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

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 $(d).$

As, $\text{x}:\text{y}=1:1$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac{1}{1}$
$\Rightarrow\text{x}=\text{y}$
Now,
$\frac{\text{3x}+4\text{y}}{5\text{x}+6\text{y}}$
$=\frac{3\text{x}+4\text{x}}{5\text{x}+6\text{x}} (As, x = y)$
$=\frac{7\text{x}}{11\text{x}}$
$=\frac{7}{11}$
Hence, the correct alternative is option $(a).$

$6$ dozen or $6 \times 12 = 72$ eggs
Cost of $72$ eggs is $= Rs. 108$
Cost of $1$ egg will be $\text{Rs. }\frac{108}{72}$
and cost of $132$ eggs will be $108$
$\text{Rs. }\frac{108}{72}\times132=\text{Rs. }198$

We know, simple interest $=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
It is given that,
$T = 2$ years
$R = 5%$
$I = Rs. 4000$
Then,
$4000=\frac{\text{P}\ \times\ 5\ \times\ 2}{100}$
$\Rightarrow4000=\frac{10\text{P}}{100}$
$\Rightarrow\text{P}=40000$
Thus,$ P = Rs. 40000$
Hence, the correct option is $(c).$

Let sum $= P$ and Interest rate $= R$
Now Simple Interest $= P$
Simple Interest $=\frac{\big(\text{sum}\times\text{interest rate}\times\text{time}\big)}{100}$
$\Rightarrow\text{P}=\frac{(\text{P}\times8\times\text{R})}{100}$
$\Rightarrow2=\frac{8\text{R}}{100}$
$\Rightarrow8\text{R}\times100$
$\Rightarrow\text{R}=\frac{100}{8}$
$\Rightarrow\text{R}=\frac{25}{2}$ $(100$ and $8$ is divided by $4)$
$\Rightarrow\text{R}=12.5\%$

Marks scored in first test $= 180$
Marks scored in second test $= 150$
Maximum marks $= 200$
Decrease in marks $= 180 - 150 = 30$
Decrease in performance in percentage $=\frac{30}{200}\times100=15\%$
Hence, the correct option is $(b).$

Let the sum be $Rs x$
$\text{A}=\frac{49\text{x}}{40}$
We know:
$A = P + S.I.$
$S.I. = A - P$
$=\Big(\frac{\text{49x}}{40}-\text{x}\Big)$
$=\frac{49\text{x}-40\text{x}}{40}$
$=\frac{9\text{x}}{40}$
Let the rate be R% per annum.
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\frac{9\text{x}}{40}=\frac{\text{x}\times\text{R}\times5}{100\times2}$
$\Rightarrow\text{R}=\frac{9\times20\times2}{10}$
$\Rightarrow\text{R}=9$

$15\ kg$ in grams are $15000$ grams. Ratio of $15000$ grams to $210$ grams is $500 : 7.$

Rate $= 10\%$ per annum
Simple interest $= 0.125 \times $ Principal
$\Rightarrow\frac{\text{Princepal}\times\text{Rate}\times\text{Time}}{100}=0.125\times\text{Principal}$
$\Rightarrow\frac{\text{Time}}{10}=0.125$
$\Rightarrow\text{Time}=1.25=1\frac14\text{ years}$

Cost price of $12$ bananas $= Rs. 25$
Cost price of one banana $=\text{Rs. }\frac{25}{12}$
Cost price of five bananas $=5\times\frac{25}{12}$
$=\frac{125}{12}=10.42$
He shells five bananas at the cost $(SP)$ of $Rs. 12.$
Gain $= SP - CP = Rs. (12 - 10.42) = Rs. 1.58$
$\text{Gain%}=\Big(\frac{\text{Gain}}{\text{C.P.}}\times100\Big)\%$
$=\Big(\frac{1.58}{10.4}\times100\Big)\%$
$=15.2\%$

Percentage of defective television $=\frac{5}{50}\times100\%$
$=10\%$

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 $(a).$

$140\ m$

Let the sum be $Rs. x$ and the rate be $r\%$
Amount $= 2x$
$\Rightarrow\text{P}+\text{S.I.}=2\text{x}$
$\Rightarrow\text{P}+\frac{\text{P}\times\text{r}\times\text{t}}{100}=\text{2x}$
$\Rightarrow\text{x}\Big(1+\frac{\text{r}\times100}{100}\Big)=\text{2x}$
$\Rightarrow\frac{100+10\text{r}}{100}=2$
$\Rightarrow10\text{r}=200-100$
$\Rightarrow10\text{r}=100$
$\Rightarrow\text{r}=\frac{100}{10}$
$\Rightarrow\text{r}=10$

As $3$ parts out of $4$ are shaded.

Given, Amount $= Rs. 220$, Principle $= Rs. 200$, Time $= 1$ year
$1$ year simple interest and compound interest are equal.
So Interest
$= 220 - 200$
$= Rs. 20$
$\text{S.I}=\frac{\text{PRT}}{100}$
$\text{R}=\frac{\text{PRT}\times100}{\text{P}\times\text{T}}$
$\text{R}=\frac{20\times100}{200\times1}$
$\text{R}=10\%$

$CP = Rs. 2500$
$SP = Rs. 2700$
Since, $SP$ is more than$ CP.$
Therefore, Profit $= SP - CP$
$= Rs. 2700 - Rs. 2500$
$= Rs. 200$
Profit Percent $=\frac{\text{Profit}}{\text{CP}}\times100$
$=\frac{200}{2500}\times100$
$=8\%$
Hence, the correct option is $(a).$

$1\frac{1}{4}\%=\frac{5}{4}\%$
$=\frac{5}{4}\times\frac{1}{100}$
$\frac{1}{4\times20}$
$=\frac{1}{80}$
Hence, the correct option is $(c).$

Required percentage
$=\frac{52-48}{48}\times100\%=\frac{25}{3}\%\text{ or }8.33\%$

Let the required number be $x.$
$8\%\ \text{of x}=6$
$\Rightarrow\frac{8}{100}\times\text{x}=6$
$\Rightarrow\frac{2\text{x}}{25}=6$
$\Rightarrow\frac{2\text{x}\times25}{25}=6\times25$
$\Rightarrow2\text{x}=150$
$\Rightarrow\text{x}=\frac{150}{2}$
$\Rightarrow\text{x}=75$
Hence, the correct option is $(d).$

Let the sum be $Rs. x$
Amount $Rs. 2x$
Time $= 10$ years
$S.I. = A - P = 2x - x = Rs. x$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\text{x}=\frac{\text{x}\times\text{R}\times10}{100}$
$\Rightarrow\text{R}=\frac{\text{x}\times100}{\text{x}\times100}$
$\Rightarrow\text{R}=10\%$