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 ?

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

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}\$

Solve Study Textbooks Guides

Use app

>>Class 6
>>Maths
>>Fractions
>>Operations on Fractions
>> $Adding and Subtracting Like...$

Adding and Subtracting Like Fractions

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

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}]\$

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

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)]\$

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\$