Exercise 6.1
Question 1:
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
[viii] 99880
(ix) 12796
[x] 55555
Answer 1:
(i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1.
Hence, unit’s digit of square of 81 is 1.
(ii) The number 272 contains its unit’s place digit 2.
So, square of 2 is 4. Hence, unit’s digit of square of 272 is 4.
(iii) The number 799 contains its unit’s place digit 9.
So, square of 9 is 81. Hence, unit’s digit of square of 799 is 1.
(iv) The number 3853 contains its unit’s place digit 3.
So, square of 3 is 9. Hence, unit’s digit of square of 3853 is 9.
(v) The number 1234 contains its unit’s place digit 4.
So, square of 4 is 16. Hence, unit’s digit of square of 1234 is 6.
(vi) The number 26387 contains its unit’s place digit 7.
So, square of 7 is 49. Hence, unit’s digit of square of 26387 is 9.
(vii) The number 52698 contains its unit’s place digit 8.
So, square of 8 is 64. Hence, unit’s digit of square of 52698 is 4.
(viii) The number 99880 contains its unit’s place digit 0.
So, square of 0 is 0. Hence, unit’s digit of square of 99880 is 0.
(ix) The number 12796 contains its unit’s place digit 6.
So, square of 6 is 36. Hence, unit’s digit of square of 12796 is 6.
(x) The number 55555 contains its unit’s place digit 5.
So, square of 5 is 25. Hence, unit’s digit of square of 55555 is 5.
Question 2:
The following numbers are obviously not perfect squares. Give reasons.
[i] 1057
[ii] 23453
(iii] 7928
(iv] 222222
[v] 64000
(vi) 89722
(vii) 222000
(viii) 505050
Answer 2:
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0.
Therefore 505050 is not a perfect square because its unit’s place digit is 0.
Question 3:
The squares of which of the following would be odd number:
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Answer 3: (i) 431 – Unit’s digit of given number is 1 and square of 1 is 1.
Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36.
Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81.
Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16.
Therefore, square of 82004 would not be an odd number.
Question 4:
Observe the following pattern and find the missing digits:
Answer 4:
Question 5:
Observe the following pattern and supply the missing numbers:
Answer 5:
Question 6:
Using the given pattern, find the missing numbers:
Answer 6:
Question 7:
Without adding, find the sum:
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Answer 7:
[i]Here,there are five odd numbers. Therefore square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 5^{2} = 25
[ii] Here, there are ten odd numbers. Therefore square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10^{2} = 100
[iii]
Here, there are twelve odd numbers. Therefore square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^{2} = 144
Question 8:
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer 8:
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Question 9:
How many numbers lie between squares of the following numbers:
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Answer 9:
(i)
Since, nonperfect square numbers between n^{2} and (n + 1)"^{2 }are 2n
Here, n = 12
Therefore, nonperfect square numbers between 12 and 13 = 2 n = 2x12 = 24
[ii]
Since, nonperfect square numbers between n^{2} and (n + 1)^{2} are 2n.
Here, n  25
Therefore, nonper feet square numbers between 25 and 26 = 2 n = 2x25 = 50
[iii]
Since, nonperfect square numbers between n and (n + 1) are 2n.
Here, n = 99
Therefore, nonperfect square numbers between 99 and 100 = 2n = 2 x 99 = 198
Exercise 6.2
Question 1: Find the squares of the following numbers:
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Answer 1:
(i) (32)^{2 }= (30 + 2)^{2} = (30)^{2} + 2x 30x 2 + (2)^{2} [∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 900+ 120 + 4 = 1024
(ii)
(35)^{2} = (30 + 5)^{2 }= (30)^{2} + 2x 30x 5 + (5)^{2 }[∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 900 + 300 + 25 = 1225
(iii)
(86)^{2 }= (80 + 6)^{2} = (80)^{2}+ 2 x 80 x 6 + (6)^{2} [∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 1600 + 960 + 36 = 7386
(iv)
(93)^{2} =(90 +3)^{2} =(90)^{2} +2x90x3 + (3)^{2 } [∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 8100 + 540 + 9 = 8649
(v)
(71)^{2} = (70+1)^{2} = 70^{2 } + 2 x 70 x 1 + 1^{2} [∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 4900 + 140 + 1 = 5041
[vi]
(46)^{2} =(40 +6)^{2} = (40)^{2}+2x40x6 +(6)^{2} [∵ (a+b)^{2} = a^{2} + b^{2} + 2ab]
= 1600 + 480 + 36 = 2116
Question 2:
Write a Pythagoras triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18
Answer 2:
(i)
There are three numbers 2m,m ^{2} 1 and m^{2} + 1 in a Pythagorean Triplet
Here, 2m = 6 => m = 6/2 = 3
There fore,
Second number (m^{2} 1) = (3)^{2} 1 = 9 1 = 8
Third number m^{2}+1 = (3)^{2} +1 = 9 + 1 = 10
Hence, Pythagorean triplet is (6,8,10).
(ii)
There are three numbers 2m, m^{2 }1 and nr +1 in a Pythagorean Triplet
Here, 2m = 6, m = 6/2 = 3
Therefore,
Second number (m^{2}  1) = (7)^{2} 1 = 49  1 = 48
Third number m^{2} +1 = (7)^{2} + l =49 + l = 50
Hence, Pythagorean triplet is (14,48, 50).
(iii)
There are three numbers 2m,m^{2} 1 and m^{2} + 1 in a Pythagorean Triplet
Here, 2m =16 , m = 8
Therefore,
Second number (m^{2}  1) = (8)^{2} 1 =64 1 = 63
Third number m^{2} +1 = (8)^{2} +1 = 64 +1 = 65
Hence, Pythagorean triplet is (16, 63, 65).
(iv)
There are three numbers 2m,m^{2} 1 and m^{2} + 1 in a Pythagorean Triplet
Here, 2m =18 , m = 9
Therefore,
Second number (m^{2}  1) = (9)^{2} 1 = 81 1 = 80
Third number m^{2} +1 = (9)^{2} +1 = 81 +1 = 62
Hence, Pythagorean triplet is (18, 80, 82).
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 