Miscellaneous

PFA the solution

40 rs. and 10 paise?

discriminant cannot be a prime no, hence option A

why can't it be a prime number?

Discriminant of a second degree polynomial ax² + bx + c ( say ) is b² + 4ac

now square of an integer is either of the form 4k or 4k+1 so b²+4ac will never be of the form of 4k + 3.

we know ,x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

for given values 

(log

p⁶ - p = (m² + m + 6)(p - 1)

p( p⁵ - 1) = (m² + m + 6)(p - 1) 

p ( p⁴ + p³ + p² + p + 1 )(p - 1) = (m² + m + 6)(p - 1) 

p ( p⁴ + p³ + p² + p + 1 ) = (m²+ m + 6)  

If p is a prime greater than 2 then LHS is an odd number , but RHS is even for all integer value of m 

So no solution for p> 2.

When p = 2 then m² +  m - 56 = 0 so m= -8, 7 Only one solution m=7 ; p=2 

x + 1/x = - √3 

Cubing both sides we get 

x³ + 1/x³ + 3 ( x + 1/x) = - 3√3 

x³ + 1/x³ -  3 √3 = -3√3 

x³ + 1/x³ = 0 

x^6 + 1 = 0. x³ 

x^6 = -1 

x^42 + x^48 + x^54 + x ^60 + x^66 + x ^72 

( -1)^7+ (-1)^8 + ( -1)^9 + ( -1)^10 + ( -1)^11 + ( -1)^12 

0 . 

got it (y)

x + 1/x = - √3 

x³ + 1/x³ = 0

x^42 + x^48 + x^54 + x ^60 + x^66 + x ^72 

= x^45(x³ + 1/x³) + x^57(x³ + 1/x³) + x^69(x³ + 1/x³)

= 0

16)

D = b^2- 4ac 

Any square is of the form 4k or 4k+1 

So D can't be of the form 4k+3 hence option 1

17. For each element listed, there is exactly one other element such that the two elements sum to 11. Thus, we can list all the 10 numbers above as 5 pairs of numbers, such that each pair sums to 11. The problem then can be solved as follows:

in any given subset with no two elements summing to 11, at most one element from each pair can be present. Thus, there are 3 ways in which each pair can contribute to a given subset (no element, the first element in the pair, or the second element in the pair). Since there are 5 pairs, the total number of ways to construct a subset with no two elements summing to 11 is 3⁵ = 243.

98

97th answer is 5292 ?

Yes, answer for 97 is correct. Can you explain your method?

Va at nth second will be = 2+(n-1)3         {since it forms an a.p with a=2 and d=3}

Vb at nth second will be = 3+ (n-1)d 

Vb=10Va

3+(n-1)d= 10(2+(n-1)d) 

(n-1)(d-30)= 17

since n>2 

n=18 ,d=31

now find sum upto n terms of both a.p

Sa= 9(4+17*3) =495

Sb= 9(6+17*31) =9*533 =4797

total distance = 495+4797= 5292 (answer)

4th line R.H.S is     

10(2 +(n-1)3)

Alternative  Approach: 

Volume of pyramid : 1/3 of area of base x height 

So 1/3 x 2562 x 40 - 1/3 x 144 x 30 

In order to divisible by 6. 

sum of digits should be a multiple of 3 and last digit should be even. 

so, when 0 is not included
2346 => 4! numbers subtract cases when 3 is at units place 
4!-3! = 18 numbers 

When 2 is not included : 3460 => sum 13 , not divisible by 3 

When 3 is not included : 2460=> sum 12 

Total 4! numbers subtract cases when 0 is at thousands place : 24 - 6 = 18

When 4 is not included : 2306 => sum 11 , not divisible by 3

When 6 is not included : 0234 => sum 9

14  numbers 

Total : 14+ 18 + 18 =50 numbers