Fractions — Exercise 3(E)

A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is:

(i) the length of AB to that of CD?

(ii) $\dfrac{1}{2}$ the length of AB to that of $\dfrac{1}{3}$ of CD?

(iii) $\dfrac{1}{5}$ of CD to that of AB?

Answer

Given,

Length of line AB = 6 cm

Length of line CD = 15 cm

(i) Required fraction $= \dfrac{AB}{CD} = \dfrac{6}{15} = \dfrac{2}{5}$

Hence, the required fraction is $\dfrac{2}{5}$.

(ii) Solving,

$\dfrac{1}{2} \text{ of } AB = \dfrac{1}{2} \times 6 = 3 \text { cm and }\\[1em] \dfrac{1}{3} \text{ of CD }= \dfrac{1}{3} \times 15 = 5 \text{cm}$

Required fraction $= \dfrac{3}{5}$

Hence, the required fraction is $\dfrac{3}{5}$.

(iii) $\dfrac{1}{5}\text{ of CD }= \dfrac{1}{5} \times 15 = 3$ cm

Required fraction = $\dfrac{3}{6} = \dfrac{1}{2}$

Hence, the required fraction is $\dfrac{1}{2}$.

Subtract $\left(\dfrac{2}{7} - \dfrac{5}{21}\right)$ from the sum of $\dfrac{3}{4}, \dfrac{5}{7} \text{ and } \dfrac{7}{12}$.

Answer

Solving, $\left(\dfrac{2}{7} - \dfrac{5}{21}\right)$ :

$\dfrac{2}{7} - \dfrac{5}{21} = \dfrac{6 - 5}{21} = \dfrac{1}{21}$

Sum of $\dfrac{3}{4}, \dfrac{5}{7} \text{ and } \dfrac{7}{12}$ (LCM of 4, 7 and 12 = 84):

$\Rightarrow \dfrac{3}{4} + \dfrac{5}{7} + \dfrac{7}{12}\\[1em] = \dfrac{63}{84} + \dfrac{60}{84} + \dfrac{49}{84} \\[1em] = \dfrac{172}{84}\\[1em] = \dfrac{43}{21}$

Now subtracting $\left(\dfrac{2}{7} - \dfrac{5}{21}\right)$ from the sum of $\dfrac{3}{4}, \dfrac{5}{7} \text{ and } \dfrac{7}{12}$ :

$\dfrac{43}{21} - \dfrac{1}{21} = \dfrac{42}{21} = 2$

Hence, the required value is 2.

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, $5\dfrac{1}{4} \text { kg }, 9\dfrac{1}{2} \text { kg and }9\dfrac{3}{4}$ kg respectively.

(i) How many kg did he sell?

(ii) How many kg are still left in the sack?

Answer

Given,

Total weight of the sack of potatoes = 120 kg

Portions sold = 6 kg, $5\dfrac{1}{4}$ kg, $9\dfrac{1}{2}$ kg and $9\dfrac{3}{4}$ kg

(i) Total sold = sum of all the portions sold

$\Rightarrow 6 + 5\dfrac{1}{4} + 9\dfrac{1}{2} + 9\dfrac{3}{4} \\[1em] = 6 + \dfrac{21}{4} + \dfrac{19}{2} + \dfrac{39}{4}$

LCM of 4 and 2 = 4.

$= \dfrac{24}{4} + \dfrac{21}{4} + \dfrac{38}{4} + \dfrac{39}{4} \\[1em] = \dfrac{122}{4} = \dfrac{61}{2} \\[1em] = 30\dfrac{1}{2} \text{ kg}$

Hence, he sold $30\dfrac{1}{2}$ kg.

(ii) Potatoes left

$= 120 - 30\dfrac{1}{2} \\[1em] = \dfrac{240}{2} - \dfrac{61}{2} \\[1em] = \dfrac{179}{2} = 89\dfrac{1}{2} \text{ kg}$

Hence, $89\dfrac{1}{2}$ kg are still left in the sack.

If a boy works for six consecutive days for 8 hours, $7\dfrac{1}{2}$ hours, $8\dfrac{1}{4}$ hours, $6\dfrac{1}{4}$ hours, $6\dfrac{3}{4}$ hours and 7 hours respectively, how much money will he earn at the rate of ₹ 36 per hour?

Answer

Given,

Hours worked on six consecutive days = 8 hours, $7\dfrac{1}{2}$ hours, $8\dfrac{1}{4}$ hours, $6\dfrac{1}{4}$ hours, $6\dfrac{3}{4}$ hours and 7 hours

Rate of pay = ₹ 36 per hour

Total hours worked = sum of the hours worked on the six days

$= 8 + 7\dfrac{1}{2} + 8\dfrac{1}{4} + 6\dfrac{1}{4} + 6\dfrac{3}{4} + 7 \\[1em] = 8 + \dfrac{15}{2} + \dfrac{33}{4} + \dfrac{25}{4} + \dfrac{27}{4} + 7$

LCM of 2 and 4 = 4.

$= \dfrac{32}{4} + \dfrac{30}{4} + \dfrac{33}{4} + \dfrac{25}{4} + \dfrac{27}{4} + \dfrac{28}{4} \\[1em] = \dfrac{175}{4} = 43\dfrac{3}{4} \text{ hours}$

Money earned = No. of hours worked × Rate per hour

$= 43\dfrac{3}{4} \times 36 = \dfrac{175}{4} \times 36 \\[1em] = \dfrac{175 \times 36}{4} = \dfrac{6300}{4} = 1575$

Hence, the boy will earn ₹ 1,575.

A student bought $4\dfrac{1}{3}$ m of yellow ribbon, $6\dfrac{1}{6}$ m of red ribbon and $3\dfrac{2}{9}$ m of blue ribbon for decorating a room. How many metres of ribbon did he buy?

Answer

Given,

Yellow ribbon bought = $4\dfrac{1}{3}$ m

Red ribbon bought = $6\dfrac{1}{6}$ m

Blue ribbon bought = $3\dfrac{2}{9}$ m

Total ribbon bought = Yellow ribbon + Red ribbon + Blue ribbon

Total ribbon $= 4\dfrac{1}{3} + 6\dfrac{1}{6} + 3\dfrac{2}{9} = \dfrac{13}{3} + \dfrac{37}{6} + \dfrac{29}{9}$

LCM of 3, 6 and 9 = 18.

$= \dfrac{78}{18} + \dfrac{111}{18} + \dfrac{58}{18} \\[1em] = \dfrac{247}{18} = 13\dfrac{13}{18} \text{ metres}$

Hence, the student bought $13\dfrac{13}{18}$ metres of ribbon.

In a business, Ram and Deepak invest $\dfrac{3}{5} \text{ and } \dfrac{2}{5}$ of the total investment. If ₹ 40,000 is the total investment, calculate the amount invested by each.

Answer

Given,

Total investment = ₹ 40,000

Ram's share of the investment = $\dfrac{3}{5}$ of the total

Deepak's share of the investment = $\dfrac{2}{5}$ of the total

Ram's investment = $\dfrac{3}{5}$ × Total investment

Ram's investment $= \dfrac{3}{5} \times 40000 = \dfrac{3 \times 40000}{5} = 24000$

Deepak's investment = $\dfrac{2}{5}$ × Total investment

Deepak's investment $= \dfrac{2}{5} \times 40000 = \dfrac{2 \times 40000}{5} = 16000$

Hence, Ram's investment is ₹ 24,000 and Deepak's investment is ₹ 16,000.

Geeta had 30 problems for home work. She worked out $\dfrac{2}{3}$ of them. How many problems were still left to be worked out by her?

Answer

Given,

Total problems for homework = 30

Problems worked out = $\dfrac{2}{3}$ of the total

Problems worked out = $\dfrac{2}{3}$ × Total problems

Problems worked out $= \dfrac{2}{3} \times 30 = 20$

Problems still left = Total problems - Problems worked out

Problems still left = 30 - 20 = 10

Hence, 10 problems were still left to be worked out.

A picture was marked at ₹ 90. It was sold at $\dfrac{3}{4}$ of its marked price. What was the sale price?

Answer

Given,

Marked price of the picture = ₹ 90

The picture was sold at $\dfrac{3}{4}$ of its marked price.

Sale price = $\dfrac{3}{4}$ × Marked price

Sale price = $\dfrac{3}{4} \times 90 = \dfrac{3 \times 90}{4} = \dfrac{270}{4} = 67.50$

Hence, the sale price was ₹ 67.50.

Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed $10\dfrac{1}{2}$ kg?

Answer

Given,

Number of parcels = 15

Weight of each parcel = $10\dfrac{1}{2}$ kg

Total weight = Number of parcels × Weight of each parcel

Total weight $= 15 \times 10\dfrac{1}{2} = 15 \times \dfrac{21}{2} = \dfrac{315}{2} = 157.5$ kg

Hence, the total weight of the parcels is 157.5 kg.

A rope is $25\dfrac{1}{2}$ m long. How many pieces each of $1\dfrac{1}{2}$ m length can be cut out from it?

Answer

Given,

Total length of the rope = $25\dfrac{1}{2}$ m

Length of each piece = $1\dfrac{1}{2}$ m

Number of pieces = Total length ÷ Length of each piece

Number of pieces $= 25\dfrac{1}{2} \div 1\dfrac{1}{2} = \dfrac{51}{2} \div \dfrac{3}{2}$

= $\dfrac{51}{2} \times \dfrac{2}{3} = \dfrac{51}{3} = 17$

Hence, 17 pieces can be cut out from the rope.

The heights of two vertical poles, above the earth's surface, are $14\dfrac{1}{4} \text { m and }22\dfrac{1}{3}$ m respectively. How much higher is the second pole as compared with the height of the first pole?

Answer

Given,

Height of the first pole = $14\dfrac{1}{4}$ m

Height of the second pole = $22\dfrac{1}{3}$ m

Difference in heights = Height of second pole - Height of first pole

Difference in heights $= 22\dfrac{1}{3} - 14\dfrac{1}{4} = \dfrac{67}{3} - \dfrac{57}{4}$

LCM of 3 and 4 = 12.

$= \dfrac{268}{12} - \dfrac{171}{12} = \dfrac{97}{12} = 8\dfrac{1}{12} \text{ m}$

Hence, the second pole is $8\dfrac{1}{12}$ m higher than the first pole.

Vijay weighed $65\dfrac{1}{2}$ kg. He gained $1\dfrac{2}{5}$ kg during the first week, $1\dfrac{1}{4}$ kg during the second week, but lost $\dfrac{5}{16}$ kg during the third week. What was his weight after the third week?

Answer

Given,

Vijay's initial weight = $65\dfrac{1}{2}$ kg

Weight gained in the first week = $1\dfrac{2}{5}$ kg

Weight gained in the second week = $1\dfrac{1}{4}$ kg

Weight lost in the third week = $\dfrac{5}{16}$ kg

Weight after third week = Initial weight + Gain in first week + Gain in second week - Loss in third week

Weight after third week $= 65\dfrac{1}{2} + 1\dfrac{2}{5} + 1\dfrac{1}{4} - \dfrac{5}{16}$

= $\dfrac{131}{2} + \dfrac{7}{5} + \dfrac{5}{4} - \dfrac{5}{16}$

LCM of 2, 5, 4 and 16 = 80.

$= \dfrac{5240}{80} + \dfrac{112}{80} + \dfrac{100}{80} - \dfrac{25}{80} \\[1em] = \dfrac{5240 + 112 + 100 - 25}{80} = \dfrac{5427}{80} = 67\dfrac{67}{80} \text{ kg}$

Hence, Vijay's weight after the third week was $67\dfrac{67}{80}$ kg.

A man spends $\dfrac{2}{5}$ of his salary on food and $\dfrac{3}{10}$ on house rent, electricity, etc. What fraction of his salary is still left with him?

Answer

Given,

Fraction of salary spent on food = $\dfrac{2}{5}$

Fraction of salary spent on house rent, electricity, etc. = $\dfrac{3}{10}$

Total fraction spent = Fraction on food + Fraction on house rent, electricity, etc.

Total fraction spent $= \dfrac{2}{5} + \dfrac{3}{10} = \dfrac{4}{10} + \dfrac{3}{10} = \dfrac{7}{10}$

Fraction left $= 1 - \dfrac{7}{10} = \dfrac{10 - 7}{10} = \dfrac{3}{10}$

Hence, $\dfrac{3}{10}$ of his salary is still left with him.

A man spends $\dfrac{2}{5}$ of his salary on food and $\dfrac{3}{10}$ of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?

Answer

Given,

Fraction of salary spent on food = $\dfrac{2}{5}$

Fraction of the remaining salary spent on house rent, electricity, etc. = $\dfrac{3}{10}$

Fraction spent on food = $\dfrac{2}{5}$

Remaining = 1 - $\dfrac{2}{5} = \dfrac{3}{5}$

Fraction spent on house rent, etc. $= \dfrac{3}{10} \text{ of } \dfrac{3}{5} = \dfrac{3}{10} \times \dfrac{3}{5} = \dfrac{9}{50}$

Fraction left = $\dfrac{3}{5} - \dfrac{9}{50} = \dfrac{30}{50} - \dfrac{9}{50} = \dfrac{21}{50}$

Hence, $\dfrac{21}{50}$ of his salary is still left with him.

Shyam bought a refrigerator for ₹ 5,000. He paid $\dfrac{1}{10}$ of the price in cash and the rest in 12 equal monthly instalments. How much did he have to pay each month?

Answer

Given,

Price of the refrigerator = ₹ 5,000

Fraction of the price paid in cash = $\dfrac{1}{10}$

Number of equal monthly instalments for the rest = 12

Amount paid in cash = $\dfrac{1}{10}$ × Price

Amount paid in cash = $\dfrac{1}{10} \times 5000 = 500$

Remaining amount = ₹ 5,000 - 500 = ₹ 4,500

Amount paid each month = Remaining amount ÷ Number of instalments

Amount paid each month = $\dfrac{4500}{12} = 375$

Hence, he had to pay ₹ 375 each month.

A lamp post has half of its length in mud and $\dfrac{1}{3}$ of its length in water.

(i) What fraction of its length is above the water?

(ii) If $3\dfrac{1}{3}$ m of the lamp post is above the water, find the whole length of the lamp post.

Answer

Given,

Fraction of the length in mud = $\dfrac{1}{2}$

Fraction of the length in water = $\dfrac{1}{3}$

(i) Fraction in mud and water $= \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3 + 2}{6} = \dfrac{5}{6}$

Fraction above water $= 1 - \dfrac{5}{6} = \dfrac{1}{6}$

Hence, $\dfrac{1}{6}$ of its length is above the water.

(ii) Let the whole length be $x$ m.

$\Rightarrow \dfrac{1}{6} \times x = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{x}{6} = \dfrac{10}{3} \\[1em] \Rightarrow x = \dfrac{10}{3} \times 6 = 20$

Hence, the whole length of the lamp post is 20 m.

I spent $\dfrac{3}{5}$ of my savings and still have ₹ 2,000 left. What were my savings?

Answer

Given,

Fraction of savings spent = $\dfrac{3}{5}$

Amount still left = ₹ 2,000

Fraction of savings left $= 1 - \dfrac{3}{5} = \dfrac{2}{5}$

Let the savings be $x$.

$\dfrac{2}{5} \times x = 2000 \\[1em] \Rightarrow x = 2000 \times \dfrac{5}{2} = 5000$

Hence, my savings were ₹ 5,000.

In a school $\dfrac{4}{5}$ of the children are boys. If the number of girls is 200, find the number of boys.

Answer

Given,

Fraction of children who are boys = $\dfrac{4}{5}$

Number of girls = 200

Fraction of girls $= 1 - \dfrac{4}{5} = \dfrac{1}{5}$

Let the total number of children be $x$.

$\dfrac{1}{5} \times x = 200 \\[1em] \Rightarrow x = 200 \times 5 = 1000$

Number of boys $= \dfrac{4}{5} \times 1000 = 800$

Hence, the number of boys is 800.

If $\dfrac{4}{5}$ of an estate is worth ₹ 42,000, find the worth of the whole estate. Also, find the value of $\dfrac{3}{7}$ of it.

Answer

Given,

Worth of $\dfrac{4}{5}$ of the estate = ₹ 42,000

Let the worth of the whole estate be $x$.

$\dfrac{4}{5} \times x = 42000 \\[1em] \Rightarrow x = 42000 \times \dfrac{5}{4} = 52500$

Worth of the whole estate = ₹ 52,500

Value of $\dfrac{3}{7}\text { of it }= \dfrac{3}{7} \times 52500 = 22500$

Hence, the worth of the whole estate is ₹ 52,500 and the value of $\dfrac{3}{7}$ of it is ₹ 22,500.

After going $\dfrac{3}{4}$ of my journey, I find that I have covered 16 km. How much journey is still left?

Answer

Given,

Fraction of the journey covered = $\dfrac{3}{4}$

Distance covered in that fraction = 16 km

Let the whole journey be $x$ km.

$\dfrac{3}{4} \times x = 16 \\[1em] \Rightarrow x = 16 \times \dfrac{4}{3} = \dfrac{64}{3}$

Journey still left $= x - 16 = \dfrac{64}{3} - 16 = \dfrac{64 - 48}{3} = \dfrac{16}{3} = 5\dfrac{1}{3}$ km

Hence, $5\dfrac{1}{3}$ km of journey is still left.

When Krishna travelled 25 km, he found that $\dfrac{3}{5}$ of his journey was still left. What was the length of the whole journey?

Answer

Given,

Distance travelled = 25 km

Fraction of the journey still left = $\dfrac{3}{5}$

Fraction of journey travelled $= 1 - \dfrac{3}{5} = \dfrac{2}{5}$

Let the whole journey be $x$ km.

$\dfrac{2}{5} \times x = 25 \\[1em] \Rightarrow x = 25 \times \dfrac{5}{2} = \dfrac{125}{2} = 62\dfrac{1}{2}$

Hence, the length of the whole journey is $62\dfrac{1}{2}$ km.

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.

Answer

Given,

Fraction of the land bought by Rajesh = $\dfrac{1}{3}$ of the total

Fraction of the land bought by Manoj = $\dfrac{1}{3}$ of the remaining

Land still left unsold = 600 m²

Let the total area be $x$ m².

Area bought by Rajesh $= \dfrac{1}{3}x$

Remaining $= x - \dfrac{1}{3}x = \dfrac{2}{3}x$

Area bought by Manoj $= \dfrac{1}{3} \text{ of } \dfrac{2}{3}x = \dfrac{1}{3} \times \dfrac{2}{3}x = \dfrac{2}{9}x$

Land left unsold

$= x - \dfrac{1}{3}x - \dfrac{2}{9}x \\[1em] = \dfrac{9x - 3x - 2x}{9} \\[1em] = \dfrac{4x}{9}\\[1em] \dfrac{4x}{9} = 600 \\[1em] \Rightarrow x = 600 \times \dfrac{9}{4} = 1350$

Hence, the total area of the piece of land is 1350 m².

A boy spent $\dfrac{3}{5}$ of his money on buying clothes and $\dfrac{1}{4}$ of the remaining money on buying shoes. If he initially had ₹ 2,400 how much did he spend on shoes?

Answer

Given,

Money the boy initially had = ₹ 2,400

Fraction of money spent on clothes = $\dfrac{3}{5}$

Fraction of the remaining money spent on shoes = $\dfrac{1}{4}$

Money spent on clothes = $\dfrac{3}{5}$ × Total money

Money spent on clothes $= \dfrac{3}{5} \times 2400 = 1440$

Remaining money = 2400 - 1440 = 960

Money spent on shoes = $\dfrac{1}{4}$ × Remaining money

Money spent on shoes $= \dfrac{1}{4} \times 960 = 240$

Hence, he spent ₹ 240 on shoes.

A boy spent $\dfrac{3}{5}$ of his money on buying clothes and $\dfrac{1}{4}$ of his money on buying shoes. If he initially had ₹ 2,400, how much did he spend on shoes?

Answer

Given,

Money the boy initially had = ₹ 2,400

Fraction of money spent on clothes = $\dfrac{3}{5}$

Fraction of money spent on shoes = $\dfrac{1}{4}$

Money spent on shoes $= \dfrac{1}{4}$ of his money $= \dfrac{1}{4} \times 2400 = 600$

Hence, he spent ₹ 600 on shoes.