Section - Mathematics
The value of \( \frac{3}{4} \) + \(\left(\frac{1}{1+\frac{5}{9}}\right)\) - \( \frac{2}{3} \) is
\( \frac{61}{84} \)
\( \frac{60}{83} \)
\( \frac{57}{77} \)
\( \frac{63}{92} \)
To solve this problem, we need to simplify the given expression by performing the operations in the correct order. We will first simplify the term inside the parenthesis, then perform addition and subtraction of the fractions.
The term inside the parenthesis is \[\left(\frac{1}{1+\frac{5}{9}}\right)\]. First, simplify the denominator:
\[ 1 + \frac{5}{9} = \frac{9}{9} + \frac{5}{9} = \frac{9+5}{9} = \frac{14}{9} \]Now substitute this back into the parenthesis:
\[ \frac{1}{1+\frac{5}{9}} = \frac{1}{\frac{14}{9}} = \frac{9}{14} \]The original expression becomes:
\[ \frac{3}{4} + \frac{9}{14} - \frac{2}{3} \]The denominators are 4, 14, and 3. The least common multiple (LCM) of 4, 14, and 3 is 84.
Now, combine the fractions with the common denominator:
\[ \frac{63}{84} + \frac{54}{84} - \frac{56}{84} = \frac{63 + 54 - 56}{84} \]Calculate the numerator:
\[ 63 + 54 - 56 = 117 - 56 = 61 \]So the final value is:
\[ \frac{61}{84} \]A) \( \frac{61}{84} \) is the correct value obtained after simplifying the expression.
We need to find the list price of the fan given the profit, the percentage by which the list price is above the cost price, and the discount percentage offered. We will set up equations based on the given percentages for cost price, list price, selling price, and profit, and then solve for the list price.
C) 47,436
To find the Highest Common Factor (HCF) of a set of numbers, we can use the prime factorization method. This involves finding the prime factors of each number and then multiplying the common prime factors, raised to their lowest powers.
Prime Factorization of 120:
\[ 120 = 2 \times 60 \\ 120 = 2 \times 2 \times 30 \\ 120 = 2 \times 2 \times 2 \times 15 \\ 120 = 2 \times 2 \times 2 \times 3 \times 5 \\ 120 = 2^3 \times 3^1 \times 5^1 \]Prime Factorization of 144:
\[ 144 = 2 \times 72 \\ 144 = 2 \times 2 \times 36 \\ 144 = 2 \times 2 \times 2 \times 18 \\ 144 = 2 \times 2 \times 2 \times 2 \times 9 \\ 144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \\ 144 = 2^4 \times 3^2 \]Prime Factorization of 168:
\[ 168 = 2 \times 84 \\ 168 = 2 \times 2 \times 42 \\ 168 = 2 \times 2 \times 2 \times 21 \\ 168 = 2 \times 2 \times 2 \times 3 \times 7 \\ 168 = 2^3 \times 3^1 \times 7^1 \]Identify Common Prime Factors and their Lowest Powers:
Calculate HCF:
Multiply the common prime factors raised to their lowest powers:
\[ \text{HCF} = 2^3 \times 3^1 \\ \text{HCF} = 8 \times 3 \\ \text{HCF} = 24 \]Correct Option: D) 24
The HCF of 120, 144, and 168 is 24.
The problem states that a number, when increased by 100%, results in 2750. We need to find the original number. Increasing a number by 100% means adding the number itself to the original number, effectively doubling it.
C) 1375. The number, when increased by 100%, becomes \(1375 + (100\% \text{ of } 1375) = 1375 + 1375 = 2750\).
This problem involves setting up and solving a system of two linear equations based on the given information about the ages of P and Q. We will define variables for their present ages and then translate the two statements into algebraic equations. Finally, we will solve these equations simultaneously to find the present age of Q.
Define variables: Let the present age of P be \(P\) years. Let the present age of Q be \(Q\) years.
Formulate the first equation: "Eight times the present age of P exceeds the present age of Q by 46 years."
This can be written as:
\[ 8P - Q = 46 \quad \text{(Equation 1)} \]Formulate the second equation: "After 10 years, three times Q's age will be 12 years less than eight times P's age at that time."
After 10 years:
The statement translates to:
\[ 3(Q + 10) = 8(P + 10) - 12 \]Expand and simplify this equation:
\[ 3Q + 30 = 8P + 80 - 12 \] \[ 3Q + 30 = 8P + 68 \]Rearrange to match the form of Equation 1 (i.e., \(8P - 3Q = \text{constant}\)):
\[ 8P - 3Q = 30 - 68 \] \[ 8P - 3Q = -38 \quad \text{(Equation 2)} \]Solve the system of equations: We have two equations:
Equation 1: \(8P - Q = 46\)
Equation 2: \(8P - 3Q = -38\)
Subtract Equation 2 from Equation 1 to eliminate \(P\):
\[ (8P - Q) - (8P - 3Q) = 46 - (-38) \] \[ 8P - Q - 8P + 3Q = 46 + 38 \] \[ 2Q = 84 \]Solve for \(Q\):
\[ Q = \frac{84}{2} \] \[ Q = 42 \]State the answer: The present age of Q is 42 years.
C) 42
To find the number of prime numbers between 317 and 322 (inclusive), we need to list all integers in this range and then test each one for primality. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Therefore, there is only 1 prime number in the given range.
D) 1 is the correct answer because only 317 is found to be a prime number in the specified range.
To calculate the profit percentage, we first need to find the cost price (CP) and selling price (SP) of a single pencil. The profit percentage is then calculated using the formula: \[\text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100\].
Ashok bought pencils at ₹180 a dozen. A dozen means 12 pencils.
\[ \text{CP of 12 pencils} = \text{₹}180 \]\[ \text{CP of 1 pencil} = \frac{180}{12} = \text{₹}15 \]He sold them for ₹16 each.
\[ \text{SP of 1 pencil} = \text{₹}16 \]Profit is the difference between Selling Price and Cost Price.
\[ \text{Profit} = \text{SP} - \text{CP} \]\[ \text{Profit} = 16 - 15 = \text{₹}1 \]The formula for profit percentage is:
\[ \text{Profit Percentage} = \frac{\text{Profit}}{\text{CP}} \times 100 \]\[ \text{Profit Percentage} = \frac{1}{15} \times 100 \]\[ \text{Profit Percentage} = \frac{100}{15} \]\[ \text{Profit Percentage} = \frac{20}{3} \]\[ \text{Profit Percentage} \approx 6.666... \% \]\[ \text{Profit Percentage} \approx 6.67 \% \]Correct Option: A) 6.67% is the calculated profit percentage.
Simplify the following.
1 × (5 × (3²)) ÷ 17 + 44 − 78
\( -\frac{533}{17} \)
\( -\frac{534}{17} \)
\( -\frac{535}{17} \)
\( -\frac{538}{17} \)
To simplify the given expression, 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).
The innermost part is \(3^2\).
\[ 3^2 = 3 \times 3 = 9 \]Substitute this back into the expression:
\[ 1 \times (5 \times 9) \div 17 + 44 - 78 \]Next, we calculate \(5 \times 9\).
\[ 5 \times 9 = 45 \]Substitute this back into the expression:
\[ 1 \times 45 \div 17 + 44 - 78 \]Next, we perform the multiplication \(1 \times 45\).
\[ 1 \times 45 = 45 \]The expression becomes:
\[ 45 \div 17 + 44 - 78 \]Next, we perform the division \(45 \div 17\). Since 45 is not perfectly divisible by 17, we keep it as a fraction.
\[ \frac{45}{17} + 44 - 78 \]First, combine the whole numbers:
\[ 44 - 78 = -34 \]Now the expression is:
\[ \frac{45}{17} - 34 \]To combine these, find a common denominator, which is 17.
\[ \frac{45}{17} - \frac{34 \times 17}{17} \] \[ \frac{45}{17} - \frac{578}{17} \]Now, subtract the numerators:
\[ \frac{45 - 578}{17} \] \[ \frac{-533}{17} \]Correct Option: A) \( -\frac{533}{17} \)
To find the amount the customer pays, we first need to calculate the discount amount based on the marked price and the discount percentage. Then, subtract this discount amount from the marked price.
Identify the Marked Price (MP): The marked price of the mobile phone is ₹15,000.
\[ \text{MP} = \text{₹}15,000 \]Identify the Discount Percentage: The shopkeeper allows a discount of 10%.
\[ \text{Discount Percentage} = 10\% \]Calculate the Discount Amount: The discount amount is calculated as a percentage of the marked price.
\[ \text{Discount Amount} = \text{Discount Percentage} \times \text{MP} \] \[ \text{Discount Amount} = 10\% \times \text{₹}15,000 \] \[ \text{Discount Amount} = \frac{10}{100} \times 15,000 \] \[ \text{Discount Amount} = 0.10 \times 15,000 \] \[ \text{Discount Amount} = \text{₹}1,500 \]Calculate the Selling Price (Amount Paid by Customer): The selling price is the marked price minus the discount amount.
\[ \text{Selling Price} = \text{MP} - \text{Discount Amount} \] \[ \text{Selling Price} = \text{₹}15,000 - \text{₹}1,500 \] \[ \text{Selling Price} = \text{₹}13,500 \]Therefore, the customer actually pays ₹13,500.
Correct Option: B) ₹13,500
Pipe A can fill a tank in 12 hours, pipe B can fill the same tank in 25 hours and pipe C can fill the same tank in 10 hours. The time taken by them to fill the same tank if they operate together is:
9 \( \frac{32}{67} \) hours
12 \( \frac{32}{67} \) hours
4 \( \frac{32}{67} \) hours
13 \( \frac{32}{67} \) hours
To find the time taken by multiple pipes to fill a tank together, we first determine the individual filling rates of each pipe. The rate is the reciprocal of the time taken to fill the tank. Then, we sum these individual rates to get the combined filling rate. Finally, the total time taken to fill the tank when all pipes operate together is the reciprocal of the combined rate.
Pipe A fills the tank in 12 hours. So, in 1 hour, Pipe A fills:
\[ \text{Rate A} = \frac{1}{12} \text{ of the tank} \]Pipe B fills the tank in 25 hours. So, in 1 hour, Pipe B fills:
\[ \text{Rate B} = \frac{1}{25} \text{ of the tank} \]Pipe C fills the tank in 10 hours. So, in 1 hour, Pipe C fills:
\[ \text{Rate C} = \frac{1}{10} \text{ of the tank} \]When all three pipes operate together, their combined rate in 1 hour is the sum of their individual rates:
\[ \text{Combined Rate} = \text{Rate A} + \text{Rate B} + \text{Rate C} \]\[ \text{Combined Rate} = \frac{1}{12} + \frac{1}{25} + \frac{1}{10} \]The prime factorization of the denominators are:
The LCM is \(2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 12 \times 25 = 300\).
The time taken is the reciprocal of the combined rate:
\[ \text{Time} = \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{67}{300}} = \frac{300}{67} \text{ hours} \]Divide 300 by 67:
\[ 300 \div 67 \]\[ 300 = 4 \times 67 + 32 \]\[ \text{Time} = 4 \frac{32}{67} \text{ hours} \]Correct Option: C) 4 \( \frac{32}{67} \) hours