This problem involves setting up a system of linear equations based on the given age relationships and then solving for the required sum. We will assign variables to the present ages of Manavi and Tanvi, and then express the ages of Tanvi's father and Manavi's brother in terms of these variables. Finally, we will use the given sums to form equations and solve them.
Define variables for present ages:
Formulate the first equation:
The sum of two times Manavi’s present age and Tanvi’s present age is 35 years.
Express other ages in terms of \(M\) and \(T\):
Formulate the second equation:
The sum of the present ages of Tanvi’s father and Manavi’s brother is 56 years.
Solve the system of equations:
We have two linear equations:
Equation 1: \(2M + T = 35\)
Equation 2: \(M + 3T = 50\)
From Equation 1, we can express \(T\) in terms of \(M\):
\[ T = 35 - 2M \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 2:\[ M + 3(35 - 2M) = 50 \]\[ M + 105 - 6M = 50 \]\[ 105 - 5M = 50 \]\[ 105 - 50 = 5M \]\[ 55 = 5M \]\[ M = \frac{55}{5} \]\[ M = 11 \]Find Tanvi's age (\(T\)):
Substitute the value of \(M\) back into Equation 3:
Calculate the sum of Manavi's and Tanvi's present ages:
\[ M + T = 11 + 13 \]\[ M + T = 24 \]Correct Option: A) 24
The sum of the present ages of Manavi and Tanvi is 24 years.
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
In the given reaction, which product remains in the solution?
The given reaction is a double displacement reaction, specifically a precipitation reaction. In such reactions, two aqueous ionic compounds react to form a new aqueous compound and an insoluble solid (precipitate). We need to identify which product remains dissolved in the solution.
\( \text{Na}_2\text{SO}_4 \text{(aq)} + \text{BaCl}_2 \text{(aq)} \rightarrow \text{BaSO}_4 \text{(s)} + 2\text{NaCl(aq)} \)
C) NaCl — Sodium chloride (NaCl) is indicated with \( \text{(aq)} \), meaning it is aqueous and remains dissolved in the solution.
The question asks to identify which of the given locations is NOT one of the sites where the Kumbh Mela religious festival is held. To answer this, one needs to know the four specific locations where the Kumbh Mela takes place.
Correct Option: A) Bodh Gaya is a significant Buddhist pilgrimage site and is not one of the four locations where the Kumbh Mela is celebrated.
A man travels from point A to B at a speed of 56\( \frac{1}{4} \) km/h and returns to A through the same route by increasing his speed by 33\( \frac{1}{3} \)% of his earlier speed. His average speed (in km/h) for the whole journey is:
63\( \frac{1}{2} \)
60\( \frac{5}{7} \)
62\( \frac{1}{4} \)
64\( \frac{2}{7} \)
To find the average speed for the whole journey, we need to calculate the total distance traveled and the total time taken. The journey involves traveling from A to B and then back from B to A. The distance is the same for both legs of the journey. We are given the initial speed and the percentage increase in speed for the return journey. We will use the formula for average speed: Average Speed = (Total Distance) / (Total Time).
The initial speed is given as \(56\frac{1}{4}\) km/h.
\[ \text{Speed}_1 = 56 + \frac{1}{4} = \frac{56 \times 4 + 1}{4} = \frac{224 + 1}{4} = \frac{225}{4} \text{ km/h} \]The speed is increased by \(33\frac{1}{3}\)\% of the earlier speed. First, convert the percentage to a fraction.
\[ 33\frac{1}{3}\% = \frac{100}{3}\% = \frac{100}{3 \times 100} = \frac{1}{3} \]So, the increase in speed is \(\frac{1}{3}\) of the earlier speed.
\[ \text{Increase in speed} = \frac{1}{3} \times \frac{225}{4} = \frac{75}{4} \text{ km/h} \]The speed for the return journey (\(\text{Speed}_2\)) is the initial speed plus the increase:
\[ \text{Speed}_2 = \frac{225}{4} + \frac{75}{4} = \frac{300}{4} = 75 \text{ km/h} \]Let the distance between A and B be \(D\). To simplify calculations, we can choose a distance that is a multiple of the denominators of the speeds. However, for two-way journeys with different speeds, it's often easier to use the formula for average speed directly:
\[ \text{Average Speed} = \frac{2 \times \text{Speed}_1 \times \text{Speed}_2}{\text{Speed}_1 + \text{Speed}_2} \]This formula is valid when the distance traveled in both directions is the same.
We have \(\text{Speed}_1 = \frac{225}{4}\) km/h and \(\text{Speed}_2 = 75\) km/h.
\[ \text{Average Speed} = \frac{2 \times \frac{225}{4} \times 75}{\frac{225}{4} + 75} \]Now, perform the division. Both 16875 and 525 are divisible by 25.
\[ 16875 \div 25 = 675 \]\[ 525 \div 25 = 21 \]So, the expression becomes:
\[ \text{Average Speed} = \frac{675 \times 2}{21} = \frac{1350}{21} \]Both 1350 and 21 are divisible by 3.
\[ 1350 \div 3 = 450 \]\[ 21 \div 3 = 7 \]So, the average speed is:
\[ \text{Average Speed} = \frac{450}{7} \text{ km/h} \]So, the average speed is \(64\frac{2}{7}\) km/h.
Correct Option: D) \(64\frac{2}{7}\)
The question asks to identify two cell organelles, besides the nucleus, that possess their own DNA and ribosomes. This points to organelles that are semi-autonomous, meaning they can synthesize some of their own proteins and replicate independently to some extent.
A) Plastids and Mitochondria — Both mitochondria and plastids (e.g., chloroplasts) are semi-autonomous organelles. They possess their own circular DNA, ribosomes (70S type, similar to prokaryotes), and can synthesize some of their own proteins. This is a key piece of evidence supporting the endosymbiotic theory, which suggests that these organelles originated from free-living prokaryotes that were engulfed by ancestral eukaryotic cells.
To determine which molecule has only single bonds, we need to draw the Lewis structure for each given molecule and identify the type of bonds (single, double, or triple) present between the atoms.
Correct Option: B) CH₄ has only single bonds (four C-H single bonds).
This question asks to identify a specific point on the principal axis of a spherical mirror based on its distance from the pole and focus. We need to recall the definitions and relationships between the key points (pole, principal focus, centre of curvature) and distances (focal length, radius of curvature) for a spherical mirror.
C) centre of curvature — As derived, the distance from the pole (P) to the centre of curvature (C) is the radius of curvature (\(R\)), which is equal to twice the focal length (\(2f\)). The focal length (\(f\)) is the distance between P and F. Therefore, a point at double the distance between P and F is the centre of curvature.
The expenditure of Sudha is 125% more than her savings. If her expenditure decreases by 6% and savings increase by 26.5%, then by what percent does her income increase?
We need to calculate the initial expenditure, savings, and income. Then, we will calculate the new expenditure and savings based on the given percentage changes. Finally, we will find the new income and determine the percentage increase in income.
Establish initial relationship between Expenditure and Savings:
Let Sudha's savings be \(S\).
Her expenditure (\(E\)) is 125% more than her savings. This means:
Assume a base value for Savings:
To simplify calculations, let's assume her initial savings \(S = 100\).
Then, her initial expenditure \(E = 2.25 \times 100 = 225\).
Calculate initial Income:
Income (\(I\)) = Expenditure (\(E\)) + Savings (\(S\)).
Calculate new Expenditure:
Her expenditure decreases by 6%.
Calculate new Savings:
Her savings increase by 26.5%.
Calculate new Income:
New Income (\(I_{\text{new}}\)) = New Expenditure (\(E_{\text{new}}\)) + New Savings (\(S_{\text{new}}\)).
Calculate the percentage increase in Income:
Percentage increase \[= \frac{\text{New Income} - \text{Initial Income}}{\text{Initial Income}} \times 100\]
Correct Option: D) 4%
This problem involves understanding relative directions. We need to visualize the positions of the poles based on the given statements and then determine the position of pole R with respect to pole T.
From these two points, we can infer that R is south-east of T. Let's represent this visually:
P --- T | | R
Or, more accurately, T is to the East of P, and P is to the North of R. So, R is South of P, and P is West of T.
T
|
P
|
RThis is not quite right. Let's draw it based on the first part: "P is to the west of T" and "P is north of R".
P --- T
|
|
RSo, from T, P is to the west. From P, R is to the south. This means R is south-west of T.
This statement gives information about Q, R, and W, but it does not directly relate to T. It primarily helps confirm the relative position of R if needed, but for the question about R with respect to T, the first statement is sufficient.
From step 1, we established that P is to the west of T, and R is to the south of P.
T
|
P
|
RIf we start at T, to reach P, we move west. From P, to reach R, we move south. Therefore, from T, R is located in the south-west direction.
C) South-west. Based on the analysis, pole R is located to the south-west of pole T.
Six students, J, K, L, M, N and O, decided to race during a PE lesson. Each had a different speed in the race. N ran faster than M but slower than L. M ran slower than J but faster than O. K ran faster than L. J ran slower than L. Who among the six ran the fastest?
To determine who ran the fastest, we need to establish the relative speeds of all six students (J, K, L, M, N, O) based on the given statements. We will use inequalities to represent their speeds and then combine them to find the overall fastest runner.
Correct Option: A) K
The combined sequence of speeds from slowest to fastest is \(O < M < N < J < L < K\). Therefore, K ran the fastest.