@Narci said:33 ?
no OA is D
@Pradeep_Rss said:Mr x has 3 varieties of rice ie 15 rs ,18 rs ,22 rs per kg.he mixes it and sells at 22 rs a kg and gains 10 % profit. if he mixes 4 kg of 1st variety what is the diffrence between amount of 2 nd and 3 rd variety.help me out without taking X ..... thanks in advance.
I am getting the difference as 10... solved using alligation... What is the OA?
@ankitpurohit991 said:I am getting the difference as 10... solved using alligation... What is the OA?
could u explain it?
@The_Loser said:For how many natural numbers, N the expression N ˛ + N + 1 results in a perfect square?
0 ?
My explanation if I am right:
Least perfect square would be (N + 1)^2 and on expanding it needs N^2 + 2N + 1 which will never be possible.
My explanation if I am right:
Least perfect square would be (N + 1)^2 and on expanding it needs N^2 + 2N + 1 which will never be possible.
@chintu61 said:prime number is always odd except 2 ,( odd^4 = odd )+ (15*odd^2=odd) , so finally odd+odd-1=even, vch will be never a prime !! so "0" values, check for p=2, its value is even 75 vch isn't prime
bhai (odd + odd - 1) != even
odd + odd is even
odd + odd is even
Rs 22 is the SP, so CP=Rs 20 (10%profit)
mix 1st 2 varieties..
15............?? .......................18 |_________|______________|
4kg......................................xkg
apply alligation-> we get ??=15+(3x/4+x)
now it is the same case as mixing (x+4)kg of another variety whose price is Rs15+(3x/4+x) with the 3rd variety of Rs 22 such that the resulting mixture is of Rs 20...
15+(3x/4+x)...........20 .................22
|________________|____________|
(x+4)kg .........................................ykg
applying alligation....
y/x+4= (20-15+(3x/4+x))/2
solve... u'll get y-x=10
@Dexian said:bhai iska solution post karo na
bhai just use a^3+b^3+c^3 - 3abc
where a=m, b=n and c=33
@Dexian said:bhai iska solution post karo na
We know that a³ + b³ + c³ = 3abc, if a + b + c = 0 and vice-versa
Since here, m³ + n³ + (-33)³ = 3*m*n*(-33)
=> m + n - 33 = 0
=> m + n = 33
Since, mn ≥ 0 we can say that there will be 35 ordered pairs of (m, n)
@chillfactor Sir kindly help me be clear with a concept.
3x + 2y + z = 15
how do i find the positive integral solutions, non negative integral solutions and integral solutions of this question?
3x + 2y + z = 15
how do i find the positive integral solutions, non negative integral solutions and integral solutions of this question?
@Logrhythm said:bhai just use a^3+b^3+c^3 - 3abc where a=m, b=n and c=33
can u please tell me how 35 solutions are there???
m+n=33
m can vary from 1- 32 ?????
@yudh said:canu please tell me how 35 solutions are there???m+n=33m can vary from 1- 32 ?????
solve m+n=33,
34C1 soluntions for m+n=33, + 1 for m=n=-33
@iLoveTorres said:@chillfactor Sir kindly help me be clear with a concept.3x + 2y + z = 15how do i find the positive integral solutions, non negative integral solutions and integral solutions of this question?
positive solutions=6
non negative solutions=17 ?
In a right angle triangle PQR , QS is an altitude to PR, /_PQR=90 , PS=16 cm , SR=4 cm , QS=?
@pankaj1988 said:In a right angle triangle PQR , QS is an altitude to PR, /_PQR=90 , PS=16 cm , SR=4 cm , QS=?
8
given: ax+by =7................(i)
ax2 + by2 = 49..................(ii)
ax3 +by3 = 133................(iii)
ax4 +by4 = 406................(iv)
to find: 2002(x+y +xy) + (a+b)/21
.
.
from (i) and (ii) we get b= 7(x-7)/(xy - y^2) ........(v)
from(ii) and (iii) we get b = 7(7x-19)/(xy^2 - y^3).....(vi)
from(iii) and (iv) we get b = 7(19x-58)/(xy^3 - y^4)....(vii)
.
.
equating the values of b derived above we get
.
(x-7) = (7x-19)/y = (19x-58)/y^2.................(viii)
.
=>(x-7)*(19x-58)/y^2 = (7x-19)^2/y^2 [multiply the LHS and RHS = square the middle]
=>(x-7)(19x-58) = (7x - 19)^2
=>x= 3 or x = -0.5
.
put x = 3 in (viii) to get
(3-7) = (7*3 - 19)/y
=>y= -0.5
so we put x=-0.5 and y=3 in (i) and (ii) to get
.
-0.5a+3b = 7
0.25a+9b = 49
=>21b = 63
=>b= 5 => a = 16
.
hence 2002(x+y+xy) + (a+b)/21 = 2002*(-0.5 +3 -0.5*3) + (16+5)/12 = 1
.
.
PS: COrrected for some calc. error.
ATDH.
