my best friend name is chandrakala. she is try to job.
The answer is depend on the distance between Mysore &
Bangalore which is not given in the question.Bird
travel=Distance/35*25km
Total game played= 60
%won =30%
Total won= 60*30/100 i.e. 18
now team plays x games and win all of those to increase the
average to 50%.
So,
(60+x)*50/100=18+x
(60+x)/2=18+x
60+x=36+2x
24=x
So the final answer is 24.
Iron and steel
Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.
120 (5×4!)
Ans =18
Explanation:
Assume that initial there were 3*X bullets.
So they got X bullets each after division.
All of them shot 4 bullets. So now they have (X – 4)
bullets each.
But it is given that,after they shot 4 bullets each, total
number of bullets remaining is equal to the bullets each
had after division i.e. X
Therefore, the equation is
3 * (X – 4) = X
3 * X – 12 = X
2 * X = 12
X = 6
Therefore the total bullets before division is = 3 * X = 18
39/50
19, 26, 33, 46, 59, 74, 91
Answer 33
Adding 7,9.13,15,17
Correct is 35
The phrase is usually ‘like looking for a needle in a haystack’. It describes a task that is virtually impossible because you would have to search a huge area
450÷100×30=135
450-135=315
Ans: 315
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125