Q: Solve:\$4 \frac{2}{3} + 3 \frac{1}{4}\$

\$\frac{(3 \times 4) + 2}{3} + \frac{(3 \times 4) + 1}{4}\$\$= \frac{14}{3} + \frac{13}{4}\$\$= \frac{(14 \times 4) + (13 \times 3)}{12} = \frac{56 + 39}{12} = \frac{95}{12}\$

Q: Divide the sum of \$(65/12)\$ and \$(8/3)\$ by their difference.

First we have to find the sum \$(65/12)+(8/3)\$\$LCM\$ of \$12\$ and \$3\$ is \$12\$Then,\$(65\times 1)/(12\times 1)=(65/12)\$and \$(8\times 4)/(3\times 4)=(32/12)\$Now,\$=(65+32)/12\$\$=(97/12)\$Now find the difference of \$(65/12)-(8/3)\$\$LCM\$ of \$12\$ and \$3\$ is \$12\$Then,\$(65\times 1)/(12\times 1)=(65/12)\$\$(8\times 4)/(3\times 4)=(32/12)\$Now,\$=(65-32)/12\$\$=(33/12)\$Now divide \$(97/12)\div (33/12)\$\$=(97\times 12)\times (12/33)\$\$=(97\times 12)/(12\times 33)\$\$=(97\times 1)/(1\times 33)\$\$=(97/33)\$

Q: Simplify: (2/3)+(5/6)-(1/9)

LCM of \$3,6,9=18\$Now let us change each of the given fraction into an equivalent fraction having \$18\$ as the denominator.\$(2/3)x(6/6)=(12/18)\$and \$(5/6)x(3/3)=(15/18)\$and \$(1/9)x(2/2)=(2/18)\$Now,\$=(12/18)+(15/18)-(2/18)\$\$=(12+15-2)/18\$\$=(27-2)/18\$\$=(25/18)\$\$=[1(7/18)]\$

Q: The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ?(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

GivenThe length of wingspan of Blue jay \$= \dfrac{41}{100}\,m\$The length of wingspan of Golden eagle \$= 2\dfrac{1}{2} \ m= \dfrac{5}{2} \,m\$The length of wingspan of Albatross \$= 3\dfrac{3}{5}\ m=\dfrac{18}5\ m\$The length of wingspan of Sea gull \$= 1\dfrac{7}{10}\ m=\dfrac{17}{10}\ m\$(a) Difference between the lengths of wingspan of Albatross and Sea gull\$=\left(\dfrac{18}{5} - \dfrac{17}{10}\right)\ \ m\\=\dfrac{36 - 17}{10} \ \ m\\= \dfrac{19}{10} \ m\\= 1\dfrac{9}{10}\ m\$(ii) Difference between the lengths of wingspan of Golden eagle and Blue jay\$= \left(\dfrac{5}{2} - \dfrac{41}{100}\right)\ m\\=\dfrac{250 - 41}{100} \ m\\= \dfrac{209}{100} \ m\\= 2\dfrac{9}{100}\ m\$

Q: Solve:\$\frac{4}{3} - \frac{1}{2}\$

\$\frac{(4 \times 2) - (1 \times 3)}{6} = \frac{8 - 3}{6} = \frac{5}{6}\$

Q: Solve\${\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 3}} \right)^3} - {\left( {{5 \over 6}} \right)^3}\$

\$\bigg(\dfrac{1}{2}\bigg)^{3}+\bigg(\dfrac{1}{3}\bigg)^{3}-\bigg(\dfrac{5}{6}\bigg)^{3}=\dfrac{3^{3}+2^{3}-5^{3}}{6^{3}}=\dfrac{27+8-125}{216}=-\dfrac{90}{216}=-\dfrac{5}{12}\$

Question 1

$4+\dfrac{1}{2}+3\dfrac{1}{2}+9\dfrac{6}{7}$.

Accepted Answer

Toppr Ask Answer

Question 2

Solve:$4 \frac{2}{3} + 3 \frac{1}{4}$

Accepted Answer

$\frac{(3 \times 4) + 2}{3} + \frac{(3 \times 4) + 1}{4}$$= \frac{14}{3} + \frac{13}{4}$$= \frac{(14 \times 4) + (13 \times 3)}{12} = \frac{56 + 39}{12} = \frac{95}{12}$

Question 3

Satypal walks $ \displaystyle \frac{2}{3}   $km from a place P towards east and then there  $ \displaystyle 1\frac{5}{7}   $km towards west Where will he be now from P ?

Accepted Answer

Let us denote the distance travelled towards east by positive sign so thus distance of Satypal from the point P would be  $ \displaystyle \frac{2}{3}   $+ $ \displaystyle (-1\frac{5}{7})   $ $= \displaystyle \frac{2}{3}   $+ $ \displaystyle \frac{-12}{7}   $ $= \displaystyle \frac{2\times 7}{3\times 7}   $+ $ \displaystyle \frac{(-12)\times 3}{7\times 3}   $ $= \displaystyle \frac{14-36}{21}   $ $= \displaystyle \frac{-22}{ 21 }   $ $= \displaystyle -1\frac{1}{21}   $Since it is negative it means Satypal is at a distance  $ \displaystyle1 \frac{1}{21}   $km towards west of P

Question 4

Divide the sum of $(65/12)$ and $(8/3)$ by their difference.

Accepted Answer

First we have to find the sum $(65/12)+(8/3)$$LCM$ of $12$ and $3$ is $12$Then,$(65\times 1)/(12\times 1)=(65/12)$and $(8\times 4)/(3\times 4)=(32/12)$Now,$=(65+32)/12$$=(97/12)$Now find the difference of $(65/12)-(8/3)$$LCM$ of $12$ and $3$ is $12$Then,$(65\times 1)/(12\times 1)=(65/12)$$(8\times 4)/(3\times 4)=(32/12)$Now,$=(65-32)/12$$=(33/12)$Now divide $(97/12)\div (33/12)$$=(97\times 12)\times (12/33)$$=(97\times 12)/(12\times 33)$$=(97\times 1)/(1\times 33)$$=(97/33)$

Question 5

Simplify: (2/3)+(5/6)-(1/9)

Accepted Answer

LCM of $3,6,9=18$Now let us change each of the given fraction into an equivalent fraction having $18$ as the denominator.$(2/3)x(6/6)=(12/18)$and $(5/6)x(3/3)=(15/18)$and $(1/9)x(2/2)=(2/18)$Now,$=(12/18)+(15/18)-(2/18)$$=(12+15-2)/18$$=(27-2)/18$$=(25/18)$$=[1(7/18)]$

Question 6

Find the sum(8/9)+(7/12)

Accepted Answer

For addition of two unlike fractions,first change them to the like fractions.LCM of 9,12=36Now,let us change each of the given fraction into an equivalent fraction having 36 as the denominator.$\dfrac{8}{9}\times \dfrac{4}{4}=\dfrac{32}{36}$$\dfrac{7}{12}\times \dfrac{3}{3}=\dfrac{21}{36}$Now.add the like fractions,$=\dfrac{32}{36}$$+\dfrac{21}{36}$$=\dfrac{32+21}{36}$$=\dfrac{53}{36}$$=[1\dfrac{17}{36}]$

Question 7

Simplify: $[8(5/6)]-[3(3/8)]+[1(7/12)]$

Accepted Answer

First we convert each mixed fraction into an improper fraction.$\Rightarrow [8(5/6)]=(53/6)$also $\Rightarrow [3(3/8)]=(27/8)$and $\Rightarrow [1(7/12)]=(19/12)$Then,$(53/6)-(27-8)-(19/12)$LCM of $6,8,12=24$Now, let us change each of the given fraction into an equivalent fraction having $24 $ as the denominator.$=[(53/6)\times(4/4)]=(212/24)$also $=[(27/8)\times (3/3)]=(81/24)$and $=[(19/12)\times(2/2)]=(38/24)$Now,$=(212/24)-(81/24)-(38/24)$$=[(212-81+38)/24]$$=[(250-81/24)]$$=(169/24)$$=[7(1/24)]$

Question 8

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:
(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

Accepted Answer

GivenThe length of wingspan of Blue jay $= \dfrac{41}{100}\,m$The length of wingspan of Golden eagle $= 2\dfrac{1}{2} \ m= \dfrac{5}{2} \,m$The length of wingspan of Albatross $= 3\dfrac{3}{5}\ m=\dfrac{18}5\ m$The length of wingspan of Sea gull $= 1\dfrac{7}{10}\ m=\dfrac{17}{10}\ m$(a) Difference between the lengths of wingspan of Albatross and Sea gull$=\left(\dfrac{18}{5} - \dfrac{17}{10}\right)\ \ m\=\dfrac{36 - 17}{10} \ \ m\= \dfrac{19}{10} \ m\= 1\dfrac{9}{10}\ m$(ii) Difference between the lengths of wingspan of Golden eagle and Blue jay$= \left(\dfrac{5}{2} - \dfrac{41}{100}\right)\ m\=\dfrac{250 - 41}{100} \ m\= \dfrac{209}{100} \ m\= 2\dfrac{9}{100}\ m$

Question 9

Solve:$\frac{4}{3} - \frac{1}{2}$

Accepted Answer

$\frac{(4 \times 2) - (1 \times 3)}{6} = \frac{8 - 3}{6} = \frac{5}{6}$

Question 10

Solve${\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 3}} \right)^3} - {\left( {{5 \over 6}} \right)^3}$

Accepted Answer

$\bigg(\dfrac{1}{2}\bigg)^{3}+\bigg(\dfrac{1}{3}\bigg)^{3}-\bigg(\dfrac{5}{6}\bigg)^{3}=\dfrac{3^{3}+2^{3}-5^{3}}{6^{3}}=\dfrac{27+8-125}{216}=-\dfrac{90}{216}=-\dfrac{5}{12}$

Addition and Subtraction of unlike fractions

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Adding and Subtracting Unlike Fractions

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Adding and Subtracting Unlike Fractions

$4 + \frac{1}{2} + 3 \frac{1}{2} + 9 \frac{6}{7}$ .

Solve:
$4 \frac{2}{3} + 3 \frac{1}{4}$

Satypal walks $\frac{2}{3}$ km from a place P towards east and then there $1 \frac{5}{7}$ km towards west Where will he be now from P ?

Divide the sum of $(65 / 12)$ and $(8 / 3)$ by their difference.

Simplify: (2/3)+(5/6)-(1/9)

Find the sum
(8/9)+(7/12)

Simplify: $[8 (5 / 6)] - [3 (3 / 8)] + [1 (7 / 12)]$

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:
(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

Solve:
$\frac{4}{3} - \frac{1}{2}$

Solve
$(\frac{1}{2})^{3} + (\frac{1}{3})^{3} - (\frac{5}{6})^{3}$

Addition and Subtraction of unlike fractions

Language

Rate

REVISE WITH CONCEPTS

Adding and Subtracting Unlike Fractions

QUICK SUMMARY WITH STORIES

Adding and subtracting fractions

Adding and Subtracting Unlike Fractions

4+21​+321​+976​.

Solve: 432​+341​

Satypal walks 32​km from a place P towards east and then there 175​km towards west Where will he be now from P ?

Divide the sum of (65/12) and (8/3) by their difference.

Simplify: (2/3)+(5/6)-(1/9)

Find the sum (8/9)+(7/12)

Simplify: [8(5/6)]−[3(3/8)]+[1(7/12)]

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below: (a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ? (b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

Solve: 34​−21​

Solve(21​)3+(31​)3−(65​)3

$4 + \frac{1}{2} + 3 \frac{1}{2} + 9 \frac{6}{7}$ .

Solve:
$4 \frac{2}{3} + 3 \frac{1}{4}$

Satypal walks $\frac{2}{3}$ km from a place P towards east and then there $1 \frac{5}{7}$ km towards west Where will he be now from P ?

Divide the sum of $(65 / 12)$ and $(8 / 3)$ by their difference.

Find the sum
(8/9)+(7/12)

Simplify: $[8 (5 / 6)] - [3 (3 / 8)] + [1 (7 / 12)]$

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:
(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

Solve:
$\frac{4}{3} - \frac{1}{2}$

Solve
$(\frac{1}{2})^{3} + (\frac{1}{3})^{3} - (\frac{5}{6})^{3}$