CBSE Maths Test Paper 2012-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 \(1 \) is a root of the equations \(ay^2 + ay + 3 = 0 \) and \(y^2 + y + b = 0 \), then find the value of \(ab \).
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Question 2 of 35
2. Question
1 point(s)In Fig., the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm).
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Question 3 of 35
3. Question
1 point(s)In Fig., a circle touches the side DF of \(\triangle \)EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, calculate the perimeter of \( \triangle\)EDF (in cm).
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Perimeter
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Question 4 of 35
4. Question
1 point(s)If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the larger circle (in cm).
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\(\)\pi R^2=\pi r_1^2+\pi r^2_2\\ \\ \left[\begin{matrix} r_1=\dfrac{10}{2}=5\ cm \\ r_2=\dfrac{24}{2}=12\ cm \end{matrix}\right.\\ \\ \pi R^2=\pi(r^2_1+r^2_2)\\ \\ R^2=5^2+12^2=25+144=169\\ \\ R=13\ cm\\ \\ \therefore\ \text{Diameter}=2(13)=26\ cm /latex]
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Question 5 of 35
5. Question
1 point(s)Find the sum of the first 25 terms of an A.P. whose nth term is given by \(t_n=2-3n \).
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\(\)\begin{aligned} &\begin{array}{l} \text { When } n=1, \quad t_{1}=2-3(1)=-1\quad …(i) \\ \text { When } n=25, \quad t_{25}=2-3(25)=-73\quad …(ii) \\ \text { As } S_{n}=\dfrac{n}{2}\left[t_{1}+t_{n}\right], \quad n=25 \\ \therefore S_{25}=\dfrac{25}{2}[-1+(-73)] \quad \ldots \text { [From (i) and (ii) } \end{array}\\ &=\dfrac{25 \times(-74)}{2}=25 \times(-37)=-925 \end{aligned} /latex]
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Question 6 of 35
6. Question
1 point(s)In a simultaneous toss of two coins, find the probability of getting:
(i) exactly one head, (ii) atmost one head.-
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The sample space is given by
Total events
(i) exactly one head
P(exactly one head)
(ii) almost one head
P(almost one head) -
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Question 7 of 35
7. Question
1 point(s)Find the value(s) of \(k \) so that the quadratic equation \(x^2-4kx+k=0 \) has equal roots.
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\(\)x^2-4kx+k=0\\ \\ \text{Here}\ a=1,\qquad b=-4k,\qquad c=k\\ \\ D=0\ ….\text{[Since, equal roots}\\ \\ \text{As}\ b^2-4ac=0\\ \\ \therefore\ (-4k)^2-4(1)\ (k)=0\\ \\ 16k^2-4k=0\\ \\ 4k(4k-1)=0\\ \\ 4k=0\qquad \text{or}\qquad 4k-1=0\\ \\ k=0\qquad\text{or}\qquad 4k=1\\ \\ \therefore k=\dfrac{1}{4} /latex]
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Question 8 of 35
8. Question
1 point(s)A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting the queen of diamonds.
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Queen of diamonds = 1
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Question 9 of 35
9. Question
1 point(s)Sum of first 20 terms of ar AP is -240 and its first term is 7. Find its 24th term
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Question 10 of 35
10. Question
1 point(s)A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 7 cm and height of the cone is equal to its diameter. Find the volume of the solid
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Volume of solid = Volume of hemisphere + Volume of cone
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Question 11 of 35
11. Question
2 point(s)Find the sum of all three digit natural numbers, which are multiples of 11.
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Question 12 of 35
12. Question
2 point(s)Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. If AP = 15 cm, then find the length of BP.
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Join
…. [Tangent is latex]\bot [/latex] to the radius through the point of contact
In rt.,
….[Pythagoras’ theorem
latex]=8^2+15^2\\ \\ =64+225=289\\ \\ OP=\sqrt{289}=17\ cm [/latex]
In rt. ….(i)
[Pythagoras theorem]
…[From (i)
latex]=2\sqrt{66}\ cm [/latex] -
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Question 13 of 35
13. Question
2 point(s)Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.
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Question 14 of 35
14. Question
2 point(s)Find the value(s) of \(k \) so that the quadratic equation \(2x^2+kx+3=0 \) has equal roots.
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Given: equation
Here…[Since, roots are equal
As latex]b^2-4ac=0\qquad\therefore\qquad k^2-4(2)\ (3)=0 [/latex] -
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Question 15 of 35
15. Question
2 point(s)Find the sum of all three digit natural numbers, which are multiples of 9.
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Question 16 of 35
16. Question
2 point(s)A box contains 35 blue, 25 white and 40 red marbles. If a marble is drawn at random from the box, find the probability that the drawn marble is
(i) white
(ii) not blue
(iii) neither white nor blue.-
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No. of blue marbles
No. of white marbles
No. of red marbles
Total no. of marbles
(i) P(white)
(ii) P(not a blue)
(iii) P(neither white nor blue) = P(red) -
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Question 17 of 35
17. Question
2 point(s)The 15th term of an AP is 3 more than twice its 7th term. If the 10th term of the AP is 41, then find its nth term.
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Question 18 of 35
18. Question
2 point(s)Find the sum of all three digit natural numbers, which are multiples of 7.
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Question 19 of 35
19. Question
2 point(s)Find the value(s) of \(k \) so that the quadratic equation \(3x^2- 2kx + 12 = 0\) has equal roots.
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Given equation
Here ….[Since roots are equal
As latex]b^2-4ac=0\quad\therefore\quad (-2k)^2-4(3)\ (12)=0 [/latex] -
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Question 20 of 35
20. Question
2 point(s)A kite is flying at a height of 45 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string assuming that there is no slack in the string.
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Let
be the height of the kite from the ground
In rt.
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Question 21 of 35
21. Question
3 point(s)A milkman was serving his customers using two types of mugs A and B of inner diameter 5 cm to serve the customers. The height of the mugs is 10 cm.
He decided to serve the customers in ‘B’ type of mug.
(a) Find the volume of the mugs of both types.
(b) Which mathematical concept is used in the above problem?
(c) By chasing the mug of type ‘B’, which value is being depicted Mug ‘A’ bv the milkman?-
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(a) Volume of mug of type ‘A’
Volume of mug of type
(b) Volume of solid figures (Mensuration)
(c) Honesty -
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Question 22 of 35
22. Question
3 point(s)In Fig., OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. [Use \( \pi=\dfrac{22}{7}] \)
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Area of shaded region
= Area of Square – Area of Quadrant
\(\)\mathrm{=(Side)^2-\dfrac{1}{4}\pi r^2\\ \\ =(7)^2-\dfrac{1}{4}\times\dfrac{22}{7}\times7\times7…[side=7\ cm,r=7\ cm\\ \\ =49-\dfrac{77}{2}=49-38.5=10.5\ cm^2} /latex] -
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Question 23 of 35
23. Question
3 point(s)If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p,
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\(\) \begin{array}{l} \mathrm{AB}=\mathrm{AC}\qquad…\text{[Given} \\ \mathrm{AB}^{2}=\mathrm{AC}^{2}\qquad…\text{Squaring} \\ (3-0)^{2}+(p-2)^{2}=(p-0)^{2}+(5-2)^{2} \\ 9+(p-2)^{2}=p^{2}+9 \\ (p-2)^{2}=p^{2} \\ (p-2)^{2}-p^{2}=0 \\ (p-2+p)(p-2-p)=0 \\ (2 p-2) \cdot(-2)=0 \\ 2 p-2=0 \\ 2 p=2 \quad \therefore \quad p=1 \end{array}/latex]
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Question 24 of 35
24. Question
3 point(s)Number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
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Multiples of 3 and 4 are 12, 24, 36, 48, i.e., 4 numbers
P(a multiple of 3 and 4) -
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Question 25 of 35
25. Question
3 point(s)Solve for \(x:4x^2-4ax+(a^2-b^2)=0 \)
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*** QuickLaTeX cannot compile formula: 4x^2-4ax+a^2-b^2=0\\ \\ [(2x)^2-2(2x)(a)+a^2]-b^2=0\\ \\ (2x-a)^2-b^2=0\\ \\ (2x-a+b)\ (2x-a-b)=0\\ \\ 2x=a-b\ \text{or}\ 2x=a+b\\ \\ x=\dfrac{a-b}{2}\ \text{or}\ x=\dfrac{a+b}{2} *** Error message: Missing number, treated as zero. leading text: ...2-4ax+a^2-b^2=0\\ \\ [(2x)^2-2(2x)(a)+a^2] Illegal unit of measure (pt inserted). leading text: ...2-4ax+a^2-b^2=0\\ \\ [(2x)^2-2(2x)(a)+a^2] Missing $ inserted. leading text: ...2-4ax+a^2-b^2=0\\ \\ [(2x)^2-2(2x)(a)+a^2] Extra }, or forgotten $. leading text: ...2-4ax+a^2-b^2=0\\ \\ [(2x)^2-2(2x)(a)+a^2] You can't use `\end' in internal vertical mode. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing $ inserted. leading text: \end{document} Extra \endgroup. leading text: \end{document} Emergency stop. -
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Question 26 of 35
26. Question
3 point(s)Solve for \(x:3x^2-2\sqrt6x+2=0 \)
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Question 27 of 35
27. Question
3 point(s)Prove that the parallelogram circumscribing a circle is a rhombus.
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Given.
is a .
To prove. is a rhombus.
Proof. In, opposite sides are equal
and ….(i)
….[Tangents drawn from an external point are equal in length
By adding these tangents,
latex](AP+PB)+(CR+DR)=AS+BQ+CQ+DS\\ \\ AB+CD=(AS+DS)+(BQ+CQ)\\ \\ AB+CD=AD+BC\\ \\ AB+AB=BC+BC [/latex]… [From (i)]
…(ii)
From (i) and (ii),
is a rhombus. -
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Question 28 of 35
28. Question
3 point(s)Construct a right triangle in which the sides, (other than the hypotenuse) are of length \(6\ cm \) and \(8\ cm \). Then construct another triangle, whose sides are \(\dfrac{3}{5} \) times the corresponding sides of the given triangle.
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Here
and ratio of corresponding sides
is the required triangle. -
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Question 29 of 35
29. Question
3 point(s)In Fig., PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm and centre O. If \(\angle \)POQ = 30° then find the area of the shaded region. [Use \(\pi=\dfrac{22}{7} \) ]
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Area of sector with radius
Area of sector with radius
Area of the shaded region
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Question 30 of 35
30. Question
3 point(s)From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. [Use latex]\pi=\dfrac{22}{7} [/latex] ]
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T.S. area of the remaining solid = C.S. area of cylinder + area of base + C.S. area of cone
\(\) \begin{array}{l} =2 \pi r h+\pi r^{2}+\pi r l \\ =\pi r(2 h+r+l) \\ =\dfrac{22}{7} \times 6[2(7)+6+\sqrt{85}] \\ =\dfrac{132}{7}(20+9.22) \quad \ldots[\because \sqrt{85}=9.219=9.22 \\ =\dfrac{132}{7} \times 29.22 \\ =\dfrac{3857.04}{7}=551.01 \mathrm{~cm}^{2} \end{array} /latex] -
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Question 31 of 35
31. Question
4 point(s)The angles of depression* of two ships from the top of a light house and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, find the height of the light house.
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Let
be the light house,
and are two ships and
Let
In rt.
In rt.
\(\)\begin{aligned} &\begin{aligned} \dfrac{1}{\sqrt{3}} &=\dfrac{h}{x+200} & \\ \sqrt{3} h &=x+200 \\ \sqrt{3} h &=h+200 & \ldots[\text { From } \end{aligned}\\ &\sqrt{3} h-h=200\\ &\begin{array}{l} (\sqrt{3}-1) h=200 \\ h=\dfrac{200}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}=\dfrac{200(1.73+1)}{3-1} \\ \quad=100(2.73) \quad \ldots\because \sqrt{3}=1.73 \\ h=273 \mathrm{~m} \end{array}\\ &\therefore \text { Height of the light house }=273 \mathrm{~m} \end{aligned} /latex] -
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Question 32 of 35
32. Question
4 point(s)If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p.
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Area of
sq units
Taking +ve sign,
Taking -ve sign, -
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Question 33 of 35
33. Question
4 point(s)A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card.-
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No. of red cards
No. of yellow cards
No. of blue cards
Total no. of cards
(i) P(a blue card)
(ii) P(not a yellow card)
(iii) P(neither yellow nor a blue card) -
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Question 34 of 35
34. Question
4 point(s)The 17th term of an AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, then find its nth term.
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\(\)\begin{array}{ll} a_{17}=5+2\left(a_{8}\right) & a_{11}=43 \\ a+16 d=5+2(a+7 d) & a+10 d=43 \\ a+16 d=5+2 a+14 d & 2 d-5+10 d=43 \\ 16 d-14 d-5=2 a-a & 12 d=48 \\ 2 d-5=a & \ldots(i) \quad \therefore d=4 \ldots \text { [From }(i) \\ \text { From }(i), a=2(4)-5=8-5=3 \\ \text { As } a_{n}=a+(n-1) d \\ \therefore a_{n}=3+(n-1) 4=3+4 n-4 \\ a_{n}=(4 n-1) \end{array} /latex]
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Question 35 of 35
35. Question
4 point(s)A shopkeeper buys some books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Find the number of books he bought.
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Let the number of books
Increased number of books
A/Q
Number of books -
![S_n=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \Rightarrow-240=\dfrac{20}{2}[2\times7+19d]\\ \\ \Rightarrow \dfrac{-240}{10}=14+19d\\ \\ \Rightarrow d=-2\\ \\ a_n=a+(n-1)d\\ \\ \Rightarrow a_{24}=a+23d=7+23\times(-2)\\ \\ =7-46=-39](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-8de957ea703b3304ac0ccf83b14de757_l3.png "Rendered by QuickLaTeX.com")
…. [Tangent is latex]\bot [/latex] to the radius through the point of contact
,
….[Pythagoras’ theorem
….(i)
…[From (i)![\begin{array}{l} a=10, \quad \mathrm{~S}_{14}=1505 \\ \mathrm{~S}_{14}=\dfrac{14}{2}[2(10)+(13) d]=1505 \\ \ldots\left[\because \mathrm{S}_{n}=\dfrac{n}{2}2 a+(n-1) d]\right. \\ 20+13 d=\dfrac{1505}{7}=215 \\ 13 d=215-20=195 \Rightarrow d=\dfrac{195}{13}=15 \\ \text { As } a_{n}=a+(n-1) d \\ \therefore \quad a_{25}=10+(25-1) 15=10+360=370 \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-b2a855e1c7ac1407a180c7301f452957_l3.png "Rendered by QuickLaTeX.com")
…[Since, roots are equal
![a_{15}=3+2\ (a_7)\qquad a_{10}=41\\ \\ a+14d=3+2(a+6d)\qquad a+9d=41\\ \\ a+14d=3+2a+12d\qquad 2d-3+9d=41\\ \\ 14d-12d=3+2a+12d\ ….\text{…[From (i)}\\ \\ 2d-3=a\ ….(i)\qquad 11d=44\\ \\ \therefore d=4 /latex]From (i)[latex]a=2(4)-3\\ \\ =8-3=5\\ \\ \mathrm{n^{th}\ term=a+(n-1)d\\ \\ \therefore\ n^{th}\ term=5+(n-1)4\\ \\ =5+4n-4\\ \\ =(4n+1)}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-4128898091ead27f86ec6aaa3ae2ef2f_l3.png "Rendered by QuickLaTeX.com")
….[Since roots are equal
be the height of the kite from the ground
is a 
.
and 
….(i)
….[Tangents drawn from an external point are equal in length
…(ii)
is a rhombus.
and ratio 
of corresponding sides
is the required triangle.
be the light house,
and 
are two ships and 
sq units![\begin{array}{l} \dfrac{1}{2}[1(p-7)+4(7-(-3))+(-9)(-3-p)]=\pm 15 \\ p-7+40+27+9 p=\pm 30 \\ 10 p+60=\pm 30 \\ 10 p=-60 \pm 30 \\ p=\dfrac{-60 \pm 30}{10} \end{array}](https://mathsgenii.com/wp-content/ql-cache/quicklatex.com-b588d20cc23ed3491eb6749511861606_l3.png "Rendered by QuickLaTeX.com")
