Question 13 Marks
Two numbers are in the ratio $3 : 4.$ If their $LCM$ is $180,$ find the numbers.
AnswerRatio in two numbers $= 3 : 4$
$LCM = 180$
Let first number $= 3x$
Then second number $= 4x$
Now $LCM = 3 \times 4 \times x = 12x$
$12x = 180$
$\Rightarrow x = 15$
Numbers will be $3 \times 15 = 45$ and $4 \times 15 = 60$
View full question & answer→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.
AnswerRatio 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: $3(7x + 7) = 2(11x + 7)$
$\Rightarrow 21x + 21 = 22x + 14$
$\Rightarrow 21 - 14 = 22x - 21x$
$\Rightarrow x = 7$
First number $= 7x = 7 \times 7 = 49 $ And
second number $= 11x = 11 \times 7 = 77 $
Hence numbers are $49, 77$
View full question & answer→Question 33 Marks
Two numbers are in the ratio $5 : 7.$ If the sum of the numbers is $720,$ find the numbers.
AnswerSum of two numbers $= 720$
Ratio of two numbers $= 5 : 7$
Let first number $= 5x$
Then second number $= 7x$
$5x + 7x = 720$
$\Rightarrow 12x = 720$
$\Rightarrow x = 60$
First number $= 5x$
$= 5 \times 60 = 300$ And second number $= 7x$
$= 7 \times 60 = 420$
View full question & answer→Question 43 Marks
Divide $Rs. 360$ between Kunal and Mohit in the ratio $7 : 8$
AnswerTotal 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 $=\frac{360\times8}{15}=24\times8$
$=\text{Rs. }192$
View full question & answer→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
$\text{B : C}=12 : 21$
$\therefore\text{A : B : C}=10:12 : 21$
View full question & answer→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 $(2 : 3)>(4 : 7)$
View full question & answer→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 $(4 : 7)>(1 : 2)$
View full question & answer→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.
AnswerRatio 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 $=\frac{5\text{x}}{14}=\frac{5\times75.6}{14}$
$=27\text{g}$
View full question & answer→Question 93 Marks
What number must be added to each term of the ratio $9 : 16$ to make the ratio $2 : 3?$
AnswerLet $x$ 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)
$\Rightarrow27+3\text{x}=32+2\text{x}$
$\Rightarrow3\text{x}-2\text{x}=32-27$
$\Rightarrow\text{x}=5$
View full question & answer→Question 103 Marks
Divide $Rs. 880$ between Rajan and Kamal in the ratio $\frac{1}{5}:\frac{1}{6}$
AnswerTotal 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 $=\text{Rs. }\frac{880\times5}{11}$
$=\text{Rs. }400$
View full question & answer→Question 113 Marks
Divide $Rs. 5600$ between $A, B$ and $C$ in the ratio $1 : 3 : 4.$
AnswerTotal 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 $=\text{Rs. }\frac{5600\times4}{8}=\text{Rs. }2800$
View full question & answer→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
$\frac{\text{B}}{\text{C}}=\frac{10}{13}$
$\text{A : B : C}=6:10:13$
View full question & answer→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 $5$
$40\div5:45\div5=8:9$
View full question & answer→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$?$
AnswerLet $x$ be subtracted from each of the given number, then
$10-\mathrm{x}, 12-\mathrm{x}, 19-\mathrm{x}$ and $24-\mathrm{x}$ are in proportion
$\frac{10-\mathrm{x}}{12-\mathrm{x}}=\frac{10-\mathrm{x}}{24-\mathrm{x}}$
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 $=4$
View full question & answer→Question 153 Marks
Compare $4 : 5$ and $7 : 9.$
AnswerThe given fraction are $\frac{4}{5}$ and $\frac{7}{9}$
$LCM$ of $5$ and $9 = 5 \times 9 = 45$
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, $(7 : 9)<(4:5)$
View full question & answer→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$?$
AnswerLet $x$ be added to each of the given numbers then
$5+x, 9+x, 7+x, 12+x$ are in proportion
$\frac{5+\mathrm{x}}{9+\mathrm{x}}=\frac{7+\mathrm{x}}{12+\mathrm{x}}$
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 $=3$
View full question & answer→Question 173 Marks
What number must be subtracted from each term of the ratio $17 : 33$ so that the ratio becomes $7 : 15?$
AnswerLet $x$ 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 $= 3$
View full question & answer→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: $\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{5}{8}\times\frac{16}{25}$
$\frac{\text{A}}{\text{C}}=\frac{2}{5}$
$\therefore\text{A}:\text{C}=2:5$
View full question & answer→Question 193 Marks
Divide $Rs. 1100$ among $A, B$ and $C$ in the ratio $2 : 3 : 5.$
AnswerThe 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 $=₹\ \frac{5}{10}\times1100=₹\ 550$
View full question & answer→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
$\therefore(5 : 6)>(7 : 9)$
View full question & answer→Question 213 Marks
Two numbers are in the ratio $5 : 9.$ On subtracting $3$ from each, the ratio becomes $1 : 2.$ Find the numbers.
AnswerThe 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,
$2(5x - 3) = 1(9x - 3)$
$ \Rightarrow 10x - 6 = 9x - 3 $
$\Rightarrow 10x - 9x = -3 + 6 $
$\Rightarrow x = 3$
First number $= 5x = 5 \times 3 = 15$ and
second number $= 9x = 9 \times 3 = 27$
Hence numbers are $15, 27$
View full question & answer→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.
AnswerRatio 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)
$4(8x + 6) = 9(3x + 6)$
$\Rightarrow 32x + 24 = 27x + 54$
$\Rightarrow 32x – 27x = 54 – 24$
$\Rightarrow 5x = 30$
$\Rightarrow x = 6$
$A’s$ present age $= 8x = 8 \times 6 = 48$ years
and $B’s$ age $= 3x = 3 \times 6 = 18$ years
View full question & answer→Question 233 Marks
The scale of a map is $1 : 5000000.$ What is the actual distance between two towns, if they are $4\ cm$ apart on the map$?$
AnswerScale of map $= 1 : 5000000$
Distance between two town on the map $= 4\ cm$
$\therefore\text{Actual distance}=\frac{5000000\times4}{1}\text{cm}$
$=\frac{5000000\times4}{100}\text{m}$
$=\frac{5000000\times4}{100\times1000}\text{km}=200\text{km}$
View full question & answer→Question 243 Marks
If $36, 54, x$ are in continued proportion, find the value of $x.$
Answer$36, 54, x$ are in continued proportion $36 : 54 : : 54 : x$
$\Rightarrow 36 \times x = 54 \times 54$
$\Rightarrow\text{x}=\frac{54\times54}{36}$
$\text{x}=\frac{54\times3}{2}=27\times2=81$
$\therefore\text{x}=81$
View full question & answer→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.
AnswerSuppose 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, $(4x - 8) : (3x - 8) = 10 : 7$
$\Rightarrow\frac{4\text{x}-8}{3\text{x}-8}=\frac{10}{7}$
$\Rightarrow 28x - 56 = 30x - 80 $
$\Rightarrow 2x = 24 $
$\Rightarrow x = 12(4x - 8) $
$= 4 \times 12 - 8 $
$= 40$ years $(3x - 8) $
$= 3 \times 12 - 8 $
$= 28$ years
View full question & answer→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, $\therefore\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{7}{5}\times\frac{9}{14}$
$\Rightarrow\frac{\text{A}}{\text{C}}=\frac{9}{10}$
$\therefore\text{A}:\text{C}=9:10$
View full question & answer→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.
AnswerTotal 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 $= 5x = 5 \times 75 = 375$
Number of $50\ P$ coins $= 8x = 8 \times 75 = 600$ and
number of $25\ P$ coins $= 4x = 4 \times 75 = 300$
View full question & answer→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 $(8 : 13)>(3:5)$
View full question & answer→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.
AnswerHeight of a tree $= 6\ cm$ And
its shadow at same time $= 8\ m$
Shadow of a pole $= 20\ m$
Let height of pole $= x\ m\ 6 : 8 = x : 20$
$\Rightarrow\text{x}=\frac{6\times20}{8}=15\text{m}$
Height of pole $= 15\ m$
View full question & answer→Question 303 Marks
If $27, 36, x$ are in continued proportion, find the value of $x.$
Answer$27, 36, x$ are in continued proportion $27 : 36 :: 36 : x\ 27 \times x = 36 \times 36$
$\Rightarrow\text{x}=\frac{36\times36}{27}$
$\text{x}=\frac{4\times36}{3}=4\times12=48$
$\therefore\text{x}=48$
View full question & answer→