4 Mark Question - Maths STD 8 Questions - Vidyadip

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4 Mark Question

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 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:
$\text{L.H.S.}=\frac{3\times5+1}{16}+\frac{2\times5-3}{7}=\frac{16}{16}+\frac{7}{7}=2$
$\text{R.H.S.}=\frac{5+3}{8}+\frac{3\times5-1}{14}=\frac{8}{8}+\frac{14}{14}=2$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=5$

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Question 24 Marks

The difference between the squares of two consecutive numbers is 31 . Find the numbers.

Answer

Let the requierd number $= x$
Then Four-fifth of the number $=x+1$
$\therefore$ According to the condition:
$(x+1)^2-(x)^2=31$
$\Rightarrow x^2+2 x+1-x^2=31$
$\Rightarrow 2 x=31-1=3030$
$\Rightarrow x=\frac{30}{2}=15$
$\therefore$ 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.

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Question 34 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:

Answer

Let breadth of the rectangle $= x cm$
Then length $=(x+9) cm$
$\therefore$ Area $=$ length $\times$ breadth $=x(x+9) cm ^2$
By increasing each lenght and breadth by 3 cm
The new lenght of the rectangle $=x+9+3$
$=(x+12) cm$
$\therefore \text { Area }=(x+12)(x+3)$
According to the condition:
$(x+12)(x+3)-a(x+9)=84$
$x^2+3 x+12 x+36-x^2-9 x=84$
$\Rightarrow 6 a=84-36=48$
$\Rightarrow x=\frac{48}{6}=8$
$\therefore$ Length of the rectangle $=a+9=8+9=17 cm$
and breadth $= x =8 cm$

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Question 44 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:
$\text{L.H.S.}=13(\text{y} - 4)-3(\text{y} - 9) - 5 (\text{y}+4)$ $=13(9-4)-3(9-9)-5(9+4)$ $=13\times5-3\times0-5\times13$ $=65-0-65$ $=0$ $=\text{R.H.S.}$

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Question 54 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?

Answer

Totle 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 × 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.

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Question 64 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:
$\text{L.H.S.}=0.18\Big(5\times3.8-4\Big)=0.18\times15=2.7$ $\text{R.H.S.}=0.5\times3.8+0.8=2.7$ $\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=3.8$

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Question 74 Marks

$\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}=13$

Answer

We 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:
$\text{L.H.S}=\frac{\text{x}}{2}+\frac{\text{x}}{3}+\frac{\text{x}}{4}$
$=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}=6+4+3$
$=13=\text{R.H.S}$

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Question 84 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:
$\text{L.H.S.}=\frac{7}{2}\times\frac{-30}{17}-\frac{5}{2}\times-\frac{30}{17}=-\frac{30}{17}$
$\text{R.H.S.}=\frac{20}{3}\times\frac{-30}{17}+10=-\frac{30}{17}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-\frac{30}{17}$

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Question 94 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:
$\text{L.H.S.}=\frac{2}{3\times(-10)}-\frac{3}{2\times(-10)}=\frac{2}{-30}-\frac{3}{-20}$ $=\frac{-4+9}{60}=\frac{5}{50}=\frac{1}{2}$ $\text{R.H.S.}=\frac{1}{12}$ $\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-10$

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Question 104 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?

Answer

Let 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)
⇒ 59x + 180 = 60x
⇒ 60x - 59x = 180
⇒ x = 180
Totle number of guests = 180

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Question 114 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:
$\text{L.H.S.}=\frac{2\text{x}}{3\text{x}+1}=\frac{2\times\Big(\frac{-3}{11}\Big)}{3\Big(\frac{-3}{11}\Big)+1}=\frac{\frac{-6}{11}}{\frac{-9}{11}+1}$ $=\frac{\frac{-6}{11}}{\frac{-9+11}{11}}=\frac{\frac{-6}{11}}{\frac{2}{11}}=\frac{-6}{11}\times\frac{11}{2}$ $\frac{-3}{1}=\text{R.H.S.}$

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Question 124 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:
$\text{L.H.S}=\frac{15(2-\text{x})-5(\text{x}+6)}{1-3\text{x}}$
$=\frac{15(2-1)-5(1+6)}{1-3\times1}$
$\frac{15\times1-5\times7}{1-3}-\frac{15-35}{-2}$
$=\frac{-20}{-2}=10$ $=\text{R.H.S}$

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Question 134 Marks

Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.

Answer

Let the requierd number = x5
Times of it = 5x
Twice of it = 2x
According to the condition:
⇒ 5x - 2x = 4 + 5
⇒ 3x = 9
$\Rightarrow\text{x}=\frac{9}{3}=3$
Required number = 3
Check: 3 × 5 - 5 = 2 × 3 + 4
⇒ 15 - 5 = 6 + 4
⇒ 10 = 10
Which is true. therefore our answer is correct.

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Question 144 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?

Answer

Let age of ashima = x
Then age of Sunita = 2x
According to the condition:
4(x - 6) = 2x + 4
⇒ 4x - 24 = 2x + 4
⇒ 4x - 2x = 4 + 24
⇒ 2x = 28
$\Rightarrow\text{x}=\frac{28}{2}=14$
$\therefore$ Sunita's present age = 2x = 2 × 14 = 28 years
And ashima's age = 14 years
Age of sunita = 28 - 2 = 26 years
And age of Ashima = 14 - 2 = 12 years

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Question 154 Marks

The distance between two stations is 340km. 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 5km/ hr. If the distance between the two trains after 2 hours of their start is 30km, find the speed of each train.

Answer

Distance between two stations = 340km.
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
⇒ 340 - (2x + 10 + 2x) = 30
⇒ 4x + 10 = 340 - 30
⇒ 4x = 340 - 30 - 10
⇒ 4x = 300
$\Rightarrow\text{x}=\frac{300}{4}=75$
$\therefore$ Speed of first train = 75km/ hr
and speed of second train = 75 + 5 = 80km/ hr

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Question 164 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:
$\text{L.H.S}$ $=\frac{5\text{x}}{3}+\frac{2}{5}$
$=\frac{5}{3}\times\frac{9}{25}+\frac{2}{5}$
$=\frac{3}{5}+\frac{2}{5}=\frac{3+2}{5}=\frac{5}{5}$
$=1=\text{R.H.S}$

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Question 174 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?

Answer

Totle number of days = 20
Let of days he worked = x
Then number of days he remained absent = 20 - x
According to the condition:
x × 60 - (20 - x) × 5 = 745
⇒ 60x = 745 + 5x = 745
⇒ 65x = 745 + 100 = 845
$\Rightarrow\text{x}=\frac{845}{65}=13$
$\therefore$ Number of days he worked = 13days
and number of days he remained absent = 20 - 13 = 7days.

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Question 184 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:
$\text{L.H.S.}=\frac{2\times\frac{5}{12}-3}{3\times\frac{5}{12}+2}=\frac{\frac{5}{6}-3}{\frac{5}{4}+2}=\frac{-4}{6}=\frac{-2}{3}$
$\text{R.H.S.}=\frac{-2}{3}$
$\text{L.H.S.}=\text{R.H.S.} \text{ for} \text{ x}=\frac{5}{12}$

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Question 194 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:
$\text{L.H.S.}=\frac{1-2\times(-10)}{7}-\frac{2-3\times(-10)}{8}$
$=\frac{1+20}{7}-\frac{2+30}{8}=9-4=-1$
$\text{R.H.S.}=\frac{3}{2}+\frac{-10}{4}=\frac{3}{2}+\frac{-5}{2}=\frac{3-5}{2}=-1$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=-10$

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Question 204 Marks

Find a number whose double is 45 greater than its half.

Answer

Let 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.

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Question 214 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:
$\text{L.H.S}=\frac{\text{x}+2}{\text{x}+5}=\frac{-4+2}{-4+5}=\frac{-2}{1}=-2$
$\text{R.H.S}=\frac{\text{x}}{\text{x}+6}=\frac{-4}{-4+6}=\frac{-4}{2}=-2$
$\therefore\text{L.H.S}=\text{R.H.S}$

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Question 224 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:
$\text{L.H.S}=\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{2\times3-(7-5\times3)}{9\times3-(3+4\times3)}$
$=\frac{6-(7-15)}{27-(3+12)}=\frac{6-7+15}{27-3-12}=\frac{14}{12}$
$=\frac{7}{6}$ $=\text{R.H.S}$

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Question 234 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:
$\text{L.H.S.}=\frac{24}{17}-2\times\frac{24}{17}+7-\frac{16}{3}\times\frac{24}{17}=\frac{-33}{17}$
$\text{R.H.S.}=3-\frac{7}{2}\times\frac{24}{17}=\frac{-33}{17}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{24}{17 }$

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Question 244 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:
$\text{L.H.S.}=\frac{3\times\frac{1}{3}-2}{3}+=\frac{3\times\frac{1}{3}+3}{2}$
$=\frac{-1}{3}+\frac{11}{6}=\frac{9}{6}=\frac{3}{2}$
$\text{R.H.S.}=\frac{1}{3}+\frac{7}{6}=\frac{9}{6}=\frac{3}{2}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for a}=\frac{1}{3}$

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Question 254 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:
$\text{L.H.S.}=\frac{2}{3\text{x}}-\frac{3}{2\text{x}}=\frac{2}{3(-10)}-\frac{3}{2(-10)}$ $=\frac{2}{-30}-\frac{3}{-20}=\frac{1}{-15}+\frac{3}{20}$ $=\frac{-4+9}{60}=\frac{5}{60}=\frac{1}{12}=\text{R.H.S.}$

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Question 264 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)
$=9\frac{1}{4}=\text{L.H.S}$

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Question 274 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:
$\text{L.H.S.}=\frac{1}{2}\times\frac{25}{2}+7\times\frac{25}{2}-6=\frac{351}{4}$
$\text{R.H.S.}=7\times\frac{25}{2}+\frac{1}{4}=\frac{351}{4}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{351}{2}$

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Question 284 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:
$\text{L.H.S.}=\frac{5\times\frac{-51}{73}}{3}-\frac{\frac{-51}{73}-1}{4}$
$=\frac{-255}{219}-\frac{-124}{292}=\frac{-54}{73}$
$\text{R.H.S.}=\frac{\frac{-51}{73}-3}{5}=\frac{-54}{73}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{-51}{73}$

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Question 294 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?

Answer

Let 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)
⇒ 5x + 6 = 3x + 18
⇒ 5x - 3x = 18 - 6
⇒ 2x = 12
⇒ x = 6
$\therefore$ Present my age = 5x = 5 × 6 = 30years
and my son's age = 6 years

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