@sameersapre23 said:Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. Find the probability that these three numbers are in A.P.100/21C399/21C3101/21C3none of these
100/21c3 ?
@sameersapre23 said:Ten people A, B, C, D, E, F, G, H, I and J attend a conference. F wants to speak before A, B wants to speak after A, C wants to speak after B and D wants to speak after C. In how many distinct orders can the 10 people speak?10!10c55! 10c5NONE OF THESE
10!/5! = 10c5*5!
@YouMadFellow said:Two dice are thrown simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on other is: A. 1/6 B. 1/3 C. 5/12 D. 11/36 (No OA )
11/36...
@sameersapre23 said:Sanjog and Rahul took part in a two-day maths contest. At the end both had attempted questions worth 500 marks. Sanjog scored 160 out of 300 attempted on the first day and 140 out of 200 attempted on the second day, so his two-day success ratio was 3/5. Rahul's attempted figures were different from Sanjog's (but with the same two-day total). Rahul had a positive integer marks on each day. For each day Rahul's success ratio was less than Sanjog's. What is the largest possible two-day success ratio that Rahul could have achieved?81/125349/500Cannot be determinedNone of these
349/500 ?
@YouMadFellow said:Two dice are thrown simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on other is: A. …™ B. …“ C. 5/12 D. 11/36
11/36 ?
@sameersapre23 said:i dont have OA of any of the questions , these all are from the mock i have....!! With no answers.We will dicuss them......!!
Hmm, we all seem to be in agreement so I guess OA ka tension nahin hai :)
regards
scrabbler
regards
scrabbler
@sameersapre23 said:Sanjog and Rahul took part in a two-day maths contest. At the end both had attempted questions worth 500 marks. Sanjog scored 160 out of 300 attempted on the first day and 140 out of 200 attempted on the second day, so his two-day success ratio was 3/5. Rahul's attempted figures were different from Sanjog's (but with the same two-day total). Rahul had a positive integer marks on each day. For each day Rahul's success ratio was less than Sanjog's. What is the largest possible two-day success ratio that Rahul could have achieved?81/125349/500Cannot be determinedNone of these
Success Ratios of Sanjog => 160/300 = 8/15 and 140/200 = 7/10.
For largest possible success ratio of Rahul, maximising his score on 2nd day.
Now, 8/15 is just more than 1/2. So, let us make Rahul get 1 correct out of 2 attempted on the first day.
On the second day, we need just less than 7/10.
Now, 7/10 of 498 = 348.6 => 348 questions can be taken as our required 2nd day number.
So, Total Correct = 348 + 1 = 349.
Success Ratio = 349/500
Easy
On 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)
(a)25(b) 35(c) 37(d) 39
On 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)
(a)25(b) 35(c) 37(d) 39
@sameersapre23 said:Ten people A, B, C, D, E, F, G, H, I and J attend a conference. F wants to speak before A, B wants to speak after A, C wants to speak after B and D wants to speak after C. In how many distinct orders can the 10 people speak?10!10c55! 10c5NONE OF THESE
10!/5! = 10c5*5!
@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
Sum of ages earlier = 6*A
After six years = 6*A + 36 + 1 = 6A + 37
6A + 37 = 7A => A = 37 ?
@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
6A = Sum.
After six years, Total = Sum + 6*6 + 1 = Sum + 37.
Now, (Sum + 37) / 7 = A
So, A = 37.
@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
37 
I will try easy ones
I will try easy ones@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
37
A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
37?
6A + 30 + 6 + 1 = 7A
A = 37
@Faruq said:A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
36 ..
@Faruq said:A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
8N = total toffees
N = 5*(7N)/ (x-1) => x - 1 = 35 => x = 36 ?
@Faruq said:EasyOn 1st January, 2000 the average age of a family of 6 people was 'A' years. After 5 years a child was born in the family and one year after that the average age was again found to be 'A' years. What is the value of 'A'? (Assume that there are no other deaths and births.)(a)25(b) 35(c) 37(d) 39
37
@Faruq said:A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
36
@Faruq said:A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
36...
..