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Arville owns three times as many screwdrivers as he does hammers and

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Arville owns three times as many screwdrivers as he does hammers and  [#permalink]

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New post 30 Sep 2018, 18:06
2
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

64% (01:39) correct 36% (01:42) wrong based on 134 sessions

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Arville owns three times as many screwdrivers as he does hammers and pliers put together. The number of hammers he owns is how many times the number of pliers he owns?

1. The ratio of screwdrivers to hammers owned by Arville is 21 : 4.
2. The ratio of screwdrivers to pliers owned by Arville is 7 : 1.

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Re: Arville owns three times as many screwdrivers as he does hammers and  [#permalink]

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New post 30 Sep 2018, 19:32
2
We have , Arville 3 times as many screwdrivers as hammers and pliers together.
So, for the sake of convenience lets say , Screwdrivers = S, Hammers = H, Pliers = P
So, S = 3(H+P)

From statement 1 we get,
Screwdrivers: Hammers = 21:4
21 = 3(4+P)
21 = 12+3P
9=3P
3=P
So, H:P = 4:3

Statement 1 is sufficient.
From statement 2 we get,
Screwdrivers: Pliers = 7:1
7 = 3(H+1)
7=3H+3
4=3H
H=4/3
So, H:P = 4/3: 1
So, H:P =4:3
So, statement 2 is also sufficient.

So, the correct option is D.
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Re: Arville owns three times as many screwdrivers as he does hammers and  [#permalink]

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New post 30 Sep 2018, 18:23
Screwdrivers = 3(Hammers+Pliers)

From statement 1:

Screwdrivers/Hammers = 21/4

Screwdrivers = \(\frac{21}{4}\)(Hammers)

Therefore \(\frac{21}{4}\)(Hammers) = 3(Hammers+Pliers)

Solving this we get,

H = 4/3P

Sufficient.

From statement 2:

Screwdrivers/Pliers = 7/1

Or S = 7P

Using this in Screwdrivers = \(\frac{21}{4}\)(Hammers)

We get,

H = P.

Sufficient.

D is the answer.
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Re: Arville owns three times as many screwdrivers as he does hammers and  [#permalink]

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New post 30 Sep 2018, 20:01
Arville owns three times as many screwdrivers as he does hammers and pliers put together. The number of hammers he owns is how many times the number of pliers he owns?

1. The ratio of screwdrivers to hammers owned by Arville is 21 : 4.
Ratio of S:(H+P)=3:1=7*3:7:1=21:7..
But H form 4 parts of 7, so pliers is 7-4=3
Ratio of H:P=4:3
Sufficient

2. The ratio of screwdrivers to pliers owned by Arville is 7 : 1.
S:P=7:1=21:3 .....
S:(H+P)=21:7, so H+P=7.......H=7-3=4
So H:P=4:3
Sufficient

D
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Arville owns three times as many screwdrivers as he does hammers and  [#permalink]

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New post 30 Jan 2019, 14:38
This question can easily be solved by spending a bit more time on the question stem and using some logic.

The information in the question stem allows us to build the equation: S = 3(H + P)
The question "The number of hammers he owns is how many times the number of pliers he owns?" simply implies H = xP, or P = H/x, so what is the value of x?
We can see that we can find the relationship between H & P by knowing the relationship between S and H or S and P. To prove this, see below:
S = 3(xP + P)
S = 3(H + H/x)

(1) - S:H = 21/4 --> Relationship between S & H --> Sufficient
(1) - S:P = 7:1 --> Relationship between S & P --> Sufficient
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Arville owns three times as many screwdrivers as he does hammers and   [#permalink] 30 Jan 2019, 14:38
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