In a group of 5 families, every family has a certain number of children, such that the number of children forms an arthmetic progression with a common difference of one, starting with two children in the first family. Despite the objection of their parents every child in a family has three times as many pets to look after as the number of offspring in the family. What is the total number of pets in the entire group of five families.
a. 270
b. 99
c. 165
d. 27
Answers
Answered by
20
These are the correct options for this question:
a. 99
b. 9
c. 55
d. 90
Answer: D
Explanation:
As the number of children are in arithmetic progression starting with 2, the five families have 2, 3, 4, 5, 6 kids respectively. As each children has kept the pets equal to the number of kids in the family, each family has n2 pets. So total = 22+32+42 +52 +62 = 90
ramaduraivijay:
This question is recently asked in TCS with these options
You can try answering with these options :)
Answered by
46
Answer:
Step-by-step explanation:
as there are total 5 families having number of children in arthmetic progression.
the number of children in 5 families are-
2, 3, 4, 5, 6
each children have pets to look after =3 times the no. of children in family
therefore,
for a single child in first family the number of pets to look after = 3 * 2
similarly,
3(2*2 +3*3 +4*4 +5*5 +6*6)=270
Similar questions