R.S Aggarwal Class 8 Chapter 8 Linear Equations Test Paper Solution
The 8th Chapter of RS Aggarwal book of Class 8 is on Linear Equations and in this post we have provided complete written solution for the test paper 8 . The solutions are prepared by our team of expert teachers.
Complete RS Aggarwal solution for class 8 is also available in our website.
Test Paper 8
A)
1) Subtract 4a2 + 5b2 – 6c2+ 8 from 2a2 – 3b2 – 4c2 – 5.
Ans:
(4a2 + 5b2 – 6c2+ 8) – (2a2 – 3b2 – 4c2 – 5)
= 4a2 + 5b2 – 6c2+ 8 – 2a2 + 3b2 + 4c2 + 5)
= (4 – 2) a2+ (5+3) b2 + (– 6 + 4) c2+ 13
= 2a2+ 8b2– 2c2+13
2) Find each of the following products:
i) (4a + 5b) × (5a – 6b)
Ans:
(4a×5a) – (4a×6b) + (5b×5a) – (5b×6b)
= 20a2 – 24ab + 25ab – 30b2
= 20a2 – ab – 30b2
3) Divide (5a3 – 4a2 + 3a + 18) by (a2 – 2a + 3).
Ans:
∴ We can say that the quotient is (5a + 6)
4) If, find the value of
i) ii)
Ans:
i) According to the problem
Squaring both sides we get
This is the solution of question (i)
Now if we square both sides of equation (i) we get
5) Evaluate {(83)2 – (17)2}
Ans:
(83)2 – (17)2
= (83+17) × (83 – 17) [∵ a2 – b2 = (a+b)(a – b)]
= 100 × 66
= 6600
6) Factorise:
i) x3 – 3x2+x – 3
Ans:
x3 – 3x2+x – 3
= x2(x – 3) + 1 (x – 3)
= (x – 3) (x2 + 1)
ii) 63x2y2 – 7
Ans:
63x2y2 – 7
= 7 (9x2y2 – 1)
= 7 [(3xy) 2 – 12]
= 7 (3xy – 1) (3xy+1)
iii) 1 – 6x + 9x2
Ans:
1 – 6x + 9x2
= 12 – 2×1×3x + (3x) 2
= (1 – 3x)2
iv) 7x2 – 19x – 6
Ans:
7x2 – 19x – 6
= 7x2 – 21x + 2x – 6
= 7x (x – 3) + 2(x – 3)
= (x – 3) (7x + 2)
7) Solve:
Ans:
17 (2x + 7) = 15 (3x + 5)
34x + 119 = 45x = 75
Or, (45 – 34)x = 119 – 75
Or, 11x = 44
∴ x = 4
8) 5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.
Ans:
Let the present ages of the man and his son is p and q respectively.
Five years ago
(p – 5) = 7 (q – 5)
Or, p – 7q = – 30 ——– (i)
After five years
p+5 = 3 (q + 5)
or, p – 3q = 10 ———– (ii)
by solving these equations we get
q = 10 and p = 40
So, the present ages of the father and the son is 40 years and 10 years respectively.
B) Mark (✓) against the correct answer in each of the following:
9) ab – a – b + 1 = ?
a) (1 – a) × (1 – b)
b) (1 – a) × (b – 1)
c) (a – 1) × (b – 1) (✓)`
d) (a – 1) × (1 – b)
Ans:
ab – a – b + 1
= a (b – 1) – 1 (b – 1)
= (a – 1) (b – 1)
10) 3+23x – 8x2 =?
a) (1 – 8x) × (3 + x)
b) (1 + 8x) × (3 – x) (✓)
c) (1 – 8x) × (3 – x)
d) none of these
Ans:
3+23x – 8x2
= 3 + 24x – x – 8x2
= 3 (1+8x) – x (1+8x)
= (1+8x)(3 – x)
11) 7x2 – 19x – 6=?
Ans:
a) (x – 3)(7x + 2) (✓)
b) (x+3)(7x – 2)
c) (x – 3)(7x – 2)
d) (7x – 3)(x + 2)
Ans:
7x2 – 19x – 6
= 7x2 – 21x + 2x – 6
= 7x (x – 3) + 2(x – 3)
= (x – 3) (7x + 2)
12) 12x2+ 60x+ 75 =?
a) (2x+5)(6x+5)
b) (3x+5)2
c) 3(2x+5)2 (✓)
d) None of these.
Ans:
12x2+ 60x+ 75
= 3 [4x2+20x+25]
= 3 [(2x)2 + 2 × 2x × 5+52]
= 3(2x+5)2
13) 10p2 + 11p + 3 = ?
a) (2p+3)(5p+1)
b) (5p+3)(2p+1) (✓)
c) (5p – 3)(2p – 1)
d) none of these
Ans:
10p2 + 11p + 3
= 10p2+ 5p + 6p + 3
= 5p (2p+1) + 3(2p+1)
= (5p+3) (2p+1)
14) 8x3 – 2x =?
a) (4x – 1)(2x – 1)x
b) (2x2+ 1)(2x – 1)
c) 2x (2x – 1)(2x + 1) (✓)
d) none of these
Ans:
8x3 – 2x
= 2x (4x2 – 1)
= 2x [(2x)2 – 12]
= 2x (2x – 1)(2x+1)
15) gives
a) x = 3
b) x = 4
c) x = 5
d) none of these
Ans:
C) 16) Fill in the blanks.
i) x2 – 18x + 81 = (___)
Ans:
x2 – 18x + 81
= x2 – 2.x.9 + 92
= (x – 9)2
So, x2 – 18x + 81 = (x – 9)2
ii) 4 – 36x2 = (__)(__)(__)
Ans:
4 – 36x2
= 22 – (6x)2
= (2+6x) (2 – 6x)
∴4 – 36x2 = (2+6x) (2 – 6x)
iii) x2 – 14x + 13 = (__)(__)
Ans:
x2 – 14x + 13
= x2 – 13x – x + 13
= x(x – 13) – 1(x – 13)
= (x – 13)(x – 1)
∴x2 – 14x + 13 = (x – 13)(x – 1)
iv) 9z2 – x2 – 4y + 4xy = (__)(__)
Ans:
9z2 – x2 – 4y + 4xy
= (3z)2 – (x2 + 4y – 4xy)
= (3z)2 – [x2 – 2.x.2y+ (2y)2]
= (3z)2 – (x –2y)2
= (3z+x–2y)(3z – x + 2y)
∴9z2 – x2 – 4y + 4xy = (3z+x–2y)(3z – x + 2y)
v) abc – ab – c + 1 = (__)(__)
Ans:
abc – ab – c + 1
= ab (c – 1) – 1 (c – 1)
= (c – 1) (ab – 1)
∴abc – ab – c + 1 = (c – 1) (ab – 1)
D)
17) Write ‘T’ for true and ‘F’ for false for each of the following:
i) (5 – 3x2) is a binomial ————- (T)
Explanation: Binomials terms are made of two different terms. The given term is also made of two different terms 5 and 3x2. So we can say that (5 – 3x2) is a binomial.
The statement is true
ii) – 8 is a monomial (T)
Explanation: Monomial terms are made of a single term. So we can say that (– 8) is a monomial.
iii) (5a – 9b) – (– 6a + 2b) = (– a –7b) ———– (F)
Explanation:
L.H.S
(5a – 9b) – (– 6a + 2b)
= 11a – 11b
We can say that the LHS is not equal to the RHS.
The statement is false.
iv) When x = 2 and y = 1, the value of (-8/7)x3y4 is (–64/7) ———- (T)
Ans:
(-8/7)x3y4
= (-8/7) × 23 × 14
= -8×8/7 = – 64/7
v) (x/4) + (x/6) – (x/2) = (3/4) => x = – 9 ———– (T)
Ans:
Or, – x = 9
Or, x = – 9
The statement is true
vi) 2x – 5 = 0 => x = 2/5 ———– (T)
Ans:
2x – 5 = 0
Or, x = 5/2
The statement is true