CBSE Class 7 Mathematics Number Play Assignment Set 01

Short Answer Questions

Question 1. Five children of a Grade 7 class stand one behind the other in a single line. Their heights, taken from the front of the line to the back, are 120 cm, 135 cm, 128 cm, 142 cm and 125 cm. Each child calls out the number of children standing in front of them who are taller than they are. Write the number called out by each child, in order from the front.
Answer:
For each child, count the taller children already standing ahead.
First child (120 cm): nobody ahead, so 0.
Second child (135 cm): only 120 is ahead, which is shorter, so 0.
Third child (128 cm): 135 is taller, so 1.
Fourth child (142 cm): nobody ahead is taller, so 0.
Fifth child (125 cm): 135, 128 and 142 are taller, so 3.
The numbers called out are 0, 0, 1, 0, 3.
In simple words: Each child counts only the taller children in front.

Teacher's Note:
1. Never count children standing behind.
2. The first number is always 0.

Question 2. A line of children follows the same calling-out rule as above. For each statement, write whether it is Always True, Only Sometimes True or Never True, and give a short reason.
(i) The first child in the line calls out 0.
(ii) The tallest child in the line calls out 0.
(iii) A child who calls out 0 is the tallest child in the line.
Answer:
(i) Always True. The first child has nobody ahead, so the count is 0.
(ii) Always True. Nobody is taller than the tallest child, so that child counts 0.
(iii) Only Sometimes True. A child says 0 whenever everyone ahead is shorter. For example, in heights 120, 135, 128, 150 the second child (135 cm) says 0, but the tallest child is the fourth (150 cm).
In simple words: Calling out 0 only means nobody ahead is taller; a taller child may stand behind.

Teacher's Note:
1. Give an example to show 'sometimes'.
2. The example must have a taller child behind.

Question 3. Using the idea of arranging numbers in pairs, find the parity of each of the following sums. Give a reason for each answer.
(i) The sum of 4 even numbers and 3 odd numbers.
(ii) The sum of 9 odd numbers.
Answer:
(i) Odd. The 4 even numbers pair up fully, so their total is even. The 3 odd numbers leave one over, so their total is odd. Even + odd = odd.
(ii) Odd. 8 of the odd numbers pair off and give an even total; one odd number is left over, so the sum is odd.
In simple words: Count the odd numbers: an odd count of them makes the sum odd.

Teacher's Note:
1. Even numbers never change the parity of a sum.
2. Only the count of odd numbers matters.

Question 4. Copy and complete the parity rules for subtraction, writing odd or even in each blank, and give one numerical example for each.
(i) even − even = ______
(ii) odd − odd = ______
(iii) even − odd = ______
(iv) odd − even = ______
Answer:
(i) even − even = even; example 18 − 6 = 12.
(ii) odd − odd = even; example 17 − 9 = 8.
(iii) even − odd = odd; example 20 − 7 = 13.
(iv) odd − even = odd; example 19 − 8 = 11.
Subtraction follows the same parity rules as addition, because taking away pairs never changes whether one is left over.
In simple words: Same parities give even; different parities give odd, for both + and −.

Teacher's Note:
1. Check each example with real numbers.
2. The rules match the addition rules.

Question 5. Write the formula that gives the nth even number and the formula that gives the nth odd number. Using these formulas, find the 45th even number and the 45th odd number.
Answer:
The nth even number is 2n, and the nth odd number is 2n − 1, as each odd number is one less than the even number at the same place.
45th even number = 2 × 45 = 90.
45th odd number = 2 × 45 − 1 = 89.
In simple words: Double the position for the even number; take away 1 for the odd number.

Teacher's Note:
1. Check with n = 1: 2 and 1.
2. Odd number = even number − 1.

Question 6. A 3 × 3 grid is filled with the numbers 1 to 9, each used exactly once. Explain why the three row sums must always add up to 45, and use this fact to show that the magic sum of a magic square made from 1 to 9 must be 15.
Answer:
Each number from 1 to 9 is in exactly one box, and each box lies in exactly one row.
So adding the three row sums adds every number once: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
In a magic square the three row sums are equal. If each is S, then 3S = 45, so S = 15.
Hence the magic sum must be 15.
In simple words: The rows share the total 45 equally, so each row is 15.

Teacher's Note:
1. The arrangement does not change the total 45.
2. Divide by the number of rows.

Question 7. Write the first ten numbers of the Virahāṅka sequence. State in one sentence the rule by which each new number is formed, and name the Prakrit scholar who first wrote down this rule.
Answer:
The first ten numbers are 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
Rule: from the third number onwards, each number is the sum of the two numbers just before it (for example 13 + 21 = 34).
The rule was first written down, as a Prakrit poem, by Virahāṅka around 700 CE.
In simple words: Add the last two numbers to get the next one.

Teacher's Note:
1. Count the terms carefully: ten numbers.
2. Virahāṅka gave the rule as a poem.

Question 8. In the cryptarithm NE + E = EN, each letter stands for one digit and different letters stand for different digits. NE and EN are two-digit numbers and E is a one-digit number. Find the digits N and E, and write the completed sum.
Answer:
Adding the single digit E to NE gives EN, the number with its digits swapped.
In value, (10N + E) + E = 10E + N, which simplifies to 9N = 8E.
N and E are single non-zero digits, so the only pair is N = 8 and E = 9.
Check: 89 + 9 = 98. So N = 8 and E = 9, and the sum is 89 + 9 = 98.
In simple words: Only 89 + 9 = 98 swaps the digits.

Teacher's Note:
1. Write each two-digit number as 10 × tens + units.
2. 9N = 8E needs N to be a multiple of 8.

Question 9. Ten children stand in a line and each calls out the number of children in front of them who are taller. Answer with reasons:
(i) What is the largest number that can possibly be called out, and by which child in the line?
(ii) Can two different children both call out 9?
Answer:
(i) The largest possible number is 9, called out by the tenth (last) child. Only that child has nine children ahead, and it says 9 when all nine are taller, that is, when it is the shortest.
(ii) No. A child at place k has only k − 1 children ahead, so only the tenth child can reach 9.
In simple words: Only the last child has nine children ahead, so only it can say 9.

Teacher's Note:
1. The limit depends on the place in the line.
2. Place k allows at most k − 1.

Question 10. For a line of 5 children following the taller-ahead rule, decide which of the two sequences below can actually happen and which cannot. Justify each answer.
(i) 0, 1, 1, 2, 4
(ii) 0, 2, 1, 0, 1
Answer:
(i) Possible. Heights 150, 135, 145, 140, 130 cm (front to back) give exactly 0, 1, 1, 2, 4. The last child is the shortest, so all four ahead are taller.
(ii) Not possible. The second child has only one child ahead, so it can call out at most 1, never 2.
In simple words: A sequence fails when a child calls a number bigger than the children ahead.

Teacher's Note:
1. Give heights to show a sequence is possible.
2. Check each number against its place.

Question 11. Without multiplying, find the parity of the number of small squares in each of these grids and state the rule you used.
(i) 35 × 47
(ii) 48 × 63
(iii) 124 × 250
Answer:
Rule: a product is odd only when both factors are odd; if even one factor is even, the product is even.
(i) 35 × 47: both odd, so the number is odd.
(ii) 48 × 63: 48 is even, so the number is even.
(iii) 124 × 250: both even, so the number is even.
In simple words: Odd × odd is odd; any even factor makes it even.

Teacher's Note:
1. Look only at the last digits.
2. State the rule before using it.

Question 12. Ravi's money box holds an odd number of ₹1 coins, an even number of ₹5 coins and an odd number of ₹10 coins. He claims the total value is ₹96. Use a parity argument to decide whether Ravi can be correct.
Answer:
An odd number of ₹1 coins gives an odd amount.
5 × (an even number) is even, so the ₹5 coins give an even amount.
10 × (any number) is even, so the ₹10 coins give an even amount.
Total = odd + even + even = odd. Since ₹96 is even, Ravi cannot be correct.
In simple words: The total must be odd, but 96 is even.

Teacher's Note:
1. Find the parity of each heap separately.
2. ₹10 coins always give an even amount.

Question 13. For each expression below, say whether it always gives an even number, always gives an odd number, or can give either. Support each answer with two values of n.
(i) 6n + 4
(ii) 4n + 1
(iii) 5n + 3
Answer:
(i) Always even: 6n is even and 4 is even. n = 1 gives 10, n = 2 gives 16.
(ii) Always odd: 4n is even, and one more than an even number is odd. n = 1 gives 5, n = 3 gives 13.
(iii) Can be either: n = 1 gives 8 (even), n = 2 gives 13 (odd), because the parity of 5n changes with n.
In simple words: An even multiple stays even; an odd multiple follows n.

Teacher's Note:
1. Check whether the multiplier of n is even or odd.
2. Then add the parity of the constant.

Question 14. A teacher draws a 3 × 3 grid to be filled with the numbers 1 to 9 without repetition. The circled row sums are 5, 16 and 24, and the circled column sums are 9, 14 and 22. Show that this grid can never be filled, even though the row sums and the column sums each total 45.
Answer:
Each circled number is the sum of three different numbers from 1 to 9.
The smallest such sum is 1 + 2 + 3 = 6 and the largest is 7 + 8 + 9 = 24.
The grid needs a row sum of 5, which is below 6, so the grid can never be filled.
A total of 45 is only a first test; passing it does not make the grid possible.
In simple words: A row sum of 5 is smaller than the smallest possible sum, 6.

Teacher's Note:
1. Check every circled number lies from 6 to 24.
2. Both tests must be passed.

Question 15. A drummer plays rhythms using only short beats (1 beat) and long beats (2 beats).
(i) How many different 7-beat rhythms can she play?
(ii) Explain briefly why this count is the same as the number of ways of climbing a 7-step staircase taking 1 or 2 steps at a time.
Answer:
(i) The counts for 1, 2, 3, … beats are 1, 2, 3, 5, 8, 13, 21, …, so there are 21 different 7-beat rhythms.
(ii) Both problems ask for the ways of writing 7 as an ordered sum of 1s and 2s: a short beat or a single step is 1, a long beat or a double step is 2. So both answers are 21.
In simple words: Rhythms and staircase climbs are both ordered sums of 1s and 2s.

Teacher's Note:
1. The 7th Virahāṅka number is 21.
2. Order matters in both problems.

Question 16. Solve the cryptarithm RR + SS = TST, where RR and SS are two-digit numbers, TST is a three-digit number, and different letters stand for different digits. Give the value of each letter and check your answer.
Answer:
Two two-digit numbers add to less than 200, so T = 1.
In value, 11R + 11S = 101 + 10S, which simplifies to 11R + S = 101.
S is a single digit, so 11R is between 92 and 101; hence R = 9 and S = 2.
Check: 99 + 22 = 121, which reads TST. So R = 9, S = 2, T = 1.
In simple words: Only 99 + 22 = 121 fits the pattern.

Teacher's Note:
1. RR means 11 × R.
2. Always finish with a check.

Question 17. Some loose sheets fall out of a register. Each sheet is printed on both sides, so it carries two page numbers, and 25 such sheets are collected. A student says the page numbers on these sheets add up to 3000. Using parity, explain why the student must be mistaken.
Answer:
A sheet carries two consecutive page numbers, one odd and one even, such as 17 and 18.
So the sum on each sheet is odd (17 + 18 = 35).
The total for 25 sheets is the sum of 25 odd numbers, and an odd count of odd numbers is odd.
The total must be odd, but 3000 is even, so the student is mistaken.
In simple words: 25 odd sheet-totals add to an odd number, never 3000.

Teacher's Note:
1. First find the parity for one sheet.
2. Then use the count of sheets.

Teacher's Note:
1. Check one row and one diagonal.
2. 15 + 8 + 13 = 36 and 15 + 12 + 9 = 36.

Question 19. The Virahāṅka sequence begins 1, 2, 3, 5, 8, 13, 21, 34, … Without actually calculating the terms, find the parity of the 20th term and the parity of the 25th term. Explain the pattern you used.
Answer:
The parities are O, E, O, O, E, O, O, E, … Since odd + even = odd and odd + odd = even, the block odd, odd, even keeps repeating.
Even terms are at places 2, 5, 8, 11, 14, 17, 20, …, places that leave remainder 2 when divided by 3.
20 = 3 × 6 + 2, so the 20th term is even.
25 = 3 × 8 + 1, so the 25th term is odd.
In simple words: Every third term from the 2nd is even; check the remainder on division by 3.

Teacher's Note:
1. Write O and E under the first few terms.
2. Remainder 2 means even.

Question 20. Solve the cryptarithm PQ + QR = RRP, where PQ and QR are two-digit numbers, RRP is a three-digit number, and different letters stand for different digits. Explain the reasoning that fixes the value of R first.
Answer:
PQ and QR are two-digit numbers, so their sum is less than 200. As RRP has three digits, its first digit is 1, so R = 1.
The sum is now 11P, that is 110 + P.
In value, (10P + Q) + (10Q + 1) = 110 + P, so 9P + 11Q = 109.
Only Q = 5 works: 9P = 54, so P = 6.
Check: 65 + 51 = 116. So P = 6, Q = 5, R = 1.
In simple words: The hundreds digit gives R = 1, and then only 65 + 51 = 116 fits.

Teacher's Note:
1. The hundreds digit of such a sum is always 1.
2. Try values of Q one by one.