Expressions using Letter-Numbers — NCERT Solutions

Checking:
$$7p - p + 5q - 2q = (7 - 1)p + (5 - 2)q = 6p + 3q$$

Mistake identified: The student wrote $7p$ instead of $6p$. They forgot to subtract $p$ from $7p$.

Correct simplest form: $6p + 3q$

Checking the distribution:
$$5(2w + 3x + 4w) = 5 \times 2w + 5 \times 3x + 5 \times 4w = 10w + 15x + 20w$$

The distribution is correct, but the expression is not in its simplest form since $10w$ and $20w$ are like terms.

Simplifying further:
$$10w + 15x + 20w = 30w + 15x$$

Correct simplest form: $30w + 15x$

Given: Cost of 1 plate Jowar roti = ₹30; Cost of 1 plate Pulao = ₹20.

Working:

• Amount from Jowar roti = $30 \times x = 30x$
• Amount from Pulao = $20 \times y = 20y$
• Total = $30x + 20y$

Correct option: (a) $30x + 20y$

Justification: Option (b) is wrong because it multiplies the total price by total plates, mixing up the two items. Option (c) swaps the prices. Options (d) and (e) are incorrect operations. Only (a) correctly multiplies each item's price by its respective quantity and adds them.

Given: $p$ customers bought only champak, $q$ customers bought only marigold, $r$ customers bought both.

Concept: Each customer (regardless of what they bought) gets exactly one flag.

Working: Total number of customers = customers who bought only champak + customers who bought only marigold + customers who bought both
$$= p + q + r$$

So total flags given = $p + q + r$.

Correct option: (a) $p + q + r$

Justification: Every customer gets exactly one flag. The total number of customers is $p + q + r$. Option (b) would be correct if $r$ customers each got 2 flags, which is not the case.

Given: Upward movement per day = $u$ cm; downward slip per night = $d$ cm; number of day-night cycles = 10.

Working:

• Total upward movement in 10 days = $10u$ cm
• Total downward slip in 10 nights = $10d$ cm
• Net distance from starting position = $10u - 10d$ cm

Expression: $10u - 10d$ cm (or equivalently $10(u - d)$ cm)

This can also be written as $10(u-d)$ using the distributive property.

Given: $d > u$, meaning the nightly slip is greater than the daily climb.

Analysis:

• Net movement per cycle = $u - d$
• Since $d > u$, we have $u - d < 0$
• After 10 cycles: $10(u - d) < 0$

Conclusion: The snail moves downward (below its starting position) overall. It ends up farther down than where it started. The snail is actually going deeper into the well rather than climbing out.

Given: Week 1 daily distance = 5 km; increase per week = $z$ km; total weeks = 3.

Working:

• Week 1: Daily distance = 5 km; Days in a week = 7; Distance in Week 1 = $7 \times 5 = 35$ km
• Week 2: Daily distance = $5 + z$ km; Distance in Week 2 = $7 \times (5 + z) = 35 + 7z$ km
• Week 3: Daily distance = $5 + 2z$ km; Distance in Week 3 = $7 \times (5 + 2z) = 35 + 14z$ km

Total distance after 3 weeks:
$$35 + (35 + 7z) + (35 + 14z) = 105 + 21z \text{ km}$$

Answer: Radha would have cycled $105 + 21z$ km after 3 weeks.

Given path: $w + 2 \xrightarrow{\times 4} 4(w+2) = 4w + 8 \xrightarrow{+12} 4w + 20$

Note: The figure is not fully visible. Based on the described transformation from $w+2$ to $4w+20$:

$$4w + 20 = 4(w + 5)$$

So one path could be: $w+2 \xrightarrow{+3} w+5 \xrightarrow{\times 4} 4w+20$

Another path: $w+2 \xrightarrow{\times 4} 4w+8 \xrightarrow{+12} 4w+20$

Another path: $w+2 \xrightarrow{+18} w+20 \xrightarrow{+3w} 4w+20$

General principle: Different sequences of operations can lead to the same final expression, demonstrating the flexibility of algebraic manipulation.

Given: $t = 4$ minutes (travel time between consecutive stations); 3 intermediate stations; stop time = 2 minutes at each station.

Diagram: Yahapur — Station 1 — Station 2 — Station 3 — Vahapur

There are 4 travel segments (between 5 points) and 3 stops.

Working:

• Total travel time = $4 \times t = 4 \times 4 = 16$ minutes
• Total stop time = $3 \times 2 = 6$ minutes
• Total time = $16 + 6 = 22$ minutes

Answer: 22 minutes.

Given: Travel time per segment = $t$ minutes; 3 intermediate stations; stop time = 2 minutes each.

Diagram: Yahapur — [t] — St.1 — [t] — St.2 — [t] — St.3 — [t] — Vahapur

• Number of travel segments = 4
• Number of stops = 3

Expression:
$$\text{Total time} = 4t + 3 \times 2 = 4t + 6 \text{ minutes}$$

Answer: The algebraic expression is $4t + 6$ minutes.

Grouping like terms:
$$= (3a + 8a - 7a) + (9b - 4b) + (-6 + 16)$$
$$= 4a + 5b + 10$$

First expand the bracket:
$$3(3a - 3b) = 9a - 9b$$
$$= 9a - 9b - 8a - 4b - 16$$
Grouping like terms:
$$= (9a - 8a) + (-9b - 4b) - 16$$
$$= a - 13b - 16$$

First expand the bracket:
$$2(2x - 3) = 4x - 6$$
$$= 4x - 6 + 8x + 12$$
Grouping like terms:
$$= (4x + 8x) + (-6 + 12)$$
$$= 12x + 6$$

Opening the bracket (negative sign changes signs inside):
$$= 8x - 2x + 3 + 12$$
Grouping like terms:
$$= (8x - 2x) + (3 + 12)$$
$$= 6x + 15$$

Opening the bracket:
$$= 8h - 5 - 7h + 9$$
Grouping like terms:
$$= (8h - 7h) + (-5 + 9)$$
$$= h + 4$$

First expand the bracket:
$$4(6m - 3n) = 24m - 12n$$
$$= 23 + 24m - 12n - 8n - 3m - 18$$
Grouping like terms:
$$= (24m - 3m) + (-12n - 8n) + (23 - 18)$$
$$= 21m - 20n + 5$$

Writing the sum:
$$(4d - 7c + 9) + (8c - 11 + 9d)$$
Grouping like terms:
$$= (4d + 9d) + (-7c + 8c) + (9 - 11)$$
$$= 13d + c - 2$$

Writing the sum:
$$(-6f + 19 - 8s) + (-23 + 13f + 12s)$$
Grouping like terms:
$$= (-6f + 13f) + (-8s + 12s) + (19 - 23)$$
$$= 7f + 4s - 4$$

First simplify the second expression:
$$16c - (11 + 9d) = 16c - 11 - 9d$$
Writing the sum:
$$(8d - 14c + 9) + (16c - 11 - 9d)$$
Grouping like terms:
$$= (8d - 9d) + (-14c + 16c) + (9 - 11)$$
$$= -d + 2c - 2$$

Writing the sum:
$$(6f - 20 + 8s) + (23 - 13f - 12s)$$
Grouping like terms:
$$= (6f - 13f) + (8s - 12s) + (-20 + 23)$$
$$= -7f - 4s + 3$$

Writing the sum:
$$(13m - 12n) + (12n - 13m)$$
Grouping like terms:
$$= (13m - 13m) + (-12n + 12n)$$
$$= 0 + 0 = 0$$

Writing the sum:
$$(-26m + 24n) + (26m - 24n)$$
Grouping like terms:
$$= (-26m + 26m) + (24n - 24n)$$
$$= 0 + 0 = 0$$

Working: $(6a + 9b - 18) - (9a - 6b + 14)$

Opening the bracket:
$$= 6a + 9b - 18 - 9a + 6b - 14$$
Grouping like terms:
$$= (6a - 9a) + (9b + 6b) + (-18 - 14)$$
$$= -3a + 15b - 32$$

Working: $(7y - 10 + 3x) - (-15x + 13 - 9y)$

Opening the bracket:
$$= 7y - 10 + 3x + 15x - 13 + 9y$$
Grouping like terms:
$$= (3x + 15x) + (7y + 9y) + (-10 - 13)$$
$$= 18x + 16y - 23$$

Working: $(11 - 10g + 3h) - (17g + 9 - 7h)$

Opening the bracket:
$$= 11 - 10g + 3h - 17g - 9 + 7h$$
Grouping like terms:
$$= (-10g - 17g) + (3h + 7h) + (11 - 9)$$
$$= -27g + 10h + 2$$

First simplify the expression we are subtracting from:
$$6a - (9b + 18) = 6a - 9b - 18$$

Working: $(6a - 9b - 18) - (9a - 6b + 14)$

Opening the bracket:
$$= 6a - 9b - 18 - 9a + 6b - 14$$
Grouping like terms:
$$= (6a - 9a) + (-9b + 6b) + (-18 - 14)$$
$$= -3a - 3b - 32$$

Working: $(-3y + 8 - 3x) - (10x + 2 + 10y)$

Opening the bracket:
$$= -3y + 8 - 3x - 10x - 2 - 10y$$
Grouping like terms:
$$= (-3x - 10x) + (-3y - 10y) + (8 - 2)$$
$$= -13x - 13y + 6$$

Working: $(7h - 8g + 20) - (8g + 4h - 10)$

Opening the bracket:
$$= 7h - 8g + 20 - 8g - 4h + 10$$
Grouping like terms:
$$= (-8g - 8g) + (7h - 4h) + (20 + 10)$$
$$= -16g + 3h + 30$$

One possible situation: A fruit seller sells mangoes at ₹8 each and bananas at ₹3 each. If he sells $x$ mangoes and $y$ bananas, the total amount earned is $8x + 3y$ rupees.

Another situation: A box contains $x$ packets with 8 pens each and $y$ packets with 3 pens each. The total number of pens is $8x + 3y$.

Note: $15x - 2x = 13x$, so this simplifies to $13x$.

One possible situation: A shopkeeper had $15x$ kg of rice. He sold $2x$ kg of rice. The remaining rice is $15x - 2x = 13x$ kg.

Another situation: A school has 15 rows of $x$ students each. 2 rows of $x$ students each went for a trip. The number of students remaining is $15x - 2x = 13x$.

Observing the pattern:

Pattern: When folded $r$ times, the rope has $2^r$ layers. Cutting through all layers gives $2^r$ cuts, producing $2^r + 1$ pieces.

Verification:

• $r = 0$: $2^0 + 1 = 1 + 1 = 2$ ✓
• $r = 1$: $2^1 + 1 = 2 + 1 = 3$ ✓
• $r = 2$: $2^2 + 1 = 4 + 1 = 5$ ✓

For $r = 10$ folds:
$$\text{Number of pieces} = 2^{10} + 1 = 1024 + 1 = 1025$$

General expression: Number of pieces when folded $r$ times = $2^r + 1$

Observing the pattern:

• 1 square: 4 matchsticks
• 2 squares: 7 matchsticks (each new square shares one side with the previous)
• 3 squares: 10 matchsticks
• Each additional square adds 3 matchsticks.

Pattern: For $w$ squares:
$$\text{Matchsticks} = 4 + 3(w - 1) = 4 + 3w - 3 = 3w + 1$$

Verification:

• $w = 1$: $3(1) + 1 = 4$ ✓
• $w = 2$: $3(2) + 1 = 7$ ✓
• $w = 3$: $3(3) + 1 = 10$ ✓

For 10 squares:
$$3 \times 10 + 1 = 31 \text{ matchsticks}$$

General formula: Number of matchsticks for $w$ squares $= 3w + 1$

Note: The figure is not visible. Based on standard traffic signal patterns, assuming the sequence repeats with a period of 3: Position 1 = Red, Position 2 = Yellow (or Orange), Position 3 = Green, Position 4 = Red, and so on.

Finding colours using remainder when divided by 3:

Position 90:
$$90 \div 3 = 30 \text{ remainder } 0$$
Remainder 0 corresponds to position 3 → Green

Position 190:
$$190 \div 3 = 63 \text{ remainder } 1$$
Remainder 1 corresponds to position 1 → Red

Position 343:
$$343 \div 3 = 114 \text{ remainder } 1$$
Remainder 1 corresponds to position 1 → Red

Expressions for positions of each colour (where $n$ is any whole number, $n \geq 0$):

• Red (position 1, 4, 7, ...): $3n + 1$
• Yellow (position 2, 5, 8, ...): $3n + 2$
• Green (position 3, 6, 9, ...): $3n + 3$ or $3n$ (for $n \geq 1$)

Note: The figure is not fully visible. Based on a common pattern where Step $n$ has $n^2$ squares (or $2n-1$ squares in a cross pattern), assuming the pattern is: Step 1 = 1 square, Step 2 = 4 squares, Step 3 = 9 squares (i.e., $n^2$ squares at Step $n$):

Number of squares at Step $n$: $n^2$

• Step 4: $4^2 = 16$ squares
• Step 10: $10^2 = 100$ squares
• Step 50: $50^2 = 2500$ squares

General formula: Number of squares at Step $n$ = $n^2$

Number of vertices: Each unit square has 4 vertices, but vertices are shared between adjacent squares. For a grid of $n \times n$ squares, the number of vertices (grid points) = $(n+1)^2$.

Formula for vertices: $(n+1)^2 = n^2 + 2n + 1$

Note: If the pattern in the figure is different (e.g., a staircase or cross), the formula would change accordingly. The method remains the same — observe the pattern, find the rule, and express it algebraically.

Observing the grid:

• Row 1: 1, 2, 3, 4
• Row 2: 5, 6, 7, 8
• Row 3: 9, 10, 11, 12
• Row $n$: $4n-3$, $4n-2$, $4n-1$, $4n$

Expressions for each column (where $n = 1, 2, 3, \ldots$):

• Column 1: $4n - 3$ → gives 1, 5, 9, 13, ...
• Column 2: $4n - 2$ → gives 2, 6, 10, 14, ...
• Column 3: $4n - 1$ → gives 3, 7, 11, 15, ...
• Column 4: $4n$ → gives 4, 8, 12, 16, ...

Method: Divide the number by 4.

• The quotient (rounded up) gives the row number.
• The remainder tells the column: remainder 1 → Col 1, remainder 2 → Col 2, remainder 3 → Col 3, remainder 0 → Col 4.

(i) 124:
$$124 \div 4 = 31 \text{ remainder } 0$$
Row = 31, Column = 4

(ii) 147:
$$147 \div 4 = 36 \text{ remainder } 3$$
Row = 37, Column = 3

(iii) 201:
$$201 \div 4 = 50 \text{ remainder } 1$$
Row = 51, Column = 1

Working:

• Row $r$ starts with the number $4(r-1) + 1 = 4r - 3$.
• Column $c$ is $c-1$ positions after the start of the row.

Number at row $r$, column $c$:
$$= 4(r-1) + c = 4r - 4 + c$$

Verification:

• Row 1, Col 1: $4(1) - 4 + 1 = 1$ ✓
• Row 2, Col 3: $4(2) - 4 + 3 = 7$ ✓
• Row 3, Col 4: $4(3) - 4 + 4 = 12$ ✓

Answer: The number at row $r$ and column $c$ is $4r - 4 + c$ (or equivalently $4(r-1) + c$).

Multiples of 3 in the grid:
3 (Row 1, Col 3), 6 (Row 2, Col 2), 9 (Row 3, Col 1), 12 (Row 3, Col 4), 15 (Row 4, Col 3), 18 (Row 5, Col 2), 21 (Row 6, Col 1), 24 (Row 6, Col 4)...

Pattern for multiples of 3: They appear in columns 3, 2, 1, 4, 3, 2, 1, 4, ... repeating with period 4. Every 4th multiple of 3 (i.e., multiples of 12) appears in Column 4.

Other patterns observed:

1. Column 4 contains all multiples of 4 (4, 8, 12, 16, ...).
2. Column 1 contains numbers of the form $4n+1$ (1, 5, 9, 13, ...) — numbers that give remainder 1 when divided by 4.
3. Each row contains 4 consecutive numbers.
4. Numbers in the same column differ by 4 (e.g., Column 2: 2, 6, 10, 14, ... each differing by 4).
5. Odd numbers appear in columns 1 and 3; even numbers appear in columns 2 and 4.
6. The sum of numbers in any row $r$ = $4r - 3 + 4r - 2 + 4r - 1 + 4r = 16r - 6$.