7 Erik has a set of number cards from 1 to 20. He picks four different cards.
Exactly three of his cards are multiples of 5.
Exactly three of his cards are even numbers.
All four of the numbers add up to less than 40.
What cards could Erik pick?

5,10,20

Step-by-step explanation:

multiples to 20

5

10

20

Step-by-step explanation:

the potential cards with multiples of 5 are :

5 10 15 20

the sum of all 4 cards is less than 40.

and exactly 3 of the cards are even. that means the 4th card is odd or uneven.

exactly 3 cards are a multiple of 5. that means the 4th card is something else.

what chances do we have to pick from the multiples of 5 ?

can we pick the 2 uneven ones ? no, that would violate the even card condition.

so, we must pick the 2 even ones : 10, 20

they are fixed.

now, 10+20 = 30, so, we cannot pick 15 as the third multiples of 5 card, or we violate the less than 40 rule.

so, we can only pick 5 as the third multiple of 5 card.

now, we have already 5, 10 and 20 (in sum 35).

the 4th card must be even (so that we get 3 even number cards), and it must be small, so that the total sum stays less than 40.

that means we can only pick 2 or 4 as the 4th card.

therefore, we have the 2 possibilities of picked cards :

2, 5, 10, 20

4, 5, 10, 20

15% chance.

step-by-step explanation:

3/10 times, the 5 year old will get a tablet. 5/10 times, the 13 yer old will get a tablet. multiply both fractions to get 15/100 times, both would have a tablet.

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.