3 Marks Question

Question 13 Marks

In drilling world’s deepest hole it was found that the temperature T in degree celcius, x km below the earth’s surface was given by $\text{T}=30^\circ+25^\circ(\text{x}-3), 3\leq\text{x}\leq15$At what depth will the temperature be between 155°C and 205°C?

Answer

GIven that, Temperature $\text{T}=30^\circ+25^\circ(\text{x}-3), 3\leq\text{x}\leq15$
Range of the temperature is 155ºC to 205ºC
$\Rightarrow155^{\circ}<\text{T}<205^{\circ}$
$\Rightarrow155^{\circ}<30^{\circ}<25^{\circ}(\text{x}-3)<205^{\circ}$
$\Rightarrow125^{\circ}<25^{\circ}(\text{x}-3)<175^{\circ}$
$\Rightarrow\frac{125}{25}<\text{x}-3<\frac{175}{25}$
$\Rightarrow5<\text{x}-3<7$
$\Rightarrow8<\text{x}<10$
Hence, the required temperature at the depth from 8 km to 10 km lies.

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Question 23 Marks

Solve for x, the inequalities Exercises.
$\frac{|\text{x}-2|-1}{|\text{x}-2|-2}\leq0$

Answer

Given that $\frac{|\text{x}-2|-1}{|\text{x}-2|-2}\leq0$
Put $|\text{x}-2|=\text{y}$
⇒ y - 1 > 0, y - 2 < 0
⇒ y > 1, y < 2
⇒ 1 < |x - 2|, |x - 2| < 2
⇒ x - 2 < -1 or x - 2 > 1 and - 2 < x -2 < 2
⇒ x < 1 or x > 3 and 0 < x < 4
Hence, the required solution is (0, 1) (3, 4).

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Question 33 Marks

Solve for x, the inequalities Exercises.
$-5\leq\frac{2-3\text{x}}{4}\leq9$

Answer

Given that $-5\leq\frac{2-3\text{x}}{4}\leq9$
$\Rightarrow-20\leq2-3\text{x}\leq36$
$\Rightarrow-20-2\leq-3\text{x}\leq36-2$
$\Rightarrow-22\leq-3\text{x}\leq34$
$\Rightarrow\frac{22}{3}\geq\text{x}\geq\frac{-34}{3}$
Hence $\text{x}\in\Big[\frac{-34}{3}, \frac{22}{3}\Big]$

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Question 43 Marks

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 $\text{F}=\frac{9}{5}{^{\circ}\text{C}}+32^{\circ}.$

Answer

Formula of conversion is $\text{F}=\frac{9}{5}{^{\circ}\text{C}}+32^{\circ}$
$\Rightarrow\text{F}-32^{\circ}=\frac{9}{5}{^{\circ}\text{C}}$
$\Rightarrow{^\circ}\text{C}=\frac{5}{9}(\text{F}-32^{\circ})$
$\Rightarrow40{^\circ}<{^\circ\text{C}}<45{^\circ}$
$\Rightarrow40{^\circ}<\frac{5}{9}(\text{F}-32^{\circ})<45^{\circ}$
$\Rightarrow40{^\circ}\times\frac{9}{5}<\text{F}-32^{\circ}<45^{\circ}\times\frac{9}{5}$
$\Rightarrow72^{\circ}+\text{F}-32^{\circ}<81^{\circ}$
$\Rightarrow72^{\circ}+32^{\circ}<\text{F}<81^{\circ}+32^{\circ}$
$\Rightarrow104^{\circ}<\text{F}<113^{\circ}$
Hence, the require range is 104°F to 113°F.

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Question 53 Marks

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

Answer

Let the length of shortest side be x cm.
According to the given information,
Longest side = 2 × Shortest side = 2x cm
And thrid side = 2 + Shortest side = (2 + x) cm
Perimeter of triangle = x + 2x + (x + 2) = 4x + 2
But it is given that,
Perimeter > 166cm
⇒ 4x + 2 > 166
⇒ 4x > 166 - 2
⇒ 4x > 164
$\Rightarrow\text{x}>\frac{164}{2}$
= 41cm
Hence, the minimum length of shortest side is 41cm.

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Question 63 Marks

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

Answer

Let us first condider the linear quation 3x + 2y = 48. We observe the shaded region and the origin on the same side of the graph of the line and (0, 0) satisfy the constraint 3x + 2y < 48.
Simiarly, we observe that the shaded region and the origin both on the same side of the graph of the line and (0, 0) satisfy the constraint x + y < 20.
We also observe notice that the shaded region is in the first quadrant where x > and y > 0
3x + 2y < 48, x + y < 20, x > 0, y > 0.

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