The differences between two numbers is 18. If four times the second numbers is less than three times the first number by 18, then what is the sum of these two numbers?

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

Method 1 to solve this question.
Let p be the first odd number.
Then next odd number is p + 2.
According to question,
Sum of five consecutive odd numbers 175
? p + p + 2 + p + 4 + p + 6 + p + 8 = 175
? 5p + 20 = 175
? p = (175 - 20)/5 = 31
According to question,
Sum of the second largest and square of smallest one = (31 + 6) + (31)²
Sum of the second largest and square of smallest one = 37 + 961 = 998

Method 2 to solve this question.
Let p be the even number.
Then next odd number is p + 1. so on p + 3 , p + 5................ P + 9
According to question,
Sum of five consecutive odd numbers 175
? p + 1 + p + 3 + p + 5 + p + 7 + p + 9 = 175
? 5p + 25 = 175
? p = (175 - 25)/5
? p = (150)/5
? p = 30
First odd number = p + 1 = 30 + 1 = 31
According to question,
Sum of the second largest and square of smallest one = (30 + 7) + (30 + 1)²
Sum of the second largest and square of smallest one = 37 + 961 = 998

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 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

Let Ram spends p minute on each Mathematics question.
According to the question.
50 x p + 100 x p/2 + 50 x p/2 = 3 x 60
p(50+ 50 + 25) = 180
? p = 180/125
? Required time = 50 x 180/125
? Required time = 2 x 180/5
? Required time = 2 x 36 = 72 min

Method 2
Let Ram spends a minute on each sciences and Gk questions.
According to the question,
50 x 2a + 150 x a = 3 x 60
100a + 150a = 180
250a = 180
a = 180/250
So total time spend on Mathematics question = 2a x 50 question
So total time spend on Mathematics question = 2 x 50 x 180/250 = 72 minutes

let the fixed charges = ? p for first 5 km
and the additional charges = ? q per km
according to the question
p + 5q = 350........... (1)
p + 20q = 800...........(2)
on subtracting Eq. 1 from Eq. 2 we get
15q = 450
q = 30
On putting the value of q in Eq. (1) we get
p = 200
? charge for a distance of 30 km = p + 25q
= 200 + 30 x 25 = ? 950