Accounting and Finance
Before Swapping
_ _ _ _ _ _ _ Shilpa ……..Reena _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
After Swapping
_ _ _ _ _ _ _ Reena _ _ _ _ _ Shilpa _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
—————————–>14=17 <——————————-
So total number of girls in that row are 30
Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5
So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.
4.5
17
One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
Answer: Pile II
Suppose 1 person consumes 1 unit food daily.
So, for 24 days, 1600 person will consume 1600×24 units.
So total units of food available = 1600×24
Now total number of person =1200
So, total food will be consumed by 1200 person in (1600×24)/1200 = 32 days.
Rs. 1500
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
For optimal size of a project team..
the % increase in staff size should be “Zero”
reason:
additional member directly proportional to increase in
staff size..
15 min 100 line
30 min 200 lines + 15 min rest total 45 min +5 min think +5 min write + 5 min rest
250 line code will be done
no