Q1
By using standard formulae, expand the following (1 to 9):
(i) (2x + 7y)2
(ii) (1/2 x + 2/3 y)2
solutions
(i) (2x + 7y)2
(ii) (1/2 x + 2/3 y)2
solutions
Q2
(i) (3x + 1/2x)2
(ii) (3x2 y + 5z)2
solutions
(ii) (3x2 y + 5z)2
solutions
Q3
(i) (3x – 1/2x)2
(ii) (1/2 x – 3/2 y)2
solutions
(ii) (1/2 x – 3/2 y)2
solutions
Q4
. (i) (x + 3) (x + 5)
(ii) (x + 3) (x – 5)
(iii) (x – 7) (x + 9)
(iv) (x – 2y) (x – 3y)
solutions
(ii) (x + 3) (x – 5)
(iii) (x – 7) (x + 9)
(iv) (x – 2y) (x – 3y)
solutions
Q5
(i) (x – 2y – z)2
(ii) (2x – 3y + 4z)2
solutions
(ii) (2x – 3y + 4z)2
solutions
Q6
(i) (2x + 3/x – 1)2
(ii) (2/3 x – 3/2x – 1)2
solutions
(ii) (2/3 x – 3/2x – 1)2
solutions
Q7
(i) (x + 2)3
(ii) (2a + b)3
solutions
(ii) (2a + b)3
solutions
Q8
A man has certain notes of denominations Rs. 20 and Rs. 5 which amount to Rs. 380. If the number of notes of each kind is interchanged, they amount to Rs. 60 less as before. Find the number of notes of each denomination.
solutions
solutions
Q9
(i) (3x + 1/x)3
(ii) (2x – 1)3
solutions
(ii) (2x – 1)3
solutions
Q10
Simplify the following (10 to 19):
(i) (a + b)2 + (a – b)2
(ii) (a + b)2 – (a – b)2
solutions
(i) (a + b)2 + (a – b)2
(ii) (a + b)2 – (a – b)2
solutions
Q11
(i) (a + 1/a)2 + (a – 1/a)2
(ii) (a + 1/a)2 – (a – 1/a)2
solutions
(ii) (a + 1/a)2 – (a – 1/a)2
solutions
Q12
(i) (3x – 1)2
– (3x – 2) (3x + 1)
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy
solutions
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy
solutions
Q13
(i) (7p + 9q) (7p – 9q)
(ii) (2x – 3/x) (2x + 3/x)
solutions
(ii) (2x – 3/x) (2x + 3/x)
solutions
Q14
(i) (2x – y + 3) (2x – y – 3)
(ii) (3x + y – 5) (3x – y – 5)
solutions
(ii) (3x + y – 5) (3x – y – 5)
solutions
Q15
(i) (x + 2/x – 3) (x – 2/x – 3)
(ii) (5 – 2x) (5 + 2x) (25 + 4x2 )
solutions
(ii) (5 – 2x) (5 + 2x) (25 + 4x2 )
solutions
Q16
(i) (x + 2y + 3) (x + 2y + 7)
(ii) (2x + y + 5) (2x + y – 9)
(iii) (x – 2y – 5) (x – 2y + 3)
(iv) (3x – 4y – 2) (3x – 4y – 6)
solutions
(ii) (2x + y + 5) (2x + y – 9)
(iii) (x – 2y – 5) (x – 2y + 3)
(iv) (3x – 4y – 2) (3x – 4y – 6)
solutions
Q17
(i) (2p + 3q) (4p2
– 6pq + 9q2
)
(ii) (x + 1/x) (x2 - 1 + 1/x2 )
solutions
(ii) (x + 1/x) (x2 - 1 + 1/x2 )
solutions
Q18
(i) (3p – 4q) (9p2 + 12pq + 16q2
)
(ii) (x – 3/x) (x2 + 3 + 9/x2 )
solutions
(ii) (x – 3/x) (x2 + 3 + 9/x2 )
solutions
Q19
(2x + 3y + 4z) (4x2 + 9y2 + 16z2
– 6xy – 12yz – 8zx).
solutions
solutions
Q20
Find the product of the following:
(i) (x + 1) (x + 2) (x + 3)
(ii) (x – 2) (x – 3) (x + 4)
solutions
(i) (x + 1) (x + 2) (x + 3)
(ii) (x – 2) (x – 3) (x + 4)
solutions
Q21
Find the coefficient of x2
and x in the product of (x – 3) (x + 7) (x – 4).
solutions
solutions
Q22
If a2 + 4a + x = (a + 2)2
, find the value of x.
solutions
solutions
Q23
Use (a + b)2 = a2 + 2ab + b2
to evaluate the following:
(i) (101)2
(ii) (1003)2
(iii) (10.2)2
solutions
(i) (101)2
(ii) (1003)2
(iii) (10.2)2
solutions
Q24
Use (a – b)2 = a2
– 2ab – b
2
to evaluate the following:
(i) (99)2
(ii) (997)2
(iii) (9.8)2
solutions
(i) (99)2
(ii) (997)2
(iii) (9.8)2
solutions
Q25
By using suitable identities, evaluate the following:
(i) (103)3
(ii) (99)3
(iii) (10.1)3
solutions
(i) (103)3
(ii) (99)3
(iii) (10.1)3
solutions
Q26
If 2a – b + c = 0, prove that 4a2
– b
2 + c2 + 4ac = 0.
solutions
solutions
Q27
If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc.
solutions
solutions
Q28
If a + b + c = 0, then find the value of a2
/bc + b2
/ca + c2
/ab
solutions
solutions
Q29
If x + y = 4, then find the value of x3 + y3 + 12xy – 64.
solutions
solutions
Q30
Without actually calculating the cubes, find the values of:
(i) (27)3 + (-17)3 + (-10)3
(ii) (-28)3 + (15)3 + (13)3
solutions
(i) (27)3 + (-17)3 + (-10)3
(ii) (-28)3 + (15)3 + (13)3
solutions
Q31
Using suitable identity, find the value of:
solutions
solutions