3 Marks Question

Question 513 Marks

Of $\frac{3}{4}$ and $\frac{5}{7}$, which is greater and by how much?

Answer

Let us compare $\frac{3}{4}$ and $\frac{5}{7}$
$3 \times 7 = 21$ and $4 \times 5 = 20$
Clearly, 21 > 20 Therefore, $\frac{3}{4}>\frac{5}{7}$ Required difference:$=\frac{3}{4}-\frac{5}{7}$
$L.C.M.$ of $4$ and $7 = (2 \times 2 \times 7) = 28$
$=\frac{21-20}{28}$
$\Big(\frac{28}{4}=7,7\times3=21\Big)$
and $\Big(\frac{28}{7}=4,4\times5=20\Big)$
$=\frac{1}{28}$
Hence, $\frac{3}{4}$ is greater than $\frac{5}{7}$ by $\frac{1}{28}$.

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Question 523 Marks

A piece of wire, $2\frac{3}{4}$ metres long, broke into two pieces. One piece is $\frac{5}{8}$ metre long. How long is the other piece?

Answer

The length of the other piece = (Length of the wire $-$ Length of one piece)
$=\Big(2\frac{3}{4}-\frac{5}{8}\Big)\ \text{m}$
$=\Big(\frac{11}{4}-\frac{5}{8}\Big)\ \text{m}$
$L.C.M.$ of $4$ and $8 = (2 \times 2 \times 2) = 8$
$=\Big(\frac{22-5}{8}\Big)\ \text{m}$
$\Big(\frac{8}{4}=2,2\times11=22\Big)$
and $\Big(\frac{8}{8}=1,1\times5=5\Big)$
$=\Big(\frac{17}{8}\Big)\ \text{m}$
$=2\frac{1}{8}\ \text{m}$
Hence, the other piece is $2\frac{1}{8}\ \text{m}$ long.

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Question 533 Marks

The weight of an empty gas cylinder is $16\frac{4}{5}\text{kg}$ and it contains $14\frac{2}{3}\text{kg}$ of gas. What is the weight of the cylinder filled with gas?

Answer

Weight of the cylinder filled with gas = Weight of the empty cylinder $+$ Weight of the gas inside the cylinder Thus, we have: $(L.C.M$ .of $5$ and $3 = (5 \times 3) = 15$
$\Big(16\frac{4}{5}+14\frac{2}{3}\Big)\text{kg}$
$=\Big(\frac{84}{5}+\frac{44}{3}\Big)\text{kg}$
$=\frac{(252+220)}{15}\text{kg}$
$=\frac{472}{15}\text{kg}$
$=31\frac{7}{15}\text{kg}$
Hence, the weight of the cylinder filled with gas is $31\frac{7}{15}\text{kg}$.

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Question 543 Marks

Find the sum: $2\frac{3}{4}+5\frac{5}{6}$

Answer

We have, $L.C.M$. of $4$ and $6 = (2 \times 2 \times 3) = 12$
$\begin{array}{c|c}2&4,6\\\hline2&2,3\\\hline3&1,3\\\hline&1,1\end{array} $
Therefore, $2\frac{3}{4}+5\frac{5}{6}$
$=\frac{11}{4}+\frac{35}{6}$
$=\frac{(66+140)}{24}$
$\Big(\frac{24}{4}=6,6\times11=66\Big)$ and
$\Big(\frac{24}{6}=4,4\times35=140\Big)$
$=\frac{206}{24}$
$=\frac{103}{12}$
$=8\frac{7}{12}$

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Question 553 Marks

A film show lasted for $3\frac{1}{3}$ hours. Out of his time, $1\frac{3}{4}$ hours was spent on advertisements. What was the actual duration of the film?

Answer

Actual duration of the film = Total duration of the show - Time spent on advertisements
$=\Big(3\frac{1}{3}-1\frac{3}{4}\Big)\ \text{hours}$
$=\Big(\frac{10}{3}-\frac{7}{4}\Big)\ \text{hours}$
$L.C.M$. of $3$ and $4 = (2 \times 2 \times 3) = 12$
$=\Big(\frac{40-21}{12}\Big)\ \text{hours}$
$\Big(\frac{12}{3}=4,4\times10=40\Big)$
and $\Big(\frac{12}{4}=3,3\times7=21\Big)$
$=\Big(\frac{19}{12}\Big)\ \text{hours}$
$=1\frac{7}{12}\ \text{hours}$
Thus, the actual duration of the film was $1\frac{7}{12}\ \text{hours}$.

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Question 563 Marks

Reduce the following fractions into its simplest form: $\frac{9}{15}$

Answer

Here, numerator $= 9$ and denominator $= 15$
Factors of $9$ are $1, 3$ and $9$
Factors of $15$ are $1, 3, 5$ and $15$
Common factors of $9$ and $15$ are $1$ and $3 H.C.F$. of $9$ and $15$ is $3$
$\therefore\frac{9}{15}=\frac{9\div3}{15\div3}=\frac{3}{5}$
Hence, the simplest form of $\frac{9}{15}$ is $\frac{3}{5}$.

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Question 573 Marks

Reduce the following fractions into its simplest form: $\frac{72}{90}$

Answer

Here, numerator $= 72$ and denominator $= 90$
Factors of $72$ are $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36$ and $72$
Factors of $90$ are $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45$ and $90$
Common factors of $72$ and $90$ are $1, 2, 3, 6, 9$ and $18 H.C.F.$ of $72$ and $90$ is $18$
$\therefore\frac{72}{90}=\frac{72\div18}{90\div18}=\frac{4}{5}$
Hence, the simplest form of $\frac{72}{90}$ is $\frac{4}{5}$.

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Question 583 Marks

Arrange the following fractions in descending order: $\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$

Answer

The given fractions are $\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{13},\frac{3}{4},\frac{3}{17}$
As the fractions have the same numerator,
we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator,
the fraction having the smaller denominator is the greater one.
Clearly, $\frac{3}{4} >\frac{3}{5} >\frac{3}{7} >\frac{3}{11}>\frac{3}{13}>\frac{3}{17}$
Hence, the given fractions can be arranged in the descending order as follows: $\frac{3}{4},\frac{3}{5},\frac{3}{7} ,\frac{3}{11},\frac{3}{13},\frac{3}{17}$

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Question 593 Marks

Arrange the following fractions in descending order:
$\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$

Answer

The given fractions are $\frac{1}{12},\frac{1}{23},\frac{1}{7},\frac{1}{9},\frac{1}{17},\frac{1}{50}$
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator,
the fraction having the smaller denominator is the greater one.
Clearly, $\frac{1}{7} >\frac{1}{9} >\frac{1}{12} >\frac{1}{17}>\frac{1}{23}>\frac{1}{50}$
Hence, the given fractions can be arranged in the descending order as follows:
$\frac{1}{7},\frac{1}{9},\frac{1}{12},\frac{1}{17},\frac{1}{23},\frac{1}{50}$

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Question 603 Marks

Compare the fractions given below: $\frac{11}{12},\frac{13}{15}$

Answer

$L.C.M.$ of $12$ and $15 = (2 \times 2 \times 3 \times 5) = 60$
Now, we convert $\frac{11}{12}$ and $\frac{13}{15}$ into equivalent fractions having $40$ as the denominator.
$\therefore\frac{11}{12}=\frac{11\times5}{12\times5}=\frac{55}{60}$ and
$\frac{13}{15}=\frac{13\times4}{15\times4}=\frac{52}{60}$
Clearly, $\frac{55}{60}>\frac{52}{60}$
$\therefore\frac{11}{12}>\frac{13}{15}$

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MATHS 3 Marks Question