Tricks in Maths

Hi everyone here I am going to solve a fundamental question related to quant the answer of which probably most of you know, but the method is critical that applies to other topics like time and distance, time and work compound interest etc.

Q. There are ten cats and dogs. Each cat eats five biscuits, and each dog eats six, and if there are total 56 biscuits find the number of cats and dogs separately?

First approach

Let the number of cats is c, and the number of dogs is d.

So according to question.

c+d=10; 5×c+6×d=56

Now multiply first equation with 5

5c+5d=50

And subtract the second by a third equation

5xc+6xd-5xc-5×d= 56-50

d=6

And apply this value of d in the very first equation

c= 10-6=4

So finally cats are 4 and dogs are 6.

but this approach is old and obsolete let's see

The cat is taking 5, and we can rearrange dog is taking 5+1=6 biscuits

So five is common for all ten animals.

And then 5×10=50

Subtract this from 56-50=6

Now we can see one is left in dog group so divide six by 1

We get six dogs and since total.animals are ten so 10-6=4 cats.

At least 6 marks we can get from the quant section by this method, and it applies to different topics

Stay happy and be selected.

Thanks