Class 6: Maths Chapter 21 solutions. Complete Class 6 Maths Chapter 21 Notes.
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Selina Class 6 ICSE Solutions Mathematics : Chapter 21- Framing Algebraic Expressions (Including Evaluation)
Selina 6th Maths Chapter 21, Class 6 Maths Chapter 21 solutions
Exercise 21
1. Write in the form of an algebraic expression:
(i) Perimeter (P) of a rectangle is two times the sum of its length (l) and its breadth (b).
(ii) Perimeter (P) of a square is four times its side.
(iii) Area of a square is square of its side.
(iv) Surface area of a cube is six times the square of its edge.
Solution:
(i) Let us assume the length be l, breadth be b and perimeter be P. Then the algebraic expression for the given statement is written as,
P = 2 (l + b)
(ii) Let us assume the side be s and area be A. Then the algebraic expression for the given statement is written as,
P = 4s
(iii) Let us assume the side be s and area be A. Then the algebraic expression for the given statement is written as,
A = (s)2
(iv) Let us assume the surface area be S and edge be a. Then the algebraic expression for the given statement is written as,
S = 6(a)2
2. Express each of the following as an algebraic expression:
(i) The sum of x and y minus m.
(ii) The product of x and y divided by m.
(iii) The subtraction of 5m from 3n and then adding 9p to it.
(iv) The product of 12, x, y and z minus the product of 5, m and n.
(v) Sum of p and 2r – s minus sum of a and 3n + 4x.
Solution:
(i)The algebraic expression for the given sentence is given below
x + y – m
(ii) The algebraic expression for the given sentence is given below
xy / m
(iii) The algebraic expression for the given sentence is given below
3n – 5m + 9p
(iv) The algebraic expression for the given sentence is given below
12xyz – 5mn
(v) The algebraic expression for the given statement is given below
p + 2r – s – (a + 3n + 4x)
3. Construct a formula for the following:
Total wages (Rs W) of a man whose basic wage is (Rs B) for t hours week plus (Rs R) per hour, if he works a total of T hours.
Solution:
The wages for t hours is Rs B.
The wages for overtime is R(T – t)
Hence, the total wages is calculated as given below,
W = B + R(T – t)
4. If x = 4, evaluate:
(i) 3x + 8
(ii) x2 – 2x
(iii) x2 / 2
Solution:
(i) 3x + 8
The value of 3x + 8 for x = 4 is calculated as below
Now, substituting x = 4 in the given equation, we get
3x + 8 = 3 × 4 + 8
= 12 + 8
= 20
Therefore, the value of 3x + 8 for x = 4 is 20
(ii) x2 – 2x
The value of x2 – 2x for x = 4 is calculated as below
Now, substituting x = 4 in the given equation, we get
x2 – 2x = 42 – 2 × 4
= 16 – 8
= 8
Therefore, the value of x2 – 2x for x = 4 is 8
(iii) x2 / 2
The value of x2 / 2 for x =4 is calculated as below
x2 / 2 = 42 / 2
= 16 / 2
= 8
Therefore, the value of x2 / 2 for x = 4 is 8
5. If m = 6, evaluate:
(i) 5m – 6
(ii) 2m2 + 3m
(iii) (2m)2
Solution:
(i) 5m – 6
The value of 5m – 6 for m = 6 is calculated as below
Now, substituting the value of m = 6 in the given equation, we get
5m – 6 = 5 × 6 – 6
= 30 – 6
= 24
Hence, the value of 5m – 6 for m = 6 is 24
(ii) 2m2 + 3m
The value of 2m2 + 3m for m = 6 is calculated as below
Now, substituting the value of m = 6 in the given equation, we get
2m2 + 3m = 2(6)2 + 3(6)
= 2 × 36 + 3 × 6
= 72 + 18
= 90
Hence, the value of 2m2 + 3m for m = 6 is 90
(iii) (2m)2
The value of (2m)2 for m = 6 is calculated as below
Now, substituting the value of m = 6 in the given equation, we get
(2m)2 = (2 × 6)2
= 122
= 144
Hence, the value of (2m)2 for m = 6 is 144
6. If x = 4, evaluate:
(i) 12x + 7
(ii) 5x2 + 4x
(iii) x2 / 8
Solution:
(i) 12x + 7
The value of 12x + 7 for x = 4 is calculated as follows,
Substituting the value of x = 4 in the given equation, we get
12x + 7 = 12 × 4 + 7
= 48 + 7
= 55
Therefore, the value of 12x + 7 for x = 4 is 55
(ii) 5x2 + 4x
The value of 5x2 + 4x for x = 4 is calculated as follows,
Substituting the value of x = 4 in the given equation, we get
5x2 + 4x = 5 × 42 + 4 × 4
= 5 × 16 + 16
= 80 + 16
= 96
Therefore, the value of 5x2 + 4x for x = 4 is 96
(iii) x2 / 8
The value of x2 / 8 for x = 4 is calculated as follows,
x2 / 8 = 42 / 8
We get,
= 16 / 8
= 2
Therefore, the value of x2 / 8 for x = 4 is 2
7. If m = 2, evaluate:
(i) 16m – 7
(ii) 15m2 – 10m
(iii) 1 / 4 × m3
Solution:
(i) 16m – 7
The value of 16m – 7 for m = 2 is calculated as below,
16m – 7 = 16 × 2 – 7
= 32 – 7
We get,
= 25
Hence, the value of 16m – 7 for m = 2 is 25
(ii) 15m2 – 10m
The value of 15m2 – 10m for m = 2 is calculated as below,
15m2 – 10m = 15 × 22 – 10 × 2
= 15 × 4 – 20
= 60 – 20
We get,
= 40
Hence, the value of 15m2 – 10m for m = 2 is 40
(iii) 1 / 4 × m3
The value of 1 / 4 × m3 for m = 2 is calculated as below,
1 / 4 × m3 = 1 / 4 × 23
We get,
= 1 / 4 × 8
= 2
Hence, the value of 1 / 4 × m3 for m = 2 is 2
8. If x = 10, evaluate:
(i) 100x + 225
(ii) 6x2 – 25x
(iii) 1 / 50 × x3
Solution:
(i) 100x + 225
The value of 100x + 225 for x = 10 is calculated as follows,
100x + 225 = 100 × 10 + 225
= 1000 + 225
We get,
= 1225
Therefore, the value of 100x + 225 for x = 10 is 1225
(ii) 6x2 – 25x
The value of 6x2 – 25x for x = 10 is calculated as follows,
6x2 – 25x = 6 × 102 – 25 × 10
= 6 × 100 – 250
= 600 – 250
We get,
= 350
Therefore, the value of 6x2 – 25x for x = 10 is 350
(iii) 1 / 50 × x3
The value of 1 / 50 × x3 for x = 10 is calculated as follows,
1 / 50 × x3 = 1 / 50 × 103
= 1 / 50 × 1000
We get,
= 20
Therefore, the value of 1 / 50 × x3 for x = 10 is 20
9. If a = – 10, evaluate:
(i) 5a
(ii) a2
(iii) a3
Solution:
(i) 5a
The value of 5a for a = – 10 is calculated as shown below,
5a = 5 × (- 10)
We get,
= – 50
Therefore, the value of 5a for a = -10 is – 50
(ii) a2
The value of a2 for a = -10 is calculated as shown below,
a2 = (- 10)2
= 100
Therefore, the value of a2 for a = – 10 is 100
(iii) a3
The value of a3 for a = – 10 is calculated as shown below,
a3 = (- 10)3
= – 1000
Therefore, the value of a3 for a = – 10 is – 1000
10. If x = – 6, evaluate:
(i) 11x
(ii) 4x2
(iii) 2x3
Solution:
(i) 11x
The value of 11x for x = – 6 is calculated as follows,
11x = 11 × (-6)
We get,
= – 66
Hence, the value of 11x for x = – 6 is – 66
(ii) 4x2
The value of 4x2 for x = – 6 is calculated as follows,
4x2 = 4 × (-6)2
We get,
= 4 × 36
= 144
Hence, the value of 4x2 for x = – 6 is 144
(iii) 2x3
The value of 2x3 for x = – 6 is calculated as follows,
2x3 = 2 × (-6)3
We get,
= 2 × (-216)
= – 432
Hence, the value of 2x3 for x = – 6 is – 432
11. If m = – 7, evaluate:
(i) 12m
(ii) 2m2
(iii) 2m3
Solution:
(i) 12m
The value of 12m for m = – 7 is calculated as below,
12m = 12 × (- 7)
We get,
= – 84
Therefore, the value of 12m for m = – 7 is – 84
(ii) 2m2
The value of 2m2 for m = – 7 is calculated as below,
2m2 = 2 × (- 7)2
We get,
= 2 × 49
= 98
Therefore, the value of 2m2 for m = – 7 is 98
(iii) 2m3
The value of 2m3 for m = – 7 is calculated as below,
2m3 = 2 × (-7)3
We get,
= 2 × (- 343)
= – 686
Therefore, the value of 2m3 for m = – 7 is – 686
12. Find the average (A) of four quantities p, q, r and s. If A = 6, p = 3, q = 5 and r = 7; find the value of s.
Solution:
The average of four quantities p, q, r and s is calculated as,
A = (p + q + r + s) / 4
Substituting the given values in the above expression, we get
6 = (3 + 5 + 7 + s) / 4
6 × 4 = 3 + 5 + 7 + s
24 = 15 + s
s = 24 – 15
We get,
s = 9
Hence, the value of s is 9
13. If a = 5 and b = 6, evaluate:
(i) 3ab
(ii) 6a2b
(iii) 2b2
Solution:
(i) 3ab
The value of 3ab for a = 5 and b = 6 is calculated as follows,
3ab = 3 × a × b
= 3 × 5 × 6
We get,
= 90
Therefore, the value of 3ab if a = 5 and b = 6 is 90
(ii) 6a2b
The value of 6a2b for a = 5 and b = 6 is calculated as follows,
6a2b = 6 × a × a × b
= 6 × 5 × 5 × 6
We get,
= 6 × 25 × 6
= 900
Therefore, the value of 6a2b if a = 5 and b = 6 is 900
(iii) 2b2
The value of 2b2 for a = 5 and b = 6 is calculated as follows,
2b2 = 2 × b × b
= 2 × 6 × 6
= 2 × 36
= 72
Therefore, the value of 2b2 if a = 5 and b = 6 is 72
14. If x = 8 and y = 2, evaluate:
(i) 9xy
(ii) 5x2y
(iii) (4y)2
Solution:
(i) 9xy
The value of 9xy for x = 8 and y = 2 is calculated as follows,
9xy = 9 × x × y
9xy = 9 × 8 × 2
9xy = 144
Hence, the value of 9xy for x = 8 and y = 2 is 144
(ii) 5x2y
The value of 5x2y for x = 8 and y = 2 is calculated as follows,
5x2y = 5 × x × x × y
= 5 × 8 × 8 × 2
= 640
Hence, the value of 5x2y for x = 8 and y = 2 is 640
(iii) (4y)2
The value of (4y)2 for x = 8 and y = 2 is calculated as follows,
(4y)2 = (4 × 2)2
= (8)2
= 64
Hence, the value of (4y)2 for x = 8 and y = 2 is 64
15. If x = 5 and y = 4, evaluate:
(i) 8xy
(ii) 3x2y
(iii) 3y2
Solution:
(i) 8xy
The value of 8xy for x = 5 and y = 4 is calculated as follows,
8xy = 8 × x × y
= 8 × 5 × 4
= 160
Therefore, the value of 8xy for x = 5 and y = 4 is 160
(ii) 3x2y
The value of 3x2y for x = 5 and y = 4 is calculated as follows,
3x2y = 3 × x × x × y
= 3 × 5 × 5 × 4
= 15 × 20
= 300
Therefore, the value of 3x2y for x = 5 and y = 4 is 300
(iii) 3y2
The value of 3y2 for x = 5 and y = 4 is calculated as follows,
3y2 = 3 × y × y
= 3 × 4 × 4
= 48
Therefore, the value of 3y2 for x = 5 and y = 4 is 48
16. If y = 5 and z = 2, evaluate:
(i) 100yz
(ii) 9y2z
(iii) 5y2
(iv) (5z)3
Solution:
(i) 100yz
The value of 100yz for y = 5 and z = 2 is calculated as below,
100yz = 100 × y × z
= 100 × 5 × 2
= 100 × 10
= 1000
Hence, the value of 100yz for y = 5 and z = 2 is 1000
(ii) 9y2z
The value of 9y2z for y = 5 and z = 2 is calculated as below,
9y2z = 9 × y × y × z
= 9 × 5 × 5 × 2
We get,
= 45 × 10
= 450
Hence, the value of 9y2z for y = 5 and z = 2 is 450
(iii) 5y2
The value of 5y2 for y = 5 and z = 2 is calculated as below,
5y2 = 5 × y × y
= 5 × 5 × 5
We get,
= 125
Hence, the value of 5y2 for y = 5 and z = 2 is 125
(iv) (5z)3
The value of (5z)3 for y =5 and z = 2 is calculated as below,
(5z)3 = (5 × z)3
= (5 × 2)3
= (10)3
= 1000
Hence, the value of (5z)3 for y = 5 and z = 2 is 1000
17. If x = 2 and y = 10, evaluate:
(i) 30xy
(ii) 50xy2
(iii) (10x)2
(iv) 5y2
Solution:
(i) 30xy
The value of 30xy for x = 2 and y = 10 is calculated as follows,
30xy = 30 × x × y
= 30 × 2 × 10
= 600
Therefore, the value of 30xy for x = 2 and y = 10 is 600
(ii) 50xy2
The value of 50xy2 for x = 2 and y = 10 is calculated as follows,
50xy2 = 50 × x × y × y
= 50 × 2 × 10 × 10
= 10000
Therefore, the value of 50xy2 for x = 2 and y = 10 is 10000
(iii) (10x)2
The value of (10x)2 for x = 2 and y = 10 is calculated as follows,
(10x)2 = (10 × x)2
= (10 × 2)2
We get,
= (20)2
= 400
Therefore, the value of (10x)2 for x = 2 and y = 10 is 400
(iv) 5y2
The value of 5y2 for x = 2 and y = 10 is calculated as follows,
5y2 = 5 × y × y
= 5 × 10 × 10
= 5 × 100
= 500
Therefore, the value of 5y2 for x = 2 and y = 10 is 500
18. If m = 3 and n = 7, evaluate:
(i) 12mn
(ii) 5mn2
(iii) (10m)2
(iv) 4n2
Solution:
(i) 12mn
The value of 12mn for m = 3 and n = 7 is calculated as follows,
12mn = 12 × m × n
= 12 × 3 × 7
= 252
Hence, the value of 12mn for m = 3 and n = 7 is 252
(ii) 5mn2
The value of 5mn2 for m = 3 and n = 7 is calculated as follows,
5mn2 = 5 × m × n2
= 5 × 3 × 72
= 5 × 3 × 7 × 7
= 735
Hence, the value of 5mn2 for m = 3 and n = 7 is 735
(iii) (10m)2
The value of (10m)2 for m = 3 and n = 7 is calculated as follows,
(10m)2 = (10 × m)2
= (10 × 3)2
= (30)2
= 900
Hence, the value of (10m)2 for m = 3 and n = 7 is 900
(iv) 4n2
The value of 4n2 for m = 3 and n = 7 is calculated as follows,
4n2 = 4 × n2
= 4 × 72
= 4 × 7 × 7
= 196
Hence, the value of 4n2 for m = 3 and n = 7 is 196
19. If a = – 10, evaluate:
(i) 3a – 2
(ii) a2 + 8a
(iii) 1 / 5 × a2
Solution:
(i) 3a – 2
The value of 3a – 2 for a = – 10 is calculated as follows,
3a – 2 = 3 × (- 10) – 2
We get,
= – 30 – 2
= – 32
Therefore, the value of 3a – 2 for a = – 10 is – 32
(ii) a2 + 8a
The value of a2 + 8a for a = – 10 is calculated as follows,
a2 + 8a = (- 10)2 + 8 × (- 10)
= 100 – 80
= 20
Therefore, the value of a2 + 8a for a = – 10 is 20
(iii) 1 / 5 × a2
The value of 1 / 5 × a2 for a = – 10 is calculated as follows,
1 / 5 × a2 = 1 / 5 × (- 10)2
= 1 / 5 × 100
We get,
= 20
Therefore, the value of 1 / 5 × a2 for a = – 10 is 20
20. If x = – 6, evaluate:
(i) 4x – 9
(ii) 3x2 + 8x
(iii) x2 / 2
Solution:
(i) 4x – 9
The value of 4x – 9 for x = – 6 is calculated as follows,
4x – 9 = 4 × (- 6) – 9
= – 24 – 9
= – 33
Hence, the value of 4x – 9 for x = – 6 is – 33
(ii) 3x2 + 8x
The value of 3x2 + 8x for x = – 6 is calculated as follows,
3x2 + 8x = 3 × (- 6)2 + 8 × (- 6)
We get,
= 3 × 36 – 48
= 108 – 48
= 60
Hence, the value of 3x2 + 8x for x = – 6 is 60
(iii) x2 / 2
The value of x2 / 2 for x = – 6 is calculated as follows,
x2 / 2 = (- 6)2 / 2
We get,
= 36 / 2
= 18
Hence, the value of x2 / 2 for x = – 6 is 18
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Selina Class 6 ICSE Solutions Mathematics : Chapter 21- Framing Algebraic Expressions (Including Evaluation)
Chapterwise Selina Publishers ICSE Solutions for Class 6 Mathematics :
- Chapter 1- Number System
- Chapter 2- Estimation
- Chapter 3- Numbers In Indian And International Systems
- Chapter 4- Place Value
- Chapter 5- Natural Numbers And Whole Numbers
- Chapter 6- Negative Numbers And Integers
- Chapter 7- Number Line
- Chapter 8- HCF And LCM
- Chapter 9- Playing With Numbers
- Chapter 10- Sets
- Chapter 11- Ratio
- Chapter 12- Proportion
- Chapter 13- Unitary Method
- Chapter 14- Fractions
- Chapter 15- Decimal Fractions
- Chapter 16- Percent (Percentage)
- Chapter 17- Idea of Speed, Distance and Time
- Chapter 18- Fundamental Concepts
- Chapter 19- Fundamental Operations
- Chapter 20- Substitution
- Chapter 21- Framing Algebraic Expressions (Including Evaluation)
- Chapter 22- Simple (Linear) Equations
- Chapter 23- Fundamental Concepts
- Chapter 24- Angles
- Chapter 25- Properties of Angles and Lines
- Chapter 26- Triangles
- Chapter 27- Quadrilateral
- Chapter 28- Polygons
- Chapter 29- The Circle
- Chapter 30- Revision Exercise Symmetry
- Chapter 31- Recognition of Solids
- Chapter 32- Perimeter and Area of Plane Figures
- Chapter 33- Data Handling
- Chapter 34- Mean and Median
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