Learn to Create a Robotic Device Using Arduino in the Free Webinar. Register Now
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Free webinar on Robotics (Block Chain) Learn to create a Robotic Device Using Arduino
30th Jan @ 5:00PM for Grade 1 to 10
Two water taps together can fill a tank in 75/8 hrs. The top of larger diameter takes 10 hr. less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Two water taps together can fill a tank in 75/8 hrs. The top of larger diameter takes 10 hr. less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Suppose that the faster pipe takes x hours to fill the tank slower pipe will take x+10 hours to fill the tank since faster tap (larger diameter tap) takes x hours to fill the tank, portion of tank filled in 1 hour = 1/x portion of tank filled in 75/8 hours = 75/8x --------------------(i) similarly portion of tank filled by slower pipe in x+10 hours to fill the tank is 1/(x+10) portion of tank filled in 75/8 hours = 75/[ 8* (x+10) ] -----------------------(ii) it is given that the tank is filled in 75/8 hours so 75/8x + 75/[ 8* (x+10) ] = 1 taking 75/8 common outside and moving it to the RHS we get 1/x + 1/(x+10) = 8/75 ( x + 10 + x )/ x(x+10) = 8/75 75 * (2x + 10) = 8 * (x2+10x) 150x + 750 = 8x2 + 80x 8x2 - 70x - 750 = 0 dividing by 2 4x2 - 35x - 375 = 0 now solve the equation and ignore the negative answer obtained ! you will get the time taken by the tap of larger diameter, add 10 to it and you will get the time taken by the smaller diametered tap PLEASE APPROVE !
Suppose that the faster pipe takes x hours to fill the tank
slower pipe will take x+10 hours to fill the tank
since faster tap (larger diameter tap) takes x hours to fill the tank,
portion of tank filled in 1 hour = 1/x
portion of tank filled in 75/8 hours = 75/8x --------------------(i)
similarly portion of tank filled by slower pipe in x+10 hours to fill the tank is 1/(x+10)
portion of tank filled in 75/8 hours = 75/[ 8* (x+10) ] -----------------------(ii)
it is given that the tank is filled in 75/8 hours
so
75/8x + 75/[ 8* (x+10) ] = 1
taking 75/8 common outside and moving it to the RHS we get
1/x + 1/(x+10) = 8/75
( x + 10 + x )/ x(x+10) = 8/75
75 * (2x + 10) = 8 * (x2+10x)
150x + 750 = 8x2 + 80x
8x2 - 70x - 750 = 0
dividing by 2
4x2 - 35x - 375 = 0
now solve the equation and ignore the negative answer obtained !
you will get the time taken by the tap of larger diameter, add 10 to it and you will get the time taken by the smaller diametered tap
PLEASE APPROVE !
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -