360
F+V-E =2;
F= faces;V= vertices;E = number of edges
avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27
X×x-x=272
X^2-x-272=0
(X-16) or (x-17)
X=16 or x=17
17^2-17=272
Answer Is 17
13 Kigs & 6 Libs –> 510tors in 10hrs
” –> 51 tors/hr
14 Libs –> 484 tors in 12hrs
” –> 121/3 tors/hr
1 Lib –> 121/42 tors/hr
==> 6 Libs –> 121/7 tors/hr
Now,
13 kigs and 6 Libs –> 51 tors/hr
Only 13 Kigs –> 51 – (121/7) tors/hr
= 236/91 tors/hr
cut a sphere in all 3 axis with center point as common point
of all axis.
u would be getting 2 lines and an arc
in other words
sphere is cut into 3 cuts in x,y,z directions
it is only similar
7days
let the total work to be done =1
so a+b+c=1 (say)
then a and b are supposed to do 7/11 th of work
so c= 1- (7/11)
c=4/11
therefore c gets (4/11)*550
=200 is the answer
150m
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.
answer is 28
>2+8=10
>reverse of 28 is 82
>subtract 28 from 82
>82-28=54
the number is 28
D
12.9
2+3+5+7+11+13+17+19+23+29=129
129/10=12.9
Employee Performance Standards. Employee performance measurements can determine an employee’s compensation, employment status or opportunities for advancement.
3 hours ago.
Thin candle melts 3/4 in 3 hours leaving 1/4
Where as in the same time thick candle melts 3/6 leaving 3/6 which is 1/2. Now thick candle is exactly twice than the thin candle.
Or via modeling:
We need to find time at which the length of the thin candle is half the thick candle. Let x be the time. Thin candle melts at 1/4 an hour and thick candle melts at 1/6 an hour. In x hours they melt at x/4 and x/6 respectively. What’s left will be 1 – x/4 and 1 – x/6. We need to find x at which :
2 * (1 – (x/4)) = 1 – (x/6)
This equation results in x = 3
65 KM