CBSE Maths Test Paper 2016-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 20 comprises of 20 questions of one mark each.
(iii) Section B: Q. No. 21 to 26 comprises of 6 questions of two mark each.
(iv) Section C: Q. No. 27 to 34 comprises of 8 questions of three mark each.
(v) Section D: Q. No. 35 to 40 comprises of 6 questions of four mark each.
(vi) Use of Calculators are not permitted.
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Question 1 of 40
1. Question
1 point(s)From an external point \(P \), tangents \(PA \) and \(PB \) are drawn to a circle with centre O. If \(\angle PAB=50^o \) then find \(\angle AOB \).
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…[Tangents drawn from external point are equal]
…[Angles equal to opp. sides]
In…[Angle sum property of a latex]\triangle [/latex]]
In cyclic quad.
…[Sum of opposite angles of a cyclic quadrilateral is latex]180^o [/latex]]
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Question 2 of 40
2. Question
1 point(s)In Fig. AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m. Find the length of the ladder. (Use \(\sqrt3=1.73 \))
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In rt.,
Length of the ladder,
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Question 3 of 40
3. Question
1 point(s)Find the 9th term from the end (towards the first term) of the A.P. 5, 9,13, …, 185.
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Here First term,
Common difference,
last term, term from the end term from the end -
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Question 4 of 40
4. Question
1 point(s)Cards marked with number 3,4,5,…., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
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Total no. of cards
Perfect square number cards are i.e., cards P(perfect square number) -
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Question 5 of 40
5. Question
1 point(s)If \(x=\dfrac{2}{3} \) and \(x=-3 \) are roots of the quadratic equation \(ax^2+7x+b=a \) and \(b \).
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Here and
Given,
Sum of roots
Product of roots… From (i)…(i) -
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Question 6 of 40
6. Question
1 point(s)Find the ratio in which y-axis divides the line segment joining the points A(5, -6), and B(-1, -4). Also find the coordinates of the point of division.
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Let
Coordinates of
…(i)
Point lies on -axis,
Required ratio
From (i), required point,
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Question 7 of 40
7. Question
1 point(s)How many terms of the A.P. 27, 24, 21, … should be taken so that their sum is zero?
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term,
Common difference,
But, i.e., number of terms cannot be zero. Number of terms -
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Question 8 of 40
8. Question
1 point(s)The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are 60° and 30° respectively. Find the height of the tower.
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Let the height of the tower,
In,
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Question 9 of 40
9. Question
1 point(s)A solid sphere of radius \(r \) is melted and recast into the shape of a solid cone of height \(r \), then find the radius of the cone.
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Question 10 of 40
10. Question
1 point(s)If \(x=2 \) and \(x=0 \) are the zeroes of the polynomials \(f(x)=2x^3-5x^2+ax+b \). Find the values of \(a \) and \(b \)
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Question 11 of 40
11. Question
1 point(s)The graph of a polynomial \(P(x) \) is shown here. Find the number of zeroes of \(P(x) \)
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Number of zeroes of
is 1 -
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Question 12 of 40
12. Question
1 point(s)Find the value of \(k \) for which one root of the quadratic equation \(kx^2-14x+8=0 \) is \(6 \) times the other
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…(1)…(2) -
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Question 13 of 40
13. Question
1 point(s)Find the area of \(a \) sector of circle of radius 21 cm and central angle \(90^o \) (take \(\pi=\dfrac{22}{7} \))
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Area
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Question 14 of 40
14. Question
1 point(s)Evaluate \(\sin45^o\cos60^o+\cos45^o\sin60^o \)
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Question 15 of 40
15. Question
1 point(s)Find angle A in the following figure
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In
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Question 16 of 40
16. Question
1 point(s)Find the discriminant of the quadratic equation \(6x^2-7x+2=0 \)
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-
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Question 17 of 40
17. Question
1 point(s)If the length of a tangent to a circle from a point is 0.5 cm and distance of a point from its centre is 1.5 cm. Find its radius
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Question 18 of 40
18. Question
1 point(s)A standard deck of \(\sqrt2 \) cards is shuffled Ritu draws a single card from the dick. What is the probability that the card is a Jack?
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Number of possible outcomes
Number of favourable outcomes -
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Question 19 of 40
19. Question
1 point(s)Two dice are thrown together. Find the probability that the sum of two numbers will be multiple of 4.
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Question 20 of 40
20. Question
1 point(s)The mean of \(n \) observations is \(\bar x \). If the first term is increased by 1 second by 2 and so on, find the new mean
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Question 21 of 40
21. Question
2 point(s)In Fig. a circle is inscribed in a \(\triangle \)ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.
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…[Given]
As we know,
[Tangents drawn from an external point are equal]
Let,
then,
Similarly, [Given]
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Question 22 of 40
22. Question
2 point(s)The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q (2, -5) and R(-3, 6), find the coordinates of P.
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Let the point
be
… Given
… Squaring both sides
Hence coordinates of point P are (16, 8). -
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Question 23 of 40
23. Question
2 point(s)How many terms of the A.P. 18,16,14,…. be taken so that their sum is zero?
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Here
term,
Common difference,…Given
But can not be zero.
Therefore, Number of terms,. -
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Question 24 of 40
24. Question
2 point(s)In Fig. AP and BP are tangents to a circle with centre O, such that AP = 5 cm and \(\angle \)APB = 60°. Find the length of chord AB.
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… [Tangents drawn from an external point are equal
Given: latex]\angle APB=60^o [/latex]
…(i)
[Angles opposite to equal sides]
In
Angle-sum-property of a
Hence,
All sides of an equilateral are equal -
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Question 25 of 40
25. Question
2 point(s)In Fig. ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.
Use \(\pi=\dfrac{22}{7} \)-
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Radius of circle,
Since the radius of all the semi-circles is same
Area of 4 semi-circles,
Area of Square,
Area of shaded region
(Area of square) (Area of semi-circle)
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Question 26 of 40
26. Question
2 point(s)In Fig. is a decorative block, made up of two solids—a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on .the top has a diameter of 3.5 cm. Find the total surface area of the block. Use \(\pi=\dfrac{22}{7} \)
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Total surface area of the block
= Total surface area of cube + C.S. Area of hemisphere – Area of circle
Side of cube
Radius of hemisphere
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Question 27 of 40
27. Question
3 point(s)In Fig. ABC is a triangle Coordinates of whose vertex A are (0, -1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of \(\triangle \)ABC and \(\triangle \)DEF.
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Let
and
Mid-point of AB = Coordinates of D
Mid-point of AC = Coordinate of E
Coordinate of F = Mid-point of BC
Area of,
As area of is always +ve -
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Question 28 of 40
28. Question
3 point(s)In Fig. are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semicircle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm, show that area of shaded region is \(25\left(\sqrt3-\dfrac{\pi}{6}\right)\ cm^2 \).
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…[Tangents drawn from an external point are equal]
…[sides are equal]
is an equilatral .
…Angles of an equilateral
Side
Area of the shaded region = Area of + Area of semi-circle – Area of sector
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Question 29 of 40
29. Question
3 point(s)If the sum of first 7 terms of an A.P is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
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Let
term , Common difference
Given:
Solving (i) and (ii), we get -
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Question 30 of 40
30. Question
3 point(s)Solve for \(x:\dfrac{2x}{x-3}+\dfrac{1}{2x+3}+\dfrac{3x+9}{(x-3)(2x+3)}=0,x\ne3,\dfrac{-3}{2} \)
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[Dividing both sides by 2
latex]\Rightarrow 2 x^{2}+3 x+2 x+3=0\\ \\ \Rightarrow x(2 x+3)+1(2 x+3)=0\\ \\ \Rightarrow(2 x+3)(x+1)=0\\ \\ \Rightarrow 2 x+3=0. \quad \text{or} \quad x+1=0\\ \\ \Rightarrow x=\dfrac{-3}{2} \quad \text{ or} \quad x=-1\ \\ \\ \text{But}, x \neq \dfrac{-3}{2}[/latex]… Given is the only solution. -
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Question 31 of 40
31. Question
3 point(s)A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all abound it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
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Radius of well,
Radius of embankment,
Height of the well,
Required height raised
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Question 32 of 40
32. Question
3 point(s)The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder. Use \(\pi=\dfrac{22}{7} \)
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Let the radius and height of cylinder be
and respectively.…(i)
Total surface area of cylinder
From (i),
Volume of cylinder -
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Question 33 of 40
33. Question
3 point(s)In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?
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Two dice can be thrown in
ways
(i) “a prime number on each dice” can be obtained as i.e., ways.
P(a prime no. on each dice)
(ii) “a total of 9 or 11” can be obtained as P(a total of 9 or 11) -
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Question 34 of 40
34. Question
3 point(s)A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for treatment. Due to this, the plane got delayed by half an hour. To reach the destination 1500 km away in time, so that the passengers could catch the connecting flight, the speed of the plane was increased by 250 km/hour than the usual speed. Find the usual speed of the plane. What value is depicted in this question?
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Let the usual speed of the plane
km/hr
Then the increased speed of the plane km/hr
Distance km
According to the question, Speed of plane km/hr.
Value: By giving utmost importance to the health and well-being of the passenger the pilot displays the value of compassion. By making up for the time lost due to this delay he displays concern for the passengers’ time and money and shows dedication towards his airline and job. -
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Question 35 of 40
35. Question
4 point(s)Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length.
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We have,
and
and are the required tangents.
By measurement -
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Question 36 of 40
36. Question
4 point(s)In Fig. O is the centre of a circle of radius 5 cm. T is a point such that OT 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
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… Tangents is to the radius through the point of contact
We have,
In rt.,
… Pythagoras’ theorem
… radius of the circle
Let,, then
… Tangents drawn from an external point
In rt.,
….Pythagoras’ theorem
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Question 37 of 40
37. Question
4 point(s)Find \(x \) in terms of \(a,b \) and \( c\): \(\dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2c}{x-c},x\ne a,b,c \)
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-
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Question 38 of 40
38. Question
4 point(s)A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take \(\sqrt3=1.732 \))
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Let
be the tree
In rt.
In rt.
… From (i)
Hence, speed of bird
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Question 39 of 40
39. Question
4 point(s)A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/ minute every succeeding minute. After how many minutes the policeman will catch the thief.
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Let total time be
minutes
Total distance covered by thief in minutes
= Speed Time
= metres
Total distance covered by policeman Thief runs = mins
Policeman runs = mins
Here, [Dividing both sides by 10
latex]\Rightarrow n^{2}-6 n+3 n-18=0\\ \\ \Rightarrow n(n-6)+3(n-6)=0\\ \\ \Rightarrow(n+3)(n-6)=0\\ \\ \Rightarrow n+3=0 \quad\ \text{or}\ \quad n-6=0\\ \\ \Rightarrow n=-3 \quad\ \text{or}\ \quad n=6 [/latex]
But (time) can not be negative. Time taken by policeman to catch the thief minutes. -
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Question 40 of 40
40. Question
4 point(s)Prove that the area of a triangle with vertices \((f, t. – 2),\left(\dfrac{f+2}{f+2}\right) \) and \((f + 3, f) \) is independent of \(f \).
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Let
Area of
Area of is always positive. Area of sq. units, which is independent of . -
…[Tangents drawn from external point are equal]
…[Angles equal to opp. sides]
…[Angle sum property of a latex]\triangle [/latex]]
…[Sum of opposite angles of a cyclic quadrilateral is latex]180^o [/latex]]
term from the end 
term from the end 
i.e., 
cards
P(perfect square number) 
and 
… From (i)
…(i)
…(i)
lies on 
-axis, 
term, 
![\begin{aligned} \Rightarrow \frac{n}{2}[2 a+(n-1) d]=0 \\ \Rightarrow \frac{n}{2}[2(27)+(n-1)(-3)]=0 \\ \Rightarrow n(54-3 n+3)=0 \\ \Rightarrow n(57-3 n)=0 \\ \Rightarrow n=0 \text { or } \quad (57-3 n)=0 \\ -3 n=-57 \\ \Rightarrow n=19 \end{aligned}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-9b07d7dcc51147f2077daa77392141b9_l3.png "Rendered by QuickLaTeX.com")
, i.e., number of terms cannot be zero.
,
is 1
…(1)
…(2)
![\dfrac{(x_1+1)+(x_2+2)+…(x_1+n)}{n}\\ \\ =\left(\dfrac{x_1+x_2+…+x_n}{n}\right) +\left(\dfrac{1+2+…+n}{n}\right)\\ \\ =\bar x+\dfrac{n}{2}\left[\dfrac{2\times1+(n-1)\times1}{n}\right]\\ \\ =\bar x+\dfrac{1}{2}(2+n-1)\\ \\ =\bar x+\dfrac{n+1}{2}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-30ca2afb770e87c260c27b076178f98b_l3.png "Rendered by QuickLaTeX.com")
…[Given]
[Tangents drawn from an external point are equal]
,
[Given]
be 
… Given
… Squaring both sides
term, 
…Given![\begin{array}{l} \Rightarrow \dfrac{n}{2}(2 a+(n-1) d)=0 \\ \Rightarrow \dfrac{n}{2}[36+(n-1)(-2)]=0 \\ \Rightarrow 2 \dfrac{n}{2}(18-n+1)=0 \\ \Rightarrow n(19-n)=0 \\ \Rightarrow n=0 \quad \text { or } \quad 19-n=0 \\ n=19 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-cf6b7df70b65f896b05deb03feff343c_l3.png "Rendered by QuickLaTeX.com")
can not be zero.
.
…(i)
![=4\left[\dfrac{1}{2}\pi r^2\right]\\ \\ =4\left[\dfrac{1}{2}\times\dfrac{22}{7}\times7\times7\right]=308\ cm^2](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-f294f11cc62fe6a22a9043fd5e4eae90_l3.png "Rendered by QuickLaTeX.com")
(Area of square) 
(Area of semi-circle)
and 
Coordinate of F = Mid-point of BC
![\begin{aligned} &=\frac{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]\\ &=\frac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]\\ &=\frac{1}{2}(8)=4 \text { sq. units }\\ &\mathrm{D}(1,0), \mathrm{E}(0,1), \mathrm{F}(1,2)\\ &\text { Area of } \Delta \mathrm{DEF} \text { , }\\ &=\frac{1}{2}[1(1-2)+0(2-0)+1(0-1)]\\ &=\frac{1}{2}(-1-1)=-1 \end{aligned} \\ \\ =1\ \text{sq. unit}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-373c90db31c874d3b5c407d7ac2ba15b_l3.png "Rendered by QuickLaTeX.com")
…[Tangents drawn from an external point are equal]
…[sides are equal]
is an equilatral 
…Angles of an equilateral 
+ Area of semi-circle 
– Area of sector 
, Common difference 
![\begin{array}{l} S_{n}=\dfrac{n}{2}[2 a+(n-1) d] \\ S_{7}=\dfrac{7}{2}(2 a+6 d) \\ 49 \times \dfrac{2}{7}=2 a+6 d \\ 2 a+6 d=14\quad ...(i) \\ S_{17}=\dfrac{17}{2}(2 a+16 d) \\ 289 \times \dfrac{2}{17}=2 a+16 d \\ 2 a+16 d=34\quad...(ii) \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-9a2dc95b234bb439370f0a0a52b7e04b_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{r} 2 a+6 d=14 \\ -2 a \pm 16 d=\pm 34 \\ -10 d=-20 \\ \hline \quad d=2 \\ \hline \end{array}\\ \\ \text { From }(i), 2 a+6(2)=14\\ \\ 2 a+12=14 \\ 2 a=14-12=2 \Rightarrow a=\dfrac{2}{2}=1 \\ S_{n}=\dfrac{n}{2}[2 a+(n-1) d] \\ =\dfrac{n}{2}[2(1)+(n-1) 2]=\dfrac{n}{2}(2+2 n-2) \\ =\dfrac{2 n^{2}}{2}=n^{2} \text { (Hence proved) }](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-71f560f22a800197c6fdbd5d6dd2e06f_l3.png "Rendered by QuickLaTeX.com")
[Dividing both sides by 2
is the only solution.
and 
respectively.
…(i)
ways
i.e., 
ways.
km/hr
km/hr
km![\dfrac{1500}{x}-\dfrac{1500}{(x+250)}=\dfrac{30}{60}\ ...\left[\begin{matrix}\because\ \text{Time}=\dfrac{\text{Distance}}{\text{speed}} \\ 30\ \text{mins}=\dfrac{30}{60}=\dfrac{1}{2}\ \text{hour}\end{matrix}\right.\\ \\ \begin{aligned} &\Rightarrow \dfrac{1,500(x+250-x)}{x(x+250)}=\frac{1}{2}\\ &\Rightarrow x(x+250)=3,000 \times 250\\ &\Rightarrow x^{2}+250 x-7,50,000=0\\ &\Rightarrow x^{2}+1,000 x-750 x-7,50,000=0\\ &\Rightarrow x(x+1,000)-750(x+1,000)=0\\ &\Rightarrow(x+1,000)(x-750)=0\\ &\Rightarrow x+1,000=0\quad\text { 2r } x-750=0\\ &\Rightarrow x=-1,000 \text { (reject) or } x=750 \end{aligned} /latex]But speed of plane can not be negative[latex]\therefore](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-f60b0d19d5d12a7dd7c3d8f19f31e7f9_l3.png "Rendered by QuickLaTeX.com")
Speed of plane 
km/hr.
and 
and 
are the required tangents.
… Tangents is 
to the radius through the point of contact
,
… Pythagoras’ theorem
… radius of the circle
, then 
… Tangents drawn from an external point
,
….Pythagoras’ theorem
![\dfrac{a}{x-a}+\dfrac{b}{x-b}=\dfrac{2 c}{x-c} \\ \Rightarrow \dfrac{a(x-b)+b(x-a)}{(x-a)(x-b)}=\dfrac{2 c}{x-c} \\ \Rightarrow (x-c)[a x-a b+b x-a b]=2 c(x-a)(x-b) \\ \Rightarrow (x-c)(a x+b x-2 a b)=2 c\left(x^{2}-b x-a x+a b\right) \\ \Rightarrow a x^{2}+b x^{2}-2 a b x-a c x-b c x+2 a b c \\ =2cx^2-2bcx-2cax+2abc \\ \Rightarrow a x^{2}+b x^{2}-2 a b x-a c x-b c x-2 c x^{2}+2 b c x+2cax=0 \\ \\ \Rightarrow a x^{2}+b x^{2}-2 c x^{2}-2 a b x+b c x+{cax}=0 \\ \Rightarrow x^{2}(a+b-2 c)+x(-2 a b+b c+c a)=0 \\ \Rightarrow x[x(a+b-2 c)+(-2 a b+b c+c a)]=0 \\ \Rightarrow x=0 \text { or } x(a+b-2 c)+(-2 a b+b c+c a)=0 \\ \Rightarrow x=0 \text { or } x(a+b-2 c)=2 a b-b c-c a=0 \\ \Rightarrow x=\frac{2 a b}{a+b} 2 c](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-8c3061c4e5f21a2f4d74bde9d58c927a_l3.png "Rendered by QuickLaTeX.com")
be the tree
… From (i)![\Rightarrow x=80\sqrt3-80\\ \\ \Rightarrow x=80(\sqrt3-1)\\ \\ \Rightarrow x=80(1.732-1)...[\sqrt3=1.732]\\ \\ \Rightarrow x=80(0.732)\\ \\ \therefore\ CE,\ x=58.56\ m](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-c05279f60110d68055158cae38995a00_l3.png "Rendered by QuickLaTeX.com")
Time
metres
Thief runs = 
mins
![\therefore 100 n=\dfrac{(n-1)}{2}[2(100)+(n-1-1)(10)]\\ \\ \Rightarrow(n-1)[200+10 n-20]=200 n\\ \\ \Rightarrow(n-1)[10 n+180]=200 n\\ \\ \Rightarrow 10 n^{2}+180 n-10 n-180-200 n=0\\ \\ \Rightarrow 10 n^{2}-30 n-180=0\\ \\ \Rightarrow n^{2}-3 n-18=0 \quad...](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-1567dc22da8b429aee73f65218fe68cf_l3.png "Rendered by QuickLaTeX.com")
[Dividing both sides by 10
(time) can not be negative.
Time taken by policeman to catch the thief 
minutes.
![= \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}[t(t+2-t)+(t+2)(t-(t-2)) +(t+3)((t-2)-(t+2))] \\ = \frac{1}{2}[t(2)+(t+2)(2)+(t+3)(-4)] \\ = \frac{1}{2}[2 t+2 t+4-4 t-12]=\frac{1}{2}[-8]=-4](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-864c36811d07e82e41d5ad4b3e90dabf_l3.png "Rendered by QuickLaTeX.com")
is always positive.
sq. units, which is independent of 
.