Question

Please answer the above question​

Answer 1

» Given :

time at that particular instant = t minutes past 3pm

time needed by minute hand to show 4pm = $\frac{t^2}{60} +15$

» Solution -

We know that,

1 hour = 60 minutes

So,

time left to 4pm = (60-t) minutes

$60-t = \frac{t^2}{60}+15\\ \\ \\ 60-15 = \frac{t^2}{60}+t\\ \\ \\ 45 = \frac{t^2+60t}{60}\\ \\ \\by\:cross\: multiplication\\ \\ \\ 45\times 60 = t^2 +60t \\ \\ \\t^2+60t-2700 = 0 \\ \\ \\ We\:know\:that,\\ \\ \\ x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ \\ In\:our\:case,\\ \\ \\ a= 1\\ \\ b= 60\\ \\ c= -2700\\ \\ \\ by\: putting\:values\\ \\ \\ t = \frac{-60\pm\sqrt{(60)^2-(4\times1\times -2700)}}{2}$ $=\frac{-60\pm\sqrt{3600+10800}}{2}\\ \\ \\ =\frac{-60\pm120}{2}\\ \\ \\ x = \frac{-60+120}{2}= \frac{60}{2} = 30 \\ \\ \\ x= \frac{-60-120}{2} = -90$

${\pink{\boxed{x = 30 \:or \:-90}}}$

${\pink{\boxed{\therefore\:answer\:is\: option\:(c)}}}$

Answer 2

Answer:

Option C.) 30

t = 30

Step-by-step explanation:

Since, this is a word problem. Let us solve this line - by - line.

Given,

't' minutes past 3. This means 't' lies somewhere between 3 pm to 4 pm

Then, time difference between 3 pm and 4 pm = 1 hour = 60 minutes.

Then,

➡ Time left to 4 pm = 1 hour - t = (60 - t) minutes ........ (i)

Also, it is given that, time needed by the minute hand of the clock to show 4 pm is 15 more than t^2 / 60 that is

(t^2 / 60) + 15 minutes.

Then,

➡ Time left to 4 pm = (t^2 / 60) + 15 minutes ...... (ii)

Clearly, we can see that, equation (i) and (ii) determine equal quantities.

So equating equation (i) and equation (ii) , we get,

➡ (60 - t) = (t^2 / 60) + 15

➡ 60 - t = (t^2 + 900) / 60

By cross - multiplication, we get,

➡ 60•(60 - t) = t^2 + 900

➡ 3600 - 60t = t^2 + 900

➡ t^2 + 900 - 3600 + 60t = 0

➡ t^2 + 60t - 2700 = 0

➡ P(x) = t^2 + 60t - 2700

We know that,

(90 × 30 = 2700 , 90 - 30 = 60)

Now the P(x) forms a Quadratic equation.

So simply,we can solve this using the Splitting the Middle Term Method.

➡ P(x) = t^2 + 90t - 30t - 2700

➡ t^2 + 90t - 30t - 2700 = 0

➡ t•(t + 90) - 30•(t + 90) = 0

Now, let us take like terms common,

➡ (t - 30)•(t + 90) = 0

Here, either, (t - 30) = 0 or (t + 90) = 0

Then, for the values of x :-

=> CASE I :

(t - 30) = 0

=> t = 30

(accepted value, as quantity of minute is always positive.)

=> CASE II :

(t + 90) = 0

=> t = - 90

(neglected value, as quantity of minute is not negative)

Hence, here, the required value of

t = 30

So, we get, t = 30