Exercise 1.3

Question 1: Find the Three Consecutive Integers

Given

The sum of three consecutive integers is 42.

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Formula

If three consecutive integers are:

$$x,\;x+1,\;x+2$$

then

$$x+(x+1)+(x+2)=42$$

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Solution

Combine like terms.

$$3x+3=42$$

Subtract 3 from both sides.

$$3x=39$$

Divide both sides by 3.

$$x=13$$

The three consecutive integers are:

$$13,\;14,\;15$$

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Verification

$$13+14+15=42$$

The condition is satisfied.

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Final Answer

$$\boxed{13,\;14,\;15}$$

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Question 2: Find the Length of (AB)

Given

A right-angled triangle (ABC) has:

• Right angle at A
• Base

$$AC=\sqrt3+\sqrt5\text{ cm}$$

• Area

$$1+\sqrt{15}\text{ cm}^2$$

Find the length of

$$AB$$

in the form

$$(a\sqrt3+b\sqrt5)\text{ cm}$$

where (a) and (b) are integers.

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Diagram

          B
          │\
      AB  │ \
          │  \
          │   \
          │    \
          │     \
          A------C
          ⟂
     AC = √3 + √5

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Formula

Area of a right triangle

$$\text{Area}=\frac12\times\text{Base}\times\text{Height}$$

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Solution

Substitute the given values.

$$1+\sqrt{15} = \frac12(\sqrt3+\sqrt5)(AB)$$

Multiply both sides by 2.

$$2+2\sqrt{15} = (\sqrt3+\sqrt5)AB$$

Divide both sides.

$$AB = \frac{2+2\sqrt{15}} {\sqrt3+\sqrt5}$$

Factor out 2.

$$AB = \frac{2(1+\sqrt{15})} {\sqrt3+\sqrt5}$$

Multiply numerator and denominator by the conjugate.

$$\frac{\sqrt5-\sqrt3} {\sqrt5-\sqrt3}$$

$$AB= \frac{2(1+\sqrt{15})(\sqrt5-\sqrt3)} {5-3}$$

Cancel 2.

$$AB= (1+\sqrt{15})(\sqrt5-\sqrt3)$$

Expand.

$$=\sqrt5-\sqrt3+\sqrt{75}-\sqrt{45}$$

Since

$$\sqrt{75}=5\sqrt3$$

and

$$\sqrt{45}=3\sqrt5$$

we get

$$=\sqrt5-\sqrt3+5\sqrt3-3\sqrt5$$

$$=4\sqrt3-2\sqrt5$$

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Final Answer

$$\boxed{AB=4\sqrt3-2\sqrt5\text{ cm}}$$

where

$$a=4,\qquad b=-2$$

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Question 3: Area of the Rectangle

Given

Length

$$2+\sqrt{18}\text{ m}$$

Width

$$5-\frac4{\sqrt2}\text{ m}$$

Express the area in the form

$$a+b\sqrt2$$

where (a) and (b) are integers.

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Solution

Simplify each expression.

$$\sqrt{18}=3\sqrt2$$

$$\frac4{\sqrt2} = 2\sqrt2$$

Therefore,

$$\text{Length}=2+3\sqrt2$$

$$\text{Width}=5-2\sqrt2$$

Multiply.

$$(2+3\sqrt2)(5-2\sqrt2)$$

Expand.

$$=10-4\sqrt2+15\sqrt2-12$$

Combine like terms.

$$=-2+11\sqrt2$$

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Final Answer

$$\boxed{-2+11\sqrt2\text{ m}^2}$$

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Question 4: Find Two Numbers

Given

The sum of two numbers is

$$68$$

and their difference is

$$22.$$

Find the two numbers.

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Solution

Let the two numbers be

$$x \text{ and } y$$

According to the given information,

$$x+y=68$$

$$x-y=22$$

Add the two equations.

$$2x=90$$

$$x=45$$

Substitute (x=45) into

$$x+y=68$$

$$45+y=68$$

$$y=23$$

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Verification

$$45+23=68$$

$$45-23=22$$

Both conditions are satisfied.

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Final Answer

$$\boxed{45 \text{ and } 23}$$

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Question 5: Convert Temperature to Fahrenheit

Given

The temperature in Lahore is

$$48^\circ C$$

Find the temperature in degrees Fahrenheit using

$$^\circ F=\frac{9}{5}\times{}^\circ C+32$$

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Formula

$$^\circ F=\frac95\times{}^\circ C+32$$

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Solution

Substitute

$$^\circ C=48$$

$$^\circ F=\frac95\times48+32$$

$$=\frac{432}{5}+32$$

$$=86.4+32$$

$$=118.4^\circ F$$

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Final Answer

$$\boxed{118.4^\circ F}$$

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Question 6: Father’s and Son’s Ages

Given

The sum of the ages of a father and his son is

$$72 \text{ years}$$

Six years ago, the father’s age was twice the son’s age.

Find the son’s age six years ago.

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Solution

Let the son’s present age be

$$x$$

Then the father’s present age is

$$72-x$$

Six years ago,

Son’s age:

$$x-6$$

Father’s age:

$$66-x$$

According to the question,

$$66-x=2(x-6)$$

$$66-x=2x-12$$

$$78=3x$$

$$x=26$$

Son’s age six years ago:

$$26-6=20$$

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Verification

Father’s age six years ago:

$$40$$

Since

$$40=2\times20$$

the answer is correct.

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Final Answer

$$\boxed{20\text{ years}}$$

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Question 7: Profit Percentage

Given

Cost Price

$$Rs.\,1500$$

Selling Price

$$Rs.\,1520$$

Find the profit percentage.

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Formula

$$\text{Profit}=SP-CP$$

$$\text{Profit \%}=\frac{\text{Profit}}{\text{Cost Price}}\times100$$

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Solution

Profit

$$=1520-1500$$

$$=Rs.\,20$$

Profit percentage

$$=\frac{20}{1500}\times100$$

$$=1.33\%$$

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Final Answer

$$\boxed{1.33\%}$$

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Question 8: Income Tax

Given

Annual income

$$Rs.\,960,000$$

Exempted amount

$$Rs.\,130,000$$

Tax rate

$$0.75\%$$

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Formula

$$\text{Taxable Income} = \text{Annual Income} - \text{Exempted Amount}$$

$$\text{Tax} = \frac{\text{Rate}}{100} \times \text{Taxable Income}$$

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Solution

Taxable income

$$=960,000-130,000$$

$$=Rs.\,830,000$$

Tax

$$=\frac{0.75}{100}\times830,000$$

$$=0.0075\times830,000$$

$$=Rs.\,6,225$$

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Final Answer

$$\boxed{Rs.\,6,225}$$

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Question 9: Compound Markup

Given

Principal Amount

$$Rs.\,375,000$$

Rate

$$14\%$$

Time

$$1\text{ year}$$

Compounded annually.

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Formula

$$A=P\left(1+\frac{r}{100}\right)^n$$

$$\text{Compound Markup}=A-P$$

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Solution

$$A=375,000\left(1+\frac{14}{100}\right)^1$$

$$=375,000\times1.14$$

$$=Rs.\,427,500$$

Compound markup

$$=427,500-375,000$$

$$=Rs.\,52,500$$

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Final Answer

$$\boxed{Rs.\,52,500}$$