Page 84 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

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

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$\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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Exemplar Solution Class 11th

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find

(i) The number of families which buy newspaper A only.

(ii) The number of families which buy none of A, B and C

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T o t a l n u m b e r o f f a m i l i e s = 10000 \Rightarrow n (U) = 10000 \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A = 40 % \Rightarrow n (A) = 40 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r B = 20 % \Rightarrow n (B) = 20 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r C = 10 % \Rightarrow n (C) = 10 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A a n d B = 5 % \Rightarrow n (A \cap B) = 5 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r B a n d C = 3 % \Rightarrow n (B \cap C) = 3 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A a n d C = 4 % \Rightarrow n (A \cap C) = 4 % \\ N u m b e r o f f a m i l i e s w h o b u y a l l t h e t h r e e N e w s p a p e r s = 2 % \Rightarrow n (A \cap B \cap C) = 2 % \\ (i) N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A o n l y \\ = n (A) - n (A \cap B) - n (A \cap C) + n (A \cap B \cap C) \\ \Rightarrow = \frac{40}{100} - \frac{5}{100} - \frac{4}{100} + \frac{2}{100} = \frac{33}{100} \\ \Rightarrow = 10000 \times \frac{33}{100} = 3300 f a m i l i e s \\ (i i) N u m b e r o f f a m i l i e s w h o b u y n o n e o f A, B a n d C \\ N e w s p a p e r i n p e r c e n t (%) = n (U) - n (A \cup B \cup C) \\ \Rightarrow n (U) - [n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)] \\ \Rightarrow [100 - (40 + 20 + 10 - 5 - 3 - 4 + 2) %] \\ \Rightarrow (100 - 60) % = 40 % \\ ∴ N u m b e r o f f a m i l i e s w h o b u y n o n e o f A, B a n d C n e w s p a p e r o u t o f 10000 f a m i l i e s a r e \\ = 10000 \times \frac{40}{100} = 4000 f a m i l i e s . \end{array}$

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Exemplar Solution Class 11th

In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s = 200 \Rightarrow n (U) = 200 \\ N u m b e r o f s t u d e n t s w h o s t u d y M a t h e m a t i c s = 120 \Rightarrow n (M) = 120 \\ N u m b e r o f s t u d e n t s w h o s t u d y P h y s i c s = 90 \Rightarrow n (P) = 90 \\ N u m b e r o f s t u d e n t s w h o s t u d y C h e m i s t r y = 70 \Rightarrow n (C) = 70 \\ N u m b e r o f s t u d e n t s w h o s t u d y M a t h e m a t i c s a n d P h y s i c s = 40 \Rightarrow n (M \cap P) = 40 \\ N u m b e r o f s t u d e n t s w h o s t u d y P h y s i c s a n d C h e m i s t r y = 30 \Rightarrow n (P \cap C) = 30 \\ N u m b e r o f s t u d e n t s w h o s t u d y C h e m i s t r y a n d M a t h e m a t i c s = 50 \Rightarrow n (C \cap M) = 50 \\ N u m b e r o f s t u d e n t s w h o s t u d y n o n e o f t h e s u b j e c t s = 20 \Rightarrow n (M' \cap P' \cap C') = 20 \\ ∴ n (U) - n (M' \cap P' \cap C') = n (M \cup P \cup C) = 200 - 20 = 180 \\ N o w, n (M \cup P \cup C) = n (M) + n (P) + n (C) - n (M \cap P) - n (P \cap C) - n (C \cap M) + n (M \cap P \cap C) \\ \Rightarrow 180 = 120 + 90 + 70 - 40 - 30 - 50 + n (M \cap P \cap C) \\ \Rightarrow 180 - 160 = n (M \cap P \cap C) \\ \Rightarrow n (M \cap P \cap C) = 20 \\ H e n c e, t h e n u m b e r o f s t u d e n t s w h o s t u d y a l l t h e t h r e e s u b j e c t s = 20 . \end{array}$

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Exemplar Solution Class 11th

In a class of 60 students, 25 students play cricket and 20 students play tennis, and10 students play both the games. Find the number of students who play neither?

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Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s = 60 \Rightarrow n (U) = 60 \\ N u m b e r o f s t u d e n t s w h o p l a y C r i c k e t = 25 \Rightarrow n (C) = 25 \\ N u m b e r o f s t u d e n t s w h o p l a y T e n n i s = 20 \Rightarrow n (T) = 20 \\ N u m b e r o f s t u d e n t s w h o p l a y b o t h t h e g a m e s = 10 \Rightarrow n (C \cap T) = 10 \\ ∴ n (C \cup T) = n (C) + n (T) - n (C \cap T) \\ = 25 + 20 - 10 = 35 \\ N u m b e r o f s t u d e n t s w h o p l a y n e i t h e r = n (U) - n (C \cup T) \\ = 60 - 35 = 25 \end{array}$

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Exemplar Solution Class 11th

Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in science, 6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed

(a) In English and Mathematics but not in Science

(b) In Mathematics and Science but not in English

(c) In Mathematics only

(d) In more than one subject only

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L e t t h e n u m b e r o f s t u d e n t s p a s s e d i n M a t h e m a t i c s M, E b e i n E n g l i s h a n d S b e i n S c i e n c e . \\ T h e n n (U) = 100, n (M) = 12, n (E) = 15, n (S) = 8 \\ n (E \cap M) = 6, n (M \cap S) = 7, n (E \cap S) = 4 a n d n (E \cap M \cap S) = 4 \\ L e t u s d r a w a v e n n d i a g r a m . A c c o r d i n g t o t h e v e n n d i a g r a m, w e g e t \\ a + b + d + e = 15 \dots (i) \\ b + c + e + f = 12 \dots (i i) \\ d + e + f + g = 8 \dots (i i i) \\ n (E \cap M) = 6 ∴ b + e = 6 \dots (i v) \\ n (M \cap S) = 7 ∴ e + f = 7 \dots (v) \\ n (E \cap S) = 4 ∴ d + e = 4 \dots (v i) \\ n (E \cap M \cap S) = 4 ∴ e = 4 \dots (v i i) \\ F r o m e q n . (i v) a n d (v i i) w e g e t b + 4 = 6 ∴ b = 2 \\ F r o m e q n . (v) a n d (v i i) w e g e t 4 + f = 7 ∴ f = 3 \\ F r o m e q n . (v i) a n d (v i i) w e g e t d + 4 = 4 ∴ d = 0 \\ F r o m e q n . (i) w e g e t a + b + d + e = 15 \\ \Rightarrow a + 2 + 0 + 4 = 15 \Rightarrow a = 9 \\ F r o m e q n . (i i) w e g e t b + c + e + f = 12 \\ \Rightarrow 2 + c + 4 + 3 = 12 \Rightarrow c = 3 \\ F r o m e q n . (i i i) w e g e t d + e + f + g = 8 \\ \Rightarrow 0 + 4 + 3 + g = 8 \Rightarrow g = 1 \\ ∴ (i) N u m b e r o f s t u d e n t s w h o p a s s e d i n E n g l i s h a n d M a t h e m a t i c s b u t n o t i n S c i e n c e, b = 2 \\ (i i) N u m b e r o f s t u d e n t s w h o p a s s e d i n M a t h e m a t i c s a n d S c i e n c e b u t n o t i n E n g l i s h, f = 3 \\ (i i i) N u m b e r o f s t u d e n t s w h o p a s s e d i n M a t h e m a t i c s o n l y, c = 3 . \end{array}$

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Exemplar Solution Class 11th

Let A, B and C be sets. Then show that

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

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Vishal Baghel | Contributor-Level 10

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$\begin{array}{l} L e t x \in A \cap (B \cup C) \\ \Rightarrow x \in A a n d (x \in B o r x \in C) \\ \Rightarrow (x \in A a n d x \in B) o r (x \in A a n d x \in C) \\ \Rightarrow (x \in A \cap B) o r (x \in A \cap C) \\ \Rightarrow (x \in A \cap B) \cup (x \in A \cap C) \dots (i) \\ L e t y \in (x \in A \cap B) \cup (x \in A \cap C) \\ \Rightarrow y \in (x \in A \cap B) o r (x \in A \cap C) \\ \Rightarrow (y \in A a n d y \in B) o r (y \in A a n d y \in C) \\ \Rightarrow y \in A a n d (y \in B o r y \in C) \\ \Rightarrow y \in A a n d y \subset (B \cup C) \\ \Rightarrow y \in A \cap (B \cup C) \dots (i i) \\ F r o m e q (i) a n d (i i) w e g e t \\ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

Let T = xx + 5/x – 7 –5 = 4x – 40/13 – x. Is T an empty set? Justify your answer.

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Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : T = {x | \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x}} \\ \Rightarrow \frac{x + 5}{x - 7} - 5 = \frac{4 x - 40}{13 - x} \\ \Rightarrow \frac{x + 5}{x - 7} - \frac{4 x - 40}{13 - x} = 5 \\ \Rightarrow \frac{(x + 5) (13 - x) - (x - 7) (4 x - 40)}{(x - 7) (13 - x)} = 5 \\ \Rightarrow \frac{13 x - x^{2} + 65 - 5 x - 4 x^{2} + 40 x + 28 x - 280}{13 x - x^{2} - 91 + 7 x} = 5 \\ \Rightarrow - 5 x^{2} + 76 x - 215 = 5 (13 x - x^{2} + 7 x - 91) \\ \Rightarrow - 5 x^{2} + 76 x - 215 = 65 x - 5 x^{2} + 35 x - 455 \\ \Rightarrow 76 x - 100 x = - 455 + 215 \\ \Rightarrow - 24 x = - 240 \Rightarrow x = \frac{240}{24} = 10 \\ ∴ T = 10 \\ H e n c e, T i s n o t a n e m p t y s e t . \end{array}$

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Exemplar Solution Class 11th

For all sets A and B, (A ∪ B) – B = A – B

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Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L . H . S . = (A \cup B) - B = (A \cup B) \cap B' [? A - B = A \cap B'] \\ = (A \cap B') \cup (B \cap B') [? A \cap A' = ?] \\ = (A - B) \cup ? \\ = A - B = R . H . S . H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

For all sets A and B, A – (A ∩ B) = A – B

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L . H . S . = A - (A \cap B) = A \cap (A \cap B)' \\ = A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') \\ = ? \cup (A - B) [? A - B = A \cap B'] \\ = A - B = R . H . S . \\ H e n c e, t h e g i v e n s t a t e m e n t i s p r o v e d . \end{array}$

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Exemplar Solution Class 11th

For all sets A and B, A – (A – B) = A ∩ B

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L . H . S . = A - (A - B) = A - (A \cap B') [? A - B = A \cap B'] \\ = A \cap (A \cap B') = A \cap [A' \cup (B')'] [? (A \cap B)' = A' \cup B'] \\ = A \cap [A' \cup B] [? (A')' = A] \\ = (A \cap A') \cup (A \cap B) = ? \cup (A \cap B) \\ = A \cap B = R . H . S . \\ H e n c e, t h e g i v e n s t a t e m e n t i s p r o v e d . \end{array}$

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In a class of 60 students, 25 students play cricket and 20 students play tennis, and10 students play both the games. Find the number of students who play neither?

Let A, B and C be sets. Then show that

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Let T = xx + 5/x – 7 –5 = 4x – 40/13 – x. Is T an empty set? Justify your answer.

For all sets A and B, (A ∪ B) – B = A – B

For all sets A and B, A – (A ∩ B) = A – B

For all sets A and B, A – (A – B) = A ∩ B