WBBSE Solutions For Class 7 Maths Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Solved Example Problems

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2

Solved Example Problems

Integers are all natural numbers including Zero (0) i.e., all whole numbers with a positive or negative sign.

For example 0, 3, (-3), 5, (-8), etc., are all integers.

Addition and Subtraction of the integers

You have already learned the number line and the representation of the numbers in the number line to attain the value of an expression in class 6.

Let us find the value of the expression 0+(+2)+( + 5)+(- 10)+(-6).

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The value of the expression 0 + 2 + 5-10-6 = (-9).

Step 1 >	(+ 0) + (+ 2) = + 2.
Step 2 >	(+ 2 ) + (+ 5) = + 7	(adding both integers with the same signs gives the result with that sign).
Step 3 >	(+ 7) + (- 10) = 7- 10 =-3	(By adding the integers with different signs the result will be the difference between the numbers with the sign of the greater number.)
Step 4 >	(-3) + (-6) =  (-9)	(adding both integers with the same signs gives the result with that sign).

 

Let us find the value of {( + 3) – (- 3)} – ( – 8) from the number line.

{(+ 3) – (- 3)} – (- 8) = {(+ 3) + ( 3)} + ( + 8)= ▭

 

 

So, we get {( + 3) – (- 3)} – (- 8) = + 14.

Commutative Property

Wbbse Class 7 Maths Solutions
We know 3 + 5 = 5 + 3 = 8. This infers that whole numbers can be added in any order.

So, the addition of whole numbers is commutative.

But let us see what happens in the case of integers

We know 4 + (- 6)= – 2 and (-6) + 4 = -2.

So, 4 + (-6) = (-6) + 4.

So, it is known that addition is commutative for integers, a + b = b + a

But subtraction is not commutative for whole numbers as well as integers.

Let us consider the integers 6 and ( – 3 ).

Example 1. 6-(- 3) = 6 + 3 = 9.

Example 2. – 3 – 6 = – 9.

So, we can conclude that subtraction is not commutative for integers.

Example 1. By selling each kg of mango, a fruit seller gains ₹ 5 per kg but loses ₹4 per kg of lichi. If he sells 10 kg mango and 14 kg of lichi, find his overall profit or loss.

Solution:

Given:

By selling each kg of mango, a fruit seller gains ₹ 5 per kg but loses ₹4 per kg of lichi. If he sells 10 kg mango and 14 kg of lichi,

On selling 1 kg mango, he earns ₹ 5

So, profit in 10 kg of mango = ₹ 5 x 10 = ₹50

Again, on selling 1 kg of lichi, the fruit seller losses ₹ 4, in other words, he earns – ₹ 4.

So his earning on selling 14 kg lichi = – ₹ 4 x 14 = – ₹ 56

So, his overall profit = ₹ 50 + ₹ (- 56) = – ₹ 6, that means he losses ₹ 6 after the sale.

Example 2. In an examination, 6 marks are given for correct answers, and – 3 marks are given for incorrect answers. If Rabin makes 7 correct answers and 5 incorrect answers, then what is his score?

Solution :

Given:

In an examination, 6 marks are given for correct answers, and – 3 marks are given for incorrect answers. If Rabin makes 7 correct answers and 5 incorrect answers

For 7 correct answers Rabin gets = 7 x 6 = 42 marks

For 5 incorrect answers, Rabin gets = 5 x (- 3) = – 15 marks.

So, his total marks = 42 + (-15) = 27 marks.

His total Score  = 27 marks.

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Example 3. In a quiz competition, Manika’s scores in five successive rounds were 25, – 2, – 10,15 and 10 with negative markings for wrong answers. Find her total score at the end of 5th round.

Solution:

Given:

In a quiz competition, Manika’s scores in five successive rounds were 25, – 2, – 10,15 and 10 with negative markings for wrong answers.

Her total score = 25 + (- 2) + ( – 10) + 15 + 10

= 25-2-10+15+10

= 50 – 12 = 38.

Her total score at the end of 5th round = 38

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Associative Properties

⇔ Consider the integers + 3,-5, and – 7.

Example 1. Let us find out the value of the expression by number line {(+ 3) + (- 5)} + (- 7).

Solution:

 

Example 2. If the expression be (+ 3)+{(-5)+(-7)} then find the value.

Solution:

 

We get, {( + 3) + (- 5)} + ( – 7) = ( + 3) + {(- 5) + (- 7).

We find that addition is associative for integers.

In general, for any integers a, b, and c, we can say, (a + b) + c = a + (b + c)

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Multiplication Of Integers

We know that the multiplication of whole numbers is repeated addition.

Example: 4 x 2 = 2 + 24-2 + 2 = 8.

Example 1. Let us find the value of 3 x 2 (both the integers are +ve).

Solution:

 

We get from number line 3×2 = ( + 2) + ( + 2) + ( + 2) = + 6.

Example 2. Now, find the value of 2 x 3.

Solution:

 

We get from number line =2×3=(+3)+(+3)=+6

∴ We find that in both cases the value is same.

So, 3 x 2 = 2 x 3 or a x b = b x a

Example 3. Now, let us find out the product value of 3 x (- 4) (One integer is +ve and the other is -ve).

Solution:

 

We get from number line 3x(-4)-(-4)+(- 4)+(-4)-12-(3×4)

Again, in case of 2x(- 4) = (-41) + (- 4) = (- 8) = (- 2 ) x 4

= (- 2) + (- 2)+ (-2 ) + (-2)

So, a x (-6) = (-a) x b = – (ab)

We thus find that while multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus (-) sign before the product. We thus get a negative integer.

So, we learn that when signs of integers are same, the answer is a +ve number.

( + )x( + )or(-)x(-) are always +ve.

When the signs of the integers are different, the answer is always – ve (+) x (-) or (-) x ( + ) are always – ve.

Multiplication of negative integers:

Now let us see the product of two negative integers

(-3) x (-2) = + 6

(-4) x (-5)= + 20

(- 4) x (- 3) = + 12.

So, after observing these products we understand that there is a rule prevailing over states that the product of two negative integers is always a positive integer.

So, (-a) x (-b) = a x b

Now, let us see the product of three or more negative integers:

1. (-5) x (-3) x (-2) = [-5 x -3] x (-2) = 15 x (-2) =-30.

2.( -3)x( -5)x( -2)x(-4) = [ -3 x -5] x [-2 x -4] = (+15) x ( + 8) = +120.

From the above results we observe that:

1. Product of two negative integers is a positive integer.
2. Product of three negative integers is a negative integer.
3. Product of four negative integers is a positive integer.

So, we can conclude that if the number of negative integers in a product is even then the product is always a positive integer. If the number of negative integers in a product is odd, then the product is always a negative integer.

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Commutative Property Of Multiplication

Now we observe the following results

Pattern -1	Pattern – 2	Conclusion
3x(-4) =-12, (-15) x 10 =-150, (-4)x(-5) = 20	(-4)x3= -12, 10 x (-15)=-150, (-5)x(-4)=20	3x(-4) = (-4 )x3, (-15)x10 = 10 x (-15), (-4)x(-5) = (-5)x(-4)

 

All the cases shown above confirm that multiplication is commutative in the case of integers.

So, a x b = b x a (where a & b are integers)

Distributive property of multiplication: We know that, 15 x (8 + 2) =(15 x 8) + (15 x 2) = 150

Now, let us see the cases for negative integers also: (- 3) x (2 + 5) = – 3 x 7 = – 21.

Also, [(-3)x 2] + [(-3) x5] =-6+-15=-21.

So, we can conclude that

a x (b + c) = a x b + a x c

We know, 5(9 – 3) = 5 x 6 = 30

Again, ( 5 x 9) – (5 x 3) = 45 – 15 = 30.

So, we can conclude that if a, b, c are there integers then we can write, a x ( b – c ) = a x b – a x c

Similarly,

(-5) x 3 = (- 15) > 1. (-15) + (- 5) = 3,2. (-15) + 3 = (-5)

(-3) x (-5) = 15 > 1. 15 ÷ (- 3) = (- 5), 2. 15 + (- 5) = (- 3)

Rule for division:

1. When we divide a +ve integer by a -ve integer or a -ve integer by a + ve integer, division is done considering whole numbers but putting a minus sign ( – ) before the quotient. Hence, the result is negative.

\(\frac{-15}{3}\) = – \(\frac{15}{3}\) = -5

So, a ÷ (-b) = (-a) ÷ b = – \(\frac{a}{b}\)

2. When we divide a negative integer by a negative integer, we divide the integers considering whole numbers but putting a ( + ) positive sign before the quotient.

\(\frac{-15}{5}\) = + \(\frac{15}{5}\) = +3

So, (- a) + (- b) = a ÷ b = \(\frac{a}{b}\)

3. Now let us take two integers like 24, ( – 3). If we divide one by other, then let us see the result.

24 ÷ ( – 3) = \(\frac{24}{-3}\) = – \(\frac{24}{3}\) = -8 ….(1)

Again, (-3) + 24= \(\frac{-3}{24}\) = – \(\frac{1}{8}\) …..(2)

We find (1) & (2) are not same.

So, we can write, a ÷ b ≠ b ÷ a

4. Let us take three integers like – 30, – 5, and + 2. We now set the number in two ways.

 

Set 1	Set 2
(-30) ÷ {(-5) + 2} = (-30) ÷ (-3) = 10	(-30) ÷ (-5)+ (-30) ÷ 2 = 6 +(-15) = 6 – 15 = – 9

 

∴ (-30) {(-5) + 2} ≠ (-30) ÷ (-5) + (-30) ÷ 2

So, we can write, a ÷ ( b + c) ≠ a ÷ b + a ÷ c

5. We now take three integers – 30, + 2, and 7. Let us set the numbers in two ways.

Set 1	Set 2
{(- 30) + 2 } ÷ 7 = – 28 ÷ 7 = – 4	{(-30) + 2}÷ 7 = \(\frac{-30}{7}\) + \(\frac{2}{7}\) = \(\frac{-30+2}{7}\) = \(\frac{-28}{7}\) = -4

 

So, we can write, (b + c) ÷ a = b ÷ a + c ÷ 4

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Multiplication And Division

Example 1. Find the value of the following products:

1. (- 18) x (- 9) x 9
2. (- 20) x (- 2) x { – 5) x 6
3. (- 1) x (- 5) x ( – 4) x (- 6)

Solution:

1. (- 18) x (- 9) x 9 = [(- 18) x (-9)] x 9 = 162 x 9 = 1458.
2. (- 20) x(-2)x(-5)x6 = (- 20 x – 2) x (- 5) x 6 = 40 x (- 5) x 6 = – 200 x 6 . = -1200.
3. (- 1) x (_ 5) x (- 4) x (- 6) = [(- 1) x (- 5)] x [(- 4) x (- 6)] = 5×24 = 120.

Example 2.  In a test containing 12 questions, 4 marks are given for every correct answer, and ( – 2) marks are given for every incorrect answer. Raju attempted all questions but only 8 of his answers were correct. Find his total score.

Solution:

Given:

In a test containing 12 questions, 4 marks are given for every correct answer, and ( – 2) marks are given for every incorrect answer. Raju attempted all questions but only 8 of his answers were correct.

Marks are given for each correct answer = 4

So, marks for 8 correct answers = 8 x 4 = 32

Marks given for one incorrect answer = – 2

So, marks for (12 – 8) = 4 incorrect answers = (-2) x 4 = -8.

Therefore, Raju’s total score = 32 + (- 8) = 32-8 =24.

Example 3. A shopkeeper earns a profit of ₹2 by sealing one pen and incurs a loss of ₹ 1 per pencil. In a particular month, he incurs a loss of ₹ 15, During this period, he sold 45 pens. How many pencils did he sell in this period?

Solution:

Given:

A shopkeeper earns a profit of ₹2 by sealing one pen and incurs a loss of ₹ 1 per pencil. In a particular month, he incurs a loss of ₹ 15, During this period, he sold 45 pens.

Profit earned by selling one pen = ₹ 2.

Profit earned by selling 45 pens = ₹2 x 45 = ₹ 90.

Loss incurred =₹ 15, which is denoted by – ₹ 15.

We know profit earned + loss incurred = Total loss

Therefore, Loss incurred =  Total loss – Profit earned = -₹ 15 – ₹90 = – ₹ 105.

Loss incurred by selling one pencil = ₹ 1.

So, number of pencils sold = – ₹ 105 ÷ – ₹1 = 105.

Number of pencils sold in this period = 105.