CBSE Class 7 Mathematics Finding the Unknown Worksheet Set 01

Long Answer Questions

Question 1. The weighing scale below is perfectly balanced. All the tins have the same weight t units, and the numbers written on the blocks give their weights in the same units.
(i) Frame an equation for this balance.
(ii) What should be removed from both pans so that only the tins are left on the left pan? Why does the scale stay balanced after this?
(iii) Solve the equation step by step, writing the operation used at each step.
(iv) Check your answer by substituting it back, and state the two properties of a balance that you have used.
Answer:
(i) The left pan has 4 tins and a block of 6, so it carries 4t + 6. The right pan carries a block of 26. The scale is balanced, so 4t + 6 = 26.
(ii) Remove a weight of 6 from both pans. Both sides lose the same amount, so they stay equal and the scale stays balanced. Now only the 4 tins are on the left, against 20 on the right: 4t = 20.
(iii) Subtract 6 from both sides: 4t + 6 − 6 = 26 − 6, so 4t = 20.
Divide both sides by 4: 4t ÷ 4 = 20 ÷ 4, so t = 5. Each tin weighs 5 units.
(iv) Check: left pan = 4 × 5 + 6 = 20 + 6 = 26 = right pan.
Properties used: (1) removing equal weights from both pans keeps the balance (subtracting the same number from both sides); (2) sharing equal weights equally among equal objects keeps the balance (dividing both sides by the same number).
In simple words: Take 6 off both pans, then share 20 equally among the 4 tins to get 5 units each.

Teacher's Note:
1. Write each pan as an expression before equating.
2. Remove the block first, then divide by the number of tins.
3. Name the operation at every step.

Question 2. Solve each equation below systematically, writing the operation done on both sides at every step, and check each solution by substitution.
(i) 7a + 8 = 50
(ii) 9 − 2b = 1
(iii) 5c + 4 = 3c + 16
Also state, in one line each, the three rules used: removing a term that is added, removing a factor, and bringing unknown terms to one side.
Answer:
(i) Subtract 8 from both sides: 7a = 42. Divide both sides by 7: a = 6. Check: 7 × 6 + 8 = 42 + 8 = 50 = RHS.
(ii) Subtract 9 from both sides: −2b = 1 − 9 = −8. Divide both sides by −2: b = 4. Check: 9 − 2 × 4 = 9 − 8 = 1 = RHS.
(iii) Subtract 3c from both sides: 2c + 4 = 16. Subtract 4 from both sides: 2c = 12. Divide both sides by 2: c = 6. Check: LHS = 30 + 4 = 34 and RHS = 18 + 16 = 34.
Rule 1: A term added on one side, when removed, appears on the other side with the opposite sign (its additive inverse).
Rule 2: When a factor is removed from a product on one side, the other side is divided by that factor.
Rule 3: When the unknown is on both sides, subtract the smaller unknown term from both sides so that the unknown remains on one side only.
In simple words: Undo additions first, then divide; if the unknown is on both sides, collect it on one side first.

Teacher's Note:
1. In (ii) the coefficient is −2, so divide by −2.
2. Write the operation beside every step.
3. Check each answer in the original equation.

Question 3. Write an account of the history of solving equations as given in the chapter, covering the following points.
(i) The meaning of the words bījagaṇita and bīja, and why the name suits the subject.
(ii) Brahmagupta’s contribution, the name of his book and its year.
(iii) Al-Khwarizmi, his book of 825 CE, and how the word “algebra” reached us.
(iv) The symbols yā, kā, nī and rū used by ancient Indian mathematicians, and how an equation was written down.
Answer:
(i) In ancient India, the study of expressions and equations (our algebra) was called bījagaṇita. Bīja means seed. A whole tree lies hidden in a seed; in the same way the answer to a problem lies hidden in the unknown number, and solving the problem is like letting that tree grow step by step.
(ii) In Chapter 18 of his book Brāhmasphuṭasiddhānta (628 CE), Brahmagupta showed how to add, subtract and multiply unknown numbers written with letters, much as we use x and y. It is one of the earliest known works on algebra, and the mathematician David Mumford called Brahmagupta the key person in creating algebra as we know it.
(iii) In the 8th century, Indian mathematics was translated into Arabic and reached Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE he wrote Hisab al-jabr wal-muqabala (“calculation by restoring and balancing”). By the 12th century it was translated into Latin and became popular in Europe, and the word al-jabr gave us “algebra”.
(iv) Unknowns were written as yā (from yāvat-tāvat, “as much as needed”); further unknowns were kā and nī, from the colour names kālaka (black) and nīlaka (blue). Known quantities were written rū (from rūpa). A dot over a number made it negative, and the two sides of an equation were written one below the other.
In simple words: Algebra was bījagaṇita in India; Brahmagupta (628 CE) laid its base, Al-Khwarizmi (825 CE) gave it the name, and yā, kā, nī, rū were the old symbols.

Teacher's Note:
1. Cover all four points with dates and names.
2. Explain the seed idea in your own words.
3. Mention how negatives were shown.

Question 4. A school orders identical notebooks from a shop. Each notebook costs the same amount, and the shop charges a fixed sum of ₹40 for printing the school crest on the whole order. The teacher pays ₹565 in all for the notebooks together with the printing charge, and each notebook costs ₹35.
(i) Taking the number of notebooks as n, write an expression for the total amount paid.
(ii) Frame an equation and solve it to find n.
(iii) Solve the same problem by a second method, by first removing the fixed printing charge from the total.
(iv) Check your answer, and find what the teacher would pay for 20 such notebooks with the same printing charge.
Answer:
(i) n notebooks at ₹35 each cost 35n rupees. With the fixed ₹40, the total is 35n + 40 rupees.
(ii) Equation: 35n + 40 = 565. Subtract 40 from both sides: 35n = 525. Divide both sides by 35: n = 15. So 15 notebooks were ordered.
(iii) Take out the printing charge first: ₹565 − ₹40 = ₹525 is spent on notebooks alone. Number of ₹35 notebooks in ₹525 = 525 ÷ 35 = 15, the same answer.
(iv) Check: 15 × 35 = ₹525 and ₹525 + ₹40 = ₹565, which matches.
For 20 notebooks: 20 × 35 + 40 = 700 + 40 = ₹740.
In simple words: Remove the fixed ₹40 and divide the rest by ₹35 to get 15 notebooks.

Teacher's Note:
1. The fixed charge is added only once.
2. Both methods do the same steps, one in symbols and one in words.
3. Keep the ₹40 when pricing 20 notebooks.

Question 5. Meera has ₹2400 saved and adds ₹300 to her savings every month. Arun has ₹3000 saved and adds ₹250 every month. They want to know when their savings will become equal.
(i) Taking m as the number of months, write an expression for Meera’s savings and one for Arun’s savings after m months.
(ii) Frame an equation for the month in which their savings are equal.
(iii) Solve the equation step by step.
(iv) Find how much each of them has at that time, and check that the two amounts are really equal.
Answer:
(i) Meera: 2400 + 300m rupees. Arun: 3000 + 250m rupees.
(ii) Savings are equal when 2400 + 300m = 3000 + 250m.
(iii) Subtract 250m from both sides: 2400 + 50m = 3000.
Subtract 2400 from both sides: 50m = 600.
Divide both sides by 50: m = 12. Their savings become equal after 12 months.
(iv) Meera: 2400 + 300 × 12 = 2400 + 3600 = ₹6000. Arun: 3000 + 250 × 12 = 3000 + 3000 = ₹6000. Both have ₹6000, so the answer is correct.
Meera starts ₹600 behind but saves ₹50 more each month, so she catches up in 600 ÷ 50 = 12 months.
In simple words: Meera gains ₹50 a month on Arun and needs to make up ₹600, so it takes 12 months.

Teacher's Note:
1. Starting amount + monthly saving × months.
2. Collect the m terms on the side with the larger coefficient.
3. Check both amounts at m = 12.

Question 6. Three students solved three equations. Go through each solution, say whether it is correct, describe the mistake if there is one, mend it and give the correct solution.
(i) 5x + 7 = 32; 5x = 32 + 7; 5x = 39; x = \( \frac{39}{5} \)
(ii) 3(y − 2) = 9; 3y − 2 = 9; 3y = 11; y = \( \frac{11}{3} \)
(iii) 6z − 4 = 2z + 12; 6z + 2z = 12 − 4; 8z = 8; z = 1
Answer:
(i) Incorrect. The +7 was moved to the RHS without changing its sign. It must become −7. Correct: 5x = 32 − 7 = 25, so x = 5. Check: 5 × 5 + 7 = 32.
(ii) Incorrect. The bracket was opened wrongly: 3 multiplies both y and 2, so 3(y − 2) = 3y − 6, not 3y − 2. Correct: 3y − 6 = 9 ⇒ 3y = 15 ⇒ y = 5. (Or keep the bracket: y − 2 = 3, so y = 5.) Check: 3 × (5 − 2) = 9.
(iii) Incorrect. Both moved terms kept the wrong sign: +2z moved to the LHS must become −2z, and −4 moved to the RHS must become +4. Correct: 6z − 2z = 12 + 4 ⇒ 4z = 16 ⇒ z = 4. Check: LHS = 24 − 4 = 20 and RHS = 8 + 12 = 20.
In simple words: All three are wrong: (i) x = 5, (ii) y = 5, (iii) z = 4.

Teacher's Note:
1. A term changes sign when it crosses the equal sign.
2. A factor outside a bracket multiplies every term inside.
3. Substitute to confirm each corrected answer.

Question 7. In a triangle, the three angles measure x, x + 20° and x + 40°.
(i) Write the equation you get from the angle sum property of a triangle.
(ii) Solve the equation to find x.
(iii) Find the three angles and check that they add up to 180°.
(iv) Name the type of triangle by its angles, and say whether such a triangle could ever be a right-angled one if the three angles kept this same form with 20° and 40° replaced by 30° and 60°.
Answer:
(i) The angles of a triangle add up to 180°, so x + (x + 20°) + (x + 40°) = 180°, that is 3x + 60° = 180°.
(ii) Subtract 60° from both sides: 3x = 120°. Divide both sides by 3: x = 40°.
(iii) The angles are 40°, 60° and 80°. Check: 40° + 60° + 80° = 180°.
(iv) All angles are less than 90°, so it is an acute-angled triangle (also scalene, as all angles differ). With x, x + 30° and x + 60°: 3x + 90° = 180°, so x = 30° and the angles are 30°, 60°, 90°. Yes, that triangle would be right-angled.
In simple words: x = 40° gives angles 40°, 60°, 80° (acute); the other form gives 30°, 60°, 90° (right-angled).

Teacher's Note:
1. Add like terms: the x's together and the degrees together.
2. Always check the angle sum.
3. A right angle needs one angle exactly 90°.

Question 8. Hari plays a number trick on his sister. He tells her: think of a number, add 4 to it, multiply the result by 2, subtract 6 from the product, and finally divide by 2. She tells him only the final answer, 12.
(i) Taking the number first thought of as x, write the expression obtained after each of the four steps.
(ii) Simplify the final expression.
(iii) Frame an equation using the final answer 12 and find the number she thought of.
(iv) State a simple rule by which Hari can get the starting number from any final answer, and test your rule with a number of your own.
Answer:
(i) Think of a number: x. Add 4: x + 4. Multiply by 2: 2(x + 4) = 2x + 8. Subtract 6: 2x + 8 − 6 = 2x + 2. Divide by 2: (2x + 2) ÷ 2 = x + 1.
(ii) The final expression is x + 1, always one more than the starting number.
(iii) x + 1 = 12, so x = 12 − 1 = 11. Check: 11 → 15 → 30 → 24 → 12.
(iv) Rule: subtract 1 from the final answer. Test with 7: 7 → 11 → 22 → 16 → 8, and 8 − 1 = 7. It works for every number because the result is always x + 1.
In simple words: The trick always gives one more than the number thought of, so subtract 1.

Teacher's Note:
1. Write one expression per step.
2. Divide every term by 2 in the last step.
3. Algebra shows why the trick always works.

Question 9. In a stationery shop, a pen costs ₹12 more than a pencil. Rekha buys 5 pens and 8 pencils and pays ₹346 in all.
(i) Taking the cost of one pencil as p rupees, write expressions for the cost of one pen, of 5 pens and of 8 pencils.
(ii) Frame an equation for Rekha’s bill and solve it step by step.
(iii) Write the cost of a pen and of a pencil, and check the bill of ₹346.
(iv) Find what 3 pens and 2 pencils would cost, and explain why the problem could be solved even though two costs were unknown at the start.
Answer:
(i) One pen = p + 12 rupees; 5 pens = 5(p + 12) rupees; 8 pencils = 8p rupees.
(ii) Equation: 5(p + 12) + 8p = 346.
Open the bracket: 5p + 60 + 8p = 346. Collect like terms: 13p + 60 = 346.
Subtract 60 from both sides: 13p = 286. Divide both sides by 13: p = 22.
(iii) A pencil costs ₹22 and a pen costs 22 + 12 = ₹34. Check: 5 × 34 = ₹170, 8 × 22 = ₹176, and 170 + 176 = ₹346.
(iv) 3 pens and 2 pencils cost 3 × 34 + 2 × 22 = 102 + 44 = ₹146. The two costs were linked (pen = pencil + ₹12), so the pen’s cost could be written using the pencil’s cost. That left only one unknown, which one equation can find.
In simple words: Write the pen as p + 12, solve 13p + 60 = 346 to get ₹22 and ₹34.

Teacher's Note:
1. Express one unknown using the other.
2. Multiply both terms inside the bracket by 5.
3. Check the full bill.

Question 10. In a parking area there are only two-wheelers and cars. The watchman counts 30 vehicles in all and 94 wheels in all.
(i) Taking the number of cars as c, write expressions for the number of two-wheelers, the number of wheels of the cars and the number of wheels of the two-wheelers.
(ii) Frame an equation in c and solve it step by step.
(iii) Find how many cars and how many two-wheelers are parked, and check both the count of heads and the count of wheels.
(iv) Explain why one unknown was enough here, and find how many wheels there would be if 4 of the two-wheelers were replaced by cars.
Answer:
(i) Two-wheelers = 30 − c; wheels of cars = 4c; wheels of two-wheelers = 2(30 − c).
(ii) Equation: 4c + 2(30 − c) = 94. Open the bracket: 4c + 60 − 2c = 94. Collect like terms: 2c + 60 = 94.
Subtract 60 from both sides: 2c = 34. Divide both sides by 2: c = 17.
(iii) 17 cars and 30 − 17 = 13 two-wheelers. Vehicles: 17 + 13 = 30. Wheels: 17 × 4 + 13 × 2 = 68 + 26 = 94. Both counts match.
(iv) The total of 30 vehicles ties the two numbers together: once cars are c, two-wheelers must be 30 − c, so one unknown is enough. With 4 two-wheelers replaced by cars: 21 cars and 9 two-wheelers, giving 21 × 4 + 9 × 2 = 84 + 18 = 102 wheels (8 more).
In simple words: Call the cars c, the rest are 30 − c; 2c + 60 = 94 gives 17 cars and 13 two-wheelers.

Teacher's Note:
1. Use the total to write the second number.
2. Each swap of a two-wheeler for a car adds 2 wheels.
3. Check both totals.

Case Study Based Questions

Question 11. A shopkeeper’s bill with an unknown rate.
Rani goes to a stationery shop before the new session begins. She buys 6 notebooks of the same kind and one geometry box. The shopkeeper writes the bill, but the ink smudges and the rate of a notebook can no longer be read. Rani can still read that the geometry box cost ₹45 and that the total of the bill was ₹255. She decides to find the rate of one notebook by writing an equation instead of guessing. She takes the cost of one notebook as n rupees and begins by writing an expression for the cost of the six notebooks.
(i) Write an expression for the cost of the 6 notebooks.
(ii) Frame an equation for Rani’s bill.
(iii) Solve the equation to find the cost of one notebook.
(iv) Check your answer by finding the total bill again, and state the cost of 10 such notebooks.
Answer:
(i) Cost of 6 notebooks = 6 × n = 6n rupees.
(ii) Notebooks + geometry box = total: 6n + 45 = 255.
(iii) Subtract 45 from both sides: 6n = 210. Divide both sides by 6: n = 35. One notebook costs ₹35.
(iv) Check: 6 × 35 + 45 = 210 + 45 = 255, the bill. Ten notebooks cost 10 × 35 = ₹350.
In simple words: Take ₹45 off the bill and share ₹210 among 6 notebooks: ₹35 each.

Teacher's Note:
1. The geometry box is the fixed part of the bill.
2. Remove it before dividing.
3. Check by rebuilding the bill.

Question 12. A bus with an unknown number of passengers.
A bus starts from the Mysuru bus stand carrying some passengers. The conductor does not note the starting number in his book. At the first stop, 12 more passengers get into the bus and nobody gets down. At the second stop, 7 passengers get down and nobody gets in. The conductor now counts the passengers carefully and finds 45 of them in the bus. He wants to tell the depot how many passengers the bus started with, so he takes that unknown number as p and writes down what happened at each stop.
(i) Write an expression for the number of passengers just after the first stop.
(ii) Frame an equation using the count of 45 passengers.
(iii) Solve the equation to find how many passengers the bus started with.
(iv) If 9 passengers had got down at the second stop instead of 7, how many would be in the bus then?
Answer:
(i) After the first stop: p + 12 passengers.
(ii) After the second stop 7 get down: p + 12 − 7 = 45, that is p + 5 = 45.
(iii) Subtract 5 from both sides: p = 40. The bus started with 40 passengers. Check: 40 + 12 − 7 = 45.
(iv) 40 + 12 − 9 = 43 passengers.
In simple words: Net change is +5, so the bus started with 45 − 5 = 40.

Teacher's Note:
1. Getting in adds, getting down subtracts.
2. Simplify +12 − 7 to +5 first.
3. Use the found value for part (iv).

Question 13. Sharing sweets equally.
On her birthday, Nazia brings a box of laddus to class. She decides to share them equally among the 8 children in her group and to keep 6 laddus aside for the class teacher. After doing this, the box is empty. Nazia remembers that the shopkeeper had packed exactly 70 laddus in the box, but she has forgotten how many laddus each child received. Her friend suggests that she take the share of one child as x and write an equation for the 70 laddus in the box.
(i) Write an expression for the number of laddus given to the 8 children.
(ii) Frame an equation for the 70 laddus in the box.
(iii) Solve the equation to find how many laddus each child received.
(iv) If the box had held 86 laddus and 6 were again kept for the teacher, how many would each child get?
Answer:
(i) 8 children get x each: 8x laddus.
(ii) Children’s share + teacher’s 6 = whole box: 8x + 6 = 70.
(iii) Subtract 6 from both sides: 8x = 64. Divide both sides by 8: x = 8. Each child got 8 laddus. Check: 8 × 8 + 6 = 70.
(iv) 8x + 6 = 86 ⇒ 8x = 80 ⇒ x = 10. Each child would get 10 laddus.
In simple words: Keep 6 aside, share 64 among 8 children: 8 each.

Teacher's Note:
1. The teacher’s 6 is the fixed part.
2. The box is empty, so the parts add up to the whole.
3. Part (iv) only changes the RHS.

Question 14. A rectangular garden with an unknown width.
A school has a rectangular kitchen garden behind the science room. The gardener knows that the garden is 18 m long, and he has just put a wire fence all around it. The record book shows that the fence used was exactly 58 m long, but the width of the garden is not written anywhere. The mathematics club of the school decides to find the width without measuring it again. They call the width w metres and recall that the perimeter of a rectangle is twice the sum of its length and its width.
(i) Write an expression for the perimeter of the garden in terms of w.
(ii) Frame an equation using the length of the fence.
(iii) Solve the equation to find the width of the garden.
(iv) Find the area of the garden, and check that the perimeter is really 58 m.
Answer:
(i) Perimeter = 2 × (length + width) = 2(18 + w) metres.
(ii) The fence is the perimeter: 2(18 + w) = 58.
(iii) Divide both sides by 2: 18 + w = 29. Subtract 18 from both sides: w = 11. The garden is 11 m wide.
(iv) Area = 18 × 11 = 198 m². Check: perimeter = 2 × (18 + 11) = 2 × 29 = 58 m, which matches the record book.
In simple words: Half the fence (29 m) is length + width, so the width is 29 − 18 = 11 m.

Teacher's Note:
1. Dividing by 2 first keeps the bracket simple.
2. Area is in square metres.
3. Check with the fence length.

Question 15. A mobile plan with a fixed charge and a per-minute rate.
Imran’s mobile company offers two plans. In Plan A, the customer pays a fixed monthly rental of ₹120 and, besides this, ₹2 for every minute of calls made during the month. In Plan B, there is no monthly rental at all, but every minute of calls costs ₹4. Imran is on Plan A. At the end of October his bill came to ₹364, and he wants to know for how many minutes he talked. He also wants to decide whether he should shift to Plan B next month. He takes the number of call minutes in a month as m.
(i) Write an expression for Imran’s monthly bill on Plan A for m minutes of calls.
(ii) Frame an equation for the October bill and find the number of minutes he talked.
(iii) What would Plan A charge for a month with 90 minutes of calls?
(iv) For how many minutes of calls would the two plans cost exactly the same?
Answer:
(i) Plan A bill = 120 + 2m rupees.
(ii) 120 + 2m = 364. Subtract 120: 2m = 244. Divide by 2: m = 122. Imran talked for 122 minutes. Check: 120 + 2 × 122 = 120 + 244 = ₹364.
(iii) 120 + 2 × 90 = 120 + 180 = ₹300.
(iv) Plan B costs 4m. Equal when 120 + 2m = 4m. Subtract 2m: 120 = 2m, so m = 60 minutes (both plans then cost ₹240).
In simple words: He talked 122 minutes; the plans cost the same at 60 minutes.

Teacher's Note:
1. Rental is fixed, call charge depends on minutes.
2. Set the two plan expressions equal for part (iv).
3. Check with the October bill.

Question 16. A grocer’s bill with an unknown rate.
Shalini buys 4 kg of sugar and 3 kg of rice from a grocery shop. The rate board in the shop shows that rice costs ₹52 per kg, but the rate of sugar has been rubbed out. The grocer hands her a bill for ₹332 in all. Shalini does not want to ask again, so she works out the rate of sugar from the bill itself. She takes the rate of sugar as s rupees per kilogram and writes the cost of the sugar and the cost of the rice separately before adding them.
(i) Write an expression for the total bill in terms of s.
(ii) Frame an equation and find the rate of sugar per kilogram.
(iii) What would 7 kg of sugar cost at this rate?
(iv) Next week the sugar rate goes up by ₹2 per kg while rice stays the same. What will 4 kg of sugar and 3 kg of rice cost then?
Answer:
(i) Sugar = 4s rupees; rice = 3 × 52 = ₹156. Total = 4s + 156 rupees.
(ii) 4s + 156 = 332. Subtract 156: 4s = 176. Divide by 4: s = 44. Sugar costs ₹44 per kg. Check: 4 × 44 + 156 = 176 + 156 = ₹332.
(iii) 7 × 44 = ₹308.
(iv) New rate = 44 + 2 = ₹46 per kg. Bill = 4 × 46 + 156 = 184 + 156 = ₹340.
In simple words: Rice is ₹156, so sugar is ₹176 for 4 kg, i.e. ₹44 per kg.

Teacher's Note:
1. Work out the known rice cost first.
2. Then the equation has one unknown.
3. Rice cost stays ₹156 in part (iv).

Question 17. Collecting money for a school picnic.
Class 7 of a government school is going on a picnic to a nearby lake. The bus has to be hired for a fixed amount of ₹3000 for the whole day, whoever goes in it. Besides this, the school spends ₹120 on food for each student who goes. The class teacher collects the money from the parents and finds that the total amount needed came to exactly ₹8400. She now wants to tell the principal how many students are going on the picnic. She takes the number of students as n and writes the bus charge and the food charge as two parts of the total.
(i) Write an expression for the total cost of the picnic for n students.
(ii) Frame an equation for the total of ₹8400 and find the number of students going.
(iii) What would the total cost be if 50 students went?
(iv) The principal can sanction at most ₹9600 for the picnic. What is the greatest number of students who can go?
Answer:
(i) Total cost = 3000 + 120n rupees.
(ii) 3000 + 120n = 8400. Subtract 3000: 120n = 5400. Divide by 120: n = 45. 45 students are going. Check: 3000 + 120 × 45 = 3000 + 5400 = ₹8400.
(iii) 3000 + 120 × 50 = 3000 + 6000 = ₹9000.
(iv) 3000 + 120n = 9600 gives 120n = 6600 and n = 55. At most 55 students can go (56 would cost ₹9720).
In simple words: Take the ₹3000 bus off and divide the rest by ₹120 per student.

Teacher's Note:
1. The bus charge does not depend on the number of students.
2. For the limit, solve for the exact amount.
3. One more student would cross the limit.

Question 18. A rangoli border made of tiles.
For the school’s annual day, the students lay a rangoli border made of square tiles along the corridor. The border grows step by step: position 1 of the pattern uses 7 tiles, position 2 uses 12 tiles, position 3 uses 17 tiles, and so on, each position needing 5 tiles more than the one before it. The art teacher notices that the number of tiles at position n is given by the expression 5n + 2. The students now have a large heap of tiles and want to know how far the border can grow.
(i) How many tiles are needed for the arrangement at position 6?
(ii) Frame an equation to find the position that uses exactly 77 tiles, and solve it.
(iii) Can an arrangement of this pattern be made with exactly 100 tiles? Give a reason.
(iv) Explain why the number of tiles increases by 5 from one position to the next.
Answer:
(i) 5 × 6 + 2 = 30 + 2 = 32 tiles.
(ii) 5n + 2 = 77. Subtract 2: 5n = 75. Divide by 5: n = 15. Position 15 uses 77 tiles.
(iii) No. 5n + 2 = 100 gives 5n = 98, and 98 is not divisible by 5, so n is not a whole number. No position uses exactly 100 tiles.
(iv) Position n uses 5n + 2 tiles and position n + 1 uses 5(n + 1) + 2 = 5n + 7 tiles, which is 5 more.
In simple words: Use 5n + 2: position 6 needs 32, 77 tiles is position 15, and 100 is impossible.

Teacher's Note:
1. A position must be a counting number.
2. Compare the rule at n and n + 1.
3. Check: 5 × 15 + 2 = 77.

Question 19. A problem from Bhāskarāchārya’s Bījagaṇita.
Old Indian mathematics books posed problems about two traders who were equally rich. Here is one such problem written in today’s language. One trader owns ₹200 in cash and 5 bullocks. A second trader owns 9 bullocks but also has a debt of ₹120, which must be subtracted from his wealth. Every bullock has the same price. The two traders are found to be equally rich. A debt is written with a minus sign, just as a dot over a number was used in ancient Indian notation to show that it was negative. Take the price of one bullock as x rupees.
(i) Write an expression for the wealth of each trader in terms of x.
(ii) Frame an equation from the statement that they are equally rich.
(iii) Solve the equation to find the price of one bullock.
(iv) Find the wealth of each trader, and check that the two are equal.
Answer:
(i) First trader: 200 + 5x rupees. Second trader: 9x − 120 rupees (debt subtracted).
(ii) 200 + 5x = 9x − 120.
(iii) Add 120 to both sides: 320 + 5x = 9x. Subtract 5x: 320 = 4x. Divide by 4: x = 80. One bullock costs ₹80.
(iv) First trader: 200 + 5 × 80 = 200 + 400 = ₹600. Second trader: 9 × 80 − 120 = 720 − 120 = ₹600. Both have ₹600.
In simple words: Equal wealth gives 200 + 5x = 9x − 120, so a bullock costs ₹80.

Teacher's Note:
1. A debt is a negative amount.
2. Move the unknowns to the side with 9x.
3. Check both traders’ wealth.

Question 20. Comparing two taxi services.
Imran has to travel from his home to the railway station and compares two taxi services. Service A charges a base fare of ₹60 and then ₹14 for every kilometre travelled. Service B charges a base fare of ₹100 and then ₹10 for every kilometre travelled. Imran wants to know for what distance the two services charge exactly the same fare, so that he can decide which one to call. He takes the distance as d kilometres and starts by writing the fare of each service as a fixed part plus a part that changes with the distance.
(i) Write an expression for the fare of each service for a journey of d kilometres.
(ii) Frame an equation for the distance at which the two fares are equal, and solve it.
(iii) Which service is cheaper for a journey of 6 km, and by how much?
(iv) Imran first wrote the solution as: 60 + 14d = 100 + 10d; 14d + 10d = 100 + 60; 24d = 160; d = \( \frac{20}{3} \). Find his mistake and mend it.
Answer:
(i) Service A: 60 + 14d rupees; Service B: 100 + 10d rupees.
(ii) 60 + 14d = 100 + 10d. Subtract 10d: 60 + 4d = 100. Subtract 60: 4d = 40, so d = 10 km. Both fares are then ₹200.
(iii) For 6 km: A = 60 + 84 = ₹144; B = 100 + 60 = ₹160. Service A is cheaper by ₹16.
(iv) Mistake: +10d moved to the LHS must become −10d, and +60 moved to the RHS must become −60; Imran kept both signs. Correct: 14d − 10d = 100 − 60 ⇒ 4d = 40 ⇒ d = 10 km.
In simple words: The fares match at 10 km; for 6 km, A is ₹16 cheaper.

Teacher's Note:
1. Fixed fare + rate × distance.
2. Moved terms change sign.
3. Check: 60 + 140 = 100 + 100 = 200.