ICSE Class 10 Maths Question Paper 2016 Solved

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

ICSE PAPER 2016

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) = 2jc3 + ?>x2 – 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 2x3 + 3x2 -kx + 5 by x-2 leaves a remainder 7.

‘ 2 O’ ‘ 7 O’
and I =
-17 0 1

(b) Given A =

and A2 = 9A + ml. Find m.

k ~

[3]
  1. [3]

…(1)

2(2)3 + 3(2)2 – k (2) + 5 7

26

13

Ans.

(b) Here, A =

=> A2 = A ■ A =

  1. 0 -1 71
  2. 0 -I 7 Given, A2 = 9A + ml
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.

  1. The slope of a line joining P (6, k) and Q (1- 3k, 3) is i . Find (0 k
  2. Midpoint of PQ, using the value of ‘k’ found in (i)
  3. Without using trigonometrical tables, evaluate : cosec2 57°-tan2 33° + cos 44 ° cosec 46° – /2 cos 45° – tan2 60 °
  4. 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.

  1. (i) Let (6, k) = {x\, and (1 – 3k, 3) = (x2, y2)

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

  1. cosec2 57° – tan2 33° + cos 44° cosec 46° —V2 cos 45° – tan2 60° => sec2 (90° – 57°) – tan2 33° + cos 44°’ ^ — V~2 • (V^)

= ^ jt x (2)2 x 3 ™ 4jt cm3

=> sec2 33° – tan2 33° +”^4^) -1-3 cos 44°

=* 1 + cos 44° ” 13

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

  1. Volume of metallic cone = ^ nr2h

1

cosec 0 = sec (90° – 9)] [ ‘sin 0 = cos (90° – 0)] [V sec2 0 – tan2 0 = 1] Ans.

  1. 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.

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

number line.

[3]
  1. 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

  1. If (3a + 2b) : (5a + 3b) = 18 : 29. Find a : b.
[3]

Answer 3.

  1. 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

  1. 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

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

ZODB = ZCBD (alternate interior angles)

Also, OB = OD (radii)

■■■ ZOBD = ZODB = 32° Ans.

  1. 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.

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

, s „ 3a + 2b 18

  1. Here> 5a + – 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.

  1. 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

lS: [3]

  1. A perfect square
  2. Divisible by 7
  3. 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)

  1. Reflect points A and B on the Y axis and name them A’ and Brespectively. Write down their coordinates.
  2. Name the figure OABCB A‘.
  3. 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]

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

Answer 4.

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

Clearly, total number of all possible outcomes = 30

  1. For getting a perfect square :

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

  1. 1

Required probability = 30 = jq Ans.

  1. For getting a number divisible by 7 :

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

  1. 2

Required probability = Ans.

  1. (i) Coordinates of A’ = (4,4)

Coordinates of B’ = (3,0)

  1. Hexagon
  2. The line of symmetry of this figure is Y-axis. Ans.
  3. (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 +

= 5000(1 + to)2

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.

  1. Solve the quadratic equation x2 — 3(x + 3) = 0; Give your answer correct to

two significant figures. [3]

  1. 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.

  1. Given, *2 – 3 (x + 3) = 0

x1 -3x-9 = 0 Here, a = 1, b = – 3 and c = – 9

b2 4ac = (- 3)2 – 4 x 1 x (- 9) = 9 + 36 = 45

  • Vb2 -4ac

x ——

3 ± V45 2 x 1 3 ± 6-708

x =

la

2

  1. + 6-708 , 3 – 6-708

and

~ 2

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

  1. 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 _

lnterest (D = -joo = f 100 >TT2 i69875

Amount = P + I = ? (45300+169-875) = ? 45469-875 Ans.

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

= p(, + T5o)”

1996-50 = 1500 1996-50

(to)”

1500

(™)3ar

i.e„ n i = 3

.\ Required time = 3 years Ans.

Question 6.

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

symmetry and name them. [3]

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

at T. [4]

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

(Hi)

  1. Given matrix A =

Find area of quadrilateral PQRS if area of A PTS = 27 cm2.

4 sin 30° cos 0° ‘ 4
cos 0° 4 sin 30° and B ~ c
J
[3]

If AX = B

  1. Write the order of matrix X.
  2. Find the matrix X’. Answer 6.

(a)

Steps of construction :

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

‘2n – 4>

[■m—(:

  1. 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.

  1. (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]

  1. Since A TPS ~ A TRQ

‘2×6-4

)■

90°= 120°

TR TP “

_6_

18 “

SP =

RQ

SP

4

SP

4x 18

= 12

Ans.

  1. 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 cm2

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 ‘ Xax b ~ 5
1 4 * 2

2×1

=> a = 2 and b = 1

The order of the matrix X = 2 x 1

  1. Let X =
  2. 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.

  1. 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

  1. 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 :

  1. The median.
  2. The interquartile range.
  3. 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 = 80th 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.

/«\th /i6oyh .

  1. Lower quartile (Qi) – term = 1 1 = 40th term.

= 29

/3n\th /3 x 160\th

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

= 60

Inter-quartile range = Q3 – Qi

  • 60 – 29 = 31 Ans.
  1. 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.

  1. If^=b=cshowthat~3 + l3 + ^= —
  2. 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]

  1. A circle of radius 2.5 cm, passing through A and C.
  2. Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.
  3. 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

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

<a> L«Xa=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} + b3 + c3 (ak)3 ^ (bk)3 ^ (,ck)3

a-> ijj cj

a3k3 b3k3 c3k3 a3 + b3 + c3 k3 + k3 + k3 = 3 k3

R.H.S. =

  1. xyz abc
  2. (qfc) (fcfc) (cfc) abc

3k3

R.H.S.

••• L.H.S.

  1. 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.

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

Q.

  1. Join PB and QB, which are the required tangents. Measure PB and QB.
  2. (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. =

1 +2

2x

T

x = 6

  1. =

1xy+2xO 1+2

– 1

y = -3

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

/•■’.ci y2~y\ -3-0 -3 1

  1. Slope of AB = = – ore” = -6=2

Now, slope of the line perpendicular to AB = – ^pe pf Ag

” 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.

  1. 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]

  1. The price paid by the shopkeeper including the tax.
  2. The VAT paid by the dealer.
  3. 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]

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

  1. A model of a ship is made to a scale 1 : 300 [3]
  2. The length of the model of the ship is 2m. Calculate the length of the ship.
  3. The area of the deck ship is J80,000 m?. Calculate the area of the deck of the model.
  4. The volume of the model is 6.5 m3. Calculate the volume of the ship.

Answer 9.

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

= ? TM x 6000 = ? 180°

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

– ? jqq x 5,400 = ? 324

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

= ? 5,724 Ans.

  1. 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.

  1. (i) Area of shaded part = Area of circle + Area of semicircle

r, jir2 = nrz +

_ 22 j ,22 (2)2

  • fj x (2 5) -f* ^ x ^

137-5 44 181-5

  • ? + ?7

= 25-93 « 26 cm2 Ans.

  1. Area of unshaded part = Area of kite – Area of shaded part

= ix Product of diagonal – 26

= jx 12×8-26

  • 48-26

= 22 cm2 Ans.

  1. 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 m2 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 m3 Ans.

Question 10.

  1. 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]

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

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

  1. Frame a frequency distribution table.
  2. To calculate mean.
  3. To determine the Modal class.

  1. 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.

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

n (n + 1) r

  1. = Px

2×12 A 100 24(24+1) 6

1200 – P x

  1. 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. (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.
  2. Through the point K, draw KL perpendicular to the horizontal axis.
  3. The value of point L on the horizontal axis represents the value of mode.

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

 

  1. 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

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

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

, , „ „ „ cos A , cos A sin A V ■ . n sin 01

  1. L.H.S. = r+^A + tanA = 1+sinA+STS |-,!m9=JSeJ

cos2 A + sin A + sin2 A (1 + sin A) cos A 1 + sin A _ _

(1 + sin A) cos A 1

cos A

[‘cos2 0 + sin2 0 = 1] = secA = R.H.S. Hence Proved.

(b)

Steps of Construction :

  1. (1) Draw a line BC = 6 5 cm.
  2. At B, draw BZ making an angle of 60° with BC.
  3. With B as centre, draw an arc of 5 cm. It cuts BZ at point A.
  4. Join AC.
  5. Taking A as centre and 3-5 cm as radius, draw a circle which is the required locus of points,
  6. Draw the bisector of angle of vertex ACB, which is the required locus of points equidistant from AC and BC.
  7. Length of XY = 5 cm.
  8. (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_Qf shares = m“ =? 32Ans

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