Linear Equation in One Variable Solved questions Amend Education Academy Poonam Dua

AMEND EDUCATION ACADEMY

Worked-out problems on linear equations in one variable:

1. The sum of three consecutive multiples of 4 is 444. Find these multiples.

Solution:

If x is a multiple of 4, the next multiple is x + 4, next to this is x + 8.

Their sum = 444

According to the question,

x + (x + 4) + (x + 8) = 444

⇒ x + x + 4 + x + 8 = 444

⇒ x + x + x + 4 + 8 = 444

⇒ 3x + 12 = 444

⇒ 3x = 444 - 12

⇒ x = 432/3

⇒ x = 144

Therefore, x + 4 = 144 + 4 = 148

Therefore, x + 8 - 144 + 8 – 152

Therefore, the three consecutive multiples of 4 are 144, 148, 152.
2. The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, the new number becomes 3/2. Find the original number.

Solution:

Let the numerator of a rational number = x

Then the denominator of a rational number = x + 3

When numerator is increased by 7, then new numerator = x + 7

When denominator is decreased by 1, then new denominator = x + 3 - 1

The new number formed = 3/2

According to the question,

(x + 7)/(x + 3 - 1) = 3/2

⇒ (x + 7)/(x + 2) = 3/2

⇒ 2(x + 7) = 3(x + 2)

⇒ 2x + 14 = 3x + 6

⇒ 3x - 2x = 14 - 6

⇒ x = 8

The original number i.e., x/(x + 3) = 8/(8 + 3) = 8/11


3. The sum of the digits of a two digit number is 7. If the number formed by reversing the digits is less than the original number by 27, find the original number.

Solution:

Let the units digit of the original number be x.

Then the tens digit of the original number be 7 - x

Then the number formed = 10(7 - x) + x × 1

                  = 70 - 10x + x = 70 - 9x

On reversing the digits, the number formed

                  = 10 × x + (7 - x) × 1

                  = 10x + 7 - x = 9x + 7

According to the question,

New number = original number - 27

9x + 7 = 70 - 9x - 27

9x + 7 = 43 - 9x

9x + 9x = 43 – 7

18x = 36

x = 36/18

x = 2

Therefore, 7 - x

        = 7 - 2

        = 5

The original number is 52



4. A motorboat goes downstream in river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water.

Solution:

Let the speed of the boat in still water = x km/hr.

Speed of the boat downstream = (x + 3) km/hr.

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) × 5   (D = Speed × Time)

Speed of the boat upstream = (x - 3) km/hr

Time taken to cover the distance = 6 hrs.

Therefore, distance covered in 6 hrs = 6(x - 3)

Therefore, the distance between two coastal towns is fixed, i.e., same.

According to the question,

5(x + 3) = 6(x - 3)

⇒ 5x + 15 = 6x - 18

⇒ 5x - 6x = -18 – 15

⇒ -x = -33

⇒ x = 33

Required speed of the boat is 33 km/hr.


5. Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Solution:

Let one part be x.

Then other part = 28 - x

It is given 6/5 of one part = 2/3 of the other.

⇒ 6/5x = 2/3(28 - x)

⇒ 3x/5 = 1/3(28 - x)

⇒ 9x = 5(28 - x)

⇒ 9x = 140 - 5x

⇒ 9x + 5x = 140

⇒ 14x = 140

⇒ x = 140/14

⇒ x = 10

Then the two parts are 10 and 28 - 10 = 18.


6. A total of $10000 is distributed among 150 persons as gift. A gift is either of $50 or $100. Find the number of gifts of each type.

Solution:

Total number of gifts = 150

Let the number of $50 is x

Then the number of gifts of $100 is (150 - x)

Amount spent on x gifts of $50 = $ 50x

Amount spent on (150 - x) gifts of $100 = $100(150 - x)

Total amount spent for prizes = $10000

According to the question,

50x + 100 (150 - x) = 10000

⇒ 50x + 15000 - 100x = 10000

⇒ -50x = 10000 - 15000

⇒ -50x = -5000

⇒ x = 5000/50

⇒ x = 100

⇒ 150 - x = 150 - 100 = 50

Therefore, gifts of $50 are 100 and gifts of $100 are 50.


The above step-by-step examples demonstrate the solved problems on linear equations in one variable.