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Expressions using Letter-Numbers

CBSE · Class 7 · Mathematics

NCERT Solutions for Expressions using Letter-Numbers — CBSE Class 7 Mathematics.

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Figure it Out (Perimeter Formulas and Basic Expressions)

1aWrite the formula for the perimeter of a triangle with all sides equal.Show solution
Given: A triangle with all sides equal (equilateral triangle). Let each side have length ss.

Concept: Perimeter = sum of all sides.

Working:
P=s+s+s=3×s=3sP = s + s + s = 3 \times s = 3s

Formula: P=3sP = 3s

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1bWrite the formula for the perimeter of a regular pentagon.Show solution
Given: A regular pentagon has 5 equal sides. Let each side have length ss.

Concept: Perimeter = number of sides × side length.

Working:
P=s+s+s+s+s=5×s=5sP = s + s + s + s + s = 5 \times s = 5s

Formula: P=5sP = 5s

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1cWrite the formula for the perimeter of a regular hexagon.Show solution
Given: A regular hexagon has 6 equal sides. Let each side have length ss.

Concept: Perimeter = number of sides × side length.

Working:
P=s×6=6sP = s \times 6 = 6s

Formula: P=6sP = 6s

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2Munirathna has a 20 m long pipe. He joins another pipe of some length kk metres to this one. Give the expression for the combined length of the pipe.Show solution
Given: Length of first pipe = 20 m; length of second pipe = kk m.

Concept: Combined length = sum of the two lengths.

Working:
Combined length=20+k metres\text{Combined length} = 20 + k \text{ metres}

Answer: The expression for the combined length is 20+k20 + k metres.

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3What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the table.Show solution
Concept: Total amount = (No. of ₹100 notes × 100) + (No. of ₹20 notes × 20) + (No. of ₹5 notes × 5).

Row 1: 3 notes of ₹100, 5 notes of ₹20, 6 notes of ₹5.
3×100+5×20+6×5=300+100+30=₹4303 \times 100 + 5 \times 20 + 6 \times 5 = 300 + 100 + 30 = ₹430

Row 2: Expression given: 6×100+4×20+3×5=600+80+15=₹6956 \times 100 + 4 \times 20 + 3 \times 5 = 600 + 80 + 15 = ₹695
So: 6 notes of ₹100, 4 notes of ₹20, 3 notes of ₹5.

Row 3: 8 notes of ₹100, 4 notes of ₹20, zz notes of ₹5.
8×100+4×20+z×5=800+80+5z=880+5z8 \times 100 + 4 \times 20 + z \times 5 = 800 + 80 + 5z = 880 + 5z

Row 4: xx notes of ₹100, yy notes of ₹20, zz notes of ₹5.
x×100+y×20+z×5=100x+20y+5zx \times 100 + y \times 20 + z \times 5 = 100x + 20y + 5z

Completed Table:

| 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=₹4303\times100+5\times20+6\times5 = ₹430 |
| 6 | 4 | 3 | 6×100+4×20+3×5=₹6956\times100+4\times20+3\times5 = ₹695 |
| 8 | 4 | zz | 800+80+5z=880+5z800+80+5z = 880+5z |
| xx | yy | zz | 100x+20y+5z100x+20y+5z |

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4Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once running, each kg of grain takes 8 seconds to grind. Which expression describes the time taken to grind yy kg of grain (machine is off initially)?
(a) 10+8+y10+8+y
(b) (10+8)×y(10+8)\times y
(c) 10×8×y10\times8\times y
(d) 10+8×y10+8\times y
(e) 10×y+810\times y+8Show solution
Given: Start-up time = 10 seconds; grinding time per kg = 8 seconds; quantity = yy kg.

Concept: Total time = start-up time + (time per kg × number of kg)
Total time=10+8×y\text{Total time} = 10 + 8 \times y

**Correct option: (d) 10+8×y10 + 8 \times y

Justification:** The machine first takes a fixed 10 seconds to start. Then it grinds yy kg at 8 seconds per kg, giving 8×y8 \times y seconds of grinding. Total = 10+8y10 + 8y seconds.

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5aWrite an algebraic expression for: 5 more than a number.Show solution
Let the number be nn.

Expression: n+5n + 5

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5bWrite an algebraic expression for: 4 less than a number.Show solution
Let the number be nn.

Expression: n−4n - 4

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5cWrite an algebraic expression for: 2 less than 13 times a number.Show solution
Let the number be nn.

Working: 13 times the number = 13n13n; 2 less than that = 13n−213n - 2.

Expression: 13n−213n - 2

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5dWrite an algebraic expression for: 13 less than 2 times a number.Show solution
Let the number be nn.

Working: 2 times the number = 2n2n; 13 less than that = 2n−132n - 13.

Expression: 2n−132n - 13

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6aDescribe a situation corresponding to the algebraic expression 8×x+3×y8 \times x + 3 \times y.Show solution
One possible situation: A shopkeeper sells two types of items. He sells xx items at ₹8 each and yy items at ₹3 each. The total amount earned is 8x+3y8x + 3y rupees.

Another example: A box has xx packets of 8 biscuits each and yy packets of 3 biscuits each. The total number of biscuits is 8x+3y8x + 3y.

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6bDescribe a situation corresponding to the algebraic expression 15×j−2×k15 \times j - 2 \times k.Show solution
One possible situation: A person earns ₹15 for each hour of work (jj hours worked) and pays ₹2 as transport cost for each trip (kk trips made). The net amount remaining is 15j−2k15j - 2k rupees.

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7In a calendar month, if any 2×32\times3 grid of dates is chosen, write expressions for the dates in the blank cells if the bottom middle cell has date ww.Show solution
Given: The bottom middle cell has date ww.

Concept: In a calendar, dates in the same column differ by 7 (one week). Dates in the same row differ by 1.

The 2×32\times3 grid has 2 rows and 3 columns.
- Bottom row: left cell = w−1w-1, middle cell = ww, right cell = w+1w+1
- Top row (one week earlier): left cell = w−8w-8, middle cell = w−7w-7, right cell = w−6w-6

Completed grid:

| w−8w-8 | w−7w-7 | w−6w-6 |
|---|---|---|
| w−1w-1 | ww | w+1w+1 |

The blank cell in the bottom row (right of ww) is w+1w+1, and the top row cells are w−8w-8, w−7w-7, w−6w-6 respectively.

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Revisiting Arithmetic Expressions

1Find the value of 23−10×223 - 10 \times 2.Show solution
Using order of operations (multiplication before subtraction):
23−10×2=23−20=323 - 10 \times 2 = 23 - 20 = 3

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2Find the value of 83+28−13+3283 + 28 - 13 + 32.Show solution
Working left to right:
83+28−13+32=111−13+32=98+32=13083 + 28 - 13 + 32 = 111 - 13 + 32 = 98 + 32 = 130

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3Find the value of 34−14+2034 - 14 + 20.Show solution
Working left to right:
34−14+20=20+20=4034 - 14 + 20 = 20 + 20 = 40

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4Find the value of 42+15−(8−7)42 + 15 - (8 - 7).Show solution
First evaluate the bracket:
8−7=18 - 7 = 1
42+15−1=57−1=5642 + 15 - 1 = 57 - 1 = 56

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5Find the value of 68−(18+13)68 - (18 + 13).Show solution
First evaluate the bracket:
18+13=3118 + 13 = 31
68−31=3768 - 31 = 37

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6Find the value of 7×4+9×67 \times 4 + 9 \times 6.Show solution
Using order of operations (multiplication before addition):
7×4+9×6=28+54=827 \times 4 + 9 \times 6 = 28 + 54 = 82

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7Find the value of 20+8×(16−6)20 + 8 \times (16 - 6).Show solution
First evaluate the bracket:
16−6=1016 - 6 = 10
20+8×10=20+80=10020 + 8 \times 10 = 20 + 80 = 100

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Figure it Out (Simplification of Algebraic Expressions)

1Add the numbers in each picture. Write their corresponding expressions and simplify them. (Note: The figures are not visible, but the method is described.)Show solution
Note: The actual figures are not available in the text. However, the method to solve such problems is:

1. Identify all the numbers/letter-numbers in the picture.
2. Write the expression by adding all of them.
3. Group like terms together (numbers with numbers, same letter-numbers together).
4. Simplify by combining like terms.

Example approach: If a picture contains values aa, 33, aa, 55, bb, bb:
a+3+a+5+b+b=(a+a)+(b+b)+(3+5)=2a+2b+8a + 3 + a + 5 + b + b = (a + a) + (b + b) + (3 + 5) = 2a + 2b + 8

Trying different orders of addition (row-wise, column-wise, diagonal) should give the same simplified result, demonstrating that the order of addition does not change the sum.

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2aSimplify: p+p+p+pp + p + p + p and p+p+p+qp + p + p + q.Show solution
Expression 1: p+p+p+pp + p + p + p
=4p= 4p

Expression 2: p+p+p+qp + p + p + q
=3p+q= 3p + q

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2bSimplify: p+q+p−qp + q + p - q.Show solution
Grouping like terms:
p+q+p−q=(p+p)+(q−q)=2p+0=2pp + q + p - q = (p + p) + (q - q) = 2p + 0 = 2p

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2cSimplify: p−q+p−qp - q + p - q.Show solution
Grouping like terms:
p−q+p−q=(p+p)+(−q−q)=2p−2qp - q + p - q = (p + p) + (-q - q) = 2p - 2q

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2dSimplify: p+q−p+qp + q - p + q.Show solution
Grouping like terms:
p+q−p+q=(p−p)+(q+q)=0+2q=2qp + q - p + q = (p - p) + (q + q) = 0 + 2q = 2q

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2eSimplify: p+q−(p+q)p + q - (p + q).Show solution
Opening the bracket (negative sign outside distributes):
p+q−(p+q)=p+q−p−q=(p−p)+(q−q)=0p + q - (p + q) = p + q - p - q = (p - p) + (q - q) = 0

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2fSimplify: p−q−p−qp - q - p - q.Show solution
Grouping like terms:
p−q−p−q=(p−p)+(−q−q)=0−2q=−2qp - q - p - q = (p - p) + (-q - q) = 0 - 2q = -2q

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2gSimplify: 2d−d−d−d2d - d - d - d.Show solution
Working step by step:
2d−d−d−d=(2d−d)−d−d=d−d−d=0−d=−d2d - d - d - d = (2d - d) - d - d = d - d - d = 0 - d = -d

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2hSimplify: 2d−d−d−c2d - d - d - c.Show solution
Grouping like terms:
2d−d−d−c=(2d−d−d)−c=0−c=−c2d - d - d - c = (2d - d - d) - c = 0 - c = -c

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2iSimplify: 2d−d−(d−c)2d - d - (d - c).Show solution
Opening the bracket:
2d−d−(d−c)=2d−d−d+c=(2d−d−d)+c=0+c=c2d - d - (d - c) = 2d - d - d + c = (2d - d - d) + c = 0 + c = c

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2jSimplify: 2d−(d−d)−c2d - (d - d) - c.Show solution
First evaluate the bracket:
d−d=0d - d = 0
2d−0−c=2d−c2d - 0 - c = 2d - c

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2kSimplify: 2d−d−c−c2d - d - c - c.Show solution
Grouping like terms:
2d−d−c−c=(2d−d)+(−c−c)=d−2c2d - d - c - c = (2d - d) + (-c - c) = d - 2c

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Mind the Mistake, Mend the Mistake

1Is the simplification 3a+2b=53a + 2b = 5 correct? If not, find the correct simplest form.Show solution
Mistake identified: 3a3a and 2b2b are unlike terms (different letter-numbers). They cannot be added to give a plain number like 5. The student incorrectly added the coefficients 3 and 2 and dropped the variables.

Correct simplest form: 3a+2b3a + 2b is already in its simplest form. It cannot be simplified further.

3a+2b\boxed{3a + 2b}

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2Is the simplification 3b−2b−b=03b - 2b - b = 0 correct? If not, find the correct simplest form.Show solution
Checking:
3b−2b−b=(3−2−1)b=0×b=03b - 2b - b = (3 - 2 - 1)b = 0 \times b = 0

This is correct. 3b−2b−b=03b - 2b - b = 0. ✓

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3Is the simplification 6(p+2)=6p+86(p + 2) = 6p + 8 correct? If not, find the correct simplest form.Show solution
Mistake identified: The student applied the distributive property incorrectly. They multiplied 6×p=6p6 \times p = 6p correctly but wrote 6×2=86 \times 2 = 8 instead of 1212.

Correct working:
6(p+2)=6×p+6×2=6p+126(p + 2) = 6 \times p + 6 \times 2 = 6p + 12

Correct simplest form: 6p+126p + 12

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4Is the simplification (4x+3y)−(3x+4y)=x+y(4x + 3y) - (3x + 4y) = x + y correct? If not, find the correct simplest form.Show solution
Mistake identified: When opening the bracket with a negative sign, the sign of 4y4y should change to −4y-4y, but the student wrote +y+y instead of −y-y.

Correct working:
(4x+3y)−(3x+4y)=4x+3y−3x−4y=(4x−3x)+(3y−4y)=x−y(4x + 3y) - (3x + 4y) = 4x + 3y - 3x - 4y = (4x - 3x) + (3y - 4y) = x - y

Correct simplest form: x−yx - y

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5Is the simplification 5−(2−6z)=3−6z5 - (2 - 6z) = 3 - 6z correct? If not, find the correct simplest form.Show solution
Mistake identified: When opening the bracket with a negative sign outside, −6z-6z becomes +6z+6z (negative of negative is positive). The student kept it as −6z-6z.

Correct working:
5−(2−6z)=5−2+6z=3+6z5 - (2 - 6z) = 5 - 2 + 6z = 3 + 6z

Correct simplest form: 3+6z3 + 6z

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6Is the simplification 2+(x+3)=2x−62 + (x + 3) = 2x - 6 correct? If not, find the correct simplest form.Show solution
Mistake identified: The student seems to have multiplied instead of adding, and also made sign errors. Opening a bracket with a positive sign outside does not change any signs.

Correct working:
2+(x+3)=2+x+3=x+52 + (x + 3) = 2 + x + 3 = x + 5

Correct simplest form: x+5x + 5

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7Is the simplification 2y+(3y−6)=−y+62y + (3y - 6) = -y + 6 correct? If not, find the correct simplest form.Show solution
Mistake identified: The student seems to have subtracted instead of adding, and reversed signs. Since the bracket has a positive sign outside, signs inside remain unchanged.

Correct working:
2y+(3y−6)=2y+3y−6=5y−62y + (3y - 6) = 2y + 3y - 6 = 5y - 6

Correct simplest form: 5y−65y - 6

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8Is the simplification 7p−p+5q−2q=7p+3q7p - p + 5q - 2q = 7p + 3q correct? If not, find the correct simplest form.
9Is the simplification 5(2w+3x+4w)=10w+15x+20w5(2w + 3x + 4w) = 10w + 15x + 20w correct? If not, find the correct simplest form.

Figure it Out (Word Problems and Applications)

1One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If xx plates of Jowar roti and yy plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day?
(a) 30x+20y30x + 20y
(b) (30+20)×(x+y)(30 + 20) \times (x + y)
(c) 20x+30y20x + 30y
(d) (30+20)×x+y(30 + 20) \times x + y
(e) 30x−20y30x - 20y
2Pushpita sells champak and marigold flowers. pp customers only bought champak, qq customers only bought marigold, and rr customers bought both. She gave a tiny national flag to every customer. How many flags did she give away?
(a) p+q+rp+q+r
(b) p+q+2rp+q+2r
(c) 2×(p+q+r)2\times(p+q+r)
(d) p+q+r+2p+q+r+2
(e) p+q+r+1p+q+r+1
(f) 2×(p+q)2\times(p+q)
3aA snail climbs up uu cm during the day and slips down dd cm during the night. This happens for 10 days and 10 nights. Write an expression describing how far away the snail is from its starting position.
3bWhat can we say about the snail's movement if d>ud > u?
4Radha cycles 5 km every day in the first week. Every week she increases the daily distance by zz km. How many kilometers would Radha have cycled after 3 weeks?
5In the figure, observe how the expression w+2w + 2 becomes 4w+204w + 20 along one path. Fill in the missing blanks on the remaining paths. (Figure not fully visible — solve based on the described path.)
6aA local train from Yahapur to Vahapur stops at three stations at equal distances. The travel time between consecutive stations is tt minutes. The train stops for 2 minutes at each of the three intermediate stations. If t=4t = 4, what is the time taken to travel from Yahapur to Vahapur?
6bWhat is the algebraic expression for the time taken to travel from Yahapur to Vahapur?
7aSimplify: 3a+9b−6+8a−4b−7a+163a + 9b - 6 + 8a - 4b - 7a + 16.
7bSimplify: 3(3a−3b)−8a−4b−163(3a - 3b) - 8a - 4b - 16.
7cSimplify: 2(2x−3)+8x+122(2x - 3) + 8x + 12.
7dSimplify: 8x−(2x−3)+128x - (2x - 3) + 12.
7eSimplify: 8h−(5+7h)+98h - (5 + 7h) + 9.
7fSimplify: 23+4(6m−3n)−8n−3m−1823 + 4(6m - 3n) - 8n - 3m - 18.
8aAdd: 4d−7c+94d - 7c + 9 and 8c−11+9d8c - 11 + 9d.
8bAdd: −6f+19−8s-6f + 19 - 8s and −23+13f+12s-23 + 13f + 12s.
8cAdd: 8d−14c+98d - 14c + 9 and 16c−(11+9d)16c - (11 + 9d).
8dAdd: 6f−20+8s6f - 20 + 8s and 23−13f−12s23 - 13f - 12s.
8eAdd: 13m−12n13m - 12n and 12n−13m12n - 13m.
8fAdd: −26m+24n-26m + 24n and 26m−24n26m - 24n.
9aSubtract 9a−6b+149a - 6b + 14 from 6a+9b−186a + 9b - 18.
9bSubtract −15x+13−9y-15x + 13 - 9y from 7y−10+3x7y - 10 + 3x.
9cSubtract 17g+9−7h17g + 9 - 7h from 11−10g+3h11 - 10g + 3h.
9dSubtract 9a−6b+149a - 6b + 14 from 6a−(9b+18)6a - (9b + 18).
9eSubtract 10x+2+10y10x + 2 + 10y from −3y+8−3x-3y + 8 - 3x.
9fSubtract 8g+4h−108g + 4h - 10 from 7h−8g+207h - 8g + 20.
10aDescribe a situation corresponding to the algebraic expression 8x+3y8x + 3y.
10bDescribe a situation corresponding to the algebraic expression 15x−2x15x - 2x.
11A straight rope cut once gives 2 pieces. Folded once and cut gives 3 pieces. 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 rr times and cut?
12Look at the matchstick pattern for squares. How many matchsticks are required to make 10 such squares? How many are required to make ww squares?
13A traffic signal changes colours in a sequence. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour. (Note: The figure shows a repeating sequence — assuming the standard sequence: Red, Yellow, Green repeating, i.e., period 3.)
14Observe the pattern of squares. 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?
15aNumbers are written in a 4-column grid (1,2,3,4 in row 1; 5,6,7,8 in row 2; etc.). Give expressions to generate all the numbers in each column (1, 2, 3, 4).
15bIn which row and column will the following numbers appear: (i) 124, (ii) 147, (iii) 201?
15cWhat number appears in row rr and column cc?
15dObserve the positions of multiples of 3 in the 4-column grid. Do you see any pattern? List other patterns you see.

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Frequently Asked Questions

What are the important topics in Expressions using Letter-Numbers for CBSE Class 7 Mathematics?
Expressions using Letter-Numbers covers several key topics that are frequently asked in CBSE Class 7 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Expressions using Letter-Numbers — CBSE Class 7 Mathematics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Expressions using Letter-Numbers Class 7 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Expressions using Letter-Numbers (CBSE Class 7 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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