To find the average of the first 150 odd natural numbers, we can use the formula for the sum of the first 'n' odd natural numbers and then divide by 'n'. Alternatively, we can recognize that the average of the first 'n' odd natural numbers is simply 'n'.
The sum of the first \(n\) odd natural numbers is given by the formula \(S_n = n^2\).
In this case, \(n = 150\), so the sum of the first 150 odd natural numbers is:
\[ S_{150} = (150)^2 = 22500 \]The average is the sum divided by the number of terms:
\[ \text{Average} = \frac{S_n}{n} = \frac{n^2}{n} = n \]Substitute \(n = 150\):
\[ \text{Average} = 150 \]It is a known property that the average of the first \(n\) odd natural numbers is \(n\).
Therefore, the average of the first 150 odd natural numbers is 150.
Correct Option: C) 150
If the interest earned during the 2nd year on a certain sum is ₹3,342, and the rate of interest is 20% per annum compounded annually, then the sum is:
The problem asks us to find the principal sum given the interest earned in the second year at a compound annual interest rate. We will use the formula for compound interest to express the interest for the first and second years in terms of the principal and then use the given information to solve for the principal.
Let the principal sum be \(P\). The rate of interest \(R = 20\%\) per annum.
Calculate the interest earned in the first year (\(CI_1\)).
\[ CI_1 = P \times \frac{R}{100} = P \times \frac{20}{100} = 0.2P \]Calculate the amount at the end of the first year (\(A_1\)). This amount becomes the principal for calculating the interest in the second year.
\[ A_1 = P + CI_1 = P + 0.2P = 1.2P \]Calculate the interest earned in the second year (\(CI_2\)). This is the interest on \(A_1\).
\[ CI_2 = A_1 \times \frac{R}{100} = 1.2P \times \frac{20}{100} = 1.2P \times 0.2 = 0.24P \]We are given that the interest earned during the second year is ₹3,342. So, we set up the equation:
\[ 0.24P = 3342 \]Solve for \(P\):
\[ P = \frac{3342}{0.24} \] \[ P = \frac{3342}{\frac{24}{100}} \] \[ P = \frac{3342 \times 100}{24} \] \[ P = \frac{334200}{24} \]Perform the division:
\[ P = 13925 \]So, the sum is ₹13,925.
B) ₹13,925 is the correct sum calculated based on the given interest for the second year and the compound interest rate.
This problem can be solved by setting up a system of two linear equations with two variables. We will assign variables to the father's age and the son's age, translate the given conditions into equations, and then solve the system of equations to find the father's age.
Let \(F\) be the father's age in years.
Let \(S\) be the son's age in years.
The first statement says: "If 2 times a son's age is added to his father's age, the sum is 36 years."
\[ F + 2S = 36 \quad \text{(Equation 1)} \]The second statement says: "If 2.5 times the father's age is added to the son's age, the sum is 72 years."
\[ S + 2.5F = 72 \quad \text{(Equation 2)} \]From Equation 1, we can express \(F\) in terms of \(S\):
\[ F = 36 - 2S \quad \text{(Equation 3)} \]Substitute Equation 3 into Equation 2:
\[ S + 2.5(36 - 2S) = 72 \]\[ S + 90 - 5S = 72 \]Combine like terms:
\[ 90 - 4S = 72 \]Subtract 90 from both sides:
\[ -4S = 72 - 90 \]\[ -4S = -18 \]Divide by -4 to find \(S\):
\[ S = \frac{-18}{-4} \]\[ S = 4.5 \text{ years} \]Now, substitute the value of \(S\) back into Equation 3 to find \(F\):
\[ F = 36 - 2(4.5) \]\[ F = 36 - 9 \]\[ F = 27 \text{ years} \]The father's age is 27 years.
A) 27 is the correct answer, as calculated by solving the system of linear equations.
To find the number of pairs of numbers satisfying the given conditions, we will use the relationship between HCF, LCM, and the product of two numbers. We'll express the numbers in terms of their HCF and then use the product to find the possible values for their co-prime factors.
Let the two numbers be \(A\) and \(B\). We are given that their HCF (Highest Common Factor) is 9.
If the HCF of two numbers is 9, then the numbers can be expressed as \(A = 9x\) and \(B = 9y\), where \(x\) and \(y\) are co-prime integers (i.e., their HCF is 1).
We are also given that the product of the two numbers is 8100.
So, \(A \times B = 8100\).
Substitute the expressions for \(A\) and \(B\) into the product equation:
\[ (9x) \times (9y) = 8100 \]\[ 81xy = 8100 \]Now, solve for the product \(xy\):
\[ xy = \frac{8100}{81} \]\[ xy = 100 \]We need to find pairs of co-prime integers \((x, y)\) whose product is 100. The pairs of factors of 100 are:
From these pairs, we must select only those where \(x\) and \(y\) are co-prime (HCF(x, y) = 1):
Thus, there are two pairs of co-prime integers \((x, y)\) whose product is 100: \((1, 100)\) and \((4, 25)\).
Each of these pairs corresponds to a unique pair of numbers \((A, B)\):
Both pairs satisfy the conditions: HCF is 9 and product is 8100.
Therefore, there are 2 such pairs of numbers.
C) 2
The problem asks us to find the greater of two consecutive positive 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 the approximate values of the numbers.
Alternatively, we can estimate. Since \(x(x+1) = 650\), \(x^2\) is approximately 650. We know that \(20^2 = 400\) and \(30^2 = 900\). So, \(x\) must be between 20 and 30. Let's try numbers close to \(\sqrt{650}\).
We know that \(25^2 = 625\) and \(26^2 = 676\). Since \(x(x+1) = 650\), \(x\) should be slightly less than \(\sqrt{650}\). Let's test \(x=25\):
\[ 25 \times (25+1) = 25 \times 26 \]\[ 25 \times 26 = 650 \]C) 26 — As determined from the calculation, the two consecutive positive natural numbers are 25 and 26, and the greater of these is 26.
The problem involves calculating time, speed, and distance. Since both Prashant and Nisha are traveling from Delhi to Kanpur, the distance covered by both is the same. We can first calculate the total distance using Prashant's speed and time, and then use this distance along with Nisha's speed to find the time she takes.
The fundamental formula is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]Since Prashant travels this distance, we use his speed and time:
\[ \text{Distance} = S_P \times T_P \]\[ \text{Distance} = 72 \text{ km/hr} \times 16 \text{ hours} \]\[ \text{Distance} = 1152 \text{ km} \]Nisha covers the same distance. We use the calculated distance and Nisha's speed:
\[ T_N = \frac{\text{Distance}}{S_N} \]\[ T_N = \frac{1152 \text{ km}}{64 \text{ km/hr}} \]\[ T_N = 18 \text{ hours} \]Correct Option: B) 18 hours
To find the final selling price after successive discounts, we need to apply each discount one after the other. First, calculate the price after the first discount. Then, apply the second discount to this new price.
Initial Marked Price:
\[ \text{Marked Price} = \text{₹}1,200 \]
First Discount (20%):
Calculate the amount of the first discount:
\[ \text{Discount}_1 = 20\% \text{ of } \text{₹}1,200 = \frac{20}{100} \times 1,200 = \text{₹}240 \]
Price after the first discount:
\[ \text{Price}_1 = \text{Marked Price} - \text{Discount}_1 = 1,200 - 240 = \text{₹}960 \]
Second Discount (10%):
The second discount is applied to the price after the first discount (₹960).
Calculate the amount of the second discount:
\[ \text{Discount}_2 = 10\% \text{ of } \text{₹}960 = \frac{10}{100} \times 960 = \text{₹}96 \]
Final Selling Price:
\[ \text{Final Selling Price} = \text{Price}_1 - \text{Discount}_2 = 960 - 96 = \text{₹}864 \]
When a discount of \(x\%\) is given, the price becomes \((100-x)\%\) of the original price. So, for successive discounts:
After a 20% discount, the price is \(100\% - 20\% = 80\%\) of the marked price.
\[ \text{Price}_1 = 1,200 \times \frac{80}{100} = 1,200 \times 0.8 = \text{₹}960 \]
After a 10% discount on \(\text{Price}_1\), the final price is \(100\% - 10\% = 90\%\) of \(\text{Price}_1\).
\[ \text{Final Selling Price} = 960 \times \frac{90}{100} = 960 \times 0.9 = \text{₹}864 \]
Correct Option: A) ₹864
To find the simple interest, we use the formula \(SI = \frac{P \times R \times T}{100}\), where \(P\) is the principal amount, \(R\) is the rate of interest per period, and \(T\) is the time period. It's crucial to ensure that the rate and time are consistent in their units (e.g., both per month or both per year).
The rate is given per month, and the time is given in months. Since both are consistent, we can directly use these values in the simple interest formula.
So, the simple interest is ₹252.
Correct Option: A) 252
This problem can be solved using the fundamental relationship between the H.C.F. (Highest Common Factor) and L.C.M. (Least Common Multiple) of two numbers. The product of two numbers is always equal to the product of their H.C.F. and L.C.M.
For any two positive integers, say \(a\) and \(b\), the relationship between their H.C.F. and L.C.M. is given by:
\[ a \times b = \text{H.C.F.}(a, b) \times \text{L.C.M.}(a, b) \]Let the first number be \(a = 102\) and the other number be \(b\). Then,
\[ 102 \times b = 51 \times 306 \]To find \(b\), divide the product of H.C.F. and L.C.M. by the given number:
\[ b = \frac{51 \times 306}{102} \]We can simplify the expression:
\[ b = \frac{51 \times 306}{2 \times 51} \]\[ b = \frac{306}{2} \]\[ b = 153 \]Thus, the other number is 153.
Correct Option: D) 153
To calculate the profit percentage, we first need to determine the cost price (CP) and selling price (SP) of a single sharpener. Then, we can use the formula for profit percentage: \[\text{Profit Percentage} = \frac{\text{SP} - \text{CP}}{\text{CP}} \times 100\].
Calculate the Cost Price (CP) per sharpener:
Nitin bought sharpeners at ₹180 per dozen. A dozen means 12 items.
\[ \text{CP of 12 sharpeners} = \text{₹}180 \]\[ \text{CP of 1 sharpener} = \frac{\text{₹}180}{12} = \text{₹}15 \]Identify the Selling Price (SP) per sharpener:
He sold them for ₹20 each.
\[ \text{SP of 1 sharpener} = \text{₹}20 \]Calculate the Profit per sharpener:
Profit is the difference between the selling price and the cost price.
\[ \text{Profit} = \text{SP} - \text{CP} = \text{₹}20 - \text{₹}15 = \text{₹}5 \]Calculate the Profit Percentage:
The profit percentage is calculated with respect to the cost price.
\[ \text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100 \]\[ \text{Profit Percentage} = \frac{5}{15} \times 100 \]\[ \text{Profit Percentage} = \frac{1}{3} \times 100 \]\[ \text{Profit Percentage} = 33.333... \% \]Round off to two decimal places:
The profit percentage rounded off to two decimal places is 33.33%.
A) 33.33 — As calculated, the profit percentage is 33.33%.