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
Solution: (i) The unit digit of the square of the number 81 is
(ii) The unit digit of the square of the number 272 is
(iii) The unit digit of the square of the number 799 is [9² = 81]
(iv) The unit digit of the square of the number 3853 is
(v) The unit digit of the square of the number 1234 is [ ]
(vi) The unit digit of the square of the number 26387 is []
(vii) The unit digit of the square of the number 52698 is []
(viii) The unit digit of the square of the number 99880 is
(ix) The unit digit of the square of the number 12796 is []
(x) The unit digit of the square of the number 55555 is []
2. The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
Solution: (i) The number 1057 ends with unit place 7 . So, 1057 is not the perfect number .
(ii) The number 23453 ends with unit place 3 . So, 23453 is not the perfect number .
(iii) The number 7928 ends with unit place 8 . So, 7928 is not the perfect number .
(iv) The number 222222 ends with unit place 2 . So, 222222 is not the perfect number .
(v) The number 64000 ends with unit place 0 . So, 64000 is not the perfect number .
(vi) The number 89722 ends with unit place 2 . So, 89722 is not the perfect number .
(vii) The number 222000 ends with unit place 0 . So, 222000 is not the perfect number .
(viii) The number 505050 ends with unit place 0 . So, 505050 is not the perfect number .
3. The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Solution: (i) 431² is an odd number . (i.e., 1² = 1)
(ii) 2826² is an even number (i.e., 6² = 36 ) .
(iii) 7779² is an odd number (i.e., 9² = 81)
(iv) 82004² is an even number (i.e., 4² =16)
4. Observe the following pattern and find the missing digits.
Solution: We have,
5. Observe the following pattern and supply the missing numbers.
Solution: We have ,
6. Using the given pattern, find the missing numbers.
Solution: We have,
[Note: (i)
(ii)
(iii)
and ]
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
Solution: (i) We have, 1 + 3 + 5 + 7 + 9
= 5² = 5 × 5 = 25
(ii) We have, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
= 10² = 10 × 10 = 100
(iii) We have, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 +21 +23
= 12² = 12 × 12 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution: (i) We have , 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) We have, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Solution: (i) 12 and 13
We know that , the squares of the numbers and is .
Therefore, the squares of the numbers and is .
(ii) 25 and 26
We know that , the squares of the numbers and is .
Therefore, the squares of the numbers and is .
(iii) 99 and 100
We know that , the squares of the numbers and is .
Therefore, , the squares of the numbers and is .
1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
[Note : For any natural number ,
We have, . So, the Pythagorean triplets are and ]
Solution: (i) 6
We know that , the Pythagorean triplets are and
We take,
So,
Then
Then the value of will not be integer .
We take,
Thus, and
Thus , the triplet are 6 , 8 and 10 .
(ii) 14
We know that , the Pythagorean triplets are and
We take ,
So ,
Then
Then the value of will not be integer .
We take,
Then
Again , and
Thus , the triplet are 14 , 48 and 50 .
(iii) 16
We know that, the Pythagorean triplets are and
We take
So ,
Then
Then the value of will not be integer .
We take,
Again, will not give an integer value for .
We take,
Thus, and
Therefore , the triplet are 16 , 63 and 65 .
(iv) 18
We know that , the Pythagorean triplets are and
We take
So ,
Then
Then the value of will not be integer .
We take,
Again, will not give an integer value for .
We take,
Thus, and
Therefore , the triplet are 18 , 80 and 82 .
1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025
Solution: (i) 9801
Therefore, the possible ‘one’s’ digits of the square root of 9801 are 1 and 9
(ii) 99856
Therefore, the possible ‘one’s’ digits of the square root of 99856 are 4 and 6 .
(iii) 998001
Therefore, the possible ‘one’s’ digits of the square root of 998001 are 1 and 9 .
(iv) 657666025
Therefore, the possible ‘one’s’ digits of the square root of 657666025 is 5 .
2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
Solution: Since, the unit’s place of the digits with 2 , 3 , 7 or 8 are not a perfect square number .
(i) The number 153 is the unit’s place of the digits with 3 . So, 153 is not a perfect square number .
(ii) The number 257 is the unit’s place of the digits with 7 . So, 257 is not a perfect square number .
(iii) The number 408 is the unit’s place of the digits with 8 . So, 408 is not a perfect square number .
(iv) The number 441 is the unit’s place of the digits with 1 . So, 441 is a perfect square number .
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution: (i) We have , 100 – 1 = 99 , 99 – 3 = 96 , 96 – 5 = 91 , 91 – 7 = 84 , 84 – 9 = 75 , 75 – 11 = 64 , 64 – 13 = 51 , 51 – 15 = 36 , 36 – 17 = 19 , 19 – 19 = 0
We know that, the sum of the first odd natural numbers is .There are 10 odd numbers for 100 .
Therefore, the square roots of 100 is 10 .
(ii) We have , 169 – 1 = 168 , 168 – 3 = 165 , 165 – 5 = 160 , 160 – 7 = 153 , 153 – 9 = 144 , 144 – 11 = 133 , 133 – 13 = 120 , 120 – 15 = 105 , 105 – 17 = 88 , 88 – 19 = 69 , 68 – 21 = 48 , 48 – 23 = 25 , 25 – 25 = 0
We know that, the sum of the first odd natural numbers is . There are 13 odd numbers for 169 .
Therefore, the square roots of 169 is 13 .
4. Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
Solution: (i) 729
Using prime factorisation Method , we have
729 = 3×3×3×3×3×3 = 3²×3²×3²
So,
(ii) 400
Using prime factorisation Method , we have
400 = 2×2×2×2×5×5 = 2²×2²×5²
So,
(iii) 1764
Using prime factorisation Method , we have
1764 = 2×2×3×3×7×7 = 2²×3²×7²
So,
(iv) 4096
Using prime factorisation Method , we have
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 2² × 2² × 2² × 2² × 2² × 2²
So,
(v) 7744
Using prime factorisation Method , we have
7744 = 2×2×2×2×2×2×11×11 = 2² × 2² × 2² × 11²
So,
(vi) 9604
Using prime factorisation Method , we have
9604 = 2×2×7×7×7×7 = 2²×7²×7²
So,
(vii) 5929
Using prime factorisation method , we have
5929 = 7×7×11×11 = 7² × 11²
So,
(viii) 9216
Using prime factorisation method , we have
9216 = 2×2×2×2×2×2×2×2×2×2×3×3 = 2² × 2² × 2² × 2² × 2² × 3²
So,
(ix) 529
Using prime factorisation method , we have
529 = 23×23 = 23²
So,
(x) 8100
Using prime factorisation method , we have
8100 = 2×2×3×3×3×3×5×5 = 2² × 3² × 3² × 5²
So,
5. For each of the following numbers,find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768
Solution: (i) We have,
252 = 2×2×3×3×7
The prime factors 7 has no pair , so 252 is not a perfect square .
Now, we multiply 252 by 7
Therefore, 252 × 7 = 2×2×3×3×7×7 is a perfect square .
Thus,
(ii) We have,
180 = 2×2×3×3×5
The prime factors 5 has no pair , so 180 is not a perfect square .
Now, we multiply 180 by 5
Therefore, 180 × 5 = 2×2×3×3×5×5 is a perfect square .
Thus,
(iii) We have,
1008 = 2×2×2×2×3×3×7
The prime factors 7 has no pair , so 1008 is not a perfect square .
Now, we multiply 1008 by 7
Therefore, 1008 × 7 = 2×2×2×2×3×3×7×7 is a perfect square .
Thus ,
(iv) We have,
2028 = 2×2×13×13×3
The prime factors 3 has no pair , so 2028 is not a perfect square .
Now, we multiply 2028 by 3
Therefore, 2028 × 3 = 2×2×13×13×3×3 is a perfect square .
Thus ,
(v) We have,
1458 = 2×3×3×3×3×3×3
The prime factors 2 has no pair , so 1458 is not a perfect square .
Now, we multiply 1458 by 2
Therefore, 1458 × 2 = 2×2×3×3×3×3×3×3 is a perfect square .
Thus,
(vi) We have,
768 = 2×2×2×2×2×2×2×2×3
The prime factors 3 has no pair , so 768 is not a perfect square .
Now, we multiply 768 by 3
Therefore, 768 × 3 = 2×2×2×2×2×2×2×2×3×3 is a perfect square .
Thus,
6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.
(i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620
Solution: (i) We have,
252 = 2×2×3×3×7
Now, we divide 252 by the factor 7 , then 252 7 = 2×2×3×3 is a perfect square number .
Therefore, the required smallest number is 7 .
So,
(ii) We have,
2925 =3×3×5×5×13
Now, we divide 2925 by the factor 13,then 2925 13 = 3×3×5×5 is a perfect square number .
Therefore, the required smallest number is 13 .
So,
(iii) We have,
396 = 2×2×3×3×11
Now , we divide 396 by the factor 11 , then 396 11 = 2×2×3×3 is a perfect square number .
Therefore, the required smallest number is 11 .
So,
(iv) We have,
2645 = 5×23×23
Now , we divide 2645 by the factor 5 , then 2645 ÷ 5 = 23×23 is a perfect square number .
Therefore, the required smallest number is 5 .
So,
(v) We have,
2800 = 2×2×2×2×5×5×7
Now, we divide 2800 by the factor 7 , then 2800 ÷ 7 = 2×2×2×2×5×5 is a perfect square number .
Therefore, the required smallest number is 7 .
So,
(vi) We have,
1620 = 2×2×3×3×3×3×5
Now, we divide 1620 by the factor 5 , then 1620 5 = 2×2×3×3×3×3 is a perfect square number .
Therefore, the required smallest number is 5 .
So,
7. The students of Class VIII of a school donated ` 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution: The total amount of donation = Rs 2401 .
We have,
2401 = 7×7×7×7 = 7² × 7²
Therefore, the number of students in the class
8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution: Total number of plants = 2025 .
Here, the number of rows = the number of plants in each row .
We have,
2025 = 3×3×3×3×5×5 = 3² × 3² × 5²
So, the number of rows
Therefore, the number of plants in each row is 45 .
9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution: The least number divisible by each one of 4 , 9 and 10 is their LCM .
We have, 4 = 2 × 2 , 9 = 3 × 3 , 10 = 2 × 5
So, the LCM of 8 , 15 and 20 is 2 × 2 × 3 × 3 × 5 = 180
The prime factorization of 180 is 180 = 2 × 2 × 3 × 3 × 5 is not perfect square .
Therefore, 180 should be multiplied by 5 .
Hence, the required square number is 180 × 5 = 900 .
10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution: The least number divisible by each one of 8 , 15 and 20 is their LCM .
We have, 8 = 2 × 2 × 2 , 15 = 3 × 5 , 20 = 2 × 2 × 5
So, the LCM of 8 , 15 and 20 is 2 × 2 × 2 × 3 × 5 = 120
The prime factorization of 120 is 120 = 2 × 2 × 2 × 3 × 5 is not perfect square .
Therefore, 120 should be multiplied by 2 × 3 × 5 = 30 .
Hence, the required square number is 120 × 30 = 3600 .
1. Find the square root of each of the following numbers by Division method.
(i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900
2. Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625
[Note : If a perfect square is of digits, then its square root will have digits if is even or if is odd ]
Solution: (i) 64
Here, (even)
Therefore, the number of digits in the square root
[ i.e., (One digit)]
(ii) 144
Here, (Odd)
Therefore, the number of digits in the square root
[ i.e., (Two digits)]
(iii) 4489
Here, (even)
Therefore, the number of digits in the square root
[ i.e., (Two digits)]
(iv) 27225
Here, (Odd)
Therefore, the number of digits in the square root
[ i.e., (Three digits)]
(v) 390625
Here, (Even)
Therefore, the number of digits in the square root
[ i.e., (Three digits)]
3. Find the square root of the following decimal numbers.
(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36
4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000
Solution: (i) (i) We find by long division method .
We get the remainder 2 . So, 20² is less than 402 by 2 .
Therefore, the required perfect square number = 402 – 2 = 400 and .
(ii) We find by long division method .
We get the remainder 53 . So, 44² is less than 1989 by 53 .
Therefore, the required perfect square number = 1989 – 53 = 1936 and
5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412
6. Find the length of the side of a square whose area is 441 m².
Solution: let be the length of the side of a square .
A/Q, Area of a square = 441
Therefore , the length of the side of a square is 21 cm .
7. In a right triangle ABC, .
(a) If AB = 6 cm, BC = 8 cm, find AC (b) If AC = 13 cm, BC = 5 cm, find AB .
8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.