Question

2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) One number is greater than thrice the other number by 2. If 4 times the smaller
number exceeds the greater by 5, find the numbers.
(ii) The ratio of income of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3.
If each of them manages to save ` 2000 per month, find their monthly income.
(iii) A two digit number is seven times the sum of its digits. The number formed by
reversing the digits is 18 less than the given number. Find the given number.
(iv) Three chairs and two tables cost ` 700 and five chairs and three tables cost ` 1100.
What is the total cost of 2 chairs and 3 tables?
(v) In a rectangle, if the length is increased and the breadth is reduced each by 2 cm
then the area is reduced by 28 cm
2
. If the length is reduced by 1 cm and the
breadth increased by 2 cm , then the area increases by 33 cm
2
. Find the area of the
rectangle.
(vi) A train travelled a certain distance at a uniform speed. If the train had been
6 km/hr faster, it would have taken 4 hours less than the scheduled time. If the
train were slower by 6 km/hr, then it would have taken 6 hours more than the
scheduled time. Find the distance covered by the train.

Answer 1

i)
Let one number be “x” and “y “ be another number.Here x>y
A.T.Q
x = 3 y + 2 ………………. (1)
4y = x + 5 …………………. (2)

Substitute the value of x equation 2 ,
4y = x + 5
4 y = 3y + 2 + 5
4 y - 3 y = 7
y = 7

Put y = 7 in the equation 1
x = 3 y + 2
x = 3(7) + 2
x = 21 + 2
x = 23

Hence, one number is 23 and other number is 7.

ii)
Let 9x and 7 x be the income of two persons & 4y and 3y are the expenditure of two persons.

Each person save ₹ 2000 (given)

Savings = Income – expenditure
9 x - 4 y = 2000 ……….(1)
7 x - 3 y = 2000 ……….(2)

Multiply the equation 1 by 3,
27 x - 12 y = 6000 ………..(3)

Multiply the equation 2 by 4,
28 x - 12 y = 8000 …………. (4)

On Subtracting eq 4 from 3
27 x - 12 y = 6000
28 x - 12 y = 8000
(-) (+) (-)
---------------------------
- x = -2000
x = 2000

Put x = 2000 in the equation 1 ,
9 x - 4 y = 2000
9 (2000) - 4 y = 2000
18000 – 4 y = 2000
-4y = 2000-18000
-4y= - 16000
y = 4000

Monthly income of first person = 9 x = 9 (2000) = ₹ 18000
Monthly income of second person = 7 x = 7 (2000) = ₹ 14000

Hence, the Monthly income of first person = ₹18000 & second person = ₹14000.

iii)
Let "x y" be the required two digit number & x is in the ten's digit place and y is in unit place.
A.T.Q
x y = 7 (x + y)

Write the two digit number (xy) in expanded form
(10 x + 1 y) = 7 x + 7 y
10 x - 7 x + 1 y - 7 y = 0
3 x - 6 y = 0 ………………… (1)

The number formed by reversing the digits is 18 less than the given number.
y x = x y - 18
Write the two digit number (yx) in expanded form
(10 y + 1 x) = 10 x + 1 y - 18
10 x - 1 x + 1 y - 10 y = 18
9 x - 9 y = 18
9(x - y) = 18
x - y = 18/9= 2
x - y = 2 ……………... (2)

Multiply equation 2 by 6,
x - y = 2
6 x – 6 y = 12………..,.....(3)

On Subtracting eq 3 from 1
3 x - 6 y = 0
6 x - 6 y = 12
(-) (+) (-)
------------------
- 3 x = -12
x = (-12)/(-3)
x = 4

Put x = 4 in equation 1 ,
3 x - 6 y = 0
3(4) - 6 y = 0
12 - 6 y = 0
-6 y = - 12
y = (-12)/(-6)
y = 2

Required two digit number = xy = 42
Hence, the two digit number is 42.

iv)
Let x be the cost of 1 chair & y be the cost of 1 table.
A.T.Q
3 x + 2 y = 700 ,................(1)
5 x + 3 y = 1100 ,............... (2)

Multiply equation 1 by 3 & eq 2 by 2,
9 x + 6 y = 2100 …………….(3)
10 x + 6 y = 2200 ……………(4)

On Subtracting eq 4 from 3,
9 x + 6 y = 2100
10 x + 6 y = 2200
(-) (-) (-)
---------------------------
-x = -100
x = 100
Cost of 1 chair = ₹100

Put x = 100 in equation 1,
3 x + 2 y = 700
3 (100) + 2 y = 700
300 + 2 y = 700
2 y = 700 - 300
2 y = 400
y = 400/2
y = 200
Cost of 1 table = ₹200

Hence, the cost of 1 chair = ₹ 100
Cost of one table = ₹ 200

v)
Let x be the length of the rectangle & y be the breadth of the rectangle.

Area of the rectangle = x y
GIVEN :
Length is increased & breadth is decreased by 2, so that the area is reduced by 28 cm²
(x + 2) (y -2) = x y - 28
x y - 2 x + 2 y - 4 = x y - 28
2 x - 2 y = 28 - 4
2 x - 2 y = 24
2(x -y ) = 24
x - y = 12 …………...(1)

GIVEN :
Length is reduced by 1 cm and the breadth is increased by 2 cm,then the area increases by 33cm²
(x-1) (y+2) = x y + 33
x y + 2 x - y - 2 = x y + 33
2 x - y + x y - x y = 33 + 2
2 x - y = 35 ……………. (2)

On Subtracting eq 2 from 1,
x - y = 12
2 x - y = 35
(-) (+) (-)
--------------
-x = -23
x = 23

Put x = 23 in equation 1,
x - y = 12
23 - y = 12
- y = 12 - 23
- y= - 11
y = 11

Hence, the length of the rectangle = 23 cm
Breadth of the rectangle = 11 cm

vi)
Let x km/hr be the speed of train & y be the time taken by the train

Distance = speed x time
Distance = x y

Given : If the train had been 6 km/hr faster,it would have taken 4 hours less than the scheduled time.

A.T.Q
(x + 6) (y - 4) = x y
x y - 4 x + 6 y - 24 = x y
x y - x y - 4 x + 6 y = 24
- 4 x + 6 y = 24
-2(2x -3y)= 24
2 x - 3 y = -12 ……………. (1)

Given : If the train were slower by 6 km/hr, then it would have taken 6 hours more than the scheduled time.

A.T.Q
(x - 6) (y + 6) = x y
x y + 6 x - 6 y - 36 = x y
x y - x y + 6 x - 6 y = 36
6 x - 6 y = 36
6(x - y )= 36
x - y = 6 ………………. (2)
Multiplying eq 2 by 2
2 x - 2 y = 12…………….(3)
On Subtracting eq 3 from 1
2 x - 3 y = -12
2 x - 2 y = 12
(-) (+) (-)
---------------------
- y = -24
y = 24

Put y = 24 in equation 1,
2 x - 3 y = -12
2 x - 3 (24) = -12
2 x - 72 = -12
2 x = -12 + 72
2 x = 60
x = 60/2
x = 30

Speed of the train = 30 km/hr &
Time taken by the train = 24 hours

Distance covered by the train = x y = 30 x 24 = 720 km

Hence, the distance covered by the train is 720 km.

HOPE THIS WILL HELP YOU…