@vijay_chandola said:@ All: Correct. kuchh jyada hi easy ho gayaIf 'a' is one of the roots of x^5 – 1 = 0 and a ≠ 1, then what can be the value of a^15 + a^16 + a^17 +.......a^50?(a) 1 (b) 5a (c) 35 (d) None of these
1..
@vijay_chandola said:@ All: Correct. kuchh jyada hi easy ho gayaIf 'a' is one of the roots of x^5 – 1 = 0 and a ≠ 1, then what can be the value of a^15 + a^16 + a^17 +.......a^50?(a) 1 (b) 5a (c) 35 (d) None of these
(a) 1 ?
@vijay_chandola said:If the roots of the equation ax3 + bx2 + cx + d = 0 are in Geometric Progression, then which of the following relations is true?(a) a*c^2 = b^2*d(b) a*c^3 = b^3*d (c) a^2*c = b*d^2 (d) a^3*c = b*d^3
x1+x2+x3=-b/a
x1x2+x2x3+x1x3=c/a
x1x2x3=-d/a
x1=u
x2=ur
x3=ur^2
u(1+r+r^2)=-b/a-------------------------------------------1
u^2(r+r^3+r^2)=c/a => u^2*r(1+r+r^2)=c/a------------------------2
u^3r^3=-d/a-------------------3
from 1 and 2
ur=-c/b
put in 3
c^3/b^3=d/a
@vijay_chandola said:Correct!Approach, anyone?
a^5 = 1
so a^15 =1, a^16 = a,....till a^50 = a^35
Now sum = 1 + a*(a^35 - 1)/(a-1) = 1 + 0 as a^35 = 1 as a^5 = 1
= 1 :)
@vijay_chandola said:Correct!Approach, anyone?
a^5=1
a^15 + a^16 + a^17 +.......a^50=a^15(1-a^16)/(1-a)
1(1-a)/(1-a)=1
a^15 + a^16 + a^17 +.......a^50=a^15(1-a^16)/(1-a)
1(1-a)/(1-a)=1
@vijay_chandola said:Correct!Approach, anyone?
a^5-1=0 => a^5=1
a^15+a^16 + a^17 +.......a^50=a^15(a^1+a^2+a^3+a^4+.....a^35)=a^15(a^36-1)/(a-1)=1*(a-1)/(a-1)=1
a^15+a^16 + a^17 +.......a^50=a^15(a^1+a^2+a^3+a^4+.....a^35)=a^15(a^36-1)/(a-1)=1*(a-1)/(a-1)=1
[3^32 / 50] gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The Value of x could be?
a) 2 b) 4 c) 6 d) 8 e) None of these
@padmanabhan1989 said:[3^32 / 50] gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The Value of x could be?a) 2 b) 4 c) 6 d) 8 e) None of these
2?
3^32 will end in 1 and when divided by 50 should give the fractional part as .b2
'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?
(a) 9 (b) 10 (c) 11 (d) None of these
(a) 9 (b) 10 (c) 11 (d) None of these
@padmanabhan1989 said:[3^32 / 50] gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The Value of x could be?a) 2 b) 4 c) 6 d) 8 e) None of these
2
@padmanabhan1989 said:[3^32 / 50] gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The Value of x could be?a) 2 b) 4 c) 6 d) 8 e) None of these
3^32 mod 50=41 mod 50=.82
==> x=.2
@soumitrabengeri said:2?3^32 will end in 1 and when divided by 50 should give the fractional part as .b2
Yes. Thanks. Got it.
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
24k-1
41
@padmanabhan1989 said:[3^32 / 50] gives the remainder and {.} denotes the fractional part of that. The fractional part is of the form (0.bx). The Value of x could be?a) 2 b) 4 c) 6 d) 8 e) None of these
.82, so x = 2, vaise no need to find the remainder as pointed by @soumitrabengeri , just the last digit and you are done :)
@rachit_28 said:.82, so x = 2, vaise no need to find the remainder as pointed by @soumitrabengeri , just the last digit and you are done
I am not finding the remainder..i am just finding the last digit of the value 3^32
Since that comes out to be 1 and division is by 50, the last digit of the decimal has to end in 2
Since that comes out to be 1 and division is by 50, the last digit of the decimal has to end in 2
@soumitrabengeri said:I am not finding the remainder..i am just finding the last digit of the value 3^32Since that comes out to be 1 and division is by 50, the last digit of the decimal has to end in 2
Bhai I found the memainder, and then i said better to follow the approach bu @soumitrabengeri , as he has done just by last digit 😃 :thumbsup:
@rachit_28 said:Bhai I found the memainder, and then i said better to follow the approach bu @soumitrabengeri , as he has done just by last digit
Ok..sorry 
If x, y, and z are positive real numbers, for what ratio of the values of y and z is the value of ( x/y + z/12x + 4y/x + x/3z) the minimum?
Ans is 1/4.
Found this one in the prep section. Could anyone post the solution. Thanks
Ans is 1/4.
Found this one in the prep section. Could anyone post the solution. Thanks
@vijay_chandola said:'X' is the largest sum of rupees which can never be paid using any number of coins of denominations Rs.4, Rs.8, Rs.13 and Rs.18. What is the sum of the digits of 'X'?(a) 9 (b) 10 (c) 11 (d) None of these
9
No is 27
No is 27