Expressions using Letter Numbers Class 7 Notes Maths Chapter 4

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Class 7 Maths Chapter 4 Notes Expressions using Letter Numbers

Class 7 Maths Notes Chapter 4 – Class 7 Expressions using Letter Numbers Notes

The Notion of Letter-Numbers Class 7 Notes

In the following picture, if each basket contains the same number of toffees, how can we represent the total number of toffees in 4 such baskets?

If each basket contains 8 toffees, the total number of toffees in 4 baskets = 4 × 8
If each basket contains 10 toffees, the total number of toffees in 4 baskets = 4 × 10
If each basket contains 20 toffees, the total number of toffees in 4 baskets = 4 × 20
So, if we let n be the number of toffees in each basket, then the total number of toffees in 4 baskets = 4 × n
In the above, the letter n in 4 × n may take on the value 8, 10, 20, or any whole number.
We call n a letter-number and 4 × n an algebraic expression.

Question 1.
Suntan is 14 years old. Find her age after
(i) 5 years
(ii) 8 years
(iii) x years
Solution:
Suman’s present age = 14 years
(i) Her age after 5 years = 14 + 5 = 19 years
(ii) Her age after 8 years = 14 + 8 = 22 years
(iii) Her age after x years = (14 + x) years

Note: The algebraic expression ‘14 + x’ is the sum of 14 and x.

Question 2.
The breadth of a rectangle is 4 cm shorter than half of its length. Find the breadth of the rectangle when its length is (i) 16 cm, (ii) 22 cm, (iii) y cm.
Solution:
Breadth of the rectangle = \(\frac{1}{2}\) × length – 4 cm
(i) When the length is 16 cm, then breadth = \(\frac{1}{2}\) × 16 – 4
= 8 – 4
= 4 cm

(ii) When the length is 22 cm, then breadth = \(\frac{1}{2}\) × 22 – 4
= 11 – 4
= 7 cm

(iii) When the length is y cm, then breadth = \(\frac{1}{2}\) × y – 4 = (\(\frac{y}{2}\) – 4) cm

Note: \(\frac{y}{2}\) means y ÷ 2 and \(\frac{y}{2}\) – 4 is an algebraic expression in letter-number y.

Question 3.
The price of a book is ₹ 25, and the price of a ball pen is ₹ 10. Find the total price of
(i) 4 books and 3 ball pens
(ii) x books and y ball pens
Solution:
(i) Price of 4 books = ₹ 25 × 4
Price of 3 ball pens = ₹ 10 × 3
Therefore, total price = ₹ (25 × 4 + 10 × 3)
= ₹ (100 + 30)
= ₹ 130

(ii) Price of x books = ₹ 25 × x
Price ofy ball pens = ₹ 10 × y
Therefore, total price = ₹ (25 × x + 10 × y) = ₹ (25x + 10y)

Note:
1. 25x + 10y is an algebraic expression in two letter-numbers x and y.
2. 25 × x is written as 25x, where the multiplication sign ‘×’ is omitted.

Revisiting Arithmetic Expressions Class 7 Notes

Question 1.
Simplify the following:
(i) m – n – m – n
(ii) 2x – (x – 3x) + 4x
Solution:
(i) m – n – m – n = m – m – n – n
= 0 – 2n
= -2n

(ii) 2x – (x – 3x) + 4x
= 2x – x + 3x + 4x
= 2x + 3x + 4x – x
= 9x – x
= 8x

Question 2.
Simplify:
(i) 4x × 2y + z
(ii) p ÷ 2q × 7
Solution:
(i) 4x × 2y + z = 4 × x × 2 × y + z
= 4 × 2 × x × y + z
= 8xy + z

(ii) p ÷ 2q × 7 = \(\frac{p}{2q}\) × 7 = \(\frac{7p}{2q}\)

Question 3.
Are the expressions 7x + 1 and 7(x + 1) equal to each other?
Solution:
Here, 7x + 1 means ‘1 more than 7 times x’;
and 7(x + 1) means ‘7 times (1 more than x)
Let x = 2. Then,
You may choose any other value of x.
7x + 1 = 7 × 2 + 1 = 14 + 1 = 15;
and 7(x + 1) = 7(2 + 1) = 7 × 3 = 21
Thus, 7x + 1 and 7(x + 1) are not equal to each other.

Question 4.
Simplify:
(i) 4(x + 2y) + 2x
(ii) 13p – 6q – (p + 2q)
Solution:
(i) 4(x + 2y) + 2x = 4x + 8y + 2x
= 4x + 2x + 8y
= 6x + 8y

(ii) 13p – 6q – (p + 2q) = 13p – 6q – p – 2q
= (13p – p) + (-6q – 2q)
= 12p – 8q

Pick Patterns and Reveal Relationships Class 7 Notes

In the above, you have seen how the use of letter-numbers and algebraic expressions allows us to express general mathematical relations concisely. Mathematical relations expressed this way are called Formulas.

Evaluation of Formulas
Consider the rectangle shown on the right.
Recall that the area of a rectangle is given by length × breadth

If we use l to represent the length and b to represent the breadth of the rectangle (in cm), then the algebraic expression for the area A (in sq cm) is l × b.
Thus, the rule A = l × b is called a formula for the area of a rectangle.
We also have a formula for the perimeter, P cm, of the rectangle in terms of l and b. It is P = 2(l + b)
We also have a formula for the perimeter, P cm, of a triangle of sides x cm, y cm, and z cm. It is P = x + y + z

Question 1.
The area A of the rectangle ABCD shown is given by A = 6b + 4b

Find the value of A when b = 4.
Solution:
For b = 4,
A = 6 × 4 + 4 × 4
= 24 + 16
= 40

Question 2.
The area A of the rectangle ABEF shown is given by A = 12w – 4w. Find the value of A when w = 5.

Solution:
For w = 5,
A = 12 × 5 – 4 × 5
= 60 – 20
= 40

Question 3.
Given the formula S_n = \(\frac{1}{2}\)n(n+1), find the value of S_n when (i) n = 8 (ii) n = 13.
Solution:

Question 4.
Given the formula S = ut + \(\frac{1}{2}\)at², find the value of s, when u = 10, t = 4 and a = 8.
Solution:
For u = 10, t = 4 and a = 8,
S = ut + \(\frac{1}{2}\)at²
= 10 × 4 + \(\frac{1}{2}\) × 8 × 4²
= 40 + 64
= 104

Question 5.
Given the formula C = \(\frac{5}{9}\)(F – 32), find the value of C when
(i) F = 212
(ii) F = 77
Solution:

Question 6.
Find the value of the algebraic expression xy² + 2xy – 7 when x = 5 and y = 3.
Solution:
For x = 5 and y = 3, we have
xy² + 2xy – 7 = 5 × 3² + 2 × 5 × 3 – 7
= 5 × 9 + 2 × 5 × 3 – 7
= 45 + 30 – 7
= 68

Question 7.
When p = 7, q = 1, and r = 3, find the value of the following algebraic expressions:
(i) r(8p + qr)
(ii) \(\frac{10 p q-r+2}{p q^2-3}\)
Solution:
For p = 7, q = 1 and r = 3, we have
(i) r(8p + qr) = 3 × (8 × 7 + 1 × 3)
= 3 × (56 + 3)
= 3 × 59
= 177

Deriving Expressions From Patterns

1. Look at the multiplication table of 6:
1 × 6 = 6
2 × 6 = 12
3 × 6 = 18
.
.
.
From the pattern, we find that the product is 6 times the number, i.e., 6n

2.

In the above, a pattern of triangles made with matchsticks is shown.
Number of triangles × 3 = Number of matchsticks.
Thus, t is the number of triangles and m is the number of matchsticks, then t × 3 = m
This is the rule for the pattern.

3.

In the above, a pattern of squares made with matchsticks is shown.
Number of squares × 3 + 1 = Number of matchsticks.
Thus, s is the number of squares and m is the number of matchsticks, then s × 3 + 1 = m
This is the rule for the pattern.

4. Look at the pattern and derive the rule.
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
.
.
.
From the pattern, we derive the rule ‘The sum of the first n odd numbers is n2.’

5. Study the pattern and write the rule for the pattern:

1 → 1 = \(\frac{1 \times(1+1)}{2}\);
2 → 3 = \(\frac{2 \times(2+1)}{2}\);
3 → 6 = \(\frac{3 \times(3+1)}{2}\);
4 → 10 = \(\frac{4 \times(4+1)}{2}\);
.
.
.
From the pattern, we find that the general rule is n → \(\frac{n \times(n+1)}{2}\)

6. Study the pattern and write the rule for the pattern:

1 → 5 = 4 × 1 + 1;
2 → 9 = 4 × 2 + 1
3 → 13 = 4 × 3 + 1;
4 → 17 = 4 × 4 + 1
From the pattern, we find that the general rule is n → 4 × n + 1

Question 1.
Look at the matchstick pattern given below:

(i) Write the rule to find the number of matchsticks in the next steps.
(ii) Find the number of matchsticks in step 11, step 15, and step 32.
Solution:
(i) The pattern of the number of matchsticks is 3, 5, 7, 9,……….
(ii) The number of sticks in the next step is (9 + 2 ), i.e., 11
Thus, the required rule is n → 2n + 1

Question 2.
Here is a pattern of the colour changes on a traffic signal

Study the pattern carefully and find the colour at (i) 12th position, (ii) 25th position, and (iii) 39th position.
Solution:
Here, Positions of red colour: 1, 5,………..
Rule is n → 4n – 3
Positions of yellow colour: 2, 4,……….
Rule is n → 2n
Positions of green colour: 3, 7,……….
Rule is n → 4n – 1
(i) Colour at 12th position is yellow.
(ii) Colour at the 25th position is red.
(iii) Colour at 39th position is green.

Question 3.
Observe the pattern given below:

How many circles are there in step 4? In step 7? In step 12?
Solution:
Number of circles in steps 1, 2, 3 are 5, 9, 13,……….
The rule, thus, formed is n → 4n + 1
So, (i) Number of circles in step 4 is (4 × 4 + 1), i.e., 17
(ii) Number of circles in step 7 is (4 × 7 + 1), i.e., 29
(iii) Number of circles in step 12 is (4 × 12 + 1), i.e., 49

Question 4.
Here is an endless 5-column grid.

(i) Write the algebraic expressions to generate all the numbers in each of the columns.
(ii) In which row and column will the following numbers appear?
45, 58, 115
(iii) Write the number in row r and column c.
Solution:
(i) In column 1, the number pattern is 1, 6, 11, 16, 21,………
The rule is n → 5n – 4
In Column 2, the number pattern is 2, 7, 12, 17, 22,………
The rule is n → 5n – 3
In Column 3, the number pattern is 3, 8, 13, 18, 23,…………
The rule is n → 5n – 2
In Column 4, the number pattern is 4, 9, 14, 19, 24,…………
The rule is n → 5n – 1
In Column 5, the number pattern is 5, 10, 15, 20, 25, and so on.
The rule is n → 5n

(ii) The number 45 will lie in Column 5;
The number 58 will lie in Column 3.
The number 114 will lie in Column 4.

(iii) The number in row r and column c is 5(r – 1) + c.