x is a positive integer divisible by 4; as x increases from 1824 to 18

Question

Competition Mode Question

x is a positive integer divisible by 4; as x increases from 1824 to 1896, which of the following must decrease?

I.  4x^2 - 4x + 4

II.  -10 - \frac{1}{x^2}

III.  \frac{4}{x^2}

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) None

unraveled · Answer

x is a positive integer divisible by 4; as x increases from 1824 to 1896, which of the following must decrease?

I.  4x^2−4x+4 
II.  −10−\frac{1}{x^2} 
III.  \frac{4}{x^2}

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) None

If x is in numerator the entity must increase.
If x is in denominator the entity must decrease.

In  4x^2−4x+4   x^2  would contribute more towards increase than 4x does towards decrease.
In  −10−\frac{1}{x^2}   x^2  would increase but since it is in denominator form it decreases  \frac{1}{x^2} . Thus,  −10−\frac{1}{1824^2}  <  −10−\frac{1}{1896^2} 
Here in  \frac{4}{x^2}  as  x^2  increases  \frac{4}{x^2}  decreases.

Answer D.

chetan2u · Answer

x is a positive integer divisible by 4; as x increases from 1824 to 1896, which of the following must decrease?

I.  4x^2 - 4x + 4.........4(x^2-x+1)=4(x(x-1)+1) ..As x increases as a positive integer, x(x-1) will also increase and so 4(x(x-1)+1) will also INCREASE.
II.  -10 - \frac{1}{x^2} ..As x increases as a positive integer, x^2 will also increase and so when x^2 is in the denominator,  \frac{1}{x^2}  will DECREASE, but since we are dealing with NEGATIVE  \frac{1}{x^2} , the value will INCREASE as it will be closer to 0.
III.  \frac{4}{x^2} ..As x increases as a positive integer, x^2 will also increase and so when x^2 is in the denominator,  \frac{1}{x^2}  will decrease, so  \frac{4}{x^2}  will also DECREASE

Only III

C

Leadership · Answer

x increases from 1824 to 1896, which of the following must decrease?

I.  4x^2−4x+4   
As X is a positive integer, the increasing integer values will result in increasing result

II.  −10−\frac{1}{x^2} 
As we keep on increasing x, the  \frac{1}{x^2}  will keep on decreasing. So, there will be less addition of -ve of  \frac{1}{x^2}  value to -10. Hence, the total result will increase

III.  \frac{4}{x2} 
The denominator will increase as the X value increases and hence, the total result will decrease

Only III - 'C' is the winner

eakabuah · Answer

Given that x is divisible by 4. We are to determine which of the following functions must decrease as x increases from 1824 to 1896.

I. 4x^2 - 4x + 4 = 4(x^2 - x + 1)
Since I is positive function of x^2, and a negative function of x in I, when x increases, the increasing effect of x^2 exceeds the decreasing effect of x for all values of x apart from 0. Hence increasing from x=1824 to x=1896 will lead to a net increasing effect on I. Hence I cannot decrease.

II. −10−1/x^2 = -(10+1/x^2)
With this function, when x increases, (10+1/x^2) reduces, but since every value of x is less than -10, the higher the value of x, closer the result of II will get to -10. So, the net effect of increasing x from 1824 to 1896 is that II will increase and get closer to -10.

III. 4/x^2.
The net effect of increasing x on this function is that III decreases and approaches 0. So as x increases from 1824 to 1896, III reduces.

So III is the only expression whereby x has a reducing effect on the results of the function when x increases from 1824 to 1896.

The answer is C.

Archit3110 · Answer

value of expression will decrease for ii & iii
IMO D

x is a positive integer divisible by 4; as x increases from 1824 to 1896, which of the following must decrease?

I. 4x2−4x+4

II. −10−1/x^2

III4/x2

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) None

exc4libur · Answer

I. 4x^2−4x+4: increase
II. −10−1/x^2: -1/4^2 = -1/16; -1/8^2 = −1/64; −1/64 > −1/16 = increase
III. 4/x^2: 4/16 = 1/4; 4/64 = 1/16; 1/16 < 1/4 = decrease

Ans (C)

freedom128 · Answer

x=1824, 1828,...,1896

I. 4(x^2-x+1) has a minimum value at x=1/2. As x>1/2 goes further away from x=1/2, the value of 4(x^2-x+1) further increases. 
INCREASE

II. -10 - 1/x^2
As x>0 moves further away from x=0, the value of -1/x^2 increases and that of -10-1/x^2 does too. 
INCREASE

III. 4/x^2
As x>0 moves further away from x=0, the value of 4/x^2 decreases
DECREASE

FINAL ANSWER IS (C) 
III only

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lacktutor · Answer

x is a positive integer divisible by 4; as x increases from 1824 to 1896, which of the following must decrease?

I.  4x^{2} —4x +4 =( 2x—1)^{2} +3 
( must increase)

II.  —10 —\frac{1}{x^{2}} = —(10+ \frac{1}{x^{2}}) 
As x increases,  10+ \frac{1}{x^{2}}  will decrease. But  —(10+\frac{1}{x^{2}})  will increase.

III.  \frac{4}{x^{2}}  will decrease

Only III will give the right solution

The answer is C

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Bunuel · Answer

OFFICIAL EXPLANATION

Answer: C

Although this question can be solved using algebra, it is significantly easier to solve by picking small numbers and observing the changes.

Although x is defined as a positive integer divisible by 4 from 1824 to 1896, there is no reason you cannot seek to determine whether the equations increase by using x = 2 and x = 4. The equations will behave the same for x = 2 and x = 1824. To be safe and convince yourself that the patterns between 2 and 4 will hold, you could also check x = 6, although this is not necessary and will consume extra time.

Equation f(2) f(4) f(6)

I.  f(x) = 4x2 – 4x + 4 :  12 ;  52 ;  124 .

II.  f(x) =-10 - \frac{1}{x^2} :   -10 – \frac{1}{4} ;  -10 – \frac{1}{16} ;  -10 – \frac{1}{36}

III.  f(x) =\frac{4}{x^2} :  1 ;  \frac{1}{4} ;  \frac{1}{9} .

Evaluate equation I. As x increases, equation I must increase. Any answer choice that includes I is wrong.

Evaluate equation II. The pattern that emerges is that as x increases, a smaller number is subtracted from a negative number. Subtracting a smaller number from a negative number actually makes the overall value increase (e.g., -10 – 20 = -30, which is less than -10 – 15 = -25). Consequently, as x increases, equation II must increase. Any answer choice that includes II is wrong.

Evaluate equation III. Both by looking at the equation and by observing the values, it is clear that as x increases, the value of equation III decreases since four is being divided by a larger number.

luisdicampo · Answer

Deconstructing the Question

We need to determine which expression must decrease as  x  increases from  1824  to  1896 .

The expressions are

\frac{4}{x^2} - \frac{4}{x} + 4

-10 - \frac{1}{x^2}

\frac{4}{x^2}

We analyze how each expression changes when  x  becomes larger.

Step-by-step

Expression III is

\frac{4}{x^2}

As  x  increases,  x^2  increases, so the fraction becomes smaller. Therefore the expression must decrease.

Expression II is

-10 - \frac{1}{x^2}

As  x  increases, the fraction

\frac{1}{x^2}

decreases. That means we subtract a smaller number, so the overall value increases rather than decreases.

Expression I is

\frac{4}{x^2} - \frac{4}{x} + 4

Here

\frac{4}{x^2}

decreases, but

-\frac{4}{x}

increases as  x  increases, while  4  stays constant.

Rewriting gives

4 + \frac{4}{x^2} - \frac{4}{x}

For large values of  x , the increase of  -\frac{4}{x}  dominates, so the expression increases.

Answer: C

x is a positive integer divisible by 4; as x increases from 1824 to 18

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x is a positive integer divisible by 4; as x increases from 1824 to 18

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