To find the largest of three consecutive multiples of 8 whose sum is 2424, we can represent the multiples algebraically. Let the first multiple be \(8x\). Then the next two consecutive multiples of 8 will be \(8x + 8\) and \(8x + 16\). We set up an equation by summing these three terms and equating it to 2424, then solve for \(x\). Once \(x\) is found, we can determine the largest multiple.
Correct Option: D) 816
The question asks about environmental factors influencing sex determination in some reptiles. This phenomenon is known as Temperature-dependent Sex Determination (TSD).
D) Temperature — Temperature is a well-documented environmental factor that determines the sex of offspring in many reptile species, a phenomenon called Temperature-dependent Sex Determination (TSD).
The question asks to identify which substance is NOT released by the gastric glands in the stomach wall. To answer this, we need to recall the primary secretions of the gastric glands and the origin of other digestive fluids.
A) saliva — Saliva is produced by the salivary glands in the mouth and is involved in the initial stages of digestion (mechanical and chemical breakdown of carbohydrates). It is not secreted by the gastric glands in the stomach.
To balance a chemical equation, we must ensure that the number of atoms of each element on the reactant side (left side) is equal to the number of atoms of the same element on the product side (right side). This adheres to the Law of Conservation of Mass. We will balance the equation step-by-step, starting with elements that appear in only one compound on each side.
Identify the unbalanced equation:
\[ \text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow \text{AlCl}_3 + \text{BaSO}_4 \]
Balance Aluminum (Al) atoms:
On the reactant side, there are 2 Al atoms in \(\text{Al}_2(\text{SO}_4)_3\). On the product side, there is 1 Al atom in \(\text{AlCl}_3\). To balance Al, place a coefficient of 2 in front of \(\text{AlCl}_3\) on the product side:
\[ \text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + \text{BaSO}_4 \]
Balance Sulfate (SOâ‚„) groups:
On the reactant side, there are 3 \(\text{SO}_4\) groups in \(\text{Al}_2(\text{SO}_4)_3\). On the product side, there is 1 \(\text{SO}_4\) group in \(\text{BaSO}_4\). To balance the \(\text{SO}_4\) groups, place a coefficient of 3 in front of \(\text{BaSO}_4\) on the product side:
\[ \text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4 \]
Balance Barium (Ba) atoms:
Now, let's check Barium. On the product side, there are 3 Ba atoms (from \(3\text{BaSO}_4\)). On the reactant side, there is 1 Ba atom in \(\text{BaCl}_2\). To balance Ba, place a coefficient of 3 in front of \(\text{BaCl}_2\) on the reactant side:
\[ \text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4 \]
Balance Chlorine (Cl) atoms:
Finally, let's check Chlorine. On the reactant side, there are 6 Cl atoms (from \(3\text{BaCl}_2\)). On the product side, there are 6 Cl atoms (from \(2\text{AlCl}_3\), since \(2 \times 3 = 6\)). The chlorine atoms are balanced.
Verify the balanced equation:
Reactants: Al = 2, S = 3, O = 12, Ba = 3, Cl = 6
Products: Al = 2, S = 3, O = 12, Ba = 3, Cl = 6
All atoms are balanced.
D) The balanced equation is \(\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4\). This option correctly shows 2 Al, 3 SOâ‚„, 3 Ba, and 6 Cl atoms on both sides of the equation.
A) \(\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{AlCl}_3 + 2\text{BaSO}_4\)
In this option, Al atoms are unbalanced (2 on reactant side, 3 on product side). Also, SOâ‚„ groups are unbalanced (3 on reactant side, 2 on product side).
B) \(2\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4\)
In this option, Al atoms are unbalanced (4 on reactant side, 2 on product side). Also, SOâ‚„ groups are unbalanced (6 on reactant side, 3 on product side).
C) \(\text{Al}_2(\text{SO}_4)_3 + 2\text{BaCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{BaSO}_4\)
In this option, Ba atoms are unbalanced (2 on reactant side, 3 on product side). Also, Cl atoms are unbalanced (4 on reactant side, 6 on product side).
The question asks to identify the title of a discussion paper released by NITI Aayog on January 15, 2024. This requires specific knowledge of recent reports and publications by NITI Aayog.
Correct Option: B) Multidimensional Poverty in India since 2005-06
This is the exact title of the discussion paper released by NITI Aayog on January 15, 2024. The paper stated that 24.82 crore people exited multidimensional poverty in India in the nine years leading up to 2022-23.
To find the effect on tax revenue, we need to calculate the percentage change in revenue. Tax revenue is the product of the tax rate and the consumption quantity. We will assume initial values for tax and consumption, calculate the initial revenue, then calculate the new tax and consumption after the given changes, and finally determine the new revenue and its percentage change.
Let the initial tax on the commodity be \(T = 100\) units (e.g., Rs. 100 per unit of commodity).
Let the initial consumption of the commodity be \(C = 100\) units.
Initial Tax Revenue \(= T \times C\)
Initial Tax Revenue \(= 100 \times 100 = 10000\) units.
The tax is reduced by 20%.
New Tax \(T' = T - (20\% \text{ of } T)\)
New Tax \(T' = 100 - (0.20 \times 100) = 100 - 20 = 80\) units.
The consumption is increased by 15%.
New Consumption \(C' = C + (15\% \text{ of } C)\)
New Consumption \(C' = 100 + (0.15 \times 100) = 100 + 15 = 115\) units.
New Tax Revenue \(= T' \times C'\)
New Tax Revenue \(= 80 \times 115 = 9200\) units.
Change in Revenue \(= \text{New Tax Revenue} - \text{Initial Tax Revenue}\)
Change in Revenue \(= 9200 - 10000 = -800\) units.
Since the change is negative, the revenue has decreased.
This means the tax revenue decreases by 8%.
Alternatively, using the percentage change formula for products:
If \(R = T \times C\), and \(T\) changes by \(x\%\) and \(C\) changes by \(y\%\), then the approximate percentage change in \(R\) is \((x + y + \frac{xy}{100})\%\).
Here, tax is reduced by 20%, so \(x = -20\).
Consumption is increased by 15%, so \(y = +15\).
\[ \text{Percentage Change} = \left( -20 + 15 + \frac{(-20) \times 15}{100} \right)\% \] \[ \text{Percentage Change} = \left( -5 + \frac{-300}{100} \right)\% \] \[ \text{Percentage Change} = (-5 - 3)\% = -8\% \]The tax revenue decreases by 8%.
B) It decreases by 8%
A diver rowing at the speed of 3 km/h in still water takes double the time going 50 km upstream compared to going 50 km downstream. The speed of the diver downstream is:
\( \frac{9}{2} \) km/h
4 km/h
\( \frac{7}{2} \) km/h
\( \frac{10}{3} \) km/h
This problem involves concepts of relative speed in rivers, specifically upstream and downstream motion. We are given the speed of the diver in still water and a relationship between the time taken to travel the same distance upstream and downstream. We need to find the speed of the diver downstream.
Let the speed of the diver in still water be \(v_d\) and the speed of the current be \(v_c\).
Given: \(v_d = 3\) km/h.
The speed of the diver upstream (\(v_u\)) is the speed in still water minus the speed of the current:
\[ v_u = v_d - v_c = 3 - v_c \]The speed of the diver downstream (\(v_{down}\)) is the speed in still water plus the speed of the current:
\[ v_{down} = v_d + v_c = 3 + v_c \]The distance covered in both cases is \(D = 50\) km.
The time taken to travel upstream (\(t_u\)) is:
\[ t_u = \frac{D}{v_u} = \frac{50}{3 - v_c} \]The time taken to travel downstream (\(t_{down}\)) is:
\[ t_{down} = \frac{D}{v_{down}} = \frac{50}{3 + v_c} \]According to the problem, the diver takes double the time going upstream compared to going downstream:
\[ t_u = 2 \times t_{down} \]Substitute the expressions for \(t_u\) and \(t_{down}\) into the equation:
\[ \frac{50}{3 - v_c} = 2 \times \frac{50}{3 + v_c} \]Cancel out 50 from both sides:
\[ \frac{1}{3 - v_c} = \frac{2}{3 + v_c} \]Cross-multiply to solve for \(v_c\):
\[ 3 + v_c = 2(3 - v_c) \] \[ 3 + v_c = 6 - 2v_c \] \[ v_c + 2v_c = 6 - 3 \] \[ 3v_c = 3 \] \[ v_c = 1 \text{ km/h} \]The question asks for the speed of the diver downstream. This is \(v_{down} = v_d + v_c\):
\[ v_{down} = 3 + 1 = 4 \text{ km/h} \]Correct Option: B) 4 km/h