on the earth
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…
A will finish work in 8 days (4days with B and 4 days with C).
d
HCF= 2
LCM = 2*5*7 = 70
acno is right answer
5 litre jug contains 4 litres of salt water solution
with 15% of salt, that means it has 4000*.15 = 600 ml salt
in it.
If 1.5 litres solution is spills out, remaining solution
is 2.5 litres, then the salf content is 2500*.15 = 375ml.
Now, the jug is filled to full capacity with water i.e.,
now the jug has 5 litre solution in it.
Now, the salt content is 375/5000 = 7.5%
8 cubes
8, 27, 64, 100, 125, 216, 343
100
Monday =y cheques
Tuesday =3y cheques
wednesday=4000 cheques
totaling to 16000 cheques
therefore y+3y+4000=16000 cheques
4y+4000=16000 cheques therefore 4y=16000-4000 cheques= 4y=12000cheques
therefore y=12000/4=3000 cheques
monday=3000 cheques
tuesday=3×3000=9000cheques
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
60*5/18*9=150