5 Marks Questions

Question 15 Marks

Solve the following system of inequalities graphically:
$2x+y\leq24,$
$x+y\geq11,$
$2x+5y\leq40,\quad x,y\geq0.$

Answer

Inequality $(x \geq 0)$ represents the region on the right of y - axis and y - axis itself. Inequality $(y \geq 0)$ represents the region above x - axis and x - axis itself.
$2 x+y \leq 24\quad\quad(1)$
$x+y \geq 11\quad\quad\quad(2)$
$2 x+5 y \leq 40\quad\quad(3)$
We first draw the graph of lines
$2x + y = 24, x + y = 11$ and $2x + 5y = 40$
Now, $2x + y = 24$ passes through A(12, 0) and B(0, 24)
Again, $x + y = 11$ passes through C(11,0) and D(0, 11)

Further $2x + 5y = 40$ passes through E(20, 0) and F(0, 8)
Shaded area PQAC is the solution area.

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Question 25 Marks

(i) If two pipes function simultaneously, the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?
(ii) The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shorter side.

Answer

(i) Let the reservoir be filled by first pipe in x hours.
Then, second pipe will fill it in (x + 10) hours.
$\therefore \quad \frac{1}{x}+\frac{1}{(x+10)}=\frac{1}{12}$
$\Rightarrow x^2-14 x-120=0$
$\Rightarrow (x - 20)(x + 6) = 0$
$\Rightarrow x = 20\quad\quad$[neglecting the -ve value of x]
So, the second pipe will take (20+10) hours, i.e., 30 hours to fill the reservoir.
(ii) Let shorter side be x cm.
Then, according to question,
Longest side = 3x cm and third side = (3x - 2) cm
Also, perimeter of triangle $\geq$ 61
i.e., Sum of all sides $\geq$ 61
$\Rightarrow x + 3x + (3x - 2)$ $\geq$ $61$
$\Rightarrow 7x - 2$ $\geq$ $61$
$\Rightarrow 2 + 7x - 2$ $\geq$ $61 + 2$
$\Rightarrow 7x$ $\geq$ $63$
$\Rightarrow \frac{7 x}{7} \geq \frac{63}{7}$
$\Rightarrow x \geq 9$
$\therefore$ Minimum length of the shortest side $= 9$ cm.

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Question 35 Marks

(i) A can contains a mixture of two liquids A and B in the ratio $7:5.$ When $9$ litres of mixture are drawn off and can is filled with B, the ratio of A and B becomes $7: 9.$ How many litres A was contained by the can initially?
(ii) A man can row 40 km upstream and 55 km downstream in 13 hours. Also, he can row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the man in still water and speed of the current.

Answer

(i) Suppose the can initially contains 7x and 5x litres of mixtures A and B, respectively.
Quantity of A in mixture left $=\left(7 x-\frac{7}{12} \times 9\right)$ litres
$=\left(7 x-\frac{21}{4}\right)$ litres
Quantity of B in mixture left $=\left(5 x-\frac{5}{12} \times 9\right)$ litres
$=\left(5 x-\frac{15}{4}\right)$ litres
$\therefore \quad \frac{\left(7 x-\frac{21}{4}\right)}{\left(5 x-\frac{15}{4}\right)+9}=\frac{7}{9}$
$\Rightarrow\frac{28 x-21}{20 x+21}=\frac{7}{9}$
$\Rightarrow252x - 189 = 140x + 147$
$\Rightarrow112x = 336$
$\Rightarrow x = 3$
So, the can contained 21 litres of A.
Let rate upstream = xkm/h and rate downstream = y km/h
Then, $\frac{40}{x}+\frac{55}{y}=13\quad\quad\ldots \text{(i)}$
and $\frac{30}{x}+\frac{44}{y}=10\quad\quad\ldots \text{(ii)}$
Multiplying (ii) by 4 and (i) by 3 and subtracting, we get
$\frac{11}{y}=1$ or $y=11$
Substituting $y = 11$ in (i), we get x = 5
$\therefore$ Rate in still water $=\frac{1}{2}(11+5)~ \text{km/h} =8~ \text{km/h}$
Rate of current $=\frac{1}{2}(11-5)~ \text{km/h} =3~ \text{km/h}$

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Applied Maths 5 Marks Questions

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