The problem states that a number, when increased by 100%, results in 2820. We need to find the original number. We can set up an equation where the original number plus 100% of itself equals 2820.
Correct Option: C) 1410 is the number. When 1410 is increased by 100% (which is 1410 itself), it becomes \(1410 + 1410 = 2820\).
This problem involves setting up and solving a system of linear equations based on the given information about the cost of pens and pencils. We will define variables for the original cost of a pen and a pencil, formulate two equations from the two scenarios described, solve these equations to find the original costs, and then calculate the final required cost.
Define variables:
Let the original cost of one pen be \(P\) (in ₹).
Let the original cost of one pencil be \(C\) (in ₹).
Formulate the first equation:
The cost of 8 pens and 10 pencils is ₹132.
\[8P + 10C = 132 \quad \text{(Equation 1)}\]
Formulate the second equation:
If the cost of a pen decreases by ₹2, the new cost of a pen is \((P - 2)\).
If the cost of a pencil increases by ₹7, the new cost of a pencil is \((C + 7)\).
The cost of 17 pens and 6 pencils with these new prices is ₹136.
\[17(P - 2) + 6(C + 7) = 136\]
\[17P - 34 + 6C + 42 = 136\]
\[17P + 6C + 8 = 136\]
\[17P + 6C = 136 - 8\]
\[17P + 6C = 128 \quad \text{(Equation 2)}\]
Solve the system of equations:
We have two equations:
1) \(8P + 10C = 132\)
2) \(17P + 6C = 128\)
Multiply Equation 1 by 3 and Equation 2 by 5 to eliminate \(C\):
\(3 \times (8P + 10C) = 3 \times 132 \implies 24P + 30C = 396 \quad \text{(Equation 3)}\)
\(5 \times (17P + 6C) = 5 \times 128 \implies 85P + 30C = 640 \quad \text{(Equation 4)}\)
Subtract Equation 3 from Equation 4:
\((85P + 30C) - (24P + 30C) = 640 - 396\)
\(61P = 244\)
\(P = \frac{244}{61}\)
\(P = 4\)
Substitute \(P = 4\) into Equation 1:
\(8(4) + 10C = 132\)
\(32 + 10C = 132\)
\(10C = 132 - 32\)
\(10C = 100\)
\(C = \frac{100}{10}\)
\(C = 10\)
So, the original cost of one pen is ₹4 and the original cost of one pencil is ₹10.
Calculate the original cost of 12 pens and 9 pencils:
Required cost \(= 12P + 9C\)
Required cost \(= 12(4) + 9(10)\)
Required cost \(= 48 + 90\)
Required cost \(= 138\)
The original cost of 12 pens and 9 pencils is ₹138.
A) ₹138 is the correct answer, as calculated from the original costs of pens and pencils.
The problem involves finding an angle in a triangle using properties of right-angled triangles and the angle sum property of triangles. We will use the given perpendicular condition to identify right angles and then calculate unknown angles step-by-step.
Given that \(BD \perp AC\), this means that \(\angle BDC = 90^\circ\).
Consider the right-angled triangle \(\triangle BDC\). We are given \(\angle DBC = 16^\circ\).
The sum of angles in \(\triangle BDC\) is \(180^\circ\). So, \(\angle BCD + \angle DBC + \angle BDC = 180^\circ\).
Substituting the known values:
\[ \angle BCD + 16^\circ + 90^\circ = 180^\circ \]\[ \angle BCD + 106^\circ = 180^\circ \]\[ \angle BCD = 180^\circ - 106^\circ \]\[ \angle BCD = 74^\circ \]Now consider \(\triangle AEC\). We know \(\angle ECA = \angle BCD = 74^\circ\) (since E is on BC, \(\angle ECA\) is the same as \(\angle BCD\)).
We are given \(\angle CAE = 51^\circ\).
The sum of angles in \(\triangle AEC\) is \(180^\circ\). So, \(\angle AEC + \angle ECA + \angle CAE = 180^\circ\).
Substituting the known values:
\[ \angle AEC + 74^\circ + 51^\circ = 180^\circ \]\[ \angle AEC + 125^\circ = 180^\circ \]\[ \angle AEC = 180^\circ - 125^\circ \]\[ \angle AEC = 55^\circ \]We need to find \(\angle AEB\). Notice that \(\angle AEC\) and \(\angle AEB\) are angles on a straight line (BC is a straight line, and E is a point on BC).
Therefore, \(\angle AEB + \angle AEC = 180^\circ\) (linear pair).
Substituting the value of \(\angle AEC\):
\[ \angle AEB + 55^\circ = 180^\circ \]\[ \angle AEB = 180^\circ - 55^\circ \]\[ \angle AEB = 125^\circ \]D) \(125^\circ\)
The problem involves dividing a total sum of money among three individuals (A, B, and C) based on a given relationship between their shares. We need to find the share of A. The key is to express the shares of A, B, and C as a ratio and then use this ratio to distribute the total amount.
Set up the given relationship:
We are given that 3 times the share of A = 9 times the share of B = 4 times the share of C. Let the shares of A, B, and C be denoted by \(S_A\), \(S_B\), and \(S_C\) respectively.
Find the ratio of the shares:
Let this common value be \(k\). So,
Convert the ratio to whole numbers:
To convert the fractional ratio to whole numbers, find the Least Common Multiple (LCM) of the denominators (3, 9, 4).
LCM(3, 9, 4) = 36.
Multiply each part of the ratio by 36:
\[ S_A : S_B : S_C = \left(\frac{1}{3} \times 36\right) : \left(\frac{1}{9} \times 36\right) : \left(\frac{1}{4} \times 36\right) \]\[ S_A : S_B : S_C = 12 : 4 : 9 \]Calculate the sum of the ratio parts:
Sum of the ratio parts \( = 12 + 4 + 9 = 25 \).
Calculate the share of A:
The total amount to be divided is ₹75,400. The share of A is the ratio part of A divided by the sum of ratio parts, multiplied by the total amount.
B) ₹36,192 is the correct share of A, calculated based on the given ratios and total amount.
To determine which number divides 774359327, we will use divisibility rules for each of the given options. This method is generally faster than performing long division for each option.
Correct Option: A) 7
To determine which number divides 545714151, we will use divisibility rules for each of the given options.
\(5 + 4 + 5 + 7 + 1 + 4 + 1 + 5 + 1 = 33\)
Since 33 is divisible by 3 (\(33 \div 3 = 11\)), the number 545714151 is divisible by 3.
For 545714151:
Since 52 is not divisible by 7, 545714151 is not divisible by 7.
D) 3 — The sum of the digits of 545714151 is 33, which is divisible by 3. Therefore, 545714151 is divisible by 3.
To find the perimeter of a sector, we need to calculate the length of its arc and add it to the lengths of the two radii that form the sector. The formula for the arc length of a sector is given by \(\frac{\theta}{360°} \times 2\pi r\), where \(\theta\) is the central angle in degrees and \(r\) is the radius. The perimeter of the sector will then be \( \text{Arc Length} + 2r \).
The formula for arc length (\(L\)) is:
\[ L = \frac{\theta}{360°} \times 2\pi r \]The perimeter (\(P\)) of a sector is the sum of the arc length and two radii:
\[ P = L + 2r \]\[ P = 2.826 + 2 \times 9 \]\[ P = 2.826 + 18 \]\[ P = 20.826 \text{ cm} \]Rounding \(20.826 \text{ cm}\) to two decimal places gives \(20.83 \text{ cm}\).
Correct Option: C) 20.83 cm
We need to find the total number of registered voters. We will work backward from the margin of victory, using the percentages of valid votes, polled votes, and registered voters.
Calculate the number of votes polled: 85% of registered voters cast their vote.
\[ \text{Votes Polled} = 0.85 \times R \]Calculate the number of valid votes: 4% of the votes polled were invalid, so 96% were valid.
\[ \text{Valid Votes} = 0.96 \times (0.85 \times R) \] \[ \text{Valid Votes} = 0.816 \times R \]Calculate votes for the winning candidate: The winning candidate got 65% of the valid votes.
\[ \text{Winning Candidate's Votes} = 0.65 \times (0.816 \times R) \] \[ \text{Winning Candidate's Votes} = 0.5304 \times R \]Calculate votes for the losing candidate: The losing candidate got \(100\% - 65\% = 35\%\) of the valid votes.
\[ \text{Losing Candidate's Votes} = 0.35 \times (0.816 \times R) \] \[ \text{Losing Candidate's Votes} = 0.2856 \times R \]Use the margin of victory to find \(R\): The winning candidate won by 2142 votes. The margin is the difference between the votes received by the two candidates.
\[ (0.5304 \times R) - (0.2856 \times R) = 2142 \] \[ (0.5304 - 0.2856) \times R = 2142 \] \[ 0.2448 \times R = 2142 \] \[ R = \frac{2142}{0.2448} \] \[ R = 8750 \]Thus, the total number of registered voters is 8750.
B) 8750
To find the third proportional of two numbers, say 'a' and 'b', we set up a proportion where 'b' is the mean proportional and 'x' is the third proportional. This means \(a : b :: b : x\), which can be written as a ratio \[\frac{a}{b} = \frac{b}{x}\].
Correct Option: A) 196 is the value of \(x\), which is the third proportional of 16 and 56.
Priya can do a certain piece of work in 26 days. Priya and Varsha can together do the same work in 13 days, and Priya, Varsha and Nisha can do the same work together in 10 days. In how many days can Priya and Nisha do the same work together?
\( \frac{71}{4} \)
\( \frac{65}{4} \)
\( \frac{56}{6} \)
\( \frac{73}{5} \)
This problem involves work and time. We can solve it by calculating the individual work rates of Priya, Varsha, and Nisha. The work rate is the reciprocal of the time taken to complete the work. Once we find Nisha's work rate, we can add it to Priya's work rate to find their combined work rate and then determine the time they take together.
Priya can do the work in 26 days.
\[ \text{Priya's 1-day work} = \frac{1}{26} \]Priya and Varsha together can do the work in 13 days.
\[ \text{(Priya + Varsha)'s 1-day work} = \frac{1}{13} \]Varsha's 1-day work is the difference between Priya and Varsha's combined 1-day work and Priya's 1-day work.
\[ \text{Varsha's 1-day work} = \frac{1}{13} - \frac{1}{26} = \frac{2 - 1}{26} = \frac{1}{26} \]Priya, Varsha, and Nisha together can do the work in 10 days.
\[ \text{(Priya + Varsha + Nisha)'s 1-day work} = \frac{1}{10} \]Nisha's 1-day work is the difference between the combined 1-day work of Priya, Varsha, and Nisha, and the combined 1-day work of Priya and Varsha.
\[ \text{Nisha's 1-day work} = \frac{1}{10} - \frac{1}{13} = \frac{13 - 10}{130} = \frac{3}{130} \]Add Priya's 1-day work and Nisha's 1-day work.
\[ \text{(Priya + Nisha)'s 1-day work} = \frac{1}{26} + \frac{3}{130} \]\[ = \frac{5}{130} + \frac{3}{130} = \frac{8}{130} = \frac{4}{65} \]The number of days is the reciprocal of their combined 1-day work.
\[ \text{Days (Priya + Nisha)} = \frac{1}{\frac{4}{65}} = \frac{65}{4} \]B) \( \frac{65}{4} \) is the correct answer, as calculated above.