HEY plz. sm1 help....
q no.93 lod-2
how many integer values of x and yare there such that 4x+7y=3while x,ya.144 b.141 c.143 d.142
ans is 142
ans is 142
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 the number of terms n.
1. The number of terms can be 16
2. The number of terms can be 5
3. The number of terms can be 25
4. The number of terms can be 20
What would be the correct option.
Plz, help me to solve it..Thanks in Advance..
x(x+1)(x+2)(x+3)...(x+(n-1))=1000, then which of the following cannot be true about the number of terms n.
1. The number of terms can be 16
2. The number of terms can be 5
3. The number of terms can be 25
4. The number of terms can be 20
What would be the correct option.
Plz, help me to solve it..Thanks in Advance..
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 the number of terms n.
1. The number of terms can be 16
2. The number of terms can be 5
3. The number of terms can be 25
4. The number of terms can be 20
What would be the correct option.
Plz, help me to solve it..Thanks in Advance..
is it option 2?
my approach...take x = 1, then you get 1*2*3*4...
but if no. of terms be 5 you get 120. hence it cannot be true...
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 the number of terms n.
1. The number of terms can be 16
2. The number of terms can be 5
3. The number of terms can be 25
4. The number of terms can be 20
What would be the correct option.
Plz, help me to solve it..Thanks in Advance..
the ans is option 2.
as 1000 should have 3 zero,the terms on LHS must have 3 5's.so,there cannot be 3 5's in the 5 consecutive numbers less than 125.
There are 5 blue socks, 4 red sochs and 3 green socks. In how many ways 4 socks could be selected?
$U!\!$H|!\!E SaysThere are 5 blue socks, 4 red sochs and 3 green socks. In how many ways 4 socks could be selected?
12c4 = 495 ways
$U!\!$H|!\!E SaysThere are 5 blue socks, 4 red sochs and 3 green socks. In how many ways 4 socks could be selected?
case 1 - all the four are same - 2 ways
case 2 - 3 are same and one different - 3C1 * 2 = 6 ways
case 3 - 2 are same and 2 are different - 3C1 * 2C2 = 6 ways
case 4 - 2 are same of one kind and 2 of other kind - 3C1*2C1/2 = 3
hence total ways is 2 + 6 + 6 + 3 = 17 ways.
is this the ans?
but if all are taken as different entity then thrr are 12C4 ways
case 1 - all the four are same - 2 ways
case 2 - 3 are same and one different - 3C1 * 2 = 6 ways
case 3 - 2 are same and 2 are different - 3C1 * 2C2 = 6 ways
case 4 - 2 are same of one kind and 2 of other kind - 3C1*2C1/2 = 3
hence total ways is 2 + 6 + 6 + 3 = 17 ways.
is this the ans?
but if all are taken as different entity then thrr are 12C4 ways
Yes.. U are right..this is a case of selecting entities from identical items..
Now try to solve this:(it's interesting)
In how many ways one or more of 5 different letters be posted to 4 different letter boxes?
I will post the answer later..
Good luck!!
yes the answer is 2...how did u approached .could you pls tell and where i was doing the mistake..
Hi all,
this is a problem from alligations.
a man purchased a cow and a calf for rs. 1300.He sold the calf at a profit of 20% and the cow at a profit of 25%.In this way ,his total profit was 23 1/3%(mixed fraction).find the cost price of the cow..LOD 1 prob 7.
Can anybody solve it by alligation method.Not by basic ..
I have solved it this way.
as 23 1/3 % (1300) = 300.therefore total amont after selling both the cow and calf = 1300 + 300 =1600.
therefore we can assume the avergae price of the calf and cow to be 800. (1600/2).
now let us assume the c.p. of cow be x .therefore the c.p of calf =1300 - x.
solving by the number line method explained in alligation .
1300-x --------------800 ------------x
------------------------------------------
1.20------------------------------- 1.25
therefor the equation will be
(x-800)/(800-1300+x) = 1.20/1.25
by solving for x i am getting the answer to be 8000. but the answer is 800.how..please tell me whats wrong in this approach..
Thanks
KAC8L591 SaysIs the ans 2
yes the answer is 2.But how..can you please solve...Thanks menace and kano for your efforts...
Now try to solve this:(it's interesting)
In how many ways one or more of 5 different letters be posted to 4 different letter boxes?
I will post the answer later..
Good luck!!
first letter has 4 boxes to choose from, similarly the second and so on....
so the no. ways =4*4*4*4*4
=4^5
Is this correct pls confirm....
cheers....
first letter has 4 boxes to choose from, similarly the second and so on....
so the no. ways =4*4*4*4*4
=4^5
Is this correct pls confirm....
cheers....
No dude it's not the correct one..
Carefully read the question it says "One or More" of 5 letter to be posted..
Check for possible scenarios in which he can keep the letter to himself i.e, he will not post the letter (only constraint is he has to post at least one letter)
Good Luck..
In how many ways one or more of 5 different letters be posted to 4 different letter boxes?
5^5 - 1
No dude it's not the correct one..
Carefully read the question it says "One or More" of 5 letter to be posted..
Check for possible scenarios in which he can keep the letter to himself i.e, he will not post the letter (only constraint is he has to post at least one letter)
Good Luck..
for 1 letter he has 4 options.
for 2 letters 4^2
for 3 letters 4^3
for 4 letters 4^4
for 5 letters 4^5
total cases=4 + 4^2 + 4^3 + 4^4 + 4^5
=1364
I feel this approach may be correct....
sir pls help....
Now try to solve this:(it's interesting)
In how many ways one or more of 5 different letters be posted to 4 different letter boxes?
I will post the answer later..
Good luck!!
each letter can be posted in 5 ways...i.e in 4 boxes and also 1 way not to post it,
hence total ways is 5^5 ways but in this there is 1 way in which no letter is posted so v need to subtract tht...hence the required ans is 5^5 -1
Hi all,
this is a problem from alligations.
a man purchased a cow and a calf for rs. 1300.He sold the calf at a profit of 20% and the cow at a profit of 25%.In this way ,his total profit was 23 1/3%(mixed fraction).find the cost price of the cow..LOD 1 prob 7.
Thanks
boss Average profit is 23 1/13 and nt 23 1/3
prfit from cow 25
prfit frm calf 20
let x n y be price of cow n calf respectively thn x + y = 1300
now also using alligations
x(25-23 1/13) = y(23 1/13 - 20)
x(1 12/13) = y (3 1/13)
x/y = 8/5
hence price of cow is 8/13 * 1300 = 800
each letter can be posted in 5 ways...i.e in 4 boxes and also 1 way not to post it,
hence total ways is 5^5 ways but in this there is 1 way in which no letter is posted so v need to subtract tht...hence the required ans is 5^5 -1
Perfect answer.. Really appreciate..
What is the sum of all 5 digit numbers which can be formed with the digits 4,3,2,1,0 without repetition?
Have a try..Good luck..