The problem states that a number, when decreased by 7%, results in 3720. We need to find the original number. We can set up an equation where the original number is represented by a variable, and then solve for that variable.
Let the unknown number be \(x\).
The number is decreased by 7%. This means we are subtracting 7% of \(x\) from \(x\).
7% of \(x\) can be written as \(\frac{7}{100} \times x\) or \(0.07x\).
When \(x\) is decreased by 7%, the new value is \(x - 0.07x\).
This new value is given as 3720. So, we can set up the equation:
\[ x - 0.07x = 3720 \]Combine the terms on the left side:
\[ (1 - 0.07)x = 3720 \]\[ 0.93x = 3720 \]To find \(x\), divide both sides by 0.93:
\[ x = \frac{3720}{0.93} \]To simplify the division, we can multiply the numerator and denominator by 100 to remove the decimal:
\[ x = \frac{3720 \times 100}{0.93 \times 100} \]\[ x = \frac{372000}{93} \]Perform the division:
\[ 372000 \div 93 \]\[ 372 \div 93 = 4 \]\[ \text{So, } 372000 \div 93 = 4000 \]Thus, the original number is 4000.
Correct Option: C) 4000
Simplify: \( 3\left(\frac{7}{3}x^{2} - 22x + 19\right) - 7(x^{2} + 9x - 14) \)
To simplify the given algebraic expression, we will use the distributive property to multiply the constants outside the parentheses with each term inside. After distributing, we will combine like terms (terms with the same variable and exponent) to arrive at the simplified expression.
Write down the given expression:
\[ 3\left(\frac{7}{3}x^{2} - 22x + 19\right) - 7(x^{2} + 9x - 14) \]Distribute the \(3\) into the first set of parentheses:
\[ 3 \times \frac{7}{3}x^{2} - 3 \times 22x + 3 \times 19 \]\[ 7x^{2} - 66x + 57 \]Distribute the \(-7\) into the second set of parentheses:
\[ -7 \times x^{2} - 7 \times 9x - 7 \times (-14) \]\[ -7x^{2} - 63x + 98 \]Now, combine the results from Step 2 and Step 3:
\[ (7x^{2} - 66x + 57) + (-7x^{2} - 63x + 98) \]\[ 7x^{2} - 66x + 57 - 7x^{2} - 63x + 98 \]Group the like terms together:
Combine the like terms:
Write the simplified expression:
\[ -129x + 155 \]Correct Option: B) −129x + 155
To find the average of the remaining two numbers, we first need to determine their sum. We can do this by subtracting the sum of the first three numbers from the total sum of all five numbers. We are given the total sum, the average of the first two numbers, and the third number.
A) 196.5 is the correct average of the remaining two numbers.
To find the final selling price after successive discounts, we need to apply each discount one after another. First, calculate the price after the initial 15% discount. Then, apply the second 10% discount to this new discounted price.
C) ₹15,300 is the final selling price after applying two successive discounts of 15% and 10%.
To determine if a number is divisible by 22, we need to check if it is divisible by both 2 and 11, because \(22 = 2 \times 11\). A number is divisible by 2 if its last digit is even. A number is divisible by 11 if the alternating sum of its digits is a multiple of 11 (i.e., 0, 11, -11, 22, -22, etc.).
A) 7467614
Since 7467614 is divisible by both 2 and 11, it is divisible by 22.
Manisha bought a toy for ₹940, decorated it at a cost of ₹10 and sold the toy for ₹659. Find the loss or profit percentage.
\( \frac{582}{19} \) % profit
\( \frac{582}{19} \) % loss
582 % loss
\( \frac{592}{19} \) % profit
To find the loss or profit percentage, we first need to calculate the total cost price (CP) of the toy, which includes the initial 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 cost price.
Calculate the Total Cost Price (CP):
The initial cost of the toy is ₹940.
The cost incurred for decoration is ₹10.
Total Cost Price (CP) = Initial Cost + Decoration Cost
\[ \text{CP} = \text{₹}940 + \text{₹}10 = \text{₹}950 \]Identify the Selling Price (SP):
The toy was sold for ₹659.
\[ \text{SP} = \text{₹}659 \]Determine if it's a Profit or Loss:
Compare CP and SP:
Since \( \text{CP} (\text{₹}950) > \text{SP} (\text{₹}659) \), there is a loss.
Calculate the Loss Amount:
Loss = CP - SP
\[ \text{Loss} = \text{₹}950 - \text{₹}659 = \text{₹}291 \]Calculate the Loss Percentage:
Loss Percentage = \[\left( \frac{\text{Loss}}{\text{CP}} \right) \times 100\]
\[ \text{Loss Percentage} = \left( \frac{291}{950} \right) \times 100 \] \[ \text{Loss Percentage} = \frac{29100}{950} \]Simplify the fraction:
Divide both numerator and denominator by 10:
\[ \text{Loss Percentage} = \frac{2910}{95} \]Divide both by 5:
\[ \text{Loss Percentage} = \frac{582}{19} \%\]Correct Option: B) \( \frac{582}{19} \) % loss
This problem involves setting up a system of two linear equations with two variables based on the given information about the ages of Ram and Ramya. We will then solve this system to find their individual ages and subsequently calculate the difference between them.
Define variables:
Let \(R\) be the age of Ram.
Let \(Y\) be the age of Ramya.
Formulate the first equation:
"2 times the age of Ram is 14 years more than 6 times the age of Ramya"
This translates to:
\[ 2R = 6Y + 14 \]Formulate the second equation:
"4 times the age of Ramya is 4 years less than the age of Ram"
This translates to:
\[ 4Y = R - 4 \]Solve the system of equations:
From the second equation, we can express \(R\) in terms of \(Y\):
Substitute \(R\) into the first equation:
Substitute \(R = 4Y + 4\) into \(2R = 6Y + 14\):
Solve for \(Y\):
Subtract \(6Y\) from both sides:
Solve for \(R\):
Substitute the value of \(Y\) back into the revised second equation \(R = 4Y + 4\):
Calculate the difference:
The difference between the ages of Ram and Ramya is \(R - Y\):
C) The difference between their ages is 13 years, which matches our calculation.
The problem asks us to find the third proportional of two given numbers, 4 and 22. The concept of third proportional is related to continuous proportion. If three numbers \(a\), \(b\), and \(c\) are in continuous proportion, then the ratio of the first to the second is equal to the ratio of the second to the third. Mathematically, this is expressed as \(a:b = b:c\), or \[\frac{a}{b} = \frac{b}{c}\]. Here, \(c\) is the third proportional.
Correct Option: B) 121
The value of \( \sqrt{121} + \sqrt{0.0121} - \sqrt{7.84} \) is _____.
To solve this problem, we need to calculate the square root of each term individually and then perform the addition and subtraction operations as indicated. We will convert decimal numbers into fractions to simplify finding their square roots.
Calculate the first term: \( \sqrt{121} \)
We know that \( 11 \times 11 = 121 \).
So, \( \sqrt{121} = 11 \).
Calculate the second term: \( \sqrt{0.0121} \)
We can write \( 0.0121 \) as a fraction:
\[ 0.0121 = \frac{121}{10000} \]Now, find the square root:
\[ \sqrt{0.0121} = \sqrt{\frac{121}{10000}} = \frac{\sqrt{121}}{\sqrt{10000}} = \frac{11}{100} = 0.11 \]Calculate the third term: \( \sqrt{7.84} \)
We can write \( 7.84 \) as a fraction:
\[ 7.84 = \frac{784}{100} \]Now, find the square root. We know that \( 28 \times 28 = 784 \).
So, find the square root:
\[ \sqrt{7.84} = \sqrt{\frac{784}{100}} = \frac{\sqrt{784}}{\sqrt{100}} = \frac{28}{10} = 2.8 \]Perform the final calculation:
Substitute the calculated values back into the original expression:
\[ \sqrt{121} + \sqrt{0.0121} - \sqrt{7.84} = 11 + 0.11 - 2.8 \]First, add the positive terms:
\[ 11 + 0.11 = 11.11 \]Now, subtract the last term:
\[ 11.11 - 2.8 = 8.31 \]The value of the expression is \( 8.31 \).
Correct Option: B) 8.31
To solve this problem, we need to use the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.
Identify the 6-digit number: N01M22.
Assign positions to the digits, starting from the rightmost digit as the 1st position (odd place).
Calculate the sum of digits at odd places (S_odd):
\[ S_{\text{odd}} = 2 + M + 0 = M + 2 \]Calculate the sum of digits at even places (S_even):
\[ S_{\text{even}} = 2 + 1 + N = N + 3 \]Apply the divisibility rule for 11: The difference \( S_{\text{odd}} - S_{\text{even}} \) must be 0 or a multiple of 11.
\[ (M + 2) - (N + 3) = 0 \text{ or a multiple of } 11 \]\[ M + 2 - N - 3 = 0 \text{ or a multiple of } 11 \]\[ M - N - 1 = 0 \text{ or a multiple of } 11 \]Consider the possible values for M and N. Since M and N are digits in a 6-digit number, they must be integers from 0 to 9. Also, N cannot be 0 because it's the leading digit of a 6-digit number, so \(1 \le N \le 9\). For M, \(0 \le M \le 9\).
Evaluate the possible values for \(M - N - 1\):
So, \( -10 \le M - N - 1 \le 7 \).
The only multiple of 11 within this range is 0.
Therefore, we must have:
\[ M - N - 1 = 0 \]\[ M - N = 1 \]Correct Option: A) M - N = 1