**Sample Question Paper**

**Mathematics-Basic **

**Class X – Session 2019-20**

**Section – A**

**Q1. If an event that cannot occur, then its probability is: [1]**

**Solution: **An event that cannot occur has 0 probability, such an event is called impossible event.

**Q2. Which of the following is not a zero of the polynomial p(x) = x ^{3} - 7x + 6 [1]**

**Solution: **Here, p(1) = (1)3 - 7(1) + 6 = 0

p(2) = (2)^{3} - 7(2) + 6 = 0

p(-2) = (-2)^{3} - 7(-2) + 6 ≠ 0

p(-3) = (-3)^{3} - 7(-3) + 6 = 0

So, -2 is not a zero of p(x).

**Q3. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be: [1]**

**Solution:** Area of first circular park whose diameter is 16 m,

=**(16/2) ^{2}**

= **(8) ^{2}**

=64m^{2}

**Area of second circular park whose diameter is 12 m,**

=(12/2)^{2}

= (6)^{2}

=36m^{2}

According to question,

Area of single circular park = Area of first circular park + Area of second circular park

(r)^{2} = 36m^{2} + 64m^{2}

r^{2} = 100

r = 10 m

**Q4. Which term of the AP: 4, 9, 14, ...... is 254? [1]**

**Solution: **Here, a = 4 and d = 5

Let n^{th} term of the AP be 254. Then,

a_{n} = a + (n - 1) d

⇒ 254 = 4 + (n - 1) (5)

⇒ 5(n - 1) = 250

⇒ n = 51

**Q5. A man goes 15 m due west and then 8 m due north. Now far is he from the starting point? [1]**

**Solution:**

op^{2}=15^{2}+ 8^{2}

op^{2}=225 +64

op^{2}=289

op= 17m

**Q6. In figure, a circle with center O is inscribed in a quadrilateral ABCD such that, it touches the sides BC,AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5cm, then the radius of the circle (in cm.) is: [1]**

**Solution:**

Since DS = DR = 5 cm

(tangents of a circle from same external point)

Now, AR = AD - DR = 25 - 5 = 18 cm

Similarly, AR = AQ = 18 cm (tangents)

Now, QB = AB - AQ = 29 - 18 = 11 cm

Similarly, QB = PB = 11 cm

Given, ∠B = 90°

So, ∠POQ = 90°

Hence, OQBP is square

QB = 11 cm

Side of square = 11 cm, so the radius = 11 cm.

**Q7. The sum of the first 20 natural numbers is ......... [1]**

**Solution**: Sum of first 20 natural numbers=

20 x (20+1) / 2 = 210

**Q8. The value of 3cosec ^{2}A - 3cot^{2}A is ........... [1]**

**Solution**: 3cosec^{2}A - 3cot^{2}A

= 3 (cosec^{2}A - cot^{2}A)

= 3 x 1 = 3

**Q9. Find the least number that is divisible by all the numbers from 1 to 10 (both inclusive): [1]**

**Solution:**

Required number = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

= 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7

= 2520

**Q10. Find the distance of the point (-4, -7) from the y-axis. [1]**

Points are(-4, -7) and (0, -7)

Distance = √(0 +4)^{2} ( -7 +7)^{2}

=√ 4^{2}+ 0^{2}

=√ 16

= 4 units

**Section - B**

**Q1. Explain whether 3 × 12 × 101 + 4 is a prime number or a composite number. [2]**

**Solution:**

3 × 12 × 101 + 4

= 4(3 × 3 × 101 + 1)

= 4(909 + 1)

= 4(910)

= 2 × 2 × 2 × 5 × 7 × 13

= a composite number

[∵ Product of more than two prime factors]

**Q2. Find the zeros of the polynomial p(x) = 12 + x - x ^{2}**

x^{2 }- x - 12

x^{2 }- 4x + 3x - 12

x(x-4) + 3(x-4)

(x+3)(x-4)

So, the two zeros are 4 and -3

**Q3. Find the value of k for which the roots of the quadratic equation 2x ^{2} + kx + 8 = 0 will have the equal roots?**

For equal roots, D = 0

b

^{2}- 4ac

b

^{2}= 4ac

k

^{2}= 4 x 2 x 8

k

^{2}= 64

k = +√64

k = +8 or k = -8

**Q4. Two unbiased coins are tossed. Find the probability of getting: [2]**

**(i) two heads**

**(ii) at least one head.**

**Solution:**

No. of possible outcomes = 4

(i) No. of favourable outcomes = 1

namely HH {Sample space S = {HH, HT, TH, TT}}

So, required probability = 1/4

(ii) No. of favourable outcomes = 3, namely {HH, HT, TH}

So, required probability = 3/4

**Q5. Solve for x and y: **

**3x + 2y = 11, 2x + 3y = 4**

**Solution:**

Given equations are:

3x + 2y = 11 .....(i)

2x + 3y = 4 ......(ii)

Eq. (ii) gives, y = (4-2x)/3……(iii)

Substituting this value of y in Eq. (i), we have

3x + {2(4-2x)/3} = 11

9x + 8 - 4x = 33

5x = 25

x = 5

Substituting this value of x in Eq. (iii), we have:

y = 4 - 2x5/3

y = -2

Thus, x = 5, y = -2