$$\frac{4}{3}, - \frac{2}{9}, - \frac{7}{8},\frac{5}{{12}}$$
$$\eqalign{ & \frac{4}{3} = 1.33, \cr & - \frac{2}{9} = - 0.22, \cr & - \frac{7}{8} = - 0.875, \cr & \frac{5}{{12}} = 0.416 \cr} $$
So, $$ - \frac{7}{8}, - \frac{2}{9},\frac{5}{{12}},\frac{4}{3}$$

Let us assume any such number which when divided by 5 leaves remainder as 3.
Let it be 8
So, now
$$\eqalign{ & = \frac{{{{\left( 8 \right)}^2} + {{\left( 8 \right)}^3}}}{5} \cr & = \frac{{64 + 512}}{5} \cr & = \frac{{576}}{5} = 1\,\,\left[ {{\text{Remainder}}} \right] \cr} $$

$$\eqalign{ & {\text{Let the two digit number be }} \cr & 10y + x\,\,\,\,\,{\text{where}}\,\,\,y > x \cr & 10x + y - 10y - x = 63 \cr & 9x - 9y = 63 \cr & x - y = 7 \cr & {\bf{x = 7,8,9}}{\text{ and}} \cr & y = 0,1,2 \cr} $$

Let the numbers are a, b
⇒ a + b = 75.....(i)
⇒ a - b = 25.....(ii)
⇒ After solving (i) and (ii)
⇒ a = 50,     b = 25
⇒ ab = 50 × 25 = 1250
So, their product is 1250

According to the question,
⇒ The least number = 3005
⇒ The greatest number = 3995
⇒ Difference will be = 3995 - 3005 = 990

$$2.3\overline {49} = \frac{{2349 - 23}}{{990}} = \frac{{2326}}{{990}}$$

A + B = 3(B + C)
A + B = 3B + 3C
A = 2B + 3C.....(i)
A + B + C = A + 30
B + C = 30.....(ii)
B = 5C (given)
5C + C = 30
C = Rs. 5
B = 30 - 5 = Rs. 25
A + 25 = 3(25 + 5)
A = 90 - 25
A = Rs. 65

Marbles in the 50th box will be kept by 1st, 2nd, 5th, 10th, 25th and 50th person is a factor of 50.
Number of marbles
=1 + 2 + 5 + 10 + 25 + 50
= 93

$$\eqalign{ & = \frac{{\left( {0.0539 - 0.002} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{\left( {0.0519} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{0.02076 + 0.0392}}{{0.04 \times 0.25}} \cr & = \frac{{0.05996}}{{0.01}} \cr & = 5.996 \cr} $$

∵ 13 + 23 + .....+ 103 = 3025
To find = 23 + 43 + 63 +.....+ 203 = ?
⇒ 23 (13 + 23 + .....+ 103)
⇒ 8 × 3025
⇒ 24200

$$1\frac{1}{2}$$ = $$\frac{3}{2}$$
$$1\frac{1}{3}$$ = $$\frac{4}{3}$$
Square = $$\frac{9}{4}$$   $$\frac{16}{9}$$
Reciprocal ratio
$$\frac{4}{9}$$ : $$\frac{9}{16}$$
64 : 81

Let the natural number (n) be 1
∴ n (n + 1) (n + 2)
= 1 (1 + 1) (1 + 2)
= 6

Let the number = 10x + y
Interchange number = 10y + x
Difference
= 10x + y - 10y - x
= 9 (x - y)
The difference is always divisible by 9

By checking last 3 digits
Then, last '3' digits of 1046529
529 is a square of 23
Hence, this is correct option

$$\eqalign{ & \frac{{{\text{Remainder of number}}}}{{17}} = \frac{{18}}{{17}} \cr & \Rightarrow {\text{Remainder = 1}} \cr} $$

$$\eqalign{ & {\text{Let the 3 numbers be }}x,y,\& \,\,z \cr & {\text{According to question}} \cr & x + y = 55 \to y = 55 - x \cr & y + z = 65 \to 55 - x + z = 65 \cr & z - x = 10.....(i) \cr & z + 3x = 110.....(ii) \cr & {\text{Solve (i)}}\,{\text{and (ii)}} \cr & {\bf{z = 35,}} \cr & x = 25, \cr & y = 30 \cr} $$

Let the given number be x
According to the question,
x2 - 25 = (x - 25)2
x2 - 25 = x2 + (25)2 - 50x
x = 13

Three digit number = 100x + 10y + z
To make number after changing last two digit = 100x + 10z + y
Now,
100x + 10y + z = 100x + 10z + y - 45
9z - 9y = 45
z - y = 5

$$\root 3 \of 7 ,\root 4 \of {13} ,\sqrt 5 $$
Taking L.C.M. of 3, 4 and 2.
$${7^{\frac{1}{3} \times 12}}$$ $${13^{\frac{1}{4} \times 12}}$$ $${5^{\frac{1}{2} \times 12}}$$ 74 133 56 ↓ ↓ ↓     2401         2197         15625    

We know that after 5! we get one zero at the end of the number
And after 10! → two zeros
And after 15! → three zeros
We can say that in 17! we get minimum three zeros
The Hundred's place value of 17! = 0