3 Mark Question - Maths STD 8 Questions - Vidyadip

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

Question 13 Marks

Solve:
$\text{m}-\frac{(\text{m}-1)}{2}=1-\frac{(\text{m}-2)}{3} $

Answer

$\text{m}-\frac{(\text{m}-1)}{2}=1-\frac{(\text{m}-2)}{3} $
$\Rightarrow \frac{2\text{m} - \text{m} + 1}{2} = 1 −\frac{ ( \text{m}−2 )}{3}$(L.C.M. of 1 and 2 is 2 )
$\Rightarrow\frac{\text{m} + 1}{2} = \frac{3 −\text{m} + 2}{3} $(L.C.M. of 1 and 3 is 3 )
$\Rightarrow\frac{\text{m} + 1}{2} = \frac{5 −\text{ m}}{3}$
$\Rightarrow 3 ( \text{m} + 1 ) = 2 ( 5 - \text{m} )$ (by cross multiplication)
$\Rightarrow3\text{m} + 3 = 10 - 2\text{m}$
$3\text{m} + 2\text{m} = 10 - 3$
$\Rightarrow 5\text{m} = 7$
$\Rightarrow\text{m}= 75$
$\therefore \text{m} = \frac{7}{5}$

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Question 23 Marks

Solve:
$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$

Answer

$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
$\Rightarrow (4+5\text{x})(2-7\text{x}) = (1-5\text{x})(3+7\text{x})$
$\Rightarrow 8 - 28\text{x} + 10\text{x} - 35\text{x}^2 = 3 + 7\text{x} - 15\text{x} - 35\text{x}^2$
$\Rightarrow-35\text{x}^2 - 18\text{x} +8 = -35\text{x}^2 - 8\text{x} + 3$
$\Rightarrow-35\text{x}^2 + 35\text{x}^2 -18\text{x} + 8\text{x} = -8 + 3$
$\Rightarrow-10\text{x} = -5$
$\Rightarrow \text{x} = -5-10= \frac{1}{2} $
$\therefore \text{x} = \frac{1}{2}$

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Question 33 Marks

A man is 10 times older than his grandson. He is also 54 years older than him. Find their present ages.

Answer

Let the age of the grandson be x years.
Then, his grandfather's age will be 10x.
Also, the grandfather is 54 years older than his grandson.
$\therefore$ Age of the grandson $= \text{x} + 54$
$10\text{x} = \text{x} + 54$
$\Rightarrow10\text{x} - \text{x} = 54$
$\Rightarrow9\text{x} = 54$
$\Rightarrow \text{x} = \frac{54}{9}= 6$
Therefore, the grandson's age is 6 years.
Grandfather's age = 10(x) = 10 × 6 = 60 years

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Question 43 Marks

Solve:
$3\text{x}+\frac{2\text{}}{3}=2\text{x}+1$

Answer

$3\text{x}+\frac{2\text{}}{3}=2\text{x}+1$
By cross multiplication:
$\Rightarrow 3\text{x} - 2\text{x} = 1 -\frac{ 2}{3}$
$ \Rightarrow \text{x} = \frac{1}{1} - \frac{2}{3}$ $\text{L.C.M}\text{ of }1\text{ and }3\text{ is }3$
$\Rightarrow\text{x}=\frac{3-2}{3}$
$\Rightarrow\text{x}=\frac{1}{3}$
$\therefore\text{x}=\frac{1}{3}$

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Question 53 Marks

Find three consecutive odd numbers whose sum is 147.

Answer

Let the first odd number be x.
Let the second odd number be $(\text{x}+2).$
Let the third odd number be $(\text{x}+4).$
$\therefore\text{x}+(\text{x}+2)+(\text{x}+4)=147$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=147$
$\Rightarrow3\text{x}+6=147$
$\Rightarrow3\text{x}=147-6$
$\Rightarrow3\text{x}=141$
$\Rightarrow\text{x}=\frac{141}{3}=47$
Therefore, the first odd number is 47.
Second odd number $=(\text{x}+2)=(47+2)=49$
Third odd number $=(\text{x}+4)=(47+4)$

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Question 63 Marks

The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as the younger one, find their present ages.

Answer

Let the age of the younger cousin be x.
Then, the age of the elder cousin will be (x + 10).
15 years ago:
Age of the younger cousin = (x - 15)
Age of elder cousin = (x + 10 - 15)
= (x - 5)
$\therefore$ (x - 5) = 2 (x - 15)
⇒ x - 5 = 2x - 30
⇒ x - 2x = -30 + 5
⇒ -x = -25
⇒ x = 25
Therefore, the present age of the younger cousin is 25 years.
Present age of elder cousin = (x + 10) = (25 + 10) = 35 years

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Question 73 Marks

Solve:
$\frac{15(2-\text{y})-5(\text{y}+6)}{1-3\text{y}}=10$

Answer

$\frac{15(2-\text{y})-5(\text{y}+6)}{1-3\text{y}}=10$
$\Rightarrow\frac{30 - 15\text{y} - 5\text{y} - 30}{1 - 3\text{y}} = 10$
$\Rightarrow \frac{-20\text{y}}{1 - 3\text{y}} = 10$
$\Rightarrow1 (-20\text{y}) = 10 (1 - 3\text{y})$ (by cross multiplication)
$\Rightarrow-20\text{y} = 10 - 30\text{y}$
$\Rightarrow-20\text{y} + 30\text{y} = 10$
$\Rightarrow10\text{y} = 10$
$\Rightarrow\text{y} = \frac{10}{10} = 1$
$\therefore\text{y} = 1$

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Question 83 Marks

Rakhi's mother is four times as old as Rakhi. After 5 years, her mother will be three times as old as she will be then. Find their present ages.

Answer

Let the present age of Rakhi be x.
Then, the present age of Rakhi's mother will be 4x.
After five years, Rakhi's age will be (x + 5).
After five years, her mother's age will be (4x + 5).
4x + 5 = 3 ( x + 5 )
⇒ 4x + 5 = 3x + 15
⇒ 4x - 3x = 15 - 5
⇒ x = 10
Present age of Rakhi = 10 years
Present age of Rakhi's mother = 4(x) = 4 × 10 = 40 years.

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Question 93 Marks

Solve:
$\frac{7\text{y}}{5}=\text{y}-4$

Answer

$\frac{7\text{y}}{5}=\text{y}-4$
By cross multiplication:
$\Rightarrow7\text{y} = 5(\text{y} - 4)$
$\Rightarrow7\text{y}=\text{5y}-20$
$\Rightarrow7\text{y}-5\text{y}=-20$
$\Rightarrow2\text{y}=-20$
$\Rightarrow\text{y}=\frac{-20}{2}=-10$
$\therefore\text{y}=-10$

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Question 103 Marks

Solve:
$\frac{2-9\text{z}}{17-4\text{z}}=\frac{4}{9}$

Answer

$\frac{2-9\text{z}}{17-4\text{z}}=\frac{4}{9}$
$\Rightarrow 5 ( 2-9\text{z}) = 4 ( 17-4\text{z})$
$\Rightarrow10 - 45\text{z} = 68 - 16\text{z}$
$\Rightarrow10 - 68 = 45\text{z} - 16\text{z}$
$\Rightarrow-58 = 29\text{z}$
$\Rightarrow29\text{z} = -58$ (by transposition)
$\Rightarrow\text{z}=\frac{-58}{29}=-2$
$\therefore\text{z}=-2$

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Question 113 Marks

Solve:
$\frac{6\text{y}-5}{2\text{y}}=\frac{7}{9}$

Answer

$\frac{6\text{y}-5}{2\text{y}}=\frac{7}{9}$
$\Rightarrow9(6\text{y}-5)=7(2\text{y})$ (by cross multiplication)
$\Rightarrow54\text{y} - 45 = 14\text{y}$
$\Rightarrow54\text{y} - 14\text{y} = 45$
$\Rightarrow 40\text{y} = 45$
$\Rightarrow \text{y} = \frac{45}{40}= \frac{9}{8}$
$\therefore\text{y} = \frac{9}{8}$

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Question 123 Marks

Solve:
$\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$

Answer

$\frac{9\text{x}-7}{3\text{x}+5}=\frac{3\text{x}-4}{\text{x}+6}$
$\Rightarrow (\text{x}+6) (9\text{x}-7) = (3\text{x}+5) (3\text{x}-4)$
$\Rightarrow9\text{x}^2 - 7\text{x} + 54\text{x} - 42 = 9\text{x}^2 - 12\text{x} + 15\text{x} - 20$
$\Rightarrow9\text{x}^2 + 47\text{x} - 42 = 9\text{x}^2 + 3\text{x} - 20$
$\Rightarrow 9\text{x}^2 - 9\text{x}^2 + 47\text{x} - 3\text{x} = -20 + 42$
$\Rightarrow44\text{x} = 22$
$\Rightarrow \text{x} = \frac{22}{44}= \frac{1}{2}$
$\therefore\text{x}=\frac{1}{2}$

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Question 133 Marks

Solve:
9x + 5 = 4(x - 2) + 8

Answer

$9\text{x} + 5 = 4(\text{x}-2)+8$
$\Rightarrow9\text{x}+\text{5}=4\text{x}-8+8$
$\Rightarrow9\text{x}+\text{5}=4\text{x}$
$\Rightarrow9\text{x}-4\text{x}=-5$
$\Rightarrow5\text{x}=-5$
$\Rightarrow\text{x}=\frac{-5}{5}=-1$
$\therefore\text{x}=-1$

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Question 143 Marks

Solve:
$\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$

Answer

$\frac{3\text{x}+5}{4\text{x}+2}=\frac{3\text{x}+4}{4\text{x}+7}$
$⇒ (4\text{x} + 7)(3\text{x} + 5) = (4\text{x} + 2)(3\text{x} +4)$
$\Rightarrow 12\text{x}^2 + 20\text{x} + 21\text{x} + 35=12\text{x}^2 + 16\text{x} + 6\text{x} +8$
$$$\Rightarrow12\text{x}^2 + 41\text{x }+ 35 =12\text{x}^2 + 22\text{x} +8$
$\Rightarrow12\text{x}^2 -12\text{x}^2 + 41\text{x} -22\text{x} = 8 - 35$
$\Rightarrow 19\text{x} = -27$
$\Rightarrow\text{x} = \frac{-27}{19} $
$\therefore\text{x} = \frac{-27}{19} $

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Question 153 Marks

Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.

Answer

Let the number be x.
Fifth part increased by $5=\frac{\text{x}}{5}+5$
Fourt part diminished by $5=\frac{\text{x}}{4}-5$
$\therefore\frac{\text{x}}{5}+5=\frac{\text{x}}{4}-5$
$\Rightarrow5+5=\frac{\text{x}}{4}-\frac{\text{x}}{5}$
$\Rightarrow10=\frac{5\text{x}-4\text{x}}{20}$
$\Rightarrow10=\frac{\text{x}}{20}$
$\Rightarrow10=\frac{\text{x}}{20}$
$\Rightarrow200=\text{x}$
$\Rightarrow\text{x}=200$
Therefore, the number is 200.

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Question 163 Marks

Three numbers are in the ratio of 4 : 5 : 6. If the sum of the largest and the smallest equals the sum of the third and 55, find the numbers.

Answer

Let the common multiple for the given three numbers be x.Then, the three numbers would be 4x, 5x and 6x.
$\therefore4\text{x} + 6\text{x} = 5\text{x} + 55$
$\Rightarrow10\text{x} = 5\text{x} + 55$
$\Rightarrow10\text{x} - 5\text{x} = 55$
$\Rightarrow 5\text{x} = 55$
$\Rightarrow\text{x} = \frac{55}{5}= 11$
$\therefore$ Smallest number $=4\text{x}=4(11)=44$
Largest number is $=6\text{x}=6(11)=66$
Third number $=5\text{x}=5(11)=55$
Therefore, the three numbers are 44, 55 and 66.

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Question 173 Marks

Evaluate $\left\{(83)^2-(17)^2\right\}$

Answer

$(83)^2-(17)^2$
$=(83+17)(83-17)\left[\text { according to the formula }(a)^2-(b)^2=(a+b)(a-b)\right]$
$=(100)(66)$
$=6600$

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Question 183 Marks

If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.

Answer

Let the number be x.
$\therefore10+4\text{x}=5\text{x}-5$
$\Rightarrow10+5=5\text{x}-4\text{x}$
$\Rightarrow15=\text{x}$
$\Rightarrow\text{x}=15$ (by transposition)
Therefore, the number is 15.

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Question 193 Marks

Two numbers are in the ratio 8 : 3. If the sum of the numbers is 143, find the numbers.

Answer

Let the numbers be 8x and 3x
$8\text{x} + 3\text{x} = 143$
$\Rightarrow11\text{x} = 143$
$\Rightarrow\text{x}= \frac{143}{11}$
$\Rightarrow\text{x} =13$
$\therefore$ One number $=8\text{x} = 8\times13 = 104$
Other number $= 3\text{x} = 3\times13 = 39$

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Question 203 Marks

$\frac{2}{3}$ of a number is 20 less than the original number. Find the number.

Answer

Let the original number be x $\frac{2}{3}$ of the number is 20 less than the original number.
$\therefore\frac{2}{3}\text{x}=\text{x}-20$
$\Rightarrow\frac{2\text{x}}{3}=\text{x}-20$
$\Rightarrow2\text{x}=3(\text{x}-20)$ (by cross multiplication)
$\Rightarrow2\text{x}=3\text{x}-60$
$\Rightarrow2\text{x}-3\text{x}=60$
$\Rightarrow-\text{x}=-60$
Therefore, the original number is 60.

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Question 213 Marks

Solve:
$\frac{8\text{x}-3}{3\text{x}}=2$

Answer

$\frac{8\text{x}-3}{3\text{x}}=2$ $\Rightarrow8\text{x}-3=2(3\text{x})$ (by cross multiplication) $\Rightarrow 8\text{x} - 3 = 6\text{x}$ $\Rightarrow 8\text{x} - 6\text{x}=3$ $\Rightarrow 2\text{x} =3$ $\Rightarrow \text{x}= \frac{3}{2}$$\therefore \text{x} = \frac{3}{2}$

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Question 223 Marks

Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part.

Answer

Let one part be x.
7 times the first part $=7\text{x}$
Let the other part be $(24-\text{x})$
5 times the second part $=5(24-\text{x})$
$\therefore7\text{x}+5(24-\text{x})=146$
$\Rightarrow7\text{x}+120-5\text{x}=146$
$\Rightarrow7\text{x}-5\text{x}=146-120$
$\Rightarrow2\text{x}=26$
$\Rightarrow\text{x}=\frac{26}{2}=13$
Therefore, one part is 13.
Other part $=(24-\text{x})=(24-13)=11$

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Question 233 Marks

Solve:
$\frac{9\text{x}}{7-6\text{x}}=15$

Answer

$\frac{9\text{x}}{7-6\text{x}}=15$$\Rightarrow\frac{9\text{x}}{7-6\text{x}}=\frac{15}{1}$
$\Rightarrow 1 ( 9\text{x}) = 15 ( 7-6\text{x})$ (by cross multiplication)
$\Rightarrow 9\text{x} = 105 - 90\text{x}$
$\Rightarrow99\text{x}=105$
$\Rightarrow\text{x}=\frac{105}{99}=\frac{35}{33}$
$\therefore\text{x}=\frac{35}{33}$

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Question 243 Marks

Solve:
$\frac{7\text{y}+4}{\text{y}+7}=\frac{-4}{3}$

Answer

$\frac{7\text{y}+4}{\text{y}+7}=\frac{-4}{3}$(by cross multiplication)
$\Rightarrow3 (7\text{y} + 4) = -4 (\text{y} + 2)$
$\Rightarrow21\text{y} + 12 = -4\text{y} - 8$
$\Rightarrow21\text{y} + 4\text{y} = -8 -12$
$\Rightarrow25\text{y} = -20$
$\Rightarrow\text{y} = -\frac{20}{25}= \frac{-4}{5}$
$\therefore\text{y}=\frac{-4}{5}$

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Question 253 Marks

Divide 4500 into two parts such that 5% of the first part is equal to 10% of the second part.

Answer

Let the first part be x.
Let the second part be (4500 − x).
$\therefore5\%\text{ of }\text{x}=10\%\text{ of }(4500-\text{x})$
$\Rightarrow\Big(\frac{5}{100}\Big)\text{x}=\Big(\frac{10}{100}\Big)(4500-\text{x})$
$\Rightarrow\frac{5\text{x}}{100}=\frac{45000-10\text{x}}{100}$
$\Rightarrow5\text{x}=45000-10\text{x}$
(by cancellation of same denominators from both the sides)
$\Rightarrow5\text{x}+10\text{x}=45000$

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Question 263 Marks

Solve:
$\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$

Answer

$\frac{2\text{x}-(7-5\text{x})}{9\text{x}-(3+4\text{x})}=\frac{7}{6}$
$\Rightarrow\frac{2\text{x} - 7 + 5\text{x}}{9\text{x} - 3 - 4\text{x}} = \frac{7}{6}$
$\Rightarrow\frac{7\text{x} - 7}{5\text{x} - 3} = \frac{7}{6}$
$\Rightarrow6 (7\text{x} - 7) = 7 (5\text{x} - 3)$ (by cross multiplication)
$\Rightarrow42\text{x} - 42 = 35\text{x} - 21$
$\Rightarrow 42\text{x} - 35\text{x} = 42 - 21$
$\Rightarrow7\text{x} = 21 $
$\Rightarrow\text{x}=\frac{21}{7}=3$
$\therefore \text{x} = 3$

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Question 273 Marks

Solve:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0

Answer

$15(\text{y} - 4) - 2(\text{y} - 9) + 5(\text{y} + 6) = 0$
$\Rightarrow 15\text{y} - 60 - 2\text{y} + 18 + 5\text{y} + 30 = 0$
$\Rightarrow15\text{y} - 2\text{y} + 5\text{y} -60 + 18 + 30 = 0$
$\Rightarrow18\text{y} - 12 = 0$
$\Rightarrow18\text{y} = 12 $
$\Rightarrow \text{y} = \frac{12}{18} = \frac{2}{3}$
$\therefore \text{y} = \frac{2}{3} $

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Question 283 Marks

If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$ find the value of:
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$

Answer

Given, $\Big(\text{x}-\frac{1}{\text{x}}\Big)=4$
Squaring both the sides:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=4^2$
$\Rightarrow(\text{x})^2-2\times(\text{x})\times\Big(\frac{1}{\text{x}}\Big)+\Big(\frac{1}{\text{x}}\Big)^2=16$
$\Rightarrow\text{x}^2-2+\frac{1}{\text{x}^2}=16$
$\Rightarrow\text{x}^2-2+\frac{1}{\text{x}^2}=16+2$
$\therefore\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=18$
$2^2+2\Rightarrow4^2+2\Rightarrow16+2\Rightarrow18$

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Question 293 Marks

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

Answer

Let the speed of the steamer in still water be x km/ h.
Speed (downstream) = (x + 1)km/ h
Speed (upstream) = (x - 1) km/h
Distance covered in 9 hours while going downstream = 9( x + 1 )km
Distance covered in 10 hours while going upstream = 10( x - 1 )km
But both of these distances will be same.
9 (x + 1) = 10 (x - 1)
⇒ 9x + 9 = 10x - 10
⇒ 9 + 10 = 10x − 9x
⇒ 19 = x
⇒ x = 19
Therefore, the speed of the steamer in still water is 19 km/ h.
Distance between the ports = 9(x + 1) = 9(19 + 1) = 9 × 20 = 180km

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Question 303 Marks

Solve:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17

Answer

$3(5\text{x} - 7) - 2(9\text{x} - 11) = 4(8\text{x} - 13) -17$
$\Rightarrow 5\text{x} - 21 - 18\text{x} + 22 = 32\text{x} - 52-17$
$\Rightarrow15\text{x} -18\text{x} - 21 + 22 = 32\text{x} −69$
$\Rightarrow-3\text{x} +1 = 32\text{x} - 69$
$\Rightarrow1 + 69 = 32\text{x} + 3\text{x} $
$\Rightarrow 70 = 35\text{x}$
$\Rightarrow 35\text{x}= 70 $ $(\text{by} \ \text{transposition})$
$\Rightarrow\text{x}= \frac{70}{35}= 2$
$\therefore\text{x}=2$

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Question 313 Marks

Solve:
$\frac{4\text{x}+7}{9-3\text{x}}=\frac{1}{4}$

Answer

$\frac{4\text{x}+7}{9-3\text{x}}=\frac{1}{4}$
$\Rightarrow4(4\text{x}+7)=1(9-3\text{x})$ (by cross multiplication)
$\Rightarrow16\text{x}+28=9-3\text{x}$
$\Rightarrow16\text{x}+3\text{x}=9-28$
$\Rightarrow19\text{x}=-19$
$\Rightarrow\text{x}=\frac{-19}{19}=1$
$\therefore\text{x}=-1$

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Question 323 Marks

Two angles of a triangle are in the ratio 4 : 5. If the sum of these angles is equal to the third angle, find the angles of the triangle.

Answer

Let the common multiple of all the three angles be x.
Then, the first angle will be 4x.
And the second angle will be 5x.
In a triangle, sum of all the three angles will be equal to 180°.
$\therefore$ Third angle $= 180 - ( 4 \text{x} + 5 \text{x} )= 180 - 9 \text{x}$
$\therefore4 \text{x} + 5 \text{x} = 180 - 9 $
$\Rightarrow9 \text{x}=180-9 \text{x}$
$\Rightarrow9\text{x}+ 9\text{x}=180$
$\Rightarrow18\text{x}=180$
$\Rightarrow\text{x}=\frac{180}{18}=10$
First angle $=4\text{x}=4\times10=40^{\circ}$
Second angle $=5\text{x}=5\times10=50^{\circ}$
Third angle $=4\text{x}+5\text{x}=9\text{x}=9\times10=90^{\circ}$

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Question 333 Marks

Find three consecutive even numbers whose sum is 234.

Answer

Let the first even number be x.
Let the second even number be x+2.
Let the third even number be x+4.
$\therefore\text{x}+\text{x}+2+\text{x}+4=234$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=234$
$\Rightarrow3\text{x}+6=234$
$\Rightarrow3\text{x}=234-6$
$\Rightarrow3\text{x}=228$
$\Rightarrow\text{x}=\frac{228}{3}=76$
$\therefore$ First even number $=\text{x}=76$
Second even number $=\text{x}+2=76+2=78$
Third even number $=\text{x}+4=80$

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Question 343 Marks

Divide 150 into three parts such that the second number is five-sixths the first and the third number is four-fifths the second.

Answer

Let the first number be x.
Then, the second number will be $\frac{5}{6}\text{x}.$
Third number $=\frac{4}{5}\Big(\frac{5}{6}\text{x}\Big)=\frac{2}{3}\text{x}$
$\therefore\text{x}+\frac{5\text{x}}{6}+\frac{2\text{x}}{3}=150$
$\Rightarrow\frac{6\text{x}+5\text{x}+4\text{x}}{6}=150$
(multiplying the L.H.S. by 6, which is the L.C.M. of 1, 6 and 3)
$\Rightarrow15\text{x}=150\times6$ (by cross multiplication)
$\Rightarrow15\text{x}=900$
$\Rightarrow\text{x}=\frac{900}{15}=60$
Therefore, the first number is 60.
Second number $=\frac{5}{6}\text{x}=\frac{5}{6}(60)=50$
Third number $=\frac{2}{3}\text{x}=\frac{2}{3}(60)=40$

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Question 353 Marks

Four-fifths of a number is 10 more than two-thirds of the number. Find the number.

Answer

Let the number be x.
Four fifths of the number is 10 more than two thirds of the number.
$\therefore\frac{4}{5}\text{x}=10+\frac{2}{3}\text{x}$
$\Rightarrow\frac{4\text{x}}{5}=\frac{30+2\text{x}}{3}$ (L.C.M. of 1 and 3 is 3)
$\Rightarrow3(4\text{x})=5(30+2\text{x})$ (by cross multiplication)
$\Rightarrow12\text{x}=150+10\text{x}$
$\Rightarrow12\text{x}-10\text{x}=150$
$\Rightarrow2\text{x}=150$
$\Rightarrow\text{x}=\frac{150}{2}=75$
Therefore, the number is 75.

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