Solution

9TH STADS [RATIONAL TO LOG]

Class 09 - Mathematics
√2 √2
1. i. 1
× =
2
irrational
√2 √2

–
ii. 7√5 = irrational
iii. -3 rational
– –
iv. (4 + √8) - 2√2
– –
= 4 + 2√2 - 2√2 = 4 rational
v. 7 - π irrational
as π is an irrational quantity
− − – –
vi. √24 + (7√6 − 2√6)
– – –
2√6 + 7√6 − 2√6
–
​= 7√6 irrational number
14 14 (5√3+ √5)
2. = ×
5√3− √5 (5√3− √5) (5√3+ √5)

[Using identity (a + b)(a - b) = a2 - b2]

14(5√3+ √5)
= 2 2
(5√3) −( √5)

14(5√3+ √5)
= 25×3−5

14(5√3+ √5)
= 75−5

14(5√3+ √5)
= 70

(5√3+ √5)
= 5
– – −− – 3 – −−
3. We have, 2

3
√7 −
1

2
√2 + 6√11 +
1

3
√7 +
2
√2 − √11

2 1 – −1 3 – −−
=( 3
+
3
)√7 + (
2
+
2
) √2 + (6 − 1)√11

– – −− – – −−
=( 3

3
) √7 +( 2

2
) √2 + 5 √11 = √7 + √2 + 5√11

4. i. 1.37
¯

Let x = 1.3777...
10x = 13.777... (i)
100x = 137.777... (ii)
Subtact (ii) - (i)
⇒ 90x = 124
124 62
x = =
90 45

ii. 2.23
¯

Let x = 2.232323...
100x = 223.232323...
99x = 221
221
x =
99
¯
¯¯¯¯¯¯¯¯
¯
iii. 34. 5678
Let x = 34.56785678... (i)
Multiply by 10000 on both side
10000x = 345678.56785678... (ii)
Subtract (ii) - (i)
⇒ 9999x = 345644
345644
⇒ x =
9999
¯
¯¯¯
¯
iv. 111.3235
Let x = 111.32353535...
100x = 11132.353535... (i)
10000x = 1113235.3535... (ii)
Subtract (ii) - (i)

1 / 20
 9900x = 1102103
x = ​ 1102103

9900

5. Given,
–
x = 5 - 2√6
1

x
= 1

(5−2√6)

(5+2√6)
= 1
×
(5−2√6) (5+2√6)

[Using identity (a + b)(a - b) = a2 - b2]

(5+2√6)
= 2 2
(5) −(2√6)

(5+2√6)
= 25−4×6

5+2√6
= 25−24
–
1

x
= 5 + 2√6
Now,
1 1 2
x
2
+
2
= (x + x
) -2
x
– – 2
= +5+2
(5 − 2√6 √6) -2
= (10) - 2 2

= 100 - 2
1
x
2
+
2
= 98
x

6. i. Here, 2

3
lies between 0 and 1.
2
∴ We take one-third of unit length and mark points on the number line XY on both sides of 0 (zero), then we represent 3
as
point A.

ii. Here, 1

4
lies between 0 and 1.
1
∴ We take one-fourth of unit length and mark points on the number line XY on both sides of 0 (zero), then we represent 4
as
point A.

3
iii. Here, - 5
lies between -1 and 0.
3
∴ We take one-fifth of unit length and mark points on the number line XY on both sides of 0 (zero), then we represen - 5
as
point A.

3√2−2√3 2√3
7. Prove that + = 11
3√2+2√3 √3− √2

LHS,
3√2−2√3 2√3
+
3√2+2√3 √3− √2

(3√2−2√3) 3√2−2√3 2√3 ( √3+ √2)

= × + ×
(3√2+2√3) 3√2−2√3 ( √3− √2) ( √3+ √2)
2

[Using identity (a + b)(a - b) = a2 - b2]

(3√2−2√3) 2√3( √3+ √2)
= 2 2
+ 2 2
(3√2) −(2√3) ( √3) −( √2)

2 2
(3√2) +(2√3) −2×3√2×2√3 2×3+2× √6
= 18−12
+ 3−2

18+12−12√6 6+2√6
= 6
+ 1

30−12√6 –
= 6
+ 6 + 2√6
6(5−2√6) –
= 6
+ 6 + 2√6
– –
= 5 - 2√6 + 6 + 2√6
= 11 = R.H.S
(3+ √θ) 3+ √8 –
8. Since 1
⇒
1
× = 9−8
= 3 + √8
3− √8 3− √8 (3+ √8)

( √7+ √6) √7+ √6 – –

1
⇒
1
× = 7−6
= (√7 + √6)
√7− √6 √7− √6 ( √7+ √6)

( √5+2) √5+2 –
1
⇒
1
× = 5−4
= (√5 + 2)
√5−2 √5−2 ( √5+2)

1 1 ( √8+ √7) √8+ √7 – –

⇒ × = 8−7
= (√8 + √7)
√8− √7 ( √8− √7) ( √8+ √7)

2 / 20
 ( √6+ √5) √6+ √5 – –
1
⇒
1
× = 6−5
= (√6 + √5)
√6− √5 √6− √5 ( √6+ √5)

So 1
+
1
+
1
−
1
−
1

3− √8 √7− √6 √5−2 √8− √7

√6− √5

−− – – – – – – –
⇒ (3 + √80 ) + (√7 + √6) + (√5 + 2) −(√8 + √7) − (√6 + √5)

⇒ 3+2=5
9. Rationalize
1 √7+ √3+ √2
⇒ ×
√7+ √3− √2 √7+ √3+ √2

√7+ √3+ √2
= 2 2
( √7+ √3) −( √2)

√7+ √3+ √2
=
7+3+2√21−2

√7+ √3+ √2
=
8+2√21

1 √7+ √3+ √2 4− √21

⇒ [ × ]
2 (4+ √21) 4− √21

( √7+ √3+ √2)(4− √21)]

= 1

2
[
16−21
]

1 −− – – –
⇒ (√21 − 4)(√7 + √3 + √2)
10

√147+ √63+ √42−4√7−4√3−4√2

⇒
10

7+4√3 7+4√3 –
10. a = 1
× =
49−48
= 7 + 4√3
7−4√3 7+4√3

7−4√3 7−4√3 –
and b = 1
× =
49−48
= 7 − 4√3
7+4√3 7−4√3

So,
i. a2 + b2 = (a + b)2 - 2ab
(7 + 4√3 + 7 - 4√3)2 - 2(7 + 4√3)(7 - 4√3)
– – – –
⇒

⇒ (14)2 - 2(49 - 48)

= 196 - 2
= 194
ii. a3 + b3
We know (a + b)3 = a3 + b3 + 3ab (a + b)
∴ a3 + b3 = (a + b)3 - 3ab (a + b)
and a + b = 14 & ab = 1
∴ a3 + b3 = (14)3 - 3(1)(14)
= 14(142 - 3)
= 14 × 194
= 2716​​
11. For 1st year
P = ₹4000, R = 8%, T = 1 year
P ×R×T 4000×8×1
I= 100
= 100
= ₹320
A = ₹4000 + ₹320 = ₹4320
For 2nd year
P = ₹4320, R = 10%, T = 1 year
I= = ₹432
4320×10×1

100

A = ₹4320 + ₹432 = ₹4752

For 3rd year
P = ₹4752, R = 10%, T = 1 year
4752×10×1
I= 100
= ₹475.20
∴ Total amount at the end of third year
= ₹4752 + ₹475.20
= ₹5227.20
Total compound interest = ₹5227.20 - ₹4000
= ₹1227.20

3 / 20
12. P = 20,000, r = 10% p.a.
t = 2 years 3months = 2 + 3

12

= 2
1

4
years
Now,
n
r
A = P (1 + )
100

1
2 1+ ×10
10 4
= 20, 000(1 + ) ( )
100 100

110 110 1
= 20, 000 × × × (1 + )
100 100 40

41
= 2 × 110 × 110 ×
40

A = ₹24,805
CI = A - P
= 24,805 - 20,000
CI = ₹4,805
13. Given:
C.I. = 205, n = 2 years, r = 5% p.a., p = ?
n
r
C . I . = P (1 + ) − P
100
n
r
= P [ (1 + ) − 1]
100

205 = P [(1 + 5

100
) − 1]

205 = P [ 21

20
×
21

20
− 1]

441
= P [ − 1]
400

41P
205 =
400

p =
400×205

41
= ₹ 2000
14. Given:
P = 10,000, n = 2 years
Amount at the end of 1 year
= 11,200
i. CI = A - P
= 11,200 - 10,000
CI = 1,200
for Ist year
let rate of interest be r%
C.I. for 1 year = r% of 10,000
r
1200 = × 10000
100

r = 12%
n
r
ii. A = p(1 + 100
)

2
12
= 10, 000(1 + )
100

2
112
= 10, 000( )
100

112 112
= 10, 000 × ×
100 100

A = 12,544
hence, the amount at the 2nd year = ₹ 12,544
2
pr
15. (C.I. - SI)2years = 2
100
2
4000×(8)
=
100×100
128
=
5
= 25.6
i.e. ₹25.6
16. Given, P = ₹7000, A = ₹9317, r = 10%, n = ?
n
r
A = P (1 + 100
)

4 / 20
 n

9317 = 7000(1 + 10

100
)

n
9317

7000
=( 11

10
)

n
1331

1000
=( 11

10
)

3 n
11 11
(
10
) =( 10
)

Comparing the power on both sides

n = 3 years
17. Given:
A = 2809 in 2 years
A = 2977.54 in 3 years
P = ?, r = ?
n
r
A = P (1 + )
100

2
r
2809 = P (1 + 100
) ...(i)
3

2977.54 = P (1 + r

100
) ...(ii)
Eqn (ii) ÷ (i)
3
r
P (1+ )
2977.54 100
=
2809×10 2
r
P (1+ )
100

3−2
r
1.06 = (1 + 100
)

1.06 = (1 + 100
r
)

100
= .06 × 100
r = 6%
Putting r = 6% in equation (i)
2

2809 = p(1 + 6

100
)

2809 = p × 106

100
×
106

100

P = ₹2500
18. Present Population = 2,16,000
r = 20% p.a.
i. Population after 2 year
n
r
A = P (1 + )
100

2
20
= 216, 000(1 + )
100

6 6
= 216000 × ×
5 5

A = 3,11,040
ii. Population before 2 year
P
A= n
r
(1+ )
100

2,16,000 216000
= =
2 6 6
20 ×
(1+ )
5 5
100

=
216000×5×5

6×6
= 1,50,000​​
19. i. Difference the depreciation value between the first year and second year = ₹4000 - ₹3600 = ₹400
Difference of depreciation
Rate of depreciation = depreciation of first year × time
× 100%
= 400

4000×1
× 100% = 10%
ii. Let the original cost be ₹x
Depreciation during first year
10 x
= 10% of x = x × 100
= 10
...(i)
But, depreciation during first year = ₹4000 (given) ...(ii)
From (i) and (ii), we get
x
= ₹4000
10

5 / 20
 x = ₹40,000
Hence, The original cost of machine is ₹40,000
iii. The cost of machine after two depreciation
= ₹40,000 - ₹4000 - ₹3600
= ₹32400
∴ Depreciation during third year = 10% of ₹32400

= 32400 × = ₹3240 10

100

Total depreciation during three years

= ₹4000 + ₹3600 + ₹3240 = ₹10840
∴ The cost of machine at the end of the third year

= ₹40,000 - ₹10840
= ₹29160
20. Let the sum of money = ₹x
For 2 years
P = ₹x, n = 2 years, A = ₹19360
n
r
A = P (1 + 100
)

19360 = x(1 + r

100
) ...(i)
For 4 years
P = ₹x, n = 4 years, A = ₹23425.60
n

A = P (1 + 100
r
)

23425.60 = x(1 + 100

r
) ...(ii)
by (ii) ÷ (i)
2
2342560

19360×100
= (1 + r

100
)

2
14641 r

12100
= (1 + 100
)

2 2

(
121

110
) = (1 + r

100
)

Comparing on both sides

=1+
121

110
r

100
r

100
= 121

110
-1= 11

110

r= 11

110
× 100
r = 10%
2

From (i), 19360 = x(1 + 10

100
)

=x×( 11

10
)

19360 = 121x

100

x= 19360×100

121

Thus, Sum (x) = ₹16,000

21. Consider (x + y + z)2 + (x - y - z)2
= x2 + y2 + z2 + 2xy + 2yz + 2zx + x2 + y2 + z2 - 2xy + 2yz - 2zx [∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca and (a - b - c)2
= a2 + b2 + c2 - 2ab + 2bc - 2ca]
= 2x2 + 2y2 + 2z2 + 4yz
= 2(x2 + y2 + z2 + 2yz)

22. i. (102)2
(100 + 2)2
(100)2 + (2)2 + 2(100)(2)
10000 + 4 + 400
10,404

6 / 20
 ii. (10.2)2
(10 + 0.2)2
(10)2 + 2(10)(0.2) + (0.2)2
100 + 4 + 0.04
104.04
1
23. Given: a 2
+
2
= 66 , find
a

(i) a − 1

a
, (ii) a 3
−
1

a3

2
i. (a − 1

a
) = a
2
+
1

2
− 2 ⋅ a⋅
1

a
a

= 66 - 2
1 −−
(a − ) = ± √64
a
1
(a − ) = ±8
a

ii. By using identity

(a - b)3 = a3 - b3 - 3ab (a - b)
1 3 1 1 1
(a −
a
) =a 3
−
3
− 3 ⋅ a⋅
a
(a −
a
)
a

(8)3 = a 3
−
1
− 3(8)
a3
1
a
3
−
3
= 512 + 24
a

a
3
−
1
= 536
a3

24. Consider (a - b)2 = (a + b)2 - 4ab

= 62 - 4 × 8 [∵ a + b = 6 and ab = 8]
⇒ (a - b)2 = 36 - 32 = 4
(a - b)2 = 4 ⇒ a - b = ±√4 [taking square root on both sides]
–
⇒

⇒ a - b = ±2

1
25. i. Given: a 2
+
2
= 47
a

Adding 2 on both sides

2
a + + 2 = 47 + 2
1

2
a
2
(a)
2
+ (
1

a
) + 2 × a×
1

a
= 49
[Using identity a2 + b2 + 2ab = (a + b)2]
2
1 2
(a + ) = (7)
a

a+
1

a
= ±7 ...(a)
ii. a3
+
1

3
a

a+
1

a
= ±7 [From (a)]
Taking cube on both sides
1 3
2
(a + ) = (±7)
a

a
3
+
1

3
+ 3 × a×
1

a
(a +
1

a
) = ±343 [Using identity (a + b)3 = a3 + b3 + 3ab(a + b)]
a

3 1
a + + 3 × (±7) = ±343
3
a

3 1
a + = (±343) − (±21)
3
a

3 1
a + = ±322
3
a

26. a + b + c = 2, ab + bc + ca = -1, abc = -2

find a3 + b3 + c3
∵ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(2)2 = a2 + b2 + c2 + 2(-1)
4 = a2 + b2 + c2 - 2
a2 + b2 + c2 = 6
Now,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
a2 + b3 + c3 - 3(-2) = (2) ((6) - {ab + bc + ca})

7 / 20
 a3 + b3 + c3 + 6 = 2(6 - (-1))
a3 + b3 + c3 = 8
27. According to identity
a3 + b3 + c3 = 3abc if (a + b + c) = 0
73 + 32 + (-10)3 = 3(7)(3)(-10) {∵ 7 + 3 + (-10) = 0}
= -630
28. x = 2y + 6
x - 2y - 6 = 0
a = x, b = -2y, c = -6
here a + b + c = 0
i.e. x - 2y - 6 = 0
then (x)3 + (-2y)3 + (-6)3 = 3(x)(-2y)(-6)
{by using identity, if a + b + c = 0 then a3 + b3 + c3 = 3abc}
= x3 - 8y3 - 216 = 36xy
= x3 - 8y3 - 36xy - 216 = 0
=0

29. i. Given a2- 5a - 1 = 0

a2- 1 = 5a
2
a −1
= 5
a
2
a −1
= 5
a

a−
1

a
= 5 ... (i)
[Using identity (a + b)2 = a2 + b2 + 2ab]
2
ii. (a + 1

a
) =a 2
+
1

2
+ 2 × a×
1

a
a
2
(a +
1

a
) =a 2
+
1
− 2 + 2 + 2
a2
2
(a +
1

a
) =a 2
+
1

2
− 2 × a×
1

a
+ 4
a

[Using identity a2 + b2 - 2ab = (a - b)2]

2 2
(a +
1

a
) = (a − 1

a
) + 4

1 2
(a +
a
) = (5) 2
+ 4 [From (1)]
= 25 + 4
2
(a +
1

a
) = 29
−−
a+
1

a
= ± √29 ...(ii)
[Using identity a2 - b2 = (a + b)(0 - b)]
1 1 1
iii. a 2
−
2
= (a + a
) (a −
a
)
a
−−
a
2
−
1

2
= (±√29) × 5 [From (1) and (2)]
a
1 −−
a
2
−
2
= ±5√29
a

30. Given:
x
2
=
1

2
=
25

36
(x > 0)
9x
1
Find x 3
+
3
27x

2 25
∵ x =
36
5
x = ±
6

But x > 0, given in question

So, x = 5

6
1

2
=
25

36
(Given)
9x
2
1 25
( ) =
3x 36

3x
=
5

6
(∵ x > 0)
Using identity
(a + b)3 = a2 + b3 + 3ab(a + b)
3 3

(x +
3x
1
) =x 3
+ (
1

3x
) + 3 ⋅ x ⋅
1

3x
(x +
1

3x
)

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

3
+ (
6
+
6
)
27x

8 / 20
 3
10 3 1 10
( ) = x + +
6 3 6
27x

3
3 1 10 10
x + = ( ) −
27x3 6 6

2
3 1 10 10
(x + = ( ) − 1)
27x3 6 6

3 1 5 100
x + = ( − 1)
3 3 36
27x

3 1 5 64
x + = ×
3 3 36
27x

3 1 80
x + =
3 27
27x

31. px - (px + qy)2 + p2x + pqy + qy

(px + qy) - (px + qy)2 + p(px + qy)
Let (px + qy) = m
-m2 + m + pm
-m2 + m(p + 1)
m((P + 1) - m)
Resubstitute value of m
(px + qy) ((p + 1) - (px + qy))
32. We have, 3ax - 6ay - 8by + 4bx + 9az + 12bz.
Now, arranging the given expression into groups,
= (3ax + 4bx) − 6ay − 8by +9az + 12bz

= (3ax + 4bx)− (6ay + 8by) + (9az + 12bz)

= x(3a + 4b) − 2y (3a + 4b) + 3z(3a + 4b)

Now, (3a + 46) is common factor for each group, so

3ax − 6ay − 8by + 4bx +9az + 12bz = (3a + 4b)(x − 2y + 3z)

which is the required factorisation of the given expression.

33. i. ab(a2 + b2 - c2) - bc(c2 - a2 - b2) + ca(a2 + b2 - c2)
= ab(a2 + b2 - c2) + bc(a2 + b2 - c2) + ca(a2 + b2 - c2)
= (a2 + b2 - c2) (ab + bc + ca)
ii. 2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)
= 2x(a - b) + 3y × 5(a - b) + 4z × (-2) (a - b)
= 2x(a - b) + 15y(a - b) - 8z(a - b)
= (a - b) (2x + 15y - 8z).
34. 2x7 - 128x
= 2x(x6 - 64)
= 2x((x2)3 - (4)3)
= 2x(x2 - 4)(x4 + 4x2 + 16)
= 2x(x + 2)(x - 2)(x2 + 4x2 + 16)
35. a(a + b - c) - bc
a2 + ab - ac - bc
a(a + b) - c(a + b)
(a - c)(a + b)
36. We have, l(x − y) 2
+ (mx − my) +(5x − 5y)

2
= l(x − y) +m(x − y) + 5(x − y)

Now, (x - y) is common factor across all the groups,

so l(x − y) + (mx − my) + (5x − 5y) = (x − y)[k(x − y) + m + 5]
2

which is the required factorisations of the given expression.

37. We have, 9x2 + 12xy + 4y2.
Given expression can be rewritten as
(3x)2 + 2⋅ 3x⋅ 2y+ (2y)2
Now, take 3x = a and 2y = b, then the expression becomes

9 / 20
 a2 + 2ab + b2 = (a + b)2
Hence, the expression becomes (3x + 2y)2
So, the required factors are (3x + 2y) and (3x + 2y).
38. 3(2a - b)2 - 19(2a - b) + 28
Let 2a - b = x
= 3x2 - 19x + 28
= 3x2 - 12x - 7x + 88
= 3x(x - 4) - 7(x - 4)
= (3x - 7) (x - 4)
Resubstituting the value of x
= (3(2a - b) - 7)(2a - b - 4)
39. x4 + x2 + 1 = x4 + x2 + 1 + x2 - x2
= (x2)2 + 2x2 + 1 - x2
= (x2)2 + 2 × x2 × 1 + (1)2 - (x)2 [Using identity a2 + 2ab + b2 = (a + b)2]
= (x2 + 1)2 - (x)2
= (x2 + 1 + x) (x2 + 1 - x) [Using identity a2 - b2 = (a + b) (a - b)]
= (x2 + x + 1) (x2 - x + 1)

40. i. x3 + xy(2 − 3x) − 6y2

x3 + 2xy - 3x2y - 6y2
x(x2 + 2y) - 3y(x2 + 2y)
(x - 3y) (x2 + 2y)
ii. pq(x2 + y2) - xy(p2 + q2)
pqx2 + pqy2 - p2xy - q2xy
pqx2 - p2xy + pqy2 - q2xy
px(qx - py) + qy(py - qx)
px(qx - py) - qy(qx - py)
(px - qy) (qx - py)

41. a7 - ab6
= a(a6 - b6)
= a(a2)3 - (b2)3)
= a(a2 - b2) ((a2)2 + (a2) (b2) + (b2)2)
= a(a + b)(a - b) (a4 + a2b2 + b4)
42. 9x2 − 4(y + 2x)2
= (3x)2 - (2(4 + 2x))2
= (3x + 2(y + 2x))(3x - 2(y + 2x))
= (3x + 2y + 4x) (3x - 2y - 4x)
= (7x + 2y) (-x - 2y)
= -(7x + 2y) (x + 2y)
43. i. 8x + 5y = 11 ...(i)
x + y = 4 ...(ii) {by elimination midpoint}
(i) - 8(ii)

y=7
Putting y = 7 in equation (ii) to get x

10 / 20
 x = 4 - 7 = -3
x = -3
ii. 4x - 3y = 8 ...(i)
18x - 3y = 29 ...(ii)
(i) - (ii)
-14x = -21
x= 21
=
14
i.e x = 3

2
3

Putting x = 3

2
in equation (i)
3
4( 2
) - 3(4) = 8
6 - 3y = 8
-3y = 2
−2
y= 3
y
iii. x

4
+
3
−
1

12
=0
3x+4y−1
⇒
12
=0
⇒ 3x + 4y = 1 ...(i)
x 5 7
− y =
2 4 4
2x−5y 7
=
4 4

2x - 5y = 7 ...(ii)
(i) × (ii) - (ii) × (iii)

19
y= 23

Put y = 19

23
in equation (i)
3x + 4y = 1
19
3x + 4 × 23
=1
76
3x = 1 - 23
23−76
3x = 23
−53
x= 23×3
−53
x= 69

44. Let unit digit = x

ten's digit = 6x
Required no. = 6x × 10 + x
= 60x + x
= 61x
A/c question
61x - 45 = 10x + 6x
61x - 16x = 45
45x = 45
x=1
i.e. unit digit = x = 1
ten's digit = 6x = 6
Required no. = 61x
= 61 × x
= 61 × 1
= 61
45. 43x + 67y = -24 ...(i)
67x + 43y = 24 ...(ii)
(i) + (ii)
110x + 110y = 0
x + y = 0 ...(iii)

11 / 20
 (i) - (ii)
-24x + 24y = -48
24(-x + y) = -48
-x + y = -2 ...(iv)
(iii) + (iv)
2y = -2
y = -1, x = 1​​
46. − = 0...(i)
3

x
2

x
+
5

y
= 19...(ii)
Multiplying eqn. (i) by 5 and eqn. (ii) by 2 and on adding we get

x= 19

38

x= 1

Putting this value of x in eqn. (ii),

5

1/2
+
2
= 19 y

5
4 +
y
= 19
5

y
= 15
y = 15
y= 1

From y = ax + 3
1 1

3
=a× 2
+3
a 1 3

2
= 3
−
1
−8
a

2
= 3

a=- 16

3
= −5 1

47. Given pair of linear equations is

10
+ = 4 and −
2 15 5
= -2
(x+y) (x−y) (x+y) (x−y)

On putting x+y
1
= u and 1

x−y
= v, we get
10u + 2v = 4 ...(i)
and 15u - 5v = -2 ...(ii)
On multiplying Eq. (i) by 5 and Eq. (ii) by 2 and then adding, we get
5(10u + 2v) + 2(15u - 5v) = 5 × 4 + 2 × ( -2)
⇒ 50u + 10v + 30u - 10v = 20 - 4

⇒ 80u = 16 ⇒ u =
1

On substituting u = 1

5
in Eq. (i), we get
10 × 1

5
+ 2v = 4 ⇒ 2 + 2v = 4 ⇒ 2v = 2 ⇒ v = 1
If u = 1

5
, then by 1

x+y
= u, we get

x+y
1
= 1

5
⇒ x + y = 5 ...(iii)
1
If v = 1, then by x−y
= v, we get
1

x−y
= 1 ⇒ x - y = 1 ...(iv)
On adding Eqs. (iii) and (iv), we get
2x = 6 ⇒ x = 3
On putting x = 3 in Eq. (iv), we get
3-y=1⇒y=2
Hence, x = 3 and y = 2, which is the required solution.

12 / 20
48. Let the numerator of the fraction = x
and the denominator of the fraction = y
Then, the fraction = x

According to question,
x−5

y−3
= 1

2x - 10 = y - 3
2x - y = 7 ...(i)
and y = x + 5 ...(ii)
Putting the value of y in eqn (i),
2x - (x + 5) = 7
2x - x - 5 = 7
x-5=7
x = 12
Putting the value of x in eqn. (ii), we get
y = 12 + 5
y = 17
12
Hence, the fraction = 17

49. Let the number of ₹1 coins = x

and the number of ₹2 coins = y
According to equation,
x × 1 + y × 2 = ₹256
x + 2y = 256 ...(i)
and y = 2x + 3...(ii)
Putting the value of y from (ii) in eqn. (i),
x + 2 × (2x + 3) = 256
x + 4x + 6 = 256
5x = 250
x = 50
by eqn. (ii),
y = 2 × 50 + 3
y = 100 + 3
y = 103
∴ The number of ₹1 coins = 50

the number of ₹2 coins = 103

and the value of ₹2 coins = 103 × 2
= ₹206
3x 5y
50. 2
−
3
+ 2 = 0 ...(i)
x y

3
+
2
=2 1

6
...(ii)
x 13 y

3
= 6
−
2

y
x = 3( 13

6
−
2
) ...(iii)
Putting this value of x in eqn. (i),
y 5y
3

2
× 3( 13

6
−
2
) - 3
= -2
y 5y
9

2
(
13

6
−
2
) - 3
= -2
39 9y 5y

4
−
4
- 3
= -2
9y 5y

4
+
3
= 39

4
+
2

1
27y+20y 39+8

12
= 4
47y 47

12
= 4

y=3
Putting this value of y in eqn. (iii), then we get
13 3
x = 3( - ) 6 2

13 / 20
 13−9
= 3( 6
)
= 4

x=2

51.

5x - 5y = 100

120
x= 2

x = 60 km/hr, y = 40 km/hr
i.e car starting from point A has speed 60 km/hr and car from point B has speed of 40 km/hr.
52. Let Sarita invested ₹x and y at a rate of 10% and 8% per annum.
According to question
x×10×1 y×8×1

100
+
100
= 1640
10x 84

100
+
100
= 1640
10x + 8y = 1,64,000 ...(i)
On interchanging rates
x×8×1 y×10×1

100
+
100
= 1600
8x + 10y = 1,60,000
(i) + (ii)
18x + 18y = 2,24,000
18(x + y) = 3,24,000 ...(iii)
(i) - (ii)
2x - 2y = 4000
x - y = 2,000 ...(iv)
(iii) + (iv)
2x = 20,000
x = 10,000
from equation (iv)
10,000 - y = 2000
y = 8,000
Hence, she invested ₹10,000 at 10% and 8000 at 8%.
53. Let the fixed charge be ₹x and charge form each extra day be ₹y.

y=3
From equation (i)
x = 27 - 4y
= 27 - 12
x = 15
Hence, fixed charge = ₹ 15
additional charge = ₹ 3/day​​
54. Let present age of Nisha and Samar be x and y years 5 years ago;
Nisha's age = x - 5
Samar's age = y - 5
x - 5 = 3(y - 5)
x - 3y = -10 ...(i)
Ten years Later;
Nisha's age = x + 10
Samar's age = y + 10

14 / 20
 x + 10 = 2(y + 10)
x - 2y = 10 ...(ii)
(i) - (ii)

from (ii)
x = 50
i.e. Nisha's age = 50 years
Samar's age = 20 years​​
55. 4

−2
+
1

−3
+
2

−1

(216) 3 (256) 4 (243) 5

4 1 2
= + +
−2 −3 −1

(6×6×6) 3 (4×4×4×4) 4 (3×3×3×3×3) 5

4 1 2
= + +
−2 −3 −1

3 4 5
(6 ) 3 (4 ) 4 (3 ) 5

=
4

−2
+
1

−3
+
2

−1
(∵ (am)n = amn)
(6) (4) (3)

= 4 × 62 + 1 × 43 + 2 × 31 (∵ 1

−n
n
= a )
a

= 4 × 36 + 64 + 2 × 3
= 144 + 64 + 6
= 214
56. 2x = 3y = 6-z = k
1 1 −1

2 = (k) , 3 = (k) x y
and 6 = (k) z

−1

6 = (k) z

−1

⇒ 3.2 = (k) z

1 1 −1
y
⇒ (k) ⋅ (k) x = (k) z

1 1 1
⇒ + + = 0
x y z

x-1 + y-1 + z-1 = 0

−6

57. 42m = (√16)

3 −− – 2
n = (√8)
−6

42m = 8 and (√16)

−−
=8
3
n

−6 1

= 23 and (2 = 23
2 2m 4 ×
n
(2 ) ) 3

−8
⇒ m= 3

4
and n = 3
​
a+b+c
58. −1 −1 −1 −1 −1 −1
a b +b c +c a

abc(a+b+c)
= 1
a+b+c

1 1
=
c+a+b
= abc
+ +
ab bc ca

3+x 3y

59. 27x+1 = 81y+2 and ( 1

3
) = (
1

9
)

3y
3+x 2
x+1 y+2
(3 )
3
= (3 )
4
and ( 1

3
) = ((
1

3
) )

3x + 3 = 4y + 8 and 3 + x = 6y
3x - 4y = 5 and x - 6y = -3
Solving y = 1, x = 3
−2 1

60. We have, 3 3
× (243) 3 × 9 3

−2 1

=3 3
× (3 × 3 × 3 × 3 × 3) 3 × (3 × 3) 3

−2 1

=3 3
× (3 )
5
3 × (3 )
2
3

−2 1

[∵ (am)n = am × n]
5×( ) 2×
=3 3
× 3 3 × 3 3

−10 2

=3 3
× 3 3 × 3 3

15 / 20
 10 2
3− +
=3 3 3 [∵ a
l
× a
m
× a
n
= a
l+m+n
]
(3×3)−10+2 9−10+2 11−10 1

=3 3 = 3 3 = 3 3 = 3 3

−y −2

y 2 x
9 × 3 × (3 2 ) ÷(27)

61. 3x 2
=
1

27
3 ×3
−y −2

2 y 2
(3 ) × 3 × ( 3 2 )

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

(3)2y+2+y+3 = (3)3x+2+3x
3y + 5 = 6x + 2
= 2x - y = 1
62. We have to prove
−1 −1 2
a a 2b
+ =
−1 −1 −1 −1 2 2
a +b a −b b −a
−1 −1
a a
L.H.S. = −1 −1
+
−1 −1
a +b a −b
−1 −1
a a
= +
1 1 1 1
+ −
a b a b
1 1

a a
= +
b+a b−a

ab ab

ab ab
= +
a(b+a) a(b−a)

ab(b−a)+ab(b+a)
=
a(b+a)(b−a)

ab(b−a+b+a)
=
2 2
a( b − a )

b×2b
=
2 2
b −a
2
2b
=
2 2
b −a

= R.H.S.
4 −7

3 4
√2× √3× √4 (3) 3 ×(5) 5

63. −1 3
÷
−3

(10) 5 ×(5) 5 (4) 5 ×6

1 1 1 −3

2 2
(2) 2 ×(3) 3 × (2 ) 4 (2 ) 5 ×3×2
= −1 3
×
4 −7

(5×2) 5 ×(5) 5 (3) 3 ×(5) 5

1 1 6 1 1 4 1 3 7
+ − + +1 − +1 − +
= (2) 2 2 5 5 × (3) 3 3 × (5) 5 5 5

=2 ×3 1 0
× 5
1
= 10​​
3
3 1 m 1 m

64. i. We have, 4 2 = (4 2 ) [∵ (a n ) = (a n ) ]

3 3
1 1

[∵ (am)n = amn]
2×( )
= [(2 2
) 2 ] = [2 2 ]

= (2)3 = 8
4
4 1 m 1 m

ii. We have, (27) 3 = (27 3 ) [∵ a n = (a n ) ]

4 4
1 1

[∵ (am)n = amn]
3×( )
= [(3 3
) 3 ] = (3 3 )

= 34 = 81
3

iii. We have, (243) 5

3 3

= (3 × 3 × 3 × 3 × 3) 5 = (3 )
5
5

3
5×
=3 = 3 [∵ (a 5
3 m
)
n
= a
m×n
]

= 3 ×3× 3 = 27

16 / 20
 2

iv. We have, 1

−2
= (27) 3 [∵
a−n
1
= a ]
n

(27) 3

2 2

= (3 × 3 × 3) 3 = (3 )
3 3

= 32
3×( )
=3 3

n
= 3 × 3 = 9 [∵ (a m
) = a
m×n
]

65. We have, log(x + 3) - log(x - 3) = 1

x+3 m
⇒ log( ) = 1 [log m − log n = log ]
x−3 n

x+3
⇒ log(
x−3
) = log 10 [∵ 1 = log10]
x+3
⇒ ( ) = 10
x−3

⇒ x + 3 = 10(x - 3)
⇒ x + 3 = 10x - 30

⇒ x - 10x = -30 - 3

⇒ -9x = -33
33 11 2
∴ x = = = 3
9 3 3

66. log (x - 2) + log (x - 5) = 1

⇒ log (x - 2)(x - 5) = 1

⇒ log x2 - 2x - 5x + 10 = 1
⇒ log x2 - 7x + 10 = 1
⇒ x2 - 7x + 10 = 101
⇒ x2 - 7x + 10 - 10 = 0
⇒ x2 - 7x = 0
x(x - 7) = 0
x = 0, 7
But x = 0 is not our answer because on putting x = 0, log 6n become (-ve) which is not defined.
Hence x = 7
67. Given that
a2 + b2 = 23 ab
a2 + b2 + 2ab = 23 ab + 2 ab
(a + b)2 = 25 ab [Using identity a2 + 2ab + b2 = (a + b)2]
2
(a+b)

25
= ab
2
a+b
(
5
) = ab
2
a+b
log ( 5
) = log ab

= log a + log b (∵ log mn = n log m and log m × n = log m + log n)

a+b
2 log 5
a+b
log 5
= 1

2
(log a + log b)
68. We have, log 7 - log 2 + log 16 - 2 log 3 - log 7

45
= 1 + log x
⇒ log 7 - log 2 + log 16 - log 32 - log
7

45
= 1 + log x [∵ n log m = log mn]
⇒ log 7 - log 2 + log 16 - log 9 - (log 7 - log 45) = 1 + log x [∵ log m

n
= log m − log n]

⇒ log 7 - log 2 + log 16 - log 9 - log 7 + log 45 = 1 + log x

⇒ (log 16 + log 45) - (log 2 + log 9) = 1 + log x

⇒ log(16 × 45) - log(2 × 9) = 1 + log x [∵ log m + log n = log mn]

16×45 m
⇒ log = 1 + log x [∵ log m - log n = log ]
2×9 n

⇒ log 40 = 1 + log x = log 10 + log x [∵ 1 = log 10]

⇒ log 40 = log 10x ⇒ 10x = 40

∴ x = 4

69. 2 log 15

18
− log
25

162
+ log
4

9
2
15 25
= log ( 18
) - log 162
+ log 4

17 / 20
 = log 225

324
- log 162
25
+ log 4

9
225

= log
324 4
+ log
25 9

162

= log 225×162

328×25
+ log
4

= log 9

2
+ log
4

= log 9

2
×
4

= log 2 RHS
Hence, Proved.
70. i. log4 32 = x - 4

4x - 4 = 32
(22)x - 4 = (2 × 2 × 2 × 2 × 2)
(22)x - 4 = 25
2(x - 4) = 5
2x - 8 = 5
2x = 13
13 1
x= 2
=6 2

ii. log7 (2x2 - 1) = 2

72 = 2x2 - 1
49 = 2x2 - 1
2x2 = 50
x2 = 25
x=±5
71. 3 (log 5 - log 3) - (log 5 - 2 log 6) = 2 - log x
3 log 5 - 3 log 3 - log 5 + 2 log 6 + log x = 2
log 53 - log 33 - log 5 + log 62 + log x = 2 × 1 (∵ n log m = log mn)
(log 53 + log 62 + log x) - (log 33 + log 5) = 2 log 10
log (53 × 62 × x) - log (33 × 5) = 2 log 102 (∵ log m + log n = log m × n)
log 125×36×x
= log 100 (∵ log m - log n = log )
27×5
m

125×36×x

27×5
= 100
x= 100×27×5

125×36

x=3
log(x+y) log(x−y) log 4
72. log 2
=
log 5
=
log(0.5)

log(x+y) log 9
⇒ =
log 2 log(0.5)
2
log(x+y) log
⇒ =
2 1
log log( )
2

log(x+y) 2 log 2
⇒ =
log 2 2−1
log

log(x+y) 2 log 2
⇒ =
log 2 −1 log 2

log(x+y)
⇒ = −2
log 2

⇒ log (x + y) = -2 log 2
⇒ log (x + y) = log 2-2
⇒ log (x + y) = log2
1

⇒ x+y= 1

⇒ 4x + 4y = 1
⇒ 4(x + y) = 1

x + y = ...(i)1

4
log(x−y) log 4
=
log 5 log 0.5

18 / 20
 log(x−y)
⇒
log 5
= -2
⇒ log (x - y) = -2 log 5
⇒ log (x - y) = log (5)2
⇒ log (x - y) = log
1

25

x-y= 25
1

25x - 25y = 1
25(x - y) = 1
1
x-y= 25
...(ii)
(i) and (ii)

29
2x = 100
29
x= 200

Putting above value of x in equation (i)

29 1
+ y =
200 4
29
y= 1

4
−
200
50−29
y= 200

y= 21

200

73. logx 49 - logx 7 + logx 1

343
+2=0
logx 49

7
+ logx ( 1

343
) = -2 (∵ log m - log n = log m

n
)
logx 7 + logx 1

343
= -2
logx 7 × 343
1
= -2
logx 7 × 343
1
= -2 (∵ log m + log n = log m × n)
1
logx 49
= -2

x-2 = 1

49
1 1
=
2 49
x

x2 = 49
x=±7
∴ x = 7 (at x = -7, logx 7 is not defined)

74. log (x + 2) + log(x - 2) = log 3 + 3 log 4

⇒ log (x + 2) (x - 2) = log 3 + 3 log 43
⇒ log (x2 - 22) = log 3 + log 64
⇒ log x2 - 4 = log 3 × 64
⇒ x2 - 4 = 192
⇒ x2 = 196
⇒ x = ± 14

but x ≠ -14
Hence x = 14
75. Given log3 m = x

3x = m ...(i)
log3 n = y

3y = n ...(ii)
i. 32x - 3 = 32x × 3-3
x 2
(3 )
= 3
3
2
(m)
= 27
2
m
= 27

19 / 20
ii. 31 - 2y + 3x = 31 × 3-2y × 33x
x 3
3×( 3 )
= 2y
3
x 3
3×( 3 )
= y 2
(3 )

3
3×(m)
= 2
(n)
3

= 3m

2
n

20 / 20