@pyashraj said:@mohitjainShl be Option E: 7, 12 N 17..Let the Amt of Vessels be x-5 kg, x kg n x+5 kg receptively..n let c% be the conc in each Vessel..Given, c/100*(x-5 + x + x + 5)= 5.4, or, cx = 180...(i)Again, c -(x-5) = 8, or, c-x=3, or, c = x+3...(ii)Thus, x^2 + 3x - 180 = 0=>(x+15)(x-12) = 0Thus, x will be 12 kg..Thus, V1= 7 kg, V2= 12 kg n V3= 17 kg..
@mohitjain said:@saurav205@mailtoankit@joyjitpalThe quantities of a sugar solution in three different vessels form an arithmetic progression with a common difference of 5 kg. All the three sugar solutions have the same concentration. The difference between the numerical values of the concentration of sugar in the sugar solution and the quantity of sugar solution in the vessel having the least quantity is 8. The total quantity of sugar in the three vessels is 5.4 kg. What are the quantities of sugar solution in the three vessels? a 5 kg, 10 kg and 15 kg b 6 kg, 11 kg and 16 kg c 7 kg, 11 kg and 18 kg d 8 kg, 13 kg and 18 kg e 7 kg, 12 kg and 17 kg
@ChirpiBird said:did it by options. e.see.. if C% is the conc .then c/100*(7+12+17)=5.4c=15c-7 (the least quantity)=8.15-7= 8only.for other options..a. C=18 and 18-5 = 13..not 8b. C=180/11 and (180/11)-6 is not 8.c. C=15 ...and 15-6=9 again wrongd C= 180/13 ...(180/13)-8 cannot be 8.only e fits.
@Logrhythm said:Chalo mein hi question daal deta hun...If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y isA)40 B)30 c)25 D)None of these
@mailtoankit said:323 ?323 = 17*1920^2004 + 16^2004 - 3^2004 - 1 mod 17 = 020^2004 + 16^2004 - 3^2004 - 1 mod 19 = 0
@MANJULNEOGI said:How many integer values of x & y satisfy the expression 4x+7y=3 where lxl
@raopradeep said:seven boxes numbered 1 to 7 are arranged in a row . each is to be filled by either by black or blue ball such that no two adjacent boxes contain blue ball. in how many ways these boxes be filled ?23333432
@raopradeep said:seven boxes numbered 1 to 7 are arranged in a row . each is to be filled by either by black or blue ball such that no two adjacent boxes contain blue ball. in how many ways these boxes be filled ?23333432
@Shrutim90 said:How many words can be formed of the letters in the word RAINBOW, so that the vowels may occupy only even positions?I solved it in this way: R_N_B_W_These four positions are even positions which can be occupied by 3 vowels.So Ans is 4P3 * 4!But the solution given in the book is:_A_I_O_4! * 3!While I have no beef with the solution given in the book, I don't understand why my approach is incorrect !
Is n^2+3n+5 divisible by 121?....plz share the approach
@maroof10 said:Is n^2+3n+5 divisible by 121?....plz share the approach
@pyashraj said:@maroof10I thnk it will never be divisible by 12..Here, n^2 + 3n + 5 => (n^2 + 2^2 + 2*2*n) - (n-1)=>(n+2)^2 - (n-1)Now, when n is odd...(n+2)^2 is odd and (n-1) is even..Thus, there difference will always give odd..Again, when n is even..(n+2)^2 is even and (n-1) is odd..Thus, there difference will always give odd..Since, in both the cases its odd..It wnt be divisible by 12..Hwever..I thnk n is always a +ve Integers..That's an assumption i am taking..