Tüm alıştırma soruları

Soru 1Soru →

Kemi and Chidi are partners in a accounting firm sharing profits and losses in the ratio of $3:2$ . They admit Ngozi into the partnership with a $\frac{1}{5}$ share in future profits. Ngozi pays $\text{₦}20,000$ as premium for goodwill, which is credited to the existing partners' capital accounts in their profit-sharing ratio. What is the amount of goodwill premium credited to Kemi's capital account (in $\text{₦}$ )?

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Cevap: 12000

Cevap

The amount credited to Kemi's capital account for goodwill premium is ₦12,000.

Kemi's share of the goodwill premium is calculated by taking her 3/5 profit share of the total ₦20,000 premium brought in by the incoming partner, which equals ₦12,000.

Adım Adım Çözüm

1

Determine Kemi's ratio share

3/5

The profit-sharing ratio between Kemi and Chidi is 3:2, meaning Kemi receives 3 out of 5 total parts.

2

Calculate Kemi's portion of the goodwill premium

₦12,000

Multiply total premium (₦20,000) by Kemi's share (3/5): 20,000 * 3 / 5 = 12,000.

Anahtar Kavram

Allocation of Goodwill Premium on Admission of a New Partner

Soru 2Soru →

The electricity account of Chukwuma Enterprises showed a trial balance debit balance of $\text{₦}240,000$ for the year ended 31 December 2025. At the beginning of the year on 1 January 2025, electricity accrued was $\text{₦}18,000$ and prepaid electricity was $\text{₦}12,000$ . At the end of the year on 31 December 2025, electricity accrued was $\text{₦}25,000$ while prepaid electricity was $\text{₦}15,000$ . What is the total electricity expense (in ₦) to be debited to the Profit and Loss Account for the year ended 31 December 2025?

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Cevap: 244000

Cevap

The total electricity expense to be debited to the Profit and Loss Account is

\text{₦}244,000

.

Under accrual accounting, the expense charged for the period equals cash paid during the year plus opening prepayments minus opening accruals plus closing accruals minus closing prepayments:

240,000 + 12,000 - 18,000 + 25,000 - 15,000 = \text{₦}244,000

.

Adım Adım Çözüm

1

Identify the total cash paid during the financial year.

Amount paid =

\text{₦}240,000

.

This is the starting cash outflow recorded in the trial balance.

2

Adjust for opening balances brought forward from the previous period.

Add opening prepayment of

\text{₦}12,000

and subtract opening accrual of

\text{₦}18,000

.

Prepayments from the prior year belong to the current period's expense, whereas accruals from the prior year were already expensed in that period and paid in the current period.

3

Adjust for closing balances at the end of the current period.

Add closing accrual of

\text{₦}25,000

and subtract closing prepayment of

\text{₦}15,000

.

Accrued expenses incurred in the current period must be added, and prepaid expenses paid for the next period must be deducted under accrual accounting concepts.

4

Calculate the net charge to the Profit and Loss Account.

240,000 + 12,000 - 18,000 + 25,000 - 15,000 = \text{₦}244,000

.

This represents the exact electricity expense incurred during the financial year 2025.

Anahtar Kavram

Accrual Accounting Adjustment for Expenses

Soru 3Soru →

At 31st December 2025, Chukwu Limited had a total trade debtors balance of ₦120,000. During the year, bad debts amounting to ₦8,000 were written off, and a debt of ₦3,000 previously written off was recovered in cash. What is the net amount of trade debtors (in ₦) to be presented in the Statement of Financial Position?

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Cevap: 112000

Cevap

112,000

The net trade debtors figure in the Statement of Financial Position is calculated by subtracting bad debts written off (₦8,000) from the gross trade debtors balance (₦120,000), giving ₦112,000. Bad debts recovered (₦3,000) are recorded as income in the Profit and Loss Account and do not affect the closing trade debtors balance.

Adım Adım Çözüm

1

Identify initial trade debtors balance

Initial trade debtors = ₦120,000

This is the unadjusted trade debtors balance before accounting for bad debts written off.

2

Deduct bad debts written off

Net trade debtors = ₦120,000 - ₦8,000 = ₦112,000

Bad debts written off represent debts that are irrecoverable and must be removed from total trade receivables.

3

Analyze the impact of bad debts recovered

Net trade debtors remains ₦112,000

Bad debts recovered (₦3,000) are credited to income in the Profit and Loss Account and debited to cash/bank, so they have no effect on closing trade debtors.

Anahtar Kavram

Calculation of net trade debtors after bad debts written off and recovered

Soru 4Soru →

The following financial figures were extracted from the books of Kemi Manufacturing Enterprises for the year ended 31st December 2025:

- Prime Cost: $\text{N}150,000$
- Factory Overheads: $\text{N}45,000$
- Opening Work-in-Progress: $\text{N}12,000$
- Closing Work-in-Progress: $\text{N}17,000$

What is the Cost of Production for the year?

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Cevap: 190000

Cevap

The Cost of Production is 190,000 Naira.

The Cost of Production is calculated using the formula: Prime Cost + Factory Overheads + Opening Work-in-Progress - Closing Work-in-Progress. Substituting the values:

\text{N}150,000 + \text{N}45,000 + \text{N}12,000 - \text{N}17,000 = \text{N}190,000

.

Adım Adım Çözüm

1

Add Factory Overheads to Prime Cost

\text{N}150,000 + \text{N}45,000 = \text{N}195,000

Factory overheads are added to prime cost to determine the total factory cost before work-in-progress adjustments.

2

Add Opening Work-in-Progress

\text{N}195,000 + \text{N}12,000 = \text{N}207,000

Opening work-in-progress represents unfinished goods from the previous period completed in the current period.

3

Deduct Closing Work-in-Progress

\text{N}207,000 - \text{N}17,000 = \text{N}190,000

Closing work-in-progress represents unfinished goods at the end of the period and must be deducted to find the cost of fully produced goods.

Anahtar Kavram

Calculation of Cost of Production from Prime Cost, Factory Overheads, and Work-in-Progress adjustments.

Soru 5Soru →

A U-tube open at both ends contains mercury of density $13\text{}600\text{ kg/m}^3$ . Water of density $1000\text{ kg/m}^3$ is poured into one arm until the water column reaches a height of $27.2\text{ cm}$ . What is the difference in height, in $\text{cm}$ , between the mercury surfaces in the two arms?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The difference in height between the mercury surfaces in the two arms is

2.0\text{ cm}

.

At the boundary level where water meets mercury, the pressure produced by the

27.2\text{ cm}

water column must equal the pressure of the mercury column above that same horizontal level. Using

h_w \rho_w = h_m \rho_m

, we solve for the mercury height difference:

h_m = \frac{27.2 \times 1000}{13600} = 2.0\text{ cm}

.

Adım Adım Çözüm

1

Equate the hydrostatic pressure exerted by the water column to the hydrostatic pressure exerted by the balancing mercury column at the interface level.

h_w \rho_w g = h_m \rho_m g

At the same horizontal level within a continuous fluid at rest, the pressures must be equal.

2

Cancel the acceleration due to gravity (

g

) from both sides of the equation.

h_w \rho_w = h_m \rho_m

Gravity acts equally on both liquid columns.

3

Substitute the known values (

h_w = 27.2\text{ cm}

,

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

,

\rho_m = 13600\text{ kg/m}^3

) into the pressure relation.

27.2 \times 1000 = h_m \times 13600

Inserting the physical quantities isolates the unknown mercury column height

h_m

.

4

Solve for the height difference

h_m

of the mercury levels.

h_m = \frac{27200}{13600} = 2.0\text{ cm}

Dividing the water pressure head product by the density of mercury yields the height of the mercury column.

Anahtar Kavram

Hydrostatic pressure equilibrium in immiscible fluids (U-tube manometer)

Soru 6Soru →

An electron in a hydrogen atom undergoes a transition from an energy state of $-1.51\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . What is the energy of the emitted photon in electron-volts ( $\text{eV}$ )?

Cevabı ve açıklamayı göster

Cevap: 1.89

Cevap

The energy of the emitted photon is

1.89\text{ eV}

.

The energy of an emitted photon during an atomic transition is given by

\Delta E = E_{\text{initial}} - E_{\text{final}}

. Substituting the given levels yields

\Delta E = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

.

Adım Adım Çözüm

1

Identify the initial and final energy states.

E_i = -1.51\text{ eV}

and

E_f = -3.40\text{ eV}

The energy of the photon corresponds to the difference between these two levels.

2

Apply the energy conservation formula for atomic emission.

E_{\text{photon}} = E_i - E_f

When an electron drops to a lower energy level, a photon carrying the lost energy is released.

3

Substitute the values and evaluate the difference.

E_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}

Subtracting the negative lower energy value yields a positive photon energy.

Anahtar Kavram

Photon energy from atomic level transitions

Soru 7Soru →

A binary operation $\otimes$ on the set of real numbers $\mathbb{R}$ is defined by $a \otimes b = a^2 + b^2 - ab$ . If $x \otimes 3 = 19$ and $x > 0$ , find the value of $x$ .

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Cevap: 5

Cevap

The value of

x

is

5

.

Applying the operation rule gives

x^2 + 3^2 - 3x = 19

, which simplifies to

x^2 - 3x - 10 = 0

. Factoring this equation yields

(x - 5)(x + 2) = 0

. Since

x

is constrained to be positive (

x > 0

), the unique valid answer is

5

.

Adım Adım Çözüm

1

Apply the definition of the binary operation to

x \otimes 3

x^2 + 3^2 - 3(x) = x^2 - 3x + 9

Substitute

a = x

and

b = 3

into

a \otimes b = a^2 + b^2 - ab

.

2

Equate the result to 19 and rearrange into standard quadratic form

x^2 - 3x - 10 = 0

Subtract 19 from both sides to set the quadratic equation to zero.

3

Factor the quadratic equation and solve for

x

(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2

Find two numbers that multiply to

-10

and add up to

-3

.

4

Apply the restriction

x > 0

x = 5

Reject the negative solution

x = -2

because

x

must be strictly positive.

Anahtar Kavram

Evaluation of Binary Operations and Solving Quadratic Equations

Soru 8Soru →

On a warm afternoon, the air temperature in a physics laboratory is $30^\circ\text{C}$ , where the saturated vapour pressure of water is $32.0\text{ mmHg}$ . When the air is cooled, condensation just begins to form on a metal vessel at $20^\circ\text{C}$ . Given that the saturated vapour pressure of water at $20^\circ\text{C}$ is $17.6\text{ mmHg}$ , what is the relative humidity of the air in percentage?

Cevabı ve açıklamayı göster

Cevap: 55

Cevap

The relative humidity of the air is

55\%

.

The dew point is the temperature at which condensation begins, indicating that the actual water vapour pressure present in the air equals the saturated vapour pressure at

20^\circ\text{C}

, which is

17.6\text{ mmHg}

. Dividing this actual vapour pressure by the saturated vapour pressure at the ambient air temperature of

30^\circ\text{C}

(

32.0\text{ mmHg}

) and multiplying by

100\%

yields

\frac{17.6}{32.0} \times 100\% = 55\%

.

Adım Adım Çözüm

1

Determine the actual vapour pressure in the air

Actual vapour pressure =

17.6\text{ mmHg}

Condensation starts at the dew point (

20^\circ\text{C}

), meaning the actual vapour pressure in the air equals the saturated vapour pressure at the dew point.

2

Determine the saturated vapour pressure at the air temperature

Saturated vapour pressure at

30^\circ\text{C}

=

32.0\text{ mmHg}

This is the maximum vapour pressure the air can exert at its current ambient temperature.

3

Compute the relative humidity percentage

\text{Relative Humidity} = \frac{17.6}{32.0} \times 100\% = 55\%

Relative humidity is defined as the ratio of actual vapour pressure to saturated vapour pressure at air temperature, expressed as a percentage.

Anahtar Kavram

Calculation of relative humidity from saturated vapour pressure at dew point and air temperature

Tahmini Süre:1m 30s

Soru 9Soru →

A metal $X$ forms two distinct chlorides. Quantitative analysis shows that in Chloride 1, $5.40\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. In Chloride 2, $3.60\text{ g}$ of $X$ combines with $10.65\text{ g}$ of chlorine. If the empirical formula of Chloride 1 is $XCl_2$ , calculate the subscript value $y$ in the empirical formula $XCl_y$ of Chloride 2.

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of the subscript y is 3.

Applying the Law of Multiple Proportions, when a fixed mass of chlorine (

10.65\text{ g}

) reacts with different masses of metal

X

(

5.40\text{ g}

and

3.60\text{ g}

), the mass ratio of

X

is

5.40 : 3.60 = 3 : 2

. This implies that for a fixed amount of metal

X

, the ratio of chlorine atoms in Chloride 1 to Chloride 2 is

2 : 3

. Since Chloride 1 is

XCl_2

, Chloride 2 must be

XCl_3

, yielding

y = 3

.

Adım Adım Çözüm

1

Determine the mass of chlorine per gram of metal X in Chloride 1.

\frac{10.65\text{ g}}{5.40\text{ g}} = 1.9722\text{ g } Cl / \text{g } X

This establishes the baseline quantitative relationship for the formula

XCl_2

.

2

Determine the mass of chlorine per gram of metal X in Chloride 2.

\frac{10.65\text{ g}}{3.60\text{ g}} = 2.9583\text{ g } Cl / \text{g } X

This determines the mass of chlorine per unit mass of metal in the second compound.

3

Calculate the simple multiple proportion ratio between the two compounds.

\frac{2.9583}{1.9722} = 1.5

According to the Law of Multiple Proportions, the masses of chlorine combining with a fixed mass of X stand in a simple whole-number ratio.

4

Multiply the subscript of chlorine in the first compound by the calculated ratio.

y = 2 \times 1.5 = 3

Since Chloride 1 has 2 chlorine atoms (

XCl_2

), Chloride 2 must have

2 \times 1.5 = 3

chlorine atoms (

XCl_3

).

Anahtar Kavram

Law of Multiple Proportions

Soru 10Soru →

The aerodynamic drag force $F$ acting on an object moving through a fluid of density $\rho$ with cross-sectional area $A$ at speed $v$ is modeled by the equation $F = \frac{1}{2} C_d \rho^a A^b v^c$ , where $C_d$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the exponent $c$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The numerical value of the exponent

c

is 2.

By applying the principle of dimensional homogeneity, the base dimension of time on the left side is

\text{T}^{-2}

(from force

[F] = \text{M L T}^{-2}

). On the right side, the only quantity containing time is velocity

[v] = \text{L T}^{-1}

, raised to power

c

, giving

\text{T}^{-c}

. Equating the exponents gives

-2 = -c

, so

c = 2

.

Adım Adım Çözüm

1

Identify the base dimensions of each physical quantity in the given equation.

[F] = \text{M L T}^{-2}

,

[\rho] = \text{M L}^{-3}

,

[A] = \text{L}^2

, and

[v] = \text{L T}^{-1}

.

C_d

is dimensionless (

[C_d] = 1

).

Dimensional homogeneity requires both sides of a physical equation to have identical base dimensions.

2

Substitute the base dimensions into the formula

F = \frac{1}{2} C_d \rho^a A^b v^c

and simplify.

\text{M L T}^{-2} = (\text{M L}^{-3})^a (\text{L}^2)^b (\text{L T}^{-1})^c = \text{M}^a \text{L}^{-3a + 2b + c} \text{T}^{-c}.

Combining powers of base dimensions allows direct comparison of corresponding exponents.

3

Equate the exponent of time (T) on both sides of the dimensional equation.

-2 = -c \implies c = 2.

The exponent of T on the left side is -2, which must equal the exponent of T on the right side (-c).

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

Soru 11Soru →

In a resonance tube experiment using a tuning fork of constant frequency, the first two consecutive resonant lengths of the air column above the water level are measured to be $23.5\text{ cm}$ and $73.5\text{ cm}$ respectively. What is the end correction of the tube in centimeters?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The end correction of the tube is

1.5\text{ cm}

.

In a resonance tube closed at one end by water, consecutive resonances occur when the air column length increases by half a wavelength. Subtracting the first resonant length from the second gives

\frac{\lambda}{2} = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}

, which yields

\lambda = 100.0\text{ cm}

and

\frac{\lambda}{4} = 25.0\text{ cm}

. The first resonance condition accounts for end correction through

L_1 + e = \frac{\lambda}{4}

. Substituting

L_1 = 23.5\text{ cm}

gives

e = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}

.

Adım Adım Çözüm

1

Determine the wavelength using consecutive resonant positions

$\frac{\lambda}{2} = L_2 - L_1 = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}$, so $\lambda = 100.0\text{ cm}$

For a column closed at one end, consecutive resonances occur at intervals of half a wavelength.

2

Calculate the quarter-wavelength value

$\frac{\lambda}{4} = \frac{100.0\text{ cm}}{4} = 25.0\text{ cm}$

The fundamental mode position of the displacement antinode corresponds to a distance of one quarter-wavelength from the closed end.

3

Calculate the end correction

$e = \frac{\lambda}{4} - L_1 = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}$

The effective length for the first resonance includes the physical length plus the end correction.

Anahtar Kavram

End Correction in Resonance Air Columns

Soru 12Soru →

The perimeter of a rectangular playfield is $28\text{ m}$ and its area is $40\text{ m}^2$ . What is the positive difference, in metres, between its length and width?

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Cevap: 6

Cevap

The positive difference between the length and width of the playfield is 6 metres.

Formulating the system gives

x + y = 14

and

xy = 40

. Substituting

y = 14 - x

yields the quadratic equation

x^2 - 14x + 40 = 0

, which factors into

(x - 10)(x - 4) = 0

. The dimensions are

10\text{ m}

and

4\text{ m}

, giving a positive difference of

10 - 4 = 6\text{ m}

.

Adım Adım Çözüm

1

Set up linear and quadratic equations for perimeter and area

x + y = 14

and

xy = 40

Perimeter formula is

2(x + y) = 28

which simplifies to

x + y = 14

, and area formula is

xy = 40

.

2

Substitute

y = 14 - x

into the quadratic area equation

x(14 - x) = 40 \implies x^2 - 14x + 40 = 0

Substitution reduces the simultaneous system to a single quadratic equation in terms of

x

.

3

Solve the quadratic equation by factoring

(x - 10)(x - 4) = 0 \implies x = 10 \text{ or } x = 4

The roots of the equation give the dimensions of the rectangle.

4

Calculate the positive difference between the two dimensions

10 - 4 = 6

Subtract the smaller dimension from the larger dimension.

Anahtar Kavram

Solving word problems involving simultaneous linear and quadratic equations

Tahmini Süre:1m 30s

Soru 13Soru →

Given the $3 \times 3$ matrix $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}$ , find the positive value of $x$ for which $\det(A) = -2$ .

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The positive value of

x

is 2.5.

Expanding the determinant of matrix

A

along the third row gives

1(4 - 1) - x(2x - 3) = -2x^2 + 3x + 3

. Setting this equal to

-2

yields

2x^2 - 3x - 5 = 0

. Factoring gives

(2x - 5)(x + 1) = 0

, yielding solutions

x = 2.5

and

x = -1

. Taking the positive value gives

x = 2.5

.

Adım Adım Çözüm

1

Calculate the determinant of matrix

A

in terms of

x

\det(A) = -2x^2 + 3x + 3

Expanding along the third row simplifies computation because of the zero entry.

2

Set the determinant expression equal to

-2

and rearrange terms

2x^2 - 3x - 5 = 0

Setting

-2x^2 + 3x + 3 = -2

forms a standard quadratic equation.

3

Factorize the quadratic equation to find the roots

x = 2.5

or

x = -1

Factoring

(2x - 5)(x + 1) = 0

yields two real solutions.

4

Filter for the positive value requested in the stem

x = 2.5

The question specifically requires the positive value of

x

.

Anahtar Kavram

Determinant of a 3x3 Matrix and Quadratic Equation Solving

Tahmini Süre:2m 30s

Soru 14Soru →

The table below shows the distribution of quiz scores for a class of students:

Score Interval	Frequency ( $f$ )
$1 - 5$	$3$
$6 - 10$	$5$
$11 - 15$	$7$
$16 - 20$	$5$

Calculate the mean score of the distribution.

Cevabı ve açıklamayı göster

Cevap: 11.5

Cevap

The mean score of the distribution is 11.5.

The mean for a grouped frequency table is calculated by taking the sum of the products of each midpoint (

x

) and its frequency (

f

), divided by the sum of all frequencies (

f

). Here,

\sum fx = 230

and

\sum f = 20

, yielding

\bar{x} = 11.5

.

Adım Adım Çözüm

1

Determine the class midpoints (

x

) for each interval.

Midpoints are 3, 8, 13, and 18.

Midpoints serve as the representative values for each class interval.

2

Calculate the product of each midpoint and frequency (

fx

).

Products are 9, 40, 91, and 90.

To find the total contribution of each class interval.

3

Find the sum of all frequencies (

\sum f

) and products (

\sum fx

).

\sum f = 20

and

\sum fx = 230

.

These totals are required for the grouped mean formula.

4

Divide the total product sum by total frequency.

\bar{x} = \frac{230}{20} = 11.5

.

Applying the formula for the mean of grouped data

\bar{x} = \frac{\sum fx}{\sum f}

.

Anahtar Kavram

Mean of Grouped Data using Class Midpoints

Soru 15Soru →

Find the smallest non-negative integer $x$ that satisfies the linear modular congruence $3x \equiv 8 \pmod{11}$ .

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The smallest non-negative integer

x

is 10.

Evaluating

3x \equiv 8 \pmod{11}

by testing multiples of

11

added to

8

gives

30

, which divided by

3

yields

x = 10

. Since

10 \in [0, 10]

, it is the canonical solution.

Adım Adım Çözüm

1

Convert the modular congruence into an algebraic equation

3x = 8 + 11k

for an integer

k

By definition of congruence modulo

11

,

3x - 8

must be a multiple of

11

.

2

Find the smallest integer

k \ge 0

such that

8 + 11k

is divisible by 3

When

k = 2

,

8 + 11(2) = 30

30

is divisible by

3

(

30 / 3 = 10

).

3

Divide by 3 to isolate

x

x = 10

3(10) = 30 \equiv 8 \pmod{11}

.

Anahtar Kavram

Linear Modular Congruence

Tahmini Süre:1m 15s

Soru 16Soru →

Find the real value of $x$ that satisfies the exponential equation $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$ .

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The value of

x

is

1.5

(or

\frac{3}{2}

).

By using index laws to group base-2 terms on the left side and base-3 terms on the right side, we obtain

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

. Rearranging gives

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}

. Equating indices gives

x = 1.5

.

Adım Adım Çözüm

1

Group like exponential terms with base 2 and base 3 on opposite sides of the equation.

4^x + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}

Grouping terms with common prime bases allows for factoring exponential terms.

2

Apply the product and power laws of indices:

4^x = 2^{2x}

,

2^{2x-1} = 2^{-1} \cdot 2^{2x}

,

3^{x \pm \frac{1}{2}} = 3^x \cdot 3^{\pm \frac{1}{2}}

.

2^{2x} + \frac{1}{2} \cdot 2^{2x} = 3^x \cdot \sqrt{3} + 3^x \cdot \frac{1}{\sqrt{3}}

Separating the variable exponents from constant exponents prepares each side for factoring.

3

Factor out

2^{2x}

from the left side and

3^x

from the right side, then simplify arithmetic terms.

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

Factoring isolates the variable terms

2^{2x}

and

3^x

from numerical constants.

4

Divide to form the ratio

\frac{4^x}{3^x} = \left(\frac{4}{3}\right)^x

and simplify the numerical fraction on the right.

\left(\frac{4}{3}\right)^x = \frac{8}{3\sqrt{3}}

Expressing both sides with unified variable bases facilitates solving for

x

by equating powers.

5

Rewrite

\frac{8}{3\sqrt{3}}

as a power of

\frac{4}{3}

and solve for

x

.

\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{\frac{3}{2}} \implies x = \frac{3}{2} = 1.5

Since

\frac{8}{3\sqrt{3}} = \frac{4^{3/2}}{3^{3/2}} = (\frac{4}{3})^{3/2}

, equating exponents yields

x = 1.5

.

Anahtar Kavram

Solving mixed-base exponential equations by grouping, factoring, and converting to a unified base ratio.

Tahmini Süre:3m 0s

Soru 17Soru →

A particle starts from rest and accelerates uniformly at a rate of $4\text{ m/s}^2$ for a duration $t_1$ . Immediately after reaching its maximum velocity, it decelerates uniformly at $2\text{ m/s}^2$ until coming to rest. If the total distance covered during the entire motion is $600\text{ m}$ , what is the total time of motion in seconds?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The total time of motion is 30 seconds.

For a two-stage motion starting and ending at rest, the peak velocity is

v_{\text{max}} = a_1 t_1 = a_2 t_2

, giving a time ratio

t_2 / t_1 = a_1 / a_2 = 4 / 2 = 2

. The total distance

S

is the area under the velocity-time triangle,

S = \frac{1}{2} v_{\text{max}} (t_1 + t_2) = \frac{1}{2} (4 t_1) (3 t_1) = 6 t_1^2

. Setting

6 t_1^2 = 600

yields

t_1 = 10\text{ s}

, which gives a total time

T = t_1 + t_2 = 30\text{ s}

.

Adım Adım Çözüm

1

Relate maximum velocity to the acceleration time

t_1

v_{\text{max}} = 4 t_1

Using

v = u + a t

starting from rest (

u = 0

).

2

Relate deceleration time

t_2

to

t_1

t_2 = 2 t_1

The final velocity is

0

, so

0 = v_{\text{max}} - a_2 t_2 \implies t_2 = \frac{4 t_1}{2} = 2 t_1

.

3

Express the total displacement

S

as a function of

t_1

S = 6 t_1^2

Displacement during acceleration

s_1 = \frac{1}{2}(4)t_1^2 = 2 t_1^2

. Displacement during deceleration

s_2 = \frac{1}{2}(2)(2 t_1)^2 = 4 t_1^2

. Total

S = 2 t_1^2 + 4 t_1^2 = 6 t_1^2

.

4

Solve for the acceleration time

t_1

t_1 = 10\text{ s}

Given

S = 600\text{ m}

, we have

6 t_1^2 = 600 \implies t_1^2 = 100 \implies t_1 = 10\text{ s}

.

5

Calculate the total time of motion

T

T = 30\text{ s}

Total time is the sum of both phases:

T = t_1 + t_2 = t_1 + 2 t_1 = 3 t_1 = 3(10) = 30\text{ s}

.

Anahtar Kavram

Multi-stage uniform motion and average velocity relations

Tahmini Süre:3m 0s

Soru 18Soru →

A curve is defined by the equation $y = x^2 - 5x + 6$ . What is the $x$ -intercept of the line normal to the curve at the point where $x = 1$ ?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The x-intercept of the normal line to the curve at x = 1 is -5.

Substituting

x = 1

into

y = x^2 - 5x + 6

yields

y = 2

, identifying the point

(1, 2)

. Differentiating gives

\frac{dy}{dx} = 2x - 5

, which equals

-3

at

x = 1

. The normal line gradient is the negative reciprocal,

\frac{1}{3}

. The line equation

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

simplifies to

x - 3y + 5 = 0

. Setting

y = 0

gives

x = -5

.

Adım Adım Çözüm

1

Calculate the y-coordinate at x = 1 to determine the point of tangency

Substituting

x = 1

into

y = x^2 - 5x + 6

gives

y = (1)^2 - 5(1) + 6 = 2

, yielding the point

(1, 2)

.

The normal line intersects the curve at the point of tangency.

2

Find the derivative of the curve and evaluate the tangent slope

\frac{dy}{dx} = 2x - 5

. At

x = 1

,

m_t = 2(1) - 5 = -3

.

The derivative evaluated at a specific point gives the slope of the tangent line to the curve.

3

Compute the slope of the normal line

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

.

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent gradient.

4

Construct the normal line equation and determine its x-intercept

Using point-slope form:

y - 2 = \frac{1}{3}(x - 1) \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0

. Setting

y = 0

gives

x + 5 = 0 \implies x = -5

.

The

x

-intercept occurs where the line crosses the

x

-axis, meaning

y = 0

.

Anahtar Kavram

The slope of the normal line to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative at that point,

m_n = -\frac{1}{f'(a)}

.

Soru 19Soru →

If the matrix $P = \begin{pmatrix} k & 3 & 1 \\ 2 & 1 & k \\ 4 & 2 & 0 \end{pmatrix}$ is singular, find the non-zero value of $k$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The non-zero value of

k

is 6.

For matrix P to be singular, its determinant must be 0. Expanding along row 3 yields 4(3k - 1) - 2(k^2 - 2) = 12k - 4 - 2k^2 + 4 = -2k^2 + 12k = 0. Factoring gives -2k(k - 6) = 0, which yields k = 0 or k = 6. The non-zero value is 6.

Adım Adım Çözüm

1

Set the determinant of matrix P to 0

\det(P) = 0

A matrix is singular if and only if its determinant equals zero.

2

Evaluate the 3x3 determinant by expanding along the third row

4 \cdot (3k - 1) - 2 \cdot (k^2 - 2) + 0 = 0

Expanding along the third row takes advantage of the zero entry to simplify computation.

3

Expand and combine like terms

-2k^2 + 12k = 0

12k - 4 - 2k^2 + 4 reduces to -2k^2 + 12k.

4

Factor out common factors and solve for k

-2k(k - 6) = 0 \implies k = 0 \text{ or } k = 6

Applying the zero-product property.

5

Select the required root

k = 6

The question specifies the non-zero value of k.

Anahtar Kavram

Determinant of a 3x3 matrix and singular matrix condition

Tahmini Süre:2m 0s

Soru 20Soru →

A body of mass $0.4\text{ kg}$ suspended vertically from a helical spring produces a static extension of $0.1\text{ m}$ . The body is then pulled down further and set into vertical simple harmonic motion with an amplitude of $0.05\text{ m}$ . What is the maximum velocity of the body in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The maximum velocity of the body during oscillation is

0.5\text{ m/s}

.

At static equilibrium, weight balances restoring force (

mg = ke

), giving

\frac{k}{m} = \frac{g}{e} = \frac{10}{0.1} = 100\text{ s}^{-2}

. The angular frequency is

\omega = \sqrt{\frac{k}{m}} = 10\text{ rad/s}

. In SHM, the maximum velocity occurs at the central equilibrium position and is given by

v_{\max} = \omega A = 10 \times 0.05 = 0.5\text{ m/s}

.

Adım Adım Çözüm

1

Relate spring stiffness to static extension

\frac{k}{m} = 100\text{ s}^{-2}

At vertical static equilibrium, the weight of the mass equals the restoring force:

mg = ke \implies \frac{k}{m} = \frac{g}{e} = \frac{10\text{ m/s}^2}{0.1\text{ m}} = 100\text{ s}^{-2}

.

2

Determine the angular frequency

\omega = 10\text{ rad/s}

The angular frequency of a mass-spring system is given by

\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10\text{ rad/s}

.

3

Calculate maximum velocity

v_{\max} = 0.5\text{ m/s}

The maximum speed in simple harmonic motion occurs at the equilibrium position and is computed using

v_{\max} = \omega A = 10\text{ rad/s} \times 0.05\text{ m} = 0.5\text{ m/s}

.

Anahtar Kavram

Maximum velocity and angular frequency derived from static extension in Simple Harmonic Motion

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