MATHS 3 Marks Question

Let the pH value of third reading be x.

$\therefore7.2<\frac{7.48+7.85+\text{x}}{3}<7.8$

⇒ 21.6 < 7.48 + 7.85 + x < 23.4

⇒ 21.6 < 15.33 + x < 23.4

⇒ 21.6 - 15.33 < x < 23.4 - 15.33

⇒ 6.27 < x < 8.07

$\therefore$ The range of pH value for the third reading is lies between 6.27 and 8.07.

Let x liters of 2% solution will have to be added to 640 liters of the 8% solution of acid.

Total quantity of mixture = (640 + x)

Total acid in the (640+x) liters of mixture

$\frac{2}{100}\text{x}+\frac{8}{100}640$

It is given that acid content in the resulting mixture must be more than 4% but less than 6%.

$\frac{4}{100}[640+\text{x}]<\Big(\frac{2}{100}\text{x}+\frac{8}{100}640\Big)<\frac{6}{100}[640+\text{x}]$

⇒ 4 [640 + x] < ( 2x + 8640) < 6 [640 + x]

⇒ 2560 + 4x < 2x + 8640 and 2x + 8640 < 3840 + 6x

⇒ 2650 - 8640 < 2x - 4x and 2x - 6x < 3840 - 8640

⇒ x < 1280 and x > 320

More than 320 litres but less than 1280 litres of 2% is to be added.

Let x be the smaller of the two consecutive odd positive intgers. Then the other odd integer is x + 2. It is given that both the integers are smaller than 10 and their sum is more than 11.

$\therefore$ x + 2 < 10 and, x + (x + 2) > 11

⇒ x < 10 - 2 and 2x + 2 > 11

⇒ x < 8 and 2x > 9

⇒ x < 8 and $\text{x}>\frac{9}{2}$

$\Rightarrow\frac{9}{2}<\text{x}<8$

⇒ x = 5, 7 [$\because$ is an odd integer]

Hence, the required pairs of odd integers are (5, 7) and (7, 9).

Let x be the smaller of the two consecutive odd natural numbers. Then the other odd integer İS x + 2

It is given that both the natural number are greater than 10 and their sum is less than 40.

$\therefore$ x > 10 and x + x + 2 < 40

⇒ x > 10 and 2x < 38

⇒ x > 10 and x < 19

⇒ 10< x < 19

⇒ x = 11, 13, 15, 17 [ $\because$ is an odd number]

 Hence, the required pairs of odd natural numbers are (11, 13), (13, 15) (15, 17) and (17, 19).

We have,

F₁ = 86º F

$\therefore\text{F}_1=\frac{9}{5}\text{C}_1+32$ $\Big[\because\text{F}=\frac{9}{5}\text{C}+32\Big]$

$\Rightarrow86=\frac{9}{5}\text{C}_1+32$

$\Rightarrow86-32=\frac{9}{5}\text{C}_1$

$\Rightarrow54=\frac{9}{5}\text{C}_1$

$\Rightarrow9\text{C}_1=5\times54$

$\Rightarrow\text{C}_1=\frac{5\times54}{9}$

$\Rightarrow\text{C}_1=5\times6=30^\circ\text{C}$

Now, $\text{F}_2=95^\circ\text{F}$

$\therefore\text{F}_2=\frac{9}{5}\text{C}_2+32$

$\Rightarrow95-32=\frac{9}{5}\text{C}_2$

$\Rightarrow63=\frac{9}{5}\text{C}_2$

$\Rightarrow9\text{C}_2=63\times5$

$\Rightarrow\text{C}_2=\frac{63\times5}{9}$

$\Rightarrow\text{C}_2=7\times5=35^\circ\text{C}$

The range of temperature of the solution is from 30ºC to 35ºC.

Let x be the smaller of the two consecutive even positive integers.

Then the other even integer is x + 2.

It is given that both the even integers are greater than 5 and their sum is less than 23.

$\therefore$ x > 5 and x + x + 2 < 23

⇒ x > 5 and 2x < 21

$\Rightarrow\text{x}>5$ and $\text{x}<\frac{21}{2}$

$\Rightarrow5<\text{x}<\frac{21}{2}=10.5$

⇒ x = 6, 8, 10 [ $\because$ is an even integer]

Hence, the required pairs of even positive integer are (6, 8),(8, 10) and (10, 12).

Suppose Rohit scores x mark in the third test then,

$65\leq\frac{65+70+\text{x}}{3}$

$195\leq135+\text{x}$

$\Rightarrow195-135\leq\text{x}$

$\Rightarrow60\leq\text{x}$

Hence, the minimum marks Rohit should score in the third test test is 60.

We have,

profit = Revenue - cost

therefore, to ear some profit, we must have

Revenue > Cost

$\Rightarrow2\text{x}>300+\frac{3}{2}\text{x}$

$\Rightarrow2\text{x}-\frac{3}{2}\text{x}>300$

$\Rightarrow\frac{4\text{x}-3\text{x}}{2}>300$

$\Rightarrow\text{x}>300\times2$

$\Rightarrow\text{x}>600$

Hence, then manufacture must sell more that 600 cassettes to realize some profit.

We have,

C₁ = 30ºC

$\therefore\text{F}_1=\frac{9}{5}\text{C}_1+32$$\Big[\because\text{F}=\frac{9}{5}\text{C}+32\Big]$

$\Rightarrow\text{F}_1=\frac{9}{5}\times30+32$

$\Rightarrow\text{F}_1=9\times6+32$

$\Rightarrow\text{F}_1=54+32$

$\Rightarrow\text{F}_1=86^\circ\text{F}$

Now,

C² = 35ºC

$\therefore\text{F}_2=\frac{9}{5}\text{C}_2+32$

$\Rightarrow\text{F}_2=\frac{9}{5}\times35+32$

$\Rightarrow\text{F}_2=9\times7+32$

$\Rightarrow\text{F}_2=63+32$

$\Rightarrow\text{F}_2=95^\circ\text{F}$

Hence, the temperature of the solution lies between 86ºF to 95ºF.

Let the length of the shortest side be x.

Then, the length of te longest side and third side of the triangle are 3x and 3x - 2 respectively.

According to question, perimeter of triangle > 61

perimeter of triangle $\geq61$

$\Rightarrow\text{x}+3\text{x}-2+3\text{x}\geq61$

$\Rightarrow7\text{x}\geq61+2$

$\Rightarrow7\text{x}\geq63$

$\Rightarrow\text{x}\geq\frac{63}{7}$

$\Rightarrow\text{x}\geq9$

$\therefore$ The maximum length of the shortest side is 9cm.

From (i) and (ii), $(-\infty,-3)\cup\Big(\frac{7}{3},\infty\Big)$is the solution set of the simultaneous equations.

From (i) and (ii), $(-\infty,2)$ is the solution set of the simultaneous equations.