CHAPTER 5 – EXERCISE 5.1 SOLUTIONS
Q1. What will be the unit digit of the squares of the following numbers?
(i) \(81 \Rightarrow\) unit digit 1 → \(1^2 = 1\) → Answer = 1
(ii) \(272 \Rightarrow\) unit digit 2 → \(2^2 = 4\) → Answer = 4
(iii) \(799 \Rightarrow\) unit digit 9 → \(9^2 = 81\) → Answer = 1
(iv) \(3853 \Rightarrow\) unit digit 3 → \(3^2 = 9\) → Answer = 9
(v) \(1234 \Rightarrow\) unit digit 4 → \(4^2 = 16\) → Answer = 6
(vi) \(26387 \Rightarrow\) unit digit 7 → \(7^2 = 49\) → Answer = 9
(vii) \(52698 \Rightarrow\) unit digit 8 → \(8^2 = 64\) → Answer = 4
(viii) \(99880 \Rightarrow\) unit digit 0 → \(0^2 = 0\) → Answer = 0
(ix) \(12796 \Rightarrow\) unit digit 6 → \(6^2 = 36\) → Answer = 6
(x) \(55555 \Rightarrow\) unit digit 5 → \(5^2 = 25\) → Answer = 5
Q2. The following numbers are obviously not perfect squares. Give reason.
(i) 1057 → unit digit 7 (a square never ends with 2, 3, 7, or 8)
(ii) 23453 → unit digit 3 → not a perfect square
(iii) 7928 → unit digit 8 → not a perfect square
(iv) 222222 → unit digit 2 → not a perfect square
(v) 64000 → ends with 3 zeros, but a perfect square can only have even number of zeros at the end → not square
(vi) 89722 → unit digit 2 → not square
(vii) 222000 → ends with 3 zeros → not square
(viii) 505050 → unit digit 0 but odd number of zeros at end → not square
Q3. The squares of which of the following would be odd numbers?
(i) 431 → odd → square is odd
(ii) 2826 → even → square is even
(iii) 7779 → odd → square is odd
(iv) 82004 → even → square is even
Q4. Observe the following pattern and find the missing digits.
\(11^2 = 121\)
\(101^2 = 10201\)
\(1001^2 = 1002001\)
\(10001^2 = 100020001\)
\(1000001^2 = 1000002000001\)
Q5. Observe the following pattern and supply the missing numbers.
\(11^2 = 121\)
\(101^2 = 10201\)
\(10101^2 = 102030201\)
\(1010101^2 = 102030405030201\)
\(101010101^2 = 10203040504030201\)
Q6. Using the given pattern, find the missing numbers.
\(1^2 + 2^2 + 2^2 = 3^2\)
\(2^2 + 3^2 + 6^2 = 7^2\)
\(3^2 + 4^2 + 12^2 = 13^2\)
\(4^2 + 5^2 + 20^2 = 21^2\)
\(5^2 + 6^2 + 30^2 = 31^2\)
\(6^2 + 7^2 + 42^2 = 43^2\)
Q7. Without adding, find the sum.
(i) \(1 + 3 + 5 + 7 + 9 = 25 = 5^2\)
(ii) Sum of first 10 odd numbers = \(10^2 = 100\)
(iii) Sum of first 12 odd numbers = \(12^2 = 144\)
Q8. Express numbers as sum of odd numbers.
(i) \(49 = 1+3+5+7+9+11+13\) (7 odd numbers)
(ii) \(121 = 1+3+5+7+9+11+13+15+17+19+21\) (11 odd numbers)
Q9. How many numbers lie between squares of the following numbers?
Hint: The difference between consecutive square numbers is given by the formula:
\((n+1)^2 - n^2 = 2n + 1\).
Therefore, the total numbers between \(n^2\) and \((n+1)^2\) = \((2n+1) - 1 = 2n\).
(i) 12 and 13
Here, \(n = 12\).
Total numbers = \(2n = 2 \times 12 = 24\).
Method 2: \(13^2 - 12^2 = 169 - 144 = 25\). So, 24 numbers lie between.
(ii) 25 and 26
Here, \(n = 25\).
Total numbers = \(2n = 2 \times 25 = 50\).
Method 2: \(26^2 - 25^2 = 676 - 625 = 51\). So, 50 numbers lie between.
(iii) 99 and 100
Here, \(n = 99\).
Total numbers = \(2n = 2 \times 99 = 198\).
Method 2: \(100^2 - 99^2 = 10000 - 9801 = 199\). So, 198 numbers lie between.