Skip to main content

Mathematics class 10 CBSE Board Paper 2014 solution Section A and B (code 30_2)

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)50Ö3 (B)150Ö3 (C) 50Ö2 (D)75

Ans: (B)150Ö3

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)300 (B)450 (C)600 (D) 900

Ans: (B)450

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) 5Ö2 (B)10Ö2 (C)5/Ö2 (D)10Ö3

Ans: (B)10Ö2

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

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

(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= 3pr2

462 = 3 x 22/7 x r2

r = 7cm

Volume of solid hemisphere = 2/3 x pr3 = 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 199 which are by both 2 and 5.

Solution:

a1 = 110 and d = 10 an = 190

an = a + (n-1)d

190 = 110 + (n-1)10

(190 -110 )/10 = n-1

8+1= n

n = 9

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 9x2 – 3kx + k hs equal root.

Solution:

For equal root: D = 0

b2 – 4ac = 0

(- 3k)2 – 4 x 9 x k = 0

9k2 = 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 (SET code 30/2) Download File

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

Structure of Matter class07

The language of chemistry class 7 Basic science soled questions. Q. What is chemistry? Ans: Chemistry is a branch of science in which we study the composition, structure, properties and change of matter. Q. What is matter? Ans: Anything that is around us having some mass and occupies space is called matter. Q. What are the different classifications of matter? Ans: Matters are classified into element, compound and mixture. Q. What is element? Ans: Elements are the purest form of substance (matter) made up of same kinds of particles. The smallest unit of element is an atom. eg. Hydrogen, oxygen, carbon. Q. What is compound?            See full post   The language of chemistry class 7 Basic science Class 07 Basic sciences Chapter the Language of Chemistry A. Answer these questions 1. What does the formula of a substance tell us? Answer: Formula of a compound or an element represents the name and number of atoms present ...