Question

water tap A take seven minutes more than water tap B for filling up a tank with water the tap A takes 16 minutes more than the time taken by the both the taps together to fill the tank find the time is each tap alone would take to fill the tank

Answer 1

Let the time taken by A is x minutes and B is y minutes,

Hence from the 1st statement we have,

x-y = 7 = > y = x-7..........................eq1

Now, in one minute tank filled by A = 1/x

and  in one minute tank filled by B = 1/y

So combine together both can fill = 1/x  + 1/y tank in one minute,

Hence time taken by both the taps to fill the tank

= 1/(1/x + 1/y)

= xy/x+y

Now from the second statement,

x- xy/x+y   = 16

=> x² + xy - xy = 16(x+y)

=> x² -16x - 16y = 0

Putting the value of y from eq1

x²-16x - 16(x-7) = 0

=> x² - 32x + 112 = 0.

Solving the above quadratic eqn, we get

x = 28 or 4

rejecting 4 as x can't be less than 7

x = 28

y = x-7 = 28-7 = 21

Hence time taken by both the taps are 28 and 21 minutes alone.

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.