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New answer posted

a year ago

Class 11th Exemplar Solution

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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Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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 $F = \frac{9}{5} C + 32 ?$

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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?

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

Solve: $4 x + 3 \geq 2 x + 17, 3 x - 5 < - 2$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t 4 x + 3 \geq 2 x + 17 \dots (i) \\ 3 x - 5 < - 2 \dots (i i) \\ F r o m e q n . (i) w e g e t \\ 4 x + 3 \geq 2 x + 17 \Rightarrow 4 x - 2 x \geq 17 - 3 \\ \Rightarrow 2 x \geq 14 \Rightarrow x \geq 7 \\ F r o m e q n . (i i) w e g e t \\ \Rightarrow 3 x - 5 < - 2 \Rightarrow 3 x < 3 \\ \Rightarrow x < 1 \\ W e s e t t h a t t h e s o l u t i o n x \geq 7 a n d x < 1 i s n o t p o s s i b l e . H e n c e t h e r e w i l l b e n o s o l u t i o n o f x . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Solve: $- 5 \leq \frac{2 x - 3}{4} \leq 9$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t - 5 \leq \frac{2 - 3 x}{4} \leq 9 \\ \Rightarrow - 20 \leq 2 - 3 x \leq 36 \Rightarrow - 22 \leq - 3 x \leq 34 \\ \Rightarrow 22 \geq 3 x \geq - 34 \\ \Rightarrow \frac{22}{3} \geq x \geq \frac{- 34}{3} \\ H e n c e x \in [\frac{- 34}{3}, \frac{22}{3}] \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Solve: $| x - 1 | \leq 5, x \geq 2$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t | x - 1 | \leq 5 a n d | x | \geq 2 \\ \Rightarrow - 5 \leq x - 1 \leq 5 a n d x \leq - 2 o r x \geq 2 \\ \Rightarrow - 5 + 1 \leq x \leq 5 + 1 a n d x \leq - 2 o r x \geq 2 \\ \Rightarrow - 4 \leq x \leq 6 a n d x \leq - 2 o r x \geq 2 \\ H e n c e x < [- 4, - 2] \cup [2, 6] \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Solve: $\frac{1}{x - 3} \leq \frac{1}{2}$

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t \frac{1}{| x | - 3} \leq \frac{1}{2} \\ \Rightarrow | x | - 3 \geq 2 [? \frac{1}{x} < \frac{1}{y} \Rightarrow x > y] \\ \Rightarrow | x | \geq 5 \\ \Rightarrow | x | \leq - 5 o r | x | \geq 5 \\ S o, x \in (- \infty, - 5] \cup [5, \infty) \dots (i) \\ H e r e, | x | - 3 \neq 0 \\ \Rightarrow | x | - 3 < 0 o r | x | - 3 > 0 \\ \Rightarrow | x | < 3 o r | x | > 3 \\ \Rightarrow - 3 < x < 3 o r x < - 3 o r x > 3 \dots (i i) \\ F r o m i n e q n . (i) a n d (i i) w e g e t \\ x \in (- \infty, - 5] \cup (- 3, 3) \cup [5, \infty) \end{array}$

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