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# 99,998^2 − 2^2 =

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Math Expert
Joined: 02 Sep 2009
Posts: 52433

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19 Sep 2018, 22:36
00:00

Difficulty:

55% (hard)

Question Stats:

66% (01:08) correct 34% (01:38) wrong based on 82 sessions

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$$99,998^2 − 2^2 =$$

(A) $$10^{10} − 4$$

(B) $$(10^5 − 4)^2$$

(C) $$10^4(10^5 − 4)$$

(D) $$10^5(10^4 − 4)$$

(E) $$10^5(10^5 − 4)$$

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Joined: 11 Sep 2015
Posts: 3360
Re: 99,998^2 − 2^2 =  [#permalink]

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20 Sep 2018, 05:50
1
Top Contributor
Bunuel wrote:
$$99,998^2 − 2^2 =$$

(A) $$10^{10} − 4$$

(B) $$(10^5 − 4)^2$$

(C) $$10^4(10^5 − 4)$$

(D) $$10^5(10^4 − 4)$$

(E) $$10^5(10^5 − 4)$$

Key concept: x² - y² = (x + y)(x - y)

So, 99,998² - 2² = (99,998 + 2)(99,998 - 2)
= (100,000)(99,996)
= (100,000)(100,000 - 4)
= (10⁵)(10⁵ - 4)

Cheers,
Brent
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##### General Discussion
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 952
WE: Supply Chain Management (Energy and Utilities)

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19 Sep 2018, 22:46
1
Bunuel wrote:
$$99,998^2 − 2^2 =$$

(A) $$10^{10} − 4$$

(B) $$(10^5 − 4)^2$$

(C) $$10^4(10^5 − 4)$$

(D) $$10^5(10^4 − 4)$$

(E) $$10^5(10^5 − 4)$$

$$99,998^2 − 2^2$$=$$(10^5-2)^2-2^2$$
Applying the formula: $$a^2-b^2=(a+b)(a-b)$$, wher $$a=(10^5-2)$$ and $$b=2$$

So, $$(10^5-2)^2-2^2$$=$$(10^5-2+2)*(10^5-2-2)$$=$$10^5(10^5 − 4)$$

Ans. (E)
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Re: 99,998^2 − 2^2 =  [#permalink]

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19 Sep 2018, 22:53

Sol: rewrite the expression in the form of (a - b)^2 - c^2. => (10^5 - 2)^2 - 2^2
Simplifying it = (10^5)^2 - ( 2. 10^5. 2)
= 10^5 ( 10^5 - 4 )

Henceforth option E.

Kudos please if you like the solution
Cheers
Hari
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Joined: 31 Oct 2013
Posts: 1014
Concentration: Accounting, Finance
GPA: 3.68
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20 Sep 2018, 01:20
1
Bunuel wrote:
$$99,998^2 − 2^2 =$$

(A) $$10^{10} − 4$$

(B) $$(10^5 − 4)^2$$

(C) $$10^4(10^5 − 4)$$

(D) $$10^5(10^4 − 4)$$

(E) $$10^5(10^5 − 4)$$

$$a^2-b^2 = (a+b) (a-b)$$

$$99998^2 - 2^2 = (99998 + 2) ( 99998 - 2)$$

99998 + 2 = 100000 =$$10^5$$

99998 - 2 = 99996 = 100000 - 4 = $$10^5 - 4$$

So,$$10^5(10^5 - 4)$$

Math Expert
Joined: 02 Aug 2009
Posts: 7213
Re: 99,998^2 − 2^2 =  [#permalink]

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20 Sep 2018, 06:16
Bunuel wrote:
$$99,998^2 − 2^2 =$$

(A) $$10^{10} − 4$$

(B) $$(10^5 − 4)^2$$

(C) $$10^4(10^5 − 4)$$

(D) $$10^5(10^4 − 4)$$

(E) $$10^5(10^5 − 4)$$

$$99,998^2 − 2^2 =$$ straightway is in the form of a$$^2-b^2$$ which is (a-b)(a+b)
so $$99,998^2 − 2^2 =(99,998-2)(99,998+2)=99,996*100,000=(100,000-4)*100,000=(10^5-4)*10^5$$
E

also 99,998 is just less than $$10^5$$, so 99,998^2 will be just less than $$10^{10}$$ AND the units digit will be 0 as $$8^2 -2^2$$ means $$4-4 =0$$
let us see the choices

(A) $$10^{10} − 4$$... this choice is slightly less than $$10^{10}$$ but it changes $$99,998^2$$ to $$10^2$$ so our answer has to be even lesser.. eliminate

(B) $$(10^5 − 4)^2$$..... units digit will be $$(10-6)^2=4^2$$, so 6...eliminate

(C) $$10^4(10^5 − 4)$$..... $$10^9-4*10^4$$.....this is just 1/10th of 10^10, so too less to be correct...eliminate

(D) $$10^5(10^4 − 4)$$..... $$10^9-4*10^5$$.....this is just 1/10th of 10^10, so too less to be correct...eliminate

(E) $$10^5(10^5 − 4)$$....10^10-4*10^5..correct

E
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: 99,998^2 − 2^2 = &nbs [#permalink] 20 Sep 2018, 06:16
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