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Maths 3 Mark Question

Ratio and Proportion · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 30 questions.

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Question 13 Marks
Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.
Answer
Ratio in two numbers = 3 : 4
LCM = 180
Let first number = 3x
Then second number = 4x
Now LCM = 3 × 4 × x = 12x
12x = 180
⇒ x = 15
Numbers will be 3 × 15 = 45 and 4 × 15 = 60
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Question 23 Marks
Two numbers are in the ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.
Answer
Ratio in two numbers = 7 : 11
Let first number = 7x
Then second number = 11x
Then adding 7 to each number, the ratio is 2 : 3
$\frac{7\text{x}+7}{11\text{x}+7}=\frac{2}{3}$
By cross multiplying:
3(7x + 7) = 2(11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 21 - 14 = 22x - 21x
⇒ x = 7
First number = 7x = 7 × 7 = 49
And second number = 11x = 11 × 7 = 77
Hence numbers are 49, 77
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Question 33 Marks
Two numbers are in the ratio 5 : 7. If the sum of the numbers is 720, find the numbers.
Answer
Sum of two numbers = 720
Ratio of two numbers = 5 : 7
Let first number = 5x
Then second number = 7x
5x + 7x = 720
⇒ 12x = 720
⇒ x = 60First number = 5x = 5 × 60 = 300
And second number = 7x = 7 × 60 = 420
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Question 43 Marks
Divide Rs. 360 between Kunal and Mohit in the ratio 7 : 8
Answer
Total amount = Rs. 360
Sum of ratios = 7 + 8 = 15
$\therefore$ Kunal's share $ \frac{360\times7}{15}=24\times7$
$=\text{Rs. }168$
Mohit's share $=\frac{360\times8}{15}=24\times8$
$=\text{Rs. }192$
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Question 53 Marks
If A : B = 5 : 6, and B : C = 4 : 7, find A : B : C.
Answer
$\text{A : B}=5:6$ and $\text{B}:\text{C}=4:7$
$\frac{\text{A}}{\text{B}}=\frac{5}{6},\frac{\text{B}}{\text{C}}=\frac{4}{7}$
LCM of 6, 4 = 12
$\therefore\frac{\text{A}}{\text{B}}=\frac{5}{6}=\frac{5\times2}{6\times2}=\frac{10}{12}$ or
$\text{A : B}=10:12$
And $\frac{\text{B}}{\text{C}}=\frac{4}{7}=\frac{4\times3}{7\times3}=\frac{12}{21}$ or
$\text{B : C}=12 : 21$
$\therefore\text{A : B : C}=10:12 : 21$
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Question 63 Marks
Which ratio is greater:
(2 : 3) or (4 : 7)
Answer
2 : 3 or 4 : 7
$\Rightarrow\frac{2}{3}$ or $\frac{4}{7}$
LCM of 3, 7 = 21
$\therefore\frac{2}{3}=\frac{2\times7}{3\times7}=\frac{14}{21}$
$\frac{4}{7}=\frac{4\times3}{7\times3}=\frac{12}{21}$
It is clear that $\frac{14}{21}$ is greater
$\therefore\frac{2}{3}>\frac{4}{7}$ or $(2 : 3)>(4 : 7)$
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Question 73 Marks
Which ratio is greater:
(1 : 2) or (4 : 7)
Answer
$(1 : 2)$ or $(4 : 7)$
$\Rightarrow\frac{1}{2}$ or $\frac{4}{7}$
LCM of 2, 7 = 14
$\therefore\frac{1}{2}=\frac{1\times7}{2\times7}=\frac{7}{14}$
$\frac{4}{7}=\frac{4\times2}{7\times2}=\frac{8}{14}$
It is clear that $\frac{8}{14}$ is greater
$\therefore\frac{8}{14}>\frac{7}{14}$ or $\frac{4}{7}>\frac{1}{2}$
Or $(4 : 7)>(1 : 2)$
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Question 83 Marks
The ratio of copper and zinc in an alloy is 9 : 5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.
Answer
Ratio in copper and zinc = 9 : 5
Let alloy = x gm
Copper $=\frac{\text{x}}{9+5}\times9=\frac{9\text{x}}{14}$
And zinc $=\frac{5\text{x}}{14}$
According to the condition,
$\frac{9\text{x}}{14}=48.6$
$\Rightarrow\text{x}=\frac{48.6\times14}{9}$
$\Rightarrow\text{x}=5.4\times14\text{g}=75.6\text{g}$
and weight of zinc $=\frac{5\text{x}}{14}=\frac{5\times75.6}{14}$
$=27\text{g}$
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Question 93 Marks
What number must be added to each term of the ratio 9 : 16 to make the ratio 2 : 3?
Answer
Let x be added to each term Then
9 + x : 16 + x = 2 : 3
$\frac{9+\text{x}}{16+\text{x}}=\frac{2}{3}$
$\Rightarrow3(9+\text{x})=2(16+\text{x})$
(By cross multiplication)
$\Rightarrow27+3\text{x}=32+2\text{x}$
$\Rightarrow3\text{x}-2\text{x}=32-27$
$\Rightarrow\text{x}=5$
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Question 103 Marks
Divide Rs. 880 between Rajan and Kamal in the ratio $\frac{1}{5}:\frac{1}{6}$
Answer
Total amount = Rs. 880
Ratio in Rajan and Kamal $\frac{1}{5}:\frac{1}{6}$
$=\frac{6\ :\ 5}{30}$
$\therefore$ Rajan's part $=\text{Rs.}=\frac{880\times6}{6+5}$
$=\text{Rs. }\frac{880\times6}{11}=\text{Rs. }480$
And kamal's part $=\text{Rs. }\frac{880\times5}{11}$
$=\text{Rs. }400$
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Question 113 Marks
Divide Rs. 5600 between A, B and C in the ratio 1 : 3 : 4.
Answer
Total amount = Rs. 5600
Ratio in A : B : C = 1 : 3 : 4
$\therefore$ A's share $=\text{Rs. }\frac{5600\times1}{1+3+4}$
$=\text{Rs. }\frac{5600\times1}{8}=\text{Rs. }700$
B's share $=\text{Rs. }\frac{5600\times3}{8}=\text{Rs. }2100$
And C's share $=\text{Rs. }\frac{5600\times4}{8}=\text{Rs. }2800$
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Question 123 Marks
If A : B = 3 : 5, B : C = 10 : 13, find A : B : C.
Answer
$\text{A : B} = 3 : 5, \text{B : C} = 10 : 13$
Or $\frac{\text{A}}{\text{B}}=\frac{3}{5}\times\frac{2}{2}=\frac{6}{10}$
$(\because\text{in }\frac{\text{B}}{\text{C}}=\text{B}=10)$
$\therefore\frac{\text{A}}{\text{B}}=\frac{6}{10}$ and $\frac{\text{B}}{\text{C}}=\frac{10}{13}$
$\text{A : B : C}=6:10:13$
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Question 133 Marks
Express the following ratios in the simplest form:
$6\frac{2}{3}:7\frac{1}{2}$
Answer
$6\frac{2}{3}:7\frac{1}{2}$ or $\frac{20}{3}:\frac{15}{2}$
LCM of 3, 2 = 6
$\therefore\frac{20\times2}{3\times2}=\frac{40}{6}$ and $\frac{15\times3}{2\times3}=\frac{45}{6}$
$\therefore40:45$
HCF of 40 and 45 = 5
$\therefore$ Dividing by 5
$40\div5:45\div5=8:9$
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Question 143 Marks
What number must be subtracted from each of the numbers $10,12,19,24$ to get the numbers which are in proportion?
Answer
Let $x$ be subtracted from each of the given number, then
$10- x , 12- x , 19- x$ and $24- x$ are in proportion
$\frac{10-x}{12-x}=\frac{10-x}{24-x}$
By cross multiplication :
$(10-x)(24-x)=(19-x)(12-x)$
$\Rightarrow 240-10 x-24 x+x^2=228-19 x-12 x+x^2$
$\Rightarrow 240-34 x+x^2=228-31 x+x^2$
$\Rightarrow-34 x+x^2+31 x-x^2=228-240$
$\Rightarrow-3 x=-2$
$\Rightarrow 3 x=12$
$\Rightarrow x=4$
Required number $=4$
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Question 153 Marks
Compare 4 : 5 and 7 : 9.
Answer
The given fraction are $\frac{4}{5}$ and $\frac{7}{9}$
LCM of 5 and 9 = 5 × 9 = 45
Now, We have:
$\frac{4\times9}{5\times9}=\frac{36}{45}$ and $\frac{7\times5}{9\times5}=\frac{35}{45}$
Clearly, $\frac{35}{45}<\frac{36}{45}$
Hence, $(7 : 9)<(4:5)$
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Question 163 Marks
What number must be added to each of the numbers 5, 9, 7, 12 to get the numbers which are in proportion?
Answer
Let $x$ be added to each of the given numbers then
$5+x, 9+x, 7+x, 12+x$ are in proportion $\frac{5+ x }{9+ x }=\frac{7+ x }{12+ x }$
$\frac{5+\text{x}}{9+\text{x}}=\frac{7+\text{x}}{12+\text{x}}$
By cross multiplication :
$(5+x)(12+x)=(7+x)(9+x)$
$\Rightarrow 60+5 x+12 x+x^2=63+7 x+9 x+x^2$
$\Rightarrow 60+17 x+x^2=63+16 x+x^2$
$\Rightarrow 17 x+x^2-16 x-x^2=63-60$
$\Rightarrow x=3$
Required number $=3$
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Question 173 Marks
What number must be subtracted from each term of the ratio 17 : 33 so that the ratio becomes 7 : 15?
Answer
Let x be subtracted from each term Then
(17 - x) : (33 - x) = 7 : 15
$\Rightarrow\frac{17-\text{x}}{33-\text{x}}=\frac{7}{15}$
(By cross multiplication)
$15(17-\text{x})=7(33-\text{x})$
$255-15\text{x}=231-7\text{x}$
$\Rightarrow255-231=-7\text{x}+15\text{x}$
$\Rightarrow24=8\text{x}$
$\Rightarrow\text{x}=\frac{24}{8}=3$
$\therefore$ Required number = 3
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Question 183 Marks
If A : B = 5 : 8 and B : C = 16 : 25, find A : C.
Answer
A : B = 5 : 8, B : C = 16 : 25
Or $\frac{\text{A}}{\text{B}}=\frac{5}{8},\frac{\text{B}}{\text{C}}=\frac{16}{25}$
Multiplying, We get:
$\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{5}{8}\times\frac{16}{25}$
$\frac{\text{A}}{\text{C}}=\frac{2}{5}$
$\therefore\text{A}:\text{C}=2:5$
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Question 193 Marks
Divide Rs. 1100 among A, B and C in the ratio 2 : 3 : 5.
Answer
The sum of ratio terms is = 2 + 3 + 5 = 10
Then, we have:
A’s share $=₹\ \frac{2}{10}\times1100=₹\ 220$
B’s share $=₹\ \frac{3}{10}\times1100=₹\ 330$
C’s share $=₹\ \frac{5}{10}\times1100=₹\ 550$
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Question 203 Marks
Which ratio is greater:
(5 : 6) or (7 : 9)
Answer
$(5 : 6)$ or $(7 : 9)$
$\Rightarrow\frac{5}{6}\text{ or }\frac{7}{9}$
LCM of 6, 9 = 18
$\therefore\frac{5}{6}=\frac{5\times3}{6\times3}=\frac{15}{18}$
$\frac{7}{9}=\frac{7\times2}{9\times2}=\frac{14}{18}$
It is clear that $\frac{15}{18}$ or 15 : 18 is greater
$\therefore(5 : 6)>(7 : 9)$
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Question 213 Marks
Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.
Answer
The ratio in two numbers = 5 : 9
Let the first number = 5x
Then second number = 9x
By subtracting 3 from each number the ratio is 1 : 2
$\frac{5\text{x}-3}{9\text{x}-3}=\frac{1}{2}$
By cross multiplication,
2(5x - 3) = 1(9x - 3)
⇒ 10x - 6 = 9x - 3
⇒ 10x - 9x = -3 + 6
⇒ x = 3
First number = 5x = 5 × 3 = 15
and second .number = 9x = 9 × 3 = 27
Hence numbers are 15, 27
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Question 223 Marks
The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present age.
Answer
Ratio in present ages of A and B = 8 : 3
Let A’s age = 8x
Then B’s age = 3x
6 years hence,
A’s age will be = 8x + 6
and B’s will be = 3x + 6
$\frac{8\text{x}+6}{3\text{x}+6}=\frac{9}{4}$
(By cross multiplication)
4(8x + 6) = 9(3x + 6)
⇒ 32x + 24 = 27x + 54
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s present age = 8x = 8 × 6 = 48 years
and B’s age = 3x = 3 × 6 = 18 years
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Question 233 Marks
The scale of a map is 1 : 5000000. What is the actual distance between two towns, if they are 4cm apart on the map?
Answer
Scale of map = 1 : 5000000
Distance between two town on the map = 4cm
$\therefore\text{Actual distance}=\frac{5000000\times4}{1}\text{cm}$
$=\frac{5000000\times4}{100}\text{m}$
$=\frac{5000000\times4}{100\times1000}\text{km}=200\text{km}$
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Question 243 Marks
If 36, 54, x are in continued proportion, find the value of x.
Answer
36, 54, x are in continued proportion
36 : 54 : : 54 : x
⇒ 36 × x = 54 × 54
$\Rightarrow\text{x}=\frac{54\times54}{36}$
$\text{x}=\frac{54\times3}{2}=27\times2=81$
$\therefore\text{x}=81$
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Question 253 Marks
The ages of A and B are in the ratio 4 : 3. Eight years ago, their ages were in the ratio 10 : 7. Find their present ages.
Answer
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively
Eight years ago, age of A = (4x - 8) yrs
Eight years ago, age of B = (3x - 8) yrs
Then,
(4x - 8) : (3x - 8) = 10 : 7
$\Rightarrow\frac{4\text{x}-8}{3\text{x}-8}=\frac{10}{7}$
⇒ 28x - 56 = 30x - 80
⇒ 2x = 24
⇒ x = 12
(4x - 8) = 4 × 12 - 8 = 40 years
(3x - 8) = 3 × 12 - 8 = 28 years
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Question 263 Marks
If A : B = 7 : 5 and B : C = 9 : 14, find A : C.
Answer
$\text{A}:\text{B}=7:5$
$\Rightarrow\frac{\text{A}}{\text{B}}=\frac{7}{5}$
$\text{B}:\text{C}=9:14$
$\Rightarrow\frac{\text{B}}{\text{C}}=\frac{9}{14}$
Multiplying we get,
$\therefore\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{7}{5}\times\frac{9}{14}$
$\Rightarrow\frac{\text{A}}{\text{C}}=\frac{9}{10}$
$\therefore\text{A}:\text{C}=9:10$
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Question 273 Marks
A bag contains Rs. 750 in the form of rupee, 50P and 25P coins in the ratio 5 : 8 : 4. Find the number of coins of each type.
Answer
Total amount = Rs. 750
Ratio in rupee, 50 P and 25 P coins = 5 : 8 : 4
Let number of rupees = 5x
Number of 50 P coins = 8x
And number of 25 coins = 4x
According to the condition,
$5\text{x}+\frac{8\text{x}}{2}+\frac{4\text{x}}{4}=750$
$\Rightarrow20\text{x}=16\text{x}+4\text{x}=3000$
$\Rightarrow40\text{x}=3000$
$\Rightarrow\text{x}=\frac{3000}{40}=75$
Number of 1 Re coins = 5x = 5 × 75 = 375
Number of 50 P coins = 8x = 8 × 75 = 600
and number of 25 P coins = 4x = 4 × 75 = 300
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Question 283 Marks
Which ratio is greater:
(3 : 5) or (8 : 13)
Answer
$(3:5)$ or $(8 : 13)$
$\Rightarrow\frac{3}{5}$ or $\frac{8}{13}$
LCM of 5, 13 = 65
$\therefore\frac{3}{5}=\frac{3\times13}{5\times13}=\frac{39}{65}$
$\frac{8}{13}=\frac{8\times5}{13\times5}=\frac{40}{65}$
It is clear that $\frac{40}{65}$ is greater
$\therefore\frac{40}{65}>\frac{39}{65}$ or $\frac{8}{13}>\frac{3}{5}$
Or $(8 : 13)>(3:5)$
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Question 293 Marks
At a certain time a tree 6m high casts a shadow of length 8 metres. At the same time a pole casts a shadow of length 20 metres. Find the height of the pole.
Answer
Height of a tree = 6cm
And its shadow at same time = 8m
Shadow of a pole = 20m
Let height of pole = x m
6 : 8 = x : 20
$\Rightarrow\text{x}=\frac{6\times20}{8}=15\text{m}$
Height of pole = 15m
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Question 303 Marks
If 27, 36, x are in continued proportion, find the value of x.
Answer
27, 36, x are in continued proportion
27 : 36 :: 36 : x
27 × x = 36 × 36
$\Rightarrow\text{x}=\frac{36\times36}{27}$
$\text{x}=\frac{4\times36}{3}=4\times12=48$
$\therefore\text{x}=48$
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