# ALGEBRAIC MANIPULATION PROBLEMS

Problem 1 :

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's present age.

Solution :

Let x be the present age of A.

Given : Thrice of A's age 6 years ago be subtracted from twice his present age is equal to his present age.

2x - 3(x - 6) = x

2x - 3x + 18 = x

-x + 18 = x

18 = x

9 = x

A's present age is 9 years.

Problem 2 :

A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 18 be subtracted from the number, the digits are reversed. Find the number.

Solution :

Let x be the digit in ones place.

Then the digit in tens place is 2x and the two digit number is

(2x)(x)

Given : If 18 be subtracted from the number, the digits are reversed.

(2x)(x) - 18 = (x)(2x) ----(1)

Idea to solve the problem further :

Consider the two digit number 35. We have 3 in tens place and 5 in ones place. So, we can write 35 as

35 = 3(10) + 5(1)

(1)----> 10(2x) + 1(x) - 18 = 10(x) + 1(2x)

20x + x - 18 = 10x + 2x

21x - 18 = 12x

9x = 18

x = 2

In the given number, the digit in the ones place is 2. Then the digit in tens place is 4.

Therefor the number is 42.

Justification :

42 - 18 = 24

24 = 24

Problem 3 :

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

Solution :

Let x be the length of the rectangle.

Then the width is 2x/3.

Perimeter = 80 cm

2(l + w) = 80

l + w = 40

x + 2x/3 = 40

3x/3 + 2x/3 = 40

(3x + 2x)/3 = 40

5x/3 = 40

5x = 120

x = 24

Length = 24 cm

Width = 2(24)/3 = 16 cm

Area of the rectangle :

= l ⋅ w

= 24 ⋅ 16

= 384 cm2

Problem 4 :

In a triangle, the second angle is 5° more than the first angle. And the third angle is three times of the first angle. Find the three angles of the triangle.

Solution :

From the given information, both the second angle third angle are compared to the first angle. So we can assume a variable for the first angle and solve the problem.

Let x be the first angle.

Then,

the second angle = x + 5°

the third angle = 3x

In any triangle, the three angles add up to 180°.

x + x + 5° + 3x = 180°

5x + 5° = 180°

5x = 175°

x = 35°

the first angle = 35°

the second angle = 35° + 5° = 40°

the third angle = 3(35°) = 105°

The three angles of the triangle are

35°, 40° and 105°

Problem 5 :

If a number of which the half is greater than 1/5 th of the number by 15, find the number.

Solution :

Let x be the number.

Given : Half of the number is greater than 1/5 th of the number by 15.

x/2 = x/5 + 15

x/2 - x/5 = 15

5x/10 - 2x/10 = 15

(5x - 2x)/10 = 15

3x/10 = 15

3x = 150

x = 50

The  number is 50.

Justification :

Half of the number :

= 50/2

= 25 ----(1)

1/5 th of the number increased by 15 :

= (1/5)(50) + 15

= 10 + 15

= 25 ----(2)

(1) = (2)

Problem 6 :

If x > 0 and  x2 - 2x - 35 = 0, then find the value of :

2x2 + 7x + 5

Solution :

Solve the quadratic equation x2 - 2x - 35 = 0.

x2 - 2x - 35 = 0

x2 - 7x + 5x - 35 = 0

x(x - 7) + 5(x - 7) = 0

(x - 7)(x + 5) = 0

x - 7 = 0  or  x + 5 = 0

 x - 7 = 0x = 7 x + 5 = 0x = -5

Since x > 0, x = 7.

To find the value of 2x2 + 7x + 5, plugin x = 7 into it.

2(7)2 + 7(7) + 5

= 2(49) + 49 + 5

= 98 + 49 + 5

= 152

Problem 7 :

If t < 0 and (t - 1)2 = 16, what is the value of t?

Solution :

(t - 1)2 = 16

Take square root on both sides.

(t - 1)2 = 16

t - 1 = ±4

t - 1 = 4  or  t - 1 = -4

 t - 1 = 4t = 5 t - 1 = -4 t = -3

Since t < 0, t = -3.

t2 = (-3)2

t2 = 9

Problem 8 :

Let x be a real number which satisfies the relations

2x - 5 > 2

3x + 3 < 18

Which of the values can x take?

A) 4

B) 5

C) 6

D) 7

Solution :

 2x - 5 > 22x > 7x > 7/2x > 3.5 ----(1) 3x + 3 < 183x < 15x < 5 ----(2)

From (1) and (2),

3.5 < x < 5

In the answer choices given, the value 4 in (A) is in the above interval.

The correct answer choice is (A).

Problem 9 :

Calculate the fifth term of the sequence defined as

f(0) = 2

f(n + 1) = 2f(n) - 1 for n ≥ 0

Solution :

The given sequence is a recursive sequence in which each term is generated using one or more previous terms for which the values are already known.

Since ≥ 0, the sequence starts when n = 0.

So, the first term is f(0), the second term is f(1), the third term is f(2) and so on.

The first term is f(0) = 2.

Substitute n = 0 in f(n + 1) = 2f(n) - 1 and evaluate f(1).

f(0 + 1) = 2f(0) - 1

f(1) = 2(2) - 1

f(1) = 3

The second term is f(1) = 3.

Substitute n = 1 in f(n + 1) = 2f(n) - 1 and evaluate f(2).

f(1 + 1) = 2f(1) - 1

f(2) = 2(3) - 1

f(2) = 5

The third term is f(2) = 5.

Substitute n = 2 in f(n + 1) = 2f(n) - 1 and evaluate f(3).

f(2 + 1) = 2f(2) - 1

f(3) = 2(5) - 1

f(3) = 9

The fourth term is f(3) = 9.

Substitute n = 3 in f(n + 1) = 2f(n) - 1 and evaluate f(4).

f(3 + 1) = 2f(3) - 1

f(4) = 2(9) - 1

f(4) = 17

The fifth term is f(4) = 17.

Problem 10 :

Find the domain of the function :

f(x) = 1/(x + 3)

Solution :

Domain of a function f(x) is all real values of x for which f(x) is defined.

f(x) = 1/(x + 3) is a rational function, because there is variable x in denominator.

In f(x) = 1/(x + 3), if x = -3, the denominator becomes zero and f(x) is undefined.

Except x = -3, for all real values of x, f(x) is defined.

Therefore, the domain of f(x) is all real values except -3.

Domain of f(x) = R - {-3}

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