CBSE Maths Test Paper 2014-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)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].
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Question 2 of 35
2. Question
1 point(s)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) = -
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Question 3 of 35
3. Question
1 point(s)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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-
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Question 4 of 35
4. Question
1 point(s)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 ] -
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Question 5 of 35
5. Question
1 point(s)Solve – \(2x^2+ax-a^2=0 \) for \(x \)
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-
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Question 6 of 35
6. Question
1 point(s)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 -
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Question 7 of 35
7. Question
1 point(s)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 -
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Question 8 of 35
8. Question
1 point(s)Given that \(\sin\alpha=\dfrac{\sqrt3}{2} \) and \(\cos\beta=0 \) the value of \(\beta-\alpha \) is
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-
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Question 9 of 35
9. Question
1 point(s)The distance of the point P(-3, 4) from the origin is
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-
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Question 10 of 35
10. Question
1 point(s)The HCF of the smallest composite number and the smallest prime number is
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HCF of (4, 2) = 2
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Question 11 of 35
11. Question
2 point(s)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. -
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Question 12 of 35
12. Question
2 point(s)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. -
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Question 13 of 35
13. Question
2 point(s)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) = -
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Question 14 of 35
14. Question
2 point(s)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)
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Question 15 of 35
15. Question
2 point(s)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]
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Question 16 of 35
16. Question
2 point(s)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), -
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Question 17 of 35
17. Question
2 point(s)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 -
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Question 18 of 35
18. Question
2 point(s)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 -
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Question 19 of 35
19. Question
2 point(s)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]
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Question 20 of 35
20. Question
2 point(s)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) -
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Question 21 of 35
21. Question
3 point(s)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. -
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Question 22 of 35
22. Question
3 point(s)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] -
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Question 23 of 35
23. Question
3 point(s)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] -
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Question 24 of 35
24. Question
3 point(s)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), -
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Question 25 of 35
25. Question
3 point(s)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]
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Question 26 of 35
26. Question
3 point(s)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
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Question 27 of 35
27. Question
3 point(s)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]
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Question 28 of 35
28. Question
3 point(s)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
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Question 29 of 35
29. Question
3 point(s)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 -
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Question 30 of 35
30. Question
3 point(s)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,
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Question 31 of 35
31. Question
4 point(s)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) -
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Question 32 of 35
32. Question
4 point(s)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
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Question 33 of 35
33. Question
4 point(s)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. -
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Question 34 of 35
34. Question
4 point(s)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 -
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Question 35 of 35
35. Question
4 point(s)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. -
…[Tangent is latex]\bot [/latex]to the radius through the point of contact]
…[Angle sum property of a quad].
ways
, i.e., 9 ways
….[ latex]\because [/latex] Volume of cylinder 
, Volume of sphere 
]
median 
…(i)
AOB is a straight line.
…. [Tangents PQ and PR are equally inclined to PO]
…. [Tangent is latex]\bot [/latex] to the radius through the point of contact]
, i.e., 3
….[Side of square, latex]OA=20\ cm [/latex] ]
in (i),
.
draw an angle of 
.
cutting the angle at 
.
is the required 
.
bisector of 
cutting 
.
as radius, draw a cirlce.
as radius draw an arc cutting the circle at 
. Join 
.
….[Given
….[Given
,
and 
are collinear![\therefore\ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0\\ \\ -2[b-(-1)]+a(-1-1)+4(1-b)=0\\ \\ =2b-2-2a+4-4b=0\\ \\ -2a-6b=-2](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-71fa804b641067d19c0fa871643b4a46_l3.png "Rendered by QuickLaTeX.com")
…[Dividing by (-2)
….[Given
and 
. Join 
and 
.
– ar of circle
…[ar. of circle latex]=\pi r^2 [/latex]
has been cut by 
.
.
….[c.p.c.t.
![\pi r^2_1h=\dfrac{10}{3}\pi\left(\dfrac{700}{3}\right)\\ \\ \left(\dfrac{1}{24}\right)^2.h=\dfrac{7000}{9}....\bigg[r_1=\frac{1}{12}\times\frac{1}{2}=\frac{1}{24}\ cm /latex] [latex]h=\dfrac{7000}{9}\times24\times24\\ \\ \mathrm{=4,48,000\ cm\ or\ 4,480\ m}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-4925351e5d08cba52309779b18b7aa23_l3.png "Rendered by QuickLaTeX.com")
and the smaller natural number be 
….[From (i)
…. [For equal roots
…. [For equal roots
…. [When latex] k=5[/latex]
…. [When latex]k=-3 [/latex]
,![\begin{array}{l} S_{10}=210 \\ \Rightarrow \dfrac{10}{2}(2 a+9 d)=210 \quad\left[\because \mathrm{S}_{n}=\dfrac{n}{2}2 a+(n-1) d]\right. \\ 5(2 a+9 d)=210 \\ 2 a+9 d=\dfrac{210}{5}=42 \\ \Rightarrow 2 a=42-9 d \quad \Rightarrow \quad a=\dfrac{42-9 d}{2} \ldots(i) \\ \mathrm{AP}=(1+2+3+\ldots)+\underbrace{(36+37+\ldots+50)}_{\mathrm{Sum}=2565} \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-ec428edcf8c99244cf3037d1c8f48c15_l3.png "Rendered by QuickLaTeX.com")
![\begin{aligned} &3 a+126 d=513\\ &3\left(\dfrac{42-9 d}{2}\right)+126 d=513 \quad \ldots \text { [From (i) }\\ &\frac{126-27 d+252 d}{2}=513\\ &126+225 d=1026\\ &225 d=1026-126=900 \quad \therefore d=\frac{900}{225}=4 \end{aligned} /latex]Putting the value of [latex]d](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-7eda496b6c0457d2fc971c67093eb991_l3.png "Rendered by QuickLaTeX.com")
in (i)
