Number System

GauravShah · 18-03-2005, 12:14 PM

Quote:

Originally Posted by sorabh

Quote:

Originally Posted by tackledude

Quote:

Originally Posted by GauravShah

Firstly, there is an error in the problem statement... it shud be .. "how many times will they tick together in the first hour?"

It is a simple problem of finding an LCM of 95/90 and 323/315.

Coz first oe ticks every 95/90 secs and second one every 323/315 secs.

LCM of fractions is given by (LCM of numerator/HCF of denominator)

Hence LCM = 1615/45 = 323/9. That is they tick simultaneously at every 323/9 secs

So in the first hour they will tick together = 3600/(323/9) times = 100.31

Since they ticked 100 times together at the end of one hour. Answer is option A] 100

Gaurav.

Gaurav i dont agree with u since both the clocks tick once in the beginning so the Option must be B] 101

tell me if iam wrong

i think the answer should be 102 becoz once they have clicked together at the start, and 100 times since 100.31 's integral part is 100 and as we have a fractional part of .31 which we have to count as the next tick so 1+100+1 = 102 its same as they ticked or they dont
no.. .31 tick like anyting exits.. like 4.5 people means 5 people. cant have half a guy..

This problem is simlar to the cases of ppl running around a circular track. We don't consider the initial point as the first time they meet each other.

i still stick with 100 being the possible answer.

Gaurav.

PS:
@tackledude - you had posted the question, do you have the official answer with you?

tackledude · 18-03-2005, 10:08 PM

Quote:

Originally Posted by GauravShah

This problem is simlar to the cases of ppl running around a circular track. We don't consider the initial point as the first time they meet each other.

i still stick with 100 being the possible answer.

Gaurav.

PS:
@tackledude - you had posted the question, do you have the official answer with you?

No, i dont have the answer to the above Q? It was told to me by a friend. Well the answer must be 101, and i dont agree with 102, since the approximation is 100. 100/323. thus it is 100.3095, so it cannot be taken as 101, so we have 100+1=101

Regards,
PREM

tackledude · **Find the solutions!! -** 19-03-2005, 12:21 PM

Q-1: In a seminar, the number of participants in physics, chemistry and mathematics are 96, 36, and 108. Find the minimum number of rooms required if in each room the same numbers of participants are to be seated and all of them being in the same subject.

Q-2: When 1388, 3309 and 7151 are divided by a certain number of three digits, the remainders are the same. Find the remainder.

Q-3: Find the greatest number of three digits when added to 45 is exactly divided by 6, 8, and 12

Regards,

PREM

GauravShah · 02:50 PM

Quote:

Originally Posted by tackledude

Q-1: In a seminar, the number of participants in physics, chemistry and mathematics are 96, 36, and 108. Find the minimum number of rooms required if in each room the same numbers of participants are to be seated and all of them being in the same subject.

The HCF of 96,36,108 i.e 12 is the number of students to be placed in each room.
Now for Physics, rooms req. = 96/12 = 8.
Similarly, for chem = 3 rooms and for maths = 9 rooms.

So total no. of rooms required = 8+3+9 = 20

Quote:

Originally Posted by tackledude

Q-2: When 1388, 3309 and 7151 are divided by a certain number of three digits, the remainders are the same. Find the remainder.

Let the 3 digit number be A and common remainder be B

So
1388 = Ax + B
3309 = Ay + B
7151 = Az + B

Taking difference of first 2 numbers, 3309 - 1388 = 1921. i.e A(y-x)
Taking difference of number 2 and 3, 7151 - 3309 = 3842. i.e A(z-x)

1921 = 113*17
3842 = 113*17*2

So the required 3 digit number is 113

Note:
1388 = 113*12 + 32
3309 = 113*29 + 32
7151 = 113*63 + 32

Quote:

Originally Posted by tackledude

Q-3: Find the greatest number of three digits when added to 45 is exactly divided by 6, 8, and 12

For a number to be divisible by 6,8 and 12, it shud be divisible by the LCM of the 3 numbers i.e. 24.

1000/24 given remainder as 16, so the first 4 digit number divisible by 24 is 1008 and the second 4 digit number will be 1008+24 = 1032.

Subtracting 45 from this will be the largest 3 digit number to which when 45 is added, it is divisible by 6,8 and12.

Hence answer is 1032 - 45 = 987

tackledude · **Answers!!! -** 19-03-2005, 06:49 PM

Quote:

Originally Posted by GauravShah

The HCF of 96,36,108 i.e 12 is the number of students to be placed in each room.
Now for Physics, rooms req. = 96/12 = 8.
Similarly, for chem = 3 rooms and for maths = 9 rooms.

So total no. of rooms required = 8+3+9 = 20 Let the 3 digit number be A and common remainder be B

So
1388 = Ax + B
3309 = Ay + B
7151 = Az + B

Taking difference of first 2 numbers, 3309 - 1388 = 1921. i.e A(y-x)
Taking difference of number 2 and 3, 7151 - 3309 = 3842. i.e A(z-x)

1921 = 113*17
3842 = 113*17*2

So the required 3 digit number is 113

Note:
1388 = 113*12 + 32
3309 = 113*29 + 32
7151 = 113*63 + 32For a number to be divisible by 6,8 and 12, it shud be divisible by the LCM of the 3 numbers i.e. 24.

1000/24 given remainder as 16, so the first 4 digit number divisible by 24 is 1008 and the second 4 digit number will be 1008+24 = 1032.

Subtracting 45 from this will be the largest 3 digit number to which when 45 is added, it is divisible by 6,8 and12.

Hence answer is 1032 - 45 = 987

A-1: 20

A-2: 32

A-3: 987

Regards
PREM

tackledude · **help! help! -** 19-03-2005, 06:51 PM

Guys can u suggest me a good book for QA. Iam planning to join a good institute in Chennai, any suggestions!

Regards,
PREM

tackledude · 07:37 AM

Quote:

Originally Posted by GauravShah

This problem is simlar to the cases of ppl running around a circular track. We don't consider the initial point as the first time they meet each other.

i still stick with 100 being the possible answer.

Gaurav.

PS:
@tackledude - you had posted the question, do you have the official answer with you?

Guys anybody from chennai here????????????????????????????????

tackledude · 07:00 PM

Quote:

Originally Posted by GauravShah

This problem is simlar to the cases of ppl running around a circular track. We don't consider the initial point as the first time they meet each other.

i still stick with 100 being the possible answer.

Gaurav.

PS:
@tackledude - you had posted the question, do you have the official answer with you?

A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds. If both the watches are started together, how many times will they tick in the first hour?

The answer to the problem is 101

tackledude · **Qqqqqqqqqqqqqq? -** 20-03-2005, 07:36 AM

What is the remainder of 2000^1000 divided by 13?

GauravShah · 20-03-2005, 08:17 AM

Quote:

Originally Posted by tackledude

What is the remainder of 2000^1000 divided by 13?

2000/13 gives remainder as 11.

Now 11^1000 = 121^500

121/13 gives remainder as 4

i.e to find the remainder of 4^500/13 = 2^1000/13

Remainder cycle for 2/13 is

2^1/13 = 2
2^2/13 = 4
2^3/13 = 8
2^4/13 = 3
2^5/13 = 6
2^6/13 = 12
2^7/13 = 11
2^8/13 = 9
2^9/13 = 5
2^10/13 = 10
2^11/13 = 7
2^12/13 = 1
2^13/13 = 2

i.e a cycle of 12.
1000/12, remainder is 4.

i.e. same as 2^4/13, heance remainder for 2000^1000/13 is 3

Gaurav.