NITISH NEGI Saystwo alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time?
12:20
take lcm of 50,48
12:20
take lcm of 50,48
12:20
take lcm of 50,48
i know the answer but i would like to ask why did we take l.c.m?
NITISH NEGI Saysi know the answer but i would like to ask why did we take l.c.m?
we need to find the common time when they will strike together..
There is a question in PnC..
There are 4 sections in an exam paper.Each section has a maximum of 45 marks.Find he number of ways in which a student can qualify if the qualifying marks is 90
1. 36546
2. 6296
3. 64906
4. none of these
It should 64906
Approach:
a + b + c + d = 90 and without any restrictions,he can score 90 marks in C(93,3)
But in some cases,scores above 45/subject are also taken into consideration
For example,if a gets more than 45,then number of possible ways are
a' + b + c + d = 44 => C(47,3)
This goes for everyone of b,c, and d
Therefore,the total cases are C(93,3) - 4*C(47,3) = 129766 - 64860 = 64906
There is a question in PnC..
There are 4 sections in an exam paper.Each section has a maximum of 45 marks.Find he number of ways in which a student can qualify if the qualifying marks is 90
1. 36546
2. 6296
3. 64906
4. none of these
p+q+r+s = 90.
=> C(93,3) solutions.
But p,q,r,s 45.
=> p'+46+q+r+s = 90.
=> p'+q+r+s = 44.
=> C(47,3) solutions.
Similarly for q,r and s.
Hence, answer = C(93,3) - 4*C(47,3) = 64906 (option 3).
the greatest number which will divide: 4003, 4126 and 4249:
(a) 43
(b) 41
(c) 45
(d) none of these
(hint: the answer is the hcf of three numbers)
which of the following represents the largest 4 digit number which can be added to 7249 in order to make the derived number divisible by each of 12, 14, 21, 33 and 54.
(a) 9123
(b) 9383
(c) 8727
(d) none of these
It should 64906
Approach:
a + b + c + d = 90 and without any restrictions,he can score 90 marks in C(93,3)
But in some cases,scores above 45/subject are also taken into consideration
For example,if a gets more than 45,then number of possible ways are
a' + b + c + d = 44 => C(47,3)
This goes for everyone of b,c, and d
Therefore,the total cases are C(93,3) - 4*C(47,3) = 129766 - 64860 = 64906
:banghead:quite weak in quant section..can you please explain it in a simpler way??
p+q+r+s = 90.
=> C(93,3) solutions.
But p,q,r,s 45.
=> p'+46+q+r+s = 90.
=> p'+q+r+s = 44.
=> C(47,3) solutions.
Similarly for q,r and s.
Hence, answer = C(93,3) - 4*C(47,3) = 64906 (option 3).
hey.. if the student scores over 90, eve then he qualifies.. so shouldnt the answer be none of these...
and how do we find the no. of ways in which he can qualify.. we cant repeat the process for 90, 91... and so on..
A ques 4 all...
Q- How many ordered triple (a,b,c) are there such that LCM (a,b)= 1000; LCM (b,c)= 2000; LCM (c,a)= 2000 ?
Options are :
a) 45 b) 70 c) 85 d) 90
It's 70
Approach:
From the given LCM's,we can say every one of a,b, and c is composed prime 2 and 5
Let's say a = 2^x1 * 5^y1;b = 2^x2 * 5^y2 and c = 2^x3 * 5^y3
LCM of a and b take maximum of 3 2's,max (x1,x2) = 3, and similarly LCM of b and c take max(x2,x3) = 4 and LCM of c and a take max(x2,x3) = 4
From this,we can say x2 cannot be 4; which means x3 has to be 4,and either x2 or x3 or both can take 3
then the number of possible triplets (x1,x2,x3) are (3,0,4),(3,1,4),(3,2,4),(3,3,4),(0,3,4),(1,3,4),and (2,3,4)
Likewise max (y1,y2) = max(y2,y3) = max(y1,y3) = 3
From this we can say atleast two of y1,y2,and y3 have to 3
Number of triplets = 3*(3!/2) + 1 = 10
Thus,the ordered triple (a,b,c) = 7*10 = 70
Good initiative.
Arun Sharma is a book with very good questions but very bad explanations.
This thread will help us explore the solutions in a better way.
Alos, LOD3 for profit and loss sucks, the problems are simply long with nothing much interesting.
So, it can be skipped ( i wasted one whole day on it grrrrr )
I want to know that which book will be essential in learning fast problem solving tricks since arun sharma gives long explanations on solutions but i cat faster solving techniques are required.
So which book to buy- Arun Sharma or Nishit sinha for quants..plz recommend
pls give me non algebraic method to solve below mentioned type of question
q)the average salary of 20 workers in an office is Rs1900 per month.If the manager's salary is added,the average salary becomes rs2000 per month.What is manager annual salary
a)24k b)25200 c)45600 d)46k e)none
pls give me non algebraic method to solve below mentioned type of question
q)the average salary of 20 workers in an office is Rs1900 per month.If the manager's salary is added,the average salary becomes rs2000 per month.What is manager annual salary
a)24k b)25200 c)45600 d)46k e)none
Think this way then:
Suppose the manager gets 19K/month,there won't be change in average salary of all
But he has increased everyone else salary,including himself,by 100
so,his monthly salary is 1900 + 21*100 = 4K and annual salary is 48K
pls give me non algebraic method to solve below mentioned type of question
q)the average salary of 20 workers in an office is Rs1900 per month.If the manager's salary is added,the average salary becomes rs2000 per month.What is manager annual salary
a)24k b)25200 c)45600 d)46k e)none
Total Sal. of 20 = 1900*20 = 38000/month.
Avg sal. incl. manager = 21 * 2000 = 42000/month.
Manager's Sal. = 42000-38000 = 4000/month = 48000/year.
Please help solving this....
Find maximum value of n such that 50! is perfectly divisible by 2520 to the power n........
Please help solving this....
Find maximum value of n such that 50! is perfectly divisible by 2520 to the power n........
Hi,
2520 = 5 x 2^3 x 3^2 x 7
Now 50! has 47 2s ,12 5s, 22 3s, 8 7s
So max value= 8
Sorry it's 8..was right at first time...
Regards,
Never Back Down
Please help solving this....
Find maximum value of n such that 50! is perfectly divisible by 2520 to the power n........
2520= 2^3* 3^2 *7 * 5 ...4 Prime no
we have to find the maximum powers of 3 and 7 to confirm our answer
50/3= 16
16/3= 5
5/3=1=22
power of 3^2= 22/2=11
power of 7 =
50/7=7
7/7=1
hence 8 power so
Maximum power of 2520 in 50! is 8
becoz maximum power of 7 in 50!=8 😃
Please help solving this....
Find maximum value of n such that 50! is perfectly divisible by 2520 to the power n........
2520 = 2^3 * 3^2 * 5 * 7.
we'll have to find out how many times each of the prime numbers occur in 50!:
divide 50 continuosly by 2 to get the quotients and add them.
++++ = 47.
divide this with 3 (power of 2) = 47/3 = 15.
similarly, for 3 this will be 11, for 5 this will be 12 and for 7 this will be 8.
Hence, for 50! to be divisible by 2520^n, we'll have to take the minimum value from above i.e, 8.
Thanks Sir! but I still didn't get it..
Why did you find maximum powers of 3 and 7 in 50!?
2520= 2^3* 3^2 *7 * 5 ...4 Prime no
we have to find the maximum powers of 3 and 7 to confirm our answer
50/3= 16
16/3= 5
5/3=1=22
power of 3^2= 22/2=11
power of 7 =
50/7=7
7/7=1
hence 8 power so
Maximum power of 2520 in 50! is 8
becoz maximum power of 7 in 50!=8 :)
Thanks Sir! but I still didn't get it..
Why did you find maximum powers of 3 and 7 in 50!?
here we are finding the maximum number of 7 in 50! bcoz 7 will the the deciding factor in 2520