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ICSE-X-Mathematics

Previous Year Paper year:2020

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#7-c [4]
In the given figure, ``TP `` and ``TQ `` are two tangents to the circle with center ``O ``, touching at ``A `` and ``C `` respectively. If ``\angle BCQ = 55^{\circ} `` and ``\angle BAP = 60^{\circ} ``, find:
i) ``\angle OBA `` and ``\angle OBC ``
ii) ``\angle AOC ``
iii) ``\angle ATC ``
Ans : i) ``\angle OAT = 90^{\circ} ``
``\therefore \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ} ``
``\angle OCT = 90^{\circ} ``
``\therefore \angle OCB = 90^{\circ} - 55^{\circ} = 35^{\circ} ``
Since ``OB = OC, \triangle OBC `` is an isosceles triangle
``\therefore \angle OBC = 35^{\circ} ``
Similarly, ``OA = OB, \triangle AOB `` is an isosceles triangle
``\therefore \angle OBA = 30^{\circ} ``
ii) ``\angle ABC = 30^{\circ} + 35^{\circ} = 65^{\circ} ``
Angle subtended by an arc at the center is twice that subtended on the circumference.
``\therefore \angle AOC = 2\angle ABC = 2 \times 65^{\circ} = 130^{\circ} ``
iii) ``AOCT `` is a quadrilateral. We know that the sum of all the internal angles of a quadrilateral is ``360^{\circ} ``
Hence, ``90^{\circ} + 130^{\circ} + 90^{\circ} + \angle ATC = 360^{\circ} ``
``\Rightarrow \angle ATC = 50^{\circ} ``
``\\ ``

#8-a [3]
What must be added to the polynomial ``2x^3 - 3x^2 - 8x `` , so that it leaves a remainder ``10 `` when divided by ``2x+1 ``?
Ans : Given : ``2x^3 - 3x^2 - 8x ``
When divided by ``2x+1 \Rightarrow x = - `` ``\frac{1}{2} ``
``f(- `` ``\frac{1}{2} `` ``) = 2 ( - `` ``\frac{1}{8} `` ``) - 3 ( `` ``\frac{1}{4} `` ``) - 8 (- `` ``\frac{1}{2} `` ``) + k ``
``\Rightarrow 10 = - `` ``\frac{1}{4} `` ``- `` ``\frac{3}{4} `` ``+ 4 + k ``
``\Rightarrow 10 = -1 + 4 + k ``
``\Rightarrow k = 7 ``

#8-b [3]
Mr. Sonu has a recurring deposit account and deposits Rs. ``750 `` per month for ``2 `` years. If he gets Rs. ``19125 `` at the time of maturity, find the rate of interest.
Ans : Given ``P = 750 `` Rs. ``n = 2 `` years ``= 24 `` months ``MV = 19125 `` Rs.
Interest ``= MV - ( P \times n) = 19125 - 750 \times 24 = 1125 `` Rs.
Interest ``= `` ``\frac{n(n+1)}{2} `` ``\times `` ``\frac{750}{12} `` ``\times `` ``\frac{r}{100} ``
``1125 = `` ``\frac{24 \times 25}{2} `` ``\times `` ``\frac{750}{12} `` ``\times `` ``\frac{r}{100} ``
``\Rightarrow r = `` ``\frac{45 \times 100}{750} `` ``= 6\% ``

#8-c [4]
Use a graph paper for this.
Take ``1 `` cm ``= 1 `` unit on both x and y axes.
i) Plot the following points on your graph sheet ``A( -4, 0), B ( -3, 2), C ( 0, 4), D ( 4, 1) `` and ``E(7, 3) ``
ii) Reflect point ``B, C, D `` and ``E `` on the x axis and name then ``B', C', D' ``and ``E' `` respectively.
iii) Join the points ``A, B, C, D, E, E', D', C', B' `` and ``A `` in order. Name the closed figure formed.
Ans :
The figure formed is a nonagon (nine edges). It is a FISH.
``\\ ``

#9-a [6]
``40 `` students enter for a game of shot put competition. The distance thrown ( in meters) is recorded below.

Distance in m 12-13 13-14 14-15 15-16 16-17 17-18 18-19

Number of Students 3 9 12 9 4 2 1

Use a graph paper to draw an ogive for the above distribution.
Use a scale of ``2 `` cm ``= 1 `` m on x-axis and ``2 `` cm ``= 5 `` students on the other axis.
Hence using your graph paper find:
i) the median
ii) Upper Quartile
iii) Number of students who cover a distance which is above ``16 `` ``\frac{1}{2} `` m
Ans :
i) the median ``= 14.7 ``
ii) Upper Quartile ``= 15.6 ``
iii) Number of students who cover a distance which is above ``16 `` ``\frac{1}{2} `` m ``= 5 `` students

#9-b [4]
If ``x = `` ``\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}- \sqrt{2a-1}} ``, prove that ``x^2 - 4ax + 1 = 0 ``
Ans : ``x = `` ``\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}- \sqrt{2a-1}} ``
Applying componendo and dividendo
``\frac{x+1}{x-1} `` ``= `` ``\frac{\sqrt{2a+1}+\sqrt{2a-1} + \sqrt{2a+1}- \sqrt{2a-1}}{\sqrt{2a+1}+\sqrt{2a-1} - \sqrt{2a+1} + \sqrt{2a-1}} ``
``\Rightarrow `` ``\frac{x+1}{x-1} `` ``= `` ``\frac{\sqrt{2a+1}}{\sqrt{2a-1}} ``
Squaring both sides
``\frac{(x+1)^2}{(x-1)^2} `` ``= `` ``\frac{2a+1}{2a-1} ``
``\frac{x^2 + 1 + 2x}{x^2 + 1 - 2x} `` ``= `` ``\frac{2a+1}{2a-1} ``
Applying componendo and dividendo
``\frac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - x^2 - 1 + 2x} `` ``= `` ``\frac{2a+1 + 2a - 1}{2a+1 - 2a+1} ``
``\Rightarrow `` ``\frac{2(x^2+1)}{4x} `` ``= `` ``\frac{4a}{2} ``
``\Rightarrow `` ``\frac{x^2+1}{2x} `` ``= `` ``\frac{4a}{2} ``
``x^2 + 1 = 4ax ``
``x^2 - 4ax + 1 = 0 ``. Hence proved.
``\\ ``

#10-a [3]
If the ``6^{th} `` terms of an AP is equal to four time its first term and the sum of first six terms is ``75 ``, find the first term and the common difference.
Ans : ``6^{th} term = 4a `` and ``S_6 = 75 ``
We know that ``t_n = a + ( n-1) d ``
``\Rightarrow 4a = a + 5 d `` ``\Rightarrow 3a = 5d `` ... ... ... ... ... i)
Also ``S_n = `` ``\frac{n}{2} `` ``[ 2a + ( n-1) d ] ``
``\Rightarrow 75 = `` ``\frac{6}{2} `` ``[ 2a + 5d ] `` ``\Rightarrow 75 = 6a + 15d `` ... ... ... ... ... ii)
Solving i) and ii)
``75 = 6a + 9a \Rightarrow a = 5 ``
Substituting in i)
``5d = 3a \Rightarrow d = `` ``\frac{3}{5} `` ``\times 5 = 3 ``
Therefore first term ``(a) = 3 `` and common difference ``(d) = 3 ``

#10-b [3]
The difference of two natural numbers us ``7 `` and their product is ``450 ``. Find the numbers.
Ans : Let the two natural numbers be ``x `` and ``y ``
Therefore ``x - y = 7 ``
Also ``xy = 450 \Rightarrow y = `` ``\frac{450}{x} ``
Substituting ``x - `` ``\frac{450}{x} `` ``= 7 ``
``\Rightarrow x^2 - 7x - 450 = 0 ``
``\Rightarrow x^2 - 25x + 18x - 450 = 0 ``
``\Rightarrow x( x - 25) + 18 ( x - 25) = 0 ``
``\Rightarrow (x-25) ( x + 18) = 0 ``
``\Rightarrow x = 25 \ or \ x = -18 `` ( not a natural number).
Therefore ``x = 25 ``. Substituting we get ``y = 25- 7 = 18 ``

#10-c [4]
Use ruler and compass for this question. Construct a circle of radius ``4.5 `` cm. Draw a chord ``AB = 6 `` cm.
i) Find the locus of points equidistant from ``A `` and ``B ``. Mark the point where it meets the circle as ``D ``.
ii) Join ``AD `` and find the locus of points which are equidistant from ``AD `` and ``AB ``. mark the point where it meets the circle as ``C ``.
Join ``BC `` and ``CD ``. Measure and write down the length of side ``CD `` of quadrilateral ``ABCD ``.
Ans :
Step 1:Draw a circle of radius 3.5 cm.
Step 2: Then take a point A and draw an arc 6 cm long to intersect the circle at B. Join AB. That is the chord
Step 3: Locus of equidistant points for AB is the perpendicular bisector. Draw perpendicular bisector.
Step 4: Mark point D. Join AD.
Step 5: Draw an angle bisector of angle DAB. This is the locus for equidistant point for point on line AD and AB.
Step 6: Measure CD.

#11-a
A model of a high rise building is made to a scale of ``1:50 ``.
i) if the height of the model is ``0.8 `` m, find the height of the actual building.
ii) If the floor area of a flat in the building is ``20 \ m^2 ``, find the floor are of that in the model.
Ans : Scale ``= 1:50 ``
``h_m = 0.8 `` m ``A_b = 20 \ m^2 ``
i) ``\frac{1}{50} `` ``= `` ``\frac{h_m}{h_b} ``
``\Rightarrow h_b = h_m \times 50 ``
``\Rightarrow h_b = 0.8 \times 50= 40 `` m
ii) ``\frac{1^2}{50^2} `` ``= `` ``\frac{A_m}{A_b} ``
``\Rightarrow `` ``\frac{1}{2500} `` ``= `` ``\frac{A_m}{20} ``
``\Rightarrow A_m = `` ``\frac{1}{125} `` ``\ m^2 `` or ``A_m = 0.008 \ m^2 ``

#11-b
From a solid wooden cylinder of height ``28 `` cm and diameter ``6 `` cm. Two conical cavities are hollowed out. The diameters of the cone is also ``6 `` cm and height is ``10.5 `` cm.
Taking ``\pi = `` ``\frac{22}{7} `` , find the volume of the remaining solid.
Ans : Volume of a cylinder ``= \pi r^2 H ``
Volume of a cone ``= `` ``\frac{1}{3} `` ``\pi r^2 h ``
Given : ``r = 3 `` cm ``H = 28 `` cm ``h = 10.5 `` cm
Volume of cylinder ``= \pi (3)^2 ( 28) = 252 \pi ``
Volume of cones ``= 2 [ `` ``\frac{1}{3} `` ``\pi (3)^2 (10.5) ] = 63 \pi ``
Therefore Volume of remaining solid ``= 252 \pi - 63 \pi = 189 \times `` ``\frac{22}{7} `` ``= 594 \ cm^3 ``

Distance in m	12-13	13-14	14-15	15-16	16-17	17-18	18-19
Number of Students	3	9	12	9	4	2	1