SSC Question Paper Set: Semi English (Maharashtra Board)

Edition: August 2013

9 Question Papers with Model Answers

Salient Features: ¾ 3 Model Question Papers for Algebra, Geometry and Science each (set of 9 Question papers). ¾ Prepared as per the new board paper pattern. ¾ Complete answers to every question with relevant marking scheme. ¾ Graphs and diagrams provided wherever necessary. ¾ Simple and lucid language. ¾ Self-evaluative in nature.

TargetPublications PVT. LTD. Mumbai, Maharashtra Tel: 022 – 6551 6551 Website : www.targetpublications.org | email : [email protected]

9 Question Papers with Model Answers

©

Target Publications PVT. LTD.

Second Edition : August 2013

Price : ` 80/-

Printed at: Spark Offset Nerul Navi Mumbai

Published by

Target Publications PVT. LTD. Shiv Mandir Sabhagriha, Mhatre Nagar, Near LIC Colony, Mithagar Road, Mulund (E), Mumbai - 400 081 Off.Tel: 022 – 6551 6551 email: [email protected]

PREFACE SSC is one of the most crucial stage in a students life as the path one decides henceforth ends up being the turning point in his/her career. The proficiency in the concerned subjects open the doors to many fields like engineering, medical, commerce and arts. However, to do so one needs to score high and be extremely dominant in subjects like Algebra, Geometry and Science. Just like any other examination, SSC examination would be dreadful only if one is under prepared. As is observed universally, preparation is the key. For the ones who are prepared and ready to take the challenge, we are proud to introduce a penultimate weapon to test your mettle. Target’s SSC Question Paper Set. The book is ideal for complete and effective preparation of SSC subjects - Algebra, Geometry and Science. The book includes 9 Question papers based on Algebra, Geometry and Science (set of 3 Question papers for each subject). Model answers with relevant marking scheme are provided to make sure students encounter no scope for error. The Question papers have been prepared as per the revised SSC paper pattern for Algebra, Geometry and Science. As the old adage goes, “Practice makes a man Perfect”, students would find here a goldmine of Question Papers to practice upon before they are up for their final battle. We are sure these Question Papers would prove to be extremely instrumental in achieving monumental scores in the Board Examinations. Constructive criticism and suggestion for the improvement of the book will be greatly welcomed and appreciated.

We wish you all the best for the exams and your future!

Yours faithfully Publisher

Index Page No. No.

1.

2.

3.

Subject

Algebra

Geometry

Science

Test No. Question Papers

Model Answers

1

1

27

2

3

35

3

5

44

1

7

52

2

10

62

3

13

71

1

15

83

2

18

92

3

22

100

Target Publications Pvt. Ltd.

S.S.C. Question Paper

S.S.C. Preliminary Examination − 1 ALGEBRA Time: 2 Hours

Total Marks: 40

Note: i. All questions are compulsory. ii. Use of calculator is not allowed. Q.1. Attempt any five:

[5]

6 =5 x

1.

Write the following quadratic equation in standard form: x −

2.

Write Dy for the following simultaneous equation: 5x − 2y = 10 ; 3x − y = 11

3.

Two coins are tossed. Write the sample space for the given experiment.

4.

For a certain frequency distribution, values of mean and median are 62.6 and 62.5 respectively. Find the value of mode.

5.

Determine whether the given value of x is a root of given quadratic equation: 4x2 − 9 = 0, x =

6.

If 12x + 13y = 29 and 13x + 12y = 21, then find x − y.

3 2

Q.2 Attempt any four:

[8]

1. Class

Frequency

5 − 15

7

15 − 25

10

25 − 35

20

35 − 45

13

Find the mean of given data by ‘Direct Method’. 2.

A card is drawn from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is a face card.

3.

Find the value of the following determinant:

3 6 −4 2 5 3

2 5 . 2

4.

Form the quadratic equation whose roots are −3 and

5.

Find S10, if a = 6 and d = 3.

6.

Write the first four terms of the A. P. where a = 10, d = −3. 1


Algebra

Q.3. Attempt any three: 1. Find three consecutive terms of an A.P. whose sum is −3 and the product of their cubes is 512.

2.

Solve the following quadratic equation by formula method: 5m2 − 2m = 2.

3.

An obtuse angle of a rhombus is greater than twice the acute angle by 60°. Find the measure of each angle.

4.

Two digit numbers are formed from the digits 2, 3, 5, 7, 9 without repetition. Find the probability of getting i. odd number ii. two digit number so formed is multiple of 7

5.

The following table gives information about the monetary investment by some residents in a city. Mode of Investment Shares Mutual funds Real Estate Gold Govt. Bonds

[9]

Percentage of residents 10 20 35 30 5

Draw a pie diagram to represent the data. Q.4. Attempt any two: 1. Below is given frequency distribution of Daily wages (in `) of 130 workers. Daily wages (in `) No. of workers

80 − 84 10

85 − 89 20

90 − 94 25

95 − 99 40

[8]

100 − 104 30

104 − 109 05

Find the median daily wages of workers. 2.

The following table gives the result of certain examination for 180 students. i. Find value of x. ii. Draw histogram Marks No. of student

3.

0 – 10 10

10 – 20 x

20 – 30 25

30 – 40 2x

If you spin the spinner, find the probability that it will point at i. a odd number. ii. multiples of 3. iii. an even numbered white sector. iv. number less than 4.

40 – 50 55

7 6

50 – 60 30

8 1 5 4

2 3

Q. 5. Attempt any two: [10] 1. A farmer borrows ` 1000 and agrees to repay with a total interest of ` 140 in 12 instalments, each instalment being less than the preceding instalment by ` 10. What should be his first and last instalment?

2

2.

If one root of the quadratic equation ax2 + bx + c = 0 is the square of the other, show that b3 + a2c + ac2 = 3abc.

3.

A publisher printed a certain number of copies of a book. He had printed 2000 copies more, each copy would have cost ` 5 less. If he had printed 1600 copies less, each copy would have cost him ` 10 more. Find the number of copies printed and the cost of each book.



S.S.C. Preliminary Examination − 1 ALGEBRA Model Answer Paper Q.1. Attempt any five:

6 =5 x

1.

x−

∴

x2 − 6 = 5x

∴

x2 − 5x − 6 = 0

2.

Dy =

3.

When two coins are tossed,

[1]

5 10

[1]

3 11

S = {HH, HT, TH, TT} 4.

Mean − Mode = 3 (Mean − Median)

∴

62.6 − Mode = 3 (62.6 − 62.5)

∴

Mode = 62.3

5.

By putting x =

[1]

[1] 3 in L.H.S., we get 2

L.H.S. = 4x2 − 9 2

⎛3⎞ =4 ⎜ ⎟ −9 ⎝2⎠ =4×

9 −9 4

∴

L.H.S. = 0

∴

L.H.S. = R.H.S.

∴

x=

6.

13x + 12y = 21

3 is a root of given quadratic equation. 2

[1]

12x + 13y = 29

− ∴

−

−

x −

y = −8

x − y = −8

[1] 27

Target Publications Pvt. Ltd. Q.2 Attempt any four: 1. Class 5 − 15 15 − 25 25 − 35 35 − 45 Total

x =

Algebra

Classmark (xi) 10 20 30 40 −

Frequency (fi) 7 10 20 13 ∑ fi = 50

Mean = 27.8

2. ∴

Total number of cards = 52 n(S) = 52 Let A be the event of getting a face card. n(A) = 12 n(A) 12 3 = = P(A) = n(S) 52 13

∴

3.

3 6 −4 2 5 3

[1]

(

= 3 6 × 2 − −4 2 × 5 3

2

= 6 6 + 20 6 = 26 6 5 2 5 −1 ∴ α + β = −3 + = 2 2 5 −15 = and α × β = −3 × 2 2 The required quadratic equation is x2 − (α + β) x + αβ = 0 ⎛ −1 ⎞ ⎛ −15 ⎞ ∴ x2 − ⎜ ⎟ x + ⎜ ⎟ =0 ⎝ 2 ⎠ ⎝ 2 ⎠ ∴ 2x2 + x − 15 = 0 4.

n [2a + (n − 1) d] 2 10 = [2 × 6 + (10 − 1) 3] 2 = 5 [12 + 27] = 5 × 39 = 195

Sn =

∴

S10

∴

S10

6.

t1 = a = 10 t2 = t1 + d = 10 − 3 = 7 t3 = t2 + d = 7 − 3 = 4 t4 = t3 + d = 4 − 3 = 1 First four terms of A.P. are 10, 7, 4, 1.

28

[½] [½] [1]

)

[1] [1]

Let α = −3 and β =

5.

∴

[1]

∑ f i xi 1390 = 27.8 = ∑ fi 50

∴

∴

fi x i 70 200 600 520 ∑ fi xi = 1390

[½] [½] [½]

[½] [½] [½]

[1] [½] [½] [½] [½]



Q.3. Attempt any three: 1. Let the 3 consecutive terms of A.P. be a − d, a, a + d According to 1st condition, a − d + a + a + d = −3 ∴ a = −1 ….(i) nd According to 2 condition, (a − d)3 × a3 × (a + d)3 = 512 ∴ (a − d) × a × (a + d) = 8 ∴ (a2 − d2) × a = 8 ∴ a3 − ad2 = 8 ∴ (−1)3 − (−1) d2 = 8 ….[From equation (i)] 2 ∴ −1 + d = 8 ∴ d2 = 9 ∴ d=±3 When d = 3, the required terms are −4, −1, 2 When d = −3, the required terms are 2, −1, −4

2.

∴

5m2 − 2m = 2 ∴ Here a = 5, b = −2, c = −2

∴ ∴

∴ ∴ ∴

[1] [½] [½]

5m2 − 2m − 2 = 0 [½]

−b ± b − 4ac 2a

[½]

=

−(−2) ± (−2) 2 − 4 × 5 × (−2) 2×5

[½]

=

2 ± 44 2×5

[½]

=

2 ± 2 11 10

[½]

m =

(

2 1 ± 11

3.

[½]

2

=

∴

[½]

m =

)

10 1 ± 11 5

[½]

D C Let ABCD is a rhombus. Let m ∠A = x° and m ∠B = y° According to given condition, y = 2x + 60° x° y° 2x − y = − 60° ….(i) A B In rhombus, adjacent angles are supplementary x + y = 180° ….(ii) Adding equation (i) and (ii) 2x − y = − 60° x + y = 180° = 120° 3x x = 40° Substituting x = 40° in equation (ii), we get 40° + y = 180° y = 140° The measure of each angle of rhombus is 40°, 140°, 40° and 140°.

[1] [1]

[1] 29

Target Publications Pvt. Ltd. 4. ∴

Algebra

S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97} n(S) = 20 i. Let A be the event of getting odd number ∴ A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75, 79, 93, 95, 97} ∴ n(A) = 16 n(A) 16 ∴ P(A) = = n(S) 20 4 ∴ P(A) = 5 ii. ∴ ∴

∴ ∴

Let B be the event that two digit number is a multiple of 7. B = {35} n(B) = 1 n(B) 1 P(B) = = n(S) 20 1 P(B) = 20

[1]

[½]

[½]

[½]

[½]

5. Mode of Investment Shares

Percentage 10

Mutual funds

20

Real Estate

35

Gold

30

Govt. Bonds

5

Total

100

Measure of Central angle (θ) 10 × 360 = 36° 100 20 × 360 = 72° 100 35 × 360 = 126° 100 30 × 360 = 108° 100 5 × 360 = 18° 100 360°

[1½]

Mutual Funds

72° Real Estate

126°

Shares

18° 108°

Gold

30

36°

Govt. Bonds

[1½]



Q.4. Attempt any two:

1. Original class 80 – 84 85 – 89 90 – 94 95 – 99

Class boundaries 79.5 – 84.5 84.5 − 89.5 89.5 − 94.5 L → 94.5 – 99.5 99.5 − 104.5 104.5 – 109.5

100 − 104 105 − 109

Frequency 10 20 25

40 ← f 30 5

Cum. frequency less than type 10 30

55 ← c.f. 95 125 130

[1½]

Total frequency = ∑f = N = 130

∴

N 130 = = 65 2 2

[½]

Cumulative frequency just greater than on equal to 65 is 95.

∴

Corresponding class 94.5 − 99.5 is the median class. Here L = 94.5, c.f. = 55, h = 5, f = 40

[½]

⎛N ⎞h Median = L + ⎜ − c.f . ⎟ ⎝2 ⎠f

[½]

= 94.5 + (65 − 55) = 94.5 +

5 40

50 40

= 94.5 + 1.25 = 95.75

∴

Median daily wages of workers is ` 95.75

2.

Total number of students = 180

∴

10 + x + 25 + 2x + 55 + 30 = 180

∴

3x + 120 = 180

∴

3x = 60

∴

x = 20 Marks 0 – 10 10 − 20 20 − 30 30 − 40 40 − 50 50 – 60

[1]

[½] No. of students 10 20 25 40 55 30

[½]

31


Algebra

[3] Scale: On X-axis: 1 cm = 10 marks On Y-axis: 1 cm = 5 students

Y

60

No. of students →

55 50 45 40 35 30 25 20 15 10 5 X′ 0

30 40 Marks →

20

10

Y′

50

60

3.

Since the pointer can point at any numbers from 1 to 8.

∴

n(S) = 8

32

i.

Let A be the event that the spinner points at odd number.

∴

A = {1, 3, 5, 7}

∴

n(A) = 4

∴

P(A) =

4 1 n (A) = = 8 2 n (S)

∴

P(A) =

1 2

X

[1]

Target Publications Pvt. Ltd. ii.

Let B be the event that the spinner points at multiples of 3.

∴

B = {3, 6}

∴

n(B) = 2

∴

P(B) =

2 1 n(B) = = 8 4 n(S)

∴

P(B) =

1 4

iii.

Let C be the event that the spinner points at even numbered white sector.

∴

C={}

∴

n(C) = 0

∴

P(C) = 0

iv.

Let D be the event that the spinner points at a number less than 4.

∴

D = {1, 2, 3}

∴

n(D) = 3

∴

P(D) =

3 n(D) = 8 n(S)

∴

P(D) =

3 8


[1]

[1]

[1]

Q. 5. Attempt any two:

1.

Here, n = 12, d = −10 and S12 = 1140 Sn =

n [2a + (n − 1) d] 2

∴

S12 =

12 [2 × a + (12 − 1) × (− 10)] 2

∴

1140 = 6 [2a − 110]

∴

190 = 2a − 110

∴

a = 150 tn

∴

[1] [1]

[1]

= a + (n − 1) d

[1]

t12 = 150 + (12 − 1) (− 10) = 150 − 110

∴

t12 = 40

∴

First instalment = ` 150 and last instalment = ` 40

[1]

33

Target Publications Pvt. Ltd. 2. ∴

∴ ∴

∴ ∴

Algebra

Let one root = α The other root = β = α2 −b α+β= a −b α + α2 = a −b α (1 + α) = a c αβ = a c α × α2 = a c α3 = a By taking cube of equation (ii),

∴

⎛ −b ⎞ α3 (1 + α)3 = ⎜ ⎟ ⎝ a ⎠

∴

α3 [1 + 3α (1 + α) + α3] =

∴

− b3 c⎡ ⎛ −b ⎞ c ⎤ 1 + 3 + = ⎜ ⎟ ⎥ a ⎢⎣ a3 ⎝ a ⎠ a⎦ 3 c ⎡ a − 3b + c ⎤ − b = ⎥ a ⎢⎣ a a3 ⎦

∴

− b3 c a − 3b + c ) = 3 2 ( a a

∴ ∴ ∴ ∴

∴ ∴ ∴ 34

[1]

….[From equation (i)] ….(ii)

[1]

…[From equation (i)] ….(iii)

[1]

…[From equation (ii) and (iii)]

[1]

3

∴

∴ ∴ ∴ 3. ∴

….(i)

− b3 a3

ac (a − 3b + c) = −b3 a2c − 3abc + ac2 = −b3 b3 + a2c + ac2 = 3abc Let the publisher printed x copies and cost of each copy be ` y. amount invested by publisher = ` xy According to 1st condition, (x + 2000) (y − 5) = xy xy − 5x + 2000y − 10,000 = xy −5x + 2000y = 10,000 ….(i) According to 2nd condition, (x − 1600) (y + 10) = xy xy + 10x −1600y − 16,000 = xy 10x − 1600y = 16,000 ….(ii) Multiplying equation (i) by 2 and then adding with equation (ii), we get −10x + 4000y = 20,000 10x − 1600y = 16,000 2400y = 36000 y = 15 Substituting value of y in equation (ii), we get 10x − 1600 × 15 = 16000 x = 4000 Number of copies printed = 4000 and cost of each book = ` 15.

[1] [1]

[1]

[1]

[1] [1]