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

Let the least number that is to be subtracted from each term of the ratio $15 : 19$ be $x.$
Then,
$(15-\text{x}):(19-\text{x})=3:4$
$\Rightarrow4(15-\text{x})=3(19-\text{x})$
$\Rightarrow60-\text{4x}=57-3\text{x}$
$\Rightarrow3\text{x}-4\text{x}=57-3\text{x}$
$\Rightarrow-\text{x}=-3$
$\therefore\text{x}=3$
So, $3$ is the least number to be subtracted from each term of the ratio $15 : 19$ to make the ratio $3 : 4.$
Hence, the correct alternative is option $(a).$

Out of Rs. $100$, money spent $= Rs. 80$
$\therefore$ Out of $Rs. 200,$ money spent
$=\frac{80}{100}\times200$
$= Rs. 160$
or
Money spent $= 80\%$ of $Rs. 200$
$=200\times\frac{80}{100}$
$= Rs. 60$
$\therefore$ Money left $= Rs. 200 - Rs. 160$
$= Rs. 40$

$10\% VAT$ on $Rs. 100$ will make it $Rs. 110$
So, for price including $VAT Rs. 110$, the original price is $Rs. 100$
Then, Price including $VAT Rs. 3300$, the original price
$=\text{Rs.} \big(\frac{100}{110}\times3300\big)$
$=\text{Rs. }3000$

$\text{Principal }=\Big(\frac{\text{I}\times100}{\text{R}\times\text{T}}\Big)$

It is given that,
Simple interest $(I) = Rs. n$
Rate $(R) = R\%$
Time $(T) = n$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{P}=\frac{100\ \times\ \text{I}}{\text{R}\ \times\ \text{T}}$
$\Rightarrow\text{P}=\frac{100\ \times\ \text{n}}{\text{R}\ \times\ \text{n}}$
$\Rightarrow\text{P}=\frac{100}{\text{R}}$
Hence, the correct option is $(c).$

Rate $=3\frac34\%$ per annum
$=\frac{15}{4}\%$ per annum
Time $=2\frac13$ years
$=\frac73$ years
$\text{S.I.}=\frac{\text{P}\times\frac{15}{4}\times\frac{7}{3}}{100}$
$\Rightarrow\text{P}=\frac{210\times100}{\Big(\frac{15}{4}\times\frac{7}{3}\Big)}$
$\Rightarrow\text{P}=600\times4$
$\Rightarrow\text{P}=\text{Rs. }2400$

Let x be subtracted from each of the term
$\frac{15-\text{x}}{19-\text{x}}=\frac{3}{4}$
$\Rightarrow 4 (15 - x) = 3 (19 - x)$
$\Rightarrow 60 - 4x = 57 - 3x$
$\Rightarrow -4x + 3x = 57 - 60$
$\Rightarrow -x = -3$
$x = 3$
Required number $= 3$

As, $24:36=\frac{24}{36}=\frac23=2:3$
So, the simplest from of $24 : 36 = 2 : 3$
Hence, the correct alternative is option $(d).$

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 $(b).$

$S.P. = Rs. 100$
Gain $= Rs. 20$
$C.P. = S.P$. - gain $= Rs. 100 - 20 = Rs. 80$
$\text{Gain%}=\frac{\text{Gain}\times100}{\text{C.P}}$
$=\frac{20\times100}{80}$
$=25\%$

$644\ km$
$1\ cm$ represents $8 \ km.$
$\therefore$ $80.5\ cm$ represents $8 \times 80.5 = 644\ km.$

Amount $= 3$ times the sum $= 3P$
Simple interest $(I) =$ Amount - Sum $= 3P - P = 2P$
Let the sum $(P)$ be $x.$
Then, simple interest $(I) = 2x$
Rate $(R) = R\%$
Time $(T) = 16$ years
$\text{I}=\frac{\text{P}\ \times\ \text{R}\ \times\ \text{T}}{100}$
$\Rightarrow\text{R}=\frac{100\ \times\ \text{I}}{\text{P}\ \times\ \text{T}}$
$\Rightarrow\text{R}=\frac{100\ \times\ 2\text{x}}{\text{x}\ \times\ 16}$
$\Rightarrow\text{R}=12.5\%$
Hence, the correct option is $(d).$

We have, $225\%$
In percentage, $225\times\frac{1}{100}=\frac{225}{100}=\frac{9}{4}$
$\therefore$ Required ratio $= 9 : 4$
Note: The sign of percentage $(\%)$ is removed dividing the number by $100.$

None of these $0.8\ cm$ represent the map $= 8.8\ km$
$1\ cm$ will represent on the map $=\frac{8.8}{0.8}\text{km}$
$\therefore$ $80.5\ cm$ will represent on the map
$=\frac{8.8\times80.5}{0.8}=\frac{88\times805\times10}{10\times8\times10}$
$=\frac{8855}{10}=885.5\text{km}$

Given that $60\%$ of $x$ is $1200$
$\frac{60}{100}\times\text{x}=1200$
$\text{x}=1200\times\frac{100}{60}$
$\text{x}=2000$

We have, $20\%$ of $700\text{m}=\frac{20}{100}\times700=20\times7=140\text{m}$
Hence, $20\%$ of $700 m$ is $140m.$

Let the sum be $Rs. x$
Let the rate be $R \%$ per annum
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times\text{R}\times5}{100}$
$=\frac{\text{Rx}}{20}$
$\therefore\frac{3\text{x}}{5}=\frac{\text{Rx}}{20}$
$\Rightarrow\text{R}=\frac{3\text{x}\times20}{5\text{x}}=12\%$

Given, $P = Rs. 12000,$
$R = 10\ %$
$T = 1$ month $=\frac{1}{12}\text{yr}$
$\therefore\text{I}=\frac{\text{P}\times\text{R}\times\text{T}}{100}=\frac{12000\times10\times1}{100\times12}=\text{Rs. }100$

Given, cost of $6$ books $= 90$
So, the cost of $1$ book $=\frac{90}{6}=15$
So, the cost of $8$ books $= 15 \times 8 = Rs. 120$

Required amount of protein
$\frac{25}{100}\times4=1\text{kg}$

$\frac{2}{5}=\frac{2}{5}\times\frac{100}{100}$
$=\frac{40}{100}$
$=40\%$
Hence, the correct option is $(c).$

Loss $= Rs. 316 - Rs. 300 = Rs. 16.$

Given, ratio of income to savings $= 4 : 1$
Let income $= 4x$ and savings $= x.$
$\therefore$ Percentage of money saved by her $=\frac{\text{Savings}}{(\text{Income+Savimgs)}}\times100\%$
$=-\frac{\text{x}}{4\text{x}+\text{x}}\times100=\frac{\text{x}}{5\text{x}}\times100\%$
$=\frac{1}{5}\times100\%=20\%$

$3m : 60cm$
$= 3 \times 100cm : 60cm$
$= 300 : 60$
$= 5 : 1.$

$9$ metre in cms are $900\ cm$, ratio of $900\ cm$ to $27$ is $100 : 3.$

Let the time be t years.
Principal $= Rs. 8000$
Amount $= Rs. 8360$
Rate $= 6\%$ per annum
Amount $=\text{Princepal}\Big(1+\frac{\text{Rate}\times\text{Times}}{100}\Big)$
$\frac{8360}{8000}=1+\frac{6\times\text{t}}{100}$
$\Rightarrow\frac{8360}{8000}-1=\frac{\text{6t}}{100}$
$\Rightarrow\text{t}=\Big(\frac{8360-8000}{8000}\Big)\times\frac{100}{6}$
$=\frac{360}{8000}\times\frac{100}{6}$
$=\frac68\times12\text{ months}$
$=9\text{ months}$

Given that $2cm = 1000km$
$2cm \rightarrow 1000km$
$2.5cm \rightarrow xkm$
By unitary method,
$\text{x}=\frac{2.5\times1000}{2}=1250\text{km}$

Number of cows that graze the field in $12$ days $= 45$
Number of cows that graze the field in $1$ day $= 45 \times 12$
$\therefore$ Number of cows that graze the field in $9$ days $=\frac{45\times12}{9}=60$

$7. 7 : x = 42 : 96$
$\Rightarrow\frac{7}{\text{x}}=\frac{42}{96}$
$\Rightarrow x = 16$

As, $\text{2a}=\text{3b}=4\text{c}$
$\Rightarrow\text{2a}=\text{3b}$ and $3\text{b}=\text{4c}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac32$ and $\frac{\text{b}}{\text{c}}=\frac43$
$\Rightarrow\text{a}:\text{b}=6:4$ and $\text{b}:\text{c}=4:3$
$\therefore\text{a}:\text{b}:\text{c}=6:4:3$
Hence, the correct alternative is option $(d).$

Let x be the number of bosy.
Then, $8 : 5 = x : 160$
$\Rightarrow\frac{8}{5}=\frac{\text{x}}{160}$
$\Rightarrow\text{x}=\frac{8\times160}{5}=256$
$\therefore$ Total strength of the school $= 256 + 160 = 416$