The problem asks us to find the greater of two consecutive natural numbers whose product is 650. We can solve this by setting up an algebraic equation or by estimating the square root of 650 to find numbers close to it.
D) 26 — As determined in the step-by-step solution, the two consecutive natural numbers are 25 and 26, and the greater of these is 26.
The problem involves calculating individual weights given average weights of groups of people. We will use the formula for average weight: \[\text{Average Weight} = \frac{\text{Total Weight}}{\text{Number of People}}\]. By setting up equations based on the given averages, we can solve for the unknown weights.
Let R, A, and M be the weights of Ram, Anwar, and Amit, respectively.
The average weight of Ram, Anwar, and Amit is 45 kg. This means:
\[ \frac{R + A + M}{3} = 45 \]\[ R + A + M = 45 \times 3 \]\[ R + A + M = 135 \quad \text{(Equation 1)} \]The average weight of Ram and Anwar is 44 kg. This means:
\[ \frac{R + A}{2} = 44 \]\[ R + A = 44 \times 2 \]\[ R + A = 88 \quad \text{(Equation 2)} \]The average weight of Anwar and Amit is 42 kg. This means:
\[ \frac{A + M}{2} = 42 \]\[ A + M = 42 \times 2 \]\[ A + M = 84 \quad \text{(Equation 3)} \]Now we have a system of three linear equations. We need to find the weight of Anwar (A).
Substitute Equation 2 into Equation 1:
\[ (R + A) + M = 135 \]\[ 88 + M = 135 \]\[ M = 135 - 88 \]\[ M = 47 \text{ kg} \]Now that we have the weight of Amit (M), substitute this value into Equation 3 to find the weight of Anwar (A):
\[ A + M = 84 \]\[ A + 47 = 84 \]\[ A = 84 - 47 \]\[ A = 37 \text{ kg} \]Thus, the weight of Anwar is 37 kg.
A) The weight of Anwar is calculated to be 37 kg, which matches this option.
Find the value of: 69-4 ÷ 6915 × 69-9
This problem involves simplifying an expression with a common base and different exponents. We will use the laws of exponents, specifically the rules for multiplication and division of powers with the same base: \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\).
The given expression is: \(69^{-4} \div 69^{15} \times 69^{-9}\)
First, let's handle the division part using the rule \(a^m \div a^n = a^{m-n}\):
\[ 69^{-4} \div 69^{15} = 69^{(-4) - (15)} = 69^{-4 - 15} = 69^{-19} \]Now, substitute this back into the original expression:
\[ 69^{-19} \times 69^{-9} \]Next, handle the multiplication part using the rule \(a^m \times a^n = a^{m+n}\):
\[ 69^{-19} \times 69^{-9} = 69^{(-19) + (-9)} = 69^{-19 - 9} = 69^{-28} \]Thus, the simplified value of the expression is \(69^{-28}\).
D) \(69^{-28}\) is the correct answer, as derived by applying the laws of exponents for division and multiplication.
To find the final amount the customer pays, we need to calculate the trade discount amount and then subtract it from the original price of the laptop. The trade discount is a percentage of the original price.
Original Price (Marked Price) = ₹46,000
Trade Discount Rate = 16%
The trade discount amount is 16% of the original price.
\[ \text{Discount Amount} = \text{Discount Rate} \times \text{Original Price} \] \[ \text{Discount Amount} = 16\% \times \text{₹}46,000 \] \[ \text{Discount Amount} = \frac{16}{100} \times 46000 \] \[ \text{Discount Amount} = 16 \times 460 \] \[ \text{Discount Amount} = \text{₹}7,360 \]The final amount is the original price minus the discount amount.
\[ \text{Final Amount} = \text{Original Price} - \text{Discount Amount} \] \[ \text{Final Amount} = \text{₹}46,000 - \text{₹}7,360 \] \[ \text{Final Amount} = \text{₹}38,640 \]Correct Option: A) ₹38,640
This problem involves concepts of speed, distance, and time. We are given information about the time taken by two individuals, Amit and Vinay, to cover a certain distance under different conditions. We need to set up equations based on the given information and solve for Amit's original speed.
Let Amit's original speed be \(S_A\) km/hr and Vinay's speed be \(S_V\) km/hr.
The distance to be covered is \(D = 102\) km.
Time taken by Amit is \[T_A = \frac{D}{S_A} = \frac{102}{S_A}\] hours.
Time taken by Vinay is \[T_V = \frac{D}{S_V} = \frac{102}{S_V}\] hours.
From the first condition, Amit takes 8 hours more than Vinay:
\[ T_A = T_V + 8 \]\[ \frac{102}{S_A} = \frac{102}{S_V} + 8 \quad \ldots(1) \]From the second condition, if Amit doubles his speed (\(2S_A\)), he would take 9 hours less than Vinay. Let the new time taken by Amit be \(T_A'\).
\[ T_A' = \frac{102}{2S_A} \]\[ T_A' = T_V - 9 \]\[ \frac{102}{2S_A} = \frac{102}{S_V} - 9 \quad \ldots(2) \]Now we have a system of two equations. Let's subtract equation (2) from equation (1):
\[ \left(\frac{102}{S_A}\right) - \left(\frac{102}{2S_A}\right) = \left(\frac{102}{S_V} + 8\right) - \left(\frac{102}{S_V} - 9\right) \]\[ \frac{102}{S_A} - \frac{102}{2S_A} = \frac{102}{S_V} + 8 - \frac{102}{S_V} + 9 \]\[ \frac{204 - 102}{2S_A} = 17 \]\[ \frac{102}{2S_A} = 17 \]Solve for \(S_A\):
\[ 102 = 17 \times 2S_A \]\[ 102 = 34S_A \]\[ S_A = \frac{102}{34} \]\[ S_A = 3 \text{ km/hr} \]We can also find Vinay's speed, if needed, by substituting \(S_A = 3\) into equation (1):
\[ \frac{102}{3} = \frac{102}{S_V} + 8 \]\[ 34 = \frac{102}{S_V} + 8 \]\[ 34 - 8 = \frac{102}{S_V} \]\[ 26 = \frac{102}{S_V} \]\[ S_V = \frac{102}{26} = \frac{51}{13} \approx 3.92 \text{ km/hr} \]Correct Option: B) 3 km/hr
Simplify: \(\frac{(1+\sin A)(\sec A-\tan A)}{\mathrm{cosec} A(1-\cos A)(\mathrm{cosec} A+\cot A)}\)
To simplify the given trigonometric expression, we will convert all terms into their sine and cosine equivalents. Then, we will use algebraic manipulations and fundamental trigonometric identities to reduce the expression to its simplest form.
The given expression is:
\[ \frac{(1+\sin A)(\sec A-\tan A)}{\mathrm{cosec} A(1-\cos A)(\mathrm{cosec} A+\cot A)} \]Convert all terms to sine and cosine:
Substitute these into the expression:
\[ \frac{(1+\sin A)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)}{\frac{1}{\sin A}(1-\cos A)\left(\frac{1}{\sin A}+\frac{\cos A}{\sin A}\right)} \]Simplify the terms in the parentheses:
\[ \frac{(1+\sin A)\left(\frac{1-\sin A}{\cos A}\right)}{\frac{1}{\sin A}(1-\cos A)\left(\frac{1+\cos A}{\sin A}\right)} \]Multiply the terms in the numerator and denominator:
Numerator:
\[ \frac{(1+\sin A)(1-\sin A)}{\cos A} = \frac{1-\sin^2 A}{\cos A} \]Using the identity \(\sin^2 A + \cos^2 A = 1\), we have \(1-\sin^2 A = \cos^2 A\). So, the numerator becomes:
\[ \frac{\cos^2 A}{\cos A} = \cos A \]Denominator:
\[ \frac{1}{\sin A}(1-\cos A)\left(\frac{1+\cos A}{\sin A}\right) = \frac{(1-\cos A)(1+\cos A)}{\sin^2 A} \]Using the algebraic identity \((a-b)(a+b) = a^2-b^2\), we get:
\[ \frac{1-\cos^2 A}{\sin^2 A} \]Using the identity \(\sin^2 A + \cos^2 A = 1\), we have \(1-\cos^2 A = \sin^2 A\). So, the denominator becomes:
\[ \frac{\sin^2 A}{\sin^2 A} = 1 \]Now, substitute the simplified numerator and denominator back into the main expression:
\[ \frac{\cos A}{1} = \cos A \]Thus, the simplified expression is \(\cos A\).
D) cos A — The step-by-step simplification leads directly to \(\cos A\).
This problem involves setting up a system of two linear equations with two variables based on the given age relationships. We will define variables for the mother's and daughter's present ages, translate the given statements into equations, solve the system of equations to find their ages, and then calculate the difference.
Define variables:
Let \(M\) be the mother's present age in years.
Let \(D\) be the daughter's present age in years.
Formulate the first equation:
"2 times a mother's present age is 26 years more than 6 times her daughter's present age."
This translates to:
\[ 2M = 6D + 26 \quad \text{(Equation 1)} \]Formulate the second equation:
"4 times the daughter's present age is 4 years less than the mother's present age."
This translates to:
\[ 4D = M - 4 \quad \text{(Equation 2)} \]Solve the system of equations:
From Equation 2, we can express \(M\) in terms of \(D\):
\[ M = 4D + 4 \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 1:
Substitute the expression for \(M\) from Equation 3 into Equation 1:
\[ 2(4D + 4) = 6D + 26 \]\[ 8D + 8 = 6D + 26 \]Solve for \(D\):
Subtract \(6D\) from both sides:
\[ 8D - 6D + 8 = 26 \]\[ 2D + 8 = 26 \]Subtract 8 from both sides:
\[ 2D = 26 - 8 \]\[ 2D = 18 \]Divide by 2:
\[ D = \frac{18}{2} \]\[ D = 9 \]Solve for \(M\):
Substitute the value of \(D\) back into Equation 3:
\[ M = 4(9) + 4 \]\[ M = 36 + 4 \]\[ M = 40 \]Calculate the difference in ages:
The difference between the ages of the mother and the daughter is \(M - D\):
\[ \text{Difference} = 40 - 9 \]\[ \text{Difference} = 31 \]C) 31
The simple interest on a certain sum for 2 \( \frac{7}{8} \) years at the rate of 2.9% per annum is ₹8,671. What will be the amount payable on the same sum at the rate of 4.5% per annum simple interest in 4 \( \frac{2}{5} \) years?
The problem involves calculating simple interest. We are given the simple interest for a certain sum, time, and rate. First, we need to find the principal amount using the given information. Once the principal is known, we can calculate the amount payable (Principal + Simple Interest) for a different time period and interest rate.
D) ₹1,24,592 is the correct amount payable after calculating the principal from the first set of conditions and then applying the new rate and time to find the new simple interest, which is added to the principal.
Find the value of x satisfying: 3(3x2 − 6) − 3(3x2 + 5x − 7) = 19
-\( \frac{13}{15} \)
-\( \frac{26}{15} \)
-\( \frac{16}{15} \)
-\( \frac{14}{15} \)
The problem requires us to solve a linear equation for the variable \(x\). We will simplify the given equation by distributing the constants, combining like terms, and then isolating \(x\).
\(3(3x^2 - 6) - 3(3x^2 + 5x - 7) = 19\)
Multiply \(3\) by each term in the first parenthesis and \(-3\) by each term in the second parenthesis.
\[ 3 \cdot 3x^2 - 3 \cdot 6 - 3 \cdot 3x^2 - 3 \cdot 5x - 3 \cdot (-7) = 19 \]\[ 9x^2 - 18 - 9x^2 - 15x + 21 = 19 \]Group the \(x^2\) terms, \(x\) terms, and constant terms together.
\[ (9x^2 - 9x^2) - 15x + (-18 + 21) = 19 \]\[ 0x^2 - 15x + 3 = 19 \]\[ -15x + 3 = 19 \]Subtract \(3\) from both sides of the equation.
\[ -15x = 19 - 3 \]\[ -15x = 16 \]Divide both sides by \(-15\).
\[ x = \frac{16}{-15} \]\[ x = -\frac{16}{15} \]C) \(-\frac{16}{15}\) is the correct value of \(x\) obtained after solving the equation.
When partners invest different amounts in a business, their profit shares are typically proportional to their individual investments. We will first find the ratio of their investments, then calculate the total sum of the ratio parts, and finally determine Uday's share of the profit based on this ratio.
The ratio of their investments (Gopal : Hema : Uday) is:
\[ 1130 : 1030 : 1840 \]Divide each number by 10 to simplify:
\[ 113 : 103 : 184 \]Sum of ratio parts \( = 113 + 103 + 184 = 400 \)
The total profit is ₹1,800. Uday's share will be his ratio part divided by the total sum of ratio parts, multiplied by the total profit.
\[ \text{Uday's share} = \frac{\text{Uday's ratio part}}{\text{Sum of ratio parts}} \times \text{Total profit} \]\[ \text{Uday's share} = \frac{184}{400} \times 1800 \]\[ \text{Uday's share} = 184 \times \frac{1800}{400} \]\[ \text{Uday's share} = 184 \times 4.5 \]\[ \text{Uday's share} = 828 \]Therefore, Uday's share in the profit is ₹828.
D) ₹828 is the correct share of Uday in the profit, calculated based on the proportion of his investment.