5 Marks Questions

Question 15 Marks

Divide $Rs. 10000$ in two parts, so that the simple interest on the first part for $4$ year at $12\%$ per annum may be equal to the simple interest on the second part for $4.5$ yr at $16\%$ per annum.

Answer

Given, money$= Rs. 10000$
Now, we have divide $Rs. 10000$ in two parts such that $SI$ on first part for $4$ year at $12%$
per annum may be equal to the $SI$ on second part for $4.5$ year at $16%.$
Let first part $= 7 x$ Then, second part $= Rs (10000 - x)$
For first part, we have $P_1=$ Rs. $x, T=4$ year and $R_1=16 \%$
$\therefore \mathrm{SI}_2=\frac{\mathrm{P}_1 \times \mathrm{R}_1 \times \mathrm{T}_1}{100}=\frac{\mathrm{x} \times 12 \times 4}{100}$
For second part $(10000-\mathrm{x})$, We have
$P_2=\text { Rs. }(10000-x), T_2=4.5 \text { year and } R_2=16 \%$
$\therefore S_2=\frac{P_2 \times R_2 \times T_2}{100}=\frac{(10000-x) \times 16 \times 45}{100}$
Since, $\mathrm{SI}_1$ is equal to $\mathrm{SI}_2$.
Then, according to the question,
$\frac{\mathrm{x} \times 12 \times 4}{100}=\frac{(10000-\mathrm{x}) \times 16 \times 45}{100}$
$\Rightarrow 48 \mathrm{x}=(10000-\mathrm{x}) \times 16 \times 45$
$\Rightarrow \frac{48 \mathrm{x}}{45 \times 16}=(10000-\mathrm{x})$
$\Rightarrow \frac{48 \mathrm{x} \times 10}{45 \times 16}=10000-\mathrm{x}$
$\Rightarrow \frac{2}{3} \mathrm{x}=10000-\mathrm{x}$
$\Rightarrow \frac{2}{3} \mathrm{x}+\mathrm{x}=10000$
$\Rightarrow \frac{5 \mathrm{x}}{3}=10000$
$\Rightarrow \mathrm{x}=10000 \times \frac{3}{5}=6000$
First part $=x Rs. 6000$
Second part $=10000-x=10000-6000= Rs. 4000$
Hence, two parts of the sum are $Rs. 6000$ and $Rs. 4000.$

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Question 25 Marks

Imagine that, a $10 \times 10$ grid has value $300$ and that this value is divided evenly among the small squares. In other words, each small square is worth $3.$ Use a new grid for each part of this problem and label each grid “Value $: 300”.$

$a.$ Shade $25\%$ of the grid. What is $25\%$ of $300?$ Compare the two answers.
$b.$ What is the value of $25$ squares?
$c.$ Shade $17\%$ of the grid. What is $17\%$ of $300?$ Compare the two answers.
$d.$ What is the value of $\frac{1}{10}$ of the grid?

Answer

$a.$ We have to shade $25\%$ of the grid, $i.e. \Big (\frac{1}{4}\Big)^\text{th}$ of grid.

$\Big(\frac{1}{4}\Big)^\text{th}$ of grid covers $25$ squares. Since, one square worth $3.$
So, total value of $25$ such squares $= 25 \times 3 \times = 75$
Now, $25\%$ of $300=\frac{25}{100}\times300=25\times3=75$
Hence, the above two values are equal.
$b.$ Value of $25$ squares $= 25 \times 3 = 75.$
$c. 17\%$ of gridd means $17$ squares. So, we will shade $17$ squares.

Total value of these $17$ squares $= 17 \times 3 = 51$
Now, $17\%$ of $300=\frac{17}{100}\times300=17\times3=51$
Hence, the above two values are equal.
$d.$ Value of $\frac{1}{10}$ of the grid in percentage $=\frac{1}{10}\times100\%=10\%$
So, $\frac{1}{10}$ of the grid meeans $10\%$ valuse of $300=\frac{10}{100}\times300=30.$

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Question 35 Marks

Designing a Healthy Diet
When you design your healthy diet, you want to make sure that you meet the dietary requirements to help you grow into a healthy adult. As you plan your menu, follow the following guidelines.
$1.$ Calculate your ideal weight as per your height from the table given at the end of this question.
$2.$ An active child should eat around $55.11$ calories for each kilogram desired weight.
$3. \ 55\%$ of calories should come from carbohydrates. There are $4$ calories in each gram of carbohydrates.
$4. \ 15\%$ of your calories should come from proteins. There are $4$ calories in each gram of proteins.
$5. \ 30\%$ of your calories may come from fats. There are $9$ calories in each gram of fats.
Following is an example to design your own healthy diet.
Example:
$1.$ Ideal weight $= 40\ kg.$
$2.$ The number of calories needed $= 40 \times 55.11 = 2204.4$
$3.$ Calories that should come from carbohydrates
$= 2204.4 \times 0.55 = 1212.42$ calories.
Therefore, required quantity of carbohydrates
$=\frac{1212.40}{4}=303.105\text{g}=300\text{g.}$
$4.$ Calories that should come from proteins
$= 2204.4 \times 0.15 = 330.66$ calories.
Therefore, required quantity of protein
$=\frac{330.66}{4}\text{g}=82.66\text{g}.$
$5.$ Calories that may come from fat $= 2204.4 \times 0.3$
$= 661.3$ calories.
Therefore, required quantity of fat
$=\frac{661.3}{9}\text{g}=73.47\text{g.}$
$1.$ Your ideal desired weight is $.......... kg.$
$2.$ The quantity of calories you need to eat is $........g$
$3.$ The quantity of protein needed is $.........g.$
$4.$ The quantity of fat required is $.........g.$
$5.$ The quantity of carbohydrates required is $.........g.$

		Ideal Height and Weight Proportion
	Men			Women
Height		Weight	Height		Weight
Feet	$cm$	Kilograms	Feet	$cm$	Kilograms
$5’$	$152$	$48$	$4’7”$	$140$	$34$
$5’1”$	$155$	$51$	$4’8”$	$142$	$36$
$5’2”$	$157$	$54$	$4’9”$	$145$	$39$
$5’3”$	$160$	$56$	$4’1”$	$147$	$41$
$5’4”$	$163$	$59$	$4’11”$	$150$	$43$
$5’5”$	$165$	$62$	$5’$	$152$	$45$
$5’6”$	$168$	$65$	$5’1”$	$155$	$48$
$5’7”$	$170$	$67$	$5’2”$	$157$	$50$
$5’8”$	$173$	$70$	$5’3”$	$160$	$52$
$5’9”$	$175$	$73$	$5’4”$	$163$	$55$
$5’10”$	$178$	$75$	$5’5”$	$165$	$57$
$5’11”$	$180$	$78$	$5’6”$	$168$	$59$
$6’$	$1834$	$81$	$5’7”$	$170$	$61$
$6’1”$	$185$	$84$	$5’8”4$	$173$	$64$
$6’2”$	$188$	$86$	$5’9”$	$175$	$66$
$6’3”$	$191$	$89$	$5’10”$	$178$	$68$
$6’4”$	$193$	$92$	$5’11”$	$180$	$70$

Answer

$1.$ Let my height be $5 \ ft.$
The, according to the table, my ideal weight $= 48\ kg.$
$2.$ The quantily of calories needed $= 48 \times 55.11 = 2645.28$ calories.
$3.$ Calorie that should come from proteins $= 2645.28 \times 0.15 = 396.76$ calories
Therefore, requlred qulred quantity of protein
$=\frac{396.76}{4}=99.19\text{g}$
$4.$ Calories thata may come form fats $= 2645.28 \times 0.3 = 793.5$ calories
Therefore, requied quantity of fats
$=\frac{793.5}{9}=88.17\text{g}.$
$5.$ Calorie that should come from carbohydrates $= 2645.28 \times 0.55 = 1454.90$ calories
Therefore, required quilared of carbohydrates
$=\frac{1454.90}{4}$
$=363.72\text{g}$
$=360 ($approx$).$

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Question 45 Marks

A doctor reports blood pressure in millimetres of mercury $(mm \ Hg)$ as a ratio of systolic blood pressure to diastolic blood pressure $($such as $140$ over $80)$. Systolic pressure is measured when the heart beats and diastolic pressure is measured when it rests. Refer to the table of blood pressure ranges for adults.

		Blood Perssure Ranges
	Normal	Prehypertension	Hypertension
Systolic	Under $120\ mm \ Hg$	$120 - 139\ mm \ Hg$	$140\ mm \ Hg$ and above
Diatolic	Under $80\ mm \ Hg$	$80 - 89\ mm \ Hg$	$90\ mm \ Hg$ and above

Manohar is a healthy $37$ year old man whose blood pressure is in the normal category.
$a.$ Calculate an approximate ratio of systolic to diastolic blood pressures in the normal range.
$b.$ If Manohar's systolic blood pressure is $102\ mm \ Hg,$ then use the ratio from part $(a)$ to predict his diastolic blood pressure.
$c.$ Calculate ratio of average systolic to average diastolic blood pressures in the prehypertension category.

Answer

$a.$ Systolic blood pressure in the normal range $= 120\ mm \ Hg$
Diastolic pressure in the normal range $= 80\ mm \ Hg$
Approximate ratio of systolic to diastolic blood pressure
$=\frac{\text{Systolic blood pressure in normal range}}{\text{Diastolic blood pressure in normal range}}=\frac{120}{80}=\frac{3}{2}$
$[$Dividing numerator and denominator by $40]$
Hence, approximate ratio is $3 : 2$
$b.$ Manohar's systolic pressure $= x \ mm \ Hg$
Let diastolic blood pressure $= X \ mm \ Hg$
According to the question,
$\frac{\text{Systolic blood pressure}}{\text{Diastolic blood pressure}}=\frac{3}{2}$
$\Rightarrow\frac{102}{\text{x}}=\frac{3}{2}$
$\Rightarrow\text{x}=\frac{102\times2}{3}$
$\therefore \text{x}=60\ \text{mm Hg.}$
Hence, Manohar's doastolic blood pressure is $68\ mm \ Hg.$
$c.$ Average systolic blood pressure in prehypertension cateegory
$=\frac{120+139}{2}=\frac{259}{2}\ \text{mm Hg}$
Average systolic blood pressure in prehypertension cateegory
$=\frac{80+89}{2}=\frac{169}{2}\ \text{mm Hg}$
Hence, ration of average systolic to average diastolic blood pressures
$=\frac{\text{Average systolic blood Pressure}}{\text{Average diastolic blood Pressure}}$
$=\frac{\frac{259}{2}}{\frac{169}{2}}$
$=\frac{259}{2}\times\frac{2}{169}$
$=\frac{259}{169}$
Hence, required ratio is $259 : 169.$

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Question 55 Marks

Earth Science: The table lists the world’s $10$ largest deserts.

Largest Deserts in the World
Desert	Area $(km^2)$
Sahara $($Africa$)$	$8,800,000$
Gobi $($Asia$)$	$1,300,000$
Australian Desert $($Australia$)$	$1,250,000$
Arabian Desert $($Asia$)$	$850,000$
Kalahari Desert $($Africa$)$	$580,000$
Chihuahuan Desert $($North America$)$	$370,000$
Takla Makan Desert $($Asia$)$	$320,000$
Kara Kum $($Asia$)$	$310,000$
Namib Desert $($Africa$)$	$310,000$
Thar Desert $($Asia$)$	$260,000$

$a.$ What are the mean, median and mode of the areas listed?
$b.$ How many times the size of the Gobi Desert is the Namib Desert?
$c.$ What percentage of the deserts listed are in Asia?
$d.$ What percentage of the total area of the deserts listed is in Asia?

Answer

$a. \text{Meen}=\frac{\text{Total ares of deserts}}{\text{Number of deserts}}$
$=\Bigg[\frac{8800000+1300000+1250000+850000+580000\\+370000+320000+310000+310000+260000}{10}\Bigg]$
$=\frac{14350000}{10}$
$=1435000\ \text{km}^2$
$\text{median}=\frac{\big(\frac{\text{N}}{2}\big)^{th} \ \text{term}+\big(\frac{\text{N}}{2}+1\big)^{th} \ \text{term}}{2}$
$=\frac{\big(\frac{\text{10}}{2}\big)^{th} \ \text{term}+\big(\frac{\text{10}}{2}+1\big)^{th} \ \text{term}}{2}$
$=\frac{5^{th} \text{ term} + 6^{th} \ \text{term}}{2}$
$\frac{580000+370000}{2}$
$=\frac{950000}{2}$
$=475000\ \text{km}^2$
Mode $=$ Most frquent observation $=310000 \ \mathrm{km}^2$
$b.$ Let the size of gobi desert is $x$ times the Namib desert.
$\therefore$ Gobi desert $= x \times $ Namib desert
$\Rightarrow1300000=\text{x}\times310000$
$\text{x}=\frac{1300000}{310000}\Rightarrow\text{x}=4.19$
Hence, the size of Gobi desert is $4.19$ times of Namib desert.
$c.$ Total number of deserts $= 10$
Number of desert in Asia $($Gobi, Arbian, Takla Makan, Kara Kum, Thar $)= 5$
Hence, percentage of deserts in Asia $=\frac{5}{10}\times100\%=50\%$
$d.$ Total number of deserts = $14350000 \ \mathrm{km}^2$
Total area of Asia's deserts $= 1300000 + 850000 + 320000 + 310000 + 260000$
$=3040000 \ \mathrm{km}^2$
Hence, percentage of the total area of the desertss listed in Asia
$=\frac{\text{Total area of Asia deserts}}{\text{Total area of all deserts}}$
$=\frac{3040000}{14350000}\times100\%$
$=21.1\%$

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Question 65 Marks

The pieces of Tangrams have been rearranged to make the given shape.

By observing the given shape, answer the following questions.
$a.$ What percentage of total has been colored?
$i.$ Red $(R) = ........$
$ii.$ Blue $(B) = .........$
$iii.$ Green $(G) = ..........$
$b.$ Check that the sum of all the percentages calculated above should be $100.$
$c.$ If we rearrange the same pieces to form some other shape, will the percentage of colures change?

Answer

$a.$ Total coloured shape
$=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{4}=1$
$i.$ Red coloured shapee
$=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$
$=\frac{1+1+1}{8}$
$=\frac{3}{8}$
Hence, percentage of red coloured
$=\frac{\text{Red coloured shape}}{\text{Total coloured shape}}\times100\%$
$=\frac{\frac{3}{8}}{1}\times100\%$
$=\frac{3}{8}\times100\%$
$=37.5\%$
$ii.$ Bule coloured shape
$=\frac{1}{4}+\frac{1}{4}$
$=\frac{1+1}{4}$
$=\frac{2}{4}$
$=\frac{1}{2}$
Hence, percentage of blue coloured shape
$=\frac{\text{Red coloured shape}}{\text{Total coloured shape}}\times100\%$
$=\frac{\frac{1}{2}}{1}\times100\%$
$=\frac{1}{2}\times100\%$
$=50\%$
$iii.$ Green coloured shape
$=\frac{1}{16}+\frac{1}{16}=\frac{1+1}{16}=\frac{2}{18}=\frac{1}{8}$
Hence, percentage of green coloured shape
$=\frac{\text{Green coloured shape}}{\text{Total coloured shape}}\times100\%$
$=\frac{\frac{1}{8}}{1}\times100\%$
$=\frac{1}{8}\times100\%$
$=12.5\%$
$b.$ Sum of ail percentages calculated
Percentage of red coloured $+$ Percentage of blue coloured $+$ Percentage of green coloued
$= 37.5 + 50 + 12.5$
$= 100\%$
$c.$ If we rearrange the same pieces to torn some other shape, the percentage of colours will not change, because we just rearrange the parts and not changing the percentage of coloures.

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Question 75 Marks

Language Application:Given below are few mathematical terms.

Find:
$a.$ The ratio of consonants to vowels in each of the terms.
$b.$ The percentage of consonants in each of the terms.

Answer

$a.$ In mathematcial term "Hypotenuse".
Number of consonants $= 6, i, e, (h, y, p, t, n, s,)$
Number of vowels $= 4, i, e, (o, e, u, e,)$
Ration of consonants to vowels
$=\frac{\text{Number of consonants}}{\text{Numberr of vowels}}=\frac{6}{4}=\frac{3}{2}$
Hence, ratio is $3 : 2.$
In mathematical term "Congruence",
Number of consonats $= 6, i, e, (c, n, g, rr, n, c,)$
Number of vowels $= 4, i, e, (o, u, e, e,)$
Ratio of consonants to vowels
$=\frac{\text{Number of consonanats}}{\text{Number of vowels}}=\frac{6}{4}=\frac{3}{2}$
Hence, ratio is $3 : 2.$
In mathematical term "Perpendicular"
Number of consonants $= 8, i, e, (p, r, p, n, d, c, l, r,)$
Number of vowels $= 5, i, e, (e, e, i, u, a,)$
Ratio of consonants to vowels
$=\frac{\text{Number of consonanats}}{\text{Number of vowels}}=\frac{8}{5}$
Hence, ratio is $8 : 5.$
In mathematical term "Transversal",
Number of consonants $= 8, i, e, (t, r, n, s, v, r, s, l,)$
Number of vowels $= 3, i, e, (a, e, a,)$
Ratio of consonants to vowels
$=\frac{\text{Number of consonanats}}{\text{Number of vowels}}=\frac{8}{3}$
Hence, ratio is $8 : 3$
In mathmatical term "Correspondence"
Number of consonants $= 9, i, e,(c, r, r, s, p, n, d, n, c,)$
Number of vowels $= 5, i, e, (o, e, o, e, e,)$
Ratio of consonants to vowels
$=\frac{\text{Number of consonanats}}{\text{Number of vowels}}=\frac{9}{5}$
Hence, ratio is $9 : 5.$
$b.$ In mathematical term "Hypotenuse",
Number of consonants $= 6$
Number of vowels $= 4 ($already calculated$)$
Total number of letters $=$ Number of consonants $+$ Number of $= 6 + 4 = 10$
Percentage of consonants
$=\frac{\text{Number of consonanats}}{\text{ Total Number of letters}}\times100\%$
$=\frac{6}{10}\times100\%$
$=60\%$
Hence, percentage of consonants is $60\%$
In mathematical term "Congruence".
Number of consonants $= 6$
Number of vowels $= 4 ($already calculated$)$
Total number of letters $-$ Number of consonants $+$ Number of vowels $= 6 + 4 = 10$
Percentage of consonants
$=\frac{\text{Number of consonanats}}{\text{ Total Number of letters}}\times100\%$
$=\frac{6}{10}\times100\%$
$=60\%$
Hence, percentage of consonants is $60\%.$
In mathematical term "Perpendicular",
Number of consonants $= 8$
Number of vowels $= 5 ($already calculated$)$
Total number of letters $=$ Number of consonants $+$ Number of vowels $= 8 + 5 = 13$
Percentage of consonants
$=\frac{\text{Number of consonanats}}{\text{ Total Number of letters}}\times100\%$
$=\frac{8}{13}\times100\%$
$=6153\%$
Hence, percentage of consonants is $61.53\%.$
In mathematical term "Transversal",
Number of consonants $= 8$
Number of vowels $= 3 ($already calculated$)$
percentage of consonants
$=\frac{\text{Number of consonanats}}{\text{ Total Number of letters}}\times100\%$
$=\frac{8}{11}\times100\%$
$=72.72\%$
Hence, percentage of consonants is $72.72\%$
In mathematical term "Correspondence"
Number of consonants $= 9$
Number of vowels $= 5 ($already calculated$)$
Total number of letters $=$ Number of consonants $+$ Number of vowels $= 9 + 5 = 14$
Percentage of consonants Number of consonants
$=\frac{\text{Number of consonanats}}{\text{ Total Number of letters}}\times100\%$
$=\frac{9}{14}\times100\%$
$=64.28\%$
Hence, percentage of consonants is $64.28\%.$

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