Logical Questions

[GAUTAMAMRIT24]

help me to solve this problems

1.When a screen is placed 3 meters from a projector,the picture occupies 3 square meters.How large will be when the projector is 5 meters from the screen?

2.John,Paul and Tim were truck drivers.They were very good friends and wanted a have lunch break together.They all drove routes starting from the warehouse.They could have a lunch break only when they had just returned from a journey.

John's journey always lasted 20 minutes,Paul's 35 minutes and Tim's 1 hour 5 minutes.Taking a new load always took 5 minutes.John loaded the first load at 07:00 and stated the first journey at 07:05.Paul loaded the first load at 07:15 and started the first journey at 07:20.Tim loaded the first load at 07:25 and started the journey at 07:30.

When could they finally have launch break together..??

 

[Deleted member 4993]

1.When a screen is placed 3 meters from a projector,the picture occupies 3 square meters.How large will be when the projector is 5 meters from the screen?

Hints:

Draw a picture - a rectangular based cone - with base 3 sq.m. and height 3 m. Join the diagonals of the rectangle.

Assume that the projector is at the apex of the cone.

Use theorems of similar triangles to solve for the diagonals of the base (hence the area) of the cone - when the height is 5 m.

2.John,Paul and Tim were truck drivers.They were very good friends and wanted a have lunch break together.They all drove routes starting from the warehouse.They could have a lunch break only when they had just returned from a journey.

John's journey always lasted 20 minutes,Paul's 35 minutes and Tim's 1 hour 5 minutes.Taking a new load always took 5 minutes.John loaded the first load at 07:00 and stated the first journey at 07:05.Paul loaded the first load at 07:15 and started the first journey at 07:20.Tim loaded the first load at 07:25 and started the journey at 07:30.

Hint:

Draw three parallel time-lines and plot their actions.

When could they finally have launch break together..??

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[soroban]

Hello, GAUTAMAMRIT24!

The second problem requires considerable work.
The final answer is unrealistic.
Also, the information is cluttered and confusing.
I will combine loading time and travel time.
(BTW, good luck with those parallel lines.
They will be over three thousand units long.)

2. John, Paul and Tim are truck drivers.
They were very good friends and wanted to have lunch together.
They all drove routes starting from the warehouse.
They could have a lunch break only when they had all just returned from a trip.

John started at 07:00 and each trip took 25 minutes.
Paul started at 07:15 and each trip took 40 minutes.
Tim started at 07:20 and each trip took 70 minutes.

When could they finally have lunch together?

The problem can be solved with Modulo Arithmetic,
. . but I will use straight Algebra (which, of course, takes longer).

Let: .\(\displaystyle \begin{Bmatrix}J &=& \text{no. of John's trips} \\ P &=& \text{no. of Paul's trips} \\ T &=& \text{no. of Tim's trips} \end{Bmatrix}\)

John's total time: .\(\displaystyle 25J\) minutes.
Paul's total time: .\(\displaystyle 40P\!+\!15\) minutes.
.Tim's total time: .\(\displaystyle 70T\!+\!25\) minutes.

These times are equal: .\(\displaystyle 25J \:=\:40P + 15 \:=\:70T + 15\)

Divide by 5: .\(\displaystyle \underbrace{5J}_{\color{blue}{[1]}} \:=\:\underbrace{8P+3}_{\color{blue}{[2]}} \:=\:\underbrace{14T +5}_{\color{blue}{[3]}}\)

From [1] = [2], we have: .\(\displaystyle 5J \:=\:8P+3 \quad\Rightarrow\quad J \:=\:\dfrac{8P+3}{5} \) .[4]

From [2] = [3], we have: .\(\displaystyle 8P+3 \:=\:14T + 5 \quad\Rightarrow\quad P \:=\:\dfrac{14T+2}{8}\) .[5]
. . . \(\displaystyle P \:=\:T + \dfrac{6T+2}{8} \quad\Rightarrow\quad P \:=\:T + \dfrac{3T+1}{8}\)

Since \(\displaystyle P\) is an integer, \(\displaystyle 3T+1\) must be a multiple of 8.
. . Hence: .\(\displaystyle 3T + 1 \:=\:8a\text{ for some integer }a.\)
Then: .\(\displaystyle T \:=\:\dfrac{8a-1}{3}\) .[6]
Hence: .\(\displaystyle T \:=\:2a + \dfrac{2a-1}{3}\)

Since \(\displaystyle T\) is an integer, \(\displaystyle 2a-1\) ,ust be a multiple of 3.
. . Hence: .\(\displaystyle 2a-1 \:=\:3b\text{ for some integer }b.\)
Then: .\(\displaystyle a \:=\:\dfrac{3b+1}{2}\) .[7]

We see that \(\displaystyle b\) must be odd: .\(\displaystyle b \:=\:2k-1\)

Substitute into [7]: .\(\displaystyle a \:=\:\dfrac{3(2k-1)+1}{2} \quad\Rightarrow\quad a \:=\:3k-1\)

Substitute into [6]: .\(\displaystyle T \:=\:\dfrac{8(3k-1)-1}{3} \quad\Rightarrow\quad T \:=\:8k-3\) .[8]

Substitute into [5]: .\(\displaystyle P \:=\:\dfrac{14(8k-3) + 2}{8} \quad\Rightarrow\quad P \:=\:14k-5\) .[9]

Substitute into [4]: .\(\displaystyle J \:=\:\dfrac{8(14k-5) + 3}{5} \:=\:\dfrac{112k-37}{5}\) .[10]
. . \(\displaystyle J \:=\:22k-7 + \dfrac{2k-2}{5} \:=\:22k-7 + \dfrac{2(k-1)}{5}\)

Since \(\displaystyle J\) is an integer, \(\displaystyle k-1\) must be a multiple of 5.
. . The least value is \(\displaystyle k = 6.\)

Substitute into [10]: .\(\displaystyle J \:=\:\dfrac{112(6) -37}{5} \quad\Rightarrow\quad \boxed{J \,=\,127}\)

Substitute into [9]: .\(\displaystyle P \:=\:14(6) - 5 \quad\Rightarrow\quad \boxed{P \,=\,79}\)

Substitute into [8]: .\(\displaystyle T \:=\:8(6)-3 \quad\Rightarrow\quad \boxed{T \,=\,45}\)


Therefore, it will take: .\(\displaystyle 25J \:=\:25(127) \:=\:3175\) minutes.



Think about the reality of this solution . . .

The three friends must drive for 2 days, 4 hours, 55 minutes non-stop
. . so they can have lunch together.
On the third day, they all stop at 11:55 a.m., just in time for lunch!

They are indeed very good friends . . .