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Hints & Solutions

MATHEMATICS

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Fortnightly Subjective Test-2

Integrated Classroom Course for Olympiads and Class-IX (2021-2022)

PART-A

A1. One rational number between –3 and –2 is ( ) ( )3 22

− + − [1]

i.e. 52

− .

A2. (a + b)2 = a2 + b2 + 2ab ⇒ (a2 + b2) = (a + b)2 – 2ab [½] = (10)2 – 2 × 21 = 100 – 42 = 58 [½] A3. ( )( )4 3 3 5 4 3 4 5 3 3 3 5+ − = × − × + × − ×

( )12 4 5 3 3 15= − + − [1]

OR

( )

( )( )2 3 75 5

2 3 7 2 3 7 2 3 7

+= ×

− − +

( )

( ) ( )2 2

5 2 3 7

2 3 7

+=

− [ (a – b) (a + b) = (a2 – b2)] [½]

( )( )

5 2 3 7

12 7

+=

−

( )5 2 3 7

5

+=

( )2 3 7= + [½]

03/07/2021 FST-IX Phase-II Code-D

Class-IX Hints & Solutions : Fortnightly Subjective Test-2_Mathematics (Code-D)_Phase-II

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A4. (5x – 3y)3 = (5x)3 + 3 (5x)2 (–3y) + 3(5x) (–3y)2 + (–3y)3 [½]

= 125x3 – 225x2y + 135xy2 – 27y3 [½]

A5. [½]

∴ 5 0.6258

= [½]

A6.

( )

4 5 4 5

2 23 33 3 3

(3 ) 3

(3 )(27 )

×=

20

23 33

3

3× ×

= [ (am)n = amn] [½]

20

633

=

3(20 – 6) = 314 – =

mm n

na aa

[½]

OR

125–5 × 25–4 × 512 = (53)–5 × (52)–4 × 512

= 5–15 × 5–8 × 512 [½]

= 5 (–15 – 8 + 12)

= 5–11 [½]

A7. The degree of the polynomial p(x) is 6. [1]

A8. p(2) = 11(2)3 – 8(2)2 – 4(2) + 23 [½]

= 71 [½]

A9. 22

1 194+ =xx

Adding 2 both sides,

22

1 2 194 2+ + = +xx

[½]

⇒ 21 196 + =

x

x

⇒ 1 196+ =xx

[ x > 1]

= 14 [½]

Hints & Solutions : Fortnightly Subjective Test-2_Mathematics (Code-D)_Phase-II Class-IX

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A10. (i) Answer (1) [1] (ii) Answer (2) [1] (iii) Answer (3) [1] (iv) Answer (1) [1] A11. (i) Answer (1) [1] (ii) Answer (3) [1] (iii) Answer (4) [1] (iv) Answer (4) [1]

PART-B

A12. 2 42 4

1 1 1 12 2 4 163 3 9 81

− + + +

b b b bb b b b

can be written as

⇒ ( )( )

2 2 42 2 4

1 1 12 4 169 813

− + + b b b

b bb [½]

⇒ 2 2 42 2 4

1 1 14 4 169 9 81

− + +

b b b

b b b [½]

⇒ 4 44 4

1 116 1681 81

− +

b b

b b

⇒ ( )( )

2424

11681

−bb

[½]

⇒ 88

12566561

−bb

[½]

OR

Let p(x) = 13x4 – 33x3 + 79x2 – 55mx + 82.

Since, 13x4 – 33x3 + 79x2 – 55mx + 82 is divisible by (x + 1).

Therefore, p(–1) = 0 [½]

13(–1)4 – 33(–1)3 + 79(–1)2 + 55m + 82 = 0 [½]

13 + 33 + 79 + 55m + 82 = 0 [½] 207 + 55m = 0

55m = –207

m = 20755

− [½]

A13. Let x = 1.777…….. …(i)

Multiplying both sides by 10, we get

10x = 17.777……. …(ii) [1] Subtracting equation (i) from (ii), we get

9x = 16 [½]

169

x = [½]

Class-IX Hints & Solutions : Fortnightly Subjective Test-2_Mathematics (Code-D)_Phase-II

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A14. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca [½]

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

⇒ (15)2 = 83 + 2(ab + bc + ca) [½]

⇒ 225 = 83 + 2(ab + bc + ca) [½]

⇒ 225 – 83 = 2(ab + bc + ca)

⇒ 142 712

ab bc ca+ + = = [½]

A15. Let p(x) = x3 – x2 + ax + b

Since, (x – 2) and (x – 3) are factors of p(x),

∴ p(2) and p(3) = 0 [½]

p(2) = (2)3 – (2)2 + a(2) + b = 0

⇒ 8 – 4 + 2a + b = 0 [½]

⇒ 2a + b = –4 …(i)

p(3) = (3)3 – (3)2 + a(3) + b = 0

⇒ 27 – 9 + 3a + b = 0 [½]

⇒ 3a + b = –18 …(ii)

On subtracting (i) from (ii), we get

a = –14 [½]

Put the value of ‘a’ in equation (i), we get

2(–14) + b = –4

⇒ b = – 4 + 28 [½]

⇒ b = 24 [½]

Hence, a = –14 and b = 24.

A16. (i) 64a2 – 80ab + 25b2 = (8a)2 – 2 × 8a × 5b + (5b)2 ...(i) [½]

We know that (a – b)2 = a2 + b2 – 2ab [½]

So, from the above identity, we can say that

64a2 – 80ab + 25b2 = (8a – 5b)2 = (8a – 5b) (8a – 5b) [½]

(ii) 2 2289 2259 16

−x y = 2 217 17 15 153 3 4 4

× ×−

× ×x y

= 2 217 15

3 4x y −

...(i) [½]

We know that, (a2 – b2) = (a + b)(a – b) [½]

So from above identity, we get

2 2289 225 17 15 17 159 16 3 4 3 4

− = + −

x y x y x y [½]

Hints & Solutions : Fortnightly Subjective Test-2_Mathematics (Code-D)_Phase-II Class-IX

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A17. Draw a number line as shown in the figure, mark a point O (representing zero) and a point A (representing 2) on it.

Draw a perpendicular AX at A on the number line and cut off an arc AB = 1 unit.

Using Pythagoras theorem, we have

OB2 = OA2 + AB2

⇒ OB2 = (2)2 + 12 = 5 [1½]

⇒ OB = 5 units

Taking O as centre and 5OB = units as radius, mark an arc cutting the number line at P.

Hence, P represents 5 on the number line. [1½] OR

We have,

2 3 2 32 3 2 3m n m nxm n m n

+ + −=

+ − −

Rationalise the denominator, we get

2 3 2 3 2 3 2 32 3 2 3 2 3 2 3m n m n m n m nxm n m n m n m n

+ + − + + −= ×

+ − − + + − [½]

⇒ ( )

( ) ( )

22 2

2 2

2 3 2 3 2 3 2 3 2 4 92 3 2 32 3 2 3

+ + − + + − + −= =

+ − ++ − −

m n m n m n m n m nxm n m nm n m n

[½]

⇒ 2 24 2 4 9

6+ −

=m m nx

n [½]

⇒ 2 22 4 9

3+ −

=m m nx

n

⇒ 2 23 2 4 9= + −nx m m n [½]

⇒ ( ) 2 23 2 4 9− = −nx m m n

Squaring on both sides, we get

9n2x2 + 4m2 – 12mnx = 4m2 – 9n2 [½]

⇒ 9n2x2 – 12mnx + 9n2 = 0 [½] A18. We have to find five real numbers which are not irrational, means they are rational numbers. [½] First we have to make same denominators of the given numbers, we get

2 7 143 7 21

× = and 5 3 157 3 21

×=

× [½]

Since, the difference between the numerators is 1. And we have to find five rational numbers. So we multiply by 6 in both numerators and denominators of the given numbers. [½]

14 6 8421 6 126

×=

× and 15 6 90

21 6 126×

=×

[½]

Now, the five rational numbers between 84126

and 90126

are 85 86 87 88 89, , , and .126 126 126 126 126

[½]

The numbers are also expressed as 85 43 29 44 89, , , and .126 63 42 63 126

[½]

Class-IX Hints & Solutions : Fortnightly Subjective Test-2_Mathematics (Code-D)_Phase-II

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A19. Here, 7 4 3x = +

∴ 1x

= 1 7 4 3

7 4 3 7 4 3−

×+ −

[½]

= ( ) ( )22

7 4 3

7 4 3

−

−

= 7 4 349 48

−−

1x

= ( )7 4 3− [½]

Now, 22

1xx

+ = 21 2 + −

x

x [½]

= ( )27 4 3 7 4 3 2+ + − − [½]

= 194 [½]

1xx

+ = ( )7 4 3 7 4 3+ + − [½]

= 14 [½]

Therefore, 22

1 115 100 10 + − + −

x xxx

= 15 × 194 – 100 × 14 – 10 [½]

= 2910 – 1400 – 10

= 1500 [1]

Edition: 2021-22