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^{th}Suqare And Square Roots Test paper1. 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

[ hint: 5

^{2}= 25 So, 55555^{2}will have 5 at unit’s place2. 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

[Numbers having only 0, 1, 4, 5, 6, and 9 at unit’s place and even number of zeroes at the end, are perfect squares.]

3. The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

4. (i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

5. Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

6. How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

[99

^{2}= 9801 , 100^{2}= 10000 ,Now, 10000 - 9801 = 199 So, there are 199 - 1 = 198 numbers lying between 99^{2}and 100^{2 ]}^{ }

7. Write a Pythagorean triplet whose one member is. (i) 14 (ii) 16 (iii) 18

[ As we know 2m, m

^{2}+1 and m^{2}-1 form a Pythagorean triplet for any number, m>1.]7. 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

Answer: Since 1

^{2}and 9^{2}give 1 at unit’s place, so these are the possible values of unit digit of the square root.Answer: 4

^{2 }= 16 and 6^{2 }= 36, hence, 4 and 6 are possibleAnswer: Same as question 1—(i)

Answer: 5

^{2 }= 25, hence 5 is possible.8. Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Answer: Option 1 can be a perfect square, others can’t be perfect squares because the unit digit of a perfect square can be only from 0, 1, 4, 5, 6, 9

9. Find the square roots of 100 and 169 by the method of repeated subtraction.

Answer: Repeated subtraction:

1. 100 - 1 = 99

2. 99 - 3 = 96

3. 96 - 5 = 91

4. 91 -7 = 84

5. 84 - 9 = 75

6. 75 - 11 = 64

7. 64 - 13 = 51

8. 51 - 15 = 36

9. 36 - 17 = 19

10. 19 - 19 = 0

We get 0 at 10th step √100 = 10

1. 169 - 1 = 168

2. 168 - 3 = 165

3. 165 - 5 = 160

4. 160 - 7 = 153

5. 153 - 9 = 144

6. 144 - 11 = 133

7. 133 - 13 = 120

8. 120 - 15 = 105

9. 105 - 17 = 88

10. 88 - 19 = 69

11. 69 - 21 = 48

12. 48 - 23 = 25

13. 25 - 25 = 0

We get 0 at 13th step √169=13

10. 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

Answer: 252 = 2 x 126

= 2 x 2 x 63

= 2 x 2 x 3 x 3 x 7

Here, 2 and 3 are in pairs but 7 needs a pair, so 252 will become a perfect square when multiplied by 7.

(ii) 180 = 3 x 3 x 2 x 2 x 5

180 needs to be multiplied by 5 to become a perfect square.

(iii) 1008

= 2 x 2 x 2 x 2 x 3 x 3 x 7

1008 needs to be multiplied by 7 to become a perfect square

(iv) 2028 = 2 x 2 x 3 x 13 x 13

2028 needs to be multiplied by 3 to become a perfect square.

(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3

1458 needs to be multiplied by 2 to become a perfect square.

(vi) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3

768 needs to be multiplied by 3 to become a perfect square.

11. 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 = 2 x 2 x 3 x 3 x 7

252 needs to be divided by 7 to become a perfect square

(ii) 2925 = 5 x 5 x 3 x 3 x 13

2925 needs to be divided by 13 to become a perfect square

(iii) 396 = 2 x 2 x 3 x 3 x 11

396 needs to be divided by 11 to become a perfect square

(iv) 2645 = 5 x 23 x 23

2645 needs to be divided by 5 to become a perfect square.

(v) 2800 = 2 x 2 x 7 x 10 x 10

2800 needs to be divided by 7 to become a perfect square.

(vi) 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5

1620 needs to be divided by 5 to become a perfect square

12. The students of Class VIII of a school donated Rs 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.

Answer: We need to calculate the square root of 2401 to get the solution

2401 = 7 x 7x 7x7 √2401 = 7x7=49

There are 49 students, each contributing 49 rupees

13. 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.

2025 = 5x5x3x3x3x3

√2025 = 5x3x3=45

There are 45 rows with 45 plants in each of them.

14 Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Answer: Let us find LCM of 4, 9 and 10

4 = 2 x 2

9 = 3 x 3

10 = 5 x 2

So, LCM = 2

^{2 }x 3^{2 }x 5 = 180Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square.

The Required number = 180 x 5 = 900

15. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Answer:

8 = 2 x 2 x 2

15 = 3 x 5

20 = 2 x 2 x 5

So, LCM = 2 x 2 x 2 x 3 x 5=120

As 3 and 5 are not in pair in LCM’s factor so we need to multiply 120 by 5 and three to make it a perfect square.

Required Number = 180 x 3 x 5 = 2700

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