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A business records a credit sale of Rs. 350 (no cash involved) using accrual accounting. How will this transaction be reflected in the financial statements?
Under accrual accounting, revenue is recognized when it is earned, not when cash is received. A credit sale of Rs. 350 means the business has earned the revenue, so it is recorded in the income statement as sales revenue. The receivable (Rs. 350) is recorded on the balance sheet as an asset, but the sale itself appears on the income statement. Cash-basis accounting would delay recognition until cash is received, but accrual accounting does not.
The transformation from tall to flat organizational structures primarily reflects efforts to:
Organizations are flattening (reducing management levels between CEO and production workers) in an effort to become more competitive. The traditional middle management role has been equated with cumbersome bureaucracy that prevents businesses from responding to market forces.
When dichloromethane (DCM) and water (H$_2$O) are used together for differential extraction, which statement correctly describes their behaviour in a separating funnel?
In solvent extraction, liquids separate into layers based on density.
- Density of DCM (CH$_2$Cl$_2$) $\approx 1.33\ \text{g/mL}$ โ denser than water.
- Density of water $\approx 1.00\ \text{g/mL}$.
- DCM and water are immiscible โ they do not dissolve in each other.
Since DCM is denser, it sinks to the bottom, and water stays on top. They separate cleanly โ no colloidal mixture forms.
Given that $x \neq 0$, compare the two quantities below.
Column A: $\dfrac{x}{|x|}$
Column B: $1$
The expression $\dfrac{x}{|x|}$ is the sign function of $x$:
- If $x > 0$: $\dfrac{x}{|x|} = \dfrac{x}{x} = 1$ โ equal to Column B
- If $x < 0$: $\dfrac{x}{|x|} = \dfrac{x}{-x} = -1$ โ less than Column B
Since $x$ can be positive or negative, and we get different results, the relationship cannot be determined from the information given.
If $\cos^{-1}\!\left(\dfrac{2}{3x}\right)+\cos^{-1}\!\left(\dfrac{3}{4x}\right)=\dfrac{\pi}{2}$ for $x>\dfrac{3}{4}$, find $x$.
$\cos^{-1}A+\cos^{-1}B=\pi/2 \Rightarrow \cos^{-1}A=\pi/2-\cos^{-1}B=\sin^{-1}B$, i.e., $A=\sin(\cos^{-1}B)=\sqrt{1-B^2}$.
$\dfrac{2}{3x}=\sqrt{1-\dfrac{9}{16x^2}}$
Square: $\dfrac{4}{9x^2}=1-\dfrac{9}{16x^2}$
Multiply by $144x^2$: $64=144x^2-81 \Rightarrow 144x^2=145 \Rightarrow x=\dfrac{\sqrt{145}}{12}$
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