Aptitude - Simplification - Discussion

Discussion Forum : Simplification - General Questions (Q.No. 13)
13.
A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
22
23
24
26
Answer: Option
Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

  and 2x + 4y = 140      x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

Discussion:
67 comments Page 2 of 7.

Shri said:   1 decade ago
How to solve x+y = 48?

x+y = 70, to get x = 26 and y=22. Please tell.

Sravani said:   1 decade ago
@Shri.

You have mistaken X+Y = 48.....1.

X+2Y = 70.2.

Subtract 1 from 2.

We get Y = 22.

By substituting we get X value 26.

Pardeshi said:   1 decade ago
Can anybody explain, how came the value of x and y?

SIREESHA said:   1 decade ago
Any simple shortcut for this type of problems. Please give it.

Laxmi said:   10 years ago
Please tell me clearly how came 2x and 4y?

Sheeraz said:   10 years ago
Please use common sense here not maths.

It can be like 48 cows he has it meas 48 cows and there 4 legs.

So multiply this 48*4 = 192.

In this question there is 140 legs.

So, 192-140 = 52.

Hens has 2 legs.

So 52/2 = 26.

Goverdhan said:   10 years ago
Hens have one head x and the cow has one leg y, then total heads are,
x + y = 48.

Hens have two legs 2x and the cow has four legs 4y, then total legs are,
2x + 4y = 140.

Solve both,

=> 2x + 2y = 96.
=> 2x + 4y = 140 then 2y = 44.
=> y = 22
=> x = 26

x is the number of hens, x = 26.

Franci said:   10 years ago
Super, thanks for your explanation.

Gaurav said:   9 years ago
Here we can see all have 2 legs.
So 48 * 2 = 96 have 2 legs, now 140 - 96 = 44 legs still remaining that means some of those we have counted previously have 2 more legs.

So, 44/2 = 22 have 4 legs and see that rest 48 - 22 = 26 have 2 legs only.

AJAY said:   9 years ago
L/2.
140/2 = 70
70 - 48 = 22=> THIS IS NUMBER OF COWS.
AND 48 - 22 = 26 =>THIS IS NUMBER OF HENS.


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