MHT CET · Maths · Determinants
The sum of three numbers is 6 . Thrice the third number when added to the first number given 7 . On adding three time first number to the sum of second and third number we get 12 . The product of these numbers is
- A \(20\)
- B \(3\)
- C \(\frac{20}{3}\)
- D \(\frac{5}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{20}{3}\)
Step-by-step Solution
Detailed explanation
Let the numbers be \(x, y, z\)
\(
\begin{aligned}
& x+y+z=6 ..(1)\\
& x+3 z==7 ..(2)\\
& 3 x+y+z=12..(3) \\
& \text { Eq. (3)-(1), gives } \\
& 2 x=6 \Rightarrow x=3
\end{aligned}
\)
\(
2 \mathrm{x}=6 \Rightarrow \mathrm{x}=3
\)
From eq. (2), we get
\(
3+3 z+7 \Rightarrow z=\frac{4}{3}
\)
From eq. (1), we get
\(
\begin{aligned}
& 3+y+\frac{4}{3}=6 \quad \Rightarrow \quad y=\frac{5}{3} \\
& \therefore x y z=(3)\left(\frac{5}{3}\right)\left(\frac{4}{3}\right)=\frac{20}{3}
\end{aligned}
\)
\(
\begin{aligned}
& x+y+z=6 ..(1)\\
& x+3 z==7 ..(2)\\
& 3 x+y+z=12..(3) \\
& \text { Eq. (3)-(1), gives } \\
& 2 x=6 \Rightarrow x=3
\end{aligned}
\)
\(
2 \mathrm{x}=6 \Rightarrow \mathrm{x}=3
\)
From eq. (2), we get
\(
3+3 z+7 \Rightarrow z=\frac{4}{3}
\)
From eq. (1), we get
\(
\begin{aligned}
& 3+y+\frac{4}{3}=6 \quad \Rightarrow \quad y=\frac{5}{3} \\
& \therefore x y z=(3)\left(\frac{5}{3}\right)\left(\frac{4}{3}\right)=\frac{20}{3}
\end{aligned}
\)
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