Maths Question 1- Answer

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)

Mrs. Singh has an account in a certain cooperative bank. The following table shows the entries on her passbook.

Date	Particulars	Withdrawal (in Rs)	Deposit (in Rs)	Balance (in Rs)
Jan, 1’09	B/F	-	-	5600
Jan, 20	To cheque	1500	-	4100
Feb, 14	To cash	1000	-	3100
March, 6	By cheque	-	3500	6600
March, 19	To cheque	5400	-	1200
April, 05	By cheque	-	2300	3500
April, 10	By cash	-	1000	4500
April, 28	To cheque	2000	-	2500
April 30	By cheque	-	500	3000
May 14	By cash	-	1200	4200
June, 20	To cheque	2800	-	1400

Mrs. Singh earns Rs 63.75 as interest at the end of June’09, where the interest is compounded annually. Find the rate of interest paid by the bank in her savings bank account.

Solution:

Principal for the month of January = Rs 4100

Principal for the month of February = Rs 3100

Principal for the month of March = Rs 1200

Principal for the month of April = Rs 2500

Principal for the month of May = Rs 3000

Principal for the month of June = Rs 1400

Total principal for 1 month = Rs 15300

∴ P = Rs 15300, I = 63.75, T = year

Question 1(b) ( 3.0 marks)

In a culture, the initial count of bacteria was 90000. It is observed that the count of bacteria is increasing at a constant rate. If the count of bacteria after 3 hours is 119790, then at what rate is the unit of bacteria increasing?

Solution:

P = 90000

A = 119790

n = 3 hours

Let the count of bacteria increase at the rate of r% per hour.

Now, A = P

Thus, the bacteria count increases at the rate of 10% per hour.

Question 1(c) ( 4.0 marks)

Solution:

Question 2(a) ( 4.0 marks)

Two lines 2x + 5y + 3 = 0 and x − 4y + 14 = 0 are intersected by a third line at the points (− a, b) and (a, 2a) respectively. What is the equation of third line?

Solution:

The point (−a, b) lies on the line 2x + 5y + 3 = 0

∴ − 2a + 5b + 3 = 0 … (1)

The point (a, 2a) lies on the line x − 4y + 14 = 0

⇒ a − 8a + 14 = 0

⇒ − 7a = −14

⇒ a = 2

Substituting in equation (1), we obtain

− 4 + 5b + 3 = 0

Thus, the required line passes through the points and (2, 4).

Hence, its equation is given by,

This is the equation of the required line.

Question 2(b) ( 3.0 marks)

In the given figure, RQ and PS are perpendicular to the line PQ.

The lengths of the sides PO and OQ are 15 cm and 5 cm respectively. If the area of ΔROQ is 7.5 cm², then find the area of ΔPOS.

Solution:

In ΔROQ and ΔSOP,

∠OQR = ∠OPS (RQ and SP are perpendicular to PQ)

∠ROQ = ∠SOP (Vertically opposite angles)

∴ΔROQ ∼ ΔSOP (By AA similarity criterion)

Now, we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Thus, the area of ΔSOP is 67.5 cm².

Question 2(c) ( 3.0 marks)

(i) If, then find the values of a and b.

Solution:

Thus, the values of a and b are 1 and −3 respectively.

Question 3(a) ( 3.0 marks)

The hypotenuse and one side of a right-angled triangle are 25 cm and 7 cm respectively. A circle is inscribed in it. Find the area of the remaining portion of the triangle. [Use π = 3.14]

Solution:

Let ΔABC be right-angled at B such that AB = 7 cm and AC = 25 cm. Let a circle with centre O be inscribed in ΔABC. Draw OL⊥AB, OM⊥BC, and ON⊥AC

Join OA, OB, and OC.

Using Pythagoras Theorem for ΔABC, we obtain

AB² + BC² = AC²

Let radius of inner circle be r.

i.e., OL = OM = ON = r

Now, area of ΔABC = ar (ΔOAB) - ar (ΔOBC) + ar (ΔOAC)

Now,

Area of the triangular portions those are not included in the circle = Area of triangle − Area of circle

Question 3(b) ( 3.0 marks)

PQ and RS are two parallel chords of a circle such that PQ = 30 cm and RS = 16 cm. If the chords are on the opposite sides of the centre and the distance between them is 23 cm, then find the radius of the circle.

Solution:

Let O be the centre of the given circle and let its radius be r cm. Draw OV⊥RS and OU⊥PQ

Since OV⊥ RS and OU⊥PQ,

RS||PQ

Therefore, points U, O, V are collinear.

It is given that UV = 23 cm

Let OV = x cm, therefore, OU = (23 − x) cm

Let OR = OP = r (Radius of the circle)

We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ RV = VS = 8 cm and PU = UQ = 15 cm

Using Pythagoras Theorem in ΔOPV, we obtain

OR² = OV² + RV²

⇒ x² + (8)² … (1)

Using Pythagoras Theorem in ΔPUO, we obtain

PO² = OU² + PU²

⇒ r² = (23 − x)² + (15)² … (2)

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

(23 − x)² + (15)² = x² + (8)²

⇒ 529 + x² − 46x + 225 = x² + 64

⇒ 46x = 529 + 225 − 64 = 690

⇒ x = 15

Putting the value of x in (1), we obtain

r² = (15)² + (8)² = 225 + 64

⇒ r² = 289 = (17)²

⇒ r = 17

Thus, the radius of the circle is 17 cm.

Question 3(c) ( 4.0 marks)

The radius and height of a cone are reduced by 20%. Find the percentage reduced by or increased by in surface area.

Solution:

Let r and h be the radius and height of the cone.

When the radius and height of the cone are reduced by 20%, the radius and height of the new cone will be

Now, surface area of the original cone, A = πr (l + r)

And, surface area of the new cone,

Therefore, percentage reduce in surface area

Question 4(a) ( 3.0 marks)

If sin A + cos A = p and sec A + cosec A = q, then prove that q (p² − 1) = 2p

Solution:

We are given that:

sin A + cos A = p and sec A + cosec A = q

To prove: q (p² − 1) = 2p

i.e., to prove:

Thus,

That is, q (p² − 1) = 2p

Hence, proved

Question 4(b) ( 3.0 marks)

(i) Prove that:

(ii) If mean, median, and mode of the given data are 11x + 5, 12x − 6, and 9x + 22, then find the value of x.

Solution:

(i)

Hence, proved

(ii) We know that:

3 Median = Mode + 2 Mean

⇒ 3 (12x − 6) = (9x + 22) + 2 (11x + 5)

⇒ 36x − 18 = 31x + 32

⇒ 5x = 50

Question 4(c) ( 4.0 marks)

(i) Find the value of:

(ii) When a die is thrown, what is the probability of getting:

(I) An odd number

(II) A number which is a multiple of 3

Solution:

(i)

(ii)

When a die is thrown, then the possible outcomes are

1, 2, 3, 4, 5 and 6

Out of these numbers, odd numbers are 1, 3 and 5, and multiples of 3 are 3 and 6.

(I) Probability of getting an odd number

(II) Probability of getting a number which is a multiple of 3

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)

Daman took a loan of Rs 27000 for years from a bank. The bank charged him an interest at the rate of 8% per annum compounded half yearly. Find the amount which Daman will have to repay.

Solution:

Here, principal, P = Rs 27000

Rate of interest, R = 8% annually i.e., 4% half yearly

Time = years

Also, there are 5 half years in years.

Now,

Thus, Daman will have to pay Rs 32,849.6 back to the bank at the end of years.

Question 5(b) ( 4.0 marks)

A shopkeeper marks an article 20% above its cost price. He sells the article at Rs 6804 after a discount of 10% and sales tax of 5%. How much did the article cost the shopkeeper?

Solution:

Let the C.P. of the article be Rs x.

∴ M.P. of the article = Rs

Rate of discount = 10%

∴Discount = Rs

∴Selling price of the article excluding sales tax = Rs

Sales tax levied

∴ Selling price of the article including sales tax

According to the given information,

Thus, the article cost Rs 6000 to the shopkeeper.

Question 5(c) ( 3.0 marks)

Ms. Amina has a recurring deposit account in a certain nationalised bank. She deposits a certain fixed amount of money per month for 4 years at the rate of 10% per annum. If she gets Rs 46240 on maturity, then find the monthly instalment.

Solution:

Let the money deposited by Amina per month be x.

∴ P = Rs x

r = 10%

Number of months (x) = 4 × 12 = 48

∴ Interest earned

∴Maturity value = Rs 48x +

Thus, the required monthly instalment is Rs 800.

Question 6(a) ( 3.0 marks)

Using the long division method, determine the remainder when the polynomial 4x⁵ + 2x⁴ − x³ + 4x² − 7 is divided by (x − 1).

Solution:

Hence, the remainder is 2.

Question 6(b) ( 4.0 marks)

(i) A ray of light is sent along the line x + y = 3. The x-axis acts as a mirror and the ray is reflected.

What is the equation of the reflected ray?

(ii) What is the solution set of the system of linear inequalities, 3x + 5 < x − 9 and 2x − 1 ≥ −(x + 34)?

Solution:

(i)

The slope of line, x + y = 3 is −1 i.e., m₁ = −1

Let the slope of reflected ray be m₂.

Using the perpendicularity condition, we obtain

m₁m₂ = −1 ⇒ m₂ = 1

The reflected ray is of the form, y = m₂x + c ⇒ y = x + c

The reflected ray passes through the point (3, 0).

∴ c = −3

Thus, the required equation is y = x − 3 or x − y = 3

(ii) The given system of linear inequalities is

3x + 5 < x − 9

2x − 1 ≥ −(x + 34)

From inequality (1), we have

3x + 5 < x − 9

⇒ 3x − x < −9 − 5

⇒ 2x < −14

⇒ x < −7 … (1)

2x − 1 ≥ −(x + 34)

⇒ 2x − 1 ≥ − x − 34

⇒ 2x + x ≥ − 34 + 1

⇒ 3x ≥ −33

⇒ x ≥ −11 … (2)

From (1) and (2), the solution of the system are real numbers lying between −11 and −7 including −11.

Thus, the required solution set is [−11, −7).

Question 6(c) ( 3.0 marks)

In a class of 90 students, if 26% of boys and 35% of girls got first class, then the total number of first class students is less than 30% of total students. If 25% of boys and 50% of girls got first class, then more than of the total students got first class. What will be the number of girls in the class?

Solution:

Let the number of girls be x.

Thus, number of boys = 90 − x

According to the given information,

35% of x + 26% of (90 − x) <>

From (1) and (2), 30 < x <>

Thus, the number of girls should lie between 30 and 40.

Question 7(a) ( 3.0 marks)

In the given figure, a circle is shown which touches the sides of the quadrilateral WXYZ. If the lengths of the opposite sides WX and YZ are 15 cm and 18 cm respectively, then find the perimeter of the quadrilateral WXYZ.

Solution:

We know that the tangents drawn from an exterior point to the circle are equal in length.

Therefore,

WA = WD (1)

XA = XB (2)

ZC = ZD (3)

YC = YB (4)

Adding equations (1), (2), (3), and (4), we obtain

WA + XA + ZC + YC = WD + XB + ZD + YB

⇒ (WA + XA) + (ZC + YC) = (WD + ZD) + (XB + YB)

⇒ WX + YZ = ZW + XY

Now, WX = 15 cm and YZ = 8 cm (Given)

∴ ZW + XY = WX + YZ = 15 cm + 8 cm = 23 cm

∴ Perimeter of quadrilateral WXYZ

= WX + XY + YZ + WZ

= (WX + YZ) + (XY + WZ)

= 23 cm + 23 cm

= 46 cm

Question 7(b) ( 4.0 marks)

Suppose that a circle is given to you without its centre and a point P outside it is given to you as shown in the following figure. How will you construct the tangents to this circle?

Solution:

In the given figure, the centre of the circle is not given. Therefore, first of all, we will find the centre of the circle.

For this, let us draw two non-parallel chords of the circle and draw their perpendicular bisectors. The point of intersection of these perpendicular bisectors is the centre O of the circle.

Now, we join the line segment OP and draw its perpendicular bisector. Let this perpendicular bisector intersect OP at point M.

Then we draw a circle with radius equal to OM, taking M as centre. Let this circle intersect the given circle at points T and Q. We join PT and PQ.

Thus, PT and PQ are the required tangents from a point P outside the circle.

Question 7(c) ( 3.0 marks)

Identify the line(s) of symmetry in the following figures. Justify your answer with reason.

(i)

(ii)

(iii)

Solution:

Line(s) of symmetry of a figure is a line at the middle of the figure that divides the figure into two parts such that one part is exactly the same as other.

(i) In the given figure, it can be observed that the line l₁ divides the given figure into two parts such that one part is exactly the same as other. Thus, line l₁ is the line of symmetry of the given figure.

(ii) In the given figure, it can be observed that the line l₄ divides the given figure into two parts such that one part is exactly the same as other. Thus, line l₄ is the line of symmetry of the given figure.

(iii) In the given figure, it can be observed that the line l₁ divides the given figure into two parts such that one part is exactly same as other. Thus, l₁ is the line of symmetry of the given figure.

Question 8(a) ( 3.0 marks)

In the following figure, PT is a tangent and ∠OPT = 30°. What is the area of ΔOTP, if perimeter of ΔOPT is cm.

Solution:

Let OT = x cm

Since PT is a tangent, OT⊥TP

Now, in ΔOPT, and

Therefore, perimeter of ΔOTP = OT + TP + OP

Therefore, area of ΔOTP

Question 8(b) ( 3.0 marks)

Two rectangular cardboards of dimension 44 cm × 22 cm are folded to form a cylinder. The first one is folded along the length and second one is folded along the breadth. Compare their

(i) Lateral surface areas

(ii) Volumes

Solution:

(i) The two given cardboards will be folded as shown in the following figure:

Along the length:

Along the breadth:

Clearly, from the figure, we can deduce that heights of the 1^st and 2^nd cylinder are 22 cm and 44 cm respectively i.e., h₁ = 22 cm and h₂ = 44 cm

And the circumferences of the 1^st cylinder and the 2^nd cylinder are 44 cm and 22 cm respectively.

Now, lateral surface area of 1^st cylinder = 2πr₁h₁

= 968 cm²

Lateral surface area of 2^nd cylinder 2πr₂h₂

= 968 cm²

Hence, the lateral surface areas of both the cylinders are same.

(ii)

⇒ Volume of 1^st cylinder = 2 × Volume of 2^nd cylinder

Question 8(c) ( 4.0 marks)

A toy is in the shape as shown in the figure. The bottom of the toy has a hemispherical raised portion. The height of the entire toy is cm whereas the height of the conical shape is 6 cm. The bases of the hemisphere, cone, and cylinder are identical. Find the volume of the entire toy and its surface area, if the radius of base of hemisphere is 2.5 cm.

Solution:

Radii of hemisphere, cylinder, and cone = 2.5 cm

Height of the cone = 6 cm

Therefore, height of the cylinder =

Now,

Volume of the toy = Volume of cylinder + Volume of cone − Volume of hemisphere

Now, slant height of the cone,

Therefore,

Surface area of the cone = Lateral surface area of cone + Lateral surface area of hemisphere + Lateral surface area of cylinder

Question 9(a) ( 3.0 marks)

The angles of elevation of the top of a tower from two points, distant a and b from its base in the same straight line, are complementary. Show that the height of the tower is .

Solution:

Let the two discussed points be A and B.

We have BC = a and AC = b

Here, CD represents the tower.

Let ∠DBC = θ

∴ ∠DAC = 90° − θ (the angles are given to be complementary)

In ΔDBC,

In ΔDAC,

Multiplying (1) and (2), we obtain

Thus, the height of the tower is.

Hence, proved

Question 9(b) ( 3.0 marks)

There are two malls of equal heights on a straight road exactly opposite to each other. The width of the road is 100 m. From a point somewhere lying in between them on the road, the angles of elevation of the top of the malls are 45° and 30° respectively. Find the heights of the malls and the distances of the point from these malls.

Solution:

Let PQ and ST represent the two malls. We have PQ = ST

It is given that the width of the road is 100 m i.e., the distance between the two malls is 100 m.

Let RS = x m

∴QR = (100 − x) m

In ΔPQR,

⇒ PQ = (100 − x) m (1)

In ΔRST,

∴The distances of the point (lying between the two malls) from the two malls are 63.37 m and 36.63 m.

Also, PQ = (100 − x) m = (100 − 63.37) m = 36.63 m

Thus, the height of the malls is 36.63 m.

Question 9(c) ( 4.0 marks)

At the foot of a mountain, the elevation of its peak is 45°. After ascending 2 km towards the peak at an inclination of 30°, the elevation becomes 60°. Find the height of the mountain.

Solution:

Let BC represents the height of the mountain, where C is the peak of the mountain.

We have to find the height of the mountain i.e., BC.

It is given that AD = 2 km

In ΔADF,

In ΔDCE,

∴AB = AF + FB

In ΔABC,

Thus, the height of the mountain is 2.74 km.

Question 10(a) ( 4.0 marks)

The following table represents the average daily earnings of 50 general stores in a market during a certain week. Find the mean daily earning of these stores by using step deviation method.

Daily earning (in rupees)	1000 − 1500	1500 − 2000	2000 − 2500	2500 − 3000	3000 − 3500
Number of stores	20	10	9	6	5

Solution:

Let a = 2250 be the assumed mean.

We calculate the deviations of the class marks from the assumed mean a. The calculations are shown in the following table:

Daily earning (in Rupees)	Number of stores, fi	Class marks, xi	di = xi− a		fiui
1000 − 1500	20	1250	−1000	−2	−40
1500 − 2000	10	1750	−500	−1	−10
2000 − 2500	9	2250 = a	0	0	0
2500 − 3000	6	2750	500	1	6
3000 − 3500	5	3250	1000	2	10
Total	50	-	-	-	−34

Here, h, class size = 500

We know that:

Therefore, the mean daily earning of the general stores is Rs 1910.

Question 10(b) ( 6.0 marks)

The following distribution gives the height of 100 students of a class.

Height of students (in cm)	120 − 130	130 − 140	140 − 150	150 − 160	160 − 170	170 − 180
Number of students	5	15	25	30	20	5

Convert above frequency distribution into more than type distribution and find the median height of the students by drawing an ogive.

Solution:

We can write the given distribution in more than type as follows:

Height (in cm)	Number of students (cf)
More than or equal to 120	100
More than or equal to 130	100 − 5 = 95
More than or equal to 140	95 − 15 = 80
More than or equal to 150	80 − 25 = 55
More than or equal to 160	55 − 30 = 25
More than or equal to 170	25 − 20 = 5

(Lower limit, Corresponding cumulative frequency) is the ordered pair of the points to be plotted. Thus, the points are (120, 100), (130, 95), (140, 80), (150, 55), (160, 25), (170, 5).

Now, by choosing an appropriate scale (along x-axis, 1 cm = 10 cm height and along y-axis, 1 cm = 10 students), we can draw the required ogive.

Now,

We mark the point 50 on vertical line and then draw a horizontal line through this point. Let the horizontal line intersect the ogive at point A. Now, through A, draw a vertical line that intersects the x-axis at point B.

Point B represents the point 152 (approx).

Therefore, the median height of the students is 152.

Question 11(a) ( 3.0 marks)

The line is perpendicular to the line joining the points (5, −9) and (17, −4). If k is an integer, then what is the value of k?

Solution:

Slope of the line joining the points (5, −9) and (17, −4) =

The equation of the given line is,

∴ Slope of this line

If two lines with the slope m₁ and m₂ are perpendicular, then m₁ × m₂ = −1

It is given that k is an integer. Therefore, k cannot be. Hence, k = 4

Thus, the value of k is 4.

Question 11(b) ( 3.0 marks)

Mohit goes to school daily from Monday to Friday. He also attends school on odd Saturdays. If we select a day randomly out of two continuous weeks, what is the probability that he goes to school on that day?

Solution:

Total number of days in 2 weeks = 14

Number of weekdays = 10

Number of odd Saturdays = 1

Number of days when Mohit goes to school during 2 continuous weeks = 11

If a day is selected at random from these 14 days, the probability that Mohit goes to school on that day

Question 11(c) ( 4.0 marks)

State whether the following statements are correct or incorrect. Justify your answers.

(i) When a die is rolled, it is equally likely that a multiple of 3 or a number greater than 4 will be obtained.

(ii) When a ball is drawn at random from a bag containing 4 yellow and 3 black balls, it is equally likely that a yellow or a black ball will be drawn.

(iii) When a coin is tossed, it is not equally likely that a head or a tail will occur.

(iv) When a card is drawn at random from a well-shuffled deck of 52 cards, it is equally likely to draw a face card or a queen.

Solution:

Two outcomes of an experiment are said to be equally likely if the possibility of occurrence of each outcome is the same.

(i) Correct

A die has six faces, numbered as 1, 2, 3, 4, 5 and 6. Out of these, 3 and 6 are multiples of 3, while 5 and 6 are greater than 4. Therefore, the number of multiples of 3 and the number of digits greater than 4 are the same. Thus, when a die is rolled, it is equally likely that a multiple of 3 or a number greater than 4 will be obtained.

(ii) Incorrect

The bag contains more yellow balls than black balls. Thus, when a ball is drawn at random from the bag containing 4 yellow and 3 black balls, it is more likely that a yellow ball will be drawn.

(iii) Incorrect

A coin has two faces: head and tail. When a coin is tossed, it is equally likely to land as a head or a tail.

(iv) Incorrect

A standard deck of 52 cards contains twelve face cards and four queens. Thus, the number of face cards is more than the number of queens. Hence, when a card is drawn at random from a well-shuffled deck of 52 cards, it is more likely that a face card will be drawn.