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# If the average (arithmetic mean) of p and r is 60 and the average (ari

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Joined: 02 Sep 2009
Posts: 50730
If the average (arithmetic mean) of p and r is 60 and the average (ari  [#permalink]

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21 Dec 2017, 20:28
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Difficulty:

15% (low)

Question Stats:

92% (00:48) correct 8% (00:11) wrong based on 36 sessions

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If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p?

A. 70
B. 40
C. 30
D. 20
E. 10

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If the average (arithmetic mean) of p and r is 60 and the average (ari  [#permalink]

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22 Dec 2017, 06:05
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Bunuel wrote:
If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p?

A. 70
B. 40
C. 30
D. 20
E. 10

$$\frac{p+r}{2}$$ = 60 -> p+r = 120 -> (1)
$$\frac{r+s}{2}$$ = 80 -> r+s = 160 -> (2)

Subtracting (1) from (2), we get
r+s - (p+r) = 160-120 = 40

Therefore, s - p = 40(Option B)
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Joined: 24 Nov 2016
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If the average (arithmetic mean) of p and r is 60 and the average (ari  [#permalink]

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23 Dec 2017, 07:29
Bunuel wrote:
If the average (arithmetic mean) of p and r is 60 and the average (arithmetic mean) of r and s is 80, what is the value of s – p?

A. 70
B. 40
C. 30
D. 20
E. 10

$$Average=\frac{Sum.of.Terms}{Number.of.Terms}$$

$$\frac{s+r}{2}=80;s+r=160$$

$$\frac{p+r}{2}=60;p+r=120$$

Subtract both equations to find $$s-p$$:

$$(s+r)-(p+r)=160-120;s-p=40$$

If the average (arithmetic mean) of p and r is 60 and the average (ari &nbs [#permalink] 23 Dec 2017, 07:29
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