T15 PA003

Set A

This Paper consists of 5 printed pages Turn over© Copyright reserved.

SECTION – A (40 Marks)

Attempt all questions from this Section.Question 1

(a) Factorise the expressionf(x) = 2x3 – 7x2 + 14x – 8.Hence, find all possible values of x for which f(x) = 0. [3]

(b) A circle with diameter 20 cm is drawn somewhere on a rectangularpiece of paper with length 40 cm and width 30 cm. A die very small insize, is dropped on the rectangular paper without seeing towards it. Ifthe die falls and lands on the paper only, find the probability that it willfall and land :(i) inside the circle.(ii) outside the circle. [3]

(c) A man invests ` 20,020 in buying shares of nominal value ` 26 at 10%premium. The dividend on the shares is 15% per annum.Calculate :(i) the number of shares he buys.(ii) the dividend he receives annually.(iii) the rate of interest he gets on his money. [4]

MATHEMATICS

(Two hours and a half )

Answers to this Paper must be written on the paper provided separately.

You will not be allowed to write during the first 15 minutes.

This time is to be spent in reading the question paper.

The time given at the head of this paper is the time allowed for writing the answers.

Attempt all questions from Section A and any four questions from Section B.

All working, including rough work, must be clearly shown and must be done on the

same sheet as the rest of the answer.

Omission of essential working will result in the loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ] .

Mathematical tables are provided.

MAHESH TUTORIALS I.C.S.E.GRADE - X (2015-2016)

Exam No. : MT/ICSE/PRELIM I - SET-A 003

T15 PA003

Set A... 2 ...

Turn over

... 2 ...

Turn over

Question 2(a) In the given figure, ABC is a triangle with

EDB = ACB. Prove that ABC ~ EBD.If BE = 6 cm, EC = 4 cm, BD = 5 cm andarea of BED = 9 cm2. Calculate the :(i) length of AB(ii) area of ABC [3]

(b) The interest charged on a certain sum is ` 720 for one year and` 1,497.60 for two years. Find, whether the interest is simple or compound. Also, calculate the rate percent and the sum. [3]

(c) Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ABC = 60º.Locate by construction the point P such that :(i) P is equidistant from B and C.(ii) P is equidistant from AB and BC.(iii) Measure and record the length of AB. [4]

Question 3(a) A shopkeeper marks his goods 40% above the cost price and then allows

two successive discounts of 10% each. Find how much will a customerpay for an article which costs the shopkeeper ` 700 and a Sales Tax of8% is levied on the sale price of the article. (Give your answer correct tothe nearest rupee) [3]

(b) If A =

–1 1

a b and A2 = I; find a and b. [3]

(c) For triangle ABC, show that :

(i) sinA + B

2 = cos

C2

(ii) tanB + C

2 = cot

A2

[4]

Question 4(a) A point A is at a distance of 10 units from the point (4, 3). Find the

co-ordinates of point A, if its ordinate is twice its abscissa. [3]

(b) Construct a regular hexagon, having one side = 3.0 cm. Draw all its linesof symmetry. [3]

D

B

A

CE

T15 PA003

Set A... 3 ...

(c) If x =2aba + b , find the value of :

x + a x + b+x – a x – b [4]

SECTION – B (40 Marks)Attempt any four questions from this Section.

Question 5(a) AB is a diameter of the circle with centre C.

APBR is as shown in the figure.APQ and RBQ are straight lines. Find :(i) PRB(ii) PBR(iii)BPR [3]

(b) How many balls each of radius 1 cm can be made by melting a bigger ballwhose diameter is 8 cm. [3]

(c) An area is paved with square tiles of a certain size and the numberrequired is 128. If the tiles had been 2 cm smaller each way, 200 tileswould have been needed to pave the same area. Find the size of thelarger tiles. [4]

Question 6(a) Determine the rate of interest per annum for a sum that becomes

729625

of itself in one year, compounded half-yearly. [3]

(b) Find the height of a tree when it is found that on walking away from it 20m, in a horizontal line through its base, the elevation of its top changesfrom 60o to 30o. [3]

(c) Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm andABC = 120º.(i) Construct a circle circumscribing the triangle ABC.(ii) Draw a cyclic quadrilateral ABCD. So that D is equidistant from

B and C. [4]

Question 7(a) Solve and graph the solution set of:

–223 x +

13

< 313

; x R. [3]

Turn overTurn over

A

B

CP

Q35o

25o

R

T15 PA003

Set A... 4 ...

Turn overTurn over

(b) In the given fig. PQ and PR are tangentsto the circle, with centre O.If QPR = 60o Calculate :(i) QOR (ii) OQR (iii) QSR [3]

(c) A page from the passbook of Asha is given below :

Calculate the interest for the period January to December at 5% perannum. [4]

Question 8(a) The line y = 3x – 2 bisects the join of (a, 3) and (2, –5), find the value of a. [3]

(b) Draw histogram to represent the following data: [3]

(c) The given figure shows two circles; one withdiameter AB and other with diameter OB. IfO is the center of the circle, AB isperpendicular to CD and OC = 28 cm; find :(i) area of the shaded portion(ii) area of the unshaded portion. [4]

Question 9(a) Vivek invests ` 4,500 in 8%, ` 10 shares at ` 15. He sells the shares

when the price rises to ` 30, and invests the proceeds in 12% ` 100shares at ` 125. Calculate :(i) the sale proceeds(ii) the number of ` 125 shares he buys.(iii) the change in his annual income from dividend. [4]

P

Q

R

OS 60o

Date Particulars Withdrawals Deposits Balance (in `) (in `) (in `)

Jan. 5 Balance B/F 3,750.00March 7 To Cheque 1,200.00 2,550.00April 2 By Cheque 2,300.00 4,850.00April 10 By Cheque 820.00 5,670.00Oct. 6 To Cheque 950.00 4,720.00Dec. 8 By Cash 1,700.00 6,420.00

Class mark 16 24 32 40 48 56 64

Frequency 8 12 15 18 25 19 10

A

D

C

OB

T15 PA003

Set A... 5 ...

(b) If quadratic equation x2 – (m + 1) x + 6 = 0, has one root as x = 3, find thevalue of m and the other root of the equation. [3]

(c) The weight of 50 apples were recorded as given below. Calculatethe mean weight, to the nearest gram by the Step Deviation Method. [3]

Question 10(a) Mrs. Geeta deposited ` 350 per month in a bank for 1 year and 3 months

under the Recurring Deposit Scheme. If the maturity value of herdeposits is ` 5,565; find the rate of interest per annum. [3]

(b) The line 5x + 4y + 20 = 0 meets the x-axis at point A and the y-axis atpoint B. Find :(i) the co-ordinates of A and B(ii) the length of segment AB(iii) the slope of AB and slope of perpendicular to AB; by using the

ordinates of A and B only. [3]

(c) Using properties of proportion, solve for x :+1 + –1+1 – –1

x xx x

=4 –1

2x

[4]

Question 11(a) Chords AB and CD of a circle when extended meet at point X. Given

AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD. [4]

(b) The daily wages of 80 workers in a building project are given below:

Wages in ` 30–40 40–50 50–60 60–70 70–80 80–90 90–100 100–110

No. of 6 10 15 19 12 8 6 4 workers

Using graph paper, draw an ogive for the above distribution.Use your ogive to estimate :(i) the median wages of the workers(ii) the percentage of workers who earn more than ` 75 day.(iii) the upper quartile wage of the workers(iv) the lower quartile wage of the workers.(v) Inter quartile range. [6]

All the Best

Weight in grams 80–85 85–90 90–95 95–100 100–105 105–110 110–115

No. of apples 5 8 10 12 8 4 3

MAHESH TUTORIALS I.C.S.E.

Model Answer Paper

Marks : 80

Time : 2½ hrs.

SUBJECT : MATHEMATICSICSE X

Exam No. : MT/ICSE/PA003

T15 I PA003 Turn over

SECTION – A (40 Marks)

Attempt all questions from this Section.A.1(a) Let f(x) = 2x3 – 7x2 – 3x + 18

For x = 2 the value of given expression,f(2) = 2(2)3 – 7(2)2 – 3(2) + 18

= 16 – 28 – 6 + 18= 0

x – 2 is a factor of 2x3 – 7x2 + 14x – 8Divide 2x3 – 7x2 + 14x – 8 by x – 2

2

3 2

3

2

2

2 32 2 7 3 18

2 4

3 3 183 6

9 18

9

( ) (+)

(+) ( )

xx x

x x xx x

x xx x

x9 18

(+) ( ) 0

x

2x3 – 7x2 – 3x + 18 = (x – 2) (2x2 – 3x – 9)= (x – 2) [2x2 – 6x + 3x – 9]= (x – 2) [2x (x – 3) + 3 (x – 3)]= (x – 2) (x – 3)(2x + 3)

All the possible factors of 2x3 – 7x2 – 3x + 18 are x – 2, x – 3 & 2x + 3 [3]

(b)

Area of circle = r2

=22 ×(10)×(10)7

20 cm

40 cm

30 cm

Turn overT15 PA003

Set A... 2 ...... 2 ...

= 22200 cm7

and area of rectangle = 40 × 30.= 1200 cm2

(i) Probability that the die falls and lands inside the circle

Favourable outcome 2200

7and n(S) = 1200

P (die inside the circle) =2200 =

7×12001142

(ii) Probability that the die falls and lands outside the circle P (die outside the circle) = 1 – P (inside the circle)

=111–42

=3142 [3]

(c) Sum Invested = ` 20,020Nominal Value = Face Value = ` 26Premium = 10%

Market Value = 26 +10100

× 26 = ` 28.60

Rate of Dividend = 15%

(i) No. of shares =Sum Invested

Market Value of 1 share

=2002028.60

No. of shares = 700(ii) Dividend = No. of shares × Rate of Dividend × Face Value

= 700 ×15100

× 26 = ` 2730

Dividend = ` 2730

(iii) Rate of interest = R%Rate of interest × Market Value = Rate of Dividend × Face ValueRate of interest × 28.60 = 15 × 26

Rate of interest =15 2628.60

Rate of interest = 13.64% [4]

Turn overT15 PA003

Set A... 3 ...... 3 ...

A.2(a) I n ABC and EBD,

BD = 5 cm, BE = 6 cm, EC = 4 cmBC = BE + EC = 6 + 4 = 10 cm

Now, ACB = EDB (Given)ABC = DBE (Common angle)

ABC ~ EBD (AA postulate)

(i) ABBE =

BCBD (Corresponding sides of similar

triangles are proportional)

AB6 =

105

AB =10×6

5 AB = 12 cm

(ii) Since, ABC ~ EBD

Area of ABCArea of EBD

2

2

BCBD

[Ratio of areas of similar triangles is equal to squares of corresponding sides]

2

2

105

10025

Area of ABC = 4 (Area of EBD) = 4(9) = 36 cm2

Area of ABC = 36 cm2 [3]

(b) Interest charged on a certain sum is ` 720 for 1 yearInterest for 2 years I = 1,497.60Interest for 2nd year I2 = 1497.60 – 720

I2 = ` 777.60 I2 – I1 = 57.60 Given interest is compound interest. [

R =100 × IP × T

=100 × 57.60

720 × 1= 8

R = 8%

I =PRT 100

P =C.I.×100

R × T

D

B

A

CE

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Set A... 4 ...... 4 ...

=720 ×100

8 × 1= ` 9000

Interest given is compound interestR = 8 %P = ` 9000. [3]

(c) [4]

PB = 4.6 cm

A.3(a) Cost Price of shopkeeper = ` 700

shopkeeper marks his goods 40% “above” the cost price

Marked Price = Cost Price +40

100 of cost price

Marked Price = 700 +40

100 × 700

Marked Price = ` 980Two successive Discounts are given of 10% each :Discount 1 = 10% of (D1), Discount 2 = 10% of (D2)

Sales Price =

1100 - D100

2100 - D100 × Marked Price

Sales Price =

100 - 10100

100 - 10100 × 980

=90

100 ×

90100

× 980

Sales Price = ` 793.80Rate of sales Tax = 8%

Total Amount = Sales Price + R% Sales Price

BA

P

C

M

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Set A... 5 ...

= 793.80 +8

100 × 793.80

= ` 857.304

Customer will pay ` 857 for the goods [to the nearest rupee] [3]

(b) A =–1 1

a b

A2 = A.A

=–1 1

a b

–1 1

a b

=

–1 × –1 + 1 × a –1 × 1 + 1 × b

a× –1 + b × a a × 1 + b × b

=

2

1 + a –1 + b

–a + ab a + b

Since A2 = I (Given)

2

1 + a –1 + b

–a + ab a + b

=

1 0

0 1

1 + a = 1 a = 1 – 1 = 0

– 1 + b = 0 b = 1 a = 0; b = 1 [3]

(c) (i) In ABC,A + B + C = 180º

A + B = 180º – CDividing by 2,

A + B2

=0180

2 –

C2

A + B

2 = 90º –

C2

sin

A + B2 = sin

0 C90 -

2

sin

A + B2 = cos

C2

Turn overT15 PA003

Set A... 6 ...

(ii) In ABC,A + B + C = 180º

B + C = 180º – ADividing by 2,

B + C2

=0180

2 –

A2

B + C

2= 90º –

A2

tan

B + C2 = tan

0 A90 -

2

tan

B + C2 = cot

A2 [4]

A.4(a) Let, (4 , 3) (x1 , y1)

Let, the x co-ordinate of point A = x2

Its y co-ordinate = y2 = 2x2

Distance between the point (4, 3) and point A

= 2 22 1 2 1+ x x y y

10 = 2 22 24 2 3 x x

10 = (x2 – 4)2 + (2x2 – 3)2 (Squaring both sides) 10 = (x2

2 – 8x2+ 16) + (4x22 – 12x2 + 9)

10 = x22 – 8x2 + 16 + 4x2

2 – 12x2 + 9 10 = 5x2

2 – 20x2 + 25 5x2

2 – 20x2 + 15 = 0 x2

2 – 4x2 + 3 = 0 x2

2 – 3x2 – x2 + 3 = 0 [Dividing throughout by 5] x2(x2 – 3)– 1 (x2 – 3) = 0 (x2 – 3) (x2 – 1) = 0 x2 – 3 = 0 or x2 – 1 = 0 x2 = 3 or x2 = 1

y2 = 2x2

y2 = 6 or 2

Point A is (3 , 6) or (1 , 2) [3]

Turn overT15 PA003

Set A... 7 ...... 7 ...

(b)

From the figure PQ, RS, XY AD, BE and CF are the lines of symmetry [3]

(c) x =2aba b x =

2aba b

xa =

2ba b

xb =

2a

a bBy componendo dividendo By componendo dividendo

x ax a

=22b a bb a b

........(i)x bx b

=22a a ba a b

.........(ii)

x ax a

+x bx b

=22b a bb a b

+22a a ba a b

[Adding (i) and (ii)]

=3b ab a

+3a ba b

=–(3 )

b aa b +

3a ba b

=3 3b a a b

a b

=

2 – 2a ba b

=

2 –a ba b

= 2 [4]

Q

P

E D

X

F

R

BA

S

Y

OC

Turn overT15 PA003

Set A... 8 ...... 8 ...

SECTION – B (40 Marks)Attempt any four questions from this Section.

A.5(a) PAB and PRB are the angles in

the same segment hence they are equal. PAB = PRB PRB = 35o

Since AB is a diameter, angle in a semi-circleis a right angle.

APB = 90o

Since APQ is a straight line BPQ = 180º – 90º

= 90o

Also RBQ is a straight line PBR = 90º + 25ºPBR = 115º

In PBR, BPR = 180º – (35º + 115º) [Sum of all angles of a ]

= 180º – 150ºBPR= 30o [3]

(b) Let radius of big ball be ‘R’ and of each smaller ball be ‘r’.

The volume of the bigger sphere =43R3

and volume of the smaller sphere =43r3

Number of balls =Volume of bigger sphereVolume of smaller sphere

Number of balls =

3

3

4R

34

r3

=

3

3

821

=34

1= 64

No of balls that can be formed by melting the bigger ball are 64 [3]

Turn overT15 PA003

Set A... 9 ...... 9 ...

(c) Let the side of larger tile be x cm Area of 1 larger tile = x2 sq.cm Area covered by 128 larger tiles = 128x2 sq.cm

Side of smaller tile will be (x – 2) cm Area of 1 smaller tile = (x – 2)2 sq. cm Area covered by 200 smaller tiles = 200 (x – 2)2 sq. cm

As per given condition,128x2 = 200(x – 2)2

128x2 = 200(x2 – 4x + 4) 128x2 = 200x2 – 800x + 800 0 = 200x2 – 128x2 – 800x + 800 72x2 – 800x + 800 = 0

Dividing throughout by 8, we get,9x2 – 100x + 100 = 0

9x2 – 90x – 10x + 100 = 0 9x(x – 10) – 10(x – 10) = 0 (x – 10) (9x – 10) = 0

x – 10 = 0 or 9x – 10 = 0

x = 10 or x =109

If x =109

Then x – 2 = –10

29

=10 – 18

9=

– 89

cm which is not acceptable

x 109

Side of larger tile is 10 cm [4]

A.6(a) Let principal = ` P

625P

Time = 1 year n = 2 ( Interest is compound half–yearly)

A = P

R1+100

n

729625

P = P

2R

1+100

729625

=

2R

1+100

22725 =

2R

1+100

729Amount = `

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Set A... 10 ...... 10 ...

2725

= 1+R

100

108 = 100 + r

108 – 100 = R R = 8% half-yearly

Rate of interest = 16% p.a. [3]

(b) AB represents the height of the treeD represents a point 20 m away from the tree.ACB and ADB are angles of elevationACB = 600 ; ADB = 300

In ADB,

tan 60o =ABx

3 =ABx

AB = 3 x ...........(i)BD = BC + CD

BD = (x + 20) mIn ADB,

tan 30o =ABBD

13 =

AB( + 20)x

AB = + 20

3x

...........(ii)

3 x = + 20

3x

[from (i)]

3x = x + 20

3x – x = 20

2x = 20

x = 10Height of tree = AB

= 3 x= 3 (10)= 1.732 × 10

AB = 17.32 m

Height of the tree is 17.32 m [3]

20 m CD

A

Bx60o30o

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Set A... 11 ...

(c)

[4]

A.7

(a)–83

3x +13

<103

Multiplying throughout by 3,– 8 3x + 1 10– 8 3x + 1 and 3x + 1 < 10– 8 – 1 3x and 3x < 10 – 1– 9 3x and 3x < 10– 3 x and x < 3– 3 x < 3

Solution set = { x : –3 x < 3, x R }

[3]

D

A

O

P

C

D

B

120º

3 4–4 –2–3 10 2–1–5

Turn overT15 PA003

Set A... 12 ...

(b) As P is the point in the exteriorof the circle PQ and PR are thetangents at Q and R respectively

Also OQ and OR are the radiihence OQ QP and QR RP.

Thus in quadrilateral OQPR,QOR = 360º – (90º + 90º + 60º)QOR = 360º – 240º

QOR = 120o

Draw QRIn OQR,

OQ = OR [Radii of the same circle] OQR = ORQ [Base angles]

OQR =12

(180º – 120º)

OQR = 30o

Angle at the centre is double the angle at the circumference QOR = 2 × QSR 120º = 2 × QSR

QSR =120º

2QSR = 60o [3]

(c)

Total Principal = ` 59,930 P = ` 59,930

R = ` 5%

T =1

12 years

Interest (I) =P × R × T

100

Month Minimum BalanceJanuary ` 3,750February ` 3,750March ` 2,550April ` 5,670May ` 5,670June ` 5,670July ` 5,670August ` 5,670September ` 5,670October ` 4,720November ` 4,720December ` 6,420

Turn overT15 PA003

Set A... 13 ...

=59,930 5 1

100 12

= 249. 708Interest (I) = ` 249.71 [4]

A.8(a) The equation of the line is y = 3x – 2

The given line bisects the join of (a, 3) and (2, –5) The co-ordinates of the mid-point of the given segment will satisfy

the equation of the line.Now,Mid-point of segment joining (a, 3) and (2, –5)

= + 2 3 – 5

,2 2

a

= + 2

, –12

a

Substituting x = + 22

a and y = –1 in the given equation, we get :

–1 = 3 + 22

a

– 2

–1 =3 + 6

2a

– 2

–2 = 3a + 6 – 4 –2 = 3a + 2 –4 = 3a

a =– 43 [3]

(b) Difference between consecutive class marks is 8.

Lower boundary of each class = Class mark –82

= Class mark – 4

and upper boundary of each class = Class mark +82

= Class mark + 4

Turn overT15 PA003

Set A... 14 ...

The frequency distribution will be :

The required histogram is shown in the figure below.

[3]

(c) (i) Area of the shaded region = Area of the circle with OB as diameter+ Area of the semicircle with CD asdiameter – Area of the ADC

Class interval Frequency

12 – 20 8

20 – 28 12

28 – 36 15

36 – 44 18

44 – 52 25

52 – 60 19

60 – 68 10

12 20

26

28

30Y

0 28 X52 60 644436

22

24

20

16

18

14

12

10

8

6

4

2

Class Interval

Freq

uenc

y

Scale :1 cm = 8 units on x-axis1 cm = 2 units on y-axis

Turn overT15 PA003

Set A... 15...

= r2 +12R2 –

12

× b × h

= 2OB

2

+12 (OC)2 –

12

× CD × OA

=22 14 147

+1 22 28 282 7

–1 56 282

= 22 × 28 + 22 × 56 – 56 × 14= 22 × 28 + 56 (22 – 14)= 22 × 28 + 56 (8)= 22 × 28 + 28 × 2 × 8= 28 × (22 + 16)= 28 × 38= 1064 sq. cm.

Area of the shaded region is 1064 sq. cm

(ii) Area of the unshaded = Area of the circle with OA as radiusportion – Area of the shaded portion

= (28)2 – 1064

=227

× 28 × 28 – 1064

= 2464 – 1064= 1400

Area of the unshaded portion is 1400 sq. cm [4]

A.9(a) Sum Invested = ` 4500

Rate of Dividend = 8%Face Value = ` 10Market Value = ` 15Firm 1 :

No. of shares =Sum Invested

Market Value of 1 share

=450015

No. of shares = 300Dividend or Annual income in firm 1= No. of shares × Rate of Dividend × Face Value

= 300 ×8

100 × 10

= ` 240Dividend in firm 1 = ` 240Vivek sells these shares when the price rises to ` 30

Sale Proceeds = 300 × 30

Turn overT15 PA003

Set A... 16 ...

(i) Sale Proceeds = ` 9000Sale Proceeds becomes sum invested for 2nd firm

Firm 2 :Sum Invested = ` 9000Rate of Dividend = 12%Face Value = ` 100Market Value = ` 125

No. of shares =Sum Invested

Market Value of 1 share

=9000125

(ii) No. of shares = 72

Dividend in firm 2 = 72 ×12

100 × 100

= ` 864 the change in his annual income = 864 – 240

Change in his annual income = ` 624 [4]

(b) x2 – (m + 1) x + 6 = 0 x = 3 is one root of the equation (3)2 – (m + 1) 3 + 6 = 0 9 – 3m – 3 + 6 = 0 12 – 3m = 0 3m = 12 m = 4Substituting value of m in the equation, x2 – (4 + 1)x + 6 = 0 x2 – 5x + 6 = 0 x2 – 2x – 3x + 6 = 0 x(x – 2) – 3(x – 2) = 0 (x – 2) (x – 3) = 0 x = 2 or x = 3 The other root is x = 2 [3]

(c) Weight

80 – 8585 – 9090 – 9595 – 100100 – 105105 – 115110 – 115

( f )

581012843

f = 50

f × t

– 15– 16– 10

0889

ft =–16

Mid - valuex

82.587.592.597.5102.5107.5117.5

Let A = 97.5 d = x – 4

– 15– 10– 5051015

t =– Axi

– 3– 2– 10123

Turn overT15 PA003

Set A... 17 ...

Mean = A +ftf

× i = 97.5 +1650

× 5 = 95.9

Mean = 96 g (correct to the nearest gram) [3]

A.10(a) Sum deposited per month (P) = ` 350

No. of months (n) = 15 monthsMaturity value = ` 5565rate of interest (r) = ?

Maturity value = (P × n) + P ×( +1)

2 ×12n n

× 100r

Maturity value = (350 × 15) +350 15 (15 1)

2 12

×

100r

5565 = 5250 + 350 ×

15(16)2 12

×100

r

5565 – 5250 = 35r 315 = 35r

r =31535

r = 9 % Rate of interest = 9% [3]

(b) (i) The equation of line is :5x + 4y + 20 = 0The line meets x-axis at point A

y = 0Substituting y = 0 in equation 5x + 4y + 20 = 0

5x + 4(0) + 20 = 0 5x + 20 = 0 5x = –20

x =–205

x = –4 Point A (–4, 0)

The line 5x + 4y + 20 = 0 meets the y-axis at point B x = 0

Substituting x = 0 in equation 5x + 4y + 20 = 0 5(0) + 4y + 20 = 0 4y = –20

y =–204

y = –5 Point B (0, –5)

Turn overT15 PA003

Set A... 18 ...

(ii) Length of segment AB = 2 22 1 2 1( – ) + ( – )x x y y

= 2 2[0 – (–4)] + (–5 – 0)

= 2 2(4) + (–5)

= 16 + 25

= 41= 6.40 units

(iii) Slope of AB =2 1

2 1

––

y yx x

=–5 – 00 – (–4)

=–54

Slope of perpendicular to AB =–1

Slope of AB

=–1–54

=45

[3]

(c)x + + x -x + - x -

1 11 1

=x4 -12

By componendo and dividendo,

1 1 1 1

1 1 1 1

x x x x

x x x x

=4 1 24 1 2

xx

2 1

2 1

x

x

=4 14 3

xx

11

xx

=

2

2

4 1

4 3

x

x

[Squaring both sides]

11

xx

=2

216 8 1

16 24 9x x

x x

By componendo - dividendo

1 11 1

x xx x

=2 2

2 216 8 1 16 24 916 8 1 16 24 9

x x x xx x x x

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Set A... 19 ...

22x

=232 16 1032 8

x xx

x =

22 16 8 5

2 16 4

x x

x

x (16x – 4) = 16x2 – 8x + 5 16x2 – 4x = 16x2 – 8x + 5 8x – 4x = 5 4x = 5

x =54 [4]

A.11(a) AB = 4 cm

BX = 6 cmDX = 5 cm

Let,CD = x cm

AX = AB + BX [ A – B – X]AX = 4 + 6AX = 10 cm

andCX = CD + DX [ C – D – X]CX = (x + 5) cm

Now,AX × BX = CX × DX [When two chords of a circle intersect

internally/externally, then the products of the lengths of segment are equal.]

10 × 6 = (x + 5) 560 = 5x + 25

60 – 25 = 5x5x = 35

x =355

x = 7 CD = 7 cm [4]

B

DA

X

C

4 cm

x

6 cm

5 cm

Turn overT15 PA003

Set A... 20 ...

(b)

Plot the point (40, 6),(50, 16), (60, 31), (70,50), (80, 62), (90, 70),(100, 76), (110, 80) bytaking class intervalsalong x-axis and c.f.along y-axis. Scaleused is 1 cm = 1 uniton both axes. Join thepoint (40, 6) to 30 onx-axis.Join these points byfree hand curvewhich is the requiredogive.

(i) To find median N = 80N2

=802

= 40

Median is the observation corresponding to c.f. = 40 Mark it by M onthe graph.

Median = 65

(ii) On x-axis point P at 75 corresponds to point Q on y-axis at 57. No. of workers who earn more than ` 75 are = 80 – 57 = 23

% workers =2380

× 100

=115

4 = 28.75%

Class Interval No. of workers f c.f.30 – 40 6 640 – 50 10 1650 – 60 15 3160 – 70 19 5070 – 80 12 6280 – 90 8 7090 – 100 6 76100 – 110 4 80

Total = 80

100

90

90

80

80

70

70

60

60

50

50

40

40

30

30

20

10

0L P N

Class Interval

c.f

3N/4

Q

N/2

Turn overT15 PA003

Set A

(iii) To find upper quartile3N4

=3 × 80

4= 60

observation corresponding to c.f. = 60 on the graph is marked by N Upper quartile = 78

(iv) To find lower quartileN4

=804

= 20 Observation corresponding to c.f. = 20 on the graph is marked y L

and it is 53.50 Lower quartile = 53.50

Inter quartile range = 78 – 53.50

Inter quartile range = 24.50 [6]

... 21 ...