Questions NCERT Solutions Class 7 Maths Part 1 Chapter 4 Expressions using Letter-Numbers

Final answer: (a) $3a$ units, (b) $5a$ units, (c) $6a$ units.

1. Let the common side length be $a$ units.
2. A triangle with all sides equal has $3$ equal sides, so its perimeter is $3a$.
3. A regular pentagon has $5$ equal sides, so its perimeter is $5a$.
4. A regular hexagon has $6$ equal sides, so its perimeter is $6a$.

For a regular polygon, perimeter equals number of equal sides times the length of each side.

Final answer: (a) $3a$ units, (b) $5a$ units, (c) $6a$ units.

Final answer: $(20+k)$ metres.

1. The first pipe is $20$ m long.
2. The second pipe has length $k$ m.
3. Combined length means first length plus second length.
4. So the expression is $20+k$ metres.

A total length is found by adding the known length and the unknown length.

Final answer: $(20+k)$ metres.

Final answer: Rs. $430$, Rs. $695$, Rs. $(880+5z)$, and Rs. $(100x+20y+5z)$.

1. For $3,5,6$ notes, total $=3\times100+5\times20+6\times5=430$.
2. For $6,4,3$ notes, total $=6\times100+4\times20+3\times5=695$.
3. For $8,4,z$ notes, total $=8\times100+4\times20+5z=880+5z$.
4. For $x,y,z$ notes, total $=100x+20y+5z$.

Total money equals number of each note multiplied by its value, then all values are added.

Final answer: Rs. $430$, Rs. $695$, Rs. $(880+5z)$, and Rs. $(100x+20y+5z)$.

Final answer: Option (d), $10+8y$.

1. The machine takes $10$ seconds to start. This is a fixed time.
2. Each kg takes $8$ seconds, so $y$ kg takes $8y$ seconds.
3. Total time equals starting time plus grinding time.
4. Therefore, total time $=10+8y$.

A fixed starting time is added once. The grinding time depends on the number of kilograms.

Final answer: Option (d), $10+8y$.

Final answer: $(a)\ d+5$, $(b)\ d-4$, $(c)\ 13d-2$, $(d)\ 2d-13$.

1. Let the number be $d$.
2. Five more than the number is $d+5$.
3. Four less than the number is $d-4$.
4. Two less than $13$ times the number is $13d-2$.
5. Thirteen less than $2$ times the number is $2d-13$.

Use a letter to stand for the number, then translate each phrase in order.

Final answer: $(a)\ d+5$, $(b)\ d-4$, $(c)\ 13d-2$, $(d)\ 2d-13$.

Final answer: One valid pair of situations is: cost of $8$ pens and $3$ notebooks; chairs made minus chairs broken.

1. For $8x+3y$, suppose a shop sells one pen for Rs. $x$ and one notebook for Rs. $y$.
2. If Abha buys $8$ pens and $3$ notebooks, the total cost is $8x+3y$.
3. For $15j-2k$, suppose a factory makes $15$ chairs each day for $j$ days.
4. If $2$ chairs break each day for $k$ days, the good chairs remaining can be written as $15j-2k$.

An algebraic expression can describe a real situation when letters stand for quantities.

Final answer: One valid pair of situations is: cost of $8$ pens and $3$ notebooks; chairs made minus chairs broken.

Final answer: Top row: $w-8,\ w-7,\ w-6$; bottom row: $w-1,\ w,\ w+1$.

1. The bottom row is $w-1,\ w,\ w+1$.
2. The top-left cell is $7$ days before $w-1$, so it is $w-8$.
3. The top-middle cell is $7$ days before $w$, so it is $w-7$.
4. The top-right cell is $7$ days before $w+1$, so it is $w-6$.

In a calendar, moving right adds $1$ and moving one row up subtracts $7$.

Final answer: Top row: $w-8,\ w-7,\ w-6$; bottom row: $w-1,\ w,\ w+1$.

Final answer: (i) $10y+2x-4$, (ii) $8p+12q+2$, (iii) $-20g+60k$.

1. First picture: $5y+x+x+5y-6+2=10y+2x-4$.
2. Second picture: four $2p$ terms give $8p$, four $3q$ terms give $12q$, and constants add to $2$.
3. So the second total is $8p+12q+2$.
4. Third picture: four $-5g$ terms give $-20g$ and twelve $5k$ terms give $60k$.

Like terms can be grouped and added. Unlike letter terms cannot be combined.

Final answer: (i) $10y+2x-4$, (ii) $8p+12q+2$, (iii) $-20g+60k$.

Final answer: $4p, 3p+q, 2p, 2p-2q, 2q, 0, -2q, -d, -c, c, 2d-c, d-2c$.

1. $p+p+p+p=4p$ and $p+p+p+q=3p+q$.
2. $p+q+p-q=2p$ and $p-q+p-q=2p-2q$.
3. $p+q-p+q=2q$, $p+q-(p+q)=0$, and $p-q-p-q=-2q$.
4. $2d-d-d-d=-d$ and $2d-d-d-c=-c$.
5. $2d-d-(d-c)=c$, $2d-(d-d)-c=2d-c$, and $2d-d-c-c=d-2c$.

Simplification means removing brackets carefully and then combining like terms.

Final answer: $4p, 3p+q, 2p, 2p-2q, 2q, 0, -2q, -d, -c, c, 2d-c, d-2c$.

Final answer: $3a+2b, 0, 6p+12, x-y, 3+6z, x+5, 5y-6, 6p+3q, 30w+15x, 3(j+2k+3h+4), 8r+12s+20$.

1. $3a+2b$ stays $3a+2b$ because $a$ and $b$ are unlike terms.
2. $3b-2b-b=0$.
3. $6(p+2)=6p+12$.
4. $(4x+3y)-(3x+4y)=x-y$.
5. $5-(2-6z)=3+6z$ and $2+(x+3)=x+5$.
6. $2y+(3y-6)=5y-6$ and $7p-p+5q-2q=6p+3q$.
7. $5(2w+3x+4w)=30w+15x$.
8. $3j+6k+9h+12=3(j+2k+3h+4)$.
9. $4(2r+3s+5)=8r+12s+20$.

A simplest form keeps unlike terms separate, distributes multiplication correctly, and handles bracket signs carefully.

Final answer: $3a+2b, 0, 6p+12, x-y, 3+6z, x+5, 5y-6, 6p+3q, 30w+15x, 3(j+2k+3h+4), 8r+12s+20$.

Final answer: The number of letter-numbers is less than or equal to the number of terms.

1. In $3a+2b$, there are $2$ terms and $2$ letter-numbers.
2. In $6p+12$, there are $2$ terms but only $1$ letter-number because $12$ is a constant.
3. In $3j+6k+9h+12$, there are $4$ terms and $3$ letter-numbers.
4. Across the corrected forms, the number of letter-numbers is not more than the number of terms.

A term may contain one letter-number, more than one letter-number, or no letter-number. Constant terms have no letter-number.

Final answer: The number of letter-numbers is less than or equal to the number of terms.

Final answer: Option (a), $30x+20y$.

1. Cost of $x$ plates of Jowar roti is $30x$.
2. Cost of $y$ plates of Pulao is $20y$.
3. Total amount is $30x+20y$.
4. The other options mix the prices and variables incorrectly.

Total earning equals price per item multiplied by number of items, added across item types.

Final answer: Option (a), $30x+20y$.

Final answer: Option (a), $p+q+r$.

1. The group $p$ has customers who bought only champak.
2. The group $q$ has customers who bought only marigold.
3. The group $r$ has customers who bought both, but each such person is still one customer.
4. So the number of flags is $p+q+r$.

The total number of customers is found by adding the three non-overlapping groups.

Final answer: Option (a), $p+q+r$.

Final answer: $(a)\ 10(u-d)$ cm. $(b)$ If $d>u$, the snail slips down more than it climbs.

1. In one day and night, the snail gains $u$ cm and loses $d$ cm.
2. Net movement in one cycle is $u-d$ cm.
3. For $10$ cycles, the movement is $10(u-d)$ cm.
4. If $d>u$, then $u-d$ is negative, so the snail moves below its starting position.

One full day-night cycle changes the position by $u-d$ cm.

Final answer: $(a)\ 10(u-d)$ cm. $(b)$ If $d>u$, the snail slips down more than it climbs.

Final answer: $105+21z$ km.

1. Week 1 daily distance is $5$ km, so week 1 total is $7\times5=35$ km.
2. Week 2 daily distance is $5+z$ km, so week 2 total is $7(5+z)$.
3. Week 3 daily distance is $5+2z$ km, so week 3 total is $7(5+2z)$.
4. Add: $35+7(5+z)+7(5+2z)=105+21z$.

Weekly distance equals daily distance multiplied by $7$ days.

Final answer: $105+21z$ km.

Final answer: See the simplified list above: it matches the NCERT answer key for Q7 to Q15.

1. Simplifying gives: $4a+5b+10$, $a-13b-16$, $12x+6$, $6x+15$, $h+4$, and $5+21m-20n$.
2. Adding pairs gives: $13d+c-2$, $7f+4s-4$, $2c-d-2$, $-7f-4s+3$, $0$, and $0$.
3. Subtracting as asked gives: $-3a+15b-32$, $16y+18x-23$, $10h-27g+2$, $-(3a+3b+32)$, $-13y-13x+6$, and $3h-16g+30$.
4. Rope pieces after $r$ folds and one cut: $r+2$.
5. For $w$ joined squares, matchsticks $=4+3(w-1)$.
6. Traffic signal positions: red at $4n-3$, green at $4n-1$, yellow at $2n$.
7. Square-pattern formula: squares $=4n+1$ and vertices $=16n+4$.
8. In a four-column grid, the number in row $r$, column $c$ is $4(r-1)+c$.

Long expression questions use the same rules: distribute, remove brackets, collect like terms, and identify the pattern.

Final answer: See the simplified list above: it matches the NCERT answer key for Q7 to Q15.