## Arithmetic Chapter 11 Square Root

**Question 1. **

**1. The square of 7**

**Solution:**

**Given Number 7**

The square of 7 = 7^{2}

= 7 x 7

= 49.

∴ The square of 7 = 49.

2. The square root of 121

**Solution:**

**Given number 121**

The square root of 121 = √I21

= (11)^{2}

=11.

And 121 = 11 x 11 = (11)^{²}

∴ The square root of 121 = 11.

3. √144

**Solution:**

**Given Number √144**

√144 = √(12)²

= 12

∴ √144 = 12.

**Read and Learn More WBBSE Solutions For Class 6 Maths**

∴ 144 = 2 x 2 x 2 x 2 x 3 x 3

= 2² x 2² x 3²

= (2 x 2 x 3)²

= (12)²

√144 = (12)²

**4. √(3² x 2²)**

**Solution:**

= √3² x √2²

= 3 x 2

= 6.

∴ √3² x 2² = 6.

5. √(5 x 7 x 5 x 7)

**Solution:**

= √(5 x 5 x 7 x 7)

= √(5² x 7²)

= √5² x √7²

= 5 x 7

= 35.

∴ √5 x 7 x 5 x 7

= √(14)²

= 14.

∴ √14 x 14

= 14.

√(5 x 7 x 5 x 7) = 14.

**Question 2. Using factorization find the square root of the following numbers: **

1. 144

**Solution:**

∴ 144 = 2 x 2 x 2 x2 x 3 x 3

∴ √144 = 2 x 2 x 3

= 12.

∴ The required square root = 12.

2. 169

**Solution:**

∴ 169 = 13 x 13

∴ √169 = 13 (one number is taken from one pair)

∴ The required square = 13.

**3. 225**

**Solution:**

∴ 225 = 3 x 3 x 5 x 5

∴ √225 = 3 x 5

= 15.

The required square = 15.

**4. 900**

**Solution:**

∴ 900 = 3 x 3 x 5 x 5

∴ √225 = 3 x 5

= 15.

**5. 15 ^{2} + 20^{2}.**

**Solution:**

∴ 625 = 5 x 5 x 5 x 5

∴ √625 = √5 x 5 x 5 x 5

= 5 x 5

= 25.

∴ The required square root = 25.

**Question 3. Write the following numbers correctly in the room (space) given below:**

**1. 20**

**Solution:**

∴ 20 = 2 x 2 x 5

= 2² x 5

**2. 27**

**Solution:**

∴ 27 = 3 x 3 x 3

= 3² x 3.

**3. 50**

**Solution:**

∴ 50 = 20 x 5 x 5

= 2 x 5²

**4. 100**

∴ 100 = 2 x 2 x 5 x 5

= 2² x 5²

**5. 108**

∴ 108 = 2 x 2 x 3 x 3 x 3

= 2² x 3² x 3.

**6. 169**

∴ 169 = 13 x 13

= (13)².

**The table is as follows:**

**Question 4. Arrange the following number in ascending order of magnitude:**

**1. √9 + √49**

**Solution:**

√9 + √49

= √(3)² +√(7)²

= 3 + 7

= 10.

√9 + √49 = 10.

2. √25 + √36

**Solution:**

√25 + √36

= √(5)² + √(6)²

= 5 + 6

= 11.

√25 + √36 = 11.

**3. √4 + √16**

**Solution:**

√4 + √16

= √(2)² + √(4)²

= 2 + 4

= 6.

√4 + √16 = 6.

∵ 6 < 10< 11< 15,

∴ √4 + √16 < √9 + √49 < √25 + √36.

∴ Arranging the given numbers in ascending order of magnitudes,

We get,

√4 + √16 < √9 + √49 < √25 + √36.

**Question 5. Without finding the actual square root determine the units place digit of the square root of the following numbers and also determine the number of digits in the square root of the given numbers :**

1. 784

**Solution:**

1. The unit’s place digit of 784 is 4.

∴ The unit’s place digit of the square root of 784 is either 2 or 8 and the number of digits in the square root of 784 is 2.

2. 1225

**Solution:**

The unit’s place digit of 1225 is 5.

∴ The units’

The units’ place digit of the square root of 1225 is 5 and the number of digits in the square root of 1225 is 2.

3. 10201

**Solution:**

The unit’s place digit of 10201 is 1

The units’, place digit of the square root of 10201 is either 1 or 9 and the number of digits in the square root of 10201 is 3.

**4. 160000. **

**Solution:**

The unit’s place digit of 160000 is 0.

The units’ place digit of the square root of 160000 is 0 and the number of digits in the square root of 160000 is 3.

**Question 6. What is the value of N in the number 202N, so that the number becomes a perfect square number?**

**Solution:**

∴ The required value of N is 5.

**Question 7. Determine the square root of the following numbers by the division method:**

**1. 529 **

**Solution:**

∴ The required square root is 23.

**2. 1764**

**Solution:**

∴ The required root is 42

**Question 8. Find the perfect square number nearer to 1000.**

**Solution:**

∴ The remainder is 39,

∴ (1000 – 39) = 961 which is a perfect square number.

Again the quotient is 31 and the next number to it is 32.

The square of 32 is (32)^{2}

= 32 x 32

= 1024.

Now the preceding perfect square number to 1000 is 961 and the succeeding perfect square number to 1000 is 1024.

Between 961 and 1024, the number 1024 is nearer to 1000 as 1024-1000 = 24 < 1000 – 961 = 39.

∴ The required perfect square number = 1024.

**Question 9. Find the least perfect square number of 4 digits.**

** Solution**: The least number of 4 digits = 1000.

The required least perfect square number of 4 digits is the least perfect square number greater than 1000.

∴ It is seen that the least number of 4 digits i.e., 1000 is not a perfect square number.

So the least perfect square number of 4 digits will be the square of (31+1)

i.e., the square of 32.

∴ The required least perfect square number of 4 digits = (32)²

= 32 x 32

= 1024.

**Question 10. Find the greatest perfect square number of 4 digits.**

** Solution**: The greatest number of 4 digits = 9999.

The required greatest perfect square number of 4 digits is the greatest perfect square number not greater than 9999.

∴ It is seen that the greatest number of 4 digits i.e., 9999 is not a perfect square. It is 198 more than the square of 99. So the greatest perfect square number of 4 digits is the square of 999.

∴ The required greatest perfect square number of 4 digits = (99)^{2} = 9801.

This can also be 9999 – 198 = 9801 which is a perfect square number.

It is to be noted that the quotient in the square root is 99. Its next integer is 99 + 1 = 100 whose square is 100^{2} = 100 x 100 = 10000, which is a perfect square number but not a number of 4 digits.

∴ The required greatest perfect square number of 4 digits = 9801.

**Question 11. What least number is to be subtracted from 9585 so that the result of subtraction is a perfect square?**

**Solution:**

∴ The required number = is 176.

In the determination of the square root of 9585, we see that there is a remainder of 176 at the end i.e., the given number is 176 more than the square of 97. So if we subtract 176 from the given number, then the remainder will be a perfect square.

176 is the required least number that will be subtracted from 9585 so that the result of subtraction is a perfect square.

**Question 12. What least number is to be added to 5320 so that the sum is a perfect square number?**

** Solution**:

∴ The required least number = 9.

**Question 13. What is the least perfect square number (other than zero) which is exactly divisible by 15, 25, 35, and 45?**

** Solution**: Since the required number is a perfect square number and divisible by 15, 25, 35, and 45, the required number will be the L. C. M. of the given numbers 15, 25, 35, and 45 or a multiple of L. C. M. of them.

∴ L.C. M. of 15, 25, 35, 45 = 3 x 5 x 5 x 7 x 3 = 3^{2} x 5^{2} x7 = 1575.

But 1575 is not a perfect square number. (This is because factor 7 is once only).

If the L. C. M. i.e., the number 1575 is multiplied by 7, Ifren the product is a perfect square number, and also if is divisible by the given numbers.

The required least perfect square number = 1575 x 7 = 11025.

**Question 14. What is the least perfect square number which has a factor of 17?**

** Solution **: The multiples of 17 are 17 x 1, 17 x 2, 17 x 3, 17 x 4,…………….. 17 x 16, 17 x 17, 17 x 18…………

These are the numbers each of which has a factor of 17.

But the least perfect square number having 17 as a factor is 17 x 17 = 289.

∴ The required least perfect square number = 289.

**Question 15. The product of two positive numbers is 1575 and their quotient is 9/7. ****Find the numbers.**

** Solution**: Let the greater number be

*and the smaller number be*

*x**(Here both x, and y are positive).*

*y.*∴ By the given condition, we get,

xy = 1575 …………….(1)

x/y = 9/7 ……………..(2)

Multiplying (1) & (2)

or, x² = 2025

or, x = 2025

= (45)^{2}

=45 ( ∵ > 0)

From (1), we get, 45y = 1575

or, y = 1575 / 45 = 35.

∴ The numbers are 35 and 45.

**Question 16. All the students of a school on republic day can be arranged in 12, 15, or 20 rows and also they can be arranged in solid squares. What is the least number of students in that school?**

**Solution:**

Since the students of a school are arranged in 12,15 or20 rows and they can also be arranged in solid squares, the least number of students in the school will be the L. C. M. (if it is a perfect square) or the multiple of L.C.M. which is a perfect square.

∴ L.C.M. of 12, 15, 20

= 2 x 2 x 3 x 5

= 2² x 3 x 5

= 60.

But 60 is not a perfect square number (Because it contains only one 3 and one 5 as factors).

The least multiple of L. C. M. which is a perfect square number = 60 x 3 x 5 = 900.

∴ The required number of students in the school = 900.

**Question 17. Shanti Devi has plucked 441 oranges from her fruit garden. She has kept the oranges in baskets such that the number of oranges in each basket is equal to the number of baskets. Find the number of baskets.**

** Solution**: Here the number of baskets and the number of oranges in each basket are equal.

The product of the number of baskets and the number of oranges in each basket the total number of oranges plucked = 441.

The product of two equal numbers = 441.

So each number = √441

= 21.

**Question 18. There are 140 students in your class. They are arranged in rows in solid squares. While arranging them in this way it is observed that there are 4 students. What is the number of rows in the solid square?**

** Solution**: If we include these 4 fewer students in the total number of students in the class, then the total number of students would be

140 + 4 = 144.

These 144 students can be arranged in rows of solid squares.

∴ Number of rows = √144

∴ 144 = 2 x 2 x 2 x 2 x 3 x 3

= 2² x 2² x 3²

= (2 x 2 x 3)²

= (12)²

∴ √144 = √(12)^{2}

= 12.

∴ The number of rows in the arrangements

= √44

=12.

The number of rows in the solid square =12.

**Question 19. ****Each member of Sukanta Smriti Pathagar has been given a subscription in rupees as the total number of members in Pathagar. If the total amount of subscriptions is Rs. 2601, then what is the number of members in the Pathagar?**

** Solution**: Each member of the Pathagar has been given a subscription in rupees = the total number of members in the Pathagar.

The total amount of subscriptions = is Rs. 2601.

.’. Product, of two equal numbers = 2601.

∴ Each number = Number of members = √2601 =51