Question Set F

cbse 10th Standard Maths Areas Related to Circles Important Questions And Answers Study Material - QB365

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Questions:

Generic 2

6 x 2 = 12

Find the zeroes of the polynomial x² + 4x -12

If HCF of 144 and 180 is expressed in the form 13m - 3, find the value of m.

A die is thrown once. What is the probability of getting a prime number.

E and F are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether EF || QR. PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

The area of a rectangle gets reduced by 80 sq units, if its length is reduced by 5 units and the breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, then the area is increased by 50 q units. Find the length and the breadth of the rectangle.

A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be Rs. 2640 at the rate of Rs. 12 per metre. Then the field is to be thoroughly ploughed at the cost of Rs. 0.50 per \(m^2\). What is the amount required to plough the field?

Generic 5

5 x 5 = 25

A farmer has a piece of land in the shape of an equilateral triangle. He divides his entire land among his four children in equal parts as shown in following figure. P, 0 and R are mid-points of sides AB, BC and AC respectively. A student find that ratio of area of ilPORto the area of ΔABC is 2 : 3. Using properties of similar triangles, a student of class X at once said, "It is false".
(i) Doyou agree with student?
(ii) Verify PQ \(=\frac{1}{2} B C\)

The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff.

In a painting competition of a school a child made Indian national flag whose perimeter was 50 cm. Its area will be decreased by 6 square cm, if length is decreased by 3 cm and breadth is increased by 2 cm then find the dimension of flag.What does the Saffron colour in flag signify?

10.

The diagonal of a rectangular field is 16 metre more than the shorter side. If the longer side is 14 metre more than the shorter side, then find the length of the sides of the field.

11.

From the given figure, calculate:
(i) the area of the shaded region and
(ii) the length of the boundary

Multiple Choice Question

18 x 1 = 18

12.

2 cards of heart and 4 cards of spade are missing from a pack of 52 cards. What is the probability of getting a black card from the remaining pack?

(a)

\(\frac{22}{52}\)

(b)

\(\frac{22}{46}\)

(c)

\(\frac{24}{52}\)

(d)

\(\frac{24}{46}\)

13.

If sin \(\alpha=\frac{\sqrt{3}}{2} \text { and } \cos \beta=\frac{\sqrt{3}}{2}\), then tan \(\alpha\) . tan \(\beta\) is

(a)

\(\sqrt {3}\)

(b)

\(\frac{1}{\sqrt{3}}\)

(c)

(d)

14.

The solution of the pair of equations x+y=a+b and a x-b y=a²-b² is

(a)

x=b, y=a

(b)

x=a, y=b

(c)

x=-a, y=b

(d)

x=a, y=-b

15.

The smallest irrational number by which √20 should be multiplied, so as to get a rational number, Is _____.

(a)

√20

(b)

√2

(c)

(d)

√5

16.

Diagonal AC of a rectangle ABCD is produced to the point E such that AC : CE = 2 : 1, AB = 8 cm and BC = 6 m. The length of DE is

(a)

\(2 \sqrt{19}\)cm

(b)

15 cm

(c)

\(3 \sqrt{17}\)cm

(d)

13 cm

17.

If \(0<\theta<\frac{\pi}{4}\) then the simplest form of \(\sqrt{1-2 \sin \theta \cos \theta} \text { is }\)

(a)

sin \(\theta\) - cos \(\theta\)

(b)

cos \(\theta\) - sin \(\theta\)

(c)

cos \(\theta\) + sin \(\theta\)

(d)

sin \(\theta\) cos \(\theta\)

18.

If the lines given by 3x + 2ky = 2 and 2x + 5y = 1 are parallel, then the value of k is

(a)

\(-\frac{5}{4}\)

(b)

\(\frac{2}{5}\)

(c)

\(\frac{15}{4}\)

(d)

\(\frac{3}{2}\)

19.

A circular field has a circumference of 360 km. TWo cyclists Sumeet and John start together and can cycle at speeds of 12 km!h and 15 km/h respectively, round the circular field. They will meet again at the starting point after

(a)

40 h

(b)

30 h

(c)

180 h

(d)

120 h

20.

The sum of areas of two squares is 468m2. If the difference of their perimeters is 24m, then the sides of the two squares are:

(a)

12m and 18m

(b)

18m and 24m

(c)

24m and 28

(d)

6m and 12m

21.

For the following distribution the modal class is

Marks below	10	20	30	40	50	60
Number of students	2	11	25	45	57	75

(a)

20-30

(b)

40-50

(c)

30-40

(d)

10-20

22.

For a symmetrical distribution, which is correct

(a)

Mean = Median = Mode

(b)

Mean < Mode < Median

(c)

Mean > Mode > Median

(d)

Mode = Mean + Median/2

23.

If x = 3 sec² – 1, y = tan² – 2 then x – 3y is equal to

(a)

(b)

(c)

(d)

24.

In ΔABC and ΔDEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE then, the two triangles are

(a)

congruent but not similar

(b)

similar but not congruent

(c)

neither congruent nor similar

(d)

congruent as well as similar

25.

Two congruent triangles are actually similar triangles with the ratio of corresponding sides as.

(a)

1:2

(b)

1:1

(c)

1:3

(d)

2:1

26.

Polynomial will have zeroes

(a)

-1

(b)

2 and -1

(c)

-2 and -1

(d)

-5

27.

If the degree of the dividend is 5 and the degree of the divisor is 3, then the degree of the quotient will be

(a)

(b)

(c)

(d)

-2

28.

Harry tosses two coins simultaneously. The probability of getting at least one head is

(a)

3/4

(b)

1/3

(c)

2/3

(d)

1/2

29.

Which of the following is not the graph of a quadratic polynomial?

(a)

(b)

(c)

(d)

Generic 3

7 x 3 = 21

30.

A car has two wipers which do not overlap. Each wipes has a blade of length 30 cm sweeping through an angle of 105°. Find the total area cleaned at each Sweep of the blades.

31.

Find two consecutive odd natural numbers, sum of whose squares is 130.

32.

Find the value of k for which the following system of equations has a unique solution
4x-5y=k,2x-3y=1

33.

A class teacher says to the three students Sanjeev, Anjali and Paras for making greeting cards. Each person take time 15, 20, 25 minutes respectively for making these cards.
(i) If all ofthem making card together, then after what time they will prepare a new card together?
(ii) Suppose if all of they start working at same time, in how much time they work together

34.

If zeroes of the polynomial x² + 4x + 2a are \(\alpha\) and \(\frac{2}{\alpha}\), then find the value of a.

35.

The set of data given below shows the ages of participants in a certain summer camp. Draw a cumulative frequency table for the data.

Age (in years)	10	11	12	13	14	15
Frequency	3	18	13	12	7	27

36.

If sin (A – B) = \(\frac{1}{2}\) cos (A + B) = \(\frac{1}{2}\) 0° < A + B \(\leq\) 90°, A > B, find A and B.

Assertion and reason

2 x 1 = 2

37.

Assertion When two coins are tossed together, the probability of getting no tail is \(\frac{1}{4}\).
Reason The probability P(E) of an event Esatisfies \(0 \leq P(E) \leq 1\).
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Reason is incorrect.
(d) Assertion is incorrect but Reason is correct.

38.

Assertion: The equation x² + x + 4 = 0 has equal roots.
Reason: Quadratic equation
Codes:
ax² + bx + c = 0, a \(\neq \) 0 has equal roots if b² -4ac = 0.
(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If Assertion is incorrect but Reason is correct.

Case Study Questions

3 x 4 = 12

39.

"Eight Ball" is a game played on a pool table with 15 balls numbered from 1 to 15 and a "cue ball" that is solid and white. Out of the 15 balls, eight are solid (non-white) coloured and numbered from 1 to 8 and seven are striped balls numbered from 9 to 15.

The 15 numbered pool balls (no cue ball) are placed in a large bowel and mixed, then one ball is drawn out at random.
Based on the above information, answer the following questions:
(i) What is the probability that the drawn ball bears number 8?
(ii) What is the probability that the drawn ball is a solid colourf and bears an even number?
(iii) what is the probability that the drawn ball bears an even number?
Or
What is the probability that the drawn ball bears a number, which is a multiple of 3?

40.

The COVID-19 pandemic, also known as the coronavirus pandemic, is an ongoing pandemic of coronavirus disease 2019 (COVID-19) caused by severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2). It was first identified in December 2019 in Wuhan, China.
During survey, the ages of 80 patients infected by COVID and admitted in the one of the City hospital were recorded and the collected data is represented in the less than cumulative frequency distribution table

Age(in year)	Below 15	Below 25	Below 35	Below 45	Below 55	Below 65
No. of patients	6	17	38	61	75	8

Based on the above information, answer the following questions
(a) The modal class interval is :

(i) 45-55

(ii) 35-45

(iii) 25-35

(iv) 15-25

(b) The median class interval is

(i) 45-55

(ii) 35-45

(iii) 25-35

(iv) 15-25

(i) 38.6 years

(ii) 35.8 years

(iii) 36.8 years

(iv) 38.5 years

(d) Which age group was affected the most?

(i) 35-45

(ii) 25-35

(iii) 15-25

(iv) 45-55

(e) How many patients of the age 45 years and above were admitted?

(i) 61

(ii) 19

(iii) 14

(iv) 23

41.

Piyush sells? saree at 8% profit and a sweater at 10% discount, thereby, getting a sum of Rs 1008. If he had sold the saree at 10% profit and the sweater at 8% discount, he would have got Rs 1028.

Denote the cost price of the saree and the list price (price before discount) of the sweater by Rs x and Rs y respectively
and answer the following questions.
(i) The 1^st situation can be represented algebraically as

(a) 2.08x + 1.9y = 2008

(b) 1.08x + 0.9y = 1008

(c) lOx + 8y = 1008

(d) 8x + 10y = 1008

(ii) The 2^nd situation can be represented algebraically as

(a) 10x + 8y = 1028

(b) 2.1x + 1.92y = 1028

(c) 1.1x + 0.92y = 1028

(d) 8x + 10y = 1028

(iii) Linear equation represented by 1^st situation intersect the x-axis at

\((a) (2800,0)\)

\((b) (2500,0)\)

\((c) \left(\frac{2500}{3}, 0\right)\)

\((d) \left(\frac{2800}{3}, 0\right)\)

(iv) Linear equation represented by 2^nd situation intersect the y-axis at

\((a) \left(0, \frac{25700}{23}\right)\)

\((b) (0,25700)\)

\((c) \left(0, \frac{25800}{23}\right)\)

\((d) (0,26800)\)

(v) Both linear equations represented by situation 1^st and 2^nd intersect each other at

(a) (400,600)

(b) (600,400)

(c) (200,200)

(d) (800,600)

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Answers Key:

Generic 2

6 x 2 = 12

Hint Let \(p(x)=x^2+4 x-12\)
\( \text { On putting } p(x)=0 \Rightarrow x^2+4 x-12=0\)
\( \Rightarrow x^2+6 x-2 x-12=0 \Rightarrow x(x+6)-2(x+6)=0\)
\( \Rightarrow (x+6)(x-2)=0 \Rightarrow x=-6 \text { and } x=2\)
So, the zeroes are -6 and 2 .

On applying Euclid's division algorithm,
180 = 144 x 1 +36
144 = 36 x 4 + 0
ஃ HCF of 144 and 180 is 36.
HCF = 36 = 13 x 3 - 3, which is of the form 13m -3
Hence, m = 3.

Total outcomes = 6
Prime numbers = 2, 3, 5, = 3
P(prime no.) = \(\frac { 3 }{ 6 } \) = \(\frac { 1 }{ 2 } \)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

Let x and y be length and breadth of rectangle.
Then, its area=xy
According to the questions,
9x-5)(y+2)=xy-80 \(\Rightarrow\) 2x-5y=-70
(x+10)(y-5)=xy+50 \(\Rightarrow\) -5x+10y=100
Length=40 units, breadth=30 units

Cost of fencing 1 m =Rs.12
Total cost of fencing the field = Rs.2640
∴ Total length to be fenced =\(={2640\over 12}m\)
= 220 m = circumference of the circle
⇒ Let radius of the circle be r cm
⇒ 2πr=220
⇒ \(2\times{22\over 7}r=220\)
⇒ r=7x5m
=35m

Area of the field=\(\pi r^2={22\over 7}\times35\times35m^2\)
Cost of ploughing 1m²=Rs.0.5
∴ Total cost of ploughing=Rs.0.50 x 22 x 5 x 35 =Rs.1925

Generic 5

5 x 5 = 25

(i) Yes,
(ii) 1 : 4.

Height of flagstaff = CD = h m
Height of tower = BD = x m
\(\angle DAB=45°,\angle CAB=60°\)

AB = 120 m
\(\Delta ABD\) is right angled
tan 45° = 1
\(\frac{x}{AB}\)=1
x = AB = 120m
\(\Delta ACB\) to right angled
tan 60° =\(\sqrt{3}\)
\(\frac { h+x }{ 120 } =\sqrt { 3 } \)
h + 120=120\(\sqrt{3}\)
h = 120\(\sqrt{3}\)-120
h = 120(\(\sqrt{3}\)-1)
h = 120(1.73-1)
h = 120 x 0.73
h = 87.6 m

Let length of the flag = x cm and breadth of the flag = y cm .
2x + 2y = 50
\(\Rightarrow\) x + Y = 25 ....(i)
(x - 3) (y + 2) = xy - 6
\(\Rightarrow\) xy + 2x - 3yx - 6 = xy - 6
\(\Rightarrow\) 2x - 3y = 0 .... (ii)
On solving the eqns. (i) and (ii),
x = 15 cm and y = 10 cm
\(\therefore\) Length of the flag = 15 cm and Breadth of the flag = 10 cm
Significane: The saffron colour is symbol of courage and sacrifice

10.

Let the length of shorter side be x m.
Length of diagonal = (x + 16) m
and, Length of longer side = (x + 14)m
x²+(x+14)²=(x+16)²z
x²-4x-60=0
x=10 m
Length of sides are 10 m and 24 m.

11.

(i) 77 cm²
(ii) 44 cm

Multiple Choice Question

18 x 1 = 18

12.

(b)
\(\frac{22}{46}\)

13.

14.

15.

(d)
√5

16.

17.

(b)
cos \(\theta\) - sin \(\theta\)

18.

19.

(d)
120 h

20.

(a)
12m and 18m

21.

(b)
40-50

22.

(a)
Mean = Median = Mode

23.

(b)
8

24.

(b)
similar but not congruent

25.

(b)
1:1

26.

27.

(b)
2

28.

(a)
3/4

29.

Generic 3

7 x 3 = 21

30.

Length of wiper blade = 30 cm = r
\(\theta\) = 105°
\(\therefore\) Area of cleaned by two blade
= 2 \(\times\) Area of sector formed by blade
\(\begin{aligned} & =2 \times \frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 105^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(30)^2 \end{aligned}\)
= 825 \(\times\) 2 cm²
= 1650 cm²

31.

\(\frac{16}{x}-1=\frac{15}{x+1}\)
\(\Rightarrow \quad \frac{16}{x}-\frac{15}{x+1}=1 \Rightarrow \frac{16(x+1)-15 x}{x(x+1)}=1\)
\(\Rightarrow \quad 16 x+16-15 x=x^{2}+x\)
\(\Rightarrow \quad x^{2}=16 \Rightarrow x^{2}=\pm 4\)
Hence, the roots are 4 and -4.

32.

For unique solution,\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{4}{2} \neq \frac{-5}{-3}\)
which is true for any real values of k.

33.

(i) The required number of minutes after which they start preparing a new card together
= LCM of (15, 20, 25)

5	15,20,25
3	3,4,5
4	1,4,5
5	1,1,5
	1,1,1

= 5 x 3 x 4 x 5 = 300 minutes
(ii) In all three of the students, Sanjeev take minimum 15 minutes to complete the greeting card. Hence minimum 15 minutes they work together.

34.

Given, \(\alpha\) and \(\frac{2}{\alpha}\) are the zeroes of x² + 4x + 2a.
We know that,
Product of the zeroes = \(=\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha\times\frac{2}{\alpha}=\frac{2a}{1}\)
\(\Rightarrow 2=2a\)
\(\therefore \quad a=1\)

35.

The cumulative frequency of first observation is the same as its frequency before it.
Now, the cumulative frequency table is

Age (in years)	Frequency	Cumulative frequency (c.f.)
10	3	3
11	18	3+18=21
12	13	21+13=34
13	12	34+12=46
14	7	46+7=53
15	27	53+27=80

36.

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30° (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60° (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

Assertion and reason

2 x 1 = 2

37.

(b) S = {HH, HT, TH, TT}
Favourable outcomes = {HH}
Total number of outcomes = 4
\(\therefore \text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{4}\)
Reason is true but not correct explanation of Assertion.

38.

(d) If Assertion is incorrect but Reason is correct.

Case Study Questions

3 x 4 = 12

39.

(i) Number of possible outcomes = 15
Number of favourable outcomes = 1
\(\therefore\) P (drawn balls bears number S) = \(\frac{1}{15}\)
(ii) Favourable outcomes are when balls numbered as 2, 4, 6 and 8.
\(\therefore\) P (drawn ball is a solid coloured and bears an even number) = \(\frac{4}{15}\)
(iii) Favourable outcomes are when balls numbered as 2, 4,6, 8, 10, 12, 14.
\(\therefore\) P (drawn ball bears even number) = \(\frac{7}{15}\)
Or
Favourable outcomes are when balls numbered as 3, 6, 9 and 15.
\(\therefore\) P (drawn ball bears a number which is multiple of 3)
\(=\frac{5}{15}=\frac{1}{3}\)

40.

Age(in yrs)	No. of patients	cf
5 – 15	6	6
15 – 25	11	17
25 – 35	21	38
35 – 45	23	61
45 – 55	14	75
55 – 65	5	80

(a) (ii) Since the highest frequency is 23 which belongs to 35 – 45.
Therefore, modal class is 35 – 45.
(b) (ii) Here, \(n=80 \Rightarrow \frac{n}{2}=40\)
which lies in 35 – 45
Therefore, medial class is 35 – 45.
(c) (iii) Here,l = 35, f₀ = 21, f₁= 23, f₂ = 14, h = 10.
\(\text { Mode }=l+\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \times h \quad \Rightarrow \text { Mode }=35+\frac{23-21}{46-21-14} \times 10\)
(d) (i) 35-45
(e) (ii)19

41.

(i) (b):Piyush sells a saree at 8% profit + sells a sweater at 10% discount = Rs 1008
\(\Rightarrow\) (100 + 8)% of x + (100 - 10)% of y = 1008
\(\Rightarrow\) 108% of x + 90% of y = 1008
\(\Rightarrow\) 1.08x + 0.9 y = 1008 ...(i)
(ii) (c): Piyush sold the saree at 10% profit + sold the sweater at 8% discount = Rs 1028
\(\Rightarrow\) (100 + 10)% of x + (100 - 8)% of Y = 1028
\(\Rightarrow\) 110%ofx+92%ofy= 1028
\(\Rightarrow\) 1.1 x + 0.92y = 1028 ...(ii)
(iii) (d): At x-axis, y = 0
\(\Rightarrow 1.08 x=1008 \Rightarrow x=\frac{1008}{1.08}=\frac{2800}{3}\)
(iv) (a): At y-axis, x = 0
\(\Rightarrow 0.92 y=1028 \Rightarrow y=\frac{1028}{0.92}=\frac{25700}{23}\)
(v) (b): Solving equations (i) and (ii), we get x = 600 and y = 400
Hence both linear equations intersect at (600, 400).