Problem on Linear Equations to Find the Unknown Value of a Variable

Topic : Linear Equations

Problem : Solve 2x + 3y = 4 and 3x - 2y = 1

Solution :
2x + 3y = 4 --------------(1)
3x - 2y = 1 --------------(2)

Multiply equation (1) by 3 and equation (2) by 2 then,
6x + 9y = 12 ----------- (3)
6x - 4y = 2 ------------ (4)
Subtraction (4) by (3) then,
+13y = 10
y = 10/13

Substitute the value of y in equation (1)
2x + 3y = 4
2x + 3(10/13) = 4
2x + 30/13 = 4
2x = 4 - 30/13
2x = (52-30)/13
2x = 22/13
x = 11/13

A Simple Problem on Integration

Topic : Integration

Problem : Integrate 1/x² * (x+2)³

Solution :

Deriving the value of Cos 11π/4

Topic : Trignometric Value of Cos 11 π/4

Problem : Find the value of cos(11π/4)

Solution :
General Formula is Cos θ = Cos (2π - θ)
So Cos 11 π/4 = Cos (2 π - 11 π/4)
= Cos (-3 π/4)

Again by another formula
Cos θ = -Cos (π + θ)
So -Cos (-3 π/4) = -Cos (π + (-3 π/4))
= -Cos (π/4)

Now Cos (π/4) = 1/√2
So -Cos (π/4) = - 1/√2

Word Problem on Simple Interest

Topic : Investment and Simple Interest

Question : Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 11%. His total annual income from the two investments was $3990. How much was invested at 11%?

Solution :
Let the amount invested at 7% be x
then the amount invested at 11% will be $(2x+2000)
Now the total annual income from these investments = $ 3990
So, the equation will be ,
7/100 *x +11/100 * (2x + 2000) = 3990
0.07x + 0.11(2x + 2000) = 3990
0.07x + 0.22x + 220 = 3990
0.29x +220 = 3990
subtract both sides by 220
0.29x = 3770
divide both sides by 0.29
x = 3770/0.29
x =13000

So, amount invested at 7% = $ 13000
Therefore the amount invested at 11% = 2x + 2000
= 2(13000) + 2000
=26000 + 2000
= $ 28000

Question on Factorization of trinomial

Topic : Factorization

Questions : Factorization of trinomial
Factorise x^2+7x+12

Solution :
X^2 +7x +12 has two binomial factors with the product of the first term as x^2
Thus we can write X^2+7x+12 + (x +a) (x +b)
a and b are the missing terms
The product of the missing terms is 12 and their sum is 7 Let us examine the
Factors of 12
1x 12 = 12 but 1 + 12=13 ≠ 7
2x 6 =12 but 2 +6 = 8 ≠ 7
3 x 4 =12 here 3 + 4=7 as required
Thus we factorise X^2 + 7x +12 = (x+3 ) (x+4)

A Word Problem on Milkman and Domesticated Animals with Him

Topic : Word problem on Milkman and domesticated animals.
Question : A milkman has some buffaloes cows and goats. He has goats equal to 5/2 times the number of cows and cows equal to 3/2 times the number of buffaloes. If there are 150 animals with the milkman, how many of each category has he with him?

Solution : Let the Number of buffaloes = Y

Number of cows = Y x 3/2 = 3Y/2 (3/2 times of buffaloes)

Number of Goats = 3Y /2 x 5/2 = 15Y /4 (5/2 times of cows)

Total number of animals = Y + 3Y/2 + 15Y/4 (Adding all of them)

= (4Y + 6Y + 15Y)/ 4

= 25Y / 4

According to the information given total animals = 150

Therefore 25Y / 4 = 150

Y = 150 x 4/ 25 = 24

Number of buffaloes = 24 (As we assumed number of buffaloes as Y)

Number of Cows = 3(24)/2 = 36 (As Cows = 3Y/2)

Number of Goats = 15(24)/ 4 = 90 (As Goats=15Y/4)