Chapter 1 Integers

We need to find a pair of integers (two integers) for each condition.

---

(a) Sum is –3

We need two integers that add up to -3.

Let's try adding a negative number and a positive number. If the negative number has a larger absolute value, the sum will be negative.

Example: $-5 + 2 = -3$.

So, a pair of integers whose sum is -3 is -5 and 2.

(Other examples: -1 + (-2) = -3, -3 + 0 = -3, -6 + 3 = -3)

---

(b) Difference is –5

We need two integers such that when the second is subtracted from the first, the result is -5.

Let the integers be $a$ and $b$. We need $a - b = -5$.

Example: Let $a = 2$. Then $2 - b = -5$. Subtract 2 from both sides: $-b = -5 - 2 = -7$. Multiply by -1: $b = 7$.

Check: $2 - 7 = -5$.

So, a pair of integers whose difference is -5 is 2 and 7.

(Other examples: $0 - 5 = -5$, $-1 - 4 = -5$, $-3 - 2 = -5$)

---

(c) Difference is 2

We need two integers such that when the second is subtracted from the first, the result is 2.

Let the integers be $a$ and $b$. We need $a - b = 2$.

Example: Let $a = 5$. Then $5 - b = 2$. Subtract 5 from both sides: $-b = 2 - 5 = -3$. Multiply by -1: $b = 3$.

Check: $5 - 3 = 2$.

So, a pair of integers whose difference is 2 is 5 and 3.

(Other examples: $0 - (-2) = 2$, $1 - (-1) = 2$, $4 - 2 = 2$)

---

(d) Sum is 0

We need two integers that add up to 0.

This happens when the two integers are additive inverses (opposites) of each other.

Example: $4 + (-4) = 0$.

So, a pair of integers whose sum is 0 is 4 and -4.

(Other examples: $1 + (-1) = 0$, $-10 + 10 = 0$, $0 + 0 = 0$)

We need to find a pair of integers (two integers) for each given condition.

---

(a) Sum is –7

We are looking for two integers, say $a$ and $b$, such that $a + b = -7$.

One way to get a negative sum is to add two negative integers. For example, if we add -3 and -4:

$-3 + (-4) = -3 - 4 = -7$

So, a pair of integers whose sum is -7 is -3 and -4.

(Other possible pairs include $0$ and $-7$, $-1$ and $-6$, $-2$ and $-5$, etc.)

---

(b) Difference is –10

We are looking for two integers, say $a$ and $b$, such that $a - b = -10$.

One way to get a negative difference is to subtract a larger number from a smaller number. For example, let the first integer be 0 and the second integer be 10:

$0 - 10 = -10$

So, a pair of integers whose difference is -10 is 0 and 10.

(Other possible pairs include $-5$ and $5$, $-2$ and $8$, etc.)

---

(c) Sum is 0

We are looking for two integers, say $a$ and $b$, such that $a + b = 0$.

This happens when the two integers are additive inverses of each other (they have the same absolute value but opposite signs). For example, if one integer is 5, its additive inverse is -5.

$5 + (-5) = 5 - 5 = 0$

So, a pair of integers whose sum is 0 is 5 and -5.

(Other possible pairs include $1$ and $-1$, $-10$ and $10$, $0$ and $0$, etc.)

We need to find pairs of integers satisfying the given conditions.

---

(a) Write a pair of negative integers whose difference gives 8.

We need two negative integers, say $a$ and $b$, such that $a < 0$, $b < 0$, and $a - b = 8$.

For the difference to be positive, the first number must be greater than the second number ($a > b$). Since both are negative, the negative number with the smaller absolute value is greater.

Example: Let $b = -10$. We need $a - (-10) = 8$, so $a + 10 = 8$. Subtract 10 from both sides: $a = 8 - 10 = -2$. Both -2 and -10 are negative integers.

Check: $(-2) - (-10) = -2 + 10 = 8$.

So, a pair of negative integers whose difference gives 8 is -2 and -10.

(Other possible pairs include $-1$ and $-9$, $-3$ and $-11$, etc.)

---

(b) Write a negative integer and a positive integer whose sum is –5.

We need a negative integer, say $a$ ($a < 0$), and a positive integer, say $b$ ($b > 0$), such that $a + b = -5$.

For the sum of a negative and a positive integer to be negative, the absolute value of the negative integer must be greater than the positive integer ($|a| > |b|$).

Example: Let the negative integer be -7 and the positive integer be 2.

$-7 + 2 = -5$.

So, a negative integer and a positive integer whose sum is -5 is -7 and 2.

(Other possible pairs include $-6$ and $1$, $-8$ and $3$, $-10$ and $5$, etc.)

---

(c) Write a negative integer and a positive integer whose difference is –3.

We need a negative integer, say $a$ ($a < 0$), and a positive integer, say $b$ ($b > 0$), such that $a - b = -3$.

Example: Let the negative integer be -1 and the positive integer be 2.

$-1 - (+2) = -1 - 2 = -3$.

So, a negative integer and a positive integer whose difference is -3 is -1 and 2.

(Other possible pairs include $-2$ and $1$, $-4$ and $1$, $-5$ and $2$, etc.)

First, we calculate the total score for each team.

---

Team A's scores in three successive rounds are –40, 10, and 0.

Total score of Team A = $(-40) + 10 + 0$

Total score of Team A = $-40 + 10 + 0$

Total score of Team A = $-30 + 0$

Total score of Team A = $-30$

---

Team B's scores in three successive rounds are 10, 0, and –40.

Total score of Team B = $10 + 0 + (-40)$

Total score of Team B = $10 + 0 - 40$

Total score of Team B = $10 - 40$

Total score of Team B = $-30$

---

Comparing the scores:

Total score of Team A = -30

Total score of Team B = -30

Since $-30 = -30$, both teams scored the same number of points.

Therefore, neither team scored more than the other; they have equal scores.

---

The scores for Team A were -40, 10, 0, and the sum was $(-40) + 10 + 0 = -30$.

The scores for Team B were 10, 0, -40, and the sum was $10 + 0 + (-40) = -30$.

The individual scores are the same for both teams, only the order in which they were obtained is different. The final sum is the same in both cases.

This example demonstrates the commutative and associative properties of addition. The order in which integers are added does not change their sum.

Yes, we can say that we can add integers in any order.

We fill in the blanks using the properties of integer addition.

---

(i) $(–5) + (–8) = (–8) + (\dots)$

This statement illustrates the commutative property of addition, which states that the order of operands does not change the sum ($a + b = b + a$).

Comparing the left and right sides, we see that $-5$ is on the left side but missing on the right side within the blank.

$(–5) + (–8) = (–8) + (\mathbf{–5})$

The missing integer is –5.

---

(ii) $–53 + \dots = –53$

This statement illustrates the additive identity property, which states that adding zero to any number results in the same number ($a + 0 = a$).

Adding what number to -53 results in -53?

$-53 + \mathbf{0} = -53$

The missing integer is 0.

---

(iii) $17 + \dots = 0$

This statement illustrates the additive inverse property, which states that adding a number to its opposite (additive inverse) results in zero ($a + (-a) = 0$).

The opposite of 17 is -17.

$17 + (\mathbf{–17}) = 0$

The missing integer is –17.

---

(iv) $[13 + (–12)] + (\dots) = 13 + [(–12) + (–7)]$

This statement illustrates the associative property of addition, which states that the grouping of operands does not change the sum $(a + b) + c = a + (b + c)$.

Comparing the left and right sides, we have 13, -12, and -7 involved in the sum.

On the left side, (13 + (-12)) are grouped. The third number is missing.

On the right side, (–12) and (–7) are grouped. The first number is 13.

The number missing on the left side is –7.

$[13 + (–12)] + (\mathbf{–7}) = 13 + [(–12) + (–7)]$

The missing integer is –7.

---

(v) $(–4) + [15 + (–3)] = [–4 + 15] + \dots$

This statement also illustrates the associative property of addition, $(a + (b + c) = (a + b) + c)$.

Comparing the left and right sides, we have –4, 15, and –3 involved in the sum.

On the left side, (15 + (–3)) are grouped. The first number is –4.

On the right side, (–4 + 15) are grouped. The third number is missing.

The number missing on the right side is –3.

$(–4) + [15 + (–3)] = [–4 + 15] + (\mathbf{–3})$

The missing integer is –3.

We find the product for each expression using the rules of integer multiplication:

$(+) \times (+) = (+)$

$(-) \times (-) = (+)$

$(+) \times (-) = (-)$

$(-) \times (+) = (-)$

Multiplication by zero is always zero.

---

(a) $3 \times (–1)$

This is a positive number multiplied by a negative number.

$3 \times (–1) = \mathbf{–3}$

---

(b) $(–1) \times 225$

This is a negative number multiplied by a positive number.

$(–1) \times 225 = \mathbf{–225}$

---

(c) $(–21) \times (–30)$

This is a negative number multiplied by a negative number.

$(–21) \times (–30) = +(21 \times 30) = +630$

$(–21) \times (–30) = \mathbf{630}$

---

(d) $(–316) \times (–1)$

This is a negative number multiplied by a negative number.

$(–316) \times (–1) = +(316 \times 1) = +316$

$(–316) \times (–1) = \mathbf{316}$

---

(e) $(–15) \times 0 \times (–18)$

Multiplying by zero always results in zero, regardless of the other numbers.

$(–15) \times 0 \times (–18) = \mathbf{0}$

---

(f) $(–12) \times (–11) \times (10)$

First, multiply the two negative numbers: $(–12) \times (–11) = +(12 \times 11) = +132$.

Then, multiply the result by the positive number: $(+132) \times 10 = 1320$.

$(–12) \times (–11) \times (10) = \mathbf{1320}$

---

(g) $9 \times (–3) \times (–6)$

First, multiply the positive number by the negative number: $9 \times (–3) = –27$.

Then, multiply the resulting negative number by the other negative number: $(–27) \times (–6) = +(27 \times 6) = +162$.

$9 \times (–3) \times (–6) = \mathbf{162}$

---

(h) $(–18) \times (–5) \times (–4)$

First, multiply the first two negative numbers: $(–18) \times (–5) = +(18 \times 5) = +90$.

Then, multiply the resulting positive number by the negative number: $(+90) \times (–4) = –(90 \times 4) = –360$.

$(–18) \times (–5) \times (–4) = \mathbf{–360}$

---

(i) $(–1) \times (–2) \times (–3) \times 4$

Multiply pairs of numbers:

$(–1) \times (–2) = +2$

$(–3) \times 4 = –12$

Now, multiply the results: $(+2) \times (–12) = –(2 \times 12) = –24$.

$(–1) \times (–2) \times (–3) \times 4 = \mathbf{–24}$

---

(j) $(–3) \times (–6) \times (–2) \times (–1)$

Multiply pairs of numbers:

$(–3) \times (–6) = +18$

$(–2) \times (–1) = +2$

Now, multiply the results: $(+18) \times (+2) = +(18 \times 2) = +36$.

$(–3) \times (–6) \times (–2) \times (–1) = \mathbf{36}$

We verify each statement by evaluating the Left Hand Side (LHS) and the Right Hand Side (RHS) separately and checking if they are equal.

---

(a) Verify $18 \times [7 + (–3)] = [18 \times 7] + [18 \times (–3)]$

LHS: $18 \times [7 + (–3)]$

First, evaluate the expression inside the brackets:

$7 + (–3) = 7 - 3 = 4$

Now, multiply 18 by the result:

$18 \times 4 = 72$

So, LHS = 72.

---

RHS: $[18 \times 7] + [18 \times (–3)]$

First, evaluate the products inside the brackets:

$18 \times 7 = 126$

$18 \times (–3) = –(18 \times 3) = –54$

Now, add the results of the products:

$126 + (–54) = 126 - 54 = 72$

So, RHS = 72.

---

Since LHS = 72 and RHS = 72, LHS = RHS.

The statement $18 \times [7 + (–3)] = [18 \times 7] + [18 \times (–3)]$ is verified.

---

(b) Verify $(–21) \times [(– 4) + (– 6)] = [(–21) \times (– 4)] + [(–21) × (– 6)]$

LHS: $(–21) \times [(–4) + (–6)]$

First, evaluate the expression inside the brackets:

$(–4) + (–6) = -4 - 6 = -10$

Now, multiply –21 by the result:

$(–21) \times (–10)$

Multiplying two negative numbers results in a positive number:

$(–21) \times (–10) = +(21 \times 10) = +210$

So, LHS = 210.

---

RHS: $[(–21) \times (–4)] + [(–21) × (–6)]$

First, evaluate the products inside the brackets:

$(–21) \times (–4) = +(21 \times 4) = +84$

$(–21) \times (–6) = +(21 \times 6) = +126$

Now, add the results of the products:

$+84 + (+126) = 84 + 126 = 210$

So, RHS = 210.

---

Since LHS = 210 and RHS = 210, LHS = RHS.

The statement $(–21) \times [(– 4) + (– 6)] = [(–21) \times (– 4)] + [(–21) × (– 6)]$ is verified.

We answer the questions based on the properties of integer multiplication.

---

(i) For any integer $a$, what is $(–1) \times a$ equal to?

Multiplying any integer by -1 gives the additive inverse (opposite) of that integer. If $a$ is positive, $(–1) \times a$ is negative. If $a$ is negative, $(–1) \times a$ is positive. If $a$ is 0, $(–1) \times 0 = 0$, which is the opposite of 0.

So, for any integer $a$, $(–1) \times a$ is equal to $-a$.

---

(ii) Determine the integer whose product with (–1) is:

Let the unknown integer be $x$. We are looking for $x$ such that $x \times (–1) = \text{given value}$. From part (i), we know that multiplying by -1 gives the opposite of the number. So, the unknown integer $x$ is the opposite of the given value.

---

(a) Product is –22

We need to find the integer $x$ such that $x \times (-1) = -22$.

The opposite of -22 is +22.

Check: $22 \times (-1) = -22$.

The integer is 22.

---

(b) Product is 37

We need to find the integer $x$ such that $x \times (-1) = 37$.

The opposite of 37 is -37.

Check: $(-37) \times (-1) = +37 = 37$.

The integer is –37.

---

(c) Product is 0

We need to find the integer $x$ such that $x \times (-1) = 0$.

The only number that gives 0 when multiplied by any non-zero number is 0 itself. Also, the opposite of 0 is 0.

Check: $0 \times (-1) = 0$.

The integer is 0.

We start with $(-1) \times 5$ and observe the pattern as the second factor decreases by 1 each time.

---

$(–1) \times 5 = –5$

$(–1) \times 4 = –4$ (Notice the product increased by 1)

$(–1) \times 3 = –3$ (The product increased by 1 again)

$(–1) \times 2 = –2$ (Product increased by 1)

$(–1) \times 1 = –1$ (Product increased by 1)

$(–1) \times 0 = 0$ (Product increased by 1)

Continuing this pattern as the second factor becomes negative:

$(–1) \times (–1)$

Following the pattern of increasing the product by 1, the next result should be $0 + 1 = 1$.

So, the pattern suggests that $(–1) \times (–1)$ should be 1.

$(–1) \times (–1) = \mathbf{1}$

This pattern provides an intuitive way to understand why the product of two negative integers is positive.

Given:

Marks for each correct answer = +5

Marks for each incorrect answer = –2

---

To Find:

Number of incorrect answers for Radhika and Jay.

---

Solution for Radhika:

Radhika's total score = 30

Number of correct answers = 10

Marks obtained for correct answers = Number of correct answers $\times$ Marks per correct answer

Marks obtained for correct answers = $10 \times (+5) = 50$

Radhika's total score is the sum of marks from correct answers and marks from incorrect answers.

Total Score = Marks from Correct Answers + Marks from Incorrect Answers

Let the marks from incorrect answers be $M_{incorrect}$.

$30 = 50 + M_{incorrect}$

Subtract 50 from both sides:

$M_{incorrect} = 30 - 50 = -20$

The marks obtained from incorrect answers is -20.

Let the number of incorrect answers be $N_{incorrect}$.

Marks from Incorrect Answers = Number of incorrect answers $\times$ Marks per incorrect answer

$-20 = N_{incorrect} \times (-2)$

To find $N_{incorrect}$, divide both sides by -2:

$N_{incorrect} = \frac{-20}{-2}$

$N_{incorrect} = 10$

Radhika attempted 10 incorrect answers.

---

Solution for Jay:

Jay's total score = –12

Number of correct answers = 4

Marks obtained for correct answers = Number of correct answers $\times$ Marks per correct answer

Marks obtained for correct answers = $4 \times (+5) = 20$

Jay's total score is the sum of marks from correct answers and marks from incorrect answers.

Total Score = Marks from Correct Answers + Marks from Incorrect Answers

Let the marks from incorrect answers be $M_{incorrect}$.

$-12 = 20 + M_{incorrect}$

Subtract 20 from both sides:

$M_{incorrect} = -12 - 20 = -32$

The marks obtained from incorrect answers is -32.

Let the number of incorrect answers be $N_{incorrect}$.

Marks from Incorrect Answers = Number of incorrect answers $\times$ Marks per incorrect answer

$-32 = N_{incorrect} \times (-2)$

To find $N_{incorrect}$, divide both sides by -2:

$N_{incorrect} = \frac{-32}{-2}$

$N_{incorrect} = 16$

Jay attempted 16 incorrect answers.

Let's denote profit by a positive integer and loss by a negative integer. We will work with amounts in paise for consistency, noting that 1 = 100 paise.

Profit per pen = +1 = +100 paise.

Loss per pencil = 40 paise = -40 paise.

---

Part (i):

Given:

Total loss in the month = 5 = -500 paise.

Number of pens sold = 45.

To Find:

Number of pencils sold.

Solution:

Total profit from selling pens = (Number of pens sold) $\times$ (Profit per pen)

Total profit from pens = $45 \times (+100 \text{ paise})$

Total profit from pens = $4500 \text{ paise}$.

The total loss incurred in the month is the sum of the total profit from pens and the total loss from pencils.

Total Loss = Total Profit from Pens + Total Loss from Pencils

Let the total loss from selling pencils be $L_{pencils}$.

$-500 \text{ paise} = 4500 \text{ paise} + L_{pencils}$

Subtract $4500 \text{ paise}$ from both sides:

$L_{pencils} = -500 \text{ paise} - 4500 \text{ paise}$

$L_{pencils} = -500 - 4500 = -5000 \text{ paise}$.

The total loss from selling pencils was -5000 paise.

Let the number of pencils sold be $N_{pencils}$.

Total Loss from Pencils = (Number of pencils sold) $\times$ (Loss per pencil)

$-5000 \text{ paise} = N_{pencils} \times (-40 \text{ paise})$.

To find $N_{pencils}$, divide the total loss from pencils by the loss per pencil:

$N_{pencils} = \frac{-5000}{-40}$

$N_{pencils} = \frac{5000}{40} = \frac{500}{4} = 125$.

She sold 125 pencils in this period.

---

Part (ii):

Given:

Total profit nor loss = 0 = 0 paise.

Number of pens sold = 70.

To Find:

Number of pencils sold.

Solution:

Total profit from selling pens = (Number of pens sold) $\times$ (Profit per pen)

Total profit from pens = $70 \times (+100 \text{ paise})$

Total profit from pens = $7000 \text{ paise}$.

The total profit/loss in the month is the sum of the total profit from pens and the total loss from pencils.

Total Profit/Loss = Total Profit from Pens + Total Loss from Pencils

Let the total loss from selling pencils be $L_{pencils}$.

$0 \text{ paise} = 7000 \text{ paise} + L_{pencils}$

Subtract $7000 \text{ paise}$ from both sides:

$L_{pencils} = 0 \text{ paise} - 7000 \text{ paise}$

$L_{pencils} = -7000 \text{ paise}$.

The total loss from selling pencils was -7000 paise.

Let the number of pencils sold be $N_{pencils}$.

Total Loss from Pencils = (Number of pencils sold) $\times$ (Loss per pencil)

$-7000 \text{ paise} = N_{pencils} \times (-40 \text{ paise})$.

To find $N_{pencils}$, divide the total loss from pencils by the loss per pencil:

$N_{pencils} = \frac{-7000}{-40}$

$N_{pencils} = \frac{7000}{40} = \frac{700}{4} = 175$.

She sold 175 pencils in the next month.

We evaluate each division expression using the rules for dividing integers.

---

(a) $(–30) \div 10$

Dividing a negative integer by a positive integer results in a negative integer.

$(–30) \div 10 = \mathbf{–3}$

---

(b) $50 \div (–5)$

Dividing a positive integer by a negative integer results in a negative integer.

$50 \div (–5) = \mathbf{–10}$

---

(c) $(–36) \div (–9)$

Dividing a negative integer by a negative integer results in a positive integer.

$(–36) \div (–9) = \mathbf{4}$

---

(d) $(–49) \div (49)$

Dividing a negative integer by a positive integer results in a negative integer.

$(–49) \div (49) = \mathbf{–1}$

---

(e) $13 \div [(–2) + 1]$

First, evaluate the expression inside the brackets:

$(–2) + 1 = –1$

Now perform the division:

$13 \div (–1)$

Dividing a positive integer by a negative integer results in a negative integer.

$13 \div (–1) = \mathbf{–13}$

---

(f) $0 \div (–12)$

Dividing zero by any non-zero integer results in zero.

$0 \div (–12) = \mathbf{0}$

---

(g) $(–31) \div [(–30) + (–1)]$

First, evaluate the expression inside the brackets:

$(–30) + (–1) = –30 – 1 = –31$

Now perform the division:

$(–31) \div (–31)$

Dividing a negative integer by a negative integer results in a positive integer.

$(–31) \div (–31) = \mathbf{1}$

---

(h) $[(–36) \div 12] \div 3$

First, evaluate the expression inside the inner brackets:

$(–36) \div 12 = –3$

Now perform the outer division:

$(–3) \div 3$

Dividing a negative integer by a positive integer results in a negative integer.

$(–3) \div 3 = \mathbf{–1}$

---

(i) $[(–6) + 5)] \div [(–2) + 1]$

First, evaluate the expressions inside both sets of brackets:

$(–6) + 5 = –1$

$(–2) + 1 = –1$

Now perform the division:

$(–1) \div (–1)$

Dividing a negative integer by a negative integer results in a positive integer.

$(–1) \div (–1) = \mathbf{1}$

We need to verify that the expression $a \div (b + c)$ is not equal to $(a \div b) + (a \div c)$ for the given values of $a$, $b$, and $c$. This shows that division is not distributive over addition for integers.

---

(a) $a = 12, b = – 4, c = 2$

We need to verify that $12 \div (–4 + 2) \neq (12 \div –4) + (12 \div 2)$.

Evaluate the Left Hand Side (LHS):

LHS = $12 \div (–4 + 2)$

First, calculate the sum inside the brackets:

$–4 + 2 = –2$

Now, perform the division:

LHS = $12 \div (–2) = –6$

---

Evaluate the Right Hand Side (RHS):

RHS = $(12 \div –4) + (12 \div 2)$

First, perform the divisions inside the brackets:

$12 \div –4 = –3$

$12 \div 2 = 6$

Now, add the results:

RHS = $(–3) + 6 = –3 + 6 = 3$

---

Compare LHS and RHS:

LHS = –6

RHS = 3

Since –6 is not equal to 3, the statement $12 \div (–4 + 2) \neq (12 \div –4) + (12 \div 2)$ is verified for the given values.

---

(b) $a = (–10), b = 1, c = 1$

We need to verify that $(–10) \div (1 + 1) \neq (–10 \div 1) + (–10 \div 1)$.

Evaluate the Left Hand Side (LHS):

LHS = $(–10) \div (1 + 1)$

First, calculate the sum inside the brackets:

$1 + 1 = 2$

Now, perform the division:

LHS = $(–10) \div 2 = –5$

---

Evaluate the Right Hand Side (RHS):

RHS = $(–10 \div 1) + (–10 \div 1)$

First, perform the divisions inside the brackets:

$–10 \div 1 = –10$

$–10 \div 1 = –10$

Now, add the results:

RHS = $(–10) + (–10) = –10 – 10 = –20$

---

Compare LHS and RHS:

LHS = –5

RHS = –20

Since –5 is not equal to –20, the statement $(–10) \div (1 + 1) \neq (–10 \div 1) + (–10 \div 1)$ is verified for the given values.

We fill in the blanks by finding the unknown integer in each division statement.

---

(a) $369 \div \_\_\_\_\_ = 369$

Dividing a number by 1 gives the same number.

$369 \div \mathbf{1} = 369$

The missing number is 1.

---

(b) $(–75) \div \_\_\_\_\_ = –1$

Dividing a number by itself (excluding 0) gives 1. To get -1, we need to divide a number by its additive inverse (opposite).

$(–75) \div \mathbf{75} = –1$

The missing number is 75.

---

(c) $(–206) \div \_\_\_\_\_ = 1$

Dividing a number by itself (excluding 0) gives 1.

$(–206) \div (\mathbf{–206}) = 1$

The missing number is –206.

---

(d) $–87 \div \_\_\_\_\_ = 87$

Dividing by -1 gives the additive inverse (opposite) of the number.

$–87 \div (\mathbf{–1}) = 87$

The missing number is –1.

---

(e) $\_\_\_\_\_ \div 1 = –87$

Dividing a number by 1 gives the number itself. Let the missing number be $x$. $x \div 1 = -87$. So, $x = -87$.

$\mathbf{–87} \div 1 = –87$

The missing number is –87.

---

(f) $\_\_\_\_\_ \div 48 = –1$

Let the missing number be $x$. $x \div 48 = -1$. To find $x$, multiply -1 by 48.

$x = -1 \times 48 = -48$

$\mathbf{–48} \div 48 = –1$

The missing number is –48.

---

(g) $20 \div \_\_\_\_\_ = –2$

Let the missing number be $x$. $20 \div x = -2$. To find $x$, divide 20 by -2.

$x = \frac{20}{-2} = -10$

$20 \div (\mathbf{–10}) = –2$

The missing number is –10.

---

(h) $\_\_\_\_\_ \div (4) = –3$

Let the missing number be $x$. $x \div 4 = -3$. To find $x$, multiply -3 by 4.

$x = -3 \times 4 = -12$

$\mathbf{–12} \div 4 = –3$

The missing number is –12.

We need to find five pairs of integers $(a, b)$ such that when $a$ is divided by $b$, the result is $-3$. This means $a = -3 \times b$, and $b$ cannot be 0.

---

Here are five such pairs:

1. Let $b = 1$. Then $a = -3 \times 1 = -3$. The pair is $\mathbf{(–3, 1)}$. Check: $-3 \div 1 = -3$.

---

2. Let $b = 2$. Then $a = -3 \times 2 = -6$. The pair is $\mathbf{(–6, 2)}$. Check: $-6 \div 2 = -3$.

---

3. Let $b = 3$. Then $a = -3 \times 3 = -9$. The pair is $\mathbf{(–9, 3)}$. Check: $-9 \div 3 = -3$.

---

4. Let $b = -1$. Then $a = -3 \times (-1) = 3$. The pair is $\mathbf{(3, –1)}$. Check: $3 \div (-1) = -3$.

---

5. Let $b = -2$. Then $a = -3 \times (-2) = 6$. The pair is $\mathbf{(6, –2)}$. Check: $6 \div (-2) = -3$. (This is the example given in the question).

Given:

Temperature at 12 noon = $10^\circ\text{C}$ (above zero, so $+10^\circ\text{C}$).

Rate of decrease = $2^\circ\text{C}$ per hour.

The temperature decreases until midnight (12 hours after 12 noon).

---

To Find:

1. The time when the temperature would be $8^\circ\text{C}$ below zero ($-8^\circ\text{C}$).

2. The temperature at midnight.

---

Solution (Part 1): Time when temperature is –8°C

The starting temperature is $+10^\circ\text{C}$. The target temperature is $-8^\circ\text{C}$.

The total decrease in temperature required is the difference between the initial and final temperatures:

Total decrease = Initial Temperature - Target Temperature

Total decrease = $10^\circ\text{C} - (-8^\circ\text{C})$

Total decrease = $10^\circ\text{C} + 8^\circ\text{C} = 18^\circ\text{C}$.

The temperature needs to decrease by $18^\circ\text{C}$.

The rate of decrease is $2^\circ\text{C}$ per hour.

The time taken for the temperature to decrease by $18^\circ\text{C}$ is:

Time = $\frac{\text{Total decrease}}{\text{Rate of decrease}}$

Time = $\frac{18^\circ\text{C}}{2^\circ\text{C/hour}} = 9$ hours.

The temperature would be $-8^\circ\text{C}$ after 9 hours from 12 noon.

9 hours after 12 noon is 9 p.m.

The temperature would be $8^\circ\text{C}$ below zero at 9 p.m.

---

Solution (Part 2): Temperature at Midnight

Midnight is 12 hours after 12 noon.

The rate of decrease is $2^\circ\text{C}$ per hour.

The total decrease in temperature from 12 noon to midnight (12 hours) is:

Total decrease = Rate of decrease $\times$ Time

Total decrease = $2^\circ\text{C/hour} \times 12 \text{ hours} = 24^\circ\text{C}$.

The temperature at midnight would be the initial temperature minus the total decrease.

Temperature at Midnight = Temperature at 12 noon - Total decrease

Temperature at Midnight = $10^\circ\text{C} - 24^\circ\text{C}$

Temperature at Midnight = $10 - 24 = -14^\circ\text{C}$.

The temperature at midnight would be –14°C.

Given:

Marks for correct answer = +3

Marks for incorrect answer = –2

Marks for not attempted question = 0

---

Part (i) - Radhika:

Given:

Radhika's total score = 20

Number of correct answers = 12

To Find:

Number of incorrect answers Radhika attempted.

Solution:

Marks obtained by Radhika for correct answers = (Number of correct answers) $\times$ (Marks per correct answer)

Marks for correct answers = $12 \times (+3) = 36$

Radhika's total score is the sum of marks from correct answers and marks from incorrect answers (since marks for not attempted questions are 0, they don't affect the score).

Total Score = Marks from Correct Answers + Marks from Incorrect Answers

Let the marks obtained from incorrect answers be $M_{incorrect}$.

$20 = 36 + M_{incorrect}$

Subtract 36 from both sides to find $M_{incorrect}$:

$M_{incorrect} = 20 - 36 = -16$

The marks obtained from incorrect answers is -16.

Let the number of incorrect answers be $N_{incorrect}$.

Marks from Incorrect Answers = (Number of incorrect answers) $\times$ (Marks per incorrect answer)

$-16 = N_{incorrect} \times (-2)$

To find $N_{incorrect}$, divide -16 by -2:

$N_{incorrect} = \frac{-16}{-2}$

$N_{incorrect} = 8$

Radhika attempted 8 questions incorrectly.

---

Part (ii) - Mohini:

Given:

Mohini's total score = –5

Number of correct answers = 7

To Find:

Number of incorrect answers Mohini attempted.

Solution:

Marks obtained by Mohini for correct answers = (Number of correct answers) $\times$ (Marks per correct answer)

Marks for correct answers = $7 \times (+3) = 21$

Mohini's total score is the sum of marks from correct answers and marks from incorrect answers.

Total Score = Marks from Correct Answers + Marks from Incorrect Answers

Let the marks obtained from incorrect answers be $M_{incorrect}$.

$-5 = 21 + M_{incorrect}$

Subtract 21 from both sides to find $M_{incorrect}$:

$M_{incorrect} = -5 - 21 = -26$

The marks obtained from incorrect answers is -26.

Let the number of incorrect answers be $N_{incorrect}$.

Marks from Incorrect Answers = (Number of incorrect answers) $\times$ (Marks per incorrect answer)

$-26 = N_{incorrect} \times (-2)$

To find $N_{incorrect}$, divide -26 by -2:

$N_{incorrect} = \frac{-26}{-2}$

$N_{incorrect} = 13$

Mohini attempted 13 questions incorrectly.

Given:

Rate of descent = 6 meters per minute (m/min).

Starting position = 10 m above ground level (+10 m relative to ground level).

Target position = –350 m (relative to ground level).

---

To Find:

Time taken to reach –350 m.

---

Solution:

First, find the total distance the elevator needs to descend. This is the difference between the starting height and the target depth.

Starting height = +10 m

Target depth = -350 m

Distance to descend = Starting height - Target depth

Distance to descend = $10 \text{ m} - (–350 \text{ m})$

Distance to descend = $10 + 350 = 360 \text{ m}$.

The elevator needs to descend a total distance of 360 meters.

---

The elevator descends at a rate of 6 m/min.

Time taken = $\frac{\text{Total distance to descend}}{\text{Rate of descent}}$

Time taken = $\frac{360 \text{ m}}{6 \text{ m/min}}$

Time taken = $60$ minutes.

Since 60 minutes is equal to 1 hour:

Time taken = 1 hour.

It will take 60 minutes (or 1 hour) for the elevator to reach –350 m.