hard maths question really hard!!?

Question

i tried for like 2 hours cant get it!!
A tank can be filled in 6 hours using two pipes. The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone. how long would each pipe alone take?
thnx alot

Answer 1

It's a tricky problem; the usual one is the reverse of knowing the flow from two pipes separate and finding the flow of them combined. However, the equation is the same.
Volume = SUM (volume/time * time)
Let the tank volume = TV.
Then for the smaller pipe, Let x be time to fill alone
TV = (vol/time)small * x hours
and TV= (vol/time)lg * (x-5) hrs.
Now (vol/time)small= TV/x
and (vol/time)large= TV/(x-5)
Together,
TV = TV/x * 6 + TV/(x-5) * 6
So, 1 = 6/x + 6/(x-5) [TV can be divided out]
And 1 = [6(x-5) + 6x ]/ [(x-5)(x)]
Multiplying
x^2- 5x = 6x-30+ 6x= 12x-30
And x^2-17x+30=0
From which (x-2)(x-15)=0
Since one condition requires x-5 to be positive, x # 2. So x=15. So x=15 and x-5 = 10 hours

TV =

Answer 2

x = hours large pipe
y = hours small pipe
x = y + 5

(1/x)6 + (1/y)6 = 1
(1/( y+ 5))6 + (1/y)6 = 1
multiply each term by lowest denominator (s(s+5)
(y)6 + (y + 5)6 = y^2 + 5y
6y + 6y + 30 = y^2 + 5y
0 = y^6 - 7y - 30
0 = (y + 3)(y - 10)
y = small pipe = 10 hours
x = large pipe = 15 hours

Answer 3

a million+a million=2 apart from, somebody with some distance too lots time on their arms desperate to tutor it... "The evidence starts off from the Peano Postulates, which define the organic numbers N. N is the smallest set fulfilling those postulates: P1. a million is in N. P2. If x is in N, then its "successor" x' is in N. P3. there is not any x such that x' = a million. P4. If x isn't a million, then there's a y in N such that y' = x. P5. If S is a subset of N, a million is in S, and the implication (x in S => x' in S) holds, then S = N. then you quite could desire to define addition recursively: Def: permit a and b be in N. If b = a million, then define a + b = a' (utilising P1 and P2). If b isn't a million, then permit c' = b, with c in N (utilising P4), and define a + b = (a + c)'. then you quite could desire to define 2: Def: 2 = a million' 2 is in N via P1, P2, and the definition of two. Theorem: a million + a million = 2 evidence: Use the 1st part of the definition of + with a = b = a million. Then a million + a million = a million' = 2 Q.E.D. notice: there is yet another formula of the Peano Postulates which replaces a million with 0 in P1, P3, P4, and P5. then you quite could desire to alter the definition of addition to this: Def: permit a and b be in N. If b = 0, then define a + b = a. If b isn't 0, then permit c' = b, with c in N, and define a + b = (a + c)'. you apart from mght could desire to define a million = 0', and a couple of = a million'. Then the evidence of the theory above is slightly diverse: evidence: Use the 2nd part of the definition of + first: a million + a million = (a million + 0)' Now use the 1st part of the definition of + on the sum in parentheses: a million + a million = (a million)' = a million' = 2 Q.E.D." Wow, he could desire to be a actual hit with the females...!! :)

Answer 4

Yeah these rate problems can be annoying until you find the trick.

A tank can be filled in 6 hours using two pipes (call them L&S). The larger pipe alone would fill it 5 hours sooner than the smaller pipe alone.

=> Call the rate of flow from the smaller pipe R, then the rate of flow from l must be 5R.

And the rate with both S&L on is R+5R=6R

Time to fill with both S&L:
2hrs = Volume/6R
=> Volume/R = 6*2hrs = 12hrs

How long would each pipe alone take?

Time to fill with only S:
Volume/R = 12hrs as above

Time to fill with only L:
Volume/5R = 2hrs * 6/5 = 2hrs24min

Answer 5

i think i asnwered this one in your previous post