CBSE Class 7 Mathematics Expressions using Letter Numbers Worksheet Set 01

Long Answer Questions

Question 1. In a stationery shop one notebook costs ₹x and one pen costs ₹y. Ankita buys 7 notebooks and 4 pens.
(i) Write algebraic expressions for the cost of the notebooks and for the cost of the pens, and then write the total bill as a sum of terms.
(ii) Find the total bill when x = 45 and y = 12.
(iii) Can the expression for the total bill be simplified into a single term? Give a reason for your answer.
Answer:
(i) One notebook costs ₹x, so 7 notebooks cost 7 × x = ₹7x. One pen costs ₹y, so 4 pens cost 4 × y = ₹4y.
Total bill = ₹(7x + 4y). Its terms are 7x and 4y; the coefficient 7 counts the notebooks and 4 counts the pens.
(ii) Put x = 45 and y = 12: notebooks = 7 × 45 = 315; pens = 4 × 12 = 48.
Total bill = 315 + 48 = ₹363.
(iii) No. 7x and 4y are unlike terms because they have different letter-numbers. Only like terms can be added into one term, so 7x + 4y is already in its simplest form.
In simple words: Cost = number of items × price of one; notebooks and pens are added but cannot be merged.

Teacher's Note:
1. A notebook and a pen have different prices, so different letters are used.
2. Substitute only after writing the expression.

Question 2. Letter-numbers help us write a single formula that works for every figure of a given kind.
(i) Taking a as the length of one side, write formulas for the perimeter of an equilateral triangle, a regular pentagon and a regular hexagon.
(ii) Write the formula for the perimeter of a square of side q, and use it to find the side of a square whose perimeter is 36 cm.
(iii) Which has the greater perimeter: a regular hexagon of side 7 cm or a regular pentagon of side 9 cm? Show your working.
Answer:
(i) In a regular figure all sides are equal, so perimeter = number of sides × side. With side a:
Equilateral triangle: 3a; regular pentagon: 5a; regular hexagon: 6a.
(ii) A square has 4 equal sides, so its perimeter is 4q. If the perimeter is 36 cm, then 4q = 36, so q = 36 ÷ 4 = 9. The side is 9 cm.
(iii) Hexagon of side 7 cm: perimeter = 6 × 7 = 42 cm. Pentagon of side 9 cm: perimeter = 5 × 9 = 45 cm.
Since 45 > 42, the regular pentagon of side 9 cm has the greater perimeter, by 3 cm. More sides do not always mean a larger perimeter; the side length matters too.
In simple words: Perimeter of a regular figure = number of sides × side length.

Teacher's Note:
1. One formula covers every figure of that kind.
2. Compare by working out both perimeters.

Teacher's Note:
1. Count the shared stick only once.
2. Test the formula on a step you can see.

Question 4. Simplify each expression and name the idea you use at each stage.
(i) 9a + 4a − 3a
(ii) 6x + 5y − 2x + 3y
(iii) (7m + 4n) − (3m + n)
Also verify your answer to part (iii) by replacing m by 2 and n by 3 in both the original and the simplified expression.
Answer:
(i) All three are like terms, so add their coefficients: 9a + 4a − 3a = (9 + 4 − 3)a = 10a.
(ii) Swap and group the terms so that like terms stand together: 6x − 2x + 5y + 3y = (6 − 2)x + (5 + 3)y = 4x + 8y. The x and y terms are unlike, so two terms remain.
(iii) A minus sign before a bracket changes the sign of every term inside it: 7m + 4n − 3m − n = (7 − 3)m + (4 − 1)n = 4m + 3n.
Verification with m = 2, n = 3: original = (14 + 12) − (6 + 3) = 26 − 9 = 17; simplified = 8 + 9 = 17. Both give 17, so the simplification is correct.
In simple words: Combine like terms, rearranging and removing brackets first where needed.

Teacher's Note:
1. n means 1n.
2. Name the idea: combining like terms, swapping and grouping, removing brackets.

Question 5. Krithika has x notes of ₹100, y notes of ₹50 and z notes of ₹10 in her purse.
(i) Write an algebraic expression for the total amount of money in her purse.
(ii) Find the total amount when x = 7, y = 4 and z = 6.
(iii) On another day she has only 5 notes of ₹100 and some notes of ₹10, and the total is ₹650. Write the expression for this situation and find how many ₹10 notes she has.
Answer:
(i) x notes of ₹100 are worth 100x; y notes of ₹50 are worth 50y; z notes of ₹10 are worth 10z.
Total amount = ₹(100x + 50y + 10z).
(ii) 100 × 7 + 50 × 4 + 10 × 6 = 700 + 200 + 60 = ₹960.
(iii) Now x = 5 and y = 0, so the amount is 100 × 5 + 10z = 500 + 10z.
500 + 10z = 650, so 10z = 150 and z = 15. She has 15 notes of ₹10.
Check: 500 + 10 × 15 = 650.
In simple words: Multiply each note value by the number of notes and add.

Teacher's Note:
1. Each kind of note needs its own letter.
2. In (iii), find what 10z must be first.

Question 6. A metro train runs from Pathanpur to Vahapur and halts at 2 stations on the way. The three stretches between stops are of equal length and each stretch takes m minutes. The train halts for 3 minutes at each of the 2 stations.
(i) Draw a rough line diagram of the journey and write an expression for the total time taken from Pathanpur to Vahapur.
(ii) Simplify the expression and find the total time when m = 7.
(iii) On a slow day the whole journey took 36 minutes. Find the value of m on that day.
Answer:
(i) A rough line diagram of the route:

There are 3 equal stretches of m minutes each and 2 halts of 3 minutes each, so the total time in minutes is m + 3 + m + 3 + m.
(ii) In simplest form: 3m + 6 minutes. For m = 7: 3 × 7 + 6 = 21 + 6 = 27 minutes.
(iii) 3m + 6 = 36, so 3m = 30 and m = 10. Each stretch took 10 minutes that day.
Check: 3 × 10 + 6 = 36 minutes.
In simple words: Add the time for three stretches and two halts.

Teacher's Note:
1. The diagram helps you count stretches and halts correctly.
2. Halts are only at the 2 middle stations.

Question 7. The expression w + 3 is put through a chain of operations. The ovals hold expressions and the boxes hold operations.
Path 1: start with w + 3, multiply by 2, then subtract 4, then add 5.
Path 2: start with w + 3, add 7, then multiply by 3.
(i) Write the expression in each box of Path 1 in its simplest form.
(ii) Write the expression in each box of Path 2 in its simplest form.
(iii) Take w = 5 and check both your final expressions by working through the chains step by step. Then answer: is 2(w + 3) the same as 2w + 3?
Answer:
(i) Path 1: After × 2: 2(w + 3) = 2w + 6. After − 4: 2w + 6 − 4 = 2w + 2. After + 5: 2w + 2 + 5 = 2w + 7.

(ii) Path 2: After + 7: w + 3 + 7 = w + 10. After × 3: 3(w + 10) = 3w + 30.

(iii) With w = 5, the start is 5 + 3 = 8.
Path 1: 8 × 2 = 16, 16 − 4 = 12, 12 + 5 = 17. Formula: 2 × 5 + 7 = 17. They agree.
Path 2: 8 + 7 = 15, 15 × 3 = 45. Formula: 3 × 5 + 30 = 45. They agree.
Is 2(w + 3) the same as 2w + 3? No. 2(w + 3) means twice the whole of w + 3, so both terms are doubled, giving 2w + 6. For w = 5, 2(w + 3) = 16 but 2w + 3 = 13.
In simple words: Do each operation on the whole expression and simplify after every step.

Teacher's Note:
1. Multiplying a bracket multiplies every term inside.
2. Checking with a number catches slips.

Question 8. Somjit sees a border on a saree in which three designs A, B and C are printed again and again in that order, starting with A at position 1.
(i) Write expressions for the position at which each design appears for the nth time.
(ii) Which design appears at position 87, and which one appears at position 100? Show how you decide.
(iii) At which position does Design B appear for the 25th time?
Answer:
(i) The designs repeat in a cycle of 3, so Design C is at positions 3, 6, 9, …, the multiples of 3.
Design C, nth time: 3n; Design B: 3n − 1 (one place before C); Design A: 3n − 2 (two places before C).
Check: for n = 2 these give 6, 5 and 4, which is exactly the second A, B, C group.
(ii) Divide the position by 3: remainder 0 means C, remainder 2 means B and remainder 1 means A.
87 = 3 × 29, remainder 0, so position 87 has Design C.
100 = 3 × 33 + 1, remainder 1, so position 100 has Design A. Indeed 3n − 2 = 100 gives n = 34, the 34th A.
(iii) Using 3n − 1 with n = 25: 3 × 25 − 1 = 74. Design B appears for the 25th time at position 74.
In simple words: C is at multiples of 3, B one before, A two before.

Teacher's Note:
1. The remainder on dividing by 3 tells the design.
2. Check the rules with the first few positions.

Question 9. In a calendar the dates in a row increase by 1 and the dates in a column increase by 7. Consider any 2 × 2 square of dates and let the top left date be a.
(i) Write the other three dates in terms of a, explaining each one in words.
(ii) Find the two diagonal sums, simplify each of them, and hence explain why the two diagonal sums are equal for every 2 × 2 square. Verify once by taking a = 14.
(iii) Now take a plus-shaped set of five dates: a date a together with the dates immediately above, below, to the left and to the right of it. Show using letter-numbers that the sum of these five dates is always 5 times the middle date, and check it for a middle date of 15.
Answer:
(i) The date to its right is 1 more: a + 1. The date just below it is in the next week, so 7 more: a + 7. The date diagonally below to the right is 1 more than a + 7: a + 8.
(ii) One diagonal: a + (a + 8) = 2a + 8. Other diagonal: (a + 1) + (a + 7) = 2a + 8.
Both diagonals give the same expression 2a + 8. Since a stands for any date, this one piece of algebra covers every 2 × 2 square at once, so the diagonal sums are always equal.
Check with a = 14: the dates are 14, 15, 21, 22. 14 + 22 = 36, 15 + 21 = 36 and 2 × 14 + 8 = 36.
(iii) With a in the middle, the date above is a − 7, below is a + 7, left is a − 1 and right is a + 1.
Sum = (a − 7) + (a − 1) + a + (a + 1) + (a + 7) = 5a + (−7 − 1 + 1 + 7) = 5a.
So the sum is always 5 times the middle date. Check with middle date 15: 8 + 14 + 15 + 16 + 22 = 75 = 5 × 15.
In simple words: Write every date using a, then simplify; the extra numbers cancel out.

Teacher's Note:
1. Right adds 1, down adds 7.
2. Algebra proves it for every square, not just one.

Question 10. The whole numbers 1, 2, 3, 4, … are written in an endless grid of 4 columns: 1, 2, 3, 4 in row 1; 5, 6, 7, 8 in row 2; 9, 10, 11, 12 in row 3; and so on.
(i) Taking r as the row number, write an expression that generates every number of Column 1, of Column 2, of Column 3 and of Column 4. Hence write the number that appears in row r and column c.
(ii) Find the row and the column in which 86 appears, and the row and the column in which 143 appears.
(iii) Which number lies in row 30, column 3? Also state one pattern you notice about the numbers of Column 4, giving a reason using your expression.
Answer:
(i) Each row holds 4 numbers, so before row r begins, 4(r − 1) numbers are already written.
Column 1: 4(r − 1) + 1 = 4r − 3; Column 2: 4(r − 1) + 2 = 4r − 2; Column 3: 4(r − 1) + 3 = 4r − 1; Column 4: 4(r − 1) + 4 = 4r.
In general, the number in row r and column c is 4(r − 1) + c.
(ii) Divide by 4 and use the quotient and remainder.
86 = 4 × 21 + 2, so r − 1 = 21 and c = 2: row 22, column 2.
143 = 4 × 35 + 3, so r − 1 = 35 and c = 3: row 36, column 3.
(iii) 4(30 − 1) + 3 = 4 × 29 + 3 = 116 + 3 = 119.
Pattern: every number in Column 4 is 4r, so Column 4 holds only multiples of 4. Similarly, Column 2 holds 4r − 2, numbers that are even but never multiples of 4, while Columns 1 and 3 hold only odd numbers.
In simple words: Count the numbers in the earlier rows, then add the column number.

Teacher's Note:
1. Remainder 0 would mean Column 4.
2. Check each answer by working forwards.

Case Study Based Questions

Question 11. The auto-rickshaw fare in a town. An auto-rickshaw in Haripur runs on a meter. The meter starts at ₹25, which covers the first kilometre of the ride. After that, ₹12 is added for every extra kilometre travelled. Deepa takes an auto from her house to the railway station every Sunday, and her younger brother Arjun likes to guess the fare before the meter shows it. Arjun says that instead of guessing again and again, they should write one rule using a letter-number so that the fare for any ride can be worked out at once. Their teacher agrees and asks them to use the letter n for the number of extra kilometres travelled after the first kilometre.
(i) What is the fare for a ride of exactly 1 km?
(ii) Write an algebraic expression for the fare of a ride of n extra kilometres.
(iii) Use the expression to find the fare for a ride of 9 km.
(iv) One day Deepa paid ₹97. How many kilometres did she travel?
Answer:
(i) A ride of exactly 1 km has no extra kilometre, so the fare is the starting amount, ₹25.
(ii) Each extra kilometre adds ₹12, so n extra kilometres add 12n. Fare = ₹(25 + 12n).
(iii) A 9 km ride has 9 − 1 = 8 extra kilometres, so n = 8. Fare = 25 + 12 × 8 = 25 + 96 = ₹121.
(iv) 25 + 12n = 97, so 12n = 72 and n = 6. With 6 extra kilometres the ride was 6 + 1 = 7 km. Check: 25 + 12 × 6 = 97.
In simple words: Fixed ₹25 for the first km plus ₹12 for each extra km.

Teacher's Note:
1. n counts only the extra kilometres.
2. Add back the first kilometre at the end of (iv).

Question 12. The school canteen bill. In the Class 7 canteen, one samosa costs ₹15 and one glass of juice costs ₹20. On a busy afternoon the canteen keeper has to add up many bills quickly, so she writes a single rule on a card: she uses x for the number of samosas and y for the number of glasses of juice in one bill. She says that once the rule is written with letter-numbers, she only has to replace x and y by the numbers of items ordered and the bill is ready. Her helper copies the rule on a chart for everyone to see.
(i) Write an expression for the cost of x samosas.
(ii) Write an expression for the total bill of x samosas and y glasses of juice.
(iii) Find the bill for 4 samosas and 3 glasses of juice.
(iv) What is the coefficient of y in the expression, and what does it stand for in this situation?
Answer:
(i) One samosa costs ₹15, so x samosas cost 15 × x = ₹15x.
(ii) Adding the cost of the juice, the total bill is ₹(15x + 20y).
(iii) Put x = 4 and y = 3: 15 × 4 + 20 × 3 = 60 + 60 = ₹120.
(iv) The coefficient of y is 20. It is the price in rupees of one glass of juice, that is, the amount added to the bill for each extra glass.
In simple words: Bill = 15 × samosas + 20 × glasses of juice.

Teacher's Note:
1. The coefficient here is a price.
2. x and y are unlike terms, so the bill stays as two terms.

Question 13. A matchstick border. Vinay makes a border for his craft chart by joining squares in a row with matchsticks. Each square has sides of one matchstick, and two squares next to each other share one matchstick, as shown below. Vinay first counts the matchsticks one by one, but his sister tells him that a formula would be faster, especially for a long border.
(i) How many matchsticks are needed to make 4 squares?
(ii) Describe the pattern in words: how many matchsticks does the first square need, and how many does each extra square need?
(iii) Write an algebraic expression for the number of matchsticks needed to make w squares, and simplify it.
(iv) How many matchsticks are needed for a border of 20 squares?
Answer:
(i) For 4 squares: 4 + 3 + 3 + 3 = 13 matchsticks.
(ii) The first square needs 4 matchsticks. Every extra square uses one matchstick that is already there, so it needs only 3 new matchsticks.
(iii) For w squares: the starting 4 and then (w − 1) threes: 4 + 3(w − 1) = 4 + 3w − 3 = 3w + 1.
(iv) For w = 20: 3 × 20 + 1 = 61 matchsticks.
Check with the picture: w = 3 gives 3 × 3 + 1 = 10, which is what the third drawing shows.
In simple words: 4 sticks for the first square and 3 for each one after it.

Teacher's Note:
1. Shared sticks are counted once.
2. Test the formula on the drawings.

Question 14. Meena's savings plan. Meena already has ₹150 in her money box. From the first week of April she decides to put in ₹40 every week out of her pocket money, and she does not take any money out. Her father tells her that if she writes the rule with a letter-number, she can find the amount in the box at the end of any week without adding ₹40 again and again. Meena uses the letter w for the number of weeks for which she has been saving.
(i) How much money is in the box at the end of 3 weeks?
(ii) Write an algebraic expression for the amount in the box at the end of w weeks.
(iii) Use the expression to find the amount at the end of 10 weeks.
(iv) At the end of which week will the box hold ₹470?
Answer:
(i) In 3 weeks she adds 3 × 40 = 120 rupees, so the box holds 150 + 120 = ₹270.
(ii) Amount after w weeks = ₹(150 + 40w).
(iii) For w = 10: 150 + 40 × 10 = 150 + 400 = ₹550.
(iv) 150 + 40w = 470, so 40w = 320 and w = 8. The box holds ₹470 at the end of the 8th week. Check: 150 + 40 × 8 = 470.
In simple words: Start with ₹150 and add ₹40 for each week.

Teacher's Note:
1. The ₹150 is added only once.
2. Undo the steps to find w in (iv).

Question 15. Fencing a rectangular lawn. A rectangular lawn in a park has length l metres and breadth b metres. The gardener wants to put a wire fence all around it. The fencing wire with fitting costs ₹60 for every metre of the boundary. The park committee does not yet know the exact size of the lawn, so the gardener writes the cost as an algebraic expression first and plans to put in the measurements later.
(i) Write an expression in its simplest form for the perimeter of the lawn.
(ii) Write an expression for the total cost of the fence, and simplify it.
(iii) Find the perimeter and the cost when l = 15 and b = 10.
(iv) If the length of the lawn is increased by 4 metres and the breadth is unchanged, write the new perimeter as a simplified expression.
Answer:
(i) Perimeter = l + b + l + b = 2l + 2b metres.
(ii) Cost = 60 × (2l + 2b) = ₹(120l + 120b).
(iii) For l = 15 and b = 10: perimeter = 2 × 15 + 2 × 10 = 30 + 20 = 50 metres; cost = 50 × 60 = ₹3000.
The expression gives the same: 120 × 15 + 120 × 10 = 1800 + 1200 = 3000.
(iv) The new length is l + 4, so the new perimeter is 2(l + 4) + 2b = 2l + 8 + 2b = 2l + 2b + 8 metres, that is 8 metres more than before.
In simple words: Perimeter is 2l + 2b, and the cost is 60 times the perimeter.

Teacher's Note:
1. 60 multiplies both terms of the perimeter.
2. Adding 4 m to the length adds 8 m to the perimeter, since there are two lengths.

Question 16. The furniture rental shop. A shop in the market rents out chairs and tables for functions. At the time of taking the furniture, the customer pays ₹50 for every chair and ₹90 for every table. When the furniture is brought back in good condition, the shopkeeper returns ₹8 for every chair and ₹12 for every table. Ganesh rents x chairs and y tables for a wedding in his family and wants to know, before the function, what the furniture will finally cost him.
(i) Write an expression for the amount Ganesh pays at the beginning.
(ii) Write an expression for the amount the shopkeeper returns.
(iii) Write the final amount paid as a difference of these two expressions and simplify it.
(iv) Find the final amount paid for 6 chairs and 2 tables.
Answer:
(i) Amount paid at the start = ₹(50x + 90y).
(ii) Amount returned = ₹(8x + 12y).
(iii) Final amount = (50x + 90y) − (8x + 12y) = 50x + 90y − 8x − 12y = (50 − 8)x + (90 − 12)y = ₹(42x + 78y).
(iv) For x = 6 and y = 2: 42 × 6 + 78 × 2 = 252 + 156 = ₹408.
Check directly: he pays 300 + 180 = 480 and gets back 48 + 24 = 72, and 480 − 72 = 408.
In simple words: Final cost = amount paid − amount returned, for chairs and tables separately.

Teacher's Note:
1. The minus sign changes both terms of the returned amount.
2. Check by working with numbers directly.

Question 17. Scores in a quiz. In an inter-house quiz, every correct answer earns p points and every wrong answer takes away q points. Charulata takes part in three rounds. Her scores in the three rounds are written by the scorer as 9p − 2q, 7p − 4q and 10p − 3q. Her friend Krishita's total after the same three rounds is written as 24p − 5q. At the end of the quiz the organisers announce that a correct answer carried 5 points and a wrong answer carried a penalty of 2 points.
(i) What does the expression 9p − 2q tell you about Charulata's first round?
(ii) Write Charulata's total score after the three rounds in its simplest form.
(iii) Find Charulata's total score when p = 5 and q = 2.
(iv) Who scored more, Charulata or Krishita, and by how much, for these values of p and q?
Answer:
(i) In the first round Charulata gave 9 correct answers and 2 wrong answers, so she earned p points nine times and lost q points twice.
(ii) Total = (9p − 2q) + (7p − 4q) + (10p − 3q) = (9 + 7 + 10)p − (2 + 4 + 3)q = 26p − 9q.
(iii) For p = 5, q = 2: 26 × 5 − 9 × 2 = 130 − 18 = 112 points.
(iv) Krishita's total = 24 × 5 − 5 × 2 = 120 − 10 = 110 points. So Charulata scored more, by 2 points.
The difference expression gives the same: (26p − 9q) − (24p − 5q) = 2p − 4q = 10 − 8 = 2.
In simple words: Add the like terms of the three rounds, then put p = 5 and q = 2.

Teacher's Note:
1. The coefficient of p counts correct answers.
2. Compare totals only after substituting the values.

Question 18. A rangoli of coloured dots. For a school festival, the children draw a long line of coloured dots for a rangoli border. Starting from position 1, the colours are repeated again and again in the fixed order red, blue, green, yellow, red, blue, green, yellow, and so on. The line is very long, so the teacher asks the class to find a way of naming the colour at any position without counting all the dots before it. One group notices that the pattern repeats after every four dots and decides to use the letter n for the number of times a colour has appeared.
(i) Which colour is at position 12?
(ii) Write an expression for the positions at which green appears for the nth time.
(iii) Which colour is at position 26? Show how you decide.
(iv) At which position does red appear for the 15th time?
Answer:
(i) The colours repeat in groups of 4 and yellow ends each group, so yellow is at the multiples of 4. As 12 = 4 × 3, position 12 is yellow.
(ii) Green is one place before yellow, so the nth green is at position 4n − 1. (For n = 1 this gives 3, which is correct.)
(iii) 26 = 4 × 6 + 2. A remainder of 2 means the second colour of a group, so position 26 is blue.
(iv) Red is the first colour of each group, at positions 4n − 3. For n = 15: 4 × 15 − 3 = 60 − 3 = 57.
In simple words: Yellow is at multiples of 4; the other colours are 1, 2 or 3 places before.

Teacher's Note:
1. The remainder on dividing by 4 gives the colour.
2. Remainder 1 red, 2 blue, 3 green, 0 yellow.

Question 19. Choosing between two taxi services. Two taxi services work in Nandipur. Taxi A charges a fixed amount of ₹60 for booking and then ₹14 for every kilometre travelled. Taxi B charges a fixed booking amount of ₹100 but only ₹10 for every kilometre travelled. Sunita travels different distances on different days and wants to decide which taxi to call. She writes the charge of each taxi as an algebraic expression, using d for the distance travelled in kilometres, so that she can compare the two services for any journey.
(i) Write an expression for the charge of Taxi A and an expression for the charge of Taxi B.
(ii) Which taxi is cheaper for a journey of 8 km? Show the amounts.
(iii) Which taxi is cheaper for a journey of 15 km? Show the amounts.
(iv) For what distance do the two taxis charge the same amount, and what is that amount?
Answer:
(i) Taxi A: ₹(60 + 14d); Taxi B: ₹(100 + 10d).
(ii) For d = 8: Taxi A = 60 + 14 × 8 = 60 + 112 = 172; Taxi B = 100 + 10 × 8 = 180. Taxi A is cheaper, by ₹8.
(iii) For d = 15: Taxi A = 60 + 14 × 15 = 60 + 210 = 270; Taxi B = 100 + 10 × 15 = 250. Taxi B is cheaper, by ₹20.
(iv) The charges are equal when 60 + 14d = 100 + 10d. Taking 10d from both sides: 60 + 4d = 100, so 4d = 40 and d = 10.
At 10 km each taxi charges 60 + 140 = 200 and 100 + 100 = 200, that is ₹200. For shorter journeys Taxi A is cheaper, and for longer journeys Taxi B is cheaper.
In simple words: Write both charges with d, then compare them for each distance.

Teacher's Note:
1. Lower booking fee helps on short trips.
2. Lower rate per km helps on long trips.

Teacher's Note:
1. Substituting a number quickly shows a wrong simplification.
2. A plus before a bracket changes nothing inside.