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# Arithmetic Aptitude :: Simplification

Home > Arithmetic Aptitude > Simplification > General Questions

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51. Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?

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Explanation:

f(2) = f(1+1)=f(1)+f(1)+4(9*1*1 - 1 )= 0+0+4*8 = 32
f(4) = f(2+2)=f(2)+f(2)+4(9*2*2 - 1 )= 32+32+4*35 = 204
f(8) = f(4+4)=f(4)+f(4)+4(9*4*4 - 1 )= 204+204+4*143 = 980
f(16) = f(8+8)=f(8)+f(8)+4(9*8*8 - 1 )= 980+980+4*575 = 4260
f(17)= f(1+16)= f(16)+f(1)+4(9*16*1 -1)= 4260+0+ 4*143= 4832

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Tags:  TCS

52. In this question, A^B means A raised to the power B. Let f(X)=1+X+X^2+....X^6. The remainder when f(X^7) is divided by f(X) is

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Explanation:

f(x)=1+x+x^2+x^3+.........+x^6
When we, put x = 0 in the given functions, then we will have remainder as 1.
f(0) = 1.
1/1 remainder = 0.
For x =1, remainder also will be 0.
But for x >1 (x = 2,3,4.......)
Remainder would be 7.

You can find remainder for x > 1 through this formula,
f(x) = 1+x+x^2+.....+x^(n).
for x > 1,
then the remainder when f(x^(n+1)) is divided by f(x) is (n+1).
n = maximum power of the f(x).

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Tags:  TCS

53. (p/q-q/p)=21/10. Then find 4p/q+4q/p?

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Explanation:

let us take p/q=x and q/p=1/x
x-1/x=21/10
then by solving we get a quadratic equation like 10x^2-10-21x=0
10x^2-25x+4x-10=0
then by taking factors we get two values for x,x=-2/5 and x=5/2
by substituting x=5/2
4(5/2)+4(2/5)=58/5

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Tags:  TCS

54. A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. The value of f(1) is

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Explanation:

Put x =1 => f(1)+2*f(6-1) = 1 => f(1) + 2*f(5) = 1
Put x = 5 => f(5)+2*f(6-5) = 5 => f(5) + 2*f(1) = 5
Put f(5) = 5 - 2*f(1) in the first equation
=> f(1) + 2*(5 - 2*f(1)) = 1
=> f(1) + 10 - 4f(1) = 1
=> f(1) = 3

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Tags:  TCS

55. Find the number of zeroes in the expression 15*32*25*22*40*75*98*112*125

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Explanation:

Maximum power of 5 in the above expression can be calculated like this. Count all the powers of 5 in the above expression. So number of zeroes are 9.

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Tags:  TCS

56. What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3

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Explanation:

take 77!=a
and 54!=b
now the given expression will look like
(a(a-2b)^3 )/(a+b)^3 + (b(2a-b)^3 )/(a+b)^3

( a (a^3 - 6a^2 b + 12ab^2 - 8b^3) + b (8a^3 - 12a^2 b + 6ab^2 - b^3) ) / (a+b)^3
(a^4 - 6a^3 b + 12 a^2 b^2 - 8ab^3 + 8a^3 b - 12a^2 b^2 + 6ab^3 - b^4) / (a+b)^3
(a^4 + 2a^3 b - 2ab^3 - b^4) / (a+b)^3
( a^4 - b^4 + 2ab(a^2 - b^2) ) / (a+b)^3
( (a^2 + b^2)(a^2 - b^2) + 2ab(a^2 - b^2) ) / (a+b)^3
( (a^2 - b^2) (a^2 + b^2 + 2ab) ) / (a+b)^3
( (a^2 - b^2)(a+b)^2 ) / (a+b)^3
( (a+b)(a-b) (a+b)^2 ) / (a+b)^3
( (a+b)^3 (a-b) ) / (a+b)^3
a-b
i.e 77! - 54!

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57. g[0]=1,g[1]=-1,g[n]=2*g[n-1]-3*g[n-2]
Then calculate g[4]= ?

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Explanation:

From the given function g[n] =2*g [n-1]-3*g [n-2]
put values of n as first 2 then 3, then 4 and you will get the answer. G[2] = 2*g[1]-3*g[0]=-2-3 =-5 , G[3]=2*g[2]-3*g[1] =2*(-5)-3*(-1)=-7, G [4] =2*g[3]-3*g[2]=2*(-7)-3*(-5) =1

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Tags:  TCS

58. In the polynomial f(x) =2*x^4 - 49*x^2+54, what is the product of the roots, and what is the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)?

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Explanation:

sum of roots= -(coeff of x^3/coeff of x^4)= -b/a = -0/2=0
product of roots=(-1)^n*(coeff of constant term/coeff of x^4)= e/a = 54/2 =27

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59. If (x-4)/7 + (7-2x)/6 = (8-x)/2 , then x= ?

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Explanation:

(x-4)/7 + (7-2x)/6 = (8-x)/2
(6x-24+49-14x)/42 = (8-x)/2
=> 50-16x = 336-42x
=> 26x = 286
=> x = 11

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60. What should come in place of both question marks?
2.5/? = ?/1.6

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