Tag: introduction to ratio and percentages

Questions Related to introduction to ratio and percentages

A shopkeeper offers a discount of 25% on a T.V and sells it for Rs.8400. What is the cost price of the T.V?

introduction to ratio and percentages comparing quantities maths

  1. Rs.8570

  2. Rs.11200

  3. Rs.9040

  4. Rs.8960

Show answer
Correct Option: B
Explanation:

Let the cost price of the T.V.$ = x$


After give $25\%$ discount 


$x - \dfrac{25}{100} \times x =$ selling price 
                        $= 8400$

$\dfrac{75}{100} \times x = 8400$

$\dfrac{3}{4} x = 8,400$

$x = \dfrac{8,400 \times 4}{3}$

$x = 11,200$

The temperature of a metal coin is increased ny $100^0$C and its diameter by 0.15%. Its area increases by nearly

introduction to ratio and percentages comparing quantities maths

  1. 0.15%

  2. 0.60%

  3. 0.30%

  4. 0.0225%

Show answer
Correct Option: C
Explanation:

$A = \pi r^2$
$\displaystyle \frac{\Delta A}{A} = 2 \frac{\Delta A}{r}$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \displaystyle \left ( \frac{\Delta A}{r} \right ) \times 100$
$\displaystyle \frac{\Delta A}{A}$% $= 2 \times 0.15 = 0.30$%

If the volume of a sphere increases by 72.8%, then its surface area increases by

introduction to ratio and percentages comparing quantities maths

  1. 20%

  2. 44%

  3. 24.3%

  4. 48.6%

Show answer
Correct Option: B
Explanation:

$\displaystyle \frac{V'}{V} = \frac{172.8}{100} = \frac{\displaystyle \frac{4}{3} \pi R^{.3}}{\displaystyle \frac{4}{3} \pi R^3}$
$\displaystyle \frac{R'}{R} = 1.2$
Now, ratio of surface area $= \displaystyle \frac{S'}{S} = \frac{4 \pi R^{.2}}{4 \pi R^3}$
$= \displaystyle \frac{S'}{S} = 1.44$
Hence surface area increased by 44%

The given table shows the prices of 3 different types of eggs $\displaystyle \frac{1}{4}$ of the eggs Priyanka bought were chicken eggs $\displaystyle \frac{1}{8}$ of them were century eggs and the rest were quail eggs If Priyanka spent a total amount of Rs. 6.50 on the chicken and century eggs how much did she spent on the quail eggs? 

Chicken eggs 20 paise each
Century eggs 90 paise each
Quail eggs 5 paise each
introduction to ratio and percentages comparing quantities maths

  1. Rs. 1.25

  2. Rs. 1.40

  3. Rs. 1.65

  4. Rs. 1.80

Show answer
Correct Option: A
Explanation:

Let the total number of eggs=x.

So, number of chicken eggs= (1/4)*x
number of century eggs= (1/8)*x

number of quail eggs= total-(chicken eggs+century eggs)
$\begin{array}{c}x - \left( {\dfrac{1}{4}x + \dfrac{1}{8}x} \right) = x - \left( {\dfrac{3}{8}x} \right)\\ = \dfrac{5}{8}x\end{array}$
Cost of 1 chicken egg=20 paise
Cost of (1/4)*x chicken eggs= $\dfrac{1}{4} \times x \times 20 = 5x$
Cost of 1 century egg=90 paise
Cost of (1/8)*x chicken eggs= $\dfrac{1}{8} \times x \times 90 = 11.25x$
Cost of 1 quail egg=5 paise
Cost of (5/8)*x chicken eggs= $\dfrac{5}{8} \times x \times 5= 3.125x$

The total cost of chicken and century eggs=5x+11.25x
=16.25x

As for the cost of chicken and century eggs=Rs 6.5
So,
16.25x=6.5
x=0.4

Cost of quail eggs=3.125*x
=3.125*0.4
=1.25

Thus Option A


On selling $17$ balls at $Rs. 720$, there is a loss equal to the cost price of $5$ balls. The cost price of a ball is:

introduction to ratio and percentages comparing quantities maths

  1. $Rs. 45$

  2. $Rs. 50$

  3. $Rs. 55$

  4. $Rs. 60$

Show answer
Correct Option: D
Explanation:

$(C.P.\ of\ 17\ balls) - (S.P.\ of\ 17\ balls) = (C.P.\ of\ 5\ balls)$
$\Rightarrow$ C.P. of $12\ balls = S.P.\ of\ 17\ balls = Rs. 720$.
$\Rightarrow C.P.\ of\ 1\ ball = Rs. \left (\dfrac {720}{12}\right )= Rs. 60$.

If selling price is doubled, the profit triples. Find the profit percent.

introduction to ratio and percentages comparing quantities maths

  1. $66\dfrac {2}{3}$

  2. $100$

  3. $105\dfrac {1}{3}$

  4. $120$

Show answer
Correct Option: B
Explanation:

Let C.P. be $Rs. x$ and S.P. be $Rs. y$.
Then, $3(y - x) = (2y - x) \Rightarrow y = 2x$.
$Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x$.
$\therefore Profit$ % $= \left (\dfrac {x}{x}\times 100\right )$% $= 100$%

Two pipes A and B can fill a tank in $15$ minutes and $20$ minutes respectively. Both the pipes are opened together but after $4$ minutes, pipe A is turned off. What is the total time required to fill the tank?

introduction to ratio and percentages comparing quantities maths

  1. $10$ min. $20$sec.

  2. $11$ min. $45$sec.

  3. $12$ min. $30$ sec.

  4. $14$ min. $40$ sec.

Show answer
Correct Option: D
Explanation:

Part filled in $4$ minutes $=4\left(\displaystyle\frac{1}{15}+\frac{1}{20}\right)=\displaystyle \frac{7}{15}$.
Remaining part$=\left(1-\displaystyle\frac{7}{15}\right)=\displaystyle\frac{8}{15}$.
Part filled by B in $1$ minute $=\displaystyle\frac{1}{20}$
$\therefore \displaystyle\frac{1}{20}:\frac{8}{15}::1:x$
$x=\left(\displaystyle\frac{8}{15}\times 1\times 20\right)=10\displaystyle\frac{2}{3}$min$=10$ min. $40$ sec.
$\therefore$ The tank will be full in $(4$ min. $+10$ min. $+40$ sec.)$=14$min. $40$sec.

What is the percent profit earned by the shopkeeper on selling the articles in his shop?
I. Labeled price of the article sold was $130$% of the cost price.
II. Cost price of each article was $Rs. 550$.
III. A discount of $10$% on labeled price was offered.

introduction to ratio and percentages comparing quantities maths

  1. Only I

  2. Only II

  3. I and III

  4. All the three are required

  5. Question cannot be answer even with information in all the three statements.

Show answer
Correct Option: C
Explanation:

I. Let C.P. be $Rs. x$.
Then, $M.P. = 130$% of $x = Rs. \left (\dfrac {13x}{10}\right )$
III. $S.P. = 90$% of M.P.
Thus, I and III give, $S.P. = Rs. \left (\dfrac {90}{100}\times \dfrac {3x}{10}\right ) = Rs. \left (\dfrac {117x}{100}\right )$
$Gain = Rs. \left (\dfrac {117x}{100} - x\right ) = Rs. \dfrac {17x}{100}$
Thus, from I and III, gain % can be obtained.
Clearly, II is redundant.

If the diameter of a sphere is decreased by $25\%$, by what percent does its curved surface area decrease?

introduction to ratio and percentages comparing quantities maths

  1. $43.75\%$

  2. $21.88\%$

  3. $50\%$

  4. $25\%$

Show answer
Correct Option: A
Explanation:

Curved surface area of sphere$=4\pi r^2$
diametre after decreases by $25\%$
$A _D=\pi d^2$
$d _1=\displaystyle\frac{75}{100}d=\frac{3d}{4}$
$A _1=\pi \left(\displaystyle\frac{3d}{4}\right)^2=\frac{9}{16}\pi d^2$
$\%$ decrease=$\displaystyle\frac{\displaystyle\frac{9}{16}\pi d^2-\pi d^2}{\pi d^2}\times 100=-43.75\%$