Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 6 We Distribute Yet Things Multiply Class 8 Question Answer to understand textbook questions step by step.
Class 8 Maths Chapter 6 We Distribute Yet Things Multiply Solutions
Ganita Prakash Class 8 Chapter 6 Solutions
Class 8 Maths Ganita Prakash Chapter 6 Solutions We Distribute Yet Things Multiply
IS THIS A MULTIPLF OF?
Figure it Out (Page 142 – 143) :
Question 1.
Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 x 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.
Answer:
| 3 × 5 | 3 × 6 | 3 × 7 |
| 4 × 5 | 4 × 6 | 4 × 7 |
| 5 × 5 | 5 × 6 | 5 × 7 |
| (p – 1)(q – 1) | (p – 1)q | (p – 1) (q + 1) |
| P(q – 1) | pq | P(q + 1) |
| (p + 1) (q – 1) | (p + 1)q | (p + 1) (q + 1) |
Question 2.
Expand the following products.
(i) (3 + u) (v – 3)
(ii) \(\frac{2}{3}\)(15 + 6a)
(iii) (10a + b) (10c + d)
(iv) (3 – x) (x – 6)
(v) (-5a + b) (c + d)
(vi) (5 + z) (y + 9)
Answer:
(i) (3 + u) (v – 3) = 3(v – 3) + u(v – 3)
= 3v – 9 + uv – 3u = 3v – 3u + uv – 9
(ii) \(\frac{2}{3}\) (15 + 6a) = \(\frac{2}{3}\) × 15 + \(\frac{2}{3}\) × 6a = 10 + 4a
(iii) (10a + b) (10c + d)
= 10a × 10c + 10a × d + b × 10c + b × d
= 100ac + 10ad + 10bc + bd
(iv) (3 – x) (x – 6) = 3(x – 6) – x(x – 6)
= 3x – x2 – 18 + 6x = – x2 + 9x – 18.
(v) (- 5a + b) (c + d)
= (- 5a + b)c + (- 5a + b)d
= – 5ac + bc – 5ad + bd
= – 5ac – 5ad + bc + bd.
(vi) (5 + z) (y + 9)
= (5 + z)y + (5 + z)9
= 5y + zy + 45 + 9z
= 5y + 9z + zy + 45.
Question 3.
Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Answer:
Let the two numbers be x and y, then:
x × y = (x + 2) × (y – 4)
xy = (x + 2)y – (x + 2)4
xy = xy + 2y – (4x + 8)
xy = xy + 2y – 4x – 8
xy – xy = 2y – 4x – 8
0 = 2y – 4x – 8
4x + 8 = 2y
2(2x + 4) = 2y
y = 2x + 4.
Examples :
(i) x = 1, y = 6 → Product = 1 × 6 = 6
Check : (1 + 2) × (6 – 4) = 3 × 2 = 6.
(ii) x = 2, y = 8 → Product = 16
Check : (2 + 2) × (8 – 4) = 4 × 4 =16.
(iii) x = 5, y = 14 → Product = 5 × 14 = 70
Check : (5 + 2) × (14 – 4) = 7 × 10 = 70.
Therefore, (1, 6), (2, 8), and (5, 14) are three examples for the given situation.
Question 4.
Expand
(i) (a + ab – 3b2) (4 + b), and
(ii) (4y + 7)(y + 11z – 3).
Answer:
(i) (a + ab – 3b2) (4 + b)
= (a + ab – 3b2)4 + (a + ab – 3b2)b
= 4a + 4ab – 12b2 + ab + ab2 – 3b3
= – 3b3 – 12b2 + ab2 + 4ab + ab + 4a
= – 3b3 – 12b2 + ab2 + 5ab + 4a.
(ii) (4y + 7) (y + 11z – 3)
= (4y + 7)y + (4y + 7)11z – (4y + 7)3
= 4y2 + 7y + 44yz + 77z – (12y + 21)
= 4y2 + 7y + 44yz + 77z – 12y – 21
= 4y2 + 7y – 12y + 44yz + 77z – 21
= 4y2 – 5y + 44yz + 77z – 21.
Question 5.
Expand (i) (a – b) (a + b),
(ii) (a – b) (a2 + ab + b2) and
(iii) (a – b)(a3 + a2b + ab2 + b3), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
Answer:
(i) (a – b)(a + b) = (a – b)a + (a – b)b
= a2 – ab + ab – b2 = a2 – b2.
(ii) (a – b) (a2 + ab + b2)
= (a – b)a2 + (a – b)ab + (a – b)b2
= a3 – a2b + a2b – ab2 + ab2 – b3 = a3 – b3.
(iii) (a – b)(a3 + a2b + ab2 + b3)
= (a – b)a3 + (a – b)a2b + (a – b)ab2 + (a – b)b3
= a4 – a3b + a3b – a2b2 + a2b2 – ab3 + ab3 – b4
= a4 – b4.
The next identity would be : (a – b)(a4 + a3b + a2b2 + ab3 + b4) = a5 – b5.
By expanding we can check it as :
(a – b) (a4 + a3b + a2b2 + ab3 + b4)
= a(a4 + a3b + a2b2 + ab3 + b4) – b(a4 + a3b + a2b2 + ab3 + b4)
= a5 + a4b + a3b2 + a2b3 + ab4 – a4b – a3b2 – ab4 – b5 = a5 – b5
2. SPECIAL CASES OFTHE DISTRIBUTIVE PROPERTY
Figure it Out (Page 149) :
Question 1.
Which is greater: (a – b)2 or (b – a)2? Justify your answer.
Answer:
Here, (a – b)2 = a2 + b2 – 2ab ……….. (1)
and (b – a)2 = b2 + a2 – 2ba
b2 + a2 = a2 + b2 and ba = ab
(b – a)2 = a2 + b2 – 2ab ……… (2)
Comparing (1) and (2), we get
(a – b)2 = (b – a)2
Question 2.
Express 100 as the difference of two squares.
Answer:
a2 – b2 = 100
(a + b) (a – b) = 100
[100 = 1 × 100, 2 × 50, 4 × 25, 5 × 20, 10 × 10]
We can take anyone
Let us take 50 × 2 = 100
Hence, (a + b) (a – b)= 50 × 2
a + b = 50 ……… (1)
a – b = 2 …….. (2)
Adding (1) and (2)
2a = 52
⇒ a = 26
Substituting a = 26 in (1)
26 + b = 50
⇒ b = 50 – 26 = 24
Let us check 262 – 242 = 676 – 576 = 100
Hence 262 – 242 = 100
Question 3.
Find 4062, 722, 1452, 10972 and 1242 using the identities you have learnt so far.
Answer:
(i) 4062 = (400 + 6)2
= 4002 + 2 × 400 × 6 + 62
= 160000 + 4800 + 36 = 164836
(ii) 722 = (50 + 22)2
= 502 + 2 × 50 × 22 + 222
= 2500 + 2200 + 484 = 5184
(iii) 1452 = (150 – 5)2
= 1502 – 2 × 150 × 5 + 52
= 22500 – 1500 + 25
= 21025
(iv) 10972 = (1100 – 3)2
= 11002 – 2 × 1100 × 3 + 32
= 1210000 – 6600 + 9
= 1203409
(v) 1242 = (100 + 24)2
= 1002 + 2 × 100 × 24 + 242
= 10000 + 4800 + 576
= 15376
Question 4.
Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Answer:
Pattern 1
2(a2 + b2) – (a + b)2 + (a – b)2
Case-I
Let a = 4, b = 2
LHS = 2(42 + 22)
= 2 × (16 + 4) = 40
RHS = (4 + 2)2 + (4 – 2)2
= 36 + 4 = 40
∴ Pattern 1 holds for counting numbers.
Case-II
Let a = -4, b = -2
LHS = 2((-4)2 + (-2)2)
= 2 × (16 + 4) = 2 × 20 = 40
RHS = (-4 + (-2))2 + (-4 – (-2))2
= (- 4 – 2)2 + (- 4 + 2)2
= (-6)2 + (-2)2 = 36 + 4 = 40
LHS = RHS
∴ Pattern 1 holds for negative integers also.
Case-III
The pattern holds for fractions also.
Pattern 2
a2 – b2 = (a + b) (a – b)
Case – I
Let a = 5, b = 3
LHS = 52 – 32 = 25 – 9 = 16
RHS = (5 + 3) (5 – 3) = 8 × 2 = 16
∴ LHS = RHS
∴ Pattern 2 holds for counting numbers.
Case-II
Let a = -5, b = -3
Now, LHS = (-5)2 – (-3)2 = 25 – 9 = 16
and RHS = [(-5) + (-3)] [(-5) – (-3)]
= (- 5 – 3) (- 5 + 3)
= (-8)(-2) = 16
∴ LHS = RHS
∴ Pattern 2 holds for negative integers also.
Case-III
and RHS = (\(\frac{1}{2}\) + \(\frac{1}{3}\)) (\(\frac{1}{2}\) – \(\frac{1}{3}\))
= (\(\frac{3 + 2}{6}\)) (\(\frac{3 – 2}{6}\))
= \(\frac{5}{6}\) \(\frac{1}{6}\) = \(\frac{5}{36}\)
∴ LHS = RHS
∴ Pattern 2 holds for fractions also.
3. THIS WAY OR THAT WAY, ALL WAYS LEAD TO THE BAY
Figure it Out (Page 154 – 156) :
Question 1.
Compute these products using the suggested identity.
(i) 462 using Identity 1A for (a + b)2
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
(iii) 912 using Identity 1B for (a – b)2
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)
Answer:
(i) 462 = (40 + 6)2 = 402 + 2 × 40 × 6 + 62
[∵ (a + b)2 = a2 + 2ab + b2]
= 1600 + 480 + 36 = 2116
(ii) 397 × 403 = (400 – 3) (400 + 3)
[∵ (a + b) × (a – b) = a2 – b2] = 4002 – 32 = 160000 – 9 = 159991
(iii) 912 = (100 – 9)2 = 1002 – 2 × 100 × 9 + 92
[∵ (a – b)2 = a2 + b2 – 2ab] = 10000 – 1800 + 81 = 8281
(iv) 43 × 45 = (44 – 1) (44 + 1)
[∵ a2 – b2 = (a + b) × (a – b)]
= 442 – 12 = 1936 – 1 = 1935
Question 2.
Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1) (p + 11)
(ii) (3a – 9b) (3a + 9b)
(iii) -(2y + 5) (3y + 4)
(iv) (6x + 5y)2
(v) (2x – \(\frac{1}{2}\))2
(vi) (7p) × (3r) × (p + 2)
Answer:
(i) (p – 1) (p + 11) = p(p + 11) – 1(p + 11)
= p2 + 11p – p – 11 = p2 + 10p – 11
(ii) (3a – 9b) (3a + 9b) = (3a)2 – (9b)2 = 9a2 – 81b2
(iii) – (2y + 5)(3y + 4) = (- 2y – 5) (3y + 4)
= – 2y(3y + 4) – 5(3y + 4)
= – 6y2 – 8y – 15y – 20 = – 6y2 – 23y – 20
(iv) (6x + 5y)2 = (6x)2 + 2(6x) (5y) + (5y)2
= 36x2 + 60xy + 25y2
(v) (2x – \(\frac{1}{2}\))2 = (2x)2 – 2 × 2x × \(\frac{1}{2}\) + (\(\frac{1}{2}\))2
= 4x2 – 2x + \(\frac{1}{4}\)
(vi) (7p) × (3r) × (p + 2) = 7p × 3r × (p + 2)
= 21pr(p + 2) = 21pr × p + 21pr × 2
= 21p2r + 42pr
Question 3.
For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
2 + s (s + 2)2 s2 + 2 s2 + 4 2s2 22s
(ii) The sum of the squares of two consecutive numbers
m2 + n2 (m + n)2
m2 + 1 m2 + (m + 1)2
m2 + (m – 1)2
(m + (m + 1))2 (2m)2 + (2m + 1)2
Answer:
(i) For “Two more than a square number”: The correct expression is s2 + 2.
(ii) For “The sum of the squares of two consecutive numbers”: The correct expression is m2 + (m + 1)2.
Question 4.
Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.
Find products of numbers lying along each diagonal – 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Hint: Label the numbers in each 2 by 2 square as
| a | (a + 1) |
| a + 7 | (a + 8) |
Answer:
Case – I
| 6 | 7 |
| 13 | 14 |
Here, 6 × 14 = 84
13 × 7 = 91
Difference = 91 – 84 = 7
Case – II
| 9 | 10 |
| 16 | 17 |
Here, 9 × 17 = 153
16 × 10= 160
Difference = 160 – 153 = 7
We observe that the difference of the diagonal products in both cases is always 7.
Question 5.
Verify which of the following statements are true.
(i) (k + 1) (k + 2) – (k + 3) is always 2.
(ii) (2q + 1) (2q – 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)2 – (4n + 3)2 is 5 less than a square number.
Answer:
(i) (k + 1)(k + 2) – (k + 3) is a multiple of 2
Let k = 5, Then (5 + 1) (5 + 2) – (5 + 3)
= 6 × 7 – 8 = 42- 8 = 34
34 is a multiple of 2.
∴ The statement is true.
(ii) (2q + 1) (2q – 3) is a multiple of 4.
Let q = 3, Then (6 + 1) (6 – 3)
= 7 × 3 = 21
21 is not a multiple of 4
∴ The statement is false.
(iii) The square of an even number is a multiple of 4.
22 – 4 = 4 × 1
42 = 16 = 4 × 4
62 = 36 = 4 × 9
∴ The statement is true.
The square of an odd number is 1 more than a multiple of 8.
32 = 9 = 8 × 1 + 1
52 = 25 = 8 × 3 + 1
72 = 49 = 8 × 6 + 1
∴ The statement is true.
(iv) (6n + 2)2 – (4n + 3)2 is 5 less than a square number.
Let n = 2, (6 × 2 + 2)2 – (4 × 2 + 3)2
= 142 – 112 = 196 – 121 = 75 = 80 – 5
But 80 is not a square number.
∴ The statement is false.
Question 6.
A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?
Answer:
Let the numbers be x and y.
x = 7a + 3, y = 7b + 5
Sum = x + y
= 7a + 3 + 7b + 5 = 7(a + b) + 8
= 7(a + b) + 7 + 1 = 7(a + b + 1) + 1
∴ The remainder on division by 7 is 1.
Difference = x – y
= (7a + 3) – (7b + 5)
= 7a + 3 – 7b – 5 = 7(a – b) – 2
= 7(a – b) – 1 + 5 (∵ -2 = – 7 + 5)
= 7(a – b – 1) + 5
∴ The remainder on division by 7 is 5.
Product = xy
= (7a + 3) (7b + 5)
= 49ab + 35a + 21b + 15
= (49ab + 35a + 21b + 14) + 1
= 7(7ab + 5a + 3b + 2) + 1
∴ The remainder on division by 7 is 1.
Question 7.
Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
Answer:
Let us take the numbers 7, 8, 9
Now, 82 – 7 × 9 = 64 – 63 = 1
Let us take the numbers 10, 11, 12
Then 112 – 10 × 12 = 121 – 120 = 1
Generalizing:
Let the numbers be a – 1, a, a + 1
Then a2 – (a + 1) (a – 1) = 1
LHS = a2 – (a + 1)(a – 1)
= a2 – (a2 – 1)
= a2 – a2 + 1 = 1
LHS = RHS
∴ Hence, the identity is correct.
Question 8.
What is the algebraic expression describing the following steps – add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
Answer:
Let the two numbers be a and b.
Step 1: a + b
Step 2: (a + b) × \(\frac{1}{2}\) (a + b)
∴ (a + b) × \(\frac{1}{2}\) (a + b) = \(\frac{1}{2}\) (a + b)2
Question 9.
Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
Answer:
(i) Let p = 14 × 26
p’ = 16 × 24
= (14 + 2) (26 – 2)
= 14 × 26 + 2 × 26 – 14 × 2 – 2 × 2
= 14 × 26 + 2(26 – 14 – 2)
= 14 × 26 + 2 × 10
p’ = p + 2 × 10
∴ p’ > p or 16 × 24 > 14 × 26
(ii) Let p = 25 × 75
p’= 26 × 74
=(25 + 1) (75 – 1)
=25 × 75 + 75 × 1 – 25 × 1 – 1 × 1
= p + (75 – 25 – 1) = p + 49
∴ p’ > p or 26 × 74 > 25 × 75
Question 10.
A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g2 sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.
Answer:
Length = w + g + 2w + g + w = 4w + 2g
Breadth = w + g + w = 2w + g
Area of park = (4w + 2g) (2w + g)
= 8w2 + 4wg + 4wg + 2g2
= 8w2 + 8wg + 2g2
Area of path = Area of park – Area of green cover
= 8w2 + 8wg + 2g2 – 2g2
= 8w2 + 8wg
∴ (8w2 + 8wg) sq. feet area needs to be tiled.
Question 11.
For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.
Answer:
Step 1: 2 vertical strips of 3 units each + 1 vertical strip of 3 units
= 3 strips of 3 units each
= 9 units squares = (1 + 2)2 unit squares
Step 2: 4 strips of 4 units each
= 16 units squares = (2 + 2)2 unit squares
Step 3: 5 strips of 5 units each
= 25 units squares = (3 + 2)2 unit squares
Step 4: (i) 6 strips of 6 units each = 2 are vertical and 4 are horizontal
(ii) Number of unit squares in step 10
= (10 + 2)2 = 144
(iii) Number of unit squares in step y = (y + 2)2
(b) (i) 
Number of unit squares in step 1 = 5 = 22 + 1
Number of unit squares in step 2 = 11 = 32 + 2
Number of unit squares in step 3 = 19 = 42 + 3
(ii) Step 1 has (1 + 1)2 + 1 or 5 squares
Step 2 has (2 + 1 )2 + 2 or 11 squares
Step 3 has (3 + 1)2 + 3 or 19 squares
Hence step 10 has (10 + 1)2 + 10 or 131 squares
(iii) Step y has [(y + 1)2 + y] squares
We Distribute Yet Things Multiply Class 8 Extra Questions
Multiple Choice Questions
Question 1.
The product of 28 and 17 increase by which number if the value of 28 is increased by 1?
(a) 17
(b) 28
(c) 45
(d) 11
Solution:
Here, (28 + 1) × 17 = (28 × 17) + 17, which is 17 more than the product 28 × 17.
(a) 17
Question 2.
The product of 28 × 17 increases by what value if 17 is increased by 1?
(a) 17
(b) 28
(c) 45
(d) 11
Solution:
Here, 28 × (17 + 1) = 28 × 17 + 28, which is 28 more than the product 28 × 17.
(b) 28
Question 3.
The product 12 × 15 will increase by what value if both the numbers are increased by 1?
(a) 12
(b) 15
(c) 27
(d) 28
Solution:
Here, (12 + 1) (15 + 1)
= (12 × 15) + 12 × 1 + 1 × 15 + 1 × 1
= (12 × 15) + 12 + 15 + 1
= (12 × 15) + 28
(d) 28
Question 4.
Let a, b, and c be three numbers.
a × (b + c) = a × b + a × c
The propery by which the above happens is :
(a) Commutative
(b) Associative
(c) Distributive
(d) Closure
Solution:
a × (b + c) = a × b + a × c is by distributive property.
(c) Distributive
Question 5.
Expanded form of a (b + c – 1) is :
(a) ab + bc – 1
(b) ab + ab – 1
(c) ab + bc – a
(d) ab + ac – a
Solution:
Here, a (b + c – 1) = ab + ac – a
Answer:
(d) ab + ac – a
Assertion and Reasoning
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).
(b) Assertion (A) and Reason (R) both are true but Reason (R) is not the correct explanation for the Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Question 1.
Assertion (A) : (a + 4) (b + 4) = ab + 4
Reason (R) : (x + y) + (u) = x + y + u
Solution:
∵ (a + 4) (b + 4) = ab + 4a + 4b + 16, So,
Assertion (A) is false.
Answer:
(d) Assertion (A) is false but Reason (R) is true.
Question 2.
Assertion (A) : A number divisible by 4 and another number divisible by 3, have sum which is always divisible by 12.
Reason (R) : A number divisible by 12 can be algebraically represented by ‘12k’, where ‘k’ is an integer.
Solution:
Let the number represented by 4 be represented by 4 m and the number represented by 3 be represented by 3n, where m and n be integers.
Now, 4m + 3n may not be divisible by 12 for all values of m and n.
Answer:
(d) Assertion (A) is false but Reason (R) is true.
Case Based Questions
Question 1.
Consider any 2 × 2 square numbers (grid) in a calender, as shown in the figure.
Answer the following questions based on the above assumptions:
(i) Write the numbers in the highlighted box as they appear.
(ii) Find the products of numbers lying along each diagonal. Find their difference.
(iii) Take any other 2 × 2 square of numbers and write it as the numbers appear- in it.
(iv) Find the product of numbers lying along each diagonal. Find their difference. Are the difference obtained in (ii) and now same?
Answer:
(i)
| 6 | 7 |
| 13 | 14 |
(ii) 6 × 14 = 84, 13 × 7 = 91
Difference = 91 – 84 = 7
(iii) Let us consider
| 2 | 3 |
| 9 | 10 |
(iv) 2 × 10 = 20, 9 × 3 = 27
Difference = 27 – 20 = 7