3 Marks Question - Maths STD 8 Questions - Vidyadip

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3 Marks Question

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

Question 13 Marks

I have Rs $1000$ in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Answer

Total amount $= Rs. 1000$
Let the number of five rupee note $= x$
$\therefore$ Ten rupees notes $= x + 10$
According to the condition, $(x + 10) \times 10 + 5x \times x = 1000$
$\Rightarrow 10a + 100 + 5a = 1000$
$\Rightarrow 15a = 1000 - 100 = 900$
$\Rightarrow\text{x}=\frac{900}{15}=60$
$\therefore$ Numbe of five rupees notes $= 60$ and number of ten rupees notes $= 60 + 10 = 70$

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Question 23 Marks

Solve the following equation and also check your result in case:
$\frac{7\text{y}+2}{5}=\frac{6\text{y}-5}{11}$

Answer

$\frac{7\text{y}+2}{5}=\frac{6\text{y}-5}{11}$
$77\text{y}-22=30\text{y}-25$
$77\text{y}-30\text{y}=-25-22$
$47\text{y}=-47$
$\text{y}=\frac{-47}{47}=-1$
Verification:
$\text{L.H.S.}=\frac{-7+2}{5}=\frac{-5}{5}=-1$
$\text{R.H.S.}=\frac{-6-5}{11}=\frac{-11}{11}=-1$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for y}=-1$

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Question 33 Marks

Solve the following equation and also check your result in case:
$\frac{3\text{x}}{4}-\frac{(\text{x}-1)}{2}=\frac{(\text{x}-2)}{3}$

Answer

$\frac{3\text{x}}{4}-\frac{(\text{x}-1)}{2}=\frac{(\text{x}-2)}{3}$
$\frac{3\text{x}-2\text{x}{+2}}{4}=\frac{\text{x}-2}{3}$
$4\text{x}-8=3\text{x}+6$
$\text{x}=14$
Check:
$\text{L.H.S.}=\frac{3\times14}{4}-\frac{14-1}{2}=\frac{21}{2}-\frac{13}{2}=\frac{8}{2}=4$
$\text{R.H.S.}=\frac{14-2}{3}=\frac{12}{3}=4$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=14$

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Question 43 Marks

Four-fifth of a number is more than three-fourth of the number by $4$. Find the number.

Answer

Let the requierd number $= x$
Then Four-fifth of the number $=\frac{4}{5}\text{x}$
And tree - fourth $=\frac{3}{4}\text{x}$
$\therefore \frac{4}{5}\text{x}-\frac{3}{47}\text{x}=4$
$\Rightarrow\frac{16\text{x}-15\text{x}=8}{20}$
$(L.C.M$.of $5, 4 = 20)$
$\therefore$ Required number $= 80$
Chake : $\frac{4}{5}$ of 80 $-\frac{3}{4}$ of $80 = 64 - 60 = 4$
wich is given $\therefore$ Our answer is correct.

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Question 53 Marks

Solve the following equation and also check your result in case: $\text{x}-\frac{(\text{x}-1)}{2}=1-\frac{(\text{x}-2)}{3}$

Answer

$\text{x}-\frac{(\text{x}-1)}{2}=1-\frac{(\text{x}-2)}{3}$
$\frac{2\text{x}-\text{x+1}}{2}=\frac{3-\text{x}+2}{3}$
$\frac{\text{x}+1}{2}=\frac{5-\text{x}}{3}$
$3\text{x}+3=10-2\text{x}$
$5\text{x}=10-3$
$\text{x}=\frac{7}{5}$ Check: $\text{L.H.S.}=\frac{7}{5}-\frac{\frac{7}{5}-1}{2}=\frac{7}{5}-\frac{1}{5}=\frac{6}{5}$
$\text{R.H.S.}=1-\frac{\frac{7}{5}-2}{3}=1-\frac{-3}{15}=\frac{6}{5}$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{7}{5}$

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Question 63 Marks

There are $180$ multiple choice questions in a test. If a candidate gets $4$ marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored $450$ marks in the test, how many questions did he answer correctly?

Answer

Number of total quations $=180$
Let the candidate answers questions correctly $= x$
$\therefore$ Uncorrect or unttended quastions $=180-x$ Total score he got $=450$
According to the condition $x \times 4-(180-x) \times 1=450 $
$\Rightarrow 4 x-180+x-= 450 $
$\Rightarrow 5 x=450+180=630 $
$\Rightarrow x=\frac{630}{5}=126$
Number of question which answered correctly $=126$

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Question 73 Marks

Find a positive value of $x$ for which the given equation is satisfied. $\frac{\text{y}^2+4}{3\text{y}^2+7}=\frac{1}{2}$

Answer

$\frac{\text{y}^2+4}{3\text{y}^2+7}=\frac{1}{2}$By cross multiplication:
$2(\text{y}^2+4)=3\text{y}^2+7$
$\Rightarrow2\text{y}^2+8=3\text{y}^2+7$
$\Rightarrow3\text{y}^2-2\text{y}^2=8-7$
$\Rightarrow\text{y}^2=1$
$\therefore\text{y}=\underline{+}\sqrt{1}=\underline{+}1$
$\because $ we have to take only positive value of $x$
$\therefore\text{y}=1$

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Question 83 Marks

Solve the following equation and verify your answer: $\frac{2\text{y}+5}{\text{y}+4}=1$

Answer

$\frac{2\text{y}+5}{\text{y}+4}=1$By cross multiplication:
$2\text{y}+5=\text{y}+4$
$\Rightarrow2\text{y}-\text{y}=4-5$ (By transposition) $\Rightarrow\text{y}=-1$
$\therefore\text{y}=-1$Verification:
$\text{L.H.S.}=\frac{2\text{y}+5}{\text{y}+4}=\frac{2(-1)+5}{-1+4}=\frac{-2+5}{-1+4}$
$=\frac{3}{3}=1=\text{R.H.S.}$

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Question 93 Marks

In a rational number, twice the numerator is $2$ more than the denominator. If $3$ is added to each, the numerator and the denominator, the new fraction is $\frac{2}{3}.$ Find the original number.

Answer

Let numerator $= x$
Then denominator $= 2x - 2$ Fraction $=\frac{\text{x}}{2\text{x}-2}$
According to the condition: $\frac{\text{x}+3}{2\text{x}-2+3}=\frac{2}{3}$
$\Rightarrow\frac{\text{x}+3}{2\text{x}+1}=\frac{2}{3}$
By cross multiplication $3(x + 3) = 2(2x + 1)$
$\Rightarrow 3x + 9 = 4x + 2$
$\Rightarrow 9 - 2 = 4x - 3x$
$\Rightarrow x = 7$
$\therefore$ Number $=\frac{\text{x}}{2\text{x}-2}$
$=\frac{7}{2\times7-2}$
$=\frac{7}{14-2}$
$=\frac{7}{12}$

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Question 103 Marks

A steamer goes downstream from one point to another in $9$ hours. It covers the same distance upstream in $10$ hours. If the speed of the stream be $1km/ hr$, find the speed of the steamer in still water and the distance between the ports.

Answer

Time taken by a steamer downstream $= 9$
hours and upstream $= 10$
hours speed of steamer $= 1km/ hr.$
Let speed of the condition: $9(x + 1) = 10(x - 1) 9x + 9 = 10x - 10$
$\Rightarrow 10x - 9x = 9 + 10$
$\Rightarrow x = 19$
$\therefore $ Speed of the steamer in still water $= 19 km/ h$ and
distance between two ports $= 9(a + 1) = 9(19 + 1) = 9 \times 20 = 180km.$

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Question 113 Marks

Find a positive value of x for which the given equation is satisfied.
$\frac{\text{x}^2-9}{5+\text{x}^2}=\frac{-5}{9}$

Answer

$\frac{\text{x}^2-9}{5+\text{x}^2}=\frac{-5}{9}$By cross multiplication:
$9(\text{x}^2-9)=-5(5+\text{x}^2)$
$\Rightarrow9\text{x}^2-81=-25-5\text{x}^2$
$\Rightarrow9\text{x}^2+5\text{x}^2=-25+81$
$\Rightarrow14\text{x}^2=56$
$\Rightarrow\text{x}^2=\frac{56}{14}=4$
$\therefore\text{x}=\underline{+}\sqrt{4}=\underline{+}2$
$\because $ we have to take only positive value of $x$
$\therefore\text{x}=2$

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Question 123 Marks

The numerator of a rational number is $3$ less than the denominator. If the denominator is increased by $5$ and the numerator by $2$, we get the rational number $\frac{1}{2}.$ Find the rational number.

Answer

Let denominator of the given rational number $= x$
Then numerator $= x - 3$
$\therefore$ Rational number $=\frac{\text{x}-3}{\text{x}}$
According to the condition: $\frac{\text{x}-3+2}{\text{x}+5}=\frac{1}{2}$
$\Rightarrow\frac{\text{x}-1}{\text{x}+5}=\frac{1}{2}$
By cross multiolication $2(x - 1) = x + 5$
$\Rightarrow 2x - 2 = x + 5$
$\Rightarrow 2x - x = 5 + 2$
$\Rightarrow x = 7$
$\therefore$ Rational number $=\frac{\text{x}-3}{\text{x}}=\frac{7-3}{7}=\frac{4}{7}$

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Question 133 Marks

Solve the following equation and also check your result in case: $\frac{\text{a}-8}{3}=\frac{\text{a}-3}{2}$

Answer

$\frac{\text{a}-8}{3}=\frac{\text{a}-3}{2}$
$2\text{a}-16=3\text{a}-9$
$3\text{a}-2\text{a}=-16+9$
$\text{a}=-7$ Verification: $\text{L.H.S.}=\frac{-7-8}{3}=\frac{-15}{3}=-5$
$\text{R.H.S.}=\frac{-7-3}{2}=\frac{-10}{2}=-5$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for a}=-7$

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Question 143 Marks

Solve the following equation and verify your answer: $\frac{5\text{x}-7}{3\text{x}}=2$

Answer

$\frac{5\text{x}-7}{3\text{x}}=\frac{2}{1}$By cross multiplication:
$2\times3\text{x}=5\text{x}-7$
$\Rightarrow6\text{x}-5\text{x}=-7$
$\Rightarrow\text{x}=-7$ (By transposition) $\therefore\text{x}=-7$Verification:
$\text{L.H.S.}=\frac{5\text{x}-7}{3\text{x}}=\frac{5\times(-7)-7}{3(-7)}$
$=\frac{-35-7}{-21}=\frac{-42}{-21}=\frac{2}{1}=\text{R.H.S.}$

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Question 153 Marks

Solve the following equation and also check your result in case:
$\frac{2\text{x}+5}{3}=3\text{x}-10$

Answer

$\frac{2\text{x}+5}{3}=3\text{x}-10$
$2\text{x}+5=9\text{x}-30$
$9\text{x}-2\text{x}=5+30$
$7\text{x}=35$
$\text{x}=\frac{35}{7}$
$\text{x}=5$
Verification:
$\text{L.H.S.}=\frac{10+5}{3}=\frac{15}{3}=5$
$\text{R.H.S.}=15-10=5$
$\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=5$

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