could u explain?? my method.. since they meet 0.4D at their first meet, the 1st person will travel 0.4D and 2nd 0.6D.. so ratio of their speed is 2:3... when they meet for 4th time, they together cover a distance of 4D.. so 1st person must have travelled 1.6D and 2nd person 2.4D.. so it is 0.4D from P.. where am i gng wrong???:shocked:
The total distance covered would not be 4D..it wud be 7D.. after their first meeting,they wud cover a total distance of 2D together..to meet again..and so in the fourth meet they wud have covered..D+ 3*2D =7D distance..Now divide it in ratio of speeds..u'll get 2.8 D from P that wud mean 2 complete rounds and .8D from P.
The total distance covered would not be 4D..it wud be 7D.. after their first meeting,they wud cover a total distance of 2D together..to meet again..and so in the fourth meet they wud have covered..D+ 3*2D =7D distance..Now divide it in ratio of speeds..u'll get 2.8 D from P that wud mean 2 complete rounds and .8D from P.
I don't know how to go about it algebraically, but I did manage to solve it.
Break down 456 into its factors: 2*2*2*3*19
From these, the 4th and 5th terms could be any of the following combinations: 2*228, 4*114, 6*76, 8*57, 12*38 and 19*24. The order signifies the 4th and 5th number, respectively.
Since we know that it's a positive AP, the other combinations need not be considered.
Now, it's given that the 9th number is 10 more than 11 times than the 4th number. So out of the combinations which form 456 as their product, the first four are immediately ruled out.
2*228 = 9th digit should be 32 4*114 = should be 54 6*76 = should be 76 8*57 = should be 98
Even the last combination is ruled out, as 19*24 = 9th digit should be 219, and there's no way we can reach that number by using d = 5
So the only one remaining:
12, 38, 64, 90, 116, 142 (from 4th to 9th)
4th * 5th = 456 9th/4th = 11x + 10
So the 1st term: 12 - 26*3 = -66
the one given its indeed a quick solution..also 456= -38 * -12 giving D=26; now the second statement simplifies to 2A + 5D = -2 ==> A= -66 simple:)
the one given above, its indeed a quick solution..also 456= -38 * -12 giving D=26; now the second statement simplifies to 2A + 5D = -2 ==> A= -66 simple
Clock A, B and C strikes every hour..B slows down and takes 2 min longer than A per hour while C becomes faster and takes 2 min less than A per hour .If they strike together at 12 midnight,when will they strike together again..
gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The value of x could be a) 2 b) 4 c) 6 d) 8 e) none of these
hi, lets look at this way..in 1 hour clock A can reach 58 mins, B reaches 60 and C reaches 62. now for them to tick again the numbers shall meet up, that is LCM (58, 60, 62) or 2*LCM (29,30,31). now it comes out to be 899 hours. so it is 37 days and 11 hours (by dividing with 24). you can leave 37 days and for time calculation just see after 11 hours; that is, 11 am. please correct me if you find something amiss.
hi, lets look at this way..in 1 hour clock A can reach 58 mins, B reaches 60 and C reaches 62. now for them to tick again the numbers shall meet up, that is LCM (58, 60, 62) or 2*LCM (29,30,31). now it comes out to be 899 hours. so it is 37 days and 11 hours (by dividing with 24). you can leave 37 days and for time calculation just see after 11 hours; that is, 11 am. please correct me if you find something amiss.
hi, lets look at this way..in 1 hour clock A can reach 58 mins, B reaches 60 and C reaches 62. now for them to tick again the numbers shall meet up, that is LCM (58, 60, 62) or 2*LCM (29,30,31). now it comes out to be 899 hours. so it is 37 days and 11 hours (by dividing with 24). you can leave 37 days and for time calculation just see after 11 hours; that is, 11 am. please correct me if you find something amiss.
I could not get the answers for these pre assesment test questions. Could some one PLEASE explain step wise to get to the answers?
1.A bag contains 3 white and 2 red balls. One by one two balls are drawn without replacement. Find the probability that the second ball is red. a.2/5 b.1/10 c.2/10 d. 3/5
2. The following table gives the probability of india's outcome and points awarded in cricket Outcome Win Loss Draw Probability 0.5 0.45 0.05 Points 2 0 1
What is the probability that India got atleast 7 points in the contest. Assume that india played 4 matches a. 0.025 b. 0.0875 c. 0.0625 d.Cant be det
3. How many triangles can be formed from N points on a circle? a. N! b.3! c N!/3! d(N-1)N(N-2)/6
4. A student is allowed to select at most n books from a collection of 2n+1 books. If total no of ways at least one book can be selected is 63,find n. a.4 b3 c 5 d6
1)2 balls are drawn one-by-one ie. without replacement. Now, 2 different cases are possible: CASE 1:First ball drwn is red. Now, probability of second ball being red too:2/5X1/4=2/20. CASE 2:First ball drawn is black. Now, probability of second ball being red:3/5X2/4=6/20. Either case is likely. So, probability of second ball being red: 2/20+6/20=8/20=2/5.
2)India can get 7 points by the following combination: 3wins+1draw (can happen in 4!/3!=4 ways). Probability = (0.5X0.5X0.5X0.05X4)= 0.125 X 0.2 = 0.025 India can get 8 points by winning all matches. can happen in only 1 way. Probability = 0.5X0.5X0.5X0.5 = 0.0625 Either event can happen. So, required probability = 0.0625+0.025 = 0.0875.
3)Since the N points are on a circle, they are not collinear. So, joining any three of them will form a triangle. Hence, no. of triangles formed will be just selection of 3 points out of N points, which is done in NC3 ways = N(N-1)(N-2)/3! = N(N-1)(N-2)/6.
Thanks a lot man..... The book had done a mistake in giving the solutions for the first and second. Wrong options marked as answers. Thanks for clarifying
gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The value of x could be a) 2 b) 4 c) 6 d) 8 e) none of these
Remainder is coming out to be 9 So , when divided by 50, it equals 0.18 Answer should be d) 8
I hope i understood the question correctly.. Please correct me if i am wrong..
1. see we have 5 balls..now if in c the possiblities, they are drawing a while first and red later OR two consecutive reds. both contains the favorable case of 2nd draw as red. so according to the law of total probability: WR + RR ==> 3/5 * 2/4 + 2/5 * 1/4 = 8/20 = 2/5. please lemme knw if the answer is smthngelse. 😃
