Tag: cartesian product of two sets

Questions Related to cartesian product of two 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$