Question

A person threw a standard dice 3 times. He obtained
two distinct odd prime numbers in two throws and an
even number which is not a factor of 18 in the third
throw. The sum of all the numbers on the opposite
faces of numbers obtained in the three throws is

Answer 1

$\huge{\boxed{\mathcal{ANSWER\::-}}}$

♦ Before we start solving we must know about a standard dice .

>> A standard dice is a cube with face numbered from 1 to 6 and Sum of the face on opposite side to the face in the front is always equal to 7 .

♦ Now provided

• In first throw :- a odd prime number

• In second throw : a odd prime number (distinct)

• In third throw :- a even number not a factor of 18

♦ Now for first two throw possible out come = 5 , 3 (odd prime)

♦ Third throw's possible outcome = 4 (not a factor of 18)

♦ Now we will find out the opposite faces

• As from the above we know that sum of face in front to its opposite = 7

Then

Opposite of 5 (let it be x) :-

$5 + x = 7$

$\implies x = 7 - 5$

$\implies x = 2$

Opposite of 3 (let it be y ) :-

$3 + y = 7$

$\implies y = 7 - 3$

$\implies y = 4$

Opposite of 3 (let it be z ) :-

$4 + z = 7$

$\implies z = 7 - 4$

$\implies z = 3$

♦ Now sum of

$x + y + z \\\\ = 2 + 3 + 4 \\\\ = 9$

♦ So sum of opposite faces = 9