Question 15 Marks
Solve the following equation and verify your answer:
$\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$
Answer$\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$
By cross multiplication:
$(3\text{x}+5)(4\text{x}+7)=(3\text{x}+4)(4\text{x}+2)$
$\Rightarrow12\text{x}^2+21\text{x}+20\text{x}+35=12\text{x}^2+6\text{x}+16\text{x}+8$
$\Rightarrow12\text{x}^2+41\text{x}-12\text{x}^2-22\text{x}=8-35$
$\Rightarrow19\text{x}=-27$
$\Rightarrow\text{x}=\frac{-27}{19}$
$\therefore\text{x}=\frac{-27}{19}$
Verification:
$\text{L.H.S.}=\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\Big(\frac{-27}{19}\Big)+5}{4\Big(\frac{-27}{19}\Big)+2}$
$=\frac{\frac{-81}{19}+5}{\frac{-108}{19}+2}=\frac{\frac{-81+95}{19}}{\frac{-108+38}{19}}=\frac{\frac{14}{19}}{\frac{-70}{19}}$
$\text{R.H.S.}=\frac{3\text{x}+4}{4\text{x}+7}$
$=\frac{3\Big(\frac{-27}{19}\Big)+4}{4\Big(\frac{-27}{19}\Big)+7}=\frac{\frac{-81}{19}+4}{\frac{-108}{19}+7}$
$=\frac{\frac{-81+76}{19}}{\frac{-108+133}{19}}=\frac{\frac{-5}{19}}{\frac{25}{19}}=\frac{-5}{19}\times\frac{19}{25}$
$=\frac{-1}{5}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 25 Marks
$(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)=0$
Answer $(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)=0$
$\Rightarrow[\text{x}^2+(2+3)\text{x}+2\times3]+[\text{x}^2+(-3-2)\text{x}\\+(-3)(-2)]-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+5\text{x}+6+\text{x}^2-5\text{x}+6-2\text{x}^2-2\text{x}=0$
$\Rightarrow\text{x}^2+\text{x}^2-2\text{x}^2+5\text{x}-5\text{x}-2\text{x}+6+6=0$
$\Rightarrow-2\text{x}+12=0$
Subtracting 12 from both sides,
$-2\text{x}+12-12=0-12$
$\Rightarrow-2\text{x}=-12$
Dividing by - 2,
$\text{x}=6$
Verification:
$\text{L.H.S}=(\text{x}+2)(\text{x}+3)+(\text{x}-3)(\text{x}-2)-2\text{x}(\text{x}+1)$
$=(6+2)(6+3)+(6-3)(6-2)-2\times6\ (6+1)$
$=8\times9+3\times4-12 \times7$
$=84-84$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 35 Marks
Solve the following equation and verify your answer: $\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{2}{3}$
Answer$\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{2}{3}$
$\Rightarrow\frac{\text{y}-7+8\text{y}}{9\text{y}-3-4\text{y}}=\frac{2}{3}$
$\Rightarrow\frac{9\text{y}-7}{5\text{y}-3}=\frac{2}{3}$
By cross multiplication:
$\Rightarrow27\text{y}-21=10\text{y}-6$
$\Rightarrow27\text{y}-21=10\text{y}-6+21$ (By transposition)
$\Rightarrow17\text{y}=15$
$\Rightarrow\text{y}=\frac{15}{17}$
$\therefore\text{y}=\frac{15}{17}$Verification:
$\text{L.H.S.}=\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{\text{y-7+8}\text{y}}{9\text{y}-3-4\text{y}}=\frac{9\text{y}-7}{5\text{y}-3}$
$=\frac{9\Big(\frac{15}{17}\Big)-7}{5\Big(\frac{15}{17}\Big)-3}=\frac{\frac{135}{17}-7}{\frac{75}{17}-3}$
$=\frac{\frac{135-199}{17}}{\frac{75-51}{17}}=\frac{\frac{16}{17}}{\frac{24}{17}}=\frac{16}{17}\times\frac{17}{24}$
$=\frac{2}{3}=\text{R.H.S.}$
View full question & answer→Question 45 Marks
The sum of the age of Anup and his father is $100$. When Anup is as his father now, he will be five times as old as his son Anuj is now. will be years older then Anup is now, when Anup is as old as his father. what are their ages now?
AnswerSum of ages of anup and his father $= 100$ years Let present age of Anup $= x$ years
$\therefore$ Age of his father $= (100 - x)$ years
$\therefore$ Age of Anuj $=\frac{100-\text{x}}{5}$ years and also Anuj's age $= (x + 8)$ years ....$I$ Anup becomes as old as his father is now after $(100 - 2x)$ years
$\therefore$ After $(100 - 2x)$ years Anuj's age $=\Big(\frac{100-\text{x}}{25}+100-2\text{x}\Big)$
$=\frac{60 -11\text{x}}{5}$ $.....II$ From $I$ and $I$ $\frac{6000-11\text{x}}{5}=\text{x}+8$
$\Rightarrow 600 - 11x = 5x + 40$
$\Rightarrow\text{x}=\frac{560}{16}=35$
$\therefore$ present age of Anup $= 35$ years and his father's age $= 100 - 35 = 65$ years and
his son's age $= \frac{100-35}{5} $ years $=\frac{65}{5}= 13$ years
View full question & answer→Question 55 Marks
A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?
AnswerLet the Amount, a lady has in the beginning $= Rs.\ x$
In first case, amount spent $=\frac{\text{x}}{2}$
Given to begger $= Rs.\ 1$
$\therefore$ Balance $=\text{x}-\Big(\frac{\text{x}}{2}+1\Big)=\frac{\text{x}}{2}-1$
$=\frac{\text{x}-2}{2}$
Amount spent on lunch $=\frac{\text{x}-2}{2\times2}=\frac{\text{x}-2}{4}$
Give to tip $= Rs.\ 2$
$\therefore$ Balance $=\frac{\text{x}-2}{2}-\Big(\frac{\text{x}-2}{4}+2\Big)$
$=\frac{\text{x}-2}{2}-\frac{\text{x}-2}{4}-2$
$=\frac{2(\text{x}-2)-(\text{x}-2)-2\times4}{4}$
$=\frac{2\text{x}-4-\text{x}+2-8}{4}$
$=\frac{\text{x}-2-8}{4}$
$=\frac{\text{x}-10}{4}$
Amount spent on purchase of a book
$=\frac{\text{x}-10}{4\times2}=\frac{\text{x}-10}{8}$
Given as bus fare $= Rs.\ 3$
$\therefore$ Blance $=\frac{\text{x}-10}{4}-\Big(\frac{\text{x}-10}{8}+3\Big)$
$=\frac{\text{x}-10}{4}-\frac{\text{x}+10}{8}-3=\frac{\text{x}-10}{8}-3$
According to the condition:
$=\frac{\text{x}-10}{8}-3=1$
$\Rightarrow\frac{\text{x}-10}{8}=1+3$
$\Rightarrow\frac{\text{x}-10}{8}=4$
$\Rightarrow\text{x} - 10 = 32$
$\Rightarrow\text{x} = 32 + 10 = 42$
The lady has an amount in the beginning
$= Rs.\ 42$
View full question & answer→Question 65 Marks
Solve the following equation and verify your answer: $\frac{1-9\text{y}}{19-3\text{y}}=\frac{5}{8}$
Answer$\frac{1-9\text{y}}{19-3\text{y}}=\frac{5}{8}$
By cross multiplication:
$5(19-3\text{y})=8(1-9\text{y})$
$\Rightarrow95-15\text{y}=8-72\text{y}$
$\Rightarrow15\text{y}+72\text{y}=8-95$
$\Rightarrow57\text{y}=-87$
$\Rightarrow\text{y}=\frac{-87}{57}=\frac{-29}{19}$
$\therefore\text{y}=\frac{-29}{19}$
Verification:
$\text{L.H.S.}=\frac{1-9\text{y}}{19-3\text{y}}=\frac{1-9\Big(\frac{-29}{19}\Big)}{19-3\Big(\frac{-29}{19}\Big)}=\frac{1+\frac{261}{19}}{91+\frac{87}{19}}$
$=\frac{\frac{19+261}{19}}{\frac{361+87}{19}}=\frac{\frac{280}{19}}{\frac{448}{19}}\times\frac{19}{448}$
$=\frac{280}{448}=\frac{280\div56}{448\div56}=\frac{5}{8}=\text{R.H.S.}$
View full question & answer→Question 75 Marks
Solve the following equation and verify your answer: $\Big(\frac{\text{x}+1}{\text{x}-4}\Big)^2=\frac{\text{x}+8}{\text{x}-2}$
Answer$\Big(\frac{\text{x}+1}{\text{x}-4}\Big)^2=\frac{\text{x}+8}{\text{x}-2}$
$\Rightarrow\frac{(\text{x}+1)^2}{(\text{x}-4)^2}=\frac{(\text{x}+8)}{(\text{x}-2)}$
By cross multiplication, $=(\text{x}+1)^2(\text{x}-2)=(\text{x}-4)^2(\text{x}+8)$
$\Rightarrow(\text{x}^2+2\text{x}+1)(\text{x}-2)=(\text{x}^2-8\text{x}+16)(\text{x}+8)$
$\Rightarrow\text{x}^3-2\text{x}^2+2\text{x}^2-4\text{x}+\text{x}-2$
$=\text{x}^3+8\text{x}^2-8\text{x}^2-64\text{x}+16\text{x}+128$
$\Rightarrow\text{x}^3-3\text{x}-2=\text{x}^3-48\text{x}+128$
$\Rightarrow\text{x}^3 -3\text{x}-\text{x}^3+48\text{x}-128+2$ (By transposition)
$\Rightarrow45\text{x}$
$=130$
$\Rightarrow\text{x}=\frac{130}{45}$
$=\frac{26}{9}$
$\therefore\text{x}=\frac{26}{9}$
Verification: $\text{L.H.S}=\bigg(\frac{\text{x}+1}{\text{x}-4}\bigg)^2=\Bigg(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\Bigg)^2$
$=\Bigg(\frac{\frac{26+9}{9}}{\frac{26-36}{9}}\Bigg)^2=\Bigg(\frac{\frac{35}{9}}{\frac{-10}{9}}\Bigg)^2$
$=\bigg(\frac{35}{9}\times\frac{9}{-10}\bigg)^2=\bigg(\frac{-7}{2}\bigg)^2=\frac{49}{4}$
$\text{R.H.S}=\frac{\text{x}+8}{\text{x}-2}=\frac{\frac{26}{9}+8}{\frac{26}{9}-2}=\frac{\frac{26+72}{9}}{\frac{26-18}{9}}=\frac{\frac{98}{9}}{\frac{8}{9}}$
$=\frac{98}{9}\times\frac{9}{8}=\frac{49}{4}$
$\therefore\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 85 Marks
The ages of sonu and Monu are in the ratio $7 : 5$. Ten years hence, the ratio of their ages will be $9 : 7$. Find their present ages.
AnswerRetio in the present ages of sonu and monu $= 7 : 5$
Let age of sonu $= 7x$ years
And age of monu $= 5x$ years
10 years hence,
The age of sonu $= 7x + 10$ years
And age of monu $= 5a + 10$ years
According to the condition:
$\frac{7\text{x}+10}{5\text{x}+10}=\frac{9}{7}$
By cross multiplication,
$7(7x + 10) = 9(5x + 10)$
$\Rightarrow 49x + 70 = 45x + 90$
$\Rightarrow 49x - 45x = 90 - 70$
$\Rightarrow 4x = 20$
$\Rightarrow\text{x}=\frac{20}{4}=5$
$\therefore$ Sonu's present age $= 7x = 7 \times 5 = 35$
and Monu's present age $= 5x = 5 \times 5 = 25$
Years.
Check: $10$ years hence, their ages will be
Sonu's $= 35 + 10 = 45$ years
Monu's $= 25 + 10 = 35$ years
$\therefore$ Ratio $45 : 35 = 9 : 7$ (Dividing by $5$)
which is given. Therefore our answer is
Correct.
View full question & answer→Question 95 Marks
Solve the following equation and also check your result in case:
$\frac{4\text{x}}{9}+\frac{1}{3}+\frac{13}{108}\text{x}=\frac{8\text{x}+19}{18}$
Answer $\frac{4\text{x}}{9}+\frac{1}{3}+\frac{13}{108}\text{x}=\frac{8\text{x}+19}{18}$ $\frac{48\text{x}+36+13\text{x}}{108}=\frac{8\text{x}+19}{18}$ $\frac{61\text{x}+36}{108}=\frac{8\text{x}+19}{18}$ $61\text{x}+36=6(8\text{x}+19)$ [Multiplying both sides by 108] $61\text{x}+36=48\text{x}+144$ $61\text{x}-48\text{x}=144-36$ $13\text{x}=78$ $\text{x}=\frac{78}{13}$ $\text{x}=6$ Thus, $\text{x}=6$ is the solution of the given equation. Check: Substituting $\text{x}=6$ in the given equation, we get:
$\text{L.H.S.}=\frac{4\times6}{9}+\frac{1}{3}+\frac{13}{108}\times6$ $=\frac{24}{9}+\frac{1}{3}+\frac{13}{18}=\frac{48+6+13}{18}=\frac{67}{18}$ $\text{R.H.S.}=\frac{8\times6+19}{18}=\frac{67}{13}$ $\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=6$ View full question & answer→Question 105 Marks
Solve the following equation and verify your answer: $\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}=\frac{-8}{3}$
Answer$\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}=\frac{-8}{3}$
$\Rightarrow\frac{2\text{x}+3-5\text{x}+7}{6\text{x}+11}=\frac{-8}{3}$
$\Rightarrow\frac{-3\text{x}+10}{6\text{x}+11}=\frac{-8}{3}$
By cross multiplication: $3(-3\text{x}+10)=-8(6\text{x}+11)$
$\Rightarrow-9\text{x}+30=-48\text{x}-88$
$\Rightarrow-9\text{x}+48\text{x}=-88-30$
$\Rightarrow39\text{x}$
$=-118$
$\Rightarrow\text{x}=\frac{-118}{39}$
$\therefore\text{x}=\frac{-118}{39}$ Verification, $\text{L.H.S.}=\frac{(2\text{x}+3)-(5\text{x}-7)}{6\text{x}+11}$
$=\frac{\Big(2\times\frac{-188}{39}+3\Big)-\Big(5\times\frac{-118}{39}-7\Big)}{6\times\frac{-188}{39}+11}$
$=\frac{\Big(\frac{-236}{39}+3\Big)-\Big(\frac{-590}{39}-7\Big)}{\frac{-708}{39}+11}$
$=\frac{\Big(\frac{-199}{39}\Big)-\Big(\frac{-863}{39}\Big)}{\frac{-279}{39}}$
$=\frac{\frac{-119}{39}+\frac{863}{39}}{\frac{-279}{39}}=\frac{\frac{-119+863}{39}}{\frac{-279}{39}}=\frac{\frac{744}{39}}{\frac{-279}{39}}$
$=\frac{744}{39}\times\frac{39}{-279}$
$=\frac{-744}{279}=\frac{744\div93}{279\div93}=\frac{-8}{3}=\text{R.H.S.}$
View full question & answer→Question 115 Marks
Solve the following equation and also check your result in case:
$\Big[(2\text{x}+3)+(\text{x+5})\Big]^2+\Big[(2\text{x}+3)-(\text{x}+5)\Big]^2=10\text{x}^2+92$
Answer$\Big[(2\text{x}+3)+(\text{x+5})\Big]^2+\Big[(2\text{x}+3)-(\text{x}+5)\Big]^2=10\text{x}^2+92$
$(3\text{x}+8)^2+(\text{x}-2)^2=10\text{x}^2+92$
$9\text{x}^2+48\text{x}+64+\text{x}^2-4\text{x}+4=10\text{x}^2+92$
$\Big[(\text{a+b})^2=\text{a}^2+\text{b}^2+2\text{ab and}(\text{a-b)}^2=\text{a}^2+\text{b}^2-2\text{ab}\Big]$
$10\text{x}^2-10\text{x}^2+44\text{x}=92-68$
$\text{x}=\frac{24}{44}$
$\text{x}=\frac{6}{11}$
Thus, $\text{x}=\frac{6}{11}$ is the solution of the given equation.
Check: Substituting $\text{x}=\frac{6}{11}$ in the given equation, we get:
$\text{L.H.S.}=\Big[(2\times\frac{6}{11}+3)+(\frac{6}{11}+5)\Big]^2$
$\ +\Big[(2\times\frac{6}{11}+3)-(\frac{6}{11}+5)\Big]^2$
$=\Big[(\frac{45}{11})+(\frac{61}{11})\Big]^2+\Big[(\frac{45}{11})-(\frac{16}{11})\Big]^2$
$=\Big(\frac{106}{11}\Big)^2+\Big(\frac{-16}{11}\Big)^2$
$\text{R.H.S.}=10\times\Big(\frac{6}{11}\Big)+92$
$=\frac{360}{121}+92=\frac{11492}{121}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{6}{11}$
View full question & answer→Question 125 Marks
$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$
Answer$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$
$=\frac{5(2\text{x}-1)-3(6\text{x}-2)}{15}=\frac{1}{3}$
$=\frac{10\text{x}-5-18\text{x}+6}{15}=\frac{1}{3}$
$(L.C.M.$ of $3, 5 = 15)$
$=\frac{-8\text{x}+1}{15}=\frac{1}{3}$
$\Rightarrow(-8\text{x}+1)\times3=1\times15$ (By cross multiplication)
$\Rightarrow$ Dividing by 3 $\frac{(-8\text{x}+1)\times3}{3}=\frac{1\times15}{3}$
$\Rightarrow-8\text{x}+1=5$
Subtracting $1$ from both side, $-8\text{x}+1-1=5-1$
$\Rightarrow-8\text{x}=4$
Dividing by $-8$, $\frac{-8\text{x}}{-8}=\frac{4}{-8}$
$\Rightarrow\text{x}=\frac{1}{-2}$
$\therefore\text{x}=\frac{-1}{2}$
Verification: $\text{L.H.S.}=\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}$
$=\frac{2\Big(\frac{-1}{2}\Big)-1}{3}-\frac{6\Big(\frac{-1}{2}\Big)-2}{5}$
$=\frac{-1-1}{3}-\frac{-3-2}{5}$
$=\frac{-2}{3}-\frac{-5}{5}=\frac{-2}{3}+1$
$=\frac{1}{3}=\text{R.H.S.}$
View full question & answer→Question 135 Marks
Solve the following equation and verify your answer: $\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=2$
Answer$\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=2$
$=\frac{(\text{x}+3)(\text{x}-2)+(\text{x}+2)(\text{x}-3)}{(\text{x}-3)(\text{x}-2)}=2$
$=\frac{\text{x}^2-2\text{x}+3\text{x}-6+\text{x}^2-3\text{x}+2\text{x}-6}{\text{x}^2-2\text{x}-3\text{x}+6}=2$
$\Rightarrow\frac{2\text{x}^2-12}{\text{x}^2-5\text{x}+6}=\frac{2}{1}$
By cross multiplication: $2\text{x}^2-12=2\text{x}^2-10\text{x}+12$
$\Rightarrow2\text{x}^2-2\text{x}^2+10\text{x}=12+12$ (By transposition)
$\Rightarrow10\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{10}=\frac{12}{5}$
$\therefore\text{x}=\frac{12}{5}$
Verification: $\text{L.H.S.}=\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=\frac{\frac{12}{5}+3}{\frac{12}{5}-3}+\frac{\frac{12}{5}+2}{\frac{12}{5}-2}$
$=\frac{\frac{12+12}{5}}{\frac{12-15}{5}}+\frac{\frac{12+10}{5}}{\frac{12-10}{5}}=\frac{\frac{27}{5}}{\frac{-3}{5}}+\frac{\frac{22}{5}}{\frac{2}{5}}$
$=\frac{27}{5}\times\frac{5}{-3}+\frac{22}{5}\times\frac{5}{2}$
$=-9+11=2=\text{R.H.S.}$
View full question & answer→Question 145 Marks
Solve the following equation and also check your result in case:
$\frac{(45-2\text{x)}}{15}-\frac{(4\text{x}+10)}{5}=\frac{(15-14\text{x})}{9}$
Answer$\frac{45-2\text{x}}{15}-\frac{4\text{x}+10}{5}=\frac{15-14\text{x}}{9}$
$\frac{45-2\text{x}-12\text{x}-30}{15}=\frac{15-14\text{x}}{9}$
$\frac{15-14\text{x}}{5}=\frac{15-14\text{x}}{3}$ [Multiplying both sides by $3$]
$45-42\text{x}=75-70\text{x}$ [After cross multiplication] $70\text{x}+42\text{x}=75-45$
$28\text{x}=30$
$\text{x}=\frac{30}{28}$
$\text{x}=\frac{15}{14}$
Thus, $\text{x}=\frac{15}{14}$ is the solution of the given equation.
Check:Substituting $\text{x}=\frac{15}{14}$ in the given equation,
we get:
$\text{L.H.S.}=\frac{45-2\times\frac{15}{14}}{15}-\frac{4\times\frac{15}{14}+10}{5}=\frac{45\times7-15}{105}-\frac{30+70}{35}$
$=\frac{300}{105}-\frac{100}{35}=0$
$\text{R.H.S.}=\frac{15-14\times\frac{15}{14}}{9}=0$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{15}{14}$
View full question & answer→Question 155 Marks
$\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)-\frac{9}{2}$
Answer$\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)-\frac{9}{2}$
$\Rightarrow\frac{2}{3}\text{x}-\frac{10}{3}-\frac{1}{4}\text{x}+\frac{1}{2}=\frac{9}{2}$
$\Rightarrow\frac{2}{3}\text{x}-\frac{1}{4}\text{x}-\frac{10}{3}+\frac{1}{2}=\frac{9}{2}$
$\frac{8\text{x} - 3\text{x}-40+6=54}{12}=\frac{9}{2}$
$(L.C.M.$ of $3, 4, 2 = 12)$$5\text{x}=-34=\frac{9\times12}{2}=9\times6=54$
$5\text{x}-34=54$ (Adding $34$ to both sides)
$5\text{x}-34+34=54+34$
$\Rightarrow5\text{x}=88$
$\Rightarrow\text{x}=\frac{88}{5}$
Verification:
$\text{L.H.S.}=\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)$
$=\frac{2}{3}\Big(\frac{88}{5}-5\Big)-\frac{1}{4}\Big(\frac{88}{5}-2\Big)$
$=\frac{2}{3}\Big(\frac{88 - 25}{5}\Big)-\frac{1}{4}\Big(\frac{88-10}{5}\Big)$
$=\frac{2}{3}\times\frac{63}{5}-\frac{1}{4}\times\frac{78}{5}$
$=\frac{42}{5}-\frac{39}{10}$
$=\frac{84-39}{10}=\frac{45}{10}=\frac{9}{2}=\text{R.H.S.}$
View full question & answer→Question 165 Marks
Solve the following equation and verify your answer: $\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$
Answer$\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$
By cross multiplication, $(9\text{x}-7)(\text{x}+6)=(3\text{x}-4)(3\text{x}+5)$
$\Rightarrow9\text{x}^2+54\text{x}-7\text{x}-42=9\text{x}^2+15\text{x}-12\text{x}-20$
$\Rightarrow9\text{x}^2+47\text{x}-42=9\text{x}^2+3\text{x}-20$
$\Rightarrow9\text{x}^2+47\text{x}-9\text{x}^2-3\text{x}=-20+42$ (By transposition)
$\Rightarrow44\text{x}$
$=22$
$\Rightarrow\text{x}=\frac{22}{44}$
$=\frac{1}{2}$
$\therefore\text{x}=\frac{1}{2}$
Verification: $\text{L.H.S}=\frac{9\text{x}-7}{3\text{x}+5}=\frac{9\times\frac{1}{2}-7}{3\times\frac{1}{2}+5}=\frac{\frac{9}{2}-7}{\frac{3}{2}+5}$
$=\frac{\frac{9-14}{2}}{\frac{3+10}{2}}=\frac{\frac{-5}{2}}{\frac{13}{2}}=\frac{-5}{2}\times\frac{2}{13}=\frac{-5}{13}$
$\text{R.H.S}=\frac{3\text{x}-4}{\text{x+6}}=\frac{3\times\frac{1}{2}-4}{\frac{1}{2}+6}=\frac{\frac{3}{2}-4}{\frac{1}{2}+6}$
$=\frac{\frac{3-8}{2}}{\frac{1+12}{2}}=\frac{\frac{-5}{2}}{\frac{13}{2}}=\frac{-5}{2}\times\frac{2}{13}=\frac{-5}{13}$
$\therefore\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 175 Marks
Solve the following equation and verify your answer: $\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=6$
Answer$\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-(\text{x}^2+2\text{x}+\text{x}+2)}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{\text{x}^2-\text{x}^2-2\text{x}-\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
$\Rightarrow\frac{-3\text{x}-2}{5\text{x}+1}=\frac{6}{1}$
By cross multiplication: $-3\text{x}-2=6(5\text{x}+1)$
$\Rightarrow-3\text{x}-2=30\text{x}+6$
$\Rightarrow-3\text{x}-30=6+2$ (By transposition)
$\Rightarrow-33\text{x}=8$
$\Rightarrow\text{x}=\frac{8}{-33}$
$=\frac{-8}{33}$
$\therefore\text{x}=\frac{-8}{33}$ Verification: $\text{L.H.S.}=\frac{\text{x}^2-(\text{x}+1)(\text{x}+2)}{5\text{x}+1}$
$=\frac{\Big(\frac{-8}{33}\Big)^2-\Big(\frac{-8}{33}+1\Big)\Big(\frac{-8}{33}+2\Big)}{\Big(\frac{-8}{33}\Big)+1}$
$=\frac{\frac{64}{1089}-\Big(\frac{25}{33}\times\frac{58}{33}\Big)}{\frac{-40+33}{33}}$
$=\frac{\frac{64}{1089}-\frac{1450}{1089}}{\frac{-7}{33}}=\frac{\frac{64-1450}{1089}}{\frac{-7}{33}}$
$=\frac{-1386}{1089}\times\frac{33}{-7}=6=\text{R.H.S.}$
View full question & answer→Question 185 Marks
Solve the following equation and verify your answer: $\Big(\frac{\text{x}+1}{\text{x}+2}\Big)^2=\frac{\text{x}+2}{\text{x}+4}$
Answer$\Big(\frac{\text{x}+1}{\text{x}+2}\Big)^2=\frac{\text{x}+2}{\text{x}+4}$
$\Rightarrow\frac{(\text{x}+1)^2}{(\text{x}+2)^2}$
$=\frac{(\text{x}+1)}{(\text{x}+4)}$
By cross multiplication,
$=(\text{x}+1)^2\times(\text{x}+4)=(\text{x}+2)^2(\text{x}+2)$
$=(\text{x}^2+2\text{x}+1)(\text{x}+4)=(\text{x}^2+4\text{x}+4)(\text{x}+2)$
$\Rightarrow\text{x}^3+4\text{x}^2+2\text{x}^2+8\text{x}+\text{x}+ 4$
$=\text{x}^3+2\text{x}^2+4\text{x}^2+8\text{x}+4\text{x}+8$
$\Rightarrow\text{x}^3+6\text{x}^2+9\text{x}+4=\text{x}^3+6\text{x}^2+12\text{x}+8$
$\Rightarrow\text{x}^3+6\text{x}^2+9\text{x}-\text{x}^3-6\text{x}^2-12\text{x}=8-4$ (By transposition)
$\Rightarrow-3\text{x}$
$=4$
$\Rightarrow\text{x}=\frac{4}{-3}$
$\therefore\text{x}=\frac{-4}{3}$
Verification: $\text{L.H.S}=\bigg(\frac{\text{x}+1}{\text{x}+2}\bigg)^2=\Bigg(\frac{\frac{-4}{3}+1}{\frac{-4}{3}+2}\Bigg)$
$=\Bigg(\frac{\frac{-4+3}{3}}{\frac{-4+6}{3}}\Bigg)^2=\Bigg(\frac{\frac{-1}{3}}{\frac{2}{3}}\Bigg)^2$
$=\bigg(\frac{\frac{-1}{3}}{3}\times\frac{3}{2}\bigg)^2=\bigg(\frac{-1}{2}\bigg)^2=\frac{1}{4}$
$\text{R.H.S}=\frac{\text{x}+2}{\text{x}+4}=\frac{\frac{-4}{3}+2}{\frac{-4}{3}+2}=\frac{\frac{-4+6}{3}}{\frac{-4+12}{3}}$
$=\frac{\frac{2}{3}}{\frac{8}{3}}=\frac{2}{3}\times\frac{3}{8}=\frac{1}{4}$
$\therefore\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 195 Marks
Solve the following equation and also check your result in case: $\frac{9\text{x}+7}{2}-\Big(\text{x}-\frac{\text{x-2}}{7}\Big)=36$
Answer$\frac{9\text{x}+7}{2}-\Big(\text{x}-\frac{\text{x-2}}{7}\Big)=36$
$\frac{63\text{x}+49-14\text{x}+2\text{x}-4}{14}=36$
$\frac{51\text{x}+45}{14}=36$
$51\text{x}+45=504$
$51\text{x}=504-45$
$\text{x}=\frac{159}{51}=9$
Thus, $\text{x}=9$ is the solution of the given equation.
Check:Substituting $\text{x}=9$ in the given equation,
we get:
$\text{L.H.S.}=\frac{9\times9+7}{2}-\Big(9-\frac{9-2}{7}\Big)$
$=\frac{88}{2}-9 \ +\frac{7}{7}=44-9+1=36$
$\text{R.H.S.}=36$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-9$
View full question & answer→Question 205 Marks
A number whose fifth part increased by $5$ is equal to its fourth part diminished by $5$. Find the number.
AnswerLet the requierd number $= x$
Then fifth part increased by $5=\frac{\text{x}}{5}+5$
fourth part diminished by $5=\frac{\text{x}}{4}+5$
According to the condition: $\frac{\text{x}}{5}+5=\frac{\text{x}}{4}-5$
$\Rightarrow\frac{\text{x}}{5}-\frac{\text{x}}{4}=-5-5$
$\Rightarrow\text{x}=\frac{4\text{x}-5\text{x}}{20}=-10$
$\Rightarrow-\frac{\text{x}}{20}=-10$
$\Rightarrow\text{x}=-10\times-20=200$
Required number $= 200$
Check: $\frac{200}{5}+5=\frac{200}{4}-5$
$\Rightarrow40 + 5 = 50 - 5$
$\Rightarrow 45 = 45$
Which is true. therefore our answer is correct.
View full question & answer→Question 215 Marks
Solve the following equation and also check your result in case: $6.5\text{x}+\frac{19.5\text{x}-32.5}{2}=6.5\text{x}+13+\Big(\frac{13\text{x}-26}{2}\Big)$
Answer$6.5\text{x}+\frac{19.5\text{x}-32.5}{2}=6.5\text{x}+13+\frac{13\text{x}-26}{2}$
$\frac{19.5\text{x}-32.5}{2}-\frac{13\text{x}-26}{2}=13$
$\frac{19.5\text{x}-32.5-13\text{x}+26}{2}=13$
$6.5\text{x}-6.5=26$ [After cross miltiplication]
$6.5\text{x}=26+6.5$
$\text{x}=\frac{32.5}{6.5}=5$
Thus, $\text{x}=5$ is the solution of the given equation.
Check:Substituting $\text{x}=5$ in the given equation,
we get:
$\text{L.H.S.}=6.5\times5+\frac{19.5\times5-32.5}{2}=65$
$\text{R.H.S.}=6.5\times5+13+\frac{13\times5-26}{2}=65$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=5$
View full question & answer→Question 225 Marks
Solve the following equation and also check your result in case: $\frac{7\text{x}-1}{4}-\frac{1}{3}\Big(2\text{x}-\frac{1-\text{x}}{2}\Big)=\frac{10}{3}$
Answer$\frac{7\text{x}-1}{4}-\frac{1}{3}\Big(2\text{x}-\frac{1-\text{x}}{2}\Big)=\frac{10}{3}$
$\frac{7\text{x}-1}{4}-\frac{2\text{x}}{3}+\frac{1-\text{x}}{6}=\frac{10}{3}$
$\frac{21\text{x}-3-8\text{x}+2-2\text{x}}{12}=\frac{10}{3}$
$11\text{x}-1=40$ [Multiplying both sides by $12$]
$11\text{x}=40+1$
$\text{x}=\frac{41}{11}$
Thus, $\text{x}-\frac{41}{11}$ is the solution of the given equation.
Check:Substituting $\text{x}-\frac{41}{11}$ in the given equation, we get:
$\text{L.H.S.}=\frac{7\times\frac{41}{11}-1}{4}-\frac{1}{3}\Bigg(2\times\frac{41}{11}-\frac{1-\frac{41}{11}}{2}\Bigg)$
$=\frac{276}{44}-\frac{82}{33}+\frac{-30}{66}=\frac{10}{3}$
$\text{R.H.S.}=\frac{10}{3}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{41}{11}$
View full question & answer→Question 235 Marks
A sum of $Rs 800$ is in the form of denominations of $Rs 10$ and $Rs 20$. If the total number of notes be $50$, find the number of notes of each type.
AnswerTotle amount $= Rs. 800$
Totle number of notes $= 50$
Let number of Nots of $Rs. 10 = x$
Then number of notes of $Rs. 20 = 50 - x$
According to the condition, $x × 10 + (50 - x) × 20 = 800$
$\Rightarrow 10x + 1000 - 20x = 800$
$ \Rightarrow -10x = 800 - 1000 = -200$
$\Rightarrow\text{x}=\frac{-200}{-10}=20$
$\therefore$ Number of $10$- repees notes $= 20$ And number of $20$-rupees noutes $= 50 - 20 = 30$
Check:$20 \times 10 + 30 \times 20 = 200 + 600 = 800$ Which is true. threrefore our is correct.
View full question & answer→Question 245 Marks
$\frac{\text{x}}{2}-\frac{4}{5}+\frac{\text{x}}{5}+\frac{3\text{x}}{10}=\frac{1}{5}$
Answer$\frac{\text{x}}{2}-\frac{4}{5}+\frac{\text{x}}{5}+\frac{3\text{x}}{10}=\frac{1}{5}$ Adding to $\frac{4}{5}$ both sides, $\frac{\text{x}}{2}+\frac{\text{x}}5{}+\frac{3\text{x}}{10}-\frac{4}{5}+\frac{4}{5}=\frac{1}{5}+\frac{4}{5}$
$\Rightarrow\frac{\text{x}}{2}+\frac{\text{x}}{5}+\frac{3\text{x}}{10}-\frac{5}{5}=1$
$\frac{5\text{x}+2\text{x}+3\text{x}}{10}=1$ $(L.C.M$ of $2, 5, 10 = 1)$ $\frac{10\text{x}}{10}=1$
$\Rightarrow\text{x}=1$
$\therefore\text{x}=1$
Verification: $\text{L.H.S}=\frac{\text{x}}{2}-\frac{4}{5}+\frac{\text{x}}{5}+\frac{3\text{x}}{10}$
$=\frac{1}{2}-\frac{4}{5}+\frac{1}{5}+\frac{3}{10}$
$=\frac{5-8+2+3}{10}=\frac{10-8}{10}=\frac{2}{10}$
$=\frac{1}{5}= \text{R.H.S}$
View full question & answer→Question 255 Marks
Solve the following equation and also check your result in case: $\frac{0.5(\text{x} - 0.4)}{0.35}-\frac{0.6(\text{x - 2.71})}{0.42}=\text{x}+6.1$
Answer$\frac{0.5(\text{x} - 0.4)}{0.35}-\frac{0.6(\text{x - 2.71})}{0.42}=\text{x}+6.1$
$\frac{(\text{x}-0.4)}{0.7}-\frac{(\text{x}-2.71)}{0.7}=\text{x}+6.1$
$\frac{\text{x}-0.4-\text{x}+2.71}{0.7}=\text{x}+6.1$
$-0.4+2.71=0.7\text{x}+4.27$
$0.7\text{x}=2.71-0.4-4.27$
$\text{x}=\frac{-1.96}{0.7}=-2.8$
Thus, $\text{x}=-2.8$ is the solution of the given equation.
Check:Substituting $\text{x}=-2.8$ in the given equation, we get:
$\text{L.H.S.}=\frac{0.5(-2.8-0.4)}{0.35}-\frac{0.6(-2.8-2.71)}{0.42}$
$=\frac{-1.6}{0.35}+\frac{3.306}{0.42}=-4.571+7.871=3.3$
$\text{R.H.S.}=-2.8+6.1=-3.3$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-2.8$
View full question & answer→Question 265 Marks
The numerator of a fraction is $6$ less than the denominator. If $3$ is added to the numerator, the fraction is equal to $\frac{2}{3}.$ What is the original fraction equal to?
AnswerLet denominator of the originel fraction $= x$
Then numberator $= x - 6$ And fraction $=\frac{\text{x}-6}{\text{x}}$
According to the condition: $\frac{\text{x}-6+3}{\text{x}}=\frac{2}{3}$
$\Rightarrow\frac{\text{x}-3}{\text{x}}=\frac{2}{3}$
$\Rightarrow 3(x - 3) = 2x$ (By cross multiplication)
$3x - 9 = 2x$
$\Rightarrow 3x - 2x = 9$
$\Rightarrow x = 9$
$\therefore$ Original fraction $=\frac{\text{x}-6}{\text{x}}=\frac{9 -6}{9}=\frac{3}{9}$
Cheek: $\frac{3+3}{9}=\frac{6}{9}=\frac{2}{3}$ Which is given. therefore our answer is correct.
View full question & answer→Question 275 Marks
Solve the following equation and verify your answer: $\frac{3\text{x}+5}{2\text{x}+7}=4$
Answer$\frac{3\text{x}+5}{2\text{x}+7}=\frac{4}{1}$
By cross multiplication:
$4(2\text{x}+7)=1\times(3\text{x}+5)$
$8\text{x}+28=3\text{x}+5$
$8\text{x}-3\text{x}=5-28$ (By transposition)
$\Rightarrow5\text{x}=-23$
$\Rightarrow\text{x}=\frac{-23}{5}$
$\text{x}=\frac{-23}{5}$Verification:
$\text{L.H.S.}=\frac{3\text{x}+5}{2\text{x}+7}=\frac{3\Big(\frac{-23}{5}\Big)+5}{2\Big(\frac{-23}{5}\Big)+7}$
$=\frac{\frac{-69}{5}+5}{\frac{-46}{5}+7}=\frac{\frac{-69+25}{5}}{\frac{-46+35}{5}}=\frac{\frac{-44}{5}}{\frac{-11}{5}}$
$\frac{-44}{5}\times\frac{-5}{11}=4=\text{R.H.S.}$
View full question & answer→Question 285 Marks
Solve the following equation and verify your answer: $\frac{2-\text{y}}{\text{y}+7}=\frac{3}{5}$
Answer$\frac{2\text{x}-3}{3\text{x}+2}=-\frac{2}{3}$By cross multiplication:
$3(\text{y}+7)=5(2-\text{y})$
$\Rightarrow3\text{y}+21=10-5\text{y}$
$\Rightarrow3\text{y}+5\text{y}=10-21$ (By transposition)
$\Rightarrow8\text{y}=-11$
$\Rightarrow\text{y}=\frac{-11}{8}$
$\therefore\text{y}=\frac{-11}{8}$Verification:
$\text{L.H.S.}=\frac{2-\text{y}}{\text{y}+7}=\frac{2+\frac{11}{8}}{\frac{-11}{8}+7}$
$=\frac{\frac{16+11}{8}}{\frac{-11+56}{8}}=\frac{\frac{27}{8}}{\frac{45}{8}}$
$=\frac{27}{8}\times\frac{8}{45}=\frac{27}{45}=\frac{3}{5}=\text{R.H.S.}$
View full question & answer→Question 295 Marks
Bhagwanti inherited $Rs. 12000.00$. She invested part of it as $10\%$ and the rest at $12\%$. Her annual income from these investments is $Rs. 1280.00$. How much did she invest at each rate?
AnswerTotal investment $= Rs. 12000.00$
Rate of interest for first part $= 10\%$ and for second part $= 12\%$
Annual income $= Rs. 1280.00$
Let the inverstment for the firest part $= Rs. x$ and
second part $= Rs. (12000 - x)$
According to the condition: $\frac{\text{x}\times10}{100}+(12000-\text{x})\times\frac{12}{100}=1280$
$\Rightarrow\frac{10\text{x}}{100}+\frac{10\text{x}+144000-12\text{x}}{100}=1280$
$\Rightarrow 144000 - 2x = 128000$
$ \Rightarrow 144000 - 128000 = 2x$
$\Rightarrow2\text{x}=16000$
$\Rightarrow\text{x}=\frac{160000}{2}=8000$
$\therefore$ $x = 8000$
$\therefore$ Amount invested on $10\% = Rs. 8000$ and amount invested on $12\% = Rs. 12000 - 8000$ $Rs. 4000$
View full question & answer→Question 305 Marks
Solve the following equation and also check your result in case: $\frac{6\text{x}+1}{2}+1=\frac{7\text{x}-3}{3}$
Answer$\frac{6\text{x}+1}{2}+1=\frac{7\text{x}-3}{3}$
$\frac{6\text{x}+1+2}{2}=\frac{7\text{x-3}}{3}$
$18\text{x}+9 =14\text{x}-6$
$18\text{x}-14\text{x}=-6-9$
$4\text{x}=-15$
$\text{x}=\frac{-15}{4}$
Check: $\text{L.H.S.}=\frac{6\times\frac{-15}{4}+1}{2}+1=\frac{-45+2+4}{4}=\frac{-39}{4}$
$\text{R.H.S.}=\frac{7\times\frac{-15}{4}-3}{2}\frac{-105-12}{12}=\frac{-39}{4}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{-15}{4}$
View full question & answer→Question 315 Marks
Ravish has three boxes whose total weight is $60\frac{1}{2}$ $kg$. Box $B$ weighs $3\frac{1}{2}$ $kg$. more then box $A$ and box $C$ weighs $5\frac{1}{3}$ $kg$. more than box $B$. Find the weight of box $A$.
AnswerTotal weight of three boxes $=60\frac{1}{2}\text{kg.}$
Let weight of box $A = x\ kg$.
Then weight of box $\text{B}=\Big(\text{x}+3\frac{1}{2}\Big)\text{kg.}$
and weight of box $\text{C}=\Big(\text{x}+3\frac{1}{2}+5\frac{1}{3}\Big)\text{kg.}$
$\therefore$ According to the condition, $\text{x}+\Big(\text{x}+3\frac{1}{2}\Big)+\Big(\text{x}+3\frac{1}{2}+5\frac{1}{3}\Big)=60\frac{1}{2}$
$\Rightarrow\text{x}+\text{x}+\frac{7}{2}+\text{x}+\frac{7}{2}+\frac{16}{3}=\frac{121}{2}$
$\Rightarrow3\text{x}+\Big(\frac{7}{2}+\frac{7}{2}+\frac{16}{3}\Big)=\frac{121}{2}$
$\Rightarrow3\text{x}=\frac{121}{2}-\Big(\frac{7}{2}+\frac{7}{2}+\frac{16}{3}\Big)$
$\Rightarrow3\text{x}=\frac{363-(21+21+32)}{6}=\frac{363-74}{6}$
$\Rightarrow3\text{x}=\frac{289}{6}$
$\Rightarrow\text{x}=\frac{289}{6\times3}=\frac{289}{18}\text{ kg.}$
$\therefore$ Weight to box $\text{A}=\frac{289}{18}\text{kg.}$
View full question & answer→Question 325 Marks
Solve the following equation and also check your result in case: $(3\text{x}-8)(3\text{x+2})-(4\text{x}-11)(2\text{x}+1)=(\text{x}-3)(\text{x}+7)$
Answer$(3\text{x}-8)(3\text{x+2})-(4\text{x}-11)(2\text{x}+1)=(\text{x}-3)(\text{x}+7)$
$9\text{x}^2+6\text{x}-24\text{x}-16-8\text{x}^2-4\text{x}+22\text{x}+11$
$=\text{x}^2+7\text{x}-3\text{x}-21$
$\text{x}^2-5=\text{x}^2+4\text{x}-21$
$4\text{x}=-5+21$
$\text{x}=\frac{16}{4}=4$
Thus, $\text{x}=4$ is the solution of the given equation.
Check:Substituting $\text{x}=4$ in the given equation, we get:
$\text{L.H.S.}=(3\times4-8)(3\times4+2)-(4\times4-11)(2\times4+1)$
$=4\times14-5\times9=11$
$\text{R.H.S.}=(4-3)(4+7)=11$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=4$
View full question & answer→Question 335 Marks
Solve the following equation and also check your result in case: $\frac{3}{4}\text{x}+4\text{x}=\frac{7}{8}+6\text{x}-6$
Answer$\frac{3}{4}\text{x}+4\text{x}=\frac{7}{8}+6\text{x}-6$
$\frac{3}{4}\text{x}-2\text{x}=\frac{7}{8}-6$
$\frac{3\text{x}-8\text{x}}{4}=\frac{7-48}{8}$
$-40\text{x}=-164$
$\text{x}=\frac{164}{40}=\frac{41}{10}$
Check: $\text{L.H.S.}=\frac{3}{4}\times\frac{41}{10}+4\times\frac{41}{10}$
$=\frac{123}{40}+\frac{164}{10}=\frac{123+656}{40}=\frac{779}{40}$
$\text{R.H.S.}=\frac{7}{8}+6\times\frac{41}{10}-6$
$=\frac{7}{8}+\frac{246}{10}-6=\frac{35+984-240}{40}=\frac{779}{40}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{41}{10}$
View full question & answer→Question 345 Marks
$\frac{\text{x}}{2}+\frac{\text{x}}{8}=\frac{1}{8}$
Answer$\frac{\text{x}}{2}+\frac{\text{x}}{8}=\frac{1}{8}$
$\frac{4\text{x}+\text{x}}{8}=\frac{1}{8}$ $(L.C.M$ of $2, 8 = 8)$
$\frac{5\text{x}}{8}=\frac{1}{8}$
Dividing by $\frac{5}{8}$
$\frac{5}{8}\text{x}+\frac{5}{8}=\frac{1}{8}+\frac{5}{8}$
$\Rightarrow\frac{5}{8}\text{x}\times\frac{8}{5}=\frac{1}{8}\times\frac{5}{8}$
$\Rightarrow\text{x}=\frac{1}{5}$
$\therefore\text{x}=\frac{1}{5}$
Verification:
$\text{L.H.S}$ $=\frac{\text{x}}{2}+\frac{\text{x}}{8}$
$=\frac{\frac{1}{5}}{2}+\frac{\frac{1}{5}}{8}=\frac{1}{5\times2}+\frac{1}{5\times8}$
$=\frac{1}{10}+\frac{1}{40}=\frac{4+1}{40}=\frac{5}{40}$
$=\frac{1}{8}=\text{R.H.S}$
View full question & answer→Question 355 Marks
Solve the following equation and verify your answer: $\frac{6}{2\text{x}-(3-4\text{x})}=\frac{2}{3}$
Answer$\frac{6}{2\text{x}-(3-4\text{x})}=\frac{2}{3}$
$\Rightarrow\frac{6}{2\text{x}-3+4\text{x}}=\frac{2}{3}$
$\Rightarrow\frac{6}{6\text{x}-3}=\frac{2}{3}$
By cross multiplication:
$2(6\text{x}-3)=6\times3$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{12}=2$
$\therefore\text{x}=2$
Verification:
$\text{L.H.S.}=\frac{6}{2\text{x}-(3-4\text{x})}=\frac{6}{2\times2-(3-4\times2)}$
$=\frac{6}{4-(3-8)}=\frac{6}{4+5}=\frac{6}{9}$
$=\frac{2}{3}=\text{R.H.S.}$
View full question & answer→Question 365 Marks
Divide $184$ into two parts such that one-third of one part may exceed one-seventh of another part by $8$.
AnswerSum of two parts $= 184$
Let first part $= x$ Then second part $= 184 - x$
According to the condition: $\frac{\text{x}}{3}=\frac{184-\text{x}}{7}+8$
$\Rightarrow\frac{\text{x}}{3}=\frac{(184-\text{x})}{7}=8$
$\frac{7\text{x}-552+3\text{x}=168}{21}$ ($\because$ $L.C.M.$ of $3, 7 = 21)$
$10\text{x} = 168 + 552 = 720$
$\text{x}=\frac{720}{10}=72$
$\therefore$ First part $= 72$ And second part $= 184 - 72 = 112$
Cheek: $\frac{72}{3}=\frac{112}{7}+8$
$\Rightarrow 24 = 16 + 8$
$\Rightarrow 24 = 24$ Which is true.
Therefore our answer is correct.
View full question & answer→Question 375 Marks
Solve the following equation and verify your answer: $\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=2$
Answer$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=\frac{2}{1}$
By cross multiplication: $(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6=2(\text{x}-5)$
$2\text{x}^2-3\text{x}+4\text{x}-6-2\text{x}^2+6=2\text{x}-10$
$\Rightarrow-\text{x}=-10$
$\Rightarrow\text{x}=10$
$\therefore\text{x}=10$
Verification: $\text{L.H.S.}=\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}$
$=\frac{(10+2)(2\times10-3)-2(10)^2+6}{10-5}$
$=\frac{12(20-3)-2\times(100)+6}{5}$
$=\frac{12\times17-200+6}{5}$
$=\frac{204-200+6}{5}$
$=\frac{10}{5}=2=\text{R.H.S.}$
View full question & answer→Question 385 Marks
$\frac{7}{\text{x}}+35=\frac{1}{10}$
Answer$\frac{7}{\text{x}}+35=\frac{1}{10}$
Subtracting $35$ from both sides, $\frac{7}{\text{x}}+35-35=\frac{1}{10}-35$
$\frac{7}{\text{x}}=\frac{1-350}{10}$
$\Rightarrow\frac{7}{\text{x}}=\frac{-349}{10}$
$\Rightarrow-349\times\text{x}=7\times10$
$\Rightarrow\text{x}=\frac{7\times10}{-349}=\frac{70}{-349}=\frac{-70}{349}$ (By cross multiplication)
$\therefore\text{x}=\frac{-70}{349}$
Verification: $\text{L.H.S}=\frac{7}{\text{x}}+35$
$=\frac{7}{\frac{-70}{349}}+35=\frac{-7\times349}{70}+35$
$=\frac{-349}{10}+35$
$=\frac{-349+350}{10}=\frac{1}{10}=\text{R.H.S}$
View full question & answer→Question 395 Marks
$\frac{2\text{x}}{3}-\frac{3\text{x}}{8}=\frac{7}{12}$
Answer$\frac{2\text{x}}{3}-\frac{3\text{x}}{8}=\frac{7}{12}$
$\frac{16\text{x}-9\text{x}}{24}=\frac{7}{12}$ $(L.C.M$ of $3, 8 = 24)$
$\frac{7\text{x}}{24}=\frac{7}{12}$ Dividing by $\frac{7}{24}$
$=\frac{7\text{x}}{24}+\frac{7}{24}=\frac{7}{12}+\frac{7}{24}$
$\Rightarrow\frac{7\text{x}}{24}\times\frac{24}{7}=\frac{7}{12}\times\frac{24}{12}$
$\Rightarrow\text{x}=2$
$\therefore\text{x}=2$ Verification: $\text{L.H.S}$
$=\frac{2\text{x}}{3}-\frac{\text{3x}}{8}=\frac{3\text{x}}{8}$
$=\frac{2\times2}{3}-\frac{3\times2}{8}=\frac{4}{3}-\frac{6}{8}$
$=\frac{4}{3}-\frac{3}{4}=\frac{16-9}{12}=\frac{7}{12}$
$=\frac{7}{12}=\text{R.H.S}$
View full question & answer→Question 405 Marks
Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Answer$5$ years ago, Let age of son = years Then, age of father $= 7\ a$ years
Presenting age of sob $= x + 5$ years and age of father $= 7x + 5$ years $5$ years hance,
age of son $= x + 5 + 5 = x + 10$ and age of father $= 7x + 5 + 5 = 7x + 10$
According to the condition: $\Rightarrow 7x + 10 = 3x + 30$
$\Rightarrow 7x - 3x = 30 - 10 = 20$
$\Rightarrow4\text{x}=20$
$\Rightarrow\text{x}=\frac{20}{4}=5$
$\therefore $ Present age of son $= x + 5 = 5 + 5 =10$ years and age of father $= 7x + 5 = 7 \times 5 + 5 =35 + 5 = 40$years.
View full question & answer→Question 415 Marks
Solve the following equation and also check your result in case: $5\Big(\frac{7\text{x}+5}{3}\Big)-\frac{23}{3}=13-\frac{4\text{x}-2}{3}$
Answer$5\Big(\frac{7\text{x}+5}{3}\Big)-\frac{23}{3}=13-\frac{4\text{x}-2}{3}$
$\frac{35\text{x}+25}{3}+\frac{4\text{x}-2}{3}=13+\frac{23}{3}$
$\frac{35\text{x}+25+4\text{x}-2}{3}=\frac{398+23}{3}$
$39\text{x}=62+23=62$ [Multiplying both sides by $3$]
$39\text{x}=62-23$
$\text{x}=\frac{39}{39}$
$\text{x}=1$
Thus, $\text{x}=1$ is the solution of the given equation.
Check:Substituting $\text{x}=1$ in the given equation, we get:
$\text{L.H.S.}=5\Big(\frac{7\times1+5}{3}\Big)-\frac{23}{3}=\frac{60}{3}-\frac{23}{3}=\frac{37}{3}$
$\text{R.H.S.}=13-\frac{4\times1-2}{3}=\frac{39-2}{3}=\frac{37}{3}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=1$
View full question & answer→Question 425 Marks
A number consists of two digits whose sum is $9$. If $27$ is subtracted from the number, its digits are reversed. Find the number.
AnswerSum of two digits $= 9$
Let units digit $= x$ Then tens digits $= 9 - x$ And number $= 10(9 - x) + x = 90 - 10x + x = 90 - 9x$
On reversing the digits, Units digits $= 9 - x$ Tens digit $= x$ And number $= 10(x) + 9 - x = 10x + 9 - x = 9x + 9x$
According to the condition: $90 - 9x - 27 = 9x = + 9 $
$\Rightarrow 9x + 9x = 90 - 27 - 9$
$ \Rightarrow 18x = 90 - 36 = 54$
$\Rightarrow\text{x}=\frac{54}{18}=3$
Number $= 90 - 9x = 90 - 9 \times 3 = 90 - 27 = 63$
Check: $63 - 27 = 36$ (Whose digits are reversed) Which is true. therefore our answer is correct.
View full question & answer→Question 435 Marks
Solve the following equation and verify your answer: $\frac{2\text{x}+1}{3\text{x}-2}=\frac{5}{9}$
Answer$\frac{2\text{x}+1}{3\text{x}-2}=\frac{5}{9}$By cross multiplication:
$9(2\text{x+1})=5(3\text{x}-2)$
$\Rightarrow18\text{x}+9=15\text{x}-10$
$\Rightarrow18\text{x}-15\text{x}=-10$(By transposition)
$\Rightarrow3\text{x}=-19$
$\Rightarrow\text{x}=\frac{-19}{3}$
$\therefore\text{x}=\frac{-19}{3}$
Verification:
$\text{L.H.S.}=\frac{2\text{x}+1}{3\text{x}-2}=\frac{2\Big(\frac{-19}{3}\Big)+1}{3\Big(\frac{-19}{3}\Big)-2}$
$=\frac{\frac{-38}{3}+1}{\frac{-57}{3}-2}=\frac{\frac{-38+3}{3}}{\frac{-57-6}{3}}$
$=\frac{-35}{3}\times\frac{3}{-63}=\frac{5}{9}=\text{R.H.S.}$
View full question & answer→Question 445 Marks
Solve the following equation and verify your answer: $\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\text{x}+3}{5\text{x}+4}$
Answer$\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\text{x}+3}{5\text{x}+4}$
By cross multiplication:
$(7\text{x}-2)(5\text{x}+4)=(7\text{x}+3)(5\text{x}-1)$
$\Rightarrow35\text{x}^2+28\text{x}-10\text{x}-8=35\text{x}^2-7\text{x}+15\text{x}-$
$\Rightarrow35\text{x}^2+18\text{x}-8=35\text{x}^2+8\text{x}-3$
$\Rightarrow35\text{x}^2+18\text{x}-35\text{x}^2-8\text{x}=-3+8$ (By transpositior)
$\Rightarrow19\text{x}=-27$
$\Rightarrow\text{x}=\frac{-27}{19}$
$\therefore\text{x}=\frac{1}{2}$
Verification:
$\text{L.H.S.}=\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\times\frac{1}{2}-2}{5\times\frac{1}{2}-1}=\frac{\frac{7}{2}-\frac{2}{1}}{\frac{5}{2}-1}$
$=\frac{\frac{7-4}{2}}{\frac{5-2}{2}}=\frac{\frac{3}{2}}{\frac{3}{2}}=\frac{3}{2}\times\frac{2}{3}=1$
$\text{R.H.S.}=\frac{7\text{x}+3}{5\text{x}+4}$
$=-\frac{7\times\frac{1}{2}+3}{5\times\frac{1}{2}+4}=\frac{\frac{7}{2}+3}{\frac{5}{2}+4}$
$=\frac{\frac{7+6}{2}}{\frac{5+8}{2}}=\frac{\frac{13}{2}}{\frac{13}{2}}=\frac{13}{2}\times\frac{2}{13}=1$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 455 Marks
Solve the following equation and also check your result in case: $\frac{(3\text{x}+1)}{16}+\frac{(2\text{x}-3)}{7}=\frac{(\text{x}+3)}{8}+\frac{(3\text{x}-1)}{14}$
Answer$\frac{(3\text{x}+1)}{16}+\frac{(2\text{x}-3)}{7}=\frac{(\text{x}+3)}{8}+\frac{(3\text{x}-1)}{14}$
$\frac{3\text{x}+1}{16}-\frac{\text{x}+3}{8}=\frac{3\text{x}-1}{14}-\frac{2\text{x}-3}{7}$
$\frac{3\text{x}+1-2\text{x}-6}{16}=\frac{3\text{x}-1-4\text{x}+6}{14}$
$\frac{\text{x}-5}{8}=\frac{-\text{x}+5}{7}$
$7\text{x}-35=-8\text{x}+40$
$15\text{x}=75$
$\text{x}=\frac{75}{15}=5$
Check: $\text{L.H.S.}=\frac{3\times5+1}{16}+\frac{2\times5-3}{7}=\frac{16}{16}+\frac{7}{7}=2$
$\text{R.H.S.}=\frac{5+3}{8}+\frac{3\times5-1}{14}=\frac{8}{8}+\frac{14}{14}=2$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=5$
View full question & answer→Question 465 Marks
The difference between the squares of two consecutive numbers is $31$. Find the numbers.
AnswerLet the requierd number $= x$
Then Four-fifth of the number $= x + 1$
$\therefore$ According to the condition:
$(x + 1)^2- (x)^2= 31$
$\Rightarrow x^2+ 2x + 1 - x^2= 31$
$\Rightarrow 2x = 31 - 1 = 3030$
$\Rightarrow\text{x}=\frac{30}{2}=15$
$\therefore$ First number $= 15$
And second number $= 15 + 1 = 16$
Hence numbers are $15, 16$
Check: $(16)^2- (15)^2= 256 - 225 = 31$
Which is given
$\therefore$ Our answer is correct.
View full question & answer→Question 475 Marks
Total investment $= Rs. 12000.00$ Rate of interest for first part $= 10\%$ and for second part $= 12\%$ Annual income $= 1280.00$ Let the inverstment for the first part $= Rs. (12000 - a)$ According to the condition:
AnswerLet breadth of the rectangle $= x\ cm$
Then length $= (x + 9)cm$
$\therefore$ Area = length × breadth $= x(x + 9)cm^2$
By increasing each lenght and breadth by $3\ cm$
The new lenght of the rectangle $= x + 9 + 3$
$= (x + 12)cm$
$\therefore$ Area $= (x + 12)(x + 3)$
According to the condition:
$(x + 12)(x + 3) - a(x + 9) = 84$
$ x^2+ 3x + 12x + 36 - x^2- 9x = 84$
$ \Rightarrow 6a = 84 - 36 = 48$
$\Rightarrow\text{x}=\frac{48}{6}=8$
$\therefore$ Length of the rectangle $= a + 9 = 8 + 9 = 17\ cm$
and breadth $= x = 8\ cm$
View full question & answer→Question 485 Marks
$13 (\text{y}-4) -3 (\text{y}-9)-5(\text{y}+4)=0$
Answer$13 (\text{y}-4) -3 (\text{y}-9)-5(\text{y}+4)=0$
$\Rightarrow13\text{y}-52-3\text{y+27}-5\text{y}-20=0$
$\Rightarrow13\text{y}-3\text{y}-5\text{y}-52+27-20=0$
$\Rightarrow13\text{y}-8\text{y}-72+27=0$
$\Rightarrow5\text{y - 45 = 0}$
Dividing by $5$, $\text{y}=9$
Verification:
$\text{L.H.S.}=13(\text{y} - 4)-3(\text{y} - 9) - 5 (\text{y}+4)$
$=13(9-4)-3(9-9)-5(9+4)$
$=13\times5-3\times0-5\times13$
$=65-0-65$
$=0$
$=\text{R.H.S.}$
View full question & answer→Question 495 Marks
Seeta Devi has $Rs\ 9$ in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have?
AnswerTotle amount $= Rs. 9$ Let fifty paisae coins $= x$
Then twenty-five paise coins $= 2x$
According to the condition: $\text{x}\times\frac{50}{100}+2\text{x}\times\frac{52}{100}=9$
$\frac{\text{x}}{2}+\frac{\text{x}}{2}=9$
$\Rightarrow\text{x}=9$
$\therefore$ Number of fifty-paise cpoin $= 9$ And number of twenty five paisa coins $= 2x = 2 \times 9 = 18$
Check : Amount $=\frac{9\times50}{100}+\frac{18\times25}{100}$
$=\frac{9}{2}+\frac{9}{2}=\frac{18}{2}=9\text{rupees}$ Which is given. Hence our answer is correct.
View full question & answer→Question 505 Marks
Solve the following equation and also check your result in case:
$0.18(5\text{x}-4)=0.5\text{x}+0.8$
Answer$0.18\Big(5\text{x}-4\Big)=0.5\text{x}+0.8$
$0.9\text{x}-0.72=0.5\text{x}+0.8$
$0.9\text{x}-0.5\text{x}=0.8+0.72$
$0.4\text{x}=1.52$
$51\text{x}=504-45$
$\text{x}=\frac{159}{51}=9$
Thus, $\text{x}=3.8$ is the solution of the given equation.
Check:Substituting $\text{x}=3.8$ in the given equation, we get:
$\text{L.H.S.}=0.18\Big(5\times3.8-4\Big)=0.18\times15=2.7$
$\text{R.H.S.}=0.5\times3.8+0.8=2.7$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=3.8$
View full question & answer→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→