Other Solved Mathematics Board Papers
MATHEMATICS (ICSE – Class X Board Paper 2012)
Two and Half Hour. 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 form 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 the loss of marks.
The intended marks for questions or parts of questions are given in brackets [ ].
Mathematical tables are provided.
SECTION A [40 Marks]
(Answer all questions from this Section.)
Question 1:
(b) The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. There expenditures are in the ratio 3:5. If each saves Rs. 80 every month, find their monthly pocket money. [3]
(c) Using the Reminder Theorem factories completely the following polynomial: [4]
Answers:
(b) Let monthly pocket money be Ravi is and Sanjeev is
They both save per month.
Therefore, their expenditure would be respectively.
Ravi’s pocket money
Sanjeev’s pocket money
(c) Given
Try
is not a factor of the given function.
Try
Therefore is a factor of
To factories
Hence,
Question 2:
(a) On what sum of money will the difference between the compound interest and simple interest for 2 years be equal Rs. 25 if the rate of interest charged for both is 5% p.a.? [3]
(b)
is an is isosceles right-angled triangle with
A semi-circle is drawn with
as the diameter. If
find the area of the shaded region.
[3]
(c) Given a line segment joining the points
Find:
(i) the ratio in which is divided by the
(ii) find the coordinates of the point of intersection.
(iii) the length of [4]
Answers:
(a) Let the sum be
Amount Compound Interest
Given difference = 25 Rs.
Therefore
(b) is a right angled triangle. Therefore
Area of the shaded region = Area of the semi circle – Area of
(c) Let the required ratio be and the point of intersection
be
Question 3:
(a) In the given figure
is the central of the circle and
is a tangent at
If
Calculate the radius of the circle. [3]
(b) Evaluate without using trigonometric tables:
[3]
(c) Marks obtained by students in a short assessment is given below, where
are two missing data:
Marks | 5 | 6 | 7 | 8 | 9 |
No. of Students | 6 | A | 16 | 13 | B |
If the mean of the distribution is 7.2 find [4]
Answers:
(a) Let the radius of the circle
Here we apply intercept theorem. Therefore:
(c) Given, the total number of students
Therefore
.. … … (i)
Therefore
.. … … (ii)
Solving (i) and (ii) we get
Question 4:
(a) Kiran deposited Rs. 200 per month for 36 months in a Bank’s recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount she gets on maturity. [3]
(b) Two coins are tossed once; Find the probability of getting:
(i) 2 heads
(ii) At 1 tail [3]
(c) Using graph paper and taking 1 cm=1 unit along both ;
(i) Plot the points
(ii) Reflect in the origin to get the images
respectively.
(iii) Write down the co-ordinates of
(iv) Give the geometrical name for the figure
(v) Draw and name its lines of symmetry. [4]
Answers:
(a)
(b) Let Heads – and Tails –
If two coins are tossed once, then the total number of possibilities would be as shown: Sample Space
(i) Event: getting two heads
(ii) Events : At least one tail
(c)
(i) Please refer to the graph shown below
(ii) Please refer to the graph shown below
(iii) Reflection of in the origin are
respectively.
(iv) Name of the geometrical figure in the graph show is Rhombus
(v) Two lines of symmetry: Both diagonal
SECTION B [40 Marks]
(Answer any four questions in this Section.)
Question 5:
(a) In the given figure,
is the diameter of a circle with center
Find (i)
(ii)
[3]
(c) A page from the savings Bank Account of Mr. Prateek is given below: [4]
Date | Particular | Withdrawal | Deposit | Balance |
Jan. 1st 2006 | B/F | – | – | 1,270 |
Jan. 7th 2006 | By Cheque | – | 2310 | 3580 |
March 9th 2006 | To Self | 2000 | – | 1580 |
March 26th 2006 | By Cash | 6200 | 7780 | |
June 10th 2006 | To Cheque | 4500 | – | 3280 |
July 15th 2006 | By Clearing | – | 2630 | 5910 |
October 18th 2006 | To Cheque | 530 | – | 5380 |
October 27th 2006 | To Self | 2690 | – | 2690 |
November 3rd 2006 | By Cash | – | 1500 | 4190 |
December 6th 2006 | To Cheque | 950 | – | 3240 |
December 23rd 2006 | By Transfer | – | 2920 | 6160 |
If he receives Rs. 198 as interest on 1st January, 2007. Find the rate of interest paid by the bank.
Answers:
(a)
(i)
(ii) In
Therefore Hence the order of Matrix is
Therefore
Therefore
and
Solving we get
(c) Qualifying principal for various months:
Month | Principal (Rs.) |
January | 3580 |
February | 3580 |
March | 1580 |
April | 7780 |
May | 7780 |
June | 3280 |
July | 3280 |
August | 5910 |
September | 5910 |
October | 2690 |
November | 4190 |
December | 3240 |
Total | 52800 |
Therefore Rate
Question 6:
(a) The principal price of an article is The wholesaler allows a discount of
to the shopkeeper. The shopkeeper sells the article to the customer at the printed price. Sales tax (under VAT) is charged at the rate of
at every stage. [4]
Find:
(i) The cost of the shopkeeper inclusive of tax
(ii) VAT paid by the shopkeeper to the Government.
(iii) The cost of the customer inclusive of tax.
(b) Solve the following inequation and represent the solution set on the number line:
[3]
(c) Without solving the following quadratic equation, find the value of for which the given equation has real and equal roots.
[3]
Answers:
(a) Printed Price
The cost to the shopkeeper inclusive of the tax
VAT paid by the shopkeeper
The cost to the customer inclusive of the tax
Therefore
For roots to be equal, we should have
Question 7:
(a) A hollow sphere of internal and external radii respectively is melted and recast into small cones of base radius
and height
Fins the number of cones. [3]
(b) Solve the following equation and give your answer correct to significant figures:
[3]
(c) As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the light house in horizontal line with its base are Find the distance between the two ships. Give your answer correct to the nearest meter. [4]
Answers:
(a) Sphere: Internal radius , External radius
Cone: Radius , Height
(b) Given :
Comparing this expression with , we get
Therefore
Question 8:
(a) A man invests Rs. 9,600 on Rs. 100 shares at Rs. 80. If the company pays him 18% dividend. Find:
(i) The number of shares he buys
(ii) His total dividend.
(iii) His percentage return on the shares. [3]
(b) In the given figure
are right angled at
respectively. Given
(i) Prove
(ii) Find [3]
Answers:
(a) Investment , Dividend
Nominal Value of the share
Market Value of the share
(b) In
(common angle)
(right angles)
Therefore (AAA postulate)
Since
Given
Applying componendo and dividendo
Simplify
Now square both sides
Simplifying
or
Question 9:
(a) The line through is perpendicular to the line
Find the value of
. [3]
(c) A car covers a distance of 400 km at a certain speed. Had the speed been 12 km per hours more, the line taken for the journey would have been 1 hour and 40 minutes less. Find the original speed of the care. [4]
Answers:
Given equation is
Since they are perpendicular,
(c) Let the speed of the car be
Therefore speed of the car is 48 km/hr.
Question 10:
(a) Construct a in which base
(i) Construct a circle circumscribing the
(ii) Draw a cyclic quadrilateral so that
is equidistant from
[4]
(b) The following distribution represents the height of 160 students of a school
Height | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 | 175-180 |
Students | 12 | 20 | 30 | 38 | 24 | 16 | 12 | 8 |
Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
(i) The medium height
(ii) The interquartile range
(iii) The number of students whose height is above 172 cm. [6]
Answers:
(a)
(i) Following Step to Constructions:
- First, using a ruler, draw a line segment
- Construct
This you can do it by using a compass. Draw an arc with B as the center. The using the same arc and intersection point ‘x”, draw two arcs one at ‘y’ and then one at ‘z’. Craw a line through B and ‘z’. You will get
angle.
- The cut
from
This you can do by setting the compass for a width of 5.5 cm.
- Then join
So you get the
.
- Now construct perpendicular bisectors of
intersecting at
Join
- Taking O as the center and
as radius draw a circle, passing through
(ii)
- Extend the right bisector of
intersecting the circle at
- Join
is required cyclic quadrilateral.
(b) Following table:
Height | F | c.f. |
140-145
145-150 150-155 155-160 160-165 165-170 170-175 175-180 |
12
20 30 38 24 16 12 8 |
12
32 62 100 124 140 152 160 |
(i) Mean
(ii) Interquartile range
(iii) No. of students above
Question 11:
(a) In
Find the radius of the inscribed circle. [3]
(b) Find the mode and medium of the following frequency distribution: [3]
X | 10 | 11 | 12 | 13 | 14 | 15 |
f | 1 | 4 | 7 | 5 | 9 | 3 |
(c) The line through
intersects
at
(i) Write the slope of the line
(ii) Write the equation of the line
(iii) Find the Co-ordinates of [4]
Answers:
(a) Given
is a right triangle. Therefore using Pythagoras theorem:
Draw as shown. They will be perpendicular to the sides of the triangle.
is a square since the angels are
(Since the tangent to a circle from an exterior point are equal in length).
Therefore
(Since the tangent to a circle from an exterior point are equal in length).
(Since the tangent to a circle from an exterior point are equal in length).
Since
(b) Mode is the value of the highest frequency.
Therefore Mode
For Median, first write the data in ascending order as follows:
Since Median is the middle most value.
(c) Given points
Slope
Equation of line:
When
Hence the co-ordinates of