Algebra Class 9th Mathematics Term 2 Tamilnadu Board Solution

Class 9th Mathematics Term 2 Tamilnadu Board Solution
Exercise 1.1
  1. (5x + 2y + 3z)^2 Expand the following
  2. (2a + 3b - c)^2 Expand the following
  3. (x - 2y - 4z)^2 Expand the following
  4. (P - 2q + r)^2 Expand the following
  5. (x + 1)(x + 4)(x + 7) Find the expansion of
  6. (p + 2)(p - 4)(p + 6) Find the expansion of
  7. (x + 5)(x - 3)(x - 1) Find the expansion of
  8. (x - a)(x - 2a)(x - 4a) Find the expansion of
  9. (3x + 1)(3x + 2)(3x + 5) Find the expansion of
  10. (2x + 3)(2x - 5)(2x - 7) Find the expansion of
  11. (x + 7)(x + 3)(x + 9) Using algebraic identities find the coefficients of x^2…
  12. (x - 5)(x - 4)(x + 2) Using algebraic identities find the coefficients of x^2…
  13. (2x + 3)(2x + 5)(2x + 7) Using algebraic identities find the coefficients of…
  14. (5x + 2)(1 - 5x)(5x + 3) Using algebraic identities find the coefficients of…
  15. If (x + a)(x + b)(x + c) ≡ x^3 - 10x^2 + 45x - 15 find a + b + c, 1/a + 1/b +…
  16. Expand : (i) (3a + 5b)^3 (ii) (4x - 3y)^3 (iii) (2y - 3/y)^3
  17. Evaluate : (i) 99^3 (ii) 101^3 (iii) 98^3 (iv) (102)^3 (v) (1002)^3…
  18. Find 8x^3 + 27y^3 if 2x + 3y = 13 and xy = 6.
  19. If x - y = - 6 and xy = 4, find the value of x^3 - y^3
  20. If x + 1/x = 4 , find the value of x^3 + 1/x^3 .
  21. If, x - 1/x = 4 find the value of x^3 - 1/x^3 .
  22. Simplify: (i) (2x + y + 4z)(4x^2 + y^2 + 16z^2 - 2xy - 4yz - 8zx) (ii) (x - 3y…
  23. Evaluate using identities : (i) 6^3 - 9^3 + 3^3 (ii) 16^3 - 6^3 - 10^3…
Exercise 1.2
  1. 2a^3 - 3a^2 b + 2a^2 c Factorize the following expressions:
  2. 16x + 64x^2 y Factorize the following expressions:
  3. 10x^3 - 25x^4 y Factorize the following expressions:
  4. xy - xz + ay - az Factorize the following expressions:
  5. p^2 + pq + pr + qr Factorize the following expressions:
  6. x^2 + 2x + 1 Factorize the following expressions:
  7. 9x^2 - 24xy + 16y^2 Factorize the following expressions:
  8. b^2 - 4 Factorize the following expressions:
  9. 1 - 36x^2 Factorize the following expressions:
  10. p^2 + q^2 + r^2 + 2pq + 2qr + 2rp Factorize the following expressions:…
  11. a^2 + 4b^2 + 36 - 4ab + 24b - 12a Factorize the following expressions:…
  12. 9x^2 + y^2 + 1 - 6xy + 6x - 2y Factorize the following expressions:…
  13. 4a^2 + b^2 + 9c^2 - 4ab - 6bc + 12ca Factorize the following expressions:…
  14. 25x^2 + 4y^2 + 9z^2 - 20xy + 12yz - 30zx Factorize the following expressions:…
  15. 27x^3 + 64y^3 Factorize the following expressions:
  16. m^3 + 8 Factorize the following expressions:
  17. a^3 + 125 Factorize the following expressions:
  18. 8x^3 - 27y^3 Factorize the following expressions:
  19. x^3 - 8y^3 Factorize the following expressions:
Exercise 1.4
  1. x + 3y = 10; 2x + y = 5 Solve the following equations by substitution method.…
  2. 2x + y = 1; 3x - 4y = 18 Solve the following equations by substitution method.…
  3. 5x + 3y = 21; 2x - y = 4 Solve the following equations by substitution method.…
  4. 1/x + 2/y = 9 2/x + 1/y = 12 (x not equal 0 , y not equal 0) Solve the…
  5. 3/x + 1/y = 7 5/x - 4/y = 6 (x not equal 0 , y not equal 0) Solve the following…
  6. Find two numbers whose sum is 24 and difference is 8.
  7. A number consists of two digits whose sum is 9. The number formed by reversing…
  8. Kavi and Kural each had a number of apples. Kavi said to Kural “If you give me 4…
  9. Solve the following inequations. (i) 2x + 715 (ii) 2(x - 2) 3 (iii) 2(x + 7) ≤ 9…
Exercise 1.5
  1. The expansion of (x + 2)(x - 1) isA. x^2 - x - 2 B. x^2 + x + 2 C. x^2 + x - 2…
  2. The expansion of (x + 1)(x - 2)(x + 3) isA. x^3 + 2x^2 - 5x - 6 B. x^3 - 2x^2 +…
  3. (x - y)(x^2 + xy + y^2) is equal toA. x^3 + y^3 B. x^2 + y^2 C. x^2 - y^2 D. x^3…
  4. Factorization of x^2 + 2x - 8 isA. (x + 4)(x - 2) B. (x - 4)(x + 2) C. (x + 4)(x…
  5. If one of the factors of x^2 - 6x - 16 is (x + 2) then other factor isA. x + 5…
  6. If (2x + 1) and (x - 3) are the factors of ax^2 - 5x + c, then the values of a…
  7. If x + y = 10 and x - y = 2, then value of x isA. 4 B. - 6 C. - 4 D. 6…
  8. The solution of 2 - x 5 isA. x - 3 B. x - 3 C. x 3 D. x 3

Exercise 1.1
Question 1.

Expand the following

(5x + 2y + 3z)2


Answer:

(5x + 2y + 3z)2


The identity is


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (5x + 2y + 3z)2


= (5x)2 + (2y)2 + (3z)2 + 2(5x)(2y) + 2(2y)(3z) + 2(3z)(5x)


= 25x2 + 4y2 + 9z2 + 20xy + 12yz + 30zx



Question 2.

Expand the following

(2a + 3b – c)2


Answer:

(2a + 3b – c)2


Here the identity used is


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


(2a + 3b – c)2 = (2a)2 + (3b)2 + (– c)2 + 2(2a)(3b) + 2(3b)(– c) + 2(– c)(2a)


= 4a2 + 9b2 + c2 + 12ab – 6bc – 4ca



Question 3.

Expand the following

(x – 2y – 4z)2


Answer:

(x – 2y – 4z)2


Here the identity used is


(x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (x – 2y – 4z) 2


= x2 + (– 2y)2 + (– 4z)2 + 2(x)(– 2y) + 2(– 2y)(– 4z) + 2 (– 4z)(x)


= x2 + 4y2 + 16z2 – 4xy + 16yz – 8yz



Question 4.

Expand the following

(P – 2q + r)2


Answer:

(P – 2q + r)2


Here we use the identity of


(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx


∴ (p – 2q + r)2


= (p)2 + (– 2q)2 + (r)2 + 2(p)(– 2q) + 2(– 2q)(r) + 2(r)(p)


= p2 + 4q2 + r2 – pq – 4qr + 2rp



Question 5.

Find the expansion of

(x + 1)(x + 4)(x + 7)


Answer:

(x + 1) (x + 4) (x + 7)


we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x + 1)(x + 4)(x + 7)


= x3 + (1 + 4 + 7) x2 + (1 × 4 + 4 × 7 + 7 × 1) x + 1 × 4 × 7


= x3 + 12x2 + (4 + 28 + 7) x + 28


= x3 + 12x2 + 39x + 28



Question 6.

Find the expansion of

(p + 2)(p – 4)(p + 6)


Answer:

(p + 2)(p – 4)(p + 6)


Here the identity used is


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (p + 2)(p – 4)(p + 6)


= p3 + (2 – 4 + 6) p2 + (2(– 4) + (– 4)6 + 6(2))p + (2)(– 4)(6)


= p3 + 4p2 + (– 8 – 24 + 12)p – 48


= p3 + 4p2 – 20p – 48



Question 7.

Find the expansion of

(x + 5)(x – 3)(x – 1)


Answer:

(x + 5)(x – 3)(x – 1)


Here the identity used is


(x + a)(x + b)(x + c) = x3 + (a + b + c) x2 + (ab + bc + ca)x + abc


∴ (x + 5)(x – 3)(x – 1)


= x3 + (5 – 3 – 1) x2 + (5(– 3) + (– 3)(– 1)(– 1)5)x + 5(– 3)(– 1)


= x3 + x2 + (– 15 + 3 – 5) x + 15


= x3 + x2 – 17x + 15



Question 8.

Find the expansion of

(x – a)(x – 2a)(x – 4a)


Answer:

(x – a)(x – 2a)(x – 4a)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x – a)(x – 2a)(x – 4a)


= x3 + (– a – 2a – 3a)x2 + [(– a)(– 2a) + (– 2a)(– 3a) + (– 3a)(– a)]x + (– a)(– 2a)(– 3a)


= x3 – 6ax2 + (2a2 + 6a2 + 3a2)x – 6a3


= x3 – 6ax2 + 11a2x – 6a3



Question 9.

Find the expansion of

(3x + 1)(3x + 2)(3x + 5)


Answer:

(3x + 1)(3x + 2)((3x + 5)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (3x + 1)(3x + 2)((3x + 5)


= (3x)3 + (1 + 2 + 5)(3x)2 + (1 × 2 + 2 × 5 + 5 × 1)(3x) + 1 × 2 × 5


= 27 x3 + (8)9x2 + (2 + 10 + 5) (3x) + 1 × 2 × 5


= 27 x3 + 72x2 + 51x + 10



Question 10.

Find the expansion of

(2x + 3)(2x – 5)(2x – 7)


Answer:

(2x + 3)(2x – 5)(2x – 7)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (2x + 3)(2x – 5)(2x – 7)


= (2x)3 + (3 – 5 – 7)(2x)2 + [(3)(– 5) + (– 5)(– 7) + (– 7)(3)](2x) + (3)(– 5)(– 7)


= 8x3 + (– 9)(4x2) + (– 15 + 35 – 21)(2x) + 105


= 8x3 – 36 x2 – 2x + 105



Question 11.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(x + 7)(x + 3)(x + 9)


Answer:

Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x + 7)(x + 3)(x + 9)


= x3 + (7 + 3 + 9)x2 + (7 × 3 + 3 × 9 + 9 × 7)x + 7 × 3 × 9


= x3 + 19x2 + (21 + 27 + 63)x + 189


= x3 + 19x2 + 111x + 189


Coefficient of


x2 = 19


x = 111


Constant term = 189



Question 12.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(x – 5)(x – 4)(x + 2)


Answer:

(x – 5)(x – 4)(x + 2)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ (x – 5)(x – 4)(x + 2)


= x3 + (– 5 – 4 + 2)x2 + {(– 5) × (– 4) + (– 4) × 2 + 2 × (– 5)}x + (– 5)(– 4)2


= x3 – 7x2 + (20 – 8 – 10)x + 40


= x3 – 7x2 + 2x + 40


Coefficient of


x2 = – 7


x = 2 and


Constant term = 40



Question 13.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(2x + 3)(2x + 5)(2x + 7)


Answer:

(2x + 3)(2x + 5)(2x + 7)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)(x)2 + (ab + bc + ca)x + abc


∴ (2x + 3) (2x + 5)(2x + 7)


= (2x)3 + 4(3 + 5 + 7)x2 + (15 + 35 + 21)2x + 105


= 8x3 + 60x2 + 142x + 105


Coefficient of


x2 = 60


x = 142


Constant term = 105



Question 14.

Using algebraic identities find the coefficients of x2 term, x term and constant term.

(5x + 2)(1 – 5x)(5x + 3)


Answer:

(5x + 2)(1 – 5x)(5x + 3)


Here we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


∴ – (5x + 2)(5x – 1)(5x + 3)


here x = 5x, a = 2, b = – 1 and c = 3


= – (5x)3 – (2 – 1 + 3)(5x)2 – (– 2 – 3 + 6)5x + 6


= – 125x3 – 100x2 – 5x + 6


Here coefficient of


x2 = – 100


x = – 5 and


constant term = 6



Question 15.

If (x + a)(x + b)(x + c) ≡ x3 – 10x2 + 45x – 15 find a + b + c,  and a2 + b2 + c2.


Answer:

Here, we have to use the identity


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


Comparing this with


x3 – 10x2 + 45x – 15 = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


⇒ – 10 = a + b + c


Now, 



And (ab + bc + ca) = 45


Now (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)


⇒ (– 10)2 – 2 × 45 = a2 + b2 + c2


⇒ 100 – 90 = a2 + b2 + c2


⇒ a2 + b2 + c2 = 10



Question 16.

Expand :

(i) (3a + 5b)3

(ii) (4x – 3y)3

(iii) 


Answer:

(i) (3a + 5b)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(3a + 5b)3


= (3a)3 + 3 (3a)2(5b) + 3(3a)(5b)2 + (5b)3


= 27a3 + 135a2b + 225ab2 + 125b3


(ii) (4x – 3y)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(4x – 3y)3


= (4x)3 + 3(4x)2(– 3y) + 3 × 4x (– 3y)2 + (– 3y)3


= 64x3 – 192 x2 + 108y2 – 27y3


(iii)


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


=




Question 17.

Evaluate :

(i) 993

(ii) 1013

(iii) 983

(iv) (102)3

(v) (1002)3


Answer:

(i) (100 – 1)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 – 1)3


= (100)3 + 3(100)2 (– 1) + 3(100) (– 1)2 + (– 1)3


= 1000000 – 30000 + 300 – 1


= 970299


(ii) 1013


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(101)3 = (100 + 1)3


= 1003 + 3 × 1002 × 1 + 3 × 100 × 12 + 13


= 1000000 + 30000 + 300 + 1


= 1030301


(iii) 983


= (100 – 2)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 – 2)3


= 1003 + (– 2)3 + 3(100)2(– 2) + 3(100)(– 2)2


= 1000000 – 8 – 60000 + 1200


= 941192


(iv) (102)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(100 + 2)3


= 1003 + 3(100)2 (2) + 3(2)2(100) + (2)3


= 1000000 + 60000 + 1200 + 8


= 1061208


(vi) (1002)3


= (1000 + 2)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(1000 + 2)3


= 10003 + 3 × 10002(2) + 3 × 1000 × (2)2 + 23


= 1000000000 + 6000000 + 12000 + 8


= 1006012008



Question 18.

Find 8x3 + 27y3 if 2x + 3y = 13 and xy = 6.


Answer:

Given 2x + 3y = 13


⇒ (2x + 3y)3 = 133


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(2x + 3y)3 = (2x)3 + 3(2x)23y + 3(2x)(3y)2 + (3y)3


⇒ 8x3 + 36x2y + 54xy2 + 27y3 = 2197


⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197


⇒ 8x3 + 27y3 = 2197 – 108 (2x + 3y) (∵ xy = 6 given)


⇒ 8x3 + 27y3 = 2197 – 108 × 13


⇒ 8x3 + 27y3 = 2197 – 1404


⇒ 8x3 + 27y3 = 793



Question 19.

If x – y = – 6 and xy = 4, find the value of x3 – y3


Answer:

The identity used here is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ (x – y)3 = x3 + 3x2(– y) + 3x(– y)2 + (– y)3


⇒ x3 – 3x2y + 3xy2 – y3


⇒ (x – y)3 = x3 – 3x2y + 3xy2 – y3


⇒ (– 6)3 = x3 – y3 + 3xy(– x + y)


⇒ – 216 + 3 × 4 (– 6) = x3 – y3


⇒ – 216 – 72 = x3 – y3


⇒ – 288 = x3 – y3



Question 20.

If , find the value of .


Answer:

Here we use the identity of


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ 


⇒ 


⇒ 


⇒ 



Question 21.

If,  find the value of .


Answer:

Here the identity used


(a + b)3 = a3 + 3a2b + 3ab2 + b3


∴ 


⇒ 


⇒ 


⇒ 



Question 22.

Simplify:

(i) (2x + y + 4z)(4x2 + y2 + 16z2 – 2xy – 4yz – 8zx)

(ii) (x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yz + 5zx)


Answer:

(i) (2x + y + 4z) (4x2 + y2 + 16z2 – 2xy – 4yz – 8zx)


⇒ Here the identity used is


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


Comparing this with given condition


(2x + y + 4z) (4x2 + y2 + 16z2 – 2xy – 4yz – 8zx) = (2x)3 + y3 + (4z)3 – 3 × 2x × y × 4z


⇒ 8x3 + y3 + 64z3 – 24xyz


(ii) (x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yx + 5zx)


⇒ Here the identity used is


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


Comparing this with given condition


(x – 3y – 5z)(x2 + 9y2 + 25z2 + 3xy – 15yx + 5zx)


= x3 + (– 3y)3 + (– 5z)3 – 3(x) (– 3y)(– 5z)


= x3 – 27y3 – 125z3 – 45xyz



Question 23.

Evaluate using identities :

(i) 63 – 93 + 33

(ii) 163 – 63 – 103


Answer:

(i) 63 – 93 + 33


Here the identity used will be


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


∴ 63 – 93 + 33


⇒ (6 + (– 9) + 3)(62 + (– 9)2 + 32 – 6 × – 9 – (– 9)3 – 6 × 3) + 3 × 6 × – 9 × 3


⇒ 0 (36 + 81 + 9 + 54 + 27 – 18) – 486


⇒ – 486


(ii) 163 – 63 – 103


Here the identity used will be


a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)


∴ 163 – 63 – 103


= (16 – 6 – 10) (162 + (– 6)2 + (– 10)2 – 16 × – 6 – (– 6)(– 10) – (16)(– 10)) + 3(16) (– 6) (– 10)


= 0 (256 + 36 + 100 + 96 – 60 + 160) + 2880


= 2880




Exercise 1.2
Question 1.

Factorize the following expressions:

2a3 – 3a2b + 2a2c


Answer:


= 2a3 – 3a2b + 2a2c
Highest common factor is a2

Taking a2 common in the term – 3a2b + 2a2c


we get,
= 2a3 + a2(– 3b + 2c)
Taking a2 common in both the terms
we get,


= a2(2a + (– 3b + 2c))
= a2(2a – 3a + 2c)
2a3 – 3a2b + 2a2c = a2(2a – 3a + 2c)



Question 2.

Factorize the following expressions:

16x + 64x2y


Answer:


= 16x + 64x2y

Highest common factor is 16x


Taking x common in the term 16x + 64x2y


we get,


= x(16 + 64xy)


= x(16 + 16(4)xy)


Taking 16 common in the term (16 + 64xy)


we get,


= 16x(1 + 4xy)


16x + 64x2y = 16x(1 + 4xy)



Question 3.

Factorize the following expressions:

10x3 – 25x4y


Answer:


= 10x3 – 25x4y
Highest common factor is 5x3

Taking x3common in the term 10x3 – 25x4y


we get,


= x3 (10– 25xy)


= x3 (5(2)– 5(2)xy)
Taking 5 common in both the terms


we get,


= 5x3 (2– 5xy)
10x3 – 25x4y = 5x3 (2– 5xy)



Question 4.

Factorize the following expressions:

xy – xz + ay – az


Answer:


= xy – xz + ay – az
Taking x common in xy – xz

we get,


= x(y – z) + ay – az


Taking ‘a’ common in ay – az


we get,


= x(y – z) + a(y – z)
Taking (y – z) common in both the terms


we get,
= (y – z)(x + a)
xy – xz + ay – az = (y – z)(x + a)



Question 5.

Factorize the following expressions:

p2 + pq + pr + qr


Answer:

= p2 + pq + pr + qr
Taking p common in p2 + pq


we get,


= p(p + q) + pr + qr
Taking r common in pr + qr
we get,
= p(p + q) + r(p + q)


Taking (p + q) common in both the terms


we get,
= (p + q)(p + r)
p2 + pq + pr + qr = (p + q)(p + r)



Question 6.

Factorize the following expressions:

x2 + 2x + 1


Answer:


= x2 + 2x + 1

Method 1


It is of the form a2 + 2ab + b2
Where a = x and b = 1.


Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ x2 + 2x + 1 = (x + 1)2


Method 2
Here 2x can be elaborated as x + x


we get,


= x2 + x + x + 1


Taking x common in x2 + x


we get,


= x(x + 1) + x + 1
Taking (x + 1) common in both the terms


we get,
= (x + 1)(x + 1)
= (x + 1)2
x2 + 2x + 1 = (x + 1)2



Question 7.

Factorize the following expressions:

9x2 – 24xy + 16y2


Answer:


= 9x2 – 24xy + 16y2
Given equation can be simplified as,

= (3x)2 – 2(3x)(4y) + (4y)2


It is of the form a2 – 2ab + b2
Where a = 3x and b = 4y
∴ using the identity (a – b)2 = a2 – 2ab + b2


We get,
⇒ 9x2 – 24xy + 16y2 = (3x – 4y)2


9x2 – 24xy + 16y2 = (3x – 4y)2



Question 8.

Factorize the following expressions:

b2 – 4


Answer:

= b2 – 4
Given equation can be simplified as,


= (b)2 – (2)2


It is of the form (p)2 – (q)2
where p = b and q = 2
∴ using the identity (p)2 – (q)2 = (p + q)(p – q)


We get,
⇒ b2 – 22 = (b + 2)(b – 2)
b2 – 4 = (b + 2)(b – 2)



Question 9.

Factorize the following expressions:

1 – 36x2


Answer:


= 1 – 36x2
Given equation can be simplified as,

= (1)2 – (6x)2


It is of the form (p)2 – (q)2
Where p = 1 and q = 6x
∴ using the identity: (p)2 – (q)2 = (p + q)(p – q)


We get,
⇒ (1)2 – (6x)2 = (1 + 6x)(1 – 6x)


1 – 36x2 = (1 + 6x)(1 – 6x)



Question 10.

Factorize the following expressions:

p2 + q2 + r2 + 2pq + 2qr + 2rp


Answer:


Method 1
= p2 + q2 + r2 + 2pq + 2qr + 2rp
= (p)2 + (q)2 + (r)2 + 2(p)(q) + 2(q)(r) + 2(r)(p)
It is of the from a2 + b2 + c2 + 2ab + 2bc + 2ca

where a = p, b = q, c = r


Using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca


We get,
⇒ p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2


Method 2


= p2 + q2 + r2 + 2pq + 2qr + 2rp
The above equation can be simplified as


= p2 + 2pq + q2 + r2 + 2qr + 2rp
= (p + q)2 + r2 + 2qr + 2rp (∵ a2 + 2ab + b2 = (a + b)2)


Taking 2r common in the term 2qr + 2rp


we get,


= (p + q)2 + r2 + 2r(q + p)
Rearranging the above equation as
= (p + q)2 + 2r(q + p) + r2


This is of the from a2 + 2ab + b2


Where, a = p + q and b = r
Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ p2 + q2 + r2 + 2pq + 2qr + 2rp = ((p + q) + r)2


= (p + q + r)2
p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2



Question 11.

Factorize the following expressions:

a2 + 4b2 + 36 – 4ab + 24b – 12a


Answer:

= a2 + 4b2 + 36 – 4ab + 24b – 12a


Method 1


The above equation can be simplified as : 
= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p = – a, q = 2b, r = 6
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2


= (– a + 2b + 6)2


= (– 1)2(a – 2b – 6)2


= (a – 2b – 6)2


Method 2


The above equation can be simplified as


= (– a)2 + (2b)2 + (6)2 + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
= {(– a)2 + 2(– a)(2b) + (2b)2} + (6)2 + 2(2b)(6) + 2(– a)(6)
(∵ a2 + 2ab + b2 = (a + b)2)
(– a + 2b)2 + (6)2 + 2(2b)(6) + 2(– a)(6)
Taking 2(6) common in term 2(2b)(6) + 2(– a)(6)
= (– a + 2b)2 + 2(6)(– a + 2b) + (6)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = – a, q = 2b and r =6


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2


= (– a + 2b + 6)2


= (– 1)2(a – 2b – 6)2


= (a – 2b – 6)2


a2 + 4b2 + 36 – 4ab + 24b – 12a = (– a + 2b + 6)2




Question 12.

Factorize the following expressions:

9x2 + y2 + 1 – 6xy + 6x – 2y


Answer:

Method 1


The above equation can be simplified as : 
= (3x)2 + (– y)2 + (1)2 + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p =3x, q = – y, r = 1
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2


= (3x – y + 1)2


Method 2


The above equation can be simplified as:


= (3x)2 + (– y)2 + (1)2 + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
= {(3x)2 + 2(3x)(– y) + (– y)2} + (1)2 + 2(3x)(1) + 2(– y)(1)
(∵ p2 + 2pq + q2 = (p + q)2)
= (3x – y)2 + (1)2 + 2(3x)(1) + 2(– y)(1)
Taking 2(1) common in term 2(3x)(1) + 2(– y)(1)
= (3x – y)2 + (1)2 + 2(1)(3x – y)
= (3x – y)2 + 2(1)(3x – y) + (1)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = 3x, q = – y and r = 1


Using the identity: (a + b)2 = a2 + 2ab + b2


We get,


⇒ 9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2
9x2 + y2 + 1 – 6xy + 6x – 2y = (3x – y + 1)2



Question 13.

Factorize the following expressions:

4a2 + b2 + 9c2 – 4ab – 6bc + 12ca


Answer:


Method 1

The above equation can be simplified as : 
= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p =2a, q = – b, r = 3c
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2


= (2a – b + 3c)2


Method 2


The above equation can be simplified as:


= (2a)2 + (– b)2 + (3c)2 + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
= {(2a)2 + 2(2a)(– b) + (– b)2} + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
(∵ p2 + 2pq + q2 = (p + q)2)


= (2a – b)2 + (3c)2 + 2(– b)(3c) + 2(3c)(2a)
Taking 2(3c) common in term 2(– b)(3c) + 2(3c)(2a)
= (2a – b)2 + (3c)2 + 2(3c)(– b + 2a)
= (2a – b)2 + 2(3c)(2a – b) + (3c)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = 2a, q = – b and r = 3c


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ 4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2
4a2 + b2 + 9c2 – 4ab – 6bc + 12ca = (2a – b + 3c)2




Question 14.

Factorize the following expressions:

25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx


Answer:


Method 1

The above equation can be simplified as : 
= (– 5x)2 + (2y)2 + (3z)2 + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
This equation is of the form: p2 + q2 + r2 + 2pq + 2qr + 2rp


where p = – 5x, q = 2y, r = 3z
Using the identity: p2 + q2 + r2 + 2pq + 2qr + 2rp = (p + q + r)2
We get,
⇒ 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)2


= (– 5x + 2y + 3z)2


= (– 1)2(5x – 2y – 3z)2


= (5x – 2y – 3z)2


Method 2


The above equation can be simplified as:


= (– 5x)2 + (2y)2 + (3z)2 + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
= {(– 5x)2 + 2(– 5x)(2y) + (2y)2} + (3z)2 + 2(2y)(3z) + 2(3z)(– 5x)
(∵ p2 + 2pq + q2 = (p + q)2)
= (– 5x + 2y)2 + (3z)2 + 2(2y)(3z) + 2(3z)(– 5x)
Taking 2(3z) common in term 2(2y)(3z) + 2(3z)(– 5x)
= (– 5x + 2y)2 + (3z)2 + 2(3z)(2y – 5x)
= (– 5x + 2y)2 + 2(3z)(2y – 5x) + (3z)2


This of the form: (p + q)2 + 2r(p + q) + r2


Where, p = – 5x, q = 2y and r = 3z


Using the identity: (p + q)2 = p2 + 2pq + q2


We get,


⇒ 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)2


= (– 5x + 2y + 3z)2


= (– 1)2(5x – 2y – 3z)2


= (5x – 2y – 3z)2


25x2 + 4y2 + 9z2 – 20xy + 12yz – 30zx = (5x – 2y – 3z)2



Question 15.

Factorize the following expressions:
27x3 + 64y3


Answer:

The above equation can be simplified as


= (3x)3 + (4y)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)


Here a = 3x and b = 4y


⇒ (3x)3 + (4y)3 = (3x + 4y)((3x)2 – (3x)(4y) + (4y)2)


= (3x + 4y)(9x2 – 12xy + 16y2)
27x3 + 64y3 = (3x + 4y)(9x2 – 12xy + 16y2)



Question 16.

Factorize the following expressions:
m3 + 8


Answer:

The above equation can be simplified as


= (m)3 + (2)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)
Here a = m and b = 2


⇒ (m)3 + (2)3 = (m + 2)((m)2 – (m)(2) + (2)2)


= (m + 2)(m2 – 2m + 4)
m3 + 8 = (m + 2)(m2 – 2m + 4)



Question 17.

Factorize the following expressions:
a3 + 125


Answer:

The above equation can be simplified as


= (a)3 + (5)3
Both the given terms are perfect cubes, factor using the sum of cubes formula,


(a)3 + (b)3 = (a + b)(a2 – ab + b2)
Here a = a and b = 5


⇒ (a)3 + (5)3 = (a + 5)((a)2 – (a)(5) + (5)2)


= (a + 5)(a2 – 5a + 25)
a3 + 123 = (a + 5)(a2 – 5a + 25)



Question 18.

Factorize the following expressions:
8x3 – 27y3


Answer:

The above equation can be simplified as


= (2x)3 – (3y)3
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,


(a)3 – (b)3 = (a – b)(a2 + ab + b2)
Here a = 2x and b = 3y


⇒ (2x)3 – (3y)3 = (2x – 3y)((2x)2 + (2x)(3y) + (3y)2)


= (2x – 3y)(4x2 + 6xy + 9y2)
8x3 – 27y3 = (2x – 3y)(4x2 + 6xy + 9y2)



Question 19.

Factorize the following expressions:
x3 – 8y3


Answer:

The above equation can be simplified as


= (x)3 – (2y)3
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,


(a)3 – (b)3 = (a – b)(a2 + ab + b2)
Here a = x and b = 2y


⇒ (x)3 – (2y)3 = (x – 2y)((x)2 + (x)(2y) + (2y)2)


= (x – y)(x2 + 2xy + 4y2)
x3 – 8y3 = (x – y)(x2 + 2xy + 4y2)





Exercise 1.4
Question 1.

Solve the following equations by substitution method.

x + 3y = 10; 2x + y = 5


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


x + 3y = 10……(1)


2x + y = 5 …..(2)


From eq(1),


X + 3y = 10


⇒ x = 10 – 3y


Substituting the value of x in eq(2)


2x + y = 5


⇒ 2(10 – 3y) + y = 5


⇒ 20 – 6y + y = 5


⇒ 20 – 5y = 5


⇒ 5y = 20 – 5


⇒ 5y = 15




Substituting value of y in eq(1)


x + 3y = 10


⇒ x + 3×3 = 10


⇒ x + 9 = 10


⇒ x = 10 – 9


⇒ x = 1


Hence,


x = 1, y = 3




Question 2.

Solve the following equations by substitution method.

2x + y = 1; 3x – 4y = 18


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


2x + y = 1….(1)


3x – 4y = 18….(2)


From eq(1),


2x + y = 1….(1)


⇒ y = 1 – 2x


Substituting the value of y in eq(2)


3x – 4y = 18….(2)


⇒ 3x – 4(1 – 2x) = 18


⇒ 3x – 4 + 8x = 18


⇒ 11x + 4 = 18


⇒ 11x = 18 + 4


⇒ 11x = 22


⟹x = 


⇒ x = 2


Substituting the value of x in eq(1),


⇒ 2x + y = 1


⇒ 2×2 + y = 1


⇒ y = 1 – 4


⇒ y = – 3


Hence,


x = 2,y = – 3



Question 3.

Solve the following equations by substitution method.

5x + 3y = 21; 2x – y = 4


Answer:

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]


5x + 3y = 21……(1)


2x – y = 4…….(2)


From eq(2)


2x – y = 4


⟹y = 2x – 4…..(3)


Substituting value of y in eq(1)


5x + 3y = 21


⇒ 5x + 3(2x – 4) = 21


⇒ 5x + 6x – 12 = 21


⇒ 11x = 33


⇒ x = 


⇒ x = 3


From eq(3)


y = 2x – 4


⇒ y = 2×3 – 4


⇒ y = 2


Hence,


x = 3, y = 2



Question 4.

Solve the following equations by substitution method.



Answer:

In such questions, if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]


Let


 = u


 = v


Then,


 = 9


⇒ u + 2v = 9……(1)


And


 = 12


⇒ 2u + v = 12……(2)


From eq(1)


u + 2v = 9


⇒ u = 9 – 2v……(3)


Substituting value of u in eq(2)


2u + v = 12


⇒ 2(9 – 2v) + v = 12


⟹18 – 4v + v = 12


⇒ 18 – 3v = 12


⇒ 3v = 18 – 12


⇒ 3v = 6


⇒ v = 2


⇒  = 2



Substituting value of v in eq(3)


u = 9 – 2v


⇒ u = 9 – 2×2


⇒ u = 5


⇒  = 5


⇒ 


Hence,


x = ,y = 



Question 5.

Solve the following equations by substitution method.



Answer:

In such questions , if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]


 +  = 7


⇒ 3u + v = 7……(1)


 = 6


⇒ 5u – 4v = 6 ……..(2)


From eq(1)


v = 7 – 3u…..(3)


Substituting the value of v in eq(2)


5u – 4(7 – 3u) = 6


⇒ 5u – 28 + 12u = 6


⇒ 17u = 6 + 28


⇒ 17u = 34


⇒ u = 34/17


⇒ u = 2


⇒  = 2



Substituting the value of u in eq(3)


v = 7 – 3×2


⇒ v = 1


⇒  = 1


⇒ y = 1


Hence, x = , y = 1



Question 6.

Find two numbers whose sum is 24 and difference is 8.


Answer:

Let the two numbers be x and y.


Since, it is given that their sum is 24,


We can write, x + y = 24


For forming second equation, we will use their difference, which is 8.


So, we can write,


x – y = 8


(we can assume any number to be greater. Here, we assumed the greater number is x)


Here, coefficient of y is equal in magnitude and opposite in sign.


So, if we add the above two equations, it will be eliminated.


Adding equation (1)and(2)


We get


2x = 32


⇒ x = 


⇒ x = 16


Substituting the value of x in eq(1)


We get


x + y = 24


⇒ 16 + y = 24


⇒ y = 24 – 16


⇒ y = 8


Hence, the two required numbers are 16 and 8.



Question 7.

A number consists of two digits whose sum is 9. The number formed by reversing the digits exceeds twice the original number by 18. Find the original number.


Answer:

Let unit digit of the number be x and its ten’s digit be y.


Then the number will be 10×y + x = 10y + x


Sum of its digits is 9.


Therefore, the sum of x and y is 9


x + y = 9……(1)


On reversing the number, ten’s digit will become unit’s digit and unit’s digit will become ten’s digit.


So, after reversing the number,


Its ten’s digit = x and unit’s digit = y


Therefore, the new number formed = 10x + y


New number formed exceeds the twice the original number by 18


⇒ 10x + y = 2(10y + x) + 18


⇒ 10x + y = 20y + 2x + 18


⇒ 8x – 19y = 18…..(2)


From equation(1),


x + y = 9


⇒ x = 9 – y…..(3)


Substituting the value of x in eq(2)


8(9 – y) – 19y = 18


⇒ 72 – 8y – 19y = 18


⇒ – 27y = 18 – 72


⇒ – 27y = – 54


⇒ y = 2


Substituting the value of y in eq(3)


x = 9 – 2


⇒ x = 7


Hence, the original number is 10y + x = 10×2 + 7 = 27



Question 8.

Kavi and Kural each had a number of apples. Kavi said to Kural “If you give me 4 of your apples, my number will be thrice yours”. Kural replied, “If you give me 26, my number will be twice yours”. How many did each have with them?.


Answer:

Let Kavi has x apples and Kural has y apples.


According to Kavi’s statement,


We can write


x + 4 = 3(y – 4)


x + 4 = 3y – 12


⇒ x = 3y – 16…..(1)


According to Kural’s statement


y + 26 = 2(x – 26)


y + 26 = 2x – 52…….(2)


Substituting the value of x from eq(1) in eq(2)


y + 26 = 2(3y – 16) – 52


⇒ y + 26 = 6y – 32 – 52


⇒ y + 26 = 6y – 84


⇒ 5y = 110


⇒ y = 22


Substituting value of y in eq(1)


x = 3×22 – 16


⇒ x = 50


Hence, Kavi has 50 apples and Kural has 22 apples.



Question 9.

Solve the following inequations.

(i) 2x + 7>15

(ii) 2(x – 2) < 3

(iii) 2(x + 7) ≤ 9

(iv) 3x + 14 ≥ 8


Answer:

(i) 2x + 7>15


2x>15 – 7


⇒ 2x>8


⇒ x>


⇒ x>4


(ii) 2x – 4<3


⇒ 2x>3 + 4


⇒ 2x>7


⇒ x>7/2


(iii) 2(x + 7)≤9


⇒ 2x + 14≤9


⇒ 2x≤9 – 14


⇒ 2x≤ – 5



(iv) 3x + 14≥8


⇒ 3x≥8 – 14


⇒ 3x≥ – 6


⇒ x≥


⇒ x≥ – 2




Exercise 1.5
Question 1.

The expansion of (x + 2)(x – 1) is
A. x2 – x – 2

B. x2 + x + 2

C. x2 + x – 2

D. x2 – x + 2


Answer:

Using identity,


(x + a)(x + b) = x2 + (a + b)x + ab


We have,


(x + 2)(x – 1)


= x2 + (2 – 1) x + (2x – 1) = x2 + x – 2


Question 2.

The expansion of (x + 1)(x – 2)(x + 3) is
A. x3 + 2x2 – 5x – 6

B. x3 – 2x2 + 5x – 6

C. x3 + 2x2 + 5x – 6

D. x3 + 2x2 + 5x + 6


Answer:

Using identity,


(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc


We have,


(x + 1)(x – 2)(x + 3) = x3 + (1 – 2 + 3)x2 + (– 2 – 6 + 3)x – 6


= x3 + 2x2 – 5x – 6


Question 3.

(x – y)(x2 + xy + y2) is equal to
A. x3 + y3

B. x2 + y2

C. x2 – y2

D. x3 – y3


Answer:

(x – y)(x2 + xy + y2) is simply the expansion of x3 – y3.


we can also obtain this result by multiplying them


i.e. (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2)


= x3 + x2y + xy2 – x2y – xy2 – y3


= x3 – y3


Question 4.

Factorization of x2 + 2x – 8 is
A. (x + 4)(x – 2)

B. (x – 4)(x + 2)

C. (x + 4)(x + 2)

D. (x – 4)(x – 2)


Answer:

Suppose (x + p)(x + q) are two factors of x2 + 2x – 8.


Then, x2 + 2x – 8 = (x + p)(x + q)


= x2 + (p + q)x + pq


So, to factorize we have to find p and q, such that pq = – 8 and p + q = 2.



∴ x2 + 2x – 8 = (x + 4)(x – 2)


Question 5.

If one of the factors of x2 – 6x – 16 is (x + 2) then other factor is
A. x + 5

B. x – 5

C. x + 8

D. x – 8


Answer:

Let the other factor be (x + p)


Then, x2 – 6x – 16 = (x + p)(x + 2)


= x2 + (p + 2)x + 2p


On comparing the coefficients on both sides, we get,


p + 2 = – 6


⇒ p = – 8


∴ The other factor is (x – 8).


Question 6.

If (2x + 1) and (x – 3) are the factors of ax2 – 5x + c, then the values of a and c are respectively
A. 2, 3

B. – 2, 3

C. 2, – 3

D. 1, – 3


Answer:

ax2 – 5x + c = (2x + 1)(x – 3)


= 2x2 – 6x + x – 3


= 2x2 – 5x – 3


On comparing the coefficients on both the sides, we get,


a = 2 and c = – 3


Question 7.

If x + y = 10 and x – y = 2, then value of x is
A. 4

B. – 6

C. – 4

D. 6


Answer:

x + y = 10


⇒ y = 10 – x substituting this value in x – y = 2


We get,


x – (10 – x) = 2


⇒ 2x – 10 = 2


⇒ 2x = 12


⇒ x = 6


Question 8.

The solution of 2 – x <5 is
A. x > – 3

B. x < – 3

C. x > 3

D. x < 3


Answer:

2 – x <5


Adding x on both the sides,


2 <5 + x


⇒ – 3 < x


⇒ x > – 3


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

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

HSC Maharashtra Board Papers 2020

(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

SSC Maharashtra Board Papers 2020

(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)

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!

BUY FROM PLAY STORE

DOWNLOAD OUR APP

HOW TO PURCHASE OUR NOTES?

S.P. Important Questions For Board Exam 2021

O.C.M. Important Questions for Board Exam. 2021

Economics Important Questions for Board Exam 2021

Chemistry Important Question Bank for board exam 2021

Physics – Section I- Important Question Bank for Maharashtra Board HSC Examination

Physics – Section II – Science- Important Question Bank for Maharashtra Board HSC 2021 Examination