Maths Case study (4 Marks)

1. (c) ₹ 40000

Solution:

Combined sales of Masoor in September and October for farmer Balwan Singh = ₹ 40000

2. (d) ₹ 15000

Solution:

Combined sales of Urad in September and October for farmer Shyam = ₹ 15000

3. (a) ₹ 24000

Solution:

Change in sales from September to October is given by:

$\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\text{A-B}=\begin{bmatrix}5000&10000&24000\\30000&20000&0\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$

$\therefore$ Decrease in sales of Mung from September to October for farmer Shyam = ₹ 24000.

4. (b)  $\begin{matrix} \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 120\\400&\ \ \ \ \ \ 200&\ \ \ \ \ 200\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$

Solution:

Required profit is given by:

2% of $\text{B}=\frac{2}{100}\times\text{B}=0.02\times\text{B}$

$\begin{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\=0.02\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 120\\400&\ \ \ \ \ \ 200&\ \ \ \ \ 200\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$

$\begin{matrix} \ \ \ \ \ \text{Urad}&\text{Masoor}&\text{Mung}\end{matrix}\\=\begin{bmatrix}100&\ \ \ \ \ \ 200&\ \ \ \ \ 120\\400&\ \ \ \ \ \ 200&\ \ \ \ \ 200\end{bmatrix}\begin{matrix}\text{Shayam}\\\text{Balwan singh}\end{matrix}$

Thus, in October Shyam receives ₹ 100, ₹ 200 and ₹ 120 as profit in the sale of each variety of pulses, respectively and Balwan Singh receives a profit of ₹ 400, ₹ 200 and ₹ 200 in the sale of each variety of pulses respectively.

5. (a) Urad

1. (d) $\begin{matrix}&\text{Boys}&\text{Girls}\end{matrix}\\\begin{matrix}\text{I}\\\text{II}\\\text{III}\end{matrix}\begin{bmatrix}80&80\\70&75\\65&90\end{bmatrix}$

Solution:

In factory A, number of units of types I, II and III for boys are 80, 70, 65 respectively and for girls number of units oftypes I, II and III are 80, 75, 90 respectively.

$\begin{matrix}&&&&\text{Boys}&\text{Girls}\end{matrix}\\\therefore\text{P}=\begin{matrix}\text{I}\\\text{II}\\\text{III}\end{matrix}\begin{bmatrix}80&80\\70&75\\65&90\end{bmatrix}$

2. (a) $\begin{matrix}&\text{Boys}&\text{Girls}\end{matrix}\\\begin{matrix}\text{I}\\\text{II}\\\text{III}\end{matrix}\begin{bmatrix}85&50\\65&55\\72&80\end{bmatrix}$

Solution:

In factory B, number of units of types I, II and III for boys are 85, 65, 72 respectively and for girls number of units of types I, II and III are 50, 55, 80 respectively.$\begin{matrix}&&&&\text{Boys}&\text{Girls}\end{matrix}\\\therefore\text{Q}=\begin{matrix}\text{I}\\\text{II}\\\text{III}\end{matrix}\begin{bmatrix}85&50\\65&55\\72&80\end{bmatrix}$

3. (c) $\begin{matrix}\ \ \text{I}&\ \ \ \ \text{II}&\ \ \ \text{III}\end{matrix}\\\begin{matrix}[165&135&137]\end{matrix}\\$

Solution:

Let X be the matrix that represent the number of units of each type produced by factory A for boys, and Y be the matrix that represent the number of units of each type produced by factory B for boys.

Then, $\begin{matrix}\ \ \ \ \ \ \ \ \ \text{I}&\ \ \ \text{II}& \text{III}\end{matrix}\\\begin{matrix}\text{X}=[80&70&65]\end{matrix}\\$ and $\begin{matrix}\ \ \ \ \ \ \ \ \ \text{I}&\ \ \ \text{II}& \text{III}\end{matrix}\\\begin{matrix}\text{Y}=[85&65&72]\end{matrix}\\$

Now, required matrix = X + Y = [80 70 65] + [85 65 72] = [165 135 137]

4. (a) $\begin{matrix}\ \ \text{I}&\ \ \ \ \text{II}&\ \ \ \text{III}\end{matrix}\\\begin{matrix}[130&130&170]\end{matrix}\\$

Solution:

Required matrix = [80 75 90] + [50 55 80] = [130 130 170]

5. (a) $\begin{bmatrix}165 & 135 & 137\\130 & 130 & 170 \end{bmatrix}$

Solution:

Clearly, R = P + Q

$=\begin{bmatrix}80 & 80 \\70 & 75 \\65 & 90 \end{bmatrix}+\begin{bmatrix}85 & 50 \\65 & 55 \\72 & 80 \end{bmatrix}=\begin{bmatrix}165 & 130 \\135 & 130 \\137 & 170 \end{bmatrix}$

$\therefore\text{R'}=\begin{bmatrix}165 & 135 & 137\\130 & 130 & 170 \end{bmatrix}$

1. (a) $\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \text{Fans}&\text{Mats}&\text{Plates}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{A}\\\text{P}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix} \ \ 40 \ \ \ & 50 & \ \ \ \ \ 25\\25 & 40 & \ \ \ \ \ 30\\35& \ 50& \ \ \ \ \ 40\end{bmatrix}$

Solution:

Clearly, $\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \text{Fans}&\text{Mats}&\text{Plates}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{A}\\\text{P}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix} \ \ 40 \ \ \ & 50 & \ \ \ \ \ 25\\25 & 40 & \ \ \ \ \ 30\\35& \ 50& \ \ \ \ \ 40\end{bmatrix}$

2. (d) $\text{Q}=\begin{bmatrix}25\\100\\50\end{bmatrix}\begin{matrix}\text{Fans}\\\text{Mats}\\\text{Plates}\end{matrix}$

Solution:

Since Q is a 3 × 1 matrix, therefore

$\text{Q}=\begin{bmatrix}25\\100\\50\end{bmatrix}\begin{matrix}\text{Fans}\\\text{Mats}\\\text{Plates}\end{matrix}$

3. (a) ₹ 7000

Solution:

Clearly, total funds collected by each school is given by the matrix

 $\text{PQ}=\begin{bmatrix}40&50&20\\25&40&30\\35&50&40\end{bmatrix}\begin{bmatrix}25\\100\\50\end{bmatrix}$ 

$=\begin{bmatrix}1000+5000+1000\\625+4000+15000\\875+5000+2000\end{bmatrix}=\begin{bmatrix}7000\\6125\\7875\end{bmatrix}$

$\therefore$ Funds collected by school A is ₹ 7000

Funds collected by school B is ₹ 6125

Funds collected by school C is ₹ 7875

4. (b) ₹ 6125
5. (b) ₹ 21000

Solution:

Total funds collected for the required purpose = ₹ (7000 + 6125 + 7875) = ₹ 21000

1. (a) $\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Calorise}&\text{Proteins}\end{matrix}\\\begin{matrix}\text{Man}\\\text{Woman}\\\text{Children}\end{matrix}\begin{bmatrix} \ \ 2400 \ \ \ & 45\\1900 & 55\\1800& \ 33&\end{bmatrix}$

Solution:

Let F be the matrix representing the number of family members and R be the matrix representing the requirement of calories and proteins for each person. Then

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{matrix}\text{Men} & \text{Women} & \text{Children}\\\end{matrix}\\\text{F} =\begin{matrix}\text{Family A} \\ \text{Family B} \end{matrix} \begin{bmatrix} 4 &\ \ \ \ \ \ \ \ 4 &\ \ \ \ \ \ \ \ \ \ \ \ 4\\2 & \ \ \ \ \ \ \ \ \ 2 & \ \ \ \ \ \ \ \ \ \ \ \ 2\end{bmatrix}$

$\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Calorise}&\text{Proteins}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \text{Man}\\\text{R}\ =\text{Woman}\\\ \ \ \ \ \ \ \ \ \ \ \text{Children}\end{matrix}\begin{bmatrix} \ \ 2400 \ \ \ & 45\\1900 & 55\\1800& \ 33&\end{bmatrix}$

2. (b) 24400

Solution:

The requirement of calories and proteins for each ofthe two families is given by the product matrix FR.

$\text{FR}=\begin{bmatrix}4&4&4\\2&2&2\end{bmatrix}\begin{bmatrix}2400&45\\1900&55\\1800&33\end{bmatrix}$

$=\begin{bmatrix}4(2400+1900+1800)&4(45+55+33)\\2(2400+1900+1800)&2(45+55+33)\end{bmatrix}$

$\begin{matrix} \ \ \ \ \ \ \ \ \ \ \ \text{Calories}&\text{Proteins}\end{matrix}\\\text{FR}=\begin{bmatrix}24400&\ \ \ \ \ \ 532\\12200&\ \ \ \ \ \ 266\end{bmatrix}\begin{matrix}\text{Family A}\\\text{Family B}\end{matrix}$

3. (c) 266 grams
4. (c) A + B

Solution:

Since, AB = B ... (i)

BA = A ... (ii)

$\therefore$ A² + B² = A × A + B × B

= A(BA) + B(AB)

= (AB)A + (BA)B

= BA + AB

= A + B

5. (a) m = q

Solution:

$\text{A}=(\text{a}_\text{ij})_{\text{m}\times\text{n}},\ \ \text{B}=(\text{b}_\text{ij})_{\text{n}\times\text{p}},\ \ \text{C}=(\text{c}_\text{ij})_{\text{p}\times\text{q}}$

$\text{BC}=(\text{b}_\text{ij})_{\text{n}\times\text{p}}\times(\text{c}_\text{ij})_{\text{p}\times\text{q}}=(\text{d}_\text{ij})_{\text{n}\times\text{q}}$

$\text{(BC)A}=(\text{d}_\text{ij})_{\text{n}\times\text{q}}\times(\text{a}_\text{ij})_{\text{m}\times\text{n}}$

Hence, (BC)A is possible only when m = q

1. (c) ₹ 30000

Solution:

Let ₹ A, ₹ B and ₹ C be the cost incurred by the organisation for villages X, Y and Z respectively. Then A, B, C will be given by the following matrix equation.

$\begin{bmatrix}400 & 300 & 100\\300 & 250 & 75\\500 & 400 & 150 \end{bmatrix}\begin{bmatrix}50 \\20\\40\end{bmatrix}=\begin{bmatrix}\text{A}\\\text{B}\\\text{C}\end{bmatrix}$

$=\begin{bmatrix}\text{A}\\\text{B}\\\text{C}\end{bmatrix}=\begin{bmatrix}400\times50+300\times20100\times40\\300\times50+250\times20+75\times40\\500\times50+400\times20+150\times40 \end{bmatrix}$

$\begin{bmatrix}2000+6000+4000\\15000+5000+3000\\25000+8000+6000 \end{bmatrix}=\begin{bmatrix}30000 \\23000\\39000\end{bmatrix}$

2. (c) ₹ 23000
3. (b) ₹ 39000
4. (c) 40

Solution:

Total number of toilets that can be expected in each village is given by the following matrix

$\begin{matrix}\text{X}\\\text{Y}\\\text{Z}\end{matrix}\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}\begin{bmatrix}\frac{2}{100}\\\frac{4}{100}\\\frac{20}{100}\end{bmatrix}$

$\begin{matrix}\text{X}\\\text{Y}\\\text{Z}\end{matrix}\begin{bmatrix}8+12+20\\6+10+15\\10+16+30\end{bmatrix}=\begin{matrix}\text{X}\\\text{Y}\\\text{Z}\end{matrix}\begin{bmatrix}40\\31\\56\end{bmatrix}$

5. (d) 46

1. (b) $\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 120\ \ \ &\ \ 100&\ \ \ \ \ 20\\\ \ \ 100&\ \ 30&\ \ \ \ \ 5\\\ \ \ 90&\ \ 40&\ \ \ \ \ 2\end{bmatrix}$

Solution:

In 2019, dealer A sold 120 Hatchback, 50 Sedan and 10 SUV; dealer B sold 100 Hatchback, 30 Sedan and 5 SUV and dealer C sold 90 Hatchback, 40 Sedan and 2 SUV.

$\therefore$ Required matrix, say P, is given by,

$\begin{matrix}& \ \ \ \ \ \ \ \ \text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{A}\\\text{P}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 120\ \ \ &\ \ 100&\ \ \ \ \ 20\\\ \ \ 100&\ \ 30&\ \ \ \ \ 5\\\ \ \ 90&\ \ 40&\ \ \ \ \ 2\end{bmatrix}$

2. (a) $\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 300\ \ \ &\ \ 150&\ \ \ \ \ 20\\\ \ \ 200&\ \ 50&\ \ \ \ \ 6\\\ \ \ 100&\ \ 60&\ \ \ \ \ 5\end{bmatrix}$

Solution:

In 2020, dealer A sold 300 Hatchback, 150 Sedan, 20 SUV dealer B sold 200 Hatchback, 50 sedan, 6 SUV dealer C sold 100 Hatchback, 60 sedan, 5 SUV.

$\therefore$ Required matrix, say Q, is given by,

$\begin{matrix}& \ \ \ \ \ \ \ \ \text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \text{A}\\\text{Q}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 300\ \ \ &\ \ 150&\ \ \ \ \ 20\\\ \ \ 200&\ \ 50&\ \ \ \ \ 6\\\ \ \ 100&\ \ 60&\ \ \ \ \ 5\end{bmatrix}$

3. (c) $\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 420\ \ \ &\ \ 200&\ \ \ \ \ 30\\\ \ \ 300&\ \ 80&\ \ \ \ \ 11\\\ \ \ 190&\ \ 100&\ \ \ \ \ 7\end{bmatrix}$

Solution:

Total number of cars sold in two given years, by each dealer, is given by

$\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hatchback}&\ \ \ \ \ \ \text{Sedan}&\ \ \ \ \ \ \ \ \ \text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\\\text{P+Q}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ 120+300\ \ \ &\ \ 50+150&\ \ \ \ \ 10+20\\100+200&\ \ 30+50&\ \ \ \ \ 5+6\\90+100&\ \ 40+60&\ \ \ \ \ 2+5\end{bmatrix}$

$\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \text{A}\\=\text{B}\\\ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 420\ \ \ &\ \ 200&\ \ \ \ \ 30\\\ \ \ 300&\ \ 80&\ \ \ \ \ 11\\\ \ \ 190&\ \ 100&\ \ \ \ \ 7\end{bmatrix}$

4. (c) $\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 180\ \ \ &\ \ 100&\ \ \ \ \ 10\\\ \ \ 100&\ \ 20&\ \ \ \ \ 1\\\ \ \ 10&\ \ 20&\ \ \ \ \ 3\end{bmatrix}$

Solution:

The increase in sales from 2019 to 2020 is given by:

$\begin{matrix}& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hatchback}&\ \ \ \ \ \ \text{Sedan}&\ \ \ \ \ \ \ \ \ \text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\\\text{Q - P}\ =\text{B}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ 300-120\ \ \ &\ 150-50&\ \ \ \ \ 20-10\\200-100&\ \ 50-30&\ \ \ \ \ 6-5\\100-90&\ \ 60-40&\ \ \ \ \ 5+2\end{bmatrix}$

$\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\ \ \ \ \ \text{A}\\=\text{B}\\\ \ \ \ \text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 180\ \ \ &\ \ 100&\ \ \ \ \ 10\\\ \ \ 100&\ \ 20&\ \ \ \ \ 1\\\ \ \ 10&\ \ 20&\ \ \ \ \ 3\end{bmatrix}$

5. (c) $\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}34000000\\16200000\\12000000\end{bmatrix}$

Solution:

The amount of profit in 2020 received by each dealer is given by the matrix.

$\begin{matrix}&\text{Hatchback}&\text{Sedan}&\text{SUV}\end{matrix}\\\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}\begin{bmatrix}\ \ \ \ \ \ 300\ \ \ &\ \ 150&\ \ \ \ \ 20\\\ \ \ 200&\ \ 50&\ \ \ \ \ 6\\\ \ \ 100&\ \ 60&\ \ \ \ \ 5\end{bmatrix}\times\begin{bmatrix}50000\\100000\\200000\end{bmatrix}$

$\begin{matrix}\ \ \ \ \text{A}\\=\text{B}\\\ \ \ \ \text{C}\end{matrix}\begin{bmatrix}15000000+15000000+4000000\\10000000+5000000+1200000\\5000000+6000000+1000000\end{bmatrix}$

$\begin{matrix}\ \ \ \ \ \text{A}\\=\text{B}\\\ \ \ \ \ \text{C}\end{matrix}\begin{bmatrix}34000000\\16200000\\12000000\end{bmatrix}$

1. (a) $\begin{matrix}\text{Notebooks}&\text{Pens}&\text{Pencils}\end{matrix}\\\begin{bmatrix}144&\ \ \ \ \ \ \ \ 60&\ \ \ \ \ 72\\120&\ \ \ \ \ \ \ \ \ 720&\ \ \ \ \ 84\\132&\ \ \ \ \ \ \ \ \ 156&\ \ \ \ \ 96\end{bmatrix}\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}$

Solution:

$\begin{matrix}\ \ \ \ \ \ \ \text{Notebooks}&\text{Pens}&\text{Pencils}\end{matrix}\\\text{X}=\begin{bmatrix}144&\ \ \ \ \ \ \ \ 60&\ \ \ \ \ 72\\120&\ \ \ \ \ \ \ \ \ 720&\ \ \ \ \ 84\\132&\ \ \ \ \ \ \ \ \ 156&\ \ \ \ \ 96\end{bmatrix}\begin{matrix}\text{A}\\\text{B}\\\text{C}\end{matrix}$

2. (b) $\begin{bmatrix}6696\\5916 \\7440\end{bmatrix}$

Solution:

Since, $\text{Y}=\begin{bmatrix}40\\12\\3\end{bmatrix}\begin{matrix}\text{Notebook}\\\text{Pens}\\\text{Pencil}\end{matrix}$

$\therefore\ \ \text{XY}=\begin{bmatrix}144&60&72\\120&72&84\\132&156&96\end{bmatrix}\begin{bmatrix}40\\12\\3\end{bmatrix}$

$=\begin{bmatrix}5760+720+216\\4800+864+252\\5280+1872+288\end{bmatrix}=\begin{bmatrix}6696\\5916\\7440\end{bmatrix}$

3. (d) ₹ 6696

Solution:

Bill of A is ₹ 6696.

4. (c) I

Solution:

(A + I)² = A² + 2A + I = 3A + I

⇒ (A + 1)³ = (3A + I) (A + I)

= 3A² + 4A + I = 7A + I

$\therefore$ (A + I)³ - 7A = I

5. (d) AB = BA

Solution:

A² - B² = (A - B) (A + B) = A² + AB - BA - B²

$\therefore$ AB = BA

1. (a) $\begin{bmatrix}3&0\\43&22\end{bmatrix}$

Solution:

$\text{AB}=\begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}5&2\\7&4\end{bmatrix}$

$=\begin{bmatrix}10-7&4-4\\15+28&6+16\end{bmatrix}=\begin{bmatrix}3&0\\43&22\end{bmatrix}$

2. (c) BA may or may not exist.
3. (b) a = -191, c = 77

Solution:

We have, CD - AB = 0

$\Rightarrow\begin{bmatrix}2&5\\3&8\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}-\begin{bmatrix}3&0\\43&22\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

$\Rightarrow\begin{bmatrix}2\text{a}+5\text{c}&2\text{b}+5\text{d}\\3\text{a}+8\text{c}&3\text{b}+8\text{d}\end{bmatrix}-\begin{bmatrix}3&0\\43&22\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

$\Rightarrow\begin{bmatrix}2\text{a}+5\text{c}-3&2\text{b}+5\text{d}\\3\text{a}+8\text{c}-43&3\text{b}+8\text{d}-22\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

By equality of matrices, we get

2a + 5c - 3 = 0 .......... (i)

3a + 8c - 43 = 0 ........... (ii)

2b+ 5d = 0 ........... (iii)

3b + 8d - 22 = 0 ............. (iv)

4. (c) b = -110, d = 44

Solution:

Solving (iii) and (iv), we get b = -110, d = 44

5. (d) $\begin{bmatrix}-186&-108\\84&48\end{bmatrix}$

Solution:

We have, $\text{B+D}=\begin{bmatrix}5&2\\7&4\end{bmatrix}+\begin{bmatrix}-191&-110\\77&44\end{bmatrix}$

$=\begin{bmatrix}-186&-108\\84&48\end{bmatrix}$

1. (b) $\begin{matrix}&&&&&&&&\text{X}&\ \ \ \ \ \ \ \text{Y}\end{matrix}\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{investment}\end{matrix}\\\text{A = Investment}\begin{bmatrix}15000&20000\end{bmatrix};\ \begin{matrix}\text{B}\ =\text{X}\\\ \ \ \ \ \ \ \ \ \ \text{Y}\end{matrix}\begin{bmatrix}\ \ 0.1\ \ \\\ \ 0.08\ \end{bmatrix}$

Solution:

If ₹ 15000 is invested in bond X, then the amount invested in bond Y = ₹ (35000 - 15000) = ₹ 20000.

$\begin{matrix}&&&&&&&&\text{X}&\ \ \ \ \ \ \ \text{Y}\end{matrix}\\\text{A = Investment}\begin{bmatrix}15000&20000\end{bmatrix}$

 and $\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{investment rate}\end{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{matrix}\text{investment rate}\end{matrix}\\\begin{matrix}\text{B}\ =\text{X}\\\ \ \ \ \ \ \ \ \ \ \text{Y}\end{matrix}\ \ \ \ \ \ \ \begin{bmatrix}\ \ 10\text{%}\ \ \\\ 8\text{%}\ \end{bmatrix} \ \ \ \ \ \ \ =\begin{matrix}\text{X}\\\text{Y}\end{matrix}\ \ \ \ \ \ \ \ \begin{bmatrix}\ \ 0.1\ \ \\0.0 8\ \end{bmatrix}$

2. (c) ₹ 3100

Solution:

The amount of interest received on each bond is given by

$\text{AB}=[15000\ \ \ 20000]\times\begin{bmatrix}0.1\\0.08\end{bmatrix}$

= [15000 × 0.1 + 20000 × 0.08] = [1500 + 1600) = 3100

3. (c) ₹ 20000 in X, ₹ 15000 in Y

Solution:

Let ₹ x be invested in bond X and then ₹ (35000 - x) will be invested in bond Y. 

Now, total amount of interest is given by:

$[\text{x}\ \ \ 35000\ \ -\ \ \ \text{x}]\begin{bmatrix}0.1\\0.08\end{bmatrix}=[0.1\text{x}+(35000-\text{x})0.08]$

But, it is given that total amount of interest = ₹ 3200

$\therefore$ 0.1x + 2800 - 0.08x = 3200

⇒ 0.02x = 400 ⇒ x = 20000

Thus, ₹ 20000 invested in bond X and ₹ 35000 - ₹ 20000 = ₹ 15000 invested in bond Y.

4. (a) AB

Solution:

AB will give the total amount of interest received on both bonds.

5. (b) ₹ 30000

Solution:

Let ₹ x invested in bond X, then we have

$\text{x}\times\frac{10}{100}=500\Rightarrow\text{x}=5000$

Thus, amount invested in bond X is ₹ 5000 and so investment in bond Y be ₹ (35000 - 5000) = ₹ 30000