NCERT Solutions Class 8 Maths Chapter 6

Q.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

Ans

1. 81- Here, the number ends with 1.So its square will end with (1×1=1).Therefore, the unit digit of the square of 81 will be 1.
2. 272 – Here, the number ends with 2.So its square will end with (2×2=4).Therefore, the unit digit of the square of 272 will be 4.
3. 799 – Here, the number ends with 9.So its square will end with (9×9= 81).Therefore, the unit digit of the square of 799 will be 1.
4. 3853- Here, the number ends with 3.So its square will end with (3×3=9).Therefore, the unit digit of the square of 3853 will be 9.
5. 1234 – Here, the number ends with 4. So its square will end with (4×4=16).Therefore, the unit digit of the square of 1234 will be 6.
6. 26387- Here, the number ends with 7. So its square will end with (7×7=49).Therefore, the unit digit of the square of 23387 will be 9.
7. 52698 – Here, the number ends with 8. So its square will end with (8×8=64).Therefore, the unit digit of the square of 52698 will be 4
8. 99880- Here, the number ends with 0. So its square will end with (0×0=0).Therefore, the unit digit of the square of 99880 will be 0.
9. 12796 – Here, the number ends with 6. So its square will end with (6×6=36).Therefore, the unit digit of the square of 12796 will be 6.
10. 55555- Here, the number ends with 5. So its square will end with (5×5=25). Therefore, the unit digit of the square of 55555 will be 5.

Q.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

Ans

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

1. 1057: Since the unit digit of 1057 is 7.Therefore, it is not a perfect square.
2. 23453: Since the unit digit of 23453 is 3.Therefore, it is not a perfect square.
3. 7928: Since the unit digit of 7928 is 8.Therefore, it is not a perfect square.
4. 222222: Since the unit digit of 222222 is 2. Therefore, it is not a perfect square.
5. 64000: A perfect square should have even number of zeroes at the end but 64000 is having three zeroes at the end which is not an even number. Therefore, it is not a perfect square.
6. 89722 : Since the unit digit of 89722 is 2.Therefore,it is not a perfect square.
7. 222000 : A perfect square should have even number of zeroes at the end but 222000 is having three zeroes at the end which is not an even number. Therefore, it is not a perfect square.
8. A perfect square should have even number of zeroes at the end but 505050 is having one zero at the end which is not an even number. Therefore, it is not a perfect square.

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

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

Ans

Here, 431 and 7779 are odd numbers and 2826 and 82004 are even numbers

We know that the square of an odd number is odd and the square of an even number is even.

Thus, the squares of 431 and 7779 will be odd numbers.

Q.4 Observe the following pattern and find the missing digits.

11² = 121

101² = 10201

1001² = 1002001

100001² = 1 ……… 2 ……… 1

10000001² = ………………………

Ans

We can observe that, the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,

100001² = 10000200001

10000001² = 100000020000001

Q.5 Observe the following pattern and supply the missing numbers.

11² = 1 2 1

101² = 1 0 2 0 1

10101² = 102030201

1010101² = ………………………

…………² = 10203040504030201

Ans

We can observe that, the squares of the given numbers have the same number of zeroes after every digit as it was in the original number.

Therefore,

1010101² = 1020304030201

101010101² = 10203040504030201

Hence, the missing numbers are:

1020304030201, 101010101²

Q.6 Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² + 4² + 12² = 13²

4² + 5² + _² = 21²

5² + _² + 30² = 31²

6² + 7² + _² = __²

Ans

Clearly, the third number is the product of the first two numbers and the fourth number can be obtained by adding 1 to the third number.

Thus, the missing numbers in the pattern will be as follows.

\begin{array}{l}{\text{4}}^{\text{2}}+{\text{5}}^{\text{2}}+\underset{¯}{{20}^2}={\text{21}}^{\text{2}}\\ {\text{5}}^{\text{2}}+\underset{¯}{6^2}+\text{3}{0}^{\text{2}}={\text{31}}^{\text{2}}\\ {\text{6}}^{\text{2}}+{\text{7}}^{\text{2}}+\underset{¯}{{42}^2}=\underset{¯}{{43}^2}\end{array}

Q.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

Ans

1. Since, the sum of first n odd natural numbers is n².Therefore,1 + 3 + 5 + 7 + 9 = (5)² = 25.

2. Since, the sum of first n odd natural numbers is n².Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100.

3. Since, the sum of first n odd natural numbers is n².Therefore,1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)² = 144

Q.8 (i) Express 49 as the sum of 7 odd numbers.

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

Ans

(i) We know that the sum of first n odd natural numbers is n².Therefore, 49 is the sum of first 7 odd natural numbers.

49 = (7)²

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) We know that the sum of first n odd natural numbers is n².Therefore, 121 is the sum of first 11 odd natural numbers.

121 = (11)²

121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q.9 How many numbers lie between squares of the following numbers?

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

Ans

We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).

(i) Between 12² and 13², there will be 2 × 12 = 24 numbers

(ii) Between 25² and 26², there will be 2 × 25 = 50 numbers

(iii) Between 99² and 100², there will be 2 × 99 = 198 numbers

Q.10 Find the square of the following numbers.

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Ans

$\begin{array}{l}\left(\mathrm{i}\right){32}^2={(30+2)}^2\\ =(30+2)(30+2)\\ =30(30+2)+2(30+2)\\ =30\times 30+2\times 30+2\times 30+2\times 2\\ =900+120+4\\ =1024\\ \\ \left(\mathrm{ii}\right){35}^2={(30+5)}^2\\ =(30+5)(30+5)\\ =30(30+5)+5(30+5)\\ =30\times 30+5\times 30+5\times 30+5\times 5\\ =900+150+150+25\\ =1225\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right){86}^2={(80+6)}^2\\ =(80+6)(80+6)\\ =80(80+6)+6(80+6)\\ =80\times 80+6\times 80+6\times 80+6\times 6\\ =6400+480+480+36\\ =7396\\ \\ \left(\mathrm{iv}\right){93}^2={(90+3)}^2\\ =(90+3)(90+3)\\ =90(90+3)+3(90+3)\\ =90\times 90+3\times 90+3\times 90+3\times 3\\ =8100+270+270+9\\ =8649\\ \\ \left(\mathrm{v}\right){71}^2={(70+1)}^2\\ =(70+1)(70+1)\\ =70(70+1)+1(70+1)\\ =70\times 70+1\times 70+1\times 70+1\times 1\\ =4900+70+70+1\\ =5041\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right){46}^2={(40+6)}^2\\ =(40+6)(40+6)\\ =40(40+6)+6(40+6)\\ =40\times 40+6\times 40+6\times 40+6\times 6\\ =1600+240+240+36\\ =2116\end{array}$

Q.11 Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14 (iii) 16 (iv) 18

Ans

$\begin{array}{l}\text{For any natural number}\mathrm{m}>\text{1},\\ \text{2}\mathrm{m},{\mathrm{m}}^{\text{2}}-\text{1},{\mathrm{m}}^{\text{2}}+\text{1 forms a Pythagorean triplet}.\\ \left(\text{i}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{6},\text{then}{\mathrm{m}}^{\text{2}}=\text{5}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\end{array}$

$\begin{array}{l}\text{If we take}{m}^{\text{2}}-\text{1}=\text{6},\text{then}{m}^{\text{2}}=\text{7}\\ \text{Again the value of}m\text{is not an integer}.\\ {\text{So, we try to take m}}^{\text{2}}\text{+ 1 = 6}\text{.}\\ {\text{Again m}}^{\text{2}}\text{= 5 will not give an integer value for m}\text{.}\\ \text{Let 2}m=\text{6}\Rightarrow m=\text{3}\\ \text{Therefore},\text{the Pythagorean triplets are}\\ \text{2}\times \text{3},{\text{3}}^{\text{2}}-\text{1},{\text{3}}^{\text{2}}+\text{1 or 6},\text{8},\text{and 1}0.\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{13}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{14},\text{then}{\mathrm{m}}^{\text{2}}=\text{15.}\\ \text{Again the value of}\mathrm{m}\text{is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{14}\Rightarrow \mathrm{m}=\text{7}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{49}-\text{1}=\text{48 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{49}+\text{1}=\text{5}0\\ \text{Therefore},\text{the required triplet is 14},\text{48},\text{and 5}0.\end{array}$

$\begin{array}{l}\left(\text{iii}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{15}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{16},\text{then}{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{16}\Rightarrow \mathrm{m}=\text{8}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{64}-\text{1}=\text{63 and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{64}+\text{1}=\text{65}\\ \text{Therefore},\text{the Pythagorean triplet is 16},\text{63},\text{and 65}.\end{array}$

$\begin{array}{l}\left(\text{iv}\right)\text{If we take}{\mathrm{m}}^{\text{2}}+\text{1}=\text{18},{\mathrm{m}}^{\text{2}}=\text{17}\\ \text{The value of}\mathrm{m}\text{will not be an integer}.\\ \text{If we take}{\mathrm{m}}^{\text{2}}-\text{1}=\text{18},\text{then}{\mathrm{m}}^{\text{2}}=\text{19}\\ \text{Again the value of}\mathrm{m}\text{\hspace{0.17em}\hspace{0.17em}is not an integer}.\\ \text{Let 2}\mathrm{m}=\text{18}\Rightarrow \mathrm{m}=\text{9}\\ \text{Thus},{\mathrm{m}}^{\text{2}}-\text{1}=\text{81}-\text{1}=\text{8}0\text{and}{\mathrm{m}}^{\text{2}}+\text{1}=\text{81}+\text{1}=\text{82}\\ \text{Therefore},\text{the Pythagorean triplet is 18},\text{8}0,\text{and 82}.\end{array}$

Q.12 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

Ans

(i) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) The one’s digit of the given number is 6. If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) The one’s digit of the given number is 1. If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) The one’s digit of the given number is 5. If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

Q.13 Without doing any calculation, find the numbers which are surely not perfect squares.

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

Ans

A perfect square of any number ends with any of the digits 0, 1,4,5,6 or 9 at its unit’s place. Also, a perfect square will end with even number of zeroes.

(i) Since the number 153 has 3 in its unit’s place, so it is not a perfect square.

(ii) Since the number 257 has 7 in its unit’s place, so it is not a perfect square.

(iii) Since the number 408 has 8 in its unit’s place, so it is not a perfect square.

(iv) Since the number 441 has 1 in its unit’s place and a perfect square number may end with 1, so it is a perfect square.

Q.14 Find the square roots of 100 and 169 by the method of repeated subtraction.

Ans

\begin{array}{l}{\text{We know that the sum of the first n odd natural numbers is n}}^{\text{2}}\text{.}\\ \text{The square root of 169 can be obtained by the method of}\\ \text{repeated subtraction as follows}.\end{array}

\begin{array}{l}\text{Consider:}\\ \\ \left(\text{i}\right)\text{100}-\text{1}\text{= 99}\\ \left(\text{ii}\right)\text{99}-\text{3}\text{= 96}\\ \left(\text{iii}\right)\text{96}-\text{5}\text{= 91}\\ \left(\text{iv}\right)\text{91}-\text{7}\text{= 84}\\ \left(\text{v}\right)\text{84}-\text{9}\text{= 75}\\ \left(\text{vi}\right)\text{75}-\text{11}\text{= 64}\\ \left(\text{vii}\right)\text{64}-\text{13 = 51}\\ \left(\text{viii}\right)\text{51}-\text{15}\text{= 36}\\ \left(\text{ix}\right)\text{36}-\text{17}\text{= 19}\\ \left(\text{x}\right)\text{19}-\text{19}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting from}\\ {\text{1 to 100, and obtained 0 at 10}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{100}}\text{=10}\end{array}

$\begin{array}{l}\text{The square root of 169 can be obtained by the method}\\ \text{of repeated subtraction as follows:}\\ \\ \text{Consider:}\\ \\ \text{(i) 169}-\text{1}\text{= 168}\\ \left(\text{ii}\right)\text{168}-\text{3}\text{= 165}\\ \left(\text{iii}\right)\text{165}-\text{5}\text{= 160}\\ \left(\text{iv}\right)\text{160}-\text{7}\text{= 153}\\ \left(\text{v}\right)\text{153}-19\text{= 144}\\ \left(\text{vi}\right)\text{144}-\text{11}\text{= 133}\\ \left(\text{vii}\right)\text{133}-\text{13}\text{= 120}\\ \left(\text{viii}\right)\text{120}-\text{15 = 105}\\ \left(\text{ix}\right)\text{105}-\text{17}\text{= 88}\\ \left(\text{x}\right)\text{88}-\text{19}\text{= 69}\\ \left(\text{xi}\right)\text{69}-\text{21}\text{= 48}\\ \left(\text{xii}\right)\text{48}-\text{23}\text{= 25}\\ \left(\text{xiii}\right)\text{25}-\text{25}\text{= 0}\\ \\ \text{We have subtracted successive odd numbers starting}\\ {\text{from 1 to 169, and obtained 0 at 13}}^{\text{th}}\text{step}\text{.}\\ \text{Therefore,}\sqrt{\text{169}}\text{=13}\end{array}$

Q.15 Find the square roots of the following numbers by the Prime Factorization Method.

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Ans

\begin{array}{l}\text{(i)}729\text{can be factorised as follows:}\\ \\ \begin{array}{cc}3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ \text{729}=\underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{3}\times \text{3}}\\ \therefore \sqrt{729}=3\times 3\times 3=\text{27}\end{array}

\begin{array}{l}\text{(ii)400}\\ \\ \begin{array}{cc}2& 400\\ 2& 200\\ 2& 100\\ 5& 50\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \text{400}=\underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{5}\times \text{5}}\\ \therefore \sqrt{400}=2\times 2\times 5=\text{20}\end{array}

\begin{array}{l}\text{(iii)1764}\\ \begin{array}{cc}2& 1764\\ 2& 882\\ 3& 441\\ 3& 147\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \text{1764}=\underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{3}\times \text{3}}\times \underset{¯}{\text{7}\times \text{7}}\\ \therefore \sqrt{1764}=2\times 3\times 7=\text{42}\end{array}

\begin{array}{l}\text{(iv)}4096\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 4096\\ 2& 2048\\ 2& 1024\\ 2& 512\\ 2& 256\\ 2& 128\\ 2& 64\\ 2& 32\\ 2& 16\\ 2& 8\\ 2& 4\\ 2& 2\\ & 1\end{array}\\ \\ \text{4096}=\underset{¯}{2\times 2}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{\text{2}\times \text{2}}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\\ \therefore \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2=\text{64}\\ \end{array}

\begin{array}{l}\text{(vi)}9604\text{can be factorised as follows:}\\ \\ \begin{array}{cc}2& 9604\\ 2& 4802\\ 7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ \text{9604}=\underset{¯}{2\times 2}\times \underset{¯}{\text{7}\times \text{7}}\times \underset{¯}{\text{7}\times \text{7}}\\ \therefore \sqrt{9604}=2\times 7\times 7=\text{98}\\ \\ \text{(vii)5929}\\ \\ \begin{array}{cc}7& 5929\\ 7& 847\\ 11& 121\\ 11& 11\\ & 1\end{array}\\ \\ \text{5929}=\underset{¯}{7\times 7}\times \underset{¯}{\text{11}\times \text{11}}\\ \therefore \sqrt{5929}=7\times 11=\text{77}\end{array}

\begin{array}{l}\text{(viii) 9216}\\ \\ \begin{array}{cc}2& 9216\\ 2& 4608\\ 2& 2304\\ 2& 1152\\ 2& 576\\ 2& 288\\ 2& 144\\ 2& 72\\ 2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \text{9216}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{\text{3}\times 3}\\ \therefore \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3=\text{96}\\ \\ \text{(ix) 529}\\ \\ \begin{array}{cc}23& 529\\ 23& 23\\ & 1\end{array}\\ \\ \text{529}=\underset{¯}{23\times 23}\\ \therefore \sqrt{529}=\text{23}\end{array}

\begin{array}{l}\text{(x)}8100\\ \\ \begin{array}{cc}2& 8100\\ 2& 4050\\ 3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 8100=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{8100}=2\times 3\times 3\times 5=90\\ \end{array}

Q.16 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

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 252 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ 252\times 7=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}\\ \\ \therefore \text{252×7 = 1764 is a perfect square}\text{.}\\ \\ \therefore \sqrt{1764}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}=2\times 3\times 7=42\\ \end{array}$

$\begin{array}{l}\text{(ii)180 can be factorised as:}\\ \\ \begin{array}{cc}2& 180\\ 2& 90\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ 180=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we multiply the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 180 has to be multiplied with 5}\\ \text{to obtain a perfect square}\text{.}\\ \text{180}\times 5=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \text{180×5 = 900 is a perfect square}\text{.}\\ \therefore \sqrt{900}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}=2\times 3\times 5=30\\ \\ \text{(iii)1008 can be factorised as:}\\ \\ \begin{array}{cc}2& 1008\\ 2& 504\\ 2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\end{array}$

\begin{array}{l}1008=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we multiply the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 1008 has to be multiplied with 7}\\ \text{to obtain a perfect square}\text{.}\\ \text{1008}\times 7=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}\\ \therefore \text{1008×7 = 7056 is a perfect square}\text{.}\\ \therefore \sqrt{7056}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{7\times 7}=2\times 2\times 3\times 7=84\\ \\ \text{(iv) 2028 can be factorised as follows:}\\ \begin{array}{cc}2& 2028\\ 2& 1014\\ 3& 507\\ 13& 169\\ 13& 13\\ & 1\end{array}\\ 2028=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2028 has to be multiplied with 3}\\ \text{to obtain a perfect square}\text{.}\\ \text{2028}\times 3=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times \underset{¯}{3\times 3}\\ \therefore \text{2028}\times 3\text{= 6084 is a perfect square}\text{.}\\ \therefore \sqrt{6084}=\underset{¯}{2\times 2}\times \underset{¯}{13\times 13}\times \underset{¯}{3\times 3}=2\times 3\times 13=78\end{array}

\begin{array}{l}\text{(v) 1458 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1458\\ 3& 729\\ 3& 243\\ 3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\\ \\ 1458=2\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \text{Here, the prime factor 2 does not have its pair}\text{.}\\ \text{If we multiply the given number by 2, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 1458 has to be multiplied with 2 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1458}\times 2=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \\ \therefore \text{1458}\times 2\text{= 2916 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2916}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}=2\times 3\times 3\times 3=54\\ \\ \end{array}

\begin{array}{l}\text{(vi) 768 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 768\\ 2& 384\\ 2& 192\\ 2& 96\\ 2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}\\ \\ 768=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times 3\\ \text{Here, the prime factor 3 does not have its pair}\text{.}\\ \text{If we multiply the given number by 3, then the number}\\ \text{will become a perfect square}\text{.}\\ \\ \text{Therefore, 768 has to be multiplied with 3 to obtain}\\ \text{a perfect square}\text{.}\\ \text{768}\times 3=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \\ \therefore \text{768}\times 3\text{= 2304 is a perfect square}\text{.}\\ \\ \therefore \sqrt{2304}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 2\times 2\times 2\times 3=48\end{array}

Q.17 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

Ans

$\begin{array}{l}\text{(i) 252 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 252\\ 2& 126\\ 3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \\ 252=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 252 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{252}÷7=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{36}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 3=6\end{array}$

$\begin{array}{l}\text{(ii) 2925 can be factorised as follows:}\\ \\ \begin{array}{cc}3& 2925\\ 3& 975\\ 5& 325\\ 5& 65\\ 13& 13\\ & 1\end{array}\\ \\ 2925=\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\times 13\\ \text{Here, the prime factor 13 does not have its pair}\text{.}\\ \text{If we divide the given number by 13, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2925 has to be divided with 13 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2925}÷13=225\text{is a perfect square}\text{.}\\ \text{225 =}\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{225}=\underset{¯}{3\times 3}\times \underset{¯}{5\times 5}=3\times 5=15\\ \\ \text{(iii) 396 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 396\\ 2& 198\\ 3& 99\\ 3& 33\\ 11& 11\\ & 1\end{array}\\ \end{array}$

$\begin{array}{l}396=\underset{¯}{3\times 3}\times \underset{¯}{2\times 2}\times 11\\ \text{Here, the prime factor 11 does not have its pair}\text{.}\\ \text{If we divide the given number by 11, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,396 has to be divided with 11 to obtain}\\ \text{a perfect square}\text{.}\\ \text{396}÷11=36\text{is a perfect square}\text{.}\\ \text{36 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{36}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}=2\times 3=6\\ \\ \text{(iv) 2645 can be factorised as follows:}\\ \\ \begin{array}{cc}5& 2645\\ 23& 529\\ 23& 23\\ & 1\end{array}\\ \\ 2645=\underset{¯}{23\times 23}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,2645 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2645}÷5=529\text{is a perfect square}\text{.}\\ \text{529 =}\underset{¯}{23\times 23}\\ \therefore \sqrt{529}=23\end{array}$

$\begin{array}{l}\text{(v) 2800 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 2800\\ 2& 1400\\ 2& 700\\ 2& 350\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}\\ \\ 2800=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}\times 7\\ \text{Here, the prime factor 7 does not have its pair}\text{.}\\ \text{If we divide the given number by 7, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore, 2800 has to be divided with 7 to obtain}\\ \text{a perfect square}\text{.}\\ \text{2800}÷7=400\text{is a perfect square}\text{.}\\ \text{400 =}\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{400}=\underset{¯}{2\times 2}\times \underset{¯}{2\times 2}\times \underset{¯}{5\times 5}=2\times 2\times 5=20\end{array}$

$\begin{array}{l}\text{(vi) 1620 can be factorised as follows:}\\ \\ \begin{array}{cc}2& 1620\\ 2& 810\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ \\ 1620=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times 5\\ \text{Here, the prime factor 5 does not have its pair}\text{.}\\ \text{If we divide the given number by 5, then the number}\\ \text{will become a perfect square}\text{.}\\ \text{Therefore,1620 has to be divided with 5 to obtain}\\ \text{a perfect square}\text{.}\\ \text{1620}÷5=324\text{is a perfect square}\text{.}\\ \text{324 =}\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\\ \therefore \sqrt{324}=\underset{¯}{2\times 2}\times \underset{¯}{3\times 3}\times \underset{¯}{3\times 3}=2\times 3\times 3=18\end{array}$

Q.18 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.

Ans

According to the question, each student donated as many rupees as the number of students of the class.
Therefore, the number of students in the class will be the square root of the amount donated by the students of the class.
The total amount of donation is Rs 2401.

\begin{array}{l}\text{Number of students in the class}=\sqrt{2400}\\ \\ \begin{array}{cc}7& 2401\\ 7& 343\\ 7& 49\\ 7& 7\\ & 1\end{array}\\ \\ 2401=\underset{¯}{7\times 7}\times \underset{¯}{7\times 7}\\ \therefore \sqrt{2401}=7\times 7=49\end{array}

Hence, the number of students in the class is 49.

Q.19 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.

Ans

According to the question, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Also, total number of plants = 2025

Since, the number of rows and number of plants in each row are equal, therefore (Number of rows)² = 2025

$\begin{array}{l}\therefore \text{Number of rows =}\sqrt{2025}\\ \\ \begin{array}{cc}3& 2025\\ 3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \\ 2025=\underset{¯}{3\times 3}\times \underset{¯}{3\times 3}\times \underset{¯}{5\times 5}\\ \therefore \sqrt{2025}=3\times 3\times 5=45\end{array}$

Hence, the number of rows and the number of plants in each row is 45.

Q.20 Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Ans

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{4},\text{9},\text{1}0\\ 2& \text{2},\text{9},\text{5}\\ 3& \text{1},\text{9},\text{5}\\ 3& \text{1},\text{3},\text{5}\\ 5& \text{1},\text{1},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900.

Q.21 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Ans

The number that will be perfectly divisible by each one of 8, 15, and 20 is their LCM.

\begin{array}{l}\text{The LCM of these numbers is as follows:}\\ \\ \begin{array}{cc}2& \text{8},\text{15},\text{2}0\\ 2& \text{4},\text{15},\text{1}0\\ 2& \text{2},\text{15},\text{5}\\ 3& \text{1},\text{15},\text{5}\\ 5& \text{1},\text{5},\text{5}\\ & \text{1},\text{1},\text{1}\end{array}\\ \end{array}

Therefore, LCM of 8, 15, and 20 = 2 × 2 ×2 × 3 × 5 =120

Here, prime factors 2, 3 and 5 do not have their pairs. Therefore, 120 is not a perfect square. If we multiply 120 with 2×3×5, then the number will become a perfect square. Therefore, 120 should be multiplied with 2×3×5 to obtain a perfect square.

Hence, the required square number is 120×2×3×5 = 3600.

Q.22 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

Ans

\begin{array}{l}\text{(i) Consider}\sqrt{2304}\\ \\ \begin{array}{cc}& 48\\ 4& \begin{array}{l}\overline{23}\overline{04}\\ -16\end{array}\\ 88& \begin{array}{l}704\\ -704\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{2304}=48\\ \\ \text{(ii) Consider}\sqrt{4489}\\ \\ \begin{array}{cc}& 67\\ 6& \begin{array}{l}\overline{44}\overline{89}\\ -36\end{array}\\ 127& \begin{array}{l}889\\ -889\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{4489}=67\end{array}

\begin{array}{l}\text{(iii) Consider}\sqrt{3481}\\ \\ \begin{array}{cc}& 59\\ 5& \begin{array}{l}\overline{34}\overline{81}\\ -25\end{array}\\ 109& \begin{array}{l}981\\ -981\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3481}=59\\ \\ \text{(iv) Consider}\sqrt{529}\\ \\ \begin{array}{cc}& 23\\ 2& \begin{array}{l}\overline{5}\overline{29}\\ -4\end{array}\\ 43& \begin{array}{l}129\\ -129\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{529}=23\end{array}

\begin{array}{l}\text{(v) Consider}\sqrt{3249}\\ \\ \begin{array}{cc}& 57\\ 5& \begin{array}{l}\overline{32}\overline{49}\\ -25\end{array}\\ 107& \begin{array}{l}749\\ -749\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3249}=57\\ \\ \text{(vi) Consider}\sqrt{1369}\\ \\ \begin{array}{cc}& 37\\ 3& \begin{array}{l}\overline{13}\overline{69}\\ -9\end{array}\\ 67& \begin{array}{l}469\\ -469\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{1369}=37\end{array}

\begin{array}{l}\text{(vii) Consider}\sqrt{5776}\\ \\ \begin{array}{cc}& 76\\ 7& \begin{array}{l}\overline{57}\overline{76}\\ -49\end{array}\\ 146& \begin{array}{l}\text{876}\\ -876\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{5776}=76\\ \\ \text{(viii) Consider}\sqrt{7921}\\ \\ \begin{array}{cc}& 89\\ 8& \begin{array}{l}\overline{79}\overline{21}\\ -64\end{array}\\ 169& \begin{array}{l}\text{1521}\\ -1521\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{7921}=89\end{array}

\begin{array}{l}\text{(ix) Consider}\sqrt{576}\\ \\ \begin{array}{cc}& 24\\ 2& \begin{array}{l}\overline{5}\overline{76}\\ -4\end{array}\\ 44& \begin{array}{l}\text{176}\\ -176\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{576}=24\\ \\ \text{(x) Consider}\sqrt{1024}\\ \\ \begin{array}{cc}& 32\\ 3& \begin{array}{l}\overline{10}\overline{24}\\ -9\end{array}\\ 62& \begin{array}{l}\text{124}\\ -124\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{1024}=32\end{array}

$\begin{array}{l}\text{(xi) Consider}\sqrt{3136}\\ \\ \begin{array}{cc}& 56\\ 5& \begin{array}{l}\overline{31}\overline{36}\\ -25\end{array}\\ 106& \begin{array}{l}\text{636}\\ -636\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{3136}=56\\ \\ \text{(xii) Consider}\sqrt{900}\\ \\ \begin{array}{cc}& 30\\ 3& \begin{array}{l}\overline{9}\overline{00}\\ -9\end{array}\\ 60& \begin{array}{l}\text{00}\\ 00\end{array}\\ & 0\end{array}\\ \\ \therefore \sqrt{900}=30\end{array}$ \begin{array}{l}\end{array}