mathamatics xam idea

Based on above situation answer the following questions.

(i) What will be the distance covered by Ajay’s car in two hours?

(a) 2(x +5) km (b) (x – 5) km

(c) 2(x + 10) km (d) (2x + 5) km

(ii) The quadratic equation in terms of speed of Raj’s car is

(a) x2 – 5 x – 500 = 0 (b) x2 + 4x – 400 = 0

(c) x2 + 5x – 500 = 0 (d) x2 – 4x + 400 = 0

(iii) What is the speed of Raj’s car? (c) 25 km/hour (d) 10 km/hour
(a) 20 km/hour (b) 15 km/hour

(iv) How much time Ajay took to travel 400 km?

(a) 20 hours (b) 40 hours (c) 25 hours (d) 16 hours

(v) What is the speed of Ajay’s car?

(a) 15 km/h (b) 20 km/h (c) 25 km/h (d) 30 km/h

Sol. We have speed of Raj’s Car = x km/h

Therefore, speed of Ajay’s Car = (x + 5)km/h

Now, Time taken to complete the journey by Raj = 400
x

and Time taken to complete the journey by Ajay = 400
x+5

According to question, Raj took 4 hours more than Ajay.
x+5 – x
∴ 400 – 400 = 4 ⇒ 400e x]x +5g o = 4
x x+5

⇒ 500 = x(x + 5) = x2 + 5x

⇒ x2 + 5x – 500 = 0 ⇒ x2 + 25x – 20x – 500 = 0

⇒ x(x + 25) – 20(x + 25) = 0

⇒ (x + 25) (x – 20) = 0 ⇒ x = 20, x ≠ – 25 (Because speed can never be negative)

∴ x = 20 km/h

(i) Distance covered by Ajay’s car in two hours

= Speed × Time = (x + 5) × 2 = 2(x + 5)km

∴ Option (a) is correct.

(ii) We get the quadratic equation x2 + 5x – 500 = 0 (Described above)

for the speed of Raj’s car.

∴ Option (c) is correct.

Case Study Based Questions 543

(iii) Speed of the Raj’s car = x km/h = 20 km/h

∴ Option (a) is correct.

(iv) Time taken by Ajay to travel 400 km = 400
x+5
400 400
= 20 + 5 = 25 =16 hours

∴ Option (d) is correct.

(v) Speed of Ajay’s car = (x + 5) km/h = 20 + 5 = 25 km/h

∴ Option (c) is correct.

2. Read the following and answer any four questions from (i) to (v).

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour

more for upstream than downstream. [CBSE Question Bank]



(i) Let speed of the stream be x km/h then speed of the motorboat in upstream will be

(a) 20 km/h (b) (20 + x) km/h (c) (20 – x) km/h (d) 2 km/h

(ii) What is the relation between speed ,distance and time?

(a) speed = distance/time (b) distance = speed/time

(c) time = speed × distance (d) speed = distance × time

(iii) Which is the correct quadratic equation for the speed of the current ?

(a) x2+ 30x − 200 = 0 (b) x2+ 20x − 400 = 0

(c) x2+ 30x − 400 = 0 (d) x2− 20x − 400 = 0

(iv) What is the speed of current ? (c) 15 km/hour (d) 25 km/hour
(a) 20 km/hour (b) 10 km/hour

(v) How much time boat took in downstream?

(a) 90 minutes (b) 15 minutes (c) 30 minutes (d) 45 minutes

Sol. We have downstream speed of motor boat = (20 + x) km/h

and upstream speed of motor boat = (20 – x) km/h

Where x km/h is the speed of the stream.

We have, 15^20 + x – 20 + xh
]20 – xg^20 + xh
15 x – 15 =1 ⇒ =1
20 – 20 + x

⇒ 15 × 2x = 400 – x2 ⇒ x2+ 30x − 400 = 0

⇒ x2+ 40x – 10x − 400 = 0 ⇒ x(x + 40) – 10(x + 40) = 0
⇒ (x + 40) (x – 10) = 0 ⇒ x = 10 x ≠ – 40
∴ x = 10 km/h
 Speed can never be negative

544 Xam idea Mathematics–X

(i) Speed of the motor boat in upstream = (20 – x) km/h
∴ Option (c) is correct.
(ii) Relation between speed, distance and time is given by

Speed = distance
time

∴ Option (a) is correct.

(iii) The quadratic equation x2 + 30x – 400 = 0 is for the speed of the current (Stream)

∴ Option (c) is correct.

(iv) Speed of the current = x km/h

= 10 km/h

∴ Option (b) is correct.

(v) Time taken by boat in downstream = 15 x h.
20 +

= 15 = 15 = 1 hr = 30 ninutes
∴ Option (c) is correct. 20 +10 30 2

zzz

Chapter-5: Arithmetic Progressions

1. Read the following and answer any four questions from (i) to (v).

India is competitive manufacturing location due to the low cost of manpower and strong technical

and engineering capabilities contributing to higher quality production runs. The production of

TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in

6th year and 22600 in 9th year. [CBSE Question Bank]

Based on the above information, answer the following questions:

(i) What is the production during first year?

(a) 2200 (b) 2500 (c) 5000 (d) 8000

(ii) What is the production during 8th year?

(a) 11,000 (b) 20,400 (c) 22,000 (d) 20,000

(iii) Total production in (during) first 3 years is

(a) 21,600 (b) 22,000 (c) 25,000 (d) 30,000

(iv) In which year, the production is `29,200?

(a) 9th (b) 12th (c) 14th (d) 15th

(v) The difference of the production during 7th year and 4th year.

(a) 1100 (b) 2200 (c) 4400 (d) 6600

Case Study Based Questions 545

Sol. (i) Let a be the production of TV sets in a factory in first year and d be the fixed number of TV
sets increased every year.

Thus we have,

a + (6 – 1)d = 16000 ⇒ a + 5d = 16000 ...(i)

a + (9 – 1)d = 22600 ⇒ a + 8d = 22600 ...(ii)

Subtracting (i) from (ii), we have

3d = 6600 ⇒ d = 6600
⇒ 3

From equation (i), we have d = 2200

a + 5 × 2200 = 16000 ⇒ a = 16000 – 11000

⇒ a = 5000

We have the production during the first year = 5000

∴ Option (c) is correct.

(ii) Production during 8th year, a8 = a + (8 – 1)d

= a + 7d

= 5000 + 7 × 2200

= 5000 + 15400

= 20400

∴ Option (b) is correct.

(iii) Total production in (during) first 3 years

= n ^2a + ]n – 1gdh
2

= 3 ^2 × 5000 + 2 × 2200h
2

= 3(5000 + 2200)

= 3 × 7200 = 21600

∴ Option (a) is correct.

(iv) Let in nth year, the production is 29200.

⇒ an = 29200
⇒ a + (n – 1)d = 29200

⇒ 5000 + (n – 1) × 2200 = 29200

⇒ (n – 1) × 2200 = 24200

⇒ (n – 1) = 24200 =11
2200

⇒ n = 12

∴ In 12th year the production is 29200.
∴ Option (b) is correct.
(v) Difference of the production during 7th year and 4th year
= (a + 6d) – (a + 3d) = 3d
= 3 × 2200 = 6600
∴ Option (d) is correct.

546 Xam idea Mathematics–X

2. Read the following and answer any four questions from (i) to (v).

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in

51 seconds and with each day of practice it takes him 2 seconds less. He wants to complete it in

31 seconds. [CBSE Question Bank]

Based on the above information, answer the following questions:

(i) Which of the following terms are in AP for the given situation?

(a) 51,53,55…. (b) 51, 49, 47…. (c) –51, –53, –55…. (d) 51, 55, 59…

(ii) What is the minimum number of days he needs to practice till his goal is achieved?

(a) 10 (b) 12 (c) 11 (d) 9

(iii) Which of the following term is not in the AP of the above given situation?

(a) 41 (b) 30 (c) 37 (d) 39

(iv) If nth term of an AP is given by an = 2n + 3 then common difference of the AP is

(a) 2 (b) 3 (c) 5 (d) 1

(v) The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of the AP is

(a) 6 (b) –6 (c) 18 (d) –18

Sol. (i) Since Veer currently runs the distance in 51 seconds and with each day of practice it takes
him 2 seconds less.

Thus, AP formed is

51, 49, 47, ....

∴ Option (b) is correct.

(ii) Let n be the minimum number of days for AP. 51, 49, 47 .....

∴ an = 31 ⇒ a +(n – 1)d = 31
51 +(n – 1) × (–2) = 31
⇒

⇒ (n – 1) × (–2) = –20

⇒ n – 1 = 10

⇒ n = 11

∴ Option (c) is correct.

(iii) Since an = 30 ⇒ a +(n – 1)d = 30
51 +(n – 1) × (–2) = 30
⇒

⇒ (n – 1) × (–2) = –21

⇒ n–1= 21
2

⇒ n= 23 ! integer
2

Case Study Based Questions 547

∴ 30 is not in AP.

∴ Option (b) is correct.
(iv) Given nth term of an AP is

an = 2n + 3
∴ d = an – an – 1 = (2n + 3) – (2(n – 1) + 3)
⇒ d = 2

∴ Option (a) is correct.

(v) Since, 2x, x + 10, 3x + 2 are three consecutive terms are in AP.

∴ (x + 10) – 2x = (3x + 2) – (x + 10)

⇒ 10 – x = 2x – 8

⇒ 18 = 3x ⇒ x = 6

∴ Option (a) is correct.

3. Read the following and answer any four questions from (i) to (v).

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays

his total loan of ` 1,18,000 by paying every month starting with the first instalment of ` 1000. If

he increases the instalment by ` 100 every month. [CBSE Question Bank]

B ased on the above information, answer the following questions:
(i) The amount paid by him in 30th instalment is

(a) `3900 (b) `3500 (c) `3700 (d) `3600

(ii) The amount paid by him upto 30 instalments is

(a) `37000 (b) `73500 (c) `75300 (d) `75000
(iii) What amount does he still have to pay after 30th instalment?

(a) `45500 (b) `49000 (c) `44500 (d) `54000

(iv) If total instalments are 40 then amount paid in the last instalment is

(a) `4900 (b) `3900 (c) `5900 (d) `9400
(v) The ratio of the 1st instalment to the last instalment is

(a) `1:49 (b) `10:49 (c) `10:39 (d) `39:10

Sol. (i) Since first instalment is of `1000 and he increases the instalment by `100 every month.

Thus, AP is formed

1000, 1100, 1200, ....

∴ a = 1000, d = 100
Amount paid by him in 30th instalment

= a + 29d

= 1000 + 29 × 100 = `3900

∴ Option (a) is correct.

548 Xam idea Mathematics–X

(ii) Amount paid upto 30 instalments
n
= 2 ^2a + ]n – 1gdh

= 30 ^2 ×1000 + 29 ×100h
2

= 30 × 4900 =15 × 4900 = ` 73500
2
∴ Option (b) is correct.

(iii) The amount still has to pay after 30 instalments

= `1,18,000 – `73,500

= `44500

∴ Option (c) is correct.
(iv) Amount paid in 40th instalment

⇒ a40 = a + 39d
= 1000 + 39 × 100

= `4900

∴ Option (a) is correct.

(v) Ratio of 1st and last instalment
a1
= a40 = 1000 = 10
4900 49

∴ Ratio is 10:49.

∴ Option (b) is correct.

zzz

Chapter-6: Coordinate Geometry

1. Read the following and answer any four questions from (i) to (v).

In order to conduct Sports Day activities in your School, lines have been drawn with chalk

powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have

been placed at a distance of 1 m from each other along AD, as shown in given figure below.

Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th

distance AD on the eighth line and posts a red flag. [CBSE Question Bank]

DC

A B

Fig. 6.1 Case Study Based Questions 549

Now she asked some questions to the students as given below:

(i) Find the position of green flag.

(a) (2, 25) (b) (2, 0.25) (c) (25, 2) (d) (0, –25)

(ii) Find the position of red flag.

(a) (8, 0) (b) (20, 8) (c) (8, 20) (d) (8, 0.2)

(iii) What is the distance between both the flags?

(a) 41 (b) 11 (c) 61 (d) 51

(iv) If Rashmi has to post a blue flag exactly halfway between the line segment joining the two
flags, where should she post her flag?

(a) (5, 22.5) (b) (10, 22) (c) (2, 8.5) (d) (2.5, 20)

(v) If Joy has to post a flag at one-fourth distance from green flag in the line segment joining
the green and red flags, then where should he post his flag?

(a) (3.5, 24) (b) (0.5, 12.5) (c) (2.25, 8.5) (d) (25, 20)

Sol. (i) Niharika runs 1 th the distance AD on the second line, post a green flag.
4
1
∴ Position of the green flag is c2, 4 ×100 m , i.e., (2, 25).

∴ Option (a) is correct.

(ii) Preet runs 1 th distance AD on the eight line and posts a red flag.
5

∴ Position of the red flag is c8, 1 ×100 m , i.e., (8, 20).
5

Its position is (8, 20).

∴ Option (c) is correct.

(iii) Distance between by green flag and red flag is given by

= (20 – 25)2 + (8 – 2)2 = 25 + 36 = 61 units

∴ Option (c) is correct.

(iv) Position of the blue flag = d 2 + 8 , 25 + 20 n = (5, 22.5)
2 2

∴ Option (a) is correct.

(v) Let P (x, y) be the position of a flag posted by Joy using section formula, we have

x = 1×8 + 4×2 = 16 = 3.2 1 4
1+4 5 P
G (x, y)
y = 1×20 + 4 ×25 = 120 = 24 (2, 25) R
1+4 5 Fig. 6.2 (8, 20)

∴ Position of the flag posted by Joy is (3.2, 24).

∴ Option (a) is correct.

2. Read the following and answer any four questions from (i) to (v).

The class X students of a school in krishnagar have been allotted a rectangular plot of land for
their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m
from each other. There is triangular grassy lawn in the plot as shown in the figure. The students
are to sow seeds of flowering plants on the remaining area of the plot. [CBSE Question Bank]

550 Xam idea Mathematics–X

B C

P
R

Q

A 1 2 3 4 5 6 7 8 9 10 D

Fig. 6.3

Now she asked some questions to the students as given below:

(i) Taking A as origin, find the coordinates of P.

(a) (4, 6) (b) (6, 4) (c) (0, 6) (d) (4,0)

(ii) What will be the coordinates of R, if C is the origin?

(a) (8, 6) (b) (3, –10) (c) (–10, –3) (d) (0, 6)

(iii) What will be the coordinates of Q, if C is the origin?

(a) (6, 13) (b) (–6, 13) (c) (–13, 6) (d) (–13, –6)

(iv) Calculate the area of the triangles if A is the origin.

(a) 4.5 (b) 6 (c) 8 (d) 6.25

(v) Calculate the area of the triangles if C is the origin.

(a) 8 (b) 5 (c) 6.25 (d) 4.5

Sol. (i) Taking A as origin we have

7
P

6
5
4
3
2
1

0 1234567

Fig. 6.4

∴ Co-ordinates of P be (4, 6).
∴ Option (a) is correct.

Case Study Based Questions 551

(ii) By taking C as origin, we have

−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0
−1

−2

R (−10, −3) −3
−4

−5

−6

−7

−8

−9

Fig. 6.5

∴ Co-ordinates of R be (–10, –3)
∴ Option (c) is correct.
(iii) When C is the origin, we have

−13 −12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0
−1

−2

−3

−4

−5

Q (−13, −6) −6
−7

−8

−9

Fig. 6.6

∴ Co-ordinates of Q be (–13, –6)
∴ Option (d) is correct.
(iv) When A is origin then co-ordinates of point P, Q and R are respectively P (4, 6), Q (3, 2) and

R (6, 5).

552 Xam idea Mathematics–X

∴ Area DPQR = 1 4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)
2

= 1 4 ×(–3) + 3× (–1) + 6× 4
2

= 1 –12 – 3 + 24 = 9 = 4.5 sq. units
2 2

∴ Option (a) is correct.

(v) When C is the origin then vertices of DPQR are respectively P (–12, –2), Q (–13, –6) and

R (–10, –3)

∴ Area DPQR = 1 –12 (–6 + 3) + (–13)×(–3 + 2) + (–10) (–2 + 6)
2

= 1 36 + 13 – 40 = 9 = 4.5 sq. units
2 2

∴ Option (d) is correct.

zzz

Chapter-7: Triangles

1. Read the following and answer any four questions from (i) to (v).


Vijay’s House Tower Ajay’s House

Fig. 7.1

Vijay is trying to find the average height of a tower near his house. He is using the properties

of similar triangles. The height of Vijay’s house if 20 m when Vijay’s house casts a shadow 10 m

long on the ground. At the same time, the tower casts a shadow 50 m long on the ground and

the house of Ajay casts 20 m shadow on the ground. [CBSE Question Bank]

(i) What is the height of the tower?

(a) 20 m (b) 50 m (c) 100 m (d) 200 m

(ii) What will be the length of the shadow of the tower when Vijay’s house casts a shadow of
12m?

(a) 75 m (b) 50 m (c) 45 m (d) 60 m

(iii) What is the height of Ajay’s house?

(a) 30 m (b) 40 m (c) 50 m (d) 20 m

Case Study Based Questions 553

(iv) When the tower casts a shadow of 40m, same time what will be the length of the shadow
of Ajay’s house?

(a) 16 m (b) 32 m (c) 20 m (d) 8 m

(v) When the tower casts a shadow of 40m, same time what will be the length of the shadow
of Vijay’s house?

(a) 15 m (b) 32 m (c) 16 m (d) 8 m

Sol. (i) Let h m be the height of tower, therefore using property of similar triangle between two
triangle i.e. for Vijay’s house and for tower with shadows.

We have,

20 = 10 ⇒ h = 100 m
h 50

∴ Height of tower is 100 m.

∴ Option (c) is correct.

(ii) When Vijay’s house casts a shadow of 12 m, we have

20 = 12 , where x is the length of shadow of tower.
h x

⇒ 12000 = 12 ⇒ x = 60 m
x

∴ Option (d) is correct.

(iii) Let H m be the height of Ajay’s house

∴ 20 = H ⇒ H = 40 m
10 20

∴ Option (b) is correct.

(iv) When the tower casts a shadow of 40 m.

∴ 100 = Height of Ajay's house
40 Length of shadow of Ajay's house

⇒ 5 = 40
2 Shadow length

⇒ Length of shadow of Ajay house = 16 m.

∴ Option (a) is correct.

(v) We have,

100 = Height of Vijay's house
40 Length of its shadow

⇒ 5 = Length 20 shadow
2 of its

⇒ Length of shadow of Vijay house = 8 m ∴ Option (d) is correct.

2. Read the following and answer any four questions from (i) to (v).

Rohan wants to measure the distance of a pond during the visit to his native. He mark points A and

B on the opposite edges of a pond as shown in the figure below. To find the distance between the

points, he makes a right-angled triangle using rope connecting B with another point C at a distance

of 12 m, connecting C to point D at a distance of 40m from point C and then connecting D to the

point A which is at a distance of 30 m from D such that ∠ADC = 90° . [CBSE Question Bank]

554 Xam idea Mathematics–X

A B 12 m C

30 m 40 m

D

Fig. 7.2
(i) Which property of geometry will be used to find the distance AC?

(a) Similarity of triangles (b) Thales Theorem

(c) Pythagoras Theorem (d) Area of similar triangles

(ii) What is the distance AC?

(a) 50m (b) 12m (c) 100m (d) 70m

(iii) Which is the following does not form a Pythagorian triplet?

(a) 7,24,25 (b) 15,8,17 (c) 5,12,13 (d) 21,20,28

(iv) The length AB is

(a) 12m (b) 38m (c) 50m (d) 100m

(v) The length of the rope used

(a) 120m (b) 70m (c) 82m (d) 22m

Sol. (i) Pythagoras Theorem will be used to find the distance AC in right angle ∆ACD.

∴ Option (c) is correct.

(ii) In right angle ∆ACD, we have

AC2 = AD2 + CD2 (Pythagoras Theorem)

⇒ AC2 = (30)2 + (40)2 = 900 + 1600 = 2500

⇒ AC = 50 m

∴ Option (a) is correct.

(iii) We have,

(21, 20, 28) is not a Pythagorean triplet because

(21)2 + (20)2 ≠ (28)2

⇒ 441 + 400 ≠ 784

⇒ 841 ≠ 784

∴ Option (d) is correct.

(iv) AB = AC – BC = 50 – 12 = 38 m
∴ Option (b) is correct.
(v) Total length of rope used
= BC + CD + DA
= 12 + 40 + 30 = 82 m
∴ Option (c) is correct.

zzz

Case Study Based Questions 555

Chapter-8: Circles

1. Read the following and answer any four questions from (i) to (v).
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a
rotating upright wheel with multiple passenger-carrying components (commonly referred to as
passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that
as the wheel turns, they are kept upright, usually by gravity. After taking a ride in Ferris wheel,
Aarti came out from the crowd and was observing her friends who were enjoying the ride . She
was curious about the different angles and measures that the wheel will form. She forms the

figure as given below. [CBSE Question Bank]

SR SR

OO

30° P 30° P
Q Q

Fig. 8.1
Based on the above information, answer the following questions:

(i) In the given figure ∠ROQ is equal to

(a) 60° (b) 100° (c) 150° (d) 90°

(ii) The measurement of ∠RQP is

(a) 75° (b) 60° (c) 30° (d) 90°

(iii) The measurement of ∠RSQ is

(a) 60° (b) 75° (c) 100° (d) 30°

(iv) The measurement of ∠ORP is

(a) 90° (b) 70° (c) 100° (d) 60°

(v) Reflex angle of ∠ROQ is

(a) 180° (b) 150° (c) 210° (d) 360°

Sol. (i) In quadrilateral PROQ, we have

∠RPQ + ∠PQO + ∠QOR + ∠ORP = 360°

⇒ 30° + 90° + ∠ROQ + 90° = 360° ( OR ⊥ PR, OQ ⊥ PQ)

⇒ ∠ROQ = 360° – 210° = 150°

⇒ ∠ROQ = 150°

∴ Option (c) is correct.

(ii) In ∆OQR, we have

∠OQR = ∠ORQ ( OQ = OR, radii of a circle)

∠QOR + ∠OQR + ∠ORQ = 180°

150° + ∠OQR + ∠OQR = 180° ⇒ 2∠OQR = 30°

⇒ ∠OQR = 15°

∠RQP = 90° – 15° = 75° ( a OQ ⊥ PQ)

∴ Option (a) is correct.

(iii) We have, Angle at the centre is double
=+12R+SQQO=R12 ×15 0° = that of circumference made by
+RSQ 75° same segment.
⇒

∴ Option (b) is correct.

556 Xam idea Mathematics–X

(iv) Since, OR is radius and PR is tangent at R

∴ OR ⊥ PR ⇒ ∠ORP = 90°

∴ Option (a) is correct.

(v) Reflex of ∠ROQ = 360° – ∠ROQ

= 360° – 150° = 210°

∴ Option (c) is correct.

2. Read the following and answer any four questions from (i) to (v).

Varun has been selected by his School to design logo for Sports Day T-shirts for students and

staff. The logo design is as given in the figure and he is working on the fonts and different

colours according to the theme. In given figure, a circle with centre O is inscribed in a ΔABC,

such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of

sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively. [CBSE Question Bank]

C

FE

AD B
Fig. 8.2

Based on the above information, answer the following questions:

(i) The length of AD is

(a) 7 cm (b) 8 cm (c) 5 cm (d) 9 cm

(ii) The length of BE is

(a) 8 cm (b) 5 cm (c) 2 cm (d) 9 cm

(iii) The length of CF is

(a) 9 cm (b) 5 cm (c) 2 cm (d) 3 cm

(iv) If radius of the circle is 4cm, then the area of ΔOAB is
(a) 20 cm2 (b) 36 cm2 (c) 24 cm2 (d) 48 cm2

(v) The area of ΔABC is
(a) 50 cm2 (b) 60 cm2 (c) 100 cm2 (d) 90 cm2

Sol. (i) Let AD = x cm

∴ DB = (12 – x) cm

DB = BE = (12 – x) cm (Tangents from external point are equal)

CE = 8 – (12 – x) = (x – 4) cm

Also, CE = CF = (x – 4) cm

and AF = 10 – (x – 4) = (14 – x) cm

AF = AD ⇒ 14 – x = x ⇒ 14 = 2x ⇒ x = 7 cm (x – 4) (12 – )
(x – 4)
We have, AD = x cm = 7 cm

∴ Option (a) is correct. 10 cm

(ii) Length of BE = (12 – x) cm = 12 – 7 = 5 cm

∴ Option (b) is correct.

(iii) Length of CF = (x – 4) cm = 7 – 4 = 3 cm

∴ Option (d) is correct. (12 – )
12

Fig. 8.3

Case Study Based Questions 557

(iv) We have,

Radius of the circle is 4 cm.

∴ OD = 4 cm
∴
Area of ∆OAB = 1 × AB× OD = 1 ×12 × 4 = 24 cm2
2 2

∴ Option (c) is correct.

(v) We have,

Area of ∆ABC = ar(∆OAB) + ar(∆OBC) + ar(∆OCA)

∴ = 24 + 1 ×8×4 + 1 ×10 × 4
2 2

∴ = 24 + 16 + 20 = 60 cm2

∴ Option (b) is correct.

zzz

Chapter-11: Heights and Distances

1. Read the following and answer any four questions from (i) to (v).

A group of students of class X visited India Gate on an education trip. The teacher and students

had interest in history as well. The teacher narrated that India Gate, official name Delhi

Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi,

dedicated to the troops of British India who died in wars fought between 1914 and 1919.The

teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly

called the Kingsway), is about 138 feet (42 metres) in height. [CBSE Question Bank]

   

Based on above information answer the following.

(i) What is the angle of elevation if they are standing at a distance of 42 m away from the
monument?

(a) 30° (b) 45° (c) 60° (d) 0°

(ii) They want to see the tower at an angle of 60° . So, they want to know the distance where
they should stand and hence find the distance.

(a) 24.24 m (b) 20.12 m (c) 42 m (d) 24.64 m

(iii) If the altitude of the Sun is at 60 , then the height of the vertical tower that will cast a

shadow of length 20 m, is

(a) 20 3 m (b) 20 m (c) 15 m (d) 15 3m
3 3

(iv) The ratio of the length of a rod and its shadow is 1:1 . The angle of elevation of the Sun is

(a) 30° (b) 45° (c) 60° (d) 90°

558 Xam idea Mathematics–X

(v) The angle formed by the line of sight with the horizontal when the object view is below
the horizontal level is

(a) corresponding angle (b) angle of elevation

(c) angle of depression (d) complete angle

Sol. (i) Let θ be the angle of elevation. India Gate
42 m
∴ tan i= 42 =1
42

⇒ tan θ = 1 = tan 45° 42 m
⇒ θ = 45° Fig. 11.1

∴ Angle of elevation is 45°.

∴ Option (b) is correct.

(ii) Let x m be the required distance.

∴ tan 60° = 42 India Gate 42 m
⇒ x

⇒ 3 = 42 ⇒ x = 42
x 3

x= 42 × 3 = 42 3 =14 3 60°
3 3 3 xm

Fig. 11.2

⇒ = 14 × 1.732 = 24.24 m

∴ Option (a) is correct.

(iii) Let h m be the height of the vertical tower.

We have,

∴ tan 60° = h hm
⇒ 20

3 = h ⇒ h = 20 3 m 60°
20
20 m
∴ Option (a) is correct. Fig. 11.3

(iv) Let θ be the angle of elevation.

∴ tan i= 1 =1
1

⇒ tan θ = 1 = tan 45°

θ = 45°

∴ Option (b) is correct.

(v) Angle of depression

∴ Option (c) is correct.

zzz

Case Study Based Questions 559

Chapter-12: Areas Related to Circles

1. Read the following and answer any four questions from (i) to (v).
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress,
blouse or coat.

Designs of some broochs are shown below. Observe them carefully.

AB

Fig. 12.1

Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. A wire
used for making 4 diameters which divide the circle into 8 equal parts.

Design B: Brooch B is made of two colours gold and silver. Outer part is made with gold. The

circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.

[CBSE Question Bank]

Based on the above information, answer the following questions:

Refer to design A

(i) The total length of silver wire required is

(a) 180 mm (b) 200 mm (c) 250 mm (d) 280 mm

(ii) The area of each sector of the brooch is

(a) 44 mm2 (b) 52 mm2 (c) 77 mm2 (d) 68 mm2

Refer to design B

(iii) The circumference of outer part (golden) is

(a) 48.49 mm (b) 82.2 mm (c) 72.50 mm (d) 62.86 mm

(iv) The difference of areas of golden and silver parts is

(a) 18π (b) 44π (c) 51π (d) 64π

(v) A boy is playing with brooch B. He makes revolution with it along its edge. How many
complete revolutions must it take to cover 80π mm ?

(a) 2 (b) 3 (c) 4 (d) 5

Sol. (i) We have diameter of the wire = 28 cm

⇒ d = 28 mm
∴ Radius (r) = 14 mm
Total length of silver wire required = circumference of circle + 4 × diameter of the circle

= 2r + 4 × dD

= 2 × 22 ×14 + 4 ×28
7

= 88 + 112 = 200 mm

∴ Option (b) is correct.

560 Xam idea Mathematics–X

(ii) Angle of the each sector = 360° = 45°
8

∴ Area of each sector of the brooch = i × rr2
360°

= 45° × 22 ×14 ×14
360° 7

= 1 × 22 × 2 ×14 = 77 mm2
8

∴ Option (c) is correct.

(iii) Now, refer to design B, we have

Circumference of silver part = 44 mm

2πr = 44 (where r is the radius of inner circle)

⇒ r= 44 ⇒ r = 44 = 7 mm
2r 22
2× 7

∴ Radius of outer part (R) = (7 + 3) = 10 mm

Circumference of outer part = 2πR

= 2π × 10 = 20 π

∴ Option (d) is correct. = 20 × 22 = 440 = 62.86 mm
7 7

(iv) Difference of areas of golden and silver part = πR2 – πr2

= π [(10)2 – (7)2]

= π(100 – 49) = 51π mm2

∴ Option (c) is correct.

(v) Circumference of outer part(circular) = 2πR

= 2π × 10 = 20π

∴ Number of revolution = 80r =4 .
20r

∴ Option (c) is correct.

zzz

Chapter-13: Surface Areas and Volumes

1. Read the following and answer any four questions from (i) to (v).

Adventure camps are the perfect place for the children to practice decision making for themselves

without parents and teachers guiding their every move. Some students of a school reached for

adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in

a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below.

After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it
was time to take shelter. Each group of four students was given a canvas of area 551 m2. Each

group had to make a conical tent to accommodate all the four students. Assuming that all the
stitching and wasting incurred while cutting, would amount to 1 m2, the students put the tents.

The radius of the tent is 7 m. [CBSE Question Bank]

Case Study Based Questions 561

d = 7 cm d = 7 cm
h = 10.5 cm

Fig. 13.1

Area = 551 m2

r=7m

Fig. 13.2

(i) The volume of cylindrical cup is

(a) 295.75 cm3 (b) 7415.5 cm3 (c) 384.88 cm3 (d) 404.25 cm3

(ii) The volume of hemispherical cup is

(a) 179.67 cm3 (b) 89.83 cm3 (c) 172.25 cm3 (d) 210.60 cm3

(iii) Which container had more juice and by how much?

(a) Hemispherical cup, 195 cm3 (b) Cylindrical glass, 207 cm3

(c) Hemispherical cup, 280.85 cm3 (d) Cylindrical glass, 314.42 cm3

(iv) The height of the conical tent prepared to accommodate four students is

(a) 18 m (b) 10 m (c) 24 m (d) 14 m

(v) How much space on the ground is occupied by each student in the conical tent

(a) 54 m2 (b) 38.5 m2 (c) 86 m2 (d) 24 m2

Sol. (i) We have, radius (r) of cylindrical cup = 7 cm
2

and its height (h) = 10.5 cm h = 10.5 cm

∴ Volume of cylindrical cup = pr2h

= 22 ×c 7 2 ×10.5
7 2
m

= 404.25 cm3 r = 7 cm
2

∴ Option (d) is correct. Fig. 13.3

(ii) We have, radius (r) of hemispherical cup = 7 cm
2

∴ Volume of hemispherical cup = 2 r r3
3

562 Xam idea Mathematics–X

= 2 × 22 ×c 7 3
3 7 2
m

= 2 × 22 × 7×7×7 = 89.83 cm3
3 7 2×2×2

∴ Option (b) is correct.

(iii) Cylindrical glass has more juice by (404.25 – 89.83) cm3 = 314.42 cm3

∴ Option (d) is correct.

(iv) We have cloth given of area 551 m2 and waste is 1 m2

∴ Curved surface area of conical tent = 551 – 1 = 550 m2

prl = 550 ⇒ 22 × 7 × l = 550
⇒ 7
l = 25 cm ⇒ l2 = 625

⇒ r2 + h2 = l2

⇒ (7)2 + h2 = 625

⇒ h2 = 625 – 49 = 576

⇒ h = 576 ⇒ h = 24 m

∴ Height of the tent is 24 m.

∴ Option (c) is correct.

(v) Space occupied by each student = r r2 = 22 × (7) 2 = 38.5 m2
4 7 4

∴ Option (b) is correct.

2. Read the following and answer any four questions from (i) to (v).

The Great Stupa at Sanchi is one of the oldest stone structure in India, and an important

monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in

the 3rd century BC. Its nucleus was a simple hemispherical brick structure built over the relics

of the Buddha. .It is a perfect example of combination of solid figures. A big hemispherical dome
22
with a cuboidal structure mounted on it. c Take r = 7 m [CBSE Question Bank]

Fig. 13.4

Case Study Based Questions 563

(i) Calculate the volume of the hemispherical dome if the height of the dome is 21 m.

(a) 19404 cu. m (b) 2000 cu. m (c) 15000 cu. m (d) 19000 cu. m

(ii) The formula to find the Volume of Sphere is

(a) 2 r r3 (b) 4 r r3 (c) 4pr2 (d) 2pr2
3 3

(iii) The cloth required to cover the hemispherical dome if the radius of its base is 14 m, is

(a) 1222 sq.m (b) 1232 sq.m (c) 1200 sq.m (d) 1400 sq.m

(iv) The total surface area of the combined figure i.e. hemispherical dome with radius 14m
and cuboidal shaped top with dimensions 8 m × 6 m × 4 m is

(a) 1200 sq. m (b) 1232 sq. m (c) 1392 sq.m (d) 1932 sq. m

(v) The volume of the cuboidal shaped top with dimensions mentioned in part (iv) above, is
(a) 182.45 m3 (b) 282.45 m3 (c) 292m3 (d) 192 m3

Sol. (i) Height of the dome = Radius of the dome = 21 m

i.e., h = r = 21 m

Volume of the hemispherical dome = 2 r r3
3
2 22
= 3 × 7 21× 21× 21

= 19404 cu. m

∴ Option (a) is correct.

(ii) Volume of sphere = 4 r r3
3

∴ Option (b) is correct.

(iii) The cloth require to cover the hemispherical dome

= Curved surface area of dome

= 2pr2

= 2× 22 ×14 ×14
7

= 1232 sq. m

∴ Option (b) is correct.

(iv) Total surface area of the combined dome with cuboidal shaped top

= C.S.A of dome + 2(lb + bh + lh) – l × b

= 1232 + 2(48 + 24 + 32) – 48

= 1392 sq. m

∴ Option (c) is correct.

(v) Volume of the cuboidal shaped top = l × b × h

=8×6×4
= 192 m3

∴ Option (d) is correct.

zzz

564 Xam idea Mathematics–X

Chapter-14: Statistics

1. Read the following and answer any four questions from (i) to (v).

COVID-19 Pandemic

The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of
coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus
2 (SARS-CoV-2) among humans.

The following table shows the age distribution of case admitted during a day in two different
hospitals

Table 1:

Age (in years): 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65
No. of cases: 6 11 21 23 14 5
Table 2:

Age (in years): 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65

No. of cases: 8 16 10 42 24 12

Based on the above information, answer the following questions: [CBSE Question Bank]

Refer to table 1

(i) The average age for which maximum cases occurred is

(a) 32.24 (b) 34.36 (c) 36.82 (d) 42.24

(ii) The upper limit of modal class is

(a) 15 (b) 25 (c) 35 (d) 45

(iii) The mean of the given data is

(a) 26.2 (b) 32.4 (c) 33.5 (d) 35.4

Refer to table 2

(iv) The mode of the given data is

(a) 41.4 (b) 48.2 (c) 55.3 (d) 64.6

(v) The median of the given data is

(a) 32.7 (b) 40.2 (c) 42.3 (d) 48.6

Sol. (i) We have for table 1

Age (in years) No. of Cases

5 – 15 6

15 – 25 11

25 – 35 21

35 – 45 23

45 – 55 14

55 – 65 5

Case Study Based Questions 565

Since 23 is the maximum frequency, therefore 35 – 45 is the modal class.
f1 – f0
∴ Mode = l + 2f1 – f0 – f2 ×h

= 35+ 23 – 21 14 ×10 = 35+ 20
2×23 – 21 – 11

= 35 + 1.82 = 36.82

∴ Option (c) is correct.

(ii) Modal class is 35 – 45.
∴ Upper limit is 45.
∴ Option (c) is correct.
(iii) We have,

Age (in years) No. of cases (fi) Mid value (xi) fixi
10
5 –15 6 20 60
30 220
15 – 25 11 40 630
50 920
25 – 35 21 60 700
300
35 – 45 23 cf 2830
8
45 – 55 14 24
34
55 – 65 5 76
100
Total 80 112

∴ Average age = / fi xi = 2830
/ fi 80

= 35.38 ≈ 35.4
∴ Option (d) is correct.

(iv) We have for table 2

Age (in years) No. of cases

5 – 15 8

15 – 25 16

25 – 35 10

35 – 45 42

45 – 55 24

55 – 65 12

Total 112

We have, 42 is the maximum frequency, therefore 35 – 45 is the modal class.
f1 – f0
∴ Mode = l + 2f1 – f0 – f2 ×h

= 35+ 42 – 10 24 ×10
2× 42 – 10 –

= 35 + 320
50
= 35 + 6.4 = 41.4

∴ Option (a) is correct.

566 Xam idea Mathematics–X

(v) We have, N = 112 ⇒ N = 56
2

∴ 76 is just greater than 56 in cumulative frequency.

∴ 35 – 45 is the median class.
N
∴ Median = l+ 2 – cf × h = 56
f

= 35 + 56 – 34 ×10 = 56
42

= 35 + 220 = 40.23 . 40.2
42

∴ Option (b) is correct.

2. Read the following and answer any four questions from (i) to (v).

Electricity Energy Consumption

Electricity energy consumption is the form of energy consumption that uses electric energy.
Global electricity consumption continues to increase faster than world population, leading to
an increase in the average amount of electricity consumed per person (per capita electricity
consumption).

Tariff : LT - Residential Bill Number :384756
: Single Phase Connected Load :3 KW
Type of Supply : 31-11-13 Meter Reading :65789

Meter Reading : 31-10-13 Previous Meter Reading :65500
Date
Units Consumed :289
Previous Reading
Date

A survey is conducted for 56 families of a colony A. The following tables gives the weekly
consumption of electricity of these families.

Weekly consumption (in 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
units): 16 12 18 6 4 0
No. of families:

The similar survey is conducted for 80 families of colony B and the data is recorded as below:

Weekly consumption (in 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
units): 0 5 10 20 40 5
No. of families:

Based on the above information, answer the following questions: [CBSE Question Bank]

Refer to data received from colony A (d) None of these
(d) None of these
(i) The median weekly consumption is (d) 30-40

(a) 12 units (b) 16 units (c) 20 units
(c) 26 units
(ii) The mean weekly consumption is (c) 20-30

(a) 19.64 units (b) 22.5 units

(iii) The modal class of the above data is

(a) 0-10 (b) 10-20

Case Study Based Questions 567

Refer to data received from colony B

(iv) The modal weekly consumption is

(a) 38.2 units (b) 43.6 units (c) 26 units (d) 32 units
(c) 38.75 units (d) 48 units
(v) The mean weekly consumption is

(a) 15.65 units (b) 32.8 units

Sol. (i) From colony A, we have

Weekly consumption (in units) No. of families (fi) cf
0 –10 16 16

10 – 20 12 28

20 – 30 18 46

30 – 40 6 52

40 – 50 4 56

50 – 60 0 56

We have, N = 56 ⇒ N = 28
2
∴ 20 – 30 is the median class.

∴ Median = l+ N – Cf ×h
2 f

= 20 + 28 – 28 ×10 = 20
18
∴ Option (c) is correct.

(ii) From colony A

Class Frequency (fi) Mid value (xi) fixi
0 –10 16 5 80

10 – 20 12 15 180
20 – 30 18 25 450

30 – 40 6 35 210

40 – 50 4 45 180
50 – 60 0 55 0

Total 56 1100

∴ Mean = / fi xi = 1100 =19.64
/ fi 56

∴ Option (a) is correct.

(iii) Since 18 is the maximum frequency in the given table, therefore 20 – 30 is the modal class.
∴ Option (c) is correct.

568 Xam idea Mathematics–X

(iv) From the given table for colony B, we have

Weekly Consumption (in units) No. of families

0 –10 0

10 – 20 5

20 – 30 10

30 – 40 20

40 – 50 40

50 – 60 5

Since 40 is the maximum frequency, therefore 40 – 50 is the modal class.

∴ Mode = l + f1 – f0 f2 ×h
2f1 – f0 –

= 40 + 40 – 20 5 ×10
80 – 20 –

= 40 + 200
55

= 40 + 3.63 = 43.63 ≈ 43.6

∴ Option (b) is correct.
(v) We have,

Class Frequency (fi) Mid value (xi) fixi
0 –10 0 5 0

10 – 20 5 15 75

20 – 30 10 25 250

30 – 40 20 35 700

40 – 50 40 45 1800

50 – 60 5 55 275

Total 80 3100

∴ Mean = / fi xi = 3100 = 38.75
/ fi 80

∴ Option (c) is correct.

zzz

Case Study Based Questions 569

Chapter-15: Probability

1. Read the following and answer any four questions from (i) to (v).

On a weekend Rani was playing cards with her family .The deck has 52 cards. If her brother

drew one card. [CBSE Question Bank]




Fig. 15.1
Based on the above information, answer the following questions:

(i) The probability of getting a king of red colour is

(a) 1 (b) 1 (c) 1 (d) 1
26 13 52 4

(ii) The probability of getting a face card is

(a) 1 (b) 1 (c) 2 (d) 3
26 13 13 13

(iii) The probability of getting a jack of hearts is

(a) 1 (b) 1 (c) 3 (d) 3
26 52 52 26

(iv) The probability of getting a red face card is

(a) 3 (b) 1 (c) 1 (d) 1
26 13 52 4

(v) The probability of getting a spade is

(a) 1 (b) 1 (c) 1 (d) 1
26 13 52 4

Sol. (i) Total no. of possible outcomes = 52

and no. of favourable outcomes = 2

∴ P(getting a king of red colour) = 2 = 1
52 26

∴ Option (a) is correct.

(ii) We have, number of face card = 12

∴ P(getting a face card) = 12 = 3
52 13

∴ Option (d) is correct.

(iii) There is only one jack of hearts.

∴ P(getting a jack of hearts) = 1
52

∴ Option (b) is correct.

570 Xam idea Mathematics–X

(iv) No. of red face card = 6

∴ P(getting a red face card) = 6 = 3
52 26

∴ Option (a) is correct.

(v) No. of spade cards = 13

∴ P(getting a spade card) = 13 = 1
52 4

∴ Option (d) is correct.

2. Read the following and answer any four questions from (i) to (v).

Rahul and Ravi planned to play Business ( board game) in which they were supposed to use two

dice. [CBSE Question Bank]

Fig. 15.2
Based on the above information, answer the following questions:

(i) Ravi got first chance to roll the dice. What is the probability that he got the sum of the
two numbers appearing on the top face of the dice as 8?

(a) 1 (b) 5 (c) 1 (d) 0
26 36 18

(ii) Rahul got next chance. What is the probability that he got the sum of the two numbers
appearing on the top face of the dice as 13?

(a) 1 (b) 5 (c) 1 (d) 0
36 18

(iii) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of

the two numbers appearing on the top face of the dice less than or equal to 12?

(a) 1 (b) 5 (c) 1 (d) 0
36 18

(iv) Rahul got next chance. What is the probability that he got the sum of the two numbers

appearing on the top face of the dice equal to 7?

(a) 5 (b) 5 (c) 1 (d) 0
9 36 6

(v) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of

the two numbers appearing on the top face of the dice greater than 8?

(a) 1 (b) 5 (c) 1 (d) 5
36 18 18

Sol. (i) When two dice are rolled once, we have

Total number of possible outcomes = 36

and favourable outcomes = {(2, 6), (3, 5), (4, 4), ), (5, 3), (6, 2)}

∴ No. of favourable outcomes = 5
5
∴ Required probability = 36

∴ Option (b) is correct.

Case Study Based Questions 571

(ii) Favourable outcomes for getting sum 13 = φ

∴ No. of favourable event = 0
0
∴ Required probability = 36 =0

∴ Option (d) is correct.

(iii) No. of favourable outcomes of getting sum less than or equal to 12 = 36
36
∴ Required probability = 36 =1

∴ Option (a) is correct.

(iv) Favourable outcomes for Rahul = {(3,4), (4, 3), (1, 6), (6, 1), (5, 2), (2, 5)}

No. of favourable outcomes = 6
6 1
∴ Required probability = 36 = 6

∴ Option (c) is correct.

(v) Favourable outcomes = {(3, 6), (6, 3), (4, 5), (5, 4), (5, 5), (6, 4), (4, 6), (5, 6), (6, 5), (6, 6)}

No. of favourable outcomes = 10
10 5
∴ Required probability = 36 = 18

∴ Option (d) is correct.

zzz

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572 Xam idea Mathematics–X