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Maths 5 Marks Questions

Comparing Quantities · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 16 questions.

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Question 15 Marks
Babita bought 160 kg of mangoes at $Rs.\ 48$ per kg. She sold $70\%$ of the mangoes at $Rs.\ 70$ per kg and the remaining mangoes at $Rs.\ 40$ per kg. Find Babita’s gain or loss per cent on the whole dealing.
Answer
Babita bought $160\ kg$ of mangoes $= Rs. 48$ per kg
So, total amount she paid $= Rs. 48 \times 160 = Rs. 7680$
She sold $70%$ of mangoes at $Rs. 70$ per kg
Cost of $70\%$ mangoes$=160\times\frac{70}{100}\times70=70\times16\times7=\text{Rs.}7840$
Remaining mangoes = $=(100-70)\%=30\%$
Cost of remaining at $Rs. 40$ per kg$=160\times\frac{30}{100}\times40=70\times16\times3\times40=\text{Rs.}1920$
Total amount received after selling mangoes $= Rs. 7840 + Rs. 1920 = Rs. 9760\ SP>CP,$
so there is a gain. Gain $= SP - CP$
$Rs. 9760 - Rs. 7680 = Rs. 2080$
$\text{Gain%}=\frac{\text{Gain}}{\text{CP}}\times100$
$\text{Gain%}=\frac{\text{2080}}{\text{7680}}\times100=27.08\%$
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Question 25 Marks
A real estate agent receives $Rs. 50,000$ as commission, which is $4\%$ of the selling price. At what price does the agent sell the property?
Answer
Let the price of the property be $Rs. x$
Commission received $= Rs. 50000$
Commission percentage $= 4\%$
According to the question,$\text{x}\times\frac{4}{100}=50000$
$\text{x}=\frac{50000\times100}{4}=50000\times25=\text{Rs.} \ 1250000$
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Question 35 Marks
Devangi’s phone subscription charges for the period $17-02-09$ to $16-03-09$ were as follows :
Period
Amount (in Rs.)
Service Tax $\%$
$17-02-09$ to $23-02-09$
$199.75$
$12$
$24-02-09$ to $16-03-09$
$599.25$
$10$
Find the final bill amount if $3\%$ education cess was also charged on service tax.
Answer
On the basis of above given table,
Amount for period $17-02-09$ to $23-02-09 = Rs. 199.75$
Amount with service tax $12\%$$=199.75+\frac{12}{100}\times199.75$
$=199.75+23.97=\text{Rs}. \ 223.72$
Amount for period $24-02-09$ to $16-03-09 = Rs. 599.25$
Amount with service tax $@10\%=599.25+599.25\times\frac{10}{100}$
$=599.25+59.925=\text{Rs}. \ 659.175$
Total bill amount = $\text{Rs.} \ 223.72+\text{Rs.} \ 659.175=\text{Rs.} \ 882.895$
Total bill amount including education cess of $3\% = 882.895 + 3\%$ of $882.895$
$=882.895+\frac{3}{100}\times882.895$
$=882.895+26.486=\text{Rs.} \ 909.39$
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Question 45 Marks
Activity
Liters per person per day
Drinking
$3$
Cooking
$4$
Bathing
$20$
Sanitation
$40$
Washing cloths
$40$
Washing utensils
$20$
Gardening
$23$
Total
$150$
$a.$ What per cent of water is used for bathing and sanitation together per day?
$b.$ How much less per cent of water is used for cooking in comparison to that used for bathing?
$c.$ What per cent of water is used for drinking, cooking and gardening together?
Answer
$a.$ On the basis of given details, water used for bathing per day $= 20\ L$
Water used for sanitation $= 40\ L$
Total water used per day $= 150\ L$
Percentage of water is used for bathing $\&$ sanitation together per day
$=\frac{20+40}{150}\times100=\frac{60}{150}\times100$
$\frac{600}{15}=40\%$
$b.$ Water used for cooking per day $= 4L$
Water used for bathing per day $= 20L$
Difference $\frac{\text{b}}{\text{w}}$ water used for cooking $\&$ bathing $= 20 - 4 = 16L$
$= 16 L$ water less used for cooking in comparison of bathing
In percentage$=\frac{16}{150}\times100=\frac{160}{15}=\frac{32}{3}\%$
$c.$ Water used for drinking per day $= 3L$
Water used for cooking per day $= 4L$
Water used for gardening per day $= 23L$
Water used for drinking, cooking $\&$ gardening together $= 3 + 4 + 23 = 30L$
In percentage$=\frac{30}{150}\times100=\frac{100}{5}=20\%$
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Question 55 Marks
Arunima bought household items whose marked price and discount $\%$ is as follows:
 
 
Item
Quantity
Rate
Amount
Discount$\%$
$(a)$
Atta
$1$ packet
$200$
$200$
$16\%$
$(b)$
Detergent
$1$ packet
$371$
$371$
$22.10\%$
$(c)$
Namkeen
$1$ packet
$153$
$153$
$18.30%$
 
Find the total amount of the bill she has to pay.
Answer
On the basis of the given data,
Rate of $1$ packet of atta $= Rs. 200$
Discount$\% = 16\%$
So, price$=200-\frac{16}{100}\times200=200-32=\text{Rs}.168$
 Discount$\% = 22.10\%$
So, price$=371-371\times\frac{22.10}{100}=371-81.991=\text{Rs.}289.009$
Rate of $1$ packet of namkeen $= 153$ Discount$\% = 18.30\%$
 So, price$=153-153\times\frac{18.30}{100}=153-153\times18.30$$=153-27.999=\text{Rs.} \ 125.001$
Total bill amount = $\text{Rs}.168+\text{Rs}.289.009$$+\text{Rs}. \ 125.001=\text{Rs}. \ 582.01$
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Question 65 Marks
Find the difference between Compound Interest and Simple Interest on $Rs. 45,000$ at $12\%$ per annum for $5$ years.
Answer
Principal $(P) = Rs. 45000$
Rate of interest $(R)= 12\%$ per annum
Time period $(T) = 5$ yr
Simple interest ,$\text{SI}=\frac{\text{P}\times\text{R}\times\text{T}}{100}=\frac{45000\times12\times5}{100}$
$=450\times60=\text{Rs}. \ 27000$
Compound interest, $CI = A - P$ Where, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{T}$$\text{A}=45000\Big(1+\frac{\text{12}}{100}\Big)^\text{5}=45000\Big(\frac{28}{25}\Big)^5$
$=45000\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25}$
$=\frac{45000\times17210368}{9765625}=45000\times1.76$
Compound interest , $\text{CI}=\text{Rs.}79200$$-\text{Rs}. \ 45000=\text{Rs.} \ 34200$
Difference between $\text{SI} \ \text{and} \ \text{CI}=\text{Rs.} \ 34200$$-\text{Rs}. \ 27000=\text{Rs.} \ 72000$
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Question 75 Marks
Given the principal $= Rs. 40,000$, rate of interest $= 8\% p.a.$ compounded annually. Find
$a.$ Interest if period is one year.
$b.$ Principal for $2^{nd}$ year.
$c.$ Interest for $2^{nd}$ year.
$d.$ Amount if period is $2$ years.
Answer
Given, principal $(P)= Rs. 40000$
Rate of interest $(R) = 8\%$ per annum
$a.$ Compound Interest for $1$ year
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{40000}\Big(1+\frac{\text{8}}{100}\Big)^{\text{1}}$
$=40000\times\frac{108}{100}$
Amount, $A = 400 \times 108 = Rs. 43200$
Compound Interest, $CI = A - P = Rs. 43200 - Rs. 40000 = Rs. 3200$
$b.$ Amount of $1^{st}$ year $=$ Principal of $2^{nd}$ year $= Rs. 4320$
$c.$ Now, for $2^{nd}$ year
Principal $= Rs. 43200$
Rate of interest, $R = 8\%$ per annum
Time, $n = 1$ year
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Question 85 Marks
With the decrease in prices of tea by $15\%$ Tonu, the chaiwallah, was able to buy $2\ kg$ more of tea with the same $Rs. 45$ that he spent each month on buying tea leaves for his chai shop. What was the reduced price of tea? What was the original price of tea?
Answer
Let chaiwallah purchase $y$ kg tea.
Let price of tea per kg $= Rs. x$
Discount $15\%$ per kg$=\text{x}-\frac{15}{100}\times\text{x}=\frac{85\text{x}}{100}$
Chailwallah can buy $2\ kg$ extra with $15\%$ discount.
But without discount,$xy = 45$
$\Big(\text{x}-\frac{15}{100}\Big)\times\text{x}=\Big(\frac{85\text{x}}{100}\Big)(\text{x+2})=45$
After solving Eqs. $(i)$ and $(ii)$
$\frac{85}{100}(45)+\frac{85\times2\text{x}}{100}=45$
$\text{45}\Big[1-\frac{85}{100}\Big]=\frac{85\times2\text{x}}{100}$
$\frac{45\times15}{100}=\frac{85\times2\text{x}}{100}$
$\text{x}=\frac{45\times15}{2\times85}=\frac{135}{34} \ \text{per} \ \text{kg}$$=3.97 \ \text{per} \ \text{kg}$
Reduced price$=\frac{85}{100}\times3.97=\frac{337.45}{100}$$=3.3748=3.38 \ \text{per} \ \text{kg}$
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Question 95 Marks
Sunscreens block harmful ultraviolet $\text{(UV)}$ rays produced by the sun. Each sunscreen has a Sun Protection Factor $\text{(SPF)}$ that tells you how many minutes you can stay in the sun before you receive one minute of burning $\text{UV}$ rays. For example, if you apply sunscreen with $\text{SPF}\ 15$, you get $1$ minute of $\text{UV}$ rays for every $15$ minutes you stay in the sun.
$1.$ A sunscreen with $\text{SPF}$ $15$ allows only$\frac{1}{15}$ of the sun’s $\text{UV}$ rays. What per cent of $\text{UV}$ rays does the sunscreen abort?
$2.$ Suppose a sunscreen allows $25\%$ of the sun’s $\text{UV}$ rays.
$a.$ What fraction of $\text{UV}$ rays does this sunscreen block? Give your answer in lowest terms.
$b.$ Use your answer from Part $(a)$ to calculate this sunscreen’s $\text{SPF}$. Explain how you found your answer.
$3.$ A label on a sunscreen with $\text{SPF}\ 30$ claims that the sunscreen blocks about $97\%$ of harmful $\text{UV}$ rays. Assuming the $\text{SPF}$ factor is accurate, is this claim true? Explain.
Answer
$i.$ A sunscreen with $\text{SPF}\ \ 15$ allows only $\frac{1}{15}$ of the sun’s $\text{UV}$ rays
It means$=1-\frac{1}{15}=\frac{14}{15}$
In percentage$\frac{\frac{14}{15}}{1}\times100=\frac{1400}{15}=93.333\%$
$ii.$
$a.$ Sunscreen allows $25\%$ of the sun’s $\text{UV}$ rays
It blocks $\text{UV}$ rays$=100-25=75\%=\frac{75}{100}=\frac{3}{4}$
$b.$ Sunscreen allows $25\%$ on $\frac{3}{4}$ of $\text{UV}$ rays. It means, it protect$=1-\frac{3}{4}=\frac{1}{4}$ of $\text{UV}$ rays Hence, it’s a $\text{SPF}\ \ 4$.
$iii.$ False,
According to the claim, for $\frac{3}{100}$ affect of $\text{UV}$ rays
$1$ minute$=33\frac{1}{3}\text{SPF}$
Affect $\neq\ 30\ \text{SPF}$
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Question 105 Marks
Ashima sold two coolers for $Rs. 3,990$ each. On selling one cooler she gained $5\%$ and on selling the the other she suffered a loss of 5%. Find her overall gain or loss $\%$ in whole transaction.
Answer
Selling price of each cooler $= Rs. 3990$
Let $Rs. x$ be the cost price of both coolers for Ashima If Ashima gets $5\%$ gain on transaction of first cooler.
Then, $3990=\text{x}+\text{x}\times\frac{5}{100}$
$3990=\frac{21\text{x}}{20}$
$\text{x}=\frac{3990\times20}{21}$
$=190\times20=\text{Rs.} \ 3800$
Also, on the another transaction of other cooler she has a loss of $5\%$.
Then, $3990=\text{x}-\text{x}\times\frac{\text{x}\times5}{100}$$3990=\frac{19\text{x}}{20}$
$\text{x}=\frac{3990\times20}{19}$
So, the total cost price for Ashima $= Rs. 3800 + Rs. 4200 = Rs. 8000$
Ashima sold both coolers $= Rs. 3900 \times 2 = Rs. 7980$
Here, $CP > SP$. So, Ashima has loss on overall transaction.
$\text{Loss}=\text{Rs.} \ 8000-\text{Rs.} \ 7980=\text{Rs.} \ 7980$
$\text{Loss}\%=\frac{20}{8000}\times100=\frac{2000}{8000}$$=\frac{1}{4}\%=0.25\%$
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Question 115 Marks
If the price of petrol, diesel and $LPG$ is slashed as follows:
Fuel
Old price / Liters litre (In Rs.)
New price / litre (In Rs.)
$\%$ Decrease
Petrol/ $L$
$45.62$
$40.62$
________
Diesel/ $L$
$32.86$
$30.86$
________
$LPG/14.2\ kg$
$304.70$
$279.70$
________
Complete the above table.
Answer
$\text{Decrease}\%=\frac{5}{45.62}\times100=$$\frac{50000}{4562}=10.96\%$For Diesel/$L$
Old price $= RS. 32.86$
New price $= Rs. 30.86$
Decrement in price = $=\text{Rs.} \ 32.56-\text{Rs.} \ 30.86=\text{Rs.} \ 2$
$\text{Decrease}\%=\frac{2}{32.86}\times100$$=\frac{200}{32.86}=6.0869=6.09\%$
For $LPG$,
Old price $= Rs. 304.70$
New price $= Rs. 279.70$
Decrement in price $=\text{Rs.} \ 304.70-\text{Rs.} \ 279.70=\text{Rs.} \ 25$
$\text{Decrease}\%=\frac{25}{304.70}\times100$$=\frac{2500}{304.70}=8.20\%$
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Question 125 Marks
A lady buys some pencils for $Rs. 3$ and an equal number for $Rs. 6.$ She sells them for $Rs. 7$. Find her gain or loss $\%$.
Answer
Let the lady buys $x$ number of pencils for $Rs. 3$
Cost price for one pencil, $\text{CP}_1=\text{Rs}. \ \frac{3}{\text{x}}$
Also, she buys same number of pencils for $Rs.6$
Cost price for one pencil, $\text{CP}_2=\text{Rs}. \ \frac{6}{\text{x}}$
Now, total pencils $=\text{x}+\text{x}=\text{2x}$
She sells, $2x$ pencils $\text{Rs.} \ 7$
Selling price of $1$ pencil $=\text{Rs}. \ \frac{7}{\text{x}}$
Case 1: We know that, Gain $= SP - CP$
$\text{Gain}=\text{Rs.} \ \Big(\frac{7}{2\text{x}}-\frac{3}{\text{x}}\Big)=\text{Rs}. \ \Big(\frac{7-6}{2\text{x}}\Big)$
$\text{Rs}. \ \frac{1}{2\text{x}}$
$\text{Gain}\%=\frac{\text{Gain}}{\text{CP}}\times100=\frac{\frac{1}{2\text{x}}}{\frac{3}{\text{x}}}\times100$$=\frac{100}{3\times3}=\frac{50}{3}\%$
Case 2: Loss $= CP - SP$
$\frac{6}{\text{x}}-\frac{7}{\text{2x}}=\frac{6\times2-7}{2\text{x}}=\text{Rs.} \ \frac{5}{2\text{x}}$
$\text{Gain}\%=\frac{\text{Gain}}{\text{CP}}\times100$
$=\frac{\frac{5}{2\text{x}}}{\frac{6}{\text{x}}}\times100=\frac{250}{6}=\frac{125}{3}\%$
$\text{Net gain}\%=\frac{125}{3}-\frac{50}{3}=\frac{75}{3}=25\%$
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Question 135 Marks
The food labels given below give information about $2$ types of soup: cream of tomato and sweet corn. Use these labels to answer the given questions. $($All the servings are based on a $2000$ calorie diet.$)$
$a.$ Which can be measured more accurately : the total amount of fat in cream of tomato soup or the total amount of fat in sweet corn soup? Explain.
$b.$ One serving of cream of tomato soup contains $29\%$ of the recommended daily value of sodium for a $2000$ calorie diet. What is the recommended daily value of sodium in milligrams? Express the answer upto $2$ decimal places.
$c.$ Find the increase per cent of sugar consumed if cream of tomato soup is chosen over sweet corn soup.
$d.$ Calculate ratio of calories from fat in sweet corn soup to the calories from fat in cream of tomato soup.
Answer
$a.$ Serving size of $1$ cup of sweet corn $= 240\ mL$
Total fat $= 2\ g$ on $2\%$
$2\%$ of fat in sweet corn of $1$ cup $($with $2$ serving per container$)$
$=\frac{2}{100}\times240$
$=\frac{48}{10}=4.8\text{g}$
For one serving$=\frac{4.8}{2}=2.4\text{g}$
Serving size of $1$ cup of cream of tomato $= 240\ mL$
Total fat $= 2g$ on $3\%$
$3\ %$ of fat in cream tomato of $1$ cup$=\frac{3}{100}\times240$
$=\frac{72}{10}=7.2\text{g}$
For one serving$=\frac{7.2}{2}=3.6\text{g}$
Hence, $2.4g$ is nearest to $2g$ in comparison of $3.6g$ to $3g$.
$b.$ On the basis of given information in the question
the cream of tomato soup contains $29\%$ of the recommended daily value of sodium for a $2000$ calories diet.
where, $29\%$ of $2000$ calories $= 690\ mg$
$c.$ If cream of tomato soup is chosen over sweet corn soup, the increase in sugar
consumed $= 11g - 5g = 6g$
$\text{Increase}\%=\frac{6}{5}\times100=\frac{600}{5}=120\%$
$d.$ Fat in sweet corn soup in calories $= 9$Fat
Fat in cream tomato soup in calories $= 20$
$\text{Ratio}=\frac{9}{20}=\frac{9}{21}=\frac{3}{7}=3:7$
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Question 145 Marks
If principal $= Rs. 1,00,000$. rate of interest $= 10\%$ compounded half yearly. Find
$a.$ Interest for $6$ months.
$b.$ Amount after $6$ months.
$c.$ Interest for next $6$ months.
$d.$ Amount after one year.
Answer
Principal $(P) = Rs. 100000$ Rate of interest $(R) = 10\%$ compounded half yearly
$a.$ Interest for $6$ months
Compound Interest, $CI = A − P$
Where, $\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{n}}$
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{n}}=100000\Big(1+\frac{10}{200}\Big)^{1}$
$=100000\times\frac{21}{20}=\text{Rs.}105000$
Compound interest, $\text{CI}= \ \text{A}-\text{P}=\text{Rs.} \ 105000$$-\text{Rs.} \ 100000=\text{Rs.} \ 5000$
$b.$ Amount after $6$ months $= Rs. 105000$
$c.$ Interest for next $6$ months
Principal $=$ Amount after $6$ months
Principal $(P) = Rs. 105000$
Rate of interest $(R)= 10\%$
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{n}}$
$\text{A}=105000\Big(1+\frac{\text{10}}{200}\Big)^{\text{1}}$
$105000\times\frac{21}{20}$
$\frac{2205000}{20}=\text{Rs.}110250$
Compound interest, $\text{CI}= \ \text{A}-\text{P}=\text{Rs.} \ 110250$$-\text{Rs.} \ 105000=\text{Rs.} \ 5250$
$d.$ Amount after $1$ year = $=\text{Rs.} \ 110250$
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Question 155 Marks
What is the percentage increase or decrease in the number of seats bwon by $A, B, C$ and $D$ in the general elections of $2009$ as compared to the results of $2004$?
Political party
Number of seats won in 2004
Number of seats won in 2009
$A$
$206$
$145$
$B$
$116$
$138$
$C$
$4$
$24$
$D$
$11$
$12$
 
Answer
For political party A Number of seats won in $2004 = 206$
Number of seats won in $2009 = 145$
Decrement in the number of seats won by party $A = 206-145=61$
$\text{Decrease}\%=\frac{61}{206}\times100=29.61\%$
For political party $B$, Number of seats won in $2004= 116$
Number of seats won in $2009 = 138$
Increment in the number of seats won by party $B = 138 - 116 = 22$
$\text{Increase}\%=\frac{22}{116}\times100=18.96\%$
For political party $C$, Number of seats won in $2004 = 4$
Number of seats won in $2009 = 24$
Increment in the number of seats won by party $C = 24 - 4 = 20$
$\text{Increase}\%=\frac{20}{4}\times100=500\%$
For political party $D$,
 Number of seats won in $2004 = 11$
Number of seats won in $2009 = 12$
Increment in the number of seats won by party $D = 12 - 11 = 1$
$\text{Increase}\%=\frac{1}{11}\times100=9.09\%$
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Question 165 Marks
With the decrease in prices of tea by $15%$ Tonu, the chaiwallah, was able to buy $2\ kg$ more of tea with the same $Rs. 45$ that he spent each month on buying tea leaves for his chai shop. What was the reduced price of tea? What was the original price of tea?Assessment Report for $2009-2010$
Class: 9B Name: Vidit Atrey Date: $31$ March $2010$
Subject Internal assessment Examination Total Final%
English Literature $\frac{20}{25}$ $\frac{82}{100}$ $\frac{102}{125}$  
English Language $\frac{22}{25}$ $\frac{91}{100}$ $\frac{113}{125}$  
Hindi Literature $\frac{18}{25}$ $\frac{67}{75}$ $\frac{85}{100}$  
Hindi Language $\frac{16}{25}$ $\frac{68}{75}$ $\frac{84}{100}$  
Mathematics $\frac{42}{50}$ $\frac{88}{100}$ $\frac{130}{150}$  
Sanskrit $\frac{14}{20}$ $\frac{75}{100}$ $\frac{99}{120}$  
Physics $\frac{45}{50}$ $\frac{90}{100}$ $\frac{135}{150}$  
Chemistry $\frac{41}{50}$ $\frac{82}{100}$ $\frac{123}{150}$  
Biology $\frac{43}{50}$ $\frac{87}{100}$ $\frac{130}{150}$  
History and Civics $\frac{19}{25}$ $\frac{68}{75}$ $\frac{87}{100}$  
Geography $\frac{17}{20}$ $\frac{71.5}{80}$ $\frac{88.5}{100}$  
Answer
Calculate final percentage of each subject English literature$=\frac{102}{125}\times100=\frac{102}{5}\times4=\frac{408}{5}=81.6\%$
English language$=\frac{113}{125}\times100=\frac{113\times4}{5}=\frac{425}{5}=90.4\%$
Hindi literature$=\frac{84}{100}\times100=84\%$
Hindi Language$=\frac{84}{100}\times100=84\%$
Mathematics$=\frac{130}{150}\times100=\frac{1300}{15}=86.67\%$
Sanskrit$=\frac{89}{120}\times100=\frac{890}{12}=74.16\%$
Physics$=\frac{135}{100}\times100=\frac{1350}{15}=90\%$
Chemistry$=\frac{123}{150}\times100=\frac{1230}{15}=82\%$
Biology$=\frac{130}{150}\times100=\frac{1300}{15}=86.66\%$
History & Civics$=\frac{87}{100}\times10=87\%$
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