Question 1204660: There are a total of 20 lions, tigers, and bears at the local zoo. The number of tigers is 2 more than the number of lions. The number of bears is 3 more than the number of lions. How many lions are at the zoo?
Found 4 solutions by mananth, ikleyn, josgarithmetic, math_tutor2020:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! There are a total of 20 lions, tigers, and bears at the local zoo.
The number of tigers is 2 more than the number of lions.
Let number of lions be x
Number of tigers will be x+2
The number of bears is 3 more than the number of lions.
The number of bears= x+3
Add them up. They equal 20
x+x+2+x+3=20
3x+5 =20
subtract 5 from both sides
3x+5-5= 20-5
3x=15
divide by 3
x=5
There are 5 lions at the zoo.
Answer by ikleyn(52814)  (Show Source):
You can put this solution on YOUR website! .
Even if you are 2-nd grade student and do not know equations yet,
you can easy solve this problem using only your common sense and skills calculating whole numbers.
From 20 animals, take out 2 tigers and 3 bears, for a minute (mentally in your mind).
You will get 15 animals, lions, tigers and bears in equal amount.
What you will do next ? - But of course, you will distribute 15 animal in 3 equal groups.
In other words, you will divide 15 by 3 and will get 5 animals in each group.
After that, you will return back 2 tigers and 3 bears and will get the final answer this way.
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Nice problem for 2-nd grade 7-years old/(young) students.
Answer by josgarithmetic(39620)  (Show Source):
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Let's make a guess that there are 6 lions because 20/3 = 6.67 approximately. I rounded down to the nearest whole number.
6 lions --> 6+2 = 8 tigers and 6+3 = 9 bears
6 lions + 8 tigers + 9 bears = 23 total animals
We have 3 too many animals. Let's take away 1 of each
6-1 = 5 lions
8-1 = 7 tigers
9-1 = 8 bears
Then,
5 lions + 7 tigers + 8 bears = 20 total animals
We have found the answer.
This method is probably not as efficient as the standard algebraic approach, but it's good to have multiple pathways to any math problem.
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