# CBSE Class 9 Mathematics Solved Guess Paper SA-II 2015

Here you can find the Mathematics Guess Paper for CBSE Class 9 SA-II 2015. It includes a various set of questions with assigned score for each.

Find **CBSE Class 9 Mathematics Guess Paper** for **SA-II 2015**. This paper is a collection of questions from the previous year question papers, along with fresh new questions and has been framed keeping the Students' perspective in mind. This will help the Students by building a sound concept before their **SA-II Examination**.

**Q.**** **What is the longest pole that can be put in a room of dimensions length = 20 cm, breadth= 20 cm and height = 10 cm?

(a) 15 cm

(b) 25 cm

(c) 20 cm

(d) 30 cm

**Sol.**** **Since, area of parallelograms on the same base and between same parallels is always same,

So as parallelogram PQRS and AQRB are on the same base and between same parallels,

Option (c) is correct.

**Q.** Find the value of *k*, if *x* = –1 and *y* = 2 is a solution of *kx* + 3*y* = 7.

**Sol.**** **We have given,

* x* = –1 and *y* = 2

Substituting the value of *x* and *y* in the equation *kx* + 3*y* = 7, we get

* k* (–1) + 3 (2) = 7

– *k* + 6 = 7

– *k* = –1

* k* = –1

**Q.** Find the amount of water displaced by a solid spherical ball of diameter 0.21 m.

**Sol. **The diameter of the solid spherical ball is = 0.21 m

Amount of water displaced = Volume of spherical ball

**Q.** Two years ago, a man's age was 3 times the square of his son's age. In 3 years’ time, his age will be 4 times his son's age. Find their present ages.

**Sol. **Let the present age of son be *x* years.

Therefore, before two years son’s age was = (*x* – 2) years.

According to the question:

Before two years, father’s age = 3(*x* – 2)^{2} years.

Thus, the present age of father = [3(*x* – 2)^{2} + 2] years

After three years, the age of son will be = (*x* + 3) years

And the age of the father will be = [3 (*x* – 2)^{2} + 2 + 3] years

= [3(*x* – 2)^{2} + 5] years

Now, according to the question:

3(*x* – 2)^{2} + 5 = 4 (*x* + 3)

⇒ 3(*x*^{2} – 4*x* + 4) + 5 = 4*x* + 12

⇒ 3*x*^{2} – 16*x*+ 5 = 0

⇒ 3*x*^{2} – 15*x*– *x*+ 5 = 0

⇒ 3*x* (*x* – 5) – 1 (*x*– 5) = 0

⇒ (3*x* – 1) (*x* – 5) = 0

⇒ 3*x* – 1 = 0 or *x* – 5 = 0

⇒ *x* = 1/3 or 5

If *x* = 1/3, then age before two years = *x* – 2 = –5/3.

This means that age of son before two years was –5/3 years. This is impossible since the age of a person cannot be negative.

Therefore, present age of son = *x *= 5 years

And, the present age of the man = [3 (*x* – 2)^{2} + 2] = 3 × 9 + 2 = 29 years

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