Number Play Class 7 Solutions Ganita Prakash Maths Chapter 6✦ Step-by-step
HBSE 7th Class Maths Solutions Chapter 6 Number Play (NCERT Ganita Prakash) — every question worked out step by step. Tap Solve step by step.
Class 7 Maths Chapter 6 Number Play Solutions
Key ideas used in this chapter
Parityeven+even = even, odd+odd = even, even+odd = odd
Productsa product is odd only if both factors are odd
Magic square3×3 of 1–9 has magic sum 15 = 3 × centre
Virahankaeach term = sum of the two before it
6.2 Picking Parity — Figure It Out (Page 131)
Q1Parity of sums
(a) 2 even + 2 odd (b) 2 odd + 3 even (c) 5 even (d) 8 odd.
(a)
even+even = even; odd+odd = even; even+even = even
(b)
(odd+odd)=even; 3 even = even; even+even = even
(c)
any number of evens stays even
(d)
8 odds = 4 pairs, each pair even ⇒ even
Q2Lakpa's coins
Odd number of ₹1, odd number of ₹5, even number of ₹10 coins total ₹205. Mistake?
Step 1
₹1 total = 1×odd = odd; ₹5 total = 5×odd = odd; ₹10 total = 10×even = even
Step 2
odd + odd + even = even + even = even
Answer
the total must be even, but 205 is odd ⇒ Lakpa made a mistake
Q3, Q4Differences & grids
(Q3) parity of even−even, odd−odd, even−odd, odd−even. (Q4) parity of small squares in 27×13, 42×78, 135×654.
Q3
even−even = even; odd−odd = even; even−odd = odd; odd−even = odd
Q4 rule
a product is odd only if both sides are odd
Q4
27×13 odd×odd = odd; 42×78 = even; 135×654 (654 even) = even
6.3 Magic Squares — Figure It Out (Pages 136–137)
Q.Magic squares
How many 3×3 magic squares from 1–9? How do “+1” and “doubling” change the magic sum? With centre m, what is each line's sum? Build one with magic sum 60.
Count
8 squares (rotations & reflections of one); magic sum = 15
+1 / double
add 1 to all 9 ⇒ sum rises by 3 = 18; double all ⇒ sum doubles = 30
Centre m
every row, column and diagonal sums to 3m
Sum 60
60 ÷ 3 = 20 = centre; use the 9 numbers 16, 17, …, 24
6.4–6.5 Virahanka & Parity — Figure It Out (Pages 143–144)
Q1, Q2Toggle & pages
(Q1) A bulb is ON; the switch is toggled 77 times — lit? (Q2) 50 double-sided loose sheets — can the page-number sum be 6000?
Q1
2 toggles = back to start; 77 = 76 (even, no change) + 1 more ⇒ state flips ⇒ OFF
Q2 Step 1
a sheet's two pages are consecutive ⇒ odd + even = odd sum
Q2 answer
50 sheets = 50 odd sums; 50 is even ⇒ total is even ⇒ yes, 6000 is possible
Q5, Q6Parity fills & sum
(Q5) (a) odd count of evens (b) even count of odds (c) even count of evens (d) odd count of odds. (Q6) parity of 1 + 2 + … + 100.
Q5
(a) even (b) even (c) even (d) odd — only an odd count of odds is odd
Q6
pair as (1+2)+(3+4)+…+(99+100) = 50 odd numbers
Q6 answer
50 (even count) of odds ⇒ even (indeed 1+…+100 = 5050)
Q7, Q8Virahanka sequence
(Q7) After 987, 1597 — next two and previous two terms. (Q8) Ways to climb 8 steps using 1 or 2 steps.
Q7 next
987 + 1597 = 2584; then 1597 + 2584 = 4181
Q7 previous
each = next − the one after: 1597 − 987 = 610; 987 − 610 = 377
Q8
ways = 1, 2, 3, 5, 8, 13, 21, 34 ⇒ 34 ways
Q9, Q10Parity & expressions
(Q9) parity of the 20th Virahanka term. (Q10) true? (a) 4m−1 always odd (b) all evens are 6j−4 (c) 2p+1 and 2q−1 both give all odd numbers (d) 2f+3 gives even and odd.
Q9
even terms sit at positions 3n+2 (2,5,8,11,…); 20 = 3×6+2 ⇒ the 20th term is even
Q10 (a)
4m is even ⇒ 4m−1 is odd ⇒ true
Q10 (b),(c),(d)
all false: 6j−4 gives 2,8,14,… (misses 4,6,…); 2p+1 (p≥1) misses 1; 2f+3 is always odd