see 5N is cuming as 5pd+260 but here 4 is cuming as remaindr n we knw 5pd is divisible by d and 260 will giv remaindr as 4 only for the nos. whch are factors of 256......but it should be grtr thn 52.....factors can be 64,128,256......256=2^8.....factors grtr thn 52 will be 2^6,2^7,2^8... hope it will make u undrstnd
5N = bd + 4..(1) N = ad + 52..(2) Multiply 2 by 5 5N = 5ad + 260 Now 260 mod Any Number Must Leave 4.... So Largest Number = 256... Rest Numbers Are Factors of 256... So total Factors = 9 But 1,2,4 will not suffice... Also Value less than 52 will not suffice.... So Values of d = 3
Let a, c 2, 4, 6, 8, 10, 12 and b 22, 24, 26, where a, b and c are distinct. Find the number of equations of the form ax2 + bx + c = 0, that can be formed such that the equation has real roots.
Let a, c 2, 4, 6, 8, 10, 12 and b 22, 24, 26, where a, b and c are distinct. Find the number of equations of the form ax2 + bx + c = 0, that can be formed such that the equation has real roots.
Let a, c 2, 4, 6, 8, 10, 12 and b 22, 24, 26, where a, b and c are distinct. Find the number of equations of the form ax2 + bx + c = 0, that can be formed such that the equation has real roots.
a)15 b)18 c)45 d)90 e)36
For the roots to be real D>=0 b^2-4ac>=0 now we can see that the maximum values of a and b possible are 10,12 and the minimum value of b = 22 22 ^2 > 4* 12 *10
11^2 > 10*12
Hence we can take all possible values from the given without any eliminations.
Ways of selecting for A n B are 6C2 * 2 Ways of selecting B are 3 hence Answer should be 6C2 * 2 *3 Hence the answer is 90.
There are 4 balls to be put in five boxes where each box can accommodate any number of balls. In how many ways can one do this if 1.Balls are similar and box are different. 2.. ball are diff and boxes are similar 3.. both ball and box are similar 4.both ball and boxes are differ
Let a, c 2, 4, 6, 8, 10, 12 and b 22, 24, 26, where a, b and c are distinct. Find the number of equations of the form ax2 + bx + c = 0, that can be formed such that the equation has real roots.
a)15 b)18 c)45 d)90 e)36
answer is 90.. as for real roots b^2>=4ac..WHICH BTW MEAN D=B^2-4AC>=0 so there r 30 cases each 4 evry value of b... in al 90 values!!
Let a, c 2, 4, 6, 8, 10, 12 and b 22, 24, 26, where a, b and c are distinct. Find the number of equations of the form ax2 + bx + c = 0, that can be formed such that the equation has real roots.
a)15 b)18 c)45 d)90 e)36
answer is 90.. as for real roots b^2>=4ac.. so there r 30 cases each 4 evry value of b... in al 90 values!!
Dear ankitrajpal, For the roots to be real the Determinant of a quadratic equation has to be zero. Determinant of a quad eqn ax^2 + bx + c = 0 is b^2 - 4ac which should be greater than zero. Just put the max values of a, b and minimum values of b and you can see that the problem is just a problem based on the selection of values from a given set of values.
We have 3 possible values for b sine we have to select from 22, 24 and 26 ..............(1) Now since a and b cannot have the same value. Hence we can select two number from the given 6 numbers in 6C2 ways. ...............(2) Since a and b are different hence we can get 2 cases for each ...............(3) Multiplying Eqns 1 , 2 and 3 we get the desired result.
There are 4 balls to be put in five boxes where each box can accommodate any number of balls. In how many ways can one do this if 1.Balls are similar and box are different. 2.. ball are diff and boxes are similar 3.. both ball and box are similar 4.both ball and boxes are differ
Dear ankitrajpal, For the roots to be real the Determinant of a quadratic equation has to be zero. Determinant of a quad eqn ax^2 + bx + c = 0 is b^2 - 4ac which should be greater than zero. Just put the max values of a, b and minimum values of b and you can see that the problem is just a problem based on the selection of values from a given set of values.
We have 3 possible values for b sine we have to select from 22, 24 and 26 ..............(1) Now since a and b cannot have the same value. Hence we can select two number from the given 6 numbers in 6C2 ways. ...............(2) Since a and b are different hence we can get 2 cases for each ...............(3) Multiplying Eqns 1 , 2 and 3 we get the desired result.
I hope the solution is clear this time.:)
@freak!!! for real roots determinant,D>=0..and not just equal 2 0..as d=0 gives real but equal roots nd v r lookin for real roots may or may not b equal!!!nd for dat d>=0.. chk it out.. Quadratic equation - Wikipedia, the free encyclopedia.. hope it helps