We have the sum of the digits of 51x3
= 5+1+x+3=9+x
Since, 51x3 is divisible by 9.
∴ (9 + x) must be divisible by 9.
∴ (9 + x) must be equal to 0 or 9 or 18 or 27
or ... But x is a digit, then
9+x=9 ⇒ x=0
9 + x = 18 ⇒ x = 9
x = 27 ⇒ x = 18, which is not possible.
∴ The required value of x = 0 or 9.

\(p={-3\over 4}\)

13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 23 x 23 x 23 x 33.
Therefore,  \(\sqrt[3]{13824}=\) 2 x 2 x 2 x 3 = 24

On plotting the given points A(1, 2) and B(2, 1) on the same graph and joining them, we get a straight line AB. When we extend the line AB on both sides, we find that it intersects the x-axis at C(3, 0) and the y-axis at D(0, 3).

Since 125 = 5 \(\times\)5\(\times\)5=5³
6^-5=(2\(\times\)3)^-5 = 2^-5 \(\times\) 3^-5
\(\therefore \frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times 125 }{ { 5 }^{ -4 }\times { 6 }^{ -5 } } =\frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times { 5 }^{ 3 } }{ { 5 }^{ -4 }\times { 2 }^{ -5 }\times { 3 }^{ -5 } } \)
\(=\frac { { 2 }^{ -5 } }{ { 2 }^{ -5 } } \times \frac { { 3 }^{ -5 } }{ { 3 }^{ -5 } } \times { 5 }^{ 3+4 }=1\times { 5 }^{ 7 }\)
=5 \(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5 = 78125.

We have
A rational number between -3 and -4
= \(\frac { (-3)+(-4) }{ 2 } =\frac { -7 }{ 2 } \)
A rational number between (-3) and \(\frac { -7 }{ 2 } \)
\(\left[ \left( -3 \right) +\left( \frac { -7 }{ 2 } \right) \right] \div 2=\left[ \frac { -6+(-7) }{ 2 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -13 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -13 }{ 4 } \)
A rational number between \(\left( \frac { -7 }{ 2 } \right) \) and (-4)
= \(\left[ \frac { -7 }{ 2 } +(-4) \right] \div 2=\left[ \frac { -7+(-8) }{ 2 } \right] \div 2\)
= \(\frac { -15 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -15 }{ 4 } \)
Thus, the three rational numbers
\(\left( \frac { -7 }{ 2 } \right) ,\left( \frac { -13 }{ 4 } \right) \) and \(\left( \frac { -15 }{ 4 } \right) \) are between (-3) and (-4).

\((\frac{2}{3})^{-2}\times(\frac{2}{3})^{5}\)=\((\frac{2}{3})^{-2+5}\)=\((\frac{2}{3})^{3}=\frac{(2)^{3}}{(3)^{3}}\)
=\(\frac{2\times2\times2}{3\times3\times3}=\frac{8}{27}\)

The prime factorisation of 9216 is
9216 = 2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)3\(\times\)3
By pairing the prime factors, we get
9216 = \(\underline { 2\times 2 } \)\(\times\)\(\underline { 2\times 2 } \)\(\times\)\(\underline { 2\times 2 } \)\(\times\)\(\underline { 2\times 2 } \)\(\times\)\(\underline { 2\times 2 } \)\(\times\)\(\underline { 3\times 3 } \)
So, \(\sqrt { 9126 } \)= 2\(\times\)2\(\times\)2\(\times\)2\(\times\)2\(\times\)3 = 96

\(\begin{matrix} \quad \quad A\quad B \\ \quad \times \quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ B\quad B\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found, since one's digit of B X 6 is 8, so B must be 2 or 4 or 6 or 8.
Then, possible values of BBB are 222, 444, 666 or 888.
If we divide these numbers by 6, then quotient should be A2 or A4 or A6 or A8.
Now, 222 \(\div\) 6 = 37, remainder = 0
But the quotient is not of the form A2, so B = 2 is not possible.
444 \(\div\) 6 = 74, remainder = 0
Also, quotient is of the form A4, which clearly works well.
Then, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 7\quad 4 \\ \quad \times \quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 4\quad 4\quad 4 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 7 and B = 4

3.2F^°

We have
\({3\over7}+({-2\over21})\times({-5\over6})={3\over7}+[({-2\over21})\times({-5\over6})]\)
\( \\ ={3\over7}+({-5\over-63})={3\over7}+{5\over63}={27+5\over63}={32\over63}\)

Given number is 21².
Using the formula, the square of any number can be expressed as two integers as
\({ n }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 2 } +\frac { { n }^{ 2 }+1 }{ 2 } \)
\(\therefore \ { 21 }^{ 2 }=\frac { 2{ 1 }^{ 2 }-1 }{ 2 } +\frac { { 21 }^{ 2 }+1 }{ 2 } =\frac { 441-1 }{ 2 } +\frac { 441+1 }{ 2 } \)
\(=\frac { 440 }{ 2 } +\frac { 442 }{ 2 } =220+221\)
which is expressed as the sum of two consecutive numbers.

We have \({-3\over13}+{15\over17}\times{-3\over13}+{17\over15}\)
\(={-3\over13}[1+{15\over17}]+{17\over15}={-3\over13}[{17+15\over17}]+{17\over15}\\ ={-3\over13}\times[{32\over 17}]+{17\over15}={-96\over221}+{17\over15} \\ ={-96\times15+221\times17\over221\times15}\\ {-1440+3757\over3315} ={2317\over3315}\)

All properties of operations on rational numbers also hold in case of integers except the following property:
a \(\div\) b is a rational number if b \(\neq\) 0 but a \(\div\)b is not necessarily an integer in case ab \(\in\) J,
All properties of operations on rational numbers also hold in case of whole numbers except the following properties:
(i) If a and b are rational numbers, then (a - b) may or may not be a whole number,
(ii) If a and b are rational numbers, then a\(\div\)b (where b \(\neq\) 0) is not necessarily a whole number,

Here, lowest observation =31
and highest observation = 65
So, we have to take class intervals 30-35, 35-40, 40-45, ... in first column and corresponding frequency i.e. number of students whose weight lie in this class interval, in the second column. Then, frequency distribution table for the above data is given below:

Class Interval   [Weights (in kg)]	Tally marks	Frequency  (Number of students)
30-35	III	3
35-40	IIII	4
40-45	IIII	4
45-50		5
50-55	I	1
55-60	I	1
60-65	I	1
65-70	I	1
Total		20