The number of groups that can be made from 5 different green balls, 4 different blue balls and
3 different red balls, if at least 1 green and 1 blue ball is to be included
In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches.
The number of ways in which the series can be won if no match ends in a draw is:
20 persons are sitting in a particular arrangement around a circular table.
3 persons are to be selected for leaders.
The number of ways of selection of 3 persons such that no 2 were sitting adjacent to each other is:
100 identical coins, each with probability $p$ of showing heads, are tossed.
If $0 < p < 1$ and the probability of showing heads on 50 coins is equal to that of 51 coins, then the value of $p$ is:
Let two fair six-faced dice $A$ and $B$ be thrown simultaneously.
Let $E_1$ be the event that die $A$ shows 4,
$E_2$ the event that die $B$ shows 2,
and $E_3$ the event that the sum of the two numbers on the dice is odd.
Which statement is false?
In a sequence of $21$ terms, the first $11$ terms are in A.P. with common difference $2$ and the last $11$ terms are in G.P. with common ratio $2$. If the middle term of A.P. is equal to the middle term of G.P., then the middle term of the entire sequence is
Let $t_1$ be the first term. For the A.P.: $t_6=t_1+10$, $t_{11}=t_1+20$.
For the G.P. (terms $11$ to $21$ with ratio $2$): $t_{16}=t_{11}\cdot 2^5=32t_{11}$.
Given $t_6=t_{16}$: $t_1+10=32(t_1+20)\Rightarrow 31t_1=-630\Rightarrow t_1=-\dfrac{630}{31}$.
Middle term of entire sequence is $t_{11}=t_1+20=-\dfrac{630}{31}+20=-\dfrac{10}{31}$.
The median of a set of $9$ distinctive observations is $20.5$. If each of the largest $4$ observations of the set is increased by $2$, then the median of the new set
Numbers are $2,4,\dots,100=2\cdot{1,\dots,50}$.
$\operatorname{Var}(1,\dots,50)=\dfrac{50^2-1}{12}=\dfrac{2499}{12}=\dfrac{833}{4}$.
Scaling by $2$: $\operatorname{Var}=4\cdot\dfrac{833}{4}=833$.
If $
\begin{vmatrix}
a & p & x \\
b & q & y \\
c & r & z
\end{vmatrix}
= 16
$, then the value of
$
\begin{vmatrix}
p+q & a+x & a+p \\
q+y & b+y & b+q \\
x+z & c+z & c+r
\end{vmatrix}
$ is:
By properties of determinants,
each column in the second determinant is the **sum of two columns** of the first.
So its value remains the same (since addition of columns preserves linearity).
Hence $\boxed{16}$.
Note $f(x) = \dfrac{3x^3 + 2|x| + 1}{x^2 + |x| + 1}$
is **odd + even combination**, so integrate separately.
After simplifying using $|x|$ symmetry and limits $[-2,2]$,
only even part contributes.
Result $= 3\log 7$.
If $A,B,C$ are angles of a triangle, then the value of
$
\begin{vmatrix}
\sin 2A & \sin C & \sin B \\
\sin C & \sin 2B & \sin A \\
\sin B & \sin A & \sin 2C
\end{vmatrix}
$
is:
** Since $A+B+C=\pi$, we have
$\cos A=-\cos(B+C)=\sin B\sin C-\cos B\cos C$ and similarly for $B,C$.
Using $\sin 2A=2\sin A\cos A$ (and cyclic forms), each row becomes a
linear combination of the other two rows, so the rows are linearly
dependent. Hence the determinant is $0$.
$\boxed{0}$
Let $f(x)=\log\!\left(\frac{2-\sin x}{2+\sin x}\right)$.
Then $f(-x)=\log\!\left(\frac{2+\sin x}{2-\sin x}\right)=-f(x)$, so $f$ is odd.
Integral over $[-\pi/2,\pi/2]$ of an odd function is $0$.
In general $AB=0\nRightarrow A=0$ or $B=0$.
Example:
$A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$,\;
$B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ are non-zero but
$AB=\begin{pmatrix}0&0\\0&0\end{pmatrix}$.
So $\boxed{\text{(A)}}$
The function $f$ is defined in $[-5,5]$ as
$
f(x)=
\begin{cases}
x, & \text{if }x\text{ is rational}\\
-x, & \text{if }x\text{ is irrational}
\end{cases}
$
For rationals $f(x)=x$, for irrationals $f(x)=-x$.
At $x=0$, both give $0$, so it’s continuous there.
At any $x\neq0$, limits from rationals and irrationals differ.
Reflexive — yes (all $(a,a)$ are present).
Symmetric — no, since $(1,2)\in R$ but $(2,1)\notin R$.
Transitive — yes, because $(1,2)$ and $(2,3)$ imply $(1,3)$ (which exists).
For reflexivity: $xRx$ means $x-x+\sqrt{2}=\sqrt{2}$ (irrational) ⇒ true.
For symmetry: $xRy⇒x-y+\sqrt{2}$ irrational, but $y-x+\sqrt{2}=-(x-y)+\sqrt{2}$ may be rational. Not always true ⇒ not symmetric.
For transitivity: fails similarly.
Put $w=z-(3+2i)$ so $|w|\le2$. Then
$2z-6+5i=2w+9i$.
Set of $2w$ is a disc of radius $4$ centered at $0$, so $2w+9i$ is a disc of radius $4$ centered at $9i$.
Minimum modulus $=\big||9|-4\big|=5$.
Let common root be $r$. Then
$r^2+br-1=0$ and $r^2+r+b=0$. Subtract:
$r(1-b)+(b+1)=0\Rightarrow r=\dfrac{b+1}{b-1}$ (for $b\ne1$).
Substitute in $r^2+br-1=0$:
\[
\frac{(b+1)^2}{(b-1)^2}+b\frac{b+1}{b-1}-1=0
\;\Rightarrow\; b^3+3b=0
\;\Rightarrow\; b\,(b^2+3)=0.
\]
Hence $b=0$ or $b=\pm i\sqrt3$.
General term
$T_k=\binom{5}{k}(y^2)^{5-k}\left(\dfrac{c}{y}\right)^k
=\binom{5}{k}c^k\,y^{10-3k}$.
For power of $y^1$: $10-3k=1\Rightarrow k=3$.
Coefficient $=\binom{5}{3}c^3=10c^3$.
(interpreting “contains 1” as **exactly one** ‘1’, repetitions allowed):**
Case-1: ‘1’ in the thousand’s place → remaining $3$ places from $\{0,\dots,7\}\setminus\{1\}$ with repetition:
$7^3=343$ ways.
Case-2: ‘1’ in any one of the last three places ($3$ choices). Thousand’s place from $\{2,\dots,7\}$ ($6$ ways). Remaining two places from $\{0,\dots,7\}\setminus\{1\}$ with repetition: $7^2=49$ ways.
Total $=343+3\cdot6\cdot49=343+882
(interpreting “contains 1” as **exactly one** ‘1’, repetitions allowed):**
Case-1: ‘1’ in the thousand’s place → remaining $3$ places from $\{0,\dots,7\}\setminus\{1\}$ with repetition:
$7^3=343$ ways.
Case-2: ‘1’ in any one of the last three places ($3$ choices). Thousand’s place from $\{2,\dots,7\}$ ($6$ ways). Remaining two places from $\{0,\dots,7\}\setminus\{1\}$ with repetition: $7^2=49$ ways.
Total $=343+3\cdot6\cdot49=343+882
(interpreting “contains 1” as **exactly one** ‘1’, repetitions allowed):**
Case-1: ‘1’ in the thousand’s place → remaining $3$ places from $\{0,\dots,7\}\setminus\{1\}$ with repetition:
$7^3=343$ ways.
Case-2: ‘1’ in any one of the last three places ($3$ choices). Thousand’s place from $\{2,\dots,7\}$ ($6$ ways). Remaining two places from $\{0,\dots,7\}\setminus\{1\}$ with repetition: $7^2=49$ ways.
Total $=343+3\cdot6\cdot49=343+882
When some unidentified/unknown person/firm sends you mail in a trustworthy/lucrative way asking for sensitive bank and online payment information, this is a case of __?
I am facing South, I turn right and walk 20 m. Then I turn right again and walk 10 m. Then I turn left and walk 10 m and then turning right walk 20 m. Then I turn right again and walk ___ m. In which direction am I from the starting point?”