In how many ways can one put 7 black balls in 4 boxes colored red,orange,blue and green?
@bs0409 said:In how many ways can one put 7 black balls in 4 boxes colored red,orange,blue and green?
r + o + b + g = 7 [where r,o,b,g >=0]
=> 10C3 = 120 ways ?
If each box needs 1 ball, then 6C3 = 20 ways ?
@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?
sp of 1 orange = 80/45 . it is 20% less than cp. so cp is greater than sp by 25 % ie ,cp= 5/4 (80/45). we need to sell at 20 % gain => he need to sell each at 6/5(5/4(80/45) = each at 8/3 rs. so for 48 he sell 48 * 3/8 = 18
@somnathbhatta said:@pyashraj@YouMadFellowwhatif boxes can remain empty also ?
Yeah, so that's what I assumed. If they remain empty, it is the first case so 120 ways.
@somnathbhatta
U'll find all ur queries resolved here..
http://www.pagalguy.com/news/permutations-combinations-cat-2011-distribute-box-chocolates-a-18150 
@YouMadFellow @pyashraj 
sry . did the calculation assuming that no boxes remain empty and thought that i didn't considered it ..
sry . did the calculation assuming that no boxes remain empty and thought that i didn't considered it ..
@bs0409 said:In how many ways can one put 7 black balls in 4 boxes colored red,orange,blue and green?
C(10, 3)
@YouMadFellow : Ab to chhod do is thread ko :splat:
Interviews mai ni poochhenge quant :P
@vijay_chandola said:C(10, 3)@YouMadFellow : Ab to chhod do is thread ko Interviews mai ni poochhenge quant
LOL..........really good one..........
From a circular sheet of paper with a radius of 20 cm, four circles of radius 5cm each are cut out. What is the ratio of the uncut to the cut portion?
(1) 5 (2) 3 (3) 6 (4) 4 (5) none of these
From a circular sheet of paper with a radius of 20 cm, four circles of radius 5cm each are cut out. What is the ratio of the uncut to the cut portion?
(1) 5 (2) 3 (3) 6 (4) 4 (5) none of these
@bs0409
Should be 3.. Required Ratio= (pi*20^2 - pi*5^2*4)/pi*4*25 = 3..
Or it is more than meets the eye..
@bs0409 said:LOL..........really good one..........From a circular sheet of paper with a radius of 20 cm, four circles of radius 5cm each are cut out. What is the ratio of the uncut to the cut portion? (1) 5 (2) 3 (3) 6 (4) 4 (5) none of these
3 ??
@bs0409 said:LOL..........really good one..........From a circular sheet of paper with a radius of 20 cm, four circles of radius 5cm each are cut out. What is the ratio of the uncut to the cut portion? (1) 5 (2) 3 (3) 6 (4) 4 (5) none of these
(pi*20^2-pi*5^2*4)/pi*5^2*4=3
what annual payment will discharge a debt of Rs.580 due in 5 years , the rate being 8% per annum?
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I was stuck to this problem. Can't solve it the faster way.
Logical explanation please.
@cadmium said:what annual payment will discharge a debt of Rs.580 due in 5 years , the rate being 8% per annum? --I was stuck to this problem. Can't solve it the faster way.Logical explanation please.
145.26
Solve the following equation:
580=x/(1+0.08)+x/(1+0.08)^2+x/(1+0.08)^3+x/(1+0.08)^4+x/(1+0.08)^5
Solve the following equation:
580=x/(1+0.08)+x/(1+0.08)^2+x/(1+0.08)^3+x/(1+0.08)^4+x/(1+0.08)^5
Let S be the set of first 14 natural numbers. A special subset of S is a subset S' which satisfies the following three properties
a) S' has exactly 8 elements
b) If x belonging to S is even, then x is in S' if and only if x/2 is in S'
c) If y belonging to S is odd, then y is in S' if and only if (y+15)/2 is in S'
Let X denotes elements of S that cannot be the part of special subset. Then n(X) (i.e. number of elements in X) equals
(1)2 (2) 3 (3) 5 (4) 6 (5) none of these
a) S' has exactly 8 elements
b) If x belonging to S is even, then x is in S' if and only if x/2 is in S'
c) If y belonging to S is odd, then y is in S' if and only if (y+15)/2 is in S'
Let X denotes elements of S that cannot be the part of special subset. Then n(X) (i.e. number of elements in X) equals
(1)2 (2) 3 (3) 5 (4) 6 (5) none of these
@bs0409 said:145.26Solve the following equation:580=x/(1+0.08)+x/(1+0.08)^2+x/(1+0.08)^3+x/(1+0.08)^4+x/(1+0.08)^5
Seems it can't be solved without calculator. Right?
Or is there any faster method of solving the figures like (xxx)^5 and like that?
@cadmium said:Seems it can't be solved without calculator. Right?Or is there any faster method of solving the figures like (xxx)^5 and like that?
U r right. Cannot be solved without a calculator. but u can do some approximations and use options to solve it.
@bs0409 said:U r right. Cannot be solved without a calculator. but u can do some approximations and use options to solve it.
Thanks bro.