To find the original number, we need to understand what "increased by 100%" means. An increase of 100% means the number is doubled. We can set up an equation where the original number plus 100% of itself equals 2760.
Let the unknown number be \(x\).
When the number is increased by 100%, it means we add 100% of the number to the original number. Mathematically, 100% of \(x\) is \(1 \times x = x\).
So, the number increased by 100% can be written as:
\[ x + (100\% \text{ of } x) \]This simplifies to:
\[ x + x = 2x \]According to the problem, this increased value is 2760. So, we set up the equation:
\[ 2x = 2760 \]To find \(x\), we divide both sides of the equation by 2:
\[ x = \frac{2760}{2} \]Calculating the value:
\[ x = 1380 \]Thus, the original number is 1380.
Correct Option: D) 1380
Pipe P can fill \( \frac{2}{9} \) part of a tank in 20 hours, and pipe Q can fill \( \frac{5}{9} \) part of the same tank in 10 hours. Both P and Q were kept open for 3 hours, then both were closed. Pipe R alone was then opened and it emptied the water in the tank in 8 hours. Pipes P, Q and R together can fill the empty tank in:
To solve this problem, we need to determine the individual filling rates of pipes P and Q, and the emptying rate of pipe R. We will then calculate the amount of tank filled by P and Q in 3 hours, and use the information about pipe R to find its emptying rate. Finally, we will calculate the combined filling/emptying rate of P, Q, and R to find the total time to fill the tank.
Calculate the filling rate of pipe P:
Pipe P fills \( \frac{2}{9} \) of the tank in 20 hours.
So, the time taken by P to fill the entire tank is:
\[ \text{Time}_P = 20 \text{ hours} \times \frac{9}{2} = 90 \text{ hours} \]The filling rate of P (part filled per hour) is:
\[ \text{Rate}_P = \frac{1}{90} \text{ part/hour} \]Calculate the filling rate of pipe Q:
Pipe Q fills \( \frac{5}{9} \) of the tank in 10 hours.
So, the time taken by Q to fill the entire tank is:
\[ \text{Time}_Q = 10 \text{ hours} \times \frac{9}{5} = 18 \text{ hours} \]The filling rate of Q (part filled per hour) is:
\[ \text{Rate}_Q = \frac{1}{18} \text{ part/hour} \]Calculate the combined work of P and Q in 3 hours:
Combined filling rate of P and Q is:
\[ \text{Rate}_{P+Q} = \text{Rate}_P + \text{Rate}_Q = \frac{1}{90} + \frac{1}{18} \]To add these fractions, find a common denominator (90):
\[ \text{Rate}_{P+Q} = \frac{1}{90} + \frac{5}{90} = \frac{6}{90} = \frac{1}{15} \text{ part/hour} \]Amount filled by P and Q in 3 hours:
\[ \text{Work}_{P+Q \text{ in 3 hours}} = \frac{1}{15} \times 3 = \frac{3}{15} = \frac{1}{5} \text{ part} \]Calculate the emptying rate of pipe R:
Pipe R alone emptied the \( \frac{1}{5} \) part of the tank in 8 hours.
The emptying rate of R (part emptied per hour) is:
\[ \text{Rate}_R = \frac{\frac{1}{5}}{8} = \frac{1}{5 \times 8} = \frac{1}{40} \text{ part/hour} \]Calculate the combined filling/emptying rate of P, Q, and R:
When P and Q are filling and R is emptying, the net rate is:
\[ \text{Net Rate}_{P+Q-R} = \text{Rate}_P + \text{Rate}_Q - \text{Rate}_R \] \[ \text{Net Rate}_{P+Q-R} = \frac{1}{90} + \frac{1}{18} - \frac{1}{40} \]We already found \[\frac{1}{90} + \frac{1}{18} = \frac{1}{15}\].
\[ \text{Net Rate}_{P+Q-R} = \frac{1}{15} - \frac{1}{40} \]Find a common denominator for 15 and 40. The least common multiple (LCM) of 15 and 40 is 120.
\[ \text{Net Rate}_{P+Q-R} = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24} \text{ part/hour} \]Calculate the time taken by P, Q, and R together to fill the empty tank:
If the net rate is \( \frac{1}{24} \) part per hour, then the time to fill the entire tank (1 part) is the reciprocal of this rate.
\[ \text{Time} = \frac{1}{\text{Net Rate}_{P+Q-R}} = \frac{1}{\frac{1}{24}} = 24 \text{ hours} \]Correct Option: C) 24 hours
This problem involves profit sharing in a partnership, which is proportional to the product of capital invested and the time for which it was invested. We will set up equations based on the given information about investments, time periods, and profit distribution, and then solve for the unknown capital contributed by M.
Let N's capital be \(x\).
According to the problem, M invested ₹31,000 more than N. So, M's capital is \(x + 31000\).
M invested for 2 months.
N invested for 7 months.
The ratio of profits is equal to the ratio of (capital \(\times\) time).
M's investment product: \((x + 31000) \times 2\)
N's investment product: \(x \times 7\)
Ratio of M's profit to N's profit:
\[ \frac{\text{Profit}_{\text{M}}}{\text{Profit}_{\text{N}}} = \frac{(x + 31000) \times 2}{x \times 7} = \frac{2x + 62000}{7x} \]Total profit is ₹5,977.
M's share is ₹139 more than N's share.
Let M's profit be \(\text{P}_{\text{M}}\) and N's profit be \(\text{P}_{\text{N}}\).
We have \(\text{P}_{\text{M}} + \text{P}_{\text{N}} = 5977\).
And \(\text{P}_{\text{M}} = \text{P}_{\text{N}} + 139\).
Substitute the second equation into the first:
\((\text{P}_{\text{N}} + 139) + \text{P}_{\text{N}} = 5977\)
\(2\text{P}_{\text{N}} + 139 = 5977\)
\(2\text{P}_{\text{N}} = 5977 - 139\)
\(2\text{P}_{\text{N}} = 5838\)
\(\text{P}_{\text{N}} = \frac{5838}{2} = 2919\)
Now find M's profit:
\(\text{P}_{\text{M}} = \text{P}_{\text{N}} + 139 = 2919 + 139 = 3058\)
Now we equate the profit ratio with the investment product ratio:
\[ \frac{\text{P}_{\text{M}}}{\text{P}_{\text{N}}} = \frac{3058}{2919} \]\[ \frac{2x + 62000}{7x} = \frac{3058}{2919} \]Cross-multiply and solve for \(x\):
\(2919 \times (2x + 62000) = 3058 \times 7x\)
\(5838x + 180978000 = 21406x\)
\(180978000 = 21406x - 5838x\)
\(180978000 = 15568x\)
\(x = \frac{180978000}{15568}\)
\(x = 11625\)
So, N's capital is ₹11,625.
M's capital is \(x + 31000 = 11625 + 31000 = 42625\).
Correct Option: B) ₹42,625
To find the simple interest, we use the formula \[\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}\]. It's crucial to ensure that the time 'T' is expressed in years, as the interest rate 'R' is given per annum.
Since the rate is per annum, the time must be in years. There are 12 months in a year.
\[ \text{T} = \frac{6 \text{ months}}{12 \text{ months/year}} = \frac{1}{2} \text{ year} \]The formula for Simple Interest (SI) is:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]The simple interest is ₹24.
B) ₹24 is the correct answer, calculated using the simple interest formula with the time converted to years.
This problem involves setting up and solving a system of linear equations. We will define variables for the original cost of a pen and a pencil, form two equations based on the given conditions, solve these equations to find the individual costs, and then calculate the final required cost.
Define variables:
Let \(P\) be the original cost of one pen (in ₹).
Let \(C\) be the original cost of one pencil (in ₹).
Formulate the first equation:
The cost of 9 pens and 14 pencils is ₹95. So, we have:
Formulate the second equation:
If the cost of a pen decreases by ₹3, the new cost of a pen is \((P - 3)\).
If the cost of a pencil increases by ₹7, the new cost of a pencil is \((C + 7)\).
The cost of 9 pens and 3 pencils with these new prices is ₹78. So, we have:
Expand and simplify this equation:
\[ 9P - 27 + 3C + 21 = 78 \] \[ 9P + 3C - 6 = 78 \] \[ 9P + 3C = 78 + 6 \] \[ 9P + 3C = 84 \quad \text{(Equation 2)} \]Solve the system of equations:
We have two equations:
Subtract Equation 2 from Equation 1:
\[ (9P + 14C) - (9P + 3C) = 95 - 84 \] \[ 11C = 11 \] \[ C = \frac{11}{11} \] \[ C = 1 \]Substitute the value of \(C = 1\) into Equation 2:
\[ 9P + 3(1) = 84 \] \[ 9P + 3 = 84 \] \[ 9P = 84 - 3 \] \[ 9P = 81 \] \[ P = \frac{81}{9} \] \[ P = 9 \]So, the original cost of a pen is ₹9 and the original cost of a pencil is ₹1.
Calculate the original cost of 17 pens and 13 pencils:
Required cost \(= 17P + 13C\)
Substitute the values of \(P = 9\) and \(C = 1\):
The original cost of 17 pens and 13 pencils is ₹166.
Correct Option: B) ₹166
Simplify the following.
3((\( \frac{4}{3} \))x² - 21x + 12) - 4(x² + 4x - 14)
To simplify the given algebraic expression, we will first distribute the constants into their respective parentheses. Then, we will combine like terms (terms with \(x^2\), terms with \(x\), and constant terms) to arrive at the simplified expression.
The given expression is:
\[ 3\left( \frac{4}{3}x^2 - 21x + 12 \right) - 4(x^2 + 4x - 14) \]Group the \(x^2\) terms, the \(x\) terms, and the constant terms.
\[ (4x^2 - 4x^2) + (-63x - 16x) + (36 + 56) \]Correct Option: C) -79x + 92
We are given the sides of a right-angled triangle in terms of \(x\) and its area. We need to find the perimeter of the triangle. First, we will use the Pythagorean theorem to establish a relationship between the sides. Then, we will use the given area to find the value of \(x\). Once \(x\) is known, we can calculate the lengths of all three sides and then find the perimeter.
Identify the sides and apply the Pythagorean theorem:
The sides of the right-angled triangle are \(x\) cm, \((x + 2)\) cm, and \((x + 4)\) cm. In a right-angled triangle, the longest side is the hypotenuse. Therefore, \((x + 4)\) cm is the hypotenuse.
According to the Pythagorean theorem:
\[ \text{base}^2 + \text{height}^2 = \text{hypotenuse}^2 \]\[ x^2 + (x + 2)^2 = (x + 4)^2 \]Solve the equation for \(x\):
\[ x^2 + (x^2 + 4x + 4) = (x^2 + 8x + 16) \]\[ 2x^2 + 4x + 4 = x^2 + 8x + 16 \]\[ 2x^2 - x^2 + 4x - 8x + 4 - 16 = 0 \]\[ x^2 - 4x - 12 = 0 \]Factor the quadratic equation:
We need two numbers that multiply to \(-12\) and add up to \(-4\). These numbers are \(-6\) and \(2\).
Find possible values for \(x\):
From the factored equation, we get two possible values for \(x\):
Calculate the lengths of the sides:
The sides are:
These sides \((6, 8, 10)\) form a Pythagorean triplet, which confirms it's a right-angled triangle.
Verify the area:
The area of a right-angled triangle is given by:
Calculate the perimeter:
The perimeter of the triangle is the sum of its sides:
B) 24
We need to find the measure of \(\angle ACD\) in a trapezium ABCD where BC || AD. We are given \(\angle ABC = 70^\circ\), \(\angle BAC = 62^\circ\), and AC = CD. We will use properties of triangles (sum of angles, isosceles triangle properties) and properties of parallel lines (alternate interior angles).
Correct Option: D) 84
The value of 3² + \(\sqrt{4^2}\) - 5\(\sqrt{81}\) - 4 is
To solve this problem, we need to evaluate each term in the given expression using the order of operations (PEMDAS/BODMAS). We will calculate squares, square roots, multiplications, and then perform additions and subtractions from left to right.
Thus, the value of the expression is -36.
Correct Option: D) -36
To find the mode of a given set of data, we need to identify the value that appears most frequently in the dataset. The mode is a measure of central tendency.
Correct Option: A) 20
The mode of the data is 20 because it is the value that occurs with the highest frequency (4 times) in the given set of marks.