13. A student got 42% marks and has secured 12 marks more than the minimum passing marks. Another student got 45% has obtained 30 marks more than the minimum passing marks. The maximum marks are:
a. 1200
b. 900
c. 600
d. 800
Answer: C
Explanation:
Let the maximum mark $= m$, and pass mark $= p$.
Marks secured by first student $= 42\%(m) = p + 12 \qquad (1)$
Marks secured by second student $= 45\%(m) = p + 30 \qquad (2)$
$(2) - (1)$, $3\%(m) = 18$
$ \Rightarrow $ $m = 18 \times \dfrac{{100}}{3} = 600$
Alternative method:
Difference in percentage of marks $= 45\% - 42\% = 3\%$
Difference in marks $= 30 - 12 = 18$
$\begin{gathered}
3\% \xrightarrow{\phantom{222222}}18 \hfill \\[4px]
1\% \xrightarrow{{\phantom{222222}}}6 \hfill \\[4px]
100\% \xrightarrow{{\phantom{222222} }}600 \hfill \\
\end{gathered} $
14. A person saves 10% of his income. If his income is increased by 20% and he saves 15% of the new income, by what percent his savings will increase?
a. 80%
b. 60%
c. 50%
d. 40%
15. Sum of 5% of a number and 9% of other number is equal to sum of 8% of first number and 7% of the second number. Find ratio between the numbers.
a. 1 : 2
b. 2 : 3
c. 3 : 2
d. 9: 13
Answer: B
Explanation:
Let the numbers are $x$ and $y$.
Then $5% \;of\; x + 9%\; of\; y = 8%\; of\; x + 7% \;of\; y$
⇒ $3\% \;of\; x = 2\%\; of\; y$
⇒ $\left( {\dfrac{3}{{100}}} \right)x = \left( {\dfrac{2}{{100}}} \right)y$
⇒ $\dfrac{x}{y} = \dfrac{2}{3}$
⇒ $x : y = 2 : 3$
16. A man spends 10% of his income on food and 80% of the remaining income on clothing. If he still has a balance of Rs. 180. what is his total income?
a. 1000
b. 1200
c. 1300
d. 1400
Answer: C
Explanation:
Let his initial income $=x$
Given, $ \Rightarrow x\left( {1 - \dfrac{{10}}{{100}}} \right)\left( {1 - \dfrac{{80}}{{100}}} \right) = 180$
$ \Rightarrow x\left( {\dfrac{{90}}{{100}}} \right)\left( {\dfrac{{20}}{{100}}} \right) = 180$
$ \Rightarrow x = \dfrac{{100}}{{90}} \times \dfrac{{100}}{{20}} \times 180 = 1000$
17. The length of a rectangle is increased by 60%. By what percent would be width have to be decreased to maintain the same area ?
a. $37\displaystyle\frac{1}{2}$ %
b. 60%
c. 75%
d. None of these
Answer: A
Explanation:
Let length = $100$ m, breadth = $100$ m
New length = $160$ m, new breadth = $x$ m
Then $160 \times x = 100 \times 100$
or $x = \displaystyle\frac{{100 \times 100}}{{160}} = \displaystyle\frac{{125}}{2}$
Decrease in breadth = $\left( {100 - \displaystyle\frac{{125}}{2}} \right)\% = 37\displaystyle\frac{1}{{2 }}%$
Alternative Method 1:
Area of rectangle $= lb$
Here length got incrased by 60% but area should remain constant.
$\left( {A + B + \frac{{AB}}{{100}}} \right) = 0$
$ \Rightarrow \left( {60 - B - \dfrac{{60B}}{{100}}} \right) = 0$
$ \Rightarrow \left( {60 - \dfrac{{160B}}{{100}}} \right) = 0$
$ \Rightarrow \dfrac{{160B}}{{100}} = 60$
$ \Rightarrow B = 60 \times \dfrac{{100}}{{160}}$
$ \require{cancel}\Rightarrow B = \cancel{60}^{15} \times \dfrac{\cancel{100}^{5}}{\cancel{160}^{\cancel{8}^2}}$ $ = \dfrac{{75}}{2} = 37\dfrac{1}{2}\% $
Alternative Method 1:
Area of rectangle $A = lb$
Length has become $= l\left( {1 + \dfrac{{60}}{{100}}} \right)$ $= l\left( {\dfrac{{160}}{{100}}} \right) = \dfrac{8}{5}l$
to keep the Area constant, the breadth has to be $ = \dfrac{5}{8}b$
$ \Rightarrow A = \dfrac{8}{5}l \times \dfrac{5}{8}b$
Chance in the percentage of breadth $ = \dfrac{{b - \dfrac{5}{8}b}}{b} \times 100$ $= \dfrac{{3b}}{{8b}} \times 100$ $ = 37\dfrac{1}{2}\% $
18. If the side of a square is increased by 30% , its area is increased by :
a. 9%
b. 30%
c. 60%
d. 69%
Answer: D
Explanation:
Let, side $= 100$ cm
Area = ${(100 \times 100)}$$\,c{m^2}$ = $10000\, c{m^2}$
New area = ${(130 \times 130)}$$\,c{m^2}$ = $16900\,c{m^2}$
Increase in area = $\left( {\displaystyle\frac{{6900}}{{10000}} \times 100} \right)\% = 69\% $
