Q.1)A solution is to be kept between 40°C and 45°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is F = (9/5)C +32?
Q.2)Solve the given inequality for real x : (x/4)< [(5x-2)/3 -(7x-3)/5]
Q.3)Solve the given inequality 3(1-x)<2(x+4) and represent the solution in the number line.
Q.4)Determine all pairs of consecutive even positive integers, both of which are greater than 5 such that their sum is less than 23.
Q.5)The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
Q.6)
Prove that the following system of linear inequalities has no solution:
x + 2y ≤ 3, 3x + 4y ≥ 12, where x ≥ 0, y ≥ 1
Q.7)If x is real number and |x| < 3, then
(a) x ≥ 3
(b) -3 < x <3
(c) x ≤ -3
(d) -3 ≤ x ≤ 3
Q.8)The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit?
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Answers
Question no. 1 ans
Let the required temperature be
. <br> Given that,
<br>
<br>
<br>
<br> Since, temperature in degree calcius lies between
. <br> Therefore,
<br>
<br>
[ multiplying throughout by 9] <br>
[ adding 160 throughout] <br>
<br>
<br>
[ divide throughout by 5} <br>
<br> Hence, the range of temperature in degree fahrenheit is.
Question no. 2
Answer
4x<3(5x−2)−5(7x−3)
⇒4x<dfrac5(5x−2)−3(7x−3)15
⇒4x<1525x−10−21x+9
⇒4x<154x−1
⇒15x<4(4x−1)
⇒15x<16x−4
⇒4<16x−15x
⇒4<x
Thus, all real numbers x, which are greater than 4, are the solution of the given inequality.
Hence, the solution set of the given inequality is (4,∞).
question no. 3 ans
Step 1: The given inequality is 3(1−x)<2(x+4) => 3−3x<2x+8 Adding -8 and 3x on both sides of the inequality. =>3−8<2x+3x =>−5<5x Dividing both sides of the inequality by a positive number 5. =>−55<5x5 =>−1
The picture there given in the above is the number line of question no. 3
For question no. 4
Let x be smaller of two consecutive even positive integers. Then the other integer is x+2.
Both integers are larger than 5.
Sum of both integers is less than 23.
We conclude and and x is even integer number.
x can be 6,8,10.
The pairs of consecutive even positive integers are (6,8),(8,10),(10,12)
For question no. 5
Let the length of shortest side = ‘x’ cm According to the question, The longest side of a triangle is twice the shortest side ⇒ Length of largest side = 2x Also, the third side is 2 cm longer than the shortest side ⇒ Length of third side = (x + 2) cm Perimeter of triangle = sum of three sides = x + 2x + x + 2 = 4x + 2 cm Now, we know that, Perimeter is more than 166 cm ⇒ 4x + 2 ≥ 166 ⇒ 4x ≥ 164 ⇒ x ≥ 41 Hence, minimum length of the shortest side should be = 41 cm.Read more on Sarthaks.com - https://www.sarthaks.com/894670/the-longest-side-triangle-twice-shortest-side-the-third-side-longer-than-the-shortest-side
For question no. 6
We have x + 2y ≤ 3, 3x + 4y > 12, x > 0, y ≥ 1
Now let’s plot lines x + 2y = 3, 3x + 4y = 12, x = 0 and y = 1 in coordinate plane.
Line x + 2y = 3 passes through the points (0, 3/2) and (3, 0).
Line 3jc + 4y = 12 passes through points (4, 0) and (0, 3).
For (0, 0), 0 + 2(0) – 3 < 0.
Therefore, the region satisfying the inequality x + 2y ≤ 3 and (0,0) lie on the same side of the line x + 2y = 3.
For (0, 0), 3(0) + 4(0)- 12 ≤0.
Therefore, the region satisfying the inequality 3x + 4y ≥ 12 and (0, 0) lie on the opposite side of the line 3x + 4y = 12.
The region satisfying x > 0 lies to the right hand side of the y-axis.
The region satisfying y > 1 lies above the line y = 1.
So, solution set is null set.
For question no. 8
The goods should be sold more than 50 in number to get some gain.
Step-by-step explanation:
Given :
Cost = C(x) = 20x + 4000
Revenue functions = R(x) = 60x + 2000
x = number of items produced and sold
To find :
Number of products to be sold to get some gain
Solution :
Cost → C(x) = 20x + 4000
Revenue → R(x) = 60x + 2000
\boxed{\sf{Profit = Revenue - Cost}}Profit=Revenue−Cost
Subsitute the data provided in the formula :
\begin{gathered}\sf{\longrightarrow}\:Profit = R(x) - C(x) \\ \\ \sf{\longrightarrow}\:Profit = (60x + 2000)-(20x + 4000) \\ \\\sf{\longrightarrow}\:Profit = 60x + 2000-20x-4000 \\ \\\sf{\longrightarrow} \:60x - 20x + 2000 - 4000 \\ \\ \sf{\longrightarrow}\:Profit = 40x - 2000\end{gathered}⟶Profit=R(x)−C(x)⟶Profit=(60x+2000)−(20x+4000)⟶Profit=60x+2000−20x−4000⟶60x−20x+2000−4000⟶Profit=40x−2000
\rule{300}{1.5}
To get some gain : 40x – 2000 > 0
\begin{gathered}\sf{\longrightarrow}\:40x > 2000 \\ \\ \sf{\longrightarrow}\:x > \dfrac{2000}{40} \\ \\\sf{\longrightarrow}\:x > 50\end{gathered}⟶40x>2000⟶x>402000⟶x>50
Goods needed = More than 50
\therefore∴ The goods should be sold more than 50 in number to get some gain.
Sorry I don't know question no. 7 but I hope that others of mi answer will be helpful to u ....
