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$
⇒ 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)
⇒ 49x + 70 = 45x + 90
⇒ 49x - 45x = 90 - 70
⇒ 4x = 20
$\Rightarrow\text{x}=\frac{20}{4}=5$
$\therefore$ Sonu's present age = 7x = 7 × 5 = 35
and Monu's present age = 5x = 5 × 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
⇒ 10x + 1000 - 20x = 800
⇒ -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 × 10 + 30 × 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}$
⇒ 3(x - 3) = 2x (By cross multiplication)
3x - 9 = 2x
⇒ 3x - 2x = 9
⇒ 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$
⇒144000 - 2x = 128000
⇒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: $\frac{72}{3}=\frac{112}{7}+8$
⇒ 24 = 16 + 8
⇒ 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.
Answer5 years ago,
Let age of son = years
Then, age of father = 7a 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:
⇒ 7x + 10 = 3x + 30
⇒ 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 × 5 + 5
=35 + 5 = 40years
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
⇒ 9x + 9x = 90 - 27 - 9
⇒ 18x = 90 - 36 = 54
$\Rightarrow\text{x}=\frac{54}{18}=3$
Number = 90 - 9x = 90 - 9 × 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:
$\text{L.H.S.}=\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\times\frac{1}{2}-2}{5\times\frac{1}{2}-1}=\frac{\frac{7}{2}-\frac{2}{1}}{\frac{5}{2}-1}$ $=\frac{\frac{7-4}{2}}{\frac{5-2}{2}}=\frac{\frac{3}{2}}{\frac{3}{2}}=\frac{3}{2}\times\frac{2}{3}=1$ $\text{R.H.S.}=\frac{7\text{x}+3}{5\text{x}+4}$ $=-\frac{7\times\frac{1}{2}+3}{5\times\frac{1}{2}+4}=\frac{\frac{7}{2}+3}{\frac{5}{2}+4}$ $=\frac{\frac{7+6}{2}}{\frac{5+8}{2}}=\frac{\frac{13}{2}}{\frac{13}{2}}=\frac{13}{2}\times\frac{2}{13}=1$ $\therefore\text{L.H.S.}=\text{R.H.S.}$
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