All practice questions

Question 161Question →

Three of the interior angles of a convex polygon are each $120^\circ$ , while the remaining interior angles are each $160^\circ$ . How many sides does the polygon have?

Show answer & explanation

Answer: 12

Answer

The polygon has 12 sides.

The sum of the interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. From the problem, the sum of the angles is

3 \times 120^\circ + (n - 3) \times 160^\circ = 160^\circ n - 120^\circ

. Equating

180^\circ(n - 2)

to

160^\circ n - 120^\circ

gives

180^\circ n - 360^\circ = 160^\circ n - 120^\circ

, which simplifies to

20^\circ n = 240^\circ

, yielding

n = 12

.

Step-by-Step Solution

1

Write down the standard formula for the sum of interior angles of an

n

-sided convex polygon.

Sum of interior angles =

(n - 2) \times 180^\circ

.

Any

n

-sided convex polygon can be split into

(n-2)

triangles, each having an interior angle sum of

180^\circ

.

2

Express the total sum of interior angles using the given values.

Sum =

3(120^\circ) + (n - 3)(160^\circ) = 360^\circ + 160^\circ n - 480^\circ = 160^\circ n - 120^\circ

.

Three angles are

120^\circ

, so the remaining

(n - 3)

angles must each equal

160^\circ

.

3

Equate the theoretical sum to the calculated sum.

180^\circ(n - 2) = 160^\circ n - 120^\circ \implies 180^\circ n - 360^\circ = 160^\circ n - 120^\circ

.

Both expressions represent the total interior angle sum of the same polygon.

4

Solve the equation for the number of sides

n

.

20^\circ n = 240^\circ \implies n = 12

.

Subtracting

160^\circ n

from both sides and adding

360^\circ

yields

20^\circ n = 240^\circ

.

Key Concept

Sum of Interior Angles of a Polygon

Question 162Question →

The $3^{\text{rd}}$ , $6^{\text{th}}$ , and $11^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $15$ , what is the $4^{\text{th}}$ term of the geometric progression?

Show answer & explanation

Answer: 125

Answer

The 4th term of the geometric progression is 125.

By writing the 3rd, 6th, and 11th terms of the AP as

15+2d

,

15+5d

, and

15+10d

, we utilize the geometric mean property

(15+5d)^2 = (15+2d)(15+10d)

to find

d=6

. This yields the GP terms

27, 45, 75

, giving a common ratio of

5/3

. Multiplying the third term

75

by

5/3

gives the 4th GP term as

125

.

Step-by-Step Solution

1

Write down the AP term expressions

T_3 = 15 + 2d

,

T_6 = 15 + 5d

,

T_{11} = 15 + 10d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

with initial term

a = 15

.

2

Apply the geometric progression condition

(15 + 5d)^2 = (15 + 2d)(15 + 10d)

If three terms

A, B, C

are in GP, then

B^2 = A \cdot C

.

3

Expand and solve the quadratic equation for the common difference

d

d = 6

Expanding gives

225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6

because

d \neq 0

.

4

Find the terms and common ratio of the GP

G_1 = 27

,

G_2 = 45

,

G_3 = 75

, and common ratio

r = \frac{5}{3}

Substituting

d = 6

gives the GP terms, and dividing consecutive terms gives

r = \frac{45}{27} = \frac{5}{3}

.

5

Calculate the 4th term of the GP

G_4 = 125

Multiplying the 3rd term by the common ratio yields

75 \times \frac{5}{3} = 125

.

Key Concept

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties

Estimated Time:2m 30s

Question 163Question →

The perpendicular bisector of the line segment joining the points $P(2, -1)$ and $Q(6, 7)$ intersects the $y$ -axis at $(0, c)$ . Find the value of $c$ .

Show answer & explanation

Answer: 5

Answer

The value of

c

is 5.

The midpoint of

PQ

is

(4, 3)

and the gradient of

PQ

is

2

. The perpendicular bisector has a gradient of

-\frac{1}{2}

and passes through

(4, 3)

. Substituting these into the line equation gives

y - 3 = -\frac{1}{2}(x - 4)

, which simplifies to

y = -\frac{1}{2}x + 5

. The line intersects the

y

-axis at

(0, 5)

, so

c = 5

.

Step-by-Step Solution

1

Find the midpoint of the line segment

PQ

Midpoint

M = (4, 3)

The perpendicular bisector must pass through the midpoint of the segment.

2

Calculate the gradient of

PQ

Gradient

m_{PQ} = 2

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for points

(2, -1)

and

(6, 7)

.

3

Determine the gradient of the perpendicular line

Perpendicular gradient

m_{\perp} = -\frac{1}{2}

Perpendicular lines have gradients that are negative reciprocals (

m_1 m_2 = -1

).

4

Formulate the equation of the perpendicular bisector and solve for the

y

-intercept

y = -\frac{1}{2}x + 5

, hence

c = 5

Using point-slope form

y - y_1 = m(x - x_1)

with point

(4, 3)

and

m = -\frac{1}{2}

, setting

x = 0

gives the

y

-intercept.

Key Concept

Perpendicular Bisector and Line Equations

Question 164Question →

A body of mass $0.2\text{ kg}$ undergoes simple harmonic motion with an angular frequency of $10\text{ rad/s}$ and an amplitude of $0.04\text{ m}$ . What is the magnitude of the maximum restoring force acting on the body in newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the maximum restoring force acting on the body is

0.8\text{ N}

.

The maximum restoring force in simple harmonic motion occurs at maximum displacement (

y = A

) and is given by

F_{\text{max}} = m \omega^2 A

. Substituting

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

yields

F_{\text{max}} = 0.2 \times (10)^2 \times 0.04 = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities from the problem statement.

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

.

These parameters are required to calculate acceleration and restoring force in simple harmonic motion.

2

Calculate the maximum acceleration of the oscillating body.

a_{\text{max}} = \omega^2 A = (10)^2 \times 0.04 = 4.0\text{ m/s}^2

.

In simple harmonic motion, maximum acceleration occurs at maximum displacement (the amplitude).

3

Determine the maximum restoring force.

F_{\text{max}} = m a_{\text{max}} = 0.2 \times 4.0 = 0.8\text{ N}

.

According to Newton's second law, force is the product of mass and acceleration.

Key Concept

Maximum Restoring Force in Simple Harmonic Motion

Question 165Question →

If $\sqrt{75} + \sqrt{12} - \sqrt{27} = k\sqrt{3}$ , what is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of

k

is 4.

Simplifying each surd into its basic form yields

\sqrt{75} = 5\sqrt{3}

,

\sqrt{12} = 2\sqrt{3}

, and

\sqrt{27} = 3\sqrt{3}

. Combining the coefficients gives

(5 + 2 - 3)\sqrt{3} = 4\sqrt{3}

. Equating

4\sqrt{3}

to

k\sqrt{3}

shows that

k = 4

.

Step-by-Step Solution

1

Simplify each individual surd by factoring out perfect squares

\sqrt{75} = 5\sqrt{3}

,

\sqrt{12} = 2\sqrt{3}

,

\sqrt{27} = 3\sqrt{3}

To combine surds through addition or subtraction, they must be converted to similar surds.

2

Combine the coefficients of the like surds

5\sqrt{3} + 2\sqrt{3} - 3\sqrt{3} = (5 + 2 - 3)\sqrt{3} = 4\sqrt{3}

Like terms with the same radical factor

\sqrt{3}

can be added and subtracted directly.

3

Compare the resulting coefficient with

k\sqrt{3}

k = 4

By direct comparison of coefficients of

\sqrt{3}

,

k

equals 4.

Key Concept

Simplification and combining of similar surds

Question 166Question →

Two $2 \times 2$ matrices are given as $A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 2 & k \end{pmatrix}$ . If the determinant of the product matrix $AB$ is equal to $30$ , what is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of k is 3.

The determinant of a product of square matrices equals the product of their individual determinants:

\det(AB) = \det(A) \cdot \det(B)

. Evaluating

\det(A)

yields

(2)(3) - (-1)(4) = 10

, and

\det(B)

yields

(1)(k) - (0)(2) = k

. Substituting these into

\det(AB) = 30

gives

10k = 30

, leading directly to

k = 3

.

Step-by-Step Solution

1

Evaluate the determinant of matrix A

\det(A) = 10

Using the

2 \times 2

determinant formula

\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

, we calculate

\det(A) = (2)(3) - (-1)(4) = 6 + 4 = 10

.

2

Evaluate the determinant of matrix B

\det(B) = k

Calculating the determinant of matrix

B

yields

\det(B) = (1)(k) - (0)(2) = k

.

3

Use the product property of determinants to solve for k

k = 3

Since

\det(AB) = \det(A) \cdot \det(B)

, we have

10k = 30

. Dividing both sides by 10 yields

k = 3

.

Key Concept

Determinant of Matrix Product

Estimated Time:1m 30s

Question 167Question →

The table below shows the frequency distribution of marks obtained by a group of students in a mathematics test:

Mark ( $x$ )	2	4	6	8	10
Frequency ( $f$ )	2	1	4	1	2

Find the mean deviation of the distribution.

Show answer & explanation

Answer: 2

Answer

The mean deviation of the distribution is 2.

To find the mean deviation, first calculate the mean

\bar{x} = \frac{\sum fx}{\sum f} = \frac{60}{10} = 6

. Next, sum the absolute deviations multiplied by their frequencies:

\sum f|x - \bar{x}| = 2(4) + 1(2) + 4(0) + 1(2) + 2(4) = 20

. Dividing this total by the sum of frequencies

10

yields a mean deviation of

2

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the distribution

\bar{x} = \frac{\sum f x}{\sum f} = \frac{(2 \times 2) + (1 \times 4) + (4 \times 6) + (1 \times 8) + (2 \times 10)}{2 + 1 + 4 + 1 + 2} = \frac{60}{10} = 6

The mean is required as the central benchmark from which individual deviations are measured.

2

Calculate the sum of absolute deviations weighted by frequency

\sum f |x - \bar{x}| = 2|2 - 6| + 1|4 - 6| + 4|6 - 6| + 1|8 - 6| + 2|10 - 6| = 8 + 2 + 0 + 2 + 8 = 20

Each absolute difference from the mean must be multiplied by its frequency to account for the total deviation.

3

Divide the total absolute deviation by the total frequency

\text{Mean Deviation} = \frac{\sum f |x - \bar{x}|}{\sum f} = \frac{20}{10} = 2

The mean deviation represents the average distance of all observations from the arithmetic mean.

Key Concept

Mean Deviation of a Frequency Distribution

Estimated Time:1m 30s

Question 168Question →

In the dimensional equation for the period of oscillation of a simple pendulum, $T = k g^a l^b$ , where $T$ is the period, $g$ is the acceleration due to gravity, $l$ is the length of the pendulum, and $k$ is a dimensionless constant, what is the numerical value of the exponent $a$ ?

Show answer & explanation

Answer: -0.5

Answer

The numerical value of the exponent

a

is -0.5.

Applying dimensional analysis to

T = k g^a l^b

, the dimension of the left-hand side is

\text{T}^1

. The right-hand side has dimensions

(\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

. Equating the exponents of time

\text{T}

yields

1 = -2a

, which gives

a = -0.5

.

Step-by-Step Solution

1

Express the dimensions of all physical quantities involved in fundamental base dimensions (M, L, T).

The dimension of period

T

is

[\text{T}]

, length

l

is

[\text{L}]

, and gravitational acceleration

g

is

[\text{L T}^{-2}]

.

Dimensional analysis requires substituting each quantity with its fundamental dimensions.

2

Formulate the dimensional homogeneity equation.

\text{M}^0 \text{L}^0 \text{T}^1 = (\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}

.

The principle of dimensional homogeneity states that the exponents of base dimensions on both sides of a physically correct equation must be equal.

3

Equate exponents of time

\text{T}

and solve for

a

.

1 = -2a \implies a = -\frac{1}{2} = -0.5

.

Comparing the powers of

\text{T}

gives a linear equation in

a

.

Key Concept

Principle of Dimensional Homogeneity

Estimated Time:1m 15s

Question 169Question →

A sector of a circle of radius $14\text{ cm}$ subtends an angle of $90^\circ$ at the centre of the circle. What is the area of the sector in $\text{cm}^2$ ? (Take $\pi = \frac{22}{7}$ )

Show answer & explanation

Answer: 154

Answer

The area of the sector is

154\text{ cm}^2

.

The area of a circular sector is given by

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2

. Substituting

\theta = 90^\circ

,

r = 14\text{ cm}

, and

\pi = \frac{22}{7}

yields

\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2

.

Step-by-Step Solution

1

Identify the formula for the area of a circular sector.

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2

The area of a sector is proportional to the central angle it subtends relative to a full circle (

360^\circ

).

2

Substitute the given values into the formula.

\text{Area} = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times (14)^2

Given radius

r = 14\text{ cm}

, angle

\theta = 90^\circ

, and

\pi = \frac{22}{7}

.

3

Simplify the fraction and calculate the numerical value.

\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{1}{4} \times 22 \times 28 = 22 \times 7 = 154\text{ cm}^2

Simplifying

\frac{90}{360}

yields

\frac{1}{4}

and dividing

196

by

7

gives

28

.

Key Concept

Area of a sector of a circle

Estimated Time:45s

Question 170Question →

A curve is defined by the equation $y = x^2 - 6x + 11$ . What is the minimum value of $y$ on this curve?

Show answer & explanation

Answer: 2

Answer

The minimum value of

y

on the curve is 2.

Differentiating

y = x^2 - 6x + 11

gives

\frac{dy}{dx} = 2x - 6

. Setting this derivative to zero yields

2x - 6 = 0

, so

x = 3

. Substituting

x = 3

into the original function gives

y = (3)^2 - 6(3) + 11 = 2

. Since

\frac{d^2y}{dx^2} = 2 > 0

, the point at

x = 3

is a local minimum, making 2 the minimum value of

y

.

Step-by-Step Solution

1

Find the first derivative of the curve function.

\frac{dy}{dx} = 2x - 6

Stationary points occur where the derivative is equal to zero.

2

Solve for the

x

-coordinate at the stationary point.

2x - 6 = 0 \implies x = 3

Setting the derivative to zero determines the input value where the slope is horizontal.

3

Calculate the corresponding

y

-value at

x = 3

.

y = (3)^2 - 6(3) + 11 = 2

Evaluating the original equation at

x = 3

yields the minimum value of

y

.

Key Concept

Finding the minimum value of a function using differentiation

Question 171Question →

In an experiment to determine the density of a solid sphere, the mass is measured as $(50.0 \pm 0.5)\text{ g}$ and the radius is measured as $(2.00 \pm 0.05)\text{ cm}$ . What is the percentage error in the calculated density of the sphere?

Show answer & explanation

Answer: 8.5

Answer

The percentage error in the calculated density of the sphere is

8.5\%

.

The density formula

\rho = \frac{3m}{4\pi r^3}

dictates that maximum relative error is given by

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right)

. Evaluating the percentage errors gives

1.0\%

for mass and

2.5\%

for radius. Summing

1.0\% + 3(2.5\%)

yields

8.5\%

.

Step-by-Step Solution

1

Determine the functional dependence of density on measured quantities.

Density

\rho = \frac{m}{V} = \frac{3m}{4\pi r^3}

.

The volume of a sphere of radius

r

is

V = \frac{4}{3}\pi r^3

.

2

Formulate the maximum fractional error relationship.

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right).

When combining uncertainties, fractional errors add, and exponents act as multiplying factors.

3

Calculate the percentage error in the mass measurement.

\frac{0.5}{50.0} \times 100\% = 1.0\%.

Percentage error in mass is the absolute uncertainty divided by the measured value multiplied by

100\%

.

4

Calculate the percentage error in the radius measurement.

\frac{0.05}{2.00} \times 100\% = 2.5\%.

Percentage error in radius is the absolute uncertainty divided by the measured value multiplied by

100\%

.

5

Calculate total percentage error in density.

Percentage error = 1.0\% + 3(2.5\%) = 8.5\%.

The radius contributes three times its relative error because volume depends on

r^3

.

Key Concept

Error propagation in derived quantities involving powers

Question 172Question →

A point $P$ lies inside a circle of radius $13\text{ cm}$ at a distance of $5\text{ cm}$ from the center $O$ . A chord $AB$ passes through point $P$ such that the ratio of segment $AP$ to segment $PB$ is $1:4$ . Calculate the total length of the chord $AB$ in centimeters.

Show answer & explanation

Answer: 30

Answer

The total length of chord

AB

is

30\text{ cm}

.

Using the Power of a Point property for an interior point

P

, the product of the chord segments is

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 144

. Given

AP : PB = 1 : 4

, we write

AP = x

and

PB = 4x

, leading to

4x^2 = 144 \implies x = 6\text{ cm}

. Summing the two segments gives

AB = 6 + 24 = 30\text{ cm}

.

Step-by-Step Solution

1

Calculate the constant product of chord segments passing through interior point

P

.

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144

.

By the intersecting chords theorem, the product of segments created by an interior point

P

on any chord equals

(R - d)(R + d) = R^2 - d^2

.

2

Set up an algebraic equation using the segment ratio

AP : PB = 1 : 4

.

Let

AP = x

and

PB = 4x

, giving

(x)(4x) = 144 \implies 4x^2 = 144

.

Expressing both chord segments in terms of a single variable

x

allows direct calculation of the segment lengths.

3

Solve for

x

to find the individual segment lengths.

x^2 = 36 \implies x = 6\text{ cm}

. Therefore,

AP = 6\text{ cm}

and

PB = 24\text{ cm}

.

Taking the positive square root gives the scale factor

x

since physical distances must be positive.

4

Sum the segment lengths to find the total chord length.

AB = AP + PB = 6 + 24 = 30\text{ cm}

.

The entire chord length is the sum of its two divided parts.

Key Concept

Intersecting Chords Theorem and Power of an Interior Point

Question 173Question →

Given that the determinant of the $3 \times 3$ matrix $M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix}$ is equal to $20$ , calculate the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

Expanding the matrix

M

along its top row yields

\det(M) = 3(2x + 4) - 1(4) = 6x + 8

. Setting this expression equal to

20

gives

6x + 8 = 20

, which simplifies to

6x = 12

, yielding

x = 2

.

Step-by-Step Solution

1

Expand the 3x3 matrix along the first row

\det(M) = 3(2x - (-4)) - 1(4 - 0) + 0

Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.

2

Simplify the algebraic expression for the determinant

\det(M) = 6x + 8

Distribute the coefficients and combine like constant terms.

3

Solve the linear equation for x

x = 2

Subtract 8 from 20 to get 12, then divide by 6.

Key Concept

Determinant of a 3x3 Matrix via Cofactor Expansion

Question 174Question →

A solid object of volume $0.002\text{ m}^3$ is completely immersed in water of density $1000\text{ kg/m}^3$ . What is the magnitude of the upthrust exerted on the object by the water? [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The magnitude of the upthrust exerted on the object is

20\text{ N}

.

According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using

U = V \cdot \rho \cdot g

. Substituting

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

yields

U = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement.

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

.

These are the essential inputs required to calculate the weight of the displaced liquid.

2

Apply Archimedes' principle to find upthrust force.

U = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Key Concept

Archimedes' Principle and Upthrust

Question 175Question →

What is the numerical value of the simplified expression $\frac{4}{\sqrt{5} - 1} - \sqrt{5}$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of the expression is 1.

Multiplying the top and bottom of

\frac{4}{\sqrt{5} - 1}

by its conjugate

(\sqrt{5} + 1)

simplifies the fraction to

\frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1

. Subtracting

\sqrt{5}

from

\sqrt{5} + 1

leaves 1.

Step-by-Step Solution

1

Rationalize the denominator of the fractional term

\sqrt{5} + 1

Multiply both numerator and denominator by the conjugate

(\sqrt{5} + 1)

to apply the difference of two squares identity

(a-b)(a+b) = a^2 - b^2

in the denominator.

2

Subtract the remaining surd term

1

Subtract

\sqrt{5}

from

\sqrt{5} + 1

, leaving the integer 1.

Key Concept

Rationalization of Binomial Denominators

Question 176Question →

Calculate the area, in square units, of the triangle formed by the straight line $3x + 4y - 24 = 0$ and the coordinate axes.

Show answer & explanation

Answer: 24

Answer

The area of the triangle formed by the line and the coordinate axes is 24 square units.

To find the area of the triangle bounded by a straight line and the coordinate axes, determine the magnitude of the

x

-intercept and

y

-intercept. Setting

y = 0

in

3x + 4y - 24 = 0

gives

x = 8

. Setting

x = 0

gives

y = 6

. The vertices of the right triangle are at

(0,0)

,

(8,0)

, and

(0,6)

. The area is calculated as

\frac{1}{2} \times 8 \times 6 = 24

square units.

Step-by-Step Solution

1

Find the

x

-intercept of the straight line

x = 8

, corresponding to the point

(8, 0)

Setting

y = 0

determines where the line crosses the horizontal axis

2

Find the

y

-intercept of the straight line

y = 6

, corresponding to the point

(0, 6)

Setting

x = 0

determines where the line crosses the vertical axis

3

Compute the area of the right-angled triangle formed with the origin

(0,0)

\text{Area} = \frac{1}{2} \times 8 \times 6 = 24

The coordinate axes are perpendicular, making the triangle right-angled with base length 8 and height 6

Key Concept

Area of a triangle bounded by a straight line and the coordinate axes

Estimated Time:1m 0s

Question 177Question →

A small business recorded the number of customer inquiries received per day over six consecutive days as follows: $4$ , $7$ , $8$ , $11$ , $13$ , and $17$ . What is the variance of the daily customer inquiries?

Show answer & explanation

Answer: 18

Answer

The variance of the daily customer inquiries is

18

.

To find the variance of the data set

\{4, 7, 8, 11, 13, 17\}

, first calculate the mean:

\bar{x} = \frac{4+7+8+11+13+17}{6} = 10

. Next, compute the squared deviation of each data point from the mean:

(4-10)^2 = 36

,

(7-10)^2 = 9

,

(8-10)^2 = 4

,

(11-10)^2 = 1

,

(13-10)^2 = 9

, and

(17-10)^2 = 49

. Summing these squared deviations gives

108

. Dividing this total by the number of observations (

6

) yields a variance of

18

.

Step-by-Step Solution

1

Calculate the arithmetic mean of the given data set.

\bar{x} = 10

The mean is required as the central point from which deviations are calculated.

2

Determine the squared deviation of each data value from the mean.

(-6)^2 = 36

,

(-3)^2 = 9

,

(-2)^2 = 4

,

1^2 = 1

,

3^2 = 9

,

7^2 = 49

Variance measures the average squared distance of data points from the mean.

3

Sum all calculated squared deviations.

\sum (x - \bar{x})^2 = 36 + 9 + 4 + 1 + 9 + 49 = 108

This provides the total sum of squares for the data set.

4

Divide the total sum of squares by the number of observations (

n = 6

).

\sigma^2 = \frac{108}{6} = 18

Population variance formula is

\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}

.

Key Concept

Variance of Ungrouped Data

Estimated Time:1m 30s

Question 178Question →

Given that $F(x) = \int (3x^2 - 2\sin x) \, dx$ and $F(0) = 6$ , what is the value of the constant of integration $C$ ?

Show answer & explanation

Answer: 4

Answer

The value of the constant of integration

C

is

4

.

Integrating

3x^2 - 2\sin x

yields

F(x) = x^3 + 2\cos x + C

. Substituting

x = 0

gives

F(0) = 2(1) + C = 2 + C

. Since

F(0) = 6

, setting

2 + C = 6

yields

C = 4

.

Step-by-Step Solution

1

Integrate the function

f(x) = 3x^2 - 2\sin x

with respect to

x

F(x) = x^3 + 2\cos x + C

The antiderivative of

3x^2

is

x^3

and the antiderivative of

-2\sin x

is

2\cos x

.

2

Evaluate

F(0)

using the antiderivative expression

F(0) = 0^3 + 2\cos(0) + C = 2 + C

Since

\cos(0) = 1

, the term

2\cos(0)

simplifies to

2

.

3

Solve for the integration constant

C

using

F(0) = 6

C = 4

Subtracting

2

from both sides of

2 + C = 6

yields

C = 4

.

Key Concept

Indefinite integration of polynomial and trigonometric functions with initial conditions

Question 179Question →

A $0.50\text{ kg}$ mass attached to a horizontal spring undergoes simple harmonic motion on a frictionless surface. The total mechanical energy of the system is $0.16\text{ J}$ and the force constant of the spring is $32\text{ N/m}$ . What is the speed of the mass, in $\text{m/s}$ , at the instant when the magnitude of its acceleration is $3.84\text{ m/s}^2$ ?

Show answer & explanation

Answer: 0.64

Answer

The speed of the mass at that instant is

0.64\text{ m/s}

.

Using the relation for total mechanical energy

E = \frac{1}{2}kA^2

, the amplitude is

A = 0.10\text{ m}

. The angular frequency is

\omega = \sqrt{k/m} = 8.0\text{ rad/s}

. From

|a| = \omega^2 |x|

, the displacement magnitude when acceleration is

3.84\text{ m/s}^2

is

|x| = 0.06\text{ m}

. Substituting these values into

v = \omega \sqrt{A^2 - x^2}

yields

v = 8.0 \sqrt{0.10^2 - 0.06^2} = 0.64\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the simple harmonic motion

\omega = 8.0\text{ rad/s}

The angular frequency depends on the stiffness constant and the mass according to \omega = \sqrt{k/m}.

2

Calculate the amplitude of oscillation from total energy

A = 0.10\text{ m}

The total mechanical energy in SHM is given by E = \frac{1}{2}kA^2.

3

Find the magnitude of displacement corresponding to the given acceleration

|x| = 0.06\text{ m}

In SHM, acceleration magnitude is related to displacement magnitude by |a| = \omega^2 |x|.

4

Calculate the speed at this displacement using the SHM velocity-displacement relation

v = 0.64\text{ m/s}

Velocity in SHM is calculated using v = \omega \sqrt{A^2 - x^2}.

Key Concept

Interdependence of energy, angular frequency, acceleration, and velocity in Simple Harmonic Motion

Question 180Question →

A projectile is launched from ground level over flat terrain. At time $t = 2\text{ s}$ after launch, the projectile passes through a point located $60\text{ m}$ horizontally and $60\text{ m}$ vertically from its launch point. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the total horizontal range of the projectile in meters?

Show answer & explanation

Answer: 240

Answer

The total horizontal range of the projectile is

240\text{ m}

.

The horizontal motion occurs at a constant velocity of

30\text{ m/s}

calculated from

\frac{60\text{ m}}{2\text{ s}}

. Substituting the vertical position (

60\text{ m}

) and time (

2\text{ s}

) into

y = u_y t - 5t^2

yields an initial vertical velocity of

40\text{ m/s}

. The total duration in the air is

T = \frac{2(40)}{10} = 8\text{ s}

. The total horizontal range is therefore

30\text{ m/s} \times 8\text{ s} = 240\text{ m}

.

Step-by-Step Solution

1

Determine the horizontal component of velocity

v_x = 30\text{ m/s}

Horizontal velocity remains constant throughout flight because there is no horizontal acceleration.

2

Determine the initial vertical component of velocity

u_y = 40\text{ m/s}

Applying the vertical displacement equation

y = u_y t - \frac{1}{2}g t^2

with

y = 60\text{ m}

,

t = 2\text{ s}

, and

g = 10\text{ m/s}^2

.

3

Calculate the total time of flight

T = 8\text{ s}

The projectile completes its full parabolic trajectory when vertical displacement returns to zero, given by

T = \frac{2 u_y}{g}

.

4

Calculate the total horizontal range

R = 240\text{ m}

The total range is the product of the constant horizontal velocity component and total time of flight (

R = v_x \times T

).

Key Concept

Independence of horizontal and vertical components of projectile motion

Answer

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