we have to select atleast 3 women and atleat 1 of then has to president or VP. We can select : either 4 women and 1 man OR 3 women and 2 men.
selection of these possibilities is given by bracket 1 and 2... now, for 4 W & 1 M, there is just a single man in the team...n he cannot simultaneously be president and VP. so atleast 1 women to be pres or vp is automatically satisfied..
now, consider 3 W & 2 M, there are 2 men on the team..and these 2 men could hold the position of president and vice president...which we do not want... so...selecting 3 W & 2 M , such that men hold position of president and VP can be done in (4C2 x 4C3 x 2C2 x 2!)..in all other ways of allocating 3 W and 2 M for various positions..atleast 1 women would be president or VP. so ..total ways = all poss ways - ways you do not want... hence u subtract d 3rd bracket from sum of 1st 2... explanation may not have been good...but i hope u got d point.. i
Can Someone plz help me with this? A group of 8 man r to b chosen from 12 men for rowing out of which 3 men can row on one side only & 2 can row on the other side only. Find the numbr of ways of arranging the crew with 4 members on each side. 40320 30240 60480 10080 PS: the answer given is 60480 but can someone plz give proper explaination.:confused:
Let number of coconuts be n, let the person take from one end , then go to middle then the other end and so on. first time the distance traveled is (n-1)*10 for second n-2 *10 ..... now the total distance traveled = sum of 1 to n-1 *10
that is n(n-1)/2 =820 n(n-1)=1640=40*41 ans 3
Suppose if there are three coconuts in that row , so according to your equation the distance travelled shd be (3-1)*10 = 20
where as if he starts from one end then he has to cover 10 m till the centre ... he will keep that coconut there and then again travel 10 m to and fro = 20 m .... thus he has travelled a total of 30 m ..... if i am going wrong .. i would like to know where ...
Let number of coconuts be n, let the person take from one end , then go to middle then the other end and so on. first time the distance traveled is (n-1)*10 for second n-2 *10 ..... now the total distance traveled = sum of 1 to n-1 *10
that is n(n-1)/2 =820 n(n-1)=1640=40*41 ans 3
I dint understand as to how sum from 1 to n-1 is n(n-1)/2.
Accordingto my understanding it should be (n-1)(n-2)/2. Please correct me if I am wrong
Can Someone plz help me with this? A group of 8 man r to b chosen from 12 men for rowing out of which 3 men can row on one side only & 2 can row on the other side only. Find the numbr of ways of arranging the crew with 4 members on each side. 40320 30240 60480 10080 PS: the answer given is 60480 but can someone plz give proper explaination.:confused:
hey puys .... no takers for this one !! hey puy veterans plz help me too !!!
#1.A train moves at a constant speed of 120kmph for one kilometre and at 40kmph for the next 1km.What is the average speed of train? a)48kmph b)50kmph c)80kmph d)72kmph e)None
....???
#2.A person saves 6% of his income.Two years later, his income shoots up by 15% but his savings remain the same. Find the hike in expenditure. a)15.95% b)15% c) 14.8% d)15.5% e)None
#1.A train moves at a constant speed of 120kmph for one kilometre and at 40kmph for the next 1km.What is the average speed of train? a)48kmph b)50kmph c)80kmph d)72kmph e)None
My solution to this one
Average Speed = Total distance / total time = 2 / (1/120 + 1/40) = 2/ (4/120) = 2/(1/30) = 60 km/hr
So answer is E None of these .... plz lemme know where am i goin wrong
#2.A person saves 6% of his income.Two years later, his income shoots up by 15% but his savings remain the same. Find the hike in expenditure. a)15.95% b)15% c) 14.8% d)15.5% e)None
Suppose that a persons income is 100Rs then savings = 6 and expenditure = 94 (assuming that the remains is the expenditure) Now 15 % hike means 115 Rs and 6 % of 115 is 6.9 thus exp is 115 - 6.9 = 108.1
now 108.1/94 is 1.15 means that there was a 15 % hike in the exp .... thus option b = 15 %
Are the answers mentioned in the above post the correct ones ? ... if so then can some one point out my mistake please
[quote=Fuzon;1117172#2.A person saves 6% of his income.Two years later, his income shoots up by 15% but his savings remain the same. Find the hike in expenditure. a)15.95% b)15% c) 14.8% d)15.5% e)None
Suppose that a persons income is 100Rs then savings = 6 and expenditure = 94 (assuming that the remains is the expenditure) Now 15 % hike means 115 Rs and 6 % of 115 is 6.9 thus exp is 115 - 6.9 = 108.1
now 108.1/94 is 1.15 means that there was a 15 % hike in the exp .... thus option b = 15 %
Are the answers mentioned in the above post the correct ones ? ... if so then can some one point out my mistake please
Mistake in the above colored text.
15% hike means 115 correct!!! But savings remain same i.e. 6 Rs. So, exp = 109
and hike in exp(%age) = (109-94)*100 /94=1500/94 which is little less than 16. Hece answer is a)15.95%
hey puys .... no takers for this one !! hey puy veterans plz help me too !!!
first go for the right hand side :
fix the 3 sailors who are gonna row the boat only on the right side : and then u have 3 out of 4 places filled on the right side now you can pick the one sailor in 7C1 ways ( as 12- 3 already picked -2 who can row only on the left side =7)
Now go for the left side :-
here fix the 2 sailors who gonna row only left and then u have 2 places to fill by choosing among 6 sailors (coz 1 out of the 7 has already been selected on the other side) so number of ways = 6C2
Now each side (right and left which each have 4 rowers) can be arranged among themselves in 4! ways.
so total answer= 7C1 * 6C2 * 4! * 4! = 60,480
A tip : Always try to figure out the combinations and then when u get the combinations you can permute or arrange them.
Hi can anyone help me out with this question from Number Systems:-
Find the gcd(111...11 hundred ones ; 111...11 sixty ones) Options:- a)111..11 forty ones b) 111..11 twenty five ones c) 111..11 twenty ones d) None of these
fix the 3 sailors who are gonna row the boat only on the right side : and then u have 3 out of 4 places filled on the right side now you can pick the one sailor in 7C1 ways ( as 12- 3 already picked -2 who can row only on the left side =7)
Now go for the left side :-
here fix the 2 sailors who gonna row only left and then u have 2 places to fill by choosing among 6 sailors (coz 1 out of the 7 has already been selected on the other side) so number of ways = 6C2
Now each side (right and left which each have 4 rowers) can be arranged among themselves in 4! ways.
so total answer= 7C1 * 6C2 * 4! * 4! = 60,480
A tip : Always try to figure out the combinations and then when u get the combinations you can permute or arrange them.
good solution dude..agreed that answer is correct...but i have 1 doubt...
we know that 3 people can row only on 1 side and 2 ppl on the other side..from this i interpret that i have 9 ppl to choose from on 1 side and 10 ppl to choose from other side...and 7 ppl could be other side..
my question is that why have you fixed those 3 ppl on 1 side n 2 on other side??as in it could be any 4 ppl from 9 on side 1..meaning u could choose 1, 2 , 3 or none from those 3 and correspondingly fill up d remaining seats on side 1... n depending upon your selection on side 1...similar combinations for side 2...
actually i tried it out..sum becomes too confusing then...
Hi can anyone help me out with this question from Number Systems:-
Find the gcd(111...11 hundred ones ; 111...11 sixty ones) Options:- a)111..11 forty ones b) 111..11 twenty five ones c) 111..11 twenty ones d) None of these
ANS: c
1111..100 times can be written as (10^100-1)/9 and 1111...60 times can be written as (10^60-1)/9 ..gcd of 100 and 60 is 20..n both have (10^20-1)/9 ie 111..20 times as factor..therefore gcd is 111..20 times
Please help in solving this .My apologies if this q was posted earlier. The question is from chapter on number systems.
Q.Suppose the product of n consecutive integers is x.(x+1)(x+2)(x+3)......(x+(n-1)) =1000 then which of the following cannot be true about number of terms n
a.The number of terms can be 16 b.The number of terms can be 5 c.The number of terms can be 25 d.The number of terms can be 20
Please help in solving this .My apologies if this q was posted earlier. The question is from chapter on number systems.
Q.Suppose the product of n consecutive integers is x.(x+1)(x+2)(x+3)......(x+(n-1)) =1000 then which of the following cannot be true about number of terms n
a.The number of terms can be 16 b.The number of terms can be 5 c.The number of terms can be 25 d.The number of terms can be 20
Please help in solving this .My apologies if this q was posted earlier. The question is from chapter on number systems.
Q.Suppose the product of n consecutive integers is x.(x+1)(x+2)(x+3)......(x+(n-1)) =1000 then which of the following cannot be true about number of terms n
a.The number of terms can be 16 b.The number of terms can be 5 c.The number of terms can be 25 d.The number of terms can be 20
1000 has 16 factors ( 1000= 2^3*5^3)..but product of consecutive 3 integers or more is a multiple of 3...and 1000 is not a multiple of 3...so how can we have 1000 as product of consecutive integers.. or am i missing a trick here..
1000 has 16 factors ( 1000= 2^3*5^3)..but product of consecutive 3 integers or more is a multiple of 3...and 1000 is not a multiple of 3...so how can we have 1000 as product of consecutive integers.. or am i missing a trick here..
u r right..1000= product of consecutive integers is not possible.
