Q. I have some five rupee coins and some two rupee coins . The number of

two rupee coins is twice the number of five rupee coins. The total money I

have is 108 rupees. How many five rupee coins do I have?

Answer

Let total no. of five rupee coins be x

No of two rupee coin be 2x

5x + 2*2x =108

i. e, 9x = 108

x = 108÷9

x = 12

Therefore, five rupees coin = 12

And two rupee coin = 2×12=24

= 70

QUESTION

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.

ANSWER

Let x be A's present age.

A's age 6 years ago = x - 6.

Thrice of A's age 6 years ago = 3(x - 6).

Twice his present age = 2x.

Given : Thrice of A's age 6 years ago be subtracted from twice his present age, the

result would be equal to his present age.

2x - 3(x - 6) = x

2x - 3x + 18 = x

- x + 18 = x

18 = 2x

Divide both sides by 2.

9 = x

Hence, A's present age is 9 years.

QUESTION

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.

ANSWER

Let x be the digit in units place.

Then, the digit in the tens place = 2x.

So, the number is (2x)x.

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

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

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

Simplify.

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

21x - 18 = 12x

9x = 18

Divide both sides by 9.

x = 2

The digit at the units place is 2.

Then, the digit at the tens place is

= 2 ⋅ 2

= 4

Hence the required number is 42.

QUESTION

For a certain commodity, the demand equation giving demand "d" in kg, for a

price "p" in dollars per kg. is d = 100(10 - p). The supply equation giving the

supply "s" in kg. for a price "p" in dollars per kg is s = 75(p - 3). Find the

equilibrium price.

ANSWER

The equilibrium price is the market price where the quantity of goods

demanded is equal to the quantity of goods supplied.

d = s

100(10 - p) = 75(p - 3)

1000 - 100p = 75p - 225

1225 = 175p

Divide both sides by 175.

7 = p

Hence, the equilibrium price is $7.

QUESTION

The fourth part of a number exceeds the sixth part by 4. Find the number.

ANSWER

Let x be the required number.

Fourth part of the number = x/4

Sixth part of the number = x/6

Given : The fourth part of a number exceeds the sixth part by 4.

x/4 - x/6 = 4

L.C.M of (4, 6) is 12.

(3x/12) - (2x/12) = 4

(3x - 2x)/12 = 4

x/12 = 4

Multiply both sides by 12.

x = 48

Hence, the required number is 48.

QUESTION

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is

80 cm. Find its area.

ANSWER

Let x be the length of the rectangle.

Then, width of the rectangle is 2x/3.

Given : Perimeter is 80 cm.

Perimeter = 80 cm

2(l + w) = 80

Divide both sides by 2.

l + w = 40

Substitute l = x and w = 2x/3.

x + 2x/3 = 40

(3x + 2x)/3 = 40

5x/3 = 40

Multiply both sides by 3/5.

x = 24

The length is 24 cm.

Then, the width is

= 2x/3

= (2 ⋅ 24)/3

= 16 cm

Formula to find the area of a rectangle is

= l ⋅ w

Substitute l = 24 and w = 16.

= 24 ⋅ 16

= 384 cm²

QUESTION

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.

ANSWER

Let x° be the first angle.

Then, we have

the second angle = x° + 5°

third angle = 3x°

We know that the sum of three angle in any triangle is 180°.

x° + (x° + 5°) + (3x°) = 180°

x + x + 5 + 3x = 180

5x + 5 = 180

Subtract 5 from both sides.

5x = 175

Divide both sides by 5.

x = 35

The first angle is 35°.

The second angle is

= 35° + 5°

= 40°

The third angle is

= 3 ⋅ 35°

= 105°

Hence, the three angles of the triangle are 35°, 40° and 105°.

HOW TO SOLVE SIMPLE EQUATIONS?

SUMMARY

An equation is a statement in which two algebraic expressions are connected by the sign of equality (=). Each of the expressions on either side of the sign of equality is called a side or member of the equation.

For example, if the expressions $4x-3$ and $2x+1$ are equal in value i.e., $4x-3= 2x+1$ , then this algebraic statement is called an equation where, $4x-3$ and $2x+1$ are the members of the equation. To solve the equation $4x-3 = 2x+1$ means to find the value of the letter $x$ . This letter $x$ is called the variable or the unknown quantity or the root of the equation. Variables are usually represented by alphabets, for example, $a, b, x, y, z$ . The equation in which the variable is of the first order, i.e., an equation in which the highest power of the involved variables is 1 is called a simple or a linear equation.

Solving Equation:

Solving a linear equation is governed by the following rules:

If the same quantity, i.e., equal quantities be added to the two expressions on either side of the equality sign, then the sums are equal.
If the same quantity, i.e., equal quantities be taken away from the two expressions on either side of the equality sign, then the differences are equal.
If the two expressions on either side of the equality sign be multiplied by the same quantity, then the products are equal.
If the two expressions on either side of the equality sign be divided by the same quantity, then the quotients are equal.

From rules 1 and 2 we can deduce an important principle, i.e., any term may be transposed from one side of the equality sign to the other by simply changing its sign.

For example, let $x-a=b+c$

Adding $a$ to both sides, we get,

$x-a+a=b+c+a$ [Rule 1]

$x=b+c+a$

Again, subtracting $c$ from both sides, we get,

$x-a-c=b+c-c=b$ [Rule 2]

Thus, we see that $(-a)$ removed from the left side appears as $(+a)$ on the right side. Again, $(-c)$ removed from the right side appears as $(+c)$ on the left side.

Hence, if $x-a=b-c+d$ , we get, $x-a-b+c-d=0$

This is called Transposition.

The sign of every term of an equation may be changed without destroying the equality.

For example, let $x-a=b+c$

$\Rightarrow (x-a) \times (-1)=(b+c) \times (-1)$ [Rule 3]

$-x+a=-b-c$

Steps to solve a simple equation:

Simplify all brackets, fractions, etc if required.
Bring all the terms containing the variables on one side and all the constant terms on the other side.
Solve the equation, obtained in Step 2 to get the value of its variable.