(a) A total of ₹ 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6%. If the total simple interest for one year was ₹ 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account ? (Use Cramer's rule). OR (b) The annual production of a commodity is given as follows. Fit a straight line trend by the method of least squares. | Year | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 | 2001 | | --- | --- | --- | --- | --- | --- | --- | --- | | Production (in tones) | 155 | 162 | 171 | 182 | 158 | 180 | 178 |

(a) A total of ₹ 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6%.

Question 32 of 44

A total of ₹ 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6%. If the total simple interest for one year was ₹ 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account ? (Use Cramer's rule). OR
The annual production of a commodity is given as follows. Fit a straight line trend by the method of least squares.

Year	1995	1996	1997	1998	1999	2000	2001
Production (in tones)	155	162	171	182	158	180	178

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2024Subjective· 5mImportance★★★★★

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(a) Cramer's rule gives investments ₹250, ₹4000, ₹4250. (b) Least-squares trend $Y=169.43+3.29X$ with origin 1998.

Part (a) — Cramer's rule. Let $x,y,z$ be the amounts (₹) invested at 2%, 3%, 6%.

Equations (interest equation $\times100$ ):

$x+y+z=8500,$

$2x+3y+6z=38000,$

$x+y-z=0\quad(\text{since }z=x+y).$

Determinant $\Delta$ :

$\Delta=\begin{vmatrix}1&1&1\\2&3&6\\1&1&-1\end{vmatrix}=1(-3-6)-1(-2-6)+1(2-3)=-9+8-1=-2.$

$\Delta_x$ (replace column 1):

$\Delta_x=\begin{vmatrix}8500&1&1\\38000&3&6\\0&1&-1\end{vmatrix}=8500(-9)-1(-38000)+1(38000)=-76500+38000+38000=-500.$

$\Delta_y$ (replace column 2):

$\Delta_y=\begin{vmatrix}1&8500&1\\2&38000&6\\1&0&-1\end{vmatrix}=1(-38000)-8500(-8)+1(-38000)=-38000+68000-38000=-8000.$

$\Delta_z$ (replace column 3):

$\Delta_z=\begin{vmatrix}1&1&8500\\2&3&38000\\1&1&0\end{vmatrix}=1(-38000)-1(-38000)+8500(-1)=-8500.$

Solutions:

$x=\frac{\Delta_x}{\Delta}=\frac{-500}{-2}=250,\quad y=\frac{-8000}{-2}=4000,\quad z=\frac{-8500}{-2}=4250.$

Check: $250+4000+4250=8500$ ✓; interest $=0.02(250)+0.03(4000)+0.06(4250)=5+120+255=380$ ✓; $z=x+y$ ✓.

Part (b) — straight-line trend by least squares. Take origin at the middle year 1998, $X=\text{year}-1998$ , so $\sum X=0$ .

| Year | $X$ | $Y$ | $XY$ | $X^2$ |

| --- | --- | --- | --- | --- | …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set MARCH1 markMCQ
$A$
(a) $|A|^{n-1}$
(b) $|A|^{n}$
(c) $|A|^{1+n}$
(d) $|A|$
›Reveal solutionSolution

Using $A(\operatorname{adj}A)=|A|\,I_n$ and taking determinants gives $|\operatorname{adj}A|=|A|^{\,n-1}$ , option (a).

Key identity. For any $n\times n$ matrix,

$A\,(\operatorname{adj}A)=|A|\,I_n.$

Take determinants of both sides.

$|A|\cdot|\operatorname{adj}A|=\big|\,|A|\,I_n\big|=|A|^{\,n}.$

…

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CBSE 2024Set MARCH1 markMCQ
$\frac{a_1}{x}+\frac{b_1}{y}=c_1,\ \frac{a_2}{x}+\frac{b_2}{y}=c_2,\ \Delta_1=\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix},\ \Delta_2=\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix},\ \Delta_3=\begin{vmatrix} c_1 & a_1 \\ c_2 & a_2 \end{vmatrix}$
(a) $\left(\frac{\Delta_3}{\Delta_1},\ \frac{\Delta_2}{\Delta_1}\right)$
(b) $\left(\frac{-\Delta_1}{\Delta_2},\ -\frac{\Delta_1}{\Delta_3}\right)$
(c) $\left(\frac{\Delta_1}{\Delta_2},\ \frac{\Delta_1}{\Delta_3}\right)$
(d) $\left(\frac{\Delta_2}{\Delta_1},\ \frac{\Delta_3}{\Delta_1}\right)$
›Reveal solutionSolution

Substitute $u=1/x,\ v=1/y$ ; Cramer's rule gives $(x,y)=\left(-\frac{\Delta_1}{\Delta_2},-\frac{\Delta_1}{\Delta_3}\right)$ .

Let $u=\dfrac1x,\ v=\dfrac1y$ . The system becomes linear in $u,v$ :

$a_1u+b_1v=c_1,\qquad a_2u+b_2v=c_2.$

With $\Delta_1=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$ , Cramer's rule gives

$u=\frac{\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}}{\Delta_1}=\frac{c_1b_2-c_2b_1}{\Delta_1}=\frac{-\Delta_2}{\Delta_1},\qquad v=\frac{\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}}{\Delta_1}=\frac{a_1c_2-a_2c_1}{\Delta_1}=\frac{-\Delta_3}{\Delta_1},$

…

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