The ratio of incomes of two persons is 8 : 5 and the ratio of their expenditure i 2 : 1. If each of them manages to save ? 1000 per month, find the difference of their monthly income

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)

Given, 2^a + 3^b = 17
and 2^{a + 2} - 3^{b + 1} = 5
? 2² x 2² - 3^b x 3¹ = 5
? 4.2^a - 3.3^b = 5
Let 2^a = x amd 3^b = y
Then, x + y = 17 ...(i)
4x - 3y = 5 ...(ii)
On multiplying Eq. (i) by 3 and adding to Eq (ii), we get
3x + 3y = 51
4x - 3y = 5
------------------
7x = 56
? x = 8

On putting the value of x in Eq. (i), we get
8 + y = 17
? y = 9

Now, 2^a = x
? 2^a = 8 (2)³
? 3^b = y = 9
? 3^b = 3²
? b = 2
Hence, a = 3 and b = 2

Let the cost of one chair be ? x
and cost of one table be ? y.
By given condition,
10x + 6y = 6200 ..(i)
and 3x + 2y = 1900
? 9x + 6y = 5700 ...(ii)

On subtracting Eq. (ii), we get
x = ? 500

From Eq (i),
5000 + 6y = 6200
? 6y = 1200
? y = ? 200

The cost of 4 chair and 5 tables
= 4x + 5y
= 2000 + 1000
= ? 3000

ax + by = c and dx + ey = f
a₁/a₂ = a/d, b₁/b₂ = b/e, c₁/c₂ = c/f
? b₁/b₂ ? c₁/c₂
? b/e ? c/f
it represent a pair of parallel lines.
? a₁/a₂ ? b₁/b₂
? a/d ? b/e
Therefore, system has unique solutions and represents a pair of intersecting lines.

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

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 us assume the first number is x and second number is y.
According to question,
sum of twice the first number and thrice the second number is 100.
2x + 3y = 100 ...................(1)
and sum of thrice the first number and twice the second number is 120.
3x + 2y = 120...................(2)
Subtract equation (1) from Equation (2) after multiply 2 with Equation (1) and 3 with Equation (2), we will get.
3(3x + 2y) - 2(2x + 3y) = 3 x 120 - 2 x 100
9x + 6y - 4x - 6y = 360 - 200
5x = 160
x = 32
put the value of x in equation (1)
2 x 32 +3y = 100
3y = 100 - 64
3y = 36
y = 12
The larger Number is 32 .

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