Mr. Brown painted the first whole house in 6 days plus 1/3 of the second house in the next 2 days. Mr. Black can paint a whole house in 8 days and had 8 days to work, so he painted the equivalent of 1 whole house. That accounts for 2-1/3 houses painted. Mr. Blue only needs to paint 2/3 of the last house, so 2/3 times 12 days equals 8 days.
Speed of Stream = 1/2 (Downstream Speed – Upstream Speed)
=1/2(40-14)
=1/2(26)
=13km/h
method 2:
for down stream case :speed=Boat speed + Stream speed
for up stream case :speed=Boat speed – Stream speed
therefore, Bs + Ss =40
Bs – Ss =14
(-) (-) (-)
———————
2Ss=26
Ss=26/2 = 13Km/h
Stream speed= 13Km/hr
length of train = 150 m
In the example first we have to find the area of the field
we have given the following values ,
cost of fencing per meter = Rs 1.50
Diameter of circular field = 28
We have to find the area , a = ?
We know that,
Area = 2 π r
= 2 x 22/7 x 14
= 2 x 22 x 2
= 44 x 2
= 88 sq. m
So, area of circle is 88 sq.m
Now just multiply it with 1.50
cost of fencing = 88 x 1.50
= Rs. 132
So, the cost of fencing the circular field is Rs. 132
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
2.5 sec
Option c
Data inadequate
I am not doing my money investment in conventional manner
2380
B
6/5
10 x 27=270