This MCQ module is based on: Integer Expressions and Chapter Exercises
TOPIC 6 OF 23
Integer Expressions and Chapter Exercises
🎓 Class 7
Mathematics
CBSE
Theory
Ch 2 — Integers
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Integer Expressions and Chapter Exercises
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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Figure it Out (Expressions & Exercises)
Q1. Find the values of: (a) \((-5) \times (18 + (-3))\) (b) \((-7) \times 4 \times (-1)\) (c) \((-2) \times (-1) \times (-5) \times (-3)\)
(a) \((-5) \times (18 - 3) = (-5) \times 15 = -75\)
(b) \((-7) \times 4 \times (-1) = (-28) \times (-1) = 28\)
(c) \((-2) \times (-1) \times (-5) \times (-3) = 2 \times (-5) \times (-3) = (-10) \times (-3) = 30\) (4 negatives → even → positive)
(b) \((-7) \times 4 \times (-1) = (-28) \times (-1) = 28\)
(c) \((-2) \times (-1) \times (-5) \times (-3) = 2 \times (-5) \times (-3) = (-10) \times (-3) = 30\) (4 negatives → even → positive)
Q2. Find the values of: (a) \((-27) \div 9\) (b) \(84 \div (-4)\) (c) \((-56) \div (-2)\)
(a) \(-27 \div 9 = -3\) (b) \(84 \div (-4) = -21\) (c) \(-56 \div (-2) = 28\)
Q3. Find the integer whose product with \((-1)\) is: (a) 27 (b) −31 (c) −1 (d) 1 (e) 0
If \(n \times (-1) = k\), then \(n = -k\).
(a) \(n = -27\) (b) \(n = 31\) (c) \(n = 1\) (d) \(n = -1\) (e) \(n = 0\)
(a) \(n = -27\) (b) \(n = 31\) (c) \(n = 1\) (d) \(n = -1\) (e) \(n = 0\)
Q4. If \(47 - 56 + 14 - 8 + 2 - 8 + 5 = -4\), then find the value of \(-47 + 56 - 14 + 8 - 2 + 8 - 5\) without calculating the full expression.
Answer: The second expression is the additive inverse of the first (every sign is flipped). So if the first = −4, the second = +4.
Q5. Try a modified version of the Collatz Conjecture with integers: Start with any number; if even, take half; if odd, multiply by −3 and add 1. Try starting with −7.
Sequence starting with −7:
−7 (odd) → \((-7) \times (-3) + 1 = 21 + 1 = 22\)
22 (even) → \(22 \div 2 = 11\)
11 (odd) → \(11 \times (-3) + 1 = -33 + 1 = -32\)
−32 (even) → \(-32 \div 2 = -16\)
−16 (even) → \(-16 \div 2 = -8\)
−8 (even) → \(-8 \div 2 = -4\)
−4 (even) → \(-4 \div 2 = -2\)
−2 (even) → \(-2 \div 2 = -1\)
−1 (odd) → \((-1) \times (-3) + 1 = 3 + 1 = 4\)
4 → 2 → 1 → ... (loops in the standard Collatz cycle)
The sequence eventually reaches 1!
−7 (odd) → \((-7) \times (-3) + 1 = 21 + 1 = 22\)
22 (even) → \(22 \div 2 = 11\)
11 (odd) → \(11 \times (-3) + 1 = -33 + 1 = -32\)
−32 (even) → \(-32 \div 2 = -16\)
−16 (even) → \(-16 \div 2 = -8\)
−8 (even) → \(-8 \div 2 = -4\)
−4 (even) → \(-4 \div 2 = -2\)
−2 (even) → \(-2 \div 2 = -1\)
−1 (odd) → \((-1) \times (-3) + 1 = 3 + 1 = 4\)
4 → 2 → 1 → ... (loops in the standard Collatz cycle)
The sequence eventually reaches 1!
Q6. In a test, (+4) marks are given for every correct answer and (−2) marks for every incorrect answer. (a) Anita answered all questions. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test? (b) Anil scored (−10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?
(a) Marks from correct: \(15 \times 4 = 60\). Total = 40. Marks from incorrect: \(40 - 60 = -20\). Number incorrect: \(-20 \div (-2) = 10\). Total questions: \(15 + 10 = \mathbf{25}\).
(b) Marks from correct: \(5 \times 4 = 20\). Total = −10. Marks from incorrect: \(-10 - 20 = -30\). Number incorrect: \(-30 \div (-2) = 15\). Total answered: \(5 + 15 = 20\). If the test has 25 questions (from part a), Anil left \(25 - 20 = \mathbf{5}\) questions unanswered.
(b) Marks from correct: \(5 \times 4 = 20\). Total = −10. Marks from incorrect: \(-10 - 20 = -30\). Number incorrect: \(-30 \div (-2) = 15\). Total answered: \(5 + 15 = 20\). If the test has 25 questions (from part a), Anil left \(25 - 20 = \mathbf{5}\) questions unanswered.
Q7. Pick the pattern — find the operations done by the machine below:
| Machine | |||
|---|---|---|---|
| 4 | 9 | −3 | 29 |
| 6 | 9 | 6 | −41 |
| 2 | 3 | −2 | 9 |
| −1 | 5 | −4 | −11 |
| −14 | −6 | −1 | ? |
Finding the pattern: Testing the operation \(a \times b - c\):
Row 1: \(4 \times 9 - (-3) = 36 + 3 = 39\) ❌ (need 29)
Testing \(a + b \times c\): Row 1: \(4 + 9 \times (-3) = 4 - 27 = -23\) ❌
Testing \(a \times c - b\): Row 1: \(4 \times (-3) - 9 = -12 - 9 = -21\) ❌
Testing \(a - b \times c\): Row 1: \(4 - 9 \times (-3) = 4 + 27 = 31\) ❌
Testing \(b \times c - a\): Row 1: \(9 \times (-3) - 4 = -27 - 4 = -31\) ❌
Testing other combinations... The operation appears to involve the pattern \(a \times b + c^2\) or similar. Students should explore systematically!
Last row: Once the pattern is found, substitute \(a = -14, b = -6, c = -1\) to get the answer.
Row 1: \(4 \times 9 - (-3) = 36 + 3 = 39\) ❌ (need 29)
Testing \(a + b \times c\): Row 1: \(4 + 9 \times (-3) = 4 - 27 = -23\) ❌
Testing \(a \times c - b\): Row 1: \(4 \times (-3) - 9 = -12 - 9 = -21\) ❌
Testing \(a - b \times c\): Row 1: \(4 - 9 \times (-3) = 4 + 27 = 31\) ❌
Testing \(b \times c - a\): Row 1: \(9 \times (-3) - 4 = -27 - 4 = -31\) ❌
Testing other combinations... The operation appears to involve the pattern \(a \times b + c^2\) or similar. Students should explore systematically!
Last row: Once the pattern is found, substitute \(a = -14, b = -6, c = -1\) to get the answer.
Q8. Imagine you're in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.
Expression: \(8 + 4 \times (-5) = 8 + (-20) = 8 - 20 = \mathbf{-12°C}\)
Q9. Find 3 consecutive numbers with a product of: (a) −6 (b) 120
(a) Product = −6: Try \((-1) \times 0 \times 1 = 0\) ❌. Try \(1 \times 2 \times 3 = 6\) — need negative. Try \((-3) \times (-2) \times (-1) = -6\) ✓ The numbers are −3, −2, −1.
(b) Product = 120: Try \(4 \times 5 \times 6 = 120\) ✓ The numbers are 4, 5, 6.
(b) Product = 120: Try \(4 \times 5 \times 6 = 120\) ✓ The numbers are 4, 5, 6.
Q10. An alien society uses a peculiar currency called 'pibs' with just two denominations: +13 pibs coin and −9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs +85 pibs?
Answer: We need to find non-negative integers \(m\) and \(n\) such that \(13m + (-9)n = 85\), i.e., \(13m - 9n = 85\).
Trying: \(m = 10\): \(130 - 9n = 85\), so \(9n = 45\), \(n = 5\). ✓
Yes! Use 10 coins of +13 pibs and 5 coins of −9 pibs: \(10 \times 13 + 5 \times (-9) = 130 - 45 = 85\) ✓
Sub-parts using two denominations (+13 and −9):
(a) +20: \(13 \times 5 + (-9) \times 5 = 65 - 45 = 20\) ✓
(b) +40: \(13 \times 4 + (-9) \times 1 = 52 - 9 = 43\) — try \(13 \times 7 + (-9) \times 6 = 91 - 54 = 37\) — try \(13 \times 1 + (-9) \times (-3)\)... Note: we can only use positive numbers of coins! \(13 \times 10 + (-9) \times 10 = 130 - 90 = 40\) ✓
(c) −50: \(13 \times 1 + (-9) \times 7 = 13 - 63 = -50\) ✓
(d) +8: \(13 \times 5 + (-9) \times 3 = 65 - 27 = 38\) — try \(13 \times 2 + (-9) \times 2 = 26 - 18 = 8\) ✓
(e) +10: \(13 \times 1 + (-9) \times 0 = 13\) — try \(13 \times 4 + (-9) \times 4 = 52 - 36 = 16\) — try \(13 \times 7 + (-9) \times 7 = 91 - 63 = 28\) — pattern: \(m = n\) gives \(4m\). Need \(4m = 10\)... not integer. Try \(13 \times 10 + (-9) \times 13 = 130 - 117 = 13\). Finding exact: \(13 \times 19 + (-9) \times 25 = 247 - 225 = 22\). This one requires more exploration!
(f) −2: \(13 \times 1 + (-9) \times 1 = 4\); \(13 \times 0 + (-9) \times 1 = -9\); Try \(13 \times 5 + (-9) \times 7 = 65 - 63 = 2\)... need −2, so \(13 \times 4 + (-9) \times 6 = 52 - 54 = -2\) ✓
(g) +1: \(13 \times 7 + (-9) \times 10 = 91 - 90 = 1\) ✓
(h) 1568 pibs: Since \(13 \times 7 - 9 \times 10 = 1\), we can scale: \(13 \times (7 \times 1568) + (-9) \times (10 \times 1568)\). Or find more efficient combinations. Yes, it is possible — since gcd(13,9) = 1, every integer can be represented!
Trying: \(m = 10\): \(130 - 9n = 85\), so \(9n = 45\), \(n = 5\). ✓
Yes! Use 10 coins of +13 pibs and 5 coins of −9 pibs: \(10 \times 13 + 5 \times (-9) = 130 - 45 = 85\) ✓
Sub-parts using two denominations (+13 and −9):
(a) +20: \(13 \times 5 + (-9) \times 5 = 65 - 45 = 20\) ✓
(b) +40: \(13 \times 4 + (-9) \times 1 = 52 - 9 = 43\) — try \(13 \times 7 + (-9) \times 6 = 91 - 54 = 37\) — try \(13 \times 1 + (-9) \times (-3)\)... Note: we can only use positive numbers of coins! \(13 \times 10 + (-9) \times 10 = 130 - 90 = 40\) ✓
(c) −50: \(13 \times 1 + (-9) \times 7 = 13 - 63 = -50\) ✓
(d) +8: \(13 \times 5 + (-9) \times 3 = 65 - 27 = 38\) — try \(13 \times 2 + (-9) \times 2 = 26 - 18 = 8\) ✓
(e) +10: \(13 \times 1 + (-9) \times 0 = 13\) — try \(13 \times 4 + (-9) \times 4 = 52 - 36 = 16\) — try \(13 \times 7 + (-9) \times 7 = 91 - 63 = 28\) — pattern: \(m = n\) gives \(4m\). Need \(4m = 10\)... not integer. Try \(13 \times 10 + (-9) \times 13 = 130 - 117 = 13\). Finding exact: \(13 \times 19 + (-9) \times 25 = 247 - 225 = 22\). This one requires more exploration!
(f) −2: \(13 \times 1 + (-9) \times 1 = 4\); \(13 \times 0 + (-9) \times 1 = -9\); Try \(13 \times 5 + (-9) \times 7 = 65 - 63 = 2\)... need −2, so \(13 \times 4 + (-9) \times 6 = 52 - 54 = -2\) ✓
(g) +1: \(13 \times 7 + (-9) \times 10 = 91 - 90 = 1\) ✓
(h) 1568 pibs: Since \(13 \times 7 - 9 \times 10 = 1\), we can scale: \(13 \times (7 \times 1568) + (-9) \times (10 \times 1568)\). Or find more efficient combinations. Yes, it is possible — since gcd(13,9) = 1, every integer can be represented!
Q11. Find the values of: (a) \((32 \times (-18)) \div ((-36))\) (b) \((32) + ((-36) \times (-18))\)
(a) \(32 \times (-18) = -576\). Then \(-576 \div (-36) = 16\).
(b) \((-36) \times (-18) = 648\). Then \(32 + 648 = 680\).
(b) \((-36) \times (-18) = 648\). Then \(32 + 648 = 680\).
Q12. Arrange the expressions in increasing order: (a) \((-348) + (-1064)\) (b) \((-348) - (-1064)\) (c) \((-348) \times (-1064)\) (d) \((-348) \div (-1064)\) (approx.) (e) \(348 \times (-1064)\) (f) \(348 \times 964\)
Let's compute each:
(a) \(-348 + (-1064) = -1412\)
(b) \(-348 - (-1064) = -348 + 1064 = 716\)
(c) \((-348) \times (-1064) = 370,272\) (positive, large)
(d) \((-348) \div (-1064) ≈ 0.327\) (positive, small)
(e) \(348 \times (-1064) = -370,272\) (negative, large magnitude)
(f) \(348 \times 964 = 335,472\) (positive, large)
Increasing order: (e) < (a) < (d) < (b) < (f) < (c)
\(-370272 < -1412 < 0.327 < 716 < 335472 < 370272\)
(a) \(-348 + (-1064) = -1412\)
(b) \(-348 - (-1064) = -348 + 1064 = 716\)
(c) \((-348) \times (-1064) = 370,272\) (positive, large)
(d) \((-348) \div (-1064) ≈ 0.327\) (positive, small)
(e) \(348 \times (-1064) = -370,272\) (negative, large magnitude)
(f) \(348 \times 964 = 335,472\) (positive, large)
Increasing order: (e) < (a) < (d) < (b) < (f) < (c)
\(-370272 < -1412 < 0.327 < 716 < 335472 < 370272\)
Q13. Given that \((-548) \times 972 = -532,656\), write the values of: (a) \((-547) \times 972\) (b) \((-548) \times 971\) (c) \((-547) \times 971\)
Using the given value and adjusting:
(a) \((-547) \times 972 = (-548 + 1) \times 972 = -532656 + 972 = \mathbf{-531,684}\)
(b) \((-548) \times 971 = (-548) \times (972 - 1) = -532656 - (-548) = -532656 + 548 = \mathbf{-532,108}\)
(c) \((-547) \times 971 = -532656 + 972 + 548 - 1 \times 1 = \text{use (a)}: (-547) \times (972-1) = -531684 + 547 = \mathbf{-531,137}\)
(a) \((-547) \times 972 = (-548 + 1) \times 972 = -532656 + 972 = \mathbf{-531,684}\)
(b) \((-548) \times 971 = (-548) \times (972 - 1) = -532656 - (-548) = -532656 + 548 = \mathbf{-532,108}\)
(c) \((-547) \times 971 = -532656 + 972 + 548 - 1 \times 1 = \text{use (a)}: (-547) \times (972-1) = -531684 + 547 = \mathbf{-531,137}\)
Q14. Given that \(207 \times (-33 + 7) = -5382\), write the value of \(-207 \times (33 - 7)\) without full calculation.
\(207 \times (-33 + 7) = 207 \times (-26) = -5382\).
\(-207 \times (33 - 7) = -207 \times 26 = -(207 \times 26) = \mathbf{-5382}\).
Same result! This is because \(207 \times (-26) = (-207) \times 26\) (swapping the sign between factors doesn't change the product).
\(-207 \times (33 - 7) = -207 \times 26 = -(207 \times 26) = \mathbf{-5382}\).
Same result! This is because \(207 \times (-26) = (-207) \times 26\) (swapping the sign between factors doesn't change the product).
Q15. Use the numbers 3, −2, 5, −6 exactly once with operations +, −, × and brackets as necessary to write an expression that: (a) gives the maximum possible result (b) gives the minimum possible result.
(a) Maximum: We want large positives. \((-2) \times (-6) \times 5 + 3 = 60 + 3 = 63\). Or \((-2) \times (-6) + 5 \times 3 = 12 + 15 = 27\). The first is larger: 63.
(b) Minimum: We want large negatives. \(3 \times (-6) \times 5 + (-2) = -90 - 2 = -92\). Or \((-2 + 3) \times 5 \times (-6) = 1 \times (-30) = -30\). The first is smaller: −92.
Even larger: \(3 \times (-6) \times 5 \times (-2)\) uses only ×, not all four operations. With constraint of using + − × ÷: explore further!
(b) Minimum: We want large negatives. \(3 \times (-6) \times 5 + (-2) = -90 - 2 = -92\). Or \((-2 + 3) \times 5 \times (-6) = 1 \times (-30) = -30\). The first is smaller: −92.
Even larger: \(3 \times (-6) \times 5 \times (-2)\) uses only ×, not all four operations. With constraint of using + − × ÷: explore further!
Q16. Fill in the blanks in at least 5 different ways with integers: (a) ☐ + ☐ × ☐ = −36 (b) (☐ − ☐) × ☐ = 12
(a) ☐ + ☐ × ☐ = −36:
1) \(0 + (-4) \times 9 = -36\)
2) \(0 + 4 \times (-9) = -36\)
3) \(-6 + (-6) \times 5 = -6 - 30 = -36\)
4) \(4 + (-8) \times 5 = 4 - 40 = -36\)
5) \(-36 + 0 \times 7 = -36\)
(b) (☐ − ☐) × ☐ = 12:
1) \((5 - 3) \times 6 = 12\)
2) \((7 - 4) \times 4 = 12\)
3) \((1 - (-1)) \times 6 = 12\)
4) \((0 - (-3)) \times 4 = 12\)
5) \((-1 - (-4)) \times 4 = 3 \times 4 = 12\)
(c) (☐ − (☐ − ☐)) = −1:
1) \((0 - (1 - 0)) = -1\)
2) \((2 - (4 - 1)) = 2 - 3 = -1\)
3) \((5 - (3 - (-3))) = 5 - 6 = -1\)
4) \((-3 - (-4 - (-2))) = -3 - (-2) = -3 + 2 = -1\)
5) \((10 - (14 - 3)) = 10 - 11 = -1\)
1) \(0 + (-4) \times 9 = -36\)
2) \(0 + 4 \times (-9) = -36\)
3) \(-6 + (-6) \times 5 = -6 - 30 = -36\)
4) \(4 + (-8) \times 5 = 4 - 40 = -36\)
5) \(-36 + 0 \times 7 = -36\)
(b) (☐ − ☐) × ☐ = 12:
1) \((5 - 3) \times 6 = 12\)
2) \((7 - 4) \times 4 = 12\)
3) \((1 - (-1)) \times 6 = 12\)
4) \((0 - (-3)) \times 4 = 12\)
5) \((-1 - (-4)) \times 4 = 3 \times 4 = 12\)
(c) (☐ − (☐ − ☐)) = −1:
1) \((0 - (1 - 0)) = -1\)
2) \((2 - (4 - 1)) = 2 - 3 = -1\)
3) \((5 - (3 - (-3))) = 5 - 6 = -1\)
4) \((-3 - (-4 - (-2))) = -3 - (-2) = -3 + 2 = -1\)
5) \((10 - (14 - 3)) = 10 - 11 = -1\)
Chapter Summary
| × | −9 | −6 | −3 | 3 | 3 | 6 | 9 |
|---|---|---|---|---|---|---|---|
| −3 | 27 | 18 | 9 | 3 | −3 | −6 | −9 |
| −2 | 18 | 12 | 6 | 2 | −2 | −4 | −6 |
| −1 | 9 | 6 | 3 | 1 | −1 | −2 | −3 |
| −3 | * | −1 | * | 1 | 1 | 2 | 3 |
| 3 | −27 | −18 | −9 | −3 | 9 | 18 | 27 |
| 6 | −54 | −36 | −18 | −6 | 18 | 36 | 54 |
| 9 | −81 | −54 | −27 | −9 | 27 | 54 | 81 |
- When two integers are multiplied, the product is positive when both the multiplier and multiplicand are positive, or when both are negative. The product is negative if one of them is positive and the other is negative.
- When two integers are divided, the quotient is positive when both the dividend and divisor are positive, or both are negative. The quotient is negative when one of them is positive and the other is negative.
- Integer multiplication is commutative: \(a \times b = b \times a\)
- Integer multiplication is associative: \(a \times (b \times c) = (a \times b) \times c\)
- Integer multiplication is distributive over addition: \(a \times (b + c) = (a \times b) + (a \times c)\)
Puzzle: Terhüchü
Terhüchü? is a traditional board game played in Assam and Nagaland. The board has 16 squares and diagonals as shown. There are 2 players, each with a set of 9 coins placed at starting positions.
Terhüchü board with starting positions — Blue (Player 1) and Red (Player 2)
Rules of Terhüchü
Objective: Capture all opponent's coins.Gameplay: Players take turns moving one coin along a line to a neighbouring vacant intersection. If an opponent's coin is at a neighbouring intersection and there is a vacant intersection beyond it, you can jump over and capture it. Multiple captures in one move are allowed.
Special rule: Inside the triangular corners, a coin may skip an intersection and move straight to the one beyond it.
Activity: Integer Operations Speed Challenge
L3 Apply
Challenge: How quickly can you compute 10 integer multiplications without a calculator? Time yourself!
- Write down 10 random integer multiplication problems (mix positive × positive, positive × negative, negative × negative).
- Start a timer and solve all 10.
- Check your answers using the sign rule: same signs → positive, different signs → negative.
- Record your time. Try to beat it tomorrow!
Try these: (a) \((-8) \times 7\) (b) \((-12) \times (-5)\) (c) \(9 \times (-11)\) (d) \((-3) \times (-3) \times (-3)\) (e) \((-1) \times (-1) \times (-1) \times (-1)\)
Answers: (a) −56 (b) 60 (c) −99 (d) −27 (3 negatives = odd = negative) (e) 1 (4 negatives = even = positive)
Competency-Based Questions
Scenario: In a quiz competition, each team starts with 0 points. For correct answers, the team gets +10 points. For wrong answers, the team loses 5 points (−5). For passing (not answering), the team neither gains nor loses. Team A answered 8 correct, 4 wrong, and passed 3 questions. Team B answered 6 correct, 2 wrong, and passed 7 questions.
Q1. Which team has a higher score?
L3 ApplyAnswer: (a) and (b) — Team A is higher.
Team A: \(8 \times 10 + 4 \times (-5) = 80 - 20 = 60\)
Team B: \(6 \times 10 + 2 \times (-5) = 60 - 10 = 50\)
Team A wins with 60 vs 50.
Team A: \(8 \times 10 + 4 \times (-5) = 80 - 20 = 60\)
Team B: \(6 \times 10 + 2 \times (-5) = 60 - 10 = 50\)
Team A wins with 60 vs 50.
Q2. Team B decides to answer 3 of its passed questions. What is the minimum number of correct answers (out of 3) needed to overtake Team A?
L4 AnalyseAnswer: Team B needs score > 60. Current: 50. If \(c\) correct and \((3-c)\) wrong out of 3:
\(50 + 10c + (-5)(3-c) = 50 + 10c - 15 + 5c = 35 + 15c > 60\)
\(15c > 25\), so \(c > 1.67\). Minimum \(c = 2\). With 2 correct, 1 wrong: \(35 + 30 = 65 > 60\) ✓. At least 2 out of 3 correct.
\(50 + 10c + (-5)(3-c) = 50 + 10c - 15 + 5c = 35 + 15c > 60\)
\(15c > 25\), so \(c > 1.67\). Minimum \(c = 2\). With 2 correct, 1 wrong: \(35 + 30 = 65 > 60\) ✓. At least 2 out of 3 correct.
Q3. A third team scored exactly 0 points after answering all 15 questions. Evaluate: is this possible? If yes, how many correct and wrong answers did they have?
L5 EvaluateAnswer: Need \(10c + (-5)(15-c) = 0\) where \(c\) = correct answers.
\(10c - 75 + 5c = 0\), so \(15c = 75\), \(c = 5\). Wrong = 10.
Yes, possible: 5 correct (+50) and 10 wrong (−50) = 0. ✓
\(10c - 75 + 5c = 0\), so \(15c = 75\), \(c = 5\). Wrong = 10.
Yes, possible: 5 correct (+50) and 10 wrong (−50) = 0. ✓
Q4. Design a new scoring system where: (a) it is impossible to score exactly 0 after answering all 15 questions, and (b) passing is penalized. State your scoring rules and prove property (a).
L6 CreateOne design: Correct = +7, Wrong = −3, Pass = −1.
Proof that 0 is impossible: All 15 answered (no passes): \(7c + (-3)(15-c) = 7c - 45 + 3c = 10c - 45\). For this to be 0: \(10c = 45\), \(c = 4.5\). Since \(c\) must be a whole number, 0 is impossible. ✓
With passes: \(7c - 3w - p = 0\) where \(c + w + p = 15\). Even with passes, reaching exactly 0 requires specific combinations — but the key property is that the "all answered" case can never yield 0.
Proof that 0 is impossible: All 15 answered (no passes): \(7c + (-3)(15-c) = 7c - 45 + 3c = 10c - 45\). For this to be 0: \(10c = 45\), \(c = 4.5\). Since \(c\) must be a whole number, 0 is impossible. ✓
With passes: \(7c - 3w - p = 0\) where \(c + w + p = 15\). Even with passes, reaching exactly 0 requires specific combinations — but the key property is that the "all answered" case can never yield 0.
Assertion–Reason Questions
Assertion (A): The product of 5 negative integers is always negative.
Reason (R): A product with an odd number of negative factors is negative.
Reason (R): A product with an odd number of negative factors is negative.
Answer: (a) — 5 is odd, so the product of 5 negative numbers is negative. R explains why A is true.
Assertion (A): \((-8) \div (-2) = -4\)
Reason (R): Dividing two negative integers always gives a positive quotient.
Reason (R): Dividing two negative integers always gives a positive quotient.
Answer: (d) — A is false: \((-8) \div (-2) = +4\), not −4. R is true: negative ÷ negative = positive. So R actually contradicts A, confirming A is wrong.
Frequently Asked Questions
How do you evaluate integer expressions with BODMAS?
Follow the BODMAS order: first solve Brackets, then Division and Multiplication from left to right, then Addition and Subtraction from left to right. Apply integer sign rules at each step. NCERT Class 7 Chapter 2 exercises require BODMAS with integers.
What types of integer exercises are in Class 7 Chapter 2?
NCERT Class 7 Chapter 2 exercises include computing sums and differences of integers, applying sign rules for products and quotients, evaluating expressions with multiple operations, solving word problems involving temperatures and debts, and verifying properties of integer operations.
How do you solve word problems involving integers?
Read the problem carefully and identify positive and negative quantities. Temperatures below zero, debts and below sea level are typically negative. Set up the expression using proper signs and evaluate step by step. NCERT Class 7 Ganita Prakash Part II uses real-world contexts for integer problems.
What properties help simplify integer calculations?
The commutative property allows reordering addition. The associative property allows regrouping. The distributive property helps expand brackets. The additive inverse gives a plus minus a equals 0. These properties simplify complex integer calculations in NCERT Class 7 Chapter 2 exercises.
How do you check your answer in integer problems?
Verify your answer by substituting it back into the original expression or by solving using a different method. For addition and subtraction, reverse the operation. For multiplication, use division to check. NCERT Class 7 Maths encourages students to verify every answer.
Frequently Asked Questions — Chapter 2
What is Integer Expressions and Chapter Exercises in NCERT Class 7 Mathematics?
Integer Expressions and Chapter Exercises is a key concept covered in NCERT Class 7 Mathematics, Chapter 2: Chapter 2. 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 Integer Expressions and Chapter Exercises step by step?
To solve problems on Integer Expressions and Chapter Exercises, 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 2: Chapter 2?
The essential formulas of Chapter 2 (Chapter 2) 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 Integer Expressions and Chapter Exercises important for the Class 7 board exam?
Integer Expressions and Chapter Exercises 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 Integer Expressions and Chapter Exercises?
Common mistakes in Integer Expressions and Chapter Exercises 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 Integer Expressions and Chapter Exercises?
End-of-chapter NCERT exercises for Integer Expressions and Chapter Exercises 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 2, and solve at least one previous-year board paper to consolidate your understanding.
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