# Exercise 2.3 [Pages 55 - 56]

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices Exercise 2.3 [Pages 55 - 56]

#### QUESTION

Exercise 2.3 | Q 1.1 | Page 55

Evaluate :

#### SOLUTION

[321][2 -4  3]

= $\left[\begin{array}{ccc}3\left(2\right)& 3\left(-4\right)& 3\left(3\right)\\ 2\left(2\right)& 2\left(-4\right)& 2\left(3\right)\\ 1\left(2\right)& 1\left(-4\right)& 1\left(3\right)\end{array}\right]$

= $\left[\begin{array}{ccc}6& 12& 9\\ 4& 8& 6\\ 2& -4& 3\end{array}\right]$.

#### QUESTION

Exercise 2.3 | Q 1.2 | Page 55

Evaluate :

#### SOLUTION

[2-1  3][431]

= [2(4) –1(3) + 3(1)]
= [8 – 3 + 3]
= [8].

#### QUESTION

Exercise 2.3 | Q 2 | Page 55

If A = $\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right],\text{B}=\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]$, state whether AB = BA? Justify your answer.

#### SOLUTION

AB = $\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right]\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+3+1& -1+0+2& -4+2+1\\ 4+9+0& 2+0+0& 8+6+0\\ 2-9+1& 1+0+2& 4-6+1\end{array}\right]$

∴ AB = $\left[\begin{array}{ccc}2& 1& -1\\ 13& 2& 14\\ -6& 3& -1\end{array}\right]$     ...(i)

BA = $\left[\begin{array}{ccc}2& 1& 4\\ 3& 0& 2\\ 1& 2& 1\end{array}\right]\left[\begin{array}{ccc}-1& 1& 1\\ 2& 3& 0\\ 1& -3& 1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+2+4& 2+3-12& 2+0+4\\ -3+0+2& 3+0-6& 3+0+2\\ -1+4+1& 1+6-3& 1+0+1\end{array}\right]$

∴ BA = $\left[\begin{array}{ccc}4& -7& 6\\ -1& -3& 5\\ 4& 4& 2\end{array}\right]$      ...(ii)
From (i) and (ii), we get
AB ≠ BA.

#### QUESTION

Exercise 2.3 | Q 3 | Page 55

Show that AB = BA, where A = $\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right],\text{B}=\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]$.

#### SOLUTION

AB = $\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]$

= $\left[\begin{array}{ccc}-2+6-3& -6+6-0& 2-3+1\\ -1+4-3& -3+4-0& 1-2+1\\ -6+18-12& -18+18+0& 6-9+4\end{array}\right]$

∴ AB = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$         ...(i)

BA = $\left[\begin{array}{ccc}1& 3& -1\\ 2& 2& -1\\ 3& 0& -1\end{array}\right]\left[\begin{array}{ccc}-2& 3& -1\\ -1& 2& -1\\ -6& 9& -4\end{array}\right]$

= $\left[\begin{array}{ccc}2-3+6& 3+6-9& -1-3+4\\ -4-2+6& 6+4-9& -2-2+4\\ -6+0+6& 9+0-9& 3+0+4\end{array}\right]$

∴ BA = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$       ...(ii)

From (i) and (ii), we get
AB = BA.

#### QUESTION

Exercise 2.3 | Q 4 | Page 55

Verify A(BC) = (AB)C, if A = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right],\text{B}=\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]$

#### SOLUTION

BC = $\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]\left[\begin{array}{ccc}3& 2& -1\\ 2& 0& -2\end{array}\right]$

= $\left[\begin{array}{ccc}6-4& 4-0& -2+4\\ -3+2& -2+0& 1-2\\ 0+6& 0+0& 0-6\end{array}\right]$

= $\left[\begin{array}{ccc}2& 4& 2\\ 1& 2& -1\\ 6& 0& -6\end{array}\right]$

∴ A(BC) = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]\left[\begin{array}{ccc}2& 4& 2\\ -1& -2& -1\\ 6& 0& -6\end{array}\right]$

= $\left[\begin{array}{ccc}2-0+6& 4-0+0& 2-0-6\\ 4-3+0& 8-6+0& 4-3-0\\ 0-4+30& 0-8+0& 0-4-30\end{array}\right]$

∴ A(BC) = $\left[\begin{array}{ccc}8& 4& -4\\ 1& 2& 1\\ 26& -8& -34\end{array}\right]$     ...(i)

AB = $\left[\begin{array}{ccc}1& 0& 1\\ 2& 3& 0\\ 0& 4& 5\end{array}\right]\left[\begin{array}{cc}2& -2\\ -1& 1\\ 0& 3\end{array}\right]$

= $\left[\begin{array}{cc}2+0+0& -2+0+3\\ 4-3+0& -4+3+0\\ 0-4+0& 0+4+15\end{array}\right]$

= $\left[\begin{array}{cc}2& 1\\ 1& -1\\ -4& 19\end{array}\right]$

∴ (AB)C = $\left[\begin{array}{cc}2& 1\\ 1& -1\\ -4& 19\end{array}\right]\left[\begin{array}{ccc}3& 2& -1\\ 2& 0& -2\end{array}\right]$

= $\left[\begin{array}{ccc}6+2& 4+0& -2-2\\ 3-2& 2-0& -1+2\\ -12+38& -8+0& 4-38\end{array}\right]$

∴ (AB)C = $\left[\begin{array}{ccc}8& 4& -4\\ 1& 2& 1\\ 26& -8& 34\end{array}\right]$       ...(ii)

From (i) and (ii), we get
A(BC) = (AB)C.

#### QUESTION

Exercise 2.3 | Q 5 | Page 55

Verify that A(B + C) = AB + AC, if A = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right],\text{B}=\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]\phantom{\rule{1ex}{0ex}}\text{and C}=\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]$.

#### SOLUTION

A(B + C) = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left\{\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]+\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]\right\}$

= $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}-1+4& 1+1\\ 3+2& -2-1\end{array}\right]$

= $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}3& 2\\ 5& -3\end{array}\right]$

= $\left[\begin{array}{cc}12-10& 8+6\\ 6+15& 4-9\end{array}\right]$

= $\left[\begin{array}{cc}2& 14\\ 21& -5\end{array}\right]$          ...(i)

AB + AC = $\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]+\left[\begin{array}{cc}4& -2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}4& 1\\ 2& -1\end{array}\right]$

= $\left[\begin{array}{cc}-4-6& 4+4\\ -2+9& 2-6\end{array}\right]+\left[\begin{array}{cc}16-4& 4+2\\ 8+6& 2-3\end{array}\right]$

= $\left[\begin{array}{cc}-10& 8\\ 7& -4\end{array}\right]+\left[\begin{array}{cc}12& 6\\ 14& -1\end{array}\right]$

= $\left[\begin{array}{cc}-10+12& 8+6\\ 7+14& -4-1\end{array}\right]$

= $\left[\begin{array}{cc}2& 14\\ 21& -5\end{array}\right]$       ...(ii)
From (i) and (ii),  we get
A(B + C) = AB + AC.

#### QUESTION

Exercise 2.3 | Q 6 | Page 56

If  A = $\left[\begin{array}{ccc}4& 3& 2\\ -1& 2& 0\end{array}\right],\text{B}=\left[\begin{array}{cc}1& 2\\ -1& 0\\ 1& -2\end{array}\right]$ show that matrix AB is non singular.

#### SOLUTION

AB = $\left[\begin{array}{ccc}4& 3& 2\\ -1& 2& 0\end{array}\right]\left[\begin{array}{cc}1& 2\\ -1& 0\\ 1& -2\end{array}\right]$

= $\left[\begin{array}{cc}4-3+2& 8+0-4\\ -1-2+0& -2+0+0\end{array}\right]$

= $\left[\begin{array}{cc}3& 4\\ -3& -2\end{array}\right]$

∴ |AB| = $|\begin{array}{cc}3& 4\\ -3& -2\end{array}|$

= – 6 + 12
= 6 ≠ 0
∴ AB is a non-singular matrix.

#### QUESTION

Exercise 2.3 | Q 7 | Page 56

If A + I = $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]$, find the product (A + I)(A − I).

#### SOLUTION

A − I = A I − 2I
= $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]-2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

= $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$

= $\left[\begin{array}{ccc}1-2& 2-0& 0-0\\ 5-0& 4-2& 2-0\\ 0-0& 7-0& -3-2\end{array}\right]$

= $\left[\begin{array}{ccc}-1& 2& 0\\ 5& 2& 2\\ 0& 7& -5\end{array}\right]$

(A + I)(A − I) = $\left[\begin{array}{ccc}1& 2& 0\\ 5& 4& 2\\ 0& 7& -3\end{array}\right]\left[\begin{array}{ccc}-1& 2& 0\\ 5& 2& 2\\ 0& 7& -5\end{array}\right]$

= $\left[\begin{array}{ccc}-1+10+0& 2+4+0& 0+4+0\\ -5+20+0& 10+8+14& 0+8-10\\ 0+35-0& 0+14-21& 0+14+15\end{array}\right]$

$\left[\begin{array}{ccc}9& 6& 4\\ 15& 32& 2\\ 35& -7& 29\end{array}\right]$.

#### QUESTION

Exercise 2.3 | Q 8 | Page 56

If A = $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, show that A2 – 4A is a scalar matrix.

#### SOLUTION

A2 – 4A = A.A – 4A
= $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]-4\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$

= $\left[\begin{array}{ccc}1+4+4& 2+2+4& 2+4+2\\ 2+2+4& 4+1+4& 4+2+2\\ 2+4+2& 4+2+2& 4+4+1\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

= $\left[\begin{array}{ccc}9& 8& 8\\ 8& 9& 8\\ 8& 8& 9\end{array}\right]-\left[\begin{array}{ccc}4& 8& 8\\ 8& 4& 8\\ 8& 8& 4\end{array}\right]$

= $\left[\begin{array}{ccc}9-4& 8-8& 8-8\\ 8-8& 9-4& 8-8\\ 8-8& 8-8& 9-4\end{array}\right]$

= $\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]$, which is a scalar martix.

#### QUESTION

Exercise 2.3 | Q 9 | Page 56

If A = $\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$, find k so that A2 – 8A – kI = O, where I is a 2 × 2 unit and O is null matrix of order 2.

#### SOLUTION

A2 – 8A – kI = O
∴ A.A 8A – kI = O

$\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]-8\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]-\text{k}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]-\left[\begin{array}{cc}8& 0\\ -8& 56\end{array}\right]-\left[\begin{array}{cc}\text{k}& 0\\ 0& \text{k}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\left[\begin{array}{cc}1-8-\text{k}& 0-0-0\\ -8+8-0& 4-6-\text{k}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

∴ By equality of matrices, we get
1 – 8 – k = 0
∴ k = – 7.

#### QUESTION

Exercise 2.3 | Q 10 | Page 56

If A = $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, prove that A2 – 5A + 7I = 0, where I is a 2 x 2 unit matrix.

#### SOLUTION

A2 – 5A + 7I = A.A – 5A + 7I

= $\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

=

= $\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

= $\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]$

= $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

= 0.

#### QUESTION

Exercise 2.3 | Q 11 | Page 56

If A = $\left[\begin{array}{cc}1& 2\\ -1& -2\end{array}\right],\text{B}=\left[\begin{array}{cc}2& \text{a}\\ -1& \text{b}\end{array}\right]$ and (A + B)2  A2 + B2, find the values of a and b.

#### SOLUTION

Given (A + B)2  A2 + B2
∴ A2 + AB + BA + B2 = A2 + B2
∴ AB + BA = 0
∴ AB = – BA

$\left[\begin{array}{cc}2-2& \text{a}+2\text{b}\\ -2+2& -\text{a}-2\text{b}\end{array}\right]=-\left[\begin{array}{cc}2-\text{a}& 4-2\text{a}\\ -1-\text{b}& -2-2\text{b}\end{array}\right]$

$\left[\begin{array}{cc}0& \text{a}+2\text{b}\\ 0& -\text{a}-2\text{b}\end{array}\right]=\left[\begin{array}{cc}-2+\text{a}& -4+2\text{a}\\ 1+\text{b}& 2+2\text{b}\end{array}\right]$

∴ By equality of matrices, we get
– 2 + a = 0 and 1 + b = 0
∴ a = 2 and b = – 1.

#### QUESTION

Exercise 2.3 | Q 12 | Page 56

Find k, if A = $\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$ and A2 = kA – 2I.

#### SOLUTION

A2 = kA – 2I
∴ A.A + 2I = kA

$\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]+2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\text{k}\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$

$\left[\begin{array}{cc}9-8& -6+4\\ 12-8& -8+4\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}1+2& -2+0\\ 4+0& -4+2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

$\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]=\left[\begin{array}{cc}3\text{k}& -2\text{k}\\ 4\text{k}& -2\text{k}\end{array}\right]$

∴ By equality of matrices, we get
3k = 3
∴ k = 1.

#### QUESTION

Exercise 2.3 | Q 13 | Page 56

Find x and y, if $\left\{4\left[\begin{array}{ccc}2& -1& 3\\ 1& 0& 2\end{array}\right]-\left[\begin{array}{ccc}3& -3& 4\\ 2& 1& 1\end{array}\right]\right\}\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

#### SOLUTION

{4[2-13102]-[3-34211]}[2-11]=[xy]

$\left\{\left[\begin{array}{ccc}8& -4& 12\\ 4& 0& 8\end{array}\right]-\left[\begin{array}{ccc}3& -3& 4\\ 2& 1& 1\end{array}\right]\right\}\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{ccc}8-3& -4+3& 12-4\\ 4-2& 0-1& 8-1\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{ccc}5& -1& 8\\ 2& -1& 7\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{c}10+1+8\\ 4+1+7\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

$\left[\begin{array}{c}19\\ 12\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

∴ By equality of matrices, we get
x = 19 and y = 12.

#### QUESTION

Exercise 2.3 | Q 14 | Page 56

Find x, y, x, if $\left\{3\left[\begin{array}{cc}2& 0\\ 0& 2\\ 2& 2\end{array}\right]-4\left[\begin{array}{cc}1& 1\\ -1& 2\\ 3& 1\end{array}\right]\right\}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$.

#### SOLUTION

{3[200222]-4[11-1231]}[12]=[x-3y-12z]

$\left\{\left[\begin{array}{cc}6& 0\\ 0& 6\\ 6& 6\end{array}\right]-\left[\begin{array}{cc}4& 4\\ -4& 8\\ 12& 4\end{array}\right]\right\}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{cc}6-4& 0-4\\ 0+4& 6-8\\ 6-12& 6-4\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{cc}2& -4\\ 4& -2\\ -6& 2\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{c}2-8\\ 4-4\\ -6+4\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

$\left[\begin{array}{c}-6\\ 0\\ -2\end{array}\right]=\left[\begin{array}{c}x-3\\ y-1\\ 2z\end{array}\right]$

∴ By equality of martices, we get
x – 3 = – 6, y – 1  = 0, 2z = – 2
∴ x = – 3, y = 1, z = – 1.

#### QUESTION

Exercise 2.3 | Q 15 | Page 56

Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks, Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.

#### SOLUTION

Let A be the matrix of pens and notebooks and B be the matrix od prices of one pen and one notebook.
Pens Notebooks
∴ A = $\left[\begin{array}{cc}4& 8\\ 5& 12\end{array}\right]\frac{\text{jay}}{\text{Ram}}$

and B = $\left[\begin{array}{c}6\\ 10\end{array}\right]\frac{\text{Pen}}{\text{Notebook}}$

The total amount required for each one of them is obtained by matrix AB.

∴ AB = $\left[\begin{array}{cc}4& 8\\ 5& 12\end{array}\right]\left[\begin{array}{c}6\\ 10\end{array}\right]$

= $\left[\begin{array}{c}24+80\\ 30+120\end{array}\right]$

= $\left[\begin{array}{c}104\\ 150\end{array}\right]$
∴ Jay needs ₹ 104 and Ram needs ₹ 150.

## PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!

## 2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

## HSCMaharashtraBoardPapers2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

## SSCMaharashtraBoardPapers2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)