cbse.jagranjosh.com CBSE Class 10 th Mathematics Solved Guess Paper

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2 Solved Guess Paper 6 Mathematics Class X SA II SET I Time Allowed: hours Maximum Marks: 9 General Instructions:. All questions are compulsory..the question paper consists of questions divided into four sections A, B, C and D.. Section A contains 4 questions of mark each, section B contains 6 questions of marks each, Section C contains questions of marks each and Section D contains questions of 4 marks each. 4.Use of calculator is not permitted.. Find k if SECTION A k x k x has equal roots. k x k x Given, On comparing with a x + b x + c = we have, a k b k c For equal roots: b 4ac k k k k kk 4 8 k,

Hence the value of k is or.. In figure PA and PB are tangents to the circle with centre O such that APB = 7 o. Find OAB. As PA and PB are tangents from external point P, so they are equal.

3 Hence the value of k is or.. In figure PA and PB are tangents to the circle with centre O such that APB = 7 o. Find OAB. As PA and PB are tangents from external point P, so they are equal. Also PAB = PBA (angle opposite to equal sides of a triangle are equal) Let PAB = PBA = x Radius drawn from centre to the tangent at the point of contact is 9. Therefore OAP = 9 In Δ APB APB PAB PBA 8 5 x x8 x x 65 therefore OAB 9 65 OAB 5.. Two unbiased dice are thrown.find the probability that the total of the numbers on dice is greater than. The total greater than that can happen on two dice is, which can be obtained as,

4 (5,6),(6,5),(6,6) Total cases = 6 = 6 Therefore P total greater than 6 P total greater than 4. If the points A (4, ) and B (x, 5) are on the circle with centre O (, ), then find the value of x. Given, A (4, ) and B (x, 5) are the points on the circle. Centre of circle is O (, ). Now, OA = OB (radius of circle) By distance formula, OB = OA x 5 4 squaring both sides x x x x. Therefore, value of x is. SECTION B 5. Solve for x : x a b x a ab b Consider,

5 9x 9 a b x a 5ab b ( ) a b x a b x a b a b 9 x 9 a b x a 4 ab ab b 9 x 9 a b x a a b b a b 9 x a b x a b a b 9 x a b a b x a b a b 9x x x a b a b x a + b x a b x a b x a b or x a b x a b or x a b This implies x or x a b a b 6. If the 8 th term of AP is and the 5 th term is 6 more than th term.find A.P. Given, The 8 th term of A.P is and the 5 th term is 6 more than th term. If a is the first term and d is the common difference then, this implies T 8 a 7d... () and

6 T =T +6 5 a 4d a d 6 4d 6 d 4 Fromequation...() a7d a 47 a 8 a Therefore the A.P is a, a + d, a + d, a + d, 7,, 5 7. The perimeter of a sector of a circle of radius 5. cm is 6.4 cm. Calculate the area of sector. Let POQ be the given sector Given, radius (r) of circle = 5. cm and perimeter of sector POQ = 6.4 cm So, length of OP + length of OQ + length of arc PQ = 6.4 cm Arc PQ = 6.4 cm Length of arc PQ = 6 cm That is, l = Arc length PQ = 6 cm Therefore,

7 Area of sector POQ r l cm. 8. Find four numbers in AP whose sum is and the sum of whose squares is. Let the term of the series be a d, a d, a + d, a + d (here a is the first term and d be the common difference). According to question, addition of 4 numbers is. So, [a d] + [a d] + [a + d] + [a + d] = 4a = a = /4 a = 5.. () Now, sum of squares of the numbers is a d a d a d a d a 9 d 6 ad a d ad a d ad a 9 d 6 ad 4 a d 4 5 d {Using...()} 4 5 d d d d d d d Answer will remain same either we take + or

8 Now the numbers are (a d), (a d), (a + d), (a + d) (5 ), (5 ), (5 + ), (5 + ), 4, 6, The diameter of a metallic sphere is 6 cm. The sphere is melted and redrawn into a wire of uniform cross-section. If the length of the wire is 6 m, then find the radius. The diameter of a metallic sphere is 6 cm, so its radius (r) will be cm. The sphere is melted and redrawn into a wire of uniform cross-section. The length of the wire is 6 m. Volume of sphere is: 4 r 4 If h is the height of the wire then the volume of the wire is, r r r h 6 6 cm The two volumes must be the same, so we get 4 6 6r r 6 r r r.cm The radius of the wire is. cm.

9 . Find the number of coins.5 cm in diameter and. cm in thickness to be melted to form a right circular cylinder of height cm and diameter 4.5 cm. Given, the number of coins.5 cm in diameter and. cm in thickness to be melted to form a right circular cylinder of height cm and diameter 4.5 cm. Each one of those coins is a cylinder. The volume of each coin (radius)²(thickness) Volume of large cylinder.5 ² 45 cm³. 8 The number of coins is Volume of large cylinder Volumeof coin 45 = coins..75 ². 9 8 cm³. SECTION C. Find the roots of the equation 9x 5x 6 by the method of completing the square. We have 9x 5x 6

10 To make complete square, add and subtract (5/) in the L.H.S of equation, x x x x x 4 5 x 5 x 6 4 x or x or x or OR 4. Solve for x: ;,, 4 x x x 4 x 4 x x x 4 x x 4 x x x 4 x 4 x x x 4 4 x x x x 4 4x 8x 4x 8 x 8x x x x 4 Comparing with a x + b x + c =

11 CBSE Class th SA-II Mathematics Solved Guess Paper 6 Set-I 4% OFF Publisher : Jagran Josh Author : Jagran Josh Expert Type the URL : Get this ebook

CBSE CLASS X MATH -SOLUTION Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or equal to 1.

CBSE CLASS X MATH -SOLUTION Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or equal to 1. CBSE CLASS X MATH -SOLUTION 011 Q1 The probability of an event is always greater than or equal to zero and less than or equal to one. Here, 3 5 = 0.6 5% = 5 100 = 0.5 Therefore, 0.6, 0.5 and 0.3 are greater