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The median AD of ΔABC is bisected at E and BE is produced to meet the side AC at F. Then, AF ∶ FC is
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Solution
For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15
respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the
corrected mean and standard deviation corresponding to the corrected figures are
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Solution
Corrected Mean and Standard Deviation
Original Mean: 40, Standard Deviation: 15
Two scores were misread: 25 → 52 and 35 → 53
Corrected Mean:
\( \mu' = \frac{3955}{100} = \boxed{39.55} \)
Corrected Standard Deviation:
\( \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} \)
✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96
The system of equations
x + y + 2z = a
x + z = b
2x + y + 3z = c
has a solution if
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Solution
Given system of equations:
$x + y + 2z = a$
$x + z = b$
$2x + y + 3z = c$
Writing in matrix form $[A|B]$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{array}\right]$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & -1 & -1 & c-2a \end{array}\right]$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-2a-(b-a) \end{array}\right]$
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-a-b \end{array}\right]$
For system to have a solution:
$\rho(A) = \rho(A|B)$
$\Rightarrow c - a - b = 0$
$\therefore \boxed{a + b = c}$
$x + y + 2z = a$
$x + z = b$
$2x + y + 3z = c$
Writing in matrix form $[A|B]$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{array}\right]$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & -1 & -1 & c-2a \end{array}\right]$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-2a-(b-a) \end{array}\right]$
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-a-b \end{array}\right]$
For system to have a solution:
$\rho(A) = \rho(A|B)$
$\Rightarrow c - a - b = 0$
$\therefore \boxed{a + b = c}$
Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two forces is θ . Find the values of θ for which the resultant force
is equal to 
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Solution
If two circles
$x^{2}+y^{2}+2gx+2fy=0$ and $x^{2}+y^{2}+2g'x+2f'y=0$ touch each other then whichof the following is true?
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Solution
If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that OPQ is an equilateral triangle,
where O is the centre of the hyperbola, then which of the following is true?
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Solution
Consider the following frequency distribution table.
| Class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 180 | $f_1$ | 34 | 180 | 136 | $f_2$ | 50 |
If the total frequency is 685 & median is 42.6 then the values of $f_1$ and $f_2$ are
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Solution
Median & Frequency Table
Given: Median = 42.6, Total Frequency = 685
Using Median Formula:
\( \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h \)
- Median Class: 40–50
- Lower boundary \( L = 40 \)
- Frequency \( f = 180 \)
- Class width \( h = 10 \)
- Cumulative freq before median class \( F = 214 + f_1 \)
Substituting values:
\( 42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82} \)
Using total frequency:
\( 662 + f_2 = 685 \Rightarrow f_2 = \boxed{23} \)
✅ Final Answer: \( f_1 = 82,\quad f_2 = 23 \)
Let $f(x) = x^2 - bx + c$, b is an odd positive integer. If f(x)=0 has two prime numbers as roots and b + c = 35, then the global minimum value of f(x) is
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Solution
Given: $f(x) = x^2 - bx + c$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$
If 
are four vectors such that 
is collinear with 
and 
is collinear with 
then 
=
are four vectors such that is collinear with and is collinear with then =1
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Solution
$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to
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Solution
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