
In a survey of 100 students studying the various languages were found to be : English only 18 , English but not Hindi 23 , English and Sanskrit 8 , English 26 , Sanskrit 48 , Sanskrit and Hindi 8, no language 24. Find how many students were studying English and Hindi ?
Answer
483.9k+ views
Hint: Draw Venn diagram and fill the details and write the equations in parallel with the information provided in the question. By using a Venn diagram you can find all the values by placing all the given values at the correct place. No language means excluding students reading any subject from all students.
Complete step-by-step answer:
Given that
Total number of students in a survey i.e. n(U ) = 100 ,
The number of students studying only English =18 i.e. a = 18
The number of students studying English but not Hindi =23 i.e. a + f = 23
The number of students studying English and Sanskrit i.e. n(E $ \cap $ S) = 8,
The number of students studying English i.e. n(E) =26,
The number of students studying Sanskrit i.e. n(S) = 48 ,
The number of students studying Sanskrit and Hindi i.e. n(S$ \cap $ H) = 8 ,
The number of students who are not studying any language i.e. u – (a+b+c+d+e+f+g) = 24.
Therefore from the given data we get
$ \Rightarrow $ a + d + f + g = 26
$ \Rightarrow $ f + g = 8
$ \Rightarrow $ a = 18
$ \Rightarrow $ 18 + 8 + d = 26
$ \Rightarrow $ d = 0
$ \Rightarrow $ g + e = 8
$ \Rightarrow $ b + f + g + e = 48
$ \Rightarrow $ a + f = 26 – 3 = 23
$ \Rightarrow $ f = 5 , g = 3
$ \Rightarrow $ a + b + c + d + e + f + g = 100 – 24 = 76
$ \Rightarrow $ 66 + c = 76
$ \Rightarrow $ c = 10
from this we can say that d+ g = 0 + 3 = 3.
The number of students who were studying English and Hindi n(E$ \cap $ H) = 3.
Note: While interpreting the question , make sure that the given information is regarding union or intersection. If it is given that both works are done then we should take union and if any of them is done we should take intersection. While solving the question write the equations that you get through the Venn diagram in parallel with solving the problem.
Complete step-by-step answer:
Given that

Total number of students in a survey i.e. n(U ) = 100 ,
The number of students studying only English =18 i.e. a = 18
The number of students studying English but not Hindi =23 i.e. a + f = 23
The number of students studying English and Sanskrit i.e. n(E $ \cap $ S) = 8,
The number of students studying English i.e. n(E) =26,
The number of students studying Sanskrit i.e. n(S) = 48 ,
The number of students studying Sanskrit and Hindi i.e. n(S$ \cap $ H) = 8 ,
The number of students who are not studying any language i.e. u – (a+b+c+d+e+f+g) = 24.
Therefore from the given data we get
$ \Rightarrow $ a + d + f + g = 26
$ \Rightarrow $ f + g = 8
$ \Rightarrow $ a = 18
$ \Rightarrow $ 18 + 8 + d = 26
$ \Rightarrow $ d = 0
$ \Rightarrow $ g + e = 8
$ \Rightarrow $ b + f + g + e = 48
$ \Rightarrow $ a + f = 26 – 3 = 23
$ \Rightarrow $ f = 5 , g = 3
$ \Rightarrow $ a + b + c + d + e + f + g = 100 – 24 = 76
$ \Rightarrow $ 66 + c = 76
$ \Rightarrow $ c = 10
from this we can say that d+ g = 0 + 3 = 3.
The number of students who were studying English and Hindi n(E$ \cap $ H) = 3.
Note: While interpreting the question , make sure that the given information is regarding union or intersection. If it is given that both works are done then we should take union and if any of them is done we should take intersection. While solving the question write the equations that you get through the Venn diagram in parallel with solving the problem.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How do I convert ms to kmh Give an example class 11 physics CBSE

Describe the effects of the Second World War class 11 social science CBSE

Which of the following methods is suitable for preventing class 11 chemistry CBSE
