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

Mathematics - 1999 ( I.C.S.E)
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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

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 = 120^o .
(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.

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.

Hence the matrix X=

(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

(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 evaluateÖ( 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) Ö(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

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

(b)

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 x^o , where tan x^o =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 = tanq
8/15 = tanq
0.533 = tanq
q= tan^-1 0.533
q= 28.05

(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) ÐABC = 90^O (angle in a semi circle) therefore Ð7+Ð8 = 90^OÐ1=Ð2=Ð3=Ð4....=Ð8
Ð2+Ð8 = 180^O .....(i) (sum of the opposite angle of cyclic quadrilateral ) 2Ð1=Ð2+Ð3=Ð4+Ð5=Ð6+Ð7
Ð7+Ð5 = 180^O....... (ii) (same reason)
adding (i) and (ii) Ð1+Ð2+Ð3+Ð4+Ð5+Ð6+Ð7=540
Ð2+Ð8+Ð7+Ð5= 360^O = Let Ð2+Ð3=x
Ð2 +Ð5 + 90^O = 360^O (\Ð7+Ð8 = 90^O)
Ð2+Ð5 = 270^O = 4 x = 540⁰
                               x = 135⁰
                               \ Ð ABC + Ð CDE = 270^o



(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

Ans. (a) A = {a,  b, c,d}   B ={ 1, 2, 3, 4}
(i) [a® 1] [b® 2] [c®3] [d®4]  1 to 1 function
(ii) [a® 2] [b® 2] [c®3] [d®4] it is many to one into
(iii) [a® 1] [b® 2] [c®3] [d®4]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
= 2pr(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