Latest Elitmus Aptitude Question SOLUTION: Q. FOUR PLANE MIRROR ARE GIVEN.THEY ARE ERECTED ON THE EARTH PERPENDICULAR TO THE EARTH FORMING A RECTANGLE ITS REFLECTIVE SURFACE IS INSIDE THE RECTANGULAR BODY .LONGER BODY MIRRO

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(#M40031145) ELITMUS Logical Reasoning question Keep an EYE

Q. FOUR PLANE MIRROR ARE GIVEN.THEY ARE ERECTED ON THE EARTH PERPENDICULAR TO THE EARTH FORMING A RECTANGLE ITS REFLECTIVE SURFACE IS INSIDE THE RECTANGULAR BODY .LONGER BODY MIRRORS HAVE LENGTH 15 CM AND SMALLER ONE HAS LENGTH 4 CM. THIS RECTANGULAR BODY IS MADE IN SUCH A WAY THAT EXCEPT LASER BEAM NOTHING CAN PASS THE BODY ITS CORNERS.
1)HOW MANY TIMES WILL BE THE LASER BEAM BOUNCE OF A MIRROR SURFACE BEFORE EXITING THE RECTANGLE FROM POINT C. IF THE BEAM TRAVELS A TOTAL DISTANCE OF 25 CM INSIDE THE RECTANGLE.

OPTION
1)5
B)4
C)3
D)NONE OF THE ABOVE

2)HOW MANY TIMES WILL BE LASER BEAM BOUNCE OFF A MIRROR SURFACE BEFORE EXITING THE RECTANGLE FROM POINT A. IF THE BEAM TRAVEL A SHORTEST POSSIBLE DISTANCE INSIDE THE RECTANGLE.

OPTION
1)22
2)15
3)4
5)NONE OF THE ABOVE

Asked In Elitmus (10 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

ANJITHA BJ  (10 years ago)

Eq for distance with no reflection = 15^2 + 4^2
Eq for distance with one reflection = 2{(15/2)^2 + 4^2 }
Eq for distance with two reflection = 3{(15/3)^2 + 4^2 }
Eq for distance with n-1 reflection = n{{15/n)^2 + 4^2 }

As per Question D = 25
25 = n{{15/n)^2 + 4^2 }
n = 5
Reflections = n - 1 = 4

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swati  (10 years ago)

Correct answer is : B
Explanation


The total distance traveled by the light is 25 cm.

The light must bounce off even number of times to get out from C.
Either it will come from A and leave from C. Bounced: 0 times
Or, the light will follow: AP->PQ->QC. Bounced: 2 times(at P and Q). In this case AP = PQ = QC
Or, the light will follow: AP->PQ->QR->RS->SC. Bounced: 4 times(at P, Q. R and S). In this case AP = PQ = QR = RS = SC
and so on...

When bounced off for 0 times:
Given, total distance traveled by the light is 25 cm = AC.
AB2+BC2 = AC2 => 42+152 = 252. Which is not true. So light didn't bounce off 0 times.

When bounced off for 2 times:
Given, total distance traveled by the light is 25 cm => AP + PQ + QC = 25, but since AP = PQ = QC => AP = 25/3
and, in this case, BP = 15/3 = 5 cm
So, AB2+BP2=AP2 => 42+52=(25/3)2. Which is also not possible. So light didn't bounce off 2 times.

When bounced off for 4 times:
Given, total distance traveled by the light is 25 cm => AP + PQ + QR + RS + SC= 25, but since AP = PQ = QR = RS = SC => AP = 25/5 = 5
and, in this case, BP = 15/5 = 3 cm
So, AB2+BP2=AP2 => 42+32=52. Which is true. Hence light didn't bounce off 4 times.

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swati  (10 years ago)

Explanation
The light coming from A can get out from A in just two reflection

The distance traveled will also be minimum in this case, because the more reflection happens, the more distance is traversed.
so ans is 5)

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Shubham Saxena  (10 years ago)

what is 15/2 and 15/3. i am unable to visualise the diagram.
you can make diagram and take a picture and send me on shubham.sxn2012@hmail.com.....

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