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Nov 24th, 2021 - Daily Quiz -CSAT 2

1. How many triangles can be formed by joining the vertexes of the polygon of “n” sides having the same number of diagonals as its sides? (a) 12     (b) 11        (c) 13         (d) 10   2. A ball when dropped from a height of 5 feet, bounces to half the height and continues bounce to half the height of each previous bounce till it reaches zero height. What is the total distance travelled by the ball? (a) 15      (b) 30        (c) 10        (d) 25   3. A snail is at the bottom of a 20 meters deep pit. Every day the snail climbs 5 meters upwards, but at night it slides 4 meters back downwards. How many days does it take before the snail reaches the top of the pit? (a) 13      (b) 16        (c) 14         (d) 15   4. How many distinct 5-digit numerals can be constructed out of the digits 1,1, 1, 6, 8? (a) 20      (b) 23        (c) 24         (d) 26   5. How many words can be formed with the letters of the word ‘OMEGA’ when ‘O’ and ‘A’ occupying end places?. (a) 10       (b) 12       (c) 18        (d) 16   Answers
Solution: No. of diagonals = nc2 – n; n = nc2 – n => n=5;  No of triangles = 5C3 = 10.  
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.  
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.   
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.  
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.  






POSTED ON 24-11-2021 BY ADMIN
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