Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)
(log 2 = 0.3010)
(a) 46 (b) 47 (c) 48 (d) 49
and please explain the method
@fireatwill said:Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)(log 2 = 0.3010)(a) 46 (b) 47 (c) 48 (d) 49and please explain the method
48 zeros?
@fireatwill said:Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)(log 2 = 0.3010)(a) 46 (b) 47 (c) 48 (d) 49and please explain the method
Here,
1/20^37 = 5^37/10^74
Number of terms in 5^37 = log(5)*37 = 25.86 = 26.
Thus, No. of terms before point = 74 - 26 = 48.
@fireatwill said:Find the number of consecutive zeroes immediately after the decimal in 1/(20^37)(log 2 = 0.3010)(a) 46 (b) 47 (c) 48 (d) 49and please explain the method
log(1/20)^37
37[log 1 - log 20 ]
log 1 =0
-37 log 20 == -37 log (2 * 10 )
-37 log 2 - 37 log 10
-11.137 -37
-48.137
the log of (1/20)^37 has -48 as its characterisitic when mantissa is negative.
so it will have 48 zeros immediately after the decimal point
@pavimai said:after travelling 50km,a train is meeting with an accident and travels 3/4th of usual speed and reaches 45 min late .had the accident happened 10 km further on it would have reached 35 min late .find the usual speed??
10/(3x/4) - 10/x = 10
=> 40/3x - 10/x = 10
=> (40-30)/3x = 10
So, x = 1/3 km/min
So, 20 km/hr.
@fireatwill said:@pavimai bhai 48 is correct ..can u explain the method
Pavimai is a girl and doesnt understand hindi 😛 as for the ans:
log(1/20^37) is your answer.
which is : 0 - 37log(20) = -37(log2 + 1) = -1.3*37=48
@Estallar12 said:10/(3x/4) - 10/x = 10=> 40/3x - 10/x = 10=> (40-30)/3x = 10So, x = 1/3 km/minSo, 20 km/hr.
sir can u explain..
@Brooklyn said:Pavimai is a girl and doesnt understand hindi as for the ans:log(1/20^37) is your answer.which is : 0 - 37log(20) = -37(log2 + 1) = -1.3*37=48
nahi nahi..muje hindi bahoot maalum..
@pavimai said:after travelling 50km,a train is meeting with an accident and travels 3/4th of usual speed and reaches 45 min late .had the accident happened 10 km further on it would have reached 35 min late .find the usual speed??
20???
@Estallar12 said:10/(3x/4) - 10/x = 10=> 40/3x - 10/x = 10=> (40-30)/3x = 10So, x = 1/3 km/minSo, 20 km/hr.
@pavimai said:sir can u explain..
\This will be the figure for it.
6x for Still Water.
10x when moving perpendicular in moving water. So, his path will be deflected at some angle. That is the Hypotenuse.
So, Base = 8x.
Now, 6x = 300
So, x = 50.
Thus, 8x = 400.
Or. Speed = 40 m/min.
@Estallar12 said:\This will be the figure for it.6x for Still Water.10x when moving perpendicular in moving water. So, his path will be deflected at some angle. That is the Hypotenuse.So, Base = 8x.Now, 6x = 300So, x = 50.Thus, 8x = 400.Or. Speed = 40 m/min.
@pavimai 
I explained the wrong question. :|
I explained the wrong question. :|@Estallar12 said:10/(3x/4) - 10/x = 10=> 40/3x - 10/x = 10=> (40-30)/3x = 10So, x = 1/3 km/minSo, 20 km/hr.
@pavimai said:sircan u explain..
:nono::nono:
See, had the accident occured 10 km later, this means he would have travelled those 10 km with the original speed x.
Difference in time = 45 - 35 = 10 minutes.
So, Had he travelled with x speed for 10 km, he would take 10 minutes less.
Thus, Time taken when travelled 10 km with 3/4 th speed - Time taken when travelled 10 km with original speed = 10 minutes.
=> 10/(3x/4) - 10/x = 10 minutes.
So, x = 1/3 km/min.
Thus, 20 km/hr.
@Estallar12 said:See, had the accident occured 10 km later, this means he would have travelled those 10 km with the original speed x.Difference in time = 45 - 35 = 10 minutes.So, Had he travelled with x speed for 10 km, he would take 10 minutes less.Thus, Time taken when travelled 10 km with 3/4 th speed - Time taken when travelled 10 km with original speed = 10 minutes.=> 10/(3x/4) - 10/x = 10 minutes.So, x = 1/3 km/min.Thus, 20 km/hr.
got it 
thanks a lot
thanks a lot