Question

A man reared cows and buffaloes. the no. of buffaloes was 7 more than one third of the number of cows. What was the total number of animals, the number of buffaloes was 21
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Answer 1

B = 21

$B = 7 + \frac{1}{3} X C$

$21 = 7 + \frac{1}{3} X C$

$\frac{1}{3} X C = 21-7$

$\frac{1}{3} X C = 14$

$C = 14 X 3$

$C = 42$

Total no. of animals = C + B

= 42 + 21

= 63

Answer 2

Answer: 63

Solution:

Let the no. of cow be x

one third of cow =x/3

A.T.Q.

the no. of buffalo = 7+x/3

the no. of buffalo=21

So,. 7+x/3 = 21

x=42

,the no. of cow =42

Hence,total no. of animal =42+21

=63