{"id":31864,"date":"2026-09-17T16:01:59","date_gmt":"2026-09-17T10:31:59","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31864"},"modified":"2026-09-17T16:02:40","modified_gmt":"2026-09-17T10:32:40","slug":"expressions-using-letter-numbers-ncert-solutions-class-7-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/expressions-using-letter-numbers-ncert-solutions-class-7-maths-ganita-prakash\/","title":{"rendered":"Expressions Using Letter-Numbers – NCERT Solutions Class 7 Maths (Ganita Prakash)"},"content":{"rendered":"\n

Expressions Using Letter-Numbers<\/strong><\/strong> – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).<\/p>\n\n\n\n

NCERT Solutions Class 7<\/h2>\n\n\nEnglish Poorvi<\/a>\n\n\nHindi Malhar<\/a>\n\n\nMaths Ganita Prakash<\/a>\n\n\nScience Curiosity<\/a>\n\n\nSocial Exploring Society<\/a>\n\n\n\n

Expressions Using Letter-Numbers<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n
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Q.1: Shabnam is 3 years older than Aftab. When Aftab\u2019s age 10 years, Shabnam\u2019s age will be 13 years. Now Aftab\u2019s age is 18 years, what will Shabnam\u2019s age be?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Shabnam is 3 years older than Aftab
Difference in Shabnam\u2019s and Aftab\u2019s age = 3 years
Aftab\u2019s present age = 18 years
Shabnam\u2019s present age = 18 + 3 = 21 years.<\/p>\n\n\n\n

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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?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

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\u2019s.<\/strong> This can be written as:
Number of matchsticks = 2 {tex}\\times{\/tex} Number of L\u2019s
Now, we can use any letter to denote the number of L\u2019s. Let\u2019s 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\u2019s. To find the number of matchsticks, we just replace n by the number of L.<\/p>\n\n\n\n

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Q.3: Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is \u20b9 35 and the price of 1 kg jaggery is \u20b9 60.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Cost of 8 coconuts {tex}=8 \\times 35=\u20b9 280{\/tex}
Cost of 9 kg jaggery {tex}=9 \\times 60=\u20b9 540{\/tex}
Total money paid {tex}=\u20b9 280+\u20b9 540=\u20b9 820{\/tex}.<\/p>\n\n\n\n

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Q.4: We are familiar with calculating the perimeters of simple shapes. Write expressions for perimeters.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

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

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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.<\/p>\n\n\n\n

 <\/td>Day 1<\/strong><\/td>Day 2<\/strong><\/td>Day 3<\/strong><\/td><\/tr>
Pencils (Price ‘{tex}c{\/tex}’)<\/td>5<\/td>3<\/td>10<\/td><\/tr>
Erasers (Price ‘{tex}d{\/tex}’)<\/td>4<\/td>6<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/p>\n\n\n\n

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}.<\/p>\n\n\n\n

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Q.6: A big rectangle is split into two smaller rectangles as shown. Write an expression describing the area of the bigger rectangle.
<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

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 \u2013 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 \u2013 4n, and they are equal, since 12n \u2013 4n = 8n. This is the expression for the area of the rectangle AEFD.
Sets of terms such as (5c, c, 10c), (12n, \u2013 4n) that involve the same letter-numbers are called like terms.<\/strong> Sets of terms such as {18c, 11d} are called unlike terms<\/strong> as they have different letter-numbers. As we have seen, like terms can be added together and simplified into a single term.<\/p>\n\n\n\n

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Q.7: Simplify the expression 4 (x + y) – y<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

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}<\/p>\n\n\n\n

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Q.8: What is the sum of the numbers in the picture (unknown values are denoted by letter-numbers)?
<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

There are many ways to go about it. Here, we show some of them.<\/p>\n\n\n\n

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

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

    \n\n\n\n

    Q.9: Given Aftab\u2019s age, how will you find out Shabnam\u2019s age?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Easy: We add 3 to Aftab\u2019s age to get Shabnam\u2019s age.<\/p>\n\n\n\n

    \n\n\n\n

    Q.10: Given the age of Shabnam, write an expression to find Aftab\u2019s age.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    We know that Aftab is 3 years younger than Shabnam. So, Aftab\u2019s age will be 3 less than Shabnam\u2019s. This can be described as
    Aftab\u2019s age = Shabnam\u2019s age – 3.
    If we again use the letter a to denote Aftab\u2019s age and the letter s to denote Shabnam\u2019s age, then the algebraic expression would be: a = s – 3, meaning 3 less than s.<\/p>\n\n\n\n

    \n\n\n\n

    Q.11: Use this expression to find Aftab\u2019s age if Shabnam\u2019s age is 20. (Expression: Aftab\u2019s age = Shabnam\u2019s age – 3.)<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    If Shabnam\u2019s age = 20 years
    Then,
    Aftab\u2019s age = Shabnam\u2019s age – 3
    = 20 – 3
    = 17 years.<\/p>\n\n\n\n

    \n\n\n\n

    Q.12: How much should she pay if she buys 10 coconuts and 5 kg jaggery?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Cost of 10 coconuts {tex}=10 \\times \u20b9 35{\/tex}
    Cost of 5 kg jaggery {tex}=5 \\times \u20b9 60{\/tex}
    Total cost {tex}=10 \\times \u20b9 35+5 \\times \u20b9 60{\/tex} {tex}=\u20b9 350+\u20b9 300=\u20b9 650{\/tex}.<\/p>\n\n\n\n

    \n\n\n\n

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

    Solution:<\/p>\n\n\n\n

    Let us identify the relationships and then write the expressions.<\/p>\n\n\n\n

    Quantity needed<\/strong><\/td>Relationship<\/strong><\/td>Expression<\/strong><\/td><\/tr>
    Cost of coconuts<\/td>Number of coconuts {tex}\\times 35{\/tex}<\/td>{tex}c \\times 35{\/tex}<\/td><\/tr>
    Cost of jaggery<\/td>Number of kgs of jaggery {tex}\\times 60{\/tex}<\/td>{tex}j \\times 60{\/tex}<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Here, \u2018c\u2019 represents the number of coconuts and \u2018j\u2019 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<\/p>\n\n\n\n

    \n\n\n\n

    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 \u2018c\u2019 represents the number of coconuts and \u2018j\u2019 represents the number of kgs of jaggery)<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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} =\u20b9 485 . {\/tex}<\/p>\n\n\n\n

    \n\n\n\n

    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.)<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

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

    \n\n\n\n

    Q.16: Find an algebraic expression to get the nth term of this sequence.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Note that here \u2018n\u2019 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 \u00d7 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<\/p>\n\n\n\n

    \n\n\n\n

    Q.17:<\/p>\n\n\n\n

     <\/td>Day 1<\/strong><\/td>Day 2<\/strong><\/td>Day 3<\/strong><\/td><\/tr>
    Pencils (Price ‘{tex}c{\/tex}’)<\/td>5<\/td>3<\/td>10<\/td><\/tr>
    Erasers (Price ‘{tex}d{\/tex}’)<\/td>4<\/td>6<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    If c <\/em>= \u20b9 50, find the total amount earned by the scale of pencils.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    {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=\u20b9 50, \\text { then } {\/tex}
    {tex} 18 c=18 \\times 50=\u20b9 900 . {\/tex}
    Therefore, \u20b9 900 is the total amount earned by the scale of pencils.<\/p>\n\n\n\n

    \n\n\n\n

    Q.18: Write the expression for the total money earned by selling erasers. Then, simplify the expression.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.19: Can the expression 18c + 11d be simplified further?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.20: Describe the procedure to get these amounts.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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}<\/p>\n\n\n\n

    \n\n\n\n

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

    Solution:<\/p>\n\n\n\n

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

    \n\n\n\n

    Q.22: Could we have written the initial expression as (40x + 75y) + (-6x –\u00a010y)?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.23: What do each of the expressions mean? (Expressions: 7p \u2013 3q, 8p \u2013 4q, and 6p \u2013 2q).<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Each of these expression involve a difference between two algebraic terms.<\/p>\n\n\n\n

    \n\n\n\n

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

    Solution:<\/p>\n\n\n\n

    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}<\/p>\n\n\n\n

    \n\n\n\n

    Q.25: Can we say who scored more? Can you explain why? (Charu\u2019 score: 21p – 9q & Krishita\u2019s score: 23p – 7q)<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Penalties for Charu = 9q
    Penalities for Krishita = 7q
    Krishita scored more because she had fewer penalties than Charu.<\/p>\n\n\n\n

    \n\n\n\n

    Q.26: Simplify this expression further. [Expression: (23p\u2013 7q)\u2013 (21p\u2013 9q)]\n\n\n\n

    Solution:<\/p>\n\n\n\n

    {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}<\/p>\n\n\n\n

    \n\n\n\n

    Q.27: Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u\u00a0<\/em>and 5 + utake.
    <\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    \"\"\/<\/figure>\n\n\n\n
    \n\n\n\n

    Q.28: After filling in the two diagrams, do you think the two expressions are equal?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Filling in the two diagrams made it clear that the two expressions are not equal.<\/p>\n\n\n\n

    \n\n\n\n

    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?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Combining like terms and simplifying constants reduces the total number of terms in an expression.<\/p>\n\n\n\n

    \n\n\n\n

    Q.30: Find out the formula of this number machine.
    <\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    The formula for the number machine above is \u201ctwo times the first number minus the second number\u201d. 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.<\/p>\n\n\n\n

    \n\n\n\n

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

    <\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    For first number machines:
    Formula:<\/strong> Two subtracted from the sum of the two numbers.
    Algebraic expression: a + b – 2.
    5th expression: a + b – 2.
    For second number machines:
    Formula:<\/strong> 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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.32: Where would design C appear for the nth<\/sup> time?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    We can see that this design appears in positions that are multiples of 3. So the nth<\/sup> occurrence of Design C will be at position 3n.<\/p>\n\n\n\n

    \n\n\n\n

    Q.33: Similarly, find the formula that gives the position where the other Designs appear for the nth<\/sup> time.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.34: Given a position number can we find out the design that appears there? Which Design appears at Position 122?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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.<\/p>\n\n\n\n

    \n\n\n\n

    Q.35: Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below.<\/p>\n\n\n\n

    Position<\/strong><\/td>Quotient on division by 3<\/strong><\/td>Remainder<\/strong><\/td><\/tr>
    99<\/td>33<\/td>0<\/td><\/tr>
    122<\/td>40<\/td>2<\/td><\/tr>
    148<\/td>49<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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\u2019s Design A; and if the remainder is 2, it\u2019s Design B.<\/p>\n\n\n\n

    \n\n\n\n

    Q.36: Use this to find what design appears at positions 99, 122, and 148.<\/p>\n\n\n\n

    Position<\/strong><\/td>Quotient on division by 3<\/strong><\/td>Remainder<\/strong><\/td><\/tr>
    99<\/td>33<\/td>0<\/td><\/tr>
    122<\/td>40<\/td>2<\/td><\/tr>
    148<\/td>49<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    Design C appears at position 99.
    Design B appears at position 122.
    Design A appears at position 148.<\/p>\n\n\n\n

    \n\n\n\n

    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?<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    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 \u2018a\u2019.<\/p>\n\n\n\n

    \n\n\n\n

    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?
    <\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    As we have been doing, first let us describe the other numbers in words.<\/p>\n\n\n\n