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The average of the numbers a, b, c, 5, 15 is s and the average of th
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15 Oct 2024, 00:03
15 Oct 2024, 00:03
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The average of the numbers a, b, c, –5, –15 is s and the average of the numbers a, b, c, 5, 15 is t. What is the value of (s–t)?
A. –8
B. –5
C. –1
D. 0
E. 5
A. –8
B. –5
C. –1
D. 0
E. 5
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Re: The average of the numbers a, b, c, 5, 15 is s and the average of th
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16 Dec 2024, 13:52
16 Dec 2024, 13:52
Expert Reply
We know that the average of 5 numbers $\(a, b, c,-5 \&-15\)$ is $\(s\)$, so we get $\(\frac{a+b+c+(-5)+(-15)}{5}=s \Rightarrow a+b+c=5 s+20 \ldots . .(1)\)$
Also as the average of 5 numbers $\(a, b, c, 5 \& 15\)$ is $\(t\)$, we get $\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+5+15}{5}=\mathrm{t} \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=5 \mathrm{t}-20 \ldots\)$.
Now, using (1) \& (2), we get $\(5 \mathrm{~s}+20=5 \mathrm{t}-20\)$, which gives $\(s-t=\frac{-20-20}{5}=\frac{-40}{5}=-8\)$
Hence the answer is (A).
Also as the average of 5 numbers $\(a, b, c, 5 \& 15\)$ is $\(t\)$, we get $\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+5+15}{5}=\mathrm{t} \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=5 \mathrm{t}-20 \ldots\)$.
Now, using (1) \& (2), we get $\(5 \mathrm{~s}+20=5 \mathrm{t}-20\)$, which gives $\(s-t=\frac{-20-20}{5}=\frac{-40}{5}=-8\)$
Hence the answer is (A).
Re: The average of the numbers a, b, c, 5, 15 is s and the average of th
[#permalink]
16 Dec 2024, 13:52
16 Dec 2024, 13:52
Expert Reply
We know that the average of 5 numbers $\(a, b, c,-5 \&-15\)$ is $\(s\)$, so we get $\(\frac{a+b+c+(-5)+(-15)}{5}=s \Rightarrow a+b+c=5 s+20 \ldots . .(1)\)$
Also as the average of 5 numbers $\(a, b, c, 5 \& 15\)$ is $\(t\)$, we get $\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+5+15}{5}=\mathrm{t} \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=5 \mathrm{t}-20 \ldots\)$.
Now, using (1) \& (2), we get $\(5 \mathrm{~s}+20=5 \mathrm{t}-20\)$, which gives $\(s-t=\frac{-20-20}{5}=\frac{-40}{5}=-8\)$
Hence the answer is (A).
Also as the average of 5 numbers $\(a, b, c, 5 \& 15\)$ is $\(t\)$, we get $\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}+5+15}{5}=\mathrm{t} \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=5 \mathrm{t}-20 \ldots\)$.
Now, using (1) \& (2), we get $\(5 \mathrm{~s}+20=5 \mathrm{t}-20\)$, which gives $\(s-t=\frac{-20-20}{5}=\frac{-40}{5}=-8\)$
Hence the answer is (A).
