Maths NCERT Exemplar Solutions Class 11th Chapter One Maths Class 11th Exemplar Solution

In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:

French = 17, English = 13, Sanskrit = 15

French and English = 09, English and Sanskrit = 4

French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study

(i) French only

(ii) English only

(iii) Sanskrit only

(iv) French and English but not Sanskrit

(v) At least one of the three languages

(vi) English and Sanskrit

(vii) None of the three languages but not French

2 Views | Posted 4 months ago

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

$\begin{array}{l} L e t u s u s e v e n n d i a g r a m m e t h o d . \\ T o t a l n u m b e r o f s t u d e n t s = 50 \Rightarrow n (U) = 50 \\ N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h = 17 \Rightarrow n (F) = 17 \\ N u m b e r o f s t u d e n t s w h o s t u d y E n g l i s h = 17 \Rightarrow n (E) = 13 \\ N u m b e r o f s t u d e n t s w h o s t u d y S a n s k r i t = 15 \Rightarrow n (S) = 15 \\ N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h a n d E n g l i s h = 9 \Rightarrow n (F \cap E) = 9 \\ N u m b e r o f s t u d e n t s w h o s t u d y E n g l i s h a n d S a n s k r i t = 4 \Rightarrow n (E \cap S) = 4 \\ N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h a n d S a n s k r i t = 5 \Rightarrow n (F \cap S) = 5 \\ N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h, E n g l i s h a n d S a n s k r i t = 3 \Rightarrow n (F \cap E \cap S) = 3 \\ n (F) = 17 \Rightarrow a + b + d + e = 17 \dots (i) \\ n (E) = 13 \Rightarrow b + c + e + f = 13 \dots (i i) \\ n (S) = 15 \Rightarrow d + e + f + g = 15 \dots (i i i) \\ n (F \cap E) = 9 ∴ b + e = 9 \dots (i v) \\ n (E \cap S) = 4 ∴ e + f = 4 \dots (v) \\ n (F \cap S) = 5 ∴ d + e = 5 \dots (v i) \\ n (E \cap F \cap S) = 3 ∴ e = 3 \dots (v i i) \\ F r o m e q n . (i v) b + 3 = 9 ∴ b = 6 \\ F r o m e q n . (v) 3 + f = 4 ∴ f = 1 \\ F r o m e q n . (v i) d + 3 = 5 ∴ d = 2 \\ F r o m e q n . (i) w e g e t a + 6 + 2 + 3 = 17 \Rightarrow a = 6 \\ F r o m e q n . (i i) w e g e t 6 + c + 3 + 1 = 13 \Rightarrow c = 3 \\ F r o m e q n . (i i i) w e g e t 2 + 3 + 1 + g = 15 \Rightarrow g = 9 \\ (i) N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h o n l y, a = 6 \\ (i i) N u m b e r o f s t u d e n t s w h o s t u d y E n g l i s h o n l y, c = 3 . \\ (i i i) N u m b e r o f s t u d e n t s w h o s t u d y S a n s k r i t o n l y, g = 9 \\ (i v) N u m b e r o f s t u d e n t s w h o s t u d y E n g l i s h a n d S a n s k r i t b u t n o t F r e n c h, f = 1 . \\ (v) N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h a n d S a n s k r i t b u t n o t E n g l i s h, d = 2 . \\ (v i) N u m b e r o f s t u d e n t s w h o s t u d y F r e n c h a n d E n g l i s h b u t n o t S a n s k r i t, b = 6 . \end{array}$

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