Question

2. i)
Ifa:b=11:15 and b:c50:17, find a:b:c
ii) Express 2
2 correct upto 2 decimal places.
iii) The length of a room is twice its width and cost of paving the floor is Rs. 544 at Rs. 4.25 per
square meter. Find the perimeter.
iv) Find the sum whose 20% is Rs. 15.
2x4-8

Answer 1

Answer:

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Answer 2

Answer:

i) Let the fixed monthly hostel charges be Rs. x and the cost of food per day taken from the mess be y.

As per the given condition,

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

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

Subtracting eq(1) from eq (2), we get

⇒6y=180

⇒y=30

Substitute y=30 in eq (1), we get

⇒x+20(30)=1000

⇒x+600=1000

⇒x=400

So, the fixed monthly hostel charge is Rs. 600 and the cost of food per day is Rs. 30.

ii) Let the numerator be x and the denominator be y, so the fraction will be

y

x

.

y

x−1

=

3

1

⇒3x−y=3⇒3x−3=y ...(1)

y+8

x

=

4

1

⇒4x=y+8 ...(2)

Substituting for y in (2) we get,

4x=3x−3+8

x=5

putting x=5 in eq(1) we get y=12

∴fraction=

y

x

=

12

5

iii) Let x be the number of correct answers and y be the number of wrong answers.

So, 3x−y=40 ...(1)

4x−2y=50 ...(2)

Multiplying (1) by 2, we get

6x−2y=80 ...(3)

4x−2y=50 ...(4)

Subtracting eq(4) from eq(3), we get

⇒2x=30

⇒x=15

Putting x=15 in eq (1), we get

⇒3(15)−y=40

⇒y=5

So, Correct answers =15 and wrong answers =5.

Total number of questions in the test =x+y=20.

iv) Let the speeds of the first and second cars be x km/hr and y km/hr

respectively.

Distance between places A and B =100km.

Moving in the same direction, relative speed will be =x−y

5×(x−y)=100

⇒x−y=20 ...(1)

Moving in opposite directions, relative speed will be =x+y

⇒1×(x+y)=100⇒x+y=100 ...(2)

Adding (1) and (2), we get

⇒2x=120

⇒x=60

Substitute in eq (1), we get

⇒y=40

The speeds of the first and second cars are 60 km/hr and 40 km/hr respectively.

v) Let the length of rectangle be l and the breadth be b.

So, the area will be A=l×b.

Case 1:

l

′

=l−5

b

′

=b+3

A

′

=A−9=lb−9

⇒(l−5)(b+3)=lb−9

⇒lb+3l−5b−15=lb−9

⇒3l−5b=6....(1)

Case 2:

l

′′

=l+3

b

′′

=b+2

A

′′

=A+67=lb+67

(l+3)(b+2)=lb+67

⇒lb+2l+3b+6=lb+67

⇒2l+3b=61...(2)

Multiplying (1) by 3 and (2) by 5 and adding both equations, we get

9l−15b=18 ...(3)

10l+15b=305 ...(4)

Adding eq (3) and eq (4), we get

19l=323

⇒l=17

Putting l=17 in eq(1), we get

⇒3(17)−5b=6

⇒45=5b

⇒b=9

The length and breadth of the rectangle are 17 and 9 units respectively.