Math, asked by pushpitabhatta21, 1 day ago

In a test,5 % of the students have failed in both English and Mathematics,10% of the students have failed in English and 15% of the students have failed in Mathematics.If 300 students have passed in both subjects, find the total number of students.

Answers

Answered by milind3024

Answer:Let the total number of candidates = 100%

Now it is given 52% candidates failed in English (E) and 42 % candidates failed in Mathematics (M)

So let n(E) = 52%, n(M) = 42%

Now if 17% candidates failed in both English and Mathematics.

Therefore n(E∩M)=17 %

So first find out total candidates failed in both English and Mathematics i.e. n(E∪M).

As we know n(E∪M)=n(E)+n(M)−n(E∩M) so use this property of set we have,

⇒n(E∪M)=52+42−17=77 %

So total candidates failed in both English and Mathematics is 77%.

So the total candidates passed in both English and Mathematics is the subtraction of total number of candidates and the total candidates failed in both English and Mathematics.

Therefore total candidates passed in both English and Mathematics is

= (100 - 77) %

= 23%.

Step-by-step explanation:

Answered by Squishyoongi

Answer:

Now it is given 52% candidates failed in English (E) and 42 % candidates failed in Mathematics (M)

So let n(E) = 52%, n(M) = 42%

Now if 17% candidates failed in both English and Mathematics.

Therefore n(E∩M)=17 %

So first find out total candidates failed in both English and Mathematics i.e. n(E∪M).

As we know n(E∪M)=n(E)+n(M)−n(E∩M) so use this property of set we have,

⇒n(E∪M)=52+42−17=77 %

So total candidates failed in both English and Mathematics is 77%.

So the total candidates passed in both English and Mathematics is the subtraction of total number of candidates and the total candidates failed in both English and Mathematics.

Therefore total candidates passed in both English and Mathematics is

= (100 - 77) %

= 23%.

So this is the required answer.

Hence option (c) is correct.

Note: In this question since things were given in percentage so on direct substitution of values back into the main formula n(E∪M)=n(E)+n(M)−n(E∩M) will eventually give out the required quantity in percentage only. No need to evaluate the percentage differently.