## ICSE Class 10 MATHS Question Paper 2016

### You can use these as maths mock test for class 10 ICSE also

## ICSE Class 10 Maths Question Paper 2016 Solved

**MATHEMATICS**

*(Two hours and a half)*

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

*You will not he 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 loss of marks.*

*The intended marks for questions or parts of questions are given in brackets [ ]. Mathematical tables are provided.*

(c) *The mean of following numbers is 68. Find the value of ‘x’ 45, 52, 60, x, 69, 70,26,81 and 94.*

*Hence estimate the median.*

Answer 1.

(a) Here, f(x) = 2jc^{3} + ?>x^{2} – kx + 5 and x-2 = 0 =s> x = 2 Given, remainder is 7 => f (2) = 7 Putting x – 2 in equation (1), we get

/ (2)

16+ 12-2A + 5

*2k*

SECTION-A (40 Marks)

*(Attempt all questions from this Section)*

Question 1.

(a) *Using remainder theorem, find the value ofk if on dividing 2x ^{3}* +

*3x*+

^{2}-kx*5 by x-2 leaves a remainder*7.

‘ 2 O’ | ‘ 7 O’ | ||||

and I = | |||||

-17 | 0 1 |

(b) *Given A =*

*and A ^{2} = 9A* +

*ml. Find m.*

*k ~*

- [3]

…(1)

2(2)^{3} + 3(2)^{2} – k (2) + 5 7

26

13

Ans.

(b) Here, A =

=> A^{2} = A ■ A =

4 + 0 | 0 + 0′ | ‘ 4 | O’ | ||

– 2-7 | 0 + 49 | -9 | 49 |

*2 0 1 7*

1 | 0 ‘ | 4 | O’ | |

0 | 1 | -9 | 49 | |

m | 0 ‘ | ‘ 4 | O’ | |

0 | m | -9 | 49 |

‘ 2 | O’ | |

-1 | + m | |

7 |

18 0 -9 63

18 + m 0 | ‘ 4 -9 | |

— 9 63 + m | 0 49 |

On comparing both sides, we get

18 + m = 4 and 63 + m = 49 which gives m = – 14 Ans.

2* 45 + 52 + 60 + x + 69 + 70 + 26 + 81 + 94

(c)

Arithmetic mean = — =

*n*

497+x

68 =

=> 612 = 497 +x

=> x = 612-497

=> *x = 115*

On arranging the given terms in ascending order of magnitude, we get

26,45,52,60,69,70,81,94,115 Since number of terms («) = 9 (odd)

. w j- /« + l\^{th} /9+l\^{th}

**• ■ Median = ^ 2**~)** ^{term} = ( T y ^{term}**

= 5 ^{th} term – 69

Ans.

[3] [4]Question 2.

*The slope of a line joining P (6, k) and Q (1- 3k,*3) is i . Find*(0 k**Midpoint of PQ, using the value of ‘k’ found in (i)**Without using trigonometrical tables, evaluate : cosec*– /2^{2}57°-tan^{2}33° + cos 44 ° cosec 46°*cos 45° – tan*^{2}60 °*A certain number of metallic cones, each of radius 2 cm and height 3 cm are*

*melted and recast into a solid sphere of radius 6 cm. Find the number of cones.* [3]

Answer 2.

- (i) Let (6, k) = {x\, and (1 – 3k, 3) = (x
_{2}, y_{2})

1

Given, slope of a line PQ = ^

*1*

*2*

*1*

*2*

Ans.

Ans.

**(‘**

-5-3 k

=*■ 6-2 k = -5-3 k

=*■ k = – 11 (ii) Mid point of P (6, – 11) and Q (34, 3)

<6 + 34 -11+3-

-) = (20,-4)

*y2-y\*

*X2-X1*

3*-k*

- cosec
^{2}57° – tan^{2}33° + cos 44° cosec 46° —V2 cos 45° – tan^{2}60° => sec^{2}(90° – 57°) – tan^{2}33° + cos 44°’ ^ — V~2 • (V^)

= ^ jt x (2)^{2} x 3 ™ 4jt cm^{3}

=> sec^{2} 33° – tan^{2} 33° +”^4^) -1-3 cos 44°

=* ^{1} + cos 44° ” ^{1} “ ^{3}

=> 1 + 1-1-3 = -2

- Volume of metallic cone = ^ nr
^{2}h

1

cosec 0 = sec (90° – 9)] [ ‘sin 0 = cos (90° – 0)]
[V sec^{2} 0 – tan^{2} 0 = 1] Ans.

^{Kr3}

4

ji x (6)^{3} 288 jt

Volume of solid sphere Volume of metallic cone 288 jt

Volume of solid sphere =

Number of cones =

= 72

Ans.

4jt

Question 3.

*Solve the following inequation, write the solution set and represent it on the*

*number line.*

*In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and*Z*CBD*=*32 °. Find :*[4J

(i) | Z OBD |

(ii) | Z AOB |

(Hi) | Z BED |

*If (3a*+*2b) : (5a*+*3b)*=*18 : 29. Find a : b.*

Answer 3.

- Given, – 3 (jr – 7) a 15 – Ix > — 3 ‘

=> – 3 (jc- 7) a 15 – Ix and 15-7* >’^{V+}‘

3

- 3* + 21 a 15 – Ix and 45-21x>.r+l

Ix-‘ix^ 15-21 and —21jc — jc> 1 — 45

4x^-6 and – 22x > – 44

_{J}

x ;> – 2 and x <2

. . . 3

On simplifying, the given inequation reduces to <; * < 2 and the required number line is

^ r^{–}^ 1 ►

-2—1 0 1 2 3

- (i) Since, AD is parallel to BC.

ZODB = ZCBD (alternate interior angles)

Also, OB = OD (radii)

■■■ ZOBD = ZODB = 32° Ans.

- Now, AD is a diameter.

So, ZABD = 90°

ZABO = ZABD – ZOBD ZABO = 90° – 32° = 58°

In A AOB

ZOAB + ZAOB + ZOBA = 180°

58° +ZAOB +58° = 180°

(Y OA = OB and Z OAB = ZOBA) ZAOB = 180°-116° = 64° Ans.

- ZBED = ZBAD = 58° (angles in same segment) Ans.

, s „ 3a + 2b 18

^{Here}> 5_{a +}– 29

87a + 58 b = 90a + 54 b

- 90a + 87a = – 5Sb + 542?

-3a = -4b a _ 4 b ~ 3

i.e., a: b = 4:3 Ans.

Question 4.

*A game of numbers has cards marked with 11, 12, 13,*,*40. A card is*

*drawn at random. Find the probability that the number on the card drawn*

*A perfect square**Divisible by 7**Use graph paper for this question.*[4]

*(Take 2 cm = I unit along both X and Y axis.)*

*Plot the points O (0, 0), A (- 4, 4), B (-3, 0) and C (0,-3)*

*Reflect points A and B on the*Y*axis and name them A’ and B*‘*respectively. Write down their coordinates.**Name the figure OABCB A*‘.*State the line of symmetry of this figure.*

*(cyMr. Lalit invested* ^ *5000 at a certain rate of interest, compounded annually*

*for two years. At the end of first year it amounts to X 5325. Calculate,* [3]

*The rate of interest.**The amount at the end of second year, to the nearest rupee.*

Answer 4.

- When a card is drawn at random, the possible outcomes are 11, 12, 13 ,40

Clearly, total number of all possible outcomes = 30

- For getting a perfect square :

The favourable outcomes are : 16,25,36 No. of favourable outcomes = 3

- 1

Required probability ^{=} 30 ^{=} jq Ans.

- For getting a number divisible by 7 :

The favourable outcomes are : 14,21, 28, 35 No. of favourable outcomes = 4

- 2

Required probability = Ans.

- (i) Coordinates of A’ = (4,4)

Coordinates of B’ = (3,0)

- Hexagon
- The line of symmetry of this figure is Y-axis. Ans.
- (i) Given, P = ^ 5,000, n = 1 year and A = X 5325

Interest = ^ (5325 – 5000) = ? 325 „ I x 100 325 x 100 „ _

^{K} “ P xn ~ 5000 x 1 ” ^{6-5 /o}

i.e., rate of interest is 6-5%. Ans.

(ii) Amount = P ^ 1 +

13 \^{2} 213 213

X

= 5000 (l + 200) = ^{5000 x}

‘ 200 ^{A} 200 = ? 5671-125

i.e., amount at the end of second year is ? 5671-125. Ans.

SECTION—B (40 Marks)

*(Attempt any four questions from this Section)*

Question 5.

*Solve the quadratic equation*x^{2}*— 3(x + 3) = 0; Give your answer correct to*

*two significant figures.* [3]

*A page from the savings bank account of Mrs. Ravi is given below.*[4]

Date | Particulars | Withdrawal
a) |
Deposit
a) |
Balance
(X) |

April 3rd, 2006 | B/F | 6000 | ||

April 7th | By cash | 2300 | 8300 | |

April 15th | By cheque | 3500 | 11800 | |

May 20th | To self | 4200 | 7600 | |

June 10th | By cash | 5800 | 13400 | |

June 15th | To self | 3100 | 10300 | |

August 13th | By cheque | 1000 | 11300 | |

August 25th | To self | 7400 | 3900 | |

September 6th 2006 | By cash | 2000 | 5900 |

*She closed the account on 30th September, 2006. Calculate the interest Mrs. Ravi earned at the end of 30th September, 2006 at 4.5% per annum interest. Hence, find the amount she receives on closing the account.*

*(cyin what time will* ^ *1500 yield* ^ *1996.50 as compound interest at 10% per annum compounded annually ?* C .”\_-

Answer 5.

- Given, *
^{2}– 3 (x + 3) = 0

x^{1} -3x-9 = 0 Here, a = 1, b = – 3 and c = – 9

b^{2} 4ac = (- 3)^{2} – 4 x 1 x (- 9) = 9 + 36 = 45

*b±*V*b*-4^{2}*ac*

x ——

3 ± V45 2 x 1 3 ± 6-708

x =

*la*

**2**

- + 6-708 , 3 – 6-708

and

~ 2

= 4-854 and – 1-854 i.e„ 4-85 and – 1-85 Ans.

- Principal for the month of April = ? 8,300

Principal for the month of May = ? 7,600

Principal for the month of June = ? 10,300

Principal for the month of July = ? 10,300

Principal for the month of August = ? 2,900

Principal for the month of September = f 5,900

Total principal for 1 month = ? 45,300

Now, P = ? 45,300, T = years and R = 4-5%

. _{WT}, PxRxT , 45300×4-5 x 1 _

** ^{lnter}est (D = -joo = **f

**100 >TT2**

^{i69}‘^{875}Amount = P + I = ? (45300+169-875) = ? 45469-875 Ans.

- Given, P = ? 1500, A = ^ 1996-50 and r = 10%

1996-50 = 1500 1996-50

1500

*i.e„ n i =* 3

.\ Required time = 3 years Ans.

Question 6.

*Construct a regular hexagon of side 5 cm. Hence construct all its lines of*

*symmetry and name them*. [3]

*In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet*

*at T.* [4]

*Prove A TPS ~ A TRQ.**Find SP if TP = 18 cm, RQ*=*4 cm and TR*=*6 cm.*

*Given matrix A*=

*Find area of quadrilateral PQRS if area of A PTS* = 27 *cm ^{2}.*

4 sin 30° | cos 0° ‘ | 4 | ||||

cos 0° | 4 sin 30° | and B ~ | c | |||

J |

*If AX* = *B*

*Write the order of matrix X.**Find the matrix X’.*Answer 6.

Steps of construction :

- Draw a line AB = 5 cm.
- From A and B, draw an angle of 120°.

*‘2n –* 4>

- From A and B, draw a line of length 5 cm.

Thus, we get the regular hexagon of side 5 cm.

A regular hexagon (6 sides) has 6 lines of symmetry.

- (i) In A TPS and A TRQ

ZSTP = ZQTR (common)

zlTPS = ZTRQ

( . Exterior angle of cyclic quadrilateral = Interior opposite angle) A TPS ~ A TRQ [By AA similarity]

- Since A TPS ~ A TRQ

‘2×6-4

**)■**

90°= 120°

TR TP “

_6_

18 “

SP =

RQ

SP

4

SP

4x 18

= 12

Ans.

- We know that the ratio between the areas of two similar triangles = ratio between the squares of its corresponding sides.

Ar(APTS) (SP)^{2} (12)^{2}_9

” 1

Ar (A RTQ)

Ar (A PTS)

Ar (A PTS)-Ar (A RTQ) Ar (A PTS)

Ar (quadrilateral PQRS)

Ar (quadrilateral PQRS)

(QR)^{2} (4)^{2}

9-1

9

8

8 x 27

= 24 cm^{2}

Ans.

(c) (i) Given, A =

Ans.

4 sin 30° cos 0° | 4 | |

cos 0° 4 sin 30° | and B = | 5 |

Let the order of matrix X be a x b.

1 | ||

4 x _{2} 1 |
4 | |

, „ 1 | ‘ X_{ax} b ~ |
5 |

^{1 4} * 2 |

2×1

=> a = 2 and b = 1

The order of the matrix X = 2 x 1

- Let X =
- 1 1 2

*2x* + *y x+ 2y*

On comparing, we get

2x + y = 4 x+ 2y = 5

and

On solving, we get: x = 1 and y-2 The matrix X =

X | l | |

y. | 2 |

Ans.

Question 7.

*An aeroplane at an altitude of 1500 metres finds that two ships are sailing*

*towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.* j4j

*The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the*

*distribution. (Take 2 cm – 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis)* [6]

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

No. of shooters | 9 | 13 | 20 | 26 | 30 | 22 | 15 | 10 | 8 | 7 |

1500 m

1500

BC

1500

1 = BC =

*Use your graph to estimate the following :*

*The median.**The interquartile range.**The number of shooters who obtained a score of more than 85%.*Answer 7.

(a) Let AB be the altitude and C and D are the positions of two ships.

In right angled triangle ABC,

BC

1500 m

In right angled triangle ABD, 1500

tan 30° =

tan 45° =

BD

J_ _ 1500 V3 ”

BD

BD = 1500 V3 = 1500 x 1-732 = 2598 m Distance between the two ships = CD

= BD-BC = 2598 – 1500 = 1098 m

/160\^{th}

( 2 ) ^{term} = 80^{th} term

■
Scores |
No. of Shooters | Cumulative frequency (c. f.) |

0-10 | 9 | 9 |

10-20 | 13 | 22 |

20-30 | 20 | 42 |

30-40 | 26 | 68 |

40-50 | 30 | 98 |

50-60 | 22 | 120 |

60-70 | 15 | 135 |

70-80 | 10 | 145 |

80-90 | 8 | 153 |

90-100 | 7 | 160 |

Ans.

(b)

Median = Q)* term

= 44

Ans.

- Lower quartile (Qi) – term = 1 1 = 40
^{th}term.

= 29

/3n\^{th} /3 x 160\^{th}

Upper quartile (Q3) = (47) term = I —^ j term = 120^{th} term

= 60

Inter-quartile range = Q3 – Qi

- 60 – 29 = 31 Ans.

- Since, 85% scores = 85% of 100 = 85

Through mark for 85 on X-axis, draw a vertical line which meets the ogive at any point. Through that point, draw a horizontal line which meets the Y-axis at the mark of 149.

.\ The no. of shooters who obtained a score of more than 85%

= 160 – 149 = 11 Ans

Scale 2 cm = 10 scores on X-axis and 2 cm = 20 shooters on Y-axis

(30;42)

(20, 22)

4044

Question 8.

^{I}f^=b^{=}c^{showthat}~3^{+}l3^{+}^= —*Draw a line AB – 5 cm. Mark a point C on AB such that AC = 3 cm. Using a*

*ruler and a compass only, construct:* [4]

*A circle of radius 2.5 cm, passing through A and C.**Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.**A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio*

^{1:2}– [3]

Answer 8.

> x

*Find the coordinates of A and B.**Find the equation of the line through P and perpendicular to AB.*

<^{a}> ^{L}«^{X}a=l = l = ^{k}

which gives x = ak,y= bk and z = ck Putting the values of x, y and z, we get

^3 -y3 ^3

L.H.S. =

*a}* ^{+} *b ^{3}*

^{ +}

*c*^

^{3}(ak)^{3}*(bk)*^ (,

^{3}*ck*)

^{3}

_{a}-> ijj _{c}j

*a ^{3}k^{3} b^{3}k^{3} c^{3}k^{3} *a

^{3 +}

*b*

^{3}^{ +}

*c*3

^{3}k^{3}+ k^{3}+ k^{3}=*k*

^{3}R.H.S. =

- xyz abc
- (qfc) (fcfc) (cfc) abc

*3k ^{3}*

R.H.S.

••• L.H.S.

- Steps of construction :

1. Draw a line AB = 5 cm.

Mark a point C on AB such that AC = 3 cm.

Draw a perpendicular bisector of AC.

Mark a point O perpendicular bisector from A of length 2-5 cm.

Hence Proved

2.

3.

4.

- Taking O as centre and OA as radius draw a circle, which is the required circle.
- Join O and B.
- Draw a circle with OB as diameter which cuts the given circle at points P and

- Join PB and QB, which are the required tangents. Measure PB and QB.
- (i) Let the coordinates of A be (jc, 0) and B be (0, y)

Given, P = (4, – 1) divides AB in the ratio 1 : 2.

*m\X2 + ntjXi*

Now, x =

m i + m2 1 x 0 + 2 x x

*=*

1 +2

*2x*

**T**

x = 6

*=*

1*xy+2*xO 1+2

– 1

Coordinates of A = (6,0) and coordinates of B = (0, – 3).

**/•■’.ci y2~y\ -3-0 -3 **1

- Slope of AB = = – ore” = -6=2

Now, slope of the line perpendicular to AB = – ^_{pe} pf _{A}g

” 1/2 “ ^{z}

Which passes through P (4,- 1).

Equation of line is, y – yi = m(x~xi)

>’-(-1) = -2(jc-4)

Hence, y = – 2x +7 Ans.

Question 9.

*A dealer buys an article at a discount of 30% from the wholesaler, the marked*

*price being* ? *6,000. The dealer sells it to a shopkeeper at a discount of 10% on the marked price. If the rate of VAT is 6% find* [3]

*The price paid by the shopkeeper including the tax.**The VAT paid by the dealer.**The given figure represents a kite with a circular and a semicircular motifs stuck*

*on it. The radius of circle is 2.5 cm and the semicircle is 2 cm. If diagonals AC and BD are of lengths 12 cm and 8 cm respectively, find the area of the :* [4]

*shaded part. Give your answer correct to the nearest whole number.**unshaded part.*

*A model of a ship is made to a scale 1 : 300*[3]*The length of the model of the ship is 2m. Calculate the length of the ship.**The area of the deck ship is J80,000 m?. Calculate the area of the deck of the model.**The volume of the model is 6.5 m*^{3}. Calculate the volume of the ship.

Answer 9.

- (i) Given, marked price of an article = ? 6,000 and discount = 30% of? 6,000

^{= ?} TM ^{x 6}–^{000 = ? 1} ’^{80}°

Sale price of the article = ? (6,000 – 1,800) = ? 4,200 Dealer sells the article to a shopkeeper at discount = 10% on M. P.

= |qq x 6,000 = ? 600

Price at which the article is bought by the shopkeeper = ? (6,000 – 600) = ? 5,400 Sales tax paid by the shopkeeper = 6% of? 5,400

The price paid by the shopkeeper including the tax = ? 5,400 + ? 324

= ? 5,724 Ans.

- Tax paid by the dealer = 6% of? 4,200

- ^ Too
^{x}^>200 ? 252

Since, tax charged = ? 324

VAT paid by the dealer = Tax charged – Tax paid

= ? 324 – ^ 252

= t!2 Ans.

- (i) Area of shaded part = Area of circle + Area of semicircle

*r, jir ^{2} = nr^{z}* +

_ 22 j ,22 (2)2

- fj x (2 5) -f* ^ x ^

137-5 44 181-5

_{?}+_{?}–_{7}

= 25-93 « 26 cm^{2} Ans.

- Area of unshaded part = Area of kite – Area of shaded part

= ix Product of diagonal – 26

= jx 12×8-26

- 48-26

= 22 cm^{2} Ans.

- Given, scale factor (k) =

r\ Length of the model _

‘ Length of the ship “

Length of the ship = 2 x 300 = 600 m Ans.

Area of the deck of the model _ _{2} Area of the deck of the ship ~

**A j , r u _{J} i 180,000 „ ,**

Area of the deck of the model = 3qq _{x} 3qq = 2 m^{2} Ans.

Volume of model _ _{3}

Volume of the ship ^{–}

Volume of the ship = 6-5 x 300 x 300 x 300

= **17,55,00,000 **m^{3} Ans.

Question 10.

*Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple*

*interest. If he gets* *?1200 as interest at the time of maturity, find :* [3]

*the monthly instalment**the amount of maturity.**The histogram below represents the scores obtained by 25 students in a*

*Mathematics mental test. Use the data to :* [4]

*Frame a frequency distribution table.**To calculate mean.**To determine the Modal class.*

*A bus covers a distance of240 km at a uniform speed. Due to heavy rain its speed*

*gets reduced by 10 km/h and as such it takes two hrs longer to cover the total*

*distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x’.* [3]

**Answer 10.**

- (i) Since, number of months (n) = 24 and rate of interest (r) = 6%

*n (n* + 1) *r*

**= Px**

2×12 ^{A} 100 24(24+1) 6

1200 – P x

- x 12 * 100

1,200 x 24 x 100

**P =**

**Ans.**

(ii) (b) (i)

**Ans.**

(ii)

given above.

Marks | No. of Students (f) | Class mark (jc) | fjc |

0-10 | 2 | 5 | 10 |

10-20 | 5 | 15 | 75 |

20-30 | 8 | 25 | 200 |

30-40 | 4 | 35 | 140 |

40-50 | 6 | 45 | 270 |

n = 25 | 2 /jc = 695 |

6 x 24 x 25 Monthly instalment = X 800 Sum deposited = ? 800 x 24 = ^ 19200 Amount of maturity = ^ 19,200 + ^ 1,200 = ? 20,400 Using the given data, frequency distribution table is as given below

‘o Calculate Mean : Construct expanded table with class mark and/jc as

n =I/=25and2/x = 695

- (1) In the given histogram, inside the highest rectangle, which represents the maximum frequency (or modal class) draw two lines AC and BD diagonally from the upper comers to C and D of adjacent rectangles.
- Through the point K, draw KL perpendicular to the horizontal axis.
- The value of point L on the horizontal axis represents the value of mode.

Mode = 24 and the modal class = 20 – 30.

- Let the uniform speed be x km/h.

Time taken by it to cover 240 km = ~ hrs .

**[ ^{T}=l]**

New speed of bus = (x- 10) km/hr New time taken by the bus to cover 240 km = hrs

10

Given, it takes two hours longer 240 240

= 2

x~ 10 ~ x

On solving, we get x = 40 or x = – 30

Since, speed can not be negative.

Hence, the value of x = 40.

i.e., the uniform speed of bus = 40 km/hr.

**Question 11.**

** cos** A

*Prove that J^*_{sJnA}+ tan A = sec A._{[3]}*Use ruler and compasses only for the following question. All construction lines GHcl urcs TYiiist be clearly shown,.**Construct a A ABC in which BC = 6.5 cm,*Z*ABC = 60°, AB = 5 cm.**Construct the locus of points at a distance of 3.5 cm from A.**Construct the locus of points equidistant from AC and BC.*

*(v) Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.*

*Ashok invested*?*26,400 on 12%,*? 25*shares of a company. If he receives a dividend of*?*2,475, find the :*[3]*number of shares he bought**Market value of each share*Answer 11.

**, , „ „ „ cos A , cos A sin A **V ■ . _{n}** sin 01**

- L.H.S. =
_{r+}^A + tanA =_{1+sin}A^{+}STS |-^{,!m9=}JSeJ

**cos ^{2} A + sin A + sin^{2} A (1 + sin A) cos A 1 + sin A _ _**

**(1 + sin A) cos A **1

**cos A**

**[‘cos ^{2} 0 + sin^{2} 0 = 1] = secA = R.H.S. Hence Proved.**

**(b)**

**Steps of Construction :**

- (1) Draw a line BC = 6 5 cm.
- At B, draw BZ making an angle of 60° with BC.
- With B as centre, draw an arc of 5 cm. It cuts BZ at point A.
- Join AC.
- Taking A as centre and 3-5 cm as radius, draw a circle which is the required locus of points,
- Draw the bisector of angle of vertex ACB, which is the required locus of points equidistant from AC and BC.
- Length of XY = 5 cm.
- (i) Given, nominal value of share = if 25, Rate of dividend = 12%

Total dividend = ? 2475

Dividend on each share = Rate of dividend x N. V.

= _{]00} x25=?3

_{XT r} , Total dividend 2475

No. of shares = Dividend on each share = = ^{825 Ans}*

…. … , „ , , Sum invested 26,400 _

(n) Market value ot each share = _{No}__{Q}–_{f shares} = _{m}“ =^{? 32}‘ ^{Ans}‘