Algebra -March 2012 - Maharashtra SSC Board · mySSC.in -Algebra-March 2012 Page 13 of 18. Let xand ybe two numbers, whose Arithmetic Mean is A and Geometric Mean is G. Compare it

Algebra - March 2012

Answer Paper

Time: 2 Hr. 30 min.-------------------------------------------------------------------------------------------------------Max. Marks: 60

Note : - (i). All questions are compulsory.

(ii). Use of calculator is not allowed.

1. Attempt any six subquestions from the following. (6 marks)

(i) Find the arithmetic mean of 4 and 6.

The arithmetic mean of two numbers x and y is given by,

The arithmetic mean of 4 and 6 is 5.

(ii) Find the first term of the sequence, where tn = 2n + 1.

Here nth term i.e. tn = 2n + 1.

Substituting n = 1,

t1 = ( 2 × 1) + 1 = 2 + 1 = 3

The first term of the given sequence is 3.

(iii) In the quadratic equation a = 3, b = 6 and c = 4, then find the value of ��+ ��.

��+ � = −2

Page 1 of 18mySSC.in -Algebra-March 2012

(iv) Write the quadratic equation 5x2 + 7 = 3x in the standard form.

The standard form of a quadratic equation is ax2 + bx + c = 0,

Rearranging the given equation we get,

5x2 − 3x + 7 = 0

5x2 − 3x + 7 = 0 is the standard form of the given quadratic equation.

(v) Find the value of y in equation 2x + y = 7 if x = 2.

The given equation is,

2x + y = 7

Substituting x = 2, we get,

(2 × 2) + y = 7

� 4 + y = 7

� y = 7 − 4 = 3

For the given equation, if x = 2 then y = 3.

(vi) Write the sample space if two coins are tossed simultaneously.

When two coins are tossed simultaneously,

S = { HH, HT, TH, TT }

When two coins are tossed simultaneously, sample space is given by S = { HH, HT, TH, TT }

(vii)

Mean is 5.

2. Attempt any five sub-questions from the following: (10 marks)


(i) Write the first five terms of the following Arithmetic Progression where:

the first term a = 3, the common difference d = 4.

It is given that,

a = 3, d = 4,

Since a = 3, t1 = 3

t2 = t1 + d = 3 + 4 = 7

t3 = t2 + d = 7 + 4 = 11

t4 = t3 + d = 11 + 4 = 15

t5 = t4 + d = 15 + 4 = 19

The first five terms of the given A. P. are 3, 7, 11, 15 and 19.

(ii) Solve the following quadratic equation by factorization method.

x2 + 10x + 24 = 0

x2 + 10x + 24 = 0

Split the middle term 10x as 6x + 4x

x2 + 6x + 4x + 24 = 0

�� x(x + 6) + 4(x + 6) = 0

�� (x + 6)(x + 4) = 0

�� x + 6 = 0 or x + 4 = 0

�� x = −6 or x = −4

x = −6 or x = −4

(iii) Find the value of the following determinant:

D = 22

(iv) Find the ninth term of the A. P. : 3, 7, 11, 15, .......


The given A. P. is, 3, 7, 11, 15, .....

Here a = 3, t1 = 3, t2 = 7, t3 = 11, t4 = 15

Now d = t2 − t1 = 7 − 3 = 4

We have,

tn = a + (n − 1)d

� t9 = 3 + (9 − 1) 4

� t9 = 3 + 32

� t9 = 35

The ninth term, t9 is 35.

(v) For a certain frequency distribution, the value of Mean is 101 and Median is 100. Find the value of Mode.

For a certain frequency distribution, it is given that,

Mean = 101, Median = 100

Now,

Mean - Mode = 3 (Mean - Median)

��101 - Mode = 3(101 - 100)

��101 - Mode = 3

��Mode = 101 - 3

��Mode = 98

Mode = 98

6. If one of the roots of the quadratic equation x2 − 11x + k = 0 is 9, then find the value of k.

The given quadratic equation is,

x2 − 11x + k = 0

One of the roots is 9.

��x = 9 satisfies the given equation.

��Substituting x = 9 in the given equation.

(9)2 − 11(9) + k = 0

�� 81 − 99 + k = 0

�� k = 99 −81

�� k = 18

k = 18

3. Attempt any four sub-questions from the following: (12 marks)


(i) Solve the following quadratic equation by completing square.

x2 − 5x − 36 = 0

x2 − 5x − 36 = 0

�� x2 − 5x = 36

−4 and 9 are the roots of the given quadratic equation.

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

3x − y = 7; x + 4y = 11

The given equations are,

(1) 3x − y = 7

(2) x + 4y = 11


Therefore by Cramer's rule we get,

(3, 2) is the solution of two given simultaneous equations.

(iii) The following data give the number of students using different modes of transport.

Mode of transport Bicycle Bus Walk

Number of students 40 20 30

Represent the above data using pie diagram.

Let's calculate the central angles corresponding to mode of transport as shown in table below.

Mode of transport Number of students Measures of central

angle

Bicycle 40

Bus 20

Walk 30

Total 90 3600

With this information the pie diagram can be shown as shown below.


(iv) Solve the following quadratic equation by using formula method.

3x2 + 7x + 4 = 0

3x2 + 7x + 4 = 0

Comparing it with ax2 + bx + c = 0,

a = 3, b = 7, c = 4

(v) Find the value of k for which the given simultaneous equations have infinitely many solutions:

4x + y = 7; 16x + ky = 28


(1) 4x + y = 7

(2) 16x + ky = 28

Comparing (i) and (ii) with a1x + b1y = c1 and a2x + b2y = c2 respectively we get,

a1 = 4, b1 = 1, c1 = 7

a2 = 16, b2 = k, c2 = 28


Now,

For k = 4, the given equations have infinetly many solutions.

4. Attempt any three from the following subquestions: (12 marks)

(i) In a class of 100 students, 60 students drink tea, 50 students drink coffee and 30 students drink both. A

student from this class is selected at random, find the probability that student takes at least one of the two

drinks (i.e. tea or coffee or both).

It is given that,

(1) The total students in a class = 100

(2) Students drinking coffee = 50

(3) Students drinking tea = 60

(4) Student drinking both coffee and tea = 30

Using (1),

n(S) = 100

Let A be the event that the selected student drink tea.

Using (3),

n(A) = 60

Let B be the event that the selected student drink coffee.

Using (2),

n(B) = 50


Now A ��B is the event that selected student drinks both coffee and tea and A ��B is the event that the student

selected drinks either tea or coffee or both.

Now using (4),

n(A ��B) = 30

(ii) In a certain G.P. if S6 = 126 and S3 = 14 then find a and r.

For the given G.P. S6 = 126 and S3 = 14

We have,


Dividing equation (ii) by (i)

Substituting this in (i),

a = 2 and r = 2.

(iii) The product of four consecutive natural numbers is 5040. Find the numbers.

Its is given that,

1) The product of four consecutive natural numbers = 5040

Let first number = x

� remaining consecutive numbers = (x + 1), (x + 2) and (x + 3)

But using (1),

x (x + 1)(x + 2)(x + 3) = 5040

� (x + 1)(x + 2)(x + 3)x = 5040


� (x2 + 3x + 2)(x2 + 3x) = 5040

Now substitute x2 + 3x = b

� (b + 2)(b)= 5040

� b2 + 2b = 5040

Let's add 1 to both the sides so that LHS becomes a perfect square.

� b2 + 2b + 1 = 5040 +1

� (b + 1)2 = 5041

Taking square root on both the sides,

b + 1 = � 71

� b + 1 = 71 or b + 1 = − 71

� b = 70 or b = − 72

Now substitute back b = x2 + 3x

�x2 + 3x = 70

and x2 + 3x = − 72

i) Let's first consider x2 + 3x = 70

� x2 + 3x− 70 = 0

� x2 + 10x− 7x− 70 = 0

� x2 + 10x− 7x− 70 = 0

� x(x + 10) − 7( x + 10) = 0

� (x + 10)( x− 7) = 0

�x = − 10 or x = 7

Since given numbers are natural numbers, x can not be negative.

�x = 7

ii) Now consider x2 + 3x = − 72

� x2 + 3x + 72 = 0

here a = 1, b = 3 and c = 72

Now b2− 4ac = 32− 4(1)(72) = 9 − 288 = − 279

i.e. b2− 4ac < 0

i.e. there are no real roots for this equation

�x = 7 is the only required root

Hence the remaining numbers are 7 + 1 = 8, 7 + 2 = 9 and 7 + 3 = 10

Four consecutive natural numbers are 7, 8, 9 and 10.

(iv) Solve the following simultaneous equations using Graphical method:

x + 3y = 7, 2x + y = −1


(1) x + 3y = 7

(2) 2x + y = −1


To draw the graphs of these two equations, we find two solutions of each equation and make tables as follows.

x + 3y = 7

x 1 −2

y 2 3

(x, y) (1, 2) (−2, 3)

2x + y = −1

x 0 2

y −1 −5

(x, y) (0, −1) (2, −5)

Now plot points (1, 2) and (−2, 3) and join them to get graph of line x +3y = 7.

Similarly plot the points (0, −1) and (2, −5) and join them to get graph of line 2x + y = −1.

The lines of the two given equations intersect each other at (−2, 3)

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

(−2, 3) is the solution of given simultaneous equations.

Q. 5. Attempt any four from the following : (20 marks)

(i) The weight of coffee (in gm) in 70 packets is given below. Determine the modal weight of coffee in a packet.


Weight (in gm) No. of packets

200 - 201 12

201 - 202 16

202 - 203 20

203 - 204 9

204 - 205 2

205 - 206 1

Weight (in gm) No. of packets

200 - 201 12 ��f1

201 - 202 26 ��fm

202 - 203 20 ��f2

203 - 204 9

204 - 205 2

205 - 206 1

L = 201, fm = 26, f1 = 12, f2 = 20, h = 1

Modal weight of the coffee in a packet is 201.70 grams.

(ii) If 'A' and 'G' are the A. M. and G.M. respectively of two numbers then prove that the numbers are


Let x and y be two numbers, whose Arithmetic Mean is A and Geometric Mean is G.

Compare it with ax2 + bx + c = 0

� a = 1, b = -2A, and c = G2

(iii) A box contains 20 cards marked with the numbers 1 to 20. One card is drawn from this box at random.

What is the probability of the following events:

(i) number on the card isa prime number

(ii) number on the card isperfect square

The sample space S is given by,


S = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 }

��n(S) = 20

(i) Let A be the event of getting a prime number.

A = { 2, 3, 5, 7, 11, 13, 17, 19 }

��n(A) = 8

(ii) Let B be the event of getting a perfect square.

B = { 1, 4, 9, 16 }

��n(B) = 4

(iv) Sagar and Aakash ran 2 km race twice. Aakash completed the first round 2 minutes earlier than Sagar. In

the second round Sagar increased his speed by 2 km/hour and Aakash reduced his speed by 2km/hour. Sagar

finished 2 minutes earlier than Aakash. Find their speeds of running in the first round.


Rejecting y = −10 as speed can not be negative,

y = 12

Substituting y = 12 in x = y − 2,

x = 10

��Speed of Sagar in the first round was 10 km/hr and speed of Aakash in the first round was 12 km/hr


Speed of Sagar in the first round was 10 km/hour and speed of Aakash in the first round was 12 km/hour

(v) Draw Histogram and frequency polygon for the following frequency distribution.

Class 5 - 10 10 - 15 15 - 20 20 - 25

Frequency 20 30 50 40

Let's prepare the table as shown below:

Class Class Mark

(xi) Frequency (fi)

Point

(xi, fi)

2.5 0 (2.5, 0)

5 - 10 7.5 20 (7.5, 20)

0 - 15 12.5 30 (12.5, 30)

15 - 20 17.5 50 (17.5, 50)

20 - 25 22.5 40 (22.5, 40)

27.5 0 (27.5, 0)

Take a scale of 2 cm = 5 on X-axis.

Take a scale of 1 cm = 5 on Y-axis.

Draw the histogram and frequency polygon as shown in figure below.



Documents

Algebra -March 2012 - Maharashtra SSC Board · mySSC.in -Algebra-March 2012 Page 13 of 18. Let xand ybe two numbers, whose Arithmetic Mean is A and Geometric Mean is G. Compare it