If difference between numerator and denominator is same in all fractions and numerator is smaller than denominator in all fractions, then smaller will be smaller and larger will larger.
So, $$\frac{{15}}{{16}}$$ is the smallest value among all.
So this is the answer

$$ = \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}}$$
$$ = \frac{1}{{3 \times 5}} + \frac{1}{{5 \times 7}} + \frac{1}{{7 \times 9}} + \frac{1}{{9 \times 11}} + \frac{1}{{11 \times 13}}$$
$$ = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{13}} \right]$$
$$ = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{{13}}} \right]$$
$$ = \frac{1}{2} \times \frac{{10}}{{39}}$$
$$ = \frac{5}{{39}}$$

On dividing 534677 by 777 we get 101 as remainder.
∴ (Divisor) - (Remainder)
= (777 - 101)
= 676

$$\eqalign{ & \frac{{{\text{Remainder of number }}}}{{17}} = \frac{{36}}{{17}} \cr & \Rightarrow {\text{Remainder = 2}} \cr} $$

A + B + C + D = 60 -----------> (1)

Now According to Question,
A = $$\frac{1}{2}$$ (B + C + D)
2A = B + C + D
Put the value of B + C + D in Equation 1
A + 2A = 60
3A = 60
A = 20

Now again according to Question
B = $$\frac{1}{3}$$ (A + C + D)
A + C + D = 3B
By Putting the Value of A + C + D in Equation 1
we get,
B + 3B = 60
4B = 60
B = 15

Now again according to Question
C = $$\frac{1}{4}$$ (A + B + D)
A + B + D = 4C
By Putting the Value of A + B + D in equation 1
We get C + 4C = 60
5C = 60
C = 12

Since A + B + C + D = 60
20 + 15 + 12 + D = 60
D = 60 - 47 = 13
Amount of D = Rs. 13

Let the number be x
⇒ According to the question,
$$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{x}{6} = 7 \cr & \Rightarrow \frac{{9x - 2x}}{{12}} = 7 \cr & \Rightarrow 7x = 7 \times 12 \cr & \Rightarrow x = 12 \cr} $$
⇒ Then $$\frac{5}{3}$$ of the number will be
$$\eqalign{ & \Rightarrow \frac{{x \times 5}}{3} \cr & = \frac{{12 \times 5}}{3} \cr & = 20 \cr} $$

Total students = z
Total boys = x
Total girls = z - x
⇒ $$\frac{{{\text{Number of girls}}}}{{{\text{Total students}}}}$$
= $$\frac{z - x}{{z}}$$
= 1 - $$\frac{x}{{z}}$$

The greatest 4 digit number = 9999
9999 is divided by 307
then, remainder = 175
∴ Number to be added = 307 - 175 = 132

$$\eqalign{ & \Leftrightarrow \frac{1}{5} + 999\frac{{494}}{{495}} \times 99\, \cr & \Rightarrow \frac{1}{5} + \left[ {999 + \frac{{494}}{{495}}} \right] \times 99\, \cr & \Rightarrow \frac{1}{5} + \left[ {999 + 1 - \frac{1}{{495}}} \right] \times 99 \cr & \Rightarrow \frac{1}{5} + \left[ {1000 - \frac{1}{{495}}} \right] \times 99 \cr & \Rightarrow \frac{1}{5} + 99000 - \frac{1}{5} \cr & \Rightarrow 99000\, \cr} $$

Note : In such type of questions assume any values which satisfy the question condition
(i) Let m = 3,   n = 1
m - n = 3 - 1 = 2 (Even)
m2 - n2 = (3)2 - (1)2 = 8
(ii) Let m = 4,   n = 2
m - n = 4 - 2 = 2 (Even)
m2 - n2 = 16 - 4 = 12
(iii) Let m = 5,   n = 3
m - n = 5 - 3 = 2 (Even)
m2 - n2 = 25 - 9 = 16
H.C.F. of 8, 12, 16 = 4
So, the number is divisible by 4

a = (0.4)2 = 0.16
b = 0.04
c = 0.4
c > a > b

On dividing 5000 by 73, we get 36 remainder
Required number = 5000 + (73 - 36) = 5037

$$\eqalign{ & \Rightarrow \frac{{13}}{1} = \frac{{13w}}{{\left( {1 - w} \right)}} \cr & \Rightarrow \frac{w}{{\left( {1 - w} \right)}} = 1 \cr & \Rightarrow w = 1 - w \cr & \Rightarrow 2w = 1 \cr & \Rightarrow w = \frac{1}{2} \cr & \therefore {\left( {2w} \right)^2} \cr & = 4{w^2} \cr & = 4 \times \frac{1}{4} \cr & = 1 \cr} $$

12x - 66 $$ \leqslant $$ 6
12x $$ \leqslant $$ 72
x $$ \leqslant $$ 6

38950 is completely divisible by 410 only.
Hence, Rs. 410 is the correct answer.

$$\eqalign{ & \Rightarrow 348 \times 265 \cr & = \left( {350 - 2} \right) \times 265 \cr & = \left( {350 \times 265} \right) - \left( {2 \times 265} \right) \cr & = \left\{ {\left( {300 + 50} \right) \times 265} \right\} - 530 \cr & = \left( {300 \times 265} \right) + \left( {50 \times 265} \right) - 530 \cr & = 79500 + 13250 - 530 \cr & = 92750 - 530 \cr & = 92220 \cr} $$

Let the property is 'P' units
According to the question,
Given part,
$$\eqalign{ & = \frac{1}{4}P + \frac{1}{2}P + \frac{1}{5}P \cr & = \frac{{5P + 10P + 4P}}{{20}} \cr & = \frac{{19}}{{20}}P \cr} $$
Part of property given away $$\frac{19}{{20}}$$

In such type of question always go through option to save your valuable time.
By option (D) 4, 2
4 + 2 × 2 = 8
4 - 2 = 2 (Satisfied)

$$\frac{1}{{24 \times 7}} = 0.0059$$

I. We have (195 - 195) = 0
∴ 195195 is divisible by 7

II. We have (181 - 181) = 0
∴ 181181 is divisible by 7

III. We have (120 - 120) = 0
∴ 120120 is divisible by 7

IV. We have (891 - 891) = 0
∴ 891891 is divisible by 7

Hence, all are divisible by 7