The most important belief is belief in yourself. Believing in who you are and your abilities frees you to pursue goals and dreams without reservation and drives you forward with confidence, independence, autonomy and direction. Combine this conviction in yourself with a solid framework of inherent beliefs from the list above, and you have an unbeatable combination.
4
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.
Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
Placing three trees in triangle and placing the fourth tree in center
99/1020
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
1, 2, 4, 13, 31, 112,?
This series is written in base 5 numbers for decimal nos of 1,2,4,8,16,32,64
so next number is 224 .
224 in base 5 = 64 in base 10.
so 224 is next number.
let x be sum.
SI of 18% for 1 year= x*18/100=0.18x;
SI of 18% for 2 years is 0.36x;
SI of 12% for 1 year= x*12/100=0.12x;
SI of 12% for 2 years is 0.24x;
Given, 0.36x-0.24x=840
0.12x=840
x=840/0.12=7000.
451 times.
Explanation: There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will
flashes 451 times in ¾ of an hour.
1+0^3=1
1+1^3=2
2+2^3=10
10+3^3=37
37+4^3=101
101+5^3=226
1,2,10,37,101,226……..
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