(i) (a): For student Anu: Fixed charge + cost of food for 25 days = Rs 4500
i.e., x + 25y = 4500 For student Bindu:
Fixed charges + cost of food for 30 days = Rs 5200
i.e., x + 30y = 5200
(ii) (b): From above, we have a₁ = 1, b₁, = 25
\(c_{1}=-4500 \text { and } a_{2}=1, b_{2}=30, c_{2}=-5200 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{25}{30}=\frac{5}{6}, \frac{c_{1}}{c_{2}}=\frac{-4500}{-5200}=\frac{45}{52} \)
\(\Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution
(iii) (c) : We have x + 25y = 4500 .......(i)
and x + 30y = 5200                      ........(ii)
Subtracting (i) from (ii), we get
5y=700 \(\Rightarrow\) y=140
\(\therefore\) Cost of food per day is Rs 140
(iv) (c): We have, x + 25y = 4500
\(\Rightarrow\) x = 4500 - 25 x 140
\(\Rightarrow\) x = 4500 - 3500 = 1000
\(\therefore\) Fixed charges per month for the hostel is Rs 1000
(v) (d): We have, x = 1000, Y = 140 and Bindu takes food for 20 days.
\(\therefore\) Amount that Bindu has to pay = Rs (1000 + 20 x 140) = Rs 3800

(i) (d): 1^st situation can be represented algebraically as 2x + 3y = 46
(ii) (c): 2^nd situation can be represented algebraically as 3x + 5y = 74
(iii) (b): We have, 2x + 3y = 46 .........(i)
3x+5y=74........... (ii)
Multiplying (i) by 5 and (ii) by 3 and then subtracting,
we get 10x - 9x = 230 - 222 \(\Rightarrow\) x = 8
\(\therefore\) Fare from Bengaluru to Malleswaram is Rs 8.
(iv) (a): Putting the value of x in equation (i), we get
3y = 46 - 2 x 8 = 30 \(\Rightarrow\) Y = 10
\(\therefore\) Fare from Bengaluru to Yeswanthpur is Rs 10.
(v) (c): We have, a₁ = 2, b₁, = 3, c₁ = -46 and
\(a_{2}=3, b_{2}=5, c_{2}=-74 \)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{5}, \frac{c_{1}}{c_{2}}=\frac{-46}{-74}=\frac{23}{37} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system oflinear equations has unique solution.

(i) (a) : Suppose two cars meet at point Q. Then,
Distance travelled by car X = A Q,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\(\therefore\) Distance travelled by car X in 9 hours = 9x km
\(\Rightarrow\) AQ=9x
Distance travelled by car Y in 9 hours = 9y km
\(\Rightarrow\) BQ=9y

Clearly, AQ - BQ = AB
\(\Rightarrow\) 9x - 9y = 90
\(\Rightarrow\) x-y =10
(ii) (c): Suppose two cars meet at point P. Then
Distance travelled by car X = AP and
Distance travelled by car Y = BP.
In this case, two cars meet in 917 hours.
\(\therefore\) Distance travelled by car X in 9/7 hours = \(\frac {9}{7}\) x km
\(\Rightarrow A P=\frac{9}{7} x\)
Distance travelled by car Y in 9/7 hours \(\frac {9}{7}\) y km
\(\Rightarrow B P=\frac{9}{7} y\)
Clearly, AP + BP = AB
\(\Rightarrow \quad \frac{9}{7} x+\frac{9}{7} y=90 \Rightarrow \frac{9}{7}(x+y)=90 \Rightarrow x+y=70\)
(iii) (b): We have x - y = 10
\(\Rightarrow x+y=70\)
Adding equations (i) and (ii), we get
2x = 80 \(\Rightarrow\) x = 40
Hence, speed of car X is 40 km/hr.
(iv) (c): We have x - y = 10
\(\Rightarrow\) 40 - Y = 10 \(\Rightarrow\) Y = 30
Hence, speed of car y is 30 km/hr.
(v) (d)

(i) (a): 1^st situation can be represented as x + 7y = 650 ...(i) and
2^nd situation can be represented as x + 11y = 970 ...(ii)
(ii) (b): Subtracting equations (i) from (ii), we get 
\(4 y=320 \Rightarrow y=80\)
\(\therefore\) Proportional expense for each person is Rs 80.
(iii) (c): Puttingy = 80 in equation (i), we get
x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90
\(\therefore\) Fixed expense for the party is Rs 90
(iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290
(v) (a): We have a₁ = 1, b₁ = 7, c₁ = -650 and 
\(a_{2}=1, b_{2}=11, c_{2}=-970 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\)
\(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution.

(i) (d): Since 8 men and 12 women can finish the work in 10 days.
\(\therefore\left(\frac{8}{x}+\frac{12}{y}\right)=\frac{1}{10} \Rightarrow \frac{80}{x}+\frac{120}{y}=1\)
(ii) (c): Since 6 men and 8 women can finish a piece of work in 14 days
\(\therefore\left(\frac{6}{x}+\frac{8}{y}\right)=\frac{1}{14} \Rightarrow \frac{84}{x}+\frac{112}{y}=1\)
(iii) (c) : Let \(\frac{1}{x}=u, \frac{1}{y}=v\)
Thus, we have
80u + 120v = 1 and 84u + 112v = 1
Solving above two equations, we get
\(v=\frac{1}{280} \Rightarrow \frac{1}{y}=\frac{1}{280} \Rightarrow y=280\)
Thus one woman alone can finish the work in 280 days.
(iv) (a): We have \(\frac{80}{x}+\frac{120}{y}=1 \Rightarrow \frac{80}{x}+\frac{120}{280}=1\)
\(\Rightarrow \quad \frac{80}{x}=1-\frac{3}{7} \Rightarrow \frac{80}{x}=\frac{4}{7}=\Rightarrow x=140\)
Thus one man alone can finish the work in 140 days.
(v) (d): We have, x = 140 and y = 280
One day's work of 14 men and 28 women
\(=\frac{14}{140}+\frac{28}{280}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}\)
Thus, work will be finished in 5 days.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850
(ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900
(iii) (c) : We have 2x + 3y = 850 .........(i)
and 3x + 2y = 900 .........(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get
5y = 750 \(\Rightarrow\) Y = 150
Thus, price of one Physics book is Rs 150.
(iv) (d): From equation (i) we have, 2x + 3 x 150 = 850
\(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200
Hence, cost of one Mathematics book = Rs 200
(v) (a): From above, we have
\(a_{1} =2, b_{1}=3, c_{1}=-850 \)
\(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\)
\(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus system of linear equations has unique solution.

Speed of boat in upstream = (x - y)km/hr and speed of boat in downstream = (x + y)km/hr.
(i) (a): 1^st situation can be represented algebraically as \(\frac{24}{x-y}+\frac{36}{x+y}=6\)
(ii) (b): 2nd situation can be represented algebraically as \(\frac{36}{x-y}+\frac{24}{x+y}=\frac{13}{2}\)
(iii) (c) : Putting \(\frac{1}{x-y}=u \text { and } \frac{1}{x+y}=v\)
we get,
24u + 36v = 6 and 36u + 24v = 13/2
Solving the above equations, we get u \(=\frac{1}{8}, v=\frac{1}{12}\)
(iv) (d): \(\because u=\frac{1}{8}=\frac{1}{x-y} \Rightarrow x-y=8\) ........(i)
\(\text { and } v=\frac{1}{12}=\frac{1}{x+y} \Rightarrow x+y=12\) .........(ii)
Adding equations (i) from (ii), we get 2x = 20 \(\Rightarrow\) x = 10
\(\therefore\) Speed of boat in still water = 10 km/hr -.
(v) (c): From equation (i), 10 - y = 8 \(\Rightarrow\) y = 2
\(\therefore\) Speed of stream = 2 km/hr.

(i) (b):Piyush sells a saree at 8% profit + sells a sweater at 10% discount = Rs 1008
\(\Rightarrow\) (100 + 8)% of x + (100 - 10)% of y = 1008
\(\Rightarrow\) 108% of x + 90% of y = 1008
\(\Rightarrow\) 1.08x + 0.9 y = 1008 ...(i)
(ii) (c): Piyush sold the saree at 10% profit + sold the sweater at 8% discount = Rs 1028
\(\Rightarrow\) (100 + 10)% of x + (100 - 8)% of Y = 1028
\(\Rightarrow\) 110%ofx+92%ofy= 1028
\(\Rightarrow\) 1.1 x + 0.92y = 1028 ...(ii)
(iii) (d): At x-axis, y = 0
\(\Rightarrow 1.08 x=1008 \Rightarrow x=\frac{1008}{1.08}=\frac{2800}{3}\)
(iv) (a): At y-axis, x = 0
\(\Rightarrow 0.92 y=1028 \Rightarrow y=\frac{1028}{0.92}=\frac{25700}{23}\)
(v) (b): Solving equations (i) and (ii), we get x = 600 and y = 400
Hence both linear equations intersect at (600, 400).

(i) (a): At x-axis, y = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) x = 4
At y-axis, x = 0
\(\therefore\) 2x + 4y = 8 \(\Rightarrow\) Y = 2
\(\therefore\) Required coordinates are (4, 0), (0, 2).
(ii) (c): At x-axis, y = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 3x = 18 \(\Rightarrow\) x = 6
At y-axis, x = 0
\(\therefore\) 3x + 6y = 18 \(\Rightarrow\) 6y = 18 \(\Rightarrow\) Y = 3
\(\therefore\) Required coordinates are (6, 0), (0, 3).
(iii) (d): Since, lines are parallel. So, point of intersection of these lines does not exist.
(iv) (a)
(v) (a): Since the lines are parallel.
\(\therefore\) These equations have no solution i.e., the given system of linear equations is inconsistent.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600.
(ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500.
(iii) (c) : At x-axis, y = 0
\(\therefore\)  At y = 0, 2x + y = 1600 becomes 2x = 1600
\(\Rightarrow\) x = 800
\(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0).
(iv) (d) : At y-axis, x = 0
\(\therefore\) 2x + Y = 1500 \(\Rightarrow\) y = 1500
\(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500).
(v) (b): We have, 2x + y = 1600 and 2x + y = 1500
Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\)
\(\therefore\) System of equations have no solution.
\(\therefore\) Lines are parallel.

(i) (b):
Area of one bedroom = 5x sq.m
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
(ii) (d):
Length of outer boundary = 12 + 15 + 12 + 15 = 54 m
(iii) (a):
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
So, 10x + 5y = 95 \(\Rightarrow\) 2x + y = 19
Also, x + 2 + y = 15\(\Rightarrow\) x + y = 13
Solving 2x + y = 19 and x + y = 13,
we get x = 6 m and y = 7 m
Area of bedroom = 5 x 6 = 30 sq. m
(iv) (c):
Area of living room = (15 x 7) – 30
= 105 – 30 =75 sq. m
(v) (c): 
Cost of 1m² laying tiles in kitchen = Rs. 50
Total cost of laying tiles in kitchen = Rs. 50 x 35 = Rs. 1750

(i) (a):
Speed of the boat in downtream = x + y km/hr
(ii) (b):
 Speed of the boat in upstream = x - y km/hr
According to this statement \(\frac{42}{x+y}+\frac{42}{x-y}=20 \Rightarrow \frac{21}{x+y}+\frac{21}{x-y}=10\) and \(\frac{14}{x+y}=\frac{6}{x-y} \Rightarrow \frac{7}{x+y}-\frac{3}{x-y}=0\)
Let \(\frac{1}{x+y}=\mathrm{p}\) and \(\frac{1}{x-y}=\mathrm{q}\) then we have 21p + 21q = 0 and 7p - 3q = 0
Solving 21p + 21q = 0 and 7p - 3q = 0 
we get \(\mathrm{p}=\frac{1}{7}=\frac{1}{x+y} \Rightarrow \mathrm{x}+\mathrm{y}=7\) and \(\mathrm{q}=\frac{1}{3}=\frac{1}{x-y} \Rightarrow \mathrm{x}-\mathrm{y}=3\) 
Again solving x + y = 7 and x - y = 3, we get x = 5 and y = 2
(iii) (a):
 Speed of the boat = 5 km/hr  
(iv) (b): 
 Speed of the river = 2 km/hr 
(v) (c):
Speed of the boat in upstream = 5 – 2 = 3 km/hr

(i) (a): Here, the cost of one notebook be Rs. x and that of pen be Rs. y. According to the statement, we have
3x + 2y = 80 and 4x + 3y = 110
(ii) (b): Solving 3x + 2y = 80 and 4x + 3y = 110, we get x = 20 and y = 10
Cost of 1 notebook = Rs. 20 and Cost of 1 pen = Rs. 10
(iii) (d): Total cost = Rs. 15 x 20 + Rs. 12 x 10
= 300 + 120
= Rs. 420
(iv) (a): Ram said that price of each notebook could be Rs. 25.
Ajay felt that Rs. 2.50 for one pen was too little. It should be at least Rs. 16 Deepak guess the cost of one pen is Rs. 10 and
Lohith guess the cost of one notebook is Rs. 30 Therefore, estimation of Deepak is correct