Chapter 4: Fractions and Decimals

Ans: The distance covers by Katrina in the morning

$= 8\dfrac{3}{4} = \dfrac{{(8 \times 3) + 3}}{4} = \dfrac{{24 + 3}}{4} = \dfrac{{27}}{4}$

The distance covers by Katrina in the evening

$= 8\dfrac{3}{4} = \dfrac{{(8 \times 3) + 3}}{4} = \dfrac{{32 + 3}}{4} = \dfrac{{35}}{4}$

Total distance travelled by her

$= \dfrac{{13}}{2}{\text{km}} + \dfrac{{35}}{4}{\text{km}}$

$= \dfrac{{13 \times 2}}{{2 \times 2}}{\text{km}} + \dfrac{{35}}{4}{\text{km}}$

${\text{L}}{\text{.C}}{\text{.M of }}2{\text{ and }}4 = 4$

$= \dfrac{{26}}{4}{\text{km}} + \dfrac{{35}}{4}{\text{km}} = \dfrac{{(26 + 35){\text{km}}}}{4}$

$= \dfrac{{61}}{4}{\text{km}} = 15\dfrac{1}{4}{\text{km}}$

Distance travelled altogether by Katrina in a day is $15\dfrac{1}{4}{\text{km}}$.

Ans: Let, the number of parts in which the rectangle has been divided be $x$

According to the question, $\dfrac{{16}}{x} = \dfrac{1}{4}$

Cross- multiplication, $16 \times 4 = 1 \times x$

$x = 64$

64 are the parts in which the rectangle has been divided.

Ans: Given, Grip size of a tennis racquet is $11\dfrac{9}{{80}}{\text{cm}}$

Now, rewrite the mixed fraction

$\Rightarrow 11\dfrac{9}{{80}} = \dfrac{{(11 \times 80) + 9}}{{80}}$

$= \dfrac{{880 + 9}}{{80}}$

$= \dfrac{{889}}{{80}}$ 

$\dfrac{{889}}{{80}}$ this is a improper fraction.

Solution coming soon.

Ans: Given,

Mr. Rajan got a job at the age of 24 years

He got retired from the job at the age of $60$ years

Total working year $= (60 - ( - 24)) = 36$year

The required fraction $\dfrac{{36}}{{60}} = \dfrac{{36 \div 12}}{{60 \div 12}} = \dfrac{3}{5}$

The fraction of his age till retirement was he in the job is $\dfrac{3}{5}$

Ans: Given, 

The food we eat remains in the stomach for a maximum of 4 hours

Total number of hours in a day is $24.$

The required fraction 

$\dfrac{4}{{24}} = \dfrac{{4 \div 4}}{{24 \div 4}} = \dfrac{1}{6}$

The fraction of the day food remains in the stomach is $\dfrac{1}{6}$

Ans: Let ${\mathbf{x}}$ should be added to ${\mathbf{25}}.{\mathbf{5}}$ to get ${\mathbf{50}}$

Now,

$25.5 + x = 50$

$x = 50 - 25.5 = 24.5$ 

The required number is $24.5$

The number that should be added to $25.5$ to get 50 is $24.5$

Ans: Convert all the data in a signal unit, ${\text{kg}}$

$1{\kern 1pt} {\kern 1pt} kg = \dfrac{1}{{1000}}g$

Given,

Alok purchased $1{\text{kg}}200{\text{g}}$ potatoes, 

$250{\text{g}}$ dhania, 

$5{\text{kg}}300{\text{g}}$ onion, 

$500{\text{g}}$ palak,

$2{\text{kg}}600{\text{g}}$ tomatoes 

$= 1{\text{kg}}200{\text{g}} = 1{\text{kg}} + \dfrac{{200}}{{1000}}{\text{kg}} = 1.200{\text{kg}}$ of potatoes

$= 0{\text{kg}}250{\text{g}} = 0{\text{kg}} + \dfrac{{250}}{{1000}}{\text{kg}} = 0.250{\text{kg}}$ of dhania

$= 5{\text{kg}}300{\text{g}} = 5{\text{kg}} + \dfrac{{300}}{{1000}}{\text{kg}} = 5.300{\text{kg}}$ of onion

$= 0{\text{kg}}500{\text{g}} = 0{\text{kg}} + \dfrac{{500}}{{1000}}{\text{kg}} = 0.500{\text{kg}}$ Of palak

$= 2{\text{kg}}600{\text{g}} = 2{\text{kg}} + \dfrac{{600}}{{1000}}{\text{kg}} = 2.600{\text{kg}}$ Of tomatoes

Total weight is $1.200 + 0.25 + 5.300 + 0.500 + 2.600 = 9.850$

Total weight of the purchase is $9.850{\text{kg}}$

The total weight of purchase of Alok in kilograms is $9.850{\text{kg}}$.

Ans: Since, all the decimals are already given in like fractions, 

$0.011,1.001,0.101,0.110$

Arranging them in ascending order, we get $0.011,0.101,0.110,1.001$

Ans: We have, $20.02$ and $2.002$

To add the above decimals, we must convert them into like decimals first. 

Writing $20.020$ and $2.002$ in a column

So,

$20.020 + 2.002 = 22.022$

Addition of $20.020$ and $2.002$ is $22.022$