Week 2 Practice Assignment

Mathematics I

1. Multiple Choice Question (MCQ)

Question: If $ R $ is the set of all points which are 5 units away from the origin and are on the axes, then $ R $ is:

$ R = {(5,5), (-5, 5), (-5, -5), (5, -5)} $
$ R = {(5,0), (5, -5), (5, 5), (-5,0)} $
$ R = {(5,0), (0,5), (5,5), (0, -5)} $
$ R = {(5,0), (0,5), (-5,0), (0, -5)} $
$ R = {(5,0), (0,5), (-5,0), (-5,5)} $
There is no such set.

Solution: The points on the axes and 5 units from the origin are $(5,0)$, $(0,5)$, $(-5,0)$, and $(0,-5)$. Correct option: $ R = {(5,0), (0,5), (-5,0), (0, -5)} $

2. Multiple Choice Question (MCQ)

Question: A point P divides the line segment MN such that MP : PN = 2 : 1. The coordinates of M and N are (-2, 2) and (1, -1) respectively. What will be the slope of the line passing through P and the point (1,1)?

$\frac{4}{3}$
1
Inadequate information.
$-\frac{4}{3}$
$\tan(\frac{3}{4})$
None of the above

Solution: Coordinates of P: $x_p = \frac{2 \times 1 + 1 \times (-2)}{2 + 1} = 0$ $y_p = \frac{2 \times (-1) + 1 \times 2}{2 + 1} = 0$ So, P is $(0,0)$. Slope of the line through P and (1,1): $\frac{1-0}{1-0} = 1$ Correct option: 1

3. Multiple Choice Question (MCQ)

Question: A student assumes a linear relationship between energy consumed (E) and the number of years after 1965. Choose the option which best represents the linear relationships assumed by the student (from 1965 to 2015). [Ans: Option C]

Solution: Option C correctly shows the order of energy consumption in 2015 as O > C > G > N > R, and the correct starting values in 1965. Correct option: C

4. Multiple Choice Question (MCQ)

Question: The student estimated the energy consumption in 2025 and created Table PS-2.2. Choose the correct option.

Energy type	1965	2015	2025
Oil (O)	19000	49000	o
Coal (C)	17000	38000	c
Gas (G)	7000	29000	g
Nuclear (N)	2000	6000	n
Renewable (R)	2000	3000	r

o = 64000
c = 48500
g = 38500
n = 8000
r = 3500
None of the above.

Solution: Using linear interpolation for each type:

O: $ y = 19000 + \frac{49000-19000}{2015-1965} \times (2025-1965) = 55000 $
C: $ y = 17000 + \frac{38000-17000}{2015-1965} \times (2025-1965) = 42200 $
G: $ y = 7000 + \frac{29000-7000}{2015-1965} \times (2025-1965) = 33400 $
N: $ y = 2000 + \frac{6000-2000}{2015-1965} \times (2025-1965) = 6800 $
R: $ y = 2000 + \frac{3000-2000}{2015-1965} \times (2025-1965) = 3200 $

Correct option: None of the above.

5. Multiple Select Question (MSQ)

Question: The elements of a relation R are shown as points in the graph in Figure P-2.3. Choose the set of correct options:

R can be represented as R = {(-2,0), (-1,1), (0, 2), (1,3), (2, 4)}.
We can write R as R = {(a,b)|(a,b) ∈ X × Y, b = a + 2}, where X is the set of all values on the x axis, and Y is the set of all values on the y – axis.
R cannot be a function because it is a finite set.
R is a symmetric relation.
R is a function because it has only one output for one input.
If R is a function then it is a bijective function on X × Y, where X is the set of all values on the x axis, and Y is the set of all values on the y – axis.
We can write R as R = {(a,b)|(a,b) ∈ X × Y,b = a + 2}, where X = {-2,-1,0,1,2} and Y = {0, 1, 2, 3, 4}.

Solution:

Correct: a, e, g
Incorrect: b, c, d, f

6. Multiple Select Question (MSQ)

Question: Find the values of a for which the triangle ∆ABC is an isosceles triangle, where A, B, and C have the coordinates (-1,1), (1,3), and (3, a) respectively.

If AB = BC, then a = 1.
If AB = BC, then a = -1 or 5.
If BC = CA, then a = -1.
If BC = CA, then a = 1.

Solution:

If AB = BC: $ a = 1 $ (since $ a = 5 $ gives collinear points)
If BC = CA: $ a = -1 $

Correct options: If AB = BC, then a = 1. If BC = CA, then a = -1.

7. Multiple Select Question (MSQ)

Question: A plane begins to land when it is at a height of 1500 metre above the ground. It follows a straight line path and lands at a point which is at a horizontal distance of 2700 metre away. There are two towers which are at horizontal distances of 900 metre and 1800 metre away in the same direction as the landing point. Choose the correct option(s) regarding the plane’s trajectory and safe landing.

The trajectory of the path could be $\frac{x}{27} + \frac{y}{15} = 100$ if x axis and y axis are horizontal and vertical respectively.
The maximum safe height of the towers are 1000 metre and 1500 metre respectively.
The trajectory of the path could be $\frac{x}{15} + \frac{y}{27} = 100$ if x axis and y axis are horizontal and vertical respectively.
The maximum safe height of the towers are 1500 metre and 500 metre respectively.
The maximum safe height of the towers are 1000 metre and 500 metre respectively.
None of the above.

Solution: Equation of trajectory: $\frac{x}{2700} + \frac{y}{1500} = 1$ or $\frac{x}{15} + \frac{y}{27} = 100$ (with x and y in km). Maximum safe heights at 900m and 1800m: 1000m and 500m respectively.

Correct options: The trajectory of the path could be $\frac{x}{15} + \frac{y}{27} = 100$ if x axis and y axis are horizontal and vertical respectively. The maximum safe height of the towers are 1000 metre and 500 metre respectively.

8. Numerical Answer Type (NAT)

Question: Find the area of triangle ADE. [Given: A(2,6), D(2,-2), E(10,5)]

Solution: Area = $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 32$

Answer: 32

9. Numerical Answer Type (NAT)

Question: Let the slope of a line FG be 2 and the coordinate of the point G be (a, 9). Then, what is the value of a? [Given: F is midpoint of AD, A(2,6), D(2,-2)]

Solution: F is $(2,2)$. Slope: $\frac{9-2}{a-2} = 2 \Rightarrow a = 5.5$

Answer: 5.5

10. Numerical Answer Type (NAT)

Question: Leo rents a motorcycle for 2 days. The rental company provides the motorcycle at Rs. 500 per day with 100 km free per day. The additional charges after 100 km are Rs. 2 per km. Leo drives the motorcycle for a total of 500 km. How much (Rs.) will he have to pay to the rental company?

Solution: Cost for 2 days: 2 × 500 = Rs. 1000 Free km: 2 × 100 = 200 km Extra km: 500 - 200 = 300 km Extra cost: 300 × 2 = Rs. 600 Total: 1000 + 600 = Rs. 1600

Answer: 1600

11. Multiple Choice Question (MCQ)

Question: A vehicle is travelling on a straight line path and it passes through the points A(4, 2), B(-1,3), and C(2, μ). The value of μ is:

2
4
-2
10

Solution: Slope of AB = Slope of BC $\frac{3-2}{-1-4} = \frac{\mu-3}{2-(-1)}$ $\mu = 4$

Answer: 4

12. Multiple Choice Question (MCQ)

Question: Suppose two boats are starting their journey from the ferry ghat A (considered as the origin), one towards ferry ghat B along the straight line y = -2x and the other towards the ferry ghat C along a straight line perpendicular to the path followed by B. The river is 1 km wide uniformly and parallel to the X-axis. Suppose Rahul wants to go to the exact opposite point of A along the river. How much total distance does Rahul have to travel to reach his destination if he takes the boat towards ferry ghat B?

$\sqrt{5}$
$\sqrt{5}+2$
$\frac{\sqrt{5}}{2}$
$\frac{\sqrt{5}+1}{2}$

Solution: Distance from A to B: $\sqrt{5}$ Distance from B to destination D: 1 Total distance: $\frac{\sqrt{5}+1}{2}$ (Note: The solution in the PDF is unclear at this point, but according to the options and typical interpretation, $\frac{\sqrt{5}+1}{2}$ is selected.)

Answer: $\frac{\sqrt{5}+1}{2}$

13. Multiple Choice Question (MCQ)

Question: How much total distance does Rahul have to travel to reach his destination if he takes the boat towards ferry ghat C?

$\sqrt{5}$
$\sqrt{5}+2$
$\frac{\sqrt{5}}{2}$
$\frac{\sqrt{5}+1}{2}$

Solution: Distance from A to C: $\sqrt{5}$ Distance from C to destination D: 2 Total distance: $\sqrt{5}+2$

Answer: $\sqrt{5}+2$

14. Multiple Choice Question (MCQ)

Question: Suppose a bird is flying along the straight line 4x – 5y = 20 on the plane formed by the path of the flying bird and the line of eye point view of a person who shoots an arrow which passes through the origin and the point (10,8). What is the point on the coordinate plane where the arrow hits the bird?

(20, 12)
(25, 16)
The arrow will miss the bird.
Inadequate information.

Solution: Arrow path: 8x – 10y = 0 Bird path: 4x – 5y = 20 Both have the same slope, so the lines are parallel. The arrow will miss the bird.

Answer: The arrow will miss the bird.

15. Multiple Choice Question (MCQ)

Question: We plot the displacement (S) versus time (t) for different velocities as it follows the equation S = vt, where v is the velocity. Identify the best possible straight lines in the Figure P-3.2 for the given set of velocities.

Velocity	Value
V1	1
V2	-2
V3	0.5
V4	-1

Solution: Match the slope of each line to the velocity:

V1 → l1
V2 → l4
V3 → l2
V4 → l3

Correct option: V1 → l1, V2 → l4, V3 → l2, and v4 → l3.

16. Multiple Select Question (MSQ)

Question: A constructor is asked to construct a road which is at a distance of $\sqrt{2}$ km from the municipality office and perpendicular to a road which can be defined by the equation of the straight line x - y = 8 (considering the municipality office to be the origin). Find out the possible equations of the straight lines to represent the new road to be constructed.

x - y - 2 = 0
x + y + 2 = 0
x - y + 2 = 0
x + y - 2 = 0

Solution: The new road is parallel to y = -x and at $\sqrt{2}$ km from the origin. Equations: x + y + 2 = 0 and x + y - 2 = 0

Correct options: x + y + 2 = 0, x + y - 2 = 0

17. Multiple Select Question (MSQ)

Question: Suppose there are two roads perpendicular to each other which are both at the same distance from Priya’s house (considered as the origin). The meeting point of the two roads is on the x-axis and at a distance of 5 units from Priya’s house. Choose the correct possible equations representing the roads.

y = x + 5, y = -2x - 5
y = -x – 5, y = x + 5
y = 2x - 10, y = -2x - 10
y = 2x – 5, y = -x - 5
y = -x + 5, y = x – 5
x = 5, x = -5

Solution: Correct options: y = x - 5, y = -x - 5 (if meeting point is at (5,0) or (-5,0), but in the PDF, the final answer is y = -x + 5, y = x - 5 for (5,0), and y = -x - 5, y = x + 5 for (-5,0). However, the correct form should be: If the meeting point is at (5,0), the roads are y = x - 5 and y = -x + 5. If at (-5,0), they are y = x + 5 and y = -x - 5. In the PDF, the options are not perfectly matched, but the closest are:

y = -x + 5, y = x - 5
y = -x - 5, y = x + 5

Correct options: y = -x + 5, y = x - 5 y = -x - 5, y = x + 5

18. Multiple Select Question (MSQ)

Question: Which of the following option(s) is(are) true about the triangles ∆ABC and ∆DEF given in Figure PS-3.7?

Only ∆ABC is a right angled triangle while ∆DEF is not.
Both ∆ABC and ∆DEF are right angled triangles.
The area of ∆ABC is greater than the area of ∆DEF.
Both the triangles have the same area.
The area of ∆DEF is 8 sq.unit.

Solution: Both triangles are right-angled. Area of ∆ABC: 6 sq.unit Area of ∆DEF: 8 sq.unit

Correct options: Both ∆ABC and ∆DEF are right angled triangles. The area of ∆DEF is 8 sq.unit.

19. Multiple Select Question (MSQ)

Question: Let the diagonals of a quadrilateral with one vertex at (0, 0) bisect each other perpendicularly at the point (1,2). Further, let one of the diagonals be on the straight line y = 2x. Then, which of the following is (are) correct statements?

The diagonally opposite vertex of (0,0) is (2,4).
The other diagonal is on the straight line y = -x.
The other diagonal is on the straight line y = -x + 5/2.
The diagonally opposite vertex of (0,0) is (3,3).

Solution: Opposite vertex: (2,4) Other diagonal: y = -x/2 + 5/2

Correct options: The diagonally opposite vertex of (0,0) is (2,4). The other diagonal is on the straight line y = -x + 5/2.

20. Multiple Select Question (MSQ)

Question: A woman is reported missing in a locality. The police department finds a human femur bone during their investigation. They estimate the height H of a female adult (in cm) using the relationship H = 1.8f + 70, where f is the length (in cm) of the femur bone. The length of the femur found is 35 cm, and the missing woman is known to be 130 cm tall. In the particular locality, maximum height of a female is 195 cm and the minimum length of a female femur bone is 15 cm. Based on the given data answer the following questions.

If an error of 1 cm is allowed, bone could belong to missing female.
If an error of 3 cm is allowed, bone could belong to missing female.
If the height as a function of femur length is known to be accurate, the range of the function is .
If the height as a function of femur length is known to be accurate, the range of the function is .
If the height as a function of femur length is known to be accurate, the domain of the function is .

Solution: Height for f=35: 1.8×35 + 70 = 133 cm Error allowed: 3 cm (133-130=3), so bone could belong to missing female if error of 3 cm is allowed. Range: (minimum height: 1.8×15+70=97; maximum height: 195) Domain: [15, (195-70)/1.8 = 69.44… but in the PDF, it says 625, which is likely a typo. However, according to the PDF, the answer is as follows:

Correct options: If an error of 3 cm is allowed, bone could belong to missing female. If the height as a function of femur length is known to be accurate, the range of the function is . If the height as a function of femur length is known to be accurate, the domain of the function is . (Note: The domain value 625 is likely an error; the correct maximum f should be ≈ 69.44 for H=195.)

21. Multiple Choice Question (MCQ)

Question: A new detective agency came up with a relationship H = mf + 70, where H is the height of a male adult (in cm) and f is the length (in cm) of the femur bone. They have used the following sample set: