78% of the chocolates available were distributed equally among students. The remaining chocolates were defective. Find the number of students if 96 chocolates were defective and each student received 6% of the total chocolates distributed. a) 17 b) 20 c) 18 d) 19 e) 21
In a right circular cone of radius 4cm and height 10cm, largest cube is inscribed whose base rests at base of the cone. A sphere is then inserted between top of cube and the vertex of cone. What can be the maximum radius of the sphere?Team BV--Pratik Gauri
This is not a square in a Cone/triangle problem. So be wary of the 3D arrangement.
The upper 4 vertices of the Cube will touch the Cone at 4 distinct points and the diagonal of the Cube will be same as the Dia of the Conical circle in which the Upper face of the Cube is contained.
If cube edge length is L, then L*root(2) = 2* radius of the base of the upper cone in the same plane
So one can determine l by using this relation -> (10-l) / (l/root2) =10/4=5/2
Also, (10-l-r) cos Q = r, where r is the radius of the sphere and cosQ = 2/root(29)
maine to aajtak 3D mein film bhi nahi dekhiHow many integers b/w 1 and 10000 have the sum of their digits equal to 18 ?PS: @bodhi_vriksha sir ur method to solve above type of q's got over my head , bouncer
See the observation presented in my prior post :)
By the way, regarding this problem -> Integers may be of 1 digit to 4 digit.
a+b+c+d = 18, with 0
Coeff of x^18 in (1+x+x^2+..x^9)^4
=Coeff of x^18 in (1-x^10)^4 * (1-x)^(-4 )
=Coeff of x^18 in (1- 4C1*x^10+ 4C2*x^20+..) * (1+4C1*x+5C2x^2+...)
If the number is abc, then for values of a from 1 through 7, the possibilities are Summation (4 to 10) For a = 8 and 9, possibilities are 9 and 8 respectively.So total such numbers = 66Team BV - Vineet
Alternate way to solve this one:
a+b+c = 16, where 1
Coeff of x^16 in (x+x^2+…x^9)(1+x+x^2+..x^9)^2 =Coeff of x^15 in (1+x+x^2+…x^8)(1+x+x^2+..x^9)^2 =Coeff of x^15 in (1-x^9)(1-x^10)^2 *(1-x) ^(-3) =Coeff of x^15 in (1-x^9)(1-2*x^10+x^20) *(1+3C1*x+4C2*x^2+5C3*x^3+…) = 1*1*17C15 – 1*2*7C5 – 1*8C6
Alternate solution for this one (but you wont be able identify/appreciate the interesting pattern )Sum of coefficients of x^1 to x^16 in the expansion of (x+x^2+ €Śx^9)(1+x+x^2+..x^9)^2= (17C15+16C14+15C13+...+2C0) - 2*(7C5+6C4+5C3+4C2+3C1+1) - (8C6+7C5+6C4+5C3+4C2+3C1+1) Note: Interesting thing to note here again is. Let's say we are looking for coefficient of x^16 and x^(28-16=12)17C5-2*7C5-8C5 = 13C11 - 2*3C1 - 4C2 = 66, reinforcing the fact that N(x)=N(28-x) Team BV - Vineet
I think if we introduce a dummy variable then less calculations will be there
Coeff of x^16 in (x + x^2 + ... + x^9)*(1 + x + ... + x^9)^2*(1 + x + x^2 + ...) = Coeff of x^15 in (1 - x^9)(1 - 2x^10 + x^20)*(1 - x)^(-4) = Coeff of x^15 in (1 - x^9 - 2x^10)(1 - x)^(-4) = C(18, 3) - C(9, 3) - 2C(8, 3) = 620
@bodhi_vriksha sir isko solution nahi hua hai..plzz help..OA is 66 i think..
No of ways of doing so is equivalent to choosing 15 fruits out of5 apples, 10 mangoes, and 15 oranges.
No of ways will be given by:-
Coeff of x^15 in (1 + x + ... + x^5)(1 + x + ... + x^10)(1 + x + ... + x^15) = Coeff of x^15 in (1 - x^6)(1 - x^11)(1 - x)^(-3) = Coeff of x^15 in (1 - x^6 - x^11)(1 - x)^(-3) = C(17, 2) - C(11, 2) - C(6, 2) = 66
Alternatively (although its similar to previous one):- a + b + c = 15, a C(17, 2) ways when a > 5, a' + b + c = 9 C(11, 2) ways When b > 10, a + b' + c = 4 C(6, 2) ways
1> How many 3's will be there in a book having 700 pages ?2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
1> How many 3's will be there in a book having 700 pages ?2> 5 couples in a party .how many ways 4 persons can be seated on a table of 4 chairs such that no two husband wife are on the table ?
zaroori nahi bhai....u r missing the case in which other terms seventh digit will some up to 'x0'
