CBSE Class 7 Mathematics Number Play Worksheet Set 01

Long Answer Questions

Question 1. Six children stand one behind the other in a line, all facing the front. Each child calls out the number of children standing in front of them who are taller than they are.
(i) Explain the calling-out rule in your own words and say why the child at the very front must always call out 0.
(ii) The heights of the six children, taken from the front of the line to the back, are 150 cm, 148 cm, 155 cm, 147 cm, 152 cm and 145 cm. Write the number called out by each child.
(iii) Describe, in words, how the six children must be arranged so that the sequence reads 0, 0, 0, 0, 0, 0, and how they must be arranged so that it reads 0, 1, 2, 3, 4, 5.
Answer:
(i) Each child looks only at the children ahead and counts how many of them are taller. Children behind are ignored. The front child has nobody ahead, so there is nothing to count and it must call out 0.
(ii) 150: nobody ahead, so 0.
148: 150 is taller, so 1.
155: 150 and 148 are shorter, so 0.
147: 150, 148 and 155 are all taller, so 3.
152: only 155 is taller, so 1.
145: all five ahead are taller, so 5.
The sequence is 0, 1, 0, 3, 1, 5.
(iii) For 0, 0, 0, 0, 0, 0 nobody may have a taller child ahead, so they stand in increasing order of height, shortest at the front and tallest at the back.
For 0, 1, 2, 3, 4, 5 every child ahead must be taller, so they stand in decreasing order of height, tallest at the front and shortest at the back.
In simple words: Count taller children ahead; all 0s means shortest to tallest, and 0 to 5 means tallest to shortest.

Teacher's Note:
1. Work child by child from the front.
2. The two special orders are opposites of each other.

Teacher's Note:
1. Draw dots to show the pairing.
2. In (iii), the count of odd numbers decides the parity.

Teacher's Note:
1. Start with the line through the centre that has the smallest sum.
2. Check every row and column at the end.

Question 4. The sequence 1, 2, 3, 5, 8, 13, 21, 34, … is one of the most famous sequences in mathematics.
(i) Write its first twelve numbers and state the rule that produces each new number.
(ii) Give a short account of the history of this sequence: in which field of study it first arose, the Indian scholars connected with it, and why the name “Virahāṅka–Fibonacci numbers” is used today.
(iii) Give two places in nature or art where these numbers appear, with an example of the actual counts.
Answer:
(i) The first twelve numbers are 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233.
Rule: from the third number onwards, each number is the sum of the two numbers just before it, for example 55 + 89 = 144.
(ii) The sequence first arose in art, in the study of poetry. Scholars classed each syllable as short (1 beat) or long (2 beats) and asked how many rhythms a given number of beats allows.
Piṅgala (around 300 BCE) worked on such questions. Around 700 CE the Prakrit scholar Virahāṅka was the first to state the rule, in the form of a poem. Gopala (about 1135 CE) and Hemachandra (about 1150 CE) wrote about it next.
Fibonacci of Italy wrote about these numbers only in 1202 CE, about 500 years later, so the joint name Virahāṅka–Fibonacci numbers gives proper credit.
(iii) Petals on a daisy are usually 13, 21 or 34. The seed spirals of a sunflower come in such counts too, for example 34 spirals one way and 55 the other. The numbers also appear in drumming patterns, architecture and visual art.
In simple words: Add the last two numbers; the idea came from Indian poetry long before Fibonacci.

Teacher's Note:
1. Learn the names and dates in order.
2. Give actual numbers for the examples.

Question 5. In Indian poetry a short syllable lasts one beat and a long syllable lasts two beats. A rhythm of n beats is any ordered arrangement of short and long syllables whose beats add up to n.
(i) Using the systematic method of writing a “1 +” in front of every 4-beat rhythm and a “2 +” in front of every 3-beat rhythm, list all the rhythms having 5 beats as sums of 1s and 2s.
(ii) Explain clearly why the number of n-beat rhythms equals the number of (n − 1)-beat rhythms plus the number of (n − 2)-beat rhythms.
(iii) Use this rule to find how many rhythms have 9 beats.
Answer:
(i) The 4-beat rhythms are 1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 2 + 2, and the 3-beat rhythms are 1 + 1 + 1, 1 + 2, 2 + 1.
Putting “1 +” before each 4-beat rhythm and “2 +” before each 3-beat rhythm gives:
1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 2, 1 + 1 + 2 + 1, 1 + 2 + 1 + 1, 1 + 2 + 2,
2 + 1 + 1 + 1, 2 + 1 + 2, 2 + 2 + 1.
That is 5 + 3 = 8 rhythms.
(ii) Every n-beat rhythm begins with either a short or a long syllable. If it begins with a short one (1 beat), the rest is a rhythm of n − 1 beats. If it begins with a long one (2 beats), the rest is a rhythm of n − 2 beats.
No rhythm can begin both ways, so none is counted twice. Hence the count for n beats = count for n − 1 beats + count for n − 2 beats.
(iii) Building up: 1, 2, 3, 5, 8, 13, 21, 34, 55 for 1 to 9 beats. So there are 55 rhythms of 9 beats.
In simple words: Split rhythms by their first syllable; that gives the add-the-last-two rule.

Teacher's Note:
1. List in a fixed order so none is missed.
2. The first syllable decides which smaller list a rhythm comes from.

Teacher's Note:
1. Check all eight lines of your final square.
2. Corners lie on three lines; edge-middles on two.

Question 7. Letter-numbers let us describe whole families of numbers at once.
(i) Write one expression that always gives an even value, one that always gives an odd value and one that can give either, and justify each choice.
(ii) Which expression lists all the even numbers and which lists all the odd numbers? Explain why the expression 6k + 2 fails to list all even numbers.
(iii) Find the 100th odd number in two different ways, and state the formula for the nth odd number.
Answer:
(i) 8p is always even, because any multiple of 8 is a multiple of 2.
6a + 5 is always odd: 6a is even, and even + odd = odd.
3n + 4 can be either: n = 2 gives 10 (even) and n = 3 gives 13 (odd), since 3n has the same parity as n.
(ii) 2n lists all the even numbers and 2n − 1 lists all the odd numbers.
6k + 2 gives 8, 14, 20, … for k = 1, 2, 3, …; it jumps in steps of 6, so even numbers such as 10, 12, 16 and 18 are skipped.
(iii) First way: the 100th even number is 2 × 100 = 200, and the odd number at the same place is one less, so it is 199.
Second way: 2n − 1 with n = 100 gives 200 − 1 = 199. The nth odd number is 2n − 1.
In simple words: 2n gives every even number, 2n − 1 every odd number; 6k + 2 jumps over many.

Teacher's Note:
1. Test each expression with two values.
2. A list must not skip any number.

Question 8. In a cryptarithm each letter stands for a single digit, the same letter always stands for the same digit, different letters stand for different digits, and no number begins with 0.
(i) Solve the cryptarithm TO + GO = OUT, where TO and GO are two-digit numbers and OUT is a three-digit number. Write down the reasoning step by step, starting with the hundreds digit.
(ii) Check your solution by doing the addition with digits.
(iii) Explain why the letter O could not have stood for any digit other than the one you found.
Answer:
(i) TO and GO are two-digit numbers, so their sum is less than 200. OUT has three digits, so O = 1.
So TO and GO both end in 1. Units column: 1 + 1 = 2 with no carry, so T = 2.
Now 21 + G1 = 1U2. In value, 10G + 22 = 102 + 10U, which gives G = U + 8.
U = 0 gives G = 8; U = 1 is not allowed because O is already 1; larger U makes G more than 9. So U = 0 and G = 8.
(ii) 21 + 81 = 102, and OUT reads 1, 0, 2, matching O = 1, U = 0, T = 2.
(iii) The largest sum of two two-digit numbers is 99 + 99 = 198, still below 200. So a three-digit answer can only begin with 1, and O can be no other digit.
In simple words: The hundreds digit must be 1, and then 21 + 81 = 102 follows.

Teacher's Note:
1. Always start from the hundreds digit.
2. Remember different letters need different digits.

Teacher's Note:
1. Check one row, one column and one diagonal each time.
2. Adding 3 to each entry adds 9 to each line.

Teacher's Note:
1. Toggles work like adding 1 each time.
2. In (iii) other fillings are also correct; check every line.

Case Study Based Questions

Question 11. Morning assembly in the school ground
Every morning the children of Class 7B line up in the school ground, one behind the other, all facing the flag post. Their class teacher plays a quick game with them. Each child must look only at the children standing in front of them and call out how many of those children are taller than they are. Children standing behind are not counted. On Monday, seven children lined up and their heights, measured from the front of the line to the back, were 138 cm, 142 cm, 135 cm, 150 cm, 141 cm, 133 cm and 147 cm. The teacher wrote down the seven numbers that were called out and showed them to the class.
(i) What number is called out by the third child in the line?
(ii) Write the complete sequence of numbers called out, from the front of the line to the back.
(iii) Which child calls out the largest number, and what does this tell you about that child's height?
(iv) The teacher claims that in a line of seven children the largest number anyone can possibly call out is 7. Is she right? Give a reason.
Answer:
(i) The third child is 135 cm tall. Both children ahead (138 cm and 142 cm) are taller, so this child calls out 2.
(ii) 138 → 0, 142 → 0, 135 → 2, 150 → 0, 141 → 2 (142 and 150 are taller), 133 → 5 (all five ahead are taller), 147 → 1 (only 150 is taller). The sequence is 0, 0, 2, 0, 2, 5, 1.
(iii) The sixth child (133 cm) calls out the largest number, 5. So all five children ahead are taller than this child; the last child (147 cm) is also taller, so this child is the shortest in the line.
(iv) No. A child at place k has only k − 1 children ahead, so the seventh child can count at most 6. The largest possible number is 6, not 7.
In simple words: Count taller children ahead; nobody in a line of seven can say more than 6.

Teacher's Note:
1. Work through the line one child at a time.
2. The number called out can never exceed the children ahead.

Question 12. Counting petals in the school garden
The Grade 7 nature club counted petals in the school garden. They found three daisies with 13, 21 and 34 petals. The teacher told them that these are not ordinary numbers: they belong to a famous sequence that begins 1, 2, 3, 5, 8, 13, 21, 34, … The club then looked at a sunflower head and carefully traced the spirals of seeds. They counted 34 spirals curving one way and 55 spirals curving the other way. The teacher explained that this sequence was first written about in India, long before it became known in Europe, and that it turns up again and again in flowers, pine cones and even in music.
(i) Name the sequence and write the next two numbers after 34.
(ii) State the rule by which each new number of the sequence is formed.
(iii) Is 100 a member of this sequence? Justify your answer.
(iv) Which Indian scholar first wrote down the rule for forming these numbers, and roughly when?
Answer:
(i) It is the Virahāṅka–Fibonacci sequence. After 34 come 21 + 34 = 55 and 34 + 55 = 89.
(ii) From the third number onwards, each number is the sum of the two numbers just before it.
(iii) No. The sequence runs …, 55, 89, 144, … and jumps straight from 89 to 144, so 100 never appears.
(iv) The Prakrit scholar Virahāṅka, around 700 CE, gave the rule in the form of a poem, about 500 years before Fibonacci (1202 CE).
In simple words: Add the last two numbers; 100 is skipped between 89 and 144.

Teacher's Note:
1. To test membership, find the neighbours on both sides.
2. Remember the date 700 CE.

Question 13. Meena's coin box
Meena saves her pocket money in a coin box. At the end of the month she empties it and sorts the coins into three heaps. She finds that she has an even number of ₹2 coins, an odd number of ₹5 coins and an even number of ₹10 coins. Before counting the money, her brother Arun challenges her to guess whether the total will be an odd amount or an even amount. Meena says she can decide this without adding anything, just by thinking about the three heaps separately. After counting, she announces that the total comes to ₹137.
(i) Is the amount contributed by the ₹2 coins odd or even? Why?
(ii) Is the amount contributed by the ₹5 coins odd or even? Why?
(iii) Using parity, decide whether a total of ₹137 is possible for Meena's coin box.
(iv) Could Meena's total instead have been ₹140? Give a reason.
Answer:
(i) Even. The ₹2 coins give 2 × (a whole number), which is a multiple of 2, whatever the count.
(ii) Odd. Each ₹5 coin gives an odd amount, and there is an odd number of them; an odd count of odd numbers adds to an odd total.
(iii) Yes, it is possible. The ₹10 coins give an even amount, so the total is even + odd + even = odd, and ₹137 is odd. For example, 6 coins of ₹2, 5 coins of ₹5 and 10 coins of ₹10 give 12 + 25 + 100 = ₹137.
(iv) No. Meena's total is always odd, and ₹140 is even.
In simple words: Only the ₹5 heap is odd, so the total is always odd.

Teacher's Note:
1. Find the parity of each heap first.
2. An example makes the 'yes' answer convincing.

Teacher's Note:
1. Check at least one row, column and diagonal.
2. The name Chautīsā tells its magic sum.

Teacher's Note:
1. Order matters, so 1 + 2 and 2 + 1 are different.
2. Split by the first stroke to explain the rule.

Question 16. Chairs in the auditorium
The caretaker of a school auditorium arranges chairs in a neat rectangular grid for every function. For Monday's prize giving he set out 27 rows with 41 chairs in each row. For Tuesday's music evening he set out 36 rows with 45 chairs in each row, and for Wednesday's parents' meeting 48 rows with 50 chairs in each row. The principal asks him each time only one thing: whether the total number of chairs is an odd number or an even number, because the seating is always shared equally between two blocks. The caretaker has learnt to answer at once, without doing any multiplication at all.
(i) State the parity of the number of chairs on each of the three days, without multiplying.
(ii) Write the rule that lets the caretaker answer so quickly.
(iii) On another day the total number of chairs turned out to be odd. What can you say about the number of rows and the number of chairs in a row?
(iv) The caretaker says that if he writes the number of chairs in a row as 4r + 2, the total can never be odd. Is he right? Give a reason.
Answer:
(i) Monday 27 × 41: both odd, so odd. Tuesday 36 × 45: 36 is even, so even. Wednesday 48 × 50: both even, so even.
(ii) The number of chairs in a rectangular grid is odd only when both the number of rows and the number in each row are odd; if either is even, the total is even.
(iii) Both the number of rows and the number of chairs in each row must be odd.
(iv) Yes, he is right. 4r is even, so 4r + 2 is even for every r. With an even number in each row, the total is even however many rows there are.
In simple words: Odd × odd is the only way to get an odd total.

Teacher's Note:
1. Look at the parity of each factor.
2. One even factor makes the product even.

Question 17. The smudged stationery bill
Rehana buys three identical notebooks from a stationery shop. When she reaches home, a drop of water has smudged the bill and the digits can no longer be read. Luckily, the shopkeeper writes bills in a neat way, and Rehana remembers the shape of the figures. The price of one notebook was a two-digit amount, which she writes as JK, and the total for the three notebooks was a three-digit amount, which she writes as LJK. Here each letter stands for a single digit, the same letter always stands for the same digit, different letters stand for different digits, and no amount begins with 0. So her bill reads JK + JK + JK = LJK.
(i) Explain why the digit L must be 1.
(ii) Find the digits J and K.
(iii) Write out the complete bill with digits and check the addition.
(iv) If Rehana had bought four such notebooks instead, what would the total have been?
Answer:
(i) LJK is 100 × L plus the two-digit number JK, so LJK = 100L + JK. The bill says LJK = 3 × JK. So 100L = 2 × JK, that is, JK = 50L. JK has two digits, so L = 1 (L = 2 would make JK = 100). Hence L = 1.
(ii) JK = 50 × 1 = 50, so J = 5 and K = 0. The digits 1, 5, 0 are all different.
(iii) The bill reads 50 + 50 + 50 = 150. Check: 3 × 50 = 150, which reads LJK with L = 1, J = 5, K = 0.
(iv) Four notebooks would cost 4 × 50 = ₹200.
In simple words: 3 × JK must equal 100 + JK, so JK = 50 and the bill is 150.

Teacher's Note:
1. Write the three-digit number as 100L + JK.
2. Check the final bill by multiplying.

Teacher's Note:
1. Row sum 8 is small, so that row uses small numbers.
2. Always check the 6 to 24 range.

Question 19. Handshakes at the science club
At the first meeting of a school science club, the members greeted one another with handshakes. Nobody shook hands with the same person twice and nobody shook their own hand. Afterwards the secretary went round and asked each member a single question: “How many hands did you shake?” She wrote all the answers in a list and then added them up. Her teacher told her that this total can be used to check whether the members have counted correctly, because the total is related in a simple way to the number of handshakes that actually took place in the room.
(i) When two members shake hands, by how much does the total of all the written numbers increase?
(ii) Explain why the total of all the written numbers must always be an even number.
(iii) At one meeting of 7 members the list read 3, 2, 4, 1, 3, 2, 2. What does this tell you?
(iv) At the next meeting the list added up to 24. How many handshakes took place?
Answer:
(i) By 2. One handshake involves two people, and each adds 1 to their own count.
(ii) Each handshake adds exactly 2 to the total. Starting from 0 and adding 2 again and again always gives an even number, so the total is twice the number of handshakes and is always even.
(iii) 3 + 2 + 4 + 1 + 3 + 2 + 2 = 17, which is odd. A correct list must total an even number, so someone has counted or written an answer wrongly; the list cannot describe real handshakes.
(iv) The total is twice the number of handshakes, so there were 24 ÷ 2 = 12 handshakes.
In simple words: Every handshake adds 2 to the total, so the total is even and equals twice the handshakes.

Teacher's Note:
1. Each handshake is counted by two people.
2. An odd total shows a mistake at once.

Question 20. The puzzle corner in the library
The school library keeps a puzzle corner where two puzzles are pinned up each week. The first is a cryptarithm: NO + NO + NO = ONE, where NO is a two-digit number and ONE is a three-digit number. Each letter stands for a single digit, the same letter always stands for the same digit, different letters stand for different digits, and no number begins with 0. The second puzzle reads: “Two numbers that stand next to each other in the Virahāṅka sequence are 610 and 987. Can you walk both forwards and backwards along the sequence?” A prize is given to the first reader who solves both puzzles and explains the reasoning.
(i) Show that the digit O must be 1.
(ii) Find the digits N and E, giving your reasoning.
(iii) Write the cryptarithm with digits and check the addition.
(iv) Solve the second puzzle: find the two numbers that come just after 987 and the two numbers that come just before 610.
Answer:
(i) NO is at most 98, so 3 × NO is at most 294; hence O is 1 or 2.
If O = 2, then 3 × (10N + 2) = 200 + 10N + E, so 20N = 194 + E. But 20N is at most 180, which is too small. So O = 1.
(ii) Units column: 1 + 1 + 1 = 3 with no carry, so E = 3.
Then 3 × (10N + 1) = 100 + 10N + 3, so 30N + 3 = 103 + 10N, 20N = 100 and N = 5. The digits 5, 1, 3 are all different.
(iii) 51 + 51 + 51 = 153. Check: 3 × 51 = 153, which reads ONE with O = 1, N = 5, E = 3.
(iv) Forwards: 610 + 987 = 1597 and 987 + 1597 = 2584.
Backwards: 987 − 610 = 377 and 610 − 377 = 233. So the sequence runs …, 233, 377, 610, 987, 1597, 2584, …
In simple words: 51 × 3 = 153; add to move forwards and subtract to move backwards.

Teacher's Note:
1. Bound the answer first to fix O.
2. Backwards, subtract the smaller from the larger neighbour.