@krum said:40 = 2^3*5so we need to express this as product of 3 numbers - 4+2 = 630 ordered pairstaking into account -ve integers = 30*4 = 120
30 ordered pairs?? i dont understand
@ravi.theja said:If x, y, z are any real numbers then the minimum possible value of x2 + 2y2 + z2 + 2yz subject of x + 2y + z = €“6 isa)-6b)6c)12d)None of these
x^2 + 2y^2 + z^2 + 2yz
=x^2 +y^2 +(y+z)^2
also (x+2y+z) = x+y+(y+z)
so put (y+z) = k
now the problem is reduced to minimize (x^2+y^2+k^2) subject to (x+y+k)=-6
for using cauchy inequality we can take pairs as (1,x) , (1,y) and (1,k)
=>(1^2 + 1^2 +1^2)*(x^2 + y^2 +k^2) >= (1*x + 1*y + 1*k)^2
=> 3*(x^2+y^2+k^2) >= (x+y+k)^2
=> 3*(x^2+y^2+k^2) >= (-6)^2
=>(x^2 + y^2 +k^2) >= 12
ATDH.
@cadmium said:In what ratio should water be mixed with a liquid costing Rs.12 per litre,so as to make a profit of 25% by selling the diluted liquid at Rs.13.75 per litre?
say x litres is added to 1 litre of liquid
acc to ques
(13.75+13.75x-12)/12*100=25
==> 13.75x=3-1.75=1.25
==> x=1/11
so every 11 part liquid should have 1 part water
acc to ques
(13.75+13.75x-12)/12*100=25
==> 13.75x=3-1.75=1.25
==> x=1/11
so every 11 part liquid should have 1 part water
@krum said:40 = 2^3*5so we need to express this as product of 3 numbers - 4+2 = 630 ordered pairstaking into account -ve integers = 30*4 = 120
You are right......
In how many ways can 6 people queue up at 4 ticket counters?
In how many ways can 6 people queue up at 4 ticket counters?
@cadmium said:Explanation please.
Profit is 25%.. therefore cp is rs 11
now using alligation
12 0
11
11 1
here we assume cost of water to be zero.
Therefore ratio of liquid to water is 11:1 hence water to liquid is 1:11
@bs0409 said:You are right......In how many ways can 6 people queue up at 4 ticket counters?
each person can go 2 any of the 4 queues.... Therefore the answer shud b 4^6
@krum said:2304 = 2^8*3^2from 1200 - 2200 we need all those which are not multiple of 2 or 3multiples of 2 - 499multiples of 3 - 333multiples of 6 - 166499+333-166 = 666999-666=333OA?
Dont hav OA....I only created the Q to get this concept.
@bvdhananjay said:each person can go 2 any of the 4 queues.... Therefore the answer shud b 4^6
dis qstn is same as the case of 6 letters gng into 4 envelopes..wich is done in envlps ^ letters i.e., 4 ^6 ways :)
@ravi.theja said:dis qstn is same as the case of 6 letters gng into 4 envelopes..wich is done in envlps ^ letters i.e., 4 ^6 ways
Yes it is.. Bt ur comment made me think abt my soln a second time.
@bs0409
Should be 6!*9C3..
x1 + x2 + x3 + x4 = 6..Thus 9C3 ways..n in each of these ways, people can arrange themselves in 6!..Thus, 6!*9C3..
@ravi.theja said:dis qstn is same as the case of 6 letters gng into 4 envelopes..wich is done in envlps ^ letters i.e., 4 ^6 ways
It is nt this case.. M so sorry... my answer is wrong.
Q. By selling 45 oranges for Rs.80, a man loses 20%.
How many should he sell for Rs.48 to gain 20% in the transaction?
@ravi.theja in the envelopes^letters type .. only 1 letter can be placed in an envelope at a tym. so that works. here the people can use queue up in any number in any counter .
@pyashraj
@pyashraj
@cadmium said:Q. By selling 45 oranges for Rs.80, a man loses 20%. How many should he sell for Rs.48 to gain 20% in the transaction?
if SP = 80 and loss is 20% then clearly CP = 100.
Now if we want to make a profit of 20% then my SP must be 120.
so to make 20% profit he should sell 45 oranges for 120
=> for 48 rs he should sell 45*(48/120) = 18.
ATDH.
@somnathbhatta said:@ravi.theja in the envelopes^letters type .. only 1 letter can be placed in an envelope at a tym. so that works. here the people can use queue up in any number in any counter . @pyashraj
sry 😃 we can use it for postboxes nt for envelopes 😃 @anytomdickandhary
Wow!
Thanks a ton !!!
@cadmium
Given, SP of 45 Oranges = 80..Thus, SP of 1 Orange = 16/9..
Since, the shopkeper is suffering a loss of 20%..CP of 1 orange = 5/4*16/9 = 20/9..
Let, n be the required number of Oranges....Thus, SP of 1 Oranges = 48/n..
Given, 20/9*6/5 = 48/n, or, n=18..