Here you get the CBSE Class 10 Science SA-II Board Exam 2017 Marking Scheme. This marking scheme released by the CBSE itself, 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.

**Importance of CBSE Class 10 Science 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.

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

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.

**A few sample questions and their key points from the CBSE Class 10 Science Board Exam 2017: Marking Scheme, are given below:**

**Q. **What is the common difference of an A.P.in which a_{21} - a_{7}= 84 ?

**Sol.**

a_{21 }– a_{7} = 84

⟹ (a + 20d) – (a + 6d) = 84 (1/2)

⟹ 14d = 84 (1/2)

⟹ d = 6

**Q. **The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the number of rotten apples in the heap?

**Sol.**

Let the number of rotten apples in the heap be n.

⟹ n / 900 = 0.18 (1/2)

⟹ n = 162 (1/2)

**Q.** A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of Q, then find the coordinatesof P and Q.

**Sol.**

Let the coordinates of points P and Q be (0, b) and (a, 0) respectively. (1/2)

∴ a/2 = 2 ⟹ a = 4 (1/2)

And, b/2 = –5 ⟹ b = –10 (1/2)

∴ P(0, –10) and Q(4, 0) (1/2)

**Q.** If the distances of P(x, y) from A(5, 1) andB(-1, 5) are equal, then prove that 3x = 2y.

**Sol.**

PA^{2 }= PB^{2}

⟹ (x – 5)^{2} + (y – 1)^{2} = (x + 1)^{2} + (y – 5)^{2} (1)

⟹ 12x = 8y

⟹ 3x = 2y (1)

**Q.** If ad ¹ bc,then prove that the equation (a^{2} + b^{2}) x^{2} + 2(ac + bd) x + (c^{2 }+ d^{2}) = 0 has no real roots.

**Sol.**

D = 4(ac + bd) – 4(a^{2} + b^{2}) (c^{2} + d^{2}) (1)

= −4(a^{2}d^{2} + b^{2}c^{2} – 2abcd)

= –4(ad – bc)^{2 }(1)

Since ad ≠ bc,

Therefore D < 0 (1/2)

Hence the quation has no real roots. (1/2)

**Q.** The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum

**Sol.**

Here *l* = 4 cm, 2*π*r_{1} = 18 cm and 2*π*r_{2} = 6 cm

⟹ *π*r_{1} = 9, *π*r_{2 }= 3 (1)

Curved surface area of frustum = *π*(r_{1} + r_{2}) × *l *or (*π*r_{1} + *π*r_{2}) × *l *(1)

= (9 + 3) × 4 (1/2)

= 48 cm^{2 }(1/2)

**Q.** Solve for x:

**Sol.**

Here [(5x + 1) + (x + 1)3](x + 4) = 5(x + 1) (5x + 1) (1)

⟹ (8x + 4)(x + 4) = 5(5x^{2} + 6x +1)

⟹ 17x^{2} − 6x – 11 = 0 (1)

⟹ (17x + 11)(x – 1) = 0 (1)

⟹ x = –11/17, x =1 (1)

**Download CBSE Class 10 Mathematics Question Paper SA – II, 2017**

**Download CBSE Class 10 Mathematics Marking Scheme SA – II, 2017**