4 Marks Question

Question 14 Marks

Solve $\frac{|x|-1}{|x|-2} \geq 0 ~x \in R , x \neq \pm 2$

Answer

We have, $\frac{|x|-1}{|x|-2} \geq 0$
$\Rightarrow\frac{y-1}{y-2} \geq 0$, where $y=|x|$
$\Rightarrow y \leq 1$ or $y>2$

$\Rightarrow |x| \leq 1$ or $|x|>2$
$\Rightarrow(-1 \leq x \leq 1)$ or $(x<-2$ or $x>2)$
$\Rightarrow x \in[-1,1]$ or $x \in[-\infty,-2) \cup(2, \infty)$
$\Rightarrow x \in[-1,1] \cup(-\infty,-2) \cup(2, \infty)$
Hence, solution set of the given inequation is [-1,1] $\cup(-\infty,-2) \cup(2, \infty).$

View full question & answer→

Question 24 Marks

(i) A runs $1 \frac{3}{4}$ times as fast as B. If A gives B a start of 84 m, how far must winning post be so that A and B might reach it at the same time?
(ii) 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is $16: 65.$ How much wine did the cask hold originally ?
(iii) Show that graph of the solution $2x - 3 > x - 5$ on number line.

Answer

(i) Ratio of the rates of A and B $=\frac{7}{4}: 1=7: 4$
So, in a race of 7 m, A gains 3 m over B.
$\therefore$ 3 m are gained by A in a race of 7 m.
$\therefore$ 84 m are gained by A in a race of $\left(\frac{7}{3} \times 84\right) m$
= 196m
$\therefore$ Winning post must be 196 m away from the starting point.
(ii) Let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations $=\left[x\left(1-\frac{8}{x}\right)^4\right]$ litres
$\therefore$ $\frac{x\left(1-\frac{8}{x}\right)^4}{x}=\frac{16}{81}$
$\Rightarrow \left(1-\frac{8}{x}\right)^4=\left(\frac{2}{3}\right)^4$
$\Rightarrow \frac{x-8}{x}=\frac{2}{3}$
$\Rightarrow 3x - 24 = 2x$
$\Rightarrow x = 24$
Hence, cask originally hold 24 litres of wine.
(iii) We have, $2x - 3 > x - 5$
Transposing x to LHS and - 3 to RHS, we get
$2x - x > - 5 + 3$
$\Rightarrow\quad\quad x>-2\Rightarrow$ $(-2, \infty)$

View full question & answer→

Question 34 Marks

Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it?

Answer

Fill the cistern by the two pipes in 1 hour $=\left(\frac{1}{14}+\frac{1}{16}\right)$
$=\frac{15}{112}$
$\therefore$ Time taken by these pipes to fill the cistern
$=\frac{112}{15}$ hours
$= 7$ h $28$ min
Due to leakage, time taken $= 7$h $28$min $+ 32$min $= 8$h
$\therefore$ Fill the cistern by (two pipes $+$ leak) in 1 hour $=\frac{1}{8}$
empty the cistern by the leak in 1 hour = $\left(\frac{15}{112}-\frac{1}{8}\right)$
$=\frac{1}{112}$
$\therefore$ Leak will empty the full cistern in 112 hours.

View full question & answer→

Question 44 Marks

(i) The milk and water in two vessels A and B are in the ratio $4 : 3$ and $2 : 3$ respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C containing half milk and half water?
(ii) A man can row $7 \frac{1}{2}~ \text{km/h}$ in still water. If in a river running at 1.5 km in an hour, it takes him 50 minutes to row to place and back, how far off is the place ?

Answer

(i) Let the C.P. of milk be ₹ 1 per litre
Milk in 1 litre mixture $A=\frac{4}{7}$ litre
Milk in 1 litre mixture of $B=\frac{2}{5}$ litre
Milk in 1 litre mixture of $C=\frac{1}{2}$ litre

$\therefore$ Required ratio = $\frac{1}{10}: \frac{1}{14}=7: 5$
(ii) Speed downstream $= (7.5 + 1.5)$ km/h $= 9$ km/h
Speed upstream $= (7.5 - 1.5)$ km/h $= 6$ km/h
Let the required distance be x km. Then,
$\frac{x}{9}+\frac{x}{6}=\frac{50}{60}$
$\Rightarrow2 x+3 x=\left(\frac{5}{6} \times 18\right)$
$\Rightarrow 5x =15$
$\Rightarrow x=3$
Hence, the required distance is 3 km.

View full question & answer→

Applied Maths 4 Marks Question

Test yourself on this topic