PRACTICE PROBLEMS ON LINEAR EQUATIONS

Problem 1 :

If 6 = 2x + 4y, what is the value of x + 2y is 

(a)  2  (b)  3  (c) 6  (d)  8

Solution :

6 = 2x + 4y

Dividing by 2 on both sides, we get

3 = x + 2y

So, option b is correct.

Problem 2 :

Solve for y in the equation 

and the value of y is.

(a)  -1  (b)  7   (c)  1   (d) -1/7

Solution :

(y + 11)/6 - (y + 1)/9 = (y + 7)/4

By making the denominators same, the least common multiple of 6 and 9 is 18.

[3(y + 11) - 2(y + 1)]/18 = (y + 7)/4

(3y + 33 - 2y - 2)/18 = (y + 7)/4

(y + 31)/18 = (y + 7)/4

4(y + 31) = 18(y + 7)

4y + 124 = 18y + 126

4y - 18y = 126 - 124

-14y = 2

y = -2/14

y = -1/7

So, option d is correct.

Problem 3 :

Pick up the correct value x for which

(a)  x = 0   (b) x = 1   (c) x = 10   (d) None of these

Solution :

x/0.5 - 1/0.05 + x/0.005 - 1/0.0005 = 0

x/0.5 + x/0.005 = 1/0.05 + 1/0.0005

To remove the decimal which is at the denominator, based on the number of digits we have at the denominator we have to multiply both numerator and denominator by 10, 100, 1000., ..... etc

10x/5 + 1000x/5 = 100/5 + 10000/5

Now the denominators are becoming the same.

10x + 1000x = 100 + 10000

1010x = 10100

x = 10100/1010

x = 10

So, option c is correct.

Problem 4 :

The denominator of the fraction exceed the numerator by 5 and if 3 be added to both the fraction becomes 3/4. Find the fraction.

(a)  12/17   (b)  15/17   (c)  12/25   (d)  13/29

Solution :

Let x/x + 5 be the required fraction.

(x + 3)/[(x + 5) + 3] = 3/4

(x + 3)/(x + 8) = 3/4

4(x + 3) = 3(x + 8)

4x + 12 = 3x + 24

4x - 3x = 24 - 12

x = 12

x + 5 = 17

So, the required fraction is 12/17.

Problem 5 :

Three persons Mr. Roy, Mr Paul and Mr. Singh together have $51. Mr Paul has $4 less than Mr Roy and Mr. Singh has got $5 less than Mr.Roy. They have the money as

(a)  (20, 16, 15)   (b)  (15, 20, 16)  (c)  (25, 11, 15)

Solution :

Let R, P and S be the amount that Mr. Roy, Mr Paul and Mr. Singh has respectively.

R + P + S = 51

P = R - 4

S = R - 5

Applying the value of P and S, we get

R + R - 4 + R - 5 = 51

3R - 9 = 51

3R = 51 + 9

3R = 60

R = 60/3

R = 20

P = 20 - 4 ==> 16

S = 20 - 5 ==> 15

The amount Mr.Roy has is 20, Mr.Paul has 16 and Mr. Singh has 15.

So, option a is correct.

Problem 6 :

A number consists of two digits. The digits in the ten's place is 3 times the digit in the unit's place. If 54 is subtracted from the number the digits are reversed. The number is 

(a)  39   (b)  92  (c) 93  (d)  94

Solution :

Let xy be the two digit number.

x = 3y 

Expanded form of xy = x + 10y

xy - 54 = yx

10x + 1y - 54 = 10y + x

10x - x + 1y - 10y = 54

9x - 9y = 54

x - y = 6

Applying the value of x, we get

3y - y = 6

2y = 6

y = 3

x = 3(3) ==> 9

So, the required number is 93, option c is correct.

Problem 7 :

The number consists of two digits, the digit in the ten's place is twice the digit in the unit's place. If 18 be subtracted from the number the digits are reversed. Find the number

(a)  42   (b)  24  (c) 33  (d)  61

Solution :

Let xy be the two digit number.

x = 2y --------(1)

xy - 18 = yx

10x + y - 18 = 10y + x

10x - x + y - 10y = 18

9x - 9y = 18

x - y = 2 --------(2)

Applying the value of x, we get

2y - y = 2

y = 2

x = 2(2)

x = 4

So, the required number is 42, option a is correct.

Problem 8 :

Solving 4^x⋅2^y = 128 and 3^3x+2y = 9^xy, we get the following roots.

(a)  7/4, 7/2  (b)  2, 3  (c)  1, 2) (d) 1, 3

Solution :

4^x⋅2^y = 128 

(2²)^x⋅2^y = 128 

2^2x+y = 128

2^2x+y = 2⁷

2x + y = 7

y = 7 - 2x --------(1)

3^3x+2y = (3²)^xy

3^3x+2y = 3^2xy

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

Applying the value of y in (2), we get

3x + 2(7 - 2x) = 2x(7 - 2x)

3x + 14 - 4x = 14x - 4x²

4x² - 4x - 14x + 3x + 14 = 0

4x² - 15x + 14 = 0

4x² - 8x - 7x + 14 = 0

4x(x - 2) - 7(x - 2) = 0

(4x - 7)(x - 2) = 0

x = 7/4 and x = 2

The roots are 7/4, 7/2, 2 and 3. Option and b are correct.

Problem 9 :

Solving 9^x = 3^y and 5^x+y+1 = 25^xy, we get the following roots.

(a)  1, 2  (b) 0, 1  (c)  0, 3  (d)  1, 3

Solution :

9^x = 3^y and 5^x+y+1 = 25^xy

(3²)^x = 3^y

3^2x = 3^y

2x = y

y = 2x --------(1)

5^x+y+1 = 5^2xy

x + y + 1 = 2xy

Applying the value of y, we get

x + 2x + 1 = 2x(2x)

3x + 1 = 4x²

4x² - 3x - 1 = 0

(4x + 1)(x - 1) = 0

x = -1/4 and x = 1

When x = -1/4, y = 2(-1/4) ==> -1/2

When x = 1, y = 2(1) ==> 2

So, option a is correct.

Problem 10 :

Solving 9x + 3y - 4z =  3, x + y - z = 0 and 2x - 5y - 4z = -20 the following roots obtained.

(a) 2, 3, 4      (b) 1, 3, 4     (c) 1, 2, 3     (d) None

Solution :

9x + 3y - 4z =  3 -------(1)

x + y - z = 0 ---------(2)

2x - 5y - 4z = -20 -------------(3)

(1) - 4(2) ==> 9x + 3y - 4z - 4x - 4y + 4z = 3 - 0

5x - y = 3 ----(4)

(1) - (3)

9x + 3y - 4z - 2x + 5y + 4z = 3 + 20

7x + 8y = 23 ----(5)

8 (4) + (5) ==> 40x - 8y + 7x + 8y = 24 + 23

47x = 47

x = 1

Applying x = 1 in (4), we get

5(1) - y = 3

5 - y = 3

y = 5 - 3

y = 2

Applying x = 1 and y = 2 in (2), we get

1 + 2 - z = 0

z = 3

(x, y, z) ==> (1, 2, 3)

So, option c is correct.