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Exemplar Solution Class 11th

Show that the following system of linear inequalities has no solution: $x + 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1 .$

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Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

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Exemplar Solution Class 11th

Find the linear inequalities for which the shaded region in the given figure is the solution set.

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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t u s f i r s t c o n s i d e r t h e l i n e x + y = 4 \\ W e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e o r i g i n (0, 0) b o t h o n t h e o p p o s i t e s i d e a n d (0, 0) \\ s a t i s f y t h e constraint x + y \leq 4 . \\ S o, x + y \geq 4 i s t h e l i n e a r i n e q u a l i t y . \\ N o w, t a k e t h e l i n e a r e q u a t i o n x + y = 8 . \\ I n t h i s c a s e, t h e s h a d e d r e g i o n a n d t h e o r i g i n (0, 0) b o t h l i e o n t h e s a m e s i d e o f t h e g r a p h \\ a n d (0, 0) s a t i s f y t h e constraint x + y \leq 8 . \\ S o, x + y \leq 4 i s a n o t h e r l i n e a r i n e q u a l i t y . \\ C o n s i d e r, x = 5 . I t i s c l e a r t h a t t h e s h a d e d a r e a l i e s t o w a r d s t h e o r i g i n . S o, x \leq 5 i s t h e \\ constraints a n d s i m i l a r l y y \leq 5 i s a l s o t h e constraints. W e a l s o o b s e r v e t h a t t h e s h a d e d r e g i o n \\ l i e s i n f i r s t q u a d r a n t, S o x \geq 0 a n d y \geq 0 . \\ H e n c e, t h e r e q u i r e d l i n e a r i n e q u a l i t i e s a r e \\ x + y \geq 4, x + y \leq 8, x \leq 5, y \leq 5, x \geq 0 a n d y \geq 0 . \end{array}$

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Exemplar Solution Class 11th

Find the linear inequalities for which the shaded region in the given figure is the solution set.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

\begin{array}{l} L e t u s c o n s i d e r t h e l i n e a r e q u a t i o n 3 x + 2 y = 48 . W e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e \\ o r i g i n b o t h o n t h e s a m e s i d e o f t h e g r a p h o f t h e l i n e a n d (0, 0) s a t i s f y t h e constraint 3 x + 3 y \leq 48 . \\ S i m i l a r l y, w e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e o r i g i n b o t h o n t h e s a m e s i d e o f t h e g r a p h \\ o f t h e l i n e a n d (0, 0) s a t i s f y t h e constraint x + y \leq 20 . \\ W e s e e t h a t t h e s h a d e d r e g i o n i s i n t h e f i r s t q u a d r a n t w h e r e x \geq 0 a n d y \geq 0 \\ H e n c e, t h e r e q u i r e d l i n e a r i n e q u a l i t i e s a r e \\ 3 x + 3 y \leq 48, x + y \leq 20, x \geq 0 a n d y \geq 0 . \end{array}

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Exemplar Solution Class 11th

Solve the following system of inequalities: $\frac{2 x + 1}{7 x - 1} > 5, \frac{x + 7}{x - 8} > 2$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t u s t a k e t h e i n e q u a t i o n \\ \frac{2 x + 1}{7 x - 1} > 5 \Rightarrow \frac{2 x + 1}{7 x - 1} - 5 > 0 \\ \Rightarrow \frac{2 x + 1 - 5 (7 x - 1)}{7 x - 1} > 0 \Rightarrow \frac{2 x + 1 - 35 x + 5}{7 x - 1} > 0 \\ \Rightarrow \frac{- 33 x + 6}{7 x - 1} > 0 \Rightarrow \frac{33 x - 6}{7 x - 1} < 0 \\ S o, x \in (\frac{1}{7}, \frac{2}{11}) \dots (i) \\ N o w f r o m second i n e q u a l i t y, w e h a v e \\ \Rightarrow \frac{x + 7}{x - 8} > 2 \Rightarrow \frac{x + 7}{x - 8} - 2 > 0 \\ \Rightarrow \frac{x + 7 - 2 x + 16}{x - 8} > 0 \Rightarrow \frac{- x + 23}{x - 8} > 0 \\ \Rightarrow \frac{x - 23}{x - 8} < 0 \\ S o, x \in (8, 23) \dots (i i) \\ N o w n o t h i n g i s c o m m o n t o t h e t w o v a l u e s o f x i . e ., n u l l s e t . \\ H e n c e, t h e g i v e n i n e q u a l i t i e s h a s n o s o l u t i o n . \end{array}$

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Exemplar Solution Class 11th

In drilling the world’s deepest hole, it was found that the temperature T in degrees Celsius, x below the earth’s surface, was given by $T = 30 + 25 (x - 3), 3 \leq x \leq 15 .$ At what depth will the temperature be between $15 5^{\circ} C$ and $20 5^{\circ} C$ ?

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t, T e m p e r a t u r e T = 3 0^{0} + 2 5^{0} (x - 3), 3 \leq x \leq 15 \\ R a n g e o f t h e t e m p e r a t u r e i s 15 5^{0} C t o 20 5^{0} C \\ ∴ 15 5^{0} < T < 20 5^{0} \\ \Rightarrow 15 5^{0} < 3 0^{0} + 2 5^{0} (x - 3) < 20 5^{0} \Rightarrow 12 5^{0} < 2 5^{0} (x - 3) < 17 5^{0} \\ \Rightarrow \frac{125}{25} < (x - 3) < \frac{175}{25} \Rightarrow 5 < x - 3 < 7 \\ \Rightarrow 8 < x < 10 \\ H e n c e, t h e r e q u i r e d t e m p e r a t u r e a t t h e d e p t h f r o m 8 k m t o 10 k m l i e s b e t w e e n 15 5^{0} C a n d 20 5^{0} C . \end{array}$

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Exemplar Solution Class 11th

The longest side of a triangle is twice the shortest side, and the third side is 2 longer than the shortest side. If the perimeter of the triangle is more than 166, then find the minimum length of the shortest side.

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e l e n g t h o f t h e s h o r t e s t s i d e b e x c m \\ ∴ L e n g t h o f t h e l o n g e s t s i d e = 2 x c m \\ a n d l e n g t h o f t h e t h i r d s i d e b e (x + 2) c m \\ P e r i m e t e r o f t h e t r i a n g l e = x c m + 2 x c m + (x + 2) c m = (4 x + 2) c m \\ A s p e r t h e c o n d i t i o n o f t h e q u e s t i o n, \\ P e r i m e t e r > 166 c m \\ \Rightarrow 4 x + 2 > 166 \Rightarrow 4 x > 166 - 2 \\ \Rightarrow 4 x > 164 \Rightarrow x > 41 c m \\ H e n c e, t h e minimum l e n g t h o f t h e s h o r t e s t s i d e o f t h e t r i a n g l e i s 41 c m . \end{array}$

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Exemplar Solution Class 11th

A solution is to be kept between $4 0^{\circ} C$ and $4 5^{\circ} C$ . What is the range of temperature in degrees Fahrenheit if the conversion formula is $F = \frac{9}{5} C + 32 ?$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t t h e r a n g e o f t e m p e r a t u r e o f t h e s o l u t i o n i s 4 0^{0} C a n d 4 5^{0} C \\ F o r m u l a o f c o n v e r s a t i o n i s F = {\frac{9}{5}}^{0} C + 3 2^{0} \\ \Rightarrow F - 3 2^{0} = {\frac{9}{5}}^{0} C \Rightarrow^{0} C = \frac{5}{9} (F - 3 2^{0}) \\ 4 0^{0} <^{0} C < 4 5^{0} \\ \Rightarrow 4 0^{0} < \frac{5}{9} (F - 3 2^{0}) < 4 5^{0} \Rightarrow 4 0^{0} \times \frac{9}{5} < F - 3 2^{0} < 4 5^{0} \times \frac{9}{5} \\ \Rightarrow 7 2^{0} < F - 3 2^{0} < 8 1^{0} \Rightarrow 7 2^{0} + 3 2^{0} < F < 8 1^{0} + 3 2^{0} \\ \Rightarrow 10 4^{0} < F < 11 3^{0} \\ H e n c e, t h e r e q u i r e d r a n g e i s 10 4^{0} F t o 11 3^{0} F . \end{array}$

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Exemplar Solution Class 11th

A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there are 460 liters of the 9% solution, how many liters of the 3% solution will have to be added?

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Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t x l i t r e s o f 3 % s o l u t i o n b e a d d e d t o 460 l i t r e s o f 9 % . \\ ∴ T o t a l a m o u n t o f m i x t u r e = (460 + x) l i t r e s \\ G i v e n t h a t t h e a c i d c o n t e n t s i n t h e r e s u l t i n g m i x t u r e i s m o r e t h a n 5 % b u t l e s s t h a n 7 % a c i d . \\ ∴ 5 % o f (460 + x) < 460 \times \frac{9}{100} + \frac{3}{100} \times x < 7 % o f (460 + x) \\ \Rightarrow \frac{5}{100} (460 + x) < \frac{4140 + 3 x}{100} < \frac{7}{100} (460 + x) \\ \Rightarrow 5 (460 + x) < 4140 + 3 x < 7 (460 + x) \\ \Rightarrow 2300 + 5 x < 4140 + 3 x < 3220 + 7 x \\ \Rightarrow 2300 + 5 x < 4140 + 3 x a n d 4140 + 3 x < 3220 + 7 x \\ \Rightarrow 5 x - 3 x < 4140 - 2300 a n d 3 x - 7 x < 3220 - 4140 \\ \Rightarrow 2 x < 1840 a n d - 4 x < - 920 \\ \Rightarrow x < \frac{1840}{2} a n d 4 x > 920 \\ \Rightarrow x < 920 a n d x > \frac{920}{4} ∴ x > 230 \\ H e n c e, t h e r e q u i r e d a m o u n t o f a c i d s o l u t i o n i s m o r e t h a n 230 l i t r e s a n d l e s s t h a n 920 l i t r e s . \end{array}$

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Exemplar Solution Class 11th

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are  $8.48$ and $8.35$ , find the range of pH values for the third reading that will result in the acidity level being normal.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e t h i r d p H v a l u e b e x . \\ G i v e n t h a t f i r s t p H v a l u e = 8.48 \\ a n d second p H v a l u e = 8.35 \\ ∴ A v e r a g e v a l u e o f p H = \frac{8.48 + 8.35 + x}{3} \\ B u t a v e r a g e v a l u e o f p H l i e s b e t w e e n 8.2 a n d 8.5 \\ ∴ 8.2 < \frac{8.48 + 8.35 + x}{3} < 8.5 \\ \Rightarrow 24.6 < 16.83 + x < 25.5 \\ \Rightarrow 24.6 - 16.83 < x < 25.5 - 16.83 \\ \Rightarrow 7.77 < x < 8.67 \\ H e n c e, t h e t h i r d p H v a l u e l i e s b e t w e e n 7.77 a n d 8.67 . \end{array}$

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Exemplar Solution Class 11th

A company manufactures cassettes. Its cost and revenue functions are $C (x) = 26, 000 + 30 x and R (x) = 43 x,$ where $x$ is the number of cassettes produced and sold in a week? How many cassettes must be sold by the company to realize some profit?

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : Cost f u n c t i o n, C (x) = 26, 000 + 30 x \\ a n d r e v e n u e f u n c t i o n R (x) = 43 x \\ N o w f o r p r o f i t P (x), R (x) > C (x) \\ \Rightarrow 43 x > 26000 + 30 x \Rightarrow 26000 + 30 x < 43 x \\ \Rightarrow 30 x - 43 x < 26000 \Rightarrow - 13 x < - 26000 \\ \Rightarrow 13 x > 26000 \Rightarrow x > 2000 \\ H e n c e, n u m b e r o f c a s s e t t e s t o b e m a n u f a c t u r e d f o r s o m e p r o f i t m u s t b e m o r e t h a n 2000 . \end{array}$

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Show that the following system of linear inequalities has no solution: $x + 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1 .$

Find the linear inequalities for which the shaded region in the given figure is the solution set.

Find the linear inequalities for which the shaded region in the given figure is the solution set.

Solve the following system of inequalities: $\frac{2 x + 1}{7 x - 1} > 5, \frac{x + 7}{x - 8} > 2$

In drilling the world’s deepest hole, it was found that the temperature T in degrees Celsius, x below the earth’s surface, was given by $T = 30 + 25 (x - 3), 3 \leq x \leq 15 .$ At what depth will the temperature be between $15 5^{\circ} C$ and $20 5^{\circ} C$ ?

A solution is to be kept between $4 0^{\circ} C$ and $4 5^{\circ} C$ . What is the range of temperature in degrees Fahrenheit if the conversion formula is $F = \frac{9}{5} C + 32 ?$