CBSE Class 7 Mathematics A Peek Beyond the Poin Worksheet Set 01

Long Answer Questions

Question 1. A tailor needs a measurement more exact than whole centimetres.
(i) Explain why a ruler marked only in whole centimetres cannot give the exact length of a small hook.
(ii) Describe how one unit is split to get one-tenths, and state how many one-tenths make one unit.
(iii) A piece of lace is 3 units and 6 one-tenths long. Write this length in three different ways and read each of them aloud in words.
Answer:
(i) A hook is rarely exactly 2 or 3 cm long. With only whole-centimetre marks we can just say it is “between 2 cm and 3 cm”; there are no marks in between, so the exact length cannot be read. A smaller unit is needed.
(ii) The unit length between two consecutive numbers is divided into 10 equal parts; each part is one-tenth, written \( \frac{1}{10} \). Since 10 × \( \frac{1}{10} \) = 1 unit, 10 one-tenths make one unit.
(iii) Mixed form: \( 3\frac{6}{10} \) units, read “three and six-tenths”.
Tenths only: 3 units = 30 one-tenths, so the length is \( \frac{36}{10} \) units, read “thirty-six tenths”.
Decimal form: 3.6 units, read “three point six”. All three show the same length.
In simple words: Splitting a unit into 10 parts lets us write 3 and 6 tenths as 36/10 or 3.6.

Teacher's Note:
1. Whole units are too coarse for small lengths.
2. Mixed form, tenths and decimal form are three names for one length.

Teacher's Note:
1. The decimal point fixes where the units place is.
2. Moving the point one place left divides the number by 10.

Question 3. Measurements are written as decimals in daily life.
(i) Write the basic relations connecting mm, cm and m; g and kg; rupees and paise.
(ii) Convert: 134 mm into cm, 325 cm into m, 465 g into kg, and 250 paise into rupees.
(iii) A roll of tape is 2.07 m long and a piece of 45 cm is cut from it. Find the length of tape left, in metres and in centimetres.
Answer:
(i) 1 cm = 10 mm, so 1 mm = \( \frac{1}{10} \) cm = 0.1 cm. 1 m = 100 cm, so 1 cm = \( \frac{1}{100} \) m = 0.01 m; also 1 m = 1000 mm. 1 kg = 1000 g, so 1 g = \( \frac{1}{1000} \) kg = 0.001 kg. ₹1 = 100 paise, so 1 paisa = \( \frac{1}{100} \) rupee = ₹0.01.
(ii) 134 mm = \( \frac{134}{10} \) cm = 13.4 cm; 325 cm = \( \frac{325}{100} \) m = 3.25 m; 465 g = \( \frac{465}{1000} \) kg = 0.465 kg; 250 paise = \( \frac{250}{100} \) rupees = ₹2.50.
(iii) 45 cm = 0.45 m. Tape left = 2.07 − 0.45 = 1.62 m = 162 cm. Check: 1.62 + 0.45 = 2.07 m.
In simple words: Know the basic relations, then divide or multiply by 10, 100 or 1000.

Teacher's Note:
1. Bring both lengths to the same unit before subtracting.
2. 2.07 m is 207 cm, so 207 − 45 = 162 cm also works.

Question 4. Decimal numbers can be shown on a number line.
(i) Describe how you would divide the number line between 2 and 3 to mark 2.7, and name the two whole numbers between which 0.45 lies.
(ii) Explain how the portion between 2.7 and 2.8 is magnified and divided to mark 2.74.
(iii) Sonu writes 2.7, 2.70, 2.700, 2.07 and 0.27. Using place value, say which of these are equal and which are different, and explain why extra zeros written at the right end do not change the value.
Answer:
(i) 2.7 is 2 units and 7 tenths, so it lies between 2 and 3. Divide the gap from 2 to 3 into 10 equal parts of 0.1 each and count 7 parts from 2; the 7th mark is 2.7. The number 0.45 has 0 units, so it lies between 0 and 1 (in fact between 0.4 and 0.5).
(ii) Take only the small gap from 2.7 to 2.8 and draw it enlarged. This gap is one-tenth, so dividing it into 10 equal parts gives parts of 0.01. Counting 4 such parts from 2.7 gives 2.74. The same magnifying between 2.74 and 2.75 would give thousandths.
(iii) 2.7, 2.70 and 2.700 all have 2 units and 7 tenths; the extra zeros only record 0 hundredths and 0 thousandths, which add nothing. So 2.7 = 2.70 = 2.700.
But 2.07 has 0 tenths and 7 hundredths, and 0.27 has 0 units, 2 tenths and 7 hundredths, so 2.07 and 0.27 are different from 2.7 and from each other.
In simple words: Zoom in on the number line in steps of 10, and remember that end zeros change nothing.

Teacher's Note:
1. Each zoom divides a gap into 10 equal parts.
2. A zero between the point and a digit changes the value; a zero at the end does not.

Question 5. The number line below shows the part between 5.6 and 5.7 divided into 10 equal parts, with three points marked P, Q and R.
(i) What decimal number does each small division stand for, and how do you know?
(ii) Write the decimal numbers shown by P, Q and R.
(iii) Which of P, Q and R is closest to 5.7? Support your answer by finding the distance of each from 5.7, and then write one decimal number that lies between P and Q.
Answer:
(i) The end marks are 5.6 and 5.7, so the stretch shown is one-tenth. It is cut into 10 equal parts, and a tenth divided into 10 gives hundredths. So each division stands for 0.01.
(ii) Counting divisions from 5.6: P is at the 2nd mark, 5.6 + 0.02 = 5.62; Q is at the 5th mark, 5.6 + 0.05 = 5.65; R is at the 9th mark, 5.6 + 0.09 = 5.69.
(iii) Distances from 5.7: P: 5.70 − 5.62 = 0.08; Q: 5.70 − 5.65 = 0.05; R: 5.70 − 5.69 = 0.01. The smallest is 0.01, so R is closest to 5.7.
A number between P and Q is 5.63, since 5.62 < 5.63 < 5.65 (thousandths such as 5.634 also work).
In simple words: Each division is 0.01, so P = 5.62, Q = 5.65, R = 5.69, and R is nearest to 5.7.

Teacher's Note:
1. Find the value of one division before reading points.
2. The nearest point is the one with the smallest distance.

Question 6. A dairy farmer records the milk he supplies to a collection centre. In the first three days he supplies 4.35 L, 3.8 L and 5.125 L. Over all six days of the week he supplies 28 L in total.
(i) Find the quantity of milk supplied in the first three days.
(ii) Find the quantity supplied in the last three days.
(iii) In which half of the week did he supply more milk, and by how much? Check your answer by adding the two totals.
Answer:
(i) With the points in a line: 4.350 + 3.800 = 8.150, and 8.150 + 5.125 = 13.275. First three days: 13.275 L.
(ii) Last three days = 28.000 − 13.275 = 14.725 L.
(iii) Both have 1 ten, but in the units place 4 > 3, so the last three days are more. Difference = 14.725 − 13.275 = 1.45 L.
Check: 13.275 + 14.725 = 28.000 L, which matches the six-day total.
In simple words: Add the first three days, subtract from 28 L, then compare.

Teacher's Note:
1. Write all numbers with three decimal places.
2. Adding the two parts back gives a quick check.

Question 7. The decimal point tells us where the whole number part ends.
(i) Show that the digits 6, 0 and 4 written as “604” could stand for three different quantities: 6 hundreds and 4 ones; 6 tens and 4 tenths; 6 ones and 4 hundredths. Write each of them in decimal notation.
(ii) Read each of your three numbers aloud in words, and explain why 0.604 is read as “zero point six zero four” and not as “zero point six hundred and four”.
(iii) Write two of the older ways used before the modern decimal point to separate the whole part from the fractional part, and name the mathematicians connected with the point and with the comma.
Answer:
(i) 6 hundreds and 4 ones = 600 + 0 + 4 = 604. 6 tens and 4 tenths = 60 + 0 + \( \frac{4}{10} \) = 60.4. 6 ones and 4 hundredths = 6 + \( \frac{0}{10} \) + \( \frac{4}{100} \) = 6.04. The same digits give three values, so a separator, the decimal point, is needed.
(ii) 604 is “six hundred and four”; 60.4 is “sixty point four” (sixty and four-tenths); 6.04 is “six point zero four” (six and four hundredths). 0.604 is read “zero point six zero four” because its digits stand for 6 tenths, 0 hundredths and 4 thousandths, which are separate places; “six hundred and four” would describe a whole number.
(iii) Older ways included a vertical mark above or beside the last whole-number digit (used by al-Uqlidisi around 950 CE), ink of different colours, and a small raised numeral showing how many fractional places there were. In the 16th century John Napier and Christopher Clavius used the point, while François Viète used the comma, which several countries still use.
In simple words: Where the point is placed decides what each digit is worth.

Teacher's Note:
1. Without a point, 604 is ambiguous.
2. Digits after the point are read one at a time.

Question 8. A group of students is comparing decimal numbers.
(i) Arrange in descending order: 7.07, 7.7, 7.077, 7.707, 7.007.
(ii) Which of 5.6, 5.09, 4.95 is closest to 5? Find the distance of each from 5 before answering.
(iii) Raju says, “A decimal number with more digits is always greater.” Decide whether this is true, giving one example that supports your decision and explaining the correct rule for comparing decimals.
Answer:
(i) Write with three places: 7.070, 7.700, 7.077, 7.707, 7.007. All have 7 units; compare tenths, then hundredths, then thousandths. Order: 7.707 > 7.7 > 7.077 > 7.07 > 7.007.
(ii) 5.6 − 5 = 0.6; 5.09 − 5 = 0.09; 5 − 4.95 = 0.05. The smallest distance is 0.05, so 4.95 is closest to 5, even though it is less than 5.
(iii) Raju is wrong. 7.7 has fewer digits than 7.077, yet 7.7 is greater, because 7.7 has 7 tenths while 7.077 has 0 tenths.
Correct rule: compare the digits place by place from the highest place. At the first place where they differ, the number with the bigger digit is greater; later digits cannot change this.
In simple words: Compare place by place, not by counting digits.

Teacher's Note:
1. Pad with zeros to the same number of places.
2. Closest means smallest distance, on either side.

Question 9. Very small and very large measurements both use the decimal system.
(i) Show step by step how many millimetres make 1 km.
(ii) A human hair is about 0.1 mm thick. How many such hairs, placed one on top of another, would make a stack 1 cm thick, and how many would make 1 m?
(iii) A sheet of newspaper is 0.08 mm thick. Find the thickness of a pile of 500 such sheets in millimetres and in centimetres, and say whether this pile is thicker or thinner than your mathematics textbook.
Answer:
(i) 1 km = 1000 m; 1 m = 100 cm, so 1 km = 1000 × 100 = 1,00,000 cm; 1 cm = 10 mm, so 1 km = 1,00,000 × 10 = 10,00,000 mm (ten lakh millimetres).
(ii) 10 one-tenths make 1, so 10 hairs make 1 mm. 1 cm = 10 mm, so 10 × 10 = 100 hairs make 1 cm. 1 m = 1000 mm, so 10 × 1000 = 10,000 hairs make 1 m.
(iii) 0.08 mm is 8 hundredths of a millimetre, so 100 sheets make 800 hundredths = 8 mm. 500 sheets = 5 × 8 = 40 mm = \( \frac{40}{10} \) cm = 4 cm. A mathematics textbook is usually about 1 cm thick, so this pile is thicker than the textbook.
In simple words: Multiply step by step through the units; 500 sheets of 0.08 mm make 4 cm.

Teacher's Note:
1. Chain the conversions one unit at a time.
2. 500 × 0.08 = 40 can also be done directly.

Question 10. Sonu claims: “When two decimal numbers are added, the sum is always greater than the sum of their whole number parts, and always less than 2 more than that sum.”
(i) Test the claim for 23.6 + 11.45 by first writing the range Sonu predicts and then finding the exact sum.
(ii) Find a pair of decimal numbers for which the first half of his claim fails, and rewrite the claim so that it becomes correct for every pair.
(iii) Write a similar rule that tells us between which two whole numbers the difference of two decimal numbers lies, and test your rule on 23.6 − 11.45.
Answer:
(i) Whole parts: 23 + 11 = 34. Sonu predicts a sum greater than 34 and less than 34 + 2 = 36. Exact sum: 23.60 + 11.45 = 35.05. Since 34 < 35.05 < 36, the claim holds here, because each fractional part is less than 1 and together they add to less than 2.
(ii) Take 5.0 and 3.0. The whole parts add to 8 and the sum is also exactly 8, so the sum is not greater than 8. The first half fails whenever both fractional parts are zero. Corrected claim: the sum is greater than or equal to the sum of the whole number parts and always less than 2 more than it.
(iii) Taking the larger number first, the difference lies between (whole part of larger − whole part of smaller) − 1 and the same expression + 1.
Test: 23 − 11 = 12, so the difference should lie between 11 and 13. Exact: 23.60 − 11.45 = 12.15, which is indeed between 11 and 13.
In simple words: Whole parts give quick bounds for sums and differences of decimals.

Teacher's Note:
1. Estimate first, then calculate.
2. A single counter-example is enough to break a rule.

Case Study Based Questions

Teacher's Note:
1. Rupees and paise are a natural use of hundredths.
2. Check change by adding it back to the bill.

Question 12. The height chart in the classroom. A Class 7 teacher measures the height of four children with a measuring tape fixed on the wall. The tape is marked in centimetres, but the teacher writes every height in metres so that the numbers can be compared easily. She reminds the class that 100 centimetres make one metre, so each centimetre is one-hundredth of a metre. The readings she writes on the chart are: Anu 1.42 m, Bala 1.4 m, Chitra 1.405 m, Dev 1.45 m.
(i) Write Bala’s height in centimetres.
(ii) Who is the tallest child in the group?
(iii) Arrange the four heights in increasing order.
(iv) Find the difference between the tallest and the shortest height, in metres and in centimetres.
Answer:
(i) 1.4 m = 1 m + \( \frac{4}{10} \) m = 100 cm + 40 cm = 140 cm.
(ii) All four have 1 unit and 4 tenths, so compare hundredths: Dev has 5 hundredths, the largest. Dev is the tallest.
(iii) With three places: 1.400, 1.405, 1.420, 1.450. Increasing order: Bala (1.4 m), Chitra (1.405 m), Anu (1.42 m), Dev (1.45 m).
(iv) 1.45 − 1.40 = 0.05 m = 5 cm.
In simple words: Write all heights with the same number of decimal places, then compare.

Teacher's Note:
1. 1.4 m is 140 cm, not 104 cm.
2. Padding with zeros makes comparison easy.

Question 13. Screws in the workshop. A carpenter keeps screws of nearly equal size in one small box. To his eye they all look the same, so he measures each one on a ruler whose every centimetre is divided into 10 equal parts. He records three of them as: Screw A: 2.7 cm, Screw B: 2 units and 4 one-tenths of a centimetre, Screw C: 36 mm. A tiny difference in length decides whether a screw will hold two pieces of wood together or not, so the carpenter is careful to note the small markings between the whole centimetres.
(i) Why is a ruler marked only in whole centimetres not useful to this carpenter?
(ii) Write the length of Screw B in decimal form.
(iii) Write the length of Screw C in centimetres.
(iv) Which screw is the longest, and how much longer is it than Screw A?
Answer:
(i) His screws differ by less than a centimetre, and whole-centimetre marks cannot show such small differences. He needs each centimetre divided into 10 equal parts so that lengths like 2.7 cm can be read.
(ii) 2 units and 4 one-tenths = \( 2\frac{4}{10} \) cm = 2.4 cm.
(iii) 36 mm = \( \frac{36}{10} \) cm = 3.6 cm, since 1 mm = 0.1 cm.
(iv) Comparing 2.7, 2.4 and 3.6, Screw C is the longest. It is longer than Screw A by 3.6 − 2.7 = 0.9 cm (9 mm).
In simple words: Change all lengths to cm decimals, then compare and subtract.

Teacher's Note:
1. Use one unit before comparing.
2. 0.9 cm and 9 mm are the same.

Question 14. The medicine dropper. A doctor gives a bottle of syrup for a child and writes on it: “Give 2.5 mL three times a day.” The bottle holds 60 mL of syrup and comes with a small measuring cup marked in tenths of a millilitre. The doctor warns the family to read the decimal point very carefully, because another medicine in the same box has a dose of only 0.25 mL, and giving the wrong one can harm the child.
(i) How much syrup is given to the child in one full day?
(ii) How much syrup is given in two full days?
(iii) How much syrup is left in the bottle at the end of the first day?
(iv) If someone reads 0.25 mL as 2.5 mL, how many times the correct dose is given?
Answer:
(i) 2.5 + 2.5 + 2.5 = 7.5 mL in one day.
(ii) 7.5 + 7.5 = 15 mL in two days.
(iii) 60.0 − 7.5 = 52.5 mL. Check: 52.5 + 7.5 = 60.0.
(iv) Shifting the point in 0.25 one place to the right gives 2.5, which is 10 times as large. So the child would get 10 times the correct dose.
In simple words: Add the doses for the day, and remember a shifted point means ten times.

Teacher's Note:
1. Repeated addition of decimals works like whole numbers.
2. Always read medicine doses carefully.

Teacher's Note:
1. Add one number at a time to avoid mistakes.
2. mm to cm: move the point one place left.

Question 16. Ritu’s kitchen shelf. Ritu checks the packets on her kitchen shelf before going shopping. Some packets show their weight in kilograms and some in grams, so she decides to write every weight in kilograms, remembering that 1000 grams make one kilogram and so one gram is one-thousandth of a kilogram. Her list is: rice 2.5 kg, dal 750 g, sugar 1.05 kg, salt 250 g, mustard seeds 50 g.
(i) Write the weights of the dal and the salt in kilograms.
(ii) Which is heavier, the sugar or the dal?
(iii) Find the total weight of all five packets in kilograms, and also write it in grams.
(iv) Ritu cooks with 1.2 kg of the rice. How much rice is left?
Answer:
(i) Dal: 750 g = \( \frac{750}{1000} \) kg = 0.75 kg; salt: 250 g = \( \frac{250}{1000} \) kg = 0.25 kg.
(ii) The sugar is heavier: 1.05 kg has 1 in the units place, while the dal (0.75 kg) has 0.
(iii) Mustard seeds: 50 g = 0.05 kg. Total = 2.50 + 0.75 + 1.05 + 0.25 + 0.05: 2.50 + 0.75 = 3.25; 3.25 + 1.05 = 4.30; 4.30 + 0.25 = 4.55; 4.55 + 0.05 = 4.60. Total = 4.6 kg = 4600 g.
(iv) Rice left = 2.5 − 1.2 = 1.3 kg (1300 g).
In simple words: Turn all grams into kilograms, then add and subtract.

Teacher's Note:
1. 50 g is 0.05 kg, not 0.5 kg.
2. Use one unit for the whole list.

Question 17. Long jump on sports day. On sports day, the long jump pit of a school has a measuring tape laid beside it. The teacher reads each jump to the nearest thousandth of a metre and writes it on the board. The four jumps recorded in the final round are: Asha 2.45 m, Babu 2.5 m, Chetan 2.405 m, Dia 2.54 m. A jump of 2.5 m or more earns a place in the district team.
(i) Arrange the four jumps in descending order.
(ii) Leaving out Babu, whose jump is closest to 2.5 m? Show the distance of each from 2.5 m.
(iii) By how many centimetres did Asha miss the district mark?
(iv) Write one decimal number that lies between Asha’s jump and Babu’s jump.
Answer:
(i) With three places: 2.540, 2.500, 2.450, 2.405. Descending order: Dia (2.54 m), Babu (2.5 m), Asha (2.45 m), Chetan (2.405 m).
(ii) Asha: 2.500 − 2.450 = 0.05 m; Chetan: 2.500 − 2.405 = 0.095 m; Dia: 2.540 − 2.500 = 0.04 m. The smallest is 0.04 m, so Dia is closest.
(iii) 2.5 − 2.45 = 0.05 m = 5 cm.
(iv) 2.48 m, since 2.45 < 2.48 < 2.50.
In simple words: Pad the jumps to three places, then compare and subtract.

Teacher's Note:
1. 1 m = 100 cm, so 0.05 m = 5 cm.
2. Closest can be above or below the mark.

Question 18. Numbers that only look like decimals. Sarayu notices three messages on the same day. A bus notice says the bus will reach the station “3.5 hours after noon”. The television scoreboard of a cricket match shows “Overs bowled: 4.3”. At home, her sister measures an opening as 2 feet 5 inches but tells the carpenter to make a door “2.5 feet wide”, and the door that comes back does not close. Sarayu realises that a number written with a point is not always a true decimal number; it depends on whether the unit has really been split into 10 equal parts. (1 hour = 60 minutes, 1 over = 6 balls, 1 foot = 12 inches.)
(i) At what time does the bus reach the station?
(ii) What does “4.3 overs” mean?
(iii) What width of door will the carpenter make, and by how much is it wrong?
(iv) Out of “3.5 hours” and “4.3 overs”, which one is a true decimal number? Give your reason.
Answer:
(i) 0.5 hour means 5 of the 10 equal parts of an hour, each 60 ÷ 10 = 6 minutes, so 30 minutes. The bus arrives 3 hours 30 minutes after noon, at 3:30 p.m.
(ii) 4 complete overs and 3 more balls; the figure after the point counts balls, not tenths.
(iii) 1 foot = 12 inches, so 0.5 foot = 6 inches. The carpenter makes a door 2 feet 6 inches wide; the opening is 2 feet 5 inches, so the door is 1 inch too wide.
(iv) “3.5 hours” is a true decimal, because the hour is really divided into 10 equal parts of 6 minutes. “4.3 overs” is not, because an over is divided into 6 balls, not 10 equal parts.
In simple words: A point means tenths only when the unit is split into 10 equal parts.

Teacher's Note:
1. Always check what the digit after the point counts.
2. 2.5 feet is 2 feet 6 inches, not 2 feet 5 inches.

Question 19. The walkathon. A school club organises a walkathon for a tree-planting fund. Each member walks around the field and the distance walked is noted down by a teacher, some in kilometres and some in metres, exactly as the members report them. The four entries of one group are: Ravi 1.2 km, Seema 950 m, Tariq 1.05 km, Uma 875 m. The group has been asked to cover a total of 5 km among themselves. (1 km = 1000 m.)
(i) Write the distances walked by Seema and Uma in kilometres.
(ii) Who walked the longest distance and who walked the shortest?
(iii) Find the total distance walked by the group, in kilometres and in metres.
(iv) How much more must the group walk to reach its target? Give the answer in metres.
Answer:
(i) Seema: 950 m = \( \frac{950}{1000} \) km = 0.95 km; Uma: 875 m = \( \frac{875}{1000} \) km = 0.875 km.
(ii) Comparing 1.200, 0.950, 1.050 and 0.875: Ravi walked the longest (1.2 km) and Uma the shortest (0.875 km).
(iii) 1.200 + 0.950 = 2.150; 2.150 + 1.050 = 3.200; 3.200 + 0.875 = 4.075. Total = 4.075 km = 4075 m.
(iv) 5.000 − 4.075 = 0.925 km = 925 m. Check: 4.075 + 0.925 = 5.000 km.
In simple words: Change metres to kilometres, add, then subtract from 5 km.

Teacher's Note:
1. Metres are thousandths of a kilometre.
2. Pad with zeros to three places before adding.

Question 20. The juice stall log. A juice stall near a park keeps a daily sales log in litres. The owner measures the juice in a jar whose marks are 0.25 L apart, so every reading he writes must end in .00, .25, .50 or .75. His log for one week shows a steady rise: Monday 18.5 L, Tuesday 19.25 L, Wednesday 20 L, Thursday 20.75 L, Friday (same pattern continues).
(i) Find the change after each term and write Friday’s reading.
(ii) Without adding exactly, say between which two whole numbers the total of Monday and Tuesday must lie, and then check by finding the exact total.
(iii) Find the total sales from Monday to Friday.
(iv) In an old log the owner finds the reading 20.4 L and says it cannot be correct. Which two readings on his jar lie nearest to 20.4 L, and which of the two is closer to it?
Answer:
(i) 18.5 + 0.75 = 19.25, 19.25 + 0.75 = 20.00, 20.00 + 0.75 = 20.75, so the change is +0.75 L each day. Friday: 20.75 + 0.75 = 21.5 L.
(ii) Whole parts: 18 + 19 = 37. Each fractional part is less than 1, so the total is more than 37 and less than 37 + 2 = 39. Exact: 18.50 + 19.25 = 37.75 L, which lies between 37 and 39.
(iii) 18.50 + 19.25 = 37.75; 37.75 + 20.00 = 57.75; 57.75 + 20.75 = 78.50; 78.50 + 21.50 = 100.00. Total = 100 L.
(iv) The nearest jar readings are 20.25 L and 20.5 L. Distances: 20.40 − 20.25 = 0.15 and 20.50 − 20.40 = 0.10. Since 0.10 < 0.15, 20.5 L is closer.
In simple words: The log rises by 0.75 L a day, and five days add up to 100 L.

Teacher's Note:
1. Find the step of a pattern by subtracting neighbours.
2. Estimate with whole parts before exact adding.