If the position of the digit ‘2’ is interchanged with the position of the digit ‘4’, and the position of the digit ‘5’ is then changed with the position of the digit ‘3’, then what is the value of the given expression?
42 ÷ 5 + 3
The problem requires us to modify a given mathematical expression by interchanging certain digits and then evaluate the new expression. We need to perform the interchanges sequentially as described and then apply the order of operations (BODMAS/PEMDAS) to find the final value.
A) 13 is the correct value of the expression after the specified digit interchanges.
If −1 is a root of the equation 3x² + px + 4 = 0, and 7x² + px + k = 0 has equal roots, then what is the value of k?
\( \frac{2}{3} \)
\( \frac{7}{4} \)
\( \frac{7}{2} \)
\( \frac{4}{3} \)
The problem involves two quadratic equations. We are given that \(-1\) is a root of the first equation, which allows us to find the value of \(p\). Then, we use this value of \(p\) in the second equation, which is stated to have equal roots. The condition for equal roots of a quadratic equation \(ax^2 + bx + c = 0\) is that its discriminant, \(b^2 - 4ac\), must be equal to zero. This condition will help us find the value of \(k\).
The first equation is \(3x^2 + px + 4 = 0\). We are given that \(-1\) is a root. This means if we substitute \(x = -1\) into the equation, it should satisfy the equation.
\[ 3(-1)^2 + p(-1) + 4 = 0 \]\[ 3(1) - p + 4 = 0 \]\[ 3 - p + 4 = 0 \]\[ 7 - p = 0 \]\[ p = 7 \]The second equation is \(7x^2 + px + k = 0\). Substitute \(p = 7\) into this equation:
\[ 7x^2 + 7x + k = 0 \]For a quadratic equation \(Ax^2 + Bx + C = 0\) to have equal roots, its discriminant must be zero, i.e., \(B^2 - 4AC = 0\).
In our second equation, \(7x^2 + 7x + k = 0\), we have:
Apply the discriminant condition:
\[ B^2 - 4AC = 0 \]\[ (7)^2 - 4(7)(k) = 0 \]\[ 49 - 28k = 0 \]From the equation \(49 - 28k = 0\):
\[ 28k = 49 \]\[ k = \frac{49}{28} \]Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 7:
\[ k = \frac{49 \div 7}{28 \div 7} \]\[ k = \frac{7}{4} \]B) \( \frac{7}{4} \) is the correct value for \(k\) as derived from the steps.
A sum of Rs 2,500 is invested in a bank at 6% interest per annum compounded annually. On maturity, the amount payable becomes Rs 2,809. Find the time period of the investment.
This problem involves compound interest. We are given the principal amount, the annual interest rate, and the final amount. We need to find the time period of the investment. We will use the compound interest formula to solve for the time period.
The formula for the amount (\(A\)) when interest is compounded annually is:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]We need to find what power of 1.06 equals 1.1236.
Let's test small integer values for \(T\):
Since \((1.06)^2 = 1.1236\), we have:
\[ (1.06)^2 = (1.06)^T \]Therefore, \(T = 2\) years.
Correct Option: C) 2 years
To find the missing number in the series, we need to identify the pattern or rule that governs the sequence of numbers. We will examine the differences between consecutive terms to see if a consistent arithmetic progression or a pattern in differences emerges.
The series with differences is:
Series: 2 3 6 12 22 37 58
1st Diff: 1 3 6 10 15 21
2nd Diff: 2 3 4 5 6
D) 22 is the number that correctly fits the identified pattern.
Ravi spends 20% of his income on petrol, one-third of the remaining income on household expenditure, and the balance is his savings. If he spends Rs 7,500 on petrol, find his household expenditure.
We need to find Ravi's household expenditure. We are given the percentage of income spent on petrol and the actual amount spent on petrol. We can use this information to find his total income. Once the total income is known, we can calculate the remaining income after petrol expenditure and then determine the household expenditure, which is one-third of this remaining amount.
Therefore, Ravi's household expenditure is Rs 10,000.
D) Rs 10,000 is the correct answer, as calculated above.
To determine which of the given numbers is a prime number, we need to test each option for divisibility by prime numbers up to its square root. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
B) 983
Let's check 983 for divisibility by prime numbers up to 31:
Since 983 is not divisible by any prime number up to its square root, 983 is a prime number.
The problem involves setting up a system of two linear equations based on the given conditions. We will define variables for the original number of pens and pencils, form equations from the total count and the changed scenario, solve for the variables, and then find the difference between the original numbers.
Let \(P\) be the original number of pens and \(C\) be the original number of pencils.
From the first statement, "Riya has a total of 60 pens and pencils," we can write the first equation:
\[ P + C = 60 \quad \text{(Equation 1)} \]From the second statement, "If she has 25 more pens and 5 less pencils, then the number of pens becomes three times the number of pencils."
New number of pens \( = P + 25 \)
New number of pencils \( = C - 5 \)
According to the condition:
\[ P + 25 = 3(C - 5) \]\[ P + 25 = 3C - 15 \]\[ P - 3C = -15 - 25 \]\[ P - 3C = -40 \quad \text{(Equation 2)} \]Now we have a system of two linear equations:
Equation 1: \( P + C = 60 \)
Equation 2: \( P - 3C = -40 \)
Subtract Equation 2 from Equation 1 to eliminate \(P\):
\[ (P + C) - (P - 3C) = 60 - (-40) \]\[ P + C - P + 3C = 60 + 40 \]\[ 4C = 100 \]\[ C = \frac{100}{4} \]\[ C = 25 \]Substitute the value of \(C\) back into Equation 1 to find \(P\):
\[ P + 25 = 60 \]\[ P = 60 - 25 \]\[ P = 35 \]So, originally Riya had 35 pens and 25 pencils.
The question asks for the difference between the number of pens and pencils she had originally:
\[ \text{Difference} = P - C \]\[ \text{Difference} = 35 - 25 \]\[ \text{Difference} = 10 \]Correct Option: D) 10
The problem presents two letter-cluster pairs (QJD : TMG and MMT : PPW) that follow a specific logical rule. We need to identify this rule and then apply it to the given options to find the pair that adheres to the same logic.
C) RPH : USK
- R (18) \(\rightarrow\) U (21): \((+3)\)
- P (16) \(\rightarrow\) S (19): \((+3)\)
- H (8) \(\rightarrow\) K (11): \((+3)\)
This pair follows the established logic.
If 60% of A = 1.2 of B = \( \frac{2}{5} \) of C, then find A : B : C.
To find the ratio A : B : C, we need to express A, B, and C in terms of a common constant. We are given three equalities involving percentages and fractions, which we will convert into a single equation and then find the relationship between A, B, and C.
Convert percentages and decimals to fractions:
Given: \[60\% \text{ of } A = 1.2 \text{ of } B = \frac{2}{5} \text{ of } C\]
Convert \(60\%\) to a fraction: \[60\% = \frac{60}{100} = \frac{3}{5}\]
Convert \(1.2\) to a fraction: \[1.2 = \frac{12}{10} = \frac{6}{5}\]
So, the given equation becomes:
\[ \frac{3}{5} A = \frac{6}{5} B = \frac{2}{5} C \]Equate each part to a constant:
Let \(k\) be a constant such that:
\[ \frac{3}{5} A = \frac{6}{5} B = \frac{2}{5} C = k \]Express A, B, and C in terms of \(k\):
From \(\frac{3}{5} A = k\), we get \(A = \frac{5k}{3}\)
From \(\frac{6}{5} B = k\), we get \(B = \frac{5k}{6}\)
From \(\frac{2}{5} C = k\), we get \(C = \frac{5k}{2}\)
Find the ratio A : B : C:
Substitute the expressions for A, B, and C:
\[ A : B : C = \frac{5k}{3} : \frac{5k}{6} : \frac{5k}{2} \]Simplify the ratio:
Divide all terms by \(5k\) (assuming \(k \neq 0\)):
\[ A : B : C = \frac{1}{3} : \frac{1}{6} : \frac{1}{2} \]Clear the denominators:
Find the least common multiple (LCM) of the denominators \(3, 6, 2\). The LCM is \(6\).
Multiply each term in the ratio by the LCM, \(6\):
\[ A : B : C = \left(\frac{1}{3} \times 6\right) : \left(\frac{1}{6} \times 6\right) : \left(\frac{1}{2} \times 6\right) \]\[ A : B : C = 2 : 1 : 3 \]D) The calculated ratio is 2 : 1 : 3.
The question asks to identify the head of a Nyaya Panchayat. This requires knowledge of the structure and function of local self-governance bodies in India, specifically the Panchayati Raj system and its judicial component.
A) Sarpanch — The Sarpanch is the elected head of the Gram Panchayat, which is the village-level self-governing institution. In many states, the Nyaya Panchayat, which is the judicial arm at the village level, is either headed by the Sarpanch or comprises members (including the Sarpanch) elected from the Gram Panchayat, with the Sarpanch often presiding over its meetings or being the primary figure. Therefore, the Sarpanch is the most appropriate answer for the head of the Nyaya Panchayat in the context of village-level administration.