QUESTIONS ON SOLVING LINEAR EQUATIONS

Problem 11 :

One machine can seal 360 packages per hour, and an older machine can seal 140 packages per hour. How many minutes will the two machines working together take to seal a total of 700 packages?

(a) 48      (b) 72       (c) 84       (d) 90

Solution :

Per hour = 60 minute

Number of packages sealed by first machine = 360/60

= 6 package

Number of packages sealed by second machine = 140/60

= 7/3 package

x is the number of minutes

6x + 7x/3 = 700

(18x + 7x)/3 = 700

25x = 700(3)

25x = 2100

x = 2100/25

x = 84

Problem 12 :

The age of a person is 8 years more than thrice the age of the sum of his two grandsons who were twins. After 8 years his age will be 10 years more then twice the sum of the ages of his grandsons. Then the age of the person when twins born is 

(a) 86 years     (b) 73 years     (c) 68 years     (d) 63 years

Solution :

Let x be the age of his grandsons.

Let y be the age of the person.

y = 3(2x) + 8

y = 6x + 8

After 8 years, age of one grand son is x + 8 each and age of person is y + 8

6x + 8 + 8 = 2(2x + 16) + 10

6x + 16 = 4x + 32 + 10

6x + 16 = 4x + 42

6x - 4x = 42 - 16

2x = 26

x = 26/2

x = 13

y = 6(13) + 8

= 78 + 8

= 86

So, the age of the person is 86 years.

Problem 13 :

Solving

we get the roots as under.

(a) 1, 4       (b) 1, 2         (c) 1, 3       (d) 1, 5

Solution :

√(x/y) + √(y/x) - 5/2 = 0 -----(1)

x + y - 5 = 0 -----(2)

From (1),

√(x/y) + √(y/x) = 5/2

Squaring on both sides, we get

x/y + y/x + 2√(x/y)(y/x) = 25/4

x/y + y/x + 2 = 25/4

x/y + y/x = (25/4) - 2

(x² + y²)/xy = 17/4

4(x² + y²) = 17xy

y = 5 - x

4(x² + (5 - x)²) = 17x(5 - x)

4(x² + 25 - 10x + x²) = 85x - 17x²

4(2x² + 25 - 10x) = 85x - 17x²

8x² + 100 - 40x = 85x - 17x²

8x² + 17x² - 40x - 85x + 100 = 0

25x² - 125x + 100 = 0

x² - 5x + 4 = 0

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

x = 1 and x = 4

When x = 1, y = 5 - 1 ==> 4

When x = 4, y = 5 - 4 ==> 1

So, the roots are 1 and 4, option a is correct.

Problem 14 :

Solving

we get the roots

(a) 1/8, 1/5       (b) 1/2, 1/3         (c) 1/13, 1/5       (d) 1/4, 1/5

Solution :

1/x² + 1/y² = 13 ------(1)

1/x + 1/y = 5 ------(2)

1/y = 5 - (1/x)

1/x² + [5 - (1/x)]² = 13

1/x² + 25 + 1/x² - 10/x = 13

2/x² -10/x = 13 - 25

2 - 10x = -12x²

12x² - 10x + 2 = 0

6x² - 5x + 1 = 0

(3x - 1)(2x - 1) = 0

x = 1/3 and x = 1/2

So, the roots are 1/2 and 1/3, option b is correct.

Problem 15 :

4/x - 5/y = (x + y)/xy + 3/10 and 3xy = 10(y - x)

(a) (2, 5)       (b) (5, 2)         (c) (2, 7)       (d) (3, 4)

Solution :

4/x - 5/y = (x + y)/xy + 3/10

3xy = 10(y - x)

(4y - 5x)/xy = [10(x + y) + 3xy]/10xy

10(4y - 5x) = 10(x + y) + 3xy

40y - 50x = 10x + 10y + 3xy

40y - 10y - 50x - 10x = 3xy

30y - 60x = 3xy

30y - 60x = 10(y - x)

30y - 10y - 60x + 10x = 0

20y - 50x = 0

y = 5x/2

Applying the value of y in terms of x in 3xy = 10(y - x)

3x(5x/2) = 10(5x/2 - x)

15x²/2 = 10(3x/2)

15x²/2 = 15x

15x² - 30x = 0

15x(x - 2) = 0

x = 0 and x = 2

When x = 0, y = 5(0)/2 ==> 0

When x = 2, y = 5(2)/2 ==> 5

(2, 5) and (0, 0) are solutions. Option a is correct.

Problem 16 :

3x - 4y + 70z = 0, 2x + 3y - 10z = 0, x + 2y + 3 z = 13

(a) (1, 3, 7)       (b) (1, 7, 3)         (c) (-10, 10, 1)      (d) none

Solution :

3x - 4y + 70z = 0 -----(1)

2x + 3y - 10z = 0 -------(2)

x + 2y + 3 z = 13 --------(3)

(1) ⋅ 3 + (2) ⋅ 4 ==> 9x - 12y + 210z + 8x+12y - 40z = 0

17x + 170z = 0

x + 10z = 0 -----(4)

(1) ⋅ 2 - (3) ⋅ 3 ==> 4x + 6y - 20z - (3x + 6y + 9z) = 0 - 39

x - 20z - 9z = -39

x - 29z = -39 ------(5)

(4) - (5) 

x + 10z - x + 29z = 0 + 39

39z = 39

z = 1

Applying z = 1, we get

x + 10(1) = 0

x = -10

Applying the values of x and z in (3), we get

-10 + 2y + 3(1) = 13

-10 + 3 + 2y = 13

-7 + 2y = 13

2y = 13 + 7

2y = 20

y = 10

So, the solution is (-10, 10, 1). Option c is correct.

Problem 17 :

A number consist of three digits of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297, find the number

(a) 306       (b) 603         (c) 504       (d) 405

Solution :

Let x0y be the required three digit number.

x + y = 9 ------(1)

y0x = x0y + 297

100y + x = 100x + 1y + 297

x - 100x + 100y - 1y = 297

-99x + 99y = 297

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

y = 3 + x

applying the value of y in (1), we get

x + 3 + x = 9

2x = 9 - 3

2x = 6

x = 3

When x = 3, y = 3 + 3 ==> 6

So, the required number is 306. Option a is correct.

Problem 18 :

The demand and supply equations for a certain commodity are 4q + 7p = 17 and 

respectively where p is the market price and q is the quantity then the equilibrium price and quantity are. 

(a) 2, 3/4    (b) 3, 1/2         (c) 5, 3/5       (d) None

Solution :

4q + 7p = 17

p = q/3 + 7/4

Applying the value of p, we get

4q + 7(q/3 + 7/4) = 17

4q + 7q/3 + 49/4 = 17

19q/3 = 17 - 49/4

19q/3 = 19/4

q/3 = 1/4

q = 3/4

Applying q = 3/4, we get

p = (3/4)/3 + 7/4

p = 1/4 + 7/4

= 8/4

p = 2

The solution is (2, 3/4). Option a is correct.

Problem 19 :

y is older than x by 7 years 15 years back x's age was 3/4 of y's age. Their present ages are :

(a) x = 40 and y = 47         (b) x = 50 and y = 43

(c) x = 43 and y = 50       (d)  x = 36 and y = 43 

Solution :

y = x + 7

Age of y 15 years back is y - 15

Age of x 15 years back is x - 15

x - 15 = (3/4)(y - 15)

Applying the value of y, we get

x - 15 = 3/4(x + 7 - 15)

4(x - 15) = 3(x - 8)

4x - 60 = 3x - 24

4x - 3x = -24 + 60

x = 36

When x = 36, y = 36 + 7 ==> 43

So, option d is correct

Problem 20 :

If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A' present age.

(a) 6          (b) 9         (c) 12       (d) 15

Solution :

Let A be the present age of A.

2A - 3(A - 6) = A

2A - 3A + 18 = A

-A + 18 = A

18 = A + A

2A = 18

A = 9

So, the present age of A is 9. Option b is correct.