Answer :

(i) Let x be the fixed charge of the food and y be the charge for food per day. According to the given information,

x + 20y = 1000 .......... (1)


x + 26y = 1180 ........... (2)


Subtracting equation (1) from equation (2), we obtain


6y = 180


y = 30


Substituting this value in equation (1), we obtain


X + 20 × 30 = 1000


x = 1000 - 600 = 400


x = 400


Hence, fixed charge = Rs 400 And charge per day = Rs 30


(ii) Let the fraction be .


According to the given information,


.......(i)


.......(ii)


Subtracting equation (i) from equation (ii), we obtain


x = 5 ............. (iii)


Putting this value in equation (i), we obtain


15 – y = 3


y = 12


Hence, the fraction is .


(iii) Let the number of right answers and wrong answers be x and y respectively.


According to the given information,

Case I

3x –y = 40......(i)


Case II

4x – 2y = 50 

2x – y = 25 .........(ii)


Subtracting equation (ii) from equation (i),


we obtain x = 15 (iii)


Substituting this in equation (ii), we obtain


30 – y = 25


y = 5


Therefore,


number of right answers = 15


And number of wrong answers = 5


Total number of questions = 20


(iv)

Let the speed of car from A be ‘a’ and of car from B be ‘b’

Speed = distance/time

Relative speed of cars when moving in same direction = a + b

Relative speed of cars when moving in opposite direction = a – b

Given, places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour

⇒ a – b = 100/5 = 20     (1)

Also, a + b = 100/1 = 100     (2)

Adding (1) and (2)

a - b + a + b = 20 + 100

⇒ 2a = 120

⇒ a = 60 km/hr

Putting value of a in (1) we get,

Thus, b = 60 – 20 = 40 km/hr
Speed of two cars are 60 km/h and 40  km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively.


Area = xy


According to the question,

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.

(x - 5) (y + 3) = xy - 9

3x – 5y – 6 = 0 ..........(i)


and if we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units

(x + 3) (y + 2) = xy + 67


2x + 3y – 61 = 0 ........(ii)


By cross-multiplication method, we obtain




x = 17, y = 9

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