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# The product of two negative numbers is 160. If the lesser of the two

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The product of two negative numbers is 160. If the lesser of the two [#permalink]

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15 Jun 2016, 00:31
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The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4
[Reveal] Spoiler: OA

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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15 Jun 2016, 02:33
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-a*-b =160 --1

-a=(-2b)-4 --2

substituting eq 2 in 1

(-2b-4)*-b=160
2b^2+4b-160=0
b^2+2b-80=o
b^2+10b-8b-80=0
b= -10,8
therefore -b =-8

substitute the value in eq 2
-a = -16-4= -20

therefore the larger number = -8

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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15 Jun 2016, 03:27
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Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

let two numbers are a and b

a = 2b-4

Find value of b.

(2b-4)*b = 160
b*(b-2) = 80

b =-8 (Given that numbers are negative)

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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15 Jun 2016, 03:49
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Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

Consider two negative numbers as $$n_1$$ and $$n_2$$.

$$n_1 * n_2 = 160$$
$$n_2 = 2n_1 - 4$$ ..... substitute this in above equation.

$$n_1 * (2n_1 - 4) = 160$$
$$2n_1^2 - 4n_2 - 160 = 0$$
$$n_1^2 - 2n_2 - 80 = 0$$
$$(n_1 - 10) (n_1 +8) = 0$$
$$n_1 = 10$$ and $$n_1 = -8$$

as mentioned in question, numbers are negative, we'll consider $$n_1 = -8$$. substitute this in second equation.

$$n_2 = 2 * (-8) - 4$$
$$n_2 = -20$$

Hence the greater number is $$-8$$.

$$Answer = D$$

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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19 Jul 2016, 11:23
The quadratic gives us two roots: (larger #)= -10 and (larger #)= +8
Curious why we didn't choose -10 since (although it is the smaller number between the roots, it is the only negative number.. and the question said that both numbers are negative).
Say we chose -10, then -10 x -16 also equals 160 and in this case -10 is in fact the larger number.

I'm sure I'm missing something since everyone got D.. can someone please tell me why D and not C?

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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19 Jul 2016, 11:28
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nk18967 wrote:
The quadratic gives us two roots: (larger #)= -10 and (larger #)= +8
Curious why we didn't choose -10 since (although it is the smaller number between the roots, it is the only negative number.. and the question said that both numbers are negative).
Say we chose -10, then -10 x -16 also equals 160 and in this case -10 is in fact the larger number.

I'm sure I'm missing something since everyone got D.. can someone please tell me why D and not C?

Hi! There,

We get the quadratic as x^2 -2x -80= 0

(x+8) (x-10)= 0

x can either be -8 or +10.

But because it is given that the numbers are -ve, we will choose -8.

Hope it is clear.
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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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19 Jul 2016, 12:04

Just realized what a silly mistake I made!

I had let the smaller # be (-a) and let the larger # be (-b)..
(-a) = 2(-b) - 4
(-a)(-b) = 160

So I solved it as:
[2(-b) - 4] [-b] = 160
[-2b - 4] [-b] = 160
2b^2 + 4b - 160 = 0
b^2 + 2b - 80 = 0
(b-8) (b+10)
So, b= 8 or b= -10

Here is where I forgot, that my number is actually (-b) !!!! That changes things because then my number can be either -(8) or -(-10)...which is +10..
SO... the Ans is obviously -8 since we need a -ve #!!

I hope I dont make these kind of mistakes when I'm taking the real test! Ugh..!

nk18967 wrote:
The quadratic gives us two roots: (larger #)= -10 and (larger #)= +8
Curious why we didn't choose -10 since (although it is the smaller number between the roots, it is the only negative number.. and the question said that both numbers are negative).
Say we chose -10, then -10 x -16 also equals 160 and in this case -10 is in fact the larger number.

I'm sure I'm missing something since everyone got D.. can someone please tell me why D and not C?

Hi! There,

We get the quadratic as x^2 -2x -80= 0

(x+8) (x-10)= 0

x can either be -8 or +10.

But because it is given that the numbers are -ve, we will choose -8.

Hope it is clear.

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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19 Jul 2016, 21:53
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Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

Use the options. The options give you the greater number. Start from the middle option.

(C) - 10
Twice of -10 is -20 and 4 less is -24.
-24 * -10 = 240 (Not correct)

We need lower product (of 160) so try option (D).

(D) - 8
Twice of -8 is -16 and 4 less is -20.
-20 * -8 = 160
Correct

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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24 Sep 2016, 07:21
Two factors of 160 can be (1,160), (2,80), (4,40), (8, 20), (16,10), (32, 5). Considering the relation, lower = 2 higher -4, we can eliminate everything other than (8,20) and (16, 10) and the answer would be (-8, -20) i.e. -8

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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06 Dec 2016, 07:28
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Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

We are given that the product of two negative numbers is 160 and that the lesser of the two numbers is 4 less than twice the greater.

We can let x = the larger number and y = the smaller number, and we can create the following equations:

(x)(y) = 160

AND

y = 2x - 4

Substituting (2x - 4) for y in the first equation yields:

x(2x - 4) = 160

2x^2 + 4x = 160

x^2 - 2x - 80 = 0

(x - 10)(x + 8) = 0

x = 10 or x = -8

Since x is negative, x = -8 and y = 2(-8) -4 = -20.

Thus, the greater number is -8.

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The product of two negative numbers is 160. If the lesser of the two [#permalink]

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06 Dec 2016, 11:41
Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

1. xy=160
2. x=2y-4
dividing 1. by 2.➡
y^2-2y-80=0
y=-8
D

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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07 Dec 2016, 10:22
Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

$$-a*-b = 160$$

Let -a > -b

So, -b = -2a - 4

Or, -a( -2a - 4 ) = 160

Or, 2a^2 + 4a = 160

Or, 2a ( a + 2 ) = 16*10

Or, a = -8

So, b = -20

Thus, the greater number is (D) -8
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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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09 Oct 2017, 19:25
i think i did this a little different than most people above.

first, broke 160 down into pairs (below):

1, 160
2, 80
4, 40
5, 32
8, 20
10, 16

NOW, remember the question tells you that the LOWER # is 4 less than TWICE the greater? --> L=2S+4

Okay, let's see which #s (from list above) make sense...

1, 160
2, 80
4, 40
5, 32
8, 20
10, 16

Don't follow my logic? Well, let's set S=8. L=16+4. So, if S=8, L=20. Perfect fit.
> Don't fall for the trap: 10, 16. Here, if S=10, then L=24 [aka 2*10+4]

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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09 Oct 2017, 21:15

-8*2 - 4 = -20

So greater number should be -8

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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15 Oct 2017, 07:47
Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

x = lesser
y = greater

(2y-4)(y) = 160
a , b , and c are too high values, so eliminate them

For e, (-12)(-4) =/ 160

For D, (-16-4)(-8) = (-20)(-8) = 160

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Re: The product of two negative numbers is 160. If the lesser of the two [#permalink]

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18 Nov 2017, 03:26
ScottTargetTestPrep wrote:
Bunuel wrote:
The product of two negative numbers is 160. If the lesser of the two numbers is 4 less than twice the greater, what is the greater number?

A) -20
B) -16
C) -10
D) -8
E) -4

We are given that the product of two negative numbers is 160 and that the lesser of the two numbers is 4 less than twice the greater.

We can let x = the larger number and y = the smaller number, and we can create the following equations:

(x)(y) = 160

AND

y = 2x - 4

Substituting (2x - 4) for y in the first equation yields:

x(2x - 4) = 160

2x^2 + 4x = 160

x^2 - 2x - 80 = 0

(x - 10)(x + 8) = 0

x = 10 or x = -8

Since x is negative, x = -8 and y = 2(-8) -4 = -20.

Thus, the greater number is -8.

hello there it says product of two negative numbers- than why you write (x)(y) = 160 shouldn't it be (-x)(-y) = 160 Also how after this x^2 - 2x - 80 = 0 you get (x - 10)(x + 8) = 0
I know formula B^2 - 4AC but couldn't use it properly, though I didn't divide whole equation by 2. Also after this (x - 10)(x + 8) = 0

you write that x = 10 or x = -8 ---> but how can x=10? if in brackets it is -10 and how can x = -8 if in brackets 8 is positive

thanks for taking time to explain and have a great day!

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Re: The product of two negative numbers is 160. If the lesser of the two   [#permalink] 18 Nov 2017, 03:26
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