To find the product of 45 and 132, we need to perform multiplication. This can be done using standard multiplication methods or by breaking down one of the numbers for easier calculation.
The product of 45 and 132 is 5940.
Correct Option: C) 5940
To solve this problem, we need to perform the addition and subtraction of decimal numbers. It's best to perform the operations sequentially from left to right, or first add all positive numbers and then subtract the negative numbers. Aligning the decimal points is crucial for accurate calculation.
C) 20.0115 is the correct result obtained after performing the addition and subtraction of the given decimal numbers.
To solve this problem, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)). In this case, we first perform the division and then the subtraction.
Thus, \(5004 \div 139 - 6 = 30\).
Correct Option: B) 30
If 2 tables and 3 chairs cost Rs 3,500 and 3 tables and 2 chairs cost Rs 4,000, then how much does a table cost?
This problem can be solved by setting up a system of two linear equations with two variables. We will represent the cost of a table and a chair with variables and then solve for the cost of a table using methods like substitution or elimination.
Let \(T\) be the cost of one table and \(C\) be the cost of one chair.
From the first statement, "2 tables and 3 chairs cost Rs 3,500", we can form the equation:
\[ 2T + 3C = 3500 \quad \text{(Equation 1)} \]From the second statement, "3 tables and 2 chairs cost Rs 4,000", we can form the equation:
\[ 3T + 2C = 4000 \quad \text{(Equation 2)} \]To eliminate \(C\) and solve for \(T\), we can multiply Equation 1 by 2 and Equation 2 by 3:
Multiply Equation 1 by 2:
\[ 2 \times (2T + 3C) = 2 \times 3500 \]\[ 4T + 6C = 7000 \quad \text{(Equation 3)} \]Multiply Equation 2 by 3:
\[ 3 \times (3T + 2C) = 3 \times 4000 \]\[ 9T + 6C = 12000 \quad \text{(Equation 4)} \]Now, subtract Equation 3 from Equation 4 to eliminate \(C\):
\[ (9T + 6C) - (4T + 6C) = 12000 - 7000 \]\[ 9T - 4T + 6C - 6C = 5000 \]\[ 5T = 5000 \]Solve for \(T\):
\[ T = \frac{5000}{5} \]\[ T = 1000 \]Thus, the cost of one table is Rs 1,000.
C) 1,000
A) 500 — This would be the cost of a chair (Rs 500) if you substitute \(T=1000\) into one of the original equations (e.g., \(2(1000) + 3C = 3500 \implies 2000 + 3C = 3500 \implies 3C = 1500 \implies C = 500\)). It is not the cost of a table.
B) 750 — This value does not satisfy the given equations for either a table or a chair.
D) 15,000 — This value is significantly higher than the total costs given in the problem and does not satisfy the equations.
The value of \(\sqrt{0.4}\) is:
To find the value of \(\sqrt{0.4}\), we can convert the decimal to a fraction and then simplify the square root. Alternatively, we can estimate the value by considering perfect squares near 0.4.
First, convert \(0.4\) into a fraction:
\[ 0.4 = \frac{4}{10} \]Simplify the fraction \(\frac{4}{10}\) by dividing both numerator and denominator by their greatest common divisor, which is 2:
\[ \frac{4}{10} = \frac{2}{5} \]Now, we need to find the square root of \(\frac{2}{5}\):
\[ \sqrt{0.4} = \sqrt{\frac{2}{5}} \]To simplify further, we can rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{5}\):
\[ \sqrt{\frac{2}{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{2} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{10}}{5} \]We know that \(\sqrt{9} = 3\) and \(\sqrt{16} = 4\). So, \(\sqrt{10}\) will be slightly greater than 3. A common approximation for \(\sqrt{10}\) is \(3.16\).
\[ \frac{\sqrt{10}}{5} \approx \frac{3.16}{5} \]The calculated value \(0.632\) is closest to \(0.63\).
D) 0.63. The calculation \(\sqrt{0.4} \approx 0.632\) rounds to \(0.63\).
This problem involves calculating averages. We can set up an equation based on the definition of average (total runs / number of innings) and solve for the unknown initial average. Then, we can find the new average.
Let the average runs for 10 innings be \(x\).
The total runs scored in 10 innings would be \(10 \times x = 10x\).
In the eleventh inning, the cricketer scored 108 runs. So, the total runs after 11 innings will be \(10x + 108\).
The number of innings is now 11.
The new average (after 11 innings) is given by the total runs after 11 innings divided by 11 innings:
\[ \text{New Average} = \frac{10x + 108}{11} \]According to the problem, the new average increased by 6 runs compared to the old average. So, the new average is \(x + 6\).
Equating the two expressions for the new average:
\[ \frac{10x + 108}{11} = x + 6 \]Now, solve for \(x\):
\[ 10x + 108 = 11(x + 6) \]\[ 10x + 108 = 11x + 66 \]\[ 108 - 66 = 11x - 10x \]\[ 42 = x \]So, the average for 10 innings was 42 runs.
The new average (after 11 innings) is \(x + 6\).
\[ \text{New Average} = 42 + 6 = 48 \]Correct Option: A) 48 runs
45% of 750 - 25% of 480 =?
To solve this problem, we need to calculate the percentage of two different numbers and then find the difference between the two results. The key is to convert percentages to decimal or fractional form and perform the multiplication and subtraction accurately.
To find 45% of 750, we can write 45% as \(\frac{45}{100}\) and multiply it by 750.
\[ 45\% \text{ of } 750 = \frac{45}{100} \times 750 \]\[ = \frac{45 \times 75}{10} \]\[ = \frac{3375}{10} \]\[ = 337.5 \]To find 25% of 480, we can write 25% as \(\frac{25}{100}\) (or \(\frac{1}{4}\)) and multiply it by 480.
\[ 25\% \text{ of } 480 = \frac{25}{100} \times 480 \]\[ = \frac{1}{4} \times 480 \]\[ = 120 \]Now, subtract the result from step 2 from the result of step 1.
\[ 337.5 - 120 = 217.5 \]Thus, \(45\% \text{ of } 750 - 25\% \text{ of } 480 = 217.5\).
Correct Option: B) 217.50
What is the value of 15+20×2+5?
To solve this problem, we need to follow the order of operations, often remembered by the acronym PEMDAS/BODMAS. This rule dictates the sequence in which mathematical operations should be performed: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).
Thus, the value of the expression is 60.
Correct Option: B) 60
When a plot is sold for Rs 18,700, the owner loses 15%. At what price must the plot be sold in order to gain 15%?
This problem involves calculating the cost price of an item given its selling price and loss percentage, and then determining the selling price required to achieve a specific profit percentage. We will use the formulas relating selling price, cost price, profit/loss percentage.
Calculate the Cost Price (CP): Given that the plot was sold for Rs 18,700, and the owner lost 15%.
The formula for Selling Price (SP) when there is a loss is:
\[ \text{SP} = \text{CP} \times \left(1 - \frac{\text{Loss Percentage}}{100}\right) \]Substitute the given values:
\[ 18700 = \text{CP} \times \left(1 - \frac{15}{100}\right) \] \[ 18700 = \text{CP} \times \left(1 - 0.15\right) \] \[ 18700 = \text{CP} \times 0.85 \]Now, solve for CP:
\[ \text{CP} = \frac{18700}{0.85} \] \[ \text{CP} = 22000 \text{ Rs} \]Calculate the new Selling Price (SP) for a 15% gain: Now that we have the Cost Price (CP = Rs 22,000), we need to find the selling price to gain 15%.
The formula for Selling Price (SP) when there is a gain is:
\[ \text{SP} = \text{CP} \times \left(1 + \frac{\text{Gain Percentage}}{100}\right) \]Substitute the values:
\[ \text{SP} = 22000 \times \left(1 + \frac{15}{100}\right) \] \[ \text{SP} = 22000 \times \left(1 + 0.15\right) \] \[ \text{SP} = 22000 \times 1.15 \]Calculate the new SP:
\[ \text{SP} = 25300 \text{ Rs} \]C) 25,300. This is the price at which the plot must be sold to gain 15%.
The prices of a scooter and a T.V. are in the ratio 7 : 5. If the scooter costs Rs 8,000 more than a T.V. set, then the price of a T.V. set is:
This problem involves ratios and basic algebra. We are given the ratio of the prices of a scooter and a T.V., and the difference in their prices. We need to find the actual price of the T.V. set. We can represent the prices using a common multiplier based on the given ratio and then form an equation using the price difference.
Let the price of the scooter be \(P_S\) and the price of the T.V. be \(P_T\).
Given that the ratio of the prices of a scooter and a T.V. is \(7:5\), we can write:
\[ \frac{P_S}{P_T} = \frac{7}{5} \]This means we can express their prices in terms of a common variable, say \(x\):
We are given that the scooter costs Rs 8,000 more than a T.V. set. This can be written as:
\[ P_S - P_T = 8000 \]The price of the T.V. set is \(P_T = 5x\).
\[ P_T = 5 \times 4000 \] \[ P_T = 20000 \]So, the price of the T.V. set is Rs 20,000.
Correct Option: A) 20,000