This MCQ module is based on: Chapter 4 Exercises and Summary
TOPIC 12 OF 23
Chapter 4 Exercises and Summary
🎓 Class 7
Mathematics
CBSE
Theory
Ch 4 — Working with Decimals
⏱ ~35 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Chapter 4 Exercises and Summary
Targeting Class 7 level in General Mathematics, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
4.4 Exercises — Figure it Out
Below are the end-of-chapter exercises for Chapter 4: Another Peek Beyond the Point.
1. Express the following as decimals: (a) \(\frac{3}{4}\) (b) \(\frac{7}{8}\) (c) \(\frac{17}{25}\) (d) \(\frac{9}{5}\)
(a) 0.75 (b) 0.875 (c) 68/100 = 0.68 (d) 1.8.
2. Multiply: (a) 0.25 × 0.4 (b) 1.75 × 2.5 (c) 9.5 × 5 (d) 12.5 × 7.5
(a) 25 × 4 = 100; 2+1 = 3 dp → 0.100 = 0.1.
(b) 175 × 25 = 4375; 2+1 = 3 dp → 4.375.
(c) 47.5.
(d) 93.75.
(b) 175 × 25 = 4375; 2+1 = 3 dp → 4.375.
(c) 47.5.
(d) 93.75.
3. Find the area of a rectangle whose length is 6.4 cm and breadth is 3.5 cm.
Area = 6.4 × 3.5 = 64 × 35 / 100 = 2240/100 = 22.4 sq cm.
4. Divide: (a) 45.6 ÷ 4 (b) 12.56 ÷ 0.4 (c) 9.8 ÷ 0.7 (d) 0.144 ÷ 1.2
(a) 11.4. (b) 125.6/4 = 31.4. (c) 98/7 = 14. (d) 1.44/12 = 0.12.
5. Fill in the following blanks appropriately: 1 cm = 10 mm; 1 m = 100 cm; 1 km = 1000 m; 1 kg = 1000 g; 1 g = 1000 mg; 1 l = 1000 ml. Convert: 5.5 km = ____ m; 35 cm = ____ mm; 14.5 cm = ____ mm; 68 g = ____ kg; 9.02 m = ____ mm; 125.5 ml = ____ l.
5.5 km = 5500 m; 35 cm = 350 mm; 14.5 cm = 145 mm; 68 g = 0.068 kg; 9.02 m = 9020 mm; 125.5 ml = 0.1255 l.
6. The division problem was set by Sridharacharya in his book Patiganita: "\(6\tfrac{1}{4}\) is divided by \(2\tfrac{1}{2}\) and \(60\tfrac{1}{4}\) is divided by \(5\tfrac{3}{4}\). Tell the quotients separately." Can you try to solve it by converting the fractions into decimals?
(i) 6.25 ÷ 2.5 = 2.5.
(ii) 60.25 ÷ 5.75 = 6025/575 = 10 remainder 275/575 ≈ 10.478.
(ii) 60.25 ÷ 5.75 = 6025/575 = 10 remainder 275/575 ≈ 10.478.
7. Fill the boxes in at least 2 different ways: (a) □ × □ = 2.4 (b) □ ÷ □ = 14.5
(a) 1.2 × 2 = 2.4; 0.6 × 4 = 2.4; 0.8 × 3 = 2.4 (many).
(b) 29 ÷ 2 = 14.5; 145 ÷ 10 = 14.5; 14.5 ÷ 1 = 14.5 (many).
(b) 29 ÷ 2 = 14.5; 145 ÷ 10 = 14.5; 14.5 ÷ 1 = 14.5 (many).
8. Find the following quotients: (a) 756 ÷ 36 (b) 7.56 ÷ 0.36 (c) 756 ÷ 3.6 (d) 75.6 ÷ 36 (e) 7560 ÷ 3.6 (f) 7.56 ÷ 3.6
(a) 21 (b) 21 (c) 210 (d) 2.1 (e) 2100 (f) 2.1. All share the 'core' 756/36 = 21, with decimal points shifted.
9. Find the missing cells if each cell represents a ÷ b. Partial grid with entries 1517, 151.7, 15.17, 1.517, 15170.
The core division is 1517 ÷ base. Each row/column adjusts the decimal: e.g., 3.7 = 37 ÷ 10 scales the 41 → 4.1; 0.37 ratio shifts twice, etc. General pattern: multiply/divide by appropriate power of 10 to shift the decimal.
10. Using the digits 2, 4, 5, 8, and 0, fill the boxes □.□ × □.□ to get the (a) maximum product (b) minimum product
(a) To maximise, put the largest digits as whole parts: 8.4 × 5.2 = 43.68 (using 8, 4, 5, 2, leaving 0). Even better: 8.5 × 4.2 = 35.7. Best: place largest digits in tens place → 8.5 × 4.2 = 35.7 or 8.4 × 5.2 = 43.68. Maximum achievable: 8.4 × 5.2 = 43.68.
(b) To minimise (non-zero): 2.0 × 4.5 = 9.0 or 2.0 × 4.8 = 9.6. Minimum: 2.0 × 4.5 = 9.0.
(b) To minimise (non-zero): 2.0 × 4.5 = 9.0 or 2.0 × 4.8 = 9.6. Minimum: 2.0 × 4.5 = 9.0.
11. Sort the following expressions in increasing order:
(a) 245.05 × 0.942368 (b) 245.05 × 7.9682 (c) 245.05 + 7.9682 (d) 245.05 × 0.942364 (e) 245.05 (f) 7.9682
(a) 245.05 × 0.942368 (b) 245.05 × 7.9682 (c) 245.05 + 7.9682 (d) 245.05 × 0.942364 (e) 245.05 (f) 7.9682
(a) ≈ 230.9 (b) ≈ 1952.6 (c) = 253.02 (d) ≈ 230.9 (very close to (a)) (e) = 245.05 (f) = 7.9682.
Increasing order: (f) < (d) ≈ (a) < (e) < (c) < (b). More precisely: f < d < a < e < c < b.
Increasing order: (f) < (d) ≈ (a) < (e) < (c) < (b). More precisely: f < d < a < e < c < b.
Chapter Summary
- In this chapter, we learnt procedures to perform decimal multiplication and division.
- For decimal multiplication, we first multiply the multiplier and multiplicand as counting numbers. The number of decimal digits in the product is the total number of decimal digits in the multiplier and multiplicand.
- Division of decimals uses the same procedure as division of counting numbers — place value division (long division). We continue until we get two numbers that do not have any common prime factors.
- There are decimal divisions where the quotient never ends. After each regrouping and dividing the new remainder is always a remainder.
Key takeaway: decimals are just an extension of place value — multiplication and division follow the same principles as counting numbers, with careful placement of the decimal point.
Puzzle Time — Hidato
In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it is made of only one piece. Usually, in every Hidato puzzle the smallest and highest numbers are given on the grid. Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any of the adjacent cells, including diagonally adjacent cells. The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution. Try solving some Hidato puzzles with classmates!
Activity: Spot the Pattern — Terminating vs Repeating
L3 ApplyMaterials: Paper, pencil, list of fractions.
Predict: Which of 1/6, 1/7, 1/8, 1/9, 1/10 are terminating decimals?
- Compute each by long division.
- Mark each as terminating (stops) or repeating.
- Factorise the denominators: 6 = 2×3, 7 = 7, 8 = 2³, 9 = 3², 10 = 2×5.
- Which denominators have only primes 2 and 5? → 8, 10 (terminating). Others repeat.
1/6 = 0.1\(\overline{6}\) (repeating). 1/7 = 0.\(\overline{142857}\). 1/8 = 0.125 (terminating). 1/9 = 0.\(\overline{1}\). 1/10 = 0.1 (terminating).
Rule: A fraction in lowest terms is terminating if and only if its denominator's only prime factors are 2 and 5.
Competency-Based Questions
Scenario: A tailor buys 18.75 m of fabric at ₹124.80 per metre. She uses the fabric to stitch equal shirts. Each shirt needs 2.5 m and she keeps the leftover for a smaller piece.
Q1. Compute the total amount she paid and the number of full shirts she can stitch.
L3 ApplyCost = 18.75 × 124.80 = 1875 × 12480 / 10000 = 23400000/10000 = ₹2340.00. Shirts = 18.75 ÷ 2.5 = 1875/250 = 7.5 shirts → so 7 full shirts with 1.25 m leftover.
Q2. What was the effective cost per full shirt (ignore leftover)? Analyse whether the leftover fabric changes this calculation significantly.
L4 AnalyseCost per shirt (if total cost attributed to 7 shirts) = 2340 ÷ 7 = ₹334.29. If attributed to only fabric used in shirts (17.5 m): 17.5 × 124.80 = ₹2184, so ₹2184 ÷ 7 = ₹312. The leftover 1.25 m represents ₹156 — a significant 6.7% of total cost.
Q3. The tailor claims: "If I buy twice the fabric (37.5 m), my per-shirt cost stays the same." Evaluate.
L5 Evaluate37.5 ÷ 2.5 = 15 shirts exactly, no leftover. Cost = 37.5 × 124.80 = ₹4680. Per shirt = 4680 ÷ 15 = ₹312. In the original case per-shirt cost (fabric-used basis) was also ₹312. Claim is true for that interpretation. But if the leftover is considered waste, doubling the fabric eliminates the leftover and makes the effective cost cheaper.
Q4. Design a purchase plan (fabric length in metres, shirt length in metres) so that the tailor uses the entire fabric with no leftover and stitches a whole number of shirts. Provide two distinct valid plans.
L6 CreatePlan 1: 20 m of fabric at 2.5 m per shirt = 8 shirts.
Plan 2: 15.75 m at 2.25 m per shirt = 7 shirts.
Criterion: fabric length ÷ shirt length must be a whole number. Many valid designs.
Plan 2: 15.75 m at 2.25 m per shirt = 7 shirts.
Criterion: fabric length ÷ shirt length must be a whole number. Many valid designs.
Assertion–Reason Questions
A: Multiplying a number by 100 moves the decimal point 2 places to the right.
R: 100 = 10².
R: 100 = 10².
Answer: (a). Each factor of 10 shifts decimal one place right; 100 = 10² shifts two places. R explains A.
A: The decimal expansion of 22/7 is not terminating.
R: The denominator 7 is not made of only 2s and 5s.
R: The denominator 7 is not made of only 2s and 5s.
Answer: (a). Fractions in lowest terms terminate iff the denominator's only primes are 2 and 5. Since 7 fails this test, 22/7 is non-terminating. R explains A.
A: 0.1 × 0.1 × 0.1 = 0.001.
R: Multiplication of decimals uses addition of decimal place counts.
R: Multiplication of decimals uses addition of decimal place counts.
Answer: (a). 1×1×1 = 1; total decimal places = 1+1+1 = 3 → 0.001. R explains A.
Frequently Asked Questions
What exercises are in Class 7 Part 2 Chapter 4?
Chapter 4 exercises include decimal multiplication of various sizes, decimal division by whole numbers and by decimals, money and measurement word problems, and checking answers by estimation. NCERT Class 7 Ganita Prakash Part 2 provides complete practice.
How do you estimate decimal answers?
Round each decimal to a nearby whole number or simple value, compute mentally, and compare with the exact answer. If the exact result is far from the estimate, recheck. NCERT Class 7 Chapter 4 emphasises estimation.
What is the summary of Chapter 4?
Key ideas: multiply decimals by treating as whole numbers and counting total decimal places; divide by whole numbers using long division; divide by decimals by shifting decimal points; estimate to verify. NCERT Class 7 Ganita Prakash Part 2 Chapter 4.
Solve: 3.6 x 2.5.
Ignore decimals: 36 x 25 = 900. Decimal places: 1 + 1 = 2. Answer: 9.00 = 9. NCERT Class 7 Chapter 4 exercises include such checks of trailing zeros.
Why study decimal operations in Class 7?
Decimal operations are essential for money, measurement, science, and further maths. Class 7 consolidates these skills so students can handle real-world computations and advanced topics like ratios and percentages. NCERT Class 7 Part 2 Chapter 4 is foundational.
How to check a decimal division answer?
Multiply the quotient by the divisor; the result should equal the dividend (within rounding). For example, if 7.2 / 0.6 = 12, then 12 x 0.6 = 7.2. NCERT Class 7 Chapter 4 recommends this verification.
Frequently Asked Questions — Chapter 4
What is Chapter 4 Exercises and Summary in NCERT Class 7 Mathematics?
Chapter 4 Exercises and Summary is a key concept covered in NCERT Class 7 Mathematics, Chapter 4: Chapter 4. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Chapter 4 Exercises and Summary step by step?
To solve problems on Chapter 4 Exercises and Summary, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 4: Chapter 4?
The essential formulas of Chapter 4 (Chapter 4) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Chapter 4 Exercises and Summary important for the Class 7 board exam?
Chapter 4 Exercises and Summary is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Chapter 4 Exercises and Summary?
Common mistakes in Chapter 4 Exercises and Summary include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Chapter 4 Exercises and Summary?
End-of-chapter NCERT exercises for Chapter 4 Exercises and Summary cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 4, and solve at least one previous-year board paper to consolidate your understanding.
AI Tutor
Mathematics Class 7 — Ganita Prakash-II
Ready