Use the textbook context
Match, attempt, then check
Each crop extends to the next solved item, so diagrams, tables and connected instructions stay with the question.
Complete worked answers
Textbook page 1
Textbook page 1 · solved item 1
Example 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? _______
Show 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.
Complete worked answers
Textbook page 2
Textbook page 2 · solved item 2
Use this expression to find Aftab’s age if Shabnam’s age is 20. (Expression: Aftab’s age = Shabnam’s age – 3.)
Show solution
If Shabnam’s age = 20 years Then, Aftab’s age = Shabnam’s age – 3 = 20 – 3 = 17 years.
Textbook page 2 · solved item 3
Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60. How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Show solution
Cost of 8 coconuts = 8 × 35 = ₹280 Cost of 9 kg jaggery = 9 × 60 = ₹540 Total money paid = ₹280 + ₹540 = ₹820.
Complete worked answers
Textbook page 3
Textbook page 3 · solved item 4
Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery. (Expression: c × 35 + j × 60, where ‘c’ represents the number of coconuts and ‘j’ represents the number of kgs of jaggery)
Show solution
Number of coconuts (c) = 7 Number of kgs of jaggery (j) = 4kg Total amount paid = c × 35 + j × 60 = 7 × 35 + 4 × 60 = 245 + 240 = ₹485.
Textbook page 3 · solved item 5
What is the perimeter of a square with side length 7 cm? Use the expression to find out. (Expression: 4 × q, where q stands for the side length.)
Show solution
Side length of square (q) = 7 cm Perimeter = 4 × q = 4 × 7 = 28 cm. Figure it out
Complete worked answers
Textbook page 4
Textbook page 4 · solved item 6
1. Write formulas for the perimeter of: (a) triangle with all sides equal. (b) 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) (c) a regular hexagon
Show solution
(a) Triangle (3 equal sides): Perimeter = 3 × side length. (b) Regular Pentagon (5 equal sides): Perimeter = 5 × side length. (c) Regular hexagon (6 equal sides): Perimeter = 6 × side length.
Textbook page 4 · solved item 7
2. 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.
Show solution
Initial length of pipe = 20 m Length of pipe joined = ‘k’ m Expression: 20 + k
Textbook page 4 · solved item 8
3. What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:
Show solution
No. of ₹100 notes No. of ₹20 notes No. of ₹5 notes Expression and total amount 3 5 6 3 × 100 + 5 × 20 + 6 × 5 = 430 6 4 3 6 × 100 + 4 × 20 + 3 × 5 = 695 8 4 z 8 × 100 + 4 × 20 + z × 5 = 800 + 80 + 5z = 880 + 5z x y z x × 100 + y × 20 + z × 5 = 100x + 20y + 5z
Textbook page 4 · solved item 9
4. 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? (a) 10 + 8 + y (b) (10 + 8) × y (c) 10 × 8 × y (d) 10 + 8 × y (e) 10 × y + 8
Show solution
Time to start 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 = 10 + 8 × y Therefore, 10 + 8 × y is the correct answer.
Textbook page 4 · solved item 10
5. Write algebraic expressions using letters of your choice. (a) 5 more than a number (b) 4 less than a number (c) 2 less than 13 times a number (d) 13 less than 2 times a number
Show solution
(a) x + 5 (Number = x) (b) y – 4 (Number = y) (c) 13 × p – 2 = 13p – 2 (Number = p) (d) 2 × z – 13 = (Number = z)
Complete worked answers
Textbook page 5
Textbook page 5 · solved item 11
6. Describe situations corresponding to the following algebraic expressions: (a) 8 × x + 3 × y (b) 15 × j – 2 × k
Show solution
(a) Sum of 8 times x and 3 times y. (b) Subtract 2 times k from 15 times j.
Textbook page 5 · solved item 12
7. In a calendar month, if any 2 × 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’.
Show solution
w - 8 w - 7 w - 6 w - 1 w w + 1
Textbook page 5 · solved item 13
1. 23 – 10 × 2
Show solution
1. 23 – 10 × 2 = 23 – 20 = 3
Textbook page 5 · solved item 14
2. 83 + 28 – 13 + 32
Show solution
2. 83 + 28 – 13 + 32 = 83 – 13 + 28 + 32 = 70 + 60 = 130.
Textbook page 5 · solved item 15
3. 34 – 14 + 20
Show solution
3. 34 – 14 + 20 = 30 + 20 = 50.
Textbook page 5 · solved item 16
4. 42 + 15 – (8 – 7)
Show solution
4. 42 + 15 – (8 – 7) = 42 + 15 – 1 = 57 – 1 = 56.
Textbook page 5 · solved item 17
5. 68 – (18 + 13)
Show solution
5. 68 – (18 + 13) = 68 - 31 = 37.
Textbook page 5 · solved item 18
6. 7 × 4 + 9 × 6
Show solution
6. 7 × 4 + 9 × 6 = 28 + 54 = 82.
Textbook page 5 · solved item 19
7. 20 + 8 × (16 – 6)
Show solution
7. 20 + 8 × (16 – 6) = 20 + 8 × 10 = 20 + 80 = 100.
Complete worked answers
Textbook page 7
Textbook page 7 · solved item 20
1. If a = -4, then 10 - a = 6. Identify the mistake, explain it, and correct the value.
Show solution
1. Given equation: 10 – a = 6 Substitute a = -4 10 – (-4) = 6 10 + 4 = 6 14 ≠ 6 Since 14 ≠ 6, a = -4 does not satisfy the equation. Now, 10 – a = 6 10 – 6 = a a = 4. Therefore, the solution to the equation is m = 16 3 .
Textbook page 7 · solved item 21
2. If d = 6, then 3d = 36. Identify the mistake, explain it, and correct the value.
Show solution
2. Consider equation: 3d = 36 Substitute d = 6 3 × 6 = 36 18 ≠ 36 Since 18 ≠ 36, d = 6 does not satisfy the equation. Now, 3d = 36 d = 36 3 d = 12. Therefore, the solution to the equation is d = 12.
Textbook page 7 · solved item 22
3. If s = 7, then 3s - 2 = 15. Identify the mistake, explain it, and correct the value.
Show solution
3. Consider equation: 3s – 2 = 15 Substitute s = 7 3 × 7 – 2 = 15 21 -2 = 15 19 ≠ 15 Since 19 ≠ 14, s = 7 does not satisfy the equation. Now, 3s – 2 = 15 3s = 15 + 2 3s = 17 s = 17 3 . Therefore, the solution to the equation is s = 17 3 .
Textbook page 7 · solved item 23
4. If r = 8, then 2r + 1 = 29. Identify the mistake, explain it, and correct the value.
Show solution
4. Consider equation: 2r + 1 = 29 Substitute r = 8 2 × 8 + 1 = 29 16 + 1 = 29 17 ≠ 29 Since 17 ≠ 29, r = 8 does not satisfy the equation. Now, 2r + 1 = 29 2r = 29 – 1 2r = 28 r = 28 2 = 14. Therefore, the solution to the equation is r = 14.
Textbook page 7 · solved item 24
5. If j = 5, then 2j = 10. Identify the mistake, explain it, and correct the value.
Show solution
5. Consider equation: 2j = 10 Substitute j = 5 2 × 5 = 10 10 = 10 Since 10 ≠ 10, j = 5 does satisfy the given equation.
Textbook page 7 · solved item 25
6. If m = -6, then 3(m + 1) = 19. Identify the mistake, explain it, and correct the value.
Show solution
6. Given equation: 3(m + 1) Substitute m = -6, 3(-6 + 1) = 19 3(-5) = 19 -15 ≠ 19 Since -15 ≠ 19; m = -6 does not satisfy the equation. 3(m + 1) = 19 3m + 3 = 19 3m = 19 – 3 3m = 16 m = 16 3 Therefore, the solution to the equation is m = 16 3 .
Textbook page 7 · solved item 26
7. If f = 3 and g = 1, then 2f - 2g = 2. Identify the mistake, explain it, and correct the value.
Show solution
7. Given equation: 2f – 2g = 2 Substitute f =3, g = 1, 2 × 3 – 2 × 1 = 2 6 – 2 = 2 4 ≠ 2 Since 4 ≠ 2; f =3, g = 1 do not satisfy the equation. If f = 2 and g = 1, then 2 × 2 – 2 × 1 = 2 4 – 2 = 2 2 = 2 Therefore, the solution to the equation are f = 2 and g = 1.
Textbook page 7 · solved item 27
8. If t = 4 and b = 3, then 2t + b = 24. Identify the mistake, explain it, and correct the value.
Show solution
8. Given equation: 2t + b = 24 Substitute t = 4, b = 3, 2 × 4 + 3 = 24 8 + 3 = 24 11 ≠ 24 Since 11 ≠ 24, t = 4, b = 3 do not satisfy the equation. Substitute t = 8, b = 8: 2 × 8 + 8 = 24 16 + 8 = 24 24 = 24 Therefore, the solution to the equation are t = 8, b = 8.
Textbook page 7 · solved item 28
9. If h = 5 and n = 6, then h - (3 - n) = 4. Identify the mistake, explain it, and correct the value.
Show solution
9. Given equation: h – (3 - n) = 4 Substitute h = 5, n = 6, 5 – (3 – 6) = 4 5 – (-3) = 4 5 + 3 = 4 8 ≠ 4 Since 11 ≠ 24, h = 5, n = 6 does not satisfy the equation. Substitute h = 5, n = 2 5 – (3 – 2) = 4 5 – 1 = 4 4 = 4 Therefore, the solution to the equation are h = 5, n = 2.
Complete worked answers
Textbook page 9
Textbook page 9 · solved item 29
If c = ₹50, find the total amount earned by the scale of pencils.
Show solution
Total amount earned = Day 1 + Day 2 + Day 3 = 5 × c + 3 × c + 10 × c = 5c + 3c + 10c = 18c. If c = ₹50, then 18c = 18 × 50 = ₹900. Therefore, ₹900 is the total amount earned by the scale of pencils.
Textbook page 9 · solved item 30
Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Show solution
Total amount earned = Day 1 + Day 2 + Day 3 = 4 × d + 6 × d + 1 × d = 4d + 6d + d = 11d.
Complete worked answers
Textbook page 10
Textbook page 10 · solved item 31
A shop rents out chairs and tables for a day’s use. To rent them, one has to first pay the following amount per piece. When the furniture is returned, the shopkeeper pays back some amount as follows. Write an expression for the total number of rupees paid if x chairs and y tables are rented. For x chairs and y tables, let us find the total amount paid at the beginning and the amount one gets back after returning the furniture. Describe the procedure to get these amounts.
Show solution
Chairs rented = x Tables rented = y Total amount paid at the beginning = Amount paid for chairs + Amount paid for tables = 40 × (x) + 75 × y = 40x + 7y. Total amount returned = Amount returned for chairs + Amount returned for tables = 6 × (x) + 10 × y = 6x + 10y. Total amount paid = (40x + 75y) – (6x + 10y)
Textbook page 10 · solved item 32
Can we simplify this expression? If yes, how? If not, why not? [Expression: (40x + 75y) – (6x + 10y)]
Show solution
Yes, the expression can be simplified further as follows: (40x + 75y) – (6x + 10y) = 40x + 75y – 6x – 10y = (40x – 6x) + (75y – 10y) = 34x + 65y.
Complete worked answers
Textbook page 11
Textbook page 11 · solved item 33
Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)?
Show 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.
Textbook page 11 · solved item 34
What do each of the expressions mean? (Expressions: 7p – 3q, 8p – 4q, and 6p – 2q).
Show solution
Each of these expression involve a difference between two algebraic terms.
Complete worked answers
Textbook page 12
Textbook page 12 · solved item 35
Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.
Show solution
Krishita’s score in three rounds could be (8p - 3q), (6p – 2q) and (9p – 2q). Therefore, total score after three rounds = (8p - 3q) + (6p – 2q) + (9p – 2q). = 8p – 3q + 6p – 2q + 9p – 2q = 23p – 7q.
Textbook page 12 · solved item 36
Can we say who scored more? Can you explain why? (Charu’ score: 21p – 9q & Krishita’s score: 23p – 7q)
Show solution
Penalties for Charu = 9q Penalities for Krishita = 7q Krishita scored more because she had fewer penalties than Charu.
Textbook page 12 · solved item 37
Simplify this expression further. [Expression: (23p – 7q) – (21p – 9q)]
Show solution
23q – 7q - (21p – 9q) = 23q – 7q – 21p + 9q = (23q – 21q) + (9q – 7q) = 2q + 2q.
Textbook page 12 · solved item 38
Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.
Show solution
55 40 25 16 13 10
Complete worked answers
Textbook page 13
Textbook page 13 · solved item 39
Are the expressions 10y – 3 and 10(y – 3) equal? Let us compare the values that these expressions take for different values of y.
Show solution
(i) For 10y – 3: If y = 0, then 10y -3 = 10(0) – 3 = 0 – 3 = -3. If y = 7, then 10y -3 = 10(7) – 3 = 70 – 3 = 67. If y = 10, then 10y – 3 = 10(10) – 3 = 100 – 3 = 97. (ii) For 10(y – 3): If y = 0, then 10(y – 3) = 10(0 – 3) = 10(-3) = -30. if y = 7, then 10(y – 3) = 10(7 – 3) = 10(4) = 40. If y = 10, then 10(y – 3) = 10(10 – 3) = 10(7) = 70.
Textbook page 13 · solved item 40
After filling in the two diagrams, do you think the two expressions are equal?
Show solution
Filling in the two diagrams made it clear that the two expressions are not equal. -3 67 97 40 70 -30 Figure it out
Textbook page 13 · solved item 41
1. 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.
Show solution
(i) Adding like terms together gives: (5y + 5y) + (x + x) + (2 – 6) = 10y + 2x + (-4) = 2x + 10y -4. (ii) Adding like terms together gives: (2p × 4) + (3q × 4) + (2 × -2) + (3 × 2) = 8p + 12q + (-4) + 6 = 8p + 12q + 2. (iii) Adding like terms together gives: (5k × 12) + (-5g × 4) = 60k + (-20g) = 40k – 20g.
Complete worked answers
Textbook page 14
Textbook page 14 · solved item 42
2. Simplify each of the following expressions: (a) p + p + p + p, p + p + p + q, p + q + p – q, (b) p − q + p − q, p + q – p + q, (c) p + q − (p + q), p – q – p – q, (d) 2d – d – d – d, 2d – d – d – c, (e) 2d – d − (d − c), 2d − (d − d) – c, (f) 2d – d – c – c
Show solution
(a) p + p + p + p = 3p. p + p + p + q = 3p + q. p + q + p – q = 2q (b) p − q + p – q = 2p -2q. p + q – p + q = 2q. (c) p + q − (p + q) = p + q – p – q = 0. p – q – p – q = -2q. (d) 2d – d – d – d = 2d – 3d = -d. 2d – d – d – c = 2d – 2d – c = -c. (i) (ii) (iii) (e) 2d – d − (d – c) = 2d – d – d + c = 2d – 2d + c = c.
2d − (d − d) – c = 2d – 0 – c = 2d – c. (f) 2d – d – c – c = d – 2c.
Textbook page 14 · solved item 43
1. 3a + 2b = 5. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
1. Incorrect simplest form because unlike terms 3a and 2b cannot be added. Correct simplest form: 3a + 2b.
Textbook page 14 · solved item 44
2. 3b - 2b - b = 0. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
2. Correct simplest form because 3b – 2b – b = 3b – 3b = 0.
Textbook page 14 · solved item 45
3. 6(p + 2) = 6p + 8. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
3. Incorrect simplest form because of wrong multiplication done during distribution. Correct simplest form: 6(p + 2) = 6×p + 6×2 = 6p + 12.
Textbook page 14 · solved item 46
4. (4x + 3y) - (3x + 4y) = x + y. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
4. Incorrect simplest form because of the subtraction mistake. Correct simplest form: (4x + 3y) − (3x + 4y) = 4x + 3y – 3x – 4y = 4x – 3x + 3y – 4y = x – y.
Textbook page 14 · solved item 47
5. 5 - (2 - 6z) = 3 - 6z. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
5. Incorrect simplest form because of the sign error. Correct simplest form: 5 − (2 − 6z) = 5 – 2 + 6z = 3 + 6z.
Textbook page 14 · solved item 48
6. 2 + (x + 3) = 2x - 6. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
6. Incorrect simplest form because of the misapplied operations. Correct simplest form: 2 + (x + 3) = 2 + x + 3 = x + 5.
Textbook page 14 · solved item 49
7. 2y + (3y - 6) = -y + 6. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
7. Incorrect simplest form because of incorrect combining. Correct simplest form: 2y + (3y − 6) = 2y + 3y – 6 = 5y – 6.
Textbook page 14 · solved item 50
8. 7p - p + 5q - 2q = 7p + 3q. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
8. Incorrect simplest form because of addition and subtraction mistakes. Correct simplest form: 7p − p + 5q − 2q = 6p + 3q.
Textbook page 14 · solved item 51
9. 5(2w + 3x + 4w) = 10w + 15x + 20w. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
9. Incorrect simplest form because of incomplete simplification. Correct simplest form: 5(2w + 3x + 4w) = 5 × 2w + 5 × 3x + 5 × 4w = 10w + 15x + 20w = 30w + 15x.
Textbook page 14 · solved item 52
10. 3j + 6k + 9h + 12 = 3(j + 2k + 3h + 4). Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
10. Correct simplest form.
Textbook page 14 · solved item 53
11. 4(2r + 3s + 5) = -20 - 8r - 12s. Check this claimed simplification, explain any mistake, and simplify correctly.
Show solution
11. Incorrect simplest form because of multiplication done incorrectly and the signs are wrong. Correct simplest form: 4(2r + 3s + 5) = (4 × 2r) + (4 × 3s) + (4 × 5) = 8r + 12s + 20.
Complete worked answers
Textbook page 15
Textbook page 15 · solved item 54
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?
Show solution
Combining like terms and simplifying constants reduces the total number of terms in an expression.
Complete worked answers
Textbook page 16
Textbook page 16 · solved item 55
Find the formulas of the number machines below and write the expression for each set of inputs.
Show solution
(A) Formula: Two subtracted from the sum of the two numbers. Algebraic expression: a + b – 2. 5th expression: a + b - 2. (B) Formula: One added to the multiplication of two numbers. Algebraic expression: (a × b) + 1 = ab + 1. 4th expression: 10 × 3 + 1 = 30 + 1 = 31. 5th expression: a × b + 1 = ab + 1.
Complete worked answers
Textbook page 17
Textbook page 17 · solved item 56
Given a position number can we find out the design that appears there? Which Design appears at Position 122? (Position at which the design A, B and C appears for nth time are 3n-2, 3n-1 and 3n)
Show solution
Yes, given a position number can help us find out the design that appears there. Since 122 ÷ 3 gives 40 as the quotient and a remainder of 2. Therefore, Design B appears at Position 122. (A) (B)
Textbook page 17 · solved item 57
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
Show 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.
Textbook page 17 · solved item 58
Use this to find what design appears at positions 99, 122, and 148.
Show solution
Design C appears at position 99. Design B appears at position 122. Design A appears at position 148.
Complete worked answers
Textbook page 19
Textbook page 19 · solved item 59
Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’.
Show solution
Considering the given 2 × 2 square: Top left number (8) = a Number to the right 8 (9) = a + 1 Position Quotient on division by 3 Remainder 99 33 0 122 40 2 148 49 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.
Textbook page 19 · solved item 60
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?
Show 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.
28 29 30 35 36 37 42 43 44
Complete worked answers
Textbook page 20
Textbook page 20 · solved item 61
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?
Show solution
Matchsticks in Step 33 = (2 × 33) + 1 = 66 + 1 = 67. Matchsticks in Step 84 = (2 × 84) + 1 = 168 + 1 = 169. Matchsticks in Step 108 = (2 × 108) + 1 = 216 + 1 = 217.
Complete worked answers
Textbook page 21
Textbook page 21 · solved item 62
Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same? [Expression: 3 + 2 × (y – 1) and 2y + 1]
Show solution
Yes, both the expressions 3 + 2 × (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 × (y – 1) = 3 + 2y – 2 = 2y + 1.
Textbook page 21 · solved item 63
What are these numbers in Step 3 and Step 4?
Show 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.
Textbook page 21 · solved item 64
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?
Show 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. Figure it out For the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described.
If required, assume some values for the unknowns and try to find the relationship.
Complete worked answers
Textbook page 22
Textbook page 22 · solved item 65
1. 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? (a) 30x + 20y (b) (30 + 20) × (x + y) (c) 20x + 30y (d) (30+20) × x + y (e) 30x – 20y
Show 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 × 30) + (y + 20) = 30x + 20y. Therefore, (a) 30x + 20y is the required expression.
Textbook page 22 · solved item 66
2. 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? (a) p + q + r (b) p + q + 2r (c) 2 × (p + q + r) (d) p + q + r + 2 (e) p + q + r + 1 (f) 2 × (p + q)
Show 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. Therefore, (a) p + q + r is the required expression.
Textbook page 22 · solved item 67
3. 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. (a) Write an expression describing how far away the snail is from its starting position. (b) What can we say about the snail’s movement if d > u?
Show solution
(a) 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 (b) 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.
Textbook page 22 · solved item 68
4. 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?
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Distance cycled in first week = 7 × 5 = 35 km Distance cycled in second week = 7 × (5 + z) = (35 + 7z) km Distance cycled in third week = 7 × (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.
Textbook page 22 · solved item 69
5. 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.
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Top left: (w + 2) → (-5) → (w – 3) → (×3) → 3w – 9. Bottom left: (w + 2) → (-8) → (w – 6) → (-4) → (w – 10). Bottom right: (w + 2) → (+3) → (w + 5) → (×4) → (4w + 20).
Complete worked answers
Textbook page 23
Textbook page 23 · solved item 70
6. 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. (a) If t = 4, what is the time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]
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There are 4 segments of travel. Trains stops 2 minutes at each station (S1, S2, S3). Time taken to travel from one station to the next station = t (a) If t = 4, Total time taken to travel from Yahapur to Vahapur = 4 × 4 + 3 × 2 = 16 + 6 = 22 minutes. (b) Time taken to travel from one station to another = t.
There are 4 segments and 3 stops of 2 minutes. Algebraic expression = 4 × t + 3 × 2 = (4t + 6) minutes. Yahapu Vahapu S1 S2 S3
Textbook page 23 · solved item 71
7. Simplify the following expressions: (a) 3a + 9b – 6 + 8a – 4b – 7a + 16 (b) 3(3a – 3b) – 8a – 4b – 16 (c) 2(2x – 3) + 8x + 12 (d) 8x – (2x – 3) + 12 (e) 8h – (5 + 7h) + 9 (f) 23 + 4(6m – 3n) – 8n – 3m – 18
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(a) 3a + 9b – 6 + 8a – 4b – 7a + 16 = 3a + 8a – 7a + 9b – 4b – 6 + 16 = 11a – 7a + 5b + 10 = 4a + 5b + 10. (b) 3(3a – 3b) – 8a – 4b – 16 = 9a – 9b – 8a – 4b – 16 = 9a – 8a – 9b – 4b – 16 = a – 13b – 16. (c) 2(2x – 3) + 8x + 12 = 4x – 6 + 8x + 12 = 4x + 8x – 6 + 12 = 12x + 6. (d) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12 = 6x + 15.
(e) 8h – (5 + 7h) + 9 = 8h – 5 + 7h + 9 = 8h + 7h + 9 – 5 = 15h + 4. (f) 23 + 4(6m – 3n) – 8n – 3m – 18 = 23 + 24m – 12n – 8n – 3m – 18 = 24m – 3m – 12n – 8n + 23 – 18 = 21m – 20n + 5.
Textbook page 23 · solved item 72
8. Add the expressions given below: (a) 4d – 7c + 9 and 8c – 11 + 9d (b) –6f + 19 – 8s and –23 + 13f + 12s (c) 8d – 14c + 9 and 16c – (11 + 9d) (d) 6f – 20 + 8s and 23 – 13f – 12s (e) 13m – 12n and 12n – 13m (f) –26m + 24n and 26m – 24n
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(a) 4d – 7c + 9 and 8c – 11 + 9d = 4d – 7c + 9 + 8c – 11 + 9d = 4d + 9d – 7c + 8c + 9 – 11 = 13d + c – 2. (b) –6f + 19 – 8s and –23 + 13f + 12s = –6f + 19 – 8s + (–23) + 13f + 12s = -6f + 13f – 8s + 12s + 19 – 23 = 7f + 4s – 4. (c) 8d – 14c + 9 and 16c – (11 + 9d) = 8d – 14c + 9 + 16c – (11 + 9d) = 8d – 14c + 9 + 16c – 11 – 9d = 16c – 14c + 8d – 9d + 9 – 11 = 2c – d – 2.
(d) 6f – 20 + 8s and 23 – 13f – 12s = 6f – 20 + 8s + 23 – 13f – 12s = 6f – 13f + 8s – 12s – 20 + 23 = -7f – 4s + 3. (e) 13m – 12n and 12n – 13m = 13m – 12n + 12n – 13m = 13m – 13m – 12n + 12n = 0. (f) –26m + 24n and 26m – 24n = -26m + 24n + 26m – 24n = -26m + 26m + 24n – 24n = 0.
Textbook page 23 · solved item 73
9. Subtract the expressions given below: (a) 9a – 6b + 14 from 6a + 9b – 18 (b) –15x + 13 – 9y from 7y – 10 + 3x (c) 17g + 9 – 7h from 11 – 10g + 3h (d) 9a – 6b + 14 from 6a – (9b + 18) (e) 10x + 2 + 10y from –3y + 8 – 3x (f) 8g + 4h – 10 from 7h – 8g + 20
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(a) 9a – 6b + 14 from 6a + 9b – 18 = 6a + 9b – 18 – (9a – 6b + 14) = 6a + 9b – 18 – 9a + 6b – 14 = 6a – 9a + 9b + 6b – 18 – 14 = 3a + 15b – 32. (b) –15x + 13 – 9y from 7y – 10 + 3x = 7y – 10 + 3x – (–15x + 13 – 9y) = 7y – 10 + 3x + 15x – 13 + 9y = 7y + 9y + 3x + 15x – 10 – 13 = 16y + 18x – 23. (c) 17g + 9 – 7h from 11 – 10g + 3h = 11 – 10g + 3h – (17g + 9 – 7h) = 11 – 10g + 3h – 17g – 9 + 7h = -10g -17g + 3h + 7h + 11 – 9 = -27g + 10h + 2.
(d) 9a – 6b + 14 from 6a – (9b + 18) = 6a – (9b + 18) – (9a – 6b + 14) = 6a – 9b – 18 – 9a + 6b – 14 = 6a – 9a – 9b + 6b – 18 – 14 = -3a – 3b – 32. (e) 10x + 2 + 10y from –3y + 8 – 3x = -3y + 8 – 3x – (10x + 2 + 10y) = -3y + 8 – 3x – 10x – 2 – 10y = -3x – 10x – 3y – 10y + 8 – 2 = -13x – 13y + 6. (f) 8g + 4h – 10 from 7h – 8g + 20 = 7h – 8g + 20 – (8g + 4h – 10) = 7h – 8g + 20 – 8g – 4h + 10 = -8g – 8g + 7h – 4h + 20 + 10 = -16g + 3h + 30.
Textbook page 23 · solved item 74
10. Describe situations corresponding to the following algebraic expressions: (a) 8x + 3y (b) 15x – 2x
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(a) 8x + 3y Ramesh buys 8 pencils, each costing ₹x, and 3 pens, each costing ₹y. Total cost = 8x + 3y (b) 15x – 2x A shopkeeper has 15 boxes, each containing x apples. He sells 2 boxes. Apples left = 15x − 2x = 13x.
Textbook page 23 · solved item 75
11. 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?
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No fold: 1 cut → 2 pieces. 1-fold: 1 cut → 3 pieces. 2-fold: 1 cut → 4 pieces. 3-fold: 1 cut → 5 pieces. Similarly, For 10-fold: 1 cut → 11 pieces. Expression with ‘r’ times fold and cut = r + 2.
Complete worked answers
Textbook page 24
Textbook page 24 · solved item 76
12. 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?
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1 square → 4 matchsticks. 2 squares → 7 matchsticks. 3 squares → 10 matchsticks. Similarly, For 10 squares → 31 matchsticks. Matchsticks required to make ‘w’ squares = 3 × w + 1 = 3w + 1.
Textbook page 24 · solved item 77
13. 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.
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Red positions: 1, 5, 9, 13……. 6 7 8 9 10 11 12 13 Yellow positions: 2, 4, 6, 8, 10……. Green positions: 3, 7, 11, 15……. Expression for red colour = 4r - 3. Expression for yellow colour = 2y. Expression for green colour = 4g – 1. 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 → Yellow 343 ÷ 4 = 85 R3 → Remainder = 3 → Green
Textbook page 24 · solved item 78
14. 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?
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Step 1: 5 squares. Step 2: 9 squares. Step 3: 13 squares. General formula: 4n + 1 Step 4: 4(4) + 1 = 16 + 1 = 17 Step 10: 4(10) + 1 = 40 + 1 = 41 Step 50: 4(50) + 1 = 200 + 1 = 201 For vertices; Step 1 → 20 Step 2 → 36 Step 3 → 52 General formula: 16n + 4.