Maths Questions

Section A

(i) Section A carries 40 marks.

(ii) This section contains four questions. All the questions are compulsory.

Question 1(a) ( 3.0 marks)

Find the compound interest paid when a sum of Rs 15000 is invested for 1 year and 4 months at per annum compounded annually.

Solution:

Principal, P = Rs 15000

Time = 1 year 4 months =

Rate, R =

Amount for the first year can be calculated as follows.

Now, this amount would act as principal for next year.

Interest for the first year = Rs 15937.50 − Rs 15000 = Rs 937.50

∴Total compound interest = Rs 937.50 + Rs 332.03 = Rs 1269.53

Question 1(b) ( 3.0 marks)

Using property of proportion, prove that if, then

Solution:

Applying componendo and dividendo, we obtain

Squaring both sides, we obtain

Again applying componendo and dividendo, we obtain

Thus, the given result is proved.

Question 1(c) ( 4.0 marks)

(i) If, then find the values of x and y.

(ii) A real number x is greater than or equal to 2 more than one-third of the number itself and is less than or equal to 3 more than half the number itself. How this situation can be represented in form of an inequality?

Solution:

(i)

(ii) The number x is greater than or equals to 2 more than one-third of the number itself.

2 more than one-third of the number x is.

It is also given that the number x is less than or equal to 3 more than half the number itself.

3 more than half the number x is.

Therefore, from (1) and (2), we obtain

Question 2(a) ( 3.0 marks)

The distance between two points P and Q is 6 units. The coordinates of point P are (4, −6). Find the coordinates of point Q, if the ordinate of Q is thrice its abscissa.

Solution:

It is given that the ordinate of Q is thrice its abscissa.

Let the coordinates of point Q be (x, 3x).

The coordinates of point P are (4, −6).

It is given that PQ = 6 units

∴ Using distance formula, we obtain

Squaring both sides, we obtain

Thus, the coordinates of point Q are:

or (−2, −6)

Question 2(b) ( 3.0 marks)

In the given figure, a cyclic quadrilateral PQRS is shown and an isosceles triangle PST with PT = TS is shown.

The line RST is a straight line. Also, PQ = QS

If ∠PQR = 72° and ∠PSQ = 56°, then find:

(i) ∠PST

(ii) ∠POS

Solution:

(i) We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

∴ ∠PST = ∠PQR

⇒ ∠PST = 72° (∠PQR = 72°, given)

(ii) It is given that PT = TS

∴ ∠SPT = ∠PST

⇒ ∠SPT = 72°

In ΔPST,

∠PST + ∠STP + ∠SPT = 180°

⇒ ∠STP = 180° − 72° − 72° = 36°

Now, we know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠POS = 2∠STP = 2 × 36° = 72°

Question 2(c) ( 4.0 marks)

(i) If tangents XY and YZ from a point Y to a circle with centre O are inclined to each other at an angle of 64°, then find ∠YOZ.

(ii) In the given figure, a cyclic quadrilateral WXYZ is shown. Find the values of x and y.

Solution:

(i)

We are given that XY and YZ are two tangents such that ∠XYZ = 64°.

Join OX, OY, and OZ.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ XO ⊥ XY and OZ ⊥ YZ

∴ ∠OXY = ∠OZY = 90°

In quadrilateral XYZO,

∠OXY + ∠XYZ + ∠OZY + ∠XOZ = 360°

(Sum of angles of a quadrilateral is 360°)

⇒ 90° + 64° + 90° + ∠XOZ = 360°

∠XOZ = 116°

Now, in ΔXYO and ZYO,

∠OXY = ∠OZY (Each equal to 90°)

XY = YZ (tangents drawn from an external point to a circle are equal)

XO = OZ (Radii of the same circle)

∴ ΔXYO ≅ ΔZYO (By SAS congruency criterion)

∴∠XOY = ∠ZOY

Also, ∠XOZ = ∠XOY + ∠ZOY = 116°

∴ ∠XOY = ∠ZOY =

Thus, ∠YOZ= 58°

(ii) We know that the sum of measures of opposite angles of a cyclic quadrilateral is 180°.

∴ ∠W + ∠Y = 180°

⇒ 4y + 2y = 180°

⇒ 6y = 180°

⇒ y = 30°

Also, ∠X + ∠Z = 180°

⇒ 2y + 30° + 5x + 40° = 180°

⇒ 2y + 5x = 180° − 70° = 110°

⇒ 2 × 30° + 5x = 110° (y = 30°)

⇒ 60° + 5x = 110°

⇒ 5x = 110° − 60° = 50°

Thus, the respective values of x and y are 10° and 30°.

Question 3(a) ( 4.0 marks)

From a solid metallic cylinder of height 24 cm, a cone of same height and same base (as cylinder) is taken out. The density of the metal is 10 gm per cm³. If the mass of the remaining portion of the cylinder is 24.64 kg, then find the cost required to paint the remaining portion at a rate of Rs 3 per 440 cm².

Solution:

Height of the cone and cylinder = 24 cm

Let radii of the cone and cylinder be r cm.

Therefore,

Volume of the remaining portion = Volume of cylinder − Volume of cone

Now, volume of the remaining portion

Therefore, radii of the cone and cylinder = 7 cm

Now, slant height of the cone,

Surface area of the remaining portion = Lateral surface area of cone + Lateral surface area of cylinder + Area of circular base

Therefore, cost of painting the remaining portion = Rs = Rs 12

Question 3(b) ( 3.0 marks)

The minute hand of a clock swept an area of 115.5 cm² while its tip travelled a distance of 22 cm. Find the radius of the minute hand and the time of rotation.

Solution:

Let r be the radius of the minute hand and be the angle of rotation of minute hand in the given period.

It is given that area swept by the minute hand = 115.5 cm²

It is also given that the tip of the minute hand travelled at a distance of 22 cm.

Dividing equations (1) and (2), we obtain

Therefore, the radius of minute hand is 10.5 cm.

Substituting r = 10.5 cm in equation (2), we obtain

Therefore, the angle of rotation for the minute hand is 120°.

For one complete rotation, i.e. 360°, the minute hand takes 60 minutes.

For 1° rotation, the minute hand takes minutes

For 120° rotation, the minute hand takes

Therefore, the time taken by the minute hand for the given rotation is 20 minutes.

Question 3(c) ( 3.0 marks)

A cylindrical glass vessel of base radius 14 cm and 40 cm height contains water up to 30 cm height. If 12 spherical metallic balls of radius 3.5 cm are dipped into it, then find the gap between the water level and the total height of the container.

Solution:

When the spherical balls are dipped in the cylindrical container, water will rise in the container in the shape of cylinder whose radius is same as the base of the container i.e., radius = 14 cm

Let x be the raise in the height of the water level in the container after putting 12 balls.

Now,

Volume of 12 balls = Volume of water raised in the cylinder

The water level will rise up to 3.5 cm.

Height of the water level after putting the balls (30 + 3.5) cm = 33.5 cm

Therefore, gap between the final water level and the top of the container = 40 cm − 33.5 cm = 6.5 cm

Question 4(a) ( 3.0 marks)

If m = cosec θ − cot θ and n = sec θ, then prove that

Solution:

We are given that,

To prove:

i.e., to prove:

Thus,

Hence, proved

Question 4(b) ( 4.0 marks)

Convert the following more than type ogive into a less than type ogive.

Solution:

First of all, we have to convert the given more than type ogive into a less than type table and then draw its ogive.

The more than type frequency distribution of the given ogive is as follows:

Class interval	Frequency
More than 400	140
More than 410	125
More than 420	120
More than 430	110
More than 440	60
More than 450	50
More than 460	35
More than 470	10

This table can be converted into a frequency distribution table with cumulative frequency as shown below:

Class interval	Frequency	Cumulative frequency
400 − 410	140 − 125 = 15	15
410 − 420	125 − 120 = 5	15 + 5 = 20
420 − 430	120 − 110 = 10	20 + 10 = 30
430 − 440	110 − 60 = 50	30 + 50 = 80
440 − 450	60 − 50 = 10	80 + 10 = 90
450 − 460	50 − 35 = 15	90 + 15 = 105
460 − 470	35 − 10 = 25	105 + 25 = 130
470 − 480	10	130 + 10 = 140

Now, the required ogive can be drawn by taking the upper class limits of each class interval on x-axis and their corresponding cumulative frequencies on y-axis as shown below.

Question 4(c) ( 3.0 marks)

(i) Prove that:

(ii) Express cot 56° − sec 68° + cos 61° in terms of trigonometric ratios of angles between 0° and 35°.

Solution:

(i)

Hence, proved

(ii)We have,

Therefore, we have

Section B

(i) Section B also carries 40 marks.

(ii) This section contains seven questions. Answer any four questions.

Question 5(a) ( 3.0 marks)

Anju purchased a microwave for Rs 27,500. After two years, its value got depreciated by 4.5%. Find the value of the microwave after 2 years.

Solution:

Here, principal, P = Rs 27, 500

Rate of depreciation, R = 4.5%

Time, n = 2 years

∴Value of microwave at the end of two years

Thus, the value of the microwave after two years is Rs 25, 080.69.

Question 5(b) ( 3.0 marks)

Jammy ordered dinner for his family from a restaurant. The bill for the lunch was Rs 428 including a VAT of 7%. Find the bill amount before VAT was added.

Solution:

Let the bill amount before VAT be Rs x.

∴VAT = 7% of Rs x = Rs = Rs

The bill amount including VAT is given to be Rs 428.

Thus, the bill amount before VAT was added was Rs 400.

Question 5(c) ( 4.0 marks)

Mr. Ambani invests Rs 9000 in shares of a certain company. He buys Rs 50 shares at a premium of Rs 10 and earns an annual income of Rs 375.

(i) Find the rate of dividend on the shares.

(ii) If he wants to increase his annual income by Rs 100, then how many more shares should he buy?

Solution:

Face value of each share = Rs 50

Market value of each share = Rs 50 + Rs 10 = Rs 60

Number of shares bought by Ambani

(i) Total income on the shares = Rs 375

∴ Number of shares × Dividend on 1 share = Rs 375

⇒ 150 × Dividend on 1 share = Rs 375

∴Dividend on 1 share = Rs = Rs 2.50

∴Rate of dividend =

(ii) Dividend on 1 share = Rs 2.50

Mr. Ambani wants to increase his income by Rs 100.

∴ Number of more shares he should buy

Question 6(a) ( 3.0 marks)

The point X (4, 3 m − 4) divides the line segment joining the points P (2, −5) and Q (7, 5) in the ratio 2:3 internally. Find the value of m.

Solution:

It is given that X divides PQ in the ratio 2:3.

By using section formula, the coordinates of point X are given by:

Comparing the coordinates of X [i.e., (4, −1)] with the given coordinates of X [i.e., (4, 3m − 4)], we obtain

Thus, the value of m is 1.

Question 6(b) ( 4.0 marks)

(i) What is the solution of the inequality?

(ii) What are the coordinates of the foot of the perpendicular drawn from the point (3, 5) to the line, 2x + 3y + 7 = 0?

Solution:

(i) The given inequality is

Let |x| = y

⇒

Case 1:

y − 1 ≥ 0 and y − 2 > 0

⇒ y ≥ 1 and y > 2

⇒ y > 2

Case 2:

y − 1 ≤ 0 and y − 2 <>

⇒ y ≤ 1 and y <>

⇒ y ≤ 1

From Case 1 and Case 2, we obtain

|x| ≤ 1 or |x| > 2

⇒ (− 1 ≤ x ≤ 1) or (2 < x or x < − 2)

⇒ x ∈ [−1, 1] or x ∈ (−∞, − 2) ∪ (2, ∞)

∴ x ∈ [−1, 1] ∪ (−∞, − 2) ∪ (2, ∞)

(ii)The equation of the given line is 2x + 3y + 7 = 0 … (1)

This can be written as , which is of the form, y = mx + c

Therefore, slope of the given line is.

Slope of the line perpendicular to the given line =

Equation of the line passing through the point (3, 5) with slope is

The foot of the perpendicular is the intersection point of the lines (1) and (2).

Solving equations (1) and (2), we obtain

Hence, the coordinates of the foot of the perpendicular are.

Question 6(c) ( 3.0 marks)

A retailer cuts three pieces from a single piece of cloth of length 93 cm. The length of the second piece is 5 cm longer than the shortest piece and the length of the third piece is twice that of the shortest piece. If the third piece is to be at least 5 cm longer than the second, then the length of the shortest piece can vary from

Solution:

Let the length of shortest piece be x cm.

Then, length of second piece = (x + 5) cm

Length of third piece = 2x cm

Total length of cloth = 93 cm

∴ x + (x +5) + 2x ≤ 93

⇒ 4x + 5 ≤ 93 ⇒ 4x ≤ 88

⇒ x ≤ 22 …(1)

It is also given that the third piece is to be at least 5 cm longer than the second.

⇒ 2x ≥ (x + 5) + 5

⇒ 2x ≥ x + 10

⇒ x ≥10 …(2)

From (1) and (2), 10 ≤ x ≤ 20

Hence, the shortest piece must be at least 10 cm long, but not more than 22 cm long.

Question 7(a) ( 3.0 marks)

Draw the line(s) of symmetry of the following figures and state the number of lines of symmetry in each case.

(i)

(ii)

(iii)

Solution:

(i)

This figure has 4 lines of symmetry.

(ii)

The given figure does not have any line of symmetry.

(iii) The given figure shows 4 lines of symmetry as shown below.

Question 7(b) ( 4.0 marks)

Draw a circle of radius 5 cm. From a point 13 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Also measure the angles between the tangents. Give the justification of the construction.

Solution:

A pair of tangents to the given circle can be constructed as follows.

(1) Taking any point O of the given plane as centre, draw a circle of 5 cm radius. Locate a point P, 13 cm away from O. Join OP.

(2) Bisect OP. Let M be the mid-point of PO.

(3) Taking M as centre and MO as radius, draw a circle.

(4) Let this circle intersect the previous circle at points Q and R.

(5) Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 12 cm each and ∠RPQ = 50°.

Justification

The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 5 cm). For this, join OQ and OR.

∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.

∴ ∠PQO = 90°

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.

Question 7(c) ( 3.0 marks)

ΔABC is shown in the figure.

D and E are points on the sides AB and AC respectively such that DE||BC.

(i) Prove that ΔADE ∼ ΔABC

(ii) If DE = 2 cm and BC = 6 cm, find

(iii) Find

Solution:

(i) In ΔADE and ΔABC,

∠ADE = ∠ABC (DE||BC and we know that corresponding angles are equal)

∠DEA = ∠BCA (DE||BC and we know that corresponding angles are equal)

∠EAD = ∠CAB (Common)

∴ΔADE ∼ ΔABC (By AAA similarity criterion)

(ii) We know that ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(iii) Now,

Question 8(a) ( 4.0 marks)

3 spheres of radii 6 cm, 8 cm, and 10 cm are melted to form a cone whose radius is twice its height. Find the surface area of this cone in terms of π.

Solution:

Let the radius (r) and height of the cone formed by melting the three spheres are 2x and x i.e., r = 2x cm and h = x cm

Since 3 spheres of radii 6 cm, 8 cm, and 10 cm are melted to form a cone, the volume of the cone is equal to the sum of the volumes of these 3 spheres.

Therefore, radius of the cone, r = 2 × 12 cm = 24 cm and height of the cone, h = 12 cm

Now, slant height of the cone (l) is given by,

Therefore, surface area of the cone

Question 8(b) ( 6.0 marks)

A right-angled triangle ABC with AB = 8 cm, BC = 15 cm, and ∠B = 90° is rotated along its hypotenuse AC. What is the solid so obtained? Find the surface area and volume of the solid. [Use]

Solution:

When the right-angled triangle ABC with AB = 8 cm, BC = 15 cm, and B = 90° is rotated along the hypotenuse AC, the solid so formed is a double cone as shown below.

Using Pythagoras Theorem for ΔABC, we obtain

AC² = AB² + BC² = (15cm)² + (8 cm)² = 289 cm² = (17 cm)²

⇒ AC = 17 cm

Area of ΔABC AB × BC

Therefore, radius of the double cone,

Now,

Volume of the double cone = Volume of cone + Volume of cone

Now,

Surface area of the double cone = Sum of lateral surface areas of cones and

= πr. AB + π.r. AC

Question 9(a) ( 3.0 marks)

From the top of a 10 m high building, the angle of elevation of the top of a tower is θand the angle of depression of its foot is 45°. Find the angleθ, if the height of the tower is 17 m more than the height of the building. (Take)

Solution:

Let AB denote the building and CD denote the tower. It is given that the height of the building is 10 m.

∴ AB = 10 m

It is also given that the angle of depression from the top of the building to the foot of the tower is 45°.

∴∠DAE = ∠ADB = 45° (As the alternate interior angles are equal)

In ΔABD,

Now it is given that the height of the tower is 17 m more than the height of the building.

In ΔAEC,

We know that tan 60° =

∴ θ = 60°

Question 9(b) ( 3.0 marks)

A helicopter flying at an altitude of 1600 metres finds that two ships are sailing in the same direction towards it. The angles of depression of the ships as observed from the helicopter are 45° and 30° respectively. Find the distance between the two ships.

Solution:

In the figure, AB denotes the height of the helicopter and the points C and D represent the two ships.

AB = 1600 m

Now, AX and BC are parallel lines.

∴∠CAX = ∠ACB (Alternate interior angles are equal)

And, ∠DAX = ∠ADB (Alternate interior angles are equal)

∴∠ADB = 45° and ∠ACB = 30°

We have to find the distance between the two ships i.e., the distance CD.

In ΔABD,

⇒ BD = 1600 m

In ΔABC,

⇒ BC = 1600

⇒ BD + DC = 1600 (BC = BD + DC)

⇒ DC =

Thus, the distance between the two ships is 1168 m.

Question 9(c) ( 4.0 marks)

Arjun standing on a horizontal plane observes that a bird at a distance of 100 m from him is at an elevation of 60°. Aditi standing on a roof of a 30 m tall building observes that the same bird is making an angle of elevation of 45°. Both Arjun and Aditi are on the opposite sides of the bird. Find the distance of the girl from the bird.

Solution:

In the figure, C denotes the position of the bird, A denotes the position of Arjun, and F denotes the position of Aditi. It is given that the bird is at a distance of 100 m from Arjun.

∴AC = 100 m

It is given that Aditi is standing on a roof of a 30 m tall building.

∴FD = EB = 30 m

In ΔACB,

In ΔCEF,

Thus, Aditi is at a distance of 79.9 m from the bird.

Question 10(a) ( 4.0 marks)

The mean of the following frequency distribution is 31.4. If there are 100 observations in total, then find the value of f₁ and f₂.

Class interval	Frequency
0 − 10	10
10 − 20	f₁
20 − 30	20
30 − 40	f₂
40 − 50	14
50 − 60	16
Total	100

Solution:

Firstly, let us find the class marks of each class interval and then product of class marks with the corresponding frequencies for each class interval as shown in the following table.

Class interval	Frequency, f_i	Class marks, x_i	f_ix_i
0 − 10	10	5	50
10 − 20	f₁	15	15f₁
20 − 30	20	25	500
30 − 40	f₂	35	35f₂
40 − 50	14	45	630
50 − 60	16	55	880
Total	100	-	2060 + 15f₁ + 35f₂

It is given that the sum of the frequencies is 100.

It is given that mean of the frequency table is 31.4.

Substituting f₂ = 24 in equation (1), we obtain f₁ = 40 − 24 = 16

Question 10(b) ( 6.0 marks)

Draw the histogram for the following data. Hence calculate the mode.

Class interval	0 − 100	100 − 200	200 − 300	300 − 400	400 − 500
Frequency	6	10	9	18	7

Solution:

W draw the histogram by taking, class interval along the horizontal axis (x-axis) and frequency along the vertical axis (y-axis)

Now, taking the scale along x-axis as, 1 cm = 100 class intervals and along the y-axis as, 1 cm = 2 frequency, we draw a histogram.

In the highest rectangle, draw two lines AC and BD from corners of the rectangles on the either side of the rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of the lines AC and BD.

Through P, draw a vertical line to meet the x-axis at M.

It can be seen that the abscissa of the point M represent 345.

Thus, the mode of the given data is 345.

Question 11(a) ( 3.0 marks)

State true or false and give reasons in support of your answer.

(i) Probability of an event can be .

(ii) Total number of outcomes for the experiment of tossing a coin and rolling a die together is 8.

(iii) The probability of drawing a red marble from a bag containing blue marbles only is 0.

Solution:

(i) False

For any event E, 0 ≤ P (E) ≤ 1, but here, . Therefore, this statement is false.

(ii) False

The sample space for the experiment of tossing a coin and rolling a die together is as follows:

{(H 1), (H 2), (H 3), (H 4), (H 5), (H 6), (T 1), (T 2), (T 3), (T 4), (T 5), (T 6)}

∴Total number of outcomes = 12

Therefore, this statement is false.

(iii) True

A red marble cannot be drawn from a bag containing blue marbles only. Therefore, this is an impossible event and probability of an impossible event is 0.

Thus, this statement is true.

Question 11(b) ( 3.0 marks)

A box contains 4 red balls, 8 white balls, and some blue balls. The probability of drawing a ball, which is neither blue nor red, is.