CBSE Maths Test Paper 2015-Class X
General Instructions:
(i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory.
(ii) Section A: Q. No.1 to 11 comprises of 11 questions of one mark each.
(iii) Section B: Q. No. 12 to 20 comprises of 9 questions of two mark each.
(iv) Section C: Q. No. 21 to 30 comprises of 10 questions of three mark each.
(v) Section D: Q. No. 31 to 35 comprises of 5 questions of four mark each.
(vi) Use of Calculators are not permitted.
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Question 1 of 35
1. Question
1 point(s)If \(x=\dfrac{-1}{2 } \) is a solution of the quadratic equation \( 3x^2+ 2kx -3 = 0, \) find the value of \(k \).
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The given quadratic equation can be written as,
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Question 2 of 35
2. Question
1 point(s)The tops of two towers of height \(x \) and \(y \), standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find \(x:y \).
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When base is same for both towers and their heights are given, i.e.,
and respectively
Let the base of towers be.
From equations (i) and (ii), -
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Question 3 of 35
3. Question
1 point(s)A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
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Total English alphabets
Number of consonants
P(letter is a consonant) -
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Question 4 of 35
4. Question
1 point(s)In Fig., PA and PB are tangents to the circle with centre O such that \(\angle \)APB = 50°. Write the measure of \(\angle \)OAB.
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… [Tangents drawn from external point are equal]
[Angles opposite equal sides]
[Quadratic rule]
… [Triangle rule]
…. [From (i)]
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Question 5 of 35
5. Question
1 point(s)In Fig,, AB is the diameter of a circle with centre O and AT tangent. If \(\angle \)AOQ = 58°, find \(\angle \)ATQ.
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… [Tangent is latex]\bot [/latex] to the radius through the point of contact]
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Question 6 of 35
6. Question
1 point(s)Solve the following quadratic equation for \(x:4x^2-4a^2x+(a^4-b^4)=0 \).
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The given quadratic equation can be written as,
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Question 7 of 35
7. Question
1 point(s)The time (in seconds) taken by 150 atheletes to own a 110 m hurdle race are tabulated below.
Class – 13.8 – 14, 14 – 14.2, 14.2 – 14.4, 14.4 – 14.6, 14.6 – 14.8, 14.8 – 15
Frequency – 2, 4, 5, 71, 48, 20
Find the number of atheletes who have completed the race in less than 14.6 seconds is :-
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-
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Question 8 of 35
8. Question
1 point(s)In the given figure, \(AD \) is the bisector of \(\angle A \). If \(BD=4\ cm,DC=3\ cm \) and \(AB=6\ cm \) then find \(AC \)
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-
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Question 9 of 35
9. Question
1 point(s)If \(P \) is prime number then what is the LCM of \(P,P^2,P^3 \)?
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LCM of
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Question 10 of 35
10. Question
1 point(s)The rational number between \( \sqrt3\) and \(\sqrt5 \) is
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-
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Question 11 of 35
11. Question
1 point(s)From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.
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In
and
…. [Tangents drawn from an external point are equal]
….[Common]
…..(TP and TQ are equally inclined to OT)
…..[CPTC]
….[SAS]
….[Linea pair]
is the right bisector of . -
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Question 12 of 35
12. Question
2 point(s)Find the middle term of the A.P. 6,13, 20, …, 216.
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The given A.P. is 6, 13, 20, …., 216
Let be the number of terms,
Middle term term term of the A.P. -
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Question 13 of 35
13. Question
2 point(s)If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with \(\angle \)B = 90°, then find the value of t.
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is a right angled triangle,
…(i)
Using distance formula,
Putting values of and in equation (i)
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Question 14 of 35
14. Question
2 point(s)Find the ratio in which the point \(P\left(\dfrac{1}{2},\dfrac{3}{2}\right) \) and \(B(2,-5) \).
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Let divide in the ratio of
Applying division formula, Required ratio -
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Question 15 of 35
15. Question
2 point(s)Find the area of the triangle ABC with A(1, -4) and mid-points of sides through A being (2, -1) and (0, -1).
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P is the mid-point of AB
Q is the mid point of AC
Area of,
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Question 16 of 35
16. Question
2 point(s)Find that non-zero value of \(k \), for which the quadratic equation \( kx^2 + 1 – 2(k – l)x +x^2 = 0\) has equal roots. Hence find the roots of the equation.
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The given quadratic equation can be written as
For equal roots,
Putting put in equation (i)
Roots are. -
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Question 17 of 35
17. Question
2 point(s)The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 30 m high, find the height of the building.
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Let height of building be
and the distance between tower and building be .
In,
Now, In
…. [From (i)]
Height of the building is m. -
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Question 18 of 35
18. Question
2 point(s)Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice.-
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Total possible outcomes
(i) The possible outcomes are when the sum of numbers on two dice is 5, i.e, 4 Required Probability,
(ii) The possible outcomes are for even numbers on both dice; 9 Required Probability, -
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Question 19 of 35
19. Question
2 point(s)If \( S_n\), denotes the sum of first \(n \) terms of an A.P., prove that \(S_{12}=3(S_8-S_4) \).
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Let
be the first term and be the common difference of A.P. …. [From (i), (ii) & (iii)] -
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Question 20 of 35
20. Question
2 point(s)In Fig., APB and AQO are semicircles, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. [Use \(\pi=\dfrac{22}{7} \)]
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… [Given]
Let
is the diameter of small semicircle
Radius of small semicircle,
is the radius of big semicircle,
is the radius of big semicircle,
Total perimeter = Perimeter of small semicircle + Perimeter of big semicircle + OB
Area of shaded region = (Area of small semicircle) + (Area of big semicircle)
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Question 21 of 35
21. Question
3 point(s)In Fig., from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use \(\pi=\dfrac{22}{7} \) and \(\sqrt5=2.236 \))
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Let
,
Height of frustum,
Lower radius of frustum
We know,
Upper radius of frustum
Slant height of frustum
Total surface area of frustum (PQCB)
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Question 22 of 35
22. Question
3 point(s)A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is \(3.5\ cm \) and the total wood used in the making of toy is \(166\left(\dfrac{5}{6}\right)\ cm^3 \). Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of \(\mathrm{Rs.\ 10\ per\ cm^2} \). [Use \(\pi=\dfrac{22}{7} \)]
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Let the height of cone = h
Radius of cone = Radius of hemisphere = r = 3.5 cm
Volume of solid wooden toy = Volume of hemisphere + Volume of cone
Height of toy
Area of hemispherical part of toy
Cost of painting -
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Question 23 of 35
23. Question
3 point(s)In Fig., from a cuboidal solid metallic block, of dimensions \(15\ cm\times10\ cm\times5\ cm \), a cylindrical hole of diameter \(7\ cm \) is drilled out. Find the surface area of the remaining block. [Use \(\pi=\dfrac{22}{7} \)]
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Let the length, breadth, height of cuboidal block be
and respectively.
Total surface area of solid cuboidal block
Radius of cylindrical hole
Area of two circular bases
Curved surface area of cylinder
Required area = (Area of cuboidal block – Area of two circular bases + Area of cylinder)
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Question 24 of 35
24. Question
3 point(s)In Fig., find the area of the shaded region. [Use \(\pi=3.14 \)]
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Let the side of big square
Let the radius of circle,
Let the side of small square
Area of square ABCD
Area of small square PQRS
Area of 4 semicircles
Required area
= (Area of big square – Area of small square – Area of 4 semicircles)
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Question 25 of 35
25. Question
3 point(s)The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \(\dfrac{29}{20} \) . Find the original fraction.
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Let the denominator and numerator of the fraction be
and respectively.
Let the fraction be
By the given condition,
New fraction,
Now denominator then, numerator The fraction is . -
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Question 26 of 35
26. Question
3 point(s)Ramkali required Rs 2500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks. What value is generated in the above situation?
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Money required for Ramkali for admission of her daughter = ₹2500
A.P. formed by saving 100, 120, 140,…. Upto 12 terms …(i)
Let, and be the first term, common difference and number of terms respectively.
She can send her daughter to school.
Value: Small savings can fulfill our big desires. -
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Question 27 of 35
27. Question
3 point(s)Solve for \(x:\dfrac{2}{x+1}+\dfrac{3}{2(x-2)}=\dfrac{23}{5x},x\ne0,-1,2 \)
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Question 28 of 35
28. Question
3 point(s)In Fig., tangents PQ and PR are drawn from an external point P to a circle with centre O, such that \(\angle \)RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find \(\angle \)RQS.
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(Tangents drawn from external point are equal)
…. [Angles opposite equal sides are equal]
In,
… Rule]
and is a transversal
… [Alternate interior angle]
… [Tangent is latex]\bot [/latex] to the radius through the point of contact]
Similarly,
In,
…. Rule]
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Question 29 of 35
29. Question
3 point(s)From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length of the flag staff is 5m, find the height of the tower.
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In
,
(i)
In,
(ii)
… [From (i)]
Height of Tower -
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Question 30 of 35
30. Question
3 point(s)A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) divisible by 2 or 3, (ii) a prime number.
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(i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13
Total outcomes = 20
Possible outcomes = 13
P(divisible by 2 or 3) =
(ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8
Total Outcomes = 20
Possible outcomes = 8
P(a prime number) -
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Question 31 of 35
31. Question
4 point(s)If A(-4, 8), B(-3, -4), C(0, -5) and D(5, 6) are the vertices of a quadrilateral ABCD, find its area.
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Join AC to form two triangles.
Given: and
Area of,
Area of,
Area of Quadrilateral ABCD
= Area ofABC + Area of ACD
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Question 32 of 35
32. Question
4 point(s)A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment.
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Let
be the height of well and be the radius of well.
Volume of earth taken out after digging the well
Let be the width of embankment formed by using (i) Total width of well including embankment
Height of embankment
Volume of well = Volume of embankment
So, Volume of embankment
Therefore, width of embankment = 10 m -
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Question 33 of 35
33. Question
4 point(s)Water is flowing at the rate of 2.52 km/hr. through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
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Let
be the internal radius of the pipe.
Radius of base of tank,
Level of water raised in the tank
If the flow rate in an hour
Then the flow rate in half an hour
So, height of water level
Volume of tank = Volume of pipe Internal diameter of pipe Internal diameter of pipe -
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Question 34 of 35
34. Question
4 point(s)To fill a swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.
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Let the bigger pipe fill the tank in
hrs. the smaller pipe fills the tanks in hrs.
Hence, the pipe with larger diameter fills the tank in 20 hours and the pipe with smaller diameter fills the tank in 30 hours. -
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Question 35 of 35
35. Question
4 point(s)Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are \(\dfrac{3}{4} \) times the corresponding sides of the isosceles triangle.
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A’BC’ is the required triangle. -
and 
respectively
.
… [Tangents drawn from external point are equal]
[Quadratic rule]
… [Triangle rule]
…. [From (i)]
… [Tangent is latex]\bot [/latex] to the radius through the point of contact]
and 
…. [Tangents drawn from an external point are equal]
….[Common]
…..(TP and TQ are equally inclined to OT)
…..[CPTC]
….[SAS]
….[Linea pair]
is the right bisector of 
.
be the number of terms,
term
term of the A.P.
is a right angled triangle,
…(i)
and 
in equation (i)
divide 
in the ratio of 
Required ratio 
,![\begin{array}{l} =\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \\ =\dfrac{1}{2}[1(2-2)+3(2+4)-1(-4-2)] \\ =\dfrac{1}{2}[1(0)+3(6)-1(-6)]=\dfrac{1}{2}[0+18+6] \\ =\dfrac{1}{2} \times 24=12 \text { sq. units } \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-173cf4a6890c029f8c263c464d375504_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \mathrm{D}=b^{2}-4 a c \\ 0=[-2(k-1)]^{2}-4 \times(k+1) \times 1 \\ 0=4(k-1)^{2}-4(k+1) \\ 0=4 k^{2}+4-8 k-4 k-4 \\ 0=4 k^{2}-12 k \quad \Rightarrow \quad 4 k(k-3)=0 \\ \Rightarrow k=3 \text { or } k=0 \quad \therefore \quad k=3 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-ecda168b5efe33e494c39fbc9264e4b8_l3.png "Rendered by QuickLaTeX.com")
put in equation (i)
.
.
…. [From (i)]
Height of the building is 
m.
when the sum of numbers on two dice is 5, i.e, 4
for even numbers on both dice; 9
be the first term and 
be the common difference of A.P.![\begin{aligned} & S_{n}=\dfrac{n}{2}(2 a+(n-1) d) \\ \therefore &\ S_{12}=\dfrac{12}{2}(2 a+(12-1) d) \\ & S_{12}=6[2 a+11 d]=12 a+66 d \\ \therefore &\ S_{8}=\dfrac{8}{2}(2 a+(8-1) d) \\ & S_{8}=4[2 a+7 d]=8 a+28 d \\ \therefore &\ S_{4}=\dfrac{4}{2}(2 a+(4-1) d) \\ & S_{4}=2[2 a+3 d]=4 a+6 d \\ & S_{12}=3\left(S_{8}-S_{4}\right) \\ & 12 a+66 d=3(8 a+28 d-4 a-6 d) \end{aligned}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-0c0a047c04ce9799df17f275be18dda6_l3.png "Rendered by QuickLaTeX.com")
…. [From (i), (ii) & (iii)]
… [Given]
is the diameter of small semicircle
is the radius of big semicircle, 
,
,
and 
respectively.
![r=\dfrac{d}{2}=\dfrac{4}{2}=2\ cm….\ [\because\ d=\dfrac{14}{2}-3=4]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-e3d82c147712800e88c66174a65dbdf8_l3.png "Rendered by QuickLaTeX.com")
![=a=4\ cm….\ [\because d=\dfrac{14}{2}-3=4]](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-0b76a07cb0e86981d09bbbc655234ca3_l3.png "Rendered by QuickLaTeX.com")
![=4\times\dfrac{1}{2}\pi r^2\\ \\ =\left[4\times\dfrac{1}{2}3.14(2)^2\right]\ cm^2=25.12\ cm^2](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-0fc7c0c3bb8358219054ae62786d6170_l3.png "Rendered by QuickLaTeX.com")
respectively.
![\begin{array}{l} \therefore \dfrac{x-3}{x}+\dfrac{x-1}{x+2}=\dfrac{29}{20} \\ \Rightarrow \dfrac{(x-3)(x+2)+x(x-1)}{x(x+2)}=\dfrac{29}{20} \\ \Rightarrow \dfrac{(x-3)(x+2)+x(x-1)}{x^{2}+2 x}=\dfrac{29}{20} \\ \Rightarrow 20[(x-3)(x+2)+x(x-1)]=29\left(x^{2}+2 x\right) \\ \Rightarrow 20\left(x^{2}-x-6+x^{2}-x\right)=29 x^{2}+58 x \\ \Rightarrow 20\left(2 x^{2}-2 x-6\right)=29 x^{2}+58 x \\ \Rightarrow 40 x^{2}-29 x^{2}-40 x-58 x=120 \\ \Rightarrow 11 x^{2}-98 x-120=0 \\ \Rightarrow 11 x^{2}-110 x+12 x-120=0 \\ \Rightarrow 11 x(x-10)+12(x-10)=0 \\ \Rightarrow(11 x+12)(x-10)=0 \\ \Rightarrow 11 x+12=0 \text { or } x-10=0 \\ \Rightarrow x=\dfrac{-12}{11} \text { (Reject) or } x=10 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-8b52a4adca40d213ad36ee46ed7565d4_l3.png "Rendered by QuickLaTeX.com")
then, numerator 
.
and ![a=100,d=20,n=12\\ \\ S_n=\dfrac{n}{2}(2a+(n-1)d)\\ \\ \Rightarrow S_{12}=\dfrac{12}{2}(2(100)+(12-1)20)\\ \\ S_{12}=\dfrac{12}{2}[2(100)+11(20)]=6[420]=₹2520](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-5eb6b13481264edbd6b11f2d263c2ea9_l3.png "Rendered by QuickLaTeX.com")
![\begin{aligned} & \dfrac{2}{x+1}+\dfrac{3}{2(x-2)}=\dfrac{23}{5 x} \\ \Rightarrow & \dfrac{4(x-2)+3(x+1)}{2(x+1)(x-2)}=\dfrac{23}{5 x} \\ \Rightarrow & 5 x[4(x-2)+3 x+3]=46(x+1)(x-2) \\ \Rightarrow & 5 x[4 x-8+3 x+3]=46\left[x^{2}-1 x-2\right] \\ \Rightarrow & 5 x(7 x-5)=46\left(x^{2}-x-2\right) \\ \Rightarrow & 35 x^{2}-25 x=46 x^{2}-46 x-92 \\ \Rightarrow & 35 x^{2}-46 x^{2}-25 x+46 x+92=0 \\ \Rightarrow & 11 x^{2}-21 x-92=0 \\ & a=11, \quad b=-21, \quad c=-92 \\ & \begin{aligned} \mathrm{D} &=b^{2}-4 a c \\ &=(-21)^{2}-4 \times 11 \times(-92) \\ &=441+4048=4,489 \end{aligned} \\ \Rightarrow & x=\dfrac{21+67}{22} \quad \text { or } \quad x=\dfrac{21-67}{22} \\ \Rightarrow & x=\dfrac{88}{22}=4 \quad \text { or } \quad x=-\dfrac{46}{22}=-\dfrac{23}{11} \\ \Rightarrow & \text { Solutions are } x=4,-\dfrac{23}{11} \end{aligned}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-40b97df6983627d93556ea38185a6645_l3.png "Rendered by QuickLaTeX.com")
(Tangents drawn from external point are equal)
…. [Angles opposite equal sides are equal]
,
… 
Rule]
and 
is a transversal
… [Alternate interior angle]
… [Tangent is latex]\bot [/latex] to the radius through the point of contact]
,
…. 
… [From (i)]
and 
![\begin{array}{l} =\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \\ =\dfrac{1}{2}[-4(-4+5)-3(-5-8)+0(8+4)] \\ =\dfrac{1}{2}[-4+39+0]=\dfrac{35}{2} \text { sq. units } \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-23d0eef2175031af96e9d35310ac80d2_l3.png "Rendered by QuickLaTeX.com")
,![\begin{array}{l} =\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \\ =\dfrac{1}{2}[-4(-5-6)+0(6-8)+5(8+5)] \\ =\dfrac{1}{2}[-4(-11)+0(-2)+5(13)] \\ =\dfrac{1}{2}[44+0+65]=\dfrac{1}{2}[109]=\dfrac{109}{2} \text { sq. units } \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-e601708bbe3c388a5ba938393ff135b5_l3.png "Rendered by QuickLaTeX.com")
be the height of well and 
![\begin{array}{l} \quad=\pi\left(R^{2}-r^{2}\right) H=176 \\ \Rightarrow \dfrac{22}{7}\left[(2+x)^{2}-(2)^{2}\right] \times \dfrac{40}{100}=176 \\ \Rightarrow \dfrac{22}{7}\left[4+4 x+x^{2}-4\right] \times \dfrac{40}{100}=176 \\ \Rightarrow \dfrac{22}{7}\left[4 x+x^{2}\right] \times \dfrac{40}{100}=176 \\ \Rightarrow \dfrac{44}{35}\left[x^{2}+4 x\right]=176=44 x^{2}+176 x=6160 \\ \Rightarrow 44\left(x^{2}+4 x-140\right)=0=x^{2}+4 x-140=0 \\ \Rightarrow x^{2}+14 x-10 x-140=0 \\ \Rightarrow x(x+14)-10(x+14)=0 \\ \Rightarrow(x+14)(x-10)=0 \\ \Rightarrow x+14=0 \quad \text { or } \quad x-10=0 \\ \Rightarrow x=-14 \text { (Reject) or } x=10 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-df47500e7c5ddc96d81f47739880b74e_l3.png "Rendered by QuickLaTeX.com")
hrs.
