Simple Linear Equation to Find the Value of a Variable

Topic : Linear Equation
Problem : Solve 0.7n - 1.5 + 7.3n = 14.5

Solution :
0.7n - 1.5 + 7.3n = 14.5
8.0n - 1.5 = 14.5
8.0n = 14.5 + 1.5
8.0n = 16.0
n = 16.0/8.0
n = 2.0
= n = 2

Integration by Tabular Method

Topic : Integration by Tabular Method
Question : Find ∫x³ Cos 2x dx


Hints :
∫x^n Cos ax dx
Take u = x^n
dv = Cos ax dx

For patterns like
∫x^n Sin ax dx , ∫x^n Cos ax dx ,
∫x^n e^(ax) dx we use tabular method


Solution :

∫x³ Cos 2x dx
u = x³ and dv = Cos 2x











+x³(Sin 2x)/2 - 3x²(-Cos 2x)/4 + 6x(-Sin 2x)/8 - 6(Cos 2x)/16
=(x³Sin 2x)/2 + 3/4 *(x²Cos 2x) - 6/8 *(xSin 2x) - 6/16 *(Cos 2x) + c

Problem on Factorization

Topic : Factorization

Problem : Solve a^4 +3a^3+27a+81

Solution :
a^4 +3a^3+27a+81
First let's group the terms
(a^4 +3a^3)+(27a+81)
Now we take common factors
from each parenthesis.
a^3(a+3)+ 27(a+3) as 81/27 = 3
Again we have a common factor
(a+3)
So, (a+3)(a^3+27)
or (a+3)(a^3+3^3)
Applying the formula x^3+y^3 = (x+y)(x^2-xy+y^2) for a^3+b^3
we get (a+3)(a+3)(a^2-3a+3^2)
(a+3)(a+3)(a^2-3a+9)
or
(a+3)^2(a^2-3a+9)

Word Problem on Finance

Topic : Finance

Question : A car dealer offers you a choice of 0% financing for 60 months or $2500 cash back on a new vehicle. You have a pre-approved 60-month loan you can use from your credit union at a 4% interest rate. If the monthly payments at 0% are $16.67 per $1000 financed, and the monthly payments at 4% are $18.41 per $1000 financed, what is the range of new car prices for which the cash back option will cost you less? For what range of car prices should you take the 0% financing?

Solution :
Let the price at which both the options will give same result be x thousand (dollor)
Total payment for cash purchase,
= 16.67 x.60
and total payment for credit purchase
=18.41x.60 - 2500
Since these are supposed to be equal,
18.41x.60-2500=16.67x.60
1104.6x-2500=1000.2x
-1000.2x = -1000.2x
------------------------
104.4x-2500=0
+2500=+2500
-----------------
104.4x=2500
x=2500/104.4
x= 23.946
So below 23.946thousand = $23.946, the cash back option will cost less.
Above $23.946 you should take 0% financing

Question to Find the Radius of the Circle

Topic : Geometrical Figures

Question : A cone is formed with an arc length AB equal to 20 cm. As the cone is formed from a sector of a circle with angle 72 degrees, determine
a) the radius of the circle from which the sector is taken, and
b) the radius of the base of the cone formed by sector ABC

ANSWER:

(a) Step1: Given, Arc Length=20cm
Angle = 72 degrees
Required = Radius( r )
Step2: Circumference = 2Pi*r
arc length = (x/360 degrees) * 2Pi*r
Step3:Putting the values
20 = (72/360)*2Pi*r --------[pi=22/7=3.14]
r = 15.915 cm
1(b) Step1: Given, Same as 1(a)
Required =the radius of the base of the cone
Step2: circumference = 2Pi*r base of cone = circumference
Step3:Putting the values
20 cm = 2Pi*r
r = 10/Pi
r = 3.18 cm

Question to Find Arc Length of a Circle

Topic : Arc Length

Question : If a Circle has 11cm radius, Find the Length of an arc in the Circle.

Solution :

We have a Formula,
measure of arc/360 = arc length/circumference

90/360 = arc length/2πr = arc length/2π(11)

90/360 = arc length/22π


Hence the arc length = 22π * 90/360 = 11π/2 = 5.5π cm

Prealgebraic question

Topic : Word Problem on Dollors

Question : Ray had $5.12 to buy 79 cent postcards and 55 cent stamps, how many did he get?

Answer :

$5.12 = 512 cents
1 post card = 79 cents
1 stamp = 55 cents 55x + 79 y = 512
number of post cards(y) can not exceed 6 as 79 (7) = 553 > 512
and number of stamps ( x) can not exceed 9 as 55 (10) = 550 > 512
also x and y are integers > 0 .hence x can be 1,2,3,4,5,6,7,8,9 and y can be 1,2,3,4,5,6
55x ends with either 0 or 5
79 y ends with 9 ,8,7,6,5,4
a b c ---- 55x
d e f -----79y
---------
5 1 2
the sum of last digits of 55x and 79 y must be 2
hence we have only one such combination: 55x ends with 5 and 79 y ends with 7
if 79 y ends with 7 , then y must be 3 ( for 9 times 3= 27)
when y = 3 , 55x + 79(3)= 512 -> x = 5
so the number of stamps = 5
number of post cards = 3

Another way to Solve the same Problem.

Here there is another method to solve it without using algebra.
With 512 cents, he will try to buy maximum number of postcards
and maximum number of stamps.
For one postcard and one stamp, the cost is (79 + 55)= 134 centsNow 134 x 3 = 402.
So he buys 3 postcards and 3 stamps.
Now 512 - 402 = 110110 is divisible by 55. 110/55 = 2
So he can buy 2 stamps more.
Hence in total, he can buy 3 postcards and 5 stamps.