Algebra | Math | Chapter 14 | Grade 5 | With Exercises Solution

Algebra - Math Chapter-14, Grade 5:

Objectives:

Students will be able to:
Define constants and variables in algebra.
Identify different types of algebraic expressions.
Define coefficient and power.
Add or subtract algebraic expressions.
Multiply algebraic expressions by terms and other expressions.
Apply FOIL method to multiply two expressions.
Solve algebraic statements by making equations.

Outcomes:

Students will have learnt to perform basic operations over algebraic terms and expressions. Also, they will have learnt to find the value of unknown quantities given with the help of known quantities in one equations.

Materials:

Balance, weighing kits, different types of objects, beads, counters, etc.

Things To Know:

➧ Those quantities which have fixed values are called constants. For example, a denotes the zones of Nepal i.e. a = 14.
➧ Those quantities which have values over a certain range are called variables. For example, '𝑥' denotes a variable: 'y' also denotes variable.
➧ Algebraic expression is formed when two or more algebraic terms are connected by '+' or '-' sign.
➧ In 5𝑥2, 5 is a coefficient, 𝑥 is a base and 2 is a power (index) of the base. Like terms can be added and subtracted.

14.1 Constants and Variables/button

Question: What are Constants?
Answer: Those quantities which have fixed values are called constants. Like, a denotes the zones of Nepal i.e. a = 14.

When somebody ask you, how many zones are there in Nepal? The answer would be 14. This is fixed number. So, it is a constant.

Also Check:

Question: What are Variables?
Answer: Those quantities which have values over a certain range are called variables. Like, x denotes the height of students of grade 5.

Is there a fixed height for students of grade 5? No! It's not. The height of different student is different. There is no fixed height for the students.

14.2 Algebraic Expression/button

Question: What is Algebraic Expression?
Answer: The mathematical expression formed by combining the variables and constants with basic operations of arithmetic is known as Algebraic Expressions. Like, 𝑥 is a variable and 7 is a constant. So, 𝑥+7 is an algebraic expression.

14.3 Algebraic Term/button

A term is either a number or a variable, or numbers and variables multiplied. An expression is a group of terms connected with '+' or '-' signs.

An algebraic expressions may be monomial, binomial, trinomial and polynomial.

Monomial expression	Contains only one term	For example a) 𝑥, b) 3y, c) 5𝑥y, etc.
Binomial expression	Contains only two terms	a) z𝑥 + 3, b) 3𝑥 + 2y, c) 2𝑥2 + 3y3, etc
Trinomial expression	Contains three terms	a( 𝑥 + y - z), b) 2𝑥2 + 3y + z, c) a𝑥2 + b𝑥 + c, etc.
Polynominal expression	Contains more than three terms	a) 𝑥 + y + z - 3, b) 𝑥3 + x2 + x - 1, c) 3a3 + 2a2b + 3ab2 + 3b, etc.

14.3 Mathematical Evaluation of Algebraic Terms or Expressions/button

The values for given variables in an algebraic terms or expressions finally evaluate the total value for a given algebraic terms or expressions.

Example 1: If 𝑥=2 and y=3, find
i) 2𝑥
ii) 𝑥 + 2y
iii) 2𝑥/𝑥+y
Solution:
i) 2𝑥
= 2 x 𝑥 [𝑥 = 2]
= 2 x 2
= 4
ii) 𝑥 + 2y
= 2 + 2 x 3
= 2 + 6
= 8
iii) 2𝑥/𝑥+y
= 2 x 2/2+3
= 4/5.

14.5 Coefficient and Power/button

Coefficient and Power can be understand by following examples.
Suppose an algebraic term be 𝑥.
Add 𝑥 two times - 𝑥 + 𝑥 = 2𝑥
Add x three times - 𝑥 + 𝑥 + 𝑥 = 3𝑥
Here, in 2𝑥, x has been added two times. The term 𝑥 is called the base and the number 2 is the coefficient of this algebraic term.
Sometimes, the algebraic terms would be like this a𝑥, by. Here, in a𝑥, x is the base and a is the coefficient which is known as literal coefficient. And in by, y is the base and x is the coefficient.
Example:
Consider a term x.
Multiply 𝑥 two times = 𝑥 x 𝑥 = 𝑥²
Multiply 𝑥 three times = 𝑥 x 𝑥 x 𝑥 = 𝑥³
Here, in 𝑥^3, 𝑥 is the base and 3 is the power which is written in the superscript.

How to read the power?
𝑥² = 𝑥 is squared.
𝑥³ = 𝑥 is cubed.
𝑥⁴ = 𝑥 to the power 4.
𝑥⁵ = 𝑥 to the power 5.
And so on ... alert-info

14.6 Like and Unlike Terms/button

Like in fractions and decimals, we have like an unlike terms in algebra too. In 𝑥, 2𝑥, 3𝑥, 4𝑥 and 5𝑥, all terms have same base 𝑥 but different coefficients. These are like terms.
∴ Like terms have same base and power but may have different coefficient. alert-info
Some other examples of like terms are-
5a, -3a → same base a
+2𝑥y, -5𝑥y → base 𝑥y
-4𝑥², 7𝑥² → same base 𝑥 and power 2
Again, lets see few other examples,
4𝑥²and 3𝑥, 8a and 4a³, 5y²and 3z²
Here,
4𝑥²has same base 𝑥 and power 2.
3𝑥 has base 𝑥 and power 1.
And, base is same but power are different.
Similarly, in 8a and 4a³
8a has base a and power 1
4a³has base a and power 3.
∴ The base is same but the power are different.
In and 5y²and 3z²,
5y²has base y and power 2.
3z²has base z and power 2.
The base is different but the powers are same.
In all these examples, the given terms haven't same base and same power. So, they are unlike terms.
Unlike terms do not have same base and same power. alert-info
Like terms can be added or subtracted, but unlike terms cannot be added or subtracted. alert-info

Grade 5: Math Subject-Chapter:14 - Algebra: Exercise 14.1 (With Solution)/button

1. Express the following statements into algebraic form.

a. Hari is 𝑥 years old. How old was he before 3 years?
Solution:
Hari is 𝑥 years old.
Before three years, he was 𝑥-3 years old.

b. Ramesh is 'y' years old now. How old he will be after 7 years?
Solution:
Ramesh is y years old.
After 7 years, he will be y+7 years old.

c. There are 𝑥 number of boys and y number of girls in a classroom. How many students are there altogether?
Solution:
Here,
Number of boys = 𝑥
Number of boys = y
∴ Total numbers of students in a classroom = 𝑥+y.

d. There are x number of people in a village. Out of them 1,600 are men. How many women are there?
Solution:
Here,
Total number of people in a village = 𝑥
Total number of men in a village = 1,600
Total number of women in a village = ?
∴ Total number of women in a village = Total number of people in a village - Total number of people in a village = x-1,600.

e. The cost of one knife is Rs. 42. What is the cost of y number of knives?
Solution:
Here,
Cost of one knife = Rs. 42
∴ Cost of y knives = Rs. 42 x y = 42y.

f. The cost of one dozen of pencil is Rs. y. What is the cost of 3 pencils?
Solution:
Here,
Cost of 1 dozen (12) pencils = Rs. y
Cost of 1 pencil = Rs. y/12
∴ Cost of 3 pencils = Rs. y/12 x 3 = Rs. y/4.

2. State whether the following statements are true or false.

a. 6𝑥 is a monomial. True
b. 7𝑥ab is binomial. False
c. The expressions 5 + y has two terms 5 and y. True
d. The coefficient of 𝑥y in -𝑥y is -1. True
e. The coefficient of a in 7a is 7a. False
f. The coefficient of m in -6mn is -6. False
g. 2𝑥y + 3𝑥 + 5y is a trinomial. True
h. 𝑥²+ 5𝑥y -6𝑥 + 9 is a polynomial. True

3. Write the type of the following expressions:

a. 𝑥y → monomial
b. ab + d → binomial
c. -𝑥 +y = b → binomial
d. 7𝑥y → monomial
e. 3 + 𝑥 ÷ → binomial
f. 𝑥² + 𝑥y + y → trinomial
g. 𝑥 + y + 1 → trinomial
h. a𝑥² + 2𝑥 + 4 → trinomial
i. 3𝑥² + 4𝑥y + y²+ 5 → polynomial

4. Write the factors of the following monomials.

a. 3a
Solution:
∴ 3 x a

b. -2𝑥
Solution:
∴ -2 x 𝑥

c. -x
Solution:
∴ -1 x 𝑥

d. -2abc
Solution:
∴ -2 x a x b x c

e. 2/3 xy
Solution:
∴ 2/3 x 𝑥 x y

f. 3ab/5
Solution:
∴ 3/5 x a x b

g. 7𝑥yz
Solution:
∴ 7 x 𝑥 x y x z

h. 17abc
Solution:
∴ 17 x a x b x c

i. -7𝑥y²/9
Solution:
∴ 7/9 x 𝑥 x y x y

5. Write down the coefficients of 𝑥 in the following monomials.

a. 3𝑥
Solution:
∴ Coefficient of 𝑥 is 3.

b. -𝑥
Solution:
∴ Coefficient of 𝑥 is 1.

c. -7𝑥
Solution:
∴ Coefficient of 𝑥 is -7.

d. 7𝑥y
Solution:
∴ Coefficient of 𝑥 is 7y.

e. -5m𝑥
Solution:
∴ Coefficient of 𝑥 is -5m.

f. 7𝑥/9
Solution:
∴ Coefficient of 𝑥 is 7/9.

g. 11𝑥yz
Solution:
∴ Coefficient of 𝑥 is 11yz.

h. 3a𝑥 ÷ 2c
Solution:
∴ Coefficient of 𝑥 is 3a.

i. 16𝑥y ÷ 15y
Solution:
∴ Coefficient of 𝑥 is 16y.

6. Write down the numerical coefficient of the given monomials.

a. 3𝑥y
Solution: ∴ 3.

b. 2abc
Solution: ∴ 2.

c. 7pqr
Solution: ∴ 7.

d. -2𝑥y/3
Solution: ∴ -2/3.

e. 2𝑥²y/3
Solution: ∴ -2/3.

f. -10𝑥yz
Solution: ∴ -10.

g. 9𝑥 ÷ 5y
Solution: ∴ 9/5.

h. -3a ÷ 7b
Solution: ∴ -3/7.

i. 3a x 7b
Solution: ∴ 21.

7. Write down the coefficient of 𝑥y² in the given monomial.

a. -7𝑥y²z
Solution: -7z.

b. 4𝑥y²/5
Solution: 4/5.

c. 5a𝑥y²z
Solution: 5az.

d. -4𝑥y²
Solution: -4.

e. -𝑥y²
Solution: -1.

f. -4b𝑥y²/5c
Solution: -4b/5c.

8. Write down the coefficient of following algebraic expressions.

a. a in -5ab²
Solution: -5b²

b. 𝑥 in -2𝑥b
Solution: -2b.

c. y in -2y
Solution: -2.

d. a in 4a/5
Solution: 4/5.

e. ab in - 7abz
Solution: -7z.

f. ax in -6a𝑥y²
Solution: -6y².

g. a²b in -8𝑥a²b
Solution: -8𝑥.

h. ab² in 11ab²z
Solution: 11z.

i. a𝑥 in 5ba𝑥y
Solution: 5by.

9. Group the like terms together.

a. 2𝑥, -7y, -4𝑥/5, 4y/5 and y
Solution: 2𝑥, 4𝑥/5 and -7y, -4y/5, y.

b. 2ab/3, 4ba, 7bz, 1bz/2 and -ab
Solution: 2ab/3, 4ba, -ab and 7bz, 1bz/2.

c. -𝑥y², 𝑥²y², 11𝑥y², -8𝑥²y² and 7𝑥y²
Solution: 𝑥y², 7𝑥y², 11𝑥y² and 𝑥²y², -8𝑥²y²

d. 𝑥²y²z², 6𝑥yz, -8𝑥²y²z², and 4𝑥yz/5
Solution: 𝑥²y²z², -8𝑥²y²z² and 6𝑥yz, 4𝑥yz/5

e. 10a𝑥, -16by, 4𝑥a, by/6 and 4a𝑥/5
Solution: 10a𝑥, 4𝑥a, 4a𝑥/5 and by/6, -16by

f. 3𝑥y/8, 5y𝑥/4, 𝑥y², 16𝑥y² and -𝑥y
Solution: 3𝑥y/8, -𝑥y, 5y𝑥/4 and 𝑥y², 16𝑥y²

10. State true or false.

a. 𝑥y and -y𝑥 are like terms. True
b. a²b and ab²are like terms. False
c. -a and a are unlike terms. False
d. 10 and 10a are like terms. False
e. -𝑥y and 5𝑥y are like terms. True
f. 9ab and 9𝑥ab are unlike terms. True

11. Find the value when 𝑥 = 7.

a. 𝑥 + 4
Solution: 𝑥 + 4 = 7 + 4 = 11.

b. 𝑥 + 7
Solution: 𝑥 + 7 = 7 + 7 = 14.

c. 𝑥 - 3
Solution: 𝑥 - 3 = 7 - 3 = 4.

d. 2𝑥 +7
Solution: 2𝑥 + 7 = 2 x 7 + 7 = 14 + 7 = 21.

e. 3𝑥 - 12
Solution: 3𝑥 - 12 = 3 x 7 - 12 = 21 - 12 = 9.

f. 22 - 3𝑥
Solution: 22 - 3𝑥 = 22 - 3 x 7 = 22 - 21 = 1.

12. Find the value when 𝑥 = 2 and y = 5.

a. 𝑥 + y
Solution: 𝑥 + y = 2 + 5 = 7.

b. 𝑥 - y
Solution: 𝑥 - y = 2 - 5 = -3.

c. 2(𝑥 + y)
Solution: 2(𝑥 + y) = 2(2 + 5) = 2 x 7 = 14.

d. 2𝑥 + 3y
Solution: 2𝑥 + 3y = 2 x 2 + 3 x 5 = 4 + 15 = 19.

e. 8𝑥 - 3y
Solution: 8𝑥 - 3y = 8 x 2 - 3 x 5 = 16 - 15 = 1.

f. 𝑥 + y/7
Solution: 𝑥 + y/7 = 2 + 5/7 = 7/7 = 1.

13. Evaluate when a = 4 and b =5.

a. ab
Solution: ab = a x b = 4 x 5 = 20.

b. 3ab
Solution: 3ab = 3 x a x b = 3 x 4 x 5 = 60.

c. 4ab + 7
Solution: 4 x a x b + 7 = 4 x 4 x 5 + 7 = 80 + 7 = 87.

d. 6ab - 9
Solution: 6ab - 9 = 6 x a x b - 9 = 6 x 4 x 5 - 9 = 120 - 9 = 111.

e. 1/4 of ab
Solution: 1/4 of ab = 1/4 x a x b = 1 x a x b/4 = 1 x 4 x 5/4 = 20/4 = 5.

14. Evaluate when, 𝑥 = 2, y = 3, a = 4 and b =5.

a. ab + 𝑥y
Solution: ab + 𝑥y = 4 x 5 + 2 x 3 = 20 + 6 = 26.

b. 2(ab + 𝑥y)
Solution: 2(ab + 𝑥y) = 2(4 x 5 + 2 x 3) = 2(20 - 6) = 2 x 14 = 28.

c. ab - 𝑥y/7
Solution: ab - 𝑥y/7 = 4 x 5 - 2 x 3/7 = 20 - 6/7 = 14/7 = 2.

d. 3ab + 4𝑥y
Solution: 3ab + 4𝑥y = 3 x 4 x 5 + 4 x 2 x 3 = 60 + 24 = 84.

e. 3𝑥y + 4ab
Solution: 3 x 2 x 3 + 4 x 4 x 5 = 18 + 80 = 98.

15. Find the value of l x b, when

a. l = 9 and b = 5
Solution: l x b = 9 x 5 = 45.

b. l = 10 and b = 7
Solution: l x b = 10 x 7 = 70.

c. l = 12 and b = 8
Solution: l x b = 12 x 8 = 96.

16. Find 2(l + b) in each case.

a. l = 9 and b = 6
Solution: 2(l + b) = 2(9 + 6) = 2 x 15 = 30.

b. l = 10 and b = 8
Solution: 2(l + b) = 2(10 + 8) = 2 x 18 = 36.

c. l = 12 and b = 8
Solution: 2(l + b) = 2(12 + 8) = 2 x 20 = 40.

17. Write the algebraic expression to represent the perimeter of the following figures. If x = 2 cm, y = 4 cm and z = 3 cm, find the perimeter of the figures in cm.

Solution:
Here, l = 𝑥 = 2 cm, b = y = 4 cm
We know that,
P = 2(l + b) = 2( 𝑥 + y ) = 2(2 cm + 4 cm) =2 x 6 cm = 12 cm.
∴ Perimeter of ABCD (P) = 12 cm.

Solution:
Here,
𝑥 = 2 cm, y = 4 cm and z = 3 cm.
And,
PQ = 3y = 3 x 4 cm = 12 cm.
QR = 2𝑥 = 2 x 2 cm = 4 cm.
RS = 2y = 2 x 4 cm = 8 cm.
SP = 𝑥 = 2 cm.
We know that,
Perimeter of PQRS (P) = PQ + QR + RS + SP = 12 cm + 4 cm + 8 cm + 2 cm = 26 cm.
∴ Perimeter of PQRS (P) = 26 cm.

Solution:
Here,
𝑥 = 2 cm, y = 4 cm and z = 3 cm.
And,
AB = 𝑥 = 2 cm.
BC = y = 4 cm.
CD = z = 3 cm.
DE = 𝑥 = 2 cm.
EF = y = 4 cm.
FA = z = 3 cm.
We know that,
Perimeter of ABCDEF (P) = AB + BC + CD + DE + EF + FA = 2 cm + 4 cm + 3 cm + 2 cm + 4 cm + 3 cm = 18 cm.
∴ Perimeter of ABCDEF (P) = 18 cm.

Solution:
Here,
𝑥 = 2 cm.
DE = 2𝑥 = 2 x 2 cm = 4 cm.
EF = 4𝑥 = 4 x 2 cm = 8 cm.
FD = 3𝑥 = 3 x 2 cm = 6 cm.
We know that,
Perimeter of DEF (P) = DE + EF + FD = 4 cm + 8 cm + 6 cm = 18 cm.
∴ Perimeter of DEF (P) = 18 cm.

14.7 Addition and Subtraction of Algebraic Expressions/button

Generally, as of above we are clear about the addition of like terms. Thus, we can add or subtract the terms of given expressions in two ways:-
a. by vertical arrangement of like terms,
For example:

b. by horizontal arrangement of like terms,
For example:
Add 3𝑥 + 4y + 6z and 5𝑥 + 3y + 2z
= (3𝑥 + 4y + 6z) + (5𝑥 + 3y + 2z)
= 3𝑥 + 4y + 6z + 5𝑥 + 3y + 2z
= 3𝑥 + 5𝑥 + 4y + 3y + 6z + 2z
= 8𝑥 + 7y + 8z.

Grade 5: Math Subject-Chapter:14 - Algebra: Exercise 14.2 (With Solution)/button

1. Add the following expressions.

i. 3𝑥 + 2 and 2𝑥 +5
Solution:
= 3𝑥 + 2 + 2𝑥 +5
= 3𝑥 + 2𝑥 + 2 + 5
= 5𝑥 + 7

j. 2a + 1 and 5a + 4
Solution:
= 2a + 1 + 5a + 4
= 2a + 5a + 1 + 4
= 7a + 5

k. 5a - 1 and 2a + 6
Solution:
= 5a - 1 + 2a + 6
= 5a + 2a + 6 - 1
= 7a + 5

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