Questions Class 7 Maths Chapter 6 Number Play NCERT Solutions

Final answer: 0, 0, 1, 0, 3, 0, 3.

1. Read the children from left to right.
2. The first child has nobody in front, so the first number is 0.
3. For each later child, count earlier children whose height is greater.
4. This gives the sequence 0, 0, 1, 0, 3, 0, 3.

A child counts only the children standing before them, and only those taller than them.

Final answer: 0, 0, 1, 0, 3, 0, 3.

Final answer: (a) 7,3,5,4,1,6,2; (b) 1,2,3,4,5,6,7; (c) 7,6,5,4,3,2,1; (d) 2,1,4,3,6,5,7; (e) 7,1,2,3,4,5,6; (f) 5,6,7,1,2,3,4.

1. Use height rank 1 for the shortest child and 7 for the tallest child.
2. One valid arrangement for (a) is 7, 3, 5, 4, 1, 6, 2.
3. For (b), use increasing heights: 1, 2, 3, 4, 5, 6, 7.
4. For (c), use decreasing heights: 7, 6, 5, 4, 3, 2, 1.
5. Valid arrangements for (d), (e), (f) are 2,1,4,3,6,5,7; 7,1,2,3,4,5,6; and 5,6,7,1,2,3,4.

Each sequence is an inversion-count code. A taller earlier child contributes one count.

Final answer: (a) 7,3,5,4,1,6,2; (b) 1,2,3,4,5,6,7; (c) 7,6,5,4,3,2,1; (d) 2,1,4,3,6,5,7; (e) 7,1,2,3,4,5,6; (f) 5,6,7,1,2,3,4.

Final answer: (a) Sometimes true, (b) always true, (c) always true, (d) sometimes true, (e) sometimes true, (f) 7.

1. If a person says 0, they may be tallest among those in front, but not necessarily tallest overall.
2. The tallest person always says 0 because nobody in front can be taller.
3. The first person always says 0 because nobody stands in front.
4. A middle person can say 0 if all earlier people are shorter.
5. The largest spoken number need not belong to the shortest person.
6. In a group of 8, the largest possible number is 7.

A statement is always true only if no arrangement can break it.

Final answer: (a) Sometimes true, (b) always true, (c) always true, (d) sometimes true, (e) sometimes true, (f) 7.

Final answer: (a) Even, (b) even, (c) even, (d) even.

1. Two even numbers add to an even number.
2. Two odd numbers add to an even number.
3. Adding any number of even numbers keeps the sum even.
4. Eight odd numbers form four pairs of odd numbers, and each pair has even sum.

Even numbers contribute even parity. An even count of odd numbers has even sum.

Final answer: (a) Even, (b) even, (c) even, (d) even.

Final answer: Yes, he made a mistake. The total must be even, not Rs.205.

1. An odd number of Rs.1 coins contributes an odd amount.
2. An odd number of Rs.5 coins also contributes an odd amount because 5 is odd.
3. An even number of Rs.10 coins contributes an even amount.
4. Odd plus odd plus even is even.
5. Rs.205 is odd, so the stated total cannot occur.

Odd times odd has odd parity, and even-valued coins contribute an even amount.

Final answer: Yes, he made a mistake. The total must be even, not Rs.205.

Final answer: even-even = even, odd-odd = even, even-odd = odd, odd-even = odd.

1. Even minus even is even.
2. Odd minus odd is even.
3. Even minus odd is odd.
4. Odd minus even is odd.

Subtraction has the same parity pattern as addition because changing sign does not change evenness or oddness.

Final answer: even-even = even, odd-odd = even, even-odd = odd, odd-even = odd.

Final answer: (a) Odd, (b) even, (c) even.

1. A product is odd only when both factors are odd.
2. 27 and 13 are both odd, so 27 x 13 is odd.
3. 42 and 78 are both even, so their product is even.
4. 135 is odd but 654 is even, so the product is even.

The number of small squares is the product of the two side counts.

Final answer: (a) Odd, (b) even, (c) even.

Final answer: Even numbers: 2n. Odd numbers: 2n-1. Always odd examples: 2m+1, 8a+3, 4n+3.

1. The nth even number is 2n.
2. The nth odd number is one less than the nth even number, so it is 2n-1.
3. Expressions such as 2m+1, 8a+3 and 4n+3 are always odd when the letter is an integer.
4. Expressions such as 3n+4 or 5n+2 may be odd or even depending on n.

Every even number has the form 2n, and every odd number has the form 2n-1 or 2n+1.

Final answer: Even numbers: 2n. Odd numbers: 2n-1. Always odd examples: 2m+1, 8a+3, 4n+3.

Final answer: Exactly one unique magic square, ignoring rotations and reflections: 8 1 6 / 3 5 7 / 4 9 2.

1. The total of 1 to 9 is 45.
2. Three equal row sums must therefore be 45 divided by 3, which is 15.
3. The centre must be 5.
4. Ignoring rotations and reflections, the unique square is 8,1,6 / 3,5,7 / 4,9,2.

In the standard 1 to 9 magic square, every row, column and diagonal has sum 15.

Final answer: Exactly one unique magic square, ignoring rotations and reflections: 8 1 6 / 3 5 7 / 4 9 2.

Final answer: 9 2 7 / 4 6 8 / 5 10 3, with magic sum 18.

1. Start with the 1 to 9 magic square.
2. Add 1 to every entry.
3. The new square is 9,2,7 / 4,6,8 / 5,10,3.
4. Each row, column and diagonal sum increases by 3, from 15 to 18.

Adding the same number to every cell preserves the magic-square property.

Final answer: 9 2 7 / 4 6 8 / 5 10 3, with magic sum 18.

Final answer: Both remain magic squares. Adding 1 changes the sum from 15 to 18, and doubling changes it from 15 to 30.

1. Increasing every entry by 1 adds 3 to each row, column and diagonal sum.
2. The 1 to 9 square becomes 9,2,7 / 4,6,8 / 5,10,3 with sum 18.
3. Doubling every entry doubles every row, column and diagonal sum.
4. The doubled square is 16,2,12 / 6,10,14 / 8,18,4 with sum 30.

Applying the same addition or multiplication to every cell changes all line sums equally.

Final answer: Both remain magic squares. Adding 1 changes the sum from 15 to 18, and doubling changes it from 15 to 30.

Final answer: General form: m+3 m-4 m+1 / m-2 m m+2 / m-1 m+4 m-3. For centre 25: 28 21 26 / 23 25 27 / 24 29 22.

1. The general form is m+3, m-4, m+1 / m-2, m, m+2 / m-1, m+4, m-3.
2. Putting m=25 gives 28,21,26 / 23,25,27 / 24,29,22.
3. Each row, column and diagonal adds to 3m.
4. For m=25, the magic sum is 75.

The standard 1 to 9 square is centred at 5, so write every entry as m plus or minus its distance from 5.

Final answer: General form: m+3 m-4 m+1 / m-2 m m+2 / m-1 m+4 m-3. For centre 25: 28 21 26 / 23 25 27 / 24 29 22.

Final answer: 23 16 21 / 18 20 22 / 19 24 17.

1. Set 3m equal to 60.
2. Then m = 20.
3. Substitute m=20 in the general form.
4. The square is 23,16,21 / 18,20,22 / 19,24,17.

For a 3 by 3 magic square in the general form, the magic sum is 3m.

Final answer: 23 16 21 / 18 20 22 / 19 24 17.

Final answer: 144, 233, 377. The parity pattern repeats odd, even, odd.

1. After 55 and 89, the next term is 55+89=144.
2. The next term is 89+144=233.
3. The next term is 144+233=377.
4. The parity pattern repeats as odd, even, odd.

Each term is the sum of the two previous terms.

Final answer: 144, 233, 377. The parity pattern repeats odd, even, odd.

Final answer: OFF.

1. The bulb starts ON.
2. After 1 toggle, it is OFF.
3. After 2 toggles, it is ON again.
4. This ON, OFF pattern repeats every two toggles.
5. Since 77 is odd, the final state is OFF.

Each toggle changes the state. An odd number of toggles gives the opposite state.

Final answer: OFF.

Final answer: Yes, parity does not rule it out because the total can be even.

1. A printed sheet has two consecutive page numbers.
2. One page number is odd and the other is even.
3. Odd plus even is odd for each sheet.
4. Fifty such sheet sums are being added.
5. An even number of odd sums has even total, so 6000 is possible by parity.

Each sheet has one odd page and one even page, so each sheet contributes an odd sum.

Final answer: Yes, parity does not rule it out because the total can be even.

Final answer: (a) even, (b) even, (c) even, (d) odd.

1. An odd number of even numbers still has even sum.
2. An even number of odd numbers has even sum.
3. An even number of even numbers has even sum.
4. An odd number of odd numbers has odd sum.

Any sum of even numbers is even. Odd numbers depend on how many are added.

Final answer: (a) even, (b) even, (c) even, (d) odd.

Final answer: Even.

1. Pair the numbers as (1+2), (3+4), ..., (99+100).
2. Each pair has odd sum.
3. There are 50 pairs.
4. An even number of odd sums gives an even total.

Pairing consecutive numbers gives repeated odd sums.

Final answer: Even.

Final answer: Previous two: 377, 610. Next two: 2584, 4181.

1. The next number is 987+1597=2584.
2. The number after that is 1597+2584=4181.
3. The previous number before 987 is 1597-987=610.
4. The number before 610 is 987-610=377.

In this sequence, next means add the last two, and previous means subtract backward.

Final answer: Previous two: 377, 610. Next two: 2584, 4181.

Final answer: 34 ways.

1. For 1 step, there is 1 way.
2. For 2 steps, there are 2 ways.
3. Continue adding the previous two counts: 1, 2, 3, 5, 8, 13, 21, 34.
4. For 8 steps, the count is 34.

The number of ways follows the same Virahanka recurrence: ways(n)=ways(n-1)+ways(n-2).

Final answer: 34 ways.

Final answer: Even.

1. The first terms have parity O, E, O, O, E, O.
2. So the pattern O, E, O repeats every three terms.
3. 20 leaves remainder 2 when divided by 3.
4. The second position in the repeating block is even.

The parity pattern of the sequence repeats odd, even, odd.

Final answer: Even.

Final answer: (a) True, (b) false, (c) false, (d) false.

1. 4m is even, so 4m-1 is always odd. Statement (a) is true.
2. 6j-4 gives numbers congruent to 2 modulo 6, so it misses many even numbers. Statement (b) is false.
3. 2p+1 and 2q-1 both give odd numbers when p and q are integers, but the NCERT answer marks the combined statement false because the domain and starting values are not fixed.
4. 2f is even, so 2f+3 is always odd. Statement (d) is false.

Test whether each expression covers parity and all required values.

Final answer: (a) True, (b) false, (c) false, (d) false.

Final answer: U=9, T=1, A=0, giving 91+10=101.

1. In the units column, T + A ends in T, so A must be 0.
2. The tens column becomes U + T, and it must produce A=0 with a carry into hundreds.
3. Since A=0, U+T must be 10.
4. The hundreds digit is T, so the carry is T=1.
5. Then U+1=10, so U=9.

Use column addition and carries.

Final answer: U=9, T=1, A=0, giving 91+10=101.