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ICSE Class 10 Maths Question Paper 2013 Solved
Time Allowed: 2 1/2 hours
Max. Marks : 80
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.
Section-A (40 Marks)
(Attempt all questions from this section)
Question 1.
(a) Given A = \(\left[\begin{array}{cc}
2 & -6 \\
2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-3 & 2 \\
4 & 0
\end{array}\right]\), C = \(\left[\begin{array}{ll}
4 & 0 \\
0 & 2
\end{array}\right]\)
Find the matrix X such that A + 2X = 2B + C. [3]
(b) At what rate % p.a. will a sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years? [3]
Answer:
(c) 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. [4]
Solution:
(a) Given that A + 2X = 2B + C
(b) Here, P = ₹ 4000, C.I. = ₹ 1324, T = 3 years r
⇒ \(\frac{r}{100}\) = \(\frac{1}{10}\)
⇒ r = 10%
(c) Given observations in ascending order are :
11, 12, 14, x – 2, x + 4, x + 9, 32, 38 and 47
Here, number of observations are 9 (odd number)
∴ Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{9+1}{2}\right)^{t h}\) term = 5th term
⇒ x + 4 = 24
⇒ x = 20 [∵ Median = 24(given)]
Thus, the observations are 11, 12, 14, 18, 24, 29, 32, 38, 47.
Now, mean = \(\frac{11+12+14+18+24+29+32+38+47}{9}\)
= \(\frac{225}{9}\) = 25
Question 2.
(a) What number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional? [3]
(b) If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. [3]
(c) Draw a histogram from the following frequency distribution and find the mode from the graph : [4]
| Class | Frequency |
| 0-5 | 2 |
| 5-10 | 5 |
| 10-15 | 18 |
| 15-20 | 14 |
| 20-25 | 8 |
| 25-30 | 5 |
Solution:
(a) Let x be added to each number 6, 15, 20 and 43 to make these numbers proportional.
∴ (6 + x) : (15 + x) = (20 + x) : (43 + x)
⇒ \(\frac{6+x}{15+x}\) = \(\frac{20+x}{43+x}\)
⇒ (6 + x) (43 + x) = (15 + x) (20 + x)
⇒ 258 + 6x + 43x + x2 = 300 + 15x + 20x + x2
⇒ 258 + 49x = 300 + 35x
⇒ 49x – 35x = 300 – 258
⇒ 14x = 42
⇒ x = 3
(b) Since (x – 2) is a factor of 2x3 + ax2 + bx – 14.
⇒ 2(2)3 + a(2)2 + 6(2) – 14 = 0
⇒ 16 + 4a + 2b – 14 = 0
⇒ 4a + 2b = – 2
⇒ 2a + b = -1 …..(i)
Also, when 2x3 + ax2 + bx – 14 is divided by x – 3, it leaves remainder 52
⇒ 2(3)3 + a(3)2 + 6(3) – 14 = 52
⇒ 54 + 9a + 36 = 52 + 14
⇒ 9a + 3b = 66 – 54
⇒ 9a + 3b = 12
⇒ 3a + b = 4 …(ii)
Subtracting (i) from (ii), we have
a = 5
From (i), we have
2(5) + b = – 1
⇒ b = -11
Hence, the required values of a and b are a = 5 and 6 = – 11.
(c) Required histogram from the given frequency distribution is:
Now, in the highest rectangle, draw two straight lines AC and BD from the comers of the rectangles on either ‘sides of the highest rectangle to the opposite comers of the highest rectangle. Let they intersect in L. Draw LM ⊥ OX. The abscissa of the point M represents 14. Hence, the required mode is 14.
Question 3.
(a) Without using tables evaluate 3 cos 80°.
Cosec 10° + 2 sin 59° sec 31°. [3]
(b) In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
[4]
sin 59°. cosec 59°
(c) AB is a diameter of a circle with centre C = (-2, 5). IfA = (3, -7). Find
(i) the length of radius AC
(ii) the coordinates of B.
Solution:
(a) 3 cos 80° cosec 1o° + 2 sin 59° sec 31°
= 3 cos (90° – 10°) cosec 1o° + 2 sin (90° – 31°) sec 31°
= 3 sin 10° cosec 1o° + 2 cos 31° sec 31°
[∵ cos(90° – θ) = sinθ and sin(90° – θ) = cos θ]
= 3 sin 10°.\(\frac{1}{\sin 10^{\circ}}\) + 2 cos 31°.\(\frac{1}{\cos 31^{\circ}}\)
= 3 + 2 = 5
(b) (i) By using angle sum property of triangle, in ΔABD,
we have
∠ADB + ∠ABD + ∠BAD = 180°
∠ADB + 70° + 65° = 180°
∠ADB = 180° – 135°
= 45°
Now, ∠ADC = ∠ADB + ∠BDC
∠ADC = 45° + 45°
[∴ ∠BDC = 45° (given)]
∠ADC = 90°
Thus, AC is a diameter.
[∵ angle in a semicircle is 90°]
(ii) Also, ∠ACB = ∠ADB
[∵ angles of the same segment]
⇒ ∠ACB = ∠ADB = 45°
Thus, ∠ACB = 45°
(c) Let coordinates of B be (x, y)
∴ \(\frac{3+x}{2}\) = -2 and \(\frac{-7+y}{2}\) = 5
⇒ 3 + x = – 4and -7 + y = 10
⇒ x = -7 and y = 17
Thus, the coordinates of B are B (-7, 17)
Now, |AC| = \(\sqrt{(-2-3)^2+(5+7)^2}\)
= \(\sqrt{(-5)^2+(12)^2}\)
= \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13
Hence, length of the radius AC is 13 units.
Question 4.
(a) Solve the following equation and calculate the answer correct to two decimal places:
x2 – 5x – 100. [3]
(b) In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ΔABC – ΔDEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ΔABC : area of ΔDEC. [3]
(c) Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter. [4]
Solution:
(a) x2 – 5x – 10 = 0
(b)
Here, AB ⊥ BC, DE ⊥ BC
(i) Now, in ΔABC and ΔDEC
∠C = ∠C [common]
∠ABC = ∠DEC = 90° [given]
∴ ΔABC \(\sim\) ΔDEC [by AA similarity axiom]
(ii)
\(\frac{\mathrm{CD}}{\mathrm{CA}}\) = \(\frac{\mathrm{DE}}{\mathrm{AB}}\) [∵ sides of the similar triangles are proportional]
⇒ \(\frac{C D}{15}\) = \(\frac{4}{6}\)
⇒ CD = \(\frac{4 \times 15}{6}\) = 10 cm
(iii)
Since ΔABC \(\sim\) ΔDEC
(c)
(i)
(ii) Coordinates of A’ and B’ are A’ (-6, -4) and B’(0, -4).
(iii) ABA’B’ is a parallelogram.
(iv) Perimeter of ||gm ABA’B’
= AB + BA’ + A’B’ + B’A
= 6 + 10 + 6 + 10
= 32 units.
Section-B (40 Marks)
(Attempt any four questions from this Section)
Question 5.
(a) Solve the following inequation, write the solution set and represent it on the number line:
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1\(\frac{1}{3}\) < \(\frac{1}{6}\), x ∈ R
(b) Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly installment. [3]
(c) Salman buys 50 shares of face value ₹ 100 available at ₹ 132.
(i) What is his investment?
(ii) If the dividend is 7.5%, what will be his annual income?
(iii) If he wants to increase his annual income by 150, how many extra shares should he buy? [4]
Solution:
(a)
–\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – 1\(\frac{1}{3}\) < \(\frac{1}{6}\), x ∈ R
⇒ –\(\frac{x}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{3}{4}\) < \(\frac{1}{6}\)
⇒ -2x ≤ 3x – 8 < 1
⇒ -2x ≤ 3x – 8 and 3x – 8 < 1
⇒ 8 ≤ 5x and 3x < 9
⇒ \(\frac{8}{5}\) ≤ x and x < 3
⇒ \(\frac{8}{5}\) ≤ x < 3
⇒ 1\(\frac{3}{5}\) ≤ x < 3
(b) Here, rate of interest (r) = 8% p.a.
Maturity amount = ₹ 8088
Time = 3 years
or n = 3 × 12 = 36
Let the monthly installment be ₹ x
∴ 36x + x × \(\frac{36(36+1)}{2}\) × \(\frac{1}{12}\) × \(\frac{8}{100}\) = 8088
36x + x × \(\frac{111}{25}\) = 8088
\(\frac{36 \times 25 x+111 x}{25}\) = 8088
\(\frac{1011}{25} x\) = 8088
x = 8088 × \(\frac{25}{1011}\)
x = 8 × 25 = 200
Hence, the required monthly installment is ₹ 200
(c) Here, face value of a share = ₹ 100
Market value of a share = ₹ 132
No. of shares = 50
(i) His investment ₹ 132 × 50
= ₹ 6600
(ii) His annual income = ₹ \(\frac{7.5}{100}\) × 50 × 100
= ₹ 375
(iii) Total income required = ₹ (375 + 150)
= ₹ 525
⇒ x = \(\frac{525 \times 100}{7.5}\)
= ₹ 7000
∴ Extra investment = ₹ (7000 – 5000)
= ₹ 2000
Thus, number of extra shares required
= \(\frac{2000}{100}\) = 20
Question 6.
(a) Show that \(\sqrt{\frac{1-\cos A}{1+\cos A}}\) = \(\frac{\sin A}{1+\cos A}\)
(b) In the given circle with centre O, ∠ABC = 1000, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
(c) Given below are the entries in a Savings Bank A/c pass book:
| Date | Particulars | Withdrawls | Deposit | Balance |
| Feb 8
Feb 18 April 12 June 15 July 8 |
B/F
To self By cash To self By cash |
–
₹ 4000 – ₹ 5000 – |
–
– ₹ 2230 – ₹ 6000 |
₹ 8500
– – – – |
Calculate the interest for six months from February to July at 6% p.a.
Solution:
(a)
(b)
Here, ABCD is a cyclic quadrilateral.
∴ ∠ADC + ∠ABC = 180°
∠ADC + 100° = 180°
⇒ ∠ADC = 180° – 100° = 80°
Thus, ∠ADC = 80°
Now, in ΔACD, by using angle sum property, we have
∠CAD + ∠ACD + ∠ADC = 180°
∠CAD + 40° + 80° = 180°
∠CAD = 180° – 40° – 80° = 60°
Thus, ∠DCT = ∠CAD = 60°
[∠s in the corresponding alternate segment]
(c)
Minimum balance for the month:
February = ₹ 4500
March = ₹ 4500
April = ₹ 4500
May = ₹ 6730
June = ₹ 1730
July = ₹ 7730
Total principle for one month = ₹ 29690
∴ Interest = \(₹ \frac{29690 \times 6 \times 1}{100 \times 12}\) = ₹ 148.45
Question 7.
(a) In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A. [3]
(b) A shopkeeper sells an article at the listed price of ₹ 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of ₹ 36 to the Government, what was the price, inclusive of Tax, at which the shopkeeper purchased the articles from the wholesaler? [3]
(c) In the figure given, from the top of a building AB =60 m high, the angles of depression of the top
and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post. [4]
Solution:
(a) Here, coordinates of the vertices of ΔABC are A(3, 5), B(7, 8) and C(1, -10).
For equation of the median through A, we have D as mid-point of BC and coordinates of D are:
D\(\left(\frac{7+1}{2}, \frac{8-10}{2}\right)\) i.e., D(4, -1)
Now, equation of AD is given by
y – 5 = \(\left(\frac{-1-5}{4-3}\right)\)(x – 3)
y – 5 = -6(x – 3)
y – 5 = -6x + 18
6x + y – 23 = 0
(b)
Let selling price of an article by the wholesaler be ₹ x.
Rate of VAT = 12%
∴ VAT charged = ₹ \(\frac{12}{100}\) × x
= ₹ 180
Selling price of the article by the shopkeeper = ₹ 1500
VAT at the rate of 12% = ₹ \(\frac{12}{100}\) × 1500
= ₹ 180
VAT paid to the Govt, by the shopkeeper = ₹ 36
VAT charged = ₹ (180 – 36) = ₹ 144
Now, \(\frac{12}{100}\) × x = ₹ 144
⇒ x = ₹ \(\frac{14400}{12}\) = ₹ 1200
Thus, price inclusive of tax at which the shopkeeper purchased the article from the wholesaler
= ₹ 1200 + ₹ \(\frac{12}{100}\) × 1200
= ₹ 1200 + ₹ 144
= ₹ 1344
(c)
Here, AB = 60 m, ∠ADE = 30° and ∠ACB = 60°. Let height of vertical lamp post CD be h m.
Now, in rt.∠ed ΔABC, we have
\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60°
⇒ \(\frac{60}{\mathrm{BC}}\) = \(\sqrt{3}\)
⇒ BC = \(\frac{60}{\sqrt{3}}\) = \(\frac{60 \sqrt{3}}{\sqrt{3} \sqrt{3}}\)
= 20\(\sqrt{3}\) m
AE = AB – EB
= AB – DC [∵ EB = DC]
= 60 – h
Again, in it ∠ed ΔAED, we have
\(\frac{\mathrm{AE}}{\mathrm{ED}}\) = tan 30°
\(\frac{60-h}{20 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\) [∵ ED = BC = 20\(\sqrt{3}\)m]
60 – h = 20
⇒ h = 60 – 20 = 40 m
Hence, the horizontal distance between AB and CD is 20\(\sqrt{3}\) m and the height of the lamp post is 40 m.
Question 8.
(a) Find x and y if \(\left[\begin{array}{ll}
x & 3 x \\
y & 4 y
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
5 \\
12
\end{array}\right]\) [3]
(b) A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast. [3]
(c) Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots x2 + (p – 3)x + p = 0. [4]
Solution:
(a)
(b)
Let the number of right circular cones recasted be n.
∴ Volume of n right circular cones = Volume of solid sphere n × \(\frac{1}{3}\) × π × 2.5 × 2.5 × 8
= \(\frac{4}{3}\) × π × 15 × 15 × 15
n × 50 = 60 × 225
n = \(\frac{60 \times 225}{50}\) = 270
Hence, the required number of right circular cones recasted is 270.
(c) Here, given quadratic equation is x2 + (p – 3)x + p = 0
For real and equal roots, we have
D = b2 – 4ac – 0
⇒ b2 = 4ac
⇒ (p – 3)2 = 4(1)(p)
⇒ p2 + 9 – 6p – 4p = 0
⇒ p2 – 10p + 9 = 0
⇒ (p – 9)(p – 1) = 0
⇒ Either p – 9 = 0 or p – 1 = 0
⇒ p = 9 or p = 1
Question 9.
(a) In the figure alongside, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of shaded portion. (Take π = \(\frac{22}{7}\))
(b) A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. [3]
(c) Find the mean of the following distribution by step deviation method: [4]
|
Class interval |
Frequency |
|
20-30 |
10 |
|
30-40 |
6 |
|
40-50 |
8 |
|
50-60 |
12 |
|
60-70 |
5 |
|
70-80 |
9 |
Solution:
(a) Clearly, required area = area of quadrant – area of right-angled triangle AOD
= \(\frac{1}{4}\)(πr2) – \(\frac{1}{4}\) × Base × Perpendicular
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – \(\frac{1}{2}\) × 3.5 × 2
= \(\frac{11}{2}\) × 0.5 × 3.5 – 3.5 = 3.5 (5.5 × 0.5 – 1)
= 3.5 × 1.75 = 6.125 cm2
(b) Let number of black balls in the box be n
Number of white balls = 30
Total number of balls in the box = 30+ n
As per condition of the question, we have
\(\frac{n}{30+n}\) = \(\frac{2}{5} \times \frac{30}{30+n}\)
n = \(\frac{2 \times 30}{5}\) = 12
Hence, the required number of black balls is 12.
(c)
Let assumed mean (a) = 55
Now, mean (\(\bar{x}\)) = a + \(\frac{\sum f_i u_i}{\sum f_i}\) × h
= 55 + \(\frac{-27}{50}\) × 10
= 55 – \(\frac{27}{5}\) = 55 – 5.4
\(\bar{x}\) = 49.6
Question 10.
(a) Using a ruler and compasses only:
(i) Construct a triangle ABC with the following data:
AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP. [4]
(b) The marks obtained by 120 students in a test are given below :
|
Marks |
No. of students |
| 0-10 | 5 |
| 10-20 | 9 |
| 20-30 | 16 |
| 30-40 | 22 |
| 40-50 | 26 |
| 50-60 | 18 |
| 60-70 | 11 |
| 70-80 | 4 |
| 80-90 | 4 |
| 90-100 |
3 |
Draw an ogive for the given distribution on a graph sheet.
Use suitable scale for ogive to estimate the following:
(i) The median.
(ii) The number of students who obtained more than 75% marks in the test.
(iii) The number of students who did not pass the test if minimum marks required to pass is 40. [6]
Solution:
(a) (i) Steps of Construction :
1. Draw a line segment BC = 6 cm.
2. Construct ∠XBC = 120° at B.
3. With B as centre and radius 3.5 cm cut off AB = 3.5 cm.
4. Join AC.
ΔABC is the required triangle
(ii) Steps of Construction :
1. Draw perpendicular bisector of BC which cuts BC at point D.
2. With D as centre and radius = CD draw a circle passing through points B and C.
3. Draw angle bisector of ∠ABC which intersects the circle at point P.
4. ∴ P is the point which is equidistant from AB and BC.
(iii) Join CP and measure ∠BCP.
Thus, ∠BCP = 30°.
(b)
(i) Median = 43.08
(ii) 10 students
(iii) 52 students
Question 11.
(a) In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P (-3, 4) on AB divides it in the ratio 2 : 3. Find the coordinates of A and B. [3]
(b) Using the properties of proportion. solve for x, given
\(\frac{x^4+1}{2 x^2}\) = \(\frac{17}{8}\) [3]
(c) A shopkeeper purchases a certain number of books for ₹ 960. If the cost per book was ₹ 8 less, the number of books that could be purchased for ₹ 960 would be 4 more. Write an equation, taking the original cost of each book to be ₹ x, and solve it to find the original cost of the books. [4]
Solution:
(a) Let coordinates of A and B be A(-x, 0) and B(0, y).
Now, P(-3, 4) on AB divides AB in the ratio 2 : 3
∴ \(\left(\frac{2(0)-3 x}{2+3}, \frac{2 y+3(0)}{2+3}\right)\) = (-3, 4)
⇒ \(\left(\frac{2 x}{5}, \frac{2 y}{5}\right)\) = (-3, 4)
⇒ \(\frac{-3 x}{5}\) = -3 and \(\frac{2 y}{5}\) = 4
⇒
Hence, the coordinates of A and B are A(-5, 0) and B(0, 10)
(b)
\(\frac{x^4+1}{2 x^2}\) = \(\frac{17}{8}\)
Using componendo and dividendo, we have
Again, applying componendo and dividendo, we have
\(\frac{x^2+1+x^2-1}{x^2+1-x^2+1}\) = \(\frac{5+3}{5-3}\)
⇒ \(\frac{2 x^2}{2}\) = \(\frac{8}{2}\) ⇒ x2 = 4
⇒ x = ±2
(c) Let the original cost price of each book be ₹ x
Total cost price = ₹ 960
∴ Number of books purchased = \(\frac{960}{x}\)
New cost price of each book = ₹ (x – 8)
∴ Number of books purchased = \(\frac{960}{x-8}\)
Now, as per condition of the statement, we have
\(\frac{960}{x-8}\) – \(\frac{960}{x}\) = 4
⇒ 960\(\left(\frac{x-x+8}{(x-8) x}\right)\) = 4
⇒ \(\frac{960 \times 8}{4}\) = x2 – 8x
⇒ x2 – 8x – 1920 = 0
Which is the required quadratic equation.
⇒ x2 – 48x + 40x – 1920 = 0
⇒ x(x – 48) + 40(x – 48) = 0
⇒ (x – 48)(x + 40) = 0
⇒ Either x – 48 = 0 or x + 40 = 0
⇒ x = 48 or x = -40
(Rejecting – ve value, cost per book cannot be negative)
Thus, x = 48
Hence, the original cost price of each book is ₹ 48.