WORD PROBLEMS ON SOLVING EQUATIONS

Problem 1 :

One hundred and fifty students are admitted to a school. They are distributed over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.

Solution :

Let x, y and z be the number of students in section A, B and C respectively.

x + y + z  =  150 -----(1)

If 6 students are shifted from section A to section C, the sections will have equal number of students.

Then, 

x - 6  =  z + 6

x - z  =  12 -----(2)

4 times of students of section C exceeds the number of students of section A by the number of students in section B.

4z  =  x + y

x + y - 4z  =  0 -----(3)

(1) - (2) : 

5z  =  150

z  =  30

Substitute 30 for z in (2). 

(2)-----> x - 30  =  12

x  =  42

Substitute 42 for x and 30 for z in (1). 

(1)-----> 42 + y + 30  =  150

y + 72  =  150

y  =  78

So, number of students in sections A, B and C are 42, 30 and 78 respectively. 

Problem 2 :

In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.

Solution :

Let 'xyz' be the three digit number. 

When the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number.

yxz  =  3xyz + 54

100y + 10x + 1z  =  3(100x + 10y + z) + 54

100y + 10x + z  =  300x + 30y + 3z + 54

-290x + 70y - 2z  =  54

-145x + 35y - z  =  27

145x - 35y + z  =  -27-----(1)

If 198 is added to the number, the digits are reversed. 

xyz + 198  =  zyx

100x + 10y + z + 198  =  100z + 10y + x

99x - 99z  =  -198

x - z  =  -2 ------(2)

The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. 

y - x  =  2(y - z)

y - x  =  2y - 2z

-x - y + 2z  =  0

x + y - 2z  =  0 -----(3)

(1) + 35(3) :

180x - 69z  =  -27 -----(4)

69(2) - (4) :

-111x  =  -111

x  =  1

Substitute 1 for x in (2). 

(2)-----> 1 - z  =  -2

-z  =  -3

z  =  3

Substitute 1 for x and 3 for z in (3). 

(3)-----> 1 + y - 2(3)  =  0

1 + y - 6  =  0

y - 5  =  0

y  =  5

So, the required three digit number is 153.

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