Previous Year Paper year:2013 : ICSE-X-Mathematics

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ICSE-X-Mathematics

Previous Year Paper year:2013

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ICSE X | Mathematics
Board Paper . 2013
ICSE Board
Class X Mathematics
Board Question Paper 2013
(Two and a half hours)
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 answers.
Attempt all questions from Section A and any four questions from Section B.
All working, including rough work, must be clearly shown and must be done on the 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.

# [40]
Section : A
Attempt all questions from this Section.

#1

#1-a [3]
Given
$$A=\begin{bmatrix}2 &-6 \\ 2 & 0 \\ \end{bmatrix}
B=\begin{bmatrix} -3 & 2 \\ 4 & 0 \\ \end{bmatrix}
C= \begin{bmatrix} 4& 0\\0 & 2 \\ \end{bmatrix}
$$
Find the matrix X such that ``A + 2X = 2B + C``.
Ans : Given ``A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \ and \ C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} ``
Substituting these values in the given expression ``A + 2X = 2B + C`` we get,
``\therefore \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} + 2X = 2 \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} ``
``2X = \begin{bmatrix} -6+4 & 4+0 \\ 8+0 & 0+2 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} ```
``X = \frac{1}{2} \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} ``
``X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} ``

#1-b [3]
At what rate % p.a. will a sum of Rs. 4000 yield Rs. 1324 as compound interest in 3
years?
Ans : Given: Principle ``=Rs.4000,\ C.I.=Rs.1324 ``
Amount = ``P+C.I = 4000+1324= Rs. \ 5324 ``
Time ``=3 \ years ``
We know that; ``A=P(1+\frac{r}{100})^n ``
``5324=4000 (1+\frac{r}{100})^3 ``
``\frac{5324}{4000}=(1+\frac{r}{100})^3 ``
``\frac{1331}{1000}=(1+\frac{r}{100})^3 ``
``(\frac{11}{10})^3 =(1+\frac{r}{100})^3 ``
Therefore, ``1+\frac{r}{100}=\frac{11}{10}``
``\frac{r}{100}=\frac{11}{10}-1 ``
``\frac{r}{100}=\frac{1}{10} ``
``r=\frac{100}{10} ``
``r=10\% ``

#1-c [4]
The median of the following observations
11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24.
Find the value of x and hence find the mean.
Ans : Given observation are ``11,12,14, (x-2), (x+4), (x+9), 32, 38, 47 `` and mediam =24
Since n=9 which is odd, therefore
Median=``\ \frac{n+1}{2}th\ term =\frac{9+1}{2}th\ term ``
``24=5th\ term ``
x+4=24
``\Rightarrow x=24-4 \Rightarrow x=20 ``
Therefore, 11,12,14, (20-2), (20+4), (20+9),32,38,47
=11,12,14,18,24,29,32,38,47
Now, ``Mean =\frac{\sum{x}}{n} ``
``=\ \frac{11+12+14+18+24+29+32+38+47}{9} =\frac{225}{9} =25 ``

#2

#2-a [3]
What number must be added to each of the numbers 6, 15, 20 and 43 to make them
proportional?
Ans : Let the number that must be added be x , then
The new number = ``(6 + x), (15 + x),(20 + x),(43 + x) ``
Since they are proportional,
``(6+x) : (15+x) :: (20+x):(43+x) ``
``(6+x)(43+x) = (15+x)(20+x) ``
``\Rightarrow 258+6x+43x+x^2 = 300+20x+15x+x^2 ``
``\Rightarrow 49x - 35x = 300-258 ``
``\Rightarrow 14x = 42 \Rightarrow x= 3 ``

#2-b [3]
If (x - 2) is a factor of the expression ``2x^3 + ax^2 + bx - 14`` and when the expression is
divided by (x - 3), it leaves a remainder 52, find the values of a and b.
Ans : Let (x-2) is a factor of the given expression;
Since ``x-2=0 \Rightarrow x=2 ``
``2x^3 + ax^2 + bx - 14 = 0 ``
In the given expression, we substitute ``x = 2 2(2^3)+a(2^2)+b(2)-14 = 0`` we get
`
``16+4a+2b- 14 = 0 ``
``4a+2b+2 = 0 ``
``4a +2b = -2``
``2a+b = -1 . . . . . (i)``
When given expression is divided by (x - 3)
``x-3 = 0 \Rightarrow x=3 ``
Similarly, in the given expression, we substitute ``x = 3 2x^3 +ax^2+bx-14 = 52 `` we get
``2 (3)^3 + a(3)^2 + b(3) - 66 = 0 ``
``54+9a+3b-66 = 0 ``
``9a+3b = 12 ``
``3a+b = 4 . . . . . (ii)``
Solving equation (i) and (ii),
``a = 5 \ and \ b = -11 ``

#2-c [4]
Draw a histogram from the following frequency distribution and find the mode from
the graph:

Class 0-5 5-10 10-15 15-20 20-25 25-30
Frequency 2 5 18 14 8 5
Ans :

#3

#3-a [3]
Without using tables evaluate:
$$ 3\cos 80°\cdot cosec 10° + 2\sin 59° \cdot sec 31°$$
Ans : ``3 \ {\mathrm{cos \ } 80{}^\circ \ cosec\ 10{}^\circ +2 \ {\mathrm{sin \ } 59{}^\circ \ sec \ 31{}^\circ \ }\ } ``
``= 3 \ {\mathrm{cos \ } 80{}^\circ \ cosec\ \left(90{}^\circ -80{}^\circ \right)+2\ {\mathrm{sin \ } 59{}^\circ \ {\mathrm{sec \ } \left(90{}^\circ -59{}^\circ \right)\ }\ }\ } ``
``= 3 \ cos \ 80° sec \ 80° + 2 \ sin \ 59° cosec \ 59° ``
``= 3 \ {\mathrm{cos \ } 80{}^\circ \ \times \frac{1}{{\mathrm{cos \ } 80{}^\circ \ }}+2 \ {\mathrm{sin \ } 59{}^\circ \times \frac{1}{sin \ 59{}^\circ }\ }\ } ``
``= 3+2=5 ``

#3-b [3]
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
Ans : Given: ``\angle BAD=65° ,\ \angle ABD=70° ,\ \angle BDC=45°``

#3-b-i
Prove that AC is a diameter of the circle.
Ans : Since ABCD is a cyclic quadrilateral
In ``\triangle \ ABD ``
``\angle BDA+\angle DAB+\angle ABD=180°`` ( sum property of a triangle)
``\angle BDA=180° -(65° +70°) =180° -135° =45° ``
Now from ``\triangle ACD ,``
``\angle ADC=\angle ADB+\angle BDC =45° +45°=90° ``
Hence ``\angle ADC`` makes right angle belongs in semi-circle therefore AC is a diameter of the circle.

#3-b-ii
Find ACB
Ans : ``\angle ACB = \angle ADB`` (Angles in the same segment of a circle)
Therefore ``\angle ABC = 45° \ (since \ \angle ADB = 45°)``