Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2016 is the best way to boost your preparation for the board exams.
ICSE Class 10 Maths Question Paper 2016 Solved
Section – A (40 Marks)
(Attempt all questions from this Section)
Question 1.
(a) Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2, leaves a remainder 7. [3]
Solution:
Let f(x) = 2x3 + 3x2 – kx + 5
When f(x) is divided by x – 2, the remainder is 7.
∴ f(2) = 7
⇒ 2(2)3 + 3(2)2 – k(2) + 5 = 7
⇒ 16 + 12 – 2k + 5 = 7
⇒ -2k = 7 – 33
⇒ -2k = -26
⇒ k = 13
(b) Given A = \(\left[\begin{array}{ll}
2 & 0 \\
-1 & 7
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and A2 = 9A + ml. Find m. [4]
Solution:
We have A = \(\left[\begin{array}{ll}
2 & 0 \\
-1 & 7
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Now, from the statement of question, it is given that
A2 = 9A + mI
∴ \(\left[\begin{array}{rr}
4 & 0 \\
-9 & 49
\end{array}\right]\) = \(\left[\begin{array}{lr}
18+m & 0 \\
-9 & 63+m
\end{array}\right]\)
By using equality of matrices, we have
18 + m = 4
⇒ m = 4 – 18
⇒ m = -14
(c) The mean of following numbers is 68. Find the value of ‘x’. [3]
45, 52, 60, x, 69, 70, 26, 81 and 94.
Hence estimate the median.
Solution:
Here, mean = 68 and n = 9
We know that mean = \(\frac{\text { Sum of the observations }}{\text { Total number of observations }}\)
68 = \(\frac{45+52+60+x+69+70+26+81+94}{9}\)
⇒ 68 × 9 = 497 + x
⇒ 612 = 497 + x
⇒ x = 115
Now, arranging the given numbers in ascending order, we have 26, 45, 52, 60, 69, 70, 81, 94, 115
Here, n = 9 (an odd number)
So, the median is \(\left(\frac{n+1}{2}\right)^{t h}\) term, i.e., 5th term
⇒ Median = 69
Question 2.
(a) The slope of a line joining P(6, k) and Q(1 – 3k, 3) is \(\frac{1}{2}\). Find [3]
(i) k
(ii) Midpoint of PQ, using the value of ‘k’ found in (i)
Solution:
Here, given points are P(6, k), Q(1 – 3k, 3) and slope of line PQ = \(\frac{1}{2}\)
We know that slope = \(\frac{\text { difference of ordinates }}{\text { difference of abscissae }}\)
∴ Slope of \(\overline{\mathrm{PQ}}\) = \(\frac{1}{2}\) = \(\frac{3-k}{1-3 k-6}\)
2(3 – k) = -3k – 5
6 – 2k = -3k – 5
3k – 2k = -5 – 6
k = -11
Now, P(6, -11), Q(1 – 3 × (-11), 3) i.e., Q(34, 3)
Let R be the midpoint of PQ.
Therefore, midpoint of
PQ = R\(\left(\frac{6+34}{2}, \frac{-11+3}{2}\right)\) = R(20, -4)
(b) Without using trigonometrical tables, evaluate : [4]
cosec257° – tan233° + cos44° cosec46° – \(\sqrt{2}\)cos45° – tan260°
Solution:
We have cosec257° – tan233° + cos44° cosec46° – \(\sqrt{2}\)cos45° – tan260°
= cosec2 (90° – 33°) – tan233° + cos44° cosec(90° – 44°) – \(\sqrt{2}\) × \(\frac{1}{\sqrt{2}}\) = -(\(\sqrt{3}\))2
= sec233° – tan233° + cos44° sec44° – 1 – 3 [∵ cosec(90° – θ) = sec θ]
= sec233° – tan233° + cos44° × \(\frac{1}{\cos 44^{\circ}}\) – 4
= 1 + 1- 4 = -2 [∵ sec2θ – tan2θ = 1]
(c) A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones. [3]
Solution:
Required number of cones
= \(\frac{\text { Volume of the solid sphere }}{\text { Volume of acone }}\)
= \(\frac{\frac{4}{3} \pi \times 6 \times 6 \times 6}{\frac{1}{3} \pi \times 2 \times 2 \times 3}\)
= 2 × 6 × 6 = 72
Thus, the required number of cones are 72.
Question 3.
(a) Solve the following inequation, write the solution set and represent it on the number line. [3]
-3(x – 7) ≥ 15 – 7x > \(\frac{x+1}{3}\), x ∈ R
Solution:
-3(x – 7) ≥ 15 – 7x > \(\frac{x+1}{3}\), x ∈ R is equivalent to
-3(x – 7) ≥ 15 – 7x
and 15 – 7x > \(\frac{x+1}{3}\)
-3x + 21 ≥ 15 – 7x
and 45 – 21x > x + 1
7x – 3x ≥ 15 – 21
and -21x – x > 1 – 45
4x ≥ -6
and -22x > -44
2x ≥ -3
and -x > -2
x ≥ –\(\frac{3}{2}\)
and x < 2
Solution set.: {x : x ∈ R and – \(\frac{3}{2}\)< x <2}
(b) In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find : [4]
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED
Solution:
(i) Here, AD is parallel to BC and BD is a transversal
Therefore ∠ADB = ∠CBD [alt. int. angles]
⇒ ∠ADB = 32° ……………… (i)
[given ∠CBD = 32°]
Consider ∆OBD which is an isosceles triangle with OD = OB = radii
Now, ∠OBD = ∠ODB
[angles opp. to equal sides of an isosceles ∆]
⇒ ∠OBD = 32°
[∵ ∠ADB or ∠ODB = 32°]
(ii) AB subtends ∠AOB at the centre and ∠ADB at point D on the circle,
⇒ ∠AOB =2 ∠ADB
⇒ ∠AOB = 2 × 32° [from eqn. (i)]
⇒ ∠AOB = 64°
(iii) Since AD is diameter
⇒ AOD is a line with centre O.
Now, ∠AOB + ∠BOD = 180° [a linear pair]
⇒ 64° + ∠BOD = 180°
⇒ ∠BOD = 180° – 64°
⇒ ∠BOD = 116°
Also, BD subtends ∠BOD at the centre and ∠BED at point E on the circle.
Therefore ∠BED = \(\frac{1}{2}\) ∠BOD
= \(\frac{1}{2}\) × 116° = 58°
(c) If (3a + 2b) : (5a + 3b) = 18 : 29. Find a : b. [3]
Solution:
We have \(\frac{3 a+2 b}{5 a+3 b}\) = \(\frac{18}{29}\)
⇒ 87a + 58b = 90a + 54b
⇒ 58b – 54b = 90a – 87a
4b = 3 a
⇒ \(\frac{4 b}{3 a}\) = 1
⇒ \(\frac{b}{a}\) = \(\frac{4}{3}\)
⇒ a : b = 4 : 3
Question 4.
(a) A game of numbers has cards marked with 11, 12, 13, …. 40. A card is drawn at random. Find the probability that the number on the card drawn is : [3]
(i) A perfect square
(ii) Divisible by 7
Solution:
Here, total number of outcomes = 30
(i) Out of 30 possible outcomes, there are three favourable outcomes which are 16, 25 and 36 (a perfect square).
Hence, required probability = \(\frac{3}{30}\) = \(\frac{1}{10}\)
(ii) Favourable outcomes = 4
i.e., {14, 21, 28, 35}
Probability (getting a card divisible by 7) = \(\frac{4}{30}\)
= \(\frac{2}{15}\)
(b) Use graph paper for this question. [4]
(Take 2 cm = 1 unit along both x and y axis.)
Plot the points O(0, 0), A (-4, 4), B (-3, 0) and C(0, -3)
(i) Reflect points A and B on the y-axis and name them A’ and B’ respectively. Write down their coordinates.
(ii) Name the figure OABCB’A’.
(iii) State the line of symmetry of this figure.
Solution:
(i) A’ (4, 4), B’ (3, 0)
(ii) Hexagonal (arrow head)
(iii) y-axis is the line of symmetry.
(c) Mr. Lalit invested ₹ 5000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate [3]
(i) The rate of interest.
(ii) The amount at the end of second year, to the nearest rupee.
Solution:
Here, Principal (P) = ₹ 5000
Rate of interest (r) = ?
Time (n) = 1 year Amount = ₹ 5325
We know that Amount = Principal + Interest
⇒ Interest = Amount – Principal
⇒ Interest for 1 year = 5325 – 5000 = ₹ 325
By using formula
Interest for 1 year = \(\frac{\mathrm{P} \times r \times 1}{100}\)
⇒ 325 = \(\frac{5000 \times r}{100}\)
⇒ 50r = 325 ⇒ r = 6.5%
Thus, rate of interest is 6.5% per annum.
Amount at the end of 1st year becomes principal for 2nd year.
∴ Interest for second year = \(\frac{5325 \times 6.5 \times 1}{100}\)
= ₹ 346.125
Thus, the amount at the end of second year
= ₹ 5325 + ₹ 346.125
= ₹ 5671.125
≅ ₹ 5671
Section – B (40 Marks)
Attempt any four questions from this Section
Question 5.
(a) Solve the quadratic equation x2 – 3 (x + 3) = 0; Give your answer correct to two significant figures. [3]
Solution:
Here x2 – 3 (x + 3) = 0
⇒ x2 – 3x – 9 = 0
Now x2 – 3x – 9 = 0 gives
x = \(\frac{3 \pm \sqrt{(3)^2-4 \times 1 \times(-9)}}{2 \times 1}=\frac{3 \pm \sqrt{9+36}}{2}\)
⇒ x = \(\frac{3 \pm \sqrt{45}}{2}=\frac{3 \pm 3 \sqrt{5}}{2}\)
= \(\frac{3 \pm 3 \times 2.236}{2}=\frac{3 \pm 6.708}{2}\)
⇒ x = \(\frac{9.708}{2}\) or \(\frac{-3.708}{2}\)
⇒ x= 4.854 or-1.854
⇒ x = 4.9 or -1.9 (correct to two significant figures)
(b) A page from the savings bank account of Mrs. Ravi is given below.
She closed the account on 30th September, 2006. Calculate the interest Mrs. Ravi earned at the end of 30th September, 2006 at 4.5% per annum interest. Hence, find the amount she receives on closing the account.
Solution:
As the interest is earned on the minimum balance between 10th day and last day of the month, we have the principal for each month given below
| Month | Balance (₹) |
| April | 8300 |
| May | 7600 |
| June | 10300 |
| July | 10300 |
| August | 3900 |
| September | 0 |
| Total | 40400 |
Principal for 1 month = ₹ 40400
Rate of interest (R) = 4.5% p.a.
Time = \(\frac{1}{12}\) year
Interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{40400 \times 4.5 \times 1}{1200}\)
⇒ I = ₹ 151.50
Required amount
= ₹ 5900 + ₹ 151.50
= ₹ 6051.50
(c) In what time will ₹ 1500 yield ₹ 1996.50 as compound interest at 10% per annum compounded annually? [3]
Solution:
We have Principal (P) = ₹ 1500,
Amount (A) = ₹ 1996.50
Rate = 10% p.a.
and Time = ?
Let the required time be n years
⇒ n = 3
Required time = 3 years
Note : Read the statement as : In what time will ₹ 1500 yield ₹ 1996.50 at 10% per annum compounded annually.
Question 6.
(a) Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. [3]
Solution:
Each interior angle = \(\left(\frac{n-2}{n}\right)\) × 180°
= \(\left(\frac{6-2}{6}\right)\) × 180° = 120°
There are six lines of symmetry- the lines joining the midpoints of the opposite sides and the diagonals passing through the centre.
XY, TS, VU, AD, BE and CF.
(b) In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T. [4]
(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.
Solution:
(i) Since PQRS is a cyclic quadrilateral
∠PSR = ∠RQT → (ext’ ∠ of cyclic
∠SPQ = ∠TRQ → (ext’ ∠ of cyclic
quadrilateral is equal to its interior opp. angle)
Now, in ∆TPS and ∆TRQ
∠TPS = ∠TRQ
∠TSP = ∠TQR
By A. A. similarity axiom, we have ∆TPS ~ ∆TRQ
(ii) From above ∆TPS and ∆TRQ
⇒ \(\frac{\mathrm{TP}}{\mathrm{TR}}=\frac{\mathrm{PS}}{\mathrm{RQ}}\) ⇒ \(\frac{18}{6}=\frac{\text { PS }}{4}\) ⇒ PS
= 12 cm
(iii) We know that the ratio of two similar triangles are equal to the ratio of the squares of any two corresponding sides
Here, ∆TPS ~ ∆TRQ
⇒ \(\frac{\text { Area of } \Delta \mathrm{TPS}}{\text { Area of } \Delta \mathrm{TRQ}}=\frac{\mathrm{TR}^2}{\mathrm{TR}^2}\)
⇒ \(\frac{27 \mathrm{~cm}^2}{\text { Area of } \Delta T R Q}=\frac{18^2}{6^2}\)
⇒ Area ∆TRQ = \(\frac{27 \times 36}{18 \times 18}\) cm2
⇒ Area of ∆TRQ = 3 cm2
Now, Area of quadrilateral
PQRS = Area of ∆TPS – Area of ∆TQR
= 27 cm2 – 3 cm2
= 24 cm2
Hence, required area = 24 cm2
(c) Given matrix A = \(\left[\begin{array}{lr}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
If AX = B
(i) Write the order of matrix X.
(ii) Find the matrix ‘X’.
Solution:
(i) We have A = \(\left[\begin{array}{lr}
4 \sin 30^{\circ} & \cos 0^{\circ} \\
\cos 0^{\circ} & 4 \sin 30^{\circ}
\end{array}\right]\)
= \(\left[\begin{array}{lr}
4 \times \frac{1}{2} & 1 \\
1 & 4 \times \frac{1}{2}
\end{array}\right]=\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\)
B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
Now, AX = B, where A is a 2 × 2 matrix, B is a 2 × 1 matrix. Therefore AX i.e., product of A and X should be 2 × 1 matrix.
⇒ Hence AX multiplication is possible iff the number of columns in A is equal to number of rows in X.
[A]2×2 [X]2×1 = [B]2×1
Also the product should of same order as B i.e., 2 × 1
⇒ Order of Matrix X is 2 × 1
(ii) Let X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), Given AX = B
⇒ \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 x+y \\
x+2 y
\end{array}\right]=\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
⇒ 2x + y = 4 ………………. (i)
x + 2y = 5 ………………… (ii)
Multiplying (i) by 2, we have
x + 2y = 8 ………………. (iii)
Now, subtracting (ii) from (iii)
⇒ 3x = 3
⇒ x = 1 and y = 4 – 2 = 2
Hence, X = \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
Question 7.
(a) An aeroplane at an altitude of 1500 metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships. [4]
Solution:
Let OP be the altitude 1500 m
∴ Height of aeroplane = 1500 m
Let A and B be the position of two ships.
Let distance between two ships, AB be x m
Let OB be y.
Consider rt. ∆OAP, \(\frac{\mathrm{OP}}{\mathrm{OA}}\) = tan 30°
⇒ \(\frac{1500}{x+y}\) = \(\frac{1}{\sqrt{3}}\)
⇒ x + y = 1500\(\sqrt{3}\) m
Consider rt. ∆OBP, \(\frac{\mathrm{OP}}{\mathrm{OB}}\) = tan 45°
⇒ \(\frac{1500}{y}\) = tan 45°
⇒ y = 1500 m
Now, distance between two ships = x
= x + y – y
= (1500\(\sqrt{3}\) – 1500)m
= 1500(\(\sqrt{3}\) – 1)m
(b) The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis). [6]
| Scores | No. of shooters |
| 0-10 | 9 |
| 10-20 | 13 |
| 20-30 | 20 |
| 30-40 | 26 |
| 40-50 | 30 |
| 50-60 | 22 |
| 60-70 | 15 |
| 70-80 | 10 |
| 80-90 | 8 |
| 90-100 | 7 |
Use your graph to estimate the following :
(i) The median
(ii) The interquartile range
(iii) The number of shooters who obtained a score of more than 85%.
Solution:
The cumulative frequency table for the given continuous distribution is :
| Scores obtained | Number of shooters | Cumulative frequency |
| 0-10 | 9 | 9 |
| 10-20 | 13 | 22 |
| 20-30 | 20 | 42 |
| 30-40 | 26 | 68 |
| 40-50 | 30 | 98 |
| 50-60 | 22 | 120 |
| 60-70 | 15 | 135 |
| 70-80 | 10 | 145 |
| 80-90 | 8 | 153 |
| 90-100 | 7 | 160 |
| Total | 160 |
Here, take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis. Plot the points (0, 0), (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153), (100, 160).
Join these points by a freehand drawing. The required ogive is shown in the graph.
Here, n (no. of shooters) = 160
(i) To find the median :
Let A be a point on Y-axis representing frequency
= \(\frac{\mathrm{n}}{\mathrm{2}}\) = \(\frac{160}{2}\) = 80
Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet X-axis
at M. The abscissa of the point M represents the median
∴ The required median = 44
(ii) The Interquartile range = Upper quartile – Lower quartile
To find Lower quartile :
Let D be the point on Y-axis representing frequency.
= \(\frac{\mathrm{n}}{\mathrm{4}}\) = \(\frac{160}{4}\) = 40
Through D draw DE || X-axis meeting the curve in E.
Draw EN || Y-axis then asbscissa of N is 29.
∴ Lower quartile = 29
Again, mark a point F on Y-axis corresponding to
\(\frac{\mathrm{3n}}{\mathrm{4}}\) i.e., 3 × \(\frac{160}{4}\) = 120
Draw FG || X-axis, meeting the curve in G.
Draw GL || Y-axis, meeting the X-axis in L.
Then abscissa is 60.
∴ Upper quartile = 60
∴ Inter quartile range = 60 – 29 = 31
Let the point B on X-axis represents 85 scores.
Through B, draw a vertical line to meet the ogive at the point R through R, draw a horizontal line to meet the Y-axis at C. The ordinate of point C represents 150 shooters.
∴ The number of shooters who obtained more than 85% scores = 160 – 150 = 10
Question 8.
(a) If \(\frac{\mathrm{x}}{\mathrm{a}}\) = \(\frac{\mathrm{y}}{\mathrm{b}}\) = \(\frac{\mathrm{z}}{\mathrm{c}}\) show that
\(\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{3 x y z}{a b c}\) [3]
Solution:
Hence proved.
(b) Draw a line AB = 5 an. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct: [4]
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.
Solution:
Steps of construction :
- Draw a line segment AB = 5 cm.
- On line AB mark a point C such that AC = 3 cm.
- From point A and C mark an arc of radius 2.5 cm which cut at O.
- Now, construct a circle of radius 2.5 cm at point O.
- Join OB and find the perpendicular bisector of OB.
- Let it bisects OB at R. Then from R construct a circle of radius OR or RB. The point of intersection of both the circles are P and Q. Join BP and BQ. Measure the length BP and BQ = 3 cm
(c) A line AB meets X-axis at A and Y-axis at B. P(4, -1) divides AB in the ratio 1:2. [3]
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.
Solution:
Let the coordinates of the point A, B be A (a, 0), B(0, b) respectively.
Since, P divides the line segment AB in the ratio 1 : 2
∴ Coordinates of P are
P\(\left(\frac{1.0+2 . a}{1+2}, \frac{1 . b+2.0}{1+2}\right)\) i.e., P\(\left(\frac{2 a}{3}, \frac{b}{3}\right)\)
P\(\left(\frac{2 a}{3}, \frac{b}{3}\right)\) = P(4, -1)
⇒ \(\frac{2 a}{3}\) = 4 ⇒ 2a = 12 ⇒ a = 6
\(\frac{b}{3}\) = -1 ⇒ b = -3
Hence, coordinates of A are (6, 0) and coordinates of B are (0, -3)
(ii) Coordinates of A (6, 0) and B (0, -3)
⇒ Slope of line AB (m) = \(\frac{-3-0}{0-6}=\frac{3}{6}\)
= \(\frac{1}{2}\) [∵m = \(\frac{y_2-y_1}{x_2-x_1}\)]
∴ Slope of line perpendicular to line
AB = –\(\left(\frac{1}{m}\right)\) = -2
Now, equation of the line passing through P(4, -1) and having slope – 2 is :
y + 1 = -2(x – 4)
⇒ 2x + y = 7
Question 9.
(a) A dealer buys an article at a discount of 30% from the wholesaler, the marked price being ₹ 6,000. The dealer sells it to a shopkeeper at a discount of 10% on the marked price. If the rate of VAT is 6%, find [3]
(i) The price paid by the shopkeeper including the tax.
(ii) The VAT paid by the dealer.
Solution:
Price at which the article bought by the dealer
= (100 – 30)% of ₹ 6000
= \(\frac{70 \times 6000}{100}\) = ₹ 4200
Tax paid = ₹ \(\frac{6}{100}\) × 4200 = ₹ 252
Selling price of the article = 90% of ₹ 6000 = ₹ 5400
Tax charged = 6% ₹ 5400
= \(\frac{6}{100}\) × 5400 = ₹ 324
(i) The price paid by shopkeeper including tax
= ₹ 5400 + ₹ 324
= ₹ 5724
(ii) VAT paid by the dealer = Tax recovered on sale to shopkeeper – Tax paid on purchase
= ₹ 324 – ₹ 252
= ₹ 72
(b) The given figure represents a kite with a circular and a semicircular motifs stuck on it. The radius of circle is 2.5 cm and the semicircle B is 2 cm. If diagonals AC and BD are of lengths 12 cm and 8 cm respectively, find the area of the: [4]
(i) shaded part. Give your answer correct to the nearest whole number.
(ii) unshaded part.
Solution:
Here, the radius (r) of semicircle is 2 cm.
Hence, area or semicircle = \(\frac{\pi r^2}{2}=\frac{\pi \times 2 \times 2}{2}\)
= 2 π cm2
Also, the radius (R) of circle is 2.5 cm.
Area of circle = πR2 = n × 2.5 × 2.5 cm2
= 6.25 π cm2
Since only a semicircle and a circle is shaded.
∴ Area of shaded part = Area of semicircle + Area of circle
= (2π + 6.25 π) cm2
= 8.25 π cm2
= 8.25 × \(\frac{22}{7}\) cm2
= 25.93 cm2
= 26 cm2 (nearest whole number)
(ii) Area of kite ABCD = 2 (area of AABC)
= 2(\(\frac{1}{2}\) × AC × \(\frac{BD}{2}\))
= 12 × 4 = 48 cm2
Area of unshaded part = Area of kite – Area of shaded part
= 48 cm2 – 26 cm2
= 22 cm2
(c) A model of a ship is made to a scale 1 : 300 [3]
(i) The length of a model of the ship is 2 m. Calculate the length of the ship.
(ii) The area of the deck ship is 180,000 m2. Calculate the area of the deck of the model.
(iii) The volume of the model is 6.5 m3. Calculate the volume of the ship.
Solution:
(i) Here, the scale factor k = \(\frac{1}{300}\)
Length of model = k × length of the ship 2 m
= \(\frac{1}{300}\)length of the ship
⇒ Length of the ship = 2 × 300 = 600 m
(ii) Area of deck of model = k2 × area of the deck of ship
= \(\frac{1}{300}\) × \(\frac{1}{300}\) × 180000 m2 = 2 m2
⇒ Area of deck of model is 2 m2.
(iii) Volume of model of ship = k3 × volume of the ship
⇒ 6.5 m3 = \(\frac{1}{300}\) × \(\frac{1}{300}\) × \(\frac{1}{300}\) × Volume of the ship
⇒ Volume of the ship = 175500000 m3
Question 10.
(a) Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find : [3]
(i) the monthly instalment
(ii) the amount of maturity.
Solution:
(i) Let monthly instalment be of ₹ x.
R = 6% p.a.
T = 2 years
⇒ n = 2 × 12 = 24
Interest = ₹ 1200
Using the formula
I = x × \(\frac{n}{2}\) (n + 1) × \(\frac{1}{12}\) × \(\frac{R}{100}\)
⇒ 1200 = x × \(\frac{24 \times 25}{2}\) × \(\frac{1}{12}\) × \(\frac{6}{100}\)
⇒ x = \(\frac{1200 \times 2}{3}\) = 800
Hence, the monthly instalment is ₹ 800
(ii) Amount of maturity = 24
x + Interest = 24 × 800 + 1200
= ₹ 20400
(b) The histogram represents the scores obtained by 25 students in a Mathematics mental test. Use the data to: [4]
(i) Frame a frequency distribution table.
(ii) To calculate mean.
(iii) To determine the Modal class.
Solution:
(i) Frequency distribution table is :
| Class interval | Frequency |
| 0-10 | 2 |
| 10-20 | 5 |
| 20-30 | 8 |
| 30-40 | 4 |
| 40-50 | 6 |
| Total | 25 |
(ii)
| Class interval | Midpoint (xi) | Frequency (fi) | fixi |
| 0-10 | 5 | 2 | 10 |
| 10-20 | 15 | 5 | 75 |
| 20-30 | 25 | 8 | 200 |
| 30-40 | 35 | 4 | 140 |
| 40-50 | 45 | 6 | 270 |
| ∑fi = 25 | ∑fixi = 695 |
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)
⇒ Mean = \(\frac{695}{25}\) = 27.8 marks
(iii) Maximum frequency shown in histogram is 8 and it falls on 20 – 30
⇒ 20 – 30 is modal class.
(c) A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate ‘x ’. [3]
Solution:
Let uniform speed of the bus be x km/h.
∴ Speed of bus (during rain)= (x – 10) km/h
Now, using the statement of question, we have
\(\frac{240}{x-10}-\frac{240}{x}\) = 2
⇒ \(\frac{240(x-x+10)}{(x-10) x}\) = 2
⇒ 2400 = 2x2 – 20x
⇒ 1200 = x2 – 10x
⇒ x2 – 10x – 1200 = 0
⇒ x2 – 40x + 30x – 1200 = 0
⇒ x(x – 40) + 30(x – 40) =0
⇒ (x + 30) (x – 40) =0
⇒ x = – 30 or x = 40
⇒ x = 40 (Rejecting x = -30, -ve value)
Hence, uniform speed of Bus is 40 km/h.
Question 11.
(a) Prove that \(\frac{\cos A}{1+\sin A}\) + tanA = secA. [3]
Solution:
(b) Use ruler and compasses only for the following question.
All construction lines and arcs must be clearly shown. [4]
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from A C and BC.
(iv) Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC arid BC. Measure XY.
Solution:
Steps of constructions :
- Construct a ∆ABC with BC = 6.5 cm, ∠ABC = 60° and AB 5 cm.
- Draw a circle with A as centre and radius 3.5 cm, which is the required locus of points at a distance of 3.5 cm from A.
- Draw the angle bisector of ∠BCA, which is the required locus of points equidistant from AC and BC.
- Mark two points, where angle bisector intersects the circle as X and Y, which are equidistant from AC and BC and are at a distance of 3.5 cm from A. On measurement XY = 5.2 cm.
(c) Ashok invested ₹ 26,400 on 12%, ₹ 25 shares of a company. If he receives a dividend of ₹ 2,475, find the: [3]
(i) Number of shares he bought
(ii) Market value of each share
Solution:
(i) Let the number of shares bought be x.
Total face value of shares = ₹ 25 x
Dividend earned = ₹ 2475
Dividend = 12% of 12
⇒ \(\frac{12}{100}\) × 25 x = 2475
⇒ 3x = 2475
⇒ x = 825.
Therefore, number of shares bought by Ashok is 825
(ii) Total investment on purchasing 825 shares
= ₹ 26400
Now, market value of each share = ₹ \(\frac{26400}{825}\)
= ₹ 32