Collins Maths Solutions Class 6 Chapter 11 Fundamentals of Algebra
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Collins Maths Class 6 Mathematics, Chapter 11, Fundamentals of Algebra. Here students can easily find Exercise wise chapter 11 solutions, Fundamentals of Algebra. Students will find proper solutions for Exercise 11.1, 11.2, 11.3 and 11.4 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.
Fundamentals of Algebra Exercise 11.1 Solution :
Question no – (1)
Solution :
(a) 7 added to the product of 5 and x
Algebraic expression,
= 5 × x + 7
= 5x + 7
(b) 13 added to the product of x and 4
Algebraic expression :
= x × 4 + 13
= 4x + 13 + 13
(c) 10 times a number taken away from 7
Algebraic expression,
= 7 – 10x
(d) Half of the sum of 2x and 2y
Algebraic expression,
= 1/2 (2x + 2y)
(e) A number t decreased by 3
Algebraic expression,
= t – 3
(f) 4 divide by a
Algebraic expression,
= 4/a
Question no – (2)
Solution :
(a) 25p + 10 = 60
Now in statement form,
= 10 added to the product of 25 and p gives 60
(b) 16 m ÷ 2 = 24
Now, in statement form :
The quotient when the product of 16 and m divided by 2 gives 24.
(c) 10x + 7y – 15
Now, in statement form :
= 15 subtracted from the sum of 10x and 7y
(d) 100 – 15y
Now in statement form :
= The product of 15 and y taken away from 100.
Question no – (3)
Solution :
(a) Given, Reshmi earns ₹500 a day working at a supermarket. How much money will she earn in p days
Algebraic expression,
Reshmi earn in p days 500 p
(b) Given, A hotel manager has to distribute ₹10,000 equally among the workers of the hotel as bonus. How much money will each worker get if there are k workers?
Algebraic expression,
= Worker will get 10000/k
Fundamentals of Algebra Exercise 11.2 Solution :
Question no – (1)
Solution :
| Algebraic Expressions |
Terms | Numeral coefficient Coefficient |
Power of the variable |
| (a) – 25×3 + 2x2 + x | – 25x3 = 2x2 = x |
= – 25 = 2 = 1 |
3 2 1 |
| (b) 4a3b2 – 2b2 + 3a – 4 | 4a3b3 = – 2b2 = 3a = – 4 |
= 4 = 2 = 3 Constant |
3 – 2 2 1 Constant |
| (c) 52z2 + 19zx | 52z2 = 19zx |
52 = 19 |
2 11 |
| (d) 6x4 + 4x5 + 20x3 – 6x2 + 2 | = 6x4 = 4x5 = 20x3 = – 6x2 = 2 |
6 4 20 -6 Constant |
4 5 3 2 Constant |
| (e) 2β + 24 | 2 β3 = 2r |
2lb = 24 |
3 Constant |
| (f) (a + 4b + 5) ÷ 2 | a/2 = 2b = 5/2 |
a/2 2b 5/2 |
1 =1 Constant |
Question no – (2)
Solution :
(a) 10p2q2 – 4p12 + 12q2p – 6qp2
Like terms :
= 10p2q – 4pq2, 12q2p – 6qp2
(b) 5a – 8a + 12a2 + 9a – 10a2
Like terms :
= 5a – 8a + 9a, 12a2 – 10a2
(c) 2x2y3 + 5x3y2 – 10x3y2 + 11x3y3 – 3x2y3
Like terms :
= 2x2y3 – 3x2y3, 5x3y2 – 10x3y2 + 11x3y3
(d) 24y10 + 6y – 4y10 + 4y2 + 7y2 – 26y6
Like terms :
= 24y10 – 4y10
= y6 – 26y6
= 4y2 + 7y2
Question no – (3)
Solution :
(a) -5x3, 5x2, 2x
= Here, Variable are like though there are different so that terms are unlike.
(b) 2a, 2b, 2c
= Here, variables are unlike
(c) 14m3, 5m2n, 10
= Here, variable are different
(d) 12p2q, 11p2q
= Here, variables are different and their power are different.
Question no – (4)
Solution :
(a) 42b
= Monomials
(b) 0.5x2
= Monomials
(c) 2p2 + 2q2
= Binomial
(d) a2 + 2ab + b2
= Triangles
(e) 4y10 + 6y3x + 72y3 + 20
= Polynomials
(f) 12abc
= Monomials
(g) 24x2 + 12y2
= Binomial
Question no – (5)
Solution :
(a) 2x + 1
= 1
(b) 5
= Constant
(c) 2
= Constant
(d) 24x2 + 12y2
= Degree = 3
(e) 2xy + 5x + 1
= 4
Fundamentals of Algebra Exercise 11.3 Solution :
Question no – (1)
Solution :
(a) 2 + 3a + 6 – 7b when a = – 1 and b = 6
= 2 + 3a + 6 – 7b
= 2 + 3 × (- 1) + 6 – 7 (6)
= 2 – 3 + 6 – 42
= 8 – 45
= – 37
(b) 3 (2x + 9) when = 10
= 3 (2x + 9)
= 3 (2 × 10 + 9)
= 2 (20 + 9)
= 3 × 29
= 87
(c) 2xy + 6x – 9y when x = – 1 and y = 2
= 2xy + 6x -9y
= 2 × (- 1) × 2 + 6 (- 1) – 9 (2)
= (- 2) × 2 – 6 – 18
= – 4 – 6 – 18
= – 28
(d) 1/x2 × 125 × y/125 × 0 when = 5 and y = – 4
= 1/x2 × 125 × y/125 × 0
When x = 5 and y = – 4
= 1/(5)2 × 125 × y/125 × 0
= 0
Question no – (2)
Solution :
(a) -5p3 + 2p2 + p when p = 6
= – 5(6)3 + 2 (6)2 + 6
= – 5 × 216 + 2 × 36 + 6
= – 1080 + 2 × 36 + 6
= 1080 + 72 + 6
= 1080 + 78
= – 1002
(b) 52z2 + 19zq when z = 3 and q = 2
= 52(3)2 + 19 × 2 × 3
= 52 × 9 + 114
= 468 + 114
= 582
(c) 4a3b2 – 2b2 + 3a – 4 when a = 4 and b = 3
= 4 × (4)3 × (3)2 – 2 × (3)2 + 3 × (3) – 4
= 4 × 89 × 64 – 2 × 9 + 9 – 4
= 2304 – 18 + 9 – 4
= 2313 – 22
= 2292
(d) 25 + x2 + 5x/5 when x = 5
= 25 + (5)2 + 5 × 5/5
= 75/5
= 15
Question no – (3)
Solution :
Value of 10 + 64/x2+ 6x when x = 2
= 10 + 64/4 + 6 × 2
= 10 + 16 + 12
= 38
Value of 10 + 64/x2+ 6x when x = 4
= 10 + 64/x2 + 6x
= 10 + 64/16 + 6 × 4
= 10 + 4 + 24
= 38
Value of 10 + 64/x2+ 6x when x = 8
= 10 + 64/82 + 6x × 8
= 10 + 64/64 + 48
= 10 + 1 + 48
= 59
Value of 10 + 64/x2+ 6x when x = 10
= 10 + 64/(10)2 + 6 × 10
= 10 + 64/100 + 60
= 70 + 0.64
= 70.64
Fundamentals of Algebra Exercise 11.4 Solution :
Question no – (1)
Solution :
(a) 5k – 4
= Is an algebraic expressions.
(b) 6p = 18
= is an algebraic equation.
(c) 5m – 4 = 16
= Is algebraic equation.
(d) 13g + 10
= Is an algebraic expression.
Question no – (2)
Solution :
(a) x – 8 = 11
Here, L.H.S = x – 8
R.H.S = 11
(b) 10z + 2 = 18
Here, L.H.S = 10z + 2
= R.H.S = 18
(c) 5p – 7 = 23
Here, L.H.S = 5p – 7
R.H.S = 23
(d) 5/4 (m – 7) = 34
Here, L.H/S = 5/4 (m – 7)
= R.H.S = 34
Question no – (3)
Solution :
Given, Three multiplied by a number gives 36.
Let, the number be x
= 3 × x = 36
= 3x = 36
Question no – (4)
Solution :
Given, Five times a number subtracted from 9 gives -11.
Let, the number BP
= 9 – p × 5 = -11
= 9 – 5p + 11 = 0
Question no – (5)
Solution :
Let, number of boys is = m
= m + 22 = 50
= m = 50 – 22
= 28
Therefore, the number of boys will be 28
Question no – (6)
Solution :
As per the given question,
Sukanya bought pencils worth = Rs 15.
She spent Rs 5 less than what Sudeshna spent.
let, Sudeshna a spent x
= x – 5 = 15
Hence, the equation will be x – 5 = 15
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