NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Chapter 4 of Ganita Prakash Part I introduces letter-numbers — letters such as a, n or w that stand for numbers. With them we write short algebraic expressions and formulas that capture a whole pattern in one line, and we learn to simplify them.
NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
| Chapter | Download | |
|---|---|---|
| Solutions PDF | Expressions Using Letter Numbers | View PDF |
NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers – Questions and Answers
All 107 questions of the chapter (textbook pages 81 – 105) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
In-text Questions — Page 81Section 4.1 The Notion of Letter-Numbers
Shabnam is always 3 years older, so add 3 to Aftab’s age.
= 18 + 3
= 21 years
Add 3 to Aftab’s age.
| Aftab’s age | Shabnam’s age |
|---|---|
| 4 | 4 + 3 = 7 |
| 10 | 10 + 3 = 13 |
| 23 | 23 + 3 = 26 |
Yes. Give Aftab’s age a short name and the rule becomes one line.
Shabnam’s age = Aftab’s age + 3
s = a + 3
Replace a by 23 in the expression a + 3.
= 23 + 3
= 26 years
In-text Questions — Page 82Section 4.1 The Notion of Letter-Numbers
| Aftab’s age | Expression for Shabnam’s age |
|---|---|
| 4 | 4 + 3 |
| 10 | 10 + 3 |
| 23 | 23 + 3 |
| ? | ? + 3 |
| a | a + 3 |
Aftab is 3 years younger, so subtract 3.
a = s – 3
Replace s by 20.
= 20 – 3
= 17 years
Every L uses exactly 2 matchsticks, so the count is always double the number of Ls.
7 Ls → 7 × 2 = 14 matchsticks
45 Ls → 45 × 2 = 90 matchsticks
In words: Number of matchsticks = 2 × Number of L’s. Taking n for the number of Ls, the expression is
In-text Questions — Page 83Section 4.1 The Notion of Letter-Numbers
Find the two costs separately and add them.
Cost of 5 kg jaggery = 5 × ₹60 = ₹300
Total cost = ₹350 + ₹300 = ₹650
Find the two costs and add them.
Cost of 9 kg jaggery = 9 × ₹60 = ₹540
Total = ₹280 + ₹540 = ₹820
Let c be the number of coconuts and j the number of kilograms of jaggery.
| Quantity needed | Relationship | Expression |
|---|---|---|
| Cost of coconuts | Number of coconuts × 35 | c × 35 |
| Cost of jaggery | Number of kgs of jaggery × 60 | j × 60 |
c × 35 + j × 60
Put c = 7 and j = 4.
= 7 × 35 + 4 × 60
= 245 + 240
= ₹485
Put q = 7 in 4 × q.
= 4 × 7
= 28 cm
Figure it Out — Pages 84–85Section 4.1 The Notion of Letter-Numbers
In each case the perimeter is the sidelength added as many times as there are sides. Let the sidelength be a units.
| Shape | Number of equal sides | Formula for perimeter |
|---|---|---|
| Equilateral triangle | 3 | 3a |
| Regular pentagon | 5 | 5a |
| Regular hexagon | 6 | 6a |
(b) Perimeter = a + a + a + a + a = 5a units
(c) Perimeter = 6a units
Joining two pipes end to end adds their lengths.
= (20 + k) metres
| No. of ₹100 notes | No. of ₹20 notes | No. of ₹5 notes | Expression and total amount |
|---|---|---|---|
| 3 | 5 | 6 | |
| 6 × 100 + 4 × 20 + 3 × 5 = 695 | |||
| 8 | 4 | z | |
| x | y | z |
Each row follows the same rule: (number of ₹100 notes × 100) + (number of ₹20 notes × 20) + (number of ₹5 notes × 5).
| 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 = 880 + 5z |
| x | y | z | x × 100 + y × 20 + z × 5 = 100x + 20y + 5z |
Row 2: 600 + 80 + 15 = ₹695
Row 3: 800 + 80 + 5z = ₹(880 + 5z)
Row 4: ₹(100x + 20y + 5z)
The answer is (d) 10 + 8 × y.
Grinding time = 8 seconds for each kg = 8 × y
Total = 10 + 8y seconds
Let the number be n.
| In words | Expression |
|---|---|
| (a) 5 more than a number | n + 5 |
| (b) 4 less than a number | n – 4 |
| (c) 2 less than 13 times a number | 13n – 2 |
| (d) 13 less than 2 times a number | 2n – 13 |
(a) 8 × x + 3 × y
A stationery shop sells a pen for ₹x and a notebook for ₹y. Abha buys 8 pens and 3 notebooks. Her bill is 8x + 3y rupees.
(b) 15 × j – 2 × k
A carpenter makes 15 stools every day and works for j days, so he makes 15j stools. On k of those days, 2 stools got damaged. The number of good stools left is 15j – 2k.
In a calendar, moving one step right adds 1 and moving one step up subtracts 7. Starting from w in the bottom-middle cell:
| w – 8 | w – 7 | w – 6 |
| w – 1 | w | w + 1 |
Directly above w = w – 7
Above-left = w – 7 – 1 = w – 8
Above-right = w – 7 + 1 = w – 6
In-text Questions — Pages 85–86Section 4.2 Revisiting Arithmetic Expressions
Write each expression as a sum of terms, turn each term into a number, then add.
2. 83 + 28 – 13 + 32 = (83 – 13) + (28 + 32) = 70 + 60 = 130
3. 34 – 14 + 20 = (34 – 14) + 20 = 20 + 20 = 40
4. 42 + 15 – (8 – 7) = 57 – 1 = 56
5. 68 – (18 + 13) = 68 – 31 = 37
6. 7 × 4 + 9 × 6 = 28 + 54 = 82
7. 20 + 8 × (16 – 6) = 20 + 8 × 10 = 20 + 80 = 100
These are the multiples of 4 in increasing order, so the term at position n is 4 times n.
29th term = 4 × 29 = 116
nth term = 4 × n = 4n
Remember 7k means 7 × k and 5m means 5 × m.
5m + 3 when m = 2 → 5 × 2 + 3 = 10 + 3 = 13
Mind the Mistake, Mend the Mistake — Page 87Section 4.3 Omission of the Multiplication Symbol
Mistake. The negative sign of a was ignored.
= 10 + 4
= 14
Mistake. 3d means 3 × d. The 3 and the 6 were written side by side as “36”.
Mistake. The 2 was subtracted before multiplying.
= 21 – 2
= 19
Mistake. The 2 and the 8 were written side by side as 28.
= 16 + 1
= 17
No mistake. This one is correct.
Mistake. Only m was multiplied by 3 (not the 1), and the minus sign of m was dropped: 3 × 6 + 1 = 19.
= 3 × (–5)
= –15
Mistake. 2f and 2g were read as the two-digit numbers 23 and 21, and 23 – 21 = 2.
= 6 – 2
= 4
Mistake. The + was treated as a ×.
= 8 + 3
= 11
Mistake. The bracket was handled wrongly.
= 5 – (–3)
= 5 + 3
= 8
In-text Questions — Pages 88–90Section 4.4 Simplification of Algebraic Expressions
| Day 1 | Day 2 | Day 3 | |
|---|---|---|---|
| Pencils (Price ‘c’) | 5 | 3 | 10 |
| Erasers (Price ‘d’) | 4 | 6 | 1 |
Multiply the number sold that day by the price per pencil.
Day 3: 10 pencils → 10c
| Day 1 | Day 2 | Day 3 | |
|---|---|---|---|
| Pencils (price c) | 5 | 3 | 10 |
| Money from pencils | 5c | 3c | 10c |
= (5c + 3c + 10c) + (4d + 6d + d)
= 18c + 11d (pencils 18c, erasers 11d)
| Day 1 | Day 2 | Day 3 | |
|---|---|---|---|
| Pencils (Price ‘c’) | 5 | 3 | 10 |
| Erasers (Price ‘d’) | 4 | 6 | 1 |
The total money from pencils is 5c + 3c + 10c, which simplifies to 18c.
18c = 18 × ₹50
= ₹900
| Day 1 | Day 2 | Day 3 | |
|---|---|---|---|
| Pencils (Price ‘c’) | 5 | 3 | 10 |
| Erasers (Price ‘d’) | 4 | 6 | 1 |
Erasers sold: 4 on Day 1, 6 on Day 2 and 1 on Day 3, each priced d.
= (4 + 6 + 1) × d
= 11d
No. It is already in its simplest form.
Try three different values of c.
| c | 5c + 3c + 10c | 18c |
|---|---|---|
| 2 | 10 + 6 + 20 = 36 | 36 |
| 5 | 25 + 15 + 50 = 90 | 90 |
| 50 | 250 + 150 + 500 = 900 | 900 |
Area of the bigger rectangle = 7v square units.
Way 2 — add the two smaller areas: 4v + 3v = 7v
So 4v + 3v = 7v
AEFD is what is left of ABCD after the strip EBCF is removed.
AE = 12 – 4 = 8, AD = n
Area of AEFD = 8 × n = 8n
Method 2 — subtract:
Area of ABCD = 12n, Area of EBCF = 4n
Area of AEFD = 12n – 4n = 8n
Amount paid at the beginning: multiply each rate by how many pieces are taken and add.
Tables: y × ₹75 = 75y
Paid at the start = 40x + 75y
Amount returned: the same procedure with the return rates.
Tables: y × ₹10 = 10y
Returned = 6x + 10y
Yes. Open the brackets, then collect like terms.
= 40x + 75y – 6x – 10y
= 40x + (–6x) + 75y + (–10y) (swapping and grouping)
= (40 – 6)x + (75 – 10)y
= 34x + 65y
In-text Questions — Pages 91–93Section 4.4 Simplification of Algebraic Expressions
Yes. Subtracting an expression is the same as adding its opposite.
So (40x + 75y) – (6x + 10y) = (40x + 75y) + (– 6x – 10y)
= 34x + 65y either way
| Round | Expression | What it says |
|---|---|---|
| 1 | 7p – 3q | 7 correct answers and 3 wrong answers |
| 2 | 8p – 4q | 8 correct answers and 4 wrong answers |
| 3 | 6p – 2q | 6 correct answers and 2 wrong answers |
= 7 × 4 + (–3 × 1)
= 28 + (–3)
= 25
Use p = 4 and q = 1 in each expression.
Round 3: 6p – 2q = 6 × 4 – 2 × 1 = 24 – 2 = 22
No penalty means nothing is taken away for a wrong answer.
Round 1 score becomes 7p – 3 × 0 = 7p = 28
Total becomes 21p – 9 × 0 = 21p = 84
Add the three round scores and simplify.
= 7p + (–3q) + 8p + (–4q) + 6p + (–2q)
= 7p + 8p + 6p + (–3q) + (–4q) + (–2q) (swapping and grouping)
= (7 + 8 + 6)p – (3 + 4 + 2)q
= 21p – 9q
Split 23 into three parts and 7 into three parts.
Round 1: 8p – 3q
Round 2: 7p – 2q
Round 3: 8p – 2q
Sum = (8 + 7 + 8)p – (3 + 2 + 2)q = 23p – 7q ✔
Krishita scored more.
Krishita loses 7q, Charu loses 9q → Krishita loses 2q less
= 23p – 7q – 21p + 9q
= 23p – 21p + (–7q) + 9q
= (23 – 21)p + (9 – 7)q
= 2p + 2q
= 4x + 4y – y (distributive property)
= 4x + 4y + (– y)
= 4x + (4 – 1)y
= 4x + 3y
No. They describe two different operations.
5 + u means 5 more than the number u
| u | 5u | 5 + u |
|---|---|---|
| 11 | 55 | 16 |
| 2 | 10 | 7 |
| 8 | 40 | 13 |
| 5 | 25 | 10 |
5 × 2 = 10 but 5 + 2 = 7
5 × 8 = 40 but 5 + 8 = 13
5 × 5 = 25 but 5 + 5 = 10
The two columns never match, so 5u ≠ 5 + u.
10y – 3 means ‘3 less than 10 times y’, while 10(y – 3) means ‘10 times (3 less than y)’.
| y | 10y – 3 | 10 (y – 3) |
|---|---|---|
| 2 | 17 | –10 |
| 0 | –3 | –30 |
| 7 | 67 | 40 |
| 10 | 97 | 70 |
y = 0: 0 – 3 = –3, but 10 × (–3) = –30
y = 7: 70 – 3 = 67, but 10 × 4 = 40
y = 10: 100 – 3 = 97, but 10 × 7 = 70
No, the two expressions are not equal.
The sum is 2r + 2s + 24. Three ways of adding:
= 12 + 12 + 2r + 2s = 2r + 2s + 24
2. Like terms together: (8 × 3) + (r + r) + (s + s)
= 24 + 2r + 2s = 2r + 2s + 24
3. Upper half, doubled: 2 × (4 × 3 + r + s)
= 2 × (12 + r + s) = 24 + 2r + 2s = 2r + 2s + 24
Figure it Out — Pages 93–94Section 4.4 Simplification of Algebraic Expressions
Picture (i) — two rows: 5y, –6, x and x, 2, 5y.
Way 2 (like terms): (5y + 5y) + (x + x) + (–6 + 2)
= 10y + 2x + (–4)
= 2x + 10y – 4
Picture (ii) — 2p and 3q appear 4 times each; the number tiles are –2, 3, 3, –2.
= 8p + 12q + 2
Way 2 (column-wise): (2p + 3q) + (3q + 2p) + (–2 + 3 + 2p + 3q) + (3 – 2 + 3q + 2p)
= 8p + 12q + 2
Picture (iii) — the four corners are –5g and every other circle is 5k (12 of them).
= –20g + 60k
Way 2 (columns): first column (–5g + 5k + 5k – 5g) = –10g + 10k
middle two columns = 8 × 5k = 40k
last column = –10g + 10k
Total = 60k – 20g
| Expression | Working | Simplest form |
|---|---|---|
| (a) p + p + p + p | 4 lots of p | 4p |
| (a) p + p + p + q | 3 lots of p, one q | 3p + q |
| (b) p + q + p – q | (p + p) + (q – q) | 2p |
| (c) p – q + p – q | (p + p) + (–q – q) | 2p – 2q |
| (d) p + q – p + q | (p – p) + (q + q) | 2q |
| (e) p + q – (p + q) | p + q – p – q | 0 |
| (f) p – q – p – q | (p – p) + (–q – q) | –2q |
| (g) 2d – d – d – d | (2 – 1 – 1 – 1)d | –d |
| (h) 2d – d – d – c | (2 – 1 – 1)d – c = 0 – c | –c |
| (i) 2d – d – (d – c) | 2d – d – d + c | c |
| (j) 2d – (d – d) – c | 2d – 0 – c | 2d – c |
| (k) 2d – d – c – c | (2 – 1)d – 2c | d – 2c |
Mind the Mistake, Mend the Mistake — Pages 94–95Section 4.4 Simplification of Algebraic Expressions
Mistake. Unlike terms were added.
No mistake.
Mistake. The 6 multiplied p, but it was added to the 2 (6 + 2 = 8) instead of multiplying it.
= 6p + 12
Mistake. 3y – 4y was worked out as 4y – 3y = y; the minus sign before the bracket was not applied to 4y.
= 4x + 3y – 3x – 4y
= (4 – 3)x + (3 – 4)y
= x – y
Mistake. Only the first term inside the bracket had its sign changed.
= 3 + 6z
Mistake. The + before the bracket was treated as ×, and the sign of 3 was also changed: 2(x – 3) = 2x – 6.
= x + 5
Mistake. The signs inside were flipped even though the bracket has a plus in front.
= 5y – 6
Mistake. The – p was left out.
= (7 – 1)p + (5 – 2)q
= 6p + 3q
Mistake. The distribution was done correctly, but like terms were not collected — so it is not the simplest form.
= 30w + 15x
Mistake. The answer is correct as an equal expression, but it is not in simplest form because the bracket is still there.
Mistake. Every term was given a minus sign, though there is no minus anywhere.
= 8r + 12s + 20
| No. | Correct simplest form | Number of terms | Number of letter-numbers |
|---|---|---|---|
| 1 | 3a + 2b | 2 | 2 |
| 2 | 0 | 1 | 0 |
| 3 | 6p + 12 | 2 | 1 |
| 4 | x – y | 2 | 2 |
| 5 | 3 + 6z | 2 | 1 |
| 6 | x + 5 | 2 | 1 |
| 7 | 5y – 6 | 2 | 1 |
| 8 | 6p + 3q | 2 | 2 |
| 9 | 30w + 15x | 2 | 2 |
| 10 | 3j + 6k + 9h + 12 | 4 | 3 |
| 11 | 8r + 12s + 20 | 3 | 2 |
Yes. In every row, the number of letter-numbers is never more than the number of terms:
Formula Detective — Pages 95–96Section 4.5 Pick Patterns and Reveal Relationships
The formula is “two times the first number minus the second number”, that is 2a – b.
| Inputs (a, b) | Expression | Output |
|---|---|---|
| 5, 2 | 2 × 5 – 2 | 8 |
| 8, 1 | 2 × 8 – 1 | 15 |
| 9, 11 | 2 × 9 – 11 | 7 |
| 10, 10 | 2 × 10 – 10 | 10 |
| 6, 4 | 2 × 6 – 4 | 8 |
Machine 1 — “add the two numbers, then subtract 2”, that is a + b – 2.
| Inputs | Expression | Output |
|---|---|---|
| 5, 2 | 5 + 2 – 2 | 5 |
| 8, 1 | 8 + 1 – 2 | 7 |
| 9, 11 | 9 + 11 – 2 | 18 |
| 10, 10 | 10 + 10 – 2 | 18 |
| a, b | a + b – 2 | a + b – 2 |
Machine 2 — “multiply the two numbers, then add 1”, that is ab + 1.
| Inputs | Expression | Output |
|---|---|---|
| 4, 1 | 4 × 1 + 1 | 5 |
| 6, 0 | 6 × 0 + 1 | 1 |
| 3, 2 | 3 × 2 + 1 | 7 |
| 10, 3 | 10 × 3 + 1 | 31 |
| a, b | a × b + 1 | ab + 1 |
Here is one you can use. Formula: 3a – 2b — “three times the first number minus twice the second”.
| Inputs (a, b) | Working | Output |
|---|---|---|
| 4, 1 | 3 × 4 – 2 × 1 | 10 |
| 5, 5 | 3 × 5 – 2 × 5 | 5 |
| 2, 3 | 3 × 2 – 2 × 3 | 0 |
| 7, 2 | 3 × 7 – 2 × 2 | 17 |
| 1, 4 | 3 × 1 – 2 × 4 | –5 |
In-text Questions — Pages 96–97Algebraic Expressions to Describe Patterns
The border repeats in blocks of three: A, B, C, A, B, C, …
| Design | Positions where it occurs | Formula for the nth time |
|---|---|---|
| A | 1, 4, 7, 10, 13, … | 3n – 2 |
| B | 2, 5, 8, 11, 14, … | 3n – 1 |
| C | 3, 6, 9, 12, 15, … | 3n |
Design C first appears at position 3, then 6, then 9 — the multiples of 3.
B occurs at 2, 5, 8, 11, 14, … — always one less than a multiple of 3.
nth occurrence of Design A: 3n – 2
Yes, and the design at position 122 is Design B.
So 122 is one less than 123, a multiple of 3
Positions of the form 3n – 1 carry Design B
| Position no. | Quotient on division by 3 | Remainder |
|---|---|---|
| 99 | 33 | 0 |
| 122 | 40 | 2 |
| 148 | 49 | 1 |
Yes. The remainder alone decides the design.
| Remainder on dividing by 3 | Form of the position | Design |
|---|---|---|
| 0 | 3n | C |
| 1 | 3n + 1, i.e. 3(n+1) – 2 | A |
| 2 | 3n + 2, i.e. 3(n+1) – 1 | B |
| Position no. | Quotient on division by 3 | Remainder |
|---|---|---|
| 99 | 33 | 0 |
| 122 | 40 | 2 |
| 148 | 49 | 1 |
| Position | Division by 3 | Remainder | Design |
|---|---|---|---|
| 99 | 99 = 3 × 33 + 0 | 0 | C |
| 122 | 122 = 3 × 40 + 2 | 2 | B |
| 148 | 148 = 3 × 49 + 1 | 1 | A |
122 = 3 × 41 – 1 → Design B
148 = 3 × 50 – 2 → Design A
In-text Questions — Pages 97–99Patterns in a Calendar
| 12 | 13 |
| 19 | 20 |
Both diagonals add up to the same number.
13 + 19 = 32
The two diagonal sums are equal.
Yes — but checking examples can never make us sure, because there are unlimited 2 × 2 squares.
Describe them in words first, then in symbols.
- the number to the right of a is 1 more than it
- the number below a is 7 more than it
- the number diagonal to a is 8 more than it
| a | a + 1 |
| a + 7 | a + 8 |
Diagonal 2: (a + 1) + (a + 7) = a + a + 1 + 7 = 2a + 8
Both diagonal sums are equal — for every value of a.
Take the square with top-left number 25 (so a = 25): the square is 25, 26 over 32, 33.
Diagonal 2: 26 + 32 = 58
Formula: 2a + 8 = 2 × 25 + 8 = 58 ✔
Now take a = 40: the square is 40, 41 over 47, 48.
2a + 8 = 80 + 8 = 88 ✔
= (8 + 22) + (14 + 16) + 15
= 30 + 30 + 15
= 75
And 75 = 5 × 15, five times the centre.
Another plus shape, centred at 26: the numbers are 19 / 25, 26, 27 / 33.
Observation: the total is always 5 times the number in the centre.
Yes, always. Let the centre be a.
| a – 7 | ||
| a – 1 | a | a + 1 |
| a + 7 |
= a + a + a + a + a + (–7 + 7) + (–1 + 1)
= 5a + 0 + 0
= 5a
Any shape that is symmetric about one cell works. Taking the centre as a:
| Shape | Numbers in terms of a | Sum |
|---|---|---|
| Row of 3 (horizontal) | a – 1, a, a + 1 | 3a |
| Column of 3 (vertical) | a – 7, a, a + 7 | 3a |
| Diagonal of 3 (↘) | a – 8, a, a + 8 | 3a |
| The X of 5 (four corners + centre) | a – 8, a – 6, a, a + 6, a + 8 | 5a |
| Full 3 × 3 block | nine numbers around a | 9a |
= 9a
In-text Questions — Pages 100–101Matchstick Patterns
Each step adds one more triangle to the row, using only 2 extra matchsticks each time, because the new triangle shares one side with the previous one.
| Step number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Triangles | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of matchsticks | 3 | 5 | 7 | 9 | 11 | 13 |
Step 4 uses 9 matchsticks, and every step adds 2.
Yes — count how many 2s are added. Step 1 starts with 3, and each later step adds one more 2.
Step 84 = 3 + 2 × 83 = 3 + 166 = 169
Step 108 = 3 + 2 × 107 = 3 + 214 = 217
Step 33 needs 67 matchsticks.
| Step | Number of 2s added | Working | Matchsticks |
|---|---|---|---|
| 33 | 32 | 3 + 2 + 2 + … (32 twos) | 67 |
| 84 | 83 | 3 + 2 × 83 | 169 |
| 108 | 107 | 3 + 2 × 107 | 217 |
Step y has one 3 and (y – 1) twos.
Noticing that the first step also contains a 2 (since 3 = 1 + 2) gives a second form:
Yes to both. Simplify the first expression and it turns into the second.
= 2y + (3 – 2)
= 2y + 1
| y | 3 + 2(y – 1) | 2y + 1 |
|---|---|---|
| 1 | 3 | 3 |
| 5 | 11 | 11 |
| 33 | 67 | 67 |
| Step | Horizontal | Diagonal | Total |
|---|---|---|---|
| 1 | 1 | 2 | 3 |
| 2 | 2 | 3 | 5 |
| 3 | 3 | 4 | 7 |
| 4 | 4 | 5 | 9 |
Step 4: 4 horizontal and 5 diagonal (4 + 5 = 9 ✔)
Horizontal sticks go 1, 2, 3, 4, … and diagonal sticks go 2, 3, 4, 5, … Each set grows by exactly 1 per step.
Diagonal at Step y = y + 1
Sum = y + (y + 1) = 2y + 1 ✔
Figure it Out — Pages 102–105Section 4.5 Pick Patterns and Reveal Relationships
The correct expression is (a) 30x + 20y.
Money from pulao = 20 × y = 20y
Total = 30x + 20y
The correct expression is (a) p + q + r.
One flag per customer
Flags given away = p + q + r
(a) In one full day-and-night the snail gains (u – d) cm. This repeats 10 times.
= 10(u – d) cm = 10u – 10d cm
(b) If d > u, then u – d is negative.
It slips down more each night than it climbs during the day, so it keeps losing ground and will never reach the top.
Work out each week separately — a week has 7 days.
| Week | Daily distance | Distance that week |
|---|---|---|
| 1 | 5 km | 7 × 5 = 35 |
| 2 | (5 + z) km | 7(5 + z) = 35 + 7z |
| 3 | (5 + 2z) km | 7(5 + 2z) = 35 + 14z |
= (35 + 35 + 35) + (7z + 14z)
= 105 + 21z km
Follow each arrow away from the centre oval w + 2, applying one box at a time.
| Path | Steps | Result |
|---|---|---|
| Top right (given) | w + 2 → +3 → w + 5 → ×4 | 4w + 20 |
| Top left | w + 2 → –5 → w – 3 → ×3 | 3w – 9 |
| Bottom right | w + 2 → –4 → w – 2 → ×3 | 3w – 6 |
| Bottom left | w + 2 → –8 → w – 6 → –4 | w – 10 |
Bottom right: (w + 2) – 4 = w – 2, and (w – 2) × 3 = 3w – 6 ✔
Bottom left: (w + 2) – 8 = w – 6, and (w – 6) – 4 = w – 10
Three stations lie between the two towns, so the journey is cut into 4 stretches, with 3 halts.
(b) Total time = 4 × t + 3 × 2 = 4t + 6 minutes
= 4a + 5b + 10
(b) 3(3a – 3b) = 9a – 9b
9a – 8a = a; –9b – 4b = –13b
= a – 13b – 16
(c) 2(2x – 3) = 4x – 6
4x + 8x = 12x; –6 + 12 = 6
= 12x + 6
(d) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12
= 6x + 15
(e) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9
= h + 4
(f) 4(6m – 3n) = 24m – 12n
24m – 3m = 21m; –12n – 8n = –20n; 23 – 18 = 5
= 21m – 20n + 5
(b) (–6f + 13f) + (–8s + 12s) + (19 – 23) = 7f + 4s – 4
(c) 16c – (11 + 9d) = 16c – 11 – 9d
(8d – 9d) + (–14c + 16c) + (9 – 11) = –d + 2c – 2
(d) (6f – 13f) + (8s – 12s) + (–20 + 23) = –7f – 4s + 3
(e) (13m – 13m) + (–12n + 12n) = 0
(f) (–26m + 26m) + (24n – 24n) = 0
“Subtract A from B” means work out B – A. Change the sign of every term of A, then add.
= 6a + 9b – 18 – 9a + 6b – 14 = –3a + 15b – 32
(b) (7y – 10 + 3x) – (–15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y = 18x + 16y – 23
(c) (11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h = –27g + 10h + 2
(d) 6a – (9b + 18) = 6a – 9b – 18
(6a – 9b – 18) – (9a – 6b + 14) = –3a – 3b – 32
(e) (–3y + 8 – 3x) – (10x + 2 + 10y)
= –3y + 8 – 3x – 10x – 2 – 10y = –13x – 13y + 6
(f) (7h – 8g + 20) – (8g + 4h – 10)
= 7h – 8g + 20 – 8g – 4h + 10 = 3h – 16g + 30
(a) 8x + 3y
A fruit seller sells mangoes at ₹x each and bananas at ₹y each. A customer buys 8 mangoes and 3 bananas, so the bill is 8x + 3y rupees.
(b) 15x – 2x
A shopkeeper has a packet of 15 pencils, each costing ₹x. Two of them are damaged and cannot be sold. The money received for the packet is 15x – 2x = 13x rupees.
Each extra fold gives one extra piece.
| Number of folds | 0 | 1 | 2 | 3 | … | 10 |
|---|---|---|---|---|---|---|
| Pieces after one cut | 2 | 3 | 4 | 5 | … | 12 |
Folded r times and cut → r + 2 pieces
The first square needs 4 matchsticks; every square after that shares one side, so it needs only 3 more.
| Squares | 1 | 2 | 3 | 4 | … | 10 |
|---|---|---|---|---|---|---|
| Matchsticks | 4 | 7 | 10 | 13 | … | 31 |
w squares = 4 + 3 × (w – 1)
= 4 + 3w – 3
= 3w + 1 matchsticks
The printed sequence is red, yellow, green, yellow, red, … so the block red – yellow – green – yellow repeats every 4 positions.
| Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Colour | Red | Yellow | Green | Yellow | Red | Yellow | Green | Yellow |
190 = 4 × 47 + 2 → remainder 2 → Yellow
343 = 4 × 85 + 3 → remainder 3 → Green
Expressions for the positions:
| Colour | Positions | Expression |
|---|---|---|
| Red | 1, 5, 9, 13, … | 4n – 3 |
| Green | 3, 7, 11, 15, … | 4n – 1 |
| Yellow | 2, 4, 6, 8, … (every even position) | 2n |
Each step is an X with one centre square and four arms. Step 1 has arms of length 1, Step 2 arms of length 2, and so on.
| Step | Centre | Four arms | Squares |
|---|---|---|---|
| 1 | 1 | 4 × 1 | 5 |
| 2 | 1 | 4 × 2 | 9 |
| 3 | 1 | 4 × 3 | 13 |
| 4 | 1 | 4 × 4 | 17 |
| 10 | 1 | 4 × 10 | 41 |
| 50 | 1 | 4 × 50 | 201 |
Step 4 → 17, Step 10 → 41, Step 50 → 201
Counting vertices: every square has 4 corners, so
= 16n + 4
(This counts 4 corners for every square. Neighbouring squares touch at one corner; if each shared corner point is counted only once, there are 4n fewer: 16n + 4 – 4n = 12n + 4.)
(a) Column expressions (with r the row number):
| Column | Numbers | Expression |
|---|---|---|
| 1 | 1, 5, 9, 13, … | 4(r – 1) + 1 = 4r – 3 |
| 2 | 2, 6, 10, 14, … | 4(r – 1) + 2 = 4r – 2 |
| 3 | 3, 7, 11, 15, … | 4(r – 1) + 3 = 4r – 1 |
| 4 | 4, 8, 12, 16, … | 4(r – 1) + 4 = 4r |
(b) Divide by 4: the remainder is the column (remainder 0 means column 4).
(ii) 147 = 4 × 36 + 3 → row 37, column 3
(iii) 201 = 4 × 50 + 1 → row 51, column 1
(c) The number in row r and column c is
(d) Multiples of 3. Their columns run 3, 2, 1, 4, 3, 2, 1, 4, … repeating.
| Multiple of 3 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 |
|---|---|---|---|---|---|---|---|---|
| Column | 3 | 2 | 1 | 4 | 3 | 2 | 1 | 4 |
Other patterns:
- Every number in column 4 is a multiple of 4.
- Even numbers appear only in columns 2 and 4; odd numbers only in columns 1 and 3.
- Going down any column, the numbers rise by 4 each time.
- Along a row the numbers rise by 1; along a downward diagonal they rise by 5.
Chapter 4 of 8 – Maths
- Previous: Chapter 3 a Peek Beyond the Point
- Next: Chapter 5 Parallel and Intersecting Lines
- All chapters: NCERT Solutions Class 7 Maths
Also for This Chapter
NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter Numbers – Overview
| Class | Class 7 |
| Subject | Maths |
| Chapter | 4 – Expressions Using Letter Numbers |
| Textbook pages | 81 – 105 |
| Questions solved | 107 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 7 Maths Chapter 4 – FAQs
Where can I download NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter Numbers PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 7 Maths book for 2027.
How many questions are solved in NCERT Class 7 Maths Chapter 4?
All 107 questions of the chapter (textbook pages 81 – 105) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.