This problem involves relating distance, speed, and time. We will set up equations based on the given information for Anmol and Nikhil's travel times under different conditions and then solve for Anmol's original speed.
Let Anmol's original speed be \(S_A\) km/hr and Nikhil's speed be \(S_N\) km/hr. The distance covered is \(D = 162\) km.
The time taken by Anmol is \[T_A = \frac{D}{S_A} = \frac{162}{S_A}\] hours.
The time taken by Nikhil is \[T_N = \frac{D}{S_N} = \frac{162}{S_N}\] hours.
According to the first condition, Anmol takes 3 hours more than Nikhil:
\[ T_A = T_N + 3 \]\[ \frac{162}{S_A} = \frac{162}{S_N} + 3 \quad \text{(Equation 1)} \]According to the second condition, if Anmol doubles his speed, his new speed is \(2S_A\). The new time taken by Anmol would be \(T_A' = \frac{162}{2S_A}\) hours.
This new time \(T_A'\) is 6 hours less than Nikhil's time \(T_N\):
\[ T_A' = T_N - 6 \]\[ \frac{162}{2S_A} = \frac{162}{S_N} - 6 \quad \text{(Equation 2)} \]We have a system of two equations. Let's subtract Equation 2 from Equation 1:
\[ \left( \frac{162}{S_A} \right) - \left( \frac{162}{2S_A} \right) = \left( \frac{162}{S_N} + 3 \right) - \left( \frac{162}{S_N} - 6 \right) \]\[ \frac{162}{S_A} - \frac{81}{S_A} = \frac{162}{S_N} + 3 - \frac{162}{S_N} + 6 \]\[ \frac{162 - 81}{S_A} = 9 \]\[ \frac{81}{S_A} = 9 \]Now, solve for \(S_A\):
\[ S_A = \frac{81}{9} \]\[ S_A = 9 \text{ km/hr} \]We can also find Nikhil's speed if needed. Substitute \(S_A = 9\) into Equation 1:
\[ \frac{162}{9} = \frac{162}{S_N} + 3 \]\[ 18 = \frac{162}{S_N} + 3 \]\[ 15 = \frac{162}{S_N} \]\[ S_N = \frac{162}{15} = 10.8 \text{ km/hr} \]Correct Option: A) 9 km/hr
Given that 170.24 = x, 170.51 = y and xz = y4, then the value of z is close to:
The problem involves solving for an unknown variable 'z' using given equations that relate powers of a base number. We will use the properties of exponents, specifically that \((a^m)^n = a^{mn}\) and if \(a^m = a^n\), then \(m=n\). We will express all terms with the same base and then equate the exponents.
We are given the following equations:
Substitute the expressions for \(x\) and \(y\) from the first two equations into the third equation:
\[ (17^{0.24})^z = (17^{0.51})^4 \]Apply the exponent rule \((a^m)^n = a^{mn}\) to both sides of the equation:
\[ 17^{(0.24) \times z} = 17^{(0.51) \times 4} \]Simplify the exponents:
\[ 17^{0.24z} = 17^{2.04} \]Since the bases are the same (17), we can equate the exponents:
\[ 0.24z = 2.04 \]Solve for \(z\):
\[ z = \frac{2.04}{0.24} \]To simplify the division, multiply the numerator and denominator by 100 to remove the decimals:
\[ z = \frac{204}{24} \]Perform the division:
\[ z = 8.5 \]To find the median of a given set of data, the first step is to arrange the data in ascending or descending order. Once ordered, the median is the middle value if the number of data points is odd. If the number of data points is even, the median is the average of the two middle values.
Correct Option: A) 57 — This is the correct median value after arranging the data in ascending order and finding the middle term.
If the longest diagonal of a cube measures 15 cm, then what is the total surface area (in sq. cm) of the cube?
540 sq.cm
180\(\sqrt{3}\) sq.cm
450 sq.cm
337.50 sq.cm
To find the total surface area of a cube, we first need to determine the length of its side. We are given the length of the longest diagonal of the cube. We can use the formula relating the diagonal and the side length of a cube to find the side length, and then use the side length to calculate the total surface area.
Correct Option: C) The total surface area of the cube is 450 sq.cm.
To find the amount of the trade discount, we need to calculate the specified percentage of the original price of the bicycle. The discount amount is simply the percentage discount multiplied by the original price.
The discount amount is 12% of Rs. 4,300.
\[ \text{Discount Amount} = \text{Discount Percentage} \times \text{Original Price} \] \[ \text{Discount Amount} = 12\% \times 4300 \]Convert the percentage to a decimal:
\[ 12\% = \frac{12}{100} = 0.12 \]Now, multiply this decimal by the original price:
\[ \text{Discount Amount} = 0.12 \times 4300 \] \[ \text{Discount Amount} = 516 \]The amount of the discount is Rs. 516.
Correct Option: D) Rs.516
To find the population at the end of the third year, we need to apply the percentage changes sequentially to the initial population. First, calculate the population after a 25% increase, then apply a 60% increase to that new population, and finally apply a 50% decrease to the population after the second year.
The current population of the town is \(P_0 = 16,362\).
An increase of 25% means the population becomes \(100\% + 25\% = 125\%\) of the original. This can be calculated as \(P_0 \times (1 + 0.25)\).
\[ P_1 = 16,362 \times (1 + 0.25) = 16,362 \times 1.25 \] \[ P_1 = 20,452.5 \]Since population must be a whole number, we can consider it as 20,453 or proceed with the decimal for intermediate calculations to maintain accuracy, and round at the end if necessary. For competitive exams, it's often better to keep fractions or decimals until the final step if possible, or round logically.
An increase of 60% means the population becomes \(100\% + 60\% = 160\%\) of the population at the end of the first year. This can be calculated as \(P_1 \times (1 + 0.60)\).
\[ P_2 = 20,452.5 \times (1 + 0.60) = 20,452.5 \times 1.60 \] \[ P_2 = 32,724 \]A decrease of 50% means the population becomes \(100\% - 50\% = 50\%\) of the population at the end of the second year. This can be calculated as \(P_2 \times (1 - 0.50)\).
\[ P_3 = 32,724 \times (1 - 0.50) = 32,724 \times 0.50 \] \[ P_3 = 16,362 \]Alternatively, we can use a single multiplicative factor:
\[ P_{\text{final}} = P_{\text{initial}} \times (1 + \text{rate}_1) \times (1 + \text{rate}_2) \times (1 - \text{rate}_3) \] \[ P_{\text{final}} = 16,362 \times (1 + 0.25) \times (1 + 0.60) \times (1 - 0.50) \] \[ P_{\text{final}} = 16,362 \times (1.25) \times (1.60) \times (0.50) \] \[ P_{\text{final}} = 16,362 \times (2.00) \times (0.50) \] \[ P_{\text{final}} = 16,362 \times 1 \] \[ P_{\text{final}} = 16,362 \]B) 16,362. The population at the end of the third year is 16,362, which is the same as the initial population.
To calculate the profit percentage, we first need to determine the cost price (CP) and selling price (SP) of a single pencil. Then, we can use the formula for profit percentage: \[\text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100\].
Arun bought pencils at the rate of Rs. 180 a dozen. A dozen means 12 pencils.
\[ \text{CP of 12 pencils} = \text{Rs. } 180 \]\[ \text{CP of 1 pencil} = \frac{180}{12} = \text{Rs. } 15 \]He sold them for Rs. 18 each.
\[ \text{SP of 1 pencil} = \text{Rs. } 18 \]Profit is the difference between Selling Price and Cost Price.
\[ \text{Profit} = \text{SP} - \text{CP} \]\[ \text{Profit} = 18 - 15 = \text{Rs. } 3 \]The formula for profit percentage is:
\[ \text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100 \]\[ \text{Profit Percentage} = \frac{3}{15} \times 100 \]\[ \text{Profit Percentage} = \frac{1}{5} \times 100 \]\[ \text{Profit Percentage} = 20\% \]A) 20% is the correct profit percentage.
To solve this problem, we first need to find the total number of parts in the given ratio. Then, we can determine the value of one part. Once we have the value of one part, we can calculate the individual shares of each person and subsequently find the difference between the largest and smallest shares.
Calculate the sum of the ratio parts:
The given ratio is \(19 : 11 : 18\).
\[ \text{Sum of ratio parts} = 19 + 11 + 18 = 48 \]Determine the value of one ratio part:
The total amount to be divided is Rs. 552.
\[ \text{Value of one part} = \frac{\text{Total amount}}{\text{Sum of ratio parts}} = \frac{552}{48} \]\[ \text{Value of one part} = 11.5 \]Calculate the individual shares:
First person's share \(= 19 \times 11.5 = \text{Rs. } 218.5\)
Second person's share \(= 11 \times 11.5 = \text{Rs. } 126.5\)
Third person's share \(= 18 \times 11.5 = \text{Rs. } 207\)
Identify the largest and smallest shares:
Largest share \(= \text{Rs. } 218.5\)
Smallest share \(= \text{Rs. } 126.5\)
Calculate the difference between the largest and smallest shares:
\[ \text{Difference} = \text{Largest share} - \text{Smallest share} \]\[ \text{Difference} = 218.5 - 126.5 = \text{Rs. } 92 \]Correct Option: C) The difference between the largest and the smallest shares is Rs. 92.
If X = \(\frac{\sin^2\theta}{1+\cos\theta}-\frac{\sin\theta}{\cot\theta-\mathrm{cosec} \theta}\), then the value of (X − 1) is:
cos²θ − cot²θ
cosec²θ + cot²θ
sec²θ − tan²θ
sin²θ + tan²θ
To find the value of \((X-1)\), we first need to simplify the expression for \(X\) using fundamental trigonometric identities. We will convert all terms to sine and cosine where possible and then simplify the algebraic expressions.
C) \(\sec^2\theta - \tan^2\theta\)
As calculated, \(X-1 = 1\). The identity \(\sec^2\theta - \tan^2\theta = 1\) matches this result.
To solve this problem, we need to evaluate the expression \((3^0 + 3^{-2} + 3^{-1}) \times 27\). This involves applying the rules of exponents, specifically that any non-zero number raised to the power of zero is 1, and a number raised to a negative exponent is the reciprocal of the number raised to the positive exponent. After simplifying the terms within the parenthesis, we will multiply the result by 27.
The expression becomes:
\[ \left(1 + \frac{1}{9} + \frac{1}{3}\right) \times 27 \]To add the fractions, find a common denominator, which is 9.
\[ 1 + \frac{1}{9} + \frac{1}{3} = \frac{9}{9} + \frac{1}{9} + \frac{3}{9} \]\[ = \frac{9 + 1 + 3}{9} \]\[ = \frac{13}{9} \]Now, multiply the sum by 27:
\[ \frac{13}{9} \times 27 \]\[ = 13 \times \frac{27}{9} \]\[ = 13 \times 3 \]\[ = 39 \]Correct Option: A) 39 is the correct value of the given expression.