Show that x^2 + y^2 + z^2
Let x; y and z be real numbers in the interval [ ˆ'2;1] such that x+y+z = 0.
@bodhi_vriksha said:1, 0, 9 are already fixed for G, H and D.And as sum of all the digits is 45 which needs to be divided among two parts (on both sides of =), but they can't be equal as 45 is odd.So clearly there difference (of the form 9k) has to be odd. That is, the carry 'k' has to be odd i.e. 1 or 3.Carry of '1' means, sum of digits of LHS = 27 and that of RHS = 18.Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 17 which can be achieved only by 9 + 8. But '9' has already been used, so this case is not possible.Carry of '3' means, sum of digits of LHS = 36 and that of RHS = 9.Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 8 which can be achieved by 2 + 6 or 3 + 5 alternately.And in each addition step, we need '1' carry as hundred's place 9 is getting '1' carry. And by adding ten's place digits, we need exactly '1' carry to be added in 9. So, certainly, by adding unit digits we must get a carry of '1' so that total requirement of carry of '3' is met.Thus with all four possible values (2, 3, 5, 6) of unit digit of sum i.e. J, we can have(i) A + C + F + J = 12 + 2 = 14 (12 can be achieved by 3 + 4 + 5)(ii) A + C + F + J = 13 + 3 = 16 (13 can be achieved by 2 + 4 + 7 or 2 + 5 + 6)(iii) A + C + F + J = 15 + 5 = 20 (15 can be achieved by 2 + 6 + 7 or 3 + 4 + 8)(iv) A + C + F + J = 16 + 6 = 22 (16 can be achieved by 3 + 5 + 8 or 4 + 5 + 7)So basically there are many cases but all possible values of A + C + F + J are 14, 16, 20 and 22 only. Team BV - Kamal Lohia
Hi Kamal,
Sorry I didn't follow the approach. Why the values of G, H and D are already fixed ??? and how can we say that the carry is odd ??
Sorry I didn't follow the approach. Why the values of G, H and D are already fixed ??? and how can we say that the carry is odd ??
@gautam22 said:every even number in dis case is of d form 3*2nnow remove all even number 3*2n /2 = 3n(including number)now remove all multiples of 6 = 3*2n/6 = n(not the number) 6n-4n = 2n6n/3 = 2n therefore all even numbersnow wen its of d form 3*(2n-1)the only number possible is 3 @ScareCrow28
Bhai check karo.. n=6
here the numbers that are less than 6 and co-prime = 1 and 5 = 2 numbers = n/3 numbers
Directly co-prime wala formula bhi use kar sakte ho
@ScareCrow28 said:Let x; y and z be real numbers in the interval [ ˆ'2;1] such that x+y+z = 0. Show that x^2 + y^2 + z^2
take the maximum and minimum end points
x=-2 and y=z=1
x^2+y^2+z^2 = 4+1+1 = 6
lo bhai ho gaya show 
@Subhashdec2 said:(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)A=-2(xy+yz+zx)xy+yz+zx should be maximum with a negative sign to get the highest value of Aif i put x as -2 y=2 and z=0we will get A as 8...is the question correct?
check the bold part... (y=2 is not available)
@Subhashdec2 said:(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)A=-2(xy+yz+zx)xy+yz+zx should be maximum with a negative sign to get the highest value of Aif i put x as -2 y=2 and z=0we will get A as 8...is the question correct?
Range [-2,1] hai..can't put y=2
@gautam22 said:pata nahi kya method dikhaoon par -2,1,1 se ho jayega
yehi method hoga yaar
see, we need to maximize the sum of three squares....for that take the maximum magnitude wala number (which is -2 here) and iske sath sum ko bhi zero karna hai so 1 and 1 lena padega sath mein obviously...
so this is by default the maximum value of x^2+y^2+z^2.....baaki saari isse kam hi hongi...
@pavanp1990 said:Hi Kamal,Sorry I didn't follow the approach. Why the values of G, H and D are already fixed ??? and how can we say that the carry is odd ??
Hi Pavan
First try to get the logic please.
"If we add 2 and 3, the sum is 5 i.e. 2 + 3 = 5, which has same sum for LHS as well as RHS as 5, right.
Now when we add 5 and 7, the sum is 12. i.e. 5 + 7 = 12
Let's compare 'sum of digits'. It is (5 + 7) i.e. 12 for LHS but (1 + 2) i.e. 3 for RHS.
Now this is not same.
I hope you can easily gauge the reason for this.
Yes..reason is the carry which we have taken while adding.
We have taken carry of 1 because of 10...i.e. it was counted as 10 in LHS but it is being counted as 1 only in RHS, creating a decrease of (10 - 1) i.e. 9 in 'sum of digits' of RHS in coparison with LHS.
Now question is, whether there will always be decrease of 9 in 'sum of digits' whenever there is a carry in the addition process.
Answer is YES. Whenever there is a carry of '1' in the addition process, it'll create a decrease of '9' in the 'sum of digits' of RHS in comparison with LHS.
Or we can state it now as "the difference in 'sum of digits' of LHS and RHS is always going to be multiple of 9 i.e. of the form 9n where 'n' gives the number of carries being taken in the addition process"
And we can also infer that LHS and RHS are always equal modulo-9"
First try to get the logic please.
"If we add 2 and 3, the sum is 5 i.e. 2 + 3 = 5, which has same sum for LHS as well as RHS as 5, right.
Now when we add 5 and 7, the sum is 12. i.e. 5 + 7 = 12
Let's compare 'sum of digits'. It is (5 + 7) i.e. 12 for LHS but (1 + 2) i.e. 3 for RHS.
Now this is not same.
I hope you can easily gauge the reason for this.
Yes..reason is the carry which we have taken while adding.
We have taken carry of 1 because of 10...i.e. it was counted as 10 in LHS but it is being counted as 1 only in RHS, creating a decrease of (10 - 1) i.e. 9 in 'sum of digits' of RHS in coparison with LHS.
Now question is, whether there will always be decrease of 9 in 'sum of digits' whenever there is a carry in the addition process.
Answer is YES. Whenever there is a carry of '1' in the addition process, it'll create a decrease of '9' in the 'sum of digits' of RHS in comparison with LHS.
Or we can state it now as "the difference in 'sum of digits' of LHS and RHS is always going to be multiple of 9 i.e. of the form 9n where 'n' gives the number of carries being taken in the addition process"
And we can also infer that LHS and RHS are always equal modulo-9"
@ScareCrow28 said:Range [-2,1] hai..can't put y=2
a+b/2 >= rt(ab)
a^2 + b^2 + 2ab >= ab
so a^2 + b^2 similarly
b^2 + c^2 c^2 + a^2
Now add 1 , 2 and 3.
2(a^2 + b^2 + c^2 )
now substitute -2 , 1 , 1 in a, b and c
a^2 + b^2 + 2ab >= ab
so a^2 + b^2 similarly
b^2 + c^2 c^2 + a^2
Now add 1 , 2 and 3.
2(a^2 + b^2 + c^2 )
now substitute -2 , 1 , 1 in a, b and c
mera mistake,
concept simple , arithmetic mean greater than or equal to geometric mean.
a^2 + b^2 + 2ab >= 4ab
a^2 + b^2 >= 2ab
There is a certain mixture of milk & water. 12 liters is drawn out from a mixture and replaced with water.after 2 such cycles the ratio of milk:water is 9:7 what is the volume of liquid with which he started .
@Pradeep_Rss said:There is a certain mixture of milk & water. 12 liters is drawn out from a mixture and replaced with water.after 2 such cycles the ratio of milk:water is 9:7 what is the volume of liquid with which he started .
You must know the initial concentration of milk in the mixture, otherwise it can't be done.
Assuming initial solution to be X liters of 100% Milk,
9/16 = (1 - 12/X)^2
i.e. (1 - 12/X) = 3/4 = 1 - 1/4 = 1 - 12/48.
i.e. X = 48.
Team BV - Kamal Lohia
Assuming initial solution to be X liters of 100% Milk,
9/16 = (1 - 12/X)^2
i.e. (1 - 12/X) = 3/4 = 1 - 1/4 = 1 - 12/48.
i.e. X = 48.
Team BV - Kamal Lohia
@Pradeep_Rss said:There is a certain mixture of milk & water. 12 liters is drawn out from a mixture and replaced with water.after 2 such cycles the ratio of milk:water is 9:7 what is the volume of liquid with which he started .
Cannot be determined.
There are two unknowns
Let the quantity of the milk be x and total quantity of the liquid be t.
The ratio of the milk after two such cycles is
(x-12/t)^2 = 9 / 16
x-12 / t = 3 / 4
4x - 48 = 3t
A Baba went to a Jungle with a certain number of toffees. He distributed the toffees equally among the Monkeys; he was left with 7 toffees. Next day he carried the same number of toffees and found that there was twice the number of Monkeys present on the first day. On dividing them equally among the Monkeys, he was left with 47 toffees. Find the number of Monkeys present on the first day
@pavanp1990 said:Cannot be determined.There are two unknownsLet the quantity of the milk be x and total quantity of the liquid be t.The ratio of the milk after two such cycles is (x-12/t)^2 = 9 / 16x-12 / t = 3 / 44x - 48 = 3t
Sorry, My mistake .
If C is the concentration after a dilutions, V is the original volume and x is the volume of liquid. Replaced each time then
C = ( V - X / V ) ^ n.
so 3t is nothing but 3x , Hence 48.
If C is the concentration after a dilutions, V is the original volume and x is the volume of liquid. Replaced each time then
C = ( V - X / V ) ^ n.
so 3t is nothing but 3x , Hence 48.
@vijay_chandola said:Cross-check is required I guess.. Here we are getting only one value of m. And for that values of y are -2,2,4.
This is incorrect. For m=-1, there will be only 2 different roots.
Hint: Two cases are possible
1> one of the equations has a double root while the other doesn't
2> both the equations have a common root
2> both the equations have a common root
Team BV - Vineet
@bodhi_vriksha said:Guys here is first one for today:
If the equation (y^2 - 2my -4(m^2 + 1))(y^2 - 4y -2m(m^2 + 1)) = 0 has exactly 3 different roots, then how many values can "m" take?
(a) 4 (b) 3 (c) 2 (d) 1Team BV - Vineet
Guys, this problem is yet to be solved correctly...
Team BV - Vineet
@bodhi_vriksha said:Guys, this problem is yet to be solved correctly...Team BV - Vineet
y^2-2my-4(m^2+1)=0
y^2-4y-2m(m^2+1)=0
y^2-4y-2m(m^2+1)=0
these two will have one common root
y^2-2my-4(m^2+1)=y^2-4y-2m(m^2+1)
y(4-2m)=(m^2+1)(4-2m)
y=m^2+1
since y has 3 different values so will m
hence 3??