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ICSE-X-Mathematics

Previous Year Paper year:2020

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#4-a [3]
A solid spherical ball of radius ``6 `` cm is melted and recasted in ``64 `` identical spherical marbles. Find the radius of each marble.
Ans : Let ``R `` be the radius of spherical ball and ``r `` be the radius of the spherical marble
Volume of spherical ball ``= `` ``\frac{4}{3} `` ``\pi R^3 = `` ``\frac{4}{3} `` ``\pi (6)^3 ``
Volume of spherical marble ``= `` ``\frac{4}{3} `` ``\pi r^3 ``
``\therefore `` ``\frac{4}{3} `` ``\pi (6)^3 = 64 \times `` ``\frac{4}{3} `` ``\pi r^3 ``
``r^3 = `` ``\frac{6^3}{64} `` ``= `` ``\frac{6^3}{4^3} ``
``\Rightarrow r = `` ``\frac{6}{4} `` ``= `` ``\frac{3}{2} `` cm

#4-b [3]
Each of the letters of the word ``'AUTH
i) a vowel
ii) One of the first ``9 `` letters of the English alphabet which appears in a given word
iii) One of the last ``9 `` letters of the English alphabet which appears in a given word
Ans : First 9 letters: ``A, B, C, D, E, F, G, H, I ``
Last 9 letters: ``R,S, T, U, V, W, X, Y, Z ``
Total outcomes ``= 10 ``
i) No. of vowels ``= 5 ``
Therefore probability ``P(Vowels) = `` ``\frac{5}{10} `` ``= `` ``\frac{1}{2} ``
ii) There are ``4 `` probable outcomes
Therefore probability ``P( \ first \ 9 \ letters\ ) = `` ``\frac{4}{10} `` ``= `` ``\frac{2}{5} ``
iii) There are ``5 `` probable outcomes
Therefore probability ``P( \ last \ 9 \ letters \ ) = `` ``\frac{5}{10} `` ``= `` ``\frac{1}{2} ``

#4-c [4]

Mr. Bedi visits the market and buys the following articles:
Medicines costing Rs. ``950 ``, GST @ ``5\% ``
A pair of shoes costing Rs. ``3000 ``, GST @ ``18\% ``
A laptop bag costing Rs ``1000 `` with a discount or ``30\% ``, GST @ ``18\% ``
i) Calculate the total amount of GST paid
ii) The total bill amount including GST paid by Mr. Bedi

Ans :

Article	Cost ( Rs.)	Final Cost (Rs.)	GST Rate	GST ( Rs.)	Final Price (Rs.)
Medicines	950	950	5%	47.50	997.50
Shoes	3000	3000	18%	540	3540
Laptop Bag	1000 @ 30% discount	700	18%	126	829
Total (Rs.)				713.50	5363.50

Therefore
i) Calculate the total amount of GST paid ``= 713.50 `` Rs.
ii) The total bill amount including GST paid by Mr. Bedi ``= 5363.50 `` Rs.

#
Section : B
[40 Marks]
(Attempt any
fourquestions from this Section.)

#5-a [3]
A company with ``500 `` shares of nominal value Rs. ``120 `` declares and annual dividend of ``15\% ``. Calculate:
i) the total amount of dividend paid by the company
ii) annual income of Mr. Sharma who holds ``80 `` shares of the company.
If the return percent of Mr. Sharma for his shares is ``10\% ``, find the market value of each share.
Ans : Number of shares ``= 500 ``
Nominal Value ``= 120 `` Rs.
Dividend ``= 15\% ``
i) Dividend ``= n \times NV \times `` ``\frac{div \%}{100} `` ``= 500 \times 120 \times `` ``\frac{15}{100} `` ``= 9000 `` Rs.
ii) ``n = 80 `` shares
Dividend ``= 80 \times 120 \times `` ``\frac{5}{100} `` ``= 1440 `` Rs.
Return ``\% = `` ``\frac{Annual \ Income }{Investment} `` ``\times 100 ``
``\Rightarrow 10 = `` ``\frac{1440}{I} `` ``\times 100 ``
``\Rightarrow I = `` ``\frac{1440 \times 100}{10} `` ``= 14400 `` Rs.
Therefore Market Value ``= `` ``\frac{14400}{80} `` ``= 180 `` Rs.

#5-b [3]
The mean of the following data is ``16 ``. Calculate the value of ``f ``.

Marks 5 10 15 20 25

No. of Students 3 7 ``f `` 9 6

Ans :

Marks ``( x) `` No. of Students ``(f) `` ``fx ``

5 3 15

10 7 70

15 ``f `` ``15f ``

20 9 180

25 6 150

``\Sigma f = 25 + f `` ``\Sigma fx = 415 + 15f ``

Given: Mean ``= 16 ``
We know ``\overline{x} = `` ``\frac{\Sigma fx}{\Sigma f} ``
``\Rightarrow 16 = `` ``\frac{415+15f}{25+f} ``
``\Rightarrow 400 + 16f = 415 + 15f ``
``\Rightarrow f = 15 ``

#5-c [4]
The ``4^{th}, 6^{th} `` and the last term of a geometric progression are ``10, 40 `` and ``640 `` respectively. If the common ratio is positive, find the first term, common ration and the number of terms of the series.
Ans : ``4^{th} `` term ``= 10 `` ``6^{th} `` term ``= 40 `` Last term ``= 640 ``
We know, ``t_n = ar^{n-1} ``
``\therefore 10 = a r^{4-1} `` ``\Rightarrow 10 = ar^3 `` ... ... ... ... ... i)
Similarly, ``40 = a r^{6-1} `` ``\Rightarrow 40 = ar^5 `` ... ... ... ... ... ii)
Dividing ii) by i) we get
``\frac{ar^5}{ar^3} `` ``= `` ``\frac{40}{10} ``
``\Rightarrow r^2 = 4 `` ``\Rightarrow r = 2 ``
Using i) we get ``10 = a (2)^3 ``
``\Rightarrow a = `` ``\frac{10}{8} `` ``= `` ``\frac{5}{4}``
We know that ``t_n = a r^{n-1} ``
``\therefore 640 = `` ``\frac{5}{4} `` ``2^{n-1} ``
``\Rightarrow 2^{n-1} = \frac{640 \times 4}{5} ``
``\Rightarrow 2^{n-1} = 512 ``
``\Rightarrow 2^{n-1} = 2^9 ``
``\Rightarrow n = 10 ``
``\\ ``

#6-a [3]
If ``A = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} `` and ``B = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} ``. Find: ``A^2 - 2AB + B^2 ``
Ans : Given
``A = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} `` and ``B = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} ``
``A^2 = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} ``
``B^2 = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} ``
``AB = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix} ``
``\therefore A^2 - 2AB + B^2 = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} - 2 \begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix} ``

#6-b [3]
In the given figure ``AB = 9 `` cm, ``PA = 7.5 `` cm and ``PC = 5 `` cm
Chord ``AD `` and ``BC `` intersect at ``P ``.
i) Prove that ``\triangle PAB \sim \triangle PCD ``
ii) Find the length of ``CD ``
iii) Find area of ``\triangle PAB : `` area of ``\triangle PCD ``
Ans :
i) Consider ``\triangle PAB `` and ``\triangle PCD ``
``\angle BPA = \angle DPC `` ( Vertically opposite angles)
``\angle ABC = \angle ADC `` (angles subtended by an arc on the circumference are equal)
``\therefore \triangle PAB \sim \triangle PCD `` (By AA similarity criterion)
ii) Since ``\triangle PAB \sim \triangle PCD ``
``\Rightarrow `` ``\frac{PA}{PC} `` ``= `` ``\frac{AB}{CD} `` ``= `` ``\frac{PB}{PD} ``
``\Rightarrow `` ``\frac{7.5}{5} `` ``= `` ``\frac{9}{CD} ``
``\Rightarrow CD = `` ``\frac{5 \times 9}{7.5} `` ``= 6 `` cm
iii) ``\frac{ar. of \triangle PAB}{ar. of \triangle PCD} `` ``= `` ``\frac{9^2}{6^2} `` ``= \frac{9}{4} ``

#6-c [4]
From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be ``45^{\circ} `` and ``60^{\circ} `` respectively. If the height of the tower is ``20 `` m, Find:
i) the height of the cliff
ii) the distance between the cliff and the tower.
Ans :
In ``\triangle ABC ``
``\tan 60^{\circ} = `` ``\frac{h}{x} `` ``\Rightarrow \sqrt{3} = `` ``\frac{h}{x} `` ``\Rightarrow h = \sqrt{3} x `` ... ... ... ... ... i)
In ``\triangle CDE ``
``\tan 45^{\circ} = `` ``\frac{h-20}{x} `` ``\Rightarrow 1 = `` ``\frac{h-20}{x} `` ``\Rightarrow x = h - 20 `` ... ... ... ... ... ii)
Substituting ``x = \sqrt{3}x - 20 ``
``\Rightarrow ( \sqrt{3} -1 ) x = 20 ``
``\Rightarrow x = `` ``\frac{20}{\sqrt{3} -1} `` ``\times `` ``\frac{\sqrt{3}+1}{\sqrt{3}+1} `` ``= 10 ( 1.732+1) = 27.32 `` m
``\therefore h = \sqrt{3} x = 1.732 \times 27.32 = 47.32 `` m
i) the height of the cliff ``= 47.32 `` m
ii) the distance between the cliff and the tower ``= 27.32 `` m
``\\ ``

#7-a [3]
Find the value of ``'p' `` is the lines ``5x - 3y + 2 = 0 `` and ``6x - py + 7 = 0 `` are perpendicular to each other. Hence find the equation of a line passing through ``(-2, -1) `` and parallel to ``6x - py + 7 = 0 ``.
Ans : Given ``5x - 3y + 2 = 0 \Rightarrow 3y = 5x + 2 \Rightarrow y = `` ``\frac{5}{3} `` ``x + `` ``\frac{2}{3} ``
Therefore slope ``m_1 = `` ``\frac{5}{3} ``
Similarly, ``6x - py + 7 = 0 \Rightarrow py = 6x + 7 \Rightarrow y = `` ``\frac{6}{p} `` ``x + `` ``\frac{7}{p} ``
Therefore slope ``m_2 = `` ``\frac{6}{p} ``
We know ``m_1 m_2 = - 1 ``
``\Rightarrow `` ``\frac{5}{7} `` ``\times `` ``\frac{6}{p} `` ``= -1 \Rightarrow p = -10 ``
Therefore slope ``m_2 = `` ``\frac{6}{-10} `` ``= - `` ``\frac{3}{5} ``
Therefore equation of line
``(y+1) = - `` ``\frac{3}{5} `` ``( x + 2) ``
``\Rightarrow 5y + 5 = -3x - 6 ``
``\Rightarrow 3x + 5y + 11 = 0 ``

#7-b [3]
Using properties of proportion find ``x:y `` given ``\frac{x^2 + 2x}{2x + 4} `` ``= `` ``\frac{y^2 + 3y}{3y + 9} ``
Ans : ``\frac{x^2 + 2x}{2x + 4} `` ``= `` ``\frac{y^2 + 3y}{3y + 9} ``
Applying componendo and dividendo
``\frac{x^2 + 2x + 2x + 4}{x^2 + 2x - 2x - 4} `` ``= `` ``\frac{y^2 + 3y + 3y + 9}{y^2 + 3y -3y - 9} ``
``\Rightarrow `` ``\frac{x^2 + 4x + 4}{x^2 - 4} `` ``= `` ``\frac{y^2 + 6y + 9}{y^2 - 9} ``
``\Rightarrow `` ``\frac{(x+2)^2}{(x+2)(x-2)} `` ``= `` ``\frac{(y+3)^2}{(y+3)(y-3)} ``
``\Rightarrow `` ``\frac{x+2}{x-2} `` ``= `` ``\frac{y+3}{y-3} ``
Applying componendo and dividendo
``\frac{x+2 + x - 2}{x+2 -x+2} `` ``= `` ``\frac{y+3 + y - 3}{y+3-y+3} ``
``\Rightarrow `` ``\frac{2x}{4} `` ``= `` ``\frac{2y}{6} ``
``\Rightarrow `` ``\frac{x}{y} `` ``= `` ``\frac{2}{3} ``
``\therefore x : y = 2:3 ``

Marks ``( x) ``	No. of Students ``(f) ``	``fx ``
5	3	15
10	7	70
15	``f ``	``15f ``
20	9	180
25	6	150
	``\Sigma f = 25 + f ``	``\Sigma fx = 415 + 15f ``