If \(\mathrm{a}+\mathrm{b}+\mathrm{c}=3, \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=6\) and \(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}=1\), where \(\mathrm{a}\), \(\mathrm{b}, \mathrm{c}\) are all non-zero, then ‘abc’ is equal to (a) \(\frac{2}{3}\) (b) \(\frac{3}{2}\) (c) \(\frac{1}{2}\) (d) \(\frac{1}{3}\)
Let the given number be \(x\). According to the problem: - The correct operation was to multiply \(x\) by \(\frac{7}{19}\): \(x \times \frac{7}{19}\). - By mistake, the boy divided \(x\) by \(\frac{7}{19}\), which is equivalent to multiplying \(x\) by \(\frac{19}{7}\): \(x \times \frac{19}{7}\). - TRead more
Let the given number be \(x\). According to the problem:
– The correct operation was to multiply \(x\) by \(\frac{7}{19}\): \(x \times \frac{7}{19}\).
– By mistake, the boy divided \(x\) by \(\frac{7}{19}\), which is equivalent to multiplying \(x\) by \(\frac{19}{7}\): \(x \times \frac{19}{7}\).
– The mistake led to a result that was 624 more than the correct answer.
Thus, we have:
\[
x \times \frac{19}{7} = x \times \frac{7}{19} + 624
\]
Rearrange the equation to isolate \(x\):
\[
x \times \frac{19}{7} – x \times \frac{7}{19} = 624
\]
Factoring \(x\) out:
\[
x \left(\frac{19}{7} – \frac{7}{19}\right) = 624
\]
To solve for \(x\), simplify the expression in the parentheses:
\[
x \left(\frac{19^2 – 7^2}{7 \times 19}\right) = 624
\]
Calculate the difference of squares:
\[
19^2 – 7^2 = (19 + 7)(19 – 7) = 26 \times 12 = 312
\]
Substitute back into the equation:
\[
x \left(\frac{312}{7 \times 19}\right) = 624
\]
Simplify the fraction:
\[
x \left(\frac{312}{133}\right) = 624
\]
Solving for \(x\):
\[
x = 624 \times \frac{133}{312} = 2 \times 133 = 266
\]
The sum of the digits of the given number \(266\) is \(2 + 6 + 6 = 14\).
Answer: (d) 14.
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Given the system of equations: 1. \(a + b + c = 3\) 2. \(a^2 + b^2 + c^2 = 6\) 3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\) We want to find the value of \(abc\). To solve this, let's first analyze the given information. The third equation can be written as: \[ \frac{ab + ac + bc}{abc} = 1 \] FRead more
Given the system of equations:
1. \(a + b + c = 3\)
2. \(a^2 + b^2 + c^2 = 6\)
3. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1\)
We want to find the value of \(abc\).
To solve this, let’s first analyze the given information. The third equation can be written as:
\[
\frac{ab + ac + bc}{abc} = 1
\]
From this, we get:
\[
ab + ac + bc = abc
\]
From the first equation, we know the sum of \(a\), \(b\), and \(c\). To find \(ab + ac + bc\), we can square the first equation and compare it to the second equation. Squaring the first equation gives us:
\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3^2 = 9
\]
Given \(a^2 + b^2 + c^2 = 6\) from the second equation, we can substitute that into our squared first equation to find \(ab + ac + bc\):
\[
6 + 2(ab + ac + bc) = 9
\]
Solving for \(ab + ac + bc\):
\[
2(ab + ac + bc) = 3
\]
\[
ab + ac + bc = \frac{3}{2}
\]
Given \(ab + ac + bc = abc\), we have:
\[
abc = \frac{3}{2}
\]
Therefore, the value of \(abc\) is \(\frac{3}{2}\).
Answer: (b) \(\frac{3}{2}\).
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