CBSE Class 7 Mathematics Another Peek beyond the Point Worksheet Set 01

Long Answer Questions

Question 1. Study the number 23.45 carefully and answer the following:
(i) Write the products 23.45 × 10, 23.45 × 100 and 23.45 × 1000.
(ii) Write the quotients 23.45 ÷ 10, 23.45 ÷ 100 and 23.45 ÷ 1000.
(iii) State the rule for multiplying a decimal by 1 followed by zeroes and the rule for dividing a decimal by 1 followed by zeroes, and say how the two rules are related to each other.
Answer:
(i) 23.45 × 10 = 234.5; 23.45 × 100 = 2345; 23.45 × 1000 = 23450.
(ii) 23.45 ÷ 10 = 2.345; 23.45 ÷ 100 = 0.2345; 23.45 ÷ 1000 = 0.02345.
(iii) Multiplying by 1 followed by zeroes: move the decimal point to the right by as many places as there are zeroes, writing zeroes at the end if needed (23.45 × 1000 = 23450).
Dividing by 1 followed by zeroes: move the point to the left by as many places as there are zeroes, writing zeroes in front if needed (23.45 ÷ 1000 = 0.02345).
The two rules are opposites of each other: × 100 makes a number 100 times bigger, while ÷ 100 (the same as × \( \frac{1}{100} \)) makes it 100 times smaller. Doing one after the other gives back the number: 23.45 × 100 ÷ 100 = 23.45.
In simple words: Multiply: point goes right; divide: point goes left by the same number of places.

Teacher's Note:
1. Count the zeroes in 10, 100, 1000.
2. Fill empty places with zeroes.
3. One operation undoes the other.

Question 2. Meena goes to a stationery shop. She buys 3 notebooks costing ₹24.50 each, 2 pens costing ₹8.75 each and one geometry box costing ₹65.25.
(i) Find the amount she pays for the notebooks and the amount she pays for the pens.
(ii) Find the total bill.
(iii) If she gives the shopkeeper a ₹200 note, how much money will she get back? Show the subtraction with the decimal points aligned.
Answer:
(i) Notebooks: 3 × 24.50; 3 × 2450 = 7350 with 2 decimal digits = ₹73.50. Pens: 2 × 8.75; 2 × 875 = 1750, so ₹17.50.
(ii) Writing the amounts one below the other with the points in a line: 73.50 + 17.50 + 65.25 = 156.25. Total bill = ₹156.25.
(iii) Write ₹200 as 200.00 so the points line up: 200.00 − 156.25 = 43.75. Meena gets back ₹43.75.
Check: 156.25 + 43.75 = 200.00. Adding in another order, 65.25 + 73.50 = 138.75 and 138.75 + 17.50 = 156.25, the same total.
In simple words: Find each cost, add for the bill, subtract from ₹200 for the change.

Teacher's Note:
1. Line up the decimal points when adding or subtracting.
2. Write 200 as 200.00.
3. Check by adding the change back.

Question 3. Find the value of 589 ÷ 8 by the long division (division using place value) method.
(i) Carry out the division and write the quotient.
(ii) Explain at which step the decimal point is placed in the quotient and why.
(iii) Verify your answer by writing \( \frac{589}{8} \) as an equivalent fraction with denominator 1000.
Answer:
(i) 589 = 5 hundreds + 8 tens + 9 ones.
5 hundreds cannot be shared among 8, so regroup as 50 tens; 50 + 8 = 58 tens. 58 tens ÷ 8 = 7 tens, remainder 2 tens.
2 tens = 20 ones; with 9 ones, 29 ones ÷ 8 = 3 ones, remainder 5 ones.
5 ones = 50 tenths; 50 ÷ 8 = 6 tenths, remainder 2 tenths. 2 tenths = 20 hundredths; 20 ÷ 8 = 2 hundredths, remainder 4 hundredths. 4 hundredths = 40 thousandths; 40 ÷ 8 = 5 thousandths, remainder 0.
So 589 ÷ 8 = 73.625.
(ii) The decimal point is placed when the ones are regrouped as tenths. Everything before that step is the whole-number part and everything after it is the decimal part.
(iii) \( \frac{589}{8} \) = \( \frac{589 × 125}{8 × 125} \) = \( \frac{73625}{1000} \) = 73.625. Both methods agree. ✓
In simple words: Long division gives 73.625, and the fraction method gives the same.

Teacher's Note:
1. Regroup the leftover into the next smaller place each time.
2. 8 × 125 = 1000 is useful.
3. The point comes in when ones become tenths.

Teacher's Note:
1. The key comparison is with 1.
2. Same idea as for fractions.
3. Use the table to see the pattern.

Question 5. The thickness of one coin of a certain kind is 1.6 mm. The coins are stacked one above the other to form a cylinder.
(i) Find the height of a stack of 45 such coins in millimetres, and then write this height in centimetres.
(ii) How many such coins are needed to make a stack exactly 12 cm high? Show the division clearly.
(iii) Write the thickness of one coin in centimetres, and use it to check your answer to part (i).
Answer:
(i) Height = 1.6 × 45 mm. 16 × 45 = 720, with one decimal digit: 72.0 = 72 mm. Since 10 mm = 1 cm, 72 ÷ 10 = 7.2 cm.
(ii) 12 cm = 12 × 10 = 120 mm. Number of coins = 120 ÷ 1.6 = \( \frac{120 × 10}{1.6 × 10} \) = \( \frac{1200}{16} \) = 75. So 75 coins are needed.
(iii) 1.6 mm = 1.6 ÷ 10 = 0.16 cm. Height of 45 coins = 45 × 0.16 = 7.2 cm, which matches part (i). Also 75 × 0.16 = 12 cm, so part (ii) is correct too.
In simple words: Multiply for the height, divide for the number of coins, and check in centimetres.

Teacher's Note:
1. Use the same unit before dividing.
2. 10 mm = 1 cm.
3. Checking in another unit catches mistakes.

Question 6. A farmer harvests 240 kg of wheat. He packs all of it into bags, each holding 7.5 kg, and sells the wheat at ₹28.50 per kilogram.
(i) Find the number of bags he fills.
(ii) Find the total money he gets by selling all the wheat.
(iii) Find the price of the wheat in one bag, and show that your answers to (i), (ii) and (iii) agree with one another.
Answer:
(i) Bags = 240 ÷ 7.5 = \( \frac{240 × 10}{7.5 × 10} \) = \( \frac{2400}{75} \) = 32. He fills 32 bags.
(ii) Total = 240 × 28.50. 240 × 2850 = 684000, with 2 decimal digits: 6840.00. He gets ₹6840.
(iii) One bag = 7.5 × 28.50. 75 × 2850 = 213750, with 1 + 2 = 3 decimal digits: 213.750. One bag is worth ₹213.75.
Agreement: 32 × 213.75 = 6840, the total in (ii); and 32 × 7.5 = 240 kg, the whole harvest. So all three answers fit together.
In simple words: 32 bags at ₹213.75 each make ₹6840, the same as 240 kg at ₹28.50.

Teacher's Note:
1. Make the divisor whole first.
2. Count decimal digits in each product.
3. Cross-check using a second route.

Question 7. The boundary of a park is in the shape of a regular octagon (a polygon with 8 equal sides). Its perimeter is 148.8 m.
(i) Find the length of one side of the park.
(ii) The park is to be fenced at ₹45.50 per metre. Find the total cost of fencing.
(iii) A second park is a square with the same perimeter. Find the length of its side and state which park has the longer side.
Answer:
(i) Side = 148.8 ÷ 8 = 18.6 m. Check: 8 × 18.6 = 148.8 m.
(ii) Cost = 148.8 × 45.50. 1488 × 4550 = 6770400; 1 + 2 = 3 decimal digits: 6770.400. The fencing costs ₹6770.40.
(iii) Square side = 148.8 ÷ 4 = 37.2 m. Since 37.2 > 18.6, the square park has the longer side.
In fact 37.2 is exactly twice 18.6: the same perimeter shared among 4 sides instead of 8 doubles each side.
In simple words: Divide the perimeter by the number of sides; fewer sides means longer sides.

Teacher's Note:
1. Perimeter ÷ number of sides = one side.
2. Fence the whole perimeter, not one side.
3. Half the sides, double the length.

Question 8. Look at the two divisions 32.76 ÷ 1.4 and 32.76 ÷ 0.14.
(i) Find both quotients, showing how you change the divisor into a counting number in each case.
(ii) Explain why the second quotient is exactly ten times the first one.
(iii) In the division 32.76 ÷ 0.14, the quotient is larger than the dividend. Explain when this happens, and give one more example of your own.
Answer:
(i) 32.76 ÷ 1.4: multiply both by 10: \( \frac{32.76 × 10}{1.4 × 10} \) = \( \frac{327.6}{14} \) = 23.4.
32.76 ÷ 0.14: multiply both by 100: \( \frac{32.76 × 100}{0.14 × 100} \) = \( \frac{3276}{14} \) = 234.
Check: 1.4 × 23.4 = 32.76 and 0.14 × 234 = 32.76.
(ii) The dividend is the same, but the second divisor 0.14 is one-tenth of 1.4. Sharing the same amount into pieces ten times smaller gives ten times as many pieces: 23.4 × 10 = 234.
(iii) This happens whenever the divisor is less than 1: a number smaller than 1 fits into the dividend more times than 1 does. Example: 7.5 ÷ 0.5 = 15, and 15 is greater than 7.5.
In simple words: Smaller divisor, bigger quotient; a divisor below 1 makes the quotient bigger than the dividend.

Teacher's Note:
1. Multiply both numbers by the same power of 10.
2. Divisor ÷ 10 means quotient × 10.
3. Any divisor between 0 and 1 works for the example.

Question 9. Divide 1 by 7 using long division and carry the division to at least six decimal places.
(i) Write the quotient you get and list the remainders in the order in which they appear.
(ii) Explain why this division never ends, using the remainders you listed.
(iii) The block of digits 142857 that appears in the quotient is a famous “cyclic” number. Multiply it by 2, by 3 and by 7, and describe what you notice in each case.
Answer:
(i) 1 ÷ 7 = 0.142857142857 … The remainders come in the order 1, 3, 2, 6, 4, 5, and then 1 again, after which the list repeats.
(ii) When dividing by 7 the only possible remainders are 1, 2, 3, 4, 5, 6 (a remainder 0 would end it). Here 0 never appears. After six steps remainder 1 comes back — the one we started with — so every step repeats and the digits 142857 come again and again for ever. So 1 ÷ 7 cannot be written with a finite number of decimal digits.
(iii) 142857 × 2 = 285714; 142857 × 3 = 428571; 142857 × 7 = 999999.
With 2 and 3 we get the same six digits 1, 4, 2, 8, 5, 7 in the same cyclic order, just starting at a different place. With 7 we get 999999, a string of nines — which is why the repeating block of \( \frac{1}{7} \) has exactly six digits.
In simple words: The remainders cycle through six values, so the six digits 142857 repeat for ever.

Teacher's Note:
1. List remainders as you go.
2. Only six non-zero remainders are possible.
3. Cyclic numbers rotate their digits.

Question 10. The four digits 1, 3, 6 and 8 are to be placed, one in each box, in the arrangement shown below.
(i) Find the arrangement that gives the greatest product, and write that product.
(ii) Find the arrangement that gives the smallest product, and write that product.
(iii) Explain the reasoning you used to decide which digits should go before the decimal points and which should go after them, without testing every possible arrangement.
Answer:
(i) Greatest product: 6.3 × 8.1 = 51.03 (63 × 81 = 5103, with 1 + 1 = 2 decimal digits).
(ii) Smallest product: 1.6 × 3.8 = 6.08 (16 × 38 = 608, with 2 decimal digits).
(iii) A digit before the point is worth ten times a digit after it, so the ones places decide most of the size.
For the greatest product, the two largest digits 8 and 6 go in the ones places and 3 and 1 after the points. The smaller leftover digit goes with the larger whole part: 8.1 × 6.3 = 51.03 beats 8.3 × 6.1 = 50.63.
For the smallest product, the two smallest digits 1 and 3 go in the ones places and 8 and 6 after the points. Pairing the larger tenths digit with the smaller whole part gives the smaller answer: 1.6 × 3.8 = 6.08 is less than 1.8 × 3.6 = 6.48.
In simple words: Big digits in the ones places give a big product; small digits there give a small product.

Teacher's Note:
1. Ones digits matter ten times more than tenths digits.
2. Then compare only the two remaining pairings.
3. 6.3 × 8.1 and 8.1 × 6.3 are the same product.

Case Study Based Questions

Question 11. A morning at the vegetable shop
Lakshmi goes to a vegetable shop near her house with a ₹500 note. The rates written on the shop board are: potatoes ₹24.60 per kilogram, onions ₹32.40 per kilogram and apples ₹160 per kilogram. Lakshmi asks the shopkeeper for 2.5 kg of potatoes, 1.5 kg of onions and 0.75 kg of apples. The shopkeeper weighs each item on a digital balance, writes the weight and the rate side by side in his bill book, and multiplies to get the cost of each item. He then adds the three amounts to make the bill and returns the balance money to Lakshmi.
(i) Find the cost of the potatoes.
(ii) Find the cost of the onions.
(iii) Find the total amount of the bill.
(iv) How much money does the shopkeeper return to Lakshmi?
Answer:
(i) 2.5 × 24.60: 25 × 2460 = 61500, with 1 + 2 = 3 decimal digits: ₹61.50.
(ii) 1.5 × 32.40: 15 × 3240 = 48600, with 3 decimal digits: ₹48.60.
(iii) Apples: 0.75 × 160 = ₹120. Bill = 61.50 + 48.60 + 120.00 = ₹230.10.
(iv) Balance = 500.00 − 230.10 = ₹269.90. Check: 230.10 + 269.90 = 500.00.
In simple words: Cost of each item = weight × rate; add for the bill and subtract from ₹500.

Teacher's Note:
1. Count decimal digits in each product.
2. Line up points while adding.

Question 12. Buying cloth by the metre
In a cloth shop, printed curtain cloth is sold at ₹85.50 per metre and plain lining cloth at ₹48 per metre. The shopkeeper measures the cloth with a metre rod and cuts exactly the length a customer asks for, so a customer may buy a length that is not a whole number of metres. Asha wants new curtains for one window. She buys 3.5 m of the printed curtain cloth and 1.2 m of the lining cloth. The printed cloth comes from a full roll that is 40 m long.
(i) Find the cost of the printed curtain cloth Asha buys.
(ii) Find the cost of the lining cloth she buys.
(iii) Find the total amount Asha has to pay.
(iv) How many pieces of 3.5 m can be cut from the full 40 m roll, and what length of cloth is left over?
Answer:
(i) 3.5 × 85.50: 35 × 8550 = 299250, with 3 decimal digits: ₹299.25.
(ii) 1.2 × 48: 12 × 48 = 576, with 1 decimal digit: ₹57.60.
(iii) Total = 299.25 + 57.60 = ₹356.85.
(iv) 40 ÷ 3.5 = \( \frac{400}{35} \); 35 × 11 = 385 and 35 × 12 = 420, so 11 pieces can be cut. Cloth used = 11 × 3.5 = 38.5 m, so 1.5 m is left over.
In simple words: Multiply length by rate, add the costs, and divide the roll to find the pieces.

Teacher's Note:
1. Only whole pieces count in (iv).
2. Leftover = roll − cloth used.

Question 13. Sharing a bill
Four friends go to a small eating place after a school match. They order 4 sandwiches at ₹95.50 each and 4 glasses of fresh juice at ₹62.40 each. They also share one large cake that costs ₹216 for the whole cake. When the bill comes, the friends decide that, instead of counting who ate what, they will simply divide the total amount equally among the four of them. One of them works out the total on paper, writing the rupees and paise one below the other with the decimal points in a line, and then divides the total by 4.
(i) Find the total cost of the 4 sandwiches.
(ii) Find the total cost of the 4 glasses of juice.
(iii) Find the total amount of the bill.
(iv) Find the share of each friend.
Answer:
(i) 4 × 95.50: 4 × 9550 = 38200, with 2 decimal digits: ₹382.
(ii) 4 × 62.40: 4 × 6240 = 24960, so ₹249.60.
(iii) Bill = 382.00 + 249.60 + 216.00 = ₹847.60.
(iv) Share = 847.60 ÷ 4 = ₹211.90 (84 ÷ 4 = 21, then 7.6 ÷ 4 = 1.9). Check: 4 × 211.90 = 847.60.
In simple words: Add everything, then divide the bill by 4.

Teacher's Note:
1. The cake price is already for the whole cake.
2. Check the share by multiplying by 4.

Question 14. A bottle of cough syrup
A doctor gives a child a bottle of cough syrup. The bottle holds 60 ml of syrup and costs ₹96. The label says that one dose is 2.5 ml and that the dose must be given three times a day, always with the measuring cup supplied with the bottle. The child’s mother measures each dose carefully, because a dose that is too small will not help and a dose that is too large is not safe. She wants to know in advance how many days the bottle will last, so that she can buy the next bottle in time if the cough has not stopped.
(i) How much syrup does the child take in one day?
(ii) For how many days will one full bottle last?
(iii) How much syrup is used in 5 days?
(iv) Find the cost of 1 ml of the syrup.
Answer:
(i) 3 × 2.5 = 7.5 ml a day.
(ii) 60 ÷ 7.5 = \( \frac{600}{75} \) = 8, so the bottle lasts 8 days.
(iii) 5 × 7.5 = 37.5 ml.
(iv) 96 ÷ 60 = \( \frac{96}{60} \) = \( \frac{16}{10} \) = 1.6, that is ₹1.60 per ml. Check: 60 × 1.6 = ₹96.
In simple words: Daily dose 7.5 ml lasts 8 days; each ml costs ₹1.60.

Teacher's Note:
1. Three doses a day, not one.
2. Make the divisor whole before dividing.

Question 15. At the petrol pump
At a petrol pump the display shows the rate, the number of litres filled and the amount to be paid. On a particular day petrol costs ₹104.50 per litre and diesel costs ₹91.20 per litre. The machine first multiplies the rate by the number of litres to show the amount; if a customer instead asks for petrol worth a fixed amount of money, the machine divides that amount by the rate to decide how many litres to fill. During one hour the attendant fills 12 litres of petrol in a car, petrol worth ₹2090 in a taxi, and fills the 5.6 litre tank of a scooter completely.
(i) Find the amount paid by the car owner.
(ii) How many litres of petrol are filled in the taxi?
(iii) Find the amount paid by the scooter owner.
(iv) How much less would 40 litres of diesel cost than 40 litres of petrol?
Answer:
(i) 12 × 104.50: 12 × 10450 = 125400, with 2 decimal digits: ₹1254.
(ii) 2090 ÷ 104.50 = \( \frac{209000}{10450} \) = 20, so 20 litres. Check: 20 × 104.50 = ₹2090.
(iii) 5.6 × 104.50: 56 × 10450 = 585200, with 3 decimal digits: ₹585.20.
(iv) Difference in rate = 104.50 − 91.20 = ₹13.30 per litre; for 40 litres: 40 × 13.30 = ₹532 less.
In simple words: Multiply for the amount, divide for the litres, and use the rate difference for (iv).

Teacher's Note:
1. Money ÷ rate = litres.
2. Find the difference per litre first in (iv).

Question 16. How far on one litre?
A family drives from their town to a hill station and back. Before starting, they fill the tank of their car completely and note the reading of the trip meter as zero. At the end of the whole journey the trip meter shows 357.5 km, and when they fill the tank completely again it takes exactly 27.5 litres of petrol. This tells them how many kilometres the car runs on one litre, which is called the fuel efficiency of the car. On that day petrol cost ₹102.40 per litre. A month later the same car is driven 195 km on a different road, and the family wants to work out the petrol used for that trip too.
(i) How many kilometres does the car run on one litre of petrol?
(ii) How much petrol is needed for a journey of 221 km?
(iii) Find the cost of the petrol used on the hill station trip.
(iv) Find the petrol used, and the cost of that petrol, for the 195 km trip.
Answer:
(i) 357.5 ÷ 27.5 = \( \frac{3575}{275} \) = 13, so 13 km per litre.
(ii) 221 ÷ 13 = 17 litres.
(iii) 27.5 × 102.40: 275 × 10240 = 2816000, with 3 decimal digits: ₹2816.
(iv) 195 ÷ 13 = 15 litres; cost = 15 × 102.40 = ₹1536.
Check: 27.5 × 13 = 357.5 km and 15 × 13 = 195 km.
In simple words: Distance ÷ litres gives km per litre; use it to find petrol for other trips.

Teacher's Note:
1. Fuel efficiency = distance ÷ fuel.
2. Fuel needed = distance ÷ efficiency.

Question 17. The milkman’s monthly account
A dairy sells cow’s milk at ₹58.50 per litre. The Das family takes 1.5 litres of milk every day in a measuring can. The milkman does not collect money daily; he writes the quantity in a small notebook each morning and collects the money for the whole month at the end of the month. In this way the family pays once for all the milk taken during the month. From the first day of the next month the dairy raises the rate of milk to ₹61.50 per litre, but the family decides to continue taking the same quantity of milk every day.
(i) Find the cost of the milk taken in one day.
(ii) How many litres of milk does the family take in one week of 7 days?
(iii) Find the amount the family pays for a month of 30 days at the old rate.
(iv) How much extra will the family have to pay in a 30 day month after the rate is raised?
Answer:
(i) 1.5 × 58.50: 15 × 5850 = 87750, with 3 decimal digits: ₹87.75.
(ii) 7 × 1.5 = 10.5 litres.
(iii) Milk in 30 days = 30 × 1.5 = 45 litres; bill = 45 × 58.50 = ₹2632.50 (also 30 × 87.75 = ₹2632.50).
(iv) Rise in rate = 61.50 − 58.50 = ₹3 per litre; extra for 45 litres = 45 × 3 = ₹135.
In simple words: Find the daily cost and monthly litres; the extra is the rate rise times 45 litres.

Teacher's Note:
1. Two routes for (iii) give the same answer.
2. Only the rise in rate matters in (iv).

Question 18. Copper wire at the ware shop
A ware shop sells copper wire at ₹14.50 per metre. The wire is kept on rolls, and each full roll is 25 m long. When a customer asks for a length that is not a whole number of metres, the shopkeeper measures it with a steel tape marked in centimetres and millimetres, cuts the wire and works out the price by multiplication. One customer buys 3.6 m of wire for a repair job. Another customer buys a full roll and asks the shopkeeper to cut it into 8 pieces of equal length for an electrician.
(i) Find the cost of one full roll of wire.
(ii) Find the cost of the 3.6 m piece.
(iii) Write the length 3.6 m in centimetres and in millimetres.
(iv) Find the length of each of the 8 equal pieces, in metres and in centimetres.
Answer:
(i) 25 × 14.50: 25 × 1450 = 36250, with 2 decimal digits: ₹362.50.
(ii) 3.6 × 14.50: 36 × 1450 = 52200, with 3 decimal digits: ₹52.20.
(iii) 3.6 m = 3.6 × 100 = 360 cm and 3.6 × 1000 = 3600 mm.
(iv) 25 ÷ 8 = 3.125 m (long division), which is 3.125 × 100 = 312.5 cm.
In simple words: Multiply length by rate for cost; × 100 and × 1000 change metres to cm and mm.

Teacher's Note:
1. 1 m = 100 cm = 1000 mm.
2. Continue long division until the remainder is 0.

Question 19. Why we need leap years
The Earth does not take exactly 365 days to go once around the Sun; it takes 365.2422 days. A calendar year, however, must have a whole number of days, so an ordinary calendar year is taken as 365 days. The small leftover time adds up year after year, and if nothing were done the months of the calendar would slowly drift away from the seasons. To correct this, an extra day is added in February every fourth year. A further correction says that a century year is a leap year only when it is divisible by 400, so the year 2000 was a leap year while 1900 was not.
(i) How many extra days does the Earth need after one calendar year of 365 days?
(ii) Using decimal multiplication, find the total number of such extra days after 100 calendar years.
(iii) In 4 years a calendar with one leap day has 1461 days. Find the number of days the Earth actually takes for 4 revolutions, and find the difference between the two.
(iv) Is the year 2100 a leap year? Give a reason for your answer.
Answer:
(i) 365.2422 − 365 = 0.2422 day.
(ii) 0.2422 × 100 = 24.22 days (point moves two places right).
(iii) 4 × 365.2422: 4 × 3652422 = 14609688, with 4 decimal digits: 1460.9688 days. Difference = 1461 − 1460.9688 = 0.0312 day; the calendar is slightly longer.
(iv) No. 2100 is divisible by 100 but not by 400, and a century year is a leap year only when it is divisible by 400.
In simple words: About a quarter day is left over each year, fixed by leap years with a century rule.

Teacher's Note:
1. Find the extra part first.
2. Century years need the 400 test.

Question 20. Profit on a sack of rice
A shopkeeper buys 45 kg of rice from a wholesale dealer at ₹38.60 per kilogram and plans to sell all of it in his shop at ₹44.50 per kilogram. He writes the buying price and the selling price in two columns of his account book, works out both totals by decimal multiplication, and subtracts to find his profit. He also likes to know his profit on each kilogram, because that helps him fix the rate for the next sack. During that week, however, 2.5 kg of the rice gets spoilt by damp and has to be thrown away, so he is able to sell only the remaining rice.
(i) Find the total amount the shopkeeper pays for the rice.
(ii) Find the amount he would get by selling all 45 kg, and the profit he would make.
(iii) Find his profit on one kilogram of rice in this plan.
(iv) Find the money he actually gets after 2.5 kg is thrown away, and his actual profit.
Answer:
(i) 45 × 38.60: 45 × 3860 = 173700, with 2 decimal digits: ₹1737.
(ii) 45 × 44.50 = ₹2002.50 (45 × 4450 = 200250). Profit = 2002.50 − 1737.00 = ₹265.50.
(iii) 265.50 ÷ 45 = ₹5.90; the same as 44.50 − 38.60 = ₹5.90 per kg.
(iv) Rice sold = 45 − 2.5 = 42.5 kg. Money = 42.5 × 44.50 = ₹1891.25 (425 × 4450 = 1891250, 3 decimal digits). Actual profit = 1891.25 − 1737.00 = ₹154.25, which is ₹111.25 less than planned.
In simple words: Profit = selling amount − buying amount; spoilt rice cuts the selling amount.

Teacher's Note:
1. The cost stays the same even if rice is spoilt.
2. Profit per kg can be found two ways.