The question asks about the fundamental principle of how Goods and Services Tax (GST) is levied. Understanding GST as a consumption-based tax is key to identifying the correct answer.
D) Destination where the goods or services are consumed — GST is a consumption-based tax. This means that the tax is ultimately paid by the final consumer of the goods or services, and it is levied at the point of consumption. For inter-state transactions, the tax accrues to the consuming state.
If \( \frac{a}{b} \) = \( \frac{4}{3} \), then what will be the value of \(\frac{6a+2b}{4a-2b}\)?
The problem provides a ratio between two variables, \(a\) and \(b\), and asks for the value of an expression involving these variables. We can solve this by expressing one variable in terms of the other using the given ratio, and then substituting this into the expression. Alternatively, we can divide the numerator and denominator of the expression by one of the variables (e.g., \(b\)) to convert it into an expression solely dependent on the ratio \(a/b\).
Given the ratio:
\[ \frac{a}{b} = \frac{4}{3} \]From this ratio, we can express \(a\) in terms of \(b\):
\[ a = \frac{4}{3}b \]Now, substitute this value of \(a\) into the given expression \(\frac{6a+2b}{4a-2b}\):
\[ \frac{6a+2b}{4a-2b} = \frac{6\left(\frac{4}{3}b\right)+2b}{4\left(\frac{4}{3}b\right)-2b} \]Simplify the numerator:
\[ 6\left(\frac{4}{3}b\right)+2b = \frac{24}{3}b + 2b = 8b + 2b = 10b \]Simplify the denominator:
\[ 4\left(\frac{4}{3}b\right)-2b = \frac{16}{3}b - 2b = \frac{16b - 6b}{3} = \frac{10b}{3} \]Substitute the simplified numerator and denominator back into the expression:
\[ \frac{10b}{\frac{10b}{3}} \]Simplify the fraction:
\[ \frac{10b}{1} \times \frac{3}{10b} = 3 \]Alternative Method:
Given the expression \(\frac{6a+2b}{4a-2b}\).
Divide both the numerator and the denominator by \(b\) (assuming \(b \neq 0\)):
\[ \frac{\frac{6a}{b}+\frac{2b}{b}}{\frac{4a}{b}-\frac{2b}{b}} = \frac{6\left(\frac{a}{b}\right)+2}{4\left(\frac{a}{b}\right)-2} \]Substitute the given ratio \[\frac{a}{b} = \frac{4}{3}\] into the expression:
\[ \frac{6\left(\frac{4}{3}\right)+2}{4\left(\frac{4}{3}\right)-2} \]Simplify the numerator:
\[ 6\left(\frac{4}{3}\right)+2 = 2 \times 4 + 2 = 8 + 2 = 10 \]Simplify the denominator:
\[ 4\left(\frac{4}{3}\right)-2 = \frac{16}{3} - 2 = \frac{16-6}{3} = \frac{10}{3} \]Substitute the simplified numerator and denominator back into the expression:
\[ \frac{10}{\frac{10}{3}} \]Simplify the fraction:
\[ 10 \times \frac{3}{10} = 3 \]D) 3 — Both methods yield the value 3 for the given expression.
A bus travelling with \( \frac{5}{7} \) of its actual speed covers 42 km in 1 h 40 min 48 s. Find the actual speed of the bus.
30 km/h
35 km/h
17 \( \frac{6}{7} \) km/h
25 km/h
To find the actual speed of the bus, we first need to calculate the speed at which it is currently travelling. This involves converting the given time into hours and then using the formula: Speed = Distance / Time. Once we have the current speed, we can use the information that this speed is \( \frac{5}{7} \) of its actual speed to find the actual speed.
Convert the given time into hours:
The time taken is 1 h 40 min 48 s.
Total time in hours:
\[ \text{Time} = 1 + \frac{2}{3} + \frac{1}{75} \text{ h} \]\[ \text{Time} = \frac{75 + 50 + 1}{75} \text{ h} \]\[ \text{Time} = \frac{126}{75} \text{ h} \]\[ \text{Time} = \frac{42}{25} \text{ h} \]Calculate the current speed of the bus:
The bus covers a distance of 42 km in \( \frac{42}{25} \) h.
Using the formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\[ \text{Current Speed} = \frac{42 \text{ km}}{\frac{42}{25} \text{ h}} \]\[ \text{Current Speed} = 42 \times \frac{25}{42} \text{ km/h} \]\[ \text{Current Speed} = 25 \text{ km/h} \]Find the actual speed of the bus:
We are given that the current speed is \( \frac{5}{7} \) of its actual speed.
Let the actual speed be \( S \text{ km/h} \).
\[ \text{Current Speed} = \frac{5}{7} \times \text{Actual Speed} \]\[ 25 = \frac{5}{7} \times S \]\[ S = 25 \times \frac{7}{5} \]\[ S = 5 \times 7 \]\[ S = 35 \text{ km/h} \]B) 35 km/h is the actual speed of the bus.
The question asks for the date of the Brexit referendum. This is a factual recall question related to significant international political events. Knowledge of key historical dates is important for competitive exams.
Correct Option: C) June 2016
The United Kingdom European Union membership referendum, commonly referred to as the Brexit referendum, was held on 23 June 2016. The result was 51.9% in favour of leaving the European Union and 48.1% in favour of remaining.
The mean proportional (also known as the geometric mean) between two numbers \(a\) and \(b\) is a number \(x\) such that the ratio of \(a\) to \(x\) is equal to the ratio of \(x\) to \(b\). This can be expressed as \[\frac{a}{x} = \frac{x}{b}\]. We will use this definition to find the mean proportional between 169 and 121.
Thus, the mean proportional between 169 and 121 is 143.
Correct Option: B) 143
If α = 30°, then find the value of 2sinαcosα.
\(\frac{1}{\sqrt{3}}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{\sqrt{2}}{3}\)
\(\frac{2}{\sqrt{3}}\)
The problem asks us to find the value of the trigonometric expression \(2\sin\alpha\cos\alpha\) when \(\alpha = 30^\circ\). We will substitute the given value of \(\alpha\) into the expression and then use the standard trigonometric values for \(\sin 30^\circ\) and \(\cos 30^\circ\).
Given \(\alpha = 30^\circ\), substitute this into the expression \(2\sin\alpha\cos\alpha\):
\[ 2\sin\alpha\cos\alpha = 2\sin(30^\circ)\cos(30^\circ) \]The standard trigonometric values for \(30^\circ\) are:
Now, substitute these values back into the expression from Step 1:
\[ 2\sin(30^\circ)\cos(30^\circ) = 2 \times \left(\frac{1}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right) \]Perform the multiplication:
\[ 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{2 \times 1 \times \sqrt{3}}{2 \times 2} \]\[ = \frac{2\sqrt{3}}{4} \]\[ = \frac{\sqrt{3}}{2} \]Thus, the value of \(2\sin\alpha\cos\alpha\) when \(\alpha = 30^\circ\) is \(\frac{\sqrt{3}}{2}\).
Alternatively, one could use the double angle identity \(\sin(2\alpha) = 2\sin\alpha\cos\alpha\). In this case:
\[ 2\sin\alpha\cos\alpha = \sin(2\alpha) \]\[ = \sin(2 \times 30^\circ) \]\[ = \sin(60^\circ) \]\[ = \frac{\sqrt{3}}{2} \]B) \(\frac{\sqrt{3}}{2}\) is the correct value obtained after substituting \(\alpha = 30^\circ\) and simplifying the expression.
The problem asks us to find the ratio \(x:y\) given a relationship between percentages of \(x\) and \(y\). We will translate the given percentage relationship into an algebraic equation and then rearrange it to find the ratio \(x/y\).
Translate the given information into an equation:
"50% of x is the same as 25% of y" can be written mathematically as:
Convert percentages to decimal or fractional form:
We know that \(50\% = \frac{50}{100} = 0.5\) and \(25\% = \frac{25}{100} = 0.25\). Substituting these values into the equation:
Alternatively, using fractions:
\[ \frac{50}{100}x = \frac{25}{100}y \] \[ \frac{1}{2}x = \frac{1}{4}y \]Rearrange the equation to find the ratio \(x/y\):
To find \(x:y\), we need to express it as a fraction \(x/y\). Divide both sides of the equation by \(y\) (assuming \(y \neq 0\)) and by the coefficient of \(x\).
Using the decimal form:
\[ 0.5x = 0.25y \] \[ \frac{x}{y} = \frac{0.25}{0.5} \] \[ \frac{x}{y} = \frac{1}{2} \]Using the fractional form:
\[ \frac{1}{2}x = \frac{1}{4}y \]Multiply both sides by 4 to clear denominators:
\[ 4 \times \frac{1}{2}x = 4 \times \frac{1}{4}y \] \[ 2x = y \]Now, divide both sides by \(2y\):
\[ \frac{2x}{2y} = \frac{y}{2y} \] \[ \frac{x}{y} = \frac{1}{2} \]Express the ratio:
The ratio \(x:y\) is therefore \(1:2\).
Correct Option: D) 1 : 2
The question asks to identify the first Indian movie with sound, released in 1931. This requires knowledge of Indian cinema history.
Correct Option: D) Alam Ara
Alam Ara (meaning "The Light of the World") was the first Indian talkie film. It was directed by Ardeshir Irani and released on March 14, 1931, at the Majestic Cinema in Mumbai. Its success marked the beginning of the sound era in Indian cinema.
The question asks to identify the discoverer of X-rays. This is a factual recall question from the history of science, specifically physics.
X-rays were discovered by Wilhelm Conrad Roentgen in 1895. He observed a new type of radiation that could pass through opaque materials and named them "X-rays" because their nature was unknown at the time. This discovery revolutionized medicine and led to him receiving the first Nobel Prize in Physics in 1901.
B) W. Roentgen — Wilhelm Conrad Roentgen (often abbreviated as W. Roentgen) discovered X-rays in 1895 while experimenting with cathode rays.
What is the value of the following expression?
\(\frac{5\times1.6-2\times1.4}{1.3}\)
To evaluate the given expression, we need to follow the order of operations (BODMAS/PEMDAS). First, perform the multiplications in the numerator, then the subtraction, and finally the division.
\[ 5 \times 1.6 \]
To multiply \(5\) by \(1.6\):
\[ 5 \times 16 = 80 \]
Since \(1.6\) has one decimal place, the result will also have one decimal place.
\[ 5 \times 1.6 = 8.0 = 8 \]
\[ 2 \times 1.4 \]
To multiply \(2\) by \(1.4\):
\[ 2 \times 14 = 28 \]
Since \(1.4\) has one decimal place, the result will also have one decimal place.
\[ 2 \times 1.4 = 2.8 \]
Now, substitute the values back into the numerator:
\[ 8 - 2.8 \]
Subtracting \(2.8\) from \(8\):
\[ 8.0 - 2.8 = 5.2 \]
The expression now becomes:
\[ \frac{5.2}{1.3} \]
To divide \(5.2\) by \(1.3\), we can multiply both the numerator and the denominator by \(10\) to remove the decimal points:
\[ \frac{5.2 \times 10}{1.3 \times 10} = \frac{52}{13} \]
Now, perform the division:
\[ 52 \div 13 = 4 \]
C) 4 — The step-by-step calculation correctly yields 4.