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Posted By: deepthi Member Level: Silver Posted Date: 09 May 2008
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2007 ICSE maths Question paper
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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 answer.
Answer all question in section A and any four question in section B.
All working, including rough work, must be clearly shown and must be done on
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.
Q1. (a) A trader loses 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 ? [3]
Sol. Let the C. P. of tea = Rs. 225 per kg.
loss = 10 % of x = (10/100) ? (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 = (225?? 10)/(9) = Rs. 250
therefore cost price of tea = Rs. 250
Gain = 10%
(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? [4]
S. P. tea = [(100 + 10)/(100)]? (250) = (110???250)/(100) = Rs. 275
Sol. Let the marked price = Rs. x
Discount = 20 %
Therefore S. P. = (x ? 80)/100 = 8x/10
Profit = 25% on the C.P.
Therefore C. P. = [(100)/(100+25)]? (8x/10) = (10/125)???(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) ? (100) = 56.25 %
Q2. A man invests 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; [6]
Sol. Here P = Rs. 5000; A = Rs. 5,600; T = 1 year
I = A - P = Rs. (5,600 - 5,000) = Rs. 600
(i) Using A = P( 1 + R /100)T , we have
5,600 = 5,000( 1 + R /100)1 , we have
?56 / 50 = 1 + R /100
? R /100 = 56 /50 - 1 = 6 /50
? R = 6 / 50 ??100 = 12%
(ii) P(for 2nd year) = Rs. (5,000 + 600) = Rs. 5,600
R = 12% ; T = 1 year
? Interest (accrued) in the second year = Rs. (5,600???12 ? 1) /100 = Rs. 672.
(iii) A = P( 1 + R /100)T
? A = 5,000( 1 + 12 /100)3
? = 5,000 ? 112 / 100 ? 112 / 100???112 / 100
= 1,75,616 = Rs. 7,024.64 = Rs. 7,025 nearly.
25
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 C(2,7);
(ii) Construct the locus of a points equidistant from A and B;
(iii) Construct the locus of a points 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. [5]
Sol.
(i) Plot the points A(1,1), B(5, 3) and C(2, 7) as shown.
(ii) Join points A, & B . Draw right bisector l of AB. Then, l is the locus of points
equidistant from A and B.
(iii) Join A & C. Now draw the bisector m of Ð CAB. Then, m is the locus of points
equidistant from AB and AC.
(iv) The point of intersection P of right bisector of AB and angle bisector of ? CAB
is the point such that PA = PB and P is equidistant from AB & AC.
(v) On measuring PA = 2.5 cm.
Q4. Use a ruler and compass only in this question.
(1) Construct the quadrilateral ABCD in which AB = 5cm, BC = 7cm and angle
ABC = 120° , 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. [5]
Sol. (1) (i) Draw a line AB = 5 cm.
(ii) Draw an angle ABX = 120°.
(iii) Taking A as center and radius equal to AB, draw an arc.
(iv) Now take C as center and radius equal to BC, draw another arc cutting BX at C.
(v) Join AD and CD.
Then ABCD is the required quadrilateral with AC as the line of symmetry.
(2) The geometrical name of quadrilateral is kite.
(3) On measuring BD = 6 cm.
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 ARC = 90o and in triangle PQS angle PSQ =90o. Given QS
= 6 cm. Calculate the length of AR. [6]
Sol. (i) ? BPQ ~ BAC [AA-similarity]
PQ = BP …(1)
AC BA
Also AP = 4
PB 3
??AP + 1 = 4 + 1
PB 3
? AP + PB = 4 + 3
PB 3
? AB = 7
PB 3
? PB= 3
AB = 7 …(2)
From (1) and (2), we have PQ = 3
AC 7
(ii) RAC ~ PSQ [AA-similarity]
? AR = AC
QS PQ
? AR = 7
6 3
? AR = 7???6 = 14 cm.
3
Q6. The figure shows a running track surrounding a grassed enclosure PQRSTU. The enclosure
consists of a rectangle PQST with a semicircular region at each end. PQ = 200 m.; PT = 70 m.
(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. Take ? to be 22. 7 [6]
Sol. (i) Area of grassed enclosure = PQ ??PT + 2 ????????? (35)2 (? r = 70/2 = 35m)
2
= 200 ? 70 + 22/7?? 35 ? 35
= 14000 + 3850 = 17850 m2
(ii) Outer perimeter of ABCDEF = 2 ? AE + 2 ? ED
= 2 ? 1/2???2pr + 2???200, where r = 35 m
= 2?? 22/7 ??35 + 400
= 220 + 400 = 620 m.
Q7. Use graph paper for this question:-
(i) Plot the points A (3,5) and B (-2,-4). Use 1cm = 1unit on both axes.
(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. [5]
Sol.
(i) First plot the points A(3,5) & B(-2,4) on the graph paper as shown above.
(ii) The coordinates of the image of A when reflected in the x-axis i.e. A' is (3,-5).
(iii) The coordinates of the image of B when reflected in the y-axis i.e. B' is (2,-4).
(iv) Geometrical name of figure AA `BB' is quadrilateral.
(v) Any point on the x-axis is invariant under reflection in the x-axis.
Say (-1, 0) and (3, 0).
SECTION - B (40 Marks)
Answer any four questions.
Q8. (a). Find the 2?? 2 matrix X which satisfies the equation.
[3 7] [0 2] + 2X = [1 -5] [4]
[2 4] [5 3] [-4 6]
Sol. The given equation can be written as
[3???0 + 7???5] [3 ??2 + 7 ? 3] + 2X =[1 -5]
[ 2 ? 0 + 4?? 5] [2 ??2 + 4?? 3] = [-4 6]
? [35 27] + 2X = [1 -5]
[20 16] [-4 6]
?2X = [1 -5] [35 27]
[-4 6] [20 16]
= [1 - 35 -5 - 27] =[ -34 -32]
[-4 -0 6-16 ] [-24 -10]
? X = [2 -17 -16]
[-12 -5]
(b) Find the equation of the line passing through (0,4) and parallel to the line 3x + 5y + 15 = 0 [3]
Sol. Any line parallel to 3x + 5y + 15 = 0 is
3x + 5y + k = 0 …(1)
Since (0, 4) lies on (1)
? 3(0) + 5(4) + k = 0 ? k = -20
?Required line is 3x + 5y - 20 = 0.
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