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Showing posts with label cbse 10th. Show all posts
Showing posts with label cbse 10th. Show all posts

Sunday, May 12, 2024

if 49x+51y=499 ,51x+49y=501, find the value of x and y

 if 49x+51y=499 ,51x+49y=501, find the value of x and y

 

Here note the coefficients of x and y are interchanged in the two equations

so we add and subtract the two equations and get two new equations which we will solve

 

49x + 51y = 499

51x +49y = 501

_________________add

100x+100y=1000     divide by 100

x+y=10---------------{1} 


49x + 51y = 499

51x +49y = 501

_________________subtract

-2x+2y=-2          divide by (-2) 


x-y =1 ------------{2}



now we try to solve the new equations {1} and {2}

x+y=10---------------{1}

x-y =1 ------------{2}

--------------------------------add

2x =11

x=(11/2)


substitute in x+y=10

(11/2)+y=10

y=10-(11/2)=(9/2)


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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



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