WORD PROBLEMS WITH RATIONAL NUMBERS

Problem 1 :

Mr. Peter bought a pizza, he ate two-fifth of it, his son ate one-fifth of it and his wife ate the rest. What amount of the pizza did his wife eat?

Solution :

Amount of pizza eaten by Mr. Peter and his son :

= ⅖ + ⅕

= ⁽² ⁺ ¹⁾⁄₅

= ⅗

From the fraction ⅗, we can understand that if we divide the pizza into five equal parts, three parts were eaten by Mr. Peter and his son. Then the remaining two of the five parts were eaten by his wife.

Amount of pizza eaten by Mr. Peter's wife :

= ⅖

Problem 2 :

In a recipe making, every 1½ cup of rice requires 2¾ cups of water. Express this, in the ratio of rice to water.

Solution :

Quantity of rice :

= 1½

= ³⁄₂

Quantity of water required :

= 2¾

= ¹¹⁄₄

Ratio between rice and water :

= ³⁄₂ : ¹¹⁄₄

Least common multiple of (2, 4) = 4.

Multiply both the terms of the ratio by 4 to get rid of the denominators 2 and 4. 

=(4 ⋅ ³⁄₂) : (4 ⋅ ¹¹⁄₄)

  = (2 ⋅ ³⁄₁) : (1 ⋅ ¹¹⁄₁)

= (2 ⋅ 3) : (1 ⋅ 11)

= 6 : 11

Problem 3 :

Ravi multiplied ²⁵⁄₈ and ¹⁶⁄₁₅ to obtain ⁴⁰⁰⁄₁₂₀. He says that the simplest form of this product is ¹⁰⁄₃ and  Chandru says the answer in the simplest form is 3⅓. Who is correct? (or) Are they both correct? Explain.

Solution :

By multiplying the fractions ²⁵⁄₈ and ¹⁶⁄₁₅, we get ⁴⁰⁰⁄₁₂₀.

⁴⁰⁰⁄₁₂₀ = ⁴⁰⁄₁₂ = ¹⁰⁄₃ (improper fraction)

Convert the improper fraction ¹⁰⁄₃ to a mixed number.

= 3⅓

So, both are correct.

Problem 4 :

A piece of wire is ⅘ m long. If it is cut into 8 pieces of equal length, how long will each piece be? 

Solution :

Original length of wire = ⅘

Number of pieces to be cut out = 8

Length of each piece :

= ⅘ ÷ 8

Change the division to multiplication by taking reciprocal of 8.

= ⅘ ⋅ ⅛

Simplify.

= ⅕ ⋅ ½

= ⅒

Length of each pieace of wire is ⅒ m.

Problem 5 :

Find the length of a room whose area is ¹⁵³⁄₁₀ sq.m and whose width is 2¹¹⁄₂₀ m.

Solution :

Width = 2¹¹⁄₂₀ = ⁵¹⁄₂₀ m

Area of room = ¹⁵³⁄₁₀ sq.m

Length x Width  = ¹⁵³⁄₁₀

Substitute width = ⁵¹⁄₂₀.

Length x ⁵¹⁄₂₀  = ¹⁵³⁄₁₀

Divide both sides by ⁵¹⁄₂₀.

Length = ¹⁵³⁄₁₀ ÷ ⁵¹⁄₂₀

Change the division to multiplication by taking reciprocal of ⁵¹⁄₂₀.

= ¹⁵³⁄₁₀ x ²⁰⁄₅₁

Simplify.

= ³⁄₁ x ²⁄₁

= 2 x 3

= 6

The length of the room is 6 m.

Problem 6 :

Subbu spends ⅓ of his monthly earnings on rent, ⅖ on food and ⅒ on monthly usuals. What fractional part of his earnings is left with him for other expenses?

Solution :

Amount spent for rent = ⅓

Amount spent for food = ⅖

Amount spent for monthly usuals = ⅒

Fractional part Subbu's earning spent for rent, food and monthly usuals :

= ⅓ + ⅖ + ⅒

Least common multiple of (3, 5, 10) = 30.

Make each denominator as 30 by multiplying each fraction by an appropriate number. 

= ⁽¹ˣ¹⁰⁾⁄₍₃ₓ₁₀₎ + ⁽²ˣ⁶⁾⁄₍₅ₓ₆₎ + ⁽¹ˣ³⁾⁄₍₁₀ₓ₃₎

= ¹⁰⁄₃₀ + ¹²⁄₃₀ + ³⁄₃₀

= ⁽¹⁰ ⁺ ¹² ⁺ ³⁾⁄₃₀

= ²⁵⁄₃₀

= ²⁵⁄₃₀

= ⅚

Fractional part of his earnings is left with him for other expenses :

= ⅙

Problem 7 :

In a constituency, ¹⁹⁄₂₅ of the voters had voted for candidate A whereas ⁷⁄₅₀ had voted for candidate B. Find the rational number of the voters who had voted for others.

Solution :

Fractional part of voters had voted for candidate A :

= ¹⁹⁄₂₅

Fractional part of voters had voted for candidate B :

= ⁷⁄₅₀

Fractional part of voters had voted for candidate B :

= ¹⁹⁄₂₅ + ⁷⁄₅₀

Least common multiple of (25, 50) = 50.

Make the first denominator as 50 by multiplying the first fraction by an appropriate number. 

= ⁽¹⁹ˣ²⁾⁄₍₂₅ₓ₂₎ + ⁷⁄₅₀

= ³⁸⁄₅₀ + ⁷⁄₅₀

= ⁽³⁸ ⁺ ⁷⁾⁄₅₀

  = ⁴⁵⁄₅₀

= ⁹⁄₁₀

Rational number of the voters who had voted for others :

= ⅒

Problem 8 :

If ¾ of a box of apples weighs 3 kg and 225 gm, how much does a full box of apples weigh?

Solution :

1 kilogram = 1000 grams

3 kg and 225 gm :

= 3000 gm + 225 gm

= 3225 gm

It is given that ¾ of a box of apples weighs 3 kg and 225 gm.

¾ of box weighs = 3225 gm

Multiply both sides by 4.

4(¾ of box weighs) = 4(3225 gm)

3 boxes weigh = 12900 gm

Divide both sides by 3.

(3 boxes weigh) ÷ 3 = (12900 gm) ÷ 3

1 box weighs = 4300 gm

= 4000 gm + 300 gm

= 4 kg and 300 gm

A full box of apples weighs 4 kg 300 gm.

Problem 9 :

Mangalam buys a water jug of capacity 3⅘ litres. If she buys another jug which is 2⅔ times as large as the smaller jug, how many litres can the larger one hold?

Solution :

Capacity of the smaller :

= 3⅘ liters

= ¹⁹⁄₅ liters

It is given that the capaicty of another jug bought is 2⅔ times as large as the smaller jug.

Capacity of larger jug :

= 2⅔ times the capacity of smaller jug

= 2⅔ ⋅ ¹⁹⁄₅ liters

= ⁸⁄₃ ⋅ ¹⁹⁄₅ liters

  = ¹⁵²⁄₁₅ liters

= 10²⁄₁₅ liters

Problem 10 :

When a water tank is empty, pipe A can fill it in 5 minutes. If the same tank is completely filled with water, pipe B can empty it in 15 minutes.  When the tank is empty, if both pipes are opened together, how long will it take to fill the tank?

Solution :

Part of the tank filled by pipe A in 1 minute :

= ⅕

Part of the tank emptied by pipe B in 1 minute :

= ¹⁄₁₅

Part of the tank filled in 1 minute, if both the pipes are opened together :

= ⅕ - ¹⁄₁₅

Least common multiple of (5, 15) = 15.

Make the first denominator as 15 by multiplying the first fraction by an appropriate number.

= ⁽¹ˣ³⁾⁄₍₅ₓ₃₎ - ¹⁄₁₅

= ³⁄₁₅ - ¹⁄₁₅

= ⁽³ ⁻ ¹⁾⁄₁₅

= ²⁄₁₅

Time taken to fill the tank, if both the pipes are opened together :

= ¹⁵⁄₂

= 7.5 minutes

or

= 7 minutes 30 seconds

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