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ICSE Class 10 Maths Question Paper 2012 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)
Question 1.
(a) If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A2 – 5A + 7I
(b) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 :7. The expenditures are in the ratio 3 : 5. If each save Rs. 80 every month, find their monthly pocket money. [3]
(c) Using a Remainder Theorem factorise completely the following polynomial.
3x2 + 2x2 – 19x + 6 [4]
Solution :
(a)
(b) Given monthly pocket money of Ravi and Sanjeev are in the ratio of 5 : 7
Let Ravi’s pocket money be 5x and Sanjeev’s pocket money be 7 x
Given Ravi’s and Sanjeev’s expenditures are in the ratio of 3 : 5
Let Ravi’s expenditure be 3y and Sanjeev’s expenditure be 5y
According to Question
Monthly savings of Ravi = Monthly savings of Sanjeev = ₹ 80
Saving = Monthly Pocket money – Monthly expenditure Ravi’s monthly savings = 5x – 3y
Sanjeev’s monthly savings = 7x – 5y
By the question
5x – 3y = 80 …(1)
7x – 5y = 80 …(2)
Multiply (1) by 7 and (2) by 5, we get
35x – 21y = 560 …(3)
35x – 25y = 400 …(4)
Subtract (4) from (3)
4y = 160
or y = 40
Putting the value of y in (1), we get
5x – 3 × 40 = 80
or 5x – 120 = 80
or 5x = 120 + 80
or 5x = 200
∴ x = 40
Ravi’s monthly pocket money = 5 × 40 = ₹ 200 and Sanjeev’s monthly pocket money = 7 × 40 = ₹ 280
(c) Given f(x) = 3x3 + 2x2 – 19x + 6
Put x = 2
∴ f(2) = 3(2)3 + 2(2)2 – 19(2) + 6
= 24 + 8 – 38 + 6 = 0
∴ By Remainder Theorem (x – 2) is a factor off (x) Now, (x – 2) is a factor of 3x3 + 2x2 – 19x + 6
∴ 3x3 + 2x2 – 19x + 6 = (x – 2) (3x2 + 8x – 3) … (1)
= (x – 2) (3x2 + 9x – x – 3)
= (x – 2) {3x(x + 3) – 1(x + 3)}
= (x – 2) (x + 3) (3x – 1)
Question 2.
(a) On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a. ? [3]
(b) ABC is an isosceles right angled triangle with ∠ABC = 90°. A semi-circle is drawn with AC as the diameter. If AB = BC = 7 cm, find the area of the shaded region. (Take π = \(\frac{22}{7}\)) [3]
(c) Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find : [4]
(i) the ratio in which AB is divided by the y- axis.
(ii) the coordinates of the point of intersection.
(iii) the length of AB.
Solution :
(a) Given
CI. – Si. = ₹ 25 for 2
Rate of interest = 5% p.a.
Let Principal be ₹ x, Time = 2 years,
Rate of interest 5% p.a.
(b) Given ABC is an isosceles triangle with AB = 7cm and BC = 7cm
∠ABC = 90°
Area of shaded portion = Area of semi circle – Area of ∆ABC
In ∆ABC, ∠ABC = 90°
∴ By Pythagoras Theorem
AC2 = AB2 + BC2 = 72 + 72
or AB = \(\sqrt{49+49}\)
= \(\sqrt{98}\) = 7\(\sqrt{2}\) cm
∴ Radius of semicircle = \(\frac{1}{2}\)AC
= \(\frac{1}{2}\) × 7\(\sqrt{2}\)
Now, Area of semicircle = \(\frac{1}{2}\)(Area of Circle)
Area of ∆ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 7 × 7 = \(\frac{49}{2}\) cm2
(c) Here, line segment AB is joining the points IA(-4, 6) and B(8, -3).
Let C(0, y) be the point of intersection of y-axis and the line segment AB.
Let the required ratio be k : 1
∴ c(\(\frac{8 k-4}{k+1}, \frac{-3+6}{k+1}\)) = C(0, y)
⇒ \(\frac{8 k-4}{k+1}\) = 0
⇒ 8k – 4 = 0
⇒ k = \(\frac{1}{2}\)
Thus, the required ratio is 1: 2
Also, y = \(\frac{-3 \times \frac{1}{2}+6}{\frac{1}{2}+1}\) = \(\frac{-3+12}{1+2}\) = \(\frac{9}{3}\) = 3
Thus, the coordinates of the point of intersection is C(0, 3)
Now, AB = \(\sqrt{(8+4)^2+(-3-6)^2}\)
Question 3.
(a) In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle. [3]
(b) Evaluate without using trigonometric tables
cos2 26° + cos 64°sin 26° + \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\) [3]
(c) Marks obtained by 40 students in a short assessment are given below, where a and b 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 a and b. [4]
Solution :
(a) In the given figure AB is a tangent at B
AB = 15 cm and AC = 7.5cm
O is the centre of the circle
We know that
AB2 = AC × AD
∴ 15 × 15 = 7.5 × AD
∴ AD = \(\frac{15 \times 15}{7.5}\) 30 cm
Now, CD = AD – AC = 30 cm – 7.5 cm
= 22.5 cm
∴ Radius of circle = \(\frac{1}{2}\)(diameter of the circle)
= \(\frac{1}{2}\) × 22.5 cm = 11.25 cm
(b) cos2 26° + cos 64° sin 26° + \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
= cos2 26° + cos(90° – 26°)sin 26° + \(\frac{\tan \left(90^{\circ}-54^{\circ}\right)}{\cot 54^{\circ}}\)
= cos2 26° + sin 26° sin 26° + \(\frac{\cot 54^{\circ}}{\cot 54^{\circ}}\)
[∵ cos(90° – 26°) = sin 26°
tan (90° – 54°) = cot 54°]
= cos2 26° + sin2 26° + 1 [∵ sin2A + cos2A = 1]
= 1 + 1 = 2
(c)
| Marks | 5 | 6 | 7 | 8 | 9 |
| Number of students | 6 | A | 16 | 13 | b |
Given mean of distribution = 7.2
Σf = 35 + a + b = 40(given)
⇒ a + b = 5
Mean
= 7.2 (given)
⇒ 246 + 6a + 9b = 40 × 7.2 = 288
∴ 6a + 9b = 42
Multiply (1) by 6 and subtract from (2), we have
or b = 4
From(1) a + 4 = 5 ⇒ a = 1
∴ a = 1, b = 4
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 least 1 tail.
(c) Using graph paper and taking 1 cm = 1 unit along both x-axis and y-axis :
(i) Plot the points A (-4, 4) and B (2, 2)
(ii) Reflect A and B in the origin to get the images A’ and B’ respectively.
(iii) Write down the co-ordinates of A’ and B’.
(iv) Give the geometrical name for the figure ABA’B’.
(v) Draw and name its lines of symmetry.
Solution :
(a) Monthly deposit = ₹ 200
Rate of interest = 11% p.a
Period of deposit (n) = 36 months
∴ Total deposited Amount = ₹ (200 × 36)
= ₹ 7200
Equivalent principal for 1 month
= \(\frac{n(n+1)}{2^{\prime}}\) × monthly deposit
= ₹\(\left(\frac{36 \times 37}{2} \times 200\right)\) = ₹(3600 × 37)
= ₹ 133200
Interest paid by the bank = ₹\(\frac{133200 \times 11}{100 \times 12}\)
= ₹111 × 11
= ₹ 1221
∴ Amount on maturity = Total Deposited Amount + Interest Amount
= ₹ (7200 – 1221)
= ₹ 8421
(b) The sample space of tossing 2 coins is (HH, HT, TH, T’T)
No. of outcomes in sample space = 4
(i) Probability of getting 2 heads = \(\frac{1}{4}\)
(ii) Probability of getting at least 1 tail. = \(\frac{3}{4}\)
(c) (i) and (ii)
(iii) Coordinates of A’ and B’ are A’( 4, -4) and B’(-2, -2)
(iv) AB A’B’ is a parallelogram
(v) AA’ and BB’ are the lines of symmetry.
Section-B (40 Marks)
(Answer any four questions from this Section)
Question 5.
(a) In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DRA
(b) Given \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 4
\end{array}\right]\) X = \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]\). Write:
(i) the order of the matrix X
(ii) the matrix X.
(c) A page from the Savings Bank Account of Mr. Prateek is given below :
If he receives Rs. 198 as interest on 1st January, 2007, find the rate of interest paid by the bank.
Solution :
(a) (i) Given AB = diameter of the circle, ∠BCD = 130°
Join BD
As ABCD is a cyclic quadrilateral
∴ ∠DCB + ∠DAB = 180°
or ∠DAB + 130° = 180°
or ∠DAB = 180° – 130° = 50°
Now, ∠ADB = 90° [angle in a semicircle]
(ii) ∴ In Δ ADB
∠DAB + ∠ADB + ∠DBA = 180°
∴ 50° + 90° + ∠DBA = 180°
or ∠DBA = 180° – 140° = 40°
(b) Given \(\left[\begin{array}{rr}
2 & 1 \\
-3 & 4
\end{array}\right]_{2 \times 2}\) X = \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]_{2 \times 1}\) … (1)
(i) As \(\left[\begin{array}{rr}
2 & 1 \\
-3 & 4
\end{array}\right]\) is a matrix of order 2 × 2 and \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]\) is a matrix of order 2 × 1
∴ X must have 2 rows and 1 column, the order of the Matrix X is 2 × 1
(ii)
∴ 2a + b = 7 … (3)
-3a + 4b = 6 …. (4)
Multiply (3) by 4 and subtract from (4), we have
⇒ a = 2
From (3), 2 × 2 + b = 7
⇒ b = 3
Thus, X = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
(c)
Interest for one year = ₹ 198
Let rate of interest be r% p.a.
198 = \(\frac{52800 \times r}{100 \times 12}\) [I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)]
⇒ r = \(\frac{198 \times 100 \times 12}{52800}\) = 4.5% p.a.
Question 6.
(a) The printed price of an article is Rs. 60,000. The wholesaler allows a discount of 20% 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 6% at every stage. Find:
(i) the cost to the shopkeeper inclusive of tax.
(ii) VAT paid by the shopkeeper to the Government.
(iii) the cost to the customer inclusive of tax. [3]
(b) Solve the following inequation and represent the solution set on the number line:
4x – 19 < \(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x ∈ R [3]
(c) Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots.
x2 + 2(m – 1)x + (m + 5) = 0 [4]
Solution :
(a) (i) Printed Price = ₹ 60,000
Discount given by wholesaler to shopkeeper = ₹ 20%
= ₹ \(\frac{60000 \times 20}{1000}\)
= ₹ 12000
∴ Net cost to shopkeeper
= ₹ (60000 – 12000)
= ₹ 4800
Sales Tax (VAT) changed by wholesaler
= ₹ \(\frac{48000 \times 6}{100}\)
= ₹ 2880
∴ Cost to shopkeeper inclusive of Tax
= ₹ (48000 + 2880)
= ₹ 50880
(ii) ∴ Sales Tax (VAT) charged by shopkeeper
= ₹ 60000 × \(\frac{6}{100}\)
= ₹ 3600
VAT Paid by shopkeeper to Government
= ₹ (3600 – 2880)
= ₹ 720
(iii) ∴ Cost to customer inclusive of Tax
= ₹ (60000 + 3600)
= ₹ 63600
(b)
4x – 19 < \(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x
4x – 19 < \(\frac{3 x}{5}\) – 2
20x – 95 < 3x – 10
20x – 3x < 95 – 10
17x < 85
-4 ≤ x
\(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x
3x – 10 ≤ -2 + 5x
-10 + 2 ≤ 5x – 3x
-8 ≤ 2x
x < 5
∴ Solution is -4 ≤ x < 5, x ∈ R
Representation
(c) Give equation is
x2 + 2(m – 1)x + (m + 5) = 0
For the given equation to have real and equal roots
Discriminant =0
⇒ b2 – 4ac = 0
Here, a = 1, b = 2(m – 1) = 2m – 2 and c = m + 5
∴ b2 – 4ac = (2m – 2)2 – 4 × 1 × (m + 5)
= 4m2 + 4 – 8m – 4m – 20
= 4m2 – 12m – 16
Now, b2 – 4ac = 4m2 – 12m – 16 = 0
⇒ m2 – 3m – 4 = 0
⇒ (m – 4)(m + 1) = 0
⇒ m = 4 and m = -1
Question 7.
(a) A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones. [3]
(b) Solve the following equation and give your answer correct to 3 significant figures :
5x2 – 3x – 4 = 0 [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 lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre. [4]
Solution :
(a) External radius of hollow sphere = 8 cm
Internal radius of hollow sphere = 6 cm
∴ Volume of hollow sphere = \(\frac{4}{3} \pi\)[R3 – r3]
= \(\frac{4}{3} \pi\)[83 – 63]
= \(\frac{4}{3} \pi\)[512 – 216]
= \(\frac{4}{3} \pi\)(296) cm3
A.T.Q. Volume of sphere = Volume of all small cones
Fora small cone r = 2 cm and h = 8 cm
∴ Volume of one small cone = \(\frac{1}{3} \pi r^2 h\)
= \(\frac{1}{3}\) × π × 2 × 2 × 8
= \(\frac{1}{3}\) × 32 × π cm3
= \(\frac{\frac{4}{3} \times \pi \times 296}{\frac{1}{3} \times 32 \times \pi}\) = \(\frac{296}{8}\) = 37
(b) Given equation is 5x2 – 3x -4 = 0
Compare with ax2 + bx + c = 0
Here, a = 5, b = 3 and c = -4
(c) Let A and B be the position of two ships
∴ Height of lighthouse OP = 80 m
Let distance between two ships, AB be x
Let OB be y
In Δ OAP ; tan 30° = \(\frac{\mathrm{OP}}{\mathrm{OA}}\) = \(\frac{80}{x+y}\)
∴ \(\frac{1}{\sqrt{3}}\) = \(\frac{80}{x+y}\)
or x + y = 80\(\sqrt{3}\)
= 80 × 1.732
= 138.56m
In Δ OBP
we have, tan 40° = \(\frac{\mathrm{OP}}{\mathrm{OB}}\) = \(\frac{80}{y}\)
∴ y = \(\frac{80}{\tan 40^{\circ}}\) = \(\frac{80}{0.8391}\)
= 95.34 m
Now, distance between two ships
= 138.56 m – 95.34 m = 43.22 m
Question 8.
(a) A man invests Rs. 9600 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 ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC – ΔAMP
(ii) Find AB and BC. [3]
(c) If x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\), using properties of proportion, show that x2 – 2ax + I = 0.
Solution :
(a) (i) Amount invested = ₹ 9600
M.V. of each share = ₹ 80
∴ No. of shares held = \(\frac{₹ 9600}{₹ 80}\) = 120
(ii) Dividend on 1 share = 18% of N.V.
= 18% of ₹ 100 = ₹ 18
Total Dividend = No. of shares × Dividend on 1 share = ₹ 120 × 18 = ₹ 2160
(iii) Percentage return on shares
= \(\frac{₹ 21600}{₹ 9600}\) × 100 = 22.5%
(b)
(i) In the given figure,
∠AMP = 90° and ∠ABC = 90°
In ΔABC and ΔAMP
∴ ∠ABC = ∠AMP = 90°
∠A = ∠A [common]
∴ ΔABC \(\sim\) ΔAMP, [by AA similarity axiom]
(ii) Since ΔABC \(\sim\) ΔAMP, therefore their sides are proportional
\(\frac{\mathrm{AC}}{\mathrm{AP}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)
∴ BC = \(\frac{10 \times 12}{15}\) = 8 cm
Now, in rt. ∠d ΔABC
AC2 = AB2 + BC2 (By Pythagoras Theorem)
or AB2 = AC2 – BC2
or AB2 = 102 – 82 = 100 – 64 = 36
∴ AB = 6 cm
(c)
Question 9.
(a) The line through A (-2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b. [3]
(b) Prove that \(\frac{\tan ^2 \theta}{(\sec \theta-1)^2}\) = \(\frac{1+\cos \theta}{1-\cos \theta}\)
(c) A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. [4]
Solution :
(a) Points A(-2, 3) and B(4, b)
Since the line through A and B is perpendicular to the line 2x – 4y = 5
∴ Product of their slopes = -1
\(\left(\frac{b-3}{4+1}\right)\left(\frac{2}{4}\right)\) = -1
\(\frac{b-3}{12}\) = -1
b – 3 = -12
b = -9
(b)
(c)
Case I
Distance = 400 km
Let speed of the car be x km/h
Case II
When speed of the car is increased by 12 km/h, then
New speed = (x + 12)km/h
Distañce = 400 km
Now, time taken (t2) = \(\frac{400}{x+12}\) hours …. (2)
According to question
t1 – t2 = 1 hour 40 minutes = 1\(\frac{40}{60}\) hours
= 1\(\frac{2}{3}\) hours = \(\frac{5}{3}\) hours.
∴ From (1) and (2)
\(\frac{400}{x}\) – \(\frac{400}{x+12}\) = \(\frac{5}{3}\)
or 400\(\left[\frac{1}{x}-\frac{1}{x+12}\right]\) = \(\frac{5}{3}\)
or 400\(\left[\frac{x+12-x}{x(x+12)}\right]\) = \(\frac{5}{3}\)
or \(\frac{12 \times 400}{x^2+12 x s}\) = \(\frac{5}{3}\)
or x2 + 12x = 12 × 400 × \(\frac{3}{5}\) = 2880
⇒ x2 + 60x – 48x – 2880 = 0
(x + 60) (x – 48) = 0
⇒ x – 48 = 0
or x + 60 = 0
⇒ x = 48
or x = -60
But x cannot be -ve
Hence,
x = 48 km/h
Question 10.
(a) Construct a triangle ABC in which base BC = 6 cm, AB = 5.5 cm and ∠ABC = 120°.
(i) Construct a circle circumscribing the triangle ABC.
(ii) Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
(b) The following distribution represents the height of 160 students of a school.
| Height (in cm) | No. of students |
|
140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 |
12 20 30 38 24 16 12 8 |
Draw an ogive 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 median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm.
Solution :
(a) Steps of Construction:
1. Take BC = 6 cm
2. Construct ∠ABC = 120° at point B, such that AB = 5.5 cm. Join AC.
3. Then, ABC is the required triangle.
4. Draw the perpendicular bisectors of sides BC and AB.
Let the perpendicular bisectors of AB and BC meet at point O.
5. Taking O as centre and radius equal to OA (or OB or OC), draw a circle
The circle so obtained is the required circle.
6. Let perpendicular bisector of BC intersect the circumference in D.
7. Join CD and DA, then ABCD is the required cyclic quadrilateral.
(b) The cumulative frequency table for the given continuous distribution is
| Height (in cm) | No. of students | Cumulative frequency |
|
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 |
Take 2 cm = 5 cm of height on x-axis & 2 cm = 20 students on y-axis
Plot the points (140, 0), (145, 12), (150, 32), (155, 62), (160, 100),(165, 124),(170, 140),(175, 152), (180, 160)
Join these points by a free hand drawing. The required ogive is shown in the graph.
Here n (no. of students) = 160, which is even
(i) To find median
Let A be a point on y-axis representing frequency =
\(\frac{1}{2}\left(\frac{n}{2}+\left(\frac{n}{2}+1\right)\right)\)
= \(\frac{1}{2}\)(80 + 81) = 80.5
Through A, draw a horizontal line to meet the ogive at A.
Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 157.4
∴ Required median = 157.4
(ii) To find lower quartile
Let B represent the point on y-axis representing frequency = \(\frac{\mathrm{N}}{4}\) = \(\frac{\mathrm{160}}{4}\) = 40
Through B, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 151.3
∴ Q1 = lowerquartile = 151.3
To Find upper quartile
Let C be a point on y-axis representing frequency = \(\frac{3 \mathrm{~N}}{4}\) = 3 × \(\frac{160}{4}\) = 120
Through C, draw a horizontal line to meet the ogive at S. Through S, draw a vertical line to meet the x-axis at T. The abscissa of the point T represents 164.2
∴ Q3 = Upper quartile = 164.2
∴ Interquartile Range = Q3 – Q1
= 164.2 – 151.3 = 12.9
(iii) Let D be a point on x-axis represent 172 cm. Through D, draw a vertical line to meet the ogive at the point E. Through E, draw a horizontal line to meet the y-axis at F. The ordinate of the point F represent 142
∴ No. of students whose height is above 172 cm = 160 – 142 = 18
Question 11.
(a) In triangle PQR, PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle.
(b) Find the mode and median of the following frequency distribution:
| x | 10 | 11 | 12 | 13 | 14 | 15 |
| f | 1 | 4 | 7 | 5 | 9 | 3 |
(c) The line through P (5, 3) intersects y axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Solution :
(a)
let the radius of the inscribed circle be x cm.
In rt. ∠d ΔPQR
PR2 = PQ2 + QR2 = 242 + 72 = 576 + 49 = 625
∴ PR = 25 cm
Join OA, OC and OB
Now, from the figure AOCQ, AQ = x = QC = OC = AO
[∵ QC and QA are tangents to the circle from Q]
Now, CR = QR – QC = 7 – x
[Tangents to the circle]
Also, CR = RB = 7 – x
[Tangents to the circle]
Now, PA = PQ – AQ = 24 – x
Also, PA = PB = 24 – x … (2)
[Tangents to the circle]
From (1) and (2)
PR = PB + RB
25 = 24 – x + 7 – x
2x = 24 + 7 – 25
2x = 6
∴ x = 3 cm
Hence, radius of inscribed circle = 3 cm
(b)
| x | 10 | 11 | 12 | 13 | 14 |
| f | 1 | 4 | 7 | 5 | 9 |
Now
| x | f | c.f. |
|
10 11 12 13 14 15 |
1
4 7 8 9 3 |
1 5 12 17 26 29 |
∴ Σf = 29
∴ N = 29 (odd)
∴ Median = \(\left(\frac{n+1}{2}\right)\)th term = \(\left(\frac{29+1}{2}\right)\)2th term = 15th term = 13
In the given distribution, the variate 14 has the maximum frequency i.e., 9.
∴ Mode = 14
(c)
(i) We know that slope of a line tan θ
[where θ is the angle made by the line with the positive direction of x-axis]
∴ slope of PQ = tan 45° = 1
⇒ m = 1
(ii) Equation of a line passing through P(5, 3) and slope
m = 1 is given by
y – 3 = 1(x – 5)
[y – y1 = m(x – x1)]
y – 3 = x – 5
or x – y = + 5 – 3 = 2
or x – y = 2
(iii) Since the line PQ meets y-axis at Q
∴ Coordinates of Q can be found by putting x = 0 in x – y = 2
y = -2
∴ Coordinates of Q are (0, -2)