Figure it Out (Page 25)

Q. Let us try to find a few more pairs of numbers from their sums and differences:
(a) Sum = 27, Difference = 9
(b) Sum = 4, Difference = 12
(c) Sum = 0, Difference = 10
(d) Sum = 0, Difference = – 10
(e) Sum = – 7, Difference = – 1
(f) Sum = – 7, Difference = – 13
Solution:
(a) Sum = 27, Difference = 9

Therefore, the numbers are 18 and 9.

(b) Sum = 4, Difference = 12

Therefore, the numbers are 8 and -4.

(c) Sum = 0, Difference = 10

Therefore, the numbers are 5 and -5.

(d) Sum = 0, Difference = – 10

Therefore, the numbers are -5 and 5.

(e) Sum = – 7, Difference = – 1

Therefore, the numbers are -4 and -3.

(f) Sum = – 7, Difference = – 13

Therefore, the numbers are -10 and 3.

Figure it Out (Page 31)

1. Using the token interpretation, find the values of:
(a) 3 × (– 2)
(b) (– 5) × (– 2)
(c) (– 4) × (– 1)
(d) (– 7) × 3
Solution:
(a) 3 × (– 2)
Two red tokens 3 times = (–6)
So, 3 × (– 2) = (– 6).

(b) (– 5) × (– 2)
Remove 2 red tokens from the zero pairs, 5 times.
So, (– 5) × (– 2) = 10.

(c) (– 4) × (– 1)
Remove 1 red token from the zero pair, 4 times.
So, (– 4) × (– 1) = 4.

(d) (– 7) × 3
Remove 3 green tokens from the zero pairs, 7 times.
So, (– 7) × 3 = (-21).

2. If 123 × 456 = 56088, without calculating, find the value of:
(a) (– 123) × 456
(b) (– 123) × (– 456)
(c) (123) × (– 456)
Solution:
(a) (– 123) × 456
= (-1) × 123 × 456
= (-1) × 56088
= -56088.

(b) (– 123) × (– 456)
= (-1) × 123 × (-1) × 456
= (-1) × (-1) × 123 × 456
= 1 × 56088 = 56088.

(c) (123) × (– 456)
= 123 × 456 × (-1)
= 56088 × (-1)
= -56088.

3. Try to frame a simple rule to multiply two integers.
Solution:
Rule for multiplying two integers:
(i) Multiply their absolute values.
(ii) If the integers have different signs, the product is negative.
(iii) If both integers have the same sign, the product is positive.

Figure it Out (Page 33)

Q. Find the following products.
(a) 4 × (– 3)
(b) (– 6) × (– 3)
(c) (– 5) × (– 1)
(d) (– 8) × 4
(e) (– 9) × 10
(f) 10 × (– 17)
Solution:
(a) 4 × (– 3)
Multiplier is positive, multiplicand is negative → product is negative.
∴ 4 × (–3) = –12.

(b) (– 6) × (– 3)
Both numbers are negative → product is positive.
∴ (–6) × (–3) = 18.

(c) (– 5) × (– 1)
Both numbers are negative → product is positive.
∴ (– 5) × (– 1) = 5.

(d) (– 8) × 4
Multiplier is negative, multiplicand is positive → product is negative.
∴ (– 8) × 4 = -32.

(e) (– 9) × 10
Multiplier is negative, multiplicand is positive → product is negative.
∴ (– 9) × 10 = -90.

(f) 10 × (– 17)
Multiplier is positive, multiplicand is negative → product is negative.
∴ 10 × (– 17) = -170.

Figure it Out (Page 39)

1. Find the values of:
(a) 14 × (– 15)
(b) (– 16) × (– 5)
(c) 36 ÷ (– 18)
(d) (– 46) ÷ (– 23)
Solution:
(a) 14 × (– 15)
Multiplier is positive, multiplicand is negative → product is negative.
∴ 14 × (– 15) = -210.

(b) (– 16) × (– 5)
Both numbers are negative → product is positive.
∴ (– 16) × (– 5) = 80.

(c) 36 ÷ (– 18)
or (-18) × ____ = 36
We know that (-18) × (- 2) = 36.
Therefore, 36 ÷ (– 18) = (- 2).

(d) (– 46) ÷ (– 23)
or (- 23) × ____ = (- 46)
We know that (- 23) × 2 = (- 46).
Therefore, (– 46) ÷ (– 23) = 2.

2. A freezing process requires that the room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
Initial temperature = 32°C
Temperature decreases at = 5°C per hour
Time = 10 hours
Decrease in temperature after 10 hours
= 32°C + 10 × (–5°C)
= 32°C – 50°C
= –18°C
∴ The room temperature after 10 hours will be –18°C.

3. A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5 per bag of grey cement sold. [Represent the profit/ loss as integers.]
(a)  The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
(b)  If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss.
Solution:
Profit on one white cement bag = ₹8
Loss on one grey cement bag = ₹5
(Profit is positive, loss is negative.)
(a)
White cement sold = 3000 bags
Grey cement sold = 5000 bags
Profit from white cement
= 3000 × 8 = 24,000.
Loss from grey cement
= 5000 × (–5) = –25,000.
Total profit/loss = 24,000 + (–25,000) = –1,000.
∴ The company has a loss of ₹1,000.

(b)
Grey cement sold = 6400 bags
Loss = 6400 × (–5) = –32,000
Let the number of white cement bags to be sold = x
Profit from white cement = x × 8
For no profit and no loss,
x × 8 + (–32,000) = 0
8x = 32,000
x = 32,000 ÷ 8
x = 4000
∴ The company must sell 4000 bags of white cement.

4. Replace the blank with an integer to make a true statement.
(a) (– 3) × _____ = 27
(b) 5 × _____ = (– 35)
(c) _____ × (– 8) = (– 56)
(d) _____ × (– 12) = 132
(e) _____ ÷ (– 8) = 7
(f) _____ ÷ 12 = – 11
Solution:
(a) (–3) × ___ = 27
or 27 ÷ (–3) = –9
∴ (–3) × (- 9) = 27

(b) 5 × _____ = (– 35)
or –35 ÷ 5 = –7
∴ 5 × (–7) = (– 35)

(c) _____ × (– 8) = (– 56)
or –56 ÷ (–8) = 7
∴ 7 × (– 8) = (– 56)

(d) _____ × (– 12) = 132
or 132 ÷ (–12) = –11
∴ (–11) × (– 12) = 132

(e) _____ ÷ (– 8) = 7
or 7 × (–8) = –56
∴ (– 56) ÷ (– 8) = 7

(f) _____ ÷ 12 = – 11
or –11 × 12 = –132
∴ (– 132) ÷ 12 = – 11

Figure it Out (Page 42 – 44)

1. Find the values of the following expressions:
(a) (– 5) × (18 + (– 3))
(b) (– 7) × 4 × (– 1)
(c) (– 2) × (– 1) × (– 5) × (– 3)
Solution
(a) (– 5) × (18 + (– 3))
= (– 5) × 18 + (– 5) × (– 3)
= (– 90) + 15
= (– 75).

(b) (– 7) × 4 × (– 1)
= (– 28) × (– 1)
= 28.

(c) (– 2) × (– 1) × (– 5) × (– 3)
= 2 × (– 5) × (– 3)
= 2 × 15
= 30.

2. Find the values of the following expressions:
(a) (– 27) ÷ 9
(b) 84 ÷ (– 4)
(c) (– 56) ÷ (– 2)
Solution:
(a) (– 27) ÷ 9
or 9 × ____ = (– 27)
We know that 9 × (– 3) = (– 27)
∴ (– 27) ÷ 9 = (– 3).

(b) 84 ÷ (– 4)
or (– 4) × _____ = 84
We know that (– 4) × (– 21) = 84
∴ 84 ÷ (– 4) = (– 21).

(c) (– 56) ÷ (– 2)
or (– 2) × _____ = (– 56)
We know that (– 2) × 28 = (– 56)
∴ (– 56) ÷ (– 2) = 28.

3. Find the integer whose product with (– 1) is:
(a) 27
(b) – 31
(c) – 1
(d) 1
(e) 0
Solution:
(a) 27
(– 1) × (– 27) = 27
So, (–27) is the required integer.

(b) – 31
(–1) × 31 = –31
So, 31 is the required integer.

(c) – 1
(–1) × 1 = –1
So, 1 is the required integer.

(d) 1
(–1) × –1 = 1
So, (– 1) is the required integer.

(e) 0
(–1) × 0 = 0
So, 0 is the required integer.

4. 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.
Solution:
Given,
47 − 56 + 14 − 8 + 2 − 8 + 5 = −4.
The expression,
– 47 + 56 – 14 + 8 – 2 + 8 – 5 = – (47 – 56 + 14 – 8 + 2 – 8 + 5) = –(– 4) = 4.
So, the value of the expression is 4.

5. Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by – 3 and add 1; repeat. An example sequence is shown below.

Try this with different starting numbers: (– 21), (– 6), and so on. Describe the patterns you observe.
Solution:
(a) Starting with (– 21)
– 21 → 64 → 32 → 16 → 8 → 4 → 2 → 1 → – 2 → – 1 → 4 → 2 → 1 → – 2 ……..

(b) Starting from (– 6)
– 6) → – 3 → 10 → 5 → – 14 → – 7 → → 22 → 11 → 32 → –16 → –8 → –4 → –2 → –1 → 4 → 2 → 1 → –2 …….

Pattern observed: For numbers like –21, –6, etc., the sequences eventually reach the same repeating loop: –2 → –1 → 4 → 2 → 1 → –2 …
All starting numbers end up in this cycle.

6. In a test, (+ 4) marks are given for every correct answer and (– 2) marks are given for every incorrect answer.
(a)  Anita answered all the questions in the test. 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?
Solution:
(a) Anita’s score = 40
Correct answers = 15
Marks from correct answers = 15 × 4 = 60
Marks lost = 60 – 40 = 20
These 20 marks were lost because of incorrect answers:
Each incorrect answer gives –2 marks.
Number of incorrect answers
= 20 ÷ 2 = 10
Total questions in the test = correct + incorrect = 15 + 10 = 25 questions.

(b) Anil’s score = –10
Correct answers = 5
Marks from correct answers = 5 × 4 = 20
Let incorrect answers be x.
Total score: 20 – 2x = –10
⇒ –2x = –10 – 20
⇒ –2x = –30
⇒ x = 15
So Anil had 15 incorrect answers.
Correct = 5
Incorrect = 15
Total answered = 20
Since no information suggests a fixed total number of questions, and his score matches perfectly,
He did not leave any question unanswered.

7. Pick the pattern — find the operations done by the machine shown below.

Solution:
The operation done by Machine 1 is:
(first number) – (second number) + (third number).

So, the result of the last group will be,
(– 16) – (– 6) + (– 9) = –16 + 6 – 9 = –25 + 6 = –19.

8. 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.
Solution:
Current temperature = 8°C
Temperature drop each hour = 5°C
Total drop in 4 hours = 4 × 5°C
So, the expression that denotes temperature after 4 hours is: 8 − (4 × 5)

9. Find 3 consecutive numbers with a product of (a) – 6, (b) 120.
Solution:
(a) – 6
Check three consecutive integers: (–3), (–2), (–1)
Product:
(−3) × (−2) × (−1) = −6
Therefore, (–3), (–2), and (–1) are the required numbers.

(b) 120
Check three consecutive integers: 4, 5, 6
Product: 4 × 5 × 6 = 120
Therefore, 4, 5 and 6 are the required numbers.

10. An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins — a + 13 pibs coin and a – 9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs + 85 pibs?
Yes, we can use 10 coins of +13 pibs and 5 coins of – 9 pibs to make a total of + 85. Using the two denominations, try to get the following totals:
(a) + 20 (b) + 40
(c) – 50 (d) + 8
(e) + 10 (f) – 2
(g) + 1
[Hint: Writing down a few multiples of 13 and 9 can help.]
(h) Is it possible to purchase an item that costs 1568 pibs?
Solution:
(a) +20
Take 5 coins of +13 and 5 coins of –9:
5 × 13 − 5 × 9 = 65 − 45 = + 20 pibs.

(b) + 40
Take 10 coins of +13 and 10 coins of –9:
10 × 13 − 10 × 9 = 130 − 90 = + 40 pibs.

(c) – 50
Take 10 coins of +13 and 20 coins of –9:
10 × 13 − 20 × 9 = 130 − 180 = − 50 pibs.

(d) + 8
Take 2 coins of +13 and 2 coins of –9:
2 × 13 − 2 × 9 = 26 − 18 = + 8 pibs.

(e) + 10
Take 7 coins of +13 and 9 coins of –9:
7 × 13 − 9 × 9 = 91 − 81 = + 10 pibs.

(f) – 2
Take 13 coins of +13 and 19 coins of –9:
13 × 13 − 19 × 9 = 169 − 171 = − 2 pibs.

(g) + 1
Take 7 coins of +13 and 10 coins of –9:
7 × 13 − 10 × 9 = 91 −90 = +1 pibs.

(h) Yes, it is possible.
Take 122 coins of +13 and 2 coins of –9:
122 × 13 − 2 × 9 = 1586 − 18 = 1568 pibs.

11. Find the values of:
(a) (32 × (– 18)) ÷ ((– 36))
(b) (32 ) ÷ ((– 36) × (– 18))
(c) (25 × (– 12)) ÷ ((45) × (– 27))
(d) (280 × (– 7)) ÷ ((– 8) × (– 35))
Solution:
(a) (32 × (– 18)) ÷ ((– 36))
= −576 ÷ (– 36)
= 16.

(b) (32 ) ÷ ((– 36) × (– 18))
= (32 ) ÷ 648
= 32/648
= 4/81.

(c) (25 × (– 12)) ÷ ((45) × (– 27))
= −300 ÷ −1215
= 300/1215
= 20/81.

(d) (280 × (– 7)) ÷ ((– 8) × (– 35))
= −1960 ÷ 280
= –1960/280
= –7.

12. Arrange the expressions given below in increasing order.
(a) (– 348) + (– 1064)
(b) (– 348) – (– 1064)
(c) 348 – (– 1064)
(d) (– 348) × (– 1064)
(e) 348 × (– 1064)
(f ) 348 × 964
Solution:
(a) (−348) + (−1064) = −1412.

(b) (−348) − (−1064) = −348 + 1064 = 716.

(c) 348 − (−1064) = 348 + 1064 = 1412.

(d) (−348) × (−1064) = 348 × 1064 = 370272.

(e) 348 × (−1064) = −370272.

(f) 348 × 964 = 335472.

Increasing order: (e) → (a) → (b) → (c) → (f) → (d).

13. Given that (– 548) × 972 = – 532656, write the values of:
(a) (– 547) × 972
(b) (– 548) × 971
(c) (– 547) × 971
Solution:
(a) (– 547) × 972
Since, −547 = −548 + 1
∴ (– 547) × 972 =
= (−548 + 1) × 972
= 972 × (−548) + 972 × 1
= – 532656 + 972
= ​–531684.

(b) (– 548) × 971
Since, 971 = 972 − 1
∴ (– 548) × 971 =
= (– 548) × (972 − 1)
= (– 548) × 972 – (– 548) × 1
= – 532656 – (– 548)
= – 532656 + 548
= –532108.

(c) (– 547) × 971
Using part (a): (– 547) × 972 = –531684
Since, 971 = 972 – 1
∴ (– 547) × 971 =
= (– 547) × (972 – 1)
= (– 547) × 972 – (– 547) × 1
= –531684 + 547
= –531137.

14. Given that 207 × (– 33 + 7) = – 5382, write the value of – 207 × (33 – 7) = _________.
Solution:
Given that
207 × (−33 + 7) = −5382
207 × (– 26) = −5382
or –(207 × 26) = –5382.

Now, −207 × (33 − 7)
= −207 × 26
= −(207 × 26)
= –5382.

15. Use the numbers 3, – 2, 5, – 6 exactly once and the operations ‘+’, ‘–’, and ‘×’ exactly once and brackets as necessary to write an expression such that —
(a)  the result is the maximum possible
(b)  the result is the minimum possible
Solution:
(a) Maximum possible value
Choose the expression −6 × (−2 −(3 + 5))
Workout: 3 + 5 = 8,  −2 − 8 = −10,  −6 × (−10) = 60.
Maximum value = 60.

(b) Minimum possible value
Choose the expression ((3 + 5) − (−2)) × (−6)
Workout: 3 + 5 = 8,  8 − (−2) = 10,  10 × (−6) = −60.
Minimum value = -60.

16. Fill in the blanks in at least 5 different ways with integers:

Solution:
(a)
Five correct fillings:
1. −1 + (−5) × 7 = −36
2. 0 + 6 × (−6) = −36
3. −6 + 5 × (−6) = −36
4. 12 + (−4) × 12 = −36
5. −10 + 2 × (−13) = −36

(b)
Five correct fillings:
1. (3 − (−3)) × 2 = 12
2. (5 − 1) × 3 = 12
3. (10 − 4) × 2 = 12
4. (7 − (−5)) × 1 = 12
5. (20 − 8) × 1 = 12

(c)
Five correct fillings:
1. 1 − (4 − 4) = −1
2. 2 − (5 − 2) = −1
3. 0 − (3 − 2) = −1
4. (−1) − (4 − 2) = −1
5. 3 − (6 − 2) = −1