,4\times5=20\Big)$ $=\frac{1}{28}$ Hence, $\frac{3}{4}$ is greater than $\frac{5}{7}$ by $\frac{1}{28}$.
View full question & answer→Question 523 Marks
A piece of wire, $2\frac{3}{4}$ metres long, broke into two pieces. One piece is $\frac{5}{8}$ metre long. How long is the other piece?
AnswerThe length of the other piece = (Length of the wire - Length of one piece)
$=\Big(2\frac{3}{4}-\frac{5}{8}\Big)\ \text{m}$
$=\Big(\frac{11}{4}-\frac{5}{8}\Big)\ \text{m}$
L.C.M. of 4 and 8 = (2 × 2 × 2) = 8
$=\Big(\frac{22-5}{8}\Big)\ \text{m}$
$\Big(\frac{8}{4}=2,2\times11=22\Big)$
and $\Big(\frac{8}{8}=1,1\times5=5\Big)$
$=\Big(\frac{17}{8}\Big)\ \text{m}$
$=2\frac{1}{8}\ \text{m}$
Hence, the other piece is $2\frac{1}{8}\ \text{m}$ long.
View full question & answer→Question 533 Marks
The weight of an empty gas cylinder is $16\frac{4}{5}\text{kg}$ and it contains $14\frac{2}{3}\text{kg}$ of gas. What is the weight of the cylinder filled with gas?
AnswerWeight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder Thus, we have: (L.C.M .of 5 and 3 = (5 × 3) = 15$\Big(16\frac{4}{5}+14\frac{2}{3}\Big)\text{kg}$
$=\Big(\frac{84}{5}+\frac{44}{3}\Big)\text{kg}$
$=\frac{(252+220)}{15}\text{kg}$
$=\frac{472}{15}\text{kg}$
$=31\frac{7}{15}\text{kg}$
Hence, the weight of the cylinder filled with gas is $31\frac{7}{15}\text{kg}$.
View full question & answer→Question 543 Marks
Find the sum:
$2\frac{3}{4}+5\frac{5}{6}$
AnswerWe have,
L.C.M. of 4 and 6 = (2 × 2 × 3) = 12
$\begin{array}{c|c}2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore,
$2\frac{3}{4}+5\frac{5}{6}$
$=\frac{11}{4}+\frac{35}{6}$
$=\frac{(66+140)}{24}$
$\Big(\frac{24}{4}=6,6\times11=66\Big)$
and $\Big(\frac{24}{6}=4,4\times35=140\Big)$
$=\frac{206}{24}$
$=\frac{103}{12}$
$=8\frac{7}{12}$
View full question & answer→Question 553 Marks
A film show lasted for $3\frac{1}{3}$ hours. Out of his time, $1\frac{3}{4}$ hours was spent on advertisements. What was the actual duration of the film?
AnswerActual duration of the film = Total duration of the show - Time spent on advertisements
$=\Big(3\frac{1}{3}-1\frac{3}{4}\Big)\ \text{hours}$
$=\Big(\frac{10}{3}-\frac{7}{4}\Big)\ \text{hours}$
L.C.M. of 3 and 4 = (2 × 2 × 3) = 12
$=\Big(\frac{40-21}{12}\Big)\ \text{hours}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
and $\Big(\frac{12}{4}=3,3\times7=21\Big)$
$=\Big(\frac{19}{12}\Big)\ \text{hours}$
$=1\frac{7}{12}\ \text{hours}$
Thus, the actual duration of the film was $1\frac{7}{12}\ \text{hours}$.
View full question & answer→Question 563 Marks
Reduce the following fractions into its simplest form:
$\frac{9}{15}$
AnswerHere, numerator = 9 and denominator = 15
Factors of 9 are 1, 3 and 9
Factors of 15 are 1, 3, 5 and 15
Common factors of 9 and 15 are 1 and 3
H.C.F. of 9 and 15 is 3
$\therefore\frac{9}{15}=\frac{9\div3}{15\div3}=\frac{3}{5}$
Hence, the simplest form of $\frac{9}{15}$ is $\frac{3}{5}$.
View full question & answer→Question 573 Marks
Reduce the following fractions into its simplest form:
$\frac{72}{90}$
AnswerHere, numerator = 72 and denominator = 90
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72
Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90
Common factors of 72 and 90 are 1, 2, 3, 6, 9 and 18
H.C.F. of 72 and 90 is 18
$\therefore\frac{72}{90}=\frac{72\div18}{90\div18}=\frac{4}{5}$
Hence, the simplest form of $\frac{72}{90}$ is $\frac{4}{5}$.
View full question & answer→Question 583 Marks
Arrange the following fractions in descending order:
$\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$
AnswerThe given fractions are $\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.
Clearly, $\frac{3}{4} >\frac{3}{5} >\frac{3}{7} >\frac{3}{11}>\frac{3}{13}>\frac{3}{17}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{3}{4},\frac{3}{5},\frac{3}{7} ,\frac{3}{11},\frac{3}{13},\frac{3}{17}$
View full question & answer→Question 593 Marks
Arrange the following fractions in descending order:
$\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$
AnswerThe given fractions are $\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.
Clearly, $\frac{1}{7} >\frac{1}{9} >\frac{1}{12} >\frac{1}{17}>\frac{1}{23}>\frac{1}{50}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{1}{7},\frac{1}{9},\frac{1}{12},\frac{1}{17},\frac{1}{23},\frac{1}{50}$
View full question & answer→Question 603 Marks
Compare the fractions given below:
$\frac{11}{12},\frac{13}{15}$
AnswerL.C.M. of 12 and 15 = (2 × 2 × 3 × 5) = 60
Now, we convert $\frac{11}{12}$ and $\frac{13}{15}$ into equivalent fractions having 40 as the denominator.
$\therefore\frac{11}{12}=\frac{11\times5}{12\times5}=\frac{55}{60}$
and $\frac{13}{15}=\frac{13\times4}{15\times4}=\frac{52}{60}$
Clearly, $\frac{55}{60}>\frac{52}{60}$
$\therefore\frac{11}{12}>\frac{13}{15}$
View full question & answer→