The sum of five consecutive odd numbers is equal to 175. What is the sum of the second largest number and the square of the smallest number among them together?

Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
? 2x + 4 = y+1
? 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
? 5x +5 = 3y - 6
? 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
? 2 x 2 -y = -3
? -y = -3 - 4
? y = 7
? Original fraction = x/y =2/7

Let the two digit number be 10x + y.
Now, according to the question,
x + y = 10............(i)
if digits reversed , number get decreased by 36,
and ( x +10y) = ( 10x + y ) - 36
? - 9y + 9x = 36
? x - y = 4 ..................... (ii)
On adding Eqs. (i) and (ii), we get
2x = 14
? x = 7
? x = 7 and y = 3
? Required number is 73.
So,the number is a multiple of a prime number.

The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

let the two-digit number be 10p + q
Now according to the question
p + q = 10
and (p +10q) + 36 = (q + 10p)
? -9q + 9p = 36
? -q + p = 4 ....(ii)
on adding eq (i) and (ii) we get
2p = 14 ? p =7
? p = 7 and q = 4

? Required number is the 73
So, the number is a multiple of prime number.

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
? Req. smallest number =42735.

Let the third numbers = p
Then, the first number = 3p
and second number = 3p/2
According to question,
( p + 3p + 3p/2 )/3 = 154
? ( 2p + 6p + 3p )/6 = 154
? p = 154 x 6/11 = 84
? Required difference = 3p - p = 2p
= 2 x 84 = 168

Let first number = p
and second number = q
According to the question,
p - q = 18 ................................(i)
and 3p - 4q = 18.....................(ii)
On multiplying Eq .(i) by 3 and subtracting Eq. (ii) from it,we get
3p - 3q = 54
3p - 4q = 18
q = 36
On putting the value of q in Eq, (i) we get
p = 18 + q
? p = 18 + 36 = 54
? Required sum = p + q = 54 + 36 = 90

Let total number of sweets distributed on children's day = x
According to the question,
x/250 - x/300 = 1
? ( 6x - 5x )/1500 = 1
? x = 1500
? Required number of sweets distributed on children's day = 1500

Let number of students in examination halls P and Q is x and y, respectively

Then as per the first condition,
x - 10 = y + 10
? x - y = 20 ..............(i)
As per the second condition,
? x + 20 = 2( y - 20)
? x - 2y = - 60 ...............(ii)
On subtraction Eq. (ii) from Eq. (i), we get
- y + 2y = 20 + 60
? y = 80
Putting the value of y Eq. (i) we get,
x - 80= 20 ? x = 100
Hence, number of students in examination halls P and Q is 100 and 80, respectively.

Let the two number are a and b
According to the question
sum of squares of two numbers = 97
i.e.., a ² + b ² = 97 .....................(i)
and square of their difference = 25
i.e., (a - b) ² = 25 .................(ii)
? a - b = 5...............(iii)
from Eq. (ii)
a² + b² - 2ab = 25
(a² + b²) - 2ab = 25
Now put the value of a ² + b ² from Eq. (i)
? 97 - 2ab = 25
? 2ab = 72
? ab = 36.................(iv)
Now, we have
(a + b)² = ( a² + b²) + 2ab
Now put the value of a ² + b ² from Eq. (i) and ab from Eq. (iv)
(a + b)² = 97 + 72 = 169
a + b = 13.............(v)
Now add the Eqs. (iii) and (v), we get
a - b + a + b = 5 +13
2a = 18
? a = 9
now put the value of a in Eq. (v)
a + b = 13
9 + b = 13
b = 13 - 9
b = 4
? Product of both the numbers = ab = 9 x 4 = 36