Question 14 Marks
A student bought $4 \frac{1}{3} \mathrm{~m}$ of yellow ribbon, 61 $1 / 6 m$ of red ribbon and $3 \frac{2}{9} \mathrm{~m}$ of blue ribbon for decorating a room. How many meters of ribbon did he buy?
Answer
$\begin{aligned} & \text { Length of yellow ribbon }=4 \frac{1}{3} \mathrm{~m}=\frac{13}{3} \mathrm{~m} \\ & \text { Length of red ribbon }=6 \frac{1}{6} \mathrm{~m}=\frac{37}{6} \mathrm{~m} \\ & \text { Length of blue ribbon }=3 \frac{2}{9} \mathrm{~m}=\frac{29}{9} \mathrm{~m} \\ & \text { Total length of ribbon }=\frac{13}{3}+\frac{37}{6}+\frac{29}{9} \\ & =\frac{13 \times 6}{3 \times 6}+\frac{37 \times 3}{6 \times 3}+\frac{29 \times 2}{9 \times 2} \ldots .(\mathrm{LCM} \text { of } 3,6,9=18) \\ & =\frac{78+111+58}{18} \\ & =\frac{247}{18} \\ & =13 \frac{13}{18} \text { meters }\end{aligned}$
View full question & answer→Question 24 Marks
A boy spent $\frac{3}{5}$ of his money on buying 1 cloth and $\frac{1}{4}$ of the remaining on buying shoes. If initially, he has ₹ 2400 ; how much did he spend on shoes?
AnswerMoney in hand ₹ 2400
Money spent on buying clothes
$=\frac{3}{5}$ of $\text{₹}$ 1440
$=\frac{3}{5} \times 2400=3 \times 480$ = ₹ 1440
Remaining money $==$ Total money - money spent on cloth
= ₹ 2400- ₹ 1440= ₹ 960
Now, boy spent $\frac{1}{4}$ of the remaining money on buying shoes.
$\therefore$ Money spent on buying shoes
= ₹ $960 \times \frac{1}{4}=$ ₹ 240
View full question & answer→Question 34 Marks
From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.
AnswerLet the piece by Rajesh $=1 \times \frac{1}{3}=\frac{1}{3} \mathrm{~m}$
Land bought by Rajesh = $1 \times \frac{1}{3}=\frac{1}{3} \mathrm{~m}$
Remaining land $=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3} \mathrm{~m}$
Now, land bought by Manoj $=\frac{2}{3} \times \frac{1}{3}=\frac{2}{9} \mathrm{~m}$
Land unsold $=\frac{2}{3}-\frac{2}{9}$
$ =\frac{6-2}{9}=\frac{4}{9} \mathrm{~m} $
Land of $\frac{4}{9} \mathrm{~m}$ remain unsold $=600 \mathrm{~m}^2$
$\therefore$ Total area of the land $=600 \times \frac{9}{4}$
$ =150 \times 9 \mathrm{~m}^2=1350 \mathrm{~m}^2 $
View full question & answer→Question 44 Marks
In a school, $\frac{4}{5}$ of the children are boys. If the number of girls is 200 , find the number of boys.
AnswerNo. of boys $=\frac{4}{5}$ of the total children
$\therefore$ No. of. girls $=\left(1-\frac{4}{5}\right)$ of total children
$=\frac{5-4}{5}=\frac{1}{5}$ of total children
But no. of girls $=200$
$\therefore \frac{1}{5}$ of total children $=200$
Hence total number of children $=200 \times \frac{5}{1}=1000$
$\therefore$ No. of boys $=\frac{4}{5}$ of $1000=\frac{4}{5} \times 1000=800$
View full question & answer→Question 54 Marks
A lamp post has half of its length in mud and $\frac{1}{3}$ of its length in water.
(i) What fraction of its length is above the water?
(ii) If $3 \frac{1}{3} \mathrm{~m}$ of the lamp post is above the water, find the whole length of the lamp post.
Answer(i) Let length of the post $=1 \mathrm{~m}$
then length of post in mud $=\frac{1}{2} \mathrm{~m}$
and length of post in water $=\frac{1}{3} \mathrm{~m}$
$\therefore$ Length of post above the water
$ \begin{aligned} & =1-\left(\frac{1}{2}+\frac{1}{3}\right)=1-\left(\frac{3+2}{6}\right) \\ & =1-\frac{5}{6}=\frac{6-5}{6}=\frac{1}{6} \mathrm{~m} \end{aligned} $
(ii) But length of post above water
$ =3 \frac{1}{3} \mathrm{~m}=\frac{10}{3} \mathrm{~m} $
$\therefore \frac{1}{6}$ th of total length $=\frac{10}{3} \mathrm{~m}$
$\therefore$ Total length $=\frac{10}{3} \times \frac{6}{1}=20 \mathrm{~m}$
View full question & answer→Question 64 Marks
Shyam bought a refrigerator for Rs. 5000 . He paid $\frac{1}{10}$ of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month?
AnswerTotal amount of the refrigerator $=$ Rs 5000
Amount paid in cash $=\frac{1}{10}$ of Rs. 5000
$ =\frac{1}{10} \times 5000=\text { Rs. } 500 $
Balance amount = Rs. 5000 - Rs. 500
$ =\text { Rs. } 4500 $
No.of equally instalments $=12$
$\therefore$ Amount of each instalment
$ \begin{aligned} & =\text { Rs. } 4500 \div 12=\text { Rs. } 4500 \times \frac{1}{12} \\ & =\text { Rs. } 375 \end{aligned} $
View full question & answer→Question 74 Marks
A man spends $\frac{2}{5}$ of his salary on food and $\frac{3}{10}$ of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?
AnswerLet the total amount of salary = Rs. 1
Amount spent on food $=\frac{2}{5}$ of Rs. $1=$ Rs. $\frac{2}{5}$
Remaining amount $=1-\frac{2}{5}$
$ =\frac{5-2}{5}=\text { Rs. } \frac{3}{5} $
Amount spent on house rent etc.
$ =\frac{3}{10} \text { of } \frac{3}{5}=\mathrm{Rs} \frac{9}{50} $
Remaining amount left $=\frac{3}{5}-\frac{9}{50}$
$ \begin{aligned} & =\text { Rs. } \frac{3}{5}-\frac{9}{50} \\ & =\text { Rs. } \frac{30-9}{50}=\text { Rs. } \frac{21}{50} \end{aligned} $
$\therefore$ Fraction of amount left $=\frac{21}{50}$
View full question & answer→Question 84 Marks
Vijay weighed $65 \frac{1}{2} \mathrm{~kg}$. He gained $1 \frac{2}{5} \mathrm{~kg}$ during the first week, $1 \frac{1}{4} \mathrm{~kg}$ during the second week, but lost $\frac{5}{16} \mathrm{~kg}$ during the 16 third week. What was his weight after the third week?
AnswerIn the beginning, weight of vijay $=65 \frac{1}{2} \mathrm{~kg}$
Gained in first week $=1 \frac{2}{5} \mathrm{~kg}$
Gained in second week $=1 \frac{1}{4} \mathrm{~kg}$
Lost in the third week $=\frac{5}{16} \mathrm{~kg}$
$\therefore$ Weight of Vijay after third week
$=\left(65 \frac{1}{2}+1 \frac{2}{5}+1 \frac{1}{4}-\frac{5}{16}\right) \mathrm{kg}$
$=\left(\frac{131}{2}+\frac{7}{5}+\frac{5}{4}-\frac{5}{16}\right) \mathrm{kg}$
$=\frac{5240+112+100-25}{80}($ LCM of 2, 5, 4 and $16=80)$
$=\frac{5452-25}{80}=\frac{5427}{80} \mathrm{~kg}=67 \frac{67}{80} \mathrm{~kg}$
View full question & answer→Question 94 Marks
Simplify: $\frac{3}{4}$ of $7 \frac{3}{7}-5 \frac{3}{5} \div 3 \frac{4}{15}$
Answer$ \frac{3}{4} \text { of } 7 \frac{3}{7}-5 \frac{3}{5} \div 3 \frac{4}{15} $
Using BODMAS
$ \frac{3}{4} \text { of } \frac{52}{7}-\frac{28}{5} \div \frac{49}{15} $
$ =\frac{39}{7}-\frac{28}{5} \div \frac{49}{15} $
$ =\frac{39}{7}-\frac{28}{5} \times \frac{15}{49}=\frac{39}{7}-\frac{12}{7} $
$ =\frac{39-12}{7}=\frac{27}{7}=3 \frac{6}{7} $
View full question & answer→Question 104 Marks
Simplify: $5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25}$
Answer$\begin{aligned} & 5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25} \\ & \frac{23}{4}-\frac{3}{7} \times \frac{63}{4}+\frac{72}{35} \div \frac{36}{25} \\ & =\frac{23}{4}-\frac{3}{7} \times \frac{63}{4}+\frac{72}{35} \times \frac{25}{36} \\ & =\frac{23}{4}-\frac{27}{4}+\frac{10}{7} \\ & =\frac{161-189+40}{28}=\frac{201-189}{28}=\frac{12}{28}=\frac{3}{7}\end{aligned}$
View full question & answer→Question 114 Marks
What number should be added to $8 \frac{2}{3}$ to $12 \frac{5}{6}$.
AnswerFor finding the required fraction, we have to subtract $8 \frac{2}{3}$ from $12 \frac{5}{6}$
$ \begin{aligned} & \therefore \text { Required number }=12 \frac{5}{6}-8 \frac{2}{3} \\ & =\frac{77}{6}-\frac{26}{3} \\ & =\frac{77 \times 1}{6 \times 1}-\frac{26 \times 2}{3 \times 2}(\because \text { L.C.M. of } 3 \text { and } 6=6) \\ & =\frac{77-52}{6}=\frac{25}{6}=4 \frac{1}{6} \end{aligned} $
View full question & answer→Question 124 Marks
Which is greater, $\frac{3}{5}$ or $\frac{7}{10}$ and by how much?
AnswerTaking the cross-multiplication, we get
$ 3 \times 10=30 \text { and } 7 \times 5=35 $
Since, $3 \times 10$ (i.e. 30 ) is smaller than $7 \times 5$ (i.e. 35 )
$ \therefore \frac{3}{5}<\frac{7}{10} $
Difference between $\frac{7}{10}$ and $\frac{3}{5}$
$ \begin{aligned} & \Rightarrow \frac{7}{10}-\frac{3}{5}(\because \text { L.C.M. of } 10 \text { and } 5=10) \\ & \Rightarrow \frac{7 \times 1}{10 \times 1}-\frac{3 \times 2}{5 \times 2} \\ & =\frac{7-6}{10}=\frac{1}{10} \end{aligned} $
$\therefore \frac{7}{10}$ is greater than $\frac{3}{5}$ by $\frac{1}{10}$
View full question & answer→Question 134 Marks
A bought $3 \frac{3}{4} \mathrm{~kg}$ of wheat and $2 \frac{1}{2} \mathrm{~kg}$ of rice. Find the total weight of wheat and rice bought.
AnswerWeight of wheat $=3 \frac{3}{4} \mathrm{~kg}=\frac{15}{4} \mathrm{~kg}$
Weight of rice $=2 \frac{1}{2} \mathrm{~kg}=\frac{5}{2} \mathrm{~kg}$
$\therefore$ Totak weight of wheat and rice
$ \begin{aligned} & =\frac{15}{4}+\frac{5}{2} \\ & =\frac{15 \times 1}{4 \times 1}+\frac{5 \times 2}{2 \times 2}(\because \text { L.C.M. of } 4 \text { and } 2=4) \\ & =\frac{15+10}{4}=\frac{25}{4} \mathrm{~kg} \\ & =6 \frac{1}{4} \mathrm{~kg} \end{aligned} $
View full question & answer→Question 144 Marks
Simplify and reduce to a simple fraction:
$\frac{1}{3} \times 4 \frac{2}{3} \div 3 \frac{1}{2} \times \frac{1}{2}$
Answer
$ \begin{aligned} & \frac{1}{3} \times 4 \frac{2}{3} \div 3 \frac{1}{2} \times \frac{1}{2} \\ & =\frac{1}{3} \times \frac{4 \times 3+2}{3} \div \frac{3 \times 2+1}{2} \times \frac{1}{2} \\ & =\frac{1}{3} \times \frac{14}{3} \div \frac{7}{2} \times \frac{1}{2} \\ & =\frac{1}{3} \times \frac{14}{3} \times \frac{2}{7} \times \frac{1}{2} \ldots \text { [Applying the rule of BODMAS] } \end{aligned} $
[Performing division first]
$ \begin{aligned} & =\frac{1 \times 14 \times 2 \times 1}{3 \times 3 \times 7 \times 2} \\ & =\frac{2}{9} \end{aligned} $
View full question & answer→Question 154 Marks
If $A=3 \frac{3}{8}$ and $B=6 \frac{5}{8}$ find:
(i) $A \div B$
(ii) $\mathrm{B} \div \mathrm{A}$
Answer
$ \begin{aligned} & A=3 \frac{3}{8}=\frac{27}{8} \\ & B=6 \frac{5}{8}=\frac{53}{8} \end{aligned} $
$ \begin{aligned} & \text { (i) } \mathrm{A} \div \mathbf{B} \\ & \Rightarrow \frac{27}{8} \div \frac{53}{8} \\ & \Rightarrow \frac{27}{8} \times \frac{8}{53}=\frac{27}{53} \end{aligned} $
(ii) $\mathrm{B} \div \mathrm{A}$
$ \Rightarrow \frac{53}{8} \div \frac{27}{8} \Rightarrow \frac{53}{8} \times \frac{8}{27} $
$\Rightarrow \frac{53}{27}=1 \frac{26}{27}$
View full question & answer→Question 164 Marks
A field is $16 \frac{1}{2} \mathrm{~m}$ long and $12 \frac{2}{5} \mathrm{~m}$ wide. Find the perimeter of the field.
Answer
$\begin{aligned} & \text { Length of field }=16 \frac{1}{2} \mathrm{~m} \\ & \text { Breadth of field }=12 \frac{2}{5} \mathrm{~m} \\ & \therefore \text { Perimeter of field }=2(\mathrm{l}+\mathrm{b}) \\ & =2 \times\left(16 \frac{1}{2}+12 \frac{2}{5}\right) \\ & =2 \times\left(\frac{33}{2}+\frac{62}{5}\right) \\ & =2 \times\left(\frac{33 \times 5}{2 \times 5}+\frac{62 \times 2}{5 \times 2}\right) \\ & (\because \text { L.C.M. of } 2 \text { and } 5=10) \\ & =2 \times\left(\frac{165+124}{10}\right) \\ & =2 \times \frac{289}{10}=\frac{289}{5} \mathrm{~m}=57 \frac{4}{5} \mathrm{~m}\end{aligned}$
View full question & answer→Question 174 Marks
Insert three fractions between
$\frac{4}{7} \text { and } \frac{3}{4}$
AnswerFraction between $\frac{4}{7}$ and $\frac{3}{4}$
$ =\frac{4+3}{7+4}=\frac{7}{11} $
Fraction between $\frac{4}{7}$ and $\frac{7}{11}$
$ =\frac{4+7}{7+11}=\frac{11}{18} $
and Fraction between $\frac{7}{11}$ and $\frac{3}{4}$
$ =\frac{7+3}{11+4}=\frac{10}{15}=\frac{2}{3} $
Hence, three fractions between $\frac{4}{7}$ and $\frac{3}{4}$ will be $\frac{11}{18}, \frac{7}{11}$ and $\frac{2}{3}$
View full question & answer→Question 184 Marks
Insert three fractions between $\frac{11}{12}$ and $\frac{2}{3}$
AnswerFraction between $\frac{11}{12}$ and $\frac{2}{3}$
$ =\frac{11+2}{12+3}=\frac{13}{15} $
Fraction between $\frac{11}{12}$ and $\frac{13}{15}$
$ =\frac{11+13}{12+15}=\frac{24}{27}=\frac{8}{9} $
and Fraction between $\frac{13}{15}$ and $\frac{2}{3}$
$ =\frac{13+2}{15+3}=\frac{15}{18}=\frac{5}{6} $
Hence, three fractions between $\frac{11}{12}$ and $\frac{2}{3}$ will be $\frac{8}{9}, \frac{13}{15}$ and $\frac{5}{6}$
View full question & answer→Question 194 Marks
Insert three fractions between $\frac{3}{8}$ and $\frac{6}{11}$
AnswerFraction between $\frac{3}{8}$ and $\frac{6}{11}$
$ =\frac{3+6}{8+11}=\frac{9}{19} $
Fraction between $\frac{3}{8}$ and $\frac{9}{19}$
$ =\frac{3+9}{8+19}=\frac{12}{27}=\frac{4}{9} $
and Fraction between $\frac{9}{19}$ and $\frac{6}{11}$
$ =\frac{9+6}{19+11}=\frac{15}{30}=\frac{1}{2} $
Hence, three fractions between $\frac{3}{8}$ and $\frac{6}{11}$ will be $\frac{4}{9}, \frac{9}{19}$ and $\frac{1}{2}$
View full question & answer→Question 204 Marks
Insert three fractions between $\frac{1}{2}$ and $\frac{5}{7}$
AnswerFraction between $\frac{1}{2}$ and $\frac{5}{7}$
$ =\frac{1+5}{2+7}=\frac{6}{9}=\frac{2}{3} $
Fraction between $\frac{1}{2}$ and $\frac{2}{3}$
$ =\frac{1+2}{2+3}=\frac{3}{5} $
and Fraction between $\frac{2}{3}$ and $\frac{5}{7}$
$ =\frac{2+5}{3+7}=\frac{7}{10} $
Hence, three fractions between $\frac{1}{2}$ and $\frac{5}{7}$ will be $\frac{3}{5}, \frac{2}{3}$ and $\frac{7}{10}$
View full question & answer→Question 214 Marks
Insert three fractions between
$\frac{2}{5} \text { and } \frac{4}{9}$
AnswerFraction between $\frac{2}{5}$ and $\frac{4}{9}$
$ =\frac{2+4}{5+9}=\frac{6}{14}=\frac{3}{7} $
Fraction between $\frac{2}{5}$ and $\frac{3}{7}$
$ =\frac{2+3}{5+7}=\frac{5}{12} $
and Fraction between $\frac{3}{7}$ and $\frac{4}{9}$
$ =\frac{3+4}{7+9}=\frac{7}{16} $
Hence, three fractions between $\frac{2}{5}$ and $\frac{4}{9}$ will be $\frac{5}{12}, \frac{3}{7}$ and $\frac{7}{16}$
View full question & answer→Question 224 Marks
Put the given fraction in ascending order by making denominator equal:
$\frac{1}{3}, \frac{2}{5}, \frac{3}{4} \text { and } \frac{1}{6}$
Answer$\frac{1}{3}, \frac{2}{5}, \frac{3}{4}$ and $\frac{1}{6}$
L.C.M. of denominators $3,5,4$ and $6=60$
$ \begin{aligned} & \therefore \frac{1}{3}=\frac{1 \times 20}{3 \times 20}=\frac{20}{60} \\ & \frac{2}{5}=\frac{2 \times 12}{5 \times 12}=\frac{24}{60} \\ & \frac{3}{4}=\frac{3 \times 15}{4 \times 15}=\frac{45}{60} \\ & \frac{1}{6}=\frac{1 \times 10}{6 \times 10}=\frac{10}{60} \end{aligned} $
From above we see that $ \frac{10}{60}<\frac{20}{60}<\frac{24}{60}<\frac{45}{60} $
or $\frac{1}{6}<\frac{1}{3}<\frac{2}{5}<\frac{3}{4}$
Hence, $\frac{1}{6}, \frac{1}{3}, \frac{2}{5}, \frac{3}{4}$ are in ascending order.
View full question & answer→Question 234 Marks
Convert given fraction into a fraction with equal numerators:
$\frac{15}{19}, \frac{25}{28}, \frac{9}{11} \text { and } \frac{45}{47}$
Answer$ \ln \frac{15}{19}, \frac{25}{28}, \frac{9}{11} \text { and } \frac{45}{47} \text {, } $
L.C.M. of $15,25,9$ and $45=225$ $ \therefore \frac{15}{19}=\frac{15 \times 15}{19 \times 15}=\frac{225}{285} $
$ \frac{25}{28}=\frac{25 \times 9}{28 \times 9}=\frac{225}{252} $
$ \frac{9}{11}=\frac{9 \times 25}{11 \times 25}=\frac{225}{275} $
and $\frac{45}{47}=\frac{45 \times 5}{47 \times 5}=\frac{225}{235}$
Hence the required fractions are $\frac{225}{285}, \frac{225}{252}, \frac{225}{275}$ and $\frac{225}{235}$
View full question & answer→Question 244 Marks
If a boy works for six consecutive days for 8 hours, $7 \frac{1}{2}$ hours, $8 \frac{1}{4}$ hours, $6 \frac{1}{4}$ hours $+6 \frac{3}{4}$ hrs and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour?
AnswerNo. of hours, a boy worked in 6 days
$=8 \mathrm{hrs}+7 \frac{1}{2} \mathrm{hrs}+8 \frac{1}{4} \mathrm{hrs}+6 \frac{1}{4} \mathrm{hrs}+6 \frac{3}{4} \mathrm{hrs}+7 \mathrm{hrs}$
$=\left(8+\frac{15}{2}+\frac{33}{4}+\frac{25}{4}+\frac{27}{4}+7\right)$ hours
$=\frac{32+30+33+25+27+28}{4}$ hours
$=\frac{175}{4}$ hours $=43 \frac{3}{4}$ hours
Earning per hour $=$ Rs. 36
$\therefore$ Total earnings $=$ Rs. $\frac{175}{4} \times 36$
$=$ Rs. $175 \times 9=$ Rs, 1575
View full question & answer→Question 254 Marks
From a sack of potatoes weighing $120 \mathrm{~kg}$, a merchant sells portions weighing $6 \mathrm{~kg}, 5 \frac{1}{4} \mathrm{~kg}, 9 \frac{1}{2} \mathrm{~kg}$, and $9 \frac{3}{4} \mathrm{~kg}$ respectively.
(i) How many kg did he sell?
(ii) How many kg is still left in the sack?
AnswerTotal quantities of potatoes $=120 \mathrm{~kg}$
(i) Quantity of potatoes he sold
$ \begin{aligned} & =6 \mathrm{~kg}+5 \frac{1}{4} \mathrm{~kg}+9 \frac{1}{2} \mathrm{~kg}+9 \frac{3}{4} \mathrm{~kg} \\ & =\left(6+\frac{21}{4}+\frac{19}{2}+\frac{39}{4}\right) \mathrm{kg} \\ & =\frac{24+21+38+39}{4} \mathrm{~kg} \\ & =\frac{122}{4}=\frac{61}{2} \mathrm{~kg}=30 \frac{1}{2} \mathrm{~kg} \end{aligned} $
(ii) Quantity of potatoes left
$ =120 \mathrm{~kg}-30 \frac{1}{2} \mathrm{~kg}=\left(\frac{120}{1}-\frac{61}{2}\right) \mathrm{kg} $
$ =\frac{240-61}{2}=\frac{179}{2} \mathrm{~kg}=89 \frac{1}{2} \mathrm{~kg} $
View full question & answer→Question 264 Marks
Simplify: $\frac{6}{5}$ of $\left(3 \frac{1}{3}-2 \frac{1}{2}\right)+\left(2 \frac{5}{21}-2\right)$
Answer
$\begin{aligned} & \frac{6}{5} \text { of }\left(3 \frac{1}{3}-2 \frac{1}{2}\right) \div\left(2 \frac{5}{21}-2\right) \\ & =\frac{6}{5} \text { of }\left(\frac{10}{3}-\frac{5}{2}\right) \div\left(\frac{47}{21}-\frac{2}{1}\right) \ldots(\text { Using BODMAS }) \\ & =\frac{6}{5} \text { of }\left(\frac{20-15}{6}\right) \div\left(\frac{47-42}{21}\right) \\ & =\frac{6}{5} \text { of } \frac{5}{6} \div \frac{5}{21} \\ & =1 \div \frac{5}{21} \\ & =1 \times \frac{21}{5}=\frac{21}{5}=4 \frac{1}{5}\end{aligned}$
View full question & answer→Question 274 Marks
Simplify: $1 \frac{1}{5}+\left\{2 \frac{1}{3}-(5+\overline{2-3})\right\}-3 \frac{1}{2}$
Answer
$\begin{aligned} & 1 \frac{1}{5}+\left\{2 \frac{1}{3}-(5+\overline{2-3})\right\}-3 \frac{1}{2} \\ & =\frac{6}{5} \div\left\{\frac{7}{3}-(5-1)\right\}-\frac{7}{2} \\ & =\frac{6}{5} \div\left\{\frac{7}{3} \div 4\right\}-\frac{7}{2}=\frac{6}{5} \div\left\{\frac{7-12}{3}\right\}-\frac{7}{2} \\ & =\frac{6}{5} \div \frac{-5}{3}-\frac{7}{2}=\frac{6}{5} \times \frac{3}{-5}-\frac{7}{2} \\ & =-\frac{18}{25}-\frac{7}{2}=\frac{-36-175}{50}=\frac{-211}{50} \\ & =-4 \frac{11}{50}\end{aligned}$
View full question & answer→Question 284 Marks
Simplify: $12 \frac{1}{2}-\left[8 \frac{1}{2}+\{9-(5-\overline{3-2})\}\right]$
Answer
$\begin{aligned} & 12 \frac{1}{2}-\left[8 \frac{1}{2}+\{9-(5-\overline{3-2})\}\right] \\ & =\frac{25}{2}-\left[\frac{17}{2}+\{9-(5-1)\}\right] \\ & =\frac{25}{2}-\left[\frac{17}{2}+\{9-4\}\right]=\frac{25}{2}-\left[\frac{17}{2}+5\right] \\ & =\frac{25}{2}-\frac{17}{2}-\frac{5}{1}=\frac{25-17-10}{2} \\ & =\frac{25-27}{2}=-\frac{2}{2}=-1\end{aligned}$
View full question & answer→Question 294 Marks
Simplify: $2 \frac{3}{4}-\left[3 \frac{1}{8} \div\left\{5-\left(4 \frac{2}{3}-\frac{11}{12}\right)\right\}\right]$
Answer
$\begin{aligned} & 2 \frac{3}{4}-\left[3 \frac{1}{8} \div\left\{5-\left(4 \frac{2}{3}-\frac{11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\left(\frac{14}{3}-\frac{11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\left(\frac{56-11}{12}\right)\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{5-\frac{45}{12}\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div\left\{\frac{60-45}{12}\right\}\right] \\ & =\frac{11}{4}-\left[\frac{25}{8} \div \frac{15}{12}\right]=\frac{11}{4}-\left[\frac{25}{8} \times \frac{12}{15}\right] \\ & =\frac{11}{4}-\frac{5}{2}=\frac{11-10}{4}=\frac{1}{4}\end{aligned}$
View full question & answer→Question 304 Marks
Reduce to a single fraction:
$\frac{2}{3}-\frac{3}{5}+3-\frac{1}{5}$
Answer
$\begin{aligned} & \frac{2}{3}-\frac{3}{5}+3-\frac{1}{5} \\ & \left.=\frac{2 \times 5}{3 \times 5}-\frac{3 \times 3}{5 \times 3}+\frac{3 \times 15}{15}-\frac{1 \times 3}{5 \times 3} \quad \text { (LCM of } 3 \text { and } 5=15\right.) \\ & =\frac{10}{15}-\frac{9}{15}+\frac{45}{15}-\frac{3}{15} \\ & =\frac{10-9+45-3}{15}=\frac{55-12}{15} \\ & =\frac{43}{15}=2 \frac{13}{15}\end{aligned}$
View full question & answer→Question 314 Marks
Arrange the given fraction in descending order by making numerator equal:
$\frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5}$
Answer$ \frac{1}{10}, \frac{6}{11}, \frac{8}{11} \text { and } \frac{3}{5} $
L.C.M. of numerators $1,6,8$ and $3=24$
$ \begin{aligned} & \therefore \frac{1}{10}=\frac{1 \times 24}{10 \times 24}=\frac{24}{240} \\ & \frac{6}{11}=\frac{6 \times 4}{11 \times 4}=\frac{24}{44} \\ & \frac{8}{11}=\frac{8 \times 3}{11 \times 3}=\frac{24}{33} \\ & \frac{3}{5}=\frac{3 \times 8}{5 \times 8}=\frac{24}{40} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{24}{33}>\frac{24}{40}>\frac{24}{44}>\frac{24}{240} \\ & \text { or } \frac{8}{11}>\frac{3}{5}>\frac{6}{11}>\frac{1}{10} \end{aligned} $
Hence, $\frac{8}{11}, \frac{3}{5}, \frac{6}{11}, \frac{1}{10}$ are in descending order.
View full question & answer→Question 324 Marks
Arrange the given fraction in descending order by making numerator equal:
$\frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11}$
Answer$ \frac{3}{7}, \frac{4}{8}, \frac{5}{7} \text { and } \frac{8}{11} $
L.C.M. of numerators $3,4,5$ and $8=120$
$ \begin{aligned} & \therefore \frac{3}{7}=\frac{3 \times 40}{7 \times 40}=\frac{120}{280} \\ & \frac{4}{9}=\frac{4 \times 30}{9 \times 30}=\frac{120}{270} \\ & \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{8}{11}=\frac{8 \times 15}{11 \times 15}=\frac{120}{165} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{120}{165}>\frac{120}{168}>\frac{120}{270}>\frac{120}{280} \\ & \text { or } \frac{8}{11}>\frac{5}{7}>\frac{4}{9}>\frac{3}{7} \end{aligned} $
Hence, $\frac{8}{11}, \frac{5}{7}, \frac{4}{9}, \frac{3}{7}$ are in descending order.
View full question & answer→Question 334 Marks
Arrange the given fraction in descending order by making numerator equal:
$\frac{5}{6}, \frac{4}{15}, \frac{8}{9} \text { and } \frac{1}{3}$
Answer$ \frac{5}{6}, \frac{4}{15}, \frac{8}{9} \text { and } \frac{1}{3} $
L.C.M. of numerators 5, 4, 8 and $1=40$
$ \begin{aligned} & \therefore \frac{5}{6}=\frac{5 \times 8}{6 \times 8}=\frac{40}{48} \\ & \frac{4}{15}=\frac{4 \times 10}{15 \times 10}=\frac{40}{150} \end{aligned} $
$ \frac{8}{9}=\frac{8 \times 5}{9 \times 5}=\frac{40}{45} $
$ \frac{1}{3}=\frac{1 \times 40}{3 \times 40}=\frac{40}{120} $
From above we see that
$ \begin{aligned} & \frac{40}{45}>\frac{40}{48}>\frac{40}{120}>\frac{40}{150} \\ & \therefore \frac{8}{9}>\frac{5}{6}>\frac{1}{3}>\frac{4}{15} \end{aligned} $
Hence, $\frac{8}{9}, \frac{5}{6}, \frac{1}{3}, \frac{4}{15}$ are in descending order.
View full question & answer→Question 344 Marks
Put the given fraction in ascending order by making denominator equal:
$\frac{5}{7}, \frac{3}{8}, \frac{9}{14} \text { and } \frac{20}{21}$
Answer$ \frac{5}{7}, \frac{3}{8}, \frac{9}{14} \text { and } \frac{20}{21} $
L.C.M. of denominators 7, 8, 14 and $21=168$
$ \begin{aligned} & \therefore \frac{5}{7}=\frac{5 \times 24}{7 \times 24}=\frac{120}{168} \\ & \frac{3}{8}=\frac{3 \times 21}{8 \times 21}=\frac{63}{168} \\ & \frac{9}{14}=\frac{9 \times 12}{14 \times 12}=\frac{108}{168} \\ & \frac{20}{21}=\frac{20 \times 8}{21 \times 8}=\frac{160}{168} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{63}{168}<\frac{108}{168}<\frac{120}{168}<\frac{160}{168} \\ & \text { or } \frac{3}{8}<\frac{9}{14}<\frac{5}{7}<\frac{20}{21} \end{aligned} $
Hence, $\frac{3}{8}, \frac{9}{14}, \frac{5}{7}$ and $\frac{20}{21}$ are in ascending order.
View full question & answer→Question 354 Marks
Put the given fraction in ascending order by making denominator equal:
$\frac{5}{6}, \frac{7}{8}, \frac{11}{12} \text { and } \frac{3}{10}$
Answer$ \frac{5}{6}, \frac{7}{8}, \frac{11}{12} \text { and } \frac{3}{10} $
L.C.M. of denominators $6,8,12$ and $10=240$
$ \begin{aligned} & \therefore \frac{5}{6}=\frac{5 \times 40}{6 \times 40}=\frac{200}{240} \\ & \frac{7}{8}=\frac{7 \times 30}{8 \times 30}=\frac{210}{240} \\ & \frac{11}{12}=\frac{11 \times 20}{12 \times 20}=\frac{220}{240} \\ & \frac{3}{10}=\frac{3 \times 24}{10 \times 24}=\frac{72}{240} \end{aligned} $
It is clear from the above that
$ \begin{aligned} & \frac{72}{240}<\frac{200}{240}<\frac{210}{240}<\frac{220}{240} \\ & \text { or } \frac{3}{10}<\frac{5}{6}<\frac{7}{8}<\frac{11}{12} \end{aligned} $
Hence, $\frac{3}{10}, \frac{5}{6}, \frac{7}{8}$ and $\frac{11}{12}$ are in ascending order.
View full question & answer→Question 364 Marks
Convert given fraction into a fraction with equal denominator:
$\frac{4}{5}, \frac{17}{20}, \frac{23}{40} \text { and } \frac{11}{16}$
Answer$ \ln \frac{4}{5}, \frac{17}{20}, \frac{23}{40} \text { and } \frac{11}{16} $
L.C.M of $5,20,40$ and $16=80$
$ \therefore \frac{4}{5}=\frac{4 \times 16}{5 \times 16}=\frac{64}{80} $
$ \frac{17}{20}=\frac{17 \times 4}{20 \times 4}=\frac{68}{80} $
$ \frac{23}{40}=\frac{23 \times 2}{40 \times 2}=\frac{46}{80} $
and $\frac{11}{16}=\frac{11 \times 5}{16 \times 5}=\frac{55}{80}$
Hence, the required fractions are $\frac{64}{80}, \frac{68}{80}, \frac{46}{80}$ and $\frac{55}{80}$.
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