Expressions Using Letter-Numbers - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Expressions Using Letter-Numbers – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Expressions Using Letter-Numbers – NCERT Solutions

Q.1: Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be?

Solution:

Shabnam is 3 years older than Aftab
Difference in Shabnam’s and Aftab’s age = 3 years
Aftab’s present age = 18 years
Shabnam’s present age = 18 + 3 = 21 years.

Q.2: Parthiv is making matchstick patterns. He repeatedly places Ls next to each other. Each L has two matchsticks as shown in figure.

How many matchsticks are needed to make 5 Ls? It will be 5 {tex}\times{/tex} 2.
How many matchsticks are needed to make 7 Ls? It will be 7 {tex}\times{/tex} 2.
How many matchsticks are needed to make 45 Ls? It will be 45 {tex}\times{/tex} 2.
Now, what is the relation between the number of Ls and the number of sticks?

Solution:

First, let us describe the relationship or the pattern here. Every L needs 2 matchsticks. So the number of matchsticks needed will be 2 times the number of L’s. This can be written as:
Number of matchsticks = 2 {tex}\times{/tex} Number of L’s
Now, we can use any letter to denote the number of L’s. Let’s use n. The algebraic expression for the number of matchsticks will be:
2 {tex}\times{/tex} n.
This expression tells us how many matchsticks are needed to make n L’s. To find the number of matchsticks, we just replace n by the number of L.

Q.3: Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹ 35 and the price of 1 kg jaggery is ₹ 60.

Solution:

Cost of 8 coconuts {tex}=8 \times 35=₹ 280{/tex}
Cost of 9 kg jaggery {tex}=9 \times 60=₹ 540{/tex}
Total money paid {tex}=₹ 280+₹ 540=₹ 820{/tex}.

Q.4: We are familiar with calculating the perimeters of simple shapes. Write expressions for perimeters.

Solution:

The perimeter of a square is 4 times the length of its side. This can be written as the expression: 4 {tex}\times{/tex} q, where q stands for the side length.

Q.5: Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

	Day 1	Day 2	Day 3
Pencils (Price ‘{tex}c{/tex}’)	5	3	10
Erasers (Price ‘{tex}d{/tex}’)	4	6	1

Solution:

Let us first find the money earned by the sale of pencils.
The money earned by selling pencils on Day 1 is 5c. Similarly, the money earned by selling pencils on Day 2 is ________, and Day 3 is ________.
The total money earned by the sale of pencils is 5c + 3c + 10c. Can we simplify this expression further and reduce the number of terms?
The expression means 5 times c is added to 3 times c is added to 10 times c. So in total, the letter-number c is added (5 + 3 + 10) times. This is what we have seen as the distributive property of numbers. Thus,
{tex} 5 \times c+3 \times c+10 \times c{/tex} {tex}=(5+3+10) \times c {/tex}
{tex}(5+3+10) \times c{/tex} can be simplified to {tex}18 \times c=18 c{/tex}.

Q.6: A big rectangle is split into two smaller rectangles as shown. Write an expression describing the area of the bigger rectangle.

Solution:

The areas of the smaller rectangles are 4v sq. units and 3v sq. units.
The area of the bigger rectangle can be found in two ways: (i) by directly using its side lengths v and (4 + 3), or (ii) by adding the areas of the smaller rectangles.
The first way gives 7v, and the second way gives 4v + 3v. We know that they are equal: 4v + 3v = 7v, and this is the required expression for the area of the bigger rectangle.
As earlier, a big rectangle is split into two smaller rectangles as shown below. Write an expression to find the area of the rectangle AEFD.
Even in this case, the area of rectangle AEFD can be found in two ways: (i) by directly using the side lengths n and (12 – 4), or (ii) subtracting the area of the rectangle EBCF from that of ABCD.

The first method gives us 8n, and the second method gives us 12n – 4n, and they are equal, since 12n – 4n = 8n. This is the expression for the area of the rectangle AEFD.
Sets of terms such as (5c, c, 10c), (12n, – 4n) that involve the same letter-numbers are called like terms. Sets of terms such as {18c, 11d} are called unlike terms as they have different letter-numbers. As we have seen, like terms can be added together and simplified into a single term.

Q.7: Simplify the expression 4 (x + y) – y

Solution:

Using the distributive property, this expression can be simplified to
{tex} 4(x+y)-y=4 x+4 y-y {/tex}
{tex} =4 x+4 y+-y {/tex}
{tex} =4 x+(4-1) y {/tex}
{tex} =4 x+3 y . {/tex}

Q.8: What is the sum of the numbers in the picture (unknown values are denoted by letter-numbers)?

Solution:

There are many ways to go about it. Here, we show some of them.

Adding row wise gives:
{tex} (4 \times 3)+(r+s)+(r+s)+(4 \times 3) {/tex}
Adding like terms together gives:
{tex} (8 \times 3)+(r+r)+(s+s) {/tex}
Adding the upper half and doubling gives:
{tex} 2 \times(4 \times 3+r+s) {/tex}

The three expressions might seem different. We can simplify each one and see that they all are the same: 2r + 2s + 24.

Q.9: Given Aftab’s age, how will you find out Shabnam’s age?

Solution:

Easy: We add 3 to Aftab’s age to get Shabnam’s age.

Q.10: Given the age of Shabnam, write an expression to find Aftab’s age.

Solution:

We know that Aftab is 3 years younger than Shabnam. So, Aftab’s age will be 3 less than Shabnam’s. This can be described as
Aftab’s age = Shabnam’s age – 3.
If we again use the letter a to denote Aftab’s age and the letter s to denote Shabnam’s age, then the algebraic expression would be: a = s – 3, meaning 3 less than s.

Q.11: Use this expression to find Aftab’s age if Shabnam’s age is 20. (Expression: Aftab’s age = Shabnam’s age – 3.)

Solution:

If Shabnam’s age = 20 years
Then,
Aftab’s age = Shabnam’s age – 3
= 20 – 3
= 17 years.

Q.12: How much should she pay if she buys 10 coconuts and 5 kg jaggery?

Solution:

Cost of 10 coconuts {tex}=10 \times ₹ 35{/tex}
Cost of 5 kg jaggery {tex}=5 \times ₹ 60{/tex}
Total cost {tex}=10 \times ₹ 35+5 \times ₹ 60{/tex} {tex}=₹ 350+₹ 300=₹ 650{/tex}.

Q.13: Write an algebraic expression to find the total amount to be paid for a given number of coconuts and quantity of jaggery.

Solution:

Let us identify the relationships and then write the expressions.

Quantity needed	Relationship	Expression
Cost of coconuts	Number of coconuts {tex}\times 35{/tex}	{tex}c \times 35{/tex}
Cost of jaggery	Number of kgs of jaggery {tex}\times 60{/tex}	{tex}j \times 60{/tex}

Here, ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery. The total amount to be paid will be:
Cost of coconuts + Cost of jaggery.
The corresponding algebraic expression can be written as: c {tex}\times{/tex} 35 + j {tex}\times{/tex} 60

Q.14: Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery. (Expression: c {tex}\times{/tex} 35 + j {tex}\times{/tex} 60, where ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery)

Solution:

Number of coconuts (c) = 7
Number of kgs of jaggery (j) = 4kg
{tex} \text { Total amount paid }=c \times 35+j \times 60 {/tex}
{tex} =7 \times 35+4 \times 60 {/tex}
{tex} =245+240 {/tex}
{tex} =₹ 485 . {/tex}

Q.15: What is the perimeter of a square with sidelength 7 cm? Use the expression to find out. (Expression: 4 {tex}\times{/tex} q, where q stands for the side length.)

Solution:

Side length of square (q) = 7 cm
Perimeter = 4 {tex}\times{/tex} q
= 4 {tex}\times{/tex} 7
= 28 cm.

Q.16: Find an algebraic expression to get the nth term of this sequence.

Solution:

Note that here ‘n’ is a letter-number that denotes a position in the sequence.
As it is the sequence of multiples of 4, it can be seen that the nth term will be 4 times n:
4 {tex}\times{/tex} n
As a standard practice, we shorten 4 × n to 4n by skipping the multiplication sign. We write the number first, followed by the letter(s). Find the value of the expression 7k when k = 4. The value is 7 {tex}\times{/tex} 4 = 28.
Find the value that the expression 5m + 3 takes when m = 2. As 5m stands for 5 {tex}\times{/tex} m, the value of the expression when m = 2 is 5 {tex}\times{/tex} 2 + 3 = 13

Q.17:

	Day 1	Day 2	Day 3
Pencils (Price ‘{tex}c{/tex}’)	5	3	10
Erasers (Price ‘{tex}d{/tex}’)	4	6	1

If c = ₹ 50, find the total amount earned by the scale of pencils.

Solution:

{tex} \text { Total amount earned = Day } 1+\text { Day } 2+\text { Day } 3 {/tex}
{tex} =5 \times c+3 \times c+10 \times c {/tex}
{tex} =5 c+3 c+10 c=18 c {/tex}
{tex} \text { If } c=₹ 50, \text { then } {/tex}
{tex} 18 c=18 \times 50=₹ 900 . {/tex}
Therefore, ₹ 900 is the total amount earned by the scale of pencils.

Q.18: Write the expression for the total money earned by selling erasers. Then, simplify the expression.

Solution:

Total amount earned = Day 1 + Day 2 + Day 3
= 4 {tex}\times{/tex} d + 6 {tex}\times{/tex} d + 1 {tex}\times{/tex} d
= 4d + 6d + d
= 11d.

Q.19: Can the expression 18c + 11d be simplified further?

Solution:

There is no way of further simplifying this expression as it contains different letter-numbers. It is in its simplest form.
In this problem, we saw the expression 5c + 3c + 10c getting simplified to the expression 18c.

Q.20: Describe the procedure to get these amounts.

Solution:

Chairs rented {tex}={x}{/tex}
Tables rented = y
Total amount paid at the beginning {tex}={/tex} Amount paid for chairs + Amount paid for tables
{tex} =40 \times(x)+75 \times y {/tex}
{tex} =40 x+7 y {/tex}
Total amount returned {tex}={/tex} Amount returned for chairs + Amount returned for tables
{tex} =6 \times(x)+10 \times y {/tex}
{tex} =6 x+10 y . {/tex}
Total amount paid {tex}=(40 x+75 y)-(6 x+10 y){/tex}

Q.21: Can we simplify this expression? If yes, how? If not, why not? [Expression: (40x + 75y) – (6x + 10y)]

Solution:

Yes, the expression can be simplified further as follows:
(40x + 75y) – (6x + 10y) = 40x + 75y – 6x – 10y
= (40x – 6x) + (75y – 10y)
= 34x + 65y.

Q.22: Could we have written the initial expression as (40x + 75y) + (-6x – 10y)?

Solution:

Yes, the initial expression (40x + 75y) – (6x + 10y) could also be written as
(40x + 75y) + (- 6x – 10y).
This is because subtracting a group of terms is the same as adding their negatives.

Q.23: What do each of the expressions mean? (Expressions: 7p – 3q, 8p – 4q, and 6p – 2q).

Solution:

Each of these expression involve a difference between two algebraic terms.

Q.24: Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.

Solution:

Krishita’s score in three rounds could be ({tex}8 p-3 q{/tex}), ({tex}6 p-2 q{/tex}), and ({tex}9 p-2 q{/tex}).
Therefore, total score after three rounds {tex}=(8 p-3 q)+(6 p-2 q)+(9 p-2 q){/tex}.
{tex} =8 p-3 q+6 p-2 q+9 p-2 q {/tex}
{tex} =23 p-7 q . {/tex}

Q.25: Can we say who scored more? Can you explain why? (Charu’ score: 21p – 9q & Krishita’s score: 23p – 7q)

Solution:

Penalties for Charu = 9q
Penalities for Krishita = 7q
Krishita scored more because she had fewer penalties than Charu.

Q.26: Simplify this expression further. [Expression: (23p– 7q)– (21p– 9q)]

Solution:

{tex} 23 q-7 q-(21 p-9 q){/tex} {tex}=23 q-7 q-21 p+9 q {/tex}
{tex} =(23 q-21 q)+(9 q-7 q) {/tex}
{tex} =2 q+2 q . {/tex}

Q.27: Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + utake.

Solution:

Q.28: After filling in the two diagrams, do you think the two expressions are equal?

Solution:

Filling in the two diagrams made it clear that the two expressions are not equal.

Q.29: Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have?

Solution:

Combining like terms and simplifying constants reduces the total number of terms in an expression.

Q.30: Find out the formula of this number machine.

Solution:

The formula for the number machine above is “two times the first number minus the second number”. When written as an algebraic expression, the formula is 2a – b. The expression for the first set of inputs is 2 {tex}\times{/tex} 5 – 2 = 8. Check that the formula holds true for each set of inputs.

Q.31: Find the formulas of the number machines below and write the expression for each set of inputs.

Solution:

For first number machines:
Formula: Two subtracted from the sum of the two numbers.
Algebraic expression: a + b – 2.
5th expression: a + b – 2.
For second number machines:
Formula: One added to the multiplication of two numbers.
Algebraic expression: (a {tex}\times{/tex} b) + 1 = ab + 1.
4th expression: 10 {tex}\times{/tex} 3 + 1 = 30 + 1 = 31.
5th expression: a {tex}\times{/tex} b + 1 = ab + 1.

Q.32: Where would design C appear for the n^th time?

Solution:

We can see that this design appears in positions that are multiples of 3. So the n^th occurrence of Design C will be at position 3n.

Q.33: Similarly, find the formula that gives the position where the other Designs appear for the n^th time.

Solution:

The positions where B occurs are 2, 5, 8, 11, 14, and so on.
We can see that the position of the nth appearance of Design B is one less than the position at which Design C appears for the nth time. Thus, the nth occurrence of Design B is at position:
3n – 1
Similarly, the expression describing the position at which the design A appears for the nth time is: 3n – 2.

Q.34: Given a position number can we find out the design that appears there? Which Design appears at Position 122?

Solution:

If the position is a multiple of 3, then clearly we have Design C. As seen earlier, if the position is one less than a multiple of 3, it has Design B, and if it is 2 less than a multiple of 3, then it has Design A.

Q.35: Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below.

Position	Quotient on division by 3	Remainder
99	33	0
122	40	2
148	49	1

Solution:

Yes, the remainders can be used for this. When the position number is divided by 3: if the remainder is 0, the design at that position is Design C; if the remainder is 1, it’s Design A; and if the remainder is 2, it’s Design B.

Q.36: Use this to find what design appears at positions 99, 122, and 148.

Position	Quotient on division by 3	Remainder
99	33	0
122	40	2
148	49	1

Solution:

Design C appears at position 99.
Design B appears at position 122.
Design A appears at position 148.

Q.37: Will the diagonal sums be equal in every 2 {tex}\times{/tex} 2 square in this endless grid? How can we be sure?

Solution:

To be sure of this we cannot check with all 2 {tex}\times{/tex} 2 squares as there are an unlimited number of them. Let us consider a 2 {tex}\times{/tex} 2 square. Its top left number can be any number. Let us call it ‘a’.

Q.38: Given that we know the top left number, how do we find the other numbers in this 2 {tex}\times{/tex} 2 square?

Solution:

As we have been doing, first let us describe the other numbers in words.

the number to the right of ‘a’ will be 1 more than it.
the number below ‘a’ will be 7 more than it.
the number diagonal to ‘a’ will be 8 more than it.

So the other numbers in the 2 {tex}\times{/tex} 2 square can be represented as shown in the grid. Let us find the diagonal sums; a + (a + 8), and (a + 1) + (a + 7).
Let us simplify them. Since the terms can be added in any order, the brackets can be opened.
a + (a + 8) = a + a + 8 = 2a + 8
(a + 1) + (a + 7) = a + 1 + a + 7 = a + a + 1 + 7 = 2a + 8
We see that both diagonal sums are equal to 2a + 8 (8 more than 2 times a).

Q.39: Verify this expression for diagonal sums by considering any 2 {tex}\times{/tex} 2 square and taking its top left number to be ‘a’.

Solution:

Considering the given 2 {tex}\times{/tex} 2 square:

Top left number (8) = a
Number to the right 8 (9) = a + 1
Number below 8 (15) = a + 7
Number diagonal to 8 (16) = a + 8
First diagonal sum (8 + 16 = 24) = a + (a + 8) = 2a + 8.
Second diagonal sum (9 + 15 = 24) = (a + 1) + (a + 7) = 2a + 8.
Hence, verified that both diagonal sums are equal to 2a + 8.

Q.40: Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?

Solution:

Sum of all numbers = 8 + 14 + 22 + 16 + 15 = 75.
Central number = 15.
75 is five times 15.

Repeating this for the above set of numbers:
Sum of all number = 29 + 35 + 43 + 37 + 36 = 180.
Central number = 36.
180 is five times 36.
Observation: The sum of a set of numbers forming a (+) shape is five times the central number of the shape.

Q.41:

How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?

Solution:

Matchsticks in Step 33 = (2 {tex}\times{/tex} 33) + 1 = 66 + 1 = 67.
Matchsticks in Step 84 = (2 {tex}\times{/tex} 84) + 1 = 168 + 1 = 169.
Matchsticks in Step 108 = (2 {tex}\times{/tex} 108) + 1 = 216 + 1 = 217.

Q.42: Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same? [Expression: 3 + 2 {tex}\times{/tex} (y – 1) and 2y + 1]

Solution:

Yes, both the expressions 3 + 2 {tex}\times{/tex} (y – 1) and 2y + 1 give the number of matchsticks at each step correctly. It is because both are exactly the same.
i.e., 3 + 2 {tex}\times{/tex} (y – 1) = 3 + 2y – 2 = 2y + 1.

Q.43: What are these numbers in Step 3 and Step 4?

Solution:

In Step 3, there are 3 matchsticks placed horizontally and 4 matchsticks placed diagonally.
In step 4, there are 4 matchsticks placed horizontally and 5 matchsticks placed diagonally.

Q.44: How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1?

Solution:

Number of horizontal matchsticks per step = y
Number of diagonally placed matchsticks per step = y + 1
Total matchsticks for step y:
Horizontal = y
Diagonal = y + 1
Total = y + (y + 1) = 2y + 1
Conclusion: The expressions are correct and they do indeed add up to 2y + 1.

Q.45: Write formulas for the perimeter of:

triangle with all sides equal.
a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all side lengths and angle measures are equal)
a regular hexagon

Solution:

Triangle (3 equal sides): Perimeter = 3 {tex}\times{/tex} side length.
Regular Pentagon (5 equal sides): Perimeter = 5 {tex}\times{/tex} side length.
Regular hexagon (6 equal sides): Perimeter = 6 {tex}\times{/tex} side length.

Q.46: Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.

Solution:

Initial length of pipe = 20 m
Length of pipe joined = ‘k’ m
Expression: 20 + k.

Q.47: What is the total amount Krithika has, if she has the following numbers of notes of ₹ 100, ₹ 20 and ₹ 5? Complete the following table:

No. of ₹ 100 notes	No. of ₹ 20 notes	No. of ₹ 5 notes	Expression and total amount
3	5	6
			{tex}6 \times 100+4 \times 20+3 \times 5=695{/tex}
8	4	z
x	y	z

Solution:

No. of ₹ 100 notes	No. of ₹ 20 notes	No. of ₹ 5 notes	Expression and total amount
3	5	6	{tex}3 \times 100+5 \times 20+6 \times 5=430{/tex}
6	4	3	{tex}6 \times 100+4 \times 20+3 \times 5=695{/tex}
8	4	z	{tex} 8 \times 100+4 \times 20+z \times 5 {/tex} {tex} =800+80+5 z {/tex} {tex} =880+5 z {/tex}
x	y	z	{tex} x \times 100+y \times 20+z \times 5 {/tex} {tex} =100 x+20 y+5 z {/tex}

Q.48: Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to complete grind ‘y’ kg of grain, assuming the machine is off initially?

{tex}10+8+y{/tex}
{tex}(10+8) \times y{/tex}
{tex}10 \times 8 \times y{/tex}
{tex}10+8 \times y{/tex}
{tex}10 \times y+8{/tex}

Solution:

Time to start the machine = 10 seconds
Time to grind 1 kg of grain = 8 seconds
Quantity of grain = y kg
Total time = Time to start the machine + Time to grind y kg of grain
= {tex}10+8 \times y{/tex}
Therefore, {tex}10+8 \times y{/tex} is the correct answer.

Q.49: Write algebraic expressions using letters of your choice.

5 more than a number
4 less than a number
2 less than 13 times a number
13 less than 2 times a number

Solution:

{tex} x+5(\text { Number }=x) {/tex}
{tex} y-4(\text { Number }=y) {/tex}
{tex} 13 \times p-2=13 p-2(\text { Number }=p) {/tex}
{tex} 2 \times z-13=(\text { Number }=z) {/tex}

Q.50: Describe situations corresponding to the following algebraic expressions:

{tex} 8 \times x+3 \times y {/tex}
{tex} 15 \times j-2 \times k {/tex}

Solution:

Sum of 8 times x and 3 times y.
Subtract 2 times k from 15 times j.

Q.51: In a calendar month, if any 2 {tex}\times{/tex} 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Solution:

Q.52: Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.

Solution:

Adding like terms together gives:
{tex} (5 y+5 y)+(x+x)+(2-6){/tex} {tex}=10 y+2 x+(-4)=2 x+10 y-4 {/tex}
Adding like terms together gives:
{tex} (2 p \times 4)+(3 q \times 4)+(2 \times-2)+(3 \times 2){/tex} {tex}=8 p+12 q+(-4)+6=8 p+12 q+2 . {/tex}
Adding like terms together gives:
{tex} (5 k \times 12)+(-5 g \times 4)=60 k+(-20 g){/tex} {tex}=40 k-20 g {/tex}

Q.53: Simplify each of the following expressions:

{tex} p+p+p+p, p+p+p+q, p+q+p-q {/tex}
{tex} p-q+p-q, p+q-p+q {/tex}
{tex} p+q-(p+q), p-q-p-q {/tex}

Solution:

{tex} p+p+p+p=3 p {/tex}
{tex} p+p+p+q=3 p+q {/tex}
{tex} p+q+p-q=2 q {/tex}
{tex}p-q+p-q=2 p-2 q {/tex}
{tex} p+q-p+q=2 q {/tex}
{tex}p+q-(p+q)=p+q-p-q=0{/tex}
{tex} p-q-p-q=-2 q {/tex}

Q.54: Simplify each of the following expressions:

{tex} 2 d-d-d-d, 2 d-d-d-c {/tex}
{tex} 2 d-d-(d-c), 2 d-(d-d)-c {/tex}
{tex} 2 d-d-c-c {/tex}

Solution:

{tex}2 d-d-d-d=2 d-3 d=-d{/tex}
{tex}2 d-d-d-c=2 d-2 d-c=-c{/tex}
{tex}2 d-d-(d-c)=2 d-d-d+c{/tex} {tex}=2 d-2 d+c=c{/tex}
{tex}2 d-(d-d)-c=2 d-0-c=2 d-c{/tex}
{tex}2 d-d-c-c=d-2 c{/tex}

Q.55: One plate of Jowar roti costs ₹ 30 and one plate of Pulao costs ₹ 20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?

{tex} 30 x+20 y {/tex}
{tex} (30+20) \times(x+y) {/tex}
{tex} 20 x+30 y {/tex}
{tex} (30+20) \times x+y {/tex}
30 x – 20 y

Solution:

Jowar roti plate = ₹ 30
Pulao plate = ₹ 20
Jowar roti ordered in a day = x
Pulao plate ordered in a day = y
Total amount earned in a day = ₹(x {tex}\times{/tex} 30) + (y + 20) = 30x + 20y.
{tex}\therefore{/tex} 30x + 20y is the required expression.

Q.56: Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day?

p + q + r
p + q + 2r
2 {tex}\times{/tex} (p + q + r)
p + q + r + 2
p + q + r + 1
2 {tex}\times{/tex} (p + q)

Solution:

Customers bought champak = p
Customers bought marigold = q
Customers bought both = r
A national flag was given to every customer.
Total national flags distributed = p + q + r.
{tex}\therefore{/tex} p + q + r is the required expression.

Q.57: A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.

Write an expression describing how far away the snail is from its starting position.
What can we say about the snail’s movement if d > u?

Solution:

Snail’s movement in a day and night = (u – v) cm
Snail’s movement in 10 days and 10 nights = 10(u – v) cm
Total distance moved by the snail from its starting position = 10(u – v) cm
if d > u, then the snail would slip more during night than it climbs each day and can never reach the height of the well.

Q.58: Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?

Solution:

Distance cycled in first week = 7 {tex}\times{/tex} 5 = 35 km
Distance cycled in second week = 7 {tex}\times{/tex} (5 + z) = (35 + 7z) km
Distance cycled in third week = 7 {tex}\times{/tex} (5 + z + z) = 7(5 + 2z) = (35 + 14z) km
Total distance cycled after 3 weeks = 35 + (35 + 7z) + (35 + 14z) = 35 + 35 + 7z + 35 + 14z
= (105 + 21z) km.

Q.59: In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

Solution:

Top left: {tex}(w+2) \rightarrow(-5) \rightarrow(w-3) {/tex} {tex}\rightarrow(\times 3) \rightarrow 3 w-9{/tex}.
Bottom left: {tex}(w+2) \rightarrow(-8) \rightarrow(w-6){/tex} {tex} \rightarrow(-4) \rightarrow(w-10){/tex}.
Bottom right: {tex}(w+2) \rightarrow(+3) \rightarrow(w+5){/tex} {tex} \rightarrow(\times 4) \rightarrow(4 w+20){/tex}.

Q.60: A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.

If t = 4, what is the time taken to travel from Yahapur to Vahapur?
What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]

Solution:

There are 4 segments of travel.
Trains stops 2 minutes at each station ({tex}{S} 1, {~S} 2, {~S} 3{/tex}).
Time taken to travel from one station to the next station {tex}=t{/tex}

If {tex}{t}=4{/tex},
Total time taken to travel from Yahapur to Vahapur {tex}=4 \times 4+3 \times 2=16+{/tex} {tex}6=22{/tex} minutes.
Time taken to travel from one station to another {tex}={t}{/tex}.
There are 4 segments and 3 stops of 2 minutes.
Algebraic expression {tex}=4 \times t+3 \times 2=(4 t+6){/tex} minutes.

Q.61: Simplify the following expressions:

{tex} 3 a+9 b-6+8 a-4 b-7 a+16 {/tex}
{tex} 3(3 a-3 b)-8 a-4 b-16 {/tex}
{tex} 2(2 x-3)+8 x+12 {/tex}

Solution:

{tex}3 a+9 b-6+8 a-4 b-7 a+16 {/tex}
{tex} =3 a+8 a-7 a+9 b-4 b-6+16 {/tex}
{tex} =11 a-7 a+5 b+10 {/tex}
{tex} =4 a+5 b+10 {/tex}
{tex}3(3 a-3 b)-8 a-4 b-16 {/tex}
{tex} =9 a-9 b-8 a-4 b-16 {/tex}
{tex} =9 a-8 a-9 b-4 b-16 {/tex}
{tex} =a-13 b-16 {/tex}
{tex} 2(2 x-3)+8 x+12 {/tex}
{tex} =4 x-6+8 x+12 {/tex}
{tex} =4 x+8 x-6+12 {/tex}
{tex} =12 x+6 {/tex}

Q.62: Simplify the following expressions:

{tex} 8 x-(2 x-3)+12 {/tex}
{tex} 8 h-(5+7 h)+9 {/tex}
{tex} 23+4(6 m-3 n)-8 n-3 m-18 {/tex}

Solution:

{tex} 8 x-(2 x-3)+12 {/tex}
{tex} =8 x-2 x+3+12 {/tex}
{tex} =6 x+15 {/tex}
{tex}8 h-(5+7 h)+9 {/tex}
{tex} =8 h-5+7 h+9 {/tex}
{tex} =8 h+7 h+9-5 {/tex}
{tex} =15 h+4 {/tex}
{tex}23+4(6 m-3 n)-8 n-3 m-18 {/tex}
{tex} =23+24 m-12 n-8 n-3 m-18 {/tex}
{tex} =24 m-3 m-12 n-8 n+23-18 {/tex}
{tex} =21 m-20 n+5 {/tex}

Q.63: Subtract the expressions given below:

{tex} 9 a-6 b+14 \text { from } 6 a+9 b-18 {/tex}
{tex} -15 x+13-9 \text { y from } 7 y-10+3 x {/tex}
{tex} 17 g+9-7 h \text { from } 11-10 g+3 h {/tex}

Solution:

{tex}9 a-6 b+14 \text { from } 6 a+9 b-18 {/tex}
{tex} =6 a+9 b-18-(9 a-6 b+14) {/tex}
{tex} =6 a+9 b-18-9 a+6 b-14 {/tex}
{tex} =6 a-9 a+9 b+6 b-18-14 {/tex}
{tex} =3 a+15 b-32 {/tex}
{tex}-15 x+13-9 y \text { from } 7 y-10+3 x {/tex}
{tex} =7 y-10+3 x-(-15 x+13-9 y) {/tex}
{tex} =7 y-10+3 x+15 x-13+9 y {/tex}
{tex} =7 y+9 y+3 x+15 x-10-13 {/tex}
{tex} =16 y+18 x-23 {/tex}
{tex}17 g+9-7 h \text { from } 11-10 g+3 h {/tex}
{tex} =11-10 g+3 h-(17 g+9-7 h) {/tex}
{tex} =11-10 g+3 h-17 g-9+7 h {/tex}
{tex} =-10 g-17 g+3 h+7 h+11-9 {/tex}
{tex} =-27 g+10 h+2 {/tex}

Q.64: Add the expressions given below:

{tex} 6 f-20+8 s \text { and } 23-13 f-12 s {/tex}
{tex} 13 m-12 n \text { and } 12 n-13 m {/tex}
{tex} -26 m+24 n \text { and } 26 m-24 n {/tex}

Solution:

{tex}6 f-20+8 s \text { and } 23-13 f-12 s {/tex}
{tex} =6 f-20+8 s+23-13 f-12 s {/tex}
{tex} =6 f-13 f+8 s-12 s-20+23 {/tex}
{tex} =-7 f-4 s+3 . {/tex}
{tex}13 m-12 n \text { and } 12 n-13 m {/tex}
{tex} =13 m-12 n+12 n-13 m {/tex}
{tex} =13 m-13 m-12 n+12 n {/tex}
{tex} =0 . {/tex}
{tex}-26 m+24 n \text { and } 26 m-24 n {/tex}
{tex} =-26 m+24 n+26 m-24 n {/tex}
{tex} =-26 m+26 m+24 n-24 n {/tex}
{tex} =0 . {/tex}

Q.65: Subtract the expressions given below:

9a – 6b + 14 from 6a + 9b – 18
-15x + 13 – 9y from 7y – 10 + 3x
17g + 9 – 7h from 11 – 10g + 3h

Solution:

{tex}9 a-6 b+14 \text { from } 6 a+9 b-18 {/tex}
{tex} =6 a+9 b-18-(9 a-6 b+14) {/tex}
{tex} =6 a+9 b-18-9 a+6 b-14 {/tex}
{tex} =6 a-9 a+9 b+6 b-18-14 {/tex}
{tex} =3 a+15 b-32 {/tex}
{tex}-15 x+13-9 y \text { from } 7 y-10+3 x {/tex}
{tex} =7 y-10+3 x-(-15 x+13-9 y) {/tex}
{tex} =7 y-10+3 x+15 x-13+9 y {/tex}
{tex} =7 y+9 y+3 x+15 x-10-13 {/tex}
{tex} =16 y+18 x-23 {/tex}
{tex}17 g+9-7 h \text { from } 11-10 g+3 h {/tex}
{tex} =11-10 g+3 h-(17 g+9-7 h) {/tex}
{tex} =11-10 g+3 h-17 g-9+7 h {/tex}
{tex} =-10 g-17 g+3 h+7 h+11-9 {/tex}
{tex} =-27 g+10 h+2 . {/tex}

Q.66: Subtract the expressions given below:

{tex} 9 a-6 b+14 \text { from } 6 a-(9 b+18) {/tex}
{tex} 10 x+2+10 y \text { from }-3 y+8-3 x {/tex}
{tex} 8 g+4 h-10 \text { from } 7 h-8 g+20 {/tex}

Solution:

{tex} 9 a-6 b+14 \text { from } 6 a-(9 b+18) {/tex}
{tex} =6 a-(9 b+18)-(9 a-6 b+14) {/tex}
{tex} =6 a-9 b-18-9 a+6 b-14 {/tex}
{tex} =6 a-9 a-9 b+6 b-18-14 {/tex}
{tex} =-3 a-3 b-32 . {/tex}
{tex}10 x+2+10 y \text { from }-3 y+8-3 x {/tex}
{tex} =-3 y+8-3 x-(10 x+2+10 y) {/tex}
{tex} =-3 y+8-3 x-10 x-2-10 y {/tex}
{tex} =-3 x-10 x-3 y-10 y+8-2 {/tex}
{tex} =-13 x-13 y+6 {/tex}
{tex}8 g+4 h-10 \text { from } 7 h-8 g+20 {/tex}
{tex} =7 h-8 g+20-(8 g+4 h-10) {/tex}
{tex} =7 h-8 g+20-8 g-4 h+10 {/tex}
{tex} =-8 g-8 g+7 h-4 h+20+10 {/tex}
{tex} =-16 g+3 h+30 {/tex}

Q.67: Describe situations corresponding to the following algebraic expressions:

8x + 3y
15x – 2x

Solution:

8x + 3y
Ramesh buys 8 pencils, each costing ₹ x, and 3 pens, each costing ₹ y.
Total cost = 8x + 3y
15x – 2x
A shopkeeper has 15 boxes, each containing x apples. He sells 2 boxes.
Apples left = 15x – 2x = 13x.

Q.68: Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

Solution:

No fold: 1 cut {tex}\rightarrow 2{/tex} pieces.
1-fold: 1 cut {tex}\rightarrow 3{/tex} pieces.
2-fold: 1 cut {tex}\rightarrow 4{/tex} pieces.
3-fold: 1 cut {tex}\rightarrow 5{/tex} pieces.
Similarly,
For 10-fold: 1 cut {tex}\rightarrow 11{/tex} pieces.
Expression with ‘{tex}r{/tex} ‘ times fold and cut {tex}=r+2{/tex}.

Q.69: Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

Solution:

1 square {tex}\rightarrow 4{/tex} matchsticks.
2 squares {tex}\rightarrow 7{/tex} matchsticks.
3 squares {tex}\rightarrow 10{/tex} matchsticks.
Similarly,
For 10 squares {tex}\rightarrow 31{/tex} matchsticks.
Matchsticks required to make ‘{tex}w{/tex}’ squares {tex}=3 \times w+1=3 w+1{/tex}.

Q.70: Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.

Solution:

Red positions: {tex}1,5,9,13 . . .{/tex}.
Yellow positions: 2, 4, 6, 8, 10…
Green positions: 3, 7, 11, 15…
Expression for red colour {tex}=4 {r}-3{/tex}.
Expression for yellow colour {tex}=2 {y}{/tex}.
Expression for green colour {tex}=4 {~g}-1{/tex}.
Red: Numbers that are 1 more than multiples of 4 (Remainder 1 when divided by 4)
Yellow: All even numbers (Remainder 0 when divided by 4)
Green: Numbers that are 3 more than multiples of 4 (Remainder 3 when divided by 4)
90 and 190: Both are even {tex}\rightarrow{/tex} Yellow
{tex}343 \div 4=85{/tex} R3 {tex}\rightarrow{/tex} Remainder {tex}=3 \rightarrow{/tex} Green

Q.71:

Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Solution:

Step 1: 5 squares.
Step 2: 9 squares.
Step 3: 13 squares.
General formula: {tex}4 n+1{/tex}
Step 4: {tex}4(4)+1=16+1=17{/tex}
Step 10: {tex}4(10)+1=40+1=41{/tex}
Step 50: {tex}4(50)+1=200+1=201{/tex}
For vertices;
Step {tex}1 \rightarrow 20{/tex}
Step {tex}2 \rightarrow 36{/tex}
Step {tex}3 \rightarrow 52{/tex}
General formula: {tex}16 n+4{/tex}.

Q.72: Numbers are written in a particular sequence in this endless 4-column grid.

Give expressions to generate all the numbers in a given column (1, 2, 3, 4).
In which row and column will the following numbers appear:
1. 124
2. 147
3. 201
What number appears in row r and column c?
Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.

Solution:

Expression to generate all the numbers in a given column {tex}(1,2,3,4){/tex}
Let {tex}r{/tex} be the row number.
Column {tex}1: 1,5,9,13, \ldots .{/tex}. which starts at 1 and adds 4 each row.
So, number in the rth row of column {tex}1=4 \times(r-1)+1{/tex}
Column 2: {tex}4 \times({r}-1)+2{/tex}
Column 3: {tex}4 \times({r}-1)+3{/tex}
Column 4: {tex}4 \times({r}-1)+4{/tex}
If c is the column number, then the general formula to generate all numbers is {tex}4 \times({r}-1)+{c}{/tex}.
1. {tex}147 \div 4 \Rightarrow{/tex} Quotient {tex}=36{/tex} and remainder is 3
  {tex} \therefore 147=4 \times 36+3 {/tex}
  Comparing it with {tex}4 \times({r}-1)+{c}{/tex}, we get
  {tex} r-1=36, c=3 {/tex}
  So, 147 will appear at row {tex}36+1=37{/tex} and column 3
2. {tex}201 \div 4 \Rightarrow{/tex} Quotient {tex}=50{/tex} and remainder is 1
  {tex} \therefore 201=4 \times 50+1 {/tex}
  Comparing it with {tex}4 \times({r}-1)+{c}{/tex}, we get
  {tex} r-1=50, c=1 {/tex}
  So, 201 will appear at row 51 and column 1.
We divide each number by 4 to find its row and column
{tex}124 \div 4 \Rightarrow{/tex} Quotient {tex}=31{/tex} and remainder is 0
{tex} \therefore 124=4 \times 31+0 \text { or } 4 \times 30+4 {/tex}
Comparing it with {tex}4 \times({r}-1)+{c}{/tex}, we get
{tex} r-1=30, c=4 {/tex}
So, {tex}{r}=31{/tex} and {tex}{c}=4{/tex}
So, row is 31 and column is 4
The number that appears in row r and column c is {tex}4({r}-1)+{c}{/tex}.
Every third number is a multiple of 3.
We can observe that even numbers always appear in column 2 and column 4.
Odd numbers always appear in column 1 and column 3.
Every row has 2 odd and 2 even numbers.
The sum of each row increases by 16 .
{tex} \text { (e.g., Row } 1: 1+2+3+4{/tex} {tex}=10 \text {, Row } 2: 5+6+7+8{/tex} {tex}=26 \text {, Row } 3: 9+10+11+12=42 \text { ) } {/tex}