C. 21000
Answer: 66.67 km approx.
Solution:
Let the first train A move at u km/h.
Let the second train B move at v km/h.
Let the distance between two trains be d km
Let the speed of bee be b km/h
Therefore, the time taken by trains to collide = d/(u+v)
Now putting all the known values into the above equation, we get,
u = 50 km/hr
v = 70 km/hr
d = 100 km
b = 80 km/hr
Therfore, the total distance travelled by bee
= b*d/(u+v)
= 80 * 100/(50+70)
= 66.67 km (approx)
Answer is 16
1,2,4,7,11,….
1+(1) =2
2+(2) =4
4+(3) =7
7+(4) =11
Therefore
11+(5) =16
The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.
Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)
= 158
Number of odd prime numbers = 10
We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd
prime numbers)
Average =
= 15.8
∴ The Average of first 10 prime numbers which are odd is 15.8
sample space=36
prob of one of the dice getting face 6
i.e n={(1,6)(2,6)(3,6)(4,6)(5,6)(6,1)(6,2)(6,3)(6,4)
(6,5)(6,6)}=11
p=n/s
p=11/36
ans:11/36
8men=6days
Xmen=1/2days
M1=8,D1=6
M2=X,D2=1/2
Formula:
M1*D1=M2*D2
8*6=X*(1/2)
X=8*6*2
X=96
Answer=96
only if the results are greater than my input
Ans : – 215
Explanation :-
1^3 = 1-1=0
2^3 = 8-1 = 7
3^3 = 27-1=26
4^3 = 64 – 1 = 63
5^3 = 125-1 = 124
Hence, Ans is
6^3 = 216-1=215
All books can be arranged in 10! ways. A single pair of books can be taken as a unit and arranged among the 8 others in 9! ways. The pair of books can also be interchanged and therefore rearranged in 2! ways. Thus the probability of the pair always being together is (9!*2!)/10!
slower train – 48 kmph = 40/3 m/s
say faster train, v m/s
therefore, {v-(40/3)}*180 = 600, => v= 60 kmph
3, 8, 15, 24, 34, 48, 63
8-3 = 5
15-8 = 7
24-15 = 9
34-24 = 10 this one should be 35-24 = 11
48 – 34 = 14 based on the correction will be 48-35 = 13
63-48 = 15
the difference will create a series 5,7,9,11,13,15…..etc
20