Hog
4:3
A—–>B (first train)
B——>A(second train)
A/B=Srureroot((time to B)/(time to A))
c
3*12=36
Let cost for apple be a Cost for banana be b and Orange be c
So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100
Further if you solve you get value of b:
b = 10 – 2c ———-4
put value of b from 4 in 1
17a + 13 [10 – 2c] + 9c = 130
Further if you solve you find value of a
a = c ———-5
Put 5 in 3
13c + 7c + 10b = 100
further solve you get: c = 1 ———-6
from 5 and 6
a = c = 1 ———-7
Substitute value of c in expression 4
b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8
therefore a + b + c = 10
There are 18 numbers between 100 and 300 that are divisible by 11: 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297.
underroot(1-r^2)/3
4/52 * 3/51
16
1+1=2
2+2=4
3+4-7
4+7=11
5+11=16
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
1000 cu.m
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
C is grand child of A
8days
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