
Find the square root of 4761
A. 69
B. 59
C. 49
D. 48
Answer
574.8k+ views
Hint: First of all, make pairs from the unit place of the given number. We will now multiply a number by itself such that the product is less than 47. Here, hence our first quotient is 6. After dividing by 6, we will get the remainder as 11 and now bring he next pair down. Now, multiply the quotient by 2 and find the unit digit when it is multiplied by the number, the product is less than or equal to the dividend.
Complete step by step answer:
We shall begin by making groups of 2 digits starting from the unit place.
Make the pair such as, $\overline {47} \overline {61} $, there are two pairs, first pair has 47 and second pair is 61.
Now, take the divisor as the largest number whose square is equal to or just less than the first number in the bar from the starting digits of the given number.
As ${6^2} = 36 < 47$ and ${7^2} = 49 > 47$
Therefore, our divisor is 6. Subtract the first pair from the square of 6 and then bring the next pair down.
Now, we get,
Next our dividend is 1161.
Write the twice of the first divisor, which is $6 \times 2 = 12$
We will now find the unit digit placed with 12 such that when that digit is multiplied by the new number, the result is less than or equal to dividend.
Therefore, we have
\[129 \times 9 = 1161\]
Hence, next digit in divisor is 9.
Now, the remainder is 0.
Hence, the square root of 4761 is 69.
Thus, option A is the correct answer.
Note: Many students make mistakes by making pairs from the starting of the number instead of a unit place, which gives the incorrect answer. The above question can also be done using the prime factorization method.
Hence, the square root of 4761 is $3 \times 23 = 69$
Complete step by step answer:
We shall begin by making groups of 2 digits starting from the unit place.
Make the pair such as, $\overline {47} \overline {61} $, there are two pairs, first pair has 47 and second pair is 61.
Now, take the divisor as the largest number whose square is equal to or just less than the first number in the bar from the starting digits of the given number.
As ${6^2} = 36 < 47$ and ${7^2} = 49 > 47$
Therefore, our divisor is 6. Subtract the first pair from the square of 6 and then bring the next pair down.
Now, we get,
| 6 | |
| 6 | \[\overline {47} \overline {61} \] |
| -36\[ \downarrow \] | |
| 1161 |
Next our dividend is 1161.
Write the twice of the first divisor, which is $6 \times 2 = 12$
We will now find the unit digit placed with 12 such that when that digit is multiplied by the new number, the result is less than or equal to dividend.
Therefore, we have
\[129 \times 9 = 1161\]
Hence, next digit in divisor is 9.
| 6 | |
| 6 | \[\overline {47} \overline {61} \] |
| -36\[ \downarrow \] | |
| 129 | 1161 -1161 |
| 0 |
Now, the remainder is 0.
Hence, the square root of 4761 is 69.
Thus, option A is the correct answer.
Note: Many students make mistakes by making pairs from the starting of the number instead of a unit place, which gives the incorrect answer. The above question can also be done using the prime factorization method.
| 3 | 4761 |
| 3 | 1587 |
| 23 | 529 |
| 23 | 23 |
| 1 |
Hence, the square root of 4761 is $3 \times 23 = 69$
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