word problem 5

The area of a right angle triangle is 63 sq.cm. The length of the base of the triangle is 5 cm. more than the altitute of the triangle. Find the length of the altitute of the triangle.

let the length of the altitute be x cm.
therefore length of the base is ( x + 5) cm.

area of the right angled triangle is (1/2) * base * altitude = (1/2)(x)( x + 5 )
but given that area = 63
therefore
(1/2)(x)( x + 5 ) = 63

or x(x+5) = 126

or
x ² + 5x - 126 = 0

solving
x ² + 14x -9x - 126 = 0
x( x +14 ) - 9( x + 14 ) = 0
( x - 9 ) ( x + 14 ) = 0

x = 9 or x = -14

ignoring negative value for x

x = 9 cm

therefore length of the altitude = 9 cm.


other word problems with some explanation


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word problem4

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number is equal to 3 less than four times the original number. What is the original number?

let digit in unit's place be x and that in ten's place of the original number be y

original number = 10y + x

given
x + y = 7 ----------(i) (sum of digits)

after the digits are reversed, the new number is 10x + y

given that new number = 4*(original number ) -3
therefore ( 10 x +y) = 4*(10y + x ) - 3

or 10x + y = 40y + 4x - 3
or 6x -39 y = -3 ----------(ii)

system is
x + y = 7 ----------(i)
6x - 39 y = -3 ----------(ii)

solving
eliminating x
6x + 6y = 42----------(i)
6x - 39 y = -3 ----------(ii)

subtracting
45 y = 45 or y =1
substituting x + 1 = 7 or x = 6

therefore
x = 6 , y = 1

original number = 16

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collection of word problems

collection of word problems

The area of a right angle triangle is 63 sq.cm. The length of the base of the triangle is 5 cm. more than the altitute of the triangle. Find the length of the altitute of the triangle. solution

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number is equal to 3 less than four times the original number. What is the original number? solution

A man has 10 paise and 25 paise coins in his purse. The man has a total of 60 coins which amount to Rs.8.25.How many 10 paise and 25 paise coins does the man have in his purse? solution

The ratio of the incomes of two persons is 9:7 and the ratio of their expenditures is 4:3 . If each of them saves Rs.200/- per month, find their monthly incomes. solution

X can do a piece of work in 2 hours, while Y can do the same piece of work in 4 hours.What is the time taken if X and Y do the work together? solution


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word problem2

A man has 10 paise and 25 paise coins in his purse. The man has a total of 60 coins which amount to Rs. 8.25.
How many 10 paise and 25 paise coins does the man have in his purse?

let the man have x 10ps. coins and y 25ps coins in his purse.

x + y = 60 (number of coins)
10x + 25y = 825 (value in paise)



solving x = 45 , y = 15

other word problems with some explanation

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word problems

The ratio of the incomes of two persons is 9:7 and the ratio of their expenditures is 4:3 . If each of them saves Rs.200/- per month, find their monthly incomes.

let their monthly incomes be 9x and 7x respectively
let their monthly expenditures be 4y and 3y respectively

therefore ( income - expenditure = saving )

9x - 4y = 200 -------------(i)
7x - 3y = 200 -------------(ii)

solving
(eliminating y)

27x -12y = 600 --------------(i)*3
28x -12y = 800 --------------(ii)*4

subtracting x = 200


monthly incomes be 9x and 7x respectively

therefore monthly incomes be 9*200 and 7*200 respectively
monthly incomes be 1800 and 1400 respectively
other word problems with some explanation

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uses of graph in solving equations

uses of graph in solving equations

solve (x^2)- 2x - 8 = 0 by graphing
solving a quadratic equation by graphing

graphical solution of system of two simultaneous linear equations in two unknowns
solve graphically 3x-7y+10=0 , 2x-y+3=0 graphical solution





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linear approximation for (sin x) /x

linear approximation for (sin x) /x about x=pi/4

taylor series (truncated) with the first two terms has been used


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implicit differentiation to find second derivative

if sqrt(x) + sqrt(y) = 1 find y"
differentiate the given expression w.r.t.x and do not forget to add dy/dx when differentiating
y w.r.t.x simplify and extract the value of dy/dx

differentiate that again w.r.t.x and do not forget to add dy/dx when differentiating
y w.r.t.x , use the earlier value for dy/dx for substitution

other questions
if √(xy) = x - 2y, find dy/dx ---------------->( implicit differentiation )

if xy + y^2 =1 , find dy/dx ---------------- implicit differentiation


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if (A-B) = pi/4 show that (1+tanA)(1+tanB) = 2tanA

if (A-B) = pi/4 show that (1+tanA)(1+tanB) = 2tanA

make it into A = B + (pi/4)
take tan on both sides, apply tan(A+B) identity
simplify and extract tanA
add tanA to both sides and then factorise
trigonometry identities

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geometric distribution

geometric distribution
proof of memory less property of geometric distribution ----------memory less property of geometric distribution

Binomial distribution

Binomial distribution
pdf (probability distribution function ) of the binomial distribution is
Mean of the binomial distribution = np

Variance of the binomial distribution = npq

mgf (moment generating function of the binomial distribution) is [q + p (e^t) ] ^n

derivation of mean and variance of binomial distribution :derivation of mean and variance of binomial distribution

derivation of mgf of binomial distribution and deduction of mean and variance from the mgf:
mgf of the binomial distribution

derivation of mode of binomial distribution : mode of binomial distribution

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