NCERT Solutions Class 7 Maths Chapter 8 Working with Fractions Exercise 8.4

The NCERT Solutions for Exercise 8.4 Class 7 Maths Chapter 8: Working with Fractions, is a hypothetical ultimate test or binary test of fraction literacy, where students are asked to solve multi-step word problems, involving the four operations (addition, subtraction, multiplication, and division) of both fractions, mixed fractions, as well as whole numbers.

It's a tedious exercise, which will require the students to convert everyday situations related to measurement, time, money, to consumption, into mathematical expressions which are mathematically accurate, and then the students will engage in the careful application of rules of arithmetic to solve regarding fractions.

1.0Download Class 7 Maths Chapter 8 Ex 8.4 NCERT Solutions PDF

Challenge your skills with complex fractional word problems! Download and go through the entire NCERT Solutions for Class 7 to have a full walk-through for this final learning checkpoint. Fully logical and worked out so you can see all work.

2.0Key Concepts of Exercise 8.4 (Integrated Problems) Class 7 Chapter 8

• Multi-Step Translation: Transforming word problems or real-world contexts into multi-step fractions (e.g., "half of the remainder", "equally divided among three").
• Identifying Operations: Understanding when a word/phrase like "of" is multiplication (​​1/2​×) and when to divide (÷3) in the case of a scenario.
• Unit Consistency: Identifying the type of unit throughout the problem and changing anything that is inconsistent (e.g., mixed fractions to improper fractions) before you do anything.
• Finding the Remainder/Total: This would include word problems where you're finding out "how much is left over after...", or finding how much/what the total is after accounting for spending or using a fraction of that amount.
• Consumption/Spending challenges: Finding how much is left after consecutively taking away the fractions.
• Area/Distance/Perimeter: Involving the fraciton multiplication/division to measure areas or distance.
• Unit Rate problems: How much it takes to do [this] in a unit of time (i.e., hours worked to do a fractional job per day).

3.0NCERT Solutions Class 7 Maths Chapter 8 Working with Fractions : Other Exercises

4.0Detailed CBSE Class 7 Chapter 8 Exercise 8.4 Solutions

1. Evaluate the following:

Sol. Row 1

(i) 3 ÷ 7/9 = 3 x 9/7 = 27/7

(ii) 14/4 ÷ 2 = 14/4 x 1/2 = 14/8. Simplified: 7/4

(iii) 2/3 ÷ 2/3 = 2/3 x 3/2 = 1

(iv) 14/6 ÷ 7/3 = 14/6 x 3/7 = (14 x 3) / (6 x 7) = 42/42 = 1

Row 2

(i) 4/3 ÷ 3/4 = 4/3 x 4/3 = 16/9

(ii) 7/4 ÷ 1/7 = 7/4 x 7/1 = 49/4

(iii) 8/2 ÷ 4/15 = 4 ÷ 4/15 = 4 x 15/4 = 15

Row 3

(i) 1/5 ÷ 1/9 = 1/5 x 9/1 = 9/5

(ii) 1/6 ÷ 11/12 = 1/6 x 12/11 = 12/66. Simplified: 2/11

(iii) 11/3 ÷ 11/8 = 11/3 x 8/11 = 8/3

2. For each of the questions below, choose the expression that describes the solution. Then simplify it.

(a) Maria bought 8 m of lace. She used 1/4 m for each bag. How many bags did she decorate?

Expression: (iii) 8 ÷ 1/4

Simplification: 8 ÷ 1/4 = 8 x 4/1 = 32 bags.

(b) 1/2 meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge?

Expression: (iv) 1/2 ÷ 8

Simplification: 1/2 ÷ 8 = 1/2 x 1/8 = 1/16 meter.

(c) A baker needs 1/6 kg of flour to make one loaf. He has 5 kg of flour. How many loaves can he make?

Expression: (iii) 5 ÷ 1/6

Simplification: 5 ÷ 1/6 = 5 x 6/1 = 30 loaves.

3. If 1/4 kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

Sol.

Flour for 12 rotis = 1/4 kg.

Flour for 1 roti = 1/4 ÷ 12 = 1/4 x 1/12 = 1/48 kg.

Flour for 6 rotis = 1/48 x 6 = 6/48 = 1/8 kg.

Answer: The flour used to make 6 rotis is 1/8 kg.

4. Pātiganita problem: What sum will be obtained by adding together 1 ÷ 1/6, 1 ÷ 1/10, 1 ÷ 1/13, 1 ÷ 1/9 and 1 ÷ 1/2.

Sol.

1 ÷ 1/6 = 6

1 ÷ 1/10 = 10

1 ÷ 1/13 = 13

1 ÷ 1/9 = 9

1 ÷ 1/2 = 2

Total sum = 6 + 10 + 13 + 9 + 2 = 40.

Answer: The friend should say 40.

5. Mira is reading a novel that has 400 pages. She read 1/5 of the pages yesterday and 3/10 of the pages today. How many more pages does she need to read to finish the novel?

Sol.

Total pages = 400.

Pages read yesterday = 1/5 x 400 = 80 pages.

Pages read today = 3/10 x 400 = 120 pages.

Total pages read = 80 + 120 = 200 pages.

Pages remaining = 400 - 200 = 200 pages.

Answer: Mira needs to read 200 more pages to finish the novel.

6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 3/4 litres of petrol?

Sol.

Distance per litre = 16 km.

Litres used = 2 3/4 = 11/4 litres.

Total distance = 11/4 x 16 = 176/4 = 44 km.

Answer: The car will go 44 km using 2 3/4 litres of petrol.

7. Amritpal decides on a destination. Train takes 5 1/6 hours. Plane takes 1/2 hour. How many hours does the plane save?

Sol.

Train time = 5 1/6 hours = 31/6 hours.

Plane time = 1/2 hours = 3/6 hours.

Time saved = 31/6 - 3/6 = 28/6 hours.

Simplified: 28/6 = 14/3 hours.

Answer: The plane saves 14/3 hours.

8. Mariam's grandmother baked a cake. Mariam and her cousins finished 4/5 of the cake. The remaining cake was shared equally by Mariam's three friends. How much of the cake did each friend get?

Sol.

Fraction of cake left = 1 - 4/5 = 1/5.

Cake shared by 3 friends = (1/5) ÷ 3

= 1/5 x 1/3 = 1/15.

Answer: Each friend got 1/15 of the cake.

9. Choose the option(s) describing the product of (565/465 x 707/676).

Sol.

Analyze the two factors:

Factor 1 (565/465) is an improper fraction, so it is > 1.

Factor 2 (707/676) is an improper fraction, so it is > 1.

Product P = (>1) x (>1). Therefore, the product P must be > 1.

(a) > 565/465: (Correct) Multiplying 565/465 by a number > 1 (707/676) increases its value.

(b) < 565/465: (Incorrect)

(c) > 707/676: (Correct) Multiplying 707/676 by a number > 1 (565/465) increases its value.

(d) < 707/676: (Incorrect)

(e) > 1: (Correct)

(f) < 1: (Incorrect)

Answer: (a), (c), (e) are correct options.

10. What fraction of the whole square is shaded?

Sol. The whole square is divided into 4 equal smaller squares.

Focus on the bottom right square - it's subdivided further:

It is split into 8 equal triangles using diagonal lines.

Of those 8 parts, 3 triangles are shaded.

Fraction of the whole taken by one small square =1/4​

Fraction of the small triangle that is shaded =3/8

Fraction of whole square which is shaded Now, 1/4​×3/8​=32/3​

∴323​ of the whole square is shaded.

11. A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the fig.) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Sol. From the figure, the first split i.e., at branch A , the colony divides into 2 equal parts.

Fraction of original ants reaching mango tree via main path A = 1/2.

Branch B divides into 2 parts. Fraction of ants on each part = 1/2 ÷ 2 = 1/4.

Fraction of original ants reaching mango tree via path B = 1/4.

Branch C divides into 4 parts. Fraction of ants on each part = 1/4 ÷ 4 = 1/16.

Fraction of original ants reaching mango tree via path C (2 of the 4 parts go to mango) = 2/16.

Fraction of original ants reaching sugarcane via path C (1 of the 4 parts goes to sugarcane) = 1/16.

Branch D divides into 2 parts. Fraction of ants on each part = 1/16 ÷ 2 = 1/32.

Fraction of original ants reaching mango tree via path D (1 of the 2 parts goes to mango) = 1/32.

Fraction of original ants reaching sugarcane via path D (1 of the 2 parts goes to sugarcane) = 1/32.

Total fraction of the original colony who reached mango tree:

= 1/2 + 1/4 + 2/16 + 1/32

= 16/32 + 8/32 + 4/32 + 1/32

= 29/32

Total fraction of the original colony who reached sugarcane field:

= 1/16 + 1/32

= 2/32 + 1/32

= 3/32

12. What is 1 - 1/2?

(1 - 1/2) x (1 - 1/3)?

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5)?

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6) x (1 - 1/7) x (1 - 1/8) x (1 - 1/9) x (1 - 1/10)?

Sol.

1 - 1/2 = 1/2

(1 - 1/2) x (1 - 1/3) = 1/2 x 2/3 = 1/3

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5)

= 1/2 x 2/3 x 3/4 x 4/5

= 1/5

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x ... x (1 - 1/10)

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x 9/10

= 1/10

Make a general statement and explain.

General Statement:

The product of the series of terms (1 - 1/k) from k=2 up to a final term (1 - 1/n) is equal to the reciprocal of the largest denominator, n.

General Expression:

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x ... x (1 - 1/n) = 1/n

Explanation:

When each term (1 - 1/k) is simplified, it becomes (k-1)/k.

The product then looks like a telescoping series:

(1/2) x (2/3) x (3/4) x ... x ((n-1)/n)

The numerator of each fraction cancels out the denominator of the previous fraction (the 2 in 1/2 cancels the 2 in 2/3, the 3 in 2/3 cancels the 3 in 3/4, and so on).

Only the numerator of the first term (1) and the denominator of the last term (n) remain.

Therefore, the value of the expression is 1/n.

5.0Key Features and Benefits: Class 7 Maths Chapter 8 Exercise 8.4

• Comprehensive Problem Solving: Brings together all fraction skills (addition, subtraction, multiplication, and division) in an authentic context.
• Building Math Models: Prepares students to transfer a sentence from English into a proper mathematical model using fractions.
• Completing the Topic: This is the thorough review to grasp this whole topic before moving on to Rational Numbers or Decimals.