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# Simplification Questions

Home > Quantitative Aptitude > Simplification > General Questions
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If a^x= b , b^y= c and c^z = a , then what is the value of x*y*z ?

A1

B3

C2

DCannot be determined

Explanation:

((((a)^x)^y)^z)=a
so.(a)^xyz=a
so xyz=1

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In this question A^B means A raised to the power B. If f(x) = ax^4 - bx^2 + x + 5 and f(-3) = 2, then f(3) =

A1

B-2

C3

D8

Explanation:

Putting the value of x= -3 in equation ax^4 - bx^2 + x + 5=2 we get 81a=9b

Now putting the values of x=3 again in equation

ax^4 - bx^2 + x + 5

81a-9b + 3 + 5

9b-9b + 3 + 5=8  therefore the answer is 8.

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What is the minimum value of abs(187m - 396n - 526) as m, n take all integer values? Here abs is the absolute value function (that is, if x > 0, then abs(x) = x and if x < 0, then abs(x) = - x).

A0

B9

C2

D1

Explanation:

187m-396n-526=187(m-2n-3)-22n+35
consider
m-2n-3=1
187(1)-22n+35=222-22n
the value =2 when n=10
m-2n-3=1
n=10
m=24
m,n are integers

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The remainder when (7)84 is divided by 342 is :

A0

B1

C49

D341

Explanation:

784 = (343) 28
Now (343) º 1mod (342) Þ (343)28 = 128 mod (342) = 1mod (342).

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Number of positive integers valued pairs (x, y) satisfying 4x - 17y = 1, x is <=1000 is :

A59

B57

C55

D58

Explanation:

4x - 17y = 1 and x £ 1000 Þ x = (1 + 17y) /4 Þ (1 + 17y)/4 £ 1000
Þ 1 + 17y £ 4000 Þ 17y £ 3999 Þ y £ 3999/17 Þ y £ 235.
Considering multiples of 4 (for x), we get 235/4 = 58.75. Hence 58 such pairs can exist.

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If |r - 6| = 11and |2q - 12| = 8, then what could be the minimum possible value of q / r ?

A-2/5

B3/14

C11/12

D2/17

Explanation:

Solution: |r-6|=11

Removing the mod bracket r-6= 11 and r-6= -11  so r values are 17, -5

Similarly  2q-12= 8 and 2q-12 = -8 so q values are 10, 2

So minimum possible value is -2/5.

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The remainder when m + n is divided by 12 is 8, and the remainder when m - n is divided by 12 is 6. If m > n, then what is the remainder when mn divided by 6?

A3

B4

C2

D1

Explanation:

m+n+8 and m+n+6 is completely divisible by 12.
m+n=8 ----- (1)
m-n=6 ------(2)
By, solving these two equation we get, m=7 & n=1
mn = 7
so when we divide 7 by 6 , we get remainder 1.

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1(1!)+2(2!)+3(3!)....2012(2012!) = ?

A2013!-1

B2011!-1

C2012!-1

DNone of these

Explanation:

1(1!)=1 => 2!-1
1(1!)+2(2!)=1+4=5 => 3!-1
1(1!)+2(2!)+3(3!)=1+4+18=23 => 4!-1
........................
.......................
1(1!)+2(2!)+3(3!)+........+2012(2012!)=2013!-1

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If X^Y denotes X raised to the power of Y, find out last two digits of (2957^3661)+(3081^3643)

A42

B38

C98

D22

Explanation:

(2957^4)^3660 *2957 +(3081)^3643
last 2 digits of (57)^4 is 01
so (01)^3660 yields last two digits as 01
{since the unit's digit is always 1 and tens digit is obtained by 0*0=0}
so 2957*01 gives last 2 digits 57
similarly (81)^3643 gives ten's digit as the unit digit of 8*3 ie 4
so 57 +41=98

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if f(x) is a function for all real numbers x holds good for maximum of 2x+4 and 12+3x. Then what is the value of x so that f(x) is equal to 2x+4.

A9

B-9

C-7

D0

Explanation:

maximum (2x+4 ,12+3x)
and ques says maximum is 2x+4
if x= -9 2x+4 = -14 and 12+3x= -15 maximum is -14
Value of x = -9

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