Tag: calculating and mental strategies 3

Questions Related to calculating and mental strategies 3

What is the percentage approximate for the fraction $\dfrac{12}{35}$?

  1. $34.28\%$

  2. $23.78\%$

  3. $30.12\%$

  4. $29.10\%$


Correct Option: A
Explanation:

To get the percent you will divide the numerator by the denominator.

Then multiply $100$ to the answer and add the percent sign.
therefore, the percentage approximate value if $\dfrac{12}{35}\times 100 = 34.28$ $\%$.

Which of the following statement/formulae is correct for Percentage Error?

  1. $\cfrac { \left| Approx.value-Exact\quad value \right| }{ Exact\quad value } $

  2. $\cfrac { \left| Exact\quad value-Approx.value \right| }{ Exact\quad value } $

  3. $\cfrac { \left| Approx.value-Exact\quad value \right| }{ Approx.value } \quad $

  4. None


Correct Option: A
Explanation:

The approximate values is the estimated values and the exact value is the real value.

In the half yearly exam only $70\%$ of the students were passed. Out of these(assed in half yearly) only $60\%$ students are passed in annual exam. Out of those who did not pass the half yearly exam, $80\%$ passed in annual exam. What per cent of the students passed the annual exam?

  1. $42\%$

  2. $56\%$

  3. $66\%$

  4. $38\%$


Correct Option: C
Explanation:

Let the total students are $100$

Given : $70\%$ students passed in half yearly examination 
Then, passed students in half yearly $=\dfrac{70}{100}\times 100=70$
Out of these passed in half yearly, $60\%$ students passed in yearly 
Then, passed students in yearly examination $=\dfrac{60}{100}\times 70=42$
The no. of students not passed in half yearly examination $=100-70=30$
Given out of these who did not pass the half yearly examination, $80\%$ students passed in annual exam
Then, out of these student who did not pass in half yearly examination, no. of students passed in yearly $=\dfrac{80}{100}\times 30=24$
Then, total no. of students passed annual examination $=42+24=66$
Then, $\%$ of passed students $=\dfrac{66}{100}\times 100=66\%$

I thought 70 people would turn up to the concert, but in fact 80 did. Find the error.

  1. $21.5 \%$

  2. $12.5 \%$

  3. $13.5 \%$

  4. None of these


Correct Option: B
Explanation:

$\dfrac{|Approximate Value - Exact Value|}{|Exact Value|} \times100$


$\dfrac{|70-80|}{80} \times 100$ = $\dfrac{10}{80} \times 100$

$\dfrac{100}{8}$ = $12.5$

Sam does an experiment to find how long it takes an apple to drop $2$ meters. Sam measures $0.62$ seconds, which is an approximate value. Calculate the error percentage.

  1. $4\%$

  2. $3\%$

  3. $6\%$

  4. None of the above


Correct Option: B
Explanation:

we know that ,
$s=\dfrac{at^2}{2}$
$a=g=9.81 msec^{-2}$
$2=\dfrac{9.81t^2}{2}$
$t=\sqrt{\dfrac{4}{9.81}}$
$t=0.6385$
Therefore, error is $(0.6385-0.62)\times 100%=3%$
$3%$ is the error percentage.

I estimated $260$ people, but $325$ came. Find the error as a percent of the exact value.

  1. $20\%$

  2. $40\%$

  3. $30\%$

  4. None of these


Correct Option: A
Explanation:

$\Rightarrow$  Expected people to come is $260$ and actual people came is $325$

$\Rightarrow$  Approximate value is $260$ and exact value is $325$
$\Rightarrow$  $Error=Exact\, value-Approximate\, value =325-260=65$.

$\Rightarrow$  $Percent\,\,error=\dfrac{65}{325}\times 100=20\%$

The report said the carpark held $240$ cars, but we counted only $ 200$ parking spaces. By what percent the report had error?

  1. $10 \%$

  2. $25 \%$

  3. $20 \%$

  4. None of these


Correct Option: C
Explanation:

$\Rightarrow$  Here, Approximate value is $240$

$\Rightarrow$  Exact value is $200$.
$\Rightarrow$  $Percent\, error$ = $\dfrac{|Approximate\,value-Exact\,value|}{Exact\,value}\times 100$

$\Rightarrow$  $Percent\,error=\dfrac{|240-200|}{|200|}\times 100=\dfrac{40}{200}\times 100=20\%$

You measure the plant to be $80$ cm high (to the nearest cm). If you could be up to $0.5$ cm wrong. Calculate your percentage error.

  1. $0.425\%$

  2. $0.525\%$

  3. $0.625\%$

  4. None of these


Correct Option: C
Explanation:

$\Rightarrow$   Measure of plant is $80\,cm$.

$\Rightarrow$   Error is $0.5\,cm$.
$\Rightarrow$  $\%\,Error=\dfrac{0.5}{80}\times 100$
$\therefore$    $\%\,Error=0.625\%$

What is the percentage error in using $\dfrac{22}{7}$ as an approximation for $\pi$?
(Give your answer to the nearest $0.01%$)

  1. $0.02\%$

  2. $0.03\%$

  3. $0.04\%$

  4. None of these


Correct Option: C
Explanation:

$\Rightarrow$  Here, Approximate value of $\pi$ = $\dfrac{22}{7}=3.142857$

$\Rightarrow$  Exact value of $\pi$ is $3.141592$
$\Rightarrow$  $\%\,Error=\dfrac{|Approximate\,value-Exact\,value|}{|Exact\,value|}\times 100$

$\Rightarrow$  $\%\,Error=\dfrac{|3.142857-3.141592|}{|3.141592|}\times 100$
$\Rightarrow$  $\%\,Error=0.04\%$

The population in $2002$ was $29,000$. It was expected to rise to $34,000$ by $2012$. In fact it rose to $33,000$. What was the percentage error?

  1. $25\%$

  2. $15\%$

  3. $12.5\%$

  4. None of these


Correct Option: A
Explanation:

$\Rightarrow$  Here, Approximate Population = $34000-29000=5000$

$\Rightarrow$  Exact population = $33000-29000=4000$
$\Rightarrow$  $\%\,Error=\dfrac{|Approximate\,value-Exact\,value|}{|Exact\,value|}\times 100$

$\Rightarrow$  $\%\,Error=\dfrac{|5000-4000|}{|4000|}\times 100$

$\therefore$   $\%\,Error=25\%$