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A spaceship on an inter-galactic tour has to transfer some cargo from a base camp to
a station 100 light sec away through an asteroid belt. The ship can carry a maximum of
100 kgs of cargo and, as a result of colliding against the asteroids, every 2 light sec of
travel causes it to lose 1 kg of cargo. There are 300 kgs of cargo available at the base
camp. Find the maximum amount of cargo (in kg) that the ship can transfer to the
station? Assume that the spaceship can store the cargo at any intermediate point along
the way and that stored cargo is not depleted by the asteroids.
Read Solution (Total 6)
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- i think ans is 50 +50+50=150
- 11 years agoHelpfull: Yes(8) No(0)
- 110 kg.
First it will make 3 trips to 40 light sec mark keeping 60+60+80 kg cargo there.
In next two trips to destination 60 light sec away, it will transfer 40+70=110 kg cargo. - 11 years agoHelpfull: Yes(7) No(3)
- in a journey the ship will collide with asteroids for 49 times
at 2,4,6,8....98 thus 49 times at 100th light sec it will be at its destination
so in 100 kgs it loses 49 kgs
so in 300 it loses 49*3=147 kgs
thus,delivered =300-147=153 kgs (ans) - 11 years agoHelpfull: Yes(4) No(0)
- the maximum amount the ship can carry is 100 kg,,,
when it travel for the first there is the loss of 50 kg,,( 2 light sec 1kg lose i.e 100 light year 50 kg loss) therefore for one time the max kg it will carry is 50 kg,,,
so 50+50+50=150 - 11 years agoHelpfull: Yes(3) No(0)
- plzz buddies ans to tzz question. as possible as gve rly to tzz plz.
- 11 years agoHelpfull: Yes(0) No(2)
- As it is given for every 2 light sec - 1kg is lost.
From base stn to 1 light sec - 100kg is stored and repeated to get remaining 200 to store all 300 kgs of cargo.
then again from the 1 light sec to 2nd light second the 100kg is shifted and repeated.
Till 100 light sec same method is done.
Thus maximum of all 300kgs of cargo is transported to station. - 9 years agoHelpfull: Yes(0) No(0)
TCS Other Question
1. Alok and Bhanu play the following game on arithmetic expressions. Given the
expression
N = (Θ + A)/B + (Θ + C + D)/E
where A, B, C, D and E are variables representing digits (0 to 9), Alok would like to
maximize N while Bhanu would like to minimize it. Towards this end, they take turns in
instantiating the variables. Alok starts and, at each move, proposes a value (digit 0-9)
and Bhanu substitutes the value for a variable of her choice. Assuming both play to their
optimal strategies, what is the value of N at the end of the game? Also find a sequence
of moves (digits by Alok and variables by Bhanu) that would yield this value.
Note: Moves that lead to a divide-by-zero condition are disallowed. A non-optimal
sequence of moves is (5 → B, 6 → C , 3 → D, 2 → E, 0 → A) and the expression
evaluates to Θ/5 + (Θ+9)/2.
2. The mean, unique mode, median and range of 21 positive integers is 21. What is the
largest value that can be in this sequence? Also find such a sequence.
Note: Given a sequence of numbers a(1) ≤ a(2) ≤ ... ≤ a(n),
The median of the sequence is the middlemost value in the sequence if n is
odd and the average of the two middle values if n is even.
The mode is the most occurring value in the sequence
The range is the difference between the largest and the smallest values, i.e.
a(n) - a(1).
For example, the sequence 2, 3, 4, 6, 6, 9 has mean = (2 + 3 + 4 + 6 + 6 + 9)/6 = 5,
median = (4+6)/2 =5, mode = 6, and range = 9 – 2 = 7.
The addition 641+852+973=2456 is incorrect. What is the largest digit that can be changed to make addition correct
A.4
B.5
C.6
D.7