Tag: cartesian product of sets

Questions Related to cartesian product of sets

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

Let a relation $R$ be defined by $R=\left {(4,5), (1,4), (4,6), (7,6), (3,7)\right }$. The relation $R^{-1}\circ R$ is given by

  1. $\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$
  2. $\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$
  3. $\left \{(1,5), (1,6), (3,6)\right \}$
  4. None of these

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have $R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$.
$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$
$(4,4)\in R^{-1}\circ R$ because $(4,5)\in R$ and $(5,4)\in R^{-1}$
$(1,1)\in R^{-1}\circ R$ because $(1,4)\in R$ and $(4,1)\in R^{-1}$
$(4,4)\in R^{-1}\circ R$ because $(4,6)\in R$ and $(6,4)\in R^{-1}$
$(4,7)\in R^{-1}\circ R$ because $(4,6)\in R$ and $(6,7)\in R^{-1}$
$(7,4)\in R^{-1}\circ R$ because $(7,6)\in R$ and $(6,4)\in R^{-1}$
$(7,7)\in R^{-1}\circ R$ because $(7,6)\in R$ and $(6,7)\in R^{-1}$
$(3,3)\in R^{-1}\circ R$ because $(3,7)\in R$ and $(7,3)\in R^{-1}$
$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$.
$\therefore$ The correct answer is $B$.
Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

Given $(a - 2, b + 3) = (6, 8)$, are equal ordered pair. Find the value of $a$ and $b$.

  1. $a = 8$ and $b = 5$
  2. $a = 8$ and $b = 3$
  3. $a = 5$ and $b = 5$
  4. $a = 8$ and $b = 6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By equality of ordered pairs, we have
$(a - 2, b + 3) = (6, 8)$
On equating we get
$a - 2 = 6$
$a = 8$
$b + 3 = 8$
$b = 5$
So, the value of$ a = 8 $ and $ b = 5.$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

$(x, y)$ and $(p, q)$ are two ordered pairs. Find the values of $x$ and $p$, if $(3x - 1, 9) = (11, p + 2)$

  1. $x = 4, p = 9$
  2. $x = 6, p = 7$
  3. $x = 4, p = 5$
  4. $x = 4, p = 7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given$(x,y)=(p,q)$
$(3x - 1, 9) = (11, p + 2)$
By equating 
$3x - 1 = 11$
$3x = 12$
$x = 4$
$9 = p + 2$
$p = 7$
So, the value of $x = 4, p = 7.$

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

 $(x, y)$ and $(p, q)$ are two ordered pairs. Find the values of $p$ and $y$, if $(4y + 5, 3p - 1) = (25, p + 1)$

  1. $p = 0, y = 5$
  2. $p = 1, y = 5$
  3. $p = 0, y = 1$
  4. $p = 1, y = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $(x,y)=(p,q)$
$(4y + 5, 3p - 1) = (25, p + 1)$
On equating we get
$4y + 5 = 25$
$4y = 25 - 5$
$4y = 20$
$y = 5$
$3p - 1 = p + 1$
$3p - p = 1 + 1$
$2p = 2$
$p = 1$
So, the value of$ p = 1, y = 5$