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Maths 5 Marks Questions

Rational Numbers · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 7 questions.

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7 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
The table given below shows the distances, in kilometres, between four villages of a state. To find the distance between two villages, locate the square where the row for one village and the column for the other village intersect.
$a.$ Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh?
$b.$ If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive?
Answer
$a.$ The distance between Himgaon and Rawalpur $=98\frac{3}{4}\text{km}$
And the distance between Sonaput and Ramgarh $=40\frac{2}{3}\text{km}$
Difference of the distance between Himgaon and Rawalpur to Sonapur and Ramgarh,
$=\Big(98\frac{3}{4}-40\frac{2}{3}\Big)$
$=\Big(\frac{395}{4}-\frac{122}{3}\Big)$
$=\Big(\frac{1185-488}{12}\Big)$
$=\frac{697}{12}$
$=58\frac{1}{12}\text{km}.$
$b.$ Distanc betwee Himgaon and Sonapur $=100\frac{5}{6}\text{km}$
And distance between Sonapur and Rawalpur $=16\frac{1}{2}\text{km}$
Total distance that he would drive, $=100\frac{5}{6}+16\frac{1}{2}$
$=\frac{605}{6}+\frac{33}{2}$
$=\frac{605+99}{6}$
$=\frac{704}{6}$
$=\frac{352}{3}$
$=117\frac{1}{3}\text{km}.$
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Question 25 Marks

The average life expectancies of males for several states are shown in the table. Express each decimal in the form $\frac{\text{p}}{\text{q}}$ and arrange the states from the least to the greatest male life expectancy.State-wise data are included below; more indicators can be found in the $“FACTFILE”$ section on the homepage for each state.
State
Male
$\frac{\text{p}}{\text{q}}$ form
Lowest terms
Andhra Pradesh
$61.6$
 
 
Assam
$57.1$
 
 
Bihar
$60.7$
 
 
Gujarat
$61.9$
 
 
Haryana
$64.1$
 
 
Himachal Pradesh
$65.1$
 
 
Karnataka
$62.4$
 
 
Kerala
$70.6$
 
 
Madhya Pradesh
$56.5$
 
 
Maharashtra
$64.5$
 
 
Orissa
$57.6$
 
 
Punjab
$66.9$
 
 
Rajasthan
$59.8$
 
 
Tamil Nadu
$63.7$
 
 
Uttar Pradesh
$58.9$
 
 
West Bengal
$62.8$
 
 
India
$60.8$
 
  
Source: Registrar General of India $(2003)$ $SRS$ Based Abridged Lefe Tables. $SRS$ Analytical Studies, Report No. $3$ of $2003$, New Delhi: Registrar General of India. The data are for the $1995-99$ period; states subsequently divided are therefore included in their pre-partition states (Chhatisgarh in $MP$, Uttaranchal in $UP$ and Jharkhand in Bihar).
Answer
State Male $\frac{\text{p}}{\text{q}}$ form Lowest terms
Andhra Pradesh $61.6$ $\frac{616}{10}$ $\frac{308}{5}$
Assam $57.1$ $\frac{571}{10}$ $\frac{571}{10}$
Bihar $60.7$ $\frac{607}{10}$ $\frac{607}{10}$
Gujarat $61.9$ $\frac{619}{10}$ $\frac{619}{10}$
Haryana $64.1$ $\frac{641}{10}$ $\frac{641}{10}$
Himachal Pradesh $65.1$ $\frac{651}{10}$ $\frac{651}{10}$
Karnataka $62.4$ $\frac{624}{10}$ $\frac{312}{5}$
Kerala $70.6$ $\frac{706}{10}$ $\frac{353}{5}$
Madhya Pradesh $56.5$ $\frac{565}{10}$ $\frac{113}{2}$
Maharashtra $64.5$ $\frac{645}{10}$ $\frac{129}{2}$
Orissa $57.6$ $\frac{576}{10}$ $\frac{288}{5}$
Punjab $66.9$ $\frac{669}{10}$ $\frac{669}{10}$
Rajasthan $59.8$ $\frac{558}{10}$ $\frac{299}{5}$
Tamil Nadu $63.7$ $\frac{637}{10}$ $\frac{637}{10}$
Uttar Pradesh $58.9$ $\frac{589}{10}$ $\frac{589}{10}$
West Bengal $62.8$ $\frac{628}{10}$ $\frac{314}{5}$
Arrangement of the states from the least to the greatest male life expectancy, Haryana, Tamil Nadu, West Bengal, Karnataka, Gujarat, AndhraPradesh, Bihar, Rajasthan, Uttar Pradesh, Orissa, Assam, Madhya Pradesh.
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Question 35 Marks
The overall width in cm of several wide-screen televisions are $97.28\text{cm}\ ,98\frac{4}{9}\text{cm},\ 98\frac{1}{25}\text{cm}$ and $97.94\text{cm}.$ Express these numbers as rational numbers in the form $\frac{\text{p}}{\text{q}}$ and arrange the widths in ascending order.
Answer
We have, width of televisions screen are $97.28\text{cm}\ ,98\frac{4}{9}\text{cm},\ 98\frac{1}{25}\text{cm}$ and $97.94\text{cm}.$ Then, firstly, we convert all widths in the rational number.
$i. 97.28\text{cm}=\frac{9728}{100} [$Remove decimal$]$
$\therefore\frac{\text{p}}{\text{q}}=\frac{2432}{25}\text{cm}$ [numerator and denoninator both dividing by $4$]
$ii. 98\frac{4}{9}\text{cm}=\frac{886}{9}\text{cm} [$convert mixed fraction into simple fraction$]$
$\therefore\frac{\text{p}}{\text{q}}=\frac{886}{9}\text{cm}$
$iii. 98\frac{1}{25}\text{cm}=\frac{2451}{25}\text{cm} [$convert mixed fraction into simple fraction$]$
$\therefore\frac{\text{p}}{\text{q}}=\frac{2451}{25}\text{cm}$
$iv. 97.94\text{cm}=\frac{9794}{100} [$remove decimal$]$
$\therefore\frac{\text{p}}{\text{q}}=\frac{4897}{50}\text{cm} [$numerator and denominator both dividing by $2]$
To arrange in ascending order, firstly we convert all the denominators same, then we get,
$\begin{array}{c|c}2&25,\ 9,\ 25,\ 50\ \\ \hline25&25,\ 9,\ 25,\ 25\\\hline9&\ 1,\ 9,\ \ 1,\ \ 1\ \ \\\hline&1, \ \ 1,\ 1,\ 1\end{array}$
$\therefore$ $\text{LCM}$ of $25, 9, 50 = 2 \times 25 \times 9 = 450$
So, $\frac{2432}{25}=\frac{2432\times18}{25\times18}=\frac{43776}{450},$
$\frac{886}{9}=\frac{886\times50}{9\times50}=\frac{44300}{450}$
$\frac{2451}{25}=\frac{2451\times18}{25\times18}=\frac{44118}{450},$
$\frac{4897}{50}=\frac{4897\times9}{50\times9}=\frac{44073}{450}$ In ascending order,
$\frac{43776}{450}>\frac{44073}{450}>\frac{44118}{450}>\frac{44300}{450}$
i.e. $97.28\text{cm}>97.94\text{cm}<98\frac{1}{25}\text{cm}<98\frac{4}{9}\text{cm}.$
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Question 45 Marks
Here is a table which gives the information about the total rainfall for several months compared to the average monthly rains of a town. Write each decimal in the form of rational number $\frac{\text{p}}{\text{q}}.$
Month
Above/ Below normal $($ in $cm)$
May
$2.6924$
June
$0.6096$
July
$-6.9088$
August
$-8.636$
Answer
$i.$ May $= 2.6924$ $=\frac{26924}{10000}$ [remove decimal]
$\Rightarrow\frac{\text{p}}{\text{q}}=\frac{6731}{2500}\text{cm} [$after dividing numerator and denominator by $4]$
$ii.$ June $= 0.6096$ $=\frac{6096}{10000} [$remove decimal$]$
$\Rightarrow\frac{\text{p}}{\text{q}}=\frac{381}{625}\text{cm} [$after dividing numerator and denominator by $16]$
$iii.$ July $= -6.9088$ $=-\frac{69088}{10000} [$remove decimal$]$
$\Rightarrow\frac{\text{p}}{\text{q}}=-\frac{4318}{625}\text{cm} [$after dividing numerator and denominator by $16]$
$iv.$  August $= -8.636$ $=-\frac{8636}{1000}$
$\Rightarrow\frac{\text{p}}{\text{q}}=-\frac{2159}{250}\text{cm} [$after dividing numerator and denominator by $4]$
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Question 55 Marks
The table shows the portion of some common materials that are recycled.
Material Recycled
Paper $\frac{5}{11}$
Aluminium cans $\frac{5}{8}$
Glass $\frac{2}{5}$
Scrap $\frac{3}{4}$
$a.$ Is the rational number expressing the amount of paper recycled more than $\frac{1}{2}$ or less than $\frac{1}{2}?$
$b.$ Which items have a Recycled amount less than $-\frac{1}{2}?$
$c.$ Is the quantity of aluminium cans recycled more $($ or less$)$ than half of the quantity of aluminium cans?
$d.$  Arrange the rate of recycling the materials from the greatest to the smallest.
Answer
$a.$ Here, $\frac{1}{2}=\frac{1}{2}\times\frac{11}{11}=\frac{11}{22}$
And $\frac{5}{11}=\frac{5}{11}\times\frac{2}{2}=\frac{10}{22}$
So, paper recycled is less than $\frac{1}{2}.$
$b.$ Similarly, $\frac{5}{8}$ is greater than $\frac{1}{2}\Big(=\frac{4}{8}\Big)$
Also, $\frac{2}{5}=\frac{2\times2}{5\times2}$
$\frac{4}{10}<\frac{1}{2}\Big(=\frac{5}{10}\Big)$
And $\frac{3}{4}>\frac{1}{2}\Big(=\frac{2}{4}\Big)$
So, the quantity of paper and glass recycled is less than $\frac{1}{2}.$
$c.$ Quantity of aluminium cans $=\frac{5}{8}\Big(=\frac{10}{16}\Big)$ is more than $\frac{1}{2}$ of the quantity of aluminium cans
$=\frac{5}{8}\times\frac{1}{2}$
$=\frac{5}{16}$
$d.$ Taking $\text{LCM}$ of $11, 8, 5, 4 = 440$
Now, $\frac{5}{11}=\frac{5}{11}\times\frac{40}{40}=\frac{220}{440}$
$\frac{5}{8}=\frac{5}{8}\times\frac{55}{55}=\frac{275}{440}$
$\frac{2}{5}=\frac{2}{5}\times\frac{88}{88}=\frac{176}{440}$
$\frac{3}{4}=\frac{3}{4}\times\frac{110}{110}=\frac{330}{440}$
As, $\frac{330}{440}>\frac{275}{440}>\frac{200}{440}>\frac{176}{440}$
i.e. $\frac{3}{4}>\frac{5}{8}>\frac{5}{11}>\frac{2}{5}$
That meens, scrap $ > $ Aluminium cans $ > $ Paper $ > $ Glass.
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Question 65 Marks
Manavi and Kuber each receives an equal allowance. The table shows the fraction of their allowance each deposits into his/her saving account and the fraction each spends at the mall. If allowance of each is $Rs. 1260$ find the amount left with each.
Where money goes Fraction of allowance
Manavi Kuber
Saving Account $\frac{1}{2}$ $\frac{1}{3}$
Spend at mall $\frac{1}{4}$ $\frac{3}{5}$
Left over ? ?
Answer
Let total cost be $Rs. 1.$
For Manvi,
Left over = Total cost - All spends $=1-\Big(\frac{1}{2}+\frac{1}{4}\Big)$
$=1-\frac{3}{4}$
$=\frac{1}{4}$
$\therefore\ $Amount $=1260\times\frac{1}{4}$
$=\text{Rs}.\ 315$
For Kuber,
Left over = Total cost - All spends $=1-\Big(\frac{1}{3}+\frac{3}{5}\Big)$
$=1-\frac{14}{15}$
$=\frac{1}{15}$
$\therefore\ $Amount $=1260\times\frac{1}{15}$
$=\text{Rs}.\ 84$
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Question 75 Marks
Roller Coaster at an amusement park is $\frac{2}{3}\text{m}$ high. If a new roller coaster is built that is $\frac{3}{5}$ times the height of the existing coaster, what will be the height of the new roller coaster?
Answer
Given, height of the existing roller coaster $=\frac{2}{3}\text{m}$
Height of new roller coaster $=\frac{3}{5}$ of height of the existing roller coaster $=\frac{3}{5}\times\frac{2}{3}$ $=\frac{2}{5}\text{m}$
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