We Distribute, Yet Things Multiply Class 8 Extra Questions Maths Chapter 6

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Class 8 We Distribute, Yet Things Multiply Extra Questions

Class 8 Maths Chapter 6 We Distribute, Yet Things Multiply Extra Questions

Class 8 Maths Chapter 6 Extra Questions – We Distribute, Yet Things Multiply Extra Questions Class 8

Question 1.
Find the product using identity.
(a) (3x + 5) × (5x – 2)
(b) (7 – x) × (7 + x)
(c) \(\frac {3}{7}\)(2x² + 5x – 3)
(d) (4y – 3)²
Solution:
(a) (3x + 5) × (5x – 2) = (3x + 5) × 5x – (3x + 5) × 2 [Distributive Property]
= 15x² + 25x – 6x – 10
= 15x² + 19x – 10
(b) (7 – x) × (7 + x) = 7² – x² = 49 – x²
(c) \(\frac{3}{7}\left(2 x^2+5 x-3\right)=\frac{3}{7} \times 2 x^2+\frac{3}{7} \times 5 x-\frac{3}{7} \times 3\) [Distributive property]
= \(\frac{6 x^2}{7}+\frac{15 x}{7}-\frac{9}{7}\)
(d) (4y – 3)² = (4y)² – 2 × 4y × 3 + 3² = 16y² – 24y + 9

Question 2.
Multiply:
(a) 378 × 101
(b) 732 × 1001
(c) 8932 × 11
Solution:
(a) 378 × 101

(b) 732 × 1001

(c) 8932 × 11 = 8932 × (10 + 1)
= 8932 × 10 + 8932
= 98252

Question 3.
The number of students going on a trip is (x + 2) and the cost of one ticket is ₹ (2x – 1). If 3 students cancel their trip, find the difference in the total cost before and after the cancellation.
Solution:
Price of each ticket = ₹ (2x – 1)
Number of students = (x + 2)
∴ Total cost of ticket before cancellation = ₹ (2x – 1) × (x + 2) = 2x² + 3x – 2
Now, 3 students have canceled their trip.
Now number of students = x + 2 – 3 = x – 1
Total cost of tickets after cancellation = ₹ (2x – 1) (x – 1) = 2x² – 3x + 1
Difference in total cost = (2x² + 3x – 2) – (2x² – 3x + 1)
= 2x² + 3x – 2 – 2x² + 3x – 1
= 6x – 3

Question 4.
A rectangular bedroom has a length of (x + 8) feet and a breadth of (x – 2) feet. Find its area. If both the length and breadth are increased by 1 foot, find the increase in the area of the room.
Solution:
Original Dimensions: The length and breadth of the room are (x + 8) feet and (x – 2) feet.
Original area of bedroom = (x + 8) (x – 2)
= (x + 8)x – (x + 8)2
= x² + 8x – 2x – 16
= x² + 6x – 16
After increasing both dimensions by 1 foot:
New Length = (x + 8) + 1 = x + 9
New Breadth = (x – 2) + 1 = (x – 1)
New Area = (x + 9)(x – 1)
= (x + 9)x – (x + 9)1 (Using Distributive Property)
= x² + 9x – x – 9
= x² + 8x – 9
Increase in Area = New Area – Original Area
= (x² + 8x – 9) – (x² + 6x – 16)
= x² + 8x – 9 – x² – 6x + 16
= 2x + 7
Thus, the area increases by (2x + 7) square feet.

Question 5.
Using identities, find:
(i) (2x + 3y)²
(ii) (xy + 3z)²
(iii) \(\left(\frac{2}{3} m+\frac{3}{2} n\right)^2\)
Solution:
(i) (2x + 3y)² = (2x)² + 2(2x)(3y) + (3y)² [∵ (a + b)² = a² + 2ab + b²]
= 2² x² + 12xy + (3y)² [∵ (ab)² = a² b²]
= 4x² + 12xy + 9y²
(ii) (xy + 3z)² = (xy)² + 2(xy)(3z) + (3z)² [∵ (a + b)² = a² + 2ab + b²]
= x² y² + 6xyz + 3² z² [∵ (ab)² = a² b²]
= x² y² + 6xyz + 9z²

Question 6.
Using identities, find:
(i) (4p – 3q)²
(ii) (b – 7)²
(iii) (6x² – 5y)²
(iv) (0.4p – 0.5q)²
Solution:

Question 7.
Using identities, find:
(i) \(\left(\frac{3}{2} m+\frac{2}{3} n\right)\left(\frac{3}{2} m-\frac{2}{3} n\right)\)
(ii) (2x + 3y)(2x – 3y)
(iii) \(\left(\frac{3}{4} x+\frac{5}{6} y\right)\left(\frac{3}{4} x-\frac{5}{6} y\right)\)
(iv) (a² + b²)(-a² + b²)
Solution:

Question 8.
Find the product of the following binomials:
\(\left(\frac{1}{2} x-\frac{1}{5} y\right)\left(\frac{1}{2} x-\frac{1}{5} y\right)\)
Solution:

Practice Questions

Question 1.
Given that 25 × 29 = 725. Without actual multiplication, find the product of
(a) 25 is increased by 1
(b) 29 is increased by 1
(c) Both 25 and 29 are increased by 1
Answer:
(a) 754
(b) 750
(c) 780

Question 2.
Given that 34 × 43 = 1462. Without actual multiplication, find the product of
(a) 34 is decreased by 1
(b) 43 is increased by 1
(c) Both 34 and 43 are decreased by 1.
Answer:
(a) 1419
(b) 1496
(c) 1386

Question 3.
Given that 39 × 57 = 2223. Without actual multiplication, find the product of
(a) 39 is decreased by 1, and 57 is increased by 1.
(b) 39 is increased by 1, and 57 is decreased by 1.
Answer:
(a) 2204
(b) 2240

Question 4.
Expand
(a) \(\frac {3a}{2}\)(a – b + 15)
(b) (a + b) (a + b)
(c) (a + b) (a² + 2ab + b²)
Answer:
(a) \(\frac{3}{2} a^2-a b \frac{3}{2}+\frac{45}{2} a\)
(b) a² + 2ab + b²
(c) a³ + 3a²b + 3ab² + b³

Question 5.
Using distributive law, find the following products:
(a) 278 × 101
(b) 3642 × 1001
(c) 635 × 99
(d) 4125 × 999
Answer:
(a) 28078
(b) 3645642
(c) 62865
(d) 4120875

Question 6.
Find the increment in the product of 3, 6, and 48 if (Hint: Use identity 1)
(a) 36 is increased by 2, and 48 is decreased by 3
(b) 36 is decreased by 3, and 48 is increased by 2
(c) 36 is increased by 4, and 48 is increased by 12
(d) 36 is decreased by 6, and 48 is decreased by 8
Answer:
(a) -18
(b) -78
(c) 672
(d) -528

Question 7.
Show using distributive law:
(a) (a + b)² = a² + 2ab + b²
(b) (a – b)² = a² – 2ab + b²
(c) (a + b)(a – b) = a² – b²

Question 8.
Show geometrically:
(a) (a + b)² = a² + 2ab + b²
(b) (a – b)² = a² – 2ab + b²
(c) (a + b)(a – b) = a² – b²
Answer:
(a) (a + b)² = a² + 2ab + b²

(b) (a – b)² = a² – 2ab + b²

(c) a² – b² = ar I – ar II
= a(a – b) + (a – b)b
= (a – b) (a + b)

Question 9.
Expand:
(a) (2x + 3y)²
(b) (3a + 5b)²
(c) (7m + 3n)²
(d) (4x – 3y)²
(e) (3a – 2b)²
(f) (8m – 3n)²
(g) (2x + 3y)(2x – 3y)
(h) (3a – 5b) (3a + 5b)
(i) (7m + 3n) (14m – 6n)
Answer:
(a) 4x² + 12xy + 9y²
(b) 9a² + 30ab + 25b²
(c) 49m² + 42mn + 9n²
(d) 16x² – 24xy + 9y²
(e) 9a² – 12ab + 4b²
(f) 64m² – 48mn + 9n²
(g) 4x² – 9y²
(h) 9a² – 25b²
(i) 98m² – 18n²

Question 10.
Find the value of
(a) 52²
(b) 101²
(c) 203²
(d) 38²
(e) 99²
(f) 197²
(g) 38 × 42
(h) 49 × 51
(i) 103 × 97
Answer:
(a) 2704
(b) 10201
(c) 41209
(d) 1444
(e) 9801
(f) 38809
(g) 1596
(h) 2499
(i) 9991

Question 11.
Using the identity (a + b)², find the area of a square with a side length of 65 m.
Answer:
4225 sq. m.

Question 12.
Using the identity (a-b)², find the area of a square with a side length of 59 m.
Answer:
3481 sq. m.

Question 13.
Using the identity (a + b)(a – b), find the area of a rectangle with length 52 m and breadth 48 m.
Answer:
2496 sq. m.

Question 14.
‘a’ and ‘b’ are any two integers. Check if
(a) (a + b)² is always greater than a² + b².
(b) (a-b)² is always smaller than a² + b².
Answer:
(a) No (Not true if a and b have opposite signs)
(b) No (Not true if a and b have opposite signs)

Question 15.
Take a pair of natural numbers. Calculate the sum of their squares. Write twice this sum as a sum of two squares.
Answer:
2(8² + 7²) = (8 + 7)² + (8 – 7)²

Question 16.
Check each of the simplifications and see if there is a mistake. If there is a mistake, then write the correct expression.
(a) \(\frac {1}{2}\)(10s – 6) + 3 = 5s – 3 + 3 = 5s
(b) 5w² + 6w = 11w²
(c) 2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b²
(d) (5m + 6n)² = 25m² + 36n²
(e) (-q + 2)² = q² – 4q + 4
(f) 3a(2b × 3c) = 6ab × 9ac = 54a²bc
Answer:
(a) Correct
(b) Incorrect
(c) Incorrect
(d) Incorrect
(e) Correct
(f) Incorrect

Question 17.
Compute these products using the suggested identity.
(i) 53² using identity for (a + b)²
(ii) 98² using identity for (a-b)²
(iii) 198 × 202 using identity for (a + b) (a – b)
Answer:
(i) 2809
(ii) 9604
(iii) 39996

Question 18.
Use either a suitable identity or the distributive property to find each of the following products.
(i) (x – 2)(x + 3)
(ii) (3a – 4b)(3a + 4b)
(iii) (2a + 1)(3a + 2)
(iv) (2x + 3y)²
(v) (2x – 3y)²
(vi) (5p) × (3r) × (p + 3)
Answer:
(i) x² + x – 6
(ii) 9a² – 16b²
(iii) 6a² + 7a + 2
(iv) 4x² + 12xy + 9y²
(v) 4x² – 12xy + 9y²
(vi) 15p²r + 45pr

Question 19.
For each statement, write an appropriate algebraic expression:
(i) Three more than a square number.
(ii) The sum of the squares of three consecutive numbers.
Answer:
(i) x² + 3
(ii) x² + (x + 1)² + (x + 2)²

Question 20.
A number leaves a remainder of 2 when divided by 5, and another number leaves a remainder of 3 when divided by 5. What is the remainder when their sum, difference, and product are divided by 5?
Answer:
x = 5a + 2, y = 5b + 3
x + y = 5a + 2 + 5b + 3
= 5(a + b) + 5
= 5(a + b + 1)
Remainder = 0
x – y = (5a + 2) – (5b + 3)
= 5(a – b) – 1
= 5(a – b) – 5 + 4
= 5(a – b – 1) + 4
Remainder = 4
xy = (5a + 2)(5b + 3)
= 25ab + 15a + 10b + 6
= 25ab + 15a + 10b + 5 + 1
= 5(5ab + 3a + 2b + 1) + 1
Remainder = 1

Question 21.
Which is larger? Find out without fully computing the product.
(a) 14 × 25 or 16 × 23
(b) 25 × 75 or 28 × 72
Answer:
(a) 16 × 23
(b) 28 × 72

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