algebra prelim q

Q.1. Attempt ANY SIX of the following : 6

(i) For the given sequence, find the next four terms.2, 6, 12, 20, 30, ...

(ii) Is list of numbers Arithmetic Progressions ?Justify 1, 4, 7, 10, ....

(iii) Find the values of discriminant of the following equation.x2 3x + 2 = 0

(iv) Determine whether the given value of x is a root of given quadraticequation : x2 2x + 1 = 0, x = 1

(v) Find the value of the determinant :5 2

7 4

(vi) If two coins are tossed then find the probability of the event that at leastone tail turns up

(vii) Calculate mean x when fixi = 100 and fi = 20.

Note :

(i) All questions are compulsory.

(ii) Use of calculator is not allowed.

Time : 2 Hours (Pages 4) Max. Marks : 60

Seat No.

Q.P. SET CODE

2012 ___ ___ 1100 - MT - w - MATHEMATICS (71) ALGEBRA - SET - A (E)

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Q.2. Solve ANY FIVE of the following : 10

(i) If 25 is the arithmetic mean between x and 46, then find x.

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

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

(iv) Find the indicated sum for the following Geometric Progressions :2, 6, 18, ... . Find S7.

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

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

Q.3. Solve ANY FOUR of the following : 12

(i) Solve the following quadratic equation by using formula :5m2 2m = 2

(ii) Without plotting the graphs, find the point of intersection of the lines2x + 5y = 13 and 4x 9y = 7.

(iii) Draw histogram for the data given below :

Also find the mode graphically.

Class intervals 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30

Frequency 4 6 3 2 1

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(iv) Three consecutive odd natural numbers are such that the product of thefirst and third is greater than four times the middle by 1. Find the numbers.

(v) Solve the following simultaneous equations using Cramers rule :

y = 5x 10

2; 4x + 5 = y

Q.4. Solve ANY THREE of the following : 12

(i) Two dice are thrown find the probability of getting :(a) The sum of the numbers on their upper faces is divisible by 9.(b) The sum of the numbers on their upper faces is at most 3.(c) The number on the upper face of the first die is less than the number

on the upper face of the second die.

(ii) Vijay invests some amount in National saving certificate. For the 1st yearhe invests Rs. 500, for the 2nd year he invests Rs. 700, for the 3rd year heinvests Rs. 900, and so on. How much amount he has invested in 12years ?

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

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


(i) Below is given frequency distribution of marks (out of 100) obtained bythe students. Find mean using step deviation method :

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of students

3 5 7 10 12 15 12 6 2 8

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Best Of Luck

Area in hectare 11 - 21 - 31 - 41 - 51 - 61 - 71 -20 30 40 50 60 70 80

No. of farmers 58 103 208 392 112 34 12

(ii) Find three consecutive terms in a G.P. such that the sum of the 2nd and3rd term is 60 and the product of all the three is 8000.

(iii) One lottery ticket is drawn at random from a bag containing 20 ticketsnumbered from 1 to 20. Find the probability that the number on theticket drawn is(a) either even or square of an integer(b) divisible by 3 or 5.

(iv) When the son will be as old as his father today, the sum of their agesthen will be 126. When the father was as old as his son is today, the sumof their ages then was 38. Find their present ages.

(v) Draw frequency polygon and frequency curve for the following data onland holding :

A.1. Attempt ANY SIX of the following :(i) t1 = 0 + 2 = 2

t2 = 2 + 4 = 6t3 = 6 + 6 = 12t4 = 12 + 8 = 20t5 = 20 + 10 = 30t6 = 30 + 12 = 42t7 = 42 + 14 = 56t8 = 56 + 16 = 72t9 = 72 + 18 = 90

The next four terms of the sequence are 42, 56, 72 and 90. 1(ii) t1 = 1, t2 = 4, t3 = 7, t4 = 10

t2 t1 = 4 1 = 3t3 t2 = 7 4 = 3t4 t3 = 10 7 = 3

The difference between any two consecutive terms is3 which is constant

The sequence is an A.P. 1(iii) x2 3x + 2 = 0

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

= ( 3)2 4 (1) (2)= 9 8= 1

= 1 1

(iv) x2 2x + 1 = 0, x = 1Putting x = 1 in L.H.S. we get,L.H.S.= (1)2 2 (1) + 1

= 1 2 (1) + 1= 2 2= 0= R.H.S.

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 60

A.P. SET CODE

2012 ___ __ 1100 - MT - w - MATHEMATICS (71) ALGEBRA - SET - A (E)

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L.H.S.= R.H.S.Thus equation is satisfied.So 1 is the root of the given quadratic equation. 1

(v)5 2

7 4= (5 4) (2 7)= 20 14

= 6 1

(vi) When two coins are tossedS = { HH, HT, TH, TT }n (S) = 4Let A be the event that atleast one tail turns upA = { HT, TH, TT }n (A) = 3

P (A) =n (A)

n (S)

P (A) = 34

1

(vii) 100Mean x 20 i ii if x

f x

Mean x 5 1A.2. Solve ANY FIVE of the following :(i) 25 is the arithmetic mean between x and 46

25 = x 462

1

50 = x + 46 x = 50 46 x = 4 1

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

(4)2 7(4) + k = 0 16 28 + k = 0 12 + k = 0 k = 12 1

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

(iv) 2, 6, 18, .........a = 2

r =6

2= 3

Sn =na (r 1)

r 1 1

S7 =7a (r 1)

r 1

S7 =72 (3 1)

3 1

S7 =72 (3 1)2

S7 = 37 1 S7 = 2187 1 S7 = 2186 1

(v) Mean = 101, Median = 100 [Given]We know,Mean Mode= 3 (Mean Median) 1

101 Mode = 3 (101 100) 101 Mode = 3 (1) 101 3 = Mode Mode = 98 1

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

+ = 3 + 10 = 13 . = 3 10 = 30 1

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

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

A.3. Solve ANY FOUR of the following :(i) 5m2 2m = 2

5m2 2m 2 = 0Comparing with ax2 + bx + c = 0 we have a = 5, b = 2, c = 2b2 4ac = ( 2)2 4 (5) ( 2)

= 4 + 40= 44

m = b b 4ac

2a

21

= ( 2) 44

2 (5)

=2 4 11

10

=2 2 11

10

1

= 2 1 11

10

=1 11

5

x = 1 115

or x =

1 11

5

1 115

and

1 11

5 are the roots of the given quadratic equation. 1

(ii) 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

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By Cramers rule,

x =D

Dx =

152

38 = 4 1

y =D

Dy

= 38

38 = 1

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

(iii) 2

Mode of the given data is 11.625 units. 1

(iv) Let the third consecutive odd natural number be x, x + 2 and x + 4As per the given condition,x (x + 4) = 4 (x + 2) + 1 1

x2 + 4x = 4x + 8 + 1 x2 = 9

Taking square root on both the sides we get,x = + 3

x is a natural number x 3 1Hence, x = 3x + 2 = 3 + 2 = 5 and x + 4 = 3 + 4 = 7

The 3 consecutive odd natural numbers are 3, 5 and 7 respectively. 1

Fre

qu

en

cy

1

2

3

4

5

6

7Scale : On X axis : 2 cm = 5 units

On Y axis : 1 cm = 1 unit

Classes

X

Y

X5 100

Y

15 2 0 25 30

Mode

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A.4. Solve ANY THREE of the following :(i) Two dice are thrown

S = { (1, 1), (1, 2), (1, 3) (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3) (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3) (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3) (6, 4), (6, 5), (6, 6) }

n (S) = 36 1(a) Let A be the event that sum of numbers on their upper faces is

divisible by 9A = { (3, 6), (4, 5), (5, 4), (6, 3) }n (A) = 4

P (A) =n (A)

n (S)

P (A) = 436

P (A) = 19

1

(b) Let B be the event that sum of number on their upper faces is atthe most 3B = { (1, 1), (1, 2), (2, 1) }n (B) = 3

P (B) =n (B)

n (S)

P (B) = 336

P (B) = 112

1

(c) Let C be the event that number on the upper face of the first die isless than the number on the upper face of second die.C = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) }n (C) = 15

P (C) =n (C)

n (S)

P (C) = 1536

P (C) = 512

1

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(ii) Yearly investments of Vijay are as follows 500, 700, 900, .......The yearly investments form an A.P. with first year investment 1(a) = 500Difference between investment done in two successive years(d) = 200.No. of years (n) = 12 1Total 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] S12 = 6 [3200] S12 = 19200 Total investment done in 12 years is Rs. 19200. 1

(iii) Let the five consecutive natural numbers be x, x + 1, x + 2, x + 3

and x + 4 respectively.

As per the given condition,

x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2 = 1455

x2 + x2 + 2x + 1 + x2 + 4x + 4 + x2 + 6x + 9 + x2 + 8x + 16 1455 = 0 1 5x2 + 20x + 30 1455 = 0 5x2 + 20x 1425 = 0

Dividing throughout by 5 we get,

x2 + 4x 285 = 0

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

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 15, 16, 17, 18 1

and 19 respectively.

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

x = 5 4y3

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

Scale : 1 cm = 1 uniton both the axes

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

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A.5. Solve ANY FOUR of the following :(i) Class width (h) = 10, Assumed mean (A) = 45

Class mark Class Mark di = xi A ui = id

hNo. of students fiui

(xi) (fi)

0 - 10 5 40 4 3 1210 - 20 15 30 3 5 1520 - 30 25 20 2 7 1430 - 40 35 10 1 10 1040 - 50 45 A 0 0 12 0 250 - 60 55 10 1 15 1560 - 70 65 20 2 12 2470 - 80 75 30 3 6 1880 - 90 85 40 4 2 890 - 100 95 50 5 8 40

Total 80 54

u =i i

i

f u

f

u =54

80

u = 0.675 1

Mean x = A hu= 45 + 10 (6.25)= 45 + 6.75 1= 51.75

Mean of marks obtained by students is 51.75 marks. 1

(ii) Let three consecutive terms of G.P. be a

r, a, ar.

As per first condition,a + ar = 60 ......(i) 1As per second condition,a

r a ar = 8000

a3 = 8000Taking cube roots on both side,

a = 80003

a = 20 20 20 3 1 a = 20

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Substituting a = 20 in (i),20 + 20r = 60

20r = 60 20 20r = 60 20 20r = 40 1

r = 4020

r = 2

ar

= 20

2= 10 1

ar = 20 2 ar = 40 The three consecutive terms of the G.P. are 10, 20, 40. 1

(iii) Let S be the sample spaceS = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

11, 12, 13, 14, 15, 16, 17, 18, 19, 20 }n (S) = 20

(a) Let A be the event that the number is evenA = { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 }

n (A) = 10P (A) =

n (A)

n (S)

P (A) = 1020

P (A) = 12

Let B be the event that the number is square of an integerB = { 1, 4, 9, 16 }

n (B) = 4

P (B) =n (B)

n (S)

P (B) = 420

P (B) = 15

A B is the event that the number is even and square of an integer A B = { 4, 16 } n (A B) = 2

P (A B) =n (A B)

n (S)

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=2

20

P (A B) = 110

A B is the event that the number is either even or square of anintegerP (A B) = P (A) + P (B) P (A B)

=1 1 1

2 5 10

=5 2 1

10 10 10

=6

10

P (A B) = 35

1

(b) Let C be the event that the number is divisible by 3C = { 3, 6, 9, 12, 15, 18 }n (C) = 6

P (C) =n (C)

n (S)

P (C) = 620

P (C) = 310

Let D be the event that the number is divisible by 5D = { 5, 10, 15, 20 }

n (D) = 4

P (D) =n (D)

n (S)

P (D) = 420

P (D) = 15

C D is the event that the number is divisible by 3 and divisibleby 5.

C D = { 15 } n (C D) = 1

P (C D) =n (C D)

n (S)

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P (C D) = 120

C D is the event that the ticket drawn bears a number which isdivisible by 3 or 5P (C D) = P (C) + P (D) P (C D)

=3 1 1

10 5 20

=6 4 1

20 20 20

P (C D) = 920

1

(iv) Let present age of father be x years and son be y yearsSons age when he will be as old as his father today = x years.

Fathers age at that time [x + (x y)] years.As per the first given condition,x + x y + x = 126

3x y = 126 ......(i) 1Fathers age when he was as old as son today = y years

Sons age at that time [y (x y)] years.As per the second given condition,y + y (x y) = 38

y + y x + y = 38 x + 3y = 38 .....(ii) 1

Multiplying (ii) by 3, 3x + 9y = 114 ......(iii)

Adding (i) and (iii),3x y = 1256

3x + 9y = 1148y = 240

y = 2408

y = 30Substituting y = 30 in (i),

3x 30 = 126 3x = 126 + 30 1 3x = 156

x = 1563

x = 52 1 The present age of father and son are 52 years and 30 years 1

respectively.

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(v) 1

2

0

25

50

75

100

125

150

175

200

225

Classes (area in hectres)

No

. o

f fa

rme

rs

XY

Scale : On X axis : 1 cm = 10 hectresOn Y axis : 1 cm = 25 farmers

X

Y

250

275

300

325

350

10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5

(15.5, 58)

(25.5, 103)

(35.5, 208)

(55.5, 112)

(65.5, 34)

(75.5, 12)

(45.5, 392)

(5.5,0) (85.5, 0)

Area in hectare Continuous classes Class mark No. of farmers

11 - 20 10.5 - 20.5 15.5 5821 - 30 20.5 - 30.5 25.5 10331 - 40 30.5 - 40.5 35.5 20841 - 50 40.5 - 50.5 45.5 39251 - 60 50.5 - 60.5 55.5 11261 - 70 60.5 - 70.5 65.5 3471 - 80 70.5 - 80.5 75.5 12

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Y

0

25

50

75

100

125

150

175

200

225

Classes (area in hectres)

No

. o

f fa

rme

rs

X

Scale : On X axis : 1 cm = 10 hectresOn Y axis : 1 cm = 25 farmers

X

250

275

300

325

350

10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5

375

400

(15.5, 58)

(25.5, 103)

(35.5, 208)

(55.5, 112)

(65.5, 34)

(75.5, 12)

(45.5, 392)

(5.5,0) (85.5, 0)

Y


(i) Find the first five terms of the following sequence, whose nth terms isgiven : tn = n

2 2n

(ii) Write the first five terms of the following Arithmetic Progression where,the common difference d and the first term a is given :a = 2, d = 2.5

(iii) Find the value of discriminant of the following equation :4x2 + kx + 2 = 0

(iv) Write the quadratic equation in standard from ax2 + bx + c = 0y2 = 5 y

(v) If Dy = 15 and D = 5 are the values of the determinants for certainsimultaneous equations in x and y, find y.

(vi) Two coins are tossed. Find the probability of the event that head appearson both the coins.

(vii) Calculate mean x when fixi = 100 and fi = 20

Note :




Seat No.

Q.P. SET CODE

2012 ___ ___ 1100 - MT - x - MATHEMATICS (71) ALGEBRA - SET - B (E)

SET - B2 / MT - x


(i) Find the geometric mean of 82 1 and 82 +1.

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

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

(iv) Write down the first five terms of the geometric progression which hasfirst term 1 and common ratio 4.

(v) For a certain frequency distribution the values of Median and Mode is95.75 and 95.5 respectively, find the mean.

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


(i) Solve the following quadratic equation by completing square :x2 + 3x + 1 = 0

(ii) The perimeter of an isosceles triangle is 24 cm. The length of its congruentsides is 13 cm less than twice the length of its base. Find the lengths ofall sides of the triangle.

(iii) The following pie diagram represent thenumber of valid votes obtained by fourstudent who contested for school captain.The total of valid votes polled was 720.Answer the following questions :(a) Who has won the election ?(b) What is the minimum number of votes?

Who got it?(c) By how many votes did the winner defect

the nearest contenstant ?

AlbertRaj

Nashima

Suja

80

12060

100

SET - B3 / MT - x

(iv) Form the quadratic equation if its one of the root is 3 2 5 .

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


(i) A game of chance consists of spinning an arrow which comes to restpointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equallylikely outcomes. What is the probability that it will point at :(a) 8 (b) an odd number(c) a number greater then 2 (d) a number lessthan 9

(ii) If S3 = 31 and S6 = 3906 then find a and r.

(iii) Solve the following equations : (p2 + p) (p2 + p 3) = 28.

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


(i) Below is given distribution of profit in Rs. per day of a shop in certaintown :

Calculate median profit of a shop.

(ii) Sachin, Sehwag and Dhoni together scored 228 runs. Their individualscores are in G.P. Sehwag and Dhoni together scored 12 runs more thanSachin. Find their Individual scores.

Profit (in Rs.) 500 - 1000 - 1500 - 2000 - 2500 - 3000 - 3500 -

900 1400 1900 2400 2900 3400 3900

No. of shops 8 18 27 21 20 18 8

SET - B4 / MT - x

Best Of Luck

(iii) A box contains 36 tickets numbered from 1 to 36. One tickets is drawn

at random. Find the probability that the number on the ticket is either

divisible by 3 or is a perfect square.

(iv) The weight of a bucket is 15 kg, when it is filled with water 3

5of its

capacity while it weights 19 kg, if it is filled with water 4

5 of its capacity.

Find the weight of bucket, if it is completely filled with water.

(v) Following data represent the age wise distribution of employees in office :

Draw : (a) less than type cumulative frequency curve (b) more than typecumulative frequency. On the same graph paper. Hence find the medianage of employees.

Age in years 25 - 30 - 35 - 40 - 45 - 50 - 55 -

30 35 40 45 50 55 60

Number of employees 4 16 19 28 22 8 3

A.1. Attempt ANY SIX of the following :(i) tn = n

2 2n t1 = 12 2 (1) = 1 2 = 1 t2 = 22 2 (2) = 4 4 = 0 t3 = 32 2 (3) = 9 6 = 3 t4 = 42 2 (4) = 16 8 = 8 t5 = 52 2 (5) = 25 10 = 15 The first five terms of the sequence are 1, 0, 3, 8 and 15. 1

(ii) a = 2, d = 2.5Here,t

1= a = 2

t2

= t1

+ d = 2 + 2.5 = 4.5t

3= t

2 + d = 4.5 + 2.5 = 7

t4

= t3 + d = 7 + 2.5 = 9.5

t5

= t4 + d = 9.5 + 2.5 = 12

The first five terms of the A.P. are 2, 4.5, 7, 9.5 and 12. 1

(iii) 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

(iv) y2 = 5 7 y

y2 + 7 y 5 = 0 1

(v) Dy = 15 and D = 5By Cramers rule,

y =D

Dy


A.P. SET CODE

2012 ___ __ 1100 - MT - x - MATHEMATICS (71) ALGEBRA - SET - B (E)

SET - B2 / MT - x

y = 15

5

y = 3 1

(vi) Two coins are tossedS = { HH, HT, TH, TT }n (S) = 4Let A be the event that head appears on both the coinsA = { HH }n (A) = 1

P (A) =n (A)

n (S)

P (A) = 14

1


f x

Mean x 5 1A.2. Solve ANY FIVE of the following :(i) Let G be the geometric mean of 82 1 and 82 1

G2 = 82 1 82 1 G2 = 82 2 12 G2 = 82 1 G2 = 81 1

Taking square roots on both sides,G = 81

G = +9 The geometric mean of 81 1 and 81 1 is 9 or 9 1

(ii) kx2 7x + 12 = 0x = 3 is root of given quadratic equation.So it satisfies the given equation

Substituting x = 3 in the given equation k(3)2 7(3) + 12 = 0 k(9) 21 + 12 = 0 1 9k 9 = 0 9k = 9 k = 9

9 k = 1 1

SET - B3 / MT - x

(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 1

5a + 6 = 4 5a = 4 6 5a = 10 a = 10

5

a = 2 1

(iv) For the G.P.The first term (a) = 1Common ratio (r) = 4t1 = a = 1t2 = ar = 1 4 = 4t3 = ar

2 = 1 (4)2 = 16 1t4 = ar

3 = 1 (4)3 = 64t5 = ar

4 = 1 (4)4 = 256

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

(v) Median = 95.75, Mode = 95.5 [Given]We know,Mean Mode = 3 (Mean Median)

Mean 95.5 = 3 (Mean 95.75) Mean 95.5 = 3 Mean 287.25 1 95.5 287.25 = 3 Mean Mean 191.75 = 2 Mean

191.752

= Mean

Mean = 95.875 1

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

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

x2 4x + ( 45) = 0 x2 4x 45 = 0 The required quadratic equation is x2 4x 45 = 0 1

SET - B4 / MT - x

A.3. Solve ANY FOUR of the following :(i) x2 + 3x + 1 = 0

x2 + 3x = 1 ..... (i)

Third term =2

1coefficientof x

2

=2

13

2

=2

3

2

=9

41

Adding 9

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

x2 + 3x + 9

4= 1 +

9

4

2

3x

2 =

4 9

4

2

3x

2 =

5

41

Taking square root on both the sides we get,

x + 3

2=

5

2

x = 32

+5

2

x = 3 52

x = 3 + 52

or x = 3 5

2

3 + 52

& 3 5

2 are the roots of the given quadratic equation. 1

(ii) Let length of one of the congruent sides of an isosceles triangle

be x cm and length of its base be y cm.

As per the first given condition,

x + x + y = 24

SET - B5 / MT - x

2x + y = 24 .......(i) 1As per the second given condition,x = 2y 13

x 2y = 13 .......(ii)Multiplying (i) by 2,4x + 2y = 48 ......(iii)

Adding (ii) and (iii),x 2y = 134x + 2y = 485x = 35

x = 355

x = 7 1Substituting x = 7 in (i),2 (7) + y = 24

14 + y = 24 y = 24 14 y = 10 The length of sides of the isosceles triangle are 7 cm, 7 cm and 1

10 cm.

(iii)

1

(a) Nashima has won the election.

(b) Minimum number of votes is 120 obtained by Suja. 1(c) Winner Nashima defeated the nearest contestant Albert by 40 votes. 1

(iv) If one of the root of the quadratic equation is 3 2 5 , then the

other root is 3 + 2 5

Name of the Measure of central angle () Number of votesCandidate

Raj 8080

360 720 = 160

Albert 100100

360 720 = 200

Suja 6060

360 720 = 120

Nashima 120120

360 720 = 240

Total 360 720

SET - B6 / MT - x

= 3 2 5 and = 3 + 2 5 + = 3 2 5 + 3 + 2 5 = 6 1

and . = 3 2 5 3 + 2 5= (3)2 2 5 2= 9 4 5= 9 20= 11 1

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

x2 6x + ( 11) = 0 x2 6x 11 = 0 The required quadratic equation is x2 6x 11 = 0 1

(v) 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

21

a

a1

2 =

b

b1

2

c

c1

2

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

A.4. Solve ANY THREE of the following :(i) Sample space for the game of chance of spinning an arrow

S = { 1, 2, 3, 4, 5, 6, 7, 8 }n (S) = 8

(a) Let A be the event that the arrow points at 8A = { 8 }

n (A) = 1

P (A) =n (A)

n (S)

P (A) = 18

1

SET - B7 / MT - x

(b) Let B be the event that the arrow points at an odd numberB = { 1, 3, 5, 7 }

n (B) = 4

P (B) =n (B)

n (S)

P (B) = 48

P (B) = 12

1

(c) Let C be the event that the arrow points at a number greater than 2C = { 3, 4, 5, 6, 7, 8 }

n (C) = 6

P (C) =n (C)

n (S)

P (C) = 68

P (C) = 34

1

(d) Let D be the event that the arrow points at number less than 9D = { 1, 2, 3, 4, 5, 6, 7, 8 }n (D) = 8

P (D) =n (D)

n (S)

P (D) = 88

P (D) = 1 1

(ii) For a G.P.

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) 1

SET - B8 / MT - x

Similarly, S6 =a (1 r )

1 r

6

But, S6 = 3906 [Given]

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

1 (r )

1 r

3 2

3 = 126 1

(1 r ) (1 r )

1 r

3 33 = 126

1 + r3 = 126 r3 = 126 1 r3 = 125

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 124a = 124

a =124

124 a = 1 1 a = 1 and r = 5.

(iii) (p2 + p)(p2 + p 3) = 28Substituting p2 + p = m we get,

m (m 3) = 28 m2 3m 28 = 0 m2 7m + 4m 28 = 0 m (m 7) + 4 (m 7) = 0 (m 7) (m + 4) = 0

SET - B9 / MT - x

m 7 = 0 or m + 4 = 0 m = 7 or m= 4 1

Resubstituting m = p2 + p we get,

p2 + p = 7 or p2 + p = 4

p2 + p 7= 0 .....(i) or p2 + p + 4 = 0 ........(ii)From (i), p2 + p 7 = 0

Comparing with ap2 + bp + c = 0 we have a =1, b = 1, c = 7

b2 4ac = (1)2 4 (1) (7)

= 1 + 28

= 29 1

p = b b 4ac

2a

2

=1 29

2(1)

=1 29

21

From (ii), p2 + p + 4 = 0

Comparing with ap2 + bp + c = 0 we have a = 1, b =1, c = 4

b2 4ac = (1)2 4 (1)(4)

= 1 16

= 15

b2 4ac < 0

The roots of this equation are not real hence not considered.

p =1 29

2

p = 1 292

or p =

1 29

21

(iv) x + y = 8 x y = 2 y = 8 x x = 2 + y

1

x 2 3 4

y 0 1 2

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

x 0 1 2

y 8 7 6

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

SET - B10 / MT - x

2


-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

SET - B11 / MT - x

A.5. Solve ANY FOUR of the following :(i)

1

Here total frequency = fi = N = 120

N2

= 120

2 = 60

Cumulative frequency (less than type) which is just greater than 60

is 74. Therefore corresponding class 1950 - 2450 is median class. 1L = 1950, N = 120, c.f. = 53, f = 21, h = 500

Median =N h

L 2

c.f.

f

=120 500

1950 532 21

1

= 1950 + (60 53) 500

21

= 1950 + (7) 500

21

= 1950 + 500

3= 1950 + 166.67 1= 2116.67

Median of profit is Rs. 2116.67. 1

(ii) Let individual scores of Sachin, Sehwag and Dhoni be a

r, a and ar

As per the first given condition,a

r + a + ar = 228 .....(i)

As per second given condition,

a + ar = 12 + a

r......(ii) 1

Class Continuous Frequency (fi) Cumulative frequency(Profit in Rs.) Classes (No. of shops) less than type

500 - 900 450 - 950 8 81000 - 1400 950 - 1450 18 261500 - 1900 1450 - 1950 27 53 c.f.2000 - 2400 1950 - 2450 21 f 742500 - 2900 2450 - 2950 20 943000 - 3400 2950 - 3450 18 1123500 - 3900 3450 - 3950 8 120

Total 120 N

SET - B12 / MT - x

Substituting (ii) in (i),a

r + 12 +

a

r= 228

12 + 2ar

= 228

2ar

= 228 12

2ar

= 216

ar

=216

2

ar

= 108 .....(iii) 1

a = 108rSubstituting (iii) in (i),108r

r + 108r + (108 r)r = 228

a108 , a 108r

r

Multiplying throughout by r,108 + 108r + 108r2 = 228

108r2 + 108r + 108 228 = 0 108r2 + 108r 120 = 0

Dividing throughout by 12 we get,9r2 + 9r 10 = 0

9r2 + 15r 6r 10 = 0 3r (3r + 5) 2 (3r + 5) = 0 (3r + 5) (3r + 2) = 0 3r + 5 = 0 or 3r 2 = 0 3r = 5 or 3r = 2 1 r = 5

3or r =

2

3

If r =53

If r =2

3a = 108r a = 108r

a = 108 5

3a = 108

2

3a = 36 5 a = 36 2a = 180 a = 72

ar = 72 2

3ar = 48

r = 5

3 is not acceptable because a (individual score) cannot be 1

negative.

SET - B13 / MT - x

ar = 48 The runs scored by Sachin, Sehwag and Dhoni are 108 runs, 1

72 runs and 48 runs.

(iii) When a ticket is drawn from a box containing 36 tickets numberedfrom 1 to 36.S = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 1

18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,32, 33, 34, 35, 36 }

n (S) = 36(a) Let A be the event that the ticket drawn bears a number which isdivisible by 3.

A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 }n (A) = 12

P (A) =n (A)

n (S)

P (A) = 1236

1

(b) Let B be the event that ticket drawn bears a number which is aperfect squareB = { 1, 4, 9, 16, 25, 36 }n (B) = 6

P (B) =n (B)

n (S)

P (B) = 636

1

A B is the event that the ticket draw bears a number which isdivisible by 3 and perfect square

A B = { 9, 36 } n (A B)= 2

P (A B)=n (A B)

n (S)

P (A B)= 236

1

A B is the event that ticket drawn bears a number which isdivisible by 3 or perfect squareP (A B)= P (A) + P (B) P (A B)

=12 6 2

36 36 36

= 1636

P (A B)= 49

1

SET - B14 / MT - x

(iv) Let weight of empty bucket be x kg. and let the weight of onlywater when the bucket is filled to its full capacity be y kg.As per the first given condition,

x + 3

5y = 15

Multiplying throughout by 5,5x + 3y = 75 .......(i) 1As per the second given condition,

x + 4

5y = 19

Multiplying throughout by 5,5x + 4y = 95 ......(ii) 1Subtracting (ii) from (i),5x + 3y = 755x + 4y = 95() () () y = 20

y = 20 1Substituting y = 20 in (i),5x + 3 (20) = 75

5x + 60 = 75 5x = 75 60 5x = 15 x = 3 1 Weight of bucket when it is filled with water to its full capacity

= x + y = 20 + 3

= 23

Weight of bucket when it is filled with water to its full capacity 1is 23 kg.

(v) 1

Classes No. of Cumulative Upper Cumulative Lower(Age in empl. frequency boundaries frequency boundariesyears) less than more than

type type

20 - 25 0 0 25 2025 - 30 4 4 30 100 2530 - 35 16 20 35 96 3035 - 40 19 39 40 80 3540 - 45 28 67 45 61 4045 - 50 22 89 50 33 4550 - 55 8 97 55 11 5055 - 60 3 100 60 3 5560 - 65 0 65 0 60

SET - B

3

Median of age of employees 42 years. 1

15 / MT - x

10

20

30

40

50

60

70

80

90

Cu

mu

lati

ve

fre

qu

en

cy

Scale : On X axis : 1 cm = 5 yearsOn Y axis : 1 cm = 10 employees

100

Class boundaries (Age in years)X

YX300

Y

40 5020 25 35 45 55 60 65

(25, 0)

(30,4)

(35, 20)

(40, 39)

(40, 61)

(35, 90)

(30,98)

(25, 100) (60, 100)

(55, 97)

(50, 89)

(45, 67)

(45, 33)

(50, 11)

(55,4)(60, 0)


(i) Is the given list of number an Arithmetic Progression ?Justify. 1, 3, 6, 10, .....

(ii) For the sequence, find the next four terms :192, 96, 48, 24, ....

(iii) Find the value of discriminant of each of the following equation :2x2 + x + 1 = 0

(iv) Find the values of a, b, c for following quadratic equation by comparingwith standard form : x2 + 5x 4 = 0

(v) Express the following information in mathematical form using twovariables : The cost of two tables and five chairs is Rs. 2200.

(vi) A die is thrown then find the probability of getting an odd number.

(vii) Calculate mean x when fixi = 100 and fi = 20.

Note :




Seat No.

Q.P. SET CODE

2012 ___ ___ 1100 - MT - y - MATHEMATICS (71) ALGEBRA - SET - C (E)

SET - C2 / MT - y


(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 1

2 and

3

4.

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

(iv) Find the twenty fifth term of the A. P. : 12, 16, 20, 24, .....

(v) For a certain frequency distribution the values of Mean and Mode are54.6 and 54 respectively. Find the value of median.

(vi) Solve the following equations by factorization method :y2 + 8y + 16 = 0


(i) Solve the following equations x4 3x2 + 2 = 0.

(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.

(iii) Draw a pie diagram to represent the world population given in the followingtable after determining the value of a :

Country India China Russia USA Other Total

Percentage ofworld population 15 20 a a 25 100

SET - C3 / MT - y

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

3. Find the number.

(v) Sum of two numbers is 60. The greater number is 8 more than thrice thesmaller number. Find the numbers.


(i) In a survey conducted among 400 students of X standard in Punedistrict, 187 offered to join Science faculty after X std. and 125students offered to join Commerce faculty after X, if a student isselected at random from this group. Find the probability that studentprefers Science or Commerce faculty.

(ii) In a school, tree plantation on Independence day was arranged. Everystudent from I standard will plant 2 trees, II standard students will plant4 trees each, III standard students will plant 8 trees each etc. If there are5 standard, how many trees are planted by the student of that school?

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

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


(i) Below is given frequency distribution of I.Q. (Intelligent Quotient) of 80candidates :

Find median I.Q. of a candidate.

I.Q, 7080 8090 90100 100110 110 120 120130 130 140

No. of 7 16 20 17 11 7 2candidates

SET - C4 / MT - y

Best Of Luck

(ii) If the nth, (2n)th, (3n)th terms of a G.P. are a, b, c respectively then showthat b2 = ac.

(iii) Two fair dice are thrown, find the probability that sum of the points ontheir uppermost faces is a perfect square or divisible by 4.

(iv) A person deposits Rs. x in savings bank account at the rate of 5% perannum and Rs. y in fixed deposit at 10% per annum. At the end of oneyear he gets Rs. 400 as total interest. If the deposits Rs. y in savingsbank account and Rs. x in fixed deposit he would get Rs. 350 as totalinterest. Find the total amount he deposited.

(v) Represent the following data using frequency curve :

Draw histogram and hence draw frequency curve.

Electricity bill in200 - 400 400-600 600 - 800 800 - 1000

a month (in Rs.)

No. of families 362 490 185 63

A.1. Attempt ANY SIX of the following :(i) t1 = 1, t2 = 3, t3 = 6, t4 = 10

t2 t1 = 3 1 = 2t3 t2 = 6 3 = 3t4 t3 = 10 6 = 4

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

(ii) t1 = 192

t2 =192

2 = 96

t3 =96

2 = 48

t4 =48

2 = 24

t5 =24

2 = 12

t6 =12

2 = 6

t7 = 6

2 = 3

t8 =3

2 =3

2

The next four terms of the sequence are 12, 6, 3 and 32

. 1

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

= ( 1)2 4 (2) (1)= 1 8= 7

= 7 1


A.P. SET CODE

2012 ___ __ 1100 - MT - y - MATHEMATICS (71) ALGEBRA - SET - C (E)

SET - C2 / MT - y

(iv) x2 + 5x 4 = 0Comparing with ax2 + bx + c = 0

a = 1, b = 5, c = 4 1

(v) Let the cost of one table be Rs. x and one chair be Rs. yAs per the given condition,

2x + 5y = 2200 1

(vi) When A die is thrownS = { 1, 2, 3, 4, 5, 6 }n (S) = 6Let A be the event of getting an odd numberA = { 1, 3, 5 }n (A) = 3

P (A) =n (A)

n (S)

P (A) = 36

P (A) = 12

1


f x

Mean x 5 1A.2. Solve ANY FIVE of the following :(i) For the G.P. 2, 6, 18, ........

The first term (a) = 2

Common ratio (r) = 6

2

= 3

Seventh term t7 = ?tn = ar

n 1 1 t7 = ar7 1 t7 = ar6 t

7= 3 (3)6

t7 = 2 (729) t7 = 1458 The common ratio of the G.P. is 3 and the seventh term is 1458. 1

SET - C3 / MT - y

(ii) The roots of the quadratic equation are 1

2 and

3

4

Let = 12

and = 34

+ = 12

+ 3

4

=

2 3

4 =

1

4

and . = 12

3

4=

3

81

We know that,x2 ( + )x + a.b = 0

x2 1

4

x +

3

8

= 0

x2 + x4

3

8= 0

Multiplying throughout by 8 we get,8x2 + 2x 3 = 0

The required quadratic equation is 8x2 + 2x 3 = 0 1

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

(iv) For the given A.P. 12, 16, 20, 24, .....Here, a = t1 = 12

d = t2 t1 = 16 12 = 4 1We know,tn = a + (n 1) d

t25 = a + (25 1) d t25 = 12 + 24 (4) t25 = 12 + 96 t25 = 108 The twenty fifth term of A.P. is 108. 1

(v) Mean = 54.6, Mode = 54 [Given]We know,Mean Mode = 3 (Mean Median) 1

54.6 54 = 3 (54.6 Median)

SET - C4 / MT - y

0.6 = 3 (54.6 Median) 0.6

3= 54.6 Median

0.2 = 54.6 Median Median = 54.6 0.2 Median = 54.4 1

(vi) y2 + 8y + 16 = 0 y2 + 4y + 4y + 16 = 0 y (y + 4) + 4 (y + 4) = 0 (y + 4) (y + 4) = 0 (y + 4)2 = 0 1

Taking square root on both the sides we get,y + 4 = 0

y = 4 y = 4 1

A.3. Solve ANY FOUR of the following :(i) x4 3x2 + 2 = 0

(x2)2 3x2 + 2 = 0Substituting x2 = m we get,

m2 3m + 2 = 0 1 m2 2m m + 2 = 0 m (m 2) 1 (m 2) = 0 (m 2) (m 1) = 0 m 2 = 0 or m 1 = 0 1 m = 2 or m = 1

Resubstituting m = x2 we get,x2 = 2 or x2 = 1

Taking square roots throughout,

x = 2 or x = +1 1

(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

31

SET - C5 / MT - y

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.

(iii) As per given condition,15 + 20 + a + a + 25 = 100

60 + 2a = 100 2a = 100 60 2a = 40 a = 20

1

2

CountryPercentage of

Measure of central angle ()world population

India 1515

100 360 = 54

China 2020

100 360 = 72

Russia 2020

100 360 = 72

USA 2020

100 360 = 72

Others 2525

100 360 = 90

Total 100 360

China

India

OtherU.S.A.

Russia 7254

9072

72

SET - C6 / MT - y

(iv) Let natural number be x

Its reciprocal is 1x

From 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

x 13

because x is natural number 1

x = 3 The natural number is 3. 1

(v) Let greater number be x and smaller number be y.As per first given condition,x + y = 60 ........(i)As per second given condition,x = 3y + 8

x 3y = 8 .......(ii) 1Subtracting (ii) from (i),x + y = 60x 3y = 8() (+) ()4y = 52

y =52

4 y = 13 1

Substituting y = 13 in (i),x + 13 = 60

x = 60 13 x = 47 The greater number is 47 and smaller number is 13. 1

A.4. Solve ANY THREE of the following :(i) Total students surveyed = 400

n (S) = 400

SET - C7 / MT - y

Let A be the event that student offered to join sciencen (A) = 187

P (A) =n (A)

n (S)

=187

4001

Let B be the event that student offered to join commercen (B) = 125

P (B) =n (B)

n (S)

P (B) = 125400

1

Here A and B are disjoint sets A B = n (A B) = 0 P (A B) =

n (A B)

n (S)

=0

400

P (A B) = 0 1A B is the event that student selected at random prefersscience or commerce facultyP (A B) = P (A) + P (B) P (A B)

=187 125

400 400 0

=312

400

P (A B) = 3950

1

(ii) No of trees planted by a student from 1st, 2nd, 3rd standards are2, 4, 8 respectivelyThe no. of trees planted by each student form a G.P with

a = 2 , r = 4

2 = 2 1

There are 5 standards. Total no. of trees planted by the student of that school is S5

Sn = na (r 1)

r 1

S5 = 5a (r 1)

r 1 1

SET - C8 / MT - y

S5 = 2 (32 1)

2 1

S5 = 2 31 1 S5 = 62

Total no of trees planted by student of school is 62 trees. 1

(iii) 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 . ( )

215 = (5)3 + 3 . (5) [ + = 5 and3 3 = 215]

215 = 125 + 15 . 1 215 125 = 15 . 90 = 15 . . = 90

15 . = 6Now, 2 = ( + )2 4. 1

(5)2 = ( + )2 4 (6) [ . = 5 and . = 6] 25 = ( + )2 24 25 + 24 = ( + )2 ( + )2 = 49

Taking square root on both the sides, we get;

+ = + 7 1 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 x2 + 7x + 6 = 0. 1

(iv) 4x = y 5 y = 2x + 1

4x + 5 = y y = 4x + 5

1

x 0 1 2 x 0 1 2

y 5 1 3 y 1 3 5

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

SET - C9 / MT - y

2


Scale : 1 cm = 1 uniton both the axesY

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

SET - C10 / MT - y


1

Here total frequency = fi = N = 80

N2

= 80

2 = 40

Cumulative frequency (less than type) which is just greater than40 is 43. Therefore corresponding class 90 - 100 is median class.L = 90, N = 80, c.f. = 23, f = 20, h = 10

Median =N h

L 2

c.f.

f 1

=80 10

90 232 20

= 90 + (40 23) 1

21

= 90 + (17)1

2= 90 + 8.5 1= 98.5

Median of I.Q.is 98.5. 1

(ii) For the G.P.Let first term be ACommon ratio be r

tn = Arn 1But, tn = a [Given]

Arn 1 = a ......(i) 1t2n = Ar

2n 1

But,t2n = b [Given]

Ar2n 1 = b ......(ii) 1t3n = Ar

3 1

Classes Frequency (fi) Cumulative frequency(I.Q.) (No. of candidates) less than type70 - 80 7 780 - 90 16 23 c.f.90 - 100 20 f 43100 - 110 17 60110 - 120 11 71120 - 130 7 78130 - 140 2 80

Total 80 N

SET - C11 / MT - y

But,t3n = c [Given]

Ar3n 1 = c .....(iii)b = Ar2n 1 [From (ii)] 1Squaring both sides,b2 = (Ar2n 1)2

b2 = A2 (r2n 1)2 b2 = A2 r4n 2 ......(iv) 1

Multiplying (i) and (iii) we get,ac = Arn 1 . Ar3n 1

ac = A2 rn 1 + 3n 1 ac = A2 r4n 2 .....(v) From (iv) and (v) we get, b2 = ac 1

(iii) Let S be the sample spaceTwo dice are thrown

S = { (1, 1), (1, 2), (1, 3) (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3) (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3) (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3) (6, 4), (6, 5), (6, 6) }

n (S) = 36 1(a) Let A be the event that sum of the points is a perfect square

A = { (1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3) }n (A) = 7

P (A) =n (A)

n (S)

P (A) = 736

1

Let B be the event that sum of the points is divisible by 4B = { (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3),

(6, 2), (6, 6) }n (B) = 9

P (B) =n (B)

n (S)

P (B) = 936

1

A B is the event that sum of scores is perfect square anddivisible by 4.

A B = { (1, 3), (2, 2), (3, 1) }

SET - C12 / MT - y

n (A B) = 3 P (A B) = 3

361

A B is the event that the number is either a perfect sqaure ordivisible by 4.P (A B) = P (A) + P (B) P (A B)

=7 9 3

36 36 36

P (A B) = 1336

1

(iv) Interest received on Rs. x = 5x

100

Interest received on Rs. y = 10y

100As per the first given condition,5x 10y

100 100 = 400

Multiplying throughout by 100 we get,5x + 10y = 40000 .....(i) 1As per second condition,5y 10x

100 100 = 350

Multiplying throughout by 1005y + 10x = 35000 ......(ii) 110x + 5y = 35000Multiplying (ii) by 2,20x + 10y = 70000 .....(iii)Substracting (iii) from (i), 5x + 10y = 4000020x + 10y = 70000() () ()

15x = 30000

x = 3000015

x = 2000 1Substituting x = 2000 in (i),5 (2000) + 10y = 40000

10000 + 10y = 40000 10000 + 10y = 40000 10y = 40000 10000

y = 3000010

1

y = 3000

SET - C13 / MT - y

x + y = 2000 + 3000 x + y = 5000 Total amount deposited is Rs. 5000. 1

(v) 1

4

Electricity billClass mark No. of families

in a month in Rs.

200 - 400 300 362

400 - 600 500 490

600 - 800 700 185

800 - 1000 900 63

No

. o

f fa

mil

ies

0

50

100

150

200

250

300

350

400

450

500

200 400 600 800 1000

Classes (Electricity bill in a month in Rs.)X

Y

Scale : On X axis : 1 cm = 100 unitsOn Y axis : 1 cm = 50 families

X1200

Y


(i) Write the first five terms of the following Arithmetic Progression where,the common difference d and the first term a are given :a = 2, d = 2.5

(ii) For the sequence, find the next four terms 2, 4, 8, 16, .....

(iii) Find the value of discriminant of the following equation :x2 + x + 1 = 0

(iv) Is the given equation a quadratic equation :11 = 4x2 x3

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

(vi) A box containing 3 red, 3 white, and 3 green balls, A ball is selected atrandom. Find the probability that ball picked up is a red ball.

(vii) Calculate mean x when fixi = 100 and fi = 20

Note :




Seat No.

Q.P. SET CODE

2012 ___ ___ 1100 - MT - z - MATHEMATICS (71) ALGEBRA - SET - D (E)

SET - D2 / MT - z


(i) Find the sum of first 11 positive numbers which are multiples of 6.

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

(iii) 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.

(iv) Find the eighteenth term of the A. P. : 1, 7, 13, 19, .....

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

(vi) Solve the following equation by factorization method :21x = 196 x2.


(i) Solve the following equation : 35y2 + 212

y = 44

(ii) Without actually solving the simultaneous equations given below, decide

whether simultaneous equations have unique solution, no solution or

infinitely many solutions 8y = x 10; 2x = 3y + 7.

(iii) Draw the histogram and use it to find the mode for the following frequencydistribution.

House - Rent 4000 - 6000 - 8000 - 10000 -(in ` per month) 6000 8000 10000 12000

Number of families 200 240 300 50

SET - D3 / MT - z

(iv) Solve the quadratic equation by factorization method m2 84 = 0.

(v) The perimeter of an isosceles triangle is 24 cm. The length of its congruentsides is 13 cm less than twice the length of its base. Find the lengths ofall sides of the triangle.


(i) Two dice are thrown. Find the probability of the events.(a) The product of numbers on their upper faces is 12.(b) The sum of the numbers on their faces is multiple of 7.

(ii) If the arithmetic mean and the geometric mean of two numbers are in theratio 5 : 4 and the sum of the two numbers is 30 then find these numbers.

(iii) If the difference of the roots of a quadratic equation is 4 and the differenceof their cubes is 208, find the quadratic equation.

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

x + 2y = 5; y = 2x 2


(i) Forty persons were examined for their Hemoglobin % in blood (in mg per100 ml) and the results were grouped as below :

Determine modal value of Hemoglobin % in blood of a person.

(ii) Sachin, Sehwag and Dhoni together scored 228 runs. Their individualscores are in G.P. Sehwag and Dhoni together scored 12 runs more thanSachin. Find their Individual scores.

Hemoglobin % 13.1 - 14.1- 15.1- 16.1- 17.1-(mg/100 ml) 14 15 16 17 18

No. of persons 8 12 10 6 4

SET - D4 / MT - z

Best Of Luck

(iii) A card is drawn at random from a well shuffled pack of cards. Find theprobability that the card drawn is :(a) a diamond card or a king(b) a spade or a face card.

(iv) A bus covers a certain distance with uniform speed. If the speed of the buswould have been increased by 15 km/h, it would have taken two hoursless to cover the same distance and if the speed of the bus would havebeen decreased by 5 km/h, it would have taken one hour more to coverthe same distance. Find the distance covered by the bus.

(v) Draw both the ogive curves on the same graph paper for the followingfrequency distribution. Also find the median :

Intelligence 60 - 80 - 100 - 120 - 140 - Totalquotient (IQ) 79 99 119 139 159

Number of students 2 9 15 11 3 40

A.1. Attempt ANY SIX of the following :(i) a = 2, d = 2.5

Here, t1 = a = 2t2 = t1 + d = 2 + 2.5 = 4.5t3 = t2 + d = 4.5 + 2.5 = 7t4 = t3 + d = 7 + 2.5 = 9.5t5 = t4 + d = 9.5 + 2.5 = 12

The first five terms of the A.P. are 2, 4.5, 7, 9.5 and 12. 1

(ii) t1 = 21 = 2

t2 = 22 = 4

t3 = 23 = 8

t4 = 24 = 16

t5 = 25 = 32

t6 = 26 = 64

t7 = 27 = 128

t8 = 28 = 256

The next four terms of the sequence are 32, 64, 128 and 256. 1

(iii) x2 + x + 1 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = 1, c = 1 = b2 4ac

= (1)2 4 (1) (1)= 1 4= 3

= 3 1

(iv) 11 = 4x2 x3

x3 + 4x2 + 11 = 0Here maximum index of the variable x is 3.So it is not a quadratic equation. 1


A.P. SET CODE

2012 ___ __ 1100 - MT - z - MATHEMATICS (71) ALGEBRA - SET - D (E)

SET - D2 / MT - z

(v) 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

(vi) Let the three red balls be R1, R2 and R3, the three green balls be G1,G2 and G3 and the three white balls be W1, W2 and W3 respectively.

S = { R1, R

2, R

3, G

1, G

2, G

3, W

1, W

2, W

3 }

n (S) = 9A is event that ball picked up is red ballA = { R1, R2, R3 }n (A) = 3

P (A) =n (A)

n (S)

=3

9

P (A) =1

31


f x

Mean x 5 1A.2. Solve ANY FIVE of the following :(i) The positive integers which are divisible by 6 are 6, 12, 18,

24, ........The number form an A.P. with a = 6, d = 6.The sum of first 11 positive integers divisible by 6 is (S11)

Sn =n

2[2a + (n 1) d]

S11 =11

2[2a + (11 1) d]

=11

2 [2 (6) + 10 (6)] 1

=11

2 [12 + 60]

SET - D3 / MT - z

=11

2 72

= 11 36Sn = 396

Sum of first 11 positive integers which are divisible by 6 is 396. 1

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

+ = 3 + ( 11) = 3 11 = 14and . = 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 X-axis is y = 0Let 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 1 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

(iv) For the given A.P. 1, 7, 13, 19, .....Here, a = t1 = 1 d = t2 t1 = 7 1 = 6We know,tn = a + (n 1) d 1

t18 = a + (18 1) d t18 = 1 + 17 (6) t18 = 1 + 102 t18 = 103

Eighteenth term of A.P. is 103. 1

(v) Mean = 101, Median = 100 [Given]We know,Mean Mode = 3 (Mean Median) 1

101 Mode = 3 (101 100) 101 Mode = 3 (1) 101 3 = Mode Mode = 98 1

SET - D4 / MT - z

(vi) 21x = 196 x2

x2 + 21x 196 = 0 x2 28x 7x 196 = 0 x (x + 28) 7 (x + 28) = 0 (x + 28) (x 7) = 0 1 x + 28 = 0 or x 7 = 0 x = 28 or x = 7 1

A.3. Solve ANY FOUR of the following :

(i) 35y2 + 12

y2 = 44

Multiplying throughout by y2 we get,35y4 + 12 = 44y2

35 (y2)2 44y2 + 12 = 0Substituting y2 = m we get,

35m2 44m + 12 = 0 35m2 14m 30m + 12 = 0 7m (5m 2) 6 (5m 2) = 0 (5m 2) (7m 6) = 0 1 5m 2 = 0 or 7m 6 = 0 5m = 2 or 7m = 6

m = 25

or m = 6

71

Resubstituting m = y2, we get;

y = 2

5

or y = 6

7

y = 25

or y = 6

71

(ii) 8y = x 10 x + 8y = 10

Comparing with a1x + b1y = c1 we get, a1 = 1, b1 = 8, c1 = 102x = 3y + 7

2x 3y = 7Comparing with a2x + b2y = c2 we get, a2 = 2, b2 = 3, c2 = 7

a

a1

2 =

1

21

b

b1

2 =

8

3 = 8

3

12

cc =

10

71

SET - D5 / MT - z

12

aa

1

2

bb

The simultaneous equations 8y = x 10 and 2x = 3y + 7 1have unique solution.

(iii) 2

Mode of house rent paid is ` 8400. 1

No

. o

f fa

mil

es

Y

0 X

20

40

60

80

100

120

140

160

180

200

220

240

4000 6000 8000 10000 12000

House rent per month (in Rs.)

X

260

280

300

Mode

Scale : On X axis : 1 cm = Rs. 1000On Y axis : 1 cm = 20 familes

SET - D6 / MT - z

(iv) m2 84 = 0

(m)2 84 2 = 0 1 (m)2 4 21 2 = 0 (m2 2 21 2 = 0 m 2 21 m 2 21 = 0 m 2 21 = 0 or m 2 21 = 0 1

m = 2 21 or m = 2 21 1

(v) Let length of one of the congruent sides of an isosceles trianglebe x cm and length of its base be y cm.As per the first given condition,x + x + y = 242x + y = 24 .......(i)As per the second given condition,x = 2y 13

x 2y = 13 .......(ii) 1Multiplying (i) by 2,4x + 2y = 48 ......(iii)Adding (ii) and (iii), x 2y = 134x + 2y = 485x = 35

x = 355

x = 7 1Substituting x = 7 in (i),2 (7) + y = 24

14 + y = 24 y = 24 14 y = 10 The length of sides of the isosceles triangle are 7 cm, 7 cm 1

and 10 cm.

A.4. Solve ANY THREE of the following :(i) Two dice are thrown

S = { (1, 1), (1, 2), (1, 3) (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3) (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3) (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3) (6, 4), (6, 5), (6, 6) }

n (S) = 36 1

SET - D7 / MT - z

(a) Let A be the event that product of numbers on their upper faces is 12A = { (2, 6), (3, 4), (4, 3), (6, 2) }n (A) = 4

P (A) =n (A)

n (S)

P (A) = 436

P (A) = 19

1

(b) Let B be the event that sum of the numbers on their upper facesis multiple of 7.B = { (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) }n (B) = 6

P (B) =n (B)

n (S)

P (B) = 636

P (B) = 16

1

(ii) Let the two numbers be x and yLet Arithmetic mean of x and y be denoted as A and geometricmean of x and y be G.

A = x y2

......(i)

G = + xy .....(ii) x + y = 30 .....(iii) [Given]

Substituting (iii) in (i),

A =30

2 A = 15

A

G=

5

4[Given]

15G

=5

4

4155

= G G = 12 xy = 12

Squaring both sides,xy = 144 1

y = 144x

......(iv) 1

SET - D8 / MT - z

Substituting (iv) in (iii), we get,

x + 144

x= 30

Multiplying throughout by x we get,x2 + 144 = 30x

x2 30x + 144 = 0 x2 24x 6x + 144 = 0 x (x 24) 6 (x 24) = 0 (x 24) (x 6) = 0 x 24 = 0 or x 6 = 0 x = 24 or x = 6 1

If x = 24 If x = 6

y = 144

xy =

144

x

y = 14424

y = 144

6 y = 6 y = 24 The two numbers are 6 and 24. 1

(iii) Let and be the roots of a quadratic equation = 4 and 3 3 = 208 [Given]

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

Also, 3 3 = ( )3 + 3. (.) 208 = (4)3 + 3. (4) [.=4and33= 208] 208 = 64 + 12 . 208 64 = 12 . 1 144 = 12 .

14412

= . . = 12 1

Now, ( )2 = ( + )2 4. (4)2 = ( + )2 4 (12) [= 4 and . = 12] 16 = ( + )2 48 16 + 48 = ( + )2 ( + )2 = 64 1

Taking square root on both the sides we get, + = + 8

x2 ( + )x + . = 0 [From (i)] x2 (8)x + 12 = 0 or x2 ( 8)x + 12 = 0 x2 8x + 12 = 0 or x2 + 8x + 12 = 0 The required quadratic equation is x2 8x + 12 = 0 or 1

x2 + 8x + 12 = 0.

SET - D9 / MT - z

(iv) x + 2y = 5 y = 2x 2

x = 5 2y

1

2

x = 3 and y = 4 is the solution of given simultaneous equations. 1

x 5 3 1 x 0 1 2

y 0 1 2 y 2 4 6

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

Y


Y

4

5

3

2

1

41 2 3 5X

-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

SET - D10 / MT - z


1

Here the maximum frequency fm = 12

The corresponding class 14.05 - 15.05 is the modal class.

L = 14.05, fm = 12, f1 = 8, f2 = 10, h = 1

Mode =

L + h2

f f

f f fm 1

m 1 21

=12 8

14.05 + 12 (12) 8 10

=4

14.05 +24 18

1

=4

14.05 +6

= 14.05 + 2

31

= 14.05 + 0.67= 14.72

Mode of hemoglobin is 14.72 mg/100 ml. 1

(ii) Let individual scores of Sachin, Sehwag and Dhoni be a

r, a and ar

As per the first given condition,a

r + a + ar = 228 .....(i)

As per second given condition,

a + ar = 12 + a

r ......(ii) 1

Substituting (ii) in (i),a

r + 12 +

a

r = 228

12 + 2ar

= 228

Hemoglobin %Continuous class No. of persons(mg / 100ml)

13.1 - 14 13.05 - 14.05 8 f114.1 - 15 14.05 - 15.05 12 fm15.1 - 16 15.05 - 16.05 10 f216.1 - 17 16.05 - 17.05 617.1 - 18 17.05 - 18.05 4

SET - D11 / MT - z

2ar

= 228 12

2ar

= 216

ar

= 216

2

ar

= 108 .....(iii) 1

a = 108rSubstituting (iii) in (i),

108r

r + 108r + (108 r)r = 228

a108 , a 108r

r

Multiplying throughout by r,108 + 108r + 108r2 = 228

108r2 + 108r + 108 228 = 0 108r2 + 108r 120 = 0

Dividing throughout by 12 we get,9r2 + 9r 10 = 0

9r2 + 15r 6r 10 = 0 3r (3r + 5) 2 (3r + 5) = 0 (3r + 5) (3r + 2) = 0 3r + 5 = 0 or 3r 2 = 0 3r = 5 or 3r = 2 1

r = 53

or r = 2

3

If r =53

If r =2

3a = 108r a = 108r

a = 108 5

3a = 108

2

3a = 36 5 a = 36 2a = 180 a = 72

ar = 72 2

3ar = 48

r = 5

3 is not acceptable because a (individual score) cannot be 1

negative.

ar = 48 The runs scored by Sachin, Sehwag and Dhoni are 108 runs, 1

72 runs and 48 runs.

SET - D12 / MT - z

(iii) There are 52 cards in a pack of cards n (S) = 52(a) Let A be the event that card drawn is a diamond card There are 13 diamond cards in a pack of cards

n (A) = 13

P (A) =n (A)

n (S)

P (A) = 1352

(b) Let B be the event that card drawn is a king There are 4 king cards in a pack of cards

n (B) = 4

P (B) =n (B)

n (S)

P (B) = 452

A B is the event that card drawn is diamond and king card There is one king card in diamonds n (A B) = 1

P (A B) =n (A B)

n (S)

P (A B) = 152

A B is the event that card drawn is diamond or a king cardP (A B) = P (A) + P (B) P (A B)

=13 4 1

52 52 52

=16

52

P (A B) = 413

1

(c) Let C be the event that the card drawn is a spade card There are 13 spade cards in a pack

n (C) = 13

P (C) =n (C)

n (S)

P (C) = 1352

Let D be event that the card drawn is a face card There are 3 face cards in each of the 4 type of cards Total no. of face cards in a pack are 4 3 = 12 n (D) = 12

SET - D13 / MT - z

P (D) =n (D)

n (S)

P (D) = 1252

C D is the event that the card drawn is spade and face card There are 3 face cards in spades n (C D) = 3 P (C D) =

n (C D)

n (S)

P (C D) = 352

C D is the event that the card drawn is a spade or face cardP (C D) = P (C) + P (D) P (C D)

=13 12 3

52 52 52

= 2252

P (C D) = 1126

1

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

Distance = xy km 1According 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)

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 = 15x = 45 1Substituting x = 45 in (ii),

45 5y = 5 5y = 5 45 5y = 40

y = 40

5 y = 8 1

SET - D

Distance = xy= 45 8= 360

Distance covered by bus is 360 km. 1

(v) 2

2

Median of IQ is 112.5 units. 1

Intelligence Continuous Frequency Cumulative Upper Cumulative LowerQuotient Classes No. of stud.frequency bou. frequency boun.

less than more than

39.5 - 59.5 0 0 59.5 39.560 - 79 59.5 - 79.5 2 2 79.5 40 59.580 - 99 79.5 - 99.5 9 11 99.5 38 79.5

100- 119 99.5 -119.5 15 26 119.5 29 99.5120- 139 119.5- 139.5 11 37 139.5 14 119.5140- 159 139.5- 159.5 3 40 159.5 3 139.5

159.5 -179.5 0 179.5 0 159.5

Cu

mu

lati

ve

fre

qu

en

cy

(N

o.

of

stu

de

nts

)

Y

5

10

15

20

25

30

35

40

45Scale : On X axis : 1 cm = 20 units

On Y axis : 1 cm = 5 students

Class boundariesX

YX0 39.5 59.5 79.5 99.5 119.5 139.5 159.5 179.5

(159.5, 0)

(139.5, 3)

(119.5, 14)

(119.5, 26)

(139.5, 37)

(159.5, 40)(59.5, 40)

(79.5, 38)

(99.5, 29)

(99.5, 11)

(79.5, 2)(59.5, 0)

Median

14 / MT - z


(i) For the sequence, find the next four terms :1, 3, 7, 15, 31, ....

(ii) Determine the nature of the roots of the following equation from itsdiscriminant : y2 4y 1 = 0.

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

simultaneous equations in x and y, find x.

(iv) In the following experiment write the sample space S, number ofsample points n (S), events P, Q, R using set and n (P), n (Q) andn (R). A die is thrown :P is the event of getting an odd number.Q is the event of getting an even number.R is the event of getting a prime number.

(v) In each of the following experiments, write the sample space S,number of sample point n (S), event A, B and n (A), n (B). Two coinsare tossed, A is the event of getting at most one head, B is the eventgetting both heads.

(vi) Is the following equation quadratic ?x2 + 4x = 11.

(vii) Express the following information in mathematical form using twovariables : The perimeter of a rectangle is 36 cm.

Note :(i) All questions are compulsory.(ii) Use of calculator is not allowed.


MT - MATHEMATICS (71) ALGEBRA - PRELIM II - PAPER - 1 (E)

Seat No.2012 ___ ___ 1100

PAPER - 12 / MT

Q.2. Attempt ANY FIVE of the following : 10

(i) For an A. P. if t4 = 12, and d = 10, then find its general term.

(ii) From 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, then

find a.

(iv) In the following experiment write the sample space S, number of

sample points n (S), events P, Q, R using set and n (P), n (Q) and

n (R). Find the events among the events defined above which are :

complementary events, mutually exclusive events and exhaustive

events.

There are 3 red, 3 white and 3 green balls in a bag. One ball is

drawn at random from a bag :

P is the event that ball is red.

Q is the event that ball is not green.

R is the event that ball is red or white.

(v) Without actually solving the simultaneous equations given below,decide whether simultaneous equations have unique solution, no

solution or infinitely many solutions. x 2y

= 13

, 2x 4y = 9

2

(vi) If P (A) = 3

4, P (B) = 1

3 and P (A B) = 1

2 then find P (A B).

Q.3. Attempt ANY FOUR of the following : 12

(i) How many three digit natural numbers are divisible by 4 ?

(ii) Solve the following quadratic equations by completing square :

m2 3m 1 = 0.

(iii) Solve the following simultaneous equations using Cramers rule :

3x + 2y + 11 = 0; 7x 4y = 9.

PAPER - 13 / MT

(iv) Following table gives frequency distribution of electricity consumptionof a household in certain area in a month :

No. of units of electricity 0 20 20 40 40 60 60 80 80 100

No. of households 4 16 41 65 8

Find modal no. of units of electricity consumed by a household in amonth.

(v) The following diagram represents thesectorwise loan amount in crores of Rs.distributed by a bank. From the informationanswer the following questions :(a) If the dairy sector received Rs. 20crores, then find the total loan disbursed.(b) Find the loan amount for agriculturesector and also for industrial sector.(c) How much additional amount didindustrial sector received than agriculturesector.

Q.4. Attempt ANY THREE of the following : 12

(i) Babubhai borrows Rs. 4000 and agrees to repay with a total interest

of Rs. 500. in 10 instalments, each instalment being less that the

preceding instalment by Rs. 10. What should be the first and the

last instalment?

(ii) Solve the following equation : (y2 + 5y) (y2 + 5y 2) 24 = 0

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

3x + 4y + 5 = 0; y = x + 4

(iv) Two dice are thrown find the probability of getting :

(a) The sum of the numbers on their upper faces is divisible by 9.

(b) The sum of the numbers on their upper faces is at most 3.

(c) The number on the upper face of the first die is less than the

number on the upper face of the second die.

Agriculture

DairyIndustry

12040

PAPER - 14 / MT

Best Of Luck

Q.5. Attempt ANY FOUR of the following : 20

(i) Find three consecutive terms in a G.P. such that the sum of the2nd and 3rd term is 60 and the product of all the three is 8000.

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

(iii) Sharad bought a table and a fan together for Rs. 5000. After sometimehe sold the table at the gain of 25% and the fan at a gain of 20%.Thus he gained 23% on the whole. Find the cost of the fan.

(iv) Following table gives frequency distribution of trees planted bydifferent housing societies in a particular locality.

No. of trees 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40

No. of students 2 7 9 8 6 4

Find the mean number of trees planted by housing society by usingstep deviation method.

(v) Following is a frequency distribution of marks :

Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Total

No. of students 2 a 56 b 2 100

If a and b are equal find their values. Draw less than cumulativefrequency curve and more than frequency curve on the same graphpaper and find the median.

A.1. Attempt ANY SIX of the following :(i) t1 = 0 + 2

0 = 0 + 1 = 1t2 = 1 + 2

1 = 1 + 2 = 3t3 = 3 + 2

2 = 3 + 4 = 7t4 = 7 + 2

3 = 7 + 8 = 15t5 = 15 + 2

4 = 15 + 16 = 31t6 = 31 + 2

5 = 31 + 32 = 63t7 = 63 + 2

6 = 63 + 64 = 127t8 = 127 + 2

7 = 127 + 128 = 255t9 = 255 + 2

8 = 255 + 256 = 511

The next four terms of the sequence are 63, 127, 255 and 511. 1

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

= (-4)2 4 (1) (-1)= 16 + 4= 20

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

(iii) Dx = 18 and D = 3By Cramers rule,

x =D

Dx

x = 183

x = 6 1

(iv) When a die is thrownS = { 1, 2, 3, 4, 5, 6 }n (S) = 6P is the event of getting an odd number

Time : 2 Hours Prelim II Model Answer Paper Max. Marks : 60

2012 ___ ___ 1100 MT - MATHEMATICS (71) ALGEBRA - PAPER - 1 (E)Seat No.

PAPER - 12 / MT

P = { 1, 3, 5 } n (P) = 3

Q is the event of getting an even numberQ = { 2, 4, 6 }

n (Q) = 3R is the event of getting a prime numberR = { 2, 3, 5 }

n (R) = 3 1

(v) Two coins are tossed S = { HH, HT, TH, TT } n (S) = 4

A is the event of getting at the most one head A = { HT, TH, TT } n (A) = 3

B is the event of getting both heads. B = { HH } n (B) = 1 1

(vi) x2 + 4x = 11 x3 + 4x 11 = 0

Here a = 1, b = 4, c = 11 are real numbers.Where a 0So it is a quadratic equation in variable x. 1

(vii) Let the length of rectangle be x cm and its breadth be y cm.Perimeter of the rectangle = 2 (l + b) = 2 (x + y)As per the given condition,2 (x + y) = 36

2x + 2y = 36 1

A.2. Attempt ANY FIVE of the following :(i) Given : For an A.P. t4 = 12, d = 10

Find : General term { tn }Sol. tn = a + (n 1) d

t4 = a + (4 1) d 12= a + 3 ( 10) a = 12 + 30 a = 42 1

PAPER - 13 / MT

tn = a + (n 1) d tn = 42 + (n 1) ( 10) tn = 42 10n + 10 tn = 52 10n The general term of A.P. is 52 10n. 1

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

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

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

15 + 2a = 19 1 2a = 19 15 2a = 4

a = 42

a = 2 1

(iv) Let 3 red balls, 3 white balls and 3 green balls be denoted as

R1, R2, R3, W1, W2, W3 and G1, G2, G3 respectively.

P is the event that the ball is red

P = { R1, R2, R3 }

n (P) = 3 1Q is the event that the ball is not green

Q = { R1, R2, R3, W1, W2, W3 }

n (Q) = 6R is the event that the ball is red or white

R = { R1, R2, R3, W1, W2, W3 }

n (R) = 6 1

PAPER - 14 / MT

(v)x 2y

= 13

x 2y = 3Comparing with a1x + b1y = c1 we get, a1 = 1, b1 = 2, c1 = 3

2x 4y =9

2 2 (2x 4y) = 9 4x 8y = 9

Comparing with a2x + b2y = c2 we get, a2 = 4, b2 = 8, c2 = 9 1

a

a1

2=

1

4

b

b1

2=

2

8 =1

4

c

c1

2=

3

9=

1

3

a

a1

2 =

b

b1

2

c

c1

2

The simultaneous equations x 2y = 13

and 1

2x 4y = 9

2 have no solution.

(vi) We have P (A) = 3

4, P (B) = 1

3 and P (A B) = 1

2 P (B) = 1 P (B)

= 1 1

3

=3 1

3

P (B) = 23

1

By addition theorem,P (A B) = P (A) + P (B) P (A B)

=3 2 1

4 3 2

=9 8 6

12 12 12

P (A B) = 1112

1

PAPER - 15 / MT

A.3. Attempt ANY FOUR of the following :(i) The three digit natural numbers that are divisible by 4 are as

follows 100, 104, 108, ........ 996.These numbers form an A.P. with a = t1 = 100, 1d = t2 t1 = 104 100 = 4.Let, tn = 996We know that for an A.P.tn = a + (n 1) d 1

996 = 100 + (n 1) 4 996 = 100 + 4n 4 996 = 96 + 4n 4n = 996 96 4n = 900 n = 900

4 n = 225 There are 225 three digit natural numbers that are divisible by 4. 1

(ii) 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

3

m 2

2

=4 9

4

3

m 2

2

=13

41

Taking square root on both the sides we get,

3m 2

=13

2

m = 3 132 2

PAPER - 16 / MT

m = 3 132

m = 3 132

or m =

3 13

2

3 132

and

3 13

2 are the roots of the given quadratic 1

equations.

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

7x 4y = 9

D =3 2

7 4 = (3 4) (2 7) = 12 14 = 36

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 Cramers 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

(iv) No. of units of electricity No. of household

0 - 20 420 - 40 1640 - 60 47 f160 - 80 65 fm80 - 10 8 f2Here the maximum frequency fm = 65.The corresponding class 60 - 80 is the modal class.L = 60, fm = 65, f1 = 47, f2 = 8, h = 20 1

Mode =

L + h2

f f

f f fm 1

m 1 2

=65 47

60 + 202 (65) 47 8

PAPER - 17 / MT

=18

60 + 20130 55

=18

60 + 2075

1

=18

60 + 415

=6

60 + 45

= 60 + 4.8= 64.8

Mode of electricity units is 64.8 units. 1

(v) (a) Let the total loan disbursed be x croresThe measure of central angle for dairy sector is 40.

Dairy sector received 20 crores of the total loan i.e. x

40 x360

= 20

x = 20 36040

x = 180 1 Total loan disbursed is Rs. 180 crores.(b) Measure of central angle for agriculture sector is 120

Amount disbursed for agriculture sector = 120 180360

= 60 croresMeasure of central angle for industrial sector is 200

Amount disbursed for industrial sector = 200 180360

= 100 crores. 1(c) Amount received by industrial sector = Rs. 100 crores.

Amount received by agricultural sector = Rs. 60 crores. Industrial sector received 100 60 = Rs. 40 crores more than

agricultural sector 1

A.4. Attempt ANY THREE of 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]

PAPER - 18 / MT

S10 =10

2 [2a + (10 1) d]

4500 = 5 [2a + 9 ( 10)] 1

45005

= 2a 90

900 = 2a 90 900 + 90 = 2a 990 = 2a 990

2= a

a = 495tn = a + (n 1) d 1

t10 = a + (10 1) d t10 = 495 + 9 ( 10) t10 = 495 90 t10 = 405 First instalment is Rs. 495 and last instalment is Rs.405. 1

(ii) (y2 + 5y) (y2 + 5y 2) 24 = 0Substiuting y2 + 5y = m we get,

m (m 2) 24 = 0 m2 2m 24 = 0 m2 6m + 4m 24 = 0 m (m 6) + 4 (m 6) = 0 (m 6) (m + 4) = 0 1 m 6 = 0 or m + 4 = 0 m = 6 or m = 4

Resubstituting m = y2 + 5y we get,y2 + 5y = 6 ......(i) or y2 + 5y = 4 .....(ii)

From (i), y2 + 5y = 6 y2 + 5y 6 = 0 y2 y + 6y 6 = 0 y (y 1) + 6 (y 1) = 0 (y 1) (y + 6) = 0 1 y 1 = 0 or y + 6 = 0 y = 1 or y = 6

From (ii), y2 + 5y = 4 y2 + 5y + 4 = 0 y2 + 4y + y + 4 = 0 y (y + 4) + 1 (y + 4) = 0 (y + 4) (y + 1) = 0 y + 4 = 0 or y + 1 = 0 1 y = 4 or y = 1 y = 1 or y = 6 or y = 4 or y = 1. 1

PAPER - 19 / MT

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

x = 5 4y3

x 3 1 5 x 0 1 2 1

y 1 2 5 y 4 5 6

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

2

Y


Y

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

-1

-2

-3

6

4

5

3

2

1

-5

-4

(5,

5)

(1,

2)

(3, 1)

(0, 4)

(1, 5)

y =

x +

4

(2, 6)

3x + 4y + 5 = 0

0

PAPER - 110 / MT

(iv) (a) Two dice are thrown S = { (1, 1), (1, 2), (1, 3) (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3) (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3) (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3) (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3) (6, 4), (6, 5), (6, 6) }

n (S) = 36 1(i) Let A be the event that sum of numbers on their upper faces is

divisible by 9A = { (3, 6), (4, 5), (5, 4), (6, 3) }n (A) = 4

P (A) =n (A)

n (S)

P (A) = 436

P (A) = 19

1

(b) Let B be the event that sum of number on their upper faces is atthe most 3.B = { (1, 1), (1, 2), (2, 1) }n (B)= 3

P (B) =n (B)

n (S)

P (B) = 336

P (B) = 112

1

(c) Let C be the event that number on the upper face of the firstdie is less than the number on the upper face of second die.C = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) }n (C)= 15

P (C) =n (C)

n (S)

P (C) = 1536

P (C) = 512

1

PAPER - 111 / MT

A.5. Attempt ANY FOUR of the following :

(i) Let three consecutive terms of G.P. be a

r, a, ar.

As per first condition,a + ar = 60 ......(i) 1As per second condition,

a

r a ar = 8000

a3 = 8000 1Taking cube roots on both side,

a = 80003

a = 20 20 20 3 a = 20 1

Substituting a = 20 in (i),20 + 20r = 60

20r = 60 20 20r = 60 20 20r = 40

r = 4020

r = 2 1

ar

= 20

2= 10

ar = 20 2 . ar = 40 The three consecutive terms of the G.P. are 10, 20, 40. 1

(ii) 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 1

x2 + x2 + 2x + 1 + x2 + 4x + 4 + x2 +6x +9 + x2 + 8x + 16 1455 = 0

5x2 + 20x + 30 1455 = 0 5x2 + 20x 1425 = 0 1

Dividing throughout by 5 we get,x2 + 4x 285 = 0

x2 15x + 19x 285 = 0 x (x 15) + 19 (x 15) = 0 (x 15) (x + 19) = 0 1

PAPER - 112 / MT

x 15 = 0 or x + 19 = 0 x = 15 or x = 19 x is a natural number

x 19 1Hence 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 15, 16, 17, 118 and 19 respectively.

(iii) Let the cost of table be Rs. x

Let the cost of fan be Rs. y

As per the first given condition,

x + y = 5000 ......(i) 1

Profit earnt on table =25

x100

Profit earnt on fan =20

y100

Total Profit earnt =23

5000100

1

As per the second condition,

25x 20y

100 100 =

115000

100Multiplying throughout by 100,

25x + 20y = 115000

Dividing throughout by 5,

5x + 4y = 23000 ......(ii) 1Multiplying (i) by 4,

4x + 4y = 20000 ......(iii)

Subtracting (iii) from (ii),

5x + 4y = 23000

4x + 4y = 20000

x = 3000 1Substituting x = 3000 in (i),

PAPER - 113 / MT

3000 + y = 5000 y = 2000 Cost of fan is Rs. 2000. 1

(iv) Class width (h) = 5, Assumed mean (M = 22.5

No. of trees Class Mark di = xi A ui = id

hNo. of fiui

(xi) students(fi)

10 - 15 12.5 10 2 2 415 - 20 17.5 5 1 7 7 220 - 25 22.5 A 0 0 9 025 - 30 27.5 5 1 8 830 - 35 32.5 10 2 6 1235 - 40 37.5 15 3 4 12

Total 36 21

u =i i

i

f u

f

u =21

36 u = 0.583 1

Mean x = A hu= 22.5 + 5 (0.583)= 22.5 + 2.92 1= 25.42

Mean of trees planted by societies 25.42 trees. 1

(v) As per the condition given to the table we have :2 + a + 56 + b + 2 = 100

60 + a + b = 100 a + b = 100 60 a + b = 40 .....(i)

If a = b thena + a = 40

2a = 40

a = 402

a = 20 a = b = 20 1

PAPER - 114 / MT

Marks Frequencies Cumulative Upper Cumulative Lowerfrequency boun- frequency bound-less than daries more than ariestype type

20 - 0 0 0 00 - 20 2 2 20 100 020 - 40 20 22 40 98 2040 - 60 56 78 60 78 40 160 - 80 20 98 80 22 6080 - 100 2 100 100 2 80100 - 120 0 0 100

Median of marks obtained is 47.3 marks. 1

2

Class boundaries (Marks)

Y

50

60

70

80

90

Cu

mu

lati

ve

fre

qu

en

cy

(N

o.

of

stu

de

nts

)

Scale : On X = axis : 1 cm = 10 marksOn Y = axis : 1 cm = 10 students

100

110

Y

20

30

40

10

X X0 40 60 802 0 100 120

Median= 47.3

Cumulative frequencycurve less than type.


(i) Is the following list of numbers an Arithmetic Progression ? Justify.22, 26, 28, 31, ...

(ii) Write the following quadratic equation in standard form ax2 + bx + c = 0.7 4x x2 = 0.

(iii) Find the value of the following determinant : 5 2

7 4

(iv) The probability that at least one of the events A and B occurs is 0.6. IfA and B occur simultaneously with probably 0.2, evaluate P (A) + P (B).

(

algebra prelim q

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