Question

Verify the property a x (bx c) = (a x b) x c by taking:
a =(7/5), b =(-9/4), c =(1/2)

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Answer 1

★ How to do :-

Here, we are given with some of the variables namely a, b and c. Those all variables are arranged in some ways. We are asked to verify that property. To verify means, we should solve those both separately and then, we should see that whether those both results are equal or not. This arrangement of fractions are classified as the associative property. In this property, we group the last two fractions in bracket whereas on the other side, we group the first two fractions with the brackets. We will solve the LHS and the RHS separately and then, we can compare the results of both sides and then verify the statement. In this question, always first we should solve the brackets. Then, we should solve the fraction. that are outside the brackets. So, let's solve!!

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➤ Solution :-

${\sf \leadsto a \times (b \times c) = (a \times b) \times c}$

Substitute the values of a, b and c.

${\sf \leadsto \dfrac{7}{5} \times \bigg( \dfrac{(-9)}{4} \times \dfrac{1}{2} \bigg) = \bigg( \dfrac{7}{5} \times \dfrac{(-9)}{4} \bigg) \times \dfrac{1}{2}}$

Let's solve the LHS and RHS separately.

LHS :-

${\sf \leadsto \dfrac{7}{5} \times \bigg( \dfrac{(-9)}{4} \times \dfrac{1}{2} \bigg)}$

Solve the fractions in bracket.

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{7}{5} \times \bigg( \dfrac{(-9) \times 1}{4 \times 2} \bigg)}$

Multiply the numbers.

${\sf \leadsto \dfrac{7}{5} \times \dfrac{(-9)}{8}}$

Write both numerators with a common denominator.

${\sf \leadsto \dfrac{7 \times (-9)}{5 \times 8}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-63)}{40} = -1 \dfrac{23}{40}}$

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RHS :-

${\sf \leadsto \bigg( \dfrac{7}{5} \times \dfrac{(-9)}{4} \bigg) \times \dfrac{1}{2}}$

Solve the fractions in bracket.

Write both numerators and denominators with a common fraction.

${\sf \leadsto \bigg( \dfrac{7 \times (-9)}{5 \times 4} \bigg) \times \dfrac{1}{2}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-63)}{20} \times \dfrac{1}{2}}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{(-63) \times 1}{20 \times 2}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-63)}{40} = -1 \dfrac{23}{40}}$

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Let's compare the results of both LHS and RHS.

Comparison :-

${\sf \leadsto \dfrac{(-63)}{40} \: and \: \dfrac{(-63)}{40}}$

As we can see that both are equal. So,

${\sf \leadsto \dfrac{(-63)}{40} = \dfrac{(-63)}{40}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!

Answer 2

$</p><p>➤ Given :-</p><p>Selling price of each machine :- ₹15000</p><p>Gain percent of first machine :- 25%</p><p>Loss percent of second machine :- 25%</p><p>[tex]\:$

➤ To Find :-

The profit or loss percentage of whole transaction...

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★ How to do :-

Here, we are given with that a mechanic sells two machines at a same price. But, he occurs some gain and loss percentage in his machines while selling those. Although the selling price of both machines are same, there are different gain and loss percent on both of the machines. As this is the given information, the cost price of both the machines are different. We are asked to find the total gain or loss percent that will obtain to the mechanic while he sells those both machines. So, first we should find the cost prices of both the machines separately. Then, we can fi

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➤ Solution :-

Cost price of first machine :-

${\sf \longrightarrow \underline{\boxed{\sf \dfrac{100}{100 + Gain \%} \times SP}}}$

Substitute the given values.

${\sf \leadsto \dfrac{100}{100 + 25} \times 15000}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{100}{125} \times 15000 = \dfrac{4}{5} \times 15000}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{4 \times 15000}{5} = \dfrac{60000}{5}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{60000}{5} = 12000}$

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Cost price of second machine :-

${\sf \longrightarrow \underline{\boxed{\sf \dfrac{100}{100 - Loss\%} \times SP}}}$

Substitute the given values.

${\sf \leadsto \dfrac{100}{100 - 25} \times 15000}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{100}{75} \times 15000 = \dfrac{4}{3} \times 15000}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{4 \times 15000}{3} = \dfrac{60000}{3}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{60000}{3} = 20000}$

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Now, let's solve the total cost price and the total selling price.

Total cost price :-

${\sf \leadsto 12000 + 20000}$

Add the values.

${\sf \leadsto 32000}$

Total selling price :-

${\sf \leadsto 15000 + 15000}$

Add the values.

${\sf \leadsto 30000}$

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Loss rupees :-

${\sf \leadsto 32000 - 30000}$

Subtract the values.

${\sf \leadsto 2000}$

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Now, let's find the loss percentage on the whole transaction.

Loss percentage :-

${\sf \longrightarrow \underline{\boxed{\sf \dfrac{Loss}{Cost \: price} \times 100}}}$

Substitute the given values.

${\sf \leadsto \dfrac{2000}{32000} \times 100}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{2000}{32000} \times 100 = \dfrac{1}{16} \times 100}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{1 \times 100}{16} = \dfrac{100}{16}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \cancel \dfrac{100}{16} = 6.25}$

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${\red{\underline{\boxed{\bf So, \: the \: loss \: percentage \: on \: whole \: transaction \: is \: 6.25\%.}}}}$[/tex]