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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In Fig., PQ and PR are two tangents to a circle with centre O. If \(\angle \)QPR = 46°, then calculate \(\angle \)QOR.
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…[Tangent is latex]\bot [/latex]to the radius through the point of contact]
…[Angle sum property of a quad].
If two different dice are rolled together, calculate the probability of getting an even number on both dice.
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Two dice can be thrown as ways
Even numbers on both the dice can be obtained as, i.e., 9 ways
P(even numbers) =
If the points \(A(x, 2), B(-3, -4)\) and \(C(7, -5)\) are collinear, then find the value of \(x\).
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Find the number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm.
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Number of solid spheres….[ latex]\because [/latex] Volume of cylinder
, Volume of sphere
]
Solve – \(2x^2+ax-a^2=0 \) for \(x \)
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Find the median using an emperical relation when it is given that mode and mean are 8 and 9 respectively
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median
3 median = 26
Median = 8.67
What is the range of first ten prime numbers?
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2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Range = 29 – 2 = 27
Given that \(\sin\alpha=\dfrac{\sqrt3}{2} \) and \(\cos\beta=0 \) the value of \(\beta-\alpha \) is
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The distance of the point P(-3, 4) from the origin is
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The HCF of the smallest composite number and the smallest prime number is
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HCF of (4, 2) = 2
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
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Given: CD and EF are two parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is a diameter of the circle.
Const: Join OA and OB. Draw OM || CD.
Proof: …(i)
[Tangents is latex]\bot [/latex] to the radius through the point of contact]
Similarly
AOB is a straight line.
Hence AOB is a diameter of the circle with O.
AB passes through centre O.
If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that \(\angle \)QPR = 120°, prove that 2PQ = PO.
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…. [Tangents PQ and PR are equally inclined to PO]
…. [Tangent is latex]\bot [/latex] to the radius through the point of contact]
In rt.
Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
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, i.e., 3
P(at least one tail) =
In Fig., a square OABC is inscribed in a quadrant OPBQ of a cirlce. If OA = 20 cm, find the area of the shaded region. (Use \(\pi=3.14 \))
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Diagonal of the square (OB) = Side
….[Side of square, latex]OA=20\ cm [/latex] ]
ar(Shaded region) = ar(Quad. Sector – ar(Square)
Solve the equation \(\dfrac{4}{x}-3=\dfrac{5}{2x+3};x\ne0,-\dfrac{3}{2} \), for \(x \).
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\(\)\begin{array}{l} \dfrac{4}{x}-\dfrac{3}{1}=\dfrac{5}{2 x+3} \\ \dfrac{4-3 x}{x}=\dfrac{5}{(2 x+3)} \\ 5 x=(2 x+3)(4-3 x) \\ 5 x=8 x-6 x^{2}+12-9 x \\ 5 x-8 x+6 x^{2}-12+9 x=0 \\ 6 x^{2}+6 x-12=0 \\ x^{2}+x-2=0\quad….\text{[Dividing by 6} \\ x^{2}+2 x-x-2=0 \\ x(x+2)-1(x+2)=0 \\ x-1=0 \quad \text { or } \quad x+2=0 \\ \therefore \quad x=1 \quad \text { or } \quad x=-2 \end{array} /latex]
If the seventh term of an AP is \(\dfrac{1}{2} \) and its ninth term is \(\dfrac{1}{7} \) find its 63rd term.
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On subtracting,
Putting the value of in (i),
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. [Take latex]\pi=\dfrac{22}{7} [/latex]]
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Given:
T.S. area of the remaining solid
= C.S. ar. of cyl. + area of base + C.S. ar. of cone
The first and last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
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Here
We know that
Solve the equation \(\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{2}{3x-1};x\ne-1,x\ne\dfrac{1}{3} \), for \(x \).
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\(\)\begin{array}{l} \dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{2}{3 x-1} \\ \dfrac{6-(x+1)}{2(x+1)}=\dfrac{2}{3 x-1}=\dfrac{6-x-1}{2 x+2}=\dfrac{2}{3 x-1} \\ 2(2 x+2)=(5-x)(3 x-1) \\ 4 x+4=15 x-5-3 x^{2}+x \\ 4 x+4-15 x+5+3 x^{2}-x=0 \\ 3 x^{2}-12 x+9=0 \\ x^{2}-4 x+3=0\ ….[\text{Dividing by 3} \\ x^{2}-3 x-x+3=0 \\ x(x-3)-1(x-3)=0 \\ (x-1)(x-3)=0 \\ x-1=0 \quad \text { or } \quad x-3=0 \\ \therefore \quad x=1 \quad \text { or } \quad x=3 \end{array} /latex]
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a:
(i) face card
(ii) red card
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Total cards = 52
Block face cards = 6
Remaining card = 52 – 6 = 46
(i) P(face card)
(ii) P(red card)
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and \(\angle \)B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
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Steps of Construction:
Draw .
From draw an angle of
.
Draw an arc cutting the angle at
.
Join is the required
.
Draw bisector of
cutting
at
.
Take as centre and
as radius, draw a cirlce.
Take as centre and
as radius draw an arc cutting the circle at
. Join
.
and
are the required tangents.
Justification:
….[Given
Since, OB is a radius of the circle.
latex]\therefore [/latex] AB is a tangent to the circle.
Also AE is a tangent to the circle.
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB.
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….[Given
latex]\therefore\ AB^2=AC^2 [/latex] ….[Squaring both sides
latex]\Rightarrow (3-0)^2+(p-2)^2\\ \\ =(p-0)^2+(5-2)^2\\ \\ 9+(p-2)^2=p^2+9\\ \\ p^2-4p+4-p^2=0\\ \\ -4p+4=0\\ \\ -4p=-4\\ \\ \Rightarrow p=1\\ \\ AB=\sqrt{(3-0)^2+(p-2)^2}\\ \\ =\sqrt{9+(1-2)^2}=\sqrt{9+1}=\sqrt{10}\ units [/latex]
Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. [Use latex]\sqrt3=1.73 [/latex]]
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Let AB be the light house and C and D be the two ships
In rt
In rt ,
The distance between the two ships,
\(\)CD=BC+BD\\ \\ =200+\dfrac{200\sqrt3}{3}….\text{[From (i) and (ii)}\\ \\ =200\left(1+\dfrac{\sqrt3}{3}\right)\\ \\ =200\left(\dfrac{3+1.73}{3}\right)\\ \\ =200\left(\dfrac{4.73}{3}\right)=\dfrac{946}{3}=315.\bar3\ m /latex]
If the points A(-2,1), B(a, b) and C(4, -1) are collinear and a – b = 1, find the values of a and b.
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and
are collinear
…[Dividing by (-2)
latex]a=1-3b [/latex] …(i) ….[Given
latex](1-3b)-b=1 [/latex] ….[From (i)
latex]1-3b-b=1\\ \\ -4b=1-1=0\quad\therefore b=\dfrac{0}{-4}=0 [/latex]
From (i),
In Fig., a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. [Use \(\pi=3.14 \) and \(\sqrt3=1.73 \) ]
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ar of an eq.
Construction: Draw and
. Join
and
.
Proof:
Radius,
Area of the shaded region
= ar of – ar of circle
…[ar. of circle latex]=\pi r^2 [/latex]
In Fig., PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [Use \(\pi=3.14 \) ]
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Perimeter of the shaded region
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled completely?
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\(\)\mathrm{Required\ time=\dfrac{Vol.\ of\ cylindrical\ tank}{Vol.\ of\ cylindrical\ pipe}}\\ \\ =\dfrac{\pi(5)^2\times2}{\pi\left(\dfrac{10}{100}\right)^2\times4000}….\ [\mathrm{\because\ Vol.\ of\ cyl.\ =\pi r^2h\ 10\ cm=\dfrac{10}{100}m\ ,\ 4\ km=4000\ m}\\ \\ =\dfrac{5\times5\times2}{\dfrac{1}{100}\times4000}\\ \\ =\dfrac{5\times5\times2}{40}=\dfrac{5}{4}\ \text{hours}=\dfrac{5}{4}\times60\ mins\\ \\ =75\ mins.\ or\ 1.\ hr.\ \&\ 15\ mins. /latex]
A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\dfrac{1}{2} \) cm, find the length of the wire.
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A solid cone has been cut by
.
.
….[c.p.c.t.
In rt. latex]\triangle ADE,\tan30^o=\dfrac{DE}{AD} [/latex]
In rt.
In rt.
Vol. of frastum of cone
Vol. of cyl. wire = Vol. of frustum of cone
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\dfrac{1}{10} \) Find the numbers.
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Let the larger natural number be and the smaller natural number be
A.T.Q.
Putting the value of in (i), we get
Putting these values of in (ii),
Numbers are -5, -10 or 10, 5
The angles of elevation and depression of the top and the bottom of a tower from the top of a building, 60 m high, are 30° and 60° respectively. Find the difference between the heights of the building and the tower and the distance between them.
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Let AB be the building and CE be the tower. In rt.
In rt.
Difference between the heights of the building and the tower DE = 20 m
Distance between them,
A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number
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Total number of cards = 49
(i) Odd numbers are 1, 3, 5,….,49, i.e., 25
P(an odd number) =
(ii) ‘A multiple of 5’ numbers are 5, 10, 15,…..,45, i.e., 9
P(a multiple of 5) =
(iii) “A perfect square” numbers are 1, 4, 9, …..,49, i.e., 7
P(a perfect square number)
(iv) “An even prime number” is 2, i.e., only one number
P(an even prime number)
Find the ratio in which the point \(P(x, 2)\) divides the line segment joining the points \(A(12, 5)\) and \(B(4, -3)\). Also find the value of \(x\).
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Let
Coordinates of P = Coordinates of P
….[From (i)
latex]\dfrac{\dfrac{12+60}{5}}{\dfrac{3+5}{5}}=x\\ \\ 2K+3K=5-2\\ \\ x=\dfrac{72}{8}=9\\ \\ 5K=3\\ \\ K=\dfrac{3}{5}….(i) [/latex]
Required ratio
Find the values of \(k \) for which the quadratic equation \((k+4)x^2+(k+1)x+1=0 \) has equal roots. Also find these roots.
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Here …. [For equal roots
latex] \begin{array}{l}\text { As } b^{2}-4 a c=0 \\ (k+1)^{2}-4(k+4) \cdot 1=0 \\ k^{2}+2 k+1-4 k-16=0 \\ k^{2}-2 k-15=0 \\ k^{2}-5 k+3 k-15=0 \\ k(k-5)+3(k-5)=0 \\ (k-5)(k+3)=0 \\ k=5 \text { or } k=-3 \end{array} [/latex]
Roots are …. [For equal roots
latex]=\dfrac{-(k+1)}{2(k+4)} [/latex]…. [When latex] k=5[/latex]
Roots are
Roots are …. [When latex]k=-3 [/latex]
Roots are 1 and 1.
In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.
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Here ,
Sum of its last 15 terms = 2565 ….[Given
latex]S_{50}-S_{35}=2565\\ \\ \dfrac{50}{2}(2a+49d)-\dfrac{35}{2}(2a+34d)=2565\\ \\ 100a+2450d-70a-1190d=2565\times2\\ \\ 30a+1260d=5130 [/latex]
Dividing both sides by 10 in (i)
A.P. is
Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which \(\dfrac{2}{5} \)th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?
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Number of spherical balls
Value – Sushant’s love for plants, nature and enviroment has been shown.