ssc algebra question paper mahesh tutorials

Paper - I

Q.1. Solve the following : 3

(i) For the sequence, find the next four terms :3, 9, 27, 81, .....

(ii) Write the quadratic equation in standard form ax2 + bx + c = 07 – 4x –x2 = 0

(iii) Find the value of the following determinant :

– 3 8

6 0


(i) Find the first three terms of the sequences for which Sn

is given

below : n (n 1) (2n 1)

6

(ii) Solve the following quadratic equation by factorization method :x2 – 5x + 6 = 0

S.S.C. Test - I

Batch : SB Marks : 20

Date : Time : 1 hr.ALGEBRA – Chapter : 1, 2, 3

MAHESH TUTORIALS

Paper - I... 2 ...

Best Of Luck


(i) Solve the following quadratic equation by factorization method :2m2 + 19m + 30 = 0

(ii) Find the value of k if x = 4 is the solution of the equation3x2 + kx – 2 = 0

(iii) Solve the following simultaneous equations using Cramer’s rule :3x + y = 1; 2x = 11y + 3


(i) Solve the following simultaneous equations :ax + by = a – b; bx = ay + a + b where, ‘a’ and ‘b’ are constants.

Paper - I

A.1. Solve the following :(i) t1 = 31 = 3

t2 = 32 = 9

t3 = 33 = 27

t4 = 34 = 81

t5 = 35 = 243

t6 = 36 = 729

t7 = 37 = 2187

t8 = 38 = 6561

The next four terms of the sequence are 243, 729, 2187 and 6561. 1

(ii) 7 – 4x – x2 = 0

– x2 – 4x + 7 = 0

– x2 – 4x + 7 = 0 1

(iii)– 3 8

6 0

= (– 3 × 0) – (8 × 6)= 0 – 48

= – 48 1

A.2. Solve the following :

(i) Sn =n (n 1) (2n 1)

6

S1 =1 (1 1) [2 (1) 1]

6

=

1 2 3

6

=

6

6= 1

S2 =2 (2 1) [2 (2) 1]

6

=

2 3 5

6

=

30

6= 5

MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.

ALGEBRA – Chapter : 1, 2, 3

MAHESH TUTORIALS

Paper - I... 2 ...

S3 =3 (3 1) [2 (3) 1]

6

=

3 4 7

6

= 14 1

We know that,t1 = S1 = 1t2 = S2 – S1 = 5 – 1 = 4t3 = S3 – S2 = 14 – 5 = 9

The first three terms of the sequence are 1, 4 and 9. 1

(ii) x2 – 5x + 6 = 0 x2 – 3x – 2x + 6 = 0 x(x – 3) – 2(x – 3) = 0 1 (x – 3) (x – 2) = 0 x – 3 = 0 or x – 2 = 0

x = 3 or x = 2 1


(i) 2m2 + 19m + 30 = 0

2m2 + 15m + 4m + 30 = 0

m(2m + 15) + 2(2m + 15) = 0 1 (2m + 15) (m + 2) = 0

2m + 15 = 0 or m + 2 = 0 1 2m = –15 or m = –2

m = –15

2or m = –2 1

(ii) 3x2 + kx – 2 = 0

x = 4 is the solution of given quadratic equation.

Substituting x = 4 in given quadratic equation, it will get satisfied.

3 (4)2 + k (4) – 2 = 0

3 (16)2 + 4k – 2 = 0 1

4k + 46 = 0 4k = – 46

k =46

4

1

k =23

2

1

Paper - I... 3 ...

(iii) 3x + y = 12x = 11y + 3

2x – 11y = 3

D =3 1

2 –11 = (3 × – 11) – (1 × 2) = – 33 – 2 = – 35

Dx =1 1

3 –11 = (1 × – 11) – (1 × 3) = – 11 – 3 = – 14 1

Dy =3 1

2 3 = (3 × 3) – (1 × 2) = 9 – 2 = 7

By Cramer’s rule,

x =D

Dx =

–14

–35 = 2

51

y =D

Dy

=7

–35 = –1

5

x = 2

5 and y =

–1

5 is the solution of given simultaneous equations. 1

A.4. Solve the following :(i) ax + by = a – b .....(i)

bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b

a b – a = [– a (a – b)] – [b (a + b)] = – a2 + ab – [ab + b2]

= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Dy =a a – b

b a b = a (a + b) – [b (a – b)] = a2 + ab – [ab – b2]

= a2 + ab – ab + b2 = a2 + b2 1By Cramer’s Rule,

x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1

x = 1 and y = – 1 are the solutions of the given simultaneous 1equations.

Paper - II


(i) Is the following list of numbers an Arithmetic Progression ? Justify.

1, 4, 7, 10, ....

(ii) Is the following equation quadratic ?

–5

3x2 = 2x + 9

(iii) If the value of the determinant m 2

–5 7 is 31 , find m.


(i) Find the first three terms of the sequence for which Sn

is given :

Sn = n2 (n + 1)

(ii) If one root of the quadratic equation 3y2 – ky + 8 = 0 is 2

3, then find

the value of k.

S.S.C. Test - I



MAHESH TUTORIALS

Paper - II... 2 ...

Best Of Luck


(i) Solve the following quardratic equation by factorization method :

1

9x2 –

2

3x + 1 = 0

(ii) Solve the following quadratic equation by factorization method :3x2 – x – 10 = 0

(iii) Solve the following simultaneous equations using Cramer’s rule :4x + 3y – 4 = 0; 6x = 8 – 5y



Paper - II


(i) t1 = 1, t2 = 4, t3 = 7, t4 = 10t2 – t1 = 4 – 1 = 3t3 – t2 = 7 – 4 = 3t4 – t3 = 10 – 7 = 3

The difference between any two consecutive terms is 3 which is constant. The sequence is an A.P. 1

(ii) –5

3x2 = 2x + 9

–5

3 x2 – 2x – 9 = 0

Here a = –5

3, b = – 2, c = – 9 are real numbers.

Where a 0.So it is a quadratic equation in variable x. 1

(iii)m 2

–5 7 = 31

(m × 7) – (2 × – 5) = 31 7m + 10 = 31 7m = 31 – 10 7m = 21

m =21

7 m = 3 1


(i) Sn = n2 (n + 1) S1 = 12 (1 + 1) = 1 (2) = 2 S2 = 22 (2 + 1) = 4 (3) = 12 S3 = 32 (3 + 1) = 9 (4) = 36 1

MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.


MAHESH TUTORIALS

Paper - II... 2 ...

We know that,

t1 = S1 = 2

t2 = S2 – S1 = 12 – 2 = 10

t3 = S3 – S2 = 36 – 12 = 24


(ii) 3y2 – ky + 8 = 0

y = 2

3 is the root of given quadratic equation.

So it satisfies the given equation.

32

3

2

– k2

3 + 8 = 0

3 4

9

–

2k

3 + 8 = 0

4

3 –

2k

3 + 8 = 0 1

Multiplying throughout by 3, we get,4 – 2k + 24 = 0

– 2k = –28

k =28

2

k = 14 1


(i)1

9x2 –

2

3x + 1 = 0

Multiplying throughout by 9, we get;

x2 – 6x + 9 = 0

x2 – 3x – 3x + 9 = 0 1

x (x – 3) – 3 (x – 3) = 0

(x – 3) (x – 3) = 0

(x – 3)2 = 0 1

Taking square root on both the sides, we get;

x – 3 = 0

x = 3 1

Paper - II... 3 ...

(ii) 3x2 – x – 10 = 0 3x2 – 6x + 5x – 10 = 0 1 3x(x – 2) + 5(x – 2) = 0 (x – 2) (3x + 5) = 0 x – 2 = 0 or 3x + 5 = 0 x = 2 or 3x = –5 1

x = 2 or x = –5

31

(iii) 4x + 3y – 4 = 0 4x + 3y = 4

6x = 8 – 5y 6x + 5y = 8

D =4 3

6 5 = (4 × 5) – (6 × 3) = 20 – 18 = 2

Dx =4 3

8 5 = (4 × 5) – (3 × 8) = 20 – 24 = – 4 1

Dy =4 4

6 8 = (4 × 8) – (4 × 6) = 32 – 24 = 8

By Cramer’s rule,

x =D

Dx =

– 4

2 = – 2 1

y =D

Dy

= 8

2 = 4

x = – 2 and y = 4 is the solution of given simultaneous equations. 1


(i) ax + by = a – b .....(i)bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b


= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Paper - II... 4 ...

Dy =a a – b



x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1


Paper - III


(i) Is the following list of numbers an Arithmetic Progression ? Justify.– 10, – 13, – 16, – 19, .....

(ii) Write the quadratic equation in standard from ax2 + bx + c = 08 – 3x – 4x2 = 0

(iii) Find the value of the following determinant :

3 6 – 4 2

5 3 2


(i) For the given sequence, find the next four terms :

1

2,

1

6,

1

18,

1

54

(ii) In the example given below determine whether the values givenagainst the quadratic equation are the roots of the equation or not.

4m2 – 9 = 0, m = 2, 2

3

S.S.C. Test - I



MAHESH TUTORIALS

Paper - III... 2 ...

Best Of Luck


(i) Solve the following simultaneous equations using Cramer’s rule :3x + y = 1; 2x = 11y + 3

(ii) Solve the following quadratic equations by factorization method :10x2 + 3x – 4 = 0

(iii) Solve the following simultaneous equations using Cramer’s rule :3x – y = 7; x + 4y = 11



Paper - III

A.1. Solve the following :(i) t1 = – 10, t2 = – 13, t3 = – 16, t4 = – 19

t2 – t1 = – 13 – (– 10) = – 13 + 10 = – 3t3 – t2 = – 16 – (– 13) = – 16 + 13 = – 3t4 – t3 = – 19 – (– 16) = – 19 + 16 = – 3

The difference between any two consecutive terms is – 3 which isconstant.

The sequence is an A.P. 1

(ii) 8 – 3x – 4x2 = 00 = 4x2 + 3x – 8

4x2 + 3x – 8 = 0 1

(iii)3 6 – 4 2

5 3 2

= 3 6 2 – – 4 2 5 3

= 6 6 – – 20 6

= 6 6 20 6

= 26 6 1


(i) t1 =1

2

t2 =1 1

2 3 =

1

6

t3 =1 1

6 3 =

1

18

MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.


MAHESH TUTORIALS


t4 =1 1

18 3 =

1

54

t5 =1 1

54 3 =

1

1621

t6 =1 1

162 3 =

1

486

t7 =1 1

486 3 =

1

1458

t8 =1 1

1458 3 =

1

4374

The next four terms of the sequence are 1

162,

1

486,

1

1458 &

1

4374. 1

(ii) a) By putting m = 2 in L.H.S. we getL.H.S. = 4(2)2 – 9

= 4(4) – 9= 16 – 9= 7 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.So 2 is not the root of the given quadratic equation. 1

b) By putting m = 2

3 in L.H.S. we get

L.H.S. = 42

3

2

– 9

= 4 × 4

9 – 9

=16

9 – 9

=16–81

9

=–65

9 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.

So 2

3 is not the root of the given quadratic equation. 1


A.3. Solve the following :(i) 3x + y = 1

2x = 11y + 3 2x – 11y = 3

D =3 1

2 –11 = (3 × – 11) – (1 × 2) = – 33 – 2 = – 35

Dx =1 1

3 –11 = (1 × – 11) – (1 × 3) = – 11 – 3 = – 14 1

Dy =3 1

2 3 = (3 × 3) – (1 × 2) = 9 – 2 = 7

By Cramer’s rule,

x =D

Dx =

–14

–35 = 2

51

y =D

Dy

=7

–35 = –1

5

x = 2

5 and y =

–1


(ii) 10x2 + 3x – 4 = 0 10x2 + 8x – 5x – 4 = 0 1 2x (5x + 4) – 1 (5x + 4) = 0 (5x + 4) (2x – 1) = 0 5x + 4 = 0 or 2x – 1 = 0 1 5x = – 4 or 2x = 1

x = –4

5or x =

1

21

(iii) 3x – y = 7x + 4y = 11

D =3 –1

1 4 = (3 × 4) – (– 1 × 1) = 12 + 1 = 13

Dx =7 –1

11 4 = (7 × 4) – (– 1 × 11) = 28 + 11 = 39 1

Dy =3 7

1 11 = (3 × 11) – (7 × 1) = 33 – 7 = 26

By Cramer’s rule,

x =D

Dx =

39

13 = 3 1


y =D

Dy

= 26

13= 2

x = 3 and y = 2 is the solution of given simultaneous equations. 1


bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b


= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Dy =a a – b



x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1


Paper - IV


(i) Write the first five terms of the following Arithmetic Progressionwhere, the common difference ‘d’ and the first term ‘a’ is given :a = 5, d = 2

(ii) Is the following equation quadratic ?(y – 2) (y + 2) = 0

(iii) If Dx = – 18 and D = 3 are the values of the determinants for certain

simultaneous equations in x and y, find x.



is given :S

n = n2 (n + 1)

(ii) If one root of the quadratic equation x2 – 7x + k = 0 is 4, then findthe value of k.


(i) State whether x =– k

2 is the root of the quadratic equation

2x2 + (k – 6) x – 3k = 0.

S.S.C. Test - I



MAHESH TUTORIALS

Paper - IV... 2 ...

Best Of Luck

(ii) Solve the following quadratic equations by factorization method :6x2 – 7x – 13 = 0.

(iii) Solve the following simultaneous equations using Cramer’s rule :3x + 2y + 11 = 0; 7x – 4y = 9



Paper - IV

A.1. Solve the following :(i) a = 5, d = 2

Here, t1 = a = 5t2 = t1 + d = 5 + 2 = 7t3 = t2 + d = 7 + 2 = 9t4 = t3 + d = 9 + 2 = 11t5 = t4 + d = 11 + 2 = 13

The first five terms of the A.P. are 5, 7, 9, 11 and 13. 1

(ii) (y – 2) (y + 2) = 0 (y)2 – (2)2 = 0 [ a2 – b2 = (a + b) (a – b)] y2 – 4 = 0 y2 + 0y – 4 = 0

Here a = 1, b = 0, c = – 4 are real numbers where a 0So it is a quadratic equation in variable y. 1

(iii) Dx = – 18 and D = 3By Cramer’s rule,

x = D

Dx

x = –18

3

x = – 6 1

A.2. Solve the following :(i) Sn = n2 (n + 1)

S1 = 12 (1 + 1) = 1 (2) = 2

S2 = 22 (2 + 1) = 4 (3) = 12

S3 = 32 (3 + 1) = 9 (4) = 36 1We know that,

MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.


MAHESH TUTORIALS

Paper - IV... 2 ...

t1 = S1 = 2

t2 = S2 – S1 = 12 – 2 = 10

t3 = S3 – S2 = 36 – 12 = 24


(ii) x2 – 7x + k = 0x = 4 is the root of given quadratic equation. 1

Substituting x = 4 in the given equationSo it satisfies the given equation.

(4)2 – 7(4) + k = 0 16 – 28 + k = 0 –12 + k = 0

k = 12 1


(i) 2x2 + (k – 6) x – 3k = 0; x =– k

2

Putting x = – k

2 in L.H.S. we get

L.H.S. =– k – k

2 (k – 6) – 3k2 2

2

1

=k k 6k

2 – – 3k4 2

2 2

=k (–k 6k)

– 3k2 2

2 2

=k – k 6k

– 3k2

2 2

=6k

– 3k2

1

= 3k – 3k= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.

So – k

2 is the root of the given quadratic equation. 1

Paper - IV... 3 ...

(ii) 6x2 – 7x – 13 = 0

6x2 – 13x + 6x – 13 = 0 1 x(6x – 13) + 1(6x – 13) = 0

(6x – 13) (x + 1) = 0

6x – 13 = 0 or x + 1 = 0 1 6x = 13 or x = –1

x = 13

6or x = –1 1

(iii) 3x + 2y + 11 = 0

3x + 2y = – 11

7x – 4y = 9

D =3 2

7 – 4 = (3 × – 4) – (2 × 7) = – 12 – 14 = – 26

Dx =–11 2

9 – 4 = (– 11 × – 4) – (2 × 9) = 44 – 18 = 26 1

Dy =3 –11

7 9 = (3 × 9) – (– 11 × 7) = 27 – (– 77) = 27 + 77 = 104

By Cramer’s rule,

x =D

Dx =

26

–26 = – 1 1

y =D

Dy

= 104

–26 = – 4

x = – 1 and y = – 4 is the solution of given simultaneous equations. 1


bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b


= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Dy =a a – b


= a2 + ab – ab + b2 = a2 + b2 1

Paper - IV... 4 ...

By Cramer’s Rule,

x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1


Paper - V


(i) Is the following list of numbers an Arithmetic Progression ? Justify.3, 6, 12, 24, ....

(ii) Write the following quadratic equation in standard form ax2 + bx + c = 0

n – 7

n = 4

(iii) If the value of the determinant m 2

–5 7 is 31 , find m.



is given :

Sn=n (n 1)

4

2 2

(ii) If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find thevalue of k.


.(i) Solve the following simultaneous equations using Cramer’s rule :

4x + 3y – 4 = 0; 6x = 8 – 5y

S.S.C. Test - I



MAHESH TUTORIALS

Paper - V... 2 ...

Best Of Luck

(ii) Solve the following quadratic equations by factorization method :3x2 – 10x + 8 = 0

(iii) Solve the following simultaneous equations using Cramer’s rule :

y = 5x – 10

2, 4x + 5 = – y



Paper - V

A.1. Solve the following :(i) t1 = 3, t2 = 6, t3 = 12, t4 = 24

t2 – t1 = 6 – 3 = 3t3 – t2 = 12 – 6 = 6t4 – t3 = 24 – 12 = 12

The difference between two consecutive terms is not constant. The sequence is not an A.P. 1

(ii) n – 7

n = 4

Multiplying throughout by n, we get,n2 – 7 = 4n

n2 – 4n – 7 = 0 1

(iii)m 2

–5 7 = 31

(m × 7) – (2 × – 5) = 31 7m + 10 = 31 7m = 31 – 10 7m = 21

m =21

7 m = 3 1


(i) Sn =n (n 1)

4

2 2

S1 =1 (1 1)

4

2 2

=1 (2)

4

2

=1 4

4

= 1

S2 =2 (2 1)

4

2 2

=4 (3)

4

2

= 9

S3 =3 (3 1)

4

2 2

=9 4

4

2

= 9 × 4 = 36 1

MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.


MAHESH TUTORIALS

Paper - V... 2 ...



(ii) x2 – 7x + k = 0x = 4 is the root of given quadratic equation.So it satisfies the given equation. 1

Substituting x = 4 in the given equation (4)2 – 7(4) + k = 0 16 – 28 + k = 0 –12 + k = 0

k = 12 1

A.3. Solve the following :(i) 4x + 3y – 4 = 0

4x + 3y = 46x = 8 – 5y

6x + 5y = 8

D =4 3

6 5 = (4 × 5) – (6 × 3) = 20 – 18 = 2

Dx =4 3

8 5 = (4 × 5) – (3 × 8) = 20 – 24 = – 4 1

Dy =4 4

6 8 = (4 × 8) – (4 × 6) = 32 – 24 = 8

By Cramer’s rule,

x =D

Dx =

– 4

2 = – 2 1

y =D

Dy

=8

2 = 4


(ii) 3x2 – 10x + 8 = 0 3x2 – 6x – 4x + 8 = 0 1 3x(x – 2) – 4(x – 2) = 0 (x – 2) (3x – 4) = 0 x – 2 = 0 or 3x – 4 = 0 x = 2 or 3x = 4 1

x = 2 or x = 4

31

Paper - V... 3 ...

(iii) y =5x – 10

2 2y = 5x – 10 – 5x + 2y = – 10

4x + 5 = – y 4x + y = – 5

D =–5 2

4 1 = (– 5 × 1) – (2 × 4) = – 5 – 8 = – 13

Dx =–10 2

–5 1 = (– 10 × 1) – (2 × – 5) = – 10 + 10 = 0 1

Dy =–5 –10

4 –5 = (– 5 × – 5) – (– 10 × 4 ) = 25 + 40 = 65

By Cramer’s rule,

x =D

Dx =

0

–13 = 0 1

y =D

Dy

=65

–13 = – 5

x = 0 and y = – 5 is the solution of given simultaneous equations. 1


bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b


= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Dy =a a – b



x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1


Paper - VI


(i) Write the first five terms of the following Arithmetic Progressionwhere, the common difference ‘d’ and the first term ‘a’ given :a = 10, d = – 3

(ii) Is the following equation quadratic ?n – 3 = 4n2

(iii) If Dy = – 15 and D = – 5 are the values of the determinants for certain

simultaneous equations in x and y, find y.



is given :S

n = n2 (n + 1)

(ii) If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then findthe value of k.


(i) Solve the following simultaneous equations using Cramer’s rule :3x – y = 7; x + 4y = 11

S.S.C. Test - I



MAHESH TUTORIALS

Paper - VI... 2 ...

Best Of Luck

(ii) Solve the following quardratic equation by factorization method :4y2 + 4y + 1 = 0

(iii) Solve the following simultaneous equations using Cramer’s rule :x + 18 = 2y; y = 2x – 9



Paper - VI

A.1. Solve the following :(i) a = 10, d = – 3

Here, t1 = a = 10t2 = t1 + d = 10 + (– 3) = 10 – 3 = 7t3 = t2 + d = 7 + (– 3) = 7 – 3 = 4t4 = t3 + d = 4 + (– 3) = 4 – 3 = 1t5 = t4 + d = 1 + (– 3) = 1 – 3 = – 2

The first five terms of the A.P. are 10, 7, 4, 1 and – 2. 1

(ii) n – 3 = 4n2

– 4n2 + n – 3 = 0Here a = – 4, b = 1, c = – 3 are real numbers.Where a 0So it is a quadratic equation in variable n. 1

(iii) Dy = – 15 and D = – 5By Cramer’s rule,

y = D

Dy

y = –15

–5

y = 3 1

A.2. Solve the following :(i) Sn = n2 (n + 1)

S1 = 12 (1 + 1) = 1 (2) = 2 S2 = 22 (2 + 1) = 4 (3) = 12 S3 = 32 (3 + 1) = 9 (4) = 36 1



MODEL ANSWER PAPER

S.S.C. Test - I


Date : Time : 1 hr.


MAHESH TUTORIALS

Paper - VI... 2 ...

(ii) kx2 – 7x + 12 = 0

x = 3 is root of given quadratic equation. 1So it satisfies the given equation

k(3)2 – 7(3) + 12 = 0 k(9) – 21 + 12 = 0 9k – 9 = 0 9k = 9

k =9

9

k = 1 1

A.3. Solve the following :(i) 3x – y = 7

x + 4y = 11

D =3 –1

1 4 = (3 × 4) – (– 1 × 1) = 12 + 1 = 13

Dx =7 –1

11 4 = (7 × 4) – (– 1 × 11) = 28 + 11= 39 1

Dy =3 7

1 11 = (3 × 11) – (7 × 1) = 33 – 7 = 26

By Cramer’s rule,

x =D

Dx =

39

13= 3 1

y =D

Dy

=26

13= 2


(ii) 4y2 + 4y + 1 = 0 4y2 + 2y + 2y + 1 = 0 2y (2y + 1) + 1 (2y + 1) = 0 1 (2y + 1) (2y + 1) = 0 (2y + 1)2 = 0

Taking square root on both the sides, we get;2y + 1 = 0 1

2y = – 1

y =1

2

1

Paper - VI... 3 ...

(iii) x + 18 = 2y x – 2y = – 18

y = 2x – 9 – 2x + y = – 9

D =1 – 2

– 2 1 = (1 × 1) – (– 2 × – 2) = 1 – 4 = – 3

Dx =–18 – 2

– 9 1 = (– 18 × 1) – (– 2 × – 9) = – 18 – 18 = – 36 1

Dy =1 –18

– 2 – 9 = (1 × – 9) – (– 18 × – 2) = – 9 – 36 = – 45

By Cramer’s rule,

x =D

Dx =

–36

–3 = 12 1

y =D

Dy

= – 45

–3 = 15



bx – ay = a + b .....(ii)

D =a b

b – a = (a × – a) – (b × b) = – a2 – b2 = – (a2 + b2) 1

Dx =a – b b


= – a2 + ab – ab – b2 = – a2 – b2 = – (a2 + b2) 1

Dy =a a – b



x =D

Dx =

– a b

– a b

2 2

2 2 = 1

y =D

Dy

= a b

– (a b )

2 2

2 2 = – 1


Paper - I


(i) The 9th term of an G.P. 3, 6, 12, 24 is ............. .

(ii) Determine the nature of roots of the following equation from its

discriminant : x2 + 3 2 x – 8 = 0

(iii) Examine whether the point (2, 5) lies on the graph of the equation3x – y = 1.


(i) Find S10

if a = 6 and d = 3.

(ii) Form the quadratic equation if its roots are 0 and – 4.

(iii) What is the equation of Y - axis? Hence, find the point of intersectionof Y - axis. and the line y = 3x + 2.


(i) Without plotting the graphs, find the point of intersection of thelines 2x + 5y = 13 and 4x – 9y = 7.

S.S.C. Test - II


Date : Time : 1 hr. 15 min.ALGEBRA – Chapter : 1, 2, 3

MAHESH TUTORIALS

Paper - I... 2 ...

Best Of Luck

(ii) Find the 4th and 9th terms of the G.P. with first term 4 and commonratio 2.

(iii) Solve the following quadratic equation by completing square :p2 – 10p + 5 = 0


(i) If S6 = 126 and S

3 = 14 then find a and r.

(ii) If the sum of the roots of the quadratic is 3 and sum of their cubesis 63, find the quadratic equation.

(iii) Solve the following simultaneous equations using graphical method :4x = y – 5; y = 2x + 1

Paper - I

A.1. Solve the following :(i) The 9th term of an G.P. 3, 6, 12, 24 is 768. 1

(ii) x2 + 3 2 x – 8 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = 3 2 , c = – 8 = b2 – 4ac

= (3 2 )2 – 4 (1) (– 8)= 9 × 2 + 32= 18 + 32= 50

> 0Hence roots of the quadratic equation are real and unequal. 1

(iii) Substituting x = 2 and y = 5 in the L.H.S. of the equation 3x – y =1

L.H.S. = 3x – y

= 3 (2) – 5

= 6 – 5

= 1

= R.H.S.

x = 2 and y = 5 satisfies the equation 3x – y = 1

Hence (2, 5) lies on the graph of the equation 3x – y = 1. 1

A.2. Solve the following :(i) For an A.P. a = 6, d = 3

Sn =n

2[2a + (n – 1) d]

S10 =10

2 [2 (6) + (10 – 1) 1] 1

S10 = 5 [12 + 9 (3)] S10 = 5 (12 + 27) S10 = 5 (39)

S10 = 195 1

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - I... 2 ...

(ii) The roots of the quadratic equation are 0 and – 4.

Let = 0 and = – 4

+ = 0 + (– 4) = – 4 and . = 0 × – 4 = 0 1We know that,

x2 – ( + )x + . = 0

x2 – (– 4)x + 0 = 0

x2 + 4x = 0

The required quadratic equation is x2 + 4x = 0 1

(iii) The equation of Y-axis is x = 0 1Let the point of intersection of the line y = 3x + 2 with Y-axis be

(0, k)

(0, k) lies on the line it satisfies the equation

Substituting x = 0 and y = k in the equation we get,

k = 3 (0) + 2

k = 2

The point of intersection of the line y = 3x + 2 with Y-axis is (0, 2). 1

A.3. Solve the following :(i) 2x + 5y = 13 ......(i)

4x – 9y = 7 .....(ii)

D =2 5

4 –9 = (2 × – 9) – (5 × 4) = – 18 – 20 = – 38

Dx =13 5

7 –9 = (13 × – 9) – (5 × 7)= – 117 – 35 = – 152

Dy =2 13

4 7 = (2 × 7) – (13 × 4) = 14 – 52 = – 38 1

By Cramer’s rule,

x =D

Dx =

–152

–38 = 4

y =D

Dy

= –38

–38 = 1 1

The point for intersection of the lines 2x + 5y = 13 and 14x – 9y = 7 is (4, 1).

Paper - I... 3 ...

(ii) For the G.P.

The first term (a) = 4

Common ratio (r) = 2

tn = arn – 1 1 t4 = ar4 – 1

t4 = ar3

t4 = 4 (2)3

t4

= 4 (8)

t4 = 32

t9 = ar9 – 1 1 t9 = ar8

t9

= 4 (2)8

t9 = 1024

The fourth and ninth term of the G.P. are 32 and 11024 respectively.

(iii) p2 – 10p + 5 = 0

p2 – 10p = – 5 ..... (i)

Third term =1

× coefficient of p2

2

=1

× –102

2

1

= (– 5)2

= 25

Adding 25 to both the sides of equation (i), we get;

p2 – 10p + 25 = – 5 + 25

(p – 5)2 = 20

p – 5 = + 20 [Taking square roots on both sides] 1

p – 5 = + 2 5

p = 5 2 5

p = 5 2 5 or p = 5 – 2 5

5 2 5 and 5 – 2 5 are the roots of the given quadratic equation. 1

Paper - I... 4 ...

A.4. Solve the following :(i) For a G.P.

Sn =a (1 – r )

1 – r

n

S6 =a (1 – r )

1 – r

6

But,S6 = 126 [Given]

a (1 – r )

1 – r

6

= 126 .....(i) 1

Similarly,

S3 = a (1 – r )

1 – r

3

But,S3 = 14 [Given]

a (1 – r )

1 – r

3

= 14 .....(ii) 1

Dividing (i) by (ii),

a (1 – r ) a (1 – r )

1 – r 1 – r

6 3

=126

14

a (1 – r ) 1 – r

1 – r a (1 – r )

6

3 = 9

1 – r

1 – r

6

3 = 9

1 – (r )

1 – r

2 3 2

3 = 9

(1 r ) (1 – r )

1 – r

3 3

3 = 9

1 + r3 = 9 r3 = 9 – 1 r3 = 8

Taking cube roots on both sidesr = 2 1

Substituting r = 2 in (ii), we get,

Paper - I... 5 ...

a (1 – 2 )

1 – 2

3

= 14

a (1 – 8)

–1 = 14

a (– 7)

–1 = 14

7a = 14 a = 2

a = 2, r = 2. 1

(ii) Let and be the roots of a quadratic equation. + = 3 and3 + 3 = 63 [Given]

We know that,x2 – ( + )x + . = 0 .......(i) 1

Also, 3 + 3 = ( + )3 – 3 . ( + ) 63 = (3)3 – 3 . (3)

[ + = 3 and 3 + 3 = 63] 63 = 27 – 9 . 1 9 . = 27 – 63 9 . = – 36

. =– 36

9 . = – 4 x2 – ( + )x + . = 0 [From (i)] 1 x2 – 3x + (– 4) = 0 [ + = 3 and . = – 4] x2 – 3x – 4 = 0

The required quadratic equation is x2 – 3x – 4 = 0. 1

(iii) 4x = y – 5 4x +5 = y y = 4x + 5

x 0 –1 –2

y 5 1 –3

(x, y) (0, 5) (–1, 1) (–2, –3)

y = 2x + 1

x 0 1 2

y 1 3 5

(x, y) (0, 1) (1, 3) (2, 5) 1

Paper - I

2


... 6 ...

Y

Scale : 1 cm = 1 uniton both the axes

Y

41 2 3 5 X-5 -4 -3 -2X

-1

-2

-3

4

2

1

-4

(0, 1)

y =

2x

+ 1

(1, 3)

(2, 5)

(–2, –3)

(–1, 1)

(0, 5)

0

5

3

-1

-5

-6

-7

Paper - II


(i) The 7th term of G.P. 2, – 6, 18 is ............. .

(ii) Determine the nature of roots of the following equation from itsdiscriminant 2y2 – 7y – 3 = 0.

(iii) Write the co-ordinates of the point of intersection of X-axis andY-axis.


(i) Write down the first five terms of the geometric progression whichhas first term 1 and common ratio 4.

(ii) Form the quadratic equation whose roots are 3 and 10.

(iii) If the point (3, 2)lies on the graph of the equation 5x + ay = 19, thenfind a.


(i) Without drawing the graphs, show that the following equations areof concurrent lines : y = 5x – 3; y = 4 – 2x; 2x + 3y = 8

S.S.C. Test - II



MAHESH TUTORIALS

Paper - II... 2 ...

Best Of Luck

(ii) If in a G.P. r = 2 and t8 = 64 then find a and S

6.

(iii) Solve the following quadratic equation by using formula :7p2 – 5p – 2 = 0


(i) Babubhai borrows Rs. 4000 and agrees to repay with a total interestof Rs. 500. in 10 instalments, each instalment being less that thepreceding instalment by Rs. 10. What should be the first and thelast instalment ?

(ii) If + = 5 and 3 + 3 = 35, find a quadratic equation whose rootsare and .

(iii) Solve the following simultaneous equations :

1 1 1+ =

3x 5y 15 ;1 1 1

+ =2x 3y 12

Paper - II

A.1. Solve the following :(i) The 7th term of G.P. 2, – 6, 18 is 1458. 1

(ii) 2y2 – 7y – 3 = 0Comparing with ay2 + by + c = 0 we have a = 2, b = – 7, c = – 3 = b2 – 4ac

= (– 7)2 – 4 (2) (– 3)= 49 + 24= 73


(iii) The co-ordinates of point of intersection of X-axis and Y-axis is(0, 0). 1

A.2. Solve the following :(i) For the G.P.

The first term (a) = 1Common ratio (r) = 4t1 = a = 1t2 = ar = 1 × 4 = 4 1t3 = ar2 = 1 × (4)2 = 16t4 = ar3 = 1 × (4)3 = 64t5 = ar4 = 1 × (4)4 = 256

The first five terms of the G.P. are 1, 4, 16, 64 and 256. 1

(ii) The roots of the quadratic equation are 3 and 10Let = 3 and = 10

+ = 3 + 10 = 13 1 . = 3 × 10 = 30We know that,x2 – ( + ) x + . = 0

x2 – 13x + 30 = 0

The required quadratic equation is x2 – 13x + 30 = 0 1

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - II... 2 ...

(iii) (3, 2) lies on the graph of the equation 5x + y = 19.It satisfies the equation,

Substituting x = 3 and y = 2 in the equation we get,5 (3) + a (2) = 19

15 + 2a = 19 2a = 19 – 15 1 2a = 4

a =4

2

a = 2 1

A.3. Solve the following :(i) y = 5x – 3

– 5x + y = – 3 ........(i)y = 4 – 2x

2x + y = 4 ........(ii)2x + 3y = 8 .......(iii)Solving (i) and (ii),Substracting (ii) from (i),– 5x + y = – 32x + y = 4(–) (–) (–)–7x = – 7

x =– 7

– 7

x = 1 1Substituting x = 1 in (ii),2 (1) + y = 4

2 + y = 4 y = 4 – 2 y = 2

Substituting x = 1 and y = 2 in L.H.S. of (iii),L.H.S. = 2x + 3y 1

= 2 (1) + 3 (2)= 2 + 6= 8= R.H.S.

(1, 2) is the solution for (iii)also. (1, 2) is a common solution for all the three equations (1, 2) is the common point for all the three lines.

The equations of are concurrent lines. 1

Paper - II... 3 ...

(ii) For a G.P.r = 2t8 = 64tn = arn – 1

t8 = ar8 – 1

64 = ar7

64 = a (2)7

64 = a (128)64

128= a 1

a =1

2

Sn =a (r – 1)

r – 1

n

S6 = a r – 1

r – 1

6

S6 =

1(2 – 1)

22 – 1

6

1

S6 =

1(64 – 1)

2–1

S6 =1

(63)2

S6 =63

2

a = 1

2 and S6 =

63

2. 1

(iii) 7p2 – 5p – 2 = 0Comparing with ap2 + bp + c = 0 we have a = 7, b = – 5, c = – 2b2 – 4ac = (– 5)2 – 4 (7) (– 2)

= 25 + 56= 81 1

p =– b b – 4ac

2a

2

=– (–5) 81

2 (7)

=5 9

14

1

Paper - II... 4 ...

p = 5 9

14

or p =

5 – 9

14

p = 14

14or p =

– 4

14

p = 1 or p = –2

7

1 and –2

7 are the roots of the given quadratic equation. 1

A.4. Solve the following :(i) Total money repaid by Babubhai in 10 instalments

= S10

= 4000 + 500= Rs. 4500

No. of instalments (n) = 10Difference between two consecutive instalments (d) = – 10 1First instalment = (a) = ?Last instalment (t10) = ?

Sn =n

2 [2a + (n – 1) d]

S10 =10

2 [2a + (10 – 1) d]

4500 = 5 [2a + 9 (– 10)] 1

4500

5= 2a – 90

900 = 2a – 90 900 + 90 = 2a 990 = 2a

990

2= a

a = 495tn = a + (n – 1) d

t10 = a + (10 – 1) d t10 = 495 + 9 (– 10) 1 t10 = 495 – 90 t10 = 405

First instalment is Rs. 495 and last instalment is Rs.405. 1

(ii) and are the roots of a quadratic equation + = 5 [Given]

3 + 3 = 35

Paper - II... 5 ...

We know that,x2 – ( + ) x + . = 0 ......(i)

Also, 3 + 3 = ( + )3 – 3 . ( + ) 1 35 = (5)3 – 3 . (5)

[ + = 5 and 3 + 3 = 35] 35 = 125 – 15 . 1 15 . = 125 – 35 15 . = 90

. =90

151

. = 6 x2 – ( + ) x + . = 0 [From (i)] x2 – 5x + 6 = 0

The required quadratic equation is x2 – 5x + 6 = 0. 1

(iii)1 1 1

+ =3x 5y 15

Multiplying throught by 15 we get,

15 1

3x

+ 15 1

5y

= 15 1

15

5

x +

3

y = 1 ........(i)

1

2x +

1

3y =1

12Multipying throught by 12,

12 1

2x

+ 12 1

3y

= 12 1

12

6

x +

4

y = 1 .........(ii)

Substituting 1

x = a and

1

y = b in (i) and (ii),

5a + 3b = 1 ........(iii) 16a + 4b = 1 ........(iv)Multiplying (iii) bt 4,20a + 12b = 4 .........(v)Multiplying (iv) by 318a + 12b = 3 ........(vi)

Paper - II... 6 ...

Subtracting (vi) from (v)20a + 12b= 418a + 12b= 3

(–) (–) (–)2a = 1

a =1

21

Substituting a = 1

2 in (iv),

6 1

2

+ 4b = 1

3 + 4b = 1 4b = 1– 3 4b = – 2

b =–2

4

b =–1

21

Resubstituting the values of a and b,

a =1

x

1

2=

1

x x = 2

b =1

y

–1

2=

1

y

y = –2

x = 2 and y = –2 is the solution of given simultaneous equations. 1

Paper - III


(i) Three consecutive numbers are in G.P. and their product is 1000then the second term a is ............. .

(ii) Determine the nature of roots of the following equation from itdiscriminant y2 + 6y + 9 = 0.

(iii) If (a, 3) is point lying on graph of equation 5x + 2y = – 4 then a............. .


(i) Find the common ratio and the 7th term of the G.P. 2, – 6, 18, .... .

(ii) Form the quadratic equation if its roots are 5 and – 7.



(i) Without drawing the graphs, show that the following equations areof concurrent lines : 2x + y = 6; x + 2y = 6; 7x – 4y = 6.

S.S.C. Test - II



MAHESH TUTORIALS


Best Of Luck

(ii) Find the indicated sums for the following Geometric Progressions :

1, 2 , 2, ... Find S10

.

(iii) Solve the following quadratic equation by completing square :n2 + 3n – 4 = 0


(i) A meeting hall has 20 seats in the first row, 24 seats in the secondrow, 28 seats in the third row, and so on and has in all 30 rows.How many seats are there in the meeting hall ?

(ii) If the difference of the roots of the quadratic equation is 5 and thedifference of their cubes is 215, find the quadratic equation.

(iii) Solve the following simultaneous equations using graphical method :x + y = 8, x – y = 2

Paper - III

A.1. Solve the following :(i) Three consecutive numbers are an G.P. and their product is

1000 then the second term a is 10. 1

(ii) y2 + 6y + 9 = 0Comparing with ay2 + by + c = 0 we have a = 1, b = 6, c = 9 = b2 – 4ac

= (6)2 – 4 (1) (9)= 36 – 36= 0

= 0Hence roots of the quadratic equation are real and equal. 1

(iii) If (a, 3) is point lying on graph of equation 5x + 2y = – 4 thena – 2. 1

A.2. Solve the following :(i) For the G.P. 2, – 6, 18, ........

The first term (a) = 2


–2

= – 3

Seventh term t7 = ?tn = arn – 1

t7 = ar7 – 1 1 t7 = ar6

t7

= 3 (–3)6

t7 = 2 (729) t7 = 1458

The common ratio of the G.P. is – 3 and the seventh term is 1458. 1

(ii) The roots of the quadratic equation are 5 and – 7.Let = 5 and = – 7

+ = 5 + (– 7) = 5 – 7 = – 2 and . = 5 × – 7 = – 35 1We know that,

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS


x2 – ( + )x + . = 0 x2 – (– 2)x + (– 35) = 0 x2 + 2x – 35 = 0

The required quadratic equation is x2 + 2x – 35 = 0. 1

(iii) The equation of Y-axis is x = 0 1Let the point of intersection of the line y = 3x + 2 with Y-axis be

(0, k)

(0, k) lies on the line it satisfies the equation

Substituting x = 0 and y = k in the equation we get,

k = 3 (0) + 2

k = 2


A.3. Solve the following :(i) 2x + y = 6 ......(i)

x + 2y = 6 .....(ii)

7x – 4y = 6 .....(iii)

Solving (i) and (iii),

Multiplying (i) by 4,

8x + 4y = 24 ......(iv)

Adding (iii) and (iv) we get,

7x – 4y = 6

8x + 4y = 2415x = 30

x =30

15 x = 2 1

Substituting x = 2 in (i),2 (2) + y = 6

4 + y = 6 y = 6 – 4 y = 2 1

Substituting x = 2 and y = 2 in L.H.S. of (ii),L.H.S. = x + 2y

= 2 + 2 (2)= 2 + 4= 6= R.H.S.


(2, 2) is the solution of (ii) also

(2, 2) is the common solution for all the three equations.

(2, 2) is the common point for all the three lines.

The equations are of concurrent lines. 1

(ii) For the G.P., a = 5, r = 2

Sn =a (r – 1)

r – 1

n

S10 = a r – 1

r – 1

10

S10 = 1 2 – 1

2 – 1

10

S10 =2 – 1

2 – 1

1012

1

S10 =2 – 1

2 – 1

5

S10 =32 – 1

2 – 1

S10 =

2 131

2 – 1 2 1

S10 =

31 2 1

2 – 1

2 2 1

S10 = 31 2 1

2 – 1

S10 = 31 2 1 1

S10 = 31 1 2 1


(iii) n2 + 3n – 4 = 0 n2 + 3n = 4 ......(i)

Third term =1

× coefficient of n2

2

=1

× 32

2

=3

2

2

=9

41

Adding 9

4 to both the sides of (i) we get,

n2 + 3n + 9

4 = 4 +

9

4

3

n +2

2

=16 + 9

4

3

n +2

2

=25

4Taking square root on both the sides we get,

n + 3

2=

5±

21

n =–3 5

2 2

n = 3

2

+

5

2or n =

3

2

–

5

2

n = 2

2or n = –

8

2 n = 1 or n = – 4

1 and – 4 are the roots of the given quadratic equation. 1

A.4. Solve the following :(i) The no. of seats in each row are as follows 20, 24, 28, .........

The no. of seats in each row form an A.P. with

First term (a) = 20

Difference between the no. of seats in two successive rows

(d) = 4 1


Total no. of rows (n) = 30

Total no. of seats in 30 rows (S30) = ?

Sn =n

2 [2a + (n – 1) d] 1

S30 =30

2 [2 (20) + (30 – 1) 4] 1

S30 = 15 [40 + 116]

S30 = 15 (156)

S30 = 2340

Total no. of seats in the meeting hall is 2340. 1

(ii) Let and be the roots of a quadratic equation.

– = 5 and3 – 3 = 215 [Given]

We know that,

x2 – ( + )x + . = 0 .......(i) 1Also, 3 – 3 = ( – )3 + 3 . ( – )

215 = (5)3 + 3 . (5) [ + = 5 and 3 – 3 = 215]

215 = 125 + 15 . 215 – 125 = 15 . 90 = 15 .

. =90

15

. = 6 1

Now, – 2 = ( + )2 – 4.

(5)2 = ( + )2 – 4 (6) [ . = 5 and . = 6]

25 = ( + )2 – 24

25 + 24 = ( + )2

( + )2 = 49 1

Taking square root on both the sides, we get;

+ = + 7

x2 – ( + )x + .= 0 [From (i)]

x2 – (7)x + 6 = 0 or x2 – (– 7)x + 6 = 0

x2 – 7x + 6 = 0 or x2 + 7x + 6 = 0

The required quadratic equation is x2 – 7x + 6 = 0 or 1x2 + 7x + 6 = 0.


(iii) x + y = 8 x – y = 2 1 y = 8 – x x = 2 + y

x 0 1 2 x 2 3 4

y 8 7 6 y 0 1 2

(x, y) (0, 8) (1, 7) (2, 6) (x, y) (2, 0) (3, 1) (4, 2)

2


Y

-1

-2


-3

-4 -3 -2 41 2-1 3 5

Y

7

6

4

5

3

2

1

8

X

(0, 8)

(1, 7)

(5, 3)

(4, 2)

(3, 1)

(2, 0)

x -

y = 2

0X

(2, 6)

x + y = 8

Paper - IV


(i) If a = 6 and d = 3. S10

= ?

(ii) Determine the nature of roots of the following equations from its

discriminant x2 + 3 2 x – 8 = 0

(iii) If (a, 3) is the point lying on the graph of the equation 5x + 2y = – 4,then find a.


(i) Find the 69th term of the G.P. 1, –1, 1, –1, ....

(ii) Form the quadratic equation if its roots are – 3 and – 11.



(i) Points (3, –1) and (6, 1) lie on the line represented by the equationpx + qy = 9, find the values of p and q.

S.S.C. Test - II



MAHESH TUTORIALS

Paper - IV... 2 ...

Best Of Luck

(ii) The taxi fare is Rs. 14 for the first kilometer and Rs. 2 for eachadditional kilometer. What will be fare for 10 kilometers ?

(iii) Solve the following quadratic equations by using formula :2n2 + 5n + 2 = 0


(i) Vijay invests some amount in National saving certificate. For the1st year he invests Rs. 500, for the 2nd year he invests Rs. 700, forthe 3rd year he invests Rs. 900, and so on. How much amount hehas invested in 12 years ?

(ii) If the difference of the roots of the quadratic equation is 3 anddifference between their cubes is 189, find the quadratic equation.

(iii) Solve the following simultaneous equations :

27

x – 2 +

31

y 3 = 85, 31

x – 2 +

27

y 3 = 89

Paper - IV

A.1. Solve the following :(i) If a = 6 and d = 3. S

10 = 195. 1

(ii) x2 + 3 2 x – 8 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = 3 2 , c = – 8 = b2 – 4ac

= (3 2 )2 – 4 (1) (– 8)= 9 × 2 + 32= 18 + 32= 50


(iii) (a, 3) is a point lying on the graph of the equation 5x + 2y = – 4,it is satisfies the equation.

Substituting x = a and y = 3 in the equation we get,5 (a) + 2 (3) = – 4

5a + 6 = – 4 5a = – 4 – 6 5a = – 10

a =–10

5 a = – 2 1

A.2. Solve the following :(i) For the G.P. 1, – 1, 1, – 1

First term (a) = 1

Common ratio (r) = –1

1 = – 1

tn = arn – 1 1 t69 = ar69 – 1

t69 = ar68

t69

= 1 (– 1)68

t69 = 1 × 1 t69 = 1

Sixtyninth term of G.P. is 1. 1

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - IV... 2 ...

(ii) The roots of the quadratic equation are -3 and -11.Let = – 3 and = – 11

+ = – 3 + (– 11) = – 3 – 11 = – 14 and . = – 3 × – 11 = 33 1We know that,x2 – ( + )x + .= 0

x2 – (– 14)x + 33 = 0 x2 + 14x + 33 = 0

The required quadratic equation is x2 + 14x + 33 = 0 1

(iii) The equation of Y-axis is x = 0 1Let the point of intersection of the line y = 3x + 2 with Y-axis be(0, k)

(0, k) lies on the line it satisfies the equation Substituting x = 0 and y = k in the equation we get,

k = 3 (0) + 2 k = 2

The point of intersection of the line y = 3x + 2 with Y-axis is 1(0, 2).

A.3. Solve the following :(i) (3, – 1) lies on the line px + qy = 9

It satisfies the equation. Substituting x = 3 and y = – 1 in the equation we get,

p (3) + q (– 1) = 9 3p – q = 9 ......(i) (6, 1) lies on the line px +qy = 9

It satisfies the equation.Substituting x = 6 and y = 1 in the equation we get,p (6) + q (1)= 9

6p + q = 9 ......(ii) 1Adding (i) and (ii),3p – q = 96p + q = 9

9p = 18

p =18

9 p = 2 1

Substituting p = 2 in (ii),6 (2) + q = 9

12 + q = 9 q = 9 – 12 q = – 3

p = 1 and q = – 3. 1

Paper - IV... 3 ...

(ii) Since the taxi fare increases by Rs. 2 every kilometer afterthe first, the successive taxi fares form an A.P.The taxi fare for first kilometer (a) = Rs. 14Increase in taxi fare in every kilometer after first kilometer(d) = 2 1No. of kilometers covered by taxi (n) = 10Taxi fare for 10 kilometers = t10 = ?tn = a + (n + 1) d 1

t10 = a + (10 – 1) d t10 = 14 + 9 (2) t10 = 14 + 18 t10 = 32

Taxi fare for ten kilometers is Rs. 32. 1

(iii) 2n2 + 5n + 2 = 0Comparing with ax2 + bx + c = 0 we have a = 2, b = 5, c = 2b2 – 4ac = (5)2 – 4 (2) (2)

= 25 – 16= 9 1

n =– b b – 4ac

2a

2

=–5 9

2 (2)

=–5 3

4

1

n = –5 3

4

or n =

–5 – 3

4

n = –2

4or n =

– 8

4

n = –1

2or n = – 2

–1

2 and – 2 are the roots of the given quadratic equation. 1

A.4. Solve the following :(i) Yearly investments of Vijay are as follows 500, 700, 900, .......

The yearly investments form an A.P. with first year investment(a) = 500Difference between investment done in two successive years(d) = 200. 1

Paper - IV... 4 ...

No. of years (n) = 12

Total investment done in 12 years = (S12) = ?

Sn =n

2[2a + (n – 1) d]

S12 =12

2[2 (500) + (12 – 1) 200] 1

S12 = 6 [1000 + 11 (200)]

S12 = 6 [1000 + 2200] 1 S12 = 6 [3200]

S12 = 19200

Total investment done in 12 years is Rs. 19200. 1

(ii) Let and be the roots of a quadratic equation

– = 3 and 3 – 3 = 189 [Given]

We know that,

x2 – ( + ) x + . = 0 ......(i)

Also,3 – 3 = ( – )3 + 3. ( – ) 1 189 = (3)3 + 3. (3) [ . = 3 and 3 – 3 = 189]

189 = 27 + 9 . 189 – 27 = 9 . 162 = 9 .

162

9= .

. = 18

Now, ( – )2 = ( + )2 – 4. 1 (3)2 = ( + )2 – 4 (18) [ – = 3 and . = 18]

9 = ( + )2 – 72

9 + 72 = ( + )2

( + )2 = 81

Taking square root on both the sides we get,

+ = + 9

x2– (+) x + . = 0 [From (i)] 1 x2 – (9)x + 18 = 0 or x2 – (– 9)x + 18 = 0

x2 – 9x + 18 = 0 or x2 + 9x + 18 = 0


Paper - IV... 5 ...

(iii)27

x – 2 +

31

y 3 = 85 ........(i)

31

x – 2 +

27

y 3 = 89 .......(ii)

Substituting 1

x – 2 = a and

1

y 3 = b in (i) and (ii),

27a + 31b = 85 .......(iii)

31a + 27b = 89 ........(iv)

Adding (iii) and (iv),

58a + 58b = 174

Dividing throughout by 58 we get,

a + b =174

58 a + b = 3 ........(v)

Subtracting (iv) from (iii), 27a + 31b = 81 31a + 27b = 89(–) (–) (–)– 4a + 4b = – 4Dividing throughout by – 4 we get,a – b = 1 ........(vi) 1Adding (v) and (vi),a + b = 3a – b = 12a = 4

a = 2Subtracting a = 2 in (v),2 + b = 3

b = 3 – 2 b = 1 1

Resubstituting the values of a and b

a =1

x – 2

2 =1

x – 2 2 (x – 2) = 1

2x – 4 = 1

2x = 1 + 4

2x = 5

Paper - IV... 6 ...

x =5

2

b =1

y 3 1

1 =1

y 3

y + 3 = 1 y = 1 – 3 y = – 2

x = 5

2 and y = –2 is the solution of given simultaneous equations. 1

Paper - V


(i) The ninth term of the G.P. 3, 6, 12, 24, ....

(ii) Determine the nature of roots of the following equation from itsdiscriminant y2 – 5y + 11 = 0.

(iii) Examine whether the point (2, 5) lies on the graph of the equation3x – y = 1.


(i) Find the 15th term of the G.P. 3, 12, 48, 192, ...

(ii) Form the quadratic equation whose roots are – 5 and 9.

(iii) What is the equation of X - axis? Hence, find the point of intersectionof the graph of the equation x + y = 3 with the X - axis.


(i) If (3, 1) is he point of intersection of lines ax + by = 7 and bx + ay = 5,find the values of a and b.

S.S.C. Test - II



MAHESH TUTORIALS

Paper - V... 2 ...

Best Of Luck

(ii) Mangala started doing physical exercise 10 minutes for the firstday. She will increase the time of exercise by 5 minutes per day, tillshe reaches 45 minutes. How many days are required to reach 45minutes ?

(iii) Solve the following quadratic equations by completing square :m2 – 3m – 1 = 0


(i) If S3 = 31 and S


(ii) If the sum of the roots of the quadratic is 3 and sum of their cubesis 63, find the quadratic equation.

(iii) Solve the following simultaneous equations using graphical method :3x + 4y + 5 = 0; y = x + 4

Paper - V

A.1. Solve the following :(i) The ninth term of G.P. 3, 6, 12, 24, ....... is 768. 1

(ii) y2 – 5y + 11 = 0Comparing with ay2 + by + c = 0 we have a = 1, b = – 5, c = 11 = b2 – 4ac

= (– 5)2 – 4 (1) (11)= 25 – 44= – 19

< 0Hence roots of the quadratic equation are not real. 1

(iii) Substituting x = 2 and y = 5 in the L.H.S. of the equation 3x – y =1L.H.S. = 3x – y

= 3 (2) – 5= 6 – 5= 1= R.H.S.

x = 2 and y = 5 satisfies the equation 3x – y = 1Hence (2, 5) lies on the graph of the equation 3x – y = 1. 1

A.2. Solve the following :(i) For the G.P. 3, 12, 48, 192, ........

First term (a) = 3


3 = 4 1

tn = arn – 1

t15 = ar15 – 1

t15 = ar14

t15

= 3 (4)14

t15 = 3 × 414

Fifteenth term of G.P. is 3 × 414. 1

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - V... 2 ...

(ii) The roots of the quadratic equation are – 5 and 9Let = – 5 and = 9

+ = – 5 + 9 = 4 . = – 5 × 9 = – 45We know that,x2 – ( + ) x + . = 0 1

x2 – 4x + (– 45) = 0 x2 – 4x – 45 = 0

The required quadratic equation is x2 – 4x – 45 = 0 1

(iii) The equation of X-axis is y = 0Let the point of intersection of graph x + y = 3 with X-axis be 1(h, 0)

(h, 0) lies on the graph, it satisfies the equation Substituting x = h and y = 0 in the equation we get,

h + 0 = 3 h = 3

The line x + y = 3 intersects the X-axis at (3, 0). 1


(i) (3, 1) is the point of intersection of lines ax + by = 7 and

bx + ay = 5

It lies on both the lines

Hence it satisfies both equations

Substituting x = 3 and y = 1 in the equation ax + by = 7 we get,

a (3) + b (1)= 7

3a + b = 7 ......(i)

Substituting x = 3 and y = 1 in the equation bx + ay = 5 we get,

b (3) + a (1) = 5

a + 3b = 5 .....(ii) 1

Multiplying (ii) by 3 we get,

3a + 9b = 15 .....(iii)

Substracting (iii) from (i),

3a + b = 7

3a + 9b = 15

(–) (–) (–)

– 8b = – 8

Paper - V... 3 ...

b =–8

–8 b = 1 1

Substituting b = 1 in (i),3a + 1 = 7

3a = 7 – 1 3a = 6

a =6

3 a = 2

a = 2 and b = 1 1

(ii) Since the workout time Mangala increases by 5 minutes everydayafter the first day, the successive workout times are in A.P.Workout time for first day (a) = 10 minutes.Increases in workout time (d) = 5 minutesLet No. of days required to reach workout time of 45 minutes be 1‘n’ days.tn = 45

tn = a + (n – 1) d 45 = 10 + (n – 1) 5 45 = 10 + 5n – 5 1 45 = 5 + 5n 45 – 5 = 5n 5n = 40 n = 8

8 days required to reach work out time of 45 minutes. 1

(iii) m2 – 3m – 1 = 0 m2 – 3m = 1 .... (i)

Third term =1

coefficient of m2

2

=1

– 32

2

=–3

2

2

=9

41

Adding 9

4 to both sides of (i) we get,

m2 – 3m + 9

4=

91

4

Paper - V... 4 ...

3

m –2

2

=4 9

4

3

m –2

2

=13

4Taking square root on both the sides we get,

3

m –2

=13

2 1

m =3 13

2 2

m =3 13

2

m = 3 13

2

or m =

3 – 13

2

3 13

2

and

3 – 13

2 are the roots of the given quadratic equations. 1


Sn = a (1 – r )

1 – r

n

S3 = a (1 – r )

1 – r

3

But, S3 = 31 [Given]

a (1 – r )

1 – r

3

= 31 .......(i)

Similarly,

S6 = a (1 – r )

1 – r

6


a (1 – r )

1 – r

6

= 3906 ......(ii) 1

Dividing (ii) by (i),

a (1 – r ) 1 – r

1 – r a (1 – r )

6

3 =3906

31

1 – r

1 – r

6

3 = 126

Paper - V... 5 ...

1 – (r )

1 – r

3 2

3 = 126

(1 r ) (1 – r )

1 – r

3 3

3 = 126

1 + r3 = 126 r3 = 126 – 1 r3 = 125 1

Taking cube roots on both sides,r = 5

Substituting r = 5 in (i),

a (1 – 5 )

1 – 5

3

= 31

a (1 – 125)

– 4 = 31

a (– 124) = 31 × – 4 1 – 124a = – 124

a =–124

–124 a = 1 a = 1 and r = 5. 1

(ii) Let and be the roots of a quadratic equation. + = 3 [Given]

and3 + 3 = 63We know that,x2 – ( + )x + . = 0 .......(i) 1

Also, 3 + 3 = ( + )3 – 3 . ( + ) 63 = (3)3 – 3 . (3) [ + = 3 and 3 + 3 = 63] 63 = 27 – 9 . 9 . = 27 – 63 9 . = – 36

. =– 36

91

. = – 4

x2 – ( + )x + . = 0 [From (i)]

x2 – 3x + (– 4) = 0 [ + = 3 and . = – 4] 1 x2 – 3x – 4 = 0

The required quadratic equation is x2 – 3x – 4 = 0. 1

Paper - V... 6 ...

(iii) 3x + 4y + 5 = 0 y = x + 4 3x = –5 – 4y

x = –5 – 4y

3 x –3 1 5 x 0 1 2

y 1 –2 –5 y 4 5 6 1

(x, y) (–3, 1) (1, –2) (5, –5) (x, y) (0, 4) (1, 5) (2, 6)

2

x = – 3 and y = 1 is the solution of given simultaneous equations. 1Y


Y

41 2 3 5 X-5 -4 -3 -2 -1X

(5, –5)

(1, –2)

(–3, 1)

(0, 4)

(1, 5)

y =

x +

4

(2, 6)

3x + 4y + 5 = 0

0

6

-1

-2

-3

3

-5

-4

2

1

5

4

Paper - VI


(i) Write down the first five terms of the geometric progression whichhas first term 1 and common ratio 4.

(ii) Find the value of discrimininant of the following equation :4x2 – kx + 2 = 0.

(iii) Write one solution of the equation 2x + y = 10.


(i) Find the indicated sums for the following Geometric Progression :2, 6, 18, ... Find S

7.

(ii) Form the quadratic equation whose roots are 0 and – 6.

(iii) If the point (3, 2)lies on the graph of the equation 5x + ay = 19, thenfind a.


(i) If (3, 1) is he point of intersection of lines ax + by = 7 and bx + ay = 5,find the values of a and b.

S.S.C. Test - II



MAHESH TUTORIALS

Paper - VI... 2 ...

Best Of Luck

(ii) There is an auditorium with 35 rows of seats. There are 20 seats inthe first row, 22 seats in the second row, 24 seats in the third row,and so on. Find the number of seats in the twenty fifth row.

(iii) Solve the following quadratic equations using formula :9p2 – 5p – 4 = 0


(i) If S6 = 126 and S


(ii) If the difference of the roots of the quadratic equation is 5 and thedifference of their cubes is 215, find the quadratic equation.

(iii) Solve the following simultaneous equations using graphical method :4x = y – 5; y = 2x + 1

Paper - VI

A.1. Solve the following :(i) For the G.P.

The first term (a) = 1Common ratio (r) = 4t1 = a = 1t2 = ar = 1 × 4 = 4t3 = ar2 = 1 × (4)2 = 16t4 = ar3 = 1 × (4)3 = 64t5 = ar4 = 1 × (4)4 = 256

The first five terms of the G.P. are 1, 4, 16, 64 and 256. 1

(ii) 4x2 – kx + 2 = 0Comparing with ax2 + bx + c = 0 we have a = 4, b = – k, c = 2 = b2 – 4ac

= (– k)2 – 4 (4) (2)= k2 – 32

= k2 – 32 1

(iii) Substituting x = 0 in the equation we get,2 (0) + y = 10

y = 10

x = 0 and y = 10 is one of the solution of 2x + y = 10 1

A.2. Solve the following :(i) 2, 6, 18, .........

a = 2

r =6

2= 3

Sn =a (r – 1)

r – 1

n

S7 =a (r – 1)

r – 1

7

1

MODEL ANSWER PAPER

S.S.C. Test - II



MAHESH TUTORIALS

Paper - VI... 2 ...

S7 =2 (3 – 1)

3 – 1

7

S7 =2 (3 – 1)

2

7

S7 = 37 – 1

S7 = 2187 – 1

S7 = 2186 1

(ii) The roots of the quadratic equation are 0 and – 6Let = 0 and = – 6

+ = 0 + (– 6) = 0 – 6 = – 6 1 . = 0 × – 6 = 0We know that,x2 – ( + ) x + . = 0

x2 – (– 6) x + 0 = 0 x2 + 6x = 0

The required quadratic equation is x2 + 6x = 0. 1

(iii) (3, 2) lies on the graph of the equation 5x + ay = 19.It satisfies the equation,

Substituting x = 3 and y = 2 in the equation we get,5 (3) + a (2) = 19 1

15 + 2a = 19 2a = 19 – 15 2a = 4

a =4

2 a = 2 1

A.3. Solve the following :(i) (3, 1) is the point of intersection of lines ax + by = 7 and bx + ay = 5

It lies on both the linesHence it satisfies both equationsSubstituting x = 3 and y = 1 in the equation ax + by = 7 we get,a (3) + b (1)=7

3a + b = 7 ......(i)Substituting x = 3 and y = 1 in the equation bx + ay = 5 we get,b (3) + a (1) = 5

a + 3b = 5 .....(ii)Multiplying (ii) by 3 we get,3a + 9b = 15 .....(iii) 1

Paper - VI... 3 ...

Substracting (iii) from (i),3a + b = 73a + 9b = 15(–) (–) (–)– 8b = – 8

b =–8

–8

b = 1 1Substituting b = 1 in (i),3a + 1 = 7

3a = 7 – 1 3a = 6

a =6

3 a = 2 1

a = 2 and b = 1.

(ii) Since the no. of seats in each row of the auditorium are 20, 22,24, ......The no. of seats in each row form an A.P.No. of seats in first row (a) = 20Difference in no. of seats in two successive rows is (d) = 2 1No. of seats in 25th row = t25 = ?tn = a + (n + 1) d

t25 = a + (25 – 1) d t25 = 20 + 24 (2) 1 t25 = 20 + 48 t25 = 68

There are 68 seats in 25th row. 1

(iii) 9p2 – 5p – 4 = 0Comparing with ap2 + bp + c = 0 we have a = 9, b = – 5, c = – 4b2 – 4ac = (– 5)2 – 4 (9) (– 4)

= 25 + 144= 169 1

p =– b ± b – 4ac

2a

2

= – –5 ± 169

2 (9)

=5 ± 13

18

Paper - VI... 4 ...

p = 5 + 13

18or p =

5 – 13

181

p = 18

18or p =

– 8

18

p = 1 or p = 4

9

1 and 4

9

are the roots of the given quadratic equation. 1


Sn = a (1 – r )

1 – r

n

S6 = a (1 – r )

1 – r

6


a (1 – r )

1 – r

6

= 126 .....(i) 1

Similarly, S3 = a (1 – r )

1 – r

3


a (1 – r )

1 – r

3

= 14 .....(ii)

Dividing (i) by (ii),

a (1 – r ) a (1 – r )

1 – r 1 – r

6 3

=126

14

a (1 – r ) 1 – r

1 – r a (1 – r )

6

3 = 9

1 – r

1 – r

6

3 = 9 1

1 – (r )

1 – r

2 3 2

3 = 9

(1 r ) (1 – r )

1 – r

3 3

3 = 9

1 + r3 = 9 r3 = 9 – 1 r3 = 8

Taking cube roots on both sidesr = 2 1

Paper - VI... 5 ...

Substituting r = 2 in (ii), we get,a (1 – 2 )

1 – 2

3

= 14

a (1 – 8)

–1 = 14

a (– 7)

–1 = 14

7a = 14 a = 2

a = 2, r = 2. 1

(ii) Let and be the roots of a quadratic equation. – = 5 and3 – 3 = 215 [Given]

We know that,x2 – ( + )x + . = 0 .......(i)

Also, 3 – 3 = ( – )3 + 3 . ( – ) 1 215 = (5)3 + 3 . (5) [ + = 5 and 3 – 3 = 215] 215 = 125 + 15 . 215 – 125 = 15 . 90 = 15 .

. =90

15 . = 6 1

Now, – 2 = ( + )2 – 4. (5)2 = ( + )2 – 4 (6) [ . = 5 and . = 6] 25 = ( + )2 – 24 25 + 24 = ( + )2

( + )2 = 49 1Taking square root on both the sides, we get; + = + 7

x2 – (+ )x + . = 0 [From (i)] x2 – (7)x + 6 = 0 or x2 – (– 7)x + 6 = 0 x2 – 7x + 6 = 0 or x2 + 7x + 6 = 0


(iii) 4x = y – 5 y = 2x + 1 4x +5 = y y = 4x + 5

x 0 –1 –2 x 0 1 2

y 5 1 –3 y 1 3 5 1

(x, y) (0, 5) (–1, 1) (–2, –3) (x, y) (0, 1) (1, 3) (2, 5)

Paper - VI... 6 ...

2


Y


Y

41 2 3 5 X-5 -4 -3 -2X

-1

-2

-3

6

4

2

1

-4

(0, 1)

y =

2x

+ 1

(1, 3)

(2, 5)

(–2, –3)

(–1, 1)

(0, 5)

0

5

3

-1

-5

-6

Paper - I


(i) If the point (3, 2)lies on the graph of the equation 5x + ay = 19, thenfind a.

(ii) Without actually solving the simultaneous equations given below,decide which simultaneous equations have unique solution, nosolution or infinitely many solutions. 3x + 5y = 16; 4x – y = 6


(i) Three times the square of a natural numbers is 363. Find thenumbers.

(ii) Solve the following simultaneous equations using Cramer’s rule :3x – y = 7; x + 4y = 11

(iii) Tinu is younger than Pinky by three years. The product of their agesis 180. Find their ages.

Q.3. Solve the following : (Any Three) 12

(i) The length of one diagonal of a rhombus is less than the seconddiagonal by 4 cm. The area of the rhombus is 30 sq.cm. Find thelength of the diagonals.

S.S.C. Test - III


Date : Time : 1 hr. 15 min.ALGEBRA – Chapter : 2, 3

MAHESH TUTORIALS

Paper - I... 2 ...

Best Of Luck

(ii) Solve the following simultaneous equations using graphical method :3x + 4y + 5 = 0; y = x + 4

(iii) The sum of the squares of five consecutive natural numbers is 1455.find them.

(iv) Seg AB is the diameter of a circle. C is the point on the circumferencesuch that in ABC, B is the less by 10º than A. Find the measuresof all the angles of ABC.

Q.4. Solve the following : (Any One) 5

(i) A car covers a distance of 240km with some speed, if the speed increasedby 20 km/hr, it will cover the same distance in 2 hours less. find thespeed of the car.

(ii) Some part of a journey of 555 km was completed by a car with speed60 km/hr then the speed is increased by 15 km/hr and the journey iscompleted. If it takes 8 hours to reach, find the time taken anddistance covered by 60km/hr speed.

Paper - I

A.1. Solve the following :(i) (3, 2) lies on the graph of the equation 5x + y = 19.

It satisfies the equation, Substituting x = 3 and y = 2 in the equation we get,

5 (3) + a (2) = 19 15 + 2a = 19 1 2a = 19 – 15 2a = 4

a =4

2 a = 2 1

(ii) 3x + 5y = 16Comparing with a1x + b1y = c1 we get, a1 = 3, b1 = 5, c1 = 164x – y = 6Comparing with a2x + b2y = c2 we get, a2 = 4, b2 = – 1, c2 = 6

a

a1

2=

3

4

b

b1

2=

5

–1 = – 5 1

c

c1

2=

16

6=

8

3

a

a1

2

b

b1

2

The simultaneous equations 3x + 5y = 16 and 4x – y = 6 have 1unique solution.

A.2. Solve the following :(i) Let the natural number be ‘x’

From the given condition,3x2 = 363 1

x2 = 363

3 x2 = 121 x = + 11 [Taking square roots] 1

MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS

Paper - I... 2 ...

x – 11 because x is a natural number

x = 11

The natural number is 11. 1

(ii) 3x – y = 7

x + 4y = 11

D =3 –1

1 4 = (3 × 4) – (– 1 × 1) = 12 + 1 = 13

Dx =7 –1

11 4 = (7 × 4) – (– 1 × 11) = 28 + 11 = 39

Dy =3 7

1 11 = (3 × 11) – (7 × 1) = 33 – 7 = 26 1

By Cramer’s rule,

x =D

Dx =

39

13= 3

y =D

Dy

=26

13= 2 1


(iii) Tinu’s age be ‘x’ years

Pinky’s age is (x + 3) years

As per the given condition,

x (x + 3) = 180

x2 + 3x – 180 = 0

x2 – 12x + 15x – 180 = 0 1 x (x – 12) + 15 (x – 12) = 0

(x – 12) (x + 15) = 0

x – 12 = 0 or x + 15 = 0

x = 12 or x = – 15 1 Age cannot be negative.

x – 15

x = 12

And x + 3 = 12 + 3 = 15 1

Tinu’s age is 12 years and Pinky’s is 15 years.

Paper - I... 3 ...

A.3. Solve the following : (Any Three)(i) Let the length of other diagonal of a rhombus be ‘x’ cm.

The length of first diagonal is (x + 4) cm. 1

Area of rhombus = 1

2 × Product of length of diagonals

Area of rhombus = 1

2 × x × (x + 4)


1

2x (x + 4) = 30

x (x + 4) = 60 x2 + 4x – 60 = 0 1 x2 + 10x – 6x – 60 = 0 x (x + 10) – 6 (x + 10) = 0 (x + 10) (x – 6) = 0 x + 10 = 0 or x – 6 = 0 x = – 10 or x = 6 1 The length of diagonal of the rhombus cannot be negative. x – 10

Hence x = 6And x + 4 = 6 + 4 = 10

The length of smaller diagonal of a rhombus is 6 cm and biggerdiagonal is 10 cm. 1

(ii) 3x + 4y + 5 = 0 3x = –5 – 4y

x = –5 – 4y

3

x –3 1 5

y 1 –2 –5

(x, y) (–3, 1) (1, –2) (5, –5)

y = x + 4

x 0 1 2

y 4 5 6

(x, y) (0, 4) (1, 5) (2, 6) 1

Paper - I... 4 ...

2


Y


Y

41 2 3 5 X-5 -4 -3 -2 -1X

(5, –5)

(1, –2)

(–3, 1)

(0, 4)

(1, 5)

y =

x +

4

(2, 6)

3x + 4y + 5 = 0

0

6

7

8

-1

-2

-3

3

-5

-4

2

1

5

4

Paper - I... 5 ...

(iii) Let the five consecutive natural numbers be x, x + 1, x + 2, x + 3and x + 4 respectively.As per the given condition,x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2 = 1455 1x2 + x2 + 2x + 1 + x2 + 4x + 4 + x2 + 6x +9 + x2 + 8x + 16 – 1455 = 05x2 + 20x + 30 – 1455 = 05x2 + 20x – 1425 = 0Dividing throughout by 5 we get,

x2 + 4x – 285 = 0 x2 – 15x + 19x – 285 = 0 1 x (x – 15) + 19 (x – 15) = 0 (x – 15) (x + 19) = 0 x – 15 = 0 or x + 19 = 0 x = 15 or x = – 19 1 x is a natural number x – 19

Hence x = 15 x + 1 = 15 + 1 = 16 x + 2 = 15 + 2 = 17 x + 3 = 15 + 3 = 18 x + 4 = 15 + 4 = 19

The required five consecutive natural numbers are 115, 16, 17, 18 and 19 respectively.

(iv)

Let the measures of A be xºand measure of B be yº.Seg AB is the diameter of a circleand C is a point on the circumference. m C = 900 [ Diameter subtends a right

angle at any point on the circle] 1In ABC,m A + m B + m C = 1800 [ Sum of measures of the angles

of a triangle is 1800] x + y + 90 = 180 x + y = 180 – 90 x + y = 90 ........(i)

As the given condition,y = x – 10

– x + y = – 10 .......(ii) 1Adding (i) and (ii),

x + y = 90– x + y = – 10

2y = 80

y =80

2 y = 40 1

A B

C

x0 y0

Paper - I... 6 ...

Substituting y = 40 in (i),x + 40 = 90

x = 90 – 40 x = 50

The measures of angle of ABC are 50º, 40º and 90º. 1

A.4. Solve the following : (Any One)(i) Let the original speed of car be x km/hr.

Distance covered is 240 km

Speed = Distance

Time

Time = Distance

Speed

Time taken by car =240

hrsx

1

New speed of car = (x + 20) km/hr

New time taken by car =240

hrsx 20

As per the given condition,240 240

–x x 20 = 2 1

1 1

240 –x x 20

= 2

x 20 – x

x (x 20)

=

2

240

20

x 20x2 =1

120

20 (120) = 1 (x2 + 20x) 2400 = x2 + 20x 0 = x2 + 20x – 2400 1 x2 + 20x – 2400 = 0 x2 + 60x – 40x – 2400 = 0 x (x + 60) – 40 (x + 60) = 0 (x + 60) (x – 40) = 0 x + 60 = 0 or x – 40 = 0 x = – 60 or x = 40 The speed of car can never be negative. 1 x – 60

Hence x = 40

The original speed of car is 40 km/hr. 1

Paper - I

(ii) Let the time for which the car travels at 60 km/hr be x hoursand at 75 (i.e. 60 + 15) km/hr be y hours .We know that, Distance = Speed × TimeDistance covered by car with the speed of 60 km/hr = 60 × x = 60x kmDistance covered by car with the speed of 75 km/hr = 75 × y = 75y km 1As per first condition,

60x + 75y = 55515 (4x + 5y) = 555

4x + 5y =55515

4x + 5y = 37 .............(i) 1As per second condition,

x + y = 8 ............(ii) 1Multiplying (ii) by 5, we get,

5x + 5y = 40 ............(iii)Subtracting (i) from (iii), we get

5x + 5y = 404x + 5y = 37 – – –

x = 3

Time taken by the car at the speed of 60 km/hr = 3 hours and 1Distance covered with the speed of 60km/hr = 60 × 3 = 180 km. 1

... 7 ...

Paper - II


(i) Find the value of the following determinants : 1.2 0.03

0.57 – 0.23

(ii) What is the equation of X - axis? Hence, find the point of intersection ofthe graph of the equation x + y = 3 with the X - axis.


(i) The length of the rectangle is greater than its breadth by 2 cm. The areaof the rectangle is 24 sq.cm, find its length and breadth.

(ii) Solve the following simultaneous equations using Cramer’s rule :

y = 5x – 10

2; 4x + 5 = – y

(iii) The sum of a natural number and its reciprocal is 10

3. Find the number.


(i) The sum of the areas of two squares is 400sq.m. If the differencebetween their perimeters is 16 m, find the sides of two square.

S.S.C. Test - III



MAHESH TUTORIALS

Paper - II... 2 ...

Best Of Luck

(ii) Solve the following simultaneous equations using graphical method :

4x = y – 5; y = 2x + 1

(iii) The divisor and quotient of the number 6123 are same and the remainderis half the divisor. Find the divisor.

(iv) Durga’s mother gave some 10 rupee notes and some 5 rupee notes toher, which amounts to Rs. 190. Durga said, ‘if the number of 10 rupeenotes and 5 rupee notes would have been interchanged, I would haveRs. 185 in my hand.’ So how many notes of rupee 10 and rupee 5 weregiven to Durga ?


(i) The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozenless for Rs. 840. Find the original cost of one dozen of banana.

(ii) Students of a school were made to stand in rows for drill. If 3 studentsless were standing in each row, 10 more rows were required and if 5students more were standing in each row then the number of rows wasreduced by 10. Find the number of students participating in the drill.

Paper - II


(i)1.2 0.03

0.57 – 0.23= (1.2 × – 0.23) – (0.03 × 0.57) 1= – 0.276 – 0.0171

= – 0.2931 1

(ii) The equation of X-axis is y = 0Let the point of intersection of graph x + y = 3 with X-axis be (h, 0) 1 (h, 0) lies on the graph, it satisfies the equation Substituting x = h and y = 0 in the equation we get,

h + 0= 3 h = 3 1

The line x + y = 3 intersects the X-axis at (3, 0).

A.2. Solve the following :(i) Let the breadth of the rectangle be ‘x’ cm.

It’s length is (x + 2) cm.As per the given condition.

Area of rectangle = Length × Breadth 1 24 = (x + 2) × x 24 = x2 + 2x 0 = x2 + 2x – 24 x2 + 2x – 24 = 0 x2 – 4x + 6x – 24 = 0 1 x (x – 4) + 6(x – 4) = 0 (x – 4) (x + 6) = 0 x – 4 = 0 or x + 6 = 0 x = 4 or x = -6 The breadth of rectangle cannot be negative. x -6

Hence x = 4 and x + 2 = 4 + 2 = 6

The length of rectangle is 6 cm and its breadth is 4 cm. 1

MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS

Paper - II... 2 ...

(ii) y = 5x – 10

2 2y = 5x – 10 – 5x + 2y = – 10

4x + 5 = – y 4x + y = – 5 1

D =–5 2

4 1 = (– 5 × 1) – (2 × 4) = – 5 – 8 = – 13

Dx =–10 2

–5 1 = (– 10 × 1) – (2 × – 5) = – 10 + 10 = 0

Dy =–5 –10

4 –5 = (– 5 × – 5) – (– 10 × 4 )= 25 + 40 = 65 1

By Cramer’s rule,

x =D

Dx =

0

–13 = 0

y =D

Dy

=65

–13 = – 5

x = 0 and y = – 5 is the solution of given simultaneous equations. 1

(iii) Let natural number be ‘x’

Its reciprocal is 1

xFrom the given condition,

x + 1

x =

10

31

Multiplying throughout by 3x,3x2 + 3 = 10x

3x2 – 10x + 3 = 0 3x2 – 9x – x + 3 = 0 3x (x – 3) – 1 (x – 3) = 0 (3x – 1) (x – 3) = 0 3x – 1 = 0 or x – 3 = 0 3x = 1 or x = 3

x = 1

3or x = 3 1

x 1

3 because x is natural number

x = 3


Paper - II... 3 ...

A.3. Solve the following : (Any Three)(i) Difference between the perimeters of two squares is 16 m

The difference between the sides of the same squares is 4 mLet the side of smaller square be x cm

The side of bigger square is (x + 4) cm Area of square = (side)2 1

As per the given condition,x2 + (x + 4)2 = 400

x2 + x2 + 8x + 16 – 400 = 0 2x2 + 8x – 384 = 0

Dividing throughout by 2 we get,x2 + 4x – 192 = 0 1

x2 – 12x + 16x – 192 = 0 x (x – 12) + 16 (x – 12) = 0 (x – 12) (x + 16) = 0 x – 12 = 0 or x + 16 = 0 x = 12 or x = – 16 1 The side of square cannot be negative x – 16

Hence x = 12 x + 4 = 12 + 4 = 16

The side of smaller square is 12 m and bigger square is 16 m. 1

(ii) 4x = y – 5

4x + 5 = y

y = 4x + 5

x 0 –1 –2

y 5 1 –3

(x, y) (0, 5) (–1, 1) (–2, –3)

y = 2x + 1

x 0 1 2

y 1 3 5

(x, y) (0, 1) (1, 3) (2, 5) 1

Paper - II... 4 ...

2

Y


Y

41 2 3 5 X-5 -4 -3 -2X

-1

-2

-3

6

4

2

1

-4

(0, 1)

y =

2x

+ 1

(1, 3)

(2, 5)

(–2, –3)

(–1, 1)

(0, 5)

0

5

3

-1

-5

-6

-7

-8


Paper - II... 5 ...

(iii) Let divisor of 6123 be ‘x’ Divisor = Quotient [Given] Quotient = x

Remainder = x

2[Given]

We know,Dividend = Divisor × Quotient + Remainder 1

6123 = x . x + x

2Multiplying throughout by 2,

2 (6123) = 2x2 + x 12246 = 2x2 + x 2x2 + x – 12246 = 0 2x2 + 157x – 156x – 12246 = 0 1 x (2x + 157) – 78 (2x + 157) = 0 (2x + 157) (x – 78) = 0 2x + 157 = 0 or (x – 78) = 0 2x + 157 = 0 or x – 78 = 0

x = – 157

2or x = 78 1

x = – 157

2 cannot be acceptable because divisor cannot be negative.

x = 78

The divisor of 6123 is 78. 1

(iv) Let the no. of Rs. 10 notes given to Durga be x and the no. ofRs.5 notes given to her be y.As per the first condition,

10x + 5y = 190 .......(i)As per the second condition,

5x + 10y = 185 ......(ii) 1Adding (i) and (ii),

15x + 15y = 375Dividing throughout by 15 we get,

x + y =375

15 x + y = 25 ......(iii) 1

Subtracting (ii) from (i),5x – 5y = 5

Dividing throughout by 5 we get,x – y = 1 .......(iv)

Adding (iii) and (iv),

Paper - II... 6 ...

x + y = 25x – y = 1

2x = 26 x = 13 1

Substituting x = 13 in (iii),13 + y = 25

y = 25 – 13 y = 12

Durga had 13 notes of Rs. 10 rupee and 12 notes of Rs. 5. 1

A.4. Solve the following : (Any One)(i) Let the cost of banana per dozen be Rs. x.

Amount for which bananas are bought = Rs. 840

No. of dozens of bananas for Rs 840 is 840

xNew cost of banana per dozen = Rs. (x + 1)

New No. of dozens of bananas for Rs 840 = 840

x 1 1


–x x 1 = 2

1 1

840 –x x 1

= 0

x 1 – x

840x (x 1)

= 2 1

1

840x x

2 = 2

840 = 2 (x2 + x) 2x2 + 2x – 840 = 0 1

Dividing throughout by 2, 2x2 + x – 420 = 0 x2 – 20x + 21x – 420 = 0 x (x – 20) + 21 (x – 20) = 0 x – 20 = 0 or x + 21 = 0 x = 20 or x = –21 1 The cost of bananas cannot be negative. x –21

Hence x = 20

The original cost of one dozen banana is Rs. 20. 1

Paper - II

(ii) Let the no. of students standing in each row be x and let no. ofrows be y. Total no. of students participating in the drill = xy

As per the first given condition,(x – 3) (y + 10) = xy

x (y + 10) – 3 (y + 10) = xy xy + 10x – 3y – 30 = xy 10x – 3y = 30 .......(i) 1

As per the second given condition,(x + 5) (y – 10) = xy

x (y – 10) + 5 (y – 10) = xy xy – 10x + 5y – 50 = xy – 10x + 5y = 50 ......(ii) 1

Adding (i) and (ii),10x – 3y = 30

– 10x + 5y = 502y = 80

y =80

21

y = 40Substituting y = 40 in (i),

10x – 3 (40) = 30 10x – 120 = 30 10x = 30 + 120 10x = 150

x =150

101

x = 15 xy = 15 × 40 = 600

600 students were participating in the drill. 1

... 7 ...

Paper - III


(i) What is the equation of Y - axis? Hence, find the point of intersection ofY - axis. and the line y = 3x + 2.

(ii) Without actually solving the simultaneous equations given below, decidewhich simultaneous equations have unique solution, no solution orinfinitely many solutions. 3y = 2 – x; 3x = 6 – 9y


(i) The sum of the squares of two consecutive natural numbers is 113. Findthe numbers.

(ii) Solve the following simultaneous equations using Cramer’s rule :3x + y = 1; 2x = 11y + 3

(iii) The length of the rectangle is greater than its breadth by 2 cm. The areaof the rectangle is 24 sq.cm, find its length and breadth.


(i) Around a square pool there is a footpath of width 2km. if the area of the

footpath is 5

4 times that of the pool. Find the area of the pool.

S.S.C. Test - III



MAHESH TUTORIALS


Best Of Luck

(ii) Solve the following simultaneous equations using graphical method :x + y = 8, x – y = 2

(iii) The length of one diagonal of a rhombus is less than the second diagonalby 4 cm. The area of the rhombus is 30 sq.cm. Find the length of thediagonals.

(iv) On the first day of the sale of tickets of a drama. 35 tickets in all weresold. If the rates of the tickets were Rs.20 and Rs.40 per ticket and thetotal collection was Rs. 900. Find the number of tickets sold of each rate.


(i) One tank can be filled up by two taps in 6 hours. The smaller tap alonetakes 5 hours more than the bigger tap alone. Find the time required byeach tap to fill the tank separately.

(ii) A bus covers a certain distance with uniform speed. If the speed of thebus would have been increased by 15 km/h, it would have taken twohours less to cover the same distance and if the speed of the bus wouldhave been decreased by 5 km/h, it would have taken one hour more tocover the same distance. Find the distance covered by the bus.

Paper - III

A.1. Solve the following :(i) The equation of Y-axis is x = 0

Let the point of intersection of the line y = 3x + 2 with Y-axis be (0, k) (0, k) lies on the line it satisfies the equation 1 Substituting x = 0 and y = k in the equation we get,

k = 3 (0) + 2 k = 2


(ii) 3y = 2 – x x + 3y = 2

Comparing with a1x + b1y = c1 we get, a1 = 1, b1 = 3, c1 = 23x = 6 – 9y

3x + 9y = 6Comparing with a2x + b2y = c2 we get, a2 = 3, b2 = 9, c2 = 6 1

a

a1

2=

1

3

b

b1

2=

3

9=

1

3

c

c1

2=

2

6=

1

3

a

a1

2=

b

b1

2=

c

c1

2

The simultaneous equations 3y = 2 – x and 3x = 6 – 9y have 1infinitely many solutions.

A.2. Solve the following :(i) Let the two consecutive natural numbers be x and x + 1

As per the given condition,x2 + (x + 1)2 = 113

x2 + x2 + 2x + 1 = 113 2x2 + 2x + 1 – 113 = 0 2x2 + 2x – 112 = 0 1

Dividing throughout by 2 we get,

MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS


x2 + x – 56 = 0 x2 + 8x – 7x – 56 = 0 x (x + 8) – 7 (x + 8) = 0 (x + 8) (x – 7) = 0 x + 8 = 0 or x – 7 = 0 1 x = – 8 or x = 7 x is a natural number x -8

Hence, x = 7And x + 1 = 7 + 1 = 8

The two consecutive natural numbers are 7 and 8 respectively. 1

(ii) 3x + y = 12x = 11y + 3

2x – 11y = 3

D =3 1

2 –11 = (3 × – 11) – (1 × 2) = – 33 – 2 = – 35

Dx =1 1

3 –11 = (1 × – 11) – (1 × 3) = – 11 – 3 = – 14

Dy =3 1

2 3 = (3 × 3) – (1 × 2) = 9 – 2 = 7 1

By Cramer’s rule,

x =D

Dx =

–14

–35 = 2

5

y =D

Dy

=7

–35 = –1

51

x = 2

5 and y =

–1


(iii) Let the breadth of the rectangle be ‘x’ cm. It’s length is (x + 2) cm.

As per the given condition. Area of rectangle = Length × Breadth 24 = (x + 2) × x 24 = x2 + 2x 0 = x2 + 2x – 24 1 x2 + 2x – 24 = 0 x2 – 4x + 6x – 24 = 0 x (x – 4) + 6(x – 4) = 0


(x – 4) (x + 6) = 0 x – 4 = 0 or x + 6 = 0 1 x = 4 or x = -6 The breadth of rectangle cannot be negative. x -6

Hence x = 4 and x + 2 = 4 + 2 = 6

The length of rectangle is 6 cm and its breadth is 4 cm. 1

A.3. Solve the following : (Any Three)(i) Let the side of inner square i.e. pool be x m.

The width of foot path around the pool is 2 m The side of outer square is (x + 2) m Area of square = side2

As per the given condition,(Area of outer square)= (Area of inner square) + (Area of footpath)

(x + 4)2 = x2 + 5

4 x2 1

x2 + 8x + 16 = x2 + 5

4x2

8x + 16 =5

4 x2

Multiplying throughout by 4 we get,32x + 64 = 5x2 1

5x2 – 32x – 64 = 0 5x2 – 40x + 8x – 64 = 0 5x (x – 8) + 8 (x – 8) = 0 (x – 8) (5x + 8) = 0 x – 8 = 0 or 5x + 8 = 0 x = 8 or 5x = – 8

x = 8 or x = – 8

51

x = – 8

5 is not acceptable because side of pool cannot be negative.

x = 8 x2 = 82

x2 = 64 Area of pool is 64 sq. m. 1

(ii) x + y = 8 x – y = 2 1 y = 8 – x x = 2 + y

x 0 1 2 x 2 3 4

y 8 7 6 y 0 1 2 1

(x, y) (0, 8) (1, 7) (2, 6) (x, y) (2, 0) (3, 1) (4, 2)


2


(iii) Let the length of other diagonal of a rhombus be ‘x’ cm. The length of first diagonal is (x + 4) cm.

Area of rhombus = 1

2 × Product of length of diagonals

Area of rhombus = 1

2 × x × (x + 4) 1


Y

-1

-2


-3

-4 -3 -2 41 2-1 3 5

Y

7

6

4

5

3

2

1

8

X

(0, 8)

(1, 7)

(5, 3)

(4, 2)

(3, 1)

(2, 0)

x -

y = 2

0X

(2, 6)x + y = 8


1

2x (x + 4) = 30

x (x + 4) = 60 x2 + 4x – 60 = 0 1 x2 + 10x – 6x – 60 = 0 x (x + 10) – 6 (x + 10) = 0 (x + 10) (x – 6) = 0 x + 10 = 0 or x – 6 = 0 x = – 10 or x = 6 1 The length of diagonal of the rhombus cannot be negative. x – 10

Hence x = 6And x + 4 = 6 + 4 = 10

The length of smaller diagonal of a rhombus is 6 cm and 1bigger diagonal is 10 cm.

(iv) Let the no. of tickets sold at Rs. 20 each be x and Rs. 40 each be y.As per first given condition,

x + y = 35 ......(i) 1As per second given condition,

20x + 40y = 900Dividing throughout by 20,

x + 2y = 45 ......(ii) 1Subtracting (ii) from (i),

x + y = 35x + 2y = 45(–) (–) (–)

– y = – 10 y = 10 1

Substituting y = 10 in (i),x + 10 = 35

x = 35 – 10 x = 25

The no. of tickets sold at Rs. 20 each and Rs. 40 each are 25tickets and 10 tickets respectively. 1

A.4. Solve the following : (Any One)(i) Let the time taken to fill a tank by a bigger tap alone be x hrs.

The time taken by smaller tap alone is (x + 5) hrs.Time taken by both the taps together to fill the same tank is 6 hrs.

Portion of tank filled in 1 hr by bigger tap = 1

x1

Portion of tank filled in 1 hr by smaller tap = 1

x 5


Portion of tank filled in 1 hr by both taps together = 1

61

As per the given condition,1

x +

1

x 5 =1

6

x 5 x

x (x 5)

=

1

6

2x 5

x 5x

2 =

1

6 6 (2x + 5) = 1 (x2 + 5x) 12x + 30 = x2 + 5x 0 = x2 + 5x – 12x – 30 1 x2 + 3x – 10x – 30 = 0 x (x + 3) – 10 (x + 3) = 0 (x + 3) (x – 10) = 0 x + 3 = 0 or x – 10 = 0 1 x = – 3 or x = 10 x is the time taken bybigger tap x – 3

Hence x = 10 x + 5 = 10 + 5 = 15

Time taken by bigger tap alone is 10 hrs and smaller tap alone 1is 15 hrs.

(ii) Let the speed of bus be x km/hr. and time taken be y hrs.Distance = Speed × Time

Distance = xy kmAccording to the first condition,

(x + 15) (y – 2) = xy x (y – 2) + 15 (y – 2) = xy xy – 2x + 15y – 30 = xy – 2x + 15y = 30 ......(i) 1

According to the second condition,(x – 5) (y + 1) = xy

x (y + 1) – 5 (y + 1) = xy xy + x – 5y – 5 = xy x – 5y = 5 ......(ii) 1

Multiplying (ii) by 3 we get,3x – 15y = 15 ......(iii)Adding (i) and (iii) we get,

– 2x + 15y = 303x – 15y = 15

x = 45 1Substituting x = 45 in (ii),

Paper - III

45 – 5y = 5 – 5y = 5 – 45 – 5y = – 40

y =– 40

–5 1

y = 8 Distance = xy

= 45 × 8= 360

Distance covered by bus is 360 km. 1

... 7 ...

Paper - IV


(i) If (a, 3) is the point lying on the graph of the equation 5x + 2y = – 4, thenfind a.

(ii) Without actually solving the simultaneous equations given below, decidewhich simultaneous equations have unique solution, no solution orinfinitely many solutions. 3x – 7y = 15; 6x = 14y + 10


(i) The sum of the ages of father and his son is 42 years. The product oftheir ages is 185, find their ages.

(ii) Find the value of k for which the given simultaneous equations haveinfinitely many solutions :4x + y = 7; 16x + ky = 28.

(iii) A natural number is greater then three times its square root by 4. Findthe number.


(i) A car covers a distance of 240km with some speed, if the speed is increasedby 20 km/hr, it will cover the same distance in 2 hours less. find thespeed of the car.

S.S.C. Test - III



MAHESH TUTORIALS

Paper - IV... 2 ...

Best Of Luck


x + 2y = 5; y = – 2x – 2

(iii) For doing some work Ganesh takes 10 days more than John. If bothwork together they complete the work in 12 days. Find the number ofdays if Ganesh worked alone?

(iv) AB is a segment. The point P is on the perpendicular bisector of segmentAB such that length of AP exceeds length of AB by 7 cm. If the perimeterof ABP is 38 cm. Find the sides of ABP.


(i) A man riding on a bicycle cover a distance of 60 km in a direction of windand comes back to his original position in 8 hours. If the speed of thewind is 10 km/hr. Find the speed of the bicycle.

(ii) Some part of a journey of 555 km was completed by a car with speed60 km/hr then the speed is increased by 15 km/hr and the journey iscompleted. If it takes 8 hours to reach, find the time taken and distancecovered by 60km/hr speed.

Paper - IV

A.1. Solve the following :(i) (a, 3) is a point lying on the graph of the equation 5x + 2y = – 4,

it is satisfies the equation. Substituting x = a and y = 3 in the equation we get, 1

5 (a) + 2 (3) = – 4 5a + 6 = – 4 5a = – 4 – 6 5a = – 10

a =–10

5 a = – 2 1

(ii) 3x – 7y = 15Comparing with a1x + b1y = c1 we get, a1 = 3, b1 = – 7, c1 = 156x = 14y + 10

6x – 14y = 10Comparing with a2x + b2y = c2 we get, a2 = 6, b2 = – 14, c2 = 10 1

a

a1

2=

3

6=

1

2

b

b1

2 =

–7

–14 =1

2

c

c1

2=

15

10 =

3

2

a

a1

2=

b

b1

2

c

c1

2

The simultaneous equations 3x – 7y = 15 and 6x = 14y + 10 1have no solution.

A.2. Solve the following :(i) Sum of ages of father and son = 42 years

Let father’s age be x years Son’s age = (42 – x) years The age of his son is (42 – x) years.


MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS

Paper - IV... 2 ...

x (42 – x) = 185 1 42x – x2 = 185 0 = x2 – 42x + 185 1 x2 – 42x + 185 = 0 x2 – 5x – 37x + 185 = 0 x (x – 5) – 37 (x – 5) = 0 (x – 5) (x – 37) = 0 x – 5 = 0 or x – 37 = 0 x = 5 or x = 37

If x = 5, father’s age is less than son’s age. x 5

Hence x = 37And 42 – x = 42 – 37 = 5

The father’s age is 37 years and son’s age 5 years. 1

(ii) 4x + y = 7Comparing with a1x + b1y = c1 we get, a1 = 4, b1 = 1, c1 = 716x + ky = 28Comparing with a2x + b2y = c2 we get, a2 = 16, b2 = k, c2 = 28 1

a

a1

2=

4

16 =

1

4

b

b1

2=

1

k

c

c1

2=

7

28=

1

41

The simultaneous equations have infinitely many solutions.

a

a1

2=

b

b1

2=

c

c1

2

1

4=

1

k=

1

4

1

4=

1

k

k = 4 1

(iii) Let the square root of natural number be x The natural number = x2

As per the given condition,x2 = 3x + 4 1

x2 – 3x – 4 = 0 x2 – 4x + 1x – 4 = 0 (x – 4) (x + 1) = 0 x – 4 = 0 or x + 1 = 0 x = 4 or x = – 1

Paper - IV... 3 ...

Square root of natural number cannot be negative 1 x – 1 x = 4 x2 = 42

x2 = 16


A.3. Solve the following : (Any Three)(i) Let the original speed of car be x km/hr.

Distance covered is 240 km

Speed = Dis tance

Time

Time = Dis tance

Speed 1

Time taken by car =240

hrsx

New speed of car = (x + 20) km/hr

New time taken by car =240

hrsx 20


–x x 20 = 2 1

1 1

240 –x x 20

= 2

x 20 – x

x (x 20)

=

2

240

20

x 20x2 =1

120 20 (120) = 1 (x2 + 20x) 2400 = x2 + 20x 0 = x2 + 20x – 2400 1 x2 + 20x – 2400 = 0 x2 + 60x – 40x – 2400 = 0 x (x + 60) – 40 (x + 60) = 0 (x + 60) (x – 40) = 0 x + 60 = 0 or x – 40 = 0 x = – 60 or x = 40 The speed of car can never be negative. x – 60

Hence x = 40

The original speed of car is 40 km/hr. 1

Paper - IV... 4 ...

(ii) x + 2y = 5 y = –2x – 2

x = 5 – 2y

x 5 3 1 x 0 1 2 1

y 0 1 2 y –2 –4 –6

(x, y) (5, 0) (3, 1) (1, 2) (x, y) (0, –2) (1, –4) (2, –6)

2


Y


Y

4

5

3

2

1

41 2 3 5 X-5 -4 -3 -2X

-2

-3

-5

-4

(5, 0)

(3, 1)

(1, 2)

(–3, 4)

(0, –2)

y = –2x – 2

x + 2y = 5

0

(1, –4)

(2, –6)

-1

-1

-6

Paper - IV... 5 ...

(iii) Let the number of days required by John alone to completethe work be x days and Ganesh alone is (x + 10) days No. of days requred by ganesh along is (x + 10 ) days.

Also number of days required by both to complete the samework is 12 days

Work done by John in 1 day = 1

x

Work done by Ganesh in 1 day = 1

x + 10 1

Work done by both in 1 day = 1

12As per the given condition,

1 1+

x x + 10 =1

121

x + 10 + x

x (x + 10) =1

12

2x + 10

x + 10x2 =1

12 12 (2x + 10) = 1 (x2 + 10x) 24x + 120 = x2 + 10x 0 = x2 + 10x – 24x – 120 x2 – 14x – 120 = 0 x2 – 20x + 6x – 120 = 0 x (x – 20) + 6 (x – 20) = 0 (x – 20) (x + 6) = 0 x – 20 = 0 or x + 6 = 0 x = 20 or x = – 6 The numbers of days cannot be negative 1 x – 6

Hence x = 20 x + 10 = 20 + 10 = 30

Ganesh alone worked for 30 days. 1

(iv) Let the length of seg AB be x cm and that of seg AP be y cm l (BP) = l (AP) = y [By perpendicular bisector theorem]

As per first condition,y = x + 7

– x + y = 7 .......(i) 1As per the second condition,

x + y + y = 38 x + 2y = 38 ......(ii) 1

Adding (i) and (ii),– x + y = 7x +2y = 38

3y = 45

P

BAx cm

y cm

T

Paper - IV... 6 ...

y =45

3 y = 15 1

Substituting y = 15 in (ii),x + 2 (15) = 38

x + 30 = 38x = 38 – 30

x = 8 l (AB) = 8 cm, l (BP) = l (AP) = 15 cm.

The length of sides of ABP are 8 cm, 15 cm and 15 cm. 1

A.4. Solve the following : (Any One)(i) Let the speed of bicycle be x km / hr

Speed of wind is 10 km /hr Speed of bicycle in the direction of wind = (x + 10) km/hr

Speed of the bicycle against the direction of wind = (x – 10) km/hr

Also, Speed = Dis tance

Time

Time = Dis tance

SpeedTime taken by man while riding in the direction of wind

=60

x 10

hrs

Time taken by man while riding against the direction of wind

=60

x – 10

hrs 1


60 60

x 10 x – 10

= 8 1

1 1

60x 10 x – 10

= 8

x – 10 x 10

(x 10) (x – 10)

=

8

60

2x

x – 1002 =2

15Dividing throughout by 2 we get,

x

x – 1002 =1

15 x2 – 100 = 15x 1 x2 – 15x – 100 = 0

Paper - IV

x2 – 20x + 5x – 100 = 0 x (x – 20) + 5 (x – 20) = 0 (x – 20) (x + 5) = 0 x – 20 = 0 or x + 5 = 0 x = 20 or x = – 5 1 The speed of bicycle cannot be negative x – 5

Hence x = 20

The speed of bicycle is 20 km / hr. 1

(ii) Let the time for which the car travels at 60 km/hr be x hoursand at 75 (i.e. 60 + 15) km/hr be y hoursWe know that, Distance = Speed × TimeDistance covered by car with the speed of 60 km/hr = 60 × x = 60x kmDistance covered by car with the speed of 75 km/hr = 75 × y = 75y km 1As per first condition,

60x + 75y = 55515 (4x + 5y) = 555

4x + 5y =55515

1

4x + 5y = 37 .............(i)As per second condition,

x + y = 8 ............(ii) 1Multiplying (ii) by 5, we get,

5x + 5y = 40 ............(iii)Subtracting (i) from (iii), we get

5x + 5y = 404x + 5y = 37 – – –

x = 3 1

Time taken by the car at the speed of 60 km/hr = 3 hours andDistance covered with the speed of 60km/hr = 60 × 3 = 180 km. 1

... 7 ...

Paper - V


(i) Find the value of the following determinants : 3 6 – 4 2

5 3 2

(ii) If the point (3, 2)lies on the graph of the equation 5x + ay = 19, then finda.


(i) The sum ‘S’ of the first ‘n’ natural numbers is given by S = n (n + 1)

2.

Find ‘n’, if the sum (S) is 276.

(ii) Find the value of k for which the given simultaneous equations haveinfinitely many solutions : 4y = kx – 10; 3x = 2y + 5

(iii) Tinu is younger than Pinky by three years. The product of their ages is180. Find their ages.


(i) For doing some work Ganesh takes 10 days more than John. If bothwork together they complete the work in 12 days. Find the number ofdays if Ganesh worked alone?

S.S.C. Test - III



MAHESH TUTORIALS

Paper - V... 2 ...

Best Of Luck

(ii) Solve the following simultaneous equations using graphical method :3x + 4y + 5 = 0; y = x + 4

(iii) Divide 40 into two parts such that the sum of their reciprocals is 8

75.

(iv) Monthly hostel charges in a college comprises of two parts, one fixedpart for the stay in the hostel and the varying part depending on thenumber of days one has taken food in the mess. Ram takes food for 20days and pays Rs. 1700 as hostel charges and Rahim takes food for 24days and pays Rs.1900 as hostel charges. Find the fixed charges andthe cost of the food per day.


(i) One diagonal of a rhombus is greater than other by 4 cm. If the area ofthe rhombus is 96 cm2, find the side of the rhombus.

(ii) A bus covers a certain distance with uniform speed. If the speed of thebus would have been increased by 15 km/h, it would have taken twohours less to cover the same distance and if the speed of the bus wouldhave been decreased by 5 km/h, it would have taken one hour more tocover the same distance. Find the distance covered by the bus.

Paper - V


(i)3 6 – 4 2

5 3 2

= 3 6 2 – – 4 2 5 3 1

= 6 6 – – 20 6

= 6 6 20 6

= 26 6 1

(ii) (3, 2) lies on the graph of the equation 5x + y = 19.It satisfies the equation,

Substituting x = 3 and y = 2 in the equation we get, 15 (3) + a (2) = 19

15 + 2a = 19 2a = 19 – 15 2a = 4

a =4

2 a = 2 1


(i) S = n (n + 1)

2From the given condition,

276 = n (n + 1)

2 552 = n2 + n n2 + n – 552 = 0 1 n2 + 24n – 23n – 552 = 0 n (n + 24) (n – 23) = 0 n + 24 = 0 or n – 23 = 0 1 n = – 24 or n = 23

n – 24 because ‘n’ cannot be negative

n = 23 1

MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS

Paper - V... 2 ...

(ii) 4y = kx – 10 – kx + 4y = – 10

C om paring w ith a1x + b1y = c1 we get, a1 = – k, b1 = 4, c1 = – 103x = 2y + 5

3x – 2y = 5Comparing with a2x + b2y = c2 we get, a2 = 3, b2 = – 2, c2 = 5 1

a

a1

2=

– k

3

b

b1

2=

4

– 2 =– 2

1

c

c1

2=

–10

5=

– 2

11

The simultaneous equations have infinitely many solutions.

a

a1

2=

b

b1

2=

c

c1

2

– k

3=

– 2

1=

– 2

1

– k

3=

– 2

1 – k = – 6

k = 6 1

(iii) Tinu’s age be ‘x’ yearsPinky’s age is (x + 3) yearsAs per the given condition,

x (x + 3) = 180 1 x2 + 3x – 180 = 0 x2 – 12x + 15x – 180 = 0 x (x – 12) + 15 (x – 12) = 0 (x – 12) (x + 15) = 0 x – 12 = 0 or x + 15 = 0 x = 12 or x = – 15 1 Age cannot be negative. x – 15 x = 12

And x + 3 = 12 + 3 = 15

Tinu’s age is 12 years and Pinky’s is 15 years. 1

A.3. Solve the following : (Any Three)(i) Let the number of days required by John alone to complete

the work be x days and Ganesh alone is (x + 10) days No. of days requred by ganesh along is (x + 10 ) days.

Paper - V... 3 ...

Also number of days required by both to complete the samework is 12 days

Work done by John in 1 day = 1

x

Work done by Ganesh in 1 day = 1

x + 10

Work done by both in 1 day = 1

121


1 1+

x x + 10 =1

121

x + 10 + x

x (x + 10) =1

12

2x + 10

x + 10x2 =1

12 12 (2x + 10) = 1 (x2 + 10x) 24x + 120 = x2 + 10x 0 = x2 + 10x – 24x – 120 x2 – 14x – 120 = 0 x2 – 20x + 6x – 120 = 0 x (x – 20) + 6 (x – 20) = 0 (x – 20) (x + 6) = 0 x – 20 = 0 or x + 6 = 0 1 x = 20 or x = – 6 The numbers of days cannot be negative x – 6

Hence x = 20 x + 10 = 20 + 10 = 30

Ganesh alone worked for 30 days. 1

(ii) 3x + 4y + 5 = 0 y = x + 4

3x = –5 – 4y

x = –5 – 4y

3

x –3 1 5 x 0 1 2

y 1 –2 –5 y 4 5 6

(x, y) (–3, 1) (1, –2) (5, –5) (x, y) (0, 4) (1, 5) (2, 6) 1

Paper - V... 4 ...

Y


Y

41 2 3 5 X-5 -4 -3 -2 -1X

(5, –5)

(1, –2)

(–3, 1)

(0, 4)

(1, 5)

y =

x +

4

(2, 6)

3x + 4y + 5 = 0

0

6

7

8

-1

-2

-3

3

-5

-4

2

1

5

4


(iii) Sum of two parts is 40Let one of the part is x

The other part is 40 – x 1

Paper - V... 5 ...


x 40 – x =

8

75

40 – x + x

x (40 – x) =8

751

40

40x – x2 =8

75 40 (75) = 8 (40x – x2)

40 × 75

8= 40x – x2

375 = 40x – x2 1 x2 – 40x + 375 = 0 x2 – 25x – 15x + 375 = 0 x (x – 25) – 15 (x – 25) = 0 (x – 25) (x – 15) = 0 x – 25 = 0 or x – 15 = 0 x = 25 or x = 15

If x = 25 or if x = 15then 40 – x = 40 – 25 = 15 then 40 – x = 40 – 15 = 25

The two parts are 25 and 15. 1

(iv) Let the fixed charge for stay in hostel be Rs. x and cost of foodper day be Rs. yRam’s total expenditure = Rs. (x + 20y)Rahim’s total expenditure = Rs. (x + 24y)As per the first given condition,

x + 20y = 1700 ......(i) 1As per the second given condition,

x + 24y = 1900 .....(ii)Substracting (ii) from (i),

x + 20y = 1700x + 24y = 1900

(–) (–) (–)– 4y = – 200

y =–200

–4 1

y = 50Substituting y = 50 in (i),

x + 20 (50) = 1700 x + 1000 = 1700 x = 1700 – 1000 1 x = 700

Fixed charge for stay in the hostel is Rs. 700 and cost of foodper day is Rs. 50.

Paper - V... 6 ...

A.4. Solve the following : (Any One)(i) Let the length of smaller diagonal of a rhombus be x cm

The length of bigger diagonal is (x + 4) cmAs per the given condition,

Area of rhombus =1

2 × Diagonal 1 × Diagonal 2


1

2 × x × (x + 4) = 96

x2 + 4x = 192 1 x2 + 4x – 192 = 0 x2 – 12x + 16x – 192 = 0 x (x – 12) + 16 (x – 12) = 0 (x – 12) (x + 16) = 0 x – 12 = 0 or x + 16 = 0 1 x = 12 or x = – 16 The length of diagonal of the rhombus cannot be negative x – 16

Hence x = 12 x + 4 = 12 + 4 = 16

Considering ABCD a rhombus l (AC) = 12 cm l (BD) = 16 cm 1

O is the intersection point of diagonal AC and BD

l (AO) = 1

2l (AC) =

1

2 × 12 = 6 cm

l (BO) = 1

2l (BD) =

1

2 × 16 = 8 cm 1

AOB is a right angled triangleIn right angled AOB,[l (AB)]2 = [l (AO)]2 + [l (BO)]2 [By Pythagoras theorem]

[l (AB)]2 = (6)2 + (8)2

[l (AB)]2 = 36 + 64 [l (AB)]2 = 100

Taking square root on both the sides we get,l (AB) = 10 cm

The side of a rhombus is 10 cm. 1

(ii) Let the speed of bus be x km/hr. and time taken be y hrs.Distance = Speed × Time

Distance = xy kmAccording to the first condition,

A B

CD

O

Paper - V

(x + 15) (y – 2) = xy x (y – 2) + 15 (y – 2) = xy xy – 2x + 15y – 30 = xy – 2x + 15y = 30 ......(i) 1

According to the second condition,(x – 5) (y + 1) = xy

x (y + 1) – 5 (y + 1) = xy xy + x – 5y – 5 = xy x – 5y = 5 ......(ii) 1

Multiplying (ii) by 3 we get,3x – 15y = 15 ......(iii)

Adding (i) and (iii) we get,– 2x + 15y = 30

3x – 15y = 15 1x = 45

Substituting x = 45 in (ii), 45 – 5y = 5 – 5y = 5 – 45 – 5y = – 40

y =– 40

–5

y = 8 1 Distance = xy

= 45 × 8= 360

Distance covered by bus is 360 km. 1

... 7 ...

Paper - VI


(i) What is the equation of X - axis? Hence, find the point of intersection ofthe graph of the equation x + y = 3 with the X - axis.

(ii) Find the value of the following determinants : 1.2 0.03

0.57 – 0.23


(i) A natural number is greater than twice its square root by 3. Find thenumber.

(ii) Find the value of p for which the given simultaneous equations have uniquesolution : 3x + y = 10; 9x + py = 23

(iii) In garden there are some rows and columns. The number of trees in arow is greater than that in each column by 10. Find the number of treesin each row if the total number of trees are 200.


(i) A rectangular playground is 420 sq.m. If its length is increased by 7 mand breadth is decreased by 5 metres, the area remains the same. Findthe length and breadth of the playground ?

S.S.C. Test - III



MAHESH TUTORIALS

Paper - VI... 2 ...

Best Of Luck


4x = y – 5; y = 2x + 1

(iii) The sum of the areas of two squares is 400sq.m. If the difference betweentheir perimeters is 16 m, find the sides of two square.

(iv) A man starts his job with a certain monthly salary and a fixed incrementevery year. If his salary will be Rs. 11000 after 2 years and Rs. 14000after 4 years of his service. What is his starting salary and what is theannual increment ?


(i) The product of four consecutive positive integers is 840. find thelargest number.

(ii) Students of a school were made to stand in rows for drill. If 3 students lesswere standing in each row, 10 more rows were required and if 5 studentsmore were standing in each row then the number of rows was reduced by10. Find the number of students participating in the drill.

Paper - VI

A.1. Solve the following :(i) The equation of X-axis is y = 0 1

Let the point of intersection of graph x + y = 3 with X-axis be (h, 0) (h, 0) lies on the graph, it satisfies the equation Substituting x = h and y = 0 in the equation we get,

h + 0 = 3 h = 3

The line x + y = 3 intersects the X-axis at (3, 0). 1

(ii)1.2 0.03

0.57 – 0.23= (1.2 × – 0.23) – (0.03 × 0.57) 1= – 0.276 – 0.0171= – 0.2931 1

A.2. Solve the following :(i) Let square root of the natural number be x

The natural number = x2 1As per the given condition,

x2 = 2x + 3 x2 – 2x – 3 = 0 x2 – 3x + 1x – 3 = 0 x (x – 3) + 1 (x – 3) = 0 (x – 3) (x + 1) = 0 x – 3 = 0 or x + 1 = 0 x = 3 or x = – 1 1 Square root of a natural number cannot be negative x – 1 x = 3 x2 = 9


(ii) 3x + y = 10Comparing with a1x + b1y = c1 we get, a1 = 3, b1 = 1, c1 = 10

MODEL ANSWER PAPER

S.S.C. Test - III



MAHESH TUTORIALS

Paper - VI... 2 ...

9x + py = 23Comparing with a2x + b2y = c2 we get, a2 = 9, b2 = p, c2 = 23 1

a

a1

2=

3

9=

1

3b

b1

2=

1

pFor the equations to have unique solution.a

a1

2

b

b1

2

1

3

1

p 1

p 3

The simultaneous equations will have unique solution for all 1values of p except 3.

(iii) Let the number of trees in each column be x The number of trees in each column is x + 10

As per the given condition, 1x (x + 10) = 200

x2 + 10x – 200 = 0 x2 + 20x – 10x – 200 = 0 x (x + 20) – 10 (x + 20) = 0 (x + 20) (x – 10) = 0 x + 20 = 0 or x – 10 = 0 1 x = –20 or x = 10 The number of trees cannot be negative x – 20

Hencex = 10

x + 10 = 10 + 10 = 20

The number of trees in each row is 20. 1

A.3. Solve the following : (Any Three)(i) Let the length of a rectangular playground be ‘x’ m.

The area of playground is 420 sq.m.

It’s breadth is 420

xm

New length = (x + 7) m

New breadth = 420

– 5 mx

1


Paper - VI... 3 ...

Area of new rectangle = Length × Breadth

New area = (x + 7) 420

– 5x

1

420 = (x + 7) × 420

5x

420 = 420 – 5x + 2940

x – 35

Multiplying throughout by x, we get;0 = – 5x2 + 2940 – 35x

5x2 + 35x – 2940 = 0 x2 + 7x – 588 = 0 [Dividing throughout by 5] x2 – 21x + 28x – 588 = 0 x (x – 21) + 28 (x – 21) = 0 (x – 21) (x + 28) = 0 x – 21 = 0 or x + 28 = 0 x = 21 or x = -28 1 The length of playground cannot be negative. x -28

Hence x = 21

And 420

x =

420

21 = 20

The length of a rectangular playground is 21 m andits breadth is 20 m. 1

(ii) 4x = y – 5

4x +5 = y

y = 4x + 5

x 0 –1 –2

y 5 1 –3

(x, y) (0, 5) (–1, 1) (–2, –3)

y = 2x + 1

x 0 1 2

y 1 3 5

(x, y) (0, 1) (1, 3) (2, 5) 1

Paper - VI... 4 ...

2


Y


Y

41 2 3 5 X-5 -4 -3 -2X

-1

-2

-3

6

4

2

1

-4

(0, 1)

y =

2x

+ 1

(1, 3)

(2, 5)

(–2, –3)

(–1, 1)

(0, 5)

0

5

3

-1

-5

-6

-7

-8

Paper - VI... 5 ...

(iii) Difference between the perimeters of two squares is 16 m The difference between the sides of the same squares is 4 m

Let the side of smaller square be x cm 1 The side of bigger square is (x + 4) cm Area of square = (side)2

As per the given condition,x2 + (x + 4)2 = 400 1

x2 + x2 + 8x + 16 – 400 = 0 2x2 + 8x – 384 = 0

Dividing throughout by 2 we get,x2 + 4x – 192 = 0

x2 – 12x + 16x – 192 = 0 x (x – 12) + 16 (x – 12) = 0 (x – 12) (x + 16) = 0 x – 12 = 0 or x + 16 = 0 x = 12 or x = – 16 1 The side of square cannot be negative x – 16

Hence x = 12 x + 4 = 12 + 4 = 16

The side of smaller square is 12 m and bigger square is 16 m. 1

(iv) Let starting salary of man be Rs. x and the fixed annualincrement be Rs. y.

As per the first given condition,x + 2y = 11000 ......(i) 1

As per the second given condition,x + 4y = 14000 ......(ii) 1

Subtracting (ii) from (i),x + 2y = 11000

x + 4y = 14000(–) (–) (–)

– 2y = – 3000

y =–3000

–2 y = 1500 1

Substituting y = 1500 in (i),x + 2 (1500) = 11000

x + 3000 = 11000 x = 11000 – 3000 x = 8000

The starting salary of man is Rs. 8000 and his fixed annual 1increment is Rs. 1500.

Paper - VI... 6 ...

A.4. Solve the following : (Any One)(i) Let the four consecutive positive integers be x, x + 1, x + 2 and x + 3

As per the given condition,x × (x + 1) × (x + 2) × (x + 3) = 840

x (x + 3) × (x + 1) (x + 2) = 840

(x2 + 3x) × (x2 + 2x + x + 2) = 840

(x2 + 3x) (x2 + 3x + 2) = 840 1

Substituting x2 + 3x = m we get,

m (m + 2) = 840

m2 + 2m – 840 = 0

m2 + 30m – 28m – 840 = 0

m (m + 30) – 28 (m + 30) = 0

(m + 30) (m – 28) = 0 m + 30 = 0 or m – 28 = 0 1 m = – 30 or m = 28

Resubstituting m = x2 + 3x we get,x2 + 3x = – 30 ......(i) or x2 + 3x = 28 ......(ii)

From (i), x2 + 3x = – 30 x2 + 3x + 30 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = 30b2 – 4ac = (3)2 – 4 (1) (30)

= 9 – 120= – 111

b2 – 4ac < 0 The roots of the above quadratic equation are not real. 1

Hence not considered.From (ii), x2 + 3x = 28

x2 + 3x – 28 = 0 x2 + 7x – 4x – 28 = 0 x (x + 7) – 4 (x + 7) = 0 (x + 7) (x – 4) = 0 x + 7 = 0 or x – 4 = 0 1 x = – 7 or x = 4 x is positive integer x – 7

Hence x = 4 x + 3 = 4 + 3 = 7

The largest required number is 7. 1

(ii) Let the no. of students standing in each row be x and let no. of rowsbe y.

Total no. of students participating in the drill = xy

Paper - VI

As per the first given condition,(x – 3) (y + 10) = xy

x (y + 10) – 3 (y + 10) = xy xy + 10x – 3y – 30 = xy 10x – 3y = 30 .......(i) 1

As per the second given condition,(x + 5) (y – 10) = xy

x (y – 10) + 5 (y – 10) = xy xy – 10x + 5y – 50 = xy – 10x + 5y = 50 ......(ii) 1

Adding (i) and (ii),10x – 3y = 30

– 10x + 5y = 502y = 80

y =80

2 y = 40 1

Substituting y = 40 in (i),10x – 3 (40) = 30

10x – 120 = 30 10x = 30 + 120 10x = 150

x =150

10 x = 15 1 xy = 15 × 40 = 600

600 students were participating in the drill. 1

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