Mathematics class 10_cbse_board_2014_solution_section_a_and_bcode_30_2
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This document contains a mathematics exam for Class 10 CBSE Board from 2014. It has two sections: Section A contains 7 multiple choice questions about topics like probability, geometry, and sequences. Section B contains 7 multi-part math problems to solve involving geometry topics like tangents of circles, properties of isosceles triangles, probability, surface area, natural numbers, and solving a quadratic equation for equal roots. The document provides the questions along with explanations and solutions for the problems in Section B.
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Ans: (A) 7/8
SECTION-B
9.In fig-01,common tangent AB and CD to the circles with centers O1 and O2 intersect at E. Prove that AB=CD
Solution:
AE= CE [tangent from E] and CE = ED [tangent from E]
Adding them we get AE+ CE = CE + ED
AB = CD
10. The incircle of an isosceles triangle ABC , in which AB=AC, touches side BC , CA and AB at D,E and F respectively.
Prove that BD = CD
Solution:
AB = AC
BF + AF = AE + CE --------(i)
BF = BD and CE = CD [tangent from B and C] -----------------(ii)
using (i)and (ii)
BD + AE = AE + CD
BD = CD
11. Two different dice are tossed together. Find the probability](https://image.slidesharecdn.com/mathematicsclass10cbseboard2014solutionsectionaandbcode302-150623111543-lva1-app6891/85/Mathematics-class-10_cbse_board_2014_solution_section_a_and_bcode_30_2-2-320.jpg)
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(i) that the number on each side is even
(ii)The sum of the number appear on two dice is 5
Solution:
(i) Total favorable outcomes are 22,42,62,24,44,64,26,46,66 = 9
Number of all possible outcomes = 36
P[E] = 9/36= ¼
(ii) Total favorable outcomes are 41,32,23,14 = 4
Number of all possible outcomes = 36
P[E] = 4/36= 1/9
12. The total surface area of a solid hemisphere is 462cm2 , find the volume?
Ans: Total surface area of solid hemisphere= 3r
2
462 = 3 x 22/7 x r2
r = 7cm
Volume of solid hemisphere = 2/3 x r3
= 2/3 x ( 22/7) x 7 x 7 x 7 = 2156/3 = 718.66cm3
13. Find the number of natural number between 101 and 999 which are by both 2 and 5.
Solution:
a1 = 110 and d = 10 an = 990
an = a + (n-1)d
990 = 110 + (n-1)10
(990 -110 )/10 = n-1
88+1= n
n = 99
Hence, there are 9 natural number between 101 and 199 which are by both 2 and 5.
14. Find the value of k for which the quadratic equation 9x
2
– 3kx + k hs equal root.
Solution:
For equal root: D = 0
b
2
– 4ac = 0 (- 3k)
2
– 4 x 9 x k = 0 9k
2
= 36k
k = 4 Hence, Value of k for which the quadratic equation 9x2
– 3kx + k ha equal root = 4
Maths class 10 CBSE Board Paper 2014 solution Section A and B (code 30_2) Download File](https://image.slidesharecdn.com/mathematicsclass10cbseboard2014solutionsectionaandbcode302-150623111543-lva1-app6891/85/Mathematics-class-10_cbse_board_2014_solution_section_a_and_bcode_30_2-3-320.jpg)
Mathematics class 10_cbse_board_2014_solution_section_a_and_bcode_30_2
- 1. www.jsuniltutorial.weebly.com/ Page 1 Series HRS Code-30/2 Summative Assessment –II Subject Mathematics class 10 CBSE Board 2014 SECTION-A 1. The probability that a number selected at random from the number 1,2,3,…..15. is a multiple of 4 is (A)4/15 (B) 2/15 (C)1/5 (D)1/3 Ans: 1/5 2. The angle of depression of car parked on the road from the top of a 150m high tower is 300 . The distance of the car from the tower in m meter is (A)503 (B)1503 (C) 502 (D)75 Ans: (B)1503 3. Two circle touches externally at P. AB is common tangent to the circle touching them at A and B . The value of <APB is (A)30 0 (B)45 0 (C)60 0 (D) 90 0 Ans: (D) 90 0 4. If k,2k -1 and 2k + 1 are three consecutive term of AP then value of k is (A)2 (B)3 (C)-3 (D)5 Ans: (B)3 5. A chord of a circle of radius 10 cm subtends a right angle at its centre. The length of chord is (A) 52 (B)102 (C)5/2 (D)103 Ans: (B)102 6. ABCD is a rectangle whose three vertices are B(4,0), C(4,3) and D(0,3) . The length of one of its diagonal is (A) 5 (B)4 (C)3 (D)25 Ans: (D)25 7. In a right triangle ABC , right angled at B ,BC = 12cm and AB = 5cm . The radius of circle inscribe in the triangle (in cm) is (A) 4 (B)3 (C)2 (D) 1 Ans: (C)2 8. In a family of 3 children, the probability of having at least one boy is (A) 7/8 (B)1/8 (C) 5/8 (D) ¾
- 2. www.jsuniltutorial.weebly.com/ Page 2 Ans: (A) 7/8 SECTION-B 9.In fig-01,common tangent AB and CD to the circles with centers O1 and O2 intersect at E. Prove that AB=CD Solution: AE= CE [tangent from E] and CE = ED [tangent from E] Adding them we get AE+ CE = CE + ED AB = CD 10. The incircle of an isosceles triangle ABC , in which AB=AC, touches side BC , CA and AB at D,E and F respectively. Prove that BD = CD Solution: AB = AC BF + AF = AE + CE --------(i) BF = BD and CE = CD [tangent from B and C] -----------------(ii) using (i)and (ii) BD + AE = AE + CD BD = CD 11. Two different dice are tossed together. Find the probability
- 3. www.jsuniltutorial.weebly.com/ Page 3 (i) that the number on each side is even (ii)The sum of the number appear on two dice is 5 Solution: (i) Total favorable outcomes are 22,42,62,24,44,64,26,46,66 = 9 Number of all possible outcomes = 36 P[E] = 9/36= ¼ (ii) Total favorable outcomes are 41,32,23,14 = 4 Number of all possible outcomes = 36 P[E] = 4/36= 1/9 12. The total surface area of a solid hemisphere is 462cm2 , find the volume? Ans: Total surface area of solid hemisphere= 3r 2 462 = 3 x 22/7 x r2 r = 7cm Volume of solid hemisphere = 2/3 x r3 = 2/3 x ( 22/7) x 7 x 7 x 7 = 2156/3 = 718.66cm3 13. Find the number of natural number between 101 and 999 which are by both 2 and 5. Solution: a1 = 110 and d = 10 an = 990 an = a + (n-1)d 990 = 110 + (n-1)10 (990 -110 )/10 = n-1 88+1= n n = 99 Hence, there are 9 natural number between 101 and 199 which are by both 2 and 5. 14. Find the value of k for which the quadratic equation 9x 2 – 3kx + k hs equal root. Solution: For equal root: D = 0 b 2 – 4ac = 0 (- 3k) 2 – 4 x 9 x k = 0 9k 2 = 36k k = 4 Hence, Value of k for which the quadratic equation 9x2 – 3kx + k ha equal root = 4 Maths class 10 CBSE Board Paper 2014 solution Section A and B (code 30_2) Download File