| Solution: No. of diagonals = nc2 – n; n = nc2 – n => n=5; No of triangles = 5C3 = 10.
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| Solution:
Total distance travelled = S; Initial height = 5feet.
Ist bounce = 5/2 feet; IInd bounce = (5/2) (1/2)
S = 5 + 2 [ (5/2) + ( 5/22) + + ( 5/23) + ……up to infinity.]
S /5= 1 + 2 [ (1/2) + ( 1/22) + + ( 1/23) + ……up to infinity.]
(S /5) – 1 = [ (1+(1/2) + ( 1/22) + + ( 1/23) + ……up to infinity.]
The problem given can be solved using the Concept of Geometric progression.
Here a = 1, r = 1/2;
Using the sum formula Sn =a/1-r.
(S/5) – 1= 2. => S= 15 feet.
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| Solution:
On the first day, the snail reaches a height of 5 meters and slides down 4 meters at night, and thus ends at a height of 1 meter.
On the second day, he reaches 6 m., but slides back to 2 m.
On the third day, he reaches 7 m., and slides back to 3 m.
...
On the fifteenth day, he reaches 19 m., and slides back to 15 m.
On the sixteenth day, he reaches 20 m., so now he is at the top of the pit.
The snail reaches the top of the pit on the 16th day.
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| Solution:
The digit 6 can be located in any of the 5 positions; then 8 can be located in in 4 positions.
Thus 5 x4 = 20 distinct 5-digit numerals can be constructed out of the digits 1,1, 1, 6, 8.
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| Solution:
When ‘O’ and ‘A’ occupying end-places
=> M.E.G. (OA)
Here (OA) are fixed, hence M, E, G can be arranged in 3! ways
But (O,A) can be arranged themselves is 2! Ways.
=> Total number of words = 3! x 2! = 12 ways.
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