Chapter 12 Algebraic Expressions Ex 12.3 Solutions
Question - 1 : - If m = 2, find the value of:
(i) m тАУ 2
(ii) 3m тАУ 5
(iii) 9 тАУ 5m
(iv) 3m2 тАУ 2m тАУ 7
(v) 5m/2тИТ4
Answer - 1 : -
(i) m тАУ2
Putting m = 2, we get
2 тАУ 2 = 0
(ii) 3m тАУ 5
Putting m = 2, we get
3 ├Ч 2 тАУ 5 = 6 тАУ 5 = 1
(iii) 9 тАУ 5m
Putting m = 2, we get
9 тАУ 5 ├Ч 2 = 9 тАУ 10 = -1
(iv) 3m2┬атАУ 2m тАУ 7Putting m = 2, we get
3(2)2┬атАУ 2(2) тАУ 7 = 3 ├Ч 4 тАУ 4 тАУ 7
=12 тАУ 4 тАУ 7 = 12 тАУ 11 = 1
(v)┬а5m/2тИТ4
Putting m = 2, we get
(5├Ч2)/2тИТ4=5тИТ4=1
Question - 2 : - If p = -2, find the value of:
(i) 4p + 7
(ii) -3p2┬а+ 4p + 7
(iii) -2p3┬атАУ 3p2┬а+ 4p + 7
Answer - 2 : -
(i) 4p + 7
Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1
(ii) -3p2┬а+ 4p + l
Putting p = -2, we get
-3(-2)2┬а+ 4(-2) + 7
= -3 ├Ч 4 тАУ 8 + 7 = -12 тАУ 8+ 7 = -13
(iii) -2p3┬атАУ 3p2┬а+ 4p + 7
Putting p = -2, we get
тАУ 2(-2)3┬атАУ 3(-2)2┬а+ 4(-2) + 7
= -2 ├Ч (-8) тАУ 3 ├Ч 4 тАУ 8 + 7
= 16 тАУ 12 тАУ 8 + 7 = 3
Question - 3 : - If a = 2, b = -2, find the value of:
(i) a2┬а+ b2
(ii) a2┬а+ ab + b2
(iii) a2┬атАУ b2
Answer - 3 : -
(i) a2┬а+ b2
Putting a = 2 and b = -2, we get
(2)2┬а+ (-2)2┬а= 4 + 4 = 8
(ii) a2┬а+ ab + b2
Putting a = 2 and b = -2, we get
(2)2┬а+ 2(-2) + (-2)2┬а= 4 тАУ 4 + 4 = 4
(iii) a2┬атАУ b2
Putting a = 2 and b = -2, we get
(2)2┬атАУ (-2)2┬а= 4 тАУ 4 = 0
Question - 4 : - When a = 0, b = -1, find the value of the givenexpressions:
(i) 2a + 2b
(ii) 2a2┬а+ b2┬а+ 1
(iii) 2a2b + 2ab2┬а+ ab
(iv) a2┬а+ ab + 2
Answer - 4 : -
(i) 2a + 2b = 2(0) + 2(-1)
= 0 тАУ 2 = -2 which is required.
(ii) 2a2┬а+ b2┬а+ 1
= 2(0)2┬а+ (-1)2┬а+ 1 =0 + 1 + 1 = 2 which isrequired.
(iii) 2a2b + 2ab2┬а+ ab
= 2(0)2┬а(-1) + 2(0)(-1)2┬а+ (0)(-1)
=0 + 0 + 0 = 0 which is required.
(iv) a2┬а+ ab + 2
= (0)2┬а+ (0)(-1) + 2
= 0 + 0 + 2 = 0 which is required.
Question - 5 : - Simplify the expressions and find the value if x isequal to 2.
(i) x + 7 +4(x тАУ 5)
(ii) 3(x + 2) + 5x тАУ 7
(iii) 6x + 5(x тАУ 2)
(iv) 4(2x тАУ 1) + 3x + 11
Answer - 5 : -
(i) x + 7 + 4(x тАУ 5) = x + 7 + 4x тАУ 20 = 5x тАУ 13
Putting x = 2, we get
= 5 ├Ч 2 тАУ 13 = 10 тАУ 13 = -3
which is required.
(ii) 3(x + 2) + 5x тАУ 7 = 3x + 6 + 5x -7 = 8x тАУ 1
Putting x = 2, we get
= 8 ├Ч 2 тАУ 1 = 16 тАУ 1 = 15
which is required.
(iii) 6x + 5(x тАУ 2) = 6x + 5x тАУ 10
= 11 ├Ч тАУ 10
Putting x = 2, we get
= 11 ├Ч 2 тАУ 10 = 22 тАУ 10 = 12
which is required.
(iv) 4(2x тАУ 1) + 3x + 11 = 8x тАУ 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= 11 ├Ч 2 + 7 = 22+ 7 = 29
Question - 6 : - Simplify these expressions and find their values if x= 3, a = -1, b = -2.
(i) 3x тАУ 5 тАУ x + 9
(ii) 2 тАУ 8x + 4x + 4
(iii) 3a + 5 тАУ 8a + 1
(iv) 10 тАУ 3b тАУ 4 тАУ 55
(v) 2a тАУ 2b тАУ 4 тАУ 5 + a
Answer - 6 : -
(i) 3x тАУ 5 тАУ x + 9 = 2x + 4
Putting x = 3, we get
2 ├Ч 3 + 4 = 6 + 4 = 10
which is required.
(ii) 2 тАУ 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6
Putting x = 2, we have
= -4 ├Ч 2 + 6 = -8 + 6 =-2
which is required.
(iii) 3a + 5 тАУ 8a +1 = 3a тАУ 8a + 5 + 1 = -5a + 6
Putting a = -1, we get
= -5(-1) + 6 = 5 + 6 = 11
which is required.
(iv) 10 тАУ 3b тАУ 4 тАУ 5b = -3b тАУ 5b + 10 тАУ 4
= -8b + 6
Putting b = -2, we get
= -8(-2) + 6 = 16 + 6 = 22
which is required.
(v) 2a тАУ 2b тАУ 4 тАУ 5 + a = 2a + a тАУ 2b тАУ 4 тАУ 5
= 3a тАУ 26 тАУ 9
Putting a = -1 and b = -2, we get
= 3(-1) тАУ 2(-2) тАУ 9
= -3 + 4 тАУ 9 = 1 тАУ 9 = -8
which is required.
Question - 7 : - (i) If z = 10, find the value of z2┬атАУ 3(z тАУ 10).
(ii) If p = -10, find the value of p2┬а-2pтАУ 100.
Answer - 7 : -
(i) z2┬атАУ 3(z тАУ 10)
= z2┬атАУ 3z + 30
Putting z = 10, we get
= (10)2┬атАУ 3(10) + 30
= 1000 тАУ 30 + 30 = 1000 which is required.
(ii) p2┬атАУ 2p тАУ 100
Putting p = -10, we get
(-10)2┬атАУ 2(-10) тАУ 100
= 100 + 20 тАУ 100 = 20 which is required.
Question - 8 : - What should be the value of a if the value of 2x2┬а+ x тАУ a equals to 5, when x = 0?
Answer - 8 : -
2x2┬а+ x тАУ a = 5
Putting x = 0, we get
2(0)2┬а+ (0) тАУ a = 5
0 + 0 тАУ a = 5
-a = 5
тЗТ a = -5which is required value.
Question - 9 : - Simplify the expression and find its value when a = 5and b = -3.
2(a2┬а+ ab) + 3 тАУ ab
Answer - 9 : -
2(a2┬а+ ab) + 3 тАУab = 2a2 + 2ab + 3 тАУ ab
= 2a2┬а+ 2ab тАУ ab + 3
= 2ab + ab + 3
Putting, a = 5 and b = -3, we get
= 2(5)2┬а+ (5)(-3) + 3
= 2 ├Ч 25 тАУ 15 + 3
= 50 тАУ 15 + 3
= 53 тАУ 15 = 38
Hence, the required value = 38.