let the total work to be done =1
so a+b+c=1 (say)
then a and b are supposed to do 7/11 th of work
so c= 1- (7/11)
c=4/11
therefore c gets (4/11)*550
=200 is the answer
Ratio is 3:2:1
Total=6
Reverse of ratio is 1/3:1/2:1/1
Portion of first part is=1/3*6=>2
Portion of second part is=1/2*6=>3
Portion of third part is 1*6=>6
∴So the new ratio is 2:3:6
Total distance to cover = 132 +110 = 242m
Speed = 72kn/hr = 20m/s
time = 242/20 = 12.1 sec.
D
Percentage of people passed = (passed in eng) + (passed in math) – ( passed in both)
Which is 90% of students passed and 10% failed
10% of X = 40
X = 400
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
A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69
d
HCF= 2
LCM = 2*5*7 = 70
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
Let the money borrowed by Nitin=Rs.P
According to the question,
(P×6×3)/100+(P×9×5)/100+(P×13×3)/100=Rs.8160
ans=8000
6*6=36 medals
6 days
1st day=1+35/7=6 remaining 30 medals
2nd day=2+28/7=6 remaining 24 medals
3rd day=3+21/7=6 remaining 18 medals
…
6th day 6 medals