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Here you get the CBSE Class 10 Mathematics SA-II Board Exam 2017 (Delhi): Marking Scheme that released by the board itself. This marking scheme has key answers (or values point) along with mark wise break up of every answer which gives better idea about the division of marks inside a question.

**CBSE Class 10 Mathematics Question Paper 2017:All India**

**Importance of CBSE Class 10 Mathematics Board Examination Marking Scheme:**

Every year CBSE Board releases the marking scheme which was used while checking the answer booklets of CBSE Board Examinations. CBSE does this with a purpose of helping its students get an idea to frame a perfect answer that contains every important key point.

The word limit while answering questions of CBSE board examination is the most crucial factor. With the help of key answers in marking scheme, students will easily understand the criteria of word limit of answers to be written in CBSE Board Exams.

**CBSE Class 10 Mathematics Marking Scheme 2017: All India**

**A few questions and their answers from CBSE Class 10 Mathematics SA-II Board Exam 2017 (Delhi): Marking Scheme, are given below:**

**Q. **Find how many integers between 200 and 500 are divisible by 8. (2)

**Sol.**

A.P. formed is 208, 216, 224,...., 496 (1)

a_{n} = 496

⟹ 208 + (n – 1) × 8 = 496 (½)

⟹ n = 37 (½)

**Q.** Find the value of k for which the equation x^{2 }+ k(2x + k* *− 1) + 2 = 0 has real and equal roots. (2)

**Sol.**

x^{2 }+ k(2x + k – 1) + 2 = 0

⟹ x^{2} + 2kx + (k^{2} – k + 2) = 0 (½)

For equal roots, b^{2} – 4ac = 0 (1)

⟹ 4k^{2} – 4k^{2} + 4k − 8 = 0 (1)

⟹ k = 2 (½)

**Q. **If the equation (1 + m^{2})x^{2} + 2mcx + c^{2} – a^{2} = 0 has equal roots then show that c^{2} = a^{2}(1 + m^{2}). (3)

**Sol.**

(1 + m^{2})x^{2} + 2mcx + c^{2} – a^{2} = 0

For equal roots, B^{2 }– 4AC = 0 (½)

⟹ 4m^{2}c^{2} – 4(1 + m^{2})(c^{2} – a^{2}) = 0 (1)

⟹ m^{2}c^{2} – c^{2} – m^{2}c^{2} + a^{2} + m^{2}a^{2} = 0 (1)

⟹ c^{2 }= a^{2}(1 + m^{2}) (½)

**CBSE Class 10 Mathematics Syllabus 2017-2018**

**Q.** Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7

(ii) have a product less than 16

(iii) is a doublet of odd numbers.

**Sol.**

Total number of outcomes = 36

(i) Favourable outcomes are

(1, 1,) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3)

(2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., 15

∴ P(sum less than 7) = 15/36 or 5/12 (1)

(ii) Favourable outcomes are

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3)

(2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (4, 1)

(4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25

∴ P(product less than 16) = 25/36 (1)

(iii) Favourable outcomes are

(1,1) (3,3) (5,5)

∴ P(doublet of odd number) = 3/36 = 1/12 (1)

**Click over the following links to get the complete CBSE Class 10 Mathematics (Delhi) question paper and its marking scheme: **

**CBSE Class 10 Mathematics Question Paper 2017: Delhi Region**

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