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Maths 3 Mark Question

Ratio, Proportion and Unitary Method · Maharashtra Board STD 6 (MSBSHSE - Semi English Medium). 59 questions.

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Question 13 Marks
Show that the following numbers are in continued proportion:
48, 60, 75
Answer
The given numbers 48, 60, 75 are in
Continued proportion if 48 : 60 :: 60 : 75.
Now, product of means = 60 × 60 = 3600
And, product of extremes = 48 × 75 = 3600
$\therefore$ Product of means = Product of extremes
So, 48 : 60 :: 60 : 75
Hence, the numbers 48, 60, 75 are in continued proportion.
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Question 23 Marks
Find the ratio of the price of a pencil to that of a ball pen, if pencil cost Rs. 96 per score and ball pens cost Rs. 50.40 per dozen.
Answer
Price of 20 pencils = Rs. 96
(1 score = 20 pencils)
Price of 1 pencil = Rs. (96 ÷ 20)
= Rs. 4.80
Price of 12 ball pens = Rs. 50.40
(1 dozen = 12)
Price of 1 ball pen = Rs. (50.40 ÷ 12)
= Rs. 4.20.
Ratio of the price of a pencil to that of a ball pen = Rs. 4.80 : Rs. 4.20
= 480 paise : 420 paise
= 480 : 420
= 48 : 42
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Question 33 Marks
Prove that (5 : 6) > (3 : 4).
Answer
We can write:
$(5:6)=\frac{5}6{}\text{ and }(3 : 4)=\frac{3}{4}$
By making their denominators same: (Taking the L.C.M of 6 and 4, which is 24)
Consider, 5 : 6
$\frac{5\times4}{6\times4}=\frac{20}{24}$
And $\frac{3\times6}{4\times6}=\frac{18}{24}$
As $20>18$
Clearly, $(5:6)>(3:4)$
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Question 43 Marks
Determine if the following ratios form a proportion:
25cm : 1m and Rs. 40 : Rs. 160
Answer
25cm : 1m and Rs. 40 : Rs. 160
$=\frac{25\text{cm}}{1000\text{cm}}=\frac{1}{4}$
$=\frac{\text{Rs. }40}{\text{Rs. }160}=\frac{1}{4}$
$\because\frac{3}{5}=\frac{3}{5}$ (Ratios are equal)
$\therefore$ 39 litres : 65 litres and 6 bottles : 10
Bottles are in proportion
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Question 53 Marks
40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?
Answer
Number of men needed to finish a piece of work in 26 days = 40
Number of men needed to finish it in 1 day = 26 × 40 = 1040 (less days means more men)
Number of men needed to finish it in 16 days $=\frac{1040}{16}=65$
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Question 63 Marks
Two numbers are in the ratio 11 : 12 and their sum is 460, find the numbers.
Answer
Let the numbers be 11x and 12x.
Then, 11x + 12x = 460
$\Rightarrow23\text{x}=460$
$\Rightarrow\frac{23\text{x}}{23}=\frac{460}{23}$
(Dividing both sides by 23)
$\Rightarrow\text{x}=20.$
$\therefore$ Required numbers are (11 × 20) and 12 × 20,that is 220 and 240
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Question 73 Marks
Divide Rs. 3450 among A, B and C in the ratio 3 : 5 : 7.
Answer
Total amount = Rs. 3450
Ratio in A, B and C shares = 3 : 5 : 7
Sum of share = 3 + 5 + 7 = 15
$\therefore\text{A's share}=\text{Rs. }\frac{3450\times3}{15}$
$=\text{Rs. }230\times3=\text{Rs. }690$
$\text{B's share}=\text{Rs.}\frac{3450\times5}{15}$
$=\text{Rs. }230\times5=\text{Rs. }1150$
$\text{C's }\text{share}=\text{Rs. }\frac{3450\times7}{15}$
$=\text{Rs. }230\times7=\text{Rs. }1610$
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Question 83 Marks
Find the value of x in the following proportions:
x : 92 :: 87 : 116
Answer
We have
Product of means = 85 × 57
Product of extremes = 51 × x
$\therefore 51\times\text{x}=85\times57$
$\Rightarrow\text{x}=\frac{85\times57}{51}=95$
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Question 93 Marks
Express the following ratios in the simplest form:
777 : 1147
Answer
The given ratio $=777:1147=\frac{777}{1147}$
To express it in simplest form, we divide the numerator and denominator by the HCF of 777 and 1147.
$=\frac{777\div37}{1147\div37}=\frac{21}{31}$

$(\text{HCF of 777 and 1147 is 37})$
$=21:31$
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Question 103 Marks
Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per month respectively. Find the ratio of:
Mr Sahai’s income to his wife’s income.
Answer
Earning of Sahai = Rs. 16800
And of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
Ratio in sahai's income and his wife
16800 : 10500
$=\frac{16800}{10500}=\frac{16800\div2100}{10500\div2100}$
$(\text{HCF of 16800 and 105000 is 2100})$
$=\frac{8}{5}=8:5$
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Question 113 Marks
Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of:
His income and savings.
Answer
Rohit monthly earnings = Rs. 15300
And his savings = Rs. 1224
So, his expenditure = Rs. 15300 - 1224=
= Rs. 14076
Now,
Ratio in his income and savings
$=15300:1224=\frac{15300}{1224}$

$=\frac{15300\div612}{1224\div612}$
(Dividing by HCF of 15300 and 1224)
$=\frac{25}{2}=25:2$
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Question 123 Marks
Write (T) for true and (F) for false in case of the following:
36 : 45 :: 80 : 100
Answer
True
Solution:
We have,
$36:45=\frac{36}{45}=\frac{4}{5}$
And $80:100=\frac{80}{100}=\frac{4}{5}$
$\therefore36:45=80:100$
So, given statemant is true
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Question 133 Marks
Write the following ratios in the simplest form:
Rs. 6.30 : Rs. 16.80
Answer
The given ratio = Rs. 6.30 : Rs. 16.80
$=\frac{\text{Rs. }6.30}{\text{Rs. }16.80}$
$=\frac{630}{1680}$
To express it in simplest form, we divide the numerator and denominator by the HCF of 630 and 1680.
Now, HCF of 630 and 1680 = 210

$\therefore\frac{630}{1680}=\frac{630\div210}{1680\div210}=\frac{3}{8}$
$\therefore\text{Required ratio}= 3:8$
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Question 143 Marks
Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per month respectively. Find the ratio of:
Mr Sahai’s income to the total income of the two.
Answer
Earning of Sahai = Rs. 16800
and of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
Sahai's income and total of their income
$=16800:27300=\frac{16800}{27300}$
$=\frac{16800\div2100}{27300\div2100}=\frac{8}{13}=8:13$
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Question 153 Marks
Express the following ratios in the simplest form:
480 : 384
Answer
The given ratio $=480:384=\frac{480}{384}$ To express it in simplest form, we divide the numerator and denominator by the HCF of 480 and 384. Now, HCF of 480 and 384 = 96
$\therefore\frac{480}{384}=\frac{480\div96}{384\div96}=\frac{5}{4}$ $\therefore\text{Required ratio}=5:4$
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Question 163 Marks
Determine if the following ratios form a proportion:
200mL : 2.5L and Rs. 4 : Rs. 50
Answer
200mL : 2.5L and Rs. 4 : Rs. 50
$\frac{200\text{mL}}{2.5\text{L}}=\frac{200}{2.5\times1000}$
$=\frac{200\times10}{25\times1000}=\frac{2}{25}$
and $\frac{\text{Rs. }4}{\text{Rs. }50}=\frac{2}{25}$ (Ratios are equal)
200mL : 2.5L and Rs. 4 : Rs. 50 are in proportion.
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Question 173 Marks
Express the following ratios in the simplest form:
85 : 561
Answer
The given ratio is $85:561=\frac{85}{561}$
To express in it simplest form, we divided the numerator and denominator by the HCF of 85 and 561.
By the HCF of 85 and 561.
Now, HCF of 85 and 561 = 17

$\therefore\frac{85}{561}=\frac{85\div17}{561\div17}=\frac{5}{33}$
$\therefore \text{Required ratio}=5.33$
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Question 183 Marks
Divide Rs. 1575 between Kamal and Madhu in the ratio 7 : 2.
Answer
Total amount = Rs. 1575
Ratio in Kamal and Madhu’s share = 7 : 2
Sum of ratios = 7 + 2 = 9
$\therefore$ kamal's share $=\text{Rs. }1225$
$\text{Rs. }175\times7=\text{Rs. }1225$
Madhu's share $=\text{Rs.}\frac{1575\times2}{9}$
$=\text{Rs. }175\times2=\text{Rs. }350$
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Question 193 Marks
Express the following ratios in the simplest form:
186 : 403
Answer
The given ratio $=186:403=\frac{186}{403}$ To express it in simplest form, we divide the numerator and denominator by the HCF of 186 and 403. Now, HCF of 186 and 403 = 31
$\therefore\frac{186}{403}=\frac{186\div31}{403\div31}=\frac{6}{13}$ $\therefore\text{Required ratio}=6:13$
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Question 203 Marks
Write the following ratios in the simplest form:
48 min : 2 hours 40 min.
Answer
The given ratio = 48 min : 2 hours 40 min.
= 48 min : (2 × 60 + 40) min
= 48 min : (120 + 40) min
= 48 min : 160 min
= 48 : 160
= 3 :10
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Question 213 Marks
The ratio of the number of male and female workers in a textile mill is 5 : 3. If there are 115 male workers, what is the number of female workers in the mill?
Answer
Let the number of male and female workers in the mill be 5x and 3x respectively.
Then,
5x = 115
$\Rightarrow\frac{5\text{x}}{5}=\frac{115}{5}$
(Dividing both sides by 5)
$\Rightarrow \text{x} = 23$
Number of female workers in the mill
$=3\text{x}$
$=3\times23=69$
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Question 223 Marks
Find the value of x in the following proportions:
55 : 11 :: x : 6
Answer
We have 55 : 11 :: x : 6
Product of means = 11 × x = 11x
Product of extremes = 55 × 6 = 330
$\therefore11\text{x}=330\Rightarrow\frac{11\text{x}}{11}=\frac{330}{11}$
(Dividing both sides by 11)
$\Rightarrow\text{x}=30$
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Question 233 Marks
Express the following ratios in the simplest form:
36 : 90
Answer
$36 : 90$$=\frac{36}{90}$
$=\frac{36\div18}{90\div18}$
$(\text{HCF of 36, 90 = 18})$
$=\frac{2}{5}=2:5$
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Question 243 Marks
The boys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448. find the number of girls.
Answer
Let the number of boys and girls in the school be 9x and 5x respectively. According to the question,
$9\text{x}+5\text{x}=44$
$\Rightarrow14\text{x}=448$
$\Rightarrow\frac{14\text{x}}{14}=\frac{448}{14}$
(Dividing both sides by 14)
$\Rightarrow\text{x}=32$
Number of girls $=5\text{x}$
$=5\times32$
$=160$
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Question 253 Marks
Write (T) for true and (F) for false in case of the following:
30 bags : 18 bags :: Rs. 450 : Rs. 270
Answer
True
Solution:
We have, 30 bags : 18 bags
$=30:18=\frac{30}{18}=\frac{5}{3}$
$\text{And Rs. }450 : \text{Rs. } 270$
$\therefore 30\text{ bags}:18\text{ Rs. }270$
$=450:270=\frac{450}{270}=\frac{5}{3}$
$\therefore30\text{bags}:18\text{ bags}=\text{Rs. }450:270$
So, the given statement is true.
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Question 263 Marks
If $9, x, x, 49$ are in proportion, find the value of x.
Hint. $x^2 = (9 \times 49) = (32 \times 72) = (3 \times 7)^2 = (21)^3$
Answer
It is given that $9, x, x, 49$ are in proportion, that is, $9 : x :: x : 49$
$\therefore$ Product of means = Product of extremes
$\text{x}\times\text{x}=9\times49$
$\Rightarrow\text{x}^2=9\times49$
$\Rightarrow\text{x}=\sqrt{9\times49}$
$\Rightarrow\text{x}=\sqrt{3\times3\times7\times7}$
$\Rightarrow\text{x}=3\times7=21$
$\therefore\text{x}=21$
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Question 273 Marks
Divide Rs. 1400 among A, B and C in the ratio 2 : 3 : 5
Answer
A : B : C = 2 : 3 : 5
Sum of ratio = 2 + 3 + 5
Total money = Rs. 1400
Then share of A = 2 × Rs. 1400 = Rs. 2800 = Rs. 280
Share of B = 3 × Rs. 1400 = Rs. 4200 = Rs. 420
Share of C $=\frac{5}{10}\times\text{Rs. }1400$
$=\text{Rs. }\frac{7000}{10}=\text{Rs. }700$
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Question 283 Marks
Show that the following numbers are in continued proportion:
36, 90, 225
Answer
The given numbers 36, 90, 225 are in
Continued proportion of 36 : 90 :: 90 : 225
Now, product of means = 90 × 90 = 8100
And, product of extremes = 36 × 225 = 8100
$\therefore$ Product of means = Product of extremes
So, 36 : 90 :: 90 : 225
Hence, the numbers 36, 90, 225 are in continued proportion.
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Question 293 Marks
The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs. 1520.
Answer
Ratio in income and savings of a family = 11 : 2
But Total savings = Rs. 1520
Let income = x
11 : 2 = x : 1520
$\Rightarrow\text{x}=\frac{11\times1520}{2}=11\times760$
$=\text{Rs. }8360$
Expenditure = total income - savings
$=\text{Rs. }8360-1520$
$=\text{Rs. }6840$
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Question 303 Marks
In an army camp, there were provisions for 425 men for 30 days. How long did the provisions last for 375 men?
Answer
Number of days for which provisions last for 425 men = 30 days
Number of days for which provisions last for 1 men = 30 × 425 = 12750 days. (less men means more days)
Number of days for which provisions last for 375 men $=\frac{12750}{375}=34\text{ days}$
Hence, provisions will last for 34 days for 375 men.
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Question 313 Marks
The ratio of zinc and copper in an alloy is 7 : 9 If the weight of copper in the alloy is 12.6kg find the weight of zinc in it.
Answer
Ratio of zinc and copper in an alloy is 7 : 9
Let the weight of zinc and copper in it be (7x) and (9x), respectvely.
Now, the weight of a copper = 12.6kg (given)
$\therefore9\text{x}=12.6$
$\Rightarrow\text{x}=\frac{12.6}{9}=1.4$
$\therefore$ Weight of zinc = 7x = 7 × 1.4 = 9.8kg
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Question 323 Marks
From each of the given pairs, find which ratio is larger:
(3 : 4) or (9 : 16)
Answer
(3 : 4) or (9 : 16)
LCM of 4, 16 = 16
$\therefore\frac{3}{4}=\frac{3\times4}{4\times4}=\frac{12}{16}$
We see that $\frac{12}{16}>\frac{9}{16}\text{ or }\frac{3}{4}>\frac{9}{16}$
$\therefore$ 3 : 4 the greater ratio
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Question 333 Marks
From the given pairs, find which ratio is larger:
(5 : 12) or (17 : 30)
Answer
(5 : 12) or (17 : 30)
$\frac{5}{12},\frac{17}{30}$
LCM of 12 and 30 = 60
$\therefore\frac{5}{12}=\frac{5\times5}{12\times5}=\frac{25}{60}$
$\text{and }\frac{17}{30}=\frac{17\times2}{30\times2}=\frac{34}{60}$
We see that $\frac{34}{60}>\frac{25}{60}\text{or }\frac{17}{30}>\frac{5}{12}$
$\therefore17:30$ is greater ratio
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Question 343 Marks
A 35cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.
Answer
Length of line segment = 35cm
Ratio = 4 : 3
Sum of ratio = 4 + 3 = 7
$\therefore\text{First part}=\frac{35\times4}{7}=20\text{cm}$
$\text{Second part }= \frac{35\times3}{7}=15\text{cm}$
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Question 353 Marks
Find the value of x in the following proportions:
51 : 85 :: 57 : x
Answer
We have 51 : 85 :: 57 : x
Product of means = 85 × 57
Product of extremes = 51 × x
$\therefore 51\times\text{x}=85\times57$
$\Rightarrow\text{x}=\frac{85\times57}{51}=95$
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Question 363 Marks
From the given pairs, find which ratio is larger:
(1 : 2) or (13 : 27)
Answer
(1 : 2) or (13 : 27)$\frac{1}{2},\frac{13}{27}$
$\text{LCM of 2 and 27 = 54}$
$\therefore\frac{1}{2}=\frac{1\times27}{2\times27}=\frac{27}{54}$
$\text{and }\frac{13}{27}=\frac{13\times2}{27\times2}=\frac{26}{54}$
We see that $\frac{27}{54}>\frac{26}{54}\text{ or }\frac{1}{2}>\frac{13}{27}$
$\therefore1:2$ is greater ratio
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Question 373 Marks
The 1st, 3rd and 4th terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.
Answer
1st term =12, third term = 8 and fourth term = 14 Let 2nd term = x, Then,
$12:\text{x}::9:14$ $\therefore\text{x}=\frac{12\times14}{8}$ $\Big(\text{b}=\frac{\text{ad}}{\text{c}}\Big)$ $=21$$\therefore\text{second term}=21$
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Question 383 Marks
Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of:
His income and expenditure.
Answer
Rohit monthly earnings = Rs. 15300 And his savings = Rs. 1224 So, his expenditure = Rs. 15300 - 1224= = Rs. 14076 Now, Ratio in his income and expenditure$=1500:14076=\frac{15300}{14076}$
$=\frac{15300\div612}{14076\div612}$ $(\text{HCF}=612)$
$=\frac{25}{23}=25:23$
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Question 393 Marks
Show that the following numbers are in continued proportion:
16, 84, 441
Answer
The given numbers 16, 84, 441 are in
Continued proportion if 16 : 84 :: 84 : 441.
Now, product of means = 84 × 84 = 7056
And, product of extremes = 16 × 441 = 7056
Product of means = Product of extremes.
So, 16 : 845 :: 84 : 441
Hence 16, 84, 441 are in continued proportion.
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Question 403 Marks
A bus covers 128km in 2 hours and a train covers 240km in 3 hours. Find the ratio of their speeds.
Answer
A bus covers in 2 hours = 128km
128 It will cover in 1 hour $=\frac{128}{2}=64\text{km}$
A train cover in 3 hours $= 240\text{km}$
It will cover in 1 hour $=\frac{240}{3}$
$=80\text{km}$
Ratio in their speeds = 64: 80
= 4 : 5
{Dividing by 16, the LCM of 64, 80}
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Question 413 Marks
Find the value of x when $36 : x :: x : 16$
Answer
Given:
$36 : x :: x : 16$
We know:
Product of means = Product of extremes
$x \times x = 36 \times 16$
$\Rightarrow x^2 = 576$
$\Rightarrow x^2 = 24^2$
$\Rightarrow x = 24$
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Question 423 Marks
A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.
Answer
Total bulbs produced per day = 630
Out of every 10 bulbs, defective bulb = 1
Out of every 10 bulbs, lighting bulbs = 10 - 1 = 9
$\therefore$ Ratio of defective bulbs to lighting bulbs = 1 : 9
Sum of the terms of the ratio = (1 + 9) = 10
$\therefore$ Number of defective bulbs produced each day$=\Big(\frac{1}{10}\times630\Big)=63.$
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Question 433 Marks
Rohit earns Rs. 15300 and saves Rs. 1224 per month. Find the ratio of:
His expenditure and savings.
Answer
Rohit monthly earnings = Rs. 15300
And his savings = Rs. 1224
So, his expenditure = Rs. 15300 - 1224=
= Rs. 14076
Now,
His expenditure and savings
$=14076:1224=\frac{14076}{1224}$
$=\frac{14076\div612}{1224\div612}=\frac{23}{2}$
$=23:2$
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Question 443 Marks
Find the value of x in the following proportions:
27 : x :: 63 : 84
Answer
We have 27 : x :: 63 : 84
Product of means = x × 63
Product of extremes = 27 × 84
$\therefore\text{x}\times63=27\times84$
$\Rightarrow\text{x}=\frac{27\times84}{63}=36$
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Question 453 Marks
In a proportion the 1st, 2nd and 4th terms are 51, 68 and 108 respectively. Find the 3rd term.
Answer
Let the third term be x.
Then 51 : 68 :: x : 108
Now, product of means = x × 68
And, product of extremes = 51 × 108
x × 68 = 51 × 108
$\Rightarrow\text{x}=\frac{51\times108}{68}$
$\Rightarrow3\times27=81$
$\text{x}=81$
Hence the third term of the given proportion is 81
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Question 463 Marks
Write the following ratios in the simplest form:
1L 35mL : 270mL
Answer
The given ratio = 1L 35mL : 270mL = (1 × 1000 + 35)mL : 270mL = 1035mL : 270mL = 1035 : 270 $=\frac{1035}{270}$ To express it in simplest form, we divide its numerator and denominator by the HCF of 1035 and 270.
Now, HCF and 1035 and 270 = 45$\therefore\frac{1035}{270}=\frac{1035\div45}{270\div45}=\frac{23}{6}$
$\therefore\text{Required ratio}=23:6$
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Question 473 Marks
The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 11.7kg., find the weight of zinc in it.
Answer
It is given that the ratio of zinc and copper in an alloy is 7 : 9.
If the weight of zinc in the alloy is 7kg then the weight of copper in the alloy is 9kg.
Now, if the wieght of copper is 9kg
Then weight of zinc = 7kg
If the wieght of copper is 1kg then
Weight of zinc $=\frac{7}{9}\text{kg}$
If the wieght of copper is 11.7kg then
Weight of zinc
$=\Big(\frac{7}{9}\times11.7\Big)\text{kg}$
$=(7\times1.3)\text{kg}$
$=9.1\text{kg}$
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Question 483 Marks
Determine if the following ratios form a proportion:
39 litres : 65 litres and 6 bottles : 10 bottles
Answer
39 litres : 65 litres and 6 bottles : 10 bottles$\frac{39}{65}=\frac{3}{5}$
$\text{and }\frac{6}{10}=\frac{3}{5}$
$\because\frac{3}{5}=\frac{3}{5}$ (Ratios are equal)
$\therefore$ 39 litres : 65 litres and 6 bottles : 10
Bottles are in proportion
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Question 493 Marks
The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.
Answer
It is given that the ratio of the length of a field to its width is 5 : 3.
If the width of the field is 3 metres then length = 5 metres.
If the width of the field is 1 metres than length $=\frac{5}{3}\text{metres}$
If the width of the field is 42 metres then length
$=\frac{5}{3}\times14\text{ metres}$
$=5\times14\text{ metres}$
$=70 \text{ metres}$
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Question 503 Marks
Write the following ratios in the simplest form:
3 weeks : 30 days
Answer
The given ratio = 3 weeks : 30 days
= (3 × 7) days : 30 days
= 21 days : 30 days
= 21 : 30
= 7 : 10
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Question 513 Marks
An electric pole casts a shadow of length 20 metres at a time when a tree 6 metres high casts a shadow of length 8 metres. Find the height of the pole.
Hint. (height of the tree) : (length of the shadow)
= (height of the pole) : (length of its shadow).
Answer
Let the height of the pole be x metres.
$\text{Now,}\frac{\text{height of the pole}}{\text{length of shadow of pole}}$
$=\frac{\text{height of thr tree}​​}{\text{Length of shadow of tree}}$
$\therefore\frac{\text{x}}{20}=\frac{6}{8}$
$\Rightarrow\text{x}=\frac{6}{8}\times20$
$=\frac{120}{8}=15$
$\therefore$ Height of the pole = 15 metres
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Question 523 Marks
Show that 48, 60, 75 are in continued proportion.
Answer
Consider 48 : 60 : : 60 : 75
Product of means = 60 × 60 = 3600
Product of extremes = 48 × 75 = 3600
So product of means = Product of extremes
Hence, 48, 60, 75 are in continued proportion.
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Question 533 Marks
The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs. 14000.
Answer
Ratio in income and expenditure = 7 : 6
Total income = Rs. 14000
Let expenditure = x, then
$7 : 6 : : 14000 : \text{x}$
$\Rightarrow\text{x}=\frac{6\times14000}{7}=\text{Rs. }12000$
Savings = Total income - Expenditure
$= \text{Rs. } 14000 - 12000$
$= \text{Rs. } 2000$
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Question 543 Marks
Write the following ratios in the simplest form:
3m 5cm : 35cm.
Answer
To given ratio = 3m 5cm : 35cm
= (3 × 100 + 5)cm : 35cm
= 305cm : 35cm
= 305 : 35
= 61 : 7
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Question 563 Marks
From the given pairs, find which ratio is larger:
(3 : 7) or (4 : 9)
Answer
(3 : 7) or (4 : 9)
$\Rightarrow\frac{3}{7},\frac{4}{9}$
$\text{LCM of }7,9=63$
$\therefore\frac{3}{7}=\frac{3\times9}{7\times9}=\frac{27}{63}$
$\frac{4}{9}=\frac{4\times7}{9\times7}=\frac{28}{63}$
We see that $\frac{28}{63}>\frac{27}{63}\text{ or }\frac{4}{9}>\frac{3}{7}$
$\therefore4:9$ is greater ratio.
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Question 573 Marks
Mr Sahai and his wife are both school teachers and earn Rs. 16800 and Rs. 10500 per month respectively. Find the ratio of:
Mrs Sahai’s income to her husband’s income.
Answer
Earning of Sahai = Rs. 16800
And of his wife = Rs. 10500
Then total income = Rs. 16800 + 10500
= Rs. 27300
Ratio in sahai's wife income and her husband's income.
= 10500 : Rs. 16800
= 5 : 8
(Dividing by the HCF of 10500 and 16800 = 2100)
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Question 583 Marks
Write the following ratios in the simplest form:
4kg : 2kg 500g
Answer
The given ratio = 4kg : 2kg 500g
= (4 × 1000)g : (2 × 1000 + 500)g
= 4000g : 2500g
= 4000 : 2500
= 40 : 25
= 8 : 5
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Question 593 Marks
Determine if the following ratios form a proportion:
2kg : 80kg and 25g : 625kg
Answer
2kg : 80kg and 25g : 625kg
$\frac{2\text{kg}}{80\text{kg}}=\frac{1}{40}$
And $\frac{25\text{g}}{625\text{kg}}=\frac{25}{625\times1000}$
$=\frac{1}{25000}$
$\because\frac{1}{40}\neq\frac{1}{25000}$
$\therefore$ 2kg : 80kg and 25g : 625kg are not in proportion.
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