Sakshi invested a sum of ₹8,900 at 10% per annum compound interest, compounded annually. If she received an amount of ₹10,769 after n years, the value of n is:
This problem involves calculating the number of years (n) for a given principal amount to grow to a specific amount under compound interest. We will use the compound interest formula to set up an equation and solve for 'n'.
The formula for compound interest, compounded annually, is:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]We know that \(1.1 \times 1.1 = 1.21\). Therefore,
\[ (1.1)^2 = (1.1)^n \]Since the bases are the same, the exponents must be equal.
\[ n = 2 \]A) 2 is the correct answer. The calculation shows that after 2 years, the invested sum will grow to the given amount.
The question asks about the consequences of the absence of decomposers in an ecosystem. Decomposers play a crucial role in breaking down dead organic matter and recycling nutrients. Understanding this role is key to determining the correct answer.
Correct Option: C) They would persist, and the natural replenishment of soil would be hindered.
This option accurately describes the consequences. Dead organic matter would not decompose, leading to its accumulation. The lack of decomposition would prevent nutrients from returning to the soil, thus hindering its natural replenishment and fertility.
This is a box-stacking puzzle. We need to deduce the exact order of the seven boxes (E, F, G, H, T, U, V) based on the given clues and then determine the number of boxes between H and F.
Total boxes: There are 7 boxes. Let's denote the positions from bottom to top as 1 to 7.
Clue 1: Only two boxes are kept below E.
This means E is at the 3rd position from the bottom (or 5th from the top).
\(\text{Position 7}\)
\(\text{Position 6}\)
\(\text{Position 5}\)
\(\text{Position 4}\)
\(\text{E (Position 3)}\)
\(\text{Position 2}\)
\(\text{Position 1}\)
Clue 2: Only one box is kept above T.
This means T is at the 6th position from the bottom (or 2nd from the top).
\(\text{Position 7}\)
\(\text{T (Position 6)}\)
\(\text{Position 5}\)
\(\text{Position 4}\)
\(\text{E (Position 3)}\)
\(\text{Position 2}\)
\(\text{Position 1}\)
Clue 3: Only one box is kept between T and U.
Since T is at position 6, U can be at position 4 (one box between T and U). U cannot be above T as there's only one box above T.
\(\text{Position 7}\)
\(\text{T (Position 6)}\)
\(\text{Position 5}\)
\(\text{U (Position 4)}\)
\(\text{E (Position 3)}\)
\(\text{Position 2}\)
\(\text{Position 1}\)
Clue 4: G is kept immediately below F.
This means F and G are together (F/G). We have two consecutive empty slots at positions 1 and 2, and one at position 5 and 7. Since F and G must be together, they can be either (F at 2, G at 1) or (F at 7, G at 5). Let's try the first case: F at 2, G at 1.
\(\text{Position 7}\)
\(\text{T (Position 6)}\)
\(\text{Position 5}\)
\(\text{U (Position 4)}\)
\(\text{E (Position 3)}\)
\(\text{F (Position 2)}\)
\(\text{G (Position 1)}\)
Clue 5: V is kept at some place below H.
The remaining boxes are H and V. The remaining empty slots are Position 7 and Position 5. If F and G are at 2 and 1, then H and V must be at 7 and 5. Since V is below H, H must be at Position 7 and V at Position 5.
\(\text{H (Position 7)}\)
\(\text{T (Position 6)}\)
\(\text{V (Position 5)}\)
\(\text{U (Position 4)}\)
\(\text{E (Position 3)}\)
\(\text{F (Position 2)}\)
\(\text{G (Position 1)}\)
This arrangement satisfies all conditions.
Final Question: How many boxes are kept between H and F?
H is at Position 7.
F is at Position 2.
The boxes between H and F are T (Position 6), V (Position 5), U (Position 4), and E (Position 3). There are 4 boxes between H and F.
D) Four
The question asks for the establishment year of the Indian Remote Sensing Satellite System (IRS). This requires knowledge of key milestones in India's space program.
Correct Option: B) 1988
To find the missing number in the series, we need to identify the pattern or rule governing the sequence of numbers. We will examine the differences between consecutive terms to see if there's a consistent arithmetic progression or an alternating pattern.
The differences are alternating between \(6\) and \(10\). The pattern of differences is \(+6, +10, +6, +10, \dots\).
Since the last difference was \(+10\) (from \(40\) to \(50\)), the next difference should be \(+6\).
Therefore, the missing number will be \(50 + 6 = 56\).
Correct Option: A) 56
The question asks to identify a combination reaction from the given options. A combination reaction is a type of chemical reaction in which two or more reactants combine to form a single product. It generally follows the form \(A + B \rightarrow AB\).
Based on the analysis, only Option A represents a combination reaction.
Correct Option: A) \( \text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 \). In this reaction, two reactants (calcium oxide and water) combine to form a single product (calcium hydroxide). This is the definition of a combination reaction.
To determine which combination produces a precipitate, we need to consider the solubility rules for ionic compounds. A precipitate forms when two soluble ionic compounds react to form an insoluble ionic compound. We will examine each option based on these rules.
Correct Option: A) NaOH + FeCl₃ — This combination produces iron(III) hydroxide (\(\text{Fe(OH)}_3\)), which is an insoluble compound and will precipitate.
We need to determine who among five people (A, B, C, D, E) is the heaviest. We will analyze each statement independently and then combine them to see if a definitive answer can be reached.
Analyze Statement I: A is heavier than C but lighter than D.
Analyze Statement II: B is not the heaviest and E is not the lightest.
Combine Statement I and Statement II:
Correct Option: D) Both Statements I and II put together are not sufficient to answer the question.
The average price of three items of furniture is Rs 16,455. If their prices are in the ratio 3:5:7, the price of the cheapest item (in Rs) is:
To find the price of the cheapest item, we first need to determine the total price of the three items using their average price. Then, we can use the given ratio of their prices to find the individual prices and identify the cheapest one.
The average price of three items is Rs 16,455.
\[ \text{Total Price} = \text{Average Price} \times \text{Number of Items} \] \[ \text{Total Price} = 16,455 \times 3 = 49,365 \text{ Rs} \]The prices of the three items are in the ratio 3:5:7. Let the prices be \(3x\), \(5x\), and \(7x\) respectively.
The sum of the individual prices must equal the total price:
\[ 3x + 5x + 7x = 49,365 \]The cheapest item corresponds to the smallest ratio part, which is \(3x\).
\[ \text{Price of Cheapest Item} = 3x = 3 \times 3,291 \] \[ \text{Price of Cheapest Item} = 9,873 \text{ Rs} \]Correct Option: D) 9,873
The resistance of a wire depends on its material (resistivity), length, and cross-sectional area. The formula for resistance will be used to determine how the resistance changes when the cross-sectional area is altered while keeping the material and length constant.
Recall the formula for resistance of a wire:
\[ R = \rho \frac{L}{A} \]where:
Given the initial conditions:
Thus, for the first wire:
\[ 10 \ \Omega = \rho \frac{L}{A} \quad \text{(Equation 1)} \]Consider the new wire:
Let the new resistance be \(R_2\). Using the resistance formula for the new wire:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{L}{2A} \]Rewrite the expression for \(R_2\) to relate it to \(R_1\):
\[ R_2 = \frac{1}{2} \left( \rho \frac{L}{A} \right) \]Substitute the value from Equation 1 into the expression for \(R_2\):
\[ R_2 = \frac{1}{2} (10 \ \Omega) \]\[ R_2 = 5 \ \Omega \]C) 5 \ \Omega — The resistance is inversely proportional to the cross-sectional area. When the cross-sectional area is doubled, the resistance is halved.