Profit and Loss, Interest, Work, Speed, Mixture and Mensuration

Chapter 2 · Arithmetic · 5 questions in the paper · 20 min read

Profit and Loss, Interest, Work, Speed, Mixture and Mensuration

What the syllabus asks for 6
  • Profit, loss, discount and marked price
  • Simple and compound interest
  • Time and work, pipes and cisterns
  • Speed, distance, time, trains and boats
  • Mixtures and alligation
  • Area, perimeter and volume of plane and solid figures

This is the second half of the arithmetic block: the "business and motion" problems. Five questions of the 50 come from here in a typical paper, and they are the most formula-driven in the whole mathematics section.

Each type has one setup that, once learnt, turns a 90-second question into a 30-second one.

The mensuration part is the one most students forget: the Dean's Office model asks even "how many squares are in this figure", so know your areas and volumes cold.

What the paper asks from this chapter
  • Profit and loss: percent on the cost price; discount on the marked price; successive discounts; a "false weight" dealer.
  • Interest: \text{SI}=\frac{PRT}{100}, compound amount P(1+\frac{R}{100})^n, and the two-year shortcut \text{CI}-\text{SI}=P(\frac{R}{100})^2.
  • Work: \frac{ab}{a+b} for two workers, and inlet/outlet pipes with a minus sign.
  • Speed: \text{km/h}\times\frac{5}{18}=\text{m/s}, average speed over equal distances, trains and boats.
  • Mixture: the alligation cross, and repeated replacement.
  • Mensuration: area and volume formulas, and percentage change in area/volume.

2.1 Profit, loss, discount and marked price

  • \text{Profit}=SP-CP, \text{Loss}=CP-SP. Profit % and loss % are always on the cost price.
  • SP=CP\left(1+\frac{p}{100}\right) for profit p\%, and SP=CP\left(1-\frac{l}{100}\right) for loss l\%.
  • Discount is on the marked price (MP): SP=MP\left(1-\frac{d}{100}\right). Successive discounts multiply: MP\times(1-d_1)(1-d_2).
  • Mark-up then discount: marking 25 % above cost and allowing 20 % discount gives 1.25\times0.8=1.00, so no profit and no loss.
  • False weight: a dealer who uses w grams for a kilogram gains \frac{1000-w}{w}\times100\%.
  • If the same article is sold at two prices, one gaining x\% and the other losing x\%, the net result is a loss of \frac{x^2}{100}\%.
Reselling at a higher profit

Problem. By selling an article for Rs 540 a man gains 20 %. At what price must he sell it to gain 30 %?

Solution. CP=\dfrac{540}{1.2}=450. New SP=450\times1.3=585.

diagram
Figure 2.1. Profit, loss and discount from one price tag, computed. An article marked Rs 1,000 is sold at 20 percent discount, so SP = Rs 800. It cost Rs 640, so profit = Rs 160, which is 25 percent of the COST price (160 / 640), while the discount of Rs 200 is 20 percent of the MARKED price. Profit is always on CP, discount always on MP.

2.2 Simple and compound interest

  • Simple interest: \text{SI}=\dfrac{PRT}{100} and amount A=P+\text{SI}. R is per year, T in years.
  • Compound interest: A=P\left(1+\frac{R}{100}\right)^n and \text{CI}=A-P. For half-yearly compounding use \frac{R}{2} and 2n.
  • Two years: \text{CI}-\text{SI}=P\left(\frac{R}{100}\right)^2. For two years at R\% the CI equals \left(2R+\frac{R^2}{100}\right)\% of P (21 % for 10 %).
  • Doubling at SI: a sum doubles in \frac{100}{R} years; it becomes n times in \frac{100(n-1)}{R} years.
  • If a sum amounts to A_1 in t_1 years and A_2 in t_2 years at SI, the yearly interest is \frac{A_2-A_1}{t_2-t_1}, and P=A_1-t_1\times (yearly interest).

photograph
(a) An old Nepali one-rupee note
Plate 2.1. An old Nepali paper note (the face is King Mahendra). Interest, profit and loss are all questions about money like this: Nepal Rastra Bank issues the notes, and banks, cooperatives and Fonepay wallets use the interest formulas of this section every day.Photo credits: (a) Prasna siwakoti, CC BY-SA 4.0.

diagram
Figure 2.2. Simple versus compound interest on Rs 10,000 at 10 percent a year, computed. Simple interest grows by Rs 1,000 every year in a straight line; compound interest grows by 10 percent of a bigger amount each time, so the curve bends upward. After 2 years the gap is only P x r^2 / 100^2 = Rs 100, but after 10 years CI amount is Rs 25,937.42 against Rs 20,000.

2.3 Time and work, pipes and cisterns

  • If A does a job in a days, his one-day work is \frac{1}{a}. Rates add: together \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}, so the time is \frac{ab}{a+b}.
  • If A+B take t days and A alone takes a, then B alone takes \frac{at}{a-t}.
  • Men–days–hours: \dfrac{M_1D_1H_1}{W_1}=\dfrac{M_2D_2H_2}{W_2}.
  • Pipes: an inlet gives +\frac{1}{x} per minute, an outlet -\frac{1}{y}. With both open: \frac{1}{x}-\frac{1}{y} of the tank per minute.
  • Work done in part: if A works k days, \frac{k}{a} is done and the rest is 1-\frac{k}{a}.

diagram
Figure 2.3. Rates of work add. A finishes a job in 10 days (1/10 per day), B in 15 days (1/15 per day), so together they do 1/10 + 1/15 = 1/6 of the job a day and need 6 days. For pipes the same idea: an inlet that fills a tank in 12 h and a leak that empties it in 18 h leave a net rate of 1/12 - 1/18 = 1/36, so the tank fills in 36 h. All values are computed with fractions.

2.4 Speed, distance, time, trains and boats

  • v=\dfrac{d}{t}. km/h to m/s: multiply by \frac{5}{18}; m/s to km/h: multiply by \frac{18}{5}. (36 km/h = 10 m/s.)
  • Average speed over equal distances at x and y is \dfrac{2xy}{x+y}, not \frac{x+y}{2}. For equal times it is \frac{x+y}{2}.
  • Trains. Crossing a pole or a person: distance =L. Crossing a platform or another train: distance =L+P (or L_1+L_2). Relative speed is the sum for opposite directions and the difference for the same direction.
  • Boats. With boat speed u and stream speed v: downstream u+v, upstream u-v, so u=\frac{D+U}{2} and v=\frac{D-U}{2}.
Common mistake

Convert units first. The three commonest wrong options are the unconverted answer, the answer in m/s when km/h is asked, and the arithmetic mean instead of \frac{2xy}{x+y}.

diagram
Figure 2.4. Trains and boats, computed. (a) A 150 m train at 54 km/h = 15 m/s crosses a 350 m platform by covering 150 + 350 = 500 m, which takes 500 / 15 = 33.3 s. (b) A boat of 15 km/h in still water in a 3 km/h stream goes at 18 km/h downstream and 12 km/h upstream; for the same distance each way the AVERAGE speed is 2 x 18 x 12 / 30 = 14.4 km/h, not 15.

2.5 Mixtures and alligation

  • Alligation. For a mixture of a cheaper and a dearer item with mean price m: \dfrac{\text{quantity of cheaper}}{\text{quantity of dearer}}=\dfrac{d-m}{m-c}.
  • Replacement. A vessel holds V litres of pure liquid. Remove x litres and refill with water, repeated k times: pure liquid left =V\left(1-\frac{x}{V}\right)^k.
  • Concentrations of two solutions mixed in ratio p:q give \frac{p\,c_1+q\,c_2}{p+q}.

diagram
Figure 2.5. Alligation as a cross. To mix rice at Rs 40 and Rs 60 per kg into a blend worth Rs 52, write the mean 52 in the middle and subtract diagonally: 60 - 52 = 8 parts of the cheap rice and 52 - 40 = 12 parts of the dear rice, so the ratio cheap : dear = 8 : 12 = 2 : 3. A 50 kg bag then holds 20 kg and 30 kg, and the cost 20 x 40 + 30 x 60 = 2,600 = 50 x 52 confirms it.

2.6 Area, perimeter and volume of plane and solid figures

Figure Area / surface Perimeter / volume
Rectangle l\times b lb 2(l+b)
Square a a^2 4a
Triangle \frac{1}{2}bh; equilateral \frac{\sqrt{3}}{4}a^2 sum of sides
Trapezium \frac{1}{2}(a+b)h
Rhombus \frac{1}{2}d_1d_2 4a
Circle \pi r^2 2\pi r
Cuboid 2(lb+bh+hl) V=lbh; diagonal \sqrt{l^2+b^2+h^2}
Cube 6a^2 V=a^3; diagonal a\sqrt{3}
Cylinder curved 2\pi rh; total 2\pi r(r+h) V=\pi r^2h
Cone curved \pi rl, l=\sqrt{r^2+h^2} V=\frac{1}{3}\pi r^2h
Sphere 4\pi r^2 V=\frac{4}{3}\pi r^3
Hemisphere curved 2\pi r^2; total 3\pi r^2 V=\frac{2}{3}\pi r^3
  • Scaling. If every length is multiplied by k, areas scale by k^2 and volumes by k^3. Doubling a sphere's radius multiplies the volume by 8.
  • Percentage change in area. Length up a\% and breadth up b\%: area changes by a+b+\frac{ab}{100} per cent. The MEC-style example (15 % and 25 %) gives 43.75\%.
Exam tip

Use \pi=\frac{22}{7} when the radius is a multiple of 7. Otherwise leave \pi in the answer and compare options.

photograph
(a) Square pyramid: V = (1/3) x base x height
photograph
(b) Cylinder: V = pi r^2 h
Plate 2.2. Solids you can measure. (a) The Great Pyramid of Giza, a square pyramid: with base 230.34 m and height 146.5 m its volume is one third of base area x height, about 2.59 million cubic metres. (b) A cylindrical hot-water tank: volume = pi r^2 h, curved surface = 2 pi r h.Photo credits: (a) Nina, CC BY 2.5; (b) Antti Leppänen, CC BY 4.0.

diagram
Figure 2.6. Archimedes' favourite result, computed. For a cone, a sphere and a cylinder of the SAME radius r = 7 cm, with height equal to the diameter (14 cm), the volumes are 718.67, 1,437.33 and 2,156 cm3, in the exact ratio 1 : 2 : 3. The pyramid on the left is checked in the cell: V = (1/3) x 230.34 x 230.34 x 146.5 = 2.59 million m3.

2.7 Eight traps in profit, interest, work, speed and mixture questions

These problems are all short formulas, so the trap is nearly always a base, a unit or a sign. The eight below come up again and again in the Dean's Office model and in earlier TU entrance papers.

  1. Profit is on cost price; discount is on marked price. A shopkeeper who marks Rs 1,000 and gives 20 percent off sells at Rs 800, and if the cost was Rs 640 the profit is 160/640=25 percent, not 20 (Fig 2.1). Reading the wrong base is the single most common error. Underline "on cost" or "on marked price" before you calculate.
  2. A "gain of p percent" means SP=CP\times\frac{100+p}{100}. To find the cost price from the selling price you must divide, not subtract. If a radio sold for Rs 1,320 at a 10 percent gain, then CP=1320\times\frac{100}{110}=1200, not 1320-132=1188. The same idea works for loss with 100-p.
  3. Simple interest is a straight line; compound interest is not. With SI=\frac{PRT}{100} the interest is the same every year, but with compound interest the amount is P\left(1+\frac{R}{100}\right)^n and each year's interest is bigger than the last (Fig 2.2). The question may ask for the interest, so subtract P from the amount. For 2 years the difference CI-SI equals \frac{PR^2}{100^2}, a quick check that saves time.
  4. Match the time unit to the rate. A rate of 12 percent per annum for 9 months is T=\frac{9}{12}=0.75 year, not 9. With half-yearly compounding, halve the rate and double the number of periods. Forgetting this turns a Rs 1,080 answer into a Rs 10,800 one.
  5. Work adds as rates, never as days. If A needs 10 days and B needs 15 days, together they need \frac{10\times15}{10+15}=6 days, not the average 12.5 days (Fig 2.3). A leak or outlet pipe has a negative rate, so subtract it. The answer is always smaller than the faster worker's time alone when both are helping.
  6. Convert units before using d=vt. Speed in km/h becomes m/s on multiplying by \frac{5}{18}, and m/s becomes km/h on multiplying by \frac{18}{5}. A train crossing a platform covers its own length plus the platform's, while crossing a pole it covers only its own length (Fig 2.4). Mixing metres with kilometres is how a correct method gets a wrong option.
  7. Average speed is not the average of the speeds. For equal distances the average speed is the harmonic mean \frac{2uv}{u+v}, so a boat going 18 km/h downstream and 12 km/h upstream averages 14.4 km/h. If the equal quantity is time, not distance, then the simple average is correct. Say which quantity is equal before you choose the formula.
  8. Alligation gives a ratio of quantities, and the mean sits in the middle. Subtract diagonally (Fig 2.5): the quantity ratio is (d-m):(m-c), which is the reverse of what students expect. The mean must lie between the two prices, otherwise the problem has no solution. Also keep \pi=\frac{22}{7} for mensuration when the radius is a multiple of 7, and remember that the cone, sphere and cylinder ratio is 1:2:3 when h=2r (Fig 2.6).

2.7.1 Worked checks

Q. A trader marks a watch 25 percent above cost and then allows a 20 percent discount. Find his gain percent.

Answer. Take CP =100. Marked price =125. Selling price =125\times0.80=100. So the gain is 0 percent. Using the successive-change formula, 25-20-\frac{25\times20}{100}=0, which agrees.

Q. Find the compound interest on Rs 8,000 for 2 years at 5 percent per annum.

Answer. Amount =8000\times1.05^2=8000\times1.1025=8820, so CI=8820-8000=\text{Rs }820.

Check with the shortcut SI=\frac{8000\times5\times2}{100}=800 and CI-SI=\frac{8000\times25}{10000}=20, so CI=820. Both routes give the same result.

Q. A can finish a job in 12 days and B in 18 days. They work together for 4 days and then A leaves. How many more days does B need?

Answer. Work done in 4 days =4\left(\frac1{12}+\frac1{18}\right)=4\times\frac{5}{36}=\frac{5}{9}, so \frac49 remains. B works at \frac1{18} per day, so he needs \frac49\times18=8 more days.

Check B in 12 days alone does \frac{12}{18}=\frac23, and A does \frac4{12}=\frac13, total 1.

Formula and fact sheet

Topic Result
Profit % \frac{SP-CP}{CP}\times100
Discount on MP; successive discounts multiply
SI, CI (2 yr) \frac{PRT}{100}; \text{CI}-\text{SI}=P(\frac{R}{100})^2
Two workers \frac{ab}{a+b} days
Average speed \frac{2xy}{x+y} (equal distances)
Boat u=\frac{D+U}{2}, v=\frac{D-U}{2}
Alligation cheaper : dearer =(d-m):(m-c)
Cylinder, cone, sphere \pi r^2h, \frac{1}{3}\pi r^2h, \frac{4}{3}\pi r^3

Chapter summary

  • Profit and loss use the cost price; discount uses the marked price.
  • Mark up 25 % and discount 20 % gives no gain, no loss.
  • For two years, compound interest exceeds simple interest by P(R/100)^2.
  • Two workers together take ab/(a+b) days; a leaking pipe subtracts its rate.
  • 36 km/h is 10 m/s; multiply by 5/18 to go from km/h to m/s.
  • Average speed over equal distances is the harmonic mean, never the simple mean.
  • A train crossing a platform covers its own length plus the platform.
  • Alligation gives the ratio directly as the two differences across the cross.

Practice questions

Set A — Recall (20 questions)

  1. Profit per cent is calculated on theaselling pricebmarked priceccost priceddiscount
  2. The simple interest on Rs 5,000 at 8 % per annum for 3 years isaRs 1,200bRs 1,000cRs 1,500dRs 1,600
  3. A speed of 36 km/h equalsa6 m/sb8 m/sc12 m/sd10 m/s
  4. A can do a job in 10 days and B in 15 days. Working together they finish ina5 daysb6 daysc8 daysd12 days
  5. An article bought for Rs 400 is sold for Rs 460. The profit per cent isa15 %b12 %c14 %d60 %
  6. The area of a circle of radius 7 cm (π = 22/7) isa44 cm²b49 cm²c154 cm²d308 cm²
  7. The volume of a cube of side 5 cm isa25 cm³b75 cm³c150 cm³d125 cm³
  8. The difference between compound and simple interest on Rs 1,000 at 10 % for 2 years isaRs 1bRs 10cRs 100dRs 20
  9. The area of a triangle of base 10 cm and height 6 cm isa60 cm²b16 cm²c30 cm²d15 cm²
  10. A marked price of Rs 800 with a 10 % discount gives a selling price ofaRs 720bRs 710cRs 700dRs 730
  11. A man covers equal distances at 40 km/h and 60 km/h. His average speed isa50 km/hb52 km/hc45 km/hd48 km/h
  12. The surface area of a sphere of radius r isa2\pi r^2b4\pi r^2c\frac{4}{3}\pi r^2d\pi r^2
  13. The curved surface area of a cylinder isa\pi r^2hb2\pi r(r+h)c2\pi rhd\pi rh
  14. To convert m/s into km/h we multiply bya\frac{5}{18}b\frac{3}{5}c\frac{1000}{60}d\frac{18}{5}
  15. The perimeter of a rectangle 12 m by 8 m isa40 mb96 mc20 md48 m
  16. The volume of a cone with radius 3 cm and height 7 cm (π = 22/7) isa44 cm³b66 cm³c132 cm³d198 cm³
  17. An article costing Rs 500 is sold at a 10 % loss. The selling price isaRs 400bRs 440cRs 450dRs 490
  18. If a worker finishes a job in 8 days, his one-day work isa\frac{1}{8}b8c\frac{1}{16}d\frac{1}{4}
  19. A rhombus has diagonals 10 cm and 24 cm. Its area isa240 cm²b34 cm²c60 cm²d120 cm²
  20. At simple interest a sum doubles itself in 10 years. The rate per annum isa5 %b10 %c12.5 %d20 %
Answers and explanations — Set A

1. (c) — profit and loss per cent are on the cost price; discount is on the marked price.

2. (a) — 5000 × 8 × 3 / 100 = 1200.

3. (d) — 36 × 5/18 = 10 m/s.

4. (b) — 10 × 15 / 25 = 6 days.

5. (a) — profit 60 on cost 400 is 15 %. Option (d) divides by the selling price's gap wrongly.

6. (c) — (22/7) × 49 = 154. Option (a) 44 is the circumference.

7. (d) — 5³ = 125. Option (a) 25 is a face area.

8. (b) — P(R/100)² = 1000 × 0.01 = 10.

9. (c) — ½ × 10 × 6 = 30. Forgetting the ½ gives 60.

10. (a) — 800 × 0.9 = 720.

11. (d) — 2 × 40 × 60 / 100 = 48. The simple mean 50 is the standard distractor.

12. (b) — surface area 4πr²; (c) is a volume-like expression.

13. (c) — curved surface = circumference × height = 2πrh; (b) is the total surface.

14. (d) — m/s to km/h is × 18/5 (= 3.6); the reverse is × 5/18.

15. (a) — 2(12 + 8) = 40.

16. (b) — (1/3)(22/7)(9)(7) = 66.

17. (c) — 500 × 0.9 = 450.

18. (a) — one-day work is 1 over the days taken.

19. (d) — ½ × 10 × 24 = 120. Option (a) forgets the ½.

20. (b) — time to double = 100 / R, so R = 100 / 10 = 10 %.

Set B — Understanding (12 questions)

  1. A shopkeeper allows successive discounts of 10 % and 20 % on a marked price of Rs 1,000. The selling price isaRs 700bRs 720cRs 730dRs 750
  2. A trader marks goods 25 % above cost and allows a 20 % discount. He makesa5 % profitb5 % lossc20 % profitdneither profit nor loss
  3. The compound interest on Rs 8,000 at 10 % per annum for 2 years isaRs 1,680bRs 1,600cRs 1,800dRs 1,720
  4. At simple interest a sum becomes Rs 6,600 in 2 years and Rs 7,200 in 3 years. The principal isaRs 5,000bRs 5,200cRs 5,400dRs 5,600
  5. A can finish a job in 12 days and B in 18 days. A works for 4 days and leaves; B finishes the rest ina8 daysb9 daysc10 daysd12 days
  6. A pipe fills a tank in 20 minutes and another empties it in 30 minutes. With both open the tank fills ina50 minb60 minc120 mind25 min
  7. A train 150 m long passes a pole in 10 seconds. Its speed isa54 km/hb45 km/hc60 km/hd72 km/h
  8. A boat rows at 8 km/h in still water and the stream flows at 2 km/h. It goes 30 km downstream and comes back. The total time isa6 hb7 hc8 hd9 h
  9. Sugar at Rs 40/kg is mixed with sugar at Rs 60/kg to give a mixture worth Rs 52/kg. The ratio of cheaper to dearer isa3:2b2:3c1:2d4:5
  10. The length of a rectangle is increased by 20 % and its breadth decreased by 10 %. The areaaincreases by 8 %bincreases by 10 %cincreases by 2 %dstays the same
  11. The radius of a sphere is doubled. Its volume becomesa2 timesb4 timesc6 timesd8 times
  12. The diagonal of a cuboid measuring 3 cm × 4 cm × 12 cm isa11 cmb12 cmc13 cmd19 cm
Answers and explanations — Set B

1. (b) — 1000 × 0.9 × 0.8 = 720. Adding the discounts (30 %) would wrongly give 700.

2. (d) — 1.25 × 0.8 = 1.00 of cost.

3. (a) — 8000 × (1.21 − 1) = 1,680. Simple interest would be 1,600.

4. (c) — yearly interest = 600, so P = 6600 − 2 × 600 = 5,400.

5. (d) — after 4 days 1 − 4/12 = 2/3 remains; B needs (2/3) × 18 = 12 days.

6. (b) — 1/20 − 1/30 = 1/60 per minute, so 60 min. Subtracting the times gives the wrong 10 min.

7. (a) — 150/10 = 15 m/s = 15 × 18/5 = 54 km/h.

8. (c) — 30/10 + 30/6 = 3 + 5 = 8 h.

9. (b) — cheaper : dearer = (60 − 52) : (52 − 40) = 8 : 12 = 2 : 3.

10. (a) — 1.2 × 0.9 = 1.08, an 8 % rise.

11. (d) — volume ∝ r³, so 2³ = 8.

12. (c) — √(9 + 16 + 144) = √169 = 13.

Set C — Application (10 questions)

  1. By selling an article for Rs 540 a man gains 20 %. To gain 30 % he must sell it foraRs 560bRs 570cRs 580dRs 585
  2. A sum at simple interest amounts to Rs 1,200 in 2 years and Rs 1,350 in 3 years. The rate per annum isa16⅔ %b15 %c12.5 %d20 %
  3. A and B together can do a job in 12 days and A alone in 20 days. B alone can do it ina24 daysb28 daysc30 daysd32 days
  4. A train 120 m long crosses a platform 180 m long in 20 seconds. Its speed isa45 km/hb54 km/hc60 km/hd72 km/h
  5. A man walks to a place at 4 km/h and returns at 6 km/h. The whole journey takes 5 hours. The distance isa6 kmb8 kmc10 kmd12 km
  6. From 40 L of milk, 4 L is drawn off and replaced with water; the process is repeated once more. The milk left isa32.4 Lb32 Lc33.6 Ld36 L
  7. The volume of a cylinder of radius 7 cm and height 10 cm (π = 22/7) isa440 cm³b770 cm³c1,540 cm³d3,080 cm³
  8. Two trains 100 m and 150 m long run in opposite directions at 36 km/h and 54 km/h. They cross each other ina8 sb10 sc12.5 sd25 s
  9. A dealer uses a 900 g weight in place of 1 kg. His gain per cent isa11⅑ %b10 %c9 %d12 %
  10. The side of a square is increased by 10 %. Its area increases bya10 %b11 %c20 %d21 %
Answers and explanations — Set C

1. (d) — CP = 540 / 1.2 = 450; 450 × 1.3 = 585.

2. (a) — yearly interest 150, P = 1200 − 300 = 900, rate = 150/900 = 16⅔ %.

3. (c) — 1/12 − 1/20 = 2/60 = 1/30, so 30 days.

4. (b) — distance 300 m in 20 s is 15 m/s = 54 km/h; forgetting the platform gives 6 m/s.

5. (d) — d/4 + d/6 = 5d/12 = 5, so d = 12 km.

6. (a) — 40 × (1 − 4/40)² = 40 × 0.81 = 32.4 L.

7. (c) — (22/7) × 49 × 10 = 1,540.

8. (b) — relative speed 90 km/h = 25 m/s; distance 250 m; time 10 s.

9. (a) — gain = 100/900 × 100 = 11⅑ %; the dealer gets 1000 g's price for 900 g of goods.

10. (d) — 10 + 10 + 100/100 = 21 %; the common error is 20 %.