Maths M.C.Q. [1 Marks Each]

 Number of pieces
$=\frac{\text{Total length of ribbon}}{\text{Length of one piece}}=\frac{\big(5\frac{1}{4}\big)}{\big(\frac{3}{4}\big)}$
$=\Bigg(\frac{\frac{(5\times4)+1}{4}}{\frac{3}{4}}\Bigg)=\bigg(\frac{\frac{21}{4}}{\frac{3}{4}}\bigg)$
$=\frac{21}{4}\times\frac{4}{3}=7$
$\big[\because\text{reciprocal of}\ \frac{3}{4}=\frac{4}{3}\big]$

The correct statement will be
$\frac{2}{3},\frac{3}{5},\frac{14}{15}$
$=\frac{10,9,14}{15}$
$LCM$ of $3, 5, 15, = 15$
or $\frac{3}{5}<\frac{2}{3}<\frac{14}{15}$

$3\times\frac{2}{3}$ means $3$ times the two-third part of anything.
$\therefore$ Option $(b)$ is correct.

We have
$0.1 - x = 0.04$
$\Rightarrow x = 0.1 - 0.04$
Converting the given decimals into like decimals, we get
$0.10$ and $0.04$
Thus, required number $= (0.10 - 0.04) = 0.06$
Hence, 0.06 should be subtracted from $0.1$ to get $0.04$

 Consider the fractions $\frac{4}{9},\frac{2}{5}$ and $\frac{1}{4}$
$LCM$ of $4, 5, 7$ and $9 = 1260$
Firstly, convert the fractions into equivalent fractions with denominator $1260$
$\Rightarrow\frac{4}{9}=\frac{4\times140}{9\times140}=\frac{560}{1260}$
$\Rightarrow\frac{2}{5}=\frac{2\times252}{5\times252}=\frac{504}{1260}$
$\Rightarrow\frac{3}{7}=\frac{3\times180}{7\times180}=\frac{540}{1260}$
$\Rightarrow\frac{1}{4}=\frac{1\times315}{4\times315}=\frac{315}{1260}$
Now,
$315<504<540<560$
$\therefore\ \frac{315}{1260}<\frac{504}{1260}<\frac{540}{1260}<\frac{560}{1260}$
$\Rightarrow\frac{1}{4}<\frac{2}{5}<\frac{3}{7}<\frac{4}{9}$
Thus, the smallest fraction is $\frac{1}{4}$

 Consider the fractions $\frac{3}{4},\frac{5}{6},\frac{2}{3},\frac{1}{2},\frac{4}{5}$ and $\frac{9}{10}$
$LCM$ of $2, 3, 4, 5, 6$ and $10 = 60$
Firstly, convert the fractions into equivalent fractions with denominator $60$
$\Rightarrow\frac{3}{4}=\frac{3\times15}{4\times15}=\frac{45}{60}$
$\Rightarrow\frac{5}{6}=\frac{5\times10}{6\times10}=\frac{50}{60}$
$\Rightarrow\frac{2}{3}=\frac{2\times20}{3\times20}=\frac{40}{60}$
$\Rightarrow\frac{1}{2}=\frac{1\times30}{2\times30}=\frac{30}{60}$
$\Rightarrow\frac{4}{5}=\frac{4\times12}{5\times12}=\frac{48}{60}$
$\Rightarrow\frac{9}{10}=\frac{9\times6}{10\times6}=\frac{54}{60}$
Now,
$30<40<45<48<50<54$
$\therefore\ \frac{30}{60}<\frac{40}{60}<\frac{45}{60}<\frac{48}{60}<\frac{50}{60}<\frac{54}{60}$
$\Rightarrow\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{9}{10}$
Thus, the fraction $\frac{4}{5}$ is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$

The number $0.257$ can be represented as $0.2 + 0.05 + 0.007 .$
Therefore we can see that digit $7$ represents
$ 0.007=7\times\frac{1}{1000}$

Total quantity of both ingredients in one packet of biscuits
$=$ Quantity of flour $+$ Quantity of sugar
$=2\frac{1}{2}\ \text{cups}+1\frac{2}{3}\text{cups}$
$=\frac{(2\times2)+1}{2}+\frac{(1\times3)+2}{3}$
$-\frac{4+1}{2}+\frac{3+2}{3}$
$=\frac{5}{2}+\frac{5}{3}$
$=\frac{5\times3+2\times5}{6}$ $\big[\because\ \text{LCM of 2 and 3 = 6}\big]$
$=\frac{15+10}{6}$
$=\frac{25}{6}$
$\therefore$ Total quantity of both ingredients used in $10$ packets
$= 10\ ×$ Total quantity of ingredients in one packet
$=10\times\frac{25}{6}=\frac{250}{6}$
Since, $\frac{250}{6}$ lies between $40$ and $50.$

 We have,
$3 \times 3 \times 3 = 27$
The sum of the decimal places in the given decimals is $1 + 1 + 1 = 3$
So, the product must contain 3 places of decimals.
$\therefore 0.3 \times 0.3 \times 0.3 = 0.027$

$626.235$
There are two $6s$ in this number, one at Hundreds place and other at Ones place.
$ \Rightarrow$ Place values of $ 6=6\times100$ and $ 6\times1$
Difference $= 600 - 6 = 594$

 The given fractions are $\frac{2}{3},\frac{3}{5}$ and $=\frac{11}{15}$
$[LCM$ of $5, 3$ and $15 = 15]$
Now, we have:
$\frac{3}{5}\times\frac{3}{3}=\frac{9}{15},\frac{2\times5}{3\times5}$
$=\frac{10}{15}$ and $\frac{11}{15}$
Clearly,
$\frac{9}{15}<\frac{10}{15}<\frac{11}{15}$
$\therefore\frac{3}{5}<\frac{2}{3}<\frac{11}{15}$

Given, $\frac{2}{3},\frac{6}{7},\frac{13}{21}$
$LCM$ of $(3, 7, 21) = 21$
$\therefore\frac{2}{3}=\frac{2}{3}\times\frac{7}{7}=\frac{14}{21}$
$\frac{6}{7}=\frac{6}{7}\times\frac{3}{3}=\frac{18}{21}$
and $\frac{13}{21}=\frac{13}{21}$
Now, compare $\frac{14}{21},\frac{18}{21}\text{and}\frac{13}{21}$
$\text{So},\frac{13}{21}<\frac{14}{21}<\frac{18}{21}$
Hence, $\frac{13}{21}<\frac{2}{2}<\frac{96}{7}$ (ascending order)
Note with same denominators, fraction with larger numerator is greater.

$ 0.34=\frac{34}{100}$

$0.06=\frac{06}{100}=\frac{3}{50}$

The given expression can be simplified as follows:
$ \frac{(0.0036)(2.8)}{(0.04)(0.1)(0.003)}$
$ =\frac{0.0036\times2.8}{0.04\times0.1 \times 0.003}=$
$ \frac{0.01008}{0.000012}=840$
Hence, the value of $ \frac{(0.0036)(2.8)}{(0.04)(0.1)(0.003)}$ is $ 840$

Factors of $84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$
Factors of $98: 1, 2, 7, 14, 49, 98$
Common factors of $84$ and $98: 1, 2, 14$
$\therefore HCF$ of $84$ and $98 = 14$
Now,
$\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7} ($Dividing numerator and senominator by the $HCF$ of $84$ and $98$ i.e., $14)$

$=75.57\div0.01$
$=\frac{75.57}{0.01}$
$=\frac{75.57\times100}{0.01\times100} ($Multiply numerator and denominator by $100$ to convert the divisor$)$
$=\frac{7557}{1}$
$=7557$

Given, $0.03 × 0.9$
Here, $3 × 9 = 27$
Sum of the decimal places to the right of the decimal point is $0.03$ and $0.09$ is $3$
So, $0.03 × 0.9 = 0.027$

$=0.012\div1.5$
$=\frac{0.012}{1.5}$
$=\frac{0.012\times10}{1.5\times10}($ Multiply the numberator and denominator by $10$ to convert the divison$)$
$=\frac{0.12}{15}$

$\therefore\ 0.012\div1.5=0.008$

To round to the nearest tenth, write down the number with a decimal point, and find the tenths place directly to the right of the decimal. Then, to the right of the tenths place, look at the number in the hundredths place. In $6050.287$ the number in the hundredths is $2.$
So, $6050.287$ rounded to the nearest tenths will be $6050.3$

We know that a fraction is irreducible (or is in its lowest terms) if the $HCF$ of its numerator and denominator is $1.$
Consider the fraction $\frac{91}{104}$
$HCF$ of $91$ and $104=13\neq1$
So, the fraction $\frac{91}{104}$ is reducible.
Consider the fraction $\frac{105}{112}$
$HCF$ of $105$ and $112=7\neq1$
So, the fraction $\frac{105}{112}$ is reducible.
Consider the fraction $\frac{51}{85}$
$HCF$ of $51$ and $85=17\neq1$
So, the fraction $\frac{51}{85}$ is reducible.
Now,
Consider the fraction $\frac{43}{83}$
$HCF$ of $43$ and $83 = 1$
So, the fraction $\frac{43}{83}$ is irreducible (or is in its lowest terms).

$ \text{10}\frac{2}{10} ⇒\text{10}\frac{2}{10}=\frac{102}{10}=10.2$

$2\text{km }5\text{m}=2\frac{5}{1000}\text{km}=2.005\text{km}$