Question 13 Marks
Solve: $\text{m}-\frac{(\text{m}-1)}{2}=1-\frac{(\text{m}-2)}{3} $
Answer$\text{m}-\frac{(\text{m}-1)}{2}=1-\frac{(\text{m}-2)}{3} $
$\Rightarrow \frac{2\text{m} - \text{m} + 1}{2} = 1 −\frac{ ( \text{m}−2 )}{3}(L.C.M.$ of $1$ and $2$ is $2 )$
$\Rightarrow\frac{\text{m} + 1}{2} = \frac{3 −\text{m} + 2}{3} (L.C.M.$ of $1$ and $3$ is $3 )$
$\Rightarrow\frac{\text{m} + 1}{2} = \frac{5 −\text{ m}}{3}$
$\Rightarrow 3 ( \text{m} + 1 ) = 2 ( 5 - \text{m} ) ($by cross multiplication$)$
$\Rightarrow3\text{m} + 3 = 10 - 2\text{m}$
$3\text{m} + 2\text{m} = 10 - 3$
$\Rightarrow 5\text{m} = 7$
$\Rightarrow\text{m}= 75$
$\therefore \text{m} = \frac{7}{5}$
View full question & answer→Question 23 Marks
Solve: $\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
Answer$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
$\Rightarrow (4+5\text{x})(2-7\text{x}) = (1-5\text{x})(3+7\text{x})$
$\Rightarrow 8 - 28\text{x} + 10\text{x} - 35\text{x}^2 = 3 + 7\text{x} - 15\text{x} - 35\text{x}^2$
$\Rightarrow-35\text{x}^2 - 18\text{x} +8 = -35\text{x}^2 - 8\text{x} + 3$
$\Rightarrow-35\text{x}^2 + 35\text{x}^2 -18\text{x} + 8\text{x} = -8 + 3$
$\Rightarrow-10\text{x} = -5$
$\Rightarrow \text{x} = -5-10= \frac{1}{2} $
$\therefore \text{x} = \frac{1}{2}$
View full question & answer→Question 33 Marks
A man is $10$ times older than his grandson. He is also $54$ years older than him. Find their present ages.
AnswerLet the age of the grandson be $x$ years.
Then, his grandfather's age will be $10x.$
Also, the grandfather is $54$ years older than his grandson.
$\therefore$ Age of the grandson $= \text{x} + 54$
$10\text{x} = \text{x} + 54$
$\Rightarrow10\text{x} - \text{x} = 54$
$\Rightarrow9\text{x} = 54$
$\Rightarrow \text{x} = \frac{54}{9}= 6$
Therefore, the grandson's age is $6$ years.
Grandfather's age $= 10(x) = 10 \times 6 = 60$ years
View full question & answer→Question 43 Marks
Solve: $3\text{x}+\frac{2\text{}}{3}=2\text{x}+1$
Answer$3\text{x}+\frac{2\text{}}{3}=2\text{x}+1$ By cross multiplication: $\Rightarrow 3\text{x} - 2\text{x} = 1 -\frac{ 2}{3}$ $ \Rightarrow \text{x} = \frac{1}{1} - \frac{2}{3}$ $\text{L.C.M}\text{ of }1\text{ and }3\text{ is }3$ $\Rightarrow\text{x}=\frac{3-2}{3}$ $\Rightarrow\text{x}=\frac{1}{3}$ $\therefore\text{x}=\frac{1}{3}$
View full question & answer→Question 53 Marks
Find three consecutive odd numbers whose sum is $147$.
AnswerLet the first odd number be $x.$
Let the second odd number be $(\text{x}+2).$
Let the third odd number be $(\text{x}+4).$
$\therefore\text{x}+(\text{x}+2)+(\text{x}+4)=147$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=147$
$\Rightarrow3\text{x}+6=147$
$\Rightarrow3\text{x}=147-6$
$\Rightarrow3\text{x}=141$
$\Rightarrow\text{x}=\frac{141}{3}=47$
Therefore, the first odd number is $47.$
Second odd number $=(\text{x}+2)=(47+2)=49$
Third odd number $=(\text{x}+4)=(47+4)$
View full question & answer→Question 63 Marks
The difference between the ages of two cousins is $10$ years. $15$ years ago, if the elder one was twice as old as the younger one, find their present ages.
AnswerLet the age of the younger cousin be $x.$
Then, the age of the elder cousin will be $(x + 10).$
$15$ years ago: Age of the younger cousin $= (x - 15)$
Age of elder cousin $= (x + 10 - 15) = (x - 5)$
$\therefore (x - 5) = 2 (x - 15)$
$\Rightarrow x - 5 = 2x - 30$
$\Rightarrow x - 2x = -30 + 5$
$\Rightarrow -x = -25$
$\Rightarrow x = 25$ Therefore, the present age of the younger cousin is $25$ years.
Present age of elder cousin $= (x + 10) = (25 + 10) = 35$ years
View full question & answer→Question 73 Marks
Solve: $\frac{15(2-\text{y})-5(\text{y}+6)}{1-3\text{y}}=10$
Answer$\frac{15(2-\text{y})-5(\text{y}+6)}{1-3\text{y}}=10$ $\Rightarrow\frac{30 - 15\text{y} - 5\text{y} - 30}{1 - 3\text{y}} = 10$ $\Rightarrow \frac{-20\text{y}}{1 - 3\text{y}} = 10$ $\Rightarrow1 (-20\text{y}) = 10 (1 - 3\text{y})$ (by cross multiplication) $\Rightarrow-20\text{y} = 10 - 30\text{y}$ $\Rightarrow-20\text{y} + 30\text{y} = 10$ $\Rightarrow10\text{y} = 10$ $\Rightarrow\text{y} = \frac{10}{10} = 1$ $\therefore\text{y} = 1$
View full question & answer→Question 83 Marks
Rakhi's mother is four times as old as Rakhi. After $5$ years, her mother will be three times as old as she will be then. Find their present ages.
AnswerLet the present age of Rakhi be $x.$
Then, the present age of Rakhi's mother will be $4x.$
After five years, Rakhi's age will be $(x + 5). $
After five years, her mother's age will be $(4x + 5).$
$4x + 5 = 3 ( x + 5 )$
$\Rightarrow 4x + 5 = 3x + 15$
$\Rightarrow 4x - 3x = 15 - 5$
$\Rightarrow x = 10$
Present age of Rakhi $= 10$
years Present age of Rakhi's mother $= 4(x) = 4 \times 10 = 40$ years.
View full question & answer→Question 93 Marks
Solve: $\frac{7\text{y}}{5}=\text{y}-4$
Answer$\frac{7\text{y}}{5}=\text{y}-4$ By cross multiplication: $\Rightarrow7\text{y} = 5(\text{y} - 4)$ $\Rightarrow7\text{y}=\text{5y}-20$ $\Rightarrow7\text{y}-5\text{y}=-20$ $\Rightarrow2\text{y}=-20$ $\Rightarrow\text{y}=\frac{-20}{2}=-10$ $\therefore\text{y}=-10$
View full question & answer→Question 103 Marks
Solve: $\frac{2-9\text{z}}{17-4\text{z}}=\frac{4}{9}$
Answer$\frac{2-9\text{z}}{17-4\text{z}}=\frac{4}{9}$ $\Rightarrow 5 ( 2-9\text{z}) = 4 ( 17-4\text{z})$ $\Rightarrow10 - 45\text{z} = 68 - 16\text{z}$ $\Rightarrow10 - 68 = 45\text{z} - 16\text{z}$ $\Rightarrow-58 = 29\text{z}$ $\Rightarrow29\text{z} = -58$ (by transposition) $\Rightarrow\text{z}=\frac{-58}{29}=-2$ $\therefore\text{z}=-2$
View full question & answer→Question 113 Marks
Solve:
$\frac{6\text{y}-5}{2\text{y}}=\frac{7}{9}$
Answer $\frac{6\text{y}-5}{2\text{y}}=\frac{7}{9}$
$\Rightarrow9(6\text{y}-5)=7(2\text{y})$ (by cross multiplication)
$\Rightarrow54\text{y} - 45 = 14\text{y}$
$\Rightarrow54\text{y} - 14\text{y} = 45$
$\Rightarrow 40\text{y} = 45$
$\Rightarrow \text{y} = \frac{45}{40}= \frac{9}{8}$
$\therefore\text{y} = \frac{9}{8}$
View full question & answer→Question 123 Marks
Solve: $\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$
Answer$\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$ $\Rightarrow (\text{x}+6) (9\text{x}-7) = (3\text{x}+5) (3\text{x}-4)$ $\Rightarrow9\text{x}^2 - 7\text{x} + 54\text{x} - 42 = 9\text{x}^2 - 12\text{x} + 15\text{x} - 20$ $\Rightarrow9\text{x}^2 + 47\text{x} - 42 = 9\text{x}^2 + 3\text{x} - 20$ $\Rightarrow 9\text{x}^2 - 9\text{x}^2 + 47\text{x} - 3\text{x} = -20 + 42$ $\Rightarrow44\text{x} = 22$ $\Rightarrow \text{x} = \frac{22}{44}= \frac{1}{2}$ $\therefore\text{x}=\frac{1}{2}$
View full question & answer→Question 133 Marks
Solve: $9x + 5 = 4(x - 2) + 8$
Answer$9\text{x} + 5 = 4(\text{x}-2)+8$
$\Rightarrow9\text{x}+\text{5}=4\text{x}-8+8$
$\Rightarrow9\text{x}+\text{5}=4\text{x}$
$\Rightarrow9\text{x}-4\text{x}=-5$
$\Rightarrow5\text{x}=-5$
$\Rightarrow\text{x}=\frac{-5}{5}=-1$
$\therefore\text{x}=-1$
View full question & answer→Question 143 Marks
Solve:
$\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$
Answer $\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$
$⇒ (4\text{x} + 7)(3\text{x} + 5) = (4\text{x} + 2)(3\text{x} +4)$
$\Rightarrow 12\text{x}^2 + 20\text{x} + 21\text{x} + 35=12\text{x}^2 + 16\text{x} + 6\text{x} +8$
$$$\Rightarrow12\text{x}^2 + 41\text{x }+ 35 =12\text{x}^2 + 22\text{x} +8$
$\Rightarrow12\text{x}^2 -12\text{x}^2 + 41\text{x} -22\text{x} = 8 - 35$
$\Rightarrow 19\text{x} = -27$
$\Rightarrow\text{x} = \frac{-27}{19} $
$\therefore\text{x} = \frac{-27}{19} $
View full question & answer→Question 153 Marks
Find the number whose fifth part increased by $5$ is equal to its fourth part diminished by $5.$
AnswerLet the number be $x.$
Fifth part increased by $5=\frac{\text{x}}{5}+5$
Fourt part diminished by $5=\frac{\text{x}}{4}-5$
$\therefore\frac{\text{x}}{5}+5=\frac{\text{x}}{4}-5$
$\Rightarrow5+5=\frac{\text{x}}{4}-\frac{\text{x}}{5}$
$\Rightarrow10=\frac{5\text{x}-4\text{x}}{20}$
$\Rightarrow10=\frac{\text{x}}{20}$
$\Rightarrow10=\frac{\text{x}}{20}$
$\Rightarrow200=\text{x}$
$\Rightarrow\text{x}=200$
Therefore, the number is $200.$
View full question & answer→Question 163 Marks
Three numbers are in the ratio of $4 : 5 : 6$. If the sum of the largest and the smallest equals the sum of the third and $55,$ find the numbers.
AnswerLet the common multiple for the given three numbers be $x.$Then, the three numbers would be $4x, 5x$ and $6x.$
$\therefore4\text{x} + 6\text{x} = 5\text{x} + 55$
$\Rightarrow10\text{x} = 5\text{x} + 55$
$\Rightarrow10\text{x} - 5\text{x} = 55$
$\Rightarrow 5\text{x} = 55$
$\Rightarrow\text{x} = \frac{55}{5}= 11$
$\therefore$ Smallest number $=4\text{x}=4(11)=44$
Largest number is $=6\text{x}=6(11)=66$
Third number $=5\text{x}=5(11)=55$
Therefore, the three numbers are $44, 55$ and $66.$
View full question & answer→Question 173 Marks
Evaluate $\left\{(83)^2-(17)^2\right\}$
Answer$(83)^2-(17)^2$
$=(83+17)(83-17)\left[\right.$ according to the formula $\left.(a)^2-(b)^2=(a+b)(a-b)\right]$
$=(100)(66)$
$=6600$
View full question & answer→Question 183 Marks
If $10$ be added to four times a certain number, the result is $5$ less than five times the number. Find the number.
AnswerLet the number be $x.$
$\therefore10+4\text{x}=5\text{x}-5$
$\Rightarrow10+5=5\text{x}-4\text{x}$
$\Rightarrow15=\text{x}$
$\Rightarrow\text{x}=15$ (by transposition)
Therefore, the number is $15.$
View full question & answer→Question 193 Marks
Two numbers are in the ratio $8 : 3.$ If the sum of the numbers is $143,$ find the numbers.
AnswerLet the numbers be $8x$ and $3x$
$8\text{x} + 3\text{x} = 143$
$\Rightarrow11\text{x} = 143$
$\Rightarrow\text{x}= \frac{143}{11}$
$\Rightarrow\text{x} =13$
$\therefore$ One number $=8\text{x} = 8\times13 = 104$
Other number $= 3\text{x} = 3\times13 = 39$
View full question & answer→Question 203 Marks
$\frac{2}{3}$ of a number is $20$ less than the original number. Find the number.
AnswerLet the original number be x $\frac{2}{3}$ of the number is $20$ less than the original number.
$\therefore\frac{2}{3}\text{x}=\text{x}-20$
$\Rightarrow\frac{2\text{x}}{3}=\text{x}-20$
$\Rightarrow2\text{x}=3(\text{x}-20)$ (by cross multiplication)
$\Rightarrow2\text{x}=3\text{x}-60$
$\Rightarrow2\text{x}-3\text{x}=60$
$\Rightarrow-\text{x}=-60$
Therefore, the original number is $60.$
View full question & answer→Question 213 Marks
Solve: $\frac{8\text{x}-3}{3\text{x}}=2$
Answer$\frac{8\text{x}-3}{3\text{x}}=2$ $\Rightarrow8\text{x}-3=2(3\text{x})$ (by cross multiplication) $\Rightarrow 8\text{x} - 3 = 6\text{x}$ $\Rightarrow 8\text{x} - 6\text{x}=3$ $\Rightarrow 2\text{x} =3$ $\Rightarrow \text{x}= \frac{3}{2}$$\therefore \text{x} = \frac{3}{2}$
View full question & answer→Question 223 Marks
Twenty-four is divided into two parts such that $7$ times the first part added to $5$ times the second part makes $146.$ Find each part.
AnswerLet one part be $x.$
$7$ times the first part $=7\text{x}$ Let the other part be $(24-\text{x})$
$5$ times the second part $=5(24-\text{x})$
$\therefore7\text{x}+5(24-\text{x})=146$
$\Rightarrow7\text{x}+120-5\text{x}=146$
$\Rightarrow7\text{x}-5\text{x}=146-120$
$\Rightarrow2\text{x}=26$
$\Rightarrow\text{x}=\frac{26}{2}=13$ Therefore, one part is $13.$
Other part $=(24-\text{x})=(24-13)=11$
View full question & answer→Question 233 Marks
Solve:
$\frac{9\text{x}}{7-6\text{x}}=15$
Answer $\frac{9\text{x}}{7-6\text{x}}=15$ $\Rightarrow\frac{9\text{x}}{7-6\text{x}}=\frac{15}{1}$
$\Rightarrow 1 ( 9\text{x}) = 15 ( 7-6\text{x})$ (by cross multiplication)
$\Rightarrow 9\text{x} = 105 - 90\text{x}$
$\Rightarrow99\text{x}=105$
$\Rightarrow\text{x}=\frac{105}{99}=\frac{35}{33}$
$\therefore\text{x}=\frac{35}{33}$
View full question & answer→Question 243 Marks
Solve: $\frac{7\text{y}+4}{\text{y}+7}=\frac{-4}{3}$
Answer$\frac{7\text{y}+4}{\text{y}+7}=\frac{-4}{3}$(by cross multiplication)
$\Rightarrow3 (7\text{y} + 4) = -4 (\text{y} + 2)$
$\Rightarrow21\text{y} + 12 = -4\text{y} - 8$
$\Rightarrow21\text{y} + 4\text{y} = -8 -12$
$\Rightarrow25\text{y} = -20$
$\Rightarrow\text{y} = -\frac{20}{25}= \frac{-4}{5}$
$\therefore\text{y}=\frac{-4}{5}$
View full question & answer→Question 253 Marks
Divide 4500 into two parts such that $5\%$ of the first part is equal to $10\%$ of the second part.
AnswerLet the first part be $x.$
Let the second part be $(4500 − x).$
$\therefore5\%\text{ of }\text{x}=10\%\text{ of }(4500-\text{x})$
$\Rightarrow\Big(\frac{5}{100}\Big)\text{x}=\Big(\frac{10}{100}\Big)(4500-\text{x})$
$\Rightarrow\frac{5\text{x}}{100}=\frac{45000-10\text{x}}{100}$
$\Rightarrow5\text{x}=45000-10\text{x}$ (by cancellation of same denominators from both the sides)
$\Rightarrow5\text{x}+10\text{x}=45000$
View full question & answer→Question 263 Marks
Solve: $\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
Answer$\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$ $\Rightarrow\frac{2\text{x} - 7 + 5\text{x}}{9\text{x} - 3 - 4\text{x}} = \frac{7}{6}$ $\Rightarrow\frac{7\text{x} - 7}{5\text{x} - 3} = \frac{7}{6}$ $\Rightarrow6 (7\text{x} - 7) = 7 (5\text{x} - 3)$ (by cross multiplication) $\Rightarrow42\text{x} - 42 = 35\text{x} - 21$ $\Rightarrow 42\text{x} - 35\text{x} = 42 - 21$ $\Rightarrow7\text{x} = 21 $ $\Rightarrow\text{x}=\frac{21}{7}=3$ $\therefore \text{x} = 3$
View full question & answer→Question 273 Marks
Solve: $15(y - 4) - 2(y - 9) + 5(y + 6) = 0$
Answer$15(\text{y} - 4) - 2(\text{y} - 9) + 5(\text{y} + 6) = 0$
$\Rightarrow 15\text{y} - 60 - 2\text{y} + 18 + 5\text{y} + 30 = 0$
$\Rightarrow15\text{y} - 2\text{y} + 5\text{y} -60 + 18 + 30 = 0$
$\Rightarrow18\text{y} - 12 = 0$
$\Rightarrow18\text{y} = 12 $
$\Rightarrow \text{y} = \frac{12}{18} = \frac{2}{3}$
$\therefore \text{y} = \frac{2}{3} $
View full question & answer→Question 283 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$ find the value of: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
AnswerGiven, $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$
Squaring both the sides:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=4^2$
$\Rightarrow(\text{x})^2-2\times(\text{x})\times\Big(\frac{1}{\text{x}}\Big)+\Big(\frac{1}{\text{x}}\Big)^2=16$
$\Rightarrow\text{x}^2-2+\frac{1}{\text{x}^2}=16$
$\Rightarrow\text{x}^2-2+\frac{1}{\text{x}^2}=16+2$
$\therefore\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=18$
$2^2+2\Rightarrow4^2+2\Rightarrow16+2\Rightarrow18$
View full question & answer→Question 293 Marks
A steamer goes downstream from one port to another in $9$ hours. It covers the same distance upstream in $10$ hours. If the speed of the stream be $1\ km/ h,$ find the speed of the steamer in still water and the distance between the ports.
AnswerLet the speed of the steamer in still water be $x\ km/ h.$
Speed (downstream) $= (x + 1)\ km/ h$
Speed (upstream) $= (x - 1)\ km/h$
Distance covered in $9$ hours while going downstream $= 9( x + 1 )\ km$
Distance covered in $10$ hours while going upstream $= 10( x - 1 )\ km$
But both of these distances will be same.
$9 (x + 1) = 10 (x - 1) $
$⇒ 9x + 9 = 10x - 10 $
$⇒ 9 + 10 = 10x − 9x $
$⇒ 19 = x $
$⇒ x = 19$
Therefore, the speed of the steamer in still water is $19\ km/ h.$
Distance between the ports $= 9(x + 1) = 9(19 + 1) = 9 × 20 = 180\ km$
View full question & answer→Question 303 Marks
Solve: $3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17$
Answer$3(5\text{x} - 7) - 2(9\text{x} - 11) = 4(8\text{x} - 13) -17$ $\Rightarrow 5\text{x} - 21 - 18\text{x} + 22 = 32\text{x} - 52-17$ $\Rightarrow15\text{x} -18\text{x} - 21 + 22 = 32\text{x} −69$ $\Rightarrow-3\text{x} +1 = 32\text{x} - 69$ $\Rightarrow1 + 69 = 32\text{x} + 3\text{x} $ $\Rightarrow 70 = 35\text{x}$ $\Rightarrow 35\text{x}= 70 $ $(\text{by} \ \text{transposition})$ $\Rightarrow\text{x}= \frac{70}{35}= 2$ $\therefore\text{x}=2$
View full question & answer→Question 313 Marks
Solve: $\frac{4\text{x}+7}{9-3\text{x}}=\frac{1}{4}$
Answer$\frac{4\text{x}+7}{9-3\text{x}}=\frac{1}{4}$ $\Rightarrow4(4\text{x}+7)=1(9-3\text{x})$ (by cross multiplication) $\Rightarrow16\text{x}+28=9-3\text{x}$ $\Rightarrow16\text{x}+3\text{x}=9-28$ $\Rightarrow19\text{x}=-19$ $\Rightarrow\text{x}=\frac{-19}{19}=1$ $\therefore\text{x}=-1$
View full question & answer→Question 323 Marks
Two angles of a triangle are in the ratio $4 : 5.$ If the sum of these angles is equal to the third angle, find the angles of the triangle.
AnswerLet the common multiple of all the three angles be $x.$
Then, the first angle will be $4x.$
And the second angle will be $5x.$
In a triangle, sum of all the three angles will be equal to $180^\circ .$
$\therefore$ Third angle $= 180 - ( 4 \text{x} + 5 \text{x} )= 180 - 9 \text{x}$
$\therefore4 \text{x} + 5 \text{x} = 180 - 9 $
$\Rightarrow9 \text{x}=180-9 \text{x}$
$\Rightarrow9\text{x}+ 9\text{x}=180$
$\Rightarrow18\text{x}=180$
$\Rightarrow\text{x}=\frac{180}{18}=10$
First angle $=4\text{x}=4\times10=40^{\circ}$
Second angle $=5\text{x}=5\times10=50^{\circ}$
Third angle $=4\text{x}+5\text{x}=9\text{x}=9\times10=90^{\circ}$
View full question & answer→Question 333 Marks
Find three consecutive even numbers whose sum is $234.$
AnswerLet the first even number be $x.$
Let the second even number be $x+2.$
Let the third even number be $x+4.$
$\therefore\text{x}+\text{x}+2+\text{x}+4=234$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=234$
$\Rightarrow3\text{x}+6=234$
$\Rightarrow3\text{x}=234-6$
$\Rightarrow3\text{x}=228$
$\Rightarrow\text{x}=\frac{228}{3}=76$
$\therefore$ First even number $=\text{x}=76$
Second even number $=\text{x}+2=76+2=78$
Third even number $=\text{x}+4=80$
View full question & answer→Question 343 Marks
Divide $150$ into three parts such that the second number is five-sixths the first and the third number is four-fifths the second.
AnswerLet the first number be $x$.
Then, the second number will be $\frac{5}{6}\text{x}.$
Third number $=\frac{4}{5}\Big(\frac{5}{6}\text{x}\Big)=\frac{2}{3}\text{x}$
$\therefore\text{x}+\frac{5\text{x}}{6}+\frac{2\text{x}}{3}=150$
$\Rightarrow\frac{6\text{x}+5\text{x}+4\text{x}}{6}=150 ($multiplying the $L.H.S.$ by $6,$ which is the $L.C.M.$ of $1, 6$ and $3)$
$\Rightarrow15\text{x}=150\times6$ (by cross multiplication)
$\Rightarrow15\text{x}=900$
$\Rightarrow\text{x}=\frac{900}{15}=60$
Therefore, the first number is $60$.
Second number $=\frac{5}{6}\text{x}=\frac{5}{6}(60)=50$
Third number $=\frac{2}{3}\text{x}=\frac{2}{3}(60)=40$
View full question & answer→Question 353 Marks
Four-fifths of a number is $10$ more than two-thirds of the number. Find the number.
AnswerLet the number be $x.$
Four fifths of the number is $10$ more than two thirds of the number.
$\therefore\frac{4}{5}\text{x}=10+\frac{2}{3}\text{x}$
$\Rightarrow\frac{4\text{x}}{5}=\frac{30+2\text{x}}{3} (L.C.M.$ of $1$ and $3$ is $3)$
$\Rightarrow3(4\text{x})=5(30+2\text{x})$ (by cross multiplication)
$\Rightarrow12\text{x}=150+10\text{x}$
$\Rightarrow12\text{x}-10\text{x}=150$
$\Rightarrow2\text{x}=150$
$\Rightarrow\text{x}=\frac{150}{2}=75$
Therefore, the number is $75.$
View full question & answer→