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Page No 73:
Question 1:
Can’t you find these products on your own?
(i) (p + q) (2m + 3n)
(ii) (4x + 3y) (2a + 3b)
(iii) (4a + 2b) (5c + 3d)
(iv) (m + n) (5a + b)
(v) (2x + 3y) (x + 2y)
(vi) (3a + 2b) (x + 2y)
Answer:
By the general principle, we know that:
(x + y) (u + v) = xu + xv + yu + yv
(i)
(p + q) (2m + 3n) = (p × 2m) + (p × 3n) + (q × 2m) + (q × 3n)
= 2pm + 3pn + 2qm + 3qn
(ii)
(4x + 3y) (2a + 3b) = (4x × 2a) + (4x × 3b) + (3y × 2a) + (3y × 3b)
= 8ax + 12bx + 6ay + 9by
(iii)
(4a + 2b) (5c + 3d) = (4a × 5c) + (4a × 3d) + (2b × 5c) + (2b × 3d)
= 20ac + 12ad + 10bc + 6bd
(iv)
(m + n) (5a + b) = (m × 5a) + (m × b) + (n × 5a) + (n × b)
= 5am + bm + 5an + bn
(v)
(2x + 3y) (x + 2y) = (2x × x) + (2x × 2y) + (3y × x) + (3y × 2y)
= 2x2 + 4xy + 3xy + 6y2
(vi)
(3a + 2b) (x + 2y) = (3a × x) + (3a × 2y) + (2b × x) + (2b × 2y)
= 3ax + 6ay + 2bx + 4by
Page No 74:
Question 1:
Find these products on your own:
(i) (x + 3y) (2a − b)
(ii) (3x + 5y) (3m − 2n)
(iii) (2r − 3s) (t − u)
(iv) (a − b) (4x − 3y)
(v) (3a − 5b) (2c − d)
(vi) (2p + 5q) (3r − 4s)
Answer:
By the general principle, we know that:
(x + y) (u − v) = xu − xv + yu − yv
(x − y) (u − v) = xu − xv − yu + yv
(x − y) (u + v) = xu + xv − yu − yv
(i)
(x + 3y) (2a − b) = (x × 2a) − (x × b) + (3y × 2a) − (3y × b)
= 2ax − bx + 6ay − 3by
(ii)
(3x + 5y) (3m − 2n) = (3x × 3m) − (3x × 2n) + (5y × 3m) − (5y × 2n)
= 9mx − 6nx + 15my − 10ny
(iii)
(2r − 3s) (t − u) = (2r × t) − (2r × u) − (3s × t) + (3s × u)
= 2rt − 2ru − 3st + 3su
(iv)
(a − b) (4x − 3y) = (a × 4x) − (a × 3y) − (b × 4x) + (b × 3y)
= 4ax − 3ay − 4bx + 3by
(v)
(3a − 5b) (2c − d) = (3a × 2c) − (5b × 2c) − (3a × d) + (5b × d)
= 6ac − 10bc − 3ad + 5bd
(vi)
(2p + 5q) (3r − 4s) = (2p × 3r) − (2p × 4s) + (5q × 3r) − (5q × 4s)
= 6pr − 8ps + 15qr −20qs
Page No 76:
Question 1:
Find the squares of the following algebraic expressions
(i) 2x + 3y
(ii) x + 2
(iii) 2x + 1
Answer:
We know that (x + y)2 = x2 + y2 + 2xy
(i)
Square of 2x + 3y = (2x + 3y)2
= (2x)2 + (3y)2 + 2 × 2x × 3y
= 4x2 + 9y2 + 12xy
(ii)
Square of x + 2 = (x + 2)2
= (x)2 + (ii)2 + 2 × 2 × x
= x2 + 4 + 4x
(iii)
Square of 2x + 1 = (2x + 1)2
= (2x)2 + (i)2 + 2 × 2x × 1
= 4x2 + 1+ 4x
Page No 76:
Question 2:
Compute the squares of these numbers mentally.
(i) 102
(ii) 202
(iii) 1001
(iv) 2002
(v) 205
(vi) 10.3
Answer:
We know that (x + y)2 = x2 + y2 + 2xy
(i)
102 = 100 + 2
⇒ (102)2 = (100 + 2)2
= (100)2 + 2 × 100 × 2 + 22
= 10000 + 400 + 4
=10404
(ii)
202 = 200 + 2
⇒ (202)2 = (200 +2)2
= (200)2 + 2 × 200 × 2 + 22
= 40000 + 800 + 4
= 40804
(iii)
1001 = 1000 + 1
⇒ (1001)2 = (1000 + 1)2
= (1000)2 + 2 × 1000 × 1 + 12
= 1000000 + 2000 + 1
= 1002001
(iv)
2002 = 2000 + 2
⇒ (2002)2 = (2000 + 2)2
= (2000)2 + 2 × 2000 × 2 + 22
= 4000000 + 8000 + 4
= 4008004
(v)
205 = 200 + 5
⇒ (205)2 = (200 + 5)2
= (200)2 + 2 × 200 × 5 + 52
= 40000 + 2000 + 25
= 42025
(vi)
10.3 = 10 + 0.3
⇒ (10.3)2 = (10 + 0.3)2
= (10)2 + 2 × 10 × 0.3 + (0.3)2
= 100 + 6 + 0.09
=106.09
Page No 76:
Question 3:
Prove that x2 + 6x + 9 is a perfect square for all natural numbers x. What can you say about its square root?
Answer:
To show that the expression x2 + 6x + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.
x2 + 6x + 9 = x2 + 2 × 3 × x + 32
= (x + 3)2 {(x + y)2 = x2 + y2 + 2xy}
∴ x2 + 6x + 9 = (x + 3)2
Hence, x2 + 6x + 9 is a perfect square.
Square root of x2 + 6x + 9 = 
=
= x + 3
Page No 76:
Question 4:
Prove that in the sequence 1, 2, 3, ….. of natural numbers, 1 added to the product of any two alternatives numbers perfect square.
Answer:
Let us assume a number a in the sequence of natural numbers 1, 2, 3,…
Alternate number after a = a + 2
Product of the two alternate numbers = (a + 2) × a
= a2 + 2a
If one is added to this expression, it will become a2 + 2a + 1.
Now, a2 + 2a + 1 = a2 + 2 × a × 1 + 12
= (a + 1)2 {(x + y)2 = x2 + y2 + 2xy}
Thus, when 1 is added to the product of any two alternate natural numbers, the result obtained is a perfect square.
Page No 77:
Question 1:
Find the squares of the algebraic expressions given below:
(i) 2x − 3y
(ii) x − 2
(iii) 2x − 1
Answer:
We know that (x − y)2 = x2 + y2 − 2xy
(i)
Square of 2x − 3y = (2x − 3y)2
= (2x)2 + (3y)2 − 2 × 2x × 3y
= 4x2 + 9y2 − 12xy
(ii)
Square of x − 2 = (x − 2)2
= x2 + 22 − 2 × x × 2
= x2 + 4 − 4x
(iii)
Square of 2x − 1 = (2x − 1)2
= (2x)2 + (i)2 − 2 × 2x × 1
= 4x2 + 1 − 4x
Page No 77:
Question 2:
Mentally compute the squares of the numbers given below:
(i) 98
(ii) 198
(iii) 999
(iv) 1998
(v) 195
(vi) 9.7
Answer:
We know that (x − y)2 = x2 + y2 − 2xy
(i)
98 = 100 − 2
⇒ (98)2 = (100 − 2)2
= (100)2 + 22 − 2 × 100 × 2
= 10000 + 4 − 400
= 9604
(ii)
198 = 200 − 2
⇒ (198)2 = (200 − 2)2
= (200)2 + 22 − 2 × 200 × 2
= 40000 + 4 − 800
= 39204
(iii)
999 = 1000 −1
⇒ (999)2 = (1000 −1)2
= (1000)2 + 12 − 2 × 1000 × 1
= 1000000 + 1 − 2000
= 998001
(iv)
1998 = 2000 − 2
⇒ (1998)2 = (2000 − 2)2
= (2000)2 + 22 − 2 × 2000 × 2
= 4000000 + 4 − 8000
= 3992004
(v)
195 = 200 − 5
⇒ (195)2 = (200 − 5)2
= (200)2 + 52 − 2 × 200 × 5
= 40000 + 25 − 2000
= 38025
(vi)
9.7 = 10 − 0.3
⇒ (9.7)2 = (10 − 0.3)2
= (10)2 + (0.3)2 − 2 × 10 × 0.3
= 100 + 0.09 − 6
= 94.09
Page No 77:
Question 3:
Prove that x2 − 6x + 9 is a perfect square for all natural numbers x. What can you say about its square root?
Answer:
To show that the expression x2 − 6x + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.
⇒ x2 − 6x + 9 = x2 − 2 × 3 × x + 9
= (x − 3)2 {(x − y)2 = x2 + y2 − 2xy}
∴ x2 − 6x + 9 = (x − 3)2
Hence, x2 − 6x + 9 is a perfect square.
Square root of x2 − 6x + 9 =
=
= x − 3
Page No 78:
Question 1:
Can’t you easily find the products below like this?
(i) 51 × 49
(ii) 98 × 102
(iii) 10.2 × 9.8
(iv) 7.3 × 6.7
Answer:
We know that (x − y)(x + y) = x2 − y2
(i)
The number 51 can be written as (50 + 1) and 49 can be written as (50 − 1).
∴ 51 × 49 = (50 + 1) (50 − 1)
= 502 − 12
= 2500 − 1
= 2499
(ii)
The number 98 can be written as (100 − 2) and 102 can be written as (100 + 2).
∴ 98 × 102 = (100 − 2) (100 + 2)
= (100)2 − 22
= 10000 − 4
= 9996
(iii)
The number 10.2 can be written as (10 + 0.2) and 9.8 can be written as (10 − 0.2).
∴ 10.2 × 9.8 = (10 + 0.2) (10 − 0.2)
= (10)2 − (0.2)2
= 100 − 0.04
= 99.96
(iv)
The number 7.3 can be written as (7 + 0.3) and 6.7 can be written as (7 − 0.3).
7.3 × 6.7 = (7 + 0.3) × (7 − 0.3)
= 72 − (0.3)2
= 49 − 0.09
= 48.91
Page No 79:
Question 1:
Compute the following in your head:
(i) 672 − 332
(ii) 1232 − 1222
(iii)
(iv) 0.272 − 0.232
Answer:
We know that x2 − y2 = (x + y) (x − y)
(i)
672 − 332 = (67 + 33) (67 − 33)
= 100 × 34
= 3400
(ii)
1232 − 1222 = (123 + 122) (123 − 122)
= 245 × 1
= 245
(iii)
(iv)
(0.27)2 − (0.23)2 = (0.27 + 0.23) (0.27 − 0.23)
= 0.5 × 0.04
= 0.02
Page No 79:
Question 2:
The diagonal of a rectangle is 65 centimetres long and one of its sides is 63 centimeres long. What is the length of the other side?
Answer:
Given: A rectangle ABCD with length of one side, DC = 63 cm and length of the diagonal, BD = 65 cm.
We know that each angle of a rectangle measures 90°.
In ΔBCD, ∠BCD = 90°
Using Pythagoras theorem, we get:
BD2 = BC2 + CD2
⇒ (65)2 = BC2 + (63)2
⇒ BC2 = (65)2 − (63)2
= (65 − 63) (65 + 63) {x2 − y2 = (x − y) (x + y)}
= 2 × 128
= 256
⇒ BC = 16
Thus, the length of the other side of the rectangle is 16 cm.
Page No 80:
Question 1:
Prove that the difference of the squares of two consecutive natural is equal to their sum.
Answer:
Let n be a natural number.
Its consecutive natural number = n + 1
Sum of these consecutive natural numbers = n + (n +1) = 2n +1
Difference of the squares of two consecutive natural numbers
= (n + 1)2 − n2
= (n + 1 − n) (n + 1 + n) {x2 − y2 = (x − y) (x + y)}
= 2n + 1
Thus, the difference of the squares of two consecutive natural numbers is equal to their sum.
Page No 80:
Question 2:
In how many different ways you can write 15 as the difference of the squares of two natural numbers?
Answer:
The factors of 15 are 1, 3, 5 and 15.
We can write 15 as:
15 = 1 × 15
= (8 − 7) (8 + 7) {x2 − y2 = (x − y) (x + y)}
= 82 − 72
So, 15 can be written as the difference of the squares of 8 and 7.
15 = 3 × 5
= (4 − 1) (4 + 1) {x2 − y2 = (x − y) (x + y)}
= 42 − 12
So, 15 can be written as the difference of the squares of 4 and 1.
Thus, there are two ways to write 15 as the difference of the squares of two natural numbers.
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