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Showing posts with label word problems. Show all posts
Showing posts with label word problems. Show all posts

Monday, January 18, 2021

Two water taps can fill a tank in [ 9+(3/8) ] hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank

 cbse chapter 4 quadratic equations word problems / time - work problems

exercise 4.3

 

9. Two water taps can fill a tank in [ 9+(3/8) ] hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank

 

Let x = time taken by tap of larger diameter to fill the tank (in hours)

 

given that the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately

 

so  time taken by tap of smaller diameter to fill the tank =(x+10) hours


together they take [ 9+(3/8) ] hours or (75/8) hours by converting the mixed fraction


remeber to use reciprocals in time /work problems


[1/x] +[1/(x+10)] = 1/{75/8}


[1/x] +[1/(x+10)] =[8/75]

 

multiply each term with 75x(x+10)

 

75(x+10) +75x =8x(x+10)


75x+750 +75x =8(x^2)+80x


8(x^2) -70x -750=0 dividing by 2


4(x^2) -35x -375=0

 

we search for two numbers 

whose sum is (-35)  and 

product =4*(-375) =(-1500) 


the numbers are (-60) and 25


use this to split the middle term


4(x^2) -35x -375=0

4(x^2) -60x +25x -375=0

 

4x[x-15]+25[x-15]=0

[x-15][4x+25] =0

 

x=15  or x={(-25) /4} which is rejected


 

x=15 hours for the tap of larger diameter.

 

x+10 =15+10=25hours for the tap of smaller diameter.


10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop atintermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.


x =speed of passenger train (in hours)
(x+11) =speed express train (in hours)

distance = 132 km

time = distance/speed
 
t1= 132/x [time for passenger train]

t2= 132/(x+11) [time for express train]
 
given express train takes 1 hour less than a passenger train to travel 132 km  
so
 
t1 -t2 =1 

using the above values

[132/x] -[132/(x+11)] =1
 
multiply x(x+11)
 
132(x+11)-132x = 1x(x+11)
1452= (x^2) + 11 x

 

(x^2) +11x -1452 =0
 
to factorise search for two numbers whose product is (-1452) and sum is 11 

numbers are 44 and (-33)

so (x+44)(x-33)=0

x=-44(which is rejected
x=33 km/hr

speed of passenger train =x = 33 km/hr
speed of express train =(x+11)=33+11 =44 km/hr

 

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If the equations x^2 -ax+b=0 and x^2-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f) 

 

solution

solving a quadratic equation by completing the square

solve 2x² -4x -7 = 0


solution



Friday, October 23, 2009

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

other word problems with some explanation


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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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index of math problems


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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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index of math problems


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Wednesday, November 19, 2008

time and work problem

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?

converting to rate (quantity of work per unit time)

X can do (1/2) of the work in one hour
Y can do (1/4)th of the work in one hour

X+Y combined can do (1/2)+(1/4)=(3/4)th of the work in one hour

therefore X and Y combined will need 1 / (3/4)
or (4/3) hours to complete the work

that is 1 hour and 20 minutes

try the following page for some formulae time and work

other word problems with some explanation


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