Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 1 Integers

EXERCISE 1 (A)

Question 1: Evaluate:
1. 427 x 8 + 2 x 427
2. 394 x 12 + 394 x (-2)
3. 558 x 27 + 3 x 558

Answer:
1. 427 x 8 + 2 x 427 = 427 x (8 + 2) (Distributive property)
= 427 x 10
= 4270
2. 394 x 12 + 394 x (-2) = 394 x (12-2) (Distributive property)
= 394 x 10
= 3940
3. 558 x 27 + 3 x 558 = 558 x (27 + 3) (Distributive property)
= 558 x 30
= 16740
The distributive property allows us to group common factors together to simplify large multiplications into smaller ones. This method is much faster than multiplying each part separately and then adding the results.
Teacher's Tip: Always look for the "repeating number" in the expression; that is your common factor!
Exam Tip: Mention the property used (e.g., "Distributive Property") in your steps to gain full marks for method.

Question 2: Evaluate:
1. 673 x 9 + 673
2. 1925 x 101 – 1925

Answer:
1. 673 x 9 + 673 = 673 x (9 + 1) (Distributive property) = 673 x 10 = 6730
2. 1925 x 101 – 1925 = 1925 x (101 – 1) (Distributive property) = 1925 x 100 = 192500
When a number appears alone in an expression like this, it is secretly multiplied by 1. Recognizing this allows us to pull out the common factor and multiply by a clean number like 10 or 100.
Teacher's Tip: Think of "673" as "673 \times 1" to make the distribution easier to see.
Exam Tip: Be careful with the minus sign in the second problem; it determines whether you add or subtract 1 inside the bracket.

Question 3: Verify:
1. 37 x {8 +(-3)} = 37 x 8 + 37 x – (3)
2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
3. {7 – (-7)} x 7 = 7 x 7 – (-7) x 7
4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)
Answer:

1. 37 x {8 + (-3)} = 37 x 8 + 37 x – (3)
Answer:
L.H.S. = 37 x {8 + (-3)}
= 37 x {8-3}
= 37 x {5}
= 37 x 5
= 185
R.H.S. = 37 x 8 + 37 – 3
= 37 x (8 – 3)
= 37 x 5
= 185
Hence, L.H.S. = R.H.S.

2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
Answer:
L.H.S. = (-82) x {(_4) + 19}
= (-82) x {-4 + 19}
= (-82)x {15}
= -82 x 15
=-1230
R.H.S. = (-82) x (-4) + (-82) x 19
= -82 x (-4 + 19)
= -82 x 15
=-1230
Hence, L.H.S. = R.H.S.

3. {7 – (-7)}. x 7 = 7 x 7 – (-1) x 7
Answer:
L.H.S. = {7 – (-7)} x 7
= {7 + 7} x 7
= {14} x 7
= 14 x 7
= 98
R.H.S. = 7 x 7 – (-7) x 7
=7 x 7+7 x 7 =
7 x (7 + 7)
= 7 x (14)
= 98
Hence, L.H.S. = R.H.S.

4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)
Answer:
L.H.S. = {(-15)-8} x-6
= {-15-8} x-6
= {-23} x-6
= -23 x- 6
= 138
R.H.S. = (-15) x (-6) – 8 x (-6)
= -6 x (-15-8)
= -6 x -23
= 138
Hence, L.H.S. = R.H.S.
Verification is the process of solving both sides of an equals sign independently to check if the statement is true. It proves that mathematical properties like distribution hold true for integers with negative signs.
Teacher's Tip: Work vertically and solve one side at a time to avoid getting signs mixed up.
Exam Tip: Clearly label your "L.H.S." and "R.H.S." steps to help the teacher follow your logic.

Question 4: Evaluate:
1. 15 x 8
2. 15 x (-8)
3. (-15) x 8
4. (-15) x -8

Answer:
1. 15 x 8= 120
2. 15 x (-8) = -120
3. (-15) x 8 = -120
4. (-15) x -8 = 120
(Since the number of negative integers in the product is even)
Multiplying integers involves multiplying the absolute values and then deciding the sign. If the signs are the same, the result is positive; if they are different, the result is negative.
Teacher's Tip: Count the minus signs: 1 minus is a negative answer, 2 minuses "cancel" into a positive.
Exam Tip: Write the final sign clearly; a small missing minus sign can lose you the whole mark!

Question 5: Evaluate:
1. 4 x 6 x 8
2. 4 x 6 x (-8)
3. 4 x (-6) x 8
4. (-4) x 6 x 8
5. 4 x (-6) x (-8)
6. (-4) x (-6) x 8
7. (-4) x 6 x (- 8)
8. (-4) x (-6) x (-8)

Answer:
1. 4 x 6 x 8 = 192
2. 4 x 6 x (-8) = -192
(It have one negative factor)
3. 4 x (-6) x 8 = -192
(It have one negative factor)
4. (-4 )x 6 x 8 = -192
(It have one negative factor)
5. 4 x (-6) x (-8) = 192
(It have two negative factors)
6. (-4) x (-6) x 8 = 192
(It have two negative factors)
7. (-4) x 6 x (-8) = 192
(It have two negative factors)
8. (-4) x (-6) x (-8) = -192
(It have three negative factors)
When multiplying multiple integers, count the total number of negative signs. An odd number of negative factors gives a negative product, while an even number gives a positive product.
Teacher's Tip: Treat the multiplication like a regular math problem first, then apply the "sign rule" at the end.
Exam Tip: If you have many terms, count the minuses and write "Product is negative" or "positive" as your first note.

Question 6: Evaluate:
1. 2 x 4 x 6 x 8
2. 2 x (-4) x 6 x 8
3. (-2) x 4 x (-6) x 8
4. (-2) x (-4) X 6 x (-8)
5. (-2) x (-4) x (-6) x (-8)

Answer:
1. 2 x 4 x 6 x 8 = 384
2. 2 x (-4) x 6 x 8 = -384
(Number of negative integer in the product is odd)
3. (-2) x 4 x (-6) x 8 = 384
(Number of negative integer in the product is even)
4. (-2) x (-4) x 6 x (-8) = -384
(Number of negative integer in the product is odd)
5. (-2) x (-4) x (-6) x (-8) = 384
(Number of negative integer in the product is even)
This question reinforces the rule about even and odd numbers of negative signs. Regardless of the number of positive integers, only the count of negative integers determines the final sign.
Teacher's Tip: Zero negative signs count as an even number, so the result stays positive.
Exam Tip: Show the sign of each multiplication step if you are feeling unsure about the final result.

Question 7: Determine the integer whose product with '-1' is:
1. -47
2. 63
3. -1
4. 0

Answer:
1. -1 x 47 = -47
Hence, integer is 47
2. -1 x -63 = 63
Hence, integer is -63
3. -1 x 1 = -1
Hence, integer is 1
4. -1 x 0 = 0
Hence, integer is 0
Multiplying any integer by -1 simply changes its sign to the opposite. If the result is negative, the original number must have been positive, and vice versa.
Teacher's Tip: Think of "-1" as a "mirror" that flips the sign of whatever it touches.
Exam Tip: Remember that 0 is special; it doesn't have a positive or negative sign, so it stays 0.

Question 8: Eighteen integers are multiplied together. What will be the sign of their product, if:
1. 15 of them are negative and 3 are positive?
2. 12 of them are negative and 6 are positive?
3. 9 of them are positive and the remaining are negative?
4. all are negative?

Answer:
1. Since out of eighteen integers, 15 of them are negative, which is odd number. Hence, sign of product will be negative (-).
2. Since out of eighteen integers 12 of them are negative, which is even number. Hence sign of product will be positive (+).
3. Since out of eighteen integers 9 of them are negative, which is odd number. Hence, sign of product will be negative (-).
4. Since all are negative, which is even number. Hence sign of product will be positive (+).
The sign of a product depends solely on the number of negative factors. Positive factors have no impact on the final sign of the multiplication.
Teacher's Tip: Don't get distracted by the positive numbers; only count the "minus" signs!
Exam Tip: Use the words "Even" and "Odd" in your explanation to show you understand the rule.

Question 9: Find which is greater?
1. (8 + 10) x 15 or 8 + 10 x 15
2. 12 x (6 – 8) or 12 x 6 – 8
3. {(-3) – 4} x (-5) or (-3) – 4 x (-5)

Answer:
1. (8 + 10) x 15 or 8 + 10 x 15
(8 + 10) x 15 = 18 x 15 = 270
8 + 10 x 15 = 8 + 150 = 158
∴(8 + 10) x 15 > 8 + 10 x 15
2. 12 x (6 – 8) or 12 x 6 – 8
12 x (6 – 8) = 12 (-2) = -24
12 x 6 – 8 = 72 – 8 = 64
∴12 x 6 – 8 > 12 x (6-8)
3. {(-3) – 4} x (-5) or (-3) – 4 x (-5)
{(-3) – 4} x (-5) = {-3 – 4} x (-5) = -7 x -5 = 35
(-3) – 4 x (-5) = -7 x (-5) = 35
∴{(-3) – 4} x (-5) = (-3) – 4 x (-5)
Brackets change the order of operations significantly. Without brackets, we must multiply before adding or subtracting, which completely changes the final result.
Teacher's Tip: Use BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction) every single time.
Exam Tip: Solve both expressions completely and write them side-by-side before using the > or < symbol.

Question 10: State, true or false :
1. product of two integers can be zero.
2. product of 120 negative integers and 121 positive integers is negative.
3. a x (b + c) = a x b + c
4. (b – c) x a=b – c x a

Answer:
1. False.
2. False.
Correct : Since 120 integers are even numbers, hence product will be positive and
for 121 integers are positive in numbers, hence product will be positive.
3. False.
Correct :a x (b + c) ≠ a x b + c
ab + ac ≠ ab + c
4. False.
Correct: (b – c) x a ≠ b – c x a
ab – ac ≠ b – ca
These statements test your fundamental understanding of integer properties. Statement 3 and 4 show how missing brackets lead to incorrect applications of the distributive property.
Teacher's Tip: Remember that the distributive property requires the outside number to be multiplied by *every* number inside the brackets.
Exam Tip: When a statement is false, always write the "Correct" version to show you really know why it is wrong.

EXERCISE 1 (B)

Question 1: Divide:
(i) 117 by 9
(ii) (-117) by 9
(iii) 117 by (-9)
(iv) (-117) by (-9)
(v) 225 by (-15)
(vi) (-552) ÷ 24
(vii) (-798) by (-21)
(viii) (-910) ÷ – 26

Answer:
(i) 117 by 9 = 117/9 = 13 × 9/9 = 13
(ii) (-117) by 9 = -117/9 = -13 × 9/9 = -13
(iii) 117 by (-9) = 117/-9 = 13 × 9-9 = -13
(iv) (-117) by (-9) = -117/-9 = 117/9 = 13 × 99 = 13
(v) 225 by (-15) = -225/15 = -15 × 15/15 = -15
(vi) (-552) / 24 = -552/24 = -23 × 24/24 = -23
(vii) (-798) by (-21) = -798/-21 = 798/21 = 38 × 21/21 = 38
(viii) (-910) / -26 = 910/26 = 35 × 26/26 = 35
Division of integers follows the same sign rules as multiplication. If you divide two numbers with the same sign, the answer is positive; if they have different signs, the answer is negative.
Teacher's Tip: Use long division for larger numbers but always decide the sign first.
Exam Tip: Don't forget to include the step where you cancel out common factors to make the division cleaner.

Question 2: Evaluate:
(i) (-234) ÷ 13
(ii) 234 ÷ (-13)
(iii) (-234) ÷ (-13)
(iv) 374 ÷ (-17)
(v) (-374) ÷ 17
(vi) (-374) ÷ (-17)
(vii) (-728) ÷ 14
(viii) 272 ÷ (-17)

Answer:
(i) (-234) / 13 = -234/13 = -18 × 13/13 = -18
(ii) 234 / (-13) = 234/-13 = -18
(iii) (-234) / (-13) = 234/13 = 18
(iv) 374 / (-17) = 374/-17 = 22 × 17/-17 = -22
(v) (-374) / 17 = -374/17 = -22 × 17/17 = -22
(vi) (-374) / (-17) = 374/17 = 22 × 17/17 = 22
(vii) (-728) / 14 = -728/14 = -52 × 14/14 = -52
(viii) 272 / (-17) = -272/17 = -16 × 17/17 = -16
The same digits result in the same absolute value, but the position of the negative sign determines the final sign of the quotient. Even if both numbers are negative, the final result becomes positive.
Teacher's Tip: Think: "Like signs = Positive (+) result; Unlike signs = Negative (-) result."
Exam Tip: Double-check your division tables for numbers like 13, 17, and 19 as they frequently appear in integer problems.

Question 3: Find the quotient in each of the following divisions:Find the quotient in each of the following divisions:
(i) 299 ÷ 23
(ii) 299 ÷ (-23)
(iii) (-384) ÷ 16
(iv) (-572) ÷ (-22)
(v) 408 ÷ (-17)

Answer:
(i) 299 / 23 = 299/23 = (23 × 13)/23 = 13
(ii) 299 / (-23) = -299/23 = -13
(iii) (-384) / 16 = -384/16 = (-24 × 16)/16 = -24
(iv) (-572) / (-22) = -572/-22 = 572/22 = (26 × 22)/22 = 26
(v) 408 / (-17) = -408/17 = (-24 × 17)/17 = -24
Finding the quotient is simply the result of the division process. In these problems, we use the property that dividing two negative numbers always results in a positive quotient.
Teacher's Tip: Quotients are just the "answers" to division problems.
Exam Tip: If a division doesn't result in a whole number, check your math again; textbook problems usually result in integers.

Question 4: Divide:
(i) 204 by 17
(ii) 152 by-19
(iii) 0 by 35
(iv) 0 by (-82)
(v) 5490 by 10
(vi) 762800 by 100

Answer:
(i) 204 / 17 = {204}/{17} = {12 × 17}/17 = 12
(ii) 152 / -19 = {152}/{-19} = -{8 × 19}/19 = -8
(iii) 0 / 35 = 0/35 = 0
(iv) 0 / (-82) = 0/-82 = 0
(v) 5490 / 10 = 5490/10 = {549 × 10}/10 = 549
(vi) 762800 / 100 = 762800/100 = 7628
Dividing zero by any non-zero integer always results in zero. Also, dividing by powers of 10 like 10 or 100 is as simple as removing the corresponding number of trailing zeros.
Teacher's Tip: You can have "zero slices of a pizza," which is just zero! But you can never divide something into "zero pieces."
Exam Tip: 0 / x = 0 is a common quick-answer question; don't overthink it.

Question 5: State, true or false :
1. 0 ÷ 32 = 0
2. 0 ÷ (-9) = 0
3. (-37) ÷ 0 = 0
4. 0 ÷ 0 = 0

Answer:
1. True.
2. True.
3. False. Correct: It is not meaningful (defined).
4. False. Correct: It is not defined.
In mathematics, division by zero is strictly prohibited and results in an undefined value. While you can divide zero by a number, you cannot divide a number into zero parts.
Teacher's Tip: Remember the "No Zero Below" rule; zero can't be the divisor!
Exam Tip: Use the term "Undefined" or "Not Meaningful" when explaining why division by zero is false.

Question 6: Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12+18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 x 3+4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 -12 ÷ 4 x 2
(vii) 16 + 8 ÷ 4- 2 x 3
(viii) 16 ÷ 8 + 4 – 2 x 3
(ix) 16 – 8 + 4 ÷ 2 x 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) x (-1) + 14 – 7

Answer:
(i) 42 / 7 + 4 = 42/7 + 4 = 6 + 4 = 10
(ii) 12 + 18 / 3 = 12 + 18/3 = 12 + 6 = 18
(iii) 19 - 20 / 4 = 19 - 20/4 = 19 - 5 = 14
(iv) 16 - 5 × 3 + 4 = 16 - 15 + 4 = 1 + 4 = 5 (Note: textbook says -3, likely a sign error in source)
(v) 6 - 8 - (-6) / 2 = 6 - 8 - (-6/2) = 6 - 8 - (-3) = 6 - 8 + 3 = 9 - 8 = 1
(vi) 13 - 12 / 4 × 2 = 13 - 12/4 × 2 = 13 - 3 × 2 = 13 - 6 = 7
(vii) 16 + 8 / 4 - 2 × 3 = 16 + 8/4 - 2 × 3 = 16 + 2 - 6 = 18 - 6 = 12
(viii) 16 / 8 + 4 - 2 × 3 = 16/8 + 4 - 2 × 3 = 2 + 4 - 6 = 6 - 6 = 0
(ix) 16 - 8 + 4 / 2 × 3 = 16 - 8 + 4/2 × 3 = 16 - 8 + 2 × 3 = 16 - 8 + 6 = 22 - 8 = 14
(x) (-4) + (-12) / (-6) = (-4) + (-12/-6) = (-4) + 2 = -2
(xi) (-18) + 6 / 3 + 5 = (-18) + 6/3 + 5 = (-18) + 2 + 5 = -18 + 7 = -11
(xii) (-20) × (-1) + 14 - 7 = 20 + 14/7 ×  (Wait, source says 14 - 7)
 20 + 7 = 27 (Note: textbook says 22, likely a difference in equation layout). Using layout: 20 + 14/7 = 22.
DMAS stands for Division, Multiplication, Addition, and Subtraction. You must perform operations in this exact order from left to right to ensure the answer is correct every time.
Teacher's Tip: Even if you want to add first, don't! Multiplication and division are much "stronger" and must happen before addition and subtraction.
Exam Tip: Underline the part of the expression you are solving in each step to stay organized.

EXERCISE 1 (C)

Question 1: Evaluate: 18-(20- 15 ÷ 3)

Answer:
18-(20- 15 ÷ 3)
= 18 - (20 - 15/5}) (Correction in source note: 15/3 = 5)
= 18 – (20 – 5)
= 18 – 20 + 5
= 18 + 5 – 20
= 23 – 20
= 3
The presence of parentheses means we must solve the expression inside the bracket before we can use the 18 outside. This hierarchy ensures consistency in mathematical results.
Teacher's Tip: Work from the inside out when you see brackets!
Exam Tip: Write each step on a new line to avoid losing track of signs.

Question 2: Evaluate: -15+ 24÷ (15-13)

Answer:
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3
Subtraction inside the brackets happened first, resulting in 2. Then, the division occurred before the final addition of the negative integer.
Teacher's Tip: Think of brackets as "VIP zones" that get served first.
Exam Tip: Remember that -15 + 12 means you are 15 steps left on the number line and move 12 steps right, ending at -3.

Question 3: Evaluate: 35 - [15 + \{14 - (13 2-1+3)}]

Answer:
35- [15 + {14-(13 + 2-1+3 )}]
= 35-[15+ 14-(13+4)]
= 35 — [15 + 14 – (13 + 4}]
= 35-{15 + 14-17]
= 35-15-14+ 17
= 35 + 17-15-14
= 52 – 29
= 23

Question 4: Evaluate: 27 - [13 + {4 - (8 + 4 - 1 + 3})}]

Answer:
27- [13 + {4-(8 + 4 –1 + 3 )}]
= 27-[13 +{4-(8+ 4 - 1 + 3 )}]
= 27-[13 + {4-8}]
= 27 – [13 + (-4)]
= 21 – [9]
= 27-9
= 18
A vinculum (the bar over numbers) acts as the highest priority bracket. You must solve the part under the bar before everything else, including parentheses.
Teacher's Tip: The vinculum bar is like an extra-strong magnet pulling those numbers together first.
Exam Tip: Pay close attention to the vinculum; missing it will result in the wrong order of operations.

Question 5: Evaluate: 32 - [43 - {51 - (20 - {18 - 7})\}]

Answer:
32 – [43 – {51 – (20 – 18 - 7)}]
= 32-[43 – {51 -(20- 11)}]
= 32-[43-{51 -9}]
= 32-[43 -42]
= 32-1
=31

Question 6: Evaluate: 46-[26-{14-(15-4÷ 2 x 2)}]

Answer:
46 – [26 – {14 – (15 – 4 ÷ 2 x 2)}]
= 46-[26- {14-(15-2 x 2)}]
= 46-[26- {14-(15 -4)}]
= 46-[26- {14- 11}]
= 46 – [26 – 3]
= 46 – 23
= 23
Even within the innermost bracket, the DMAS rule still applies. You must divide 4 by 2 before you can move on to multiplication or subtraction inside the parentheses.
Teacher's Tip: DMAS applies inside *every* set of brackets individually.

Question 7: Evaluate: 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]

Answer:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
= 45-[38- {60 ÷ 3-(6-3)÷ 3}]
= 45-[38 -{20-3 ÷ 3}]
= 45-[38- {20-1}]
= 45-[38- 19]
= 45-19
= 26
Solving the 9 / 3 part first is essential because it is nested inside the deepest parentheses. Once that is done, we continue with the remaining divisions before we can subtract from 45.
Teacher's Tip: If you get a fraction, you probably did the subtraction before the division.
Exam Tip: Don't lose the negative sign in front of the square bracket as you work downwards.

Question 8: Evaluate: 17 - [17 - {17 - (17 - {17 - 17})\}]

Answer:
17- [17-{17-(17 – 17 )}]
= 17-[17-{17-(17-0)}]
= 17 – [17 – {17 — 17}]
= 17 — [17 — 0]
= 17-17
= 0
This problem is a clever exercise in recognizing the repeating zero pattern caused by the brackets. Each inner subtraction results in either 17 or 0, eventually canceling everything out to zero.
Teacher's Tip: These look intimidating but are often very simple if you solve from the center.
Exam Tip: Be precise with your brackets; missing one will ruin the whole zero-cancellation effect.

Question 9: Evaluate: 2550 - [510 - {270 - (90 - {80 + 7)}]

Answer:
2550- [510-{270-(90- )}]
= 2550 – [510 – {270 – (90 – 87)}]
= 2550 -[510- {270 -3}]
= 2550-[510-267]
= 2550 – 243
= 2307
The vinculum over 80 + 7 is solved first to get 87. Subtracting this from 90 gives us the key value needed to begin stripping away the outer curly and square brackets.
Teacher's Tip: For big numbers like 2550, use the margin for scratch calculations to keep your main steps neat.
Exam Tip: Always double-check your subtraction: 510 - 267 = 243.

Question 10: Evaluate: 30 + [{-2 × (25 -{13 - 3})}]

Answer:
30+ [{-2 x (25- 13 - 3)}]
= 30 + [{-2 x (25 – 10)}]
= 30 + [{-2 x 15}]
= 30 + [-30]
= 30-30
= 0
The negative integer multiplication inside the bracket resulted in -30. When we add -30 to the positive 30 outside, the result is zero, demonstrating how additive inverses work.
Teacher's Tip: Adding a negative is the same as subtraction: 30 + (-30) is just 30 - 30.
Exam Tip: Put negative results in brackets to keep them separate from operation signs.

Question 11: Evaluate: 88-{5-(-48)+ (-16)}

Answer:
88 - {5 - (-48)+(-16)}
88 –{5 -(-48)/(-16)}
= 88 – {5-3}
= 88 – 2
= 86

Question 12: Evaluate: 9 × (8 - {3 + 2}) - 2 × (2 +{3 + 3})

Answer:
9 × (8 - {3 + 2}) - 2 × (2 +{3 + 3})
= 9 x (8 – 5) – 2(2 + 6)
= 9 x 3 – 2 x 8
= 27- 16
= 11
There are two independent parts separated by a minus sign in this expression. We solve each part following the order of brackets and vinculum first, then multiply, and finally subtract.
Teacher's Tip: Imagine there is a "wall" between the two parts until the very last step.
Exam Tip: Don't forget that multiplication comes before the final subtraction!

Question 13: Evaluate: 2 - [3 - {6 - (5 - {4 - 3})}]

Answer:
2 – [3 – {6 – (5 – {4 - 3})}]
⇒ 2 – [3 – {6 – (5 – 1)}]
⇒ 2 – [3 – {6 – 4}]
⇒2 – (3 – 2)
⇒2-1 = 1
The complexity here is the depth of the nesting. By methodically solving from vinculum to the square bracket, we simplify the problem into a simple "2 minus 1" calculation.
Teacher's Tip: Draw arrows from the brackets to show which ones you have finished.
Exam Tip: Make sure your final answer is clearly visible; underlining it helps.

EXERCISE 1 (D)

Question 1: The sum of two integers is -15. If one of them is 9, find the other.

Answer:
Sum of two integers = -15
One integer = 9
∴ Second integer = -15 – 9
= -(15 + 9)
= -24
When we know the sum and one part, we find the other part by subtracting. In the case of integers, subtracting a positive from a negative moves you further into the negative side of the number line.
Teacher's Tip: Think of it like this: "What do I add to 9 to get -15?" You need to go back 24 steps!

Question 2: The difference between an integer and -6 is -5. Find the values of x.

Answer:
The difference between an integer
= x-(-6) = -5
∴ Value of
⇒ x – (-6) = -5
⇒ x + 6 = -5
x = -5 – 6
x = -11

Question 3: The sum of two integers is 28. If one integer is -45, find the other.

Answer:
Sum of two integers = 28
One integer = -45
∴ Second integer = 28 – (-45)
= 28 + 45
= 73
Subtracting a negative number is the same as adding its absolute value. This is why the result is 73, which is larger than the original sum of 28.
Teacher's Tip: If you subtract a "debt" (negative), you are actually gaining "money" (positive)!
Exam Tip: Show the step where -(-45) becomes +45 to prove your understanding of signs.

Question 4: The sum of two integers is -56. If one integer is -42, find the other.

Answer:
Sum of two integers = -56
One integer = -42
∴Second integer = -56 – (-42)
= -56+ 42
=-14
In this scenario, both integers are negative. We find the difference by adding the absolute value of the first integer to the sum, which moves us closer to zero on the negative scale.
Teacher's Tip: -56 is your destination. You already have -42. You only need -14 more to get there.
Exam Tip: Be extremely careful with sign changes; -56 + 42 is not 98!

Question 5: The difference between an integer x and (-9) is 6. Find all possible values of x.

Answer:
The difference between an integer x – (-9) = 6 or -9 – x = 6
∴ Value of x
⇒ x – (-9) = 6 or ⇒ -9 – x = 6
⇒ x + 9 = 6 or Answer-x = 6 + 9
⇒ x = 6 – 9 or ⇒ -x = 15
⇒x = -3 or ⇒ x = -15
Hence, possible values ofx are -3 and -15.
A "difference" between two numbers can be calculated in two directions (A-B or B-A). This leads to two separate equations and two valid possible answers for x.
Teacher's Tip: "All possible values" is a clue that there is more than one answer!
Exam Tip: Split your page into two columns to solve both possible equations simultaneously.

Question 6: Evaluate:
1. (-1) x (-1) x (-1) x ….60 times.
2. (-1) x (-1) x (-1) x (-1) x …. 75 times.

Answer:
1. 1 (because (-1) is multiplied even times.)
2. -1 (because (-1) is multiplied odd times.)

Question 7: Evaluate:
1. (-2) x (-3) x (-4) x (-5) X (-6)
2. (-3) x (-6) x (-9) x (-12)
3. (-11) x (-15) + (-11) x (-25)
4. 10 x (-12) + 5 x (-12)

Answer:
1. (-2) x (-3) x (-4) x (-5) x (-6)
⇒ 6 x 20 x (-6) = 120 x (-6)
= -720
2. (-3) x (-6) x (-9) x (-12)
⇒ 18 x 108
= 1944
3. (-11) x (-15) + (-11) x (-25)
⇒ 165 + 275
= 440
4. 10 x (-12) + 5 x (-12)
⇒ -120-60
= -180

Question 8:
1. If x x (-1) = -36, is x positive or negative?
2. If x x (-1) = 36, is x positive or negative?

Answer:
1. x x (-1) = -36
-lx = -36
x = -36/-1
x = 36
∵ x = 36
∴ It is a positive integer.
2. x x (-1) = 36
-1x = 36
x = 36/-1
x = -36
∵x = -36
∴It is a negative integer.
therefore It is a negative integer.
By dividing the result by -1, we find the original value of x. This tests the rule that a negative product requires one positive and one negative factor.
Teacher's Tip: Multiplication by -1 is like a "toggle switch" for signs.
Exam Tip: Don't just guess; show the algebra to prove your reasoning.

Question 9: Write all the integers between -15 and 15, which are divisible by 2 and 3.

Answer:
The integers between -15 and 15 are :
-12, -6, 0, 6 and 12
That are divisible by 2 and 3.

Question 10: Write all the integers between -5 and 5, which are divisible by 2 or 3.

Answer:
The integers between -5 and 5 are :
-4, -3, -2, 0, 2, 3 and 4
That are divisible by 2 or 3.
The word "or" means the number can be a multiple of 2, a multiple of 3, or both. We check each integer in the range to see if it fits either category.
Teacher's Tip: "Or" is more inclusive than "and," so your list will usually be longer.
Exam Tip: List the numbers in order from smallest to largest to keep your work neat.

Question 11: Evaluate:
1. (-20) + (-8) ÷ (-2) x 3
2. (-5) – (-48) ÷ (-16) + (-2) x 6
3. 16 + 8 ÷ 4- 2 x 3
4. 16 ÷ 8 x 4 – 2 x 3
5. 27 – [5 + {28 – (29 – 7)}]
6. 48 – [18 – {16 – (5 – 4+1)}]
7. -8 – {-6 (9 – 11) + 18 = -3}
8. (24 ÷ 12-9 – 12) – (3 x 8 ÷ 4 + 1)
Answer:
We know that, if these type of expressions that has more than one fundamental operations, we use the rule of DMAS i.e., First of all we perform D (division), then M (multiplication), then A (addition) and in the last S (subtraction).
1. (-20) + (-8) ÷ (-2) x 3
⇒ -20 + 4 x 3
⇒ -20+ 12
=-8

2. (-5) – (-48) ÷ (-16) + (-2) x 6
Answer:
⇒ (-5) – 3 + (-2) x 6
⇒ -5 – 3 – 12
⇒ -8- 12
= -20

3. 16 + 8 ÷ 4 – 2 x 3
Answer:
⇒ 16 + 2 – 2 x 3
⇒16 + 2 – 6
⇒ 18-6
= 12

4. 16 ÷ 8 x 4 – 2 x 3
Answer:
⇒ 2 x 4 – 2 x 3
⇒ 8 – 6
= 2

5. 27 – [5 + {28 – (29 – 7)}]
Answer:
⇒ 27 – [5 + {28 – 22}]
⇒ 27 – [5 + 6]
⇒ 27 — 11
= 16

6. 48-[18-{16-(5 – 4+1)}]
Answer:
⇒ 48-[18-{16-(5-5)}]
⇒ 48-[18- {16-0)}]
⇒ 48-[18- 16]
⇒ 48 – 2
= 46

7. -8 – {-6 (9 – 11) + 18 ÷ -3}
Answer:
⇒ -8 – {-6 (-2) – 6}
⇒ -8- {12-6}
⇒ -8 – {6}
⇒ -8-6
= -14

8. (24 ÷ 12-9 – 12) – (3 x 8 = 4 + 1)
Answer:
⇒ (24 ÷ 3-12)-(3 x 2 + 1)
⇒ (8- 12)-(6+ 1)
⇒ —4 — 7
= —11

This collection of problems serves as a final review of the rules for BODMAS and the properties of integers. They show how small changes in signs or brackets can lead to entirely different final answers.
Teacher's Tip: If you see a minus sign outside a bracket, remember it acts like -1 multiplying everything inside when you open it!
Exam Tip: Don't rush; these are the most "scorable" questions if you just take them one step at a time.

Question 12: Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.

Answer:
Required number = (Sum of all integers between 20 and 30 – Integers between 20 and
30)
(20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 +
27 + 28 + 29 )
⇒ 20 + 30 = 50
∴ Required number = 50
The integers "from 20 to 30" include both endpoints, whereas "between 20 and 30" excludes them. When you subtract the two sums, only the endpoints (20 and 30) are left to be added.
Teacher's Tip: You don't need to add the whole long string! Just cross out the numbers that appear in both brackets.
Exam Tip: Read keywords like "from" and "between" very carefully as they define your set of numbers.

Question 13: Add the product of (-13) and (-17) to the quotient of (-187) and 11.

Answer:
(-13) x (-17)+ (-187- 11)
⇒ (-13) x (-17) + (-17)
⇒ 221 – 17 = 204
The problem requires two operations first: a multiplication and a division. Adding a negative result (-17) to a positive product (221) means we are actually performing a subtraction.
Teacher's Tip: Use BODMAS; find the product and the quotient before you do the adding.
Exam Tip: Product of two negatives is positive (+221), but the quotient of a negative and positive is negative (-17).

Question 14: The product of two integers is -180. If one of them is 12, find the other.

Answer:
The product of two integers = -180 One integer = 12
∴ Second integer = -180 – 12 = -15
Division is the inverse operation of multiplication. By dividing the total product by the known factor, we identify the missing number, which must be negative to result in a negative product.
Teacher's Tip: Think: "12 times what equals -180?" Since the result is negative, our answer must have a minus sign.
Exam Tip: Always perform the division to get the absolute value first, then attach the correct sign.

Question 15:
1. A number changes from -20 to 30. What is the increase or decrease in the number?
2. A number changes from 40 to -30. What is the increase or decrease in the number?

Answer:
1. ∵A number changes from = -20 to 30
⇒ -20 – 30 = -50
∴-50, it will be increases.
2. ∵A number changes from = 40 to -30
⇒ 40 – (-30)
40 + 30 = 70
∴70, it will be decreases"