@onlytj said:164 , values 125,5,34
Please explain how you found the three values !!
@kingsleyx said:Please explain how you found the three values !!
@krum ka solution dekh lo. Same approach.
How about this ... Question attached ...
@SufyS said:How about this ... Question attached ...
x+1.125x+1.125^2x=(200000-26400)*1.125^3
=>(1.265625+1.125+1)x=173600*1.125^3
=>x=247176.5625/3.390625
=>x=72900
=>(1.265625+1.125+1)x=173600*1.125^3
=>x=247176.5625/3.390625
=>x=72900
@Zedai said:6?? aur kuch nahi dimag mein aa raha
@krum said:i gave up on 6 too
@nick_baba said:i'm in...
(a^2 + b^2 + c^2 + d^2) = (a+b+c+d)^2 - 2(ab+ac+ad+bc+bd+cd) = 40
For d to be greatest, a,b,c must be least and equal. Let a = b = c = x
=> x = (8-d)/3 and 3[(8-d)/3]^2 + d^2 = 40
=> d^2 - 4d - 14 = 0
=> d(max) = 2 + 3sqrt(2)
.There is a cuboidical dice with each of its faces numbered with a positive integer. The product of the numbers on the faces converging at each corner is writtern at the respective corner.The sum of all the products is 1995. If the sum of all the numbers (on the six faces) is s, what is the number of possible values of s?
(A) 6 (B) 12 (C) 18 (D) 24
(A) 6 (B) 12 (C) 18 (D) 24
A circular dart board of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point X in the circle. What is the probability that X is closer to the center of the circle than the periphery?
@vijay_chandola @soumitrabengeri @Zedai @rkshtsurana @surajsrivastav @sauravd2001 @grkkrg @gs4890
here is the ans.
Let, x, y and z are in AP.
Let, x = y – d and z = y + d
Hence, the given equation becomes;
(y – d)^2 + y^2 + (y + d)^2 = (–d) × (–d) × (2d)
∴ 3y^2 + 2d^2 = 2d^3
∴ 3y^2 = 2(d – 1)d^2
So to have integer solutions, we need 2(d – 1) to be a number which is 3 times a perfect square.
This is satisfied if d – 1 = 6n^2 for some integer n.
Thus d = 6n^2 + 1 and 3y^2 = d^2 × 2(6n^2)
∴ y^2 = 4d^2n^2
Without loss of generality, we can assume;
y = 2dn = 2n × (6n^2 + 1).
∴ x = y – d = (2n – 1) × (6n^2 + 1), and z = y + d = (2n + 1) × (6n^2 + 1)
So (x, y, z) = ((2n – 1)(6n2 + 1), 2n(6n2 + 1), (2n + 1) (6n2 + 1) ) is the required solution for integer n.
I.e. for every integer n, there is at least one solution.
Hence, there are infinitely many integer solutions for the given equation.
Hence, option 4.
Let, x = y – d and z = y + d
Hence, the given equation becomes;
(y – d)^2 + y^2 + (y + d)^2 = (–d) × (–d) × (2d)
∴ 3y^2 + 2d^2 = 2d^3
∴ 3y^2 = 2(d – 1)d^2
So to have integer solutions, we need 2(d – 1) to be a number which is 3 times a perfect square.
This is satisfied if d – 1 = 6n^2 for some integer n.
Thus d = 6n^2 + 1 and 3y^2 = d^2 × 2(6n^2)
∴ y^2 = 4d^2n^2
Without loss of generality, we can assume;
y = 2dn = 2n × (6n^2 + 1).
∴ x = y – d = (2n – 1) × (6n^2 + 1), and z = y + d = (2n + 1) × (6n^2 + 1)
So (x, y, z) = ((2n – 1)(6n2 + 1), 2n(6n2 + 1), (2n + 1) (6n2 + 1) ) is the required solution for integer n.
I.e. for every integer n, there is at least one solution.
Hence, there are infinitely many integer solutions for the given equation.
Hence, option 4.
@kingsleyx said:A circular dart board of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point X in the circle. What is the probability that X is closer to the center of the circle than the periphery?
1/4 ?
@SufyS said:How about this ... Question attached ...
You can always use this formula for these type of questions:
Each Installment = (P.r/100)/[1-(100/100+r)^n]
here P = 200000 - 26400 = 173600
r = 12.5
n = 3
So, Each Installment is 72900...
@kingsleyx said:A circular dart board of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point X in the circle. What is the probability that X is closer to the center of the circle than the periphery?
1/4? whats the OA?
@kingsleyx said:A circular dart board of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point X in the circle. What is the probability that X is closer to the center of the circle than the periphery?
Consider a circle at half the radius of the dartboard.
If the point X lies inside this circle or on it, it is closer to the center than the periphery, otherwise false...
find the ratio of the area of the smaller to the larger circle.
pi(1/2)^2 / pi(1)^2 = 1/4
If the point X lies inside this circle or on it, it is closer to the center than the periphery, otherwise false...
find the ratio of the area of the smaller to the larger circle.
pi(1/2)^2 / pi(1)^2 = 1/4
@rkshtsurana i knw. mera b dimag nai chala itna to. but yeah the ans is right. i thot sum1 will post a better approach.
@kingsleyx said:.There is a cuboidical dice with each of its faces numbered with a positive integer. The product of the numbers on the faces converging at each corner is writtern at the respective corner.The sum of all the products is 1995. If the sum of all the numbers (on the six faces) is s, what is the number of possible values of s?(A) 6 (B) 12 (C) 18 (D) 24
18?