Question 515 Marks
$\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}=13$
AnswerWe have:$\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}=13$
$L.C.M.$ of $2, 3, 4 = 12$
$\therefore\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}=13$
$6\text{x}+4\text{x}+3\text{x}=13\times12$
$\Rightarrow13\text{x}=156$
$\frac{13\text{x}}{13}=\frac{156}{13}$ (Dividing both sides by $13$)
$\Rightarrow\text{x}=12$
$\therefore\text{x}=12$
Verification: $\text{L.H.S}=\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}$
$=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}=6+4+3$
$=13=\text{R.H.S}$
View full question & answer→Question 525 Marks
Solve the following equation and also check your result in case: $\frac{7}{2}\text{x}-\frac{5}{2}\text{x}=\frac{20}{3}\text{x}+10$
Answer$\frac{7}{2}\text{x}-\frac{5}{2}\text{x}=\frac{20}{3}\text{x}+10$
$\frac{7\text{x}-5\text{x}}{2}=\frac{20\text{x+30}}{3}$
$40\text{x}+60 =6\text{x}$
$\text{x}=\frac{60}{34}=-\frac{30}{17}$
Check: $\text{L.H.S.}=\frac{7}{2}\times\frac{-30}{17}-\frac{5}{2}\times-\frac{30}{17}=-\frac{30}{17}$
$\text{R.H.S.}=\frac{20}{3}\times\frac{-30}{17}+10=-\frac{30}{17}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-\frac{30}{17}$
View full question & answer→Question 535 Marks
Solve the following equation and also check your result in case: $\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{1}{12}$
Answer$\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{1}{12}$
$\frac{4-9}{6\text{x}}=\frac{1}{12}$
$\frac{-5}{6\text{x}}=\frac{1}{12}$
$6\text{x}=-60$
$\text{x}=\frac{-60}{6}$
$\text{x}=-10$
Thus, $\text{x}=-10$ is the solution of the given equation.
Check:Substituting $\text{x}=-10$ in the given equation, we get:
$\text{L.H.S.}=\frac{2}{3\times(-10)}-\frac{3}{2\times(-10)}=\frac{2}{-30}-\frac{3}{-20}$
$=\frac{-4+9}{60}=\frac{5}{50}=\frac{1}{2}$
$\text{R.H.S.}=\frac{1}{12}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-10$
View full question & answer→Question 545 Marks
At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink any thing. How many guests were in all?
AnswerLet Total number of guests $= x$
Guests who drank colas $=\frac{\text{x}}{4}$
Guests who drank squash $=\frac{\text{x}}{3}$
Guests who drank juice $=\frac{2}{3}\text{x}$
Guests who drank none of these $= 3$
According to the condition: $\frac{\text{x}}{4}+\frac{\text{x}}{3}+\frac{2}{5}\text{x}+3=\text{x}$
$\frac{15\text{x}+20\text{x}+24\text{x}+180\text{x}=60\text{x}}{60}$ $(L.C.M. 4, 3, 5 = 60)$
$\Rightarrow 59x + 180 = 60x$ $\Rightarrow 60x - 59x = 180$
$\Rightarrow x = 180$
Totle number of guests $= 180$
View full question & answer→Question 555 Marks
Solve the following equation and verify your answer: $\frac{2\text{x}}{3\text{x}+1}=-3$
Answer$\frac{2\text{x}}{3\text{x}+1}=\frac{-3}{1}$By cross multiplication:
$2\text{x}=-3(3\text{x}+1)$
$\Rightarrow2\text{x}+9\text{x}=-3$ (By transposition)
$\Rightarrow11\text{x}=-3$
$\Rightarrow\text{x}=\frac{-3}{11}$
$\therefore\text{x}=\frac{-3}{11}$
Verification:
$\text{L.H.S.}=\frac{2\text{x}}{3\text{x}+1}=\frac{2\times\Big(\frac{-3}{11}\Big)}{3\Big(\frac{-3}{11}\Big)+1}=\frac{\frac{-6}{11}}{\frac{-9}{11}+1}$
$=\frac{\frac{-6}{11}}{\frac{-9+11}{11}}=\frac{\frac{-6}{11}}{\frac{2}{11}}=\frac{-6}{11}\times\frac{11}{2}$
$\frac{-3}{1}=\text{R.H.S.}$
View full question & answer→Question 565 Marks
Solve the following equation and verify your answer: $\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=10$
Answer$\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=10$
$\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}=\frac{10}{1}$
$\Rightarrow\frac{-20\text{x}}{1-3\text{x}}$
$=\frac{10}{1}$
By cross multiplication, $-20\text{x}=10(1-3\text{x})$
$\Rightarrow-20\text{x}-10=30\text{x}$
$\Rightarrow20\text{x}+30\text{x}=10$
$\Rightarrow10\text{x}=10$
$\Rightarrow\text{x}=\frac{10}{10}=1$
$=1$
$\therefore\text{x}=1$
Verification: $\text{L.H.S}=\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}$
$=\frac{15(2-1)-5(1+6)}{1-3\times1}$
$\frac{15\times1-5\times7}{1-3}-\frac{15-35}{-2}$
$=\frac{-20}{-2}=10$
$=\text{R.H.S}$
View full question & answer→Question 575 Marks
Find a number such that when $5$ is subtracted from $5$ times the number, the result is $4$ more than twice the number.
AnswerLet the requierd number $= x5$
Times of it $= 5x$
Twice of it $= 2x$
According to the condition:
$\Rightarrow 5x - 2x = 4 + 5$
$\Rightarrow 3x = 9$
$\Rightarrow\text{x}=\frac{9}{3}=3$
Required number $= 3$
Check: $3 \times 5 - 5 = 2 \times 3 + 4$
$\Rightarrow 15 - 5 = 6 + 4$
$\Rightarrow 10 = 10$
Which is true. therefore our answer is correct.
View full question & answer→Question 585 Marks
Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times Ashima's age. How old were they two years ago?
AnswerLet age of ashima $= x$
Then age of Sunita $= 2x$
According to the condition:
$4(x - 6) = 2x + 4$
$\Rightarrow 4x - 24 = 2x + 4$
$\Rightarrow 4x - 2x = 4 + 24$
$\Rightarrow 2x = 28$
$\Rightarrow\text{x}=\frac{28}{2}=14$
$\therefore$ Sunita's present age $= 2x = 2 \times 14 = 28$ years
And ashima's age $= 14$ years
Age of sunita $= 28 - 2 = 26$ years
And age of Ashima $= 14 - 2 = 12$ years
View full question & answer→Question 595 Marks
The distance between two stations is $340\ km$. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by $5\ km/ hr$. If the distance between the two trains after $2$ hours of their start is $30\ km$, find the speed of each train.
AnswerDistance between two stations $= 340\ km.$
Let the speed of the first train $= x\ km/ hr.$
Then speed of second train $= (x + 5)\ km/ h.$
Time $= 2$ hours Distance travelled by the first train in $2$ hours =$ 2x\ km$ and
distance travelled by the first trian in $2$ hours $= 2(x + 5)\ km$
According to the condition, $340 - [2(x + 5) + 2x] = 30km$
$\Rightarrow 340 - (2x + 10 + 2x) = 30 $
$\Rightarrow 4x + 10 = 340 - 30 $
$\Rightarrow 4x = 340 - 30 - 10$
$ \Rightarrow 4x = 300$
$\Rightarrow\text{x}=\frac{300}{4}=75$
$\therefore$ Speed of first train $= 75\ km/ hr$ and speed of second train $= 75 + 5 = 80\ km/ hr$
View full question & answer→Question 605 Marks
$\frac{5\text{x}}{3}+\frac{2}{5}=1$
Answer$\frac{5\text{x}}{3}+\frac{2}{5}=1$
Subtracting $\frac{2}{5}$ from both sides: $\frac{5\text{x}}{3}+\frac{2}{5}-\frac{2}{5}=1-\frac{2}{5}$
$\Rightarrow\frac{5\text{x}}{3}=\frac{3}{5}$
$\Rightarrow$ Multiplying $\frac{3}{5}$ both sides $\frac{5}{3}\text{x}\times\frac{3}{5}=\frac{3}{5}\times\frac{3}{5}$
$\Rightarrow\text{x}=\frac{9}{25}$
Verification: $\text{L.H.S}$
$=\frac{5\text{x}}{3}+\frac{2}{5}$
$=\frac{5}{3}\times\frac{9}{25}+\frac{2}{5}$
$=\frac{3}{5}+\frac{2}{5}=\frac{3+2}{5}=\frac{5}{5}$
$=1=\text{R.H.S}$
View full question & answer→Question 615 Marks
A labourer is engaged for $20$ days on the condition that he will receive $Rs.\ 60$ for each day, he works and he will be fined $Rs.\ 5$ for each day, he is absent. If he receives $Rs.\ 745$ in all, for how many days he remained absent?
AnswerTotle number of days $= 20$
Let of days he worked $= x$
Then number of days he remained absent $= 20 - x$
According to the condition: $x \times 60 - (20 - x) \times 5 = 745$
$\Rightarrow 60x = 745 + 5x = 745$
$ \Rightarrow 65x = 745 + 100 = 845$
$\Rightarrow\text{x}=\frac{845}{65}=13$
$\therefore$ Number of days he worked$ = 13$ days and number of days he remained absent $= 20 - 13 = 7$days.
View full question & answer→Question 625 Marks
Solve the following equation and verify your answer: $\frac{2\text{x}-3}{3\text{x}+2}=-\frac{2}{3}$
Answer$\frac{2\text{x}-3}{3\text{x}+2}=-\frac{2}{3}$
$6\text{x}-9=-6\text{x}-4$ (After cross multiplication)
$6\text{x}+6\text{x}=-4+9$
$\text{x}=\frac{5}{12}$
$\therefore\text{x}=\frac{5}{12}$ is the solution of the given equation.
Chek: $\text{L.H.S.}=\frac{2\times\frac{5}{12}-3}{3\times\frac{5}{12}+2}=\frac{\frac{5}{6}-3}{\frac{5}{4}+2}=\frac{-4}{6}=\frac{-2}{3}$
$\text{R.H.S.}=\frac{-2}{3}$
$\text{L.H.S.}=\text{R.H.S.} \text{ for} \text{ x}=\frac{5}{12}$
View full question & answer→Question 635 Marks
Solve the following equation and also check your result in case: $\frac{(1-2\text{x})}{7}-\frac{(2-3\text{x})}{8}=\frac{3}{2}+\frac{\text{x}}{4}$
Answer$\frac{(1-2\text{x})}{7}-\frac{(2-3\text{x})}{8}=\frac{3}{2}+\frac{\text{x}}{4}$
$\frac{1-2\text{x}}{7}=\frac{3}{2}+\frac{\text{x}}{4}+\frac{2-3\text{x}}{8}$
$\frac{1-2\text{x}}{7}=\frac{14-\text{x}}{8}$
$8-16\text{x}=98-7\text{x}$
$-16\text{x}+7\text{x}=98-8$
$\text{x}=\frac{-90}{9}$
$\text{x}=-10$
Check: $\text{L.H.S.}=\frac{1-2\times(-10)}{7}-\frac{2-3\times(-10)}{8}$
$=\frac{1+20}{7}-\frac{2+30}{8}=9-4=-1$
$\text{R.H.S.}=\frac{3}{2}+\frac{-10}{4}=\frac{3}{2}+\frac{-5}{2}=\frac{3-5}{2}=-1$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-10$
View full question & answer→Question 645 Marks
Find a number whose double is $45$ greater than its half.
AnswerLet the requierd number $= x$
Then Four-fifth of the number $= 2x$ And half of it $=\frac{\text{x}}{2}$
$\therefore$ According to the condition:$2\text{x}-\frac{\text{x}}{2}=45$
$\Rightarrow\frac{4\text{x}-\text{x}}{2}=45$
$\Rightarrow\frac{3}{2}\text{x}=45$
$\Rightarrow\text{x}=\frac{45\times2}{3}=30$
$\therefore$ Required number $= 30$
Check: $2\times30-\frac{1}{2}\times30$ $= 6 - 15 = 45$, which is given
$\therefore$ Our answer is correct.
View full question & answer→Question 655 Marks
Solve the following equation and verify your answer: $\frac{\text{x}+2}{\text{x}+5}=\frac{\text{x}}{\text{x}+6}$
Answer$\frac{\text{x}+2}{\text{x}+5}=\frac{\text{x}}{\text{x}+6}$ By cross multiplication, $(\text{x}+2)(\text{x}+6)=\text{x}(\text{x}+5)$
$\Rightarrow\text{x}^2+6\text{x}+2\text{x}+12=\text{x}^2+5\text{x}$
$\Rightarrow\text{x}^2+8\text{x}+12=\text{x}^2+5\text{x}$
$\Rightarrow\text{x}^2+8\text{x}-\text{x}^2-5\text{x}=-12$
$\Rightarrow3\text{x}$
$=-12$
$\Rightarrow\text{x}=\frac{-12}{3}$
$=-4$
$\therefore\text{x}=-4$
Verification: $\text{L.H.S}=\frac{\text{x}+2}{\text{x}+5}=\frac{-4+2}{-4+5}=\frac{-2}{1}=-2$
$\text{R.H.S}=\frac{\text{x}}{\text{x}+6}=\frac{-4}{-4+6}=\frac{-4}{2}=-2$
$\therefore\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 665 Marks
Solve the following equation and verify your answer: $\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
Answer$\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
$\Rightarrow\frac{2\text{x}-7+5\text{x}}{9\text{x}-3-4\text{x}}=\frac{7}{6}$
$\Rightarrow\frac{7\text{x}-7}{5\text{x}-3}=\frac{7}{6}$
By cross multiplication, $6(7\text{x}-7)=7(5\text{x}-3)$
$\Rightarrow42\text{x}-42=35\text{x}-21$
$\Rightarrow42\text{x}-35\text{x}=-21+42$
$\Rightarrow7\text{x}=21$
$\Rightarrow\text{x}=\frac{21}{7}=3$
$=3$
$\therefore\text{x}=3$
Verification: $\text{L.H.S}=\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{2\times3-(7-5\times3)}{9\times3-(3+4\times3)}$
$=\frac{6-(7-15)}{27-(3+12)}=\frac{6-7+15}{27-3-12}=\frac{14}{12}$
$=\frac{7}{6}$
$=\text{R.H.S}$
View full question & answer→Question 675 Marks
Solve the following equation and also check your result in case: $\text{x}-2\text{x}+2-\frac{16}{3}\text{x}+5=3-\frac{7}{2}\text{x}$
Answer$\text{x}-2\text{x}+2-\frac{16}{3}\text{x}+5=3-\frac{7}{2}\text{x}$
$\frac{3\text{x}-6\text{x}+6-16\text{x+15}}{3}=\frac{6-7\text{x}}{2}$
$\frac{-19\text{x}+21}{3}=\frac{6-7\text{x}}{2}$
$-38\text{x}+42=18-21\text{x}$
$-21\text{x}+38\text{x}=42-18$
$=17\text{x}=24$
$\text{x}=\frac{24}{17}$
Check: $\text{L.H.S.}=\frac{24}{17}-2\times\frac{24}{17}+7-\frac{16}{3}\times\frac{24}{17}=\frac{-33}{17}$
$\text{R.H.S.}=3-\frac{7}{2}\times\frac{24}{17}=\frac{-33}{17}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{24}{17 }$
View full question & answer→Question 685 Marks
Solve the following equation and also check your result in case: $\frac{3\text{a}-2}{3}=\frac{2\text{a}+3}{2}=\text{a}+\frac{7}{6}$
Answer$\frac{3\text{a}-2}{3}=\frac{2\text{a}+3}{2}=\text{a}+\frac{7}{6}$
$\frac{6\text{a}-4+6\text{a}+9}{6}=\text{a}+\frac{7}{6}$
$12\text{a}+5 =6\text{a}+7$
$6\text{a}=7-5$
$\text{a}=\frac{2}{6}=\frac{1}{3}$
Check: $\text{L.H.S.}=\frac{3\times\frac{1}{3}-2}{3}+=\frac{3\times\frac{1}{3}+3}{2}$
$=\frac{-1}{3}+\frac{11}{6}=\frac{9}{6}=\frac{3}{2}$
$\text{R.H.S.}=\frac{1}{3}+\frac{7}{6}=\frac{9}{6}=\frac{3}{2}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for a}=\frac{1}{3}$
View full question & answer→Question 695 Marks
Solve the following equation and verify your answer: $\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{1}{12}$
Answer$\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{1}{12}$ By cross multiplication:
$\frac{4-9}{6\text{x}}=\frac{1}{12}$
$\Rightarrow\frac{-5}{6\text{x}}=\frac{1}{12}$
$\Rightarrow6\text{x}=-5\times12$
$\Rightarrow\text{x}=\frac{-5\times12}{6}=-10$
$\therefore\text{x}=-10$
Verification:
$\text{L.H.S.}=\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{2}{3(-10)}-\frac{3}{2(-10)}$
$=\frac{2}{-30}-\frac{3}{-20}=\frac{1}{-15}+\frac{3}{20}$
$=\frac{-4+9}{60}=\frac{5}{60}=\frac{1}{12}=\text{R.H.S.}$
View full question & answer→Question 705 Marks
$9\frac{1}{4}=\text{y}-1\frac{1}{3}$
Answer$9\frac{1}{4}=\text{y}-1\frac{1}{3} $
$\Rightarrow\frac{37}{4}$
$=\text{y}-\frac{4}{3}$
$\Rightarrow\frac{37}{4}+\frac{4}{3}$
$=\text{y}-\frac{4}{3}+\frac{4}{3}$ (Adding $\frac{4}{3}$ to both sides)
$\Rightarrow\text{y}=\frac{111+16}{12}$
$=\frac{127}{12}$
$=10\frac{7}{12}$
$\therefore\text{y}=10\frac{7}{12}$ Verification: $\text{R.H.S}$
$=\text{y}-1\frac{1}{3}=10\frac{7}{12}-1\frac{1}{3}$
$=\frac{127}{12}-\frac{4}{3}$
$=\frac{127-16}{12}=\frac{111}{12}=\frac{37}{4}$ (Dividing by $3$)
$=9\frac{1}{4}=\text{L.H.S}$
View full question & answer→Question 715 Marks
Solve the following equation and also check your result in case: $\frac{1}{2}\text{x}+7\text{x}-6=7\text{x}+\frac{1}{4}$
Answer$\frac{1}{2}\text{x}+7\text{x}-6=7\text{x}+\frac{1}{4}$
$\frac{1}{2}\text{x}+7\text{x}-7\text{x}=\frac{1}{4}+6$
$\frac{\text{x}}{2}=\frac{1+24}{4}$
$\frac{\text{x}}{2}=\frac{25}{4}$
$\text{x}=\frac{25}{2}$
Check: $\text{L.H.S.}=\frac{1}{2}\times\frac{25}{2}+7\times\frac{25}{2}-6=\frac{351}{4}$
$\text{R.H.S.}=7\times\frac{25}{2}+\frac{1}{4}=\frac{351}{4}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{351}{2}$
View full question & answer→Question 725 Marks
Solve the following equation and also check your result in case: $\frac{5\text{x}}{3}-\frac{(\text{x}-1)}{4}=\frac{(\text{x}-3)}{5}$
Answer$\frac{5\text{x}}{3}-\frac{(\text{x}-1)}{4}=\frac{(\text{x}-3)}{5}$
$\frac{20\text{x}-3\text{x}{+3}}{12}=\frac{\text{x}-3}{5}$
$\frac{17\text{x+3}}{12}=\frac{\text{x}-3}{5}$
$85\text{x}+15=12\text{x}-36$
$73\text{x}=-51$
$\text{x}=\frac{-51}{73}$
Check: $\text{L.H.S.}=\frac{5\times\frac{-51}{73}}{3}-\frac{\frac{-51}{73}-1}{4}$
$=\frac{-255}{219}-\frac{-124}{292}=\frac{-54}{73}$
$\text{R.H.S.}=\frac{\frac{-51}{73}-3}{5}=\frac{-54}{73}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{-51}{73}$
View full question & answer→Question 735 Marks
I am currently $5$ times as old as my son. In $6$ years time I will be three times as old as he will be then. What are our ages now?
AnswerLet present age of my son $= x$ years Then my age $= 5x$ years After $6$ years,
My age will be $= 5x + 6$ and my son's age $= x + 6$
According to the condition $5x + 6 = 3(x + 6)$
$\Rightarrow 5x + 6 = 3x + 18$
$ \Rightarrow 5x - 3x = 18 - 6 $
$\Rightarrow 2x = 12$
$ \Rightarrow x = 6$
$\therefore$ Present my age $= 5x = 5 \times 6 = 30$ years and my son's age $= 6$ years
View full question & answer→