PART – A

Section 1: (2 marks each)

1. Find the L.C.M. and H.C.F. of 25, 125, and 625 by the Prime Factorization method.

Answer:

Prime factorization of 25 = 5²

Prime factorization of 125 = 5³

Prime factorization of 625 = 5⁴

HCF = 5² = 25 (The lowest power of common prime factor)

LCM = 5⁴ = 625 (The highest power of the common prime factor)

2. If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12}, then show that n(A ∪ B) = n(A) + n(B) – n(A ∩ B).

Answer:

A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}

A ∩ B = {6}

So, n(A ∪ B) = 8, n(A) = 5, n(B) = 4, n(A ∩ B) = 1

Now, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

8 = 5 + 4 – 1, which is true.

3. Solve the pair of linear equations 3x + 2y = 6 and x + 3y = 13 by the Elimination method.

Answer:

Multiply the second equation by 3:

3x + 9y = 39

Subtract the first equation from this:

(3x + 9y) – (3x + 2y) = 39 – 6

7y = 33

y = 33 / 7

Substitute y = 33/7 in 3x + 2y = 6:

3x + 2 × 33/7 = 6

3x = 6 – 66/7 = 42/7 – 66/7 = -24/7

x = -24/21 = -8/7

So, x = -8/7 and y = 33/7.

4. Find the points of trisection of the line segment joining the points (-4, 3) and (9, 4).

Answer:

The points of trisection divide the line segment into three equal parts. Let the trisection points be P₁ and P₂.

Using the section formula, the coordinates of P₁ (1st trisection) are:

P₁ = (2x₂ + x₁) / 3, (2y₂ + y₁) / 3 = (2(9) + (-4)) / 3, (2(4) + 3) / 3 = (14/3, 11/3)

The coordinates of P₂ (2nd trisection) are:

P₂ = (x₂ + 2x₁) / 3, (y₂ + 2y₁) / 3 = (9 + 2(-4)) / 3, (4 + 2(3)) / 3 = (1/3, 10/3)

So, the points of trisection are (14/3, 11/3) and (1/3, 10/3).

5. In the given figure, if DE ∥ BC, Justify.

Answer:

If DE ∥ BC, by Basic Proportionality Theorem (also called Thales’ theorem), the sides of the triangles formed by the line DE cutting the triangle are proportional. This means the ratio of the corresponding sides of the triangles will be equal.

6. Write the formula to find the total surface area of a cone and explain each term in it.

Answer:

The total surface area of a cone is given by:

Total Surface Area = πr(r + l)

where:

• r = radius of the base of the cone,
• l = slant height of the cone,
• π = constant, approximately 3.14159.

7. Use Euclid’s division lemma to show that the square of any positive integer is of the form 5n, 5n+1, or 5n+4 where n is a whole number.

Answer:

Euclid’s division lemma states that for any integer a and any positive integer b, there exist unique integers q (quotient) and r (remainder) such that:

a = bq + r where 0 ≤ r < b

Let b = 5, and consider any integer a. By Euclid’s division lemma, we can write:

a = 5q + r where r = 0, 1, 2, 3, or 4

Now, let’s square a:

• If r = 0, then a = 5q, and a² = (5q)² = 25q² = 5(5q²), which is of the form 5n.
• If r = 1, then a = 5q + 1, and a² = (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1, which is of the form 5n + 1.
• If r = 2, then a = 5q + 2, and a² = (5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q + 1) + 4, which is of the form 5n + 4.
• If r = 3, then a = 5q + 3, and a² = (5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4, which is of the form 5n + 4.
• If r = 4, then a = 5q + 4, and a² = (5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1, which is of the form 5n + 1.

Thus, the square of any integer is always of the form 5n, 5n + 1, or 5n + 4.

8. Draw a circle of diameter 8 cm from a point 6 cm away from its centre. Construct the pair of tangents to the circle and measure their length.

Answer:

Given: The diameter of the circle is 8 cm, so the radius r = 4 cm. The distance from the external point to the center is 6 cm.

Steps to construct the tangents:

• Draw the circle with center O and radius 4 cm.
• Mark the point P outside the circle, 6 cm away from the center.
• Join OP (the line from the center to the external point), which has a length of 6 cm.
• Draw a perpendicular from the center O to the line OP, which meets the circle at point T. This point T will be the point of contact of the tangents.
• Using the right triangle OTP, the length of the tangent can be calculated using the Pythagoras theorem:

OP² = OT² + PT²

Substitute the known values:

6² = 4² + PT²

36 = 16 + PT²

PT² = 20 → PT = √20 ≈ 4.47 cm

The length of each tangent is approximately 4.47 cm.

Mathematics Problems and Solutions

10. If two dice are thrown at the same time, find the probability of getting sum of the dots on top is a composite number.

Answer:

When two dice are thrown, the possible sums of the dots on top range from 2 to 12. These sums can be classified into prime numbers and composite numbers:

• Prime numbers: 2, 3, 5, 7, 11
• Composite numbers: 4, 6, 8, 9, 10, 12

Now, let’s calculate the total possible outcomes when throwing two dice. Since each die has 6 faces, there are:

6 × 6 = 36 total outcomes.

Next, let’s count the outcomes where the sum is a composite number:

• Sum = 4: (1, 3), (2, 2), (3, 1) → 3 outcomes
• Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → 5 outcomes
• Sum = 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → 5 outcomes
• Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes
• Sum = 10: (4, 6), (5, 5), (6, 4) → 3 outcomes
• Sum = 12: (6, 6) → 1 outcome

So, the total number of outcomes where the sum is a composite number is:

3 + 5 + 5 + 4 + 3 + 1 = 21 favorable outcomes.

Therefore, the probability is:

Probability = favorable outcomes / total outcomes = 21 / 36 = 7 / 12.

11. Find the value of p and k, If 3 is a common root for the quadratic equations 2x² + px + 9 = 0 and p(x² + 2x) + k = 0.

Answer:

Given that 3 is a common root of both quadratic equations, we substitute x = 3 in both equations.

First equation:

2x² + px + 9 = 0

Substitute x = 3:

2(3)² + p(3) + 9 = 0 ⟹ 18 + 3p + 9 = 0 ⟹ 3p + 27 = 0 ⟹ p = -9.

Second equation:

p(x² + 2x) + k = 0

Substitute p = -9 and x = 3:

-9((3)² + 2(3)) + k = 0 ⟹ -9(9 + 6) + k = 0 ⟹ -9(15) + k = 0 ⟹ -135 + k = 0 ⟹ k = 135.

Thus, p = -9 and k = 135.

12. The first and the last terms of an AP are 105 and 995 respectively. If the common difference is 5, how many terms are there and what is their sum?

Answer:

The general formula for the nth term of an arithmetic progression (AP) is:

aₙ = a₁ + (n - 1) d

Where:

• a₁ is the first term,
• d is the common difference,
• aₙ is the nth term.

We are given:

• a₁ = 105
• aₙ = 995
• d = 5

Substitute these values into the nth term formula:

995 = 105 + (n - 1) × 5

Now, solve for n:

995 - 105 = (n - 1) × 5 ⟹ 890 = (n - 1) × 5 ⟹ n - 1 = 890 / 5 = 178 ⟹ n = 179.

Thus, the number of terms is n = 179.

The sum of the first n terms of an AP is given by the formula:

Sₙ = n / 2 × (a₁ + aₙ)

Substitute the known values:

S₁₇₉ = 179 / 2 × (105 + 995) = 179 / 2 × 1100 = 179 × 550 = 98450.

Thus, the sum of the terms is 98450.

SECTION – III

4 x 6 = 24

Instructions:
i. Answer any 4 of the following questions.
ii. Each question carries 6 marks.

13. Prove that \( 2\sqrt{3} – \sqrt{5} \) is an irrational number.

Answer:
We are tasked with proving that \( 2\sqrt{3} – \sqrt{5} \) is irrational. To do this, we assume the opposite—that the number is rational—and show this leads to a contradiction.

Assume that \( 2\sqrt{3} – \sqrt{5} \) is a rational number. Let:

\( 2\sqrt{3} – \sqrt{5} = \frac{p}{q} \), where \( p \) and \( q \) are integers with no common factors (i.e., \( \frac{p}{q} \) is in its simplest form).

Rearranging, we get:

\( 2\sqrt{3} = \sqrt{5} + \frac{p}{q} \)

Squaring both sides:

\( (2\sqrt{3})^2 = \left( \sqrt{5} + \frac{p}{q} \right)^2 \)

Expanding:

\( 4 \times 3 = \left( 5 \right) + 2 \times 5 \times \frac{p}{q} + \left( \frac{p}{q} \right)^2 \)

\( 12 = 5 + 2 \times 5 \times \frac{p}{q} + \frac{p^2}{q^2} \)

Simplifying:

\( 7 = 2 \times 5 \times \frac{p}{q} + \frac{p^2}{q^2} \)

Notice that the left-hand side (7) is rational, but the right-hand side involves \( \sqrt{5} \), an irrational number, multiplied by rational terms. Therefore, for both sides to be equal, we would need \( \sqrt{5} \) to be rational, which is a contradiction since \( \sqrt{5} \) is irrational.

Hence, our initial assumption that \( 2\sqrt{3} – \sqrt{5} \) is rational must be false. Therefore, \( 2\sqrt{3} – \sqrt{5} \) is irrational.

14. Draw the graph of the polynomial \( p(x) = 5x^2 + 3x – 1 \) on the graph paper. Find its zeros from the graph.

Answer:
To find the zeros of the polynomial \( p(x) = 5x^2 + 3x – 1 \) by graphing, follow these steps:

1. Plot the function: First, plot the function \( p(x) = 5x^2 + 3x – 1 \) using a graphing method, either by hand or using graphing software. The graph is a parabola opening upwards because the coefficient of \( x^2 \) (which is 5) is positive.
2. Find the x-intercepts (zeros): The x-intercepts represent the zeros of the polynomial. From the graph, observe where the curve crosses the x-axis. These points give the approximate values of \( x \) for which \( p(x) = 0 \).
3. Find the exact zeros algebraically: Use the quadratic formula:

   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

   For \( p(x) = 5x^2 + 3x – 1 \), \( a = 5 \), \( b = 3 \), and \( c = -1 \):

   x = \frac{-3 \pm \sqrt{3^2 - 4(5)(-1)}}{2(5)} = \frac{-3 \pm \sqrt{9 + 20}}{10} = \frac{-3 \pm \sqrt{29}}{10}

   Thus, the exact zeros are:

   x = \frac{-3 + \sqrt{29}}{10} and x = \frac{-3 - \sqrt{29}}{10}

   These can be approximated numerically.

15. Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours when he travels 120 km by train and the rest by car. He takes 20 minutes more if he travels 200 km by train and the rest by car. Find the speed of the train and the car.

Answer:
Let the speed of the train be \( t \) km/h and the speed of the car be \( c \) km/h.

1. First scenario: Rahim travels 120 km by train and the remaining 480 km by car, taking 8 hours.
   Time taken by train = \( \frac{120}{t} \)
   Time taken by car = \( \frac{480}{c} \)
   Total time = 8 hours, so:

   120/t + 480/c = 8

2. Second scenario: Rahim travels 200 km by train and the remaining 400 km by car, taking 20 minutes more (which is \( \frac{1}{3} \) hour more) than the first scenario.
   Time taken by train = \( \frac{200}{t} \)
   Time taken by car = \( \frac{400}{c} \)
   Total time = 8 hours + \( \frac{1}{3} \) hour, so:

   200/t + 400/c = 8 + 1/3

3. Solve the system of equations: After solving, we find:

   t = 60 km/h and c = 80 km/h

Thus, the speed of the train is 60 km/h and the speed of the car is 80 km/h.

16. A person from the top of a building of height 15 meters observes the top and the bottom (foot) of a cell tower with the angle of elevation as 60° and the angle of depression as 45° respectively. Find the height of the cell tower.

Answer:
Let the height of the cell tower be \( h \) meters. Let the height of the building be 15 meters. Let the distance from the base of the building to the foot of the tower be \( d \).

1. Using the angle of elevation (60°): In the right triangle formed by the height of the building and the line of sight to the top of the tower:

   tan(60°) = (h - 15) / d

   \[ \Rightarrow \sqrt{3} = \frac{h – 15}{d} \]
2. Using the angle of depression (45°): In the right triangle formed by the height of the building and the line of sight to the foot of the tower:

   tan(45°) = 15 / d

   \[ \Rightarrow 1 = \frac{15}{d} \quad \Rightarrow d = 15 \text{ meters}. \]
3. Substitute \( d = 15 \) into the first equation: \[ \sqrt{3} = \frac{h – 15}{15} \quad \Rightarrow h – 15 = 15\sqrt{3} \quad \Rightarrow h = 15 + 15\sqrt{3} \approx 40.98 \, \text{meters}. \]

Thus, the height of the cell tower is approximately 40.98 meters.

17. Two dice are rolled at the same time and the sum of the numbers appearing on them is noted. Find the probability of getting each sum, from 4 to 8, separately.

Answer:
When two dice are rolled, the possible sums range from 2 to 12. We will calculate the probability of getting each sum from 4 to 8.

1. Sum = 4: The possible pairs are (1,3), (2,2), (3,1). There are 3 outcomes.
2. Sum = 5: The possible pairs are (1,4), (2,3), (3,2), (4,1). There are 4 outcomes.
3. Sum = 6: The possible pairs are (1,5), (2,4), (3,3), (4,2), (5,1). There are 5 outcomes.
4. Sum = 7: The possible pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 outcomes.
5. Sum = 8: The possible pairs are (2,6), (3,5), (4,4), (5,3), (6,2). There are 5 outcomes.

The total number of possible outcomes when rolling two dice is 36 (since each die has 6 faces, and \( 6 \times 6 = 36 \)).

Thus, the probabilities are:

• Probability of getting sum 4: \( \frac{3}{36} = \frac{1}{12} \)
• Probability of getting sum 5: \( \frac{4}{36} = \frac{1}{9} \)
• Probability of getting sum 6: \( \frac{5}{36} \)
• Probability of getting sum 7: \( \frac{6}{36} = \frac{1}{6} \)
• Probability of getting sum 8: \( \frac{5}{36} \)

Question 18:

The following table gives the marks obtained by 40 students in SA – I exams in the Mathematics subject. Draw the “Less Than” and “Greater Than” graphs.

Marks Range	Number of Students
50 – 55	4
55 – 60	5
60 – 65	10
65 – 70	8
70 – 75	7
75 – 80	6

Answer:

To draw the “Less Than” and “Greater Than” graphs, we need to create cumulative frequency tables for both.

Less Than Cumulative Frequency:

• Less than 55: 4
• Less than 60: 4 + 5 = 9
• Less than 65: 9 + 10 = 19
• Less than 70: 19 + 8 = 27
• Less than 75: 27 + 7 = 34
• Less than 80: 34 + 6 = 40

Greater Than Cumulative Frequency:

• Greater than 50: 40 – 4 = 36
• Greater than 55: 36 – 5 = 31
• Greater than 60: 31 – 10 = 21
• Greater than 65: 21 – 8 = 13
• Greater than 70: 13 – 7 = 6
• Greater than 75: 6 – 6 = 0

Now you can plot these cumulative frequencies on graphs:

For the “Less Than” graph:

Plot the cumulative frequencies against the upper limits of the intervals (55, 60, 65, 70, 75, 80).

For the “Greater Than” graph:

Plot the cumulative frequencies against the lower limits of the intervals (50, 55, 60, 65, 70, 75).

Math Quiz – Part B