Maths MCQ

As, $\text{a}:\text{b}=4:5$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4}{5}$
Also, $\text{b}:\text{c}=2:\text{3}$
$\Rightarrow\frac{\text{b}}{\text{c}}=\frac{2}{3}$
So, $\text{a}:\text{c}=\frac{\text{a}}{\text{c}}$
$=\frac{\text{ab}}{\text{bc}}$
$=\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{c}}$
$=\frac45\times\frac23$
$=\frac{8}{15}$
$=8:15$
Hence, the correct alternative is option $(b).$

Let the number that must be added to each term of the ratio $9 : 16$ be $x.$
Then,
$(9+\text{x}):(16+\text{x})=2:3$
$\Rightarrow\frac{(9+\text{x})}{(16+\text{x})}=\frac{2}{3}$
$\Rightarrow3(9+\text{x})=2(16+\text{x})$
$\Rightarrow27+\text{3x}=32+\text{x}$
$\Rightarrow3\text{x}-2\text{x}=32-27$
$\therefore\text{x}=5$
So, $5$ must be added to each term of the ratio $9 : 16$ to make the ratio $2 : 3.$
Hence, the correct alternative is option $(a).$

Let $P's$ share be $Rs. x.$
Then $Q's$ share $= Rs. (840 - x)$
As, $P's$ share : $Q's$ share $= 3 : 4$
$\Rightarrow\frac{\text{P's share}}{\text{Q's share}}=\frac34$
$\Rightarrow\frac{\text{x}}{(840-\text{x})}=\frac{3}{4}$
$\Rightarrow\text{4x}=3(840-\text{x})$
$\Rightarrow\text{4x}=3\times840-\text{3x}$
$\Rightarrow\text{4x}+\text{3x}=3\times840$
$\Rightarrow7\text{x}=3\times840$
$\text{x}=\frac{3\times840}{7}$
$\Rightarrow\text{x}=3\times120$
$\therefore\text{x}=360$
So, $P's$ share is $Rs. 360.$
Hence, the correct alternative is option $(c).$

As, $\frac{1}{\text{a}}:\frac{1}{\text{b}}:\frac{1}{\text{c}}=3:4:5$
$\Rightarrow\frac{1}{\text{a}}:\frac{1}{\text{b}}=3:4$ and $\frac{1}{\text{b}}:\frac{1}{\text{c}}=4:5$
$\Rightarrow\frac{1}{\text{a}}\div\frac{1}{\text{b}}=\frac{3}{4}$ and $\frac{1}{\text{b}}\div\frac{1}{\text{c}}=\frac45$
$\Rightarrow\frac{1}{\text{a}}\times\frac{\text{b}}{1}=\frac{3}{4}$ and $\frac{1}{\text{b}}\times\frac{\text{c}}{1}=\frac{4}{5}$
$\Rightarrow\frac{\text{b}}{\text{a}}=\frac{3}{4}$ and $\frac{\text{c}}{\text{b}}=\frac{4}{5}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4}{3}$ and $\frac{\text{b}}{\text{c}}=\frac{5}{4} ($Reciprocal of both sides$)$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{4\times5}{3\times5}$ and $\frac{\text{b}}{\text{c}}=\frac{5\times3}{4\times3}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{20}{15}$ and $\frac{\text{b}}{c}=\frac{15}{12}$
$\Rightarrow\text{a}:\text{b}=20:15$ and $\text{b}:\text{c}=15:12$
$\therefore\text{a}:\text{b}:\text{c}=20:15:12$
Hence, the correct alternative is option $(b).$

As, $a : b = 3 : 4$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac34$
So, $\text{4a}:\text{3b}=\frac{4\text{a}}{3\text{b}}$
$=\frac43\times\frac{\text{a}}{\text{b}}$
$=\frac43\times\frac34$
$=\frac{12}{12}$
$=\frac11$
$=1:1$
Hence, the correct alternative is option $(c).$

Since,
$\frac{1}{12}:\frac{1}{60}$
$=\frac{1}{12}\div\frac{1}{60}$
$=\frac{1}{12}\times\frac{60}{1}$
$=\frac{60}{12}$
$=\frac{5}{1}$
$=5:1$
Hence, the correct alternative is option $(c).$

Let the fourth term be $x.$
As, $3:5: :21:\text{x}$
$\Rightarrow\frac35=\frac{21}{\text{x}}$
$\Rightarrow3\text{x}=21\times5$
$\Rightarrow\text{x}=\frac{21\times5}{3}$
$\Rightarrow\text{x}=7\times5$
$\therefore\text{x}=35$
So, the fourth term is $35.$
Hence, the correct alternative is option $(b).$

Let $x$ be the mean proportional of $6$ and $24.$
Then,
$\frac{\text{6}}{\text{x}}=\frac{\text{x}}{24}$
$\Rightarrow\text{x}^2=6\times24 ($By cross multiplication$)$
$\Rightarrow\text{x}^2=144$
$\therefore\text{x}=12$
So, the mean proportional of $6$ and $24$ is $12.$
Hence, the correct alternative is option $(b).$

As, $\text{a}:\text{b}=5:7$ and $\text{b}:\text{c}=6:11$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac57$ and $\frac{\text{b}}{\text{c}}=\frac{6}{11}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{5\times6}{7\times6}$ and $\frac{\text{b}}{\text{c}}=\frac{6\times7}{11\times7}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{30}{42}$ and $\frac{\text{b}}{\text{c}}=\frac{42}{77}$
$\Rightarrow\text{a}:\text{b}=30:42$ and $\text{b}:\text{c}=42:77$
$\therefore\text{a}:\text{b}:\text{c}=30:42:77$
Hence, the correct alternative is option $(b).$

As, $\text{2x}=3\text{y}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac32$
And, $\text{4y}=\text{5z}$
$\Rightarrow\frac{\text{y}}{\text{z}}=\frac54$
Now, $\text{x}:\text{z}=\frac{\text{x}}{\text{z}}$
$=\frac{\text{xy}}{\text{yz}}$
$=\frac{\text{x}}{\text{y}}\times\frac{\text{y}}{\text{z}}$
$=\frac32\times\frac54$
$=\frac{15}{8}$
$=15:8$
Hence, the correct alternative is option $(d).$

Let the third proportional of $3$ and $27$ be $x.$
Then,
$3:27::27:\text{x}$
$\Rightarrow3:27=27:\text{x}$
$\Rightarrow\frac{3}{27}=\frac{27}{\text{x}}$
$\Rightarrow3\text{x}=27\times27$
$\Rightarrow\text{x}=\frac{27\times27}{3}$
$\Rightarrow\text{x}=27\times9$
$\therefore\text{x}=243$
So, the third proportional of $3$ and $27$ is $243.$
Hence, the correct alternative is option $(a).$