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# Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10

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Math Expert
Joined: 02 Sep 2009
Posts: 58408
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10  [#permalink]

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22 Mar 2018, 23:56
00:00

Difficulty:

25% (medium)

Question Stats:

92% (01:10) correct 8% (01:31) wrong based on 38 sessions

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Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10^2 grams of sugar. The number of grams of sugar in solution Y is how many times the number of grams of sugar in solution X?

(A) 0.0002
(B) 0.002
(C) 200
(D) 2,000
(E) 20,000

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Current Student
Joined: 07 Jan 2016
Posts: 1088
Location: India
GMAT 1: 710 Q49 V36
Re: Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10  [#permalink]

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23 Mar 2018, 00:19
Bunuel wrote:
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10^2 grams of sugar. The number of grams of sugar in solution Y is how many times the number of grams of sugar in solution X?

(A) 0.0002
(B) 0.002
(C) 200
(D) 2,000
(E) 20,000

8x10^2/4x10^ -2

= 2 x 10^2 / 10^-2 = 2 x 10^4

(E) imo
Senior SC Moderator
Joined: 22 May 2016
Posts: 3548
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10  [#permalink]

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23 Mar 2018, 08:53
Bunuel wrote:
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10^2 grams of sugar. The number of grams of sugar in solution Y is how many times the number of grams of sugar in solution X?

(A) 0.0002
(B) 0.002
(C) 200
(D) 2,000
(E) 20,000

Y, number of grams of sugar = $$8*10^2$$
X, number of grams of sugar = $$4*10^{-2}$$

Number of grams of sugar in Y is how many times the number of grams of sugar in X?

$$\frac{Y}{X}=\frac{8*10^2}{4*10^{-2}}=$$?

1) Move $$10^{-2}$$ to the numerator.
Any number with a negative power in the denominator can be moved to the numerator.
Change the exponent's sign
$$\frac{Y}{X}=\frac{8*10^2}{4*10^{-2}}=\frac{8*10^2*10^2}{4}=$$

$$\frac{8*10^{(2+2)}}{4}=\frac{8*10^4}{4}=2*10^{4}=20,000$$

OR
2) Division of exponents (same base) =
Keep the base, subtract the exponents

$$\frac{Y}{X}=\frac{8*10^2}{4*10^{-2}}=\frac{8*10^{2-(-2)}}{4}=$$

$$\frac{8*10^{2+2}}{4}=\frac{8*10^4}{4}=2*10^4=20,000$$

OR
3) divide integers, then powers of 10.
Integers and powers of 10 are factors of the numbers. Results of separate operations are also factors -- i.e., multiply the results

Number in Y is how many times number in X?
Number in Y = $$8*10^2$$
Number in X = $$4*10^{-2}$$

$$\frac{8}{4} = 2$$
$$\frac{10^2}{10^{-2}}=10^{2-(-2)}=10^{2+2}=10^4$$
Multiply results: $$2*10^4=20,000$$

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Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10  [#permalink]

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23 Mar 2018, 09:43
Bunuel wrote:
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10^2 grams of sugar. The number of grams of sugar in solution Y is how many times the number of grams of sugar in solution X?

(A) 0.0002
(B) 0.002
(C) 200
(D) 2,000
(E) 20,000

Number of grams of sugar in solution Y:X $$= \frac{8*10^{2}}{4*10^{-2}} = \frac{8*10^{2}*10^{2}}{4} = 2*10^{4} = 20,000 (E)$$

Key point to remember: $$a^{-b} = \frac{1}{a^{b}}$$ in this case: $$4*10^{-2} = \frac{4}{10^2}$$
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2816
Re: Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10  [#permalink]

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26 Mar 2018, 17:46
Bunuel wrote:
Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10^2 grams of sugar. The number of grams of sugar in solution Y is how many times the number of grams of sugar in solution X?

(A) 0.0002
(B) 0.002
(C) 200
(D) 2,000
(E) 20,000

(8 x 10^2)/(4 x 10^(-2))

(8/4) x 10^(2 - (-2))

2 x 10^4

20,000

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Re: Solution X contains 4*10^(–2) grams of sugar. Solution Y contains 8*10   [#permalink] 26 Mar 2018, 17:46
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