Exponded farm
pont It Number is standard from
wsite coch, Nomborin standby use. The fost
problem has been completed for you
1 3.000 + 200+30+5 2.800+20+5
3
4.
5 hundreds 2 tons and Thell thausand and
I ons
your
Answers
Answer:
Step-by-step explanation: 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.