Question

A water tap a text 7 minutes more than water type b for filling up a tank with water that Tap a takes 16 minutes more than the time taken by both the taps together to fill the tank find the time is tap alone would take to fill the tank

Answer 1

hey mate....

Let Tap B takes x mins

Tap B takes x + 7 mins

Tap B takes x mins

⇒ 1 min = 1/x of the tank filled

Tap A takes (x + 7) mins

⇒ 1 min = 1/(x + 7) of the tank filled

Together:

1 min = 1/x + 1/(x + 7)

1 min = [ x + 7 + x] /x(x + 7)

1 min = (2x + 7) /x(x + 7) of the tank

Total time needed to fill the tank together:

mins needed = x(x + 7)/(2x + 7)

The tap A takes 16 minutes more than the time taken by both the taps together to fill the tank:

Tap A = Tap A and Tap B + 16 mins

(x + 7) = x(x + 7)/(2x + 7) + 16

x + 7 - 16 = x(x + 7)/(2x + 7)

x - 9 = x(x + 7)/(2x + 7)

x(x + 7) = (x - 9)(2x + 7)

x² + 7x = 2x² + 7x - 18x - 63

x² - 18x - 63 = 0

(x - 21)(x + 3) = 0

x = 21 or x = - 3 (rejected, time cannot be negative)

Find the time needed:

Tap B = x = 21 mins

Tap A = x + 7 = 21 + 7 = 28 mins

Answer: Tap A takes 28 mins and Tap B takes 21 mins


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Answer 2

ʟᴇᴛ ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ʙʏ ᴀ ɪꜱ x ᴍɪɴᴜᴛᴇꜱ ᴀɴᴅ ʙ ɪꜱ ʏ ᴍɪɴᴜᴛᴇꜱ,

ʜᴇɴᴄᴇ ꜰʀᴏᴍ ᴛʜᴇ 1ꜱᴛ ꜱᴛᴀᴛᴇᴍᴇɴᴛ ᴡᴇ ʜᴀᴠᴇ,

x-ʏ = 7 = > ʏ = x-7..........................ᴇQ1

ɴᴏᴡ, ɪɴ ᴏɴᴇ ᴍɪɴᴜᴛᴇ ᴛᴀɴᴋ ꜰɪʟʟᴇᴅ ʙʏ ᴀ = 1/x

ᴀɴᴅ  ɪɴ ᴏɴᴇ ᴍɪɴᴜᴛᴇ ᴛᴀɴᴋ ꜰɪʟʟᴇᴅ ʙʏ ʙ = 1/ʏ

ꜱᴏ ᴄᴏᴍʙɪɴᴇ ᴛᴏɢᴇᴛʜᴇʀ ʙᴏᴛʜ ᴄᴀɴ ꜰɪʟʟ = 1/x  + 1/ʏ ᴛᴀɴᴋ ɪɴ ᴏɴᴇ ᴍɪɴᴜᴛᴇ,

ʜᴇɴᴄᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ʙʏ ʙᴏᴛʜ ᴛʜᴇ ᴛᴀᴘꜱ ᴛᴏ ꜰɪʟʟ ᴛʜᴇ ᴛᴀɴᴋ

= 1/(1/x + 1/ʏ)

= xʏ/x+ʏ

ɴᴏᴡ ꜰʀᴏᴍ ᴛʜᴇ ꜱᴇᴄᴏɴᴅ ꜱᴛᴀᴛᴇᴍᴇɴᴛ,

x- xʏ/x+ʏ   = 16

=> x² + xʏ - xʏ = 16(x+ʏ)

=> x² -16x - 16ʏ = 0

ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏꜰ ʏ ꜰʀᴏᴍ ᴇQ1

x²-16x - 16(x-7) = 0

=> x² - 32x + 112 = 0.

ꜱᴏʟᴠɪɴɢ ᴛʜᴇ ᴀʙᴏᴠᴇ Qᴜᴀᴅʀᴀᴛɪᴄ ᴇQɴ, ᴡᴇ ɢᴇᴛ

x = 28 ᴏʀ 4

ʀᴇᴊᴇᴄᴛɪɴɢ 4 ᴀꜱ x ᴄᴀɴ'ᴛ ʙᴇ ʟᴇꜱꜱ ᴛʜᴀɴ 7

x = 28

ʏ = x-7 = 28-7 = 21

ʜᴇɴᴄᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ʙʏ ʙᴏᴛʜ ᴛʜᴇ ᴛᴀᴘꜱ ᴀʀᴇ 28 ᴀɴᴅ 21 ᴍɪɴᴜᴛᴇꜱ ᴀʟᴏɴᴇ.