plz give the answer
Answers
Let the time taken by the smaller diameter tap = x
larger = x-10
total time taken = 75 /8
portion filled in one hour by smaller diameter tap = 1/x
and by larger diamter tap = 1/x-10
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
2x+10/x²-10x = 8/75
8(x²-10x) = 75 ×2 (x-5)
8/2 (x²-10x) = 75 (x-5)
4x²-40x = 75x-375
4x² -40x-75x +375 = 0
4x²-115x + 375 = 0
4x²-100x-15x +375 = 0
4x(x-25)-15(x-25)=0
(x-25)(4x-15)
x= 25
x= 15/4
If x= 25
the x-10 = 25-10 = 15
if x = 15/4
x-10 = 15/4 - 10 = 15-40/4 = -25/4
since time cannot be negative therefore x = 25
Answer:
Let time taken by smaller tap = x hours
Let time taken by larger tap = ( x - 10) hr
Case 1
Tank filled by smaller tap
Portion of tank filled in one hour = 1/x
Portion of tank filled in 75/8 hour = 75/8 * 1/x
= 75/8x
Case 2
Tank filled by larger tap
Portion of tank filled in one hour = 1/x-10
Portion of tank filled in 75/8 hour = 75/8 * 1/x-10
= 75/8(x-10)
75/8x + 75/8(x-10) = 1
75/8(1/x +1/x-10). = 1
1/x + 1/x-10 = 8/75
After solving the equation you get the value of x