Rational Numbers — Exercise 1.6

In a bag, there are 20 kg of fruits. If $7\dfrac{1}{6}$ kg of these fruits are oranges, $8\dfrac{2}{3}$ kg of these are apples and the rest are grapes. Find the weight of the grapes in the bag.

Answer

Subtract the total weight of the oranges and the apples from the total weight of the fruits.

Total weight of fruits in the bag = 20 kg

Weight of oranges $= 7\dfrac{1}{6} \text{ kg} = \dfrac{43}{6} \text{ kg}$

Weight of apples $= 8\dfrac{2}{3} \text{ kg} = \dfrac{26}{3} \text{ kg}$

Finding the weight of oranges and apples together,

$\Rightarrow \dfrac{43}{6} + \dfrac{26}{3} \\[1em] [\text{ LCM of } 6 \text{ and } 3 = 6] \\[1em] \Rightarrow \dfrac{43 + 26 \times 2}{6} \\[1em] \Rightarrow \dfrac{43 + 52}{6} \\[1em] \Rightarrow \dfrac{95}{6} \text{ kg}$

Finding the weight of grapes,

$\Rightarrow 20 - \dfrac{95}{6} \\[1em] \Rightarrow \dfrac{20 \times 6 - 95}{6} \\[1em] \Rightarrow \dfrac{120 - 95}{6} \\[1em] \Rightarrow \dfrac{25}{6} = 4\dfrac{1}{6} \text{ kg}$

Hence, the weight of the grapes in the bag is $\mathbf{4\dfrac{1}{6} \text{ kg}}$

The population of a city is 6,63,432. If $\dfrac{1}{2}$ of the population are adult males and $\dfrac{1}{3}$ of the population are adult females, then find the number of children in the city.

Answer

First find the fraction of the population which are children, then find that fraction of the total population.

Total population of the city = 6,63,432

Finding the fraction of the population which are adults,

$\Rightarrow \dfrac{1}{2} + \dfrac{1}{3} \\[1em] [\text{ LCM of } 2 \text{ and } 3 = 6] \\[1em] \Rightarrow \dfrac{3 + 2}{6} \\[1em] \Rightarrow \dfrac{5}{6}$

Finding the fraction of the population which are children,

$\Rightarrow 1 - \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{6 - 5}{6} \\[1em] \Rightarrow \dfrac{1}{6}$

Finding the number of children,

$\Rightarrow \dfrac{1}{6} \times 663432 \\[1em] \Rightarrow 110572$

Hence, there are 1,10,572 children in the city.

In an examination, 400 students appeared. If $\dfrac{2}{3}$ of the boys and all 130 girls passed in examination, then find how many boys failed in the examination?

Answer

First find the number of boys, then find how many of them passed and subtract from the total number of boys.

Total number of students who appeared = 400

Number of girls = 130

Number of boys = 400 - 130 = 270

Finding the number of boys who passed,

$\dfrac{2}{3} \times 270 = 180$

Finding the number of boys who failed,

Number of boys who failed = 270 - 180 = 90

Hence, 90 boys failed in the examination.

A car is moving at the speed of $40\dfrac{2}{3}$ km/h. How much distance will it cover in $\dfrac{9}{10}$ hours?

Answer

We know that distance = speed × time.

Speed of the car $= 40\dfrac{2}{3} \text{ km/h} = \dfrac{122}{3} \text{ km/h}$

Time taken $= \dfrac{9}{10}$ hours

Finding the distance covered,

$\Rightarrow \dfrac{122}{3} \times \dfrac{9}{10} \\[1em] \Rightarrow \dfrac{122 \times 9}{3 \times 10} \\[1em] \Rightarrow \dfrac{1098}{30} \\[1em] \Rightarrow \dfrac{183}{5} = 36\dfrac{3}{5} \text{ km}$

Hence, the car will cover a distance of $\mathbf{36\dfrac{3}{5} \text{ km}}$

Find the area of a square lawn whose one side is $5\dfrac{7}{9}$ m long.

Answer

We know that area of a square = side × side.

Side of the square lawn $= 5\dfrac{7}{9} \text{ m} = \dfrac{52}{9} \text{ m}$

Finding the area of the square lawn,

$\Rightarrow \dfrac{52}{9} \times \dfrac{52}{9} \\[1em] \Rightarrow \dfrac{52 \times 52}{9 \times 9} \\[1em] \Rightarrow \dfrac{2704}{81} \\[1em] \Rightarrow 33\dfrac{31}{81} \text{ sq. m}$

Hence, the area of the square lawn is $\mathbf{33\dfrac{31}{81} \text{ sq. m}}$

Perimeter of a rectangle is $15\dfrac{3}{7}$ m. If the length is $4\dfrac{2}{7}$ m, find its breadth.

Answer

We know that perimeter of a rectangle = 2 (length + breadth).

Perimeter of the rectangle $= 15\dfrac{3}{7} \text{ m} = \dfrac{108}{7} \text{ m}$

Length of the rectangle $= 4\dfrac{2}{7} \text{ m} = \dfrac{30}{7} \text{ m}$

Let the breadth of the rectangle be $b$ metres.

$\Rightarrow \dfrac{108}{7} = 2\left(\dfrac{30}{7} + b\right) \\[1em] \Rightarrow \dfrac{30}{7} + b = \dfrac{108}{7} \div 2 \\[1em] \Rightarrow \dfrac{30}{7} + b = \dfrac{108}{7} \times \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{30}{7} + b = \dfrac{54}{7} \\[1em] \Rightarrow b = \dfrac{54}{7} - \dfrac{30}{7} \\[1em] \Rightarrow b = \dfrac{54 - 30}{7} \\[1em] \Rightarrow b = \dfrac{24}{7} = 3\dfrac{3}{7} \text{ m}$

Hence, the breadth of the rectangle is $\mathbf{3\dfrac{3}{7} \text{ m}}$

Rahul had a rope of $325\dfrac{4}{5}$ m long. He cut off a $150\dfrac{3}{5}$ m long piece, then he divided the rest of the rope into 3 parts of equal length. Find the length of each part.

Answer

First find the length of the remaining rope, then divide it by 3.

Total length of the rope $= 325\dfrac{4}{5} \text{ m} = \dfrac{1629}{5} \text{ m}$

Length of the piece cut off $= 150\dfrac{3}{5} \text{ m} = \dfrac{753}{5} \text{ m}$

Finding the length of the remaining rope,

$\Rightarrow \dfrac{1629}{5} - \dfrac{753}{5} \\[1em] \Rightarrow \dfrac{1629 - 753}{5} \\[1em] \Rightarrow \dfrac{876}{5} \text{ m}$

Finding the length of each part,

$\Rightarrow \dfrac{876}{5} \div 3 \\[1em] \Rightarrow \dfrac{876}{5} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{876}{15} \\[1em] \Rightarrow \dfrac{292}{5} = 58\dfrac{2}{5} \text{ m}$

Hence, the length of each part is $\mathbf{58\dfrac{2}{5} \text{ m}}$

If $3\dfrac{1}{2}$ litre of liquid soap costs ₹$270\dfrac{3}{8}$, then find the cost of 4 litre of liquid soap.

Answer

First find the cost of 1 litre of liquid soap by unitary method, then find the cost of 4 litre.

Quantity of liquid soap $= 3\dfrac{1}{2} \text{ litre} = \dfrac{7}{2} \text{ litre}$

Cost of $\dfrac{7}{2} \text{ litre of liquid soap} = ₹270\dfrac{3}{8} = ₹\dfrac{2163}{8}$

Finding the cost of 1 litre of liquid soap,

$\Rightarrow \dfrac{2163}{8} \div \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{2163}{8} \times \dfrac{2}{7} \\[1em] \Rightarrow \dfrac{4326}{56} \\[1em] \Rightarrow ₹\dfrac{309}{4}$

Finding the cost of 4 litre of liquid soap,

$\Rightarrow \dfrac{309}{4} \times 4 \\[1em] \Rightarrow ₹309$

Hence, the cost of 4 litre of liquid soap is ₹309.

Ramesh earns ₹40,000 per month. He spends $\dfrac{3}{8}$ of the income on food, $\dfrac{1}{5}$ of the remaining on LIC premium and then $\dfrac{1}{2}$ of the remaining on other expenses. How much money is left with him?

Answer

Find the money spent at each stage step by step and subtract it from the amount remaining at that stage.

Monthly income of Ramesh = ₹40,000

Finding the money spent on food,

$\Rightarrow \dfrac{3}{8} \times 40000 \\[1em] \Rightarrow ₹15000$

Remaining money $= ₹40000 - ₹15000 = ₹25000$

Finding the money spent on LIC premium,

$\Rightarrow \dfrac{1}{5} \times 25000 \\[1em] \Rightarrow ₹5000$

Remaining money = $₹25000 - ₹5000 = ₹20000$

Finding the money spent on other expenses,

$\Rightarrow \dfrac{1}{2} \times 20000 \\[1em] \Rightarrow ₹10000$

Money left with him = $₹20000 - ₹10000 = ₹10000$

Hence, ₹10,000 is left with Ramesh.

A, B, C, D and E went to a restaurant for dinner. A paid $\dfrac{1}{2}$ of the bill, B paid $\dfrac{1}{5}$ of the bill and rest of the bill was shared equally by C, D and E. What fraction of the bill was paid by each?

Answer

First find the fraction of the bill paid by A and B together, then divide the remaining fraction equally among C, D and E.

Finding the fraction of the bill paid by A and B together,

$\Rightarrow \dfrac{1}{2} + \dfrac{1}{5} \\[1em] [\text{ LCM of } 2 \text{ and } 5 = 10] \\[1em] \Rightarrow \dfrac{5 + 2}{10} \\[1em] \Rightarrow \dfrac{7}{10}$

Finding the remaining fraction of the bill,

$\Rightarrow 1 - \dfrac{7}{10} \\[1em] \Rightarrow \dfrac{10 - 7}{10} \\[1em] \Rightarrow \dfrac{3}{10}$

Finding the fraction of the bill paid by each of C, D and E,

$\Rightarrow \dfrac{3}{10} \div 3 \\[1em] \Rightarrow \dfrac{3}{10} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{1}{10}$

Hence, A paid $\mathbf{\dfrac{1}{2}}$, B paid $\mathbf{\dfrac{1}{5}}$ and each of C, D and E paid $\mathbf{\dfrac{1}{10}}$ of the bill.

$\dfrac{2}{5}$ of total number of students of a school come by car while $\dfrac{1}{4}$ of students come by bus. All the other students walk to school of which $\dfrac{1}{3}$ walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in that school?

Answer

Let the total number of students in the school be $x$.

Finding the fraction of students who come by car or bus,

$\Rightarrow \dfrac{2}{5} + \dfrac{1}{4} \\[1em] [\text{ LCM of } 5 \text{ and } 4 = 20] \\[1em] \Rightarrow \dfrac{8 + 5}{20} \\[1em] \Rightarrow \dfrac{13}{20}$

Finding the fraction of students who walk to school,

$\Rightarrow 1 - \dfrac{13}{20} \\[1em] \Rightarrow \dfrac{20 - 13}{20} \\[1em] \Rightarrow \dfrac{7}{20}$

So, the number of students who walk to school $= \dfrac{7x}{20}$

According to the given condition,

$\Rightarrow \dfrac{1}{3} \times \dfrac{7x}{20} = 224 \\[1em] \Rightarrow \dfrac{7x}{60} = 224 \\[1em] \Rightarrow x = \dfrac{224 \times 60}{7} \\[1em] \Rightarrow x = 32 \times 60 \\[1em] \Rightarrow x = 1920$

Hence, 1920 students study in that school.

A mother and her two sons got a room constructed for ₹60,000. The elder son contributes $\dfrac{3}{8}$ of his mother's contribution while the younger son contributes $\dfrac{1}{2}$ of his mother's share. How much do the three contribute individually?

Answer

Let the mother's contribution be ₹$x$.

Elder son's contribution $= ₹\dfrac{3x}{8}$ and younger son's contribution $= ₹\dfrac{x}{2}$

According to the given condition,

$\Rightarrow x + \dfrac{3x}{8} + \dfrac{x}{2} = 60000 \\[1em] \Rightarrow \dfrac{8x + 3x + 4x}{8} = 60000 \\[1em] \Rightarrow \dfrac{15x}{8} = 60000 \\[1em] \Rightarrow x = \dfrac{60000 \times 8}{15} \\[1em] \Rightarrow x = 32000$

Finding the elder son's contribution,

$\Rightarrow \dfrac{3}{8} \times 32000 \\[1em] \Rightarrow ₹12000$

Finding the younger son's contribution,

$\Rightarrow \dfrac{1}{2} \times 32000 \\[1em] \Rightarrow ₹16000$

Hence, the mother, the elder son and the younger son contribute ₹32,000, ₹12,000 and ₹16,000 respectively.

In a class of 56 students, the number of boys is $\dfrac{2}{5}$ th of the number of girls. Find the number of boys and girls.

Answer

Let the number of girls be $x$, so the number of boys $= \dfrac{2x}{5}$

According to the given condition,

$\Rightarrow x + \dfrac{2x}{5} = 56 \\[1em] \Rightarrow \dfrac{5x + 2x}{5} = 56 \\[1em] \Rightarrow \dfrac{7x}{5} = 56 \\[1em] \Rightarrow x = \dfrac{56 \times 5}{7} \\[1em] \Rightarrow x = 40$

Finding the number of boys,

$\Rightarrow \dfrac{2}{5} \times 40 \\[1em] \Rightarrow 16$

Hence, there are 16 boys and 40 girls in the class.

A man donated $\dfrac{1}{10}$ th of his money to a school, $\dfrac{1}{6}$ th of the remaining to an orphanage and the remaining money he distributed equally among his three children. If each child gets ₹50000, how much money did the man originally have?

Answer

Let the man originally have ₹$x$.

Money donated to the school $= ₹\dfrac{x}{10}$

Finding the remaining money,

$\Rightarrow x - \dfrac{x}{10} \\[1em] \Rightarrow \dfrac{10x - x}{10} \\[1em] \Rightarrow ₹\dfrac{9x}{10}$

Finding the money donated to the orphanage,

$\Rightarrow \dfrac{1}{6} \times \dfrac{9x}{10} \\[1em] \Rightarrow \dfrac{9x}{60} \\[1em] \Rightarrow ₹\dfrac{3x}{20}$

Finding the money left,

$\Rightarrow \dfrac{9x}{10} - \dfrac{3x}{20} \\[1em] \Rightarrow \dfrac{18x - 3x}{20} \\[1em] \Rightarrow \dfrac{15x}{20} = ₹\dfrac{3x}{4}$

This money was distributed equally among his three children.

According to the given condition,

$\Rightarrow \dfrac{3x}{4} \div 3 = 50000 \\[1em] \Rightarrow \dfrac{3x}{4} \times \dfrac{1}{3} = 50000 \\[1em] \Rightarrow \dfrac{x}{4} = 50000 \\[1em] \Rightarrow x = 50000 \times 4 \\[1em] \Rightarrow x = 200000$

Hence, the man originally had ₹2,00,000.

If $\dfrac{1}{4}$ of a number is added to $\dfrac{1}{3}$ of that number, the result is 15 greater than half of that number. Find the number.

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{1}{4}x + \dfrac{1}{3}x = \dfrac{1}{2}x + 15 \\[1em] [\text{Multiplying both sides by } 12] \\[1em] \Rightarrow 3x + 4x = 6x + 180 \\[1em] \Rightarrow 7x = 6x + 180 \\[1em] \Rightarrow 7x - 6x = 180 \\[1em] \Rightarrow x = 180$

Hence, the required number is 180.

A student was asked to multiply a given number by $\dfrac{4}{5}$. By mistake, he divided the given number by $\dfrac{4}{5}$. His answer was 36 more than the correct answer. What was the given number?

Answer

Let the given number be $x$.

$\Rightarrow \text{Correct answer} = x \times \dfrac{4}{5} = \dfrac{4x}{5} \\[1em] \Rightarrow \text{Answer obtained by the student} = x \div \dfrac{4}{5} = x \times \dfrac{5}{4} = \dfrac{5x}{4}$

According to the given condition,

$\Rightarrow \dfrac{5x}{4} - \dfrac{4x}{5} = 36 \\[1em] \Rightarrow \dfrac{25x - 16x}{20} = 36 \\[1em] \Rightarrow \dfrac{9x}{20} = 36 \\[1em] \Rightarrow x = \dfrac{36 \times 20}{9} \\[1em] \Rightarrow x = 80$

Hence, the given number is 80.