have already given them.
ans is 588...
Hi guys,
The crew of an 8 members team is to be chosen from 12 men,of which 3 must row on one side only and 2 must row on the other side only.Find the number of arrangement the crew with 4 members on each side.
Ans is 60,480....
Hi guys,
The crew of an 8 members team is to be chosen from 12 men,of which 3 must row on one side only and 2 must row on the other side only.Find the number of arrangement the crew with 4 members on each side.
Ans is 60,480....
Not sure if you are answering or asking.
But anyway here's how it goes
C(7,3)-First step
Next, out of these 3 any 2 can be on one side and one on the other, so C(3,2)
Next 4 on each side can be arranged among themselves in 4!, 4! ways each
so C(7,3)*C(3,2)*4! *4! = 60480
A young girl counted in the following way on the fingers of he left hand .she started calling the thumb 1,index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger7,index finger 8,thumb 9then back to the index finger fo 10,middle finger 11,and so on .she counted upto 1994.she ended on her.
A)thumb B)index finger c)middle finger d)ring finger
WITH SOLUTION PLEASE
hey,i m confused abt one concept,
plz chk Pg no 26(Ch 1 no system)
self practice exercise
Q3,n Q 5, Q 6
if we divide 157! by 12^n or 18^n
wat diff steps shud we take...
same goes for 50! divided by 2520^n and 12600^n
plz clarify ppl
hey,i m confused abt one concept,
plz chk Pg no 26(Ch 1 no system)
self practice exercise
Q3,n Q 5, Q 6
if we divide 157! by 12^n or 18^n
wat diff steps shud we take...
same goes for 50! divided by 2520^n and 12600^n
plz clarify ppl
157! mod 12^n
Now we have to find how many powers of 2 and 3 are there in 157!
157! = P * 2 ^ 152 * 3 ^ 75
Now, to make a 12 we need two 2s and one 3. So, we can make a maximum of seventy-five 12s from 157!. So, n = 75.
To make an eighteen, you need two 3s and one 2. So, you can make a maximum of 37 eighteens from 157!
You can try for the other two in a similar fashion 😃
A young girl counted in the following way on the fingers of he left hand .she started calling the thumb 1,index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger7,index finger 8,thumb 9then back to the index finger fo 10,middle finger 11,and so on .she counted upto 1994.she ended on her.
A)thumb B)index finger c)middle finger d)ring finger
WITH SOLUTION PLEASE
:banghead:
You can see the pattern to be
01, 02, 03, 04, 05
09, 08, 07, 06, __
__, 10, 11, 12, 13
17, 16, 15, 14, __
__, 18, 19, 20, __
So, we can see that the middle number is always of the format 4k-1. So, 1995 would be on the middle finger. Now, we have to see if the previous number would be on the ring or the index finger. We can see that for k = 1, the previous number is on the index finger. For k = 2, it is the ring finger and so on. So, for odd 'k's it is the index finger and for even 'k's it is the ring finger. 1995 is 1996 - 1 or (4 * 499) - 1. So, here, k = 499 and so, it should be the index finger.
Option (B)
Am I correct? 😃
A young girl counted in the following way on the fingers of he left hand .she started calling the thumb 1,index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger7,index finger 8,thumb 9then back to the index finger fo 10,middle finger 11,and so on .she counted upto 1994.she ended on her.
A)thumb B)index finger c)middle finger d)ring finger
WITH SOLUTION PLEASE
:banghead:
We can observe, that each count on reaching the thumb would be of the form 8k+1
1993 = 8*(249) + 1
So at 1994, she ended on the index finger.
gud one...
:)
m in tooooo...
A young girl counted in the following way on the fingers of he left hand .she started calling the thumb 1,index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger7,index finger 8,thumb 9then back to the index finger fo 10,middle finger 11,and so on .she counted upto 1994.she ended on her.
A)thumb B)index finger c)middle finger d)ring finger
WITH SOLUTION PLEASE
:banghead:
We can see that she will be at little fingers for numbers 5, 13, 21, ..., i.e, at the count of numbers of form 8k + 5
Also, she will be at thumb for numbers 1, 9, 17, 25, ..., i.e, at the count of numbers of form 8k + 1
So, we should find 1994 (mod 8 ), i.e., remainder when 1994 is divided by 8.
1994 (mod 8 ) = 2, i.e, it is of form 8k + 2
When the count is 8k + 1 she is at thumb, hence for 8k + 2 she will be at index finger.
Hence, when count is 1994, she will at index finger
hey,i m confused abt one concept,
plz chk Pg no 26(Ch 1 no system)
self practice exercise
Q3,n Q 5, Q 6
if we divide 157! by 12^n or 18^n
wat diff steps shud we take...
same goes for 50! divided by 2520^n and 12600^n
plz clarify ppl
Exponent of any prime p in n! is given by:-
E = + + + ...., where is greatest integer less than x
I don' know what the question is as I don't have the book, but I'll try to explain the concept.
Suppose 157! is divisible by i) 12 and ii) 18, then we have find the maximum value of n.
Now, 12 = 2*3
Exponent of 2 in 157! = 78 + 39 + 19 + 9 + 4 + 2 + 1 = 152
=> Exponent of 2 in 157! = 152/2 = 76
Exponent of 3 in 157! = 52 + 17 + 5 + 1 = 75
Since exponent of 3 is less than that of 4, highest value of n will be 75
In case of 18, answer will be 37, as highest exponent of 3 will be = 37 which is lower than 152(exponent of 2)
Similarly for the other case, i.e., 50! by 2520 and 12600.
here 2520 = 8*9*5*7
look for exponent of 8, i.e, 2 in 50!. exponent of 2 in 50! is 47.
=> Exponent of 8 in 50! will be = 15
Similarly for 9, 5 and 7, it comes out to be 7, 12, 8
=> highest value of n will be lowest of 15, 7, 12, 8, i.e, 7
In case of 12600 = 8*9*25*7
Exponents of 8, 9, 25, 7 are 15, 7, 6, 8 respectively. So highest value of n will be 6
NOTE:- You should always find the prime factors of the number whose exponent is required. Also, the formula mentioned at the top of the post is applicable only if we have prime numbers in the denominator, thats why its necessary to break the number in prime factors.
.............
We can see that she will be at little fingers for numbers 5, 13, 21, ..., i.e, at the count of numbers of form 8k + 5
Also, she will be at thumb for numbers 1, 9, 17, 25, ..., i.e, at the count of numbers of form 8k + 1
So, we should find 1994 (mod 8 ), i.e., remainder when 1994 is divided by 8.
1994 (mod 8 ) = 2, i.e, it is of form 8k + 2
When the count is 8k + 1 she is at thumb, hence for 8k + 2 she will be at index finger.
Hence, when count is 1994, she will at index finger
Exponent of any prime p in n! is given by:-
E = + + + ...., where is greatest integer less than x
I don' know what the question is as I don't have the book, but I'll try to explain the concept.
Suppose 157! is divisible by i) 12 and ii) 18, then we have find the maximum value of n.
Now, 12 = 2*3
Exponent of 2 in 157! = 78 + 39 + 19 + 9 + 4 + 2 + 1 = 152
=> Exponent of 2 in 157! = 152/2 = 76
Exponent of 3 in 157! = 52 + 17 + 5 + 1 = 75
Since exponent of 3 is less than that of 4, highest value of n will be 75
In case of 18, answer will be 37, as highest exponent of 3 will be = 37 which is lower than 152(exponent of 2)
Similarly for the other case, i.e., 50! by 2520 and 12600.
here 2520 = 8*9*5*7
look for exponent of 8, i.e, 2 in 50!. exponent of 2 in 50! is 47.
=> Exponent of 8 in 50! will be = 15
Similarly for 9, 5 and 7, it comes out to be 7, 12, 8
=> highest value of n will be lowest of 15, 7, 12, 8, i.e, 7
In case of 12600 = 8*9*25*7
Exponents of 8, 9, 25, 7 are 15, 7, 6, 8 respectively. So highest value of n will be 6
NOTE:- You should always find the prime factors of the number whose exponent is required. Also, the formula mentioned at the top of the post is applicable only if we have prime numbers in the denominator, thats why its necessary to break the number in prime factors.
hi Puys,plz solve this ques :
Q.Brass is an alloy of copper & zinc .Bronze is an alloy containing 80% of copper,4 % of zinc and 16% of tin.A fused mass of brass & bronze is found to contain 74% of copper ,16% of zinc,and 10% of tin.
The ratio of copper to zinc in Brass is :
a)64% & 36% b)33% & 67% c)50% & 75% d)35% & 65% e)none of these
Now we have a for fuse of alloy let it be 100 gm
so it contain..
copper 74 gm(coming from both)
zinc 16%(both)
10 gm Tin(coming from Bronze as it contain tin_)
now 100 gm Bronze gives 16gm tin..
=>100/16*10 gm will give 10 gm tin..
so 62.5 gm of bronze is used..
so it give 4%zinc=62.5*0.04=2.5gm
and 74% cu=50gm
So total Cu in fuse
74gm.(out of which 50gm from bronze)
=>24 gm from Brass
and 16-2.5=13.5gm Zinc from Brass..
so total weight used 37.5 gm of brass
hence after calutaing %gs we get
(A)A66% 36%
one silly ques:
how to convert 6(2/13)% into fraction.. pls generalise in form of x(y/z)%
reposting the question:
how to convert (400/6500)*100 into 6(2/13)%..
reposting the question:
how to convert (400/6500)*100 into 6(2/13)%..
See let say
2% of 100..
so we will write( 2/100)*100=2%
similiarly
(400/6500)*100=80/13%
Now 80 divided by 13 gives quotient 6 and reminder 2..
so we can write it as 6(2/13)%
If you want to convert it back..simply multiply 13*6(quotient)+2(rem)=80..
An A.P P consist of n terms .from the progressions three different progression P1 P2, P3are created such that P1 is obtained by 1st,4th ,7th ....terms of P ,P2 has the 2nd.5th ,8th ...terms of P and P3 has 3rd,6th ,9th ...terms of P.It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original progression P.what are the possible values of n?(with solution please)
reposting the question....
Anyone from Mumbai Central suburbs feels like donating ????............
It seems from the trailing posts that theres a huge lag between demand & supply, so donor's...... wake up, we need Ur contribution......
Anyone from Mumbai Central suburbs feels like donating ????............
It seems from the trailing posts that theres a huge lag between demand & supply, so donor's...... wake up, we need Ur contribution......
OH O !! sorry wrong place to post....
I am finding this problem as tedious maths process....how to do this with out much calculation?
in 1919 ,cricketer scored 891 runs at ana average of 34.27;in 1920 949 runs at an average of 28.75;in 1921 1329 at ana average of 42.87 and in 1922 ,1101 at aan average of 36.70 .what was the batting average in 4 years?
every year number of matches played
1919 = a= 891/34.27 = 26 (approx)
1920 = b = 949/28.75 = 33
1921 = c = 1329/42.87= 31
1922 =d 1101/36.70 = 29
number of matches played in 4 years= 26+33+31+29 = 119
so 4 year avg = (891+949+1329+1101) / 119 = 37 approx