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# ICSE Set Qa1 Year Icse Maths99 Qa1.php Mathematics Exam Paper for students online

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 Mathematics - 1999 ( I.C.S.E) You are on Answers 1 to 6 Q1. (a) A trader losses 10% on his cost price by selling tea at Rs. 225 per kg. At what price per kg should he sell it to gain 10% on his cost price ? (b) When a discount of 20% is given on the marked price of an article, a shopkeeper makes a profit of 25% on his cost price. What would be his percentage profit on cost if the article was sold at the marked price ? Ans.(a) Let the C. P. of tea=   Rs. x per kg. loss = 10 % of   x  = (10/100) x (x) = x/10 Therefore S.P. of tea = x-(x / 10) = (10x-x) /10  = 9x/10 But S. P. of tea  = Rs. 225 therefore  9x/10 = 225 x=(225x10)/(9) = Rs. 250 therefore cost price of tea = Rs. 250 Gain = 10% S. P. tea = [(100+10)/(100)]x(250) = (110x250)/(100) = Rs. 275  Ans.(b)  Let the marked price = Rs. X Discount = 20 % Therefore S. P. = (Xx80)/100 = 8X/10 Profit = 25% on the C.P. Therefore C. P. = [(100)/(100+25)]x (8X/10) = (10/125)x(8X) = Rs. 0.64X Now let S. P = Marked price = Rs X Therefore Profit = X- 0.64X = 0.36X Therefore Profit percentage = (0.36X/0.64X) x (100) = 56.25 % Q2. A man invest Rs.  5,000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 5,600. Calculate : (i) the rate of interest per annum; (ii) the interest accrued in the second year; (iii) the amount at the end of the third year; Ans(i) A= P [1 + (R/100)]n 5,600 = 5,000[1 + (R/100)]1 5,600= 5,000 +50 R 600 = 50 R R=12%  Ans (ii) A= P [1 + (R/100)]n A=Rs. 5,000[ 1 + (12/100)]2 = Rs. 5,000[112/100]2 = Rs. 5,000 x 1.12x 1.12 = Rs. 6272 C.I. = A - P = 6,272-5,000 = Rs. 1,272 Ans. (iii)A= P[1 +(R/100)]n =5,000(112/100)3          = 5,000 x (1.12)3 A= Rs. 7,024.64 Q3. Use graph paper for this question. Take 2cm=1 unit on the both axes (i) Plot the points A (1,1), B(5,3) and (2,7); (ii) Construct the locus of a point equidistant from A and B; (iii) Construct the locus of a point equidistant from AB and AC; (iv) Locate the point P such that PA = PB and P is equidistant from AB and AC; (v) Measure and record the length PA in cm. Ans. 3(i) (ii) Join points A, & B Draw the perpendicular bisector of AB cutting  it at point P. Point P is the locus of points equidistant from A & B (iii) Join AB & AC.Now draw the perpendicular bisector on both lines intersecting at a point P. Now point P is the locus of point equidistant. (iv) Point is equidistant from AB & AC such that EA= PB (v) Join PA & measure  it . Q4. Use a ruler and compass only in this question. (1) Construct the quadrilateral ABCD in which AB = 5cm, BC = 7cm and angle ABC = 1200, given that AC is its only line of symmetry. (2) Write down the geometrical name of the quadrilateral. (3) Measure and record the length of BD in cm. Ans.(1) (i)Draw a line AB = 5 cm. (ii)Draw an  angle ABX = 120o . (iii)Taking B as center and 7cm. radius mark an arc on BX at the point C . (iv)Join AC . (v)Taking A as center and 7cm radius mark an arc  . (vi)Now take C as center and 5 cm radius cut the previous marked arc it is the required point D. ABCD is the required quadrilateral. (2) The geometrical name of quadrilateral is parallelogram. (3) Try your self. Q5. In the figure given below P is a point on AB such that AP:PB = 4:3 .PQ is parallel to AC . (i) Calculate the ratio PQ:AC,giving reasons for your answer , (ii) In triangle ARC  angle =90o and in triangle PQS  angle PSQ =90o. Given QS =6 cm. Calculate the length of AR. Ans. (1)In triangle ACB and triangle PQB angle B= angle B  ..............(common) angle  BAC =angle BPQ ..............(corresponding angles) therefore triangle ACB ~ triangle PQB BQ = BP = PQ BC     BA     AC  PQ = 3 AC    7 PQ:AC = 3:7 (ii)Now in triangle ARC and triangle QSP angle R = angle S = 900      angle ACR = angle QPS .....................(corresponding angles) therefore triangle ARC ~ triangle QSR ..........(AA) AR/QS =  RC/SR  =  AC/QP AR/6 = 7/3 AR = 42/3  =  14cm. Q6.The figure shows a running track surrounding a grassed enclosure (i) Calculate the area of grassed enclosure in m2 (ii) Given that the track is of constant width 7m,calculate the outer perimeter ABCDEF of the track. Ans. (i)Area of grass enclosure =area of rectangle + 2(circular ends) = L xB + (pr2/2 +pr2/2) = 200 x 70 + (22/7 x 35 x35) = 53900000 m2 (ii) Perimeter of outer region = L+L+2(circular ends) =400+264 = 664m2

Q7.Use graph paper for this question:-
(i) Plot the point A (3,5)and B (-2,-4). Use 1cm = 1unit on both axis.
(ii) A' is the image of A when reflected in the  x axis .Write down the coordinates of A' and plot it on the graph paper.
(iii) B' is the image of B when reflected in the y axis, followed by the reflection in the origin.Write down the coordinates of B' and
plot it on the graph paper.
(iv) Write down the geometrical name of the figure AA' BB' .
(v) Name two invariant points under reflection in the X axis

Ans.(i) First plot the point A(3,5) & (-2,4) on the graph paper using 1cm =1unit.
(ii)The coordinates of the image of A when reflected in the X- axis i.e A' are (3,-5). Now plot this point on the graph paper.
(iii)The coordinates of the image of B when reflected in the Y- axis i.e B' are (2,-4).Now plot the coordinates of this point on the graph
paper.
(iv)&(v) Do your self.

Q8.(a) Find the 2x2 matrix X which satisfies the equation. (b) Find the equation of the line passing through (0,4) and parallel to the line 3x+5y +15=0
(c) In the figure given below PQRS and PXYZ  are parallelogram.Prove that they are of equal area ? Ans-(a)
 [ 3x0+7x5 3x2+7x3 ] + 2X = [ 1 -5 ] 2x0+4x5 2x2+4x3 -4 6
 [ 35 27 ] + 2X = [ 1 -5 ] 20 16 -4 6
 2x = [ 1 -5 ] - [ 35 27 ] -4 6 20 16
 2x = [ -34 -32 ] -24 -10

Hence the matrix X=

 - [ 17 16 ] 12 5

(b) Equation of a line is 3x +5y +15 =0
y - y1 = m (x- x1)
-Coefficient of x / coefficient of y = -3/5
it passing  through the points :- (0, 4)
y-4 = 13/5 (x - 0 )
5y- 20 = -3x
3x + 5y = 20

Q9.(a)Solve the inequations :-
12+11/6 x  < 5+3x,x
R .
Represent the solution on a number line.

(b) In the figure given line APB meets the x axis at A , y axis at B. P is the point (-4,2)and AP:PB = 1:2. Write down coordinates
of A and B.
(c) Use logarithm to evaluate
( 0.874 )     , correct to three significant figures.....................................0.0591

Ans(a) 12 + 11/6x < 5 + 3x
12 + 11x/6  < 5+ 3x
(11x/ 6) -(3x)  < 5 - 12
(11x- 18x)/ ( 6 )  < -7
(-7x)/ (6)  < -7
(x/ 6)  < (-7/7)
(x/6)  < 1
x  < 6
(-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 )
(b) Let quardinate of point A (e, 0 ) and (0, y)
By section formula
= (m1n2 + m2m1)/( m1+m2)
-4 = (1 x 0+2(n))/(1+2)
-12 = 2n
n= -6
and 2 (1xy+2x0)/ (3)
y = 6
So the quardinates are - 6, 0 and 6, 0
(c)
(o.874)/0.0591
1/2 log o.874 - log 0.0591
1/2(-0.0585) - (-1.2284)
- 0.0292 + 1.2284 = 1.192
antilog ( 1.192) = 15.82 ans.

Q10(a) Use graph paper for this equation .
(i) Draw the graphs of 3x-y-2=0 and 2x+y-8=0.Take 1cm=1unit on both axis and plot only three points per line .
(ii) Write down the coordinates of the point of intersection and the area of the triangle formed by the lines and the x-axis
(b) The marks obtained by the set of students in an examination are given below :
MARKS :                  5    10   15   20   25   30
NO OF STUDENTS :   6    4     6     12     x     4
Given that the mean mark of the set is 18 ,Calculate the numerical value of x.

Ans. 3x - y -2 = 0
2x +y - 8 =0
For Ist eq.
3x - 2 = y

 x : 0 1 2 y : -2 1 4

For IInd eq. 2x + y - 8 = 0
-2x + 8 = y

 x : 0 1 2 y : 0 6 4

(b)

 xi fi fi xi 5 6 30 10 4 40 15 6 90 20 12 240 25 x 25x 30 4 120 32+x 540+25x

mean = Sfixi/Sfdi
18 =  (520+25x)/ (32+x)
576 +18x = 570 +25x
56 = 7x
x = 8

Q11. (a) A trader buys x articles for a total cost of Rs. 600.
(i) Write down the cost of one article in terms of x.If the cost per article were Rs.5 more ,The number of articles that can be
brought  for Rs. 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it to find x.
(b) With reference to the figure given below, a man stands on the ground at point A ,Which is on the same horizontal plane as B ,
the foot of a vertical pole BC. The height of the pole is 10 m .The man's eye is 2m above the ground .He observes the angle of
elevation at C ,the top of the poles as xo , where tan xo =2/5,calculate : (i) the distance AB in m :

(ii) the angle of elevation of the top of the pole when he is  standing 15m from the pole. Give your answer to nearest degree.

Ans. (a) Let the number of articles = x
Total Money spend = Rs. 600
cost of 1 article = 600/x
A, T. S:- (600/x)+(5) = 600/ x-4
(ii) (600/x) - (600/x-4) = -5
(600(x-4)- 600x) /( x(x-4))= -5/1
(600x - 2400 - 600x)/(x2 - 4x) = -5/1
-5x2 +20x = - 2400
5x2 - 20x -2400 = 0
5(x2- 4x- 480 ) = 0
x2 -24x + 20x - 480 = 0
x(x- 24 )+20 (x-24 )= 0
(x+20 ) (x-24)=0
x= -20
x= 24
(b)(i) In
D DEC tan = 2/5 = CE/DE
2/5 = 8/DE
DE = 20 m
DE = AB = 20m
(ii) In
D CFE
CE/FE = tan
q
8/15 = tanq
0.533 = tanq
q= tan-1 0.533
q= 28.05

Q12. (a) In the figure given above ,AE is the diameter of the circle .Write down the numerical value of angle ABC + angle CDE.
Give reasons for your answer.
(b) Use ruler and compass only in this equation.
(i) Draw a circle , center O and radius 4cm.
(ii) Mark a point P such that OP=7cm.
Construct the two tangents to the circle from P. Measure and record the length of one of the tangents.
(c)A man invests Rs. 1680 in buying shares of nominal value Rs.24 and selling at 12% premium . The dividend on the shares os
15% per annum.
(i)Calculate the number of shares he buys;
(ii)Calculate the dividend he receives annually.

Ans. (a) ABC = 90O (angle in a semi circle) therefore 7+8 = 90O1=2=3=4....=8
2+8 = 180O .....(i) (sum of the opposite angle of cyclic quadrilateral ) 21=2+3=4+5=6+7
7+5 = 180O....... (ii) (same reason)
adding (i) and (ii)
1+2+3+4+5+6+7=540
2+8+7+5= 360O = Let 2+3=x
2 +5 + 90O = 360O (\�7+8 = 90O)
2+5 = 270O = 4 x = 5400
x = 1350

\ ABC +
CDE = 270o

(b) (i) Draw a circle of radius = 4cm
(ii) Mark a point P such that OP = 7cm
(iii) Find perpendicular bisector of OP at M
(iv) Taking m on centre and draw a circle which intersect the previous circle at R and S
(v) Join RP, SP
(vi) RP and SP are required tangents
(c) Total money invested = 1680
A.T. S.  :- (12/ 100) x (24) = (72/25)
cost for each share = (24+72)/(1+25) = (600+ 72)/(25) = 672/25
No. of shares( 1680 x 25) /( 672) =(2625/42)=62 shares
Annual dividend :- (15x1680)/(122) = Rs. 225

Q13. (a) Given A ={a,b,c,d};B={1,2,3,4},
(i) From orderd pairs showing a 1 to 1 function from A to B;
(ii)From ordered pairs showing a many to 1 function from A to B;
(iii)Explain why it is not possible to construct ordered pairs which represent a many to 1 onto function from A to B.

(b)With reference to the figure given alongside , a metal container in the form of a cylinder is surmounted by a hemisphere
of the same radius .The internal height of the cylinder is 7 m and the internal radius is 3.5m Calculate :
(i) the total area of the internal surface,excluding the base :
(ii)the internal volume of the container in m3. Ans. (a) A = {a,  b, c,d}   B ={ 1, 2, 3, 4}
(i) [a
1] [b 2] [c3] [d4]  1 to 1 function
(ii)
[a 2] [b 2] [c3] [d4] it is many to one into
(iii) [a
1] [b 2] [c3] [d4]there is no element which has pre- image is set A so to form this orders not possible

(b) (i)Total internal surface area = 2prh +3p r2
= 2
pr(h+r)
= 2 x 22/7 x3.5(7+3.5)
= 231 m2
(ii) Total internal volume =
p r2 h +2/3p r3
=
pr2 (h+2/3r)
= 359.33 m3

Q14(a) The center of a circle of radius 13 units is  the point (3,6 ).P(7,8)is a point inside the circle.APB is a chord of the circle
such that AP=PB .Calculate the length of AB.
(b)Use the graph paper for this question.The table given below shows the monthly wages of some factory workers.
(i)Using the table ,calculate the cumulative frequencies of workers.
(ii)Draw the cumulative frequency curve .Use 2cm =Rs 500,Starting the origin at Rs. 6500 on x axis ,and 2cm =10 workers on
the y-axis.
(iii)Use your graph to write down the median wage in Rs.
 WAGES IN Rs. No OF WORKERS CUMULATTIVE FREQUENCY 6500-7000 10 ........................ 7000-7500 18 ........................ 7500-8000 22 ........................ 8000-8500 25 ......................... 8500-9000 17 ......................... 9000-9500 10 ......................... 9500-10000 8 .........................
Ans. (a) Solution to find :- OP2 = (7-3)2 + (9-6)2
Op2 = 16+9
OP2 = 25
OP = 5
Now in triangle OPA :- OA2 - OP2 = AP2
(13)2 -( 5)2 = AP2
169- 25 AP2
144 = AP2
AP = 12
Therefore PO = 12
AB = 12 + 12 = 24 units
 Wages in Rs. No. of workers Cumulative Frequency 6500-7000 10 10 7000-7500 18 28 7500-8000 22 50 8000-8500 25 75 8500-9000 17 92 9000-9500 10 102 9500-10000 8 110 