This problem involves setting up and solving linear equations based on the given age conditions at different points in time. We will define variables for the current ages of the father and son, then use the information from 4 years ago to form the first equation. Finally, we will use the future condition to form a second equation and solve for the number of years required.
Define current ages:
Let the current age of the father be \(F\) years.
Let the current age of the son be \(S\) years.
Formulate the equation for 4 years ago:
4 years ago, the father's age was \(F - 4\).
4 years ago, the son's age was \(S - 4\).
According to the problem, 4 years ago, the father's age was 33 years more than twice his son's age:
\[ F - 4 = 2(S - 4) + 33 \]\[ F - 4 = 2S - 8 + 33 \]\[ F - 4 = 2S + 25 \]\[ F = 2S + 29 \quad \text{(Equation 1)} \]Formulate the equation for the future condition:
Let \(x\) be the number of years from now after which the father will be twice his son's age.
After \(x\) years, the father's age will be \(F + x\).
After \(x\) years, the son's age will be \(S + x\).
According to the problem, after \(x\) years, the father will be twice his son's age:
\[ F + x = 2(S + x) \]\[ F + x = 2S + 2x \]\[ F = 2S + x \quad \text{(Equation 2)} \]Solve for \(x\):
We have two expressions for \(F\). Equate Equation 1 and Equation 2:
\[ 2S + 29 = 2S + x \]Simplify and find \(x\):
Subtract \(2S\) from both sides:
\[ 29 = x \]So, after 29 years from now, the father will be twice his son's age.
A) 29 is the correct answer, as calculated from the given conditions.
To find the other diagonal of a rhombus, we first need to determine the length of its side using the given perimeter. Then, we can use the properties of a rhombus, specifically that its diagonals bisect each other at right angles, and apply the Pythagorean theorem to find half the length of the unknown diagonal.
B) 12 cm is the correct answer, as calculated above.
To find the average speed for the entire journey, we need to calculate the total distance traveled and the total time taken. The average speed is then the total distance divided by the total time.
Correct Option: D) 73
The value of \(\sqrt{169}\) + \(\sqrt{0.0324}\) − \(\sqrt{7.29}\) is _____.
To solve this problem, we need to calculate the square root of each term separately and then perform the indicated addition and subtraction operations. We will convert decimal numbers to fractions to simplify finding their square roots.
We know that \(13 \times 13 = 169\).
\[ \sqrt{169} = 13 \]First, convert the decimal to a fraction:
\[ 0.0324 = \frac{324}{10000} \]Now, find the square root:
\[ \sqrt{0.0324} = \sqrt{\frac{324}{10000}} = \frac{\sqrt{324}}{\sqrt{10000}} \]We know that \(18 \times 18 = 324\) and \(100 \times 100 = 10000\).
\[ \frac{\sqrt{324}}{\sqrt{10000}} = \frac{18}{100} = 0.18 \]First, convert the decimal to a fraction:
\[ 7.29 = \frac{729}{100} \]Now, find the square root:
\[ \sqrt{7.29} = \sqrt{\frac{729}{100}} = \frac{\sqrt{729}}{\sqrt{100}} \]We know that \(27 \times 27 = 729\) and \(10 \times 10 = 100\).
\[ \frac{\sqrt{729}}{\sqrt{100}} = \frac{27}{10} = 2.7 \]Substitute the calculated square root values back into the original expression:
\[ \sqrt{169} + \sqrt{0.0324} - \sqrt{7.29} = 13 + 0.18 - 2.7 \]First, add the positive terms:
\[ 13 + 0.18 = 13.18 \]Now, subtract the last term:
\[ 13.18 - 2.7 = 10.48 \]Thus, the value of the expression is \(10.48\).
D) The calculated value of the expression is \(10.48\), which matches option D.
To find the mode of a given set of data, we need to identify the value that appears most frequently in the dataset. This involves counting the occurrences of each unique number.
Correct Option: A) 13 is the mode because it appears 4 times, which is the highest frequency among all the given marks.
To find the measure of \(\angle AEB\), we will use the properties of triangles, specifically the sum of angles in a triangle and the exterior angle theorem. We will first find \(\angle C\) using \(\triangle BDC\), then find \(\angle ACB\) (which is the same as \(\angle C\)). After that, we will find \(\angle AEC\) using \(\triangle AEC\) and finally \(\angle AEB\) as it forms a linear pair with \(\angle AEC\).
Consider \(\triangle BDC\).
Given that \(BD \perp AC\) at \(D\), we have \(\angle BDC = 90^\circ\).
We are also given \(\angle DBC = 63^\circ\).
The sum of angles in a triangle is \(180^\circ\). So, in \(\triangle BDC\):
\[ \angle BDC + \angle DBC + \angle BCD = 180^\circ \] \[ 90^\circ + 63^\circ + \angle BCD = 180^\circ \] \[ 153^\circ + \angle BCD = 180^\circ \] \[ \angle BCD = 180^\circ - 153^\circ \] \[ \angle BCD = 27^\circ \]
Thus, \(\angle C = 27^\circ\).
Now consider \(\triangle AEC\).
We know \(\angle C = 27^\circ\) and we are given \(\angle CAE = 76^\circ\).
The sum of angles in \(\triangle AEC\) is \(180^\circ\):
\[ \angle CAE + \angle ACE + \angle AEC = 180^\circ \] \[ 76^\circ + 27^\circ + \angle AEC = 180^\circ \] \[ 103^\circ + \angle AEC = 180^\circ \] \[ \angle AEC = 180^\circ - 103^\circ \] \[ \angle AEC = 77^\circ \]
Angles \(\angle AEB\) and \(\angle AEC\) form a linear pair on the line \(BC\).
Therefore, their sum is \(180^\circ\):
\[ \angle AEB + \angle AEC = 180^\circ \] \[ \angle AEB + 77^\circ = 180^\circ \] \[ \angle AEB = 180^\circ - 77^\circ \] \[ \angle AEB = 103^\circ \]
Correct Option: B) 103°
Ayushi bought a toy for ₹461, decorated it at a cost of ₹15, and sold the toy for ₹918. Find the loss or gain percentage.
\( \frac{660}{7} \) % loss
\( \frac{650}{7} \) % loss
\( \frac{650}{7} \) % profit
650 % profit
To find the loss or gain percentage, we first need to calculate the total cost price (CP) of the toy, which includes the purchase price and the decoration cost. Then, we compare this total CP with the selling price (SP) to determine if there was a profit or a loss. Finally, we calculate the percentage based on the total CP.
The cost price of the toy is the sum of the initial purchase price and the cost incurred for decorating it.
\[ \text{Initial purchase price} = \text{₹}461 \]\[ \text{Decoration cost} = \text{₹}15 \]
\[ \text{Total CP} = \text{Initial purchase price} + \text{Decoration cost} \]
\[ \text{Total CP} = \text{₹}461 + \text{₹}15 = \text{₹}476 \]
The selling price of the toy is given as ₹918.
\[ \text{SP} = \text{₹}918 \]
Compare the Total CP with the SP.
Since \( \text{SP} (\text{₹}918) > \text{Total CP} (\text{₹}476) \), there is a profit.
\[ \text{Profit} = \text{SP} - \text{Total CP} \]
\[ \text{Profit} = \text{₹}918 - \text{₹}476 = \text{₹}442 \]
The profit percentage is calculated using the formula:
\[ \text{Profit Percentage} = \left( \frac{\text{Profit}}{\text{Total CP}} \right) \times 100\% \]
\[ \text{Profit Percentage} = \left( \frac{442}{476} \right) \times 100\% \]
Simplify the fraction:
Divide both numerator and denominator by their greatest common divisor. Both 442 and 476 are divisible by 2.
\[ \frac{442}{2} = 221 \]
\[ \frac{476}{2} = 238 \]
So, the fraction becomes \( \frac{221}{238} \). Now, 221 is \(13 \times 17\), and 238 is \(2 \times 7 \times 17\). Both are divisible by 17.
\[ \frac{221}{17} = 13 \]
\[ \frac{238}{17} = 14 \]
So, the simplified fraction is \( \frac{13}{14} \).
\[ \text{Profit Percentage} = \left( \frac{13}{14} \right) \times 100\% \]
\[ \text{Profit Percentage} = \frac{1300}{14}\% \]
Divide 1300 by 14:
\[ \frac{1300}{14} = \frac{650}{7}\% \]
Correct Option: C) The profit percentage is \( \frac{650}{7} \% \).
The simple interest on a certain sum for 4 \( \frac{5}{9} \) years at the rate of 3.6% per annum is ₹5,453. What will be the amount payable on the same sum at the rate of 2.7% per annum simple interest in 6 \( \frac{2}{9} \) years?
The problem involves simple interest calculations. First, we need to find the principal sum using the given simple interest, time, and rate. Once the principal is known, we can calculate the amount payable for a different time period and interest rate using the simple interest formula and then adding it to the principal.
Calculate the Principal (P) from the first scenario:
Given:
The formula for Simple Interest is: \[\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\]
Rearranging to find P:
\[ \text{P} = \frac{\text{SI} \times 100}{\text{R} \times \text{T}} \] \[ \text{P} = \frac{5453 \times 100}{\frac{3.6}{100} \times \frac{41}{9}} \]Let's re-evaluate the formula application:
\[ 5453 = \frac{\text{P} \times 3.6 \times \frac{41}{9}}{100} \] \[ \text{P} = \frac{5453 \times 100 \times 9}{3.6 \times 41} \] \[ \text{P} = \frac{545300 \times 9}{3.6 \times 41} \] \[ \text{P} = \frac{4907700}{147.6} \] \[ \text{P} = 33250 \]So, the principal sum is ₹33,250.
Calculate the Simple Interest (SI') for the second scenario:
Given:
Using the formula: \[\text{SI'} = \frac{\text{P} \times \text{R'} \times \text{T'}}{100}\]
\[ \text{SI'} = \frac{33250 \times 2.7 \times \frac{56}{9}}{100} \] \[ \text{SI'} = \frac{33250 \times 2.7 \times 56}{100 \times 9} \] \[ \text{SI'} = \frac{33250 \times 0.3 \times 56}{100} \quad \text{(since } 2.7/9 = 0.3) \] \[ \text{SI'} = \frac{33250 \times 16.8}{100} \] \[ \text{SI'} = 332.5 \times 16.8 \] \[ \text{SI'} = 5586 \]So, the simple interest in the second scenario is ₹5,586.
Calculate the total Amount Payable:
Amount = Principal + Simple Interest
\[ \text{Amount} = \text{P} + \text{SI'} \] \[ \text{Amount} = 33250 + 5586 \] \[ \text{Amount} = 38836 \]The amount payable will be ₹38,836.
Correct Option: D) ₹38,836
To find the simple interest, we use the formula for simple interest: \[\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\], where \(\text{P}\) is the principal amount, \(\text{R}\) is the rate of interest per annum, and \(\text{T}\) is the time in years.
The formula for simple interest is:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]Therefore, the simple interest is ₹176.
Correct Option: A) 176 is the correct simple interest calculated using the given principal, rate, and time.
The problem can be solved using the fundamental relationship between the Highest Common Factor (HCF), the Least Common Multiple (LCM), and the product of two numbers. This relationship states that for any two positive integers, the product of the numbers is equal to the product of their HCF and LCM.
For two numbers, say \(a\) and \(b\), the relationship is:
\[ a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b) \]Let the first number be \(a = 82\) and the other number be \(b\).
\[ 82 \times b = 2 \times 410 \]Divide both sides by 82:
\[ b = \frac{820}{82} \]\[ b = 10 \]The other number is 10.
C) 10 is the correct answer. When the HCF is 2, the LCM is 410, and one number is 82, the other number is found to be 10 by applying the HCF-LCM product rule.