There are two numbers such that the sum of twice the first number and thrice the second number is 100 and the sum of thrice the first number and twice the second number is 120. Which is the larger number?

Method 1 to solve the given question.
Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y.
According to question, after interchanging the digits, the new number is 10y + x.
Then,
New Number - 18 = Original number
10y + x = 10x + y + 18
? 9y - 9x = 18
? y - x = 2 .....................(1)
Again according to question,
Sum of digits of Original Number = 8
x + y = 8 ..............................(2)
Add the equation (1) and (2) , we will get
y - x + x + y = 2 + 8
? 2y = 10
? y = 5
Put the value of Y in equation (2) , we will get
x + 5 = 8
? x = 8 - 5 = 3
Then , the original number = 10x + y
Put the value of x and y and get the original number.
The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35

Method 2 to solve the given question.
Given that sum of digits of original number is 8.
Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.)
Now Original number = 10( 8 - x) + x
After interchanging the digits , the number = 10x + (8 - x) = 9x + 8
According to question.
New number = original number + 18
? 9x + 8 = 10(8 - x) + x + 18
? 9x + 8 = 80 - 10x + x + 18
? 9x + 8 = 98 - 9x
? 9x + 9x = 98 - 8
? 18x = 90
? x = 90/18
? x = 5
The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35

Let the fraction be x/y,
According to the question,
(x + 5) / (y + 5) = 7/8
? 8x + 40 = 7x + 35
? 8x - 7y = -5 ..(i)

Again, according to the question,
(x + 3)/(y + 3) = 6/7
? 7x + 21 = 6y + 18
? 7x - 6y = -3 ..(ii)


On multiplying Eq. (i) by 6 and Eq. (ii) by 7 and subtracting, we get
48x - 42y = -30
49x - 42y = -21
---------------------
-x = -9
? x = 9
On putting te value of x in Eq. (i) , we get
72 - 7y = -5
? -7y = -5 - 72
? y = -5 - 72
? y = (-77) / (-7) = 11
? Required fraction = 9/11

Let the incomes of two persons be 8x and 5x and their expenditure be 2y and y , respectively.
? Saving = Income - Expenditure
? 1000 = 8x - 2y ...(i)
and 1000 = 5x - y ...(ii)


On multiplying Eq. (ii) by 2 and subtracting from Eq. . (i) , we get
8x - 2y = 1000
10x - 2y = 2000
----------------------
-2x = -1000
? x = 500

? Monthly incomes are
8x = ? 4000 and
5x = ? 2500

? Difference = ? 4000 - 2500 = ? 1500

Let the capital of one is x and that of another is y .

According to the question.
x + 100 = 2(y - 100)
x + 100 = 2y - 200
or x - 2y = -300 ...(i)

Again, according to the question
y + 10 = 6(x - 10)
? y + 10 = 6x - 10
? 6x - y = 70 ..(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i) , we get
x - 2y = -300
12x - 2y = 140
----------------
-11x = -440
? x = 40

On putting the value of x in Eq. (i) , we get
40 - 2y = -300
? 2y = 340
? y = 170
So, their capital are ? 40 and ? 170.

(p/x) + (q/y) = m ..(i)
(q/x) + (p/y) = n ...(ii)

On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get
(pq/x) + (q²)/y = mq
(pq/x) + (p²)/y = np
-----------------------------------------
(q²/y) - (p²/y) = mq - np
? (q² - p²) = y(mq - np)
? y = (q² - p²)/mp - np = (p² - q²)/np - mq

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get
(p²/x) + (pq/y) = mp
(q²/x) + (pq/y) = nq
-----------------------------------------
(p²/x) - (q²/x) = mp - nq
? (p² - q²) = x (mp - nq)
? x = (p² - q²)/(mp - nq)
? x = (p² - q²)/(mp - nq)
and y = (p² - q²) / (np - mq)

Let us assume the first book published in year x.
According to question,
Books are published at seven years interval,
First edition book published in year = x
Second edition book published in year = x + 7
Third edition book published in year = x + 14
Four edition book published in year = x + 21
Five edition book published in year = x + 28
Six edition book published in year = x + 35
Seven edition book published in year = x + 42
When the seventh book was issued, the sum of the publication year was 13,524.
x + x + 7 + x +14 + x +21 + x + 28 + x + 35 + x + 42 =13524
147 + 7x = 13524
7x = 13524 -147
? x = 13377/7 = 1911
The first book published in year x = 1911

Let the ten's digits be x and unit's digit be y.
Then,(10x + y) - (10y + x)=36 ?9(x-y)=36
? x - y=4

Let us assume the two digits number of the number is a and b.
According to given question,
The original number is 10a + b.
After interchanging the digits , the Number is 10b + a.
The number formed is greater than the the original number by 45,
We will get,
New number = Original number + 45
? 10b + a = 10a + b + 45
? 10b + a - 10a - b = 45
? 9b - 9a = 45
? b - a = 5
According the question the difference between the digits is 5 which will not help us to solve this question because both condition given in question are same. So can't be determined the answer.

Now try Hit and Trail method
Try one by one all option. All 3 Options are true.
But we cannot choose all 3 option, but as per question we cannot find the answer.

Let us assume the ten's digit of the number is n. Then the unit's digit will be 3n.
According to question,
The sum of the digits is equal to 8.
sum of the digit = 8
n + 3n = 8
? 4n = 8
? n = 2
So, ten's digit is 2 and unit's digit is 6.
So the number = 10n + 3n = 10 x 2 + 3 x 2 = 20 + 6 = 26.

Suppose the number of cows = a therefore , the number of herdsmen = a
The total number of legs = Legs of cows + Legs of herdsmen ( Cow has 4 legs and herdsmen has 2 legs )
The total number of legs = a x 4 + a x 2 = 4a + 2a = 6a
The total number of heads = Heads of cows + Heads of herdsmen ( Cow has 1 head and herdsmen has also 1 head )
The total number of heads = a + a = 2a
According to question,
The total number of legs was 28 less than four times the number of heads,
6a = 4 x 2a - 28
? 8a - 28 = 6a
? 8a - 6a = 28
? a = 28/2 = 14