Previous Year Paper year:2010 : ICSE-X-Mathematics

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ICSE-X-Mathematics

Previous Year Paper year:2010

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ICSE X | Mathematics
ICSE Board
Class X Mathematics
Board Question Paper 2010
(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]
Solve the following in equation and represent the solution set on the number line.
$$R - 3 \lt -\frac12 - \frac{2x}{3} \le \frac{5}{6} ,x \epsilon R $$
Ans : ``-3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}``
``-3 < -\frac{1}{2}-\frac{2x}{3}``
``-18 < -3 -4x``
``4x < 15 or x < \frac{15}{4}``
``-\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6} or -3-4x \leq 5``
``-8 \leq 4x or -2 \leq x ``
Therefore ``\{ x: -2 \leq x < \frac{15}{4}, x \in R \}``

#1-b [3]
Tarun bought and article for Rs. 8000 and spent Rs. 1000 for transportation. He marked
the article Rs. 11,700 and sold it to a customer. If the customer had to pay 10% sales
tax, find:
Ans : Cost Price = Rs. 8000
Overheads = Rs. 1000
Listed Price = Rs. 11700
Sales Tax rate ``= 10\% ``

#1-b-i
the customer.s price
Ans : Customer Price = ``11700+11700 \times \frac{10}{100} = Rs. 12870``

#1-b-ii
Tarun.s profit percent.
Ans : Profit ``= \frac{11700-8000-1000}{9000} = 30\%``

#1-c [4]
Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs. 2500 per
month for two years. At the time of maturity he got Rs. 67,500. Find:
Ans : ``P = Rs. \ 2500, \ no \ of \ months = 24, \ \\ \\ rate = r\% \ Maturity \ Amount \ = Rs. 67500``
Maturity ``\ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}``
``67500 =2500 \times 24 +2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100} ``
``r = \frac{(67500-2500 \times 24) \times (2 \times 12) \times 100}{2500 \times 24 \times 25} \Rightarrow r=12\% ``
Interest ``=2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{12}{100} = Rs. \ 7500 ``

#1-c-i
the total interest earned by Mr. Gupta.

#1-c-ii
the rate of interest per annum.

#2

#2-a [3]
$$
\text{Given }A= \begin{bmatrix} 3 & -2 \\ -1 & 4 \\ \end{bmatrix}
,B= \begin{bmatrix} 6 \\ 1\\ \end{bmatrix}
,C= \begin{bmatrix} -4 \\5 \\ \end{bmatrix}
and D = \begin{bmatrix} 2 \\ 2 \\ \end{bmatrix}
$$
Find ``AB + 2C -4D``.
Ans : AB+2C-4D
``= \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}. \begin{bmatrix} 6 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \begin{bmatrix} 2 \\ 2 \end{bmatrix}``
``= \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}``
``= \begin{bmatrix} 0 \\ 0 \end{bmatrix}``

#2-b [3]
Nikita invests Rs. 6000 for two years at a certain rate of interest compounded annually.
At the end of first year it amounts to Rs. 6720. Calculate:
Ans : Compound Interest for 1 year
``P=6000\ Rs.; \ r=x\%; Compounded \ yearly \ n=1 \ year``
``A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{x}{100})^{1}``
Given ``6000(1+\frac{x}{100})^{1}=6720 \Rightarrow x= 12\% ``
Amount at the end of second year
``A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{12}{100})^{2} = 7526.40 \ Rs.``

#2-b-i
the rate of interest.

#2-b-ii
the amount at the end of the second year.