We are given the sum and the Least Common Multiple (LCM) of two numbers. We need to find the two numbers. A key property relating two numbers, their Highest Common Factor (HCF), and their LCM is that the product of the two numbers is equal to the product of their HCF and LCM. We can use this property along with the given sum to find the numbers. Alternatively, we can test the given options to see which pair satisfies both conditions.
Let the two numbers be \(a\) and \(b\).
Given that their sum is 57:
\[ a + b = 57 \quad \text{(Equation 1)} \]Given that their L.C.M. is 812:
\[ \text{LCM}(a, b) = 812 \]We know the property that for any two numbers \(a\) and \(b\), the product of the numbers is equal to the product of their HCF and LCM:
\[ a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b) \]Let's test the given options to see which pair satisfies both conditions.
Option A: 21, 36
Sum: \(21 + 36 = 57\). This condition is satisfied.
Now, let's find the LCM of 21 and 36.
Prime factorization of \(21 = 3 \times 7\)
Prime factorization of \(36 = 2^2 \times 3^2\)
\( \text{LCM}(21, 36) = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 36 \times 7 = 252 \).
Since \(252 \neq 812\), this option is incorrect.
Option B: 26, 31
Sum: \(26 + 31 = 57\). This condition is satisfied.
Now, let's find the LCM of 26 and 31.
Prime factorization of \(26 = 2 \times 13\)
Prime factorization of \(31 = 31\) (31 is a prime number)
\( \text{LCM}(26, 31) = 2 \times 13 \times 31 = 26 \times 31 = 806 \).
Since \(806 \neq 812\), this option is incorrect.
Option C: 28, 29
Sum: \(28 + 29 = 57\). This condition is satisfied.
Now, let's find the LCM of 28 and 29.
Prime factorization of \(28 = 2^2 \times 7\)
Prime factorization of \(29 = 29\) (29 is a prime number)
Since 28 and 29 are coprime (their HCF is 1), their LCM is their product.
\( \text{LCM}(28, 29) = 28 \times 29 \).
Let's calculate the product:
\[ 28 \times 29 = 28 \times (30 - 1) = 28 \times 30 - 28 \times 1 = 840 - 28 = 812 \]Since \( \text{LCM}(28, 29) = 812 \), this option satisfies both conditions.
Option D: 33, 24
Sum: \(33 + 24 = 57\). This condition is satisfied.
Now, let's find the LCM of 33 and 24.
Prime factorization of \(33 = 3 \times 11\)
Prime factorization of \(24 = 2^3 \times 3\)
\( \text{LCM}(33, 24) = 2^3 \times 3 \times 11 = 8 \times 3 \times 11 = 24 \times 11 = 264 \).
Since \(264 \neq 812\), this option is incorrect.
C) 28, 29 — This pair satisfies both conditions: their sum is \(28 + 29 = 57\) and their LCM is \(28 \times 29 = 812\).
To find the greatest number that divides given numbers leaving specific remainders, we first subtract the remainders from the respective numbers. The resulting numbers will be perfectly divisible by the required greatest number. Therefore, the problem reduces to finding the HCF (Highest Common Factor) of these new numbers.
Identify the given numbers and their respective remainders:
Subtract the remainders from their corresponding numbers to get the numbers that are perfectly divisible by the required greatest number:
Now, we need to find the HCF of 123, 82, and 205.
Find the prime factors of each of these new numbers:
Identify the common prime factor among 123, 82, and 205. The common prime factor is 41.
The HCF is the product of the common prime factors. In this case, the only common prime factor is 41.
Therefore, the greatest number that divides 131, 91, and 215, leaving remainders 8, 9, and 10 respectively, is 41.
D) 41 is the HCF of 123, 82, and 205, and thus the greatest number satisfying the given conditions.
By selling an article at \( \frac{2}{5} \) of its actual selling price, Nirbhay incurs a loss of 26%. If he sells it at 84% of its actual selling price, then the profit percentage is:
We are given a scenario where an article is sold at a fraction of its actual selling price, resulting in a loss. We need to find the profit percentage if it were sold at a different fraction of its actual selling price. We will use the concepts of cost price (CP), selling price (SP), loss percentage, and profit percentage.
To find the average speed for a journey where the distance covered in both directions is the same but the speeds are different, we can use the formula for average speed, which is the total distance divided by the total time. Alternatively, for a two-way journey with equal distances, a specific formula for average speed can be used: \[\text{Average Speed} = \frac{2xy}{x+y}\], where \(x\) and \(y\) are the speeds for the two legs of the journey.
Identify the given speeds:
Understand the journey: The man travels from Visakhapatnam to Nagpur and returns to Visakhapatnam via the same route. This means the distance covered in both directions is the same.
Apply the formula for average speed for equal distances: When the distance covered in each direction is the same, the average speed can be calculated using the harmonic mean of the two speeds:
\[ \text{Average Speed} = \frac{2xy}{x+y} \]Substitute the given values into the formula:
\[ \text{Average Speed} = \frac{2 \times 55 \times 66}{55 + 66} \]Calculate the sum in the denominator:
\[ 55 + 66 = 121 \]Substitute the sum back into the formula:
\[ \text{Average Speed} = \frac{2 \times 55 \times 66}{121} \]Simplify the expression:
So, we can cancel out \(11\) from the numerator and denominator:
\[ \text{Average Speed} = \frac{2 \times (5 \times 11) \times (6 \times 11)}{11 \times 11} \]\[ \text{Average Speed} = \frac{2 \times 5 \times 6 \times 11 \times 11}{11 \times 11} \]\[ \text{Average Speed} = 2 \times 5 \times 6 \]\[ \text{Average Speed} = 10 \times 6 \]\[ \text{Average Speed} = 60 \text{ km/hr} \]Correct Option: B) 60
To find the average of the remaining two numbers, we first need to determine their sum. We are given the total sum of five numbers, the average of the first two, and the value of the third number. We can use this information to calculate the sum of the first two numbers and then subtract the known values from the total sum to find the sum of the remaining two numbers. Finally, we divide this sum by 2 to get their average.
So, the sum of the first two numbers is:
\[ \text{Sum of first two numbers} = \text{Average} \times \text{Count} = 80 \times 2 = 160 \]Correct Option: D) 197
Given that 580.5 = x, 580.79 = y and xz = y10, then the value of z is close to:
The problem involves simplifying expressions with exponents and then solving for an unknown variable, \(z\). We will use the properties of exponents, specifically that \((a^m)^n = a^{mn}\) and \(a^m \cdot a^n = a^{m+n}\), and if \(a^m = a^n\), then \(m=n\).
We are given three relationships:
Substitute \(x = 58^{0.5}\) and \(y = 58^{0.79}\) into the equation \(x^z = y^{10}\):
\[ (58^{0.5})^z = (58^{0.79})^{10} \]Since the bases are the same (58), the exponents must be equal:
\[ 0.5z = 7.9 \]Divide both sides by 0.5:
\[ z = \frac{7.9}{0.5} \] \[ z = \frac{79}{5} \] \[ z = 15.8 \]Correct Option: D) 15.8
Pallavi bought a toy for ₹740, decorated it at a cost of ₹16 and sold the toy for ₹558. Find the loss or profit percentage.
550% loss
\( \frac{560}{21} \) % profit
\( \frac{550}{21} \) % profit
\( \frac{550}{21} \) % loss
To find the loss or profit 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 of this profit or loss relative to the total cost price.
Calculate the Total Cost Price (CP):
The cost price includes the initial purchase price and any additional expenses incurred to prepare the item for sale.
Identify the Selling Price (SP):
The selling price is the amount for which the toy was sold.
Determine Profit or Loss:
Compare the Total CP with the SP. If SP > Total CP, it's a profit. If SP < Total CP, it's a loss.
Here, \( \text{SP} = \text{₹}558 \) and \( \text{Total CP} = \text{₹}756 \). Since \( \text{SP} < \text{Total CP} \), there is a loss.
\[ \text{Loss} = \text{Total CP} - \text{SP} \]Calculate the Loss Percentage:
The loss percentage is calculated using the formula:
Simplify the fraction:
Both 198 and 756 are divisible by common factors. Let's simplify step by step.
Divide by 2:
\[ \frac{198}{756} = \frac{99}{378} \]Divide by 9:
\[ \frac{99}{378} = \frac{11}{42} \]So, the loss percentage is:
\[ \text{Loss Percentage} = \frac{11}{42} \times 100 \% = \frac{1100}{42} \% \]Divide by 2 again:
\[ \text{Loss Percentage} = \frac{550}{21} \% \]Since it's a loss, the answer is \( \frac{550}{21} \% \) loss.
D) \( \frac{550}{21} \% \) loss
A) 550% loss — This is incorrect because the calculation \( \frac{550}{21} \) is approximately 26.19%, not 550%.
B) \( \frac{560}{21} \% \) profit — This is incorrect because there was a loss, not a profit, and the calculated percentage is different.
C) \( \frac{550}{21} \% \) profit — This is incorrect because there was a loss, not a profit. The magnitude of the percentage is correct but the nature (profit/loss) is wrong.
To find the original number, we need to set up an equation based on the given information. The problem states that when a number is increased by 80%, it results in 2790. We can represent the unknown number as a variable, say \(x\), and then form an equation to solve for \(x\).
Thus, the number is 1550.
Correct Option: C) 1550
To find the net price after a trade discount, we first calculate the amount of the discount based on the listed price. Then, we subtract this discount amount from the listed price to get the final net price.
Correct Option: D) ₹630
A cylindrical rod has an outer curved surface area of 3300 cm². If the length of the rod is 72 cm, then the outer radius (in cm) of the rod, correct to two places of decimal, is: (Take π = \( \frac{22}{7} \))
To find the outer radius of the cylindrical rod, we will use the formula for the curved surface area of a cylinder. We are given the curved surface area and the length (height) of the rod, and we need to solve for the radius.
The curved surface area of a cylinder is given by:
\[ \text{CSA} = 2 \pi r h \]where \(r\) is the radius and \(h\) is the height (length) of the cylinder.First, simplify the right side of the equation:
\[ 3300 = \frac{44 \times 72}{7} \times r \]\[ 3300 = \frac{3168}{7} \times r \]Multiply both sides by 7 and divide by 3168:
\[ r = \frac{3300 \times 7}{3168} \]\[ r = \frac{23100}{3168} \]Correct Option: B) 7.29
The outer radius of the rod, correct to two decimal places, is 7.29 cm.