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Two water taps together can fill a tank in  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.

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Let the time taken by the smaller diameter tap be A hours Let the time taken by the larger diameter tap be A-10 hours Total time taken with both Taps together= 9 3/8 = 75 /8 hours Amount filled in one hour by smaller diameter tap = 1/A and by larger diamter tap = 1/(A-10) units As it takes 75/8 hours...
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Let the time taken by the smaller diameter tap be A hours

Let the time taken by the larger diameter tap be A-10 hours

Total time taken with both Taps together= 9  3/8 = 75 /8 hours

Amount filled in one hour by smaller diameter tap = 1/A [use concept of proportions, in A hours it fills 1 complete unit then in 1 hour it will fill 1/A units]

and by larger diamter tap = 1/(A-10) units

As it takes 75/8 hours to fill complete unit .... in 1 hour it will fill 1/(75/8) = 8/75

[1/A] + [1/(A-10)] = 8/75

Take LCM

(A-10+A)/[A*(A-10)] = 8/75

(2A-10)/(A²-10A) = 8/75

8(A²-10A) = 75 ×(2A-10){cross multiply}

8/2 (A²-10A) = 75 (A-5) {taking 2 common and dividing}

4A²-40A = 75A-375

4A² -40A-75A +375 = 0

4A²-115A + 375 = 0

4A²-100A-15A +375 = 0

4A(A-25)-15(A-25)=0

(A-25)(4A-15)

A= 25 hours

or

A= 15/4 hours

If A= 25 hours

then A-10 = 25-10 = 15 hours

if A = 15/4 hours

then A-10 = 15/4 - 10 = 15-40/4 = -25/4 hours which is not possible

since time cannot be negative therefore A = 25 hours

 

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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)...
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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.

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Tank with smaller diameter___25 hours Tank with larger diameter__15 hours
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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)...
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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

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Professional Structural Designer

Time take by larger tap is 15 hours.
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Best Tuition for CBSE 10 Class Maths!

25 hours and 15 hours
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The tap with larger diameter will fill it in 15 hours while lesser diameter will fill in 25 hours.
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