CBSE Class 10 Mathematics Solved Question Paper 2017 is available here. Students can download the complete paper from the link given at the end of this article.

Though the structure of this paper is wholly different from that released by the board for Borad Exams 2018, but the importance of questions asked in previous years’ exams cannot be ignored as the syllabus for Class 10 Mathematics remains unaltered. So, studenst must practice the previous years’ papers to get a command over the important topics and prepare accordingly for the CBSE Board Exams 2018.

**CBSE Class 10 New Exam Pattern 2017-2018**

**CBSE Class 10 Mathematics Solved Question Paper 2017, contains:**

- 31 question divided into four sections - A, B, C and D
- Simple and apt solution for each question
- Every solution framed on the basis of the CBSE Marking Scheme
- Neat and clean figures where required

Students need to practice CBSE Class 10 Mathematics Solved Question Paper 2017, to check their preparedness and build confidence for the final exam.

**A few sample questions from the CBSE Class 10 Mathematics Solved Question Paper 2017, are given below:**

**Q. **If *ad* ≠ *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.**

The given equation is (*a*^{2} + *b*^{2})*x*^{2} + 2 (*ac* + *bd*)*x* + (*c*^{2} + *d*^{2}) = 0

We know, *D *= *b*^{2}−4*ac*

Thus,

*D *= [2(*ac*+*bd*)]^{2 }− 4(*a*^{2}+*b*^{2})(*c*^{2}+*d*^{2})]

= [4(*a*^{2}*c*^{2}+*b*^{2}*d*^{2}+ 2*abcd*)] − 4(*a*^{2}+*b*^{2})(*c*^{2}+*d*^{2})

= 4[(*a*^{2}*c*^{2}+*b*^{2}*d*^{2}+ 2*abcd*) − (*a*^{2}*c*^{2}+*a*^{2}*d*^{2}+*b*^{2}*c*^{2}+*b*^{2}*d*^{2})]

= 4[*a*^{2}*c*^{2}+*b*^{2}*d*^{2 }+ 2*abcd*−*a*^{2}*c*^{2}−*a*^{2}*d*^{2}−*b*^{2}*c*^{2}−*b*^{2}*d*^{2}]

= 4[2*abcd *− *b*^{2}*c*^{2 }− *a*^{2}*d*^{2}]

= −4[*a*^{2}*d*^{2}+*b*^{2}*c*^{2}−2*abcd*]

= −4[*ad*−*bc*]^{2}

But we know that *ad* ≠ *bc*

Therefore, (*ad*−*bc*) ≠0

⟹ (*ad*−*bc*)^{2} > 0

⟹ −4(*ad*−*bc*)^{2} < 0

⟹ *D *< 0

Hence, the given equation has no real roots.

**Q. **The dimensions of a solid iron cuboid are 4·4 m × 2·6 m × 1·0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.

**Sol.**

Given, dimensions of a solid iron cuboid = 4·4 m × 2·6 m × 1·0 m

Thus, Voume of cuboid = 11.44 m^{3}

Also given, inner radius of cylindrical pipe, *r* = 30 cm = 0.30 m

And, thickness of cylindrical pipe = 5 cm

Thus, outer radius of cylindrical pipe,* R* = 30 + 5 = 35 cm = 0.35 m

Let the length of the cylindrical pipe be *h* cm.

Therefore, volume of cylindrical pipe = *πR*^{2}*h *− *πr*^{2}*h *= *πh*(*R*^{2}−*r*^{2})

= *πh* [(0.35)^{2 }− (0.30)^{2}]

= *πh*(0.35 − 0.30)(0.35 + 0.30)

= *πh*(0.05 × 0.65) = 0.0325*πh *m^{3}

According to the question,

Volume of iron cylindrical pipe = Volume solid iron cuboid

⟹ 0.0325*πh* = 11.4

⟹ *h* = 112 m

Thus, the required length of pipe = 112 m.

**Q. **If the points A(*k* + 1, 2*k*), B(3*k*, 2*k* + 3) and C(5*k* − 1, 5*k*) are collinear, then find the value of *k*.

**Sol.**

Given that points A(*k* + 1, 2*k*), B(3*k*, 2*k* + 3) and C(5*k* − 1, 5*k*) are collinear.

Therefore, the area of the triangle formed by these three points = 0

Thus the required values of *k* are 1/2 and 2.

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

(i) even sum, and

(ii) even product.

**Sol.**

On throwing the two different dice the various outcomes obtained are as follows:

(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), (3,6),

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

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

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

Thus, here total number of outcomes are 36.

(i) Even sum outcomes are as follows:

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

(2,2), (2,4), (2,6),

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

(4,2), (4,4), (4,6),

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

(6,2), (6,4), (6,6)

Thus, number of favourable outcomes = 18

P(even sum) = 18/36 = 1/2

Hence, the probability of getting an even sum is 1/2.

(ii) Even product outcomes are as follows:

(1,2), (1,4), (1,6),

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

(3,2), (3,4), (3,6),

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

(5,2), (5,4), (5,6),

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

Thus number of favourable outcomes = 27

P (even product) = 27/36 = 3/4

Hence, the probability of getting an even product is 3/4.

**Download the complete CBSE Class 10 Mathematics Solved Question Paper 2017**

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