CBSE X Mathematics 2012 Solution (SET 1) Section D

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1 Section D Q 9. 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. Let the number of books purchased by the shopkeeper be x. Cost price of x books = Rs Original cost price of one book Rs x If the shopkeeper had purchased 4 more books, then the number of books purchased by him would be (x + 4). 80 New cost price of one book Rs x 4 Given, Original cost price of one book New cost price of one book = Rs x x 4 80 x4 80x 1 x x 4 80x x = x (x + 4) x + 4x = 0 x + 4x 0 = 0 x + 0x 16x 0 = 0 x(x + 0x) 16(x + 0) = 0 (x 16) (x + 0) = 0 x 16 = 0 or x + 0 = 0 x = 16 or x = 0 x = 16 ( Number of books cannot be negative) Thus, the number of books purchased by the shopkeeper is 16. OR The sum of two numbers is 9 and the sum of their reciprocals is 1. Find the numbers. Let the first number be x. Given: First number + Second number = 9 x + Second number = 9 Second number = 9 x

2 1 1 1 Given, First Number Second Number x 9 x 9xx 1 x 9 x 9 = 9x x x 9x + 18 = 0 x 6x x + 18 = 0 x(x 6) (x 6) = 0 (x ) (x 6) = 0 x = 0 or x 6 = 0 x = or x = 6 When x =, we have 9 x = 9 = 6 When x = 6, we have 9 x = 9 6 = Thus, the two numbers are and 6. Q 0. Sum of the first 14 terms of an AP is 1505, and its first term is 10. Find its 5 th term. Given: First term, a = 10 and sum of first 14 terms, S 14 = 1505 Let d be the common difference of the A.P. n The sum of first n terms of an AP, Sn a n 1 d 14 S d 7 0 1d d 7 0 1d 15 1d d 195 d 15 5 th term of the A.P., a an a n 1 d Thus, the 5 th term of the A.P is 70.

Q 1. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Given: A circle (O, r) and a tangent l at point A.

Proof: Among all line segments joining the centre O to any point on l, the perpendicular is the shortest to l. So, in order to prove OA l we need to prove that OA is shorter than OB.

3 Q 1. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Given: A circle (O, r) and a tangent l at point A. To prove: OA l Construction: Take any point B other than A on the tangent l. Join OB. Suppose OB meets the circle at C. Proof: Among all line segments joining the centre O to any point on l, the perpendicular is the shortest to l. So, in order to prove OA l we need to prove that OA is shorter than OB. OA = OC (Radius of same circle) Now, OB = OC + BC OB > OC OB > OA OA < OB B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l. Hence, OA l. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. OR

Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure. We know that, tangents drawn from an external point to the circle are equal in length.

4 Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure. We know that, tangents drawn from an external point to the circle are equal in length. Therefore, AP = AS BP = BQ... 1 CQ = CR DR = DS AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS) [Using (1)] = (AS + DS) + (BQ + CQ) = AD + BC Hence, AB + CD = AD + BC Q. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. Use π 7 Radius of cone = Radius of hemisphere =.5 cm Total height of solid (OC) = 9.5 cm OD + DC = 9.5 OD +.5 = 9.5 [DC (radius of hemisphere) =.5 cm] OD = 6 cm Height of cone = OD = 6 cm Volume of solid = Volume of cone + Volume of hemisphere

5 1 πr h πr πr hr (.5) cm cm cm cm cm Thus, the volume of the solid is cm. Q. A bucket is in the form of a frustum of a cone and it can hold 8.49 litres of water. If the radii of its circular ends are 8 cm and 1 cm, find the height of the bucket. Use π 7 Let the height of the bucket be h cm. Suppose r 1 and r be the radii of the circular ends of the bucket. Given, r 1 = 8 cm and r = 1 cm Capacity of bucket = 8.49 litres Volume of the bucket = cm [1 litre = 1000 cm ] 1 hr1 r r1 r cm 1 h cm 8490 cm 7 h cm h cm 181 h 15cm Thus, the height of the bucket is 15 cm.

Q 4. The angle of elevation of the top of a hill at the foot of a tower is 60 and the angle of depression from the top of the tower to the foot of the hill is 0.

6 Q 4. The angle of elevation of the top of a hill at the foot of a tower is 60 and the angle of depression from the top of the tower to the foot of the hill is 0. If the tower is 50 m high, find the height of the hill. Let AB be the hill and CD be the tower. Angle of elevation of the hill at the foot of the tower is 60, i.e., ADB = 60 and the angle of depression of the foot of hill from the top of the tower is 0, i.e., CBD = 0. Now in right angled CBD: CD tan 0 BD CD BD tan 0 50 BD 1 BD 50 m In right ABD: AB tan 60 BD AB = BD tan 60 = 50 m 50 m 150 m Hence, the height of the hill is 150 m.

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1 / 24 CBSE-XII-017 EXAMINATION CBSE-X-01 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instuctions : 1. All questions are compulsory.. The question paper consists of 34 questions