4 Mark Question - Maths STD 8 Questions - Vidyadip

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

Question 14 Marks

Solve:
$4-\frac{2(\text{z}-4)}{3}=\frac{1}{2}(2\text{z}+5)$

Answer

$4-\frac{2(\text{z}-4)}{3}=\frac{1}{2}(2\text{z}+5)$
$\Rightarrow\frac{12-2(\text{z}-4)}{3}=\frac{1(2\text{z}+5)}{2}$ $\text{(L.C.M of 1 and 3 is 3)}$
$\Rightarrow\frac{12-2\text{z}+8}{3}=\frac{2\text{z}+5}{2}$
$\Rightarrow\frac{20-2\text{z}}{3}=\frac{2\text{z}+5}{2}$
$\Rightarrow2(20-2\text{z})=3(2\text{z}+5)$ $\text{(by cross multiplication)}$
$\Rightarrow40-4\text{z}=6\text{z}+15$
$\Rightarrow40-15=6\text{z}+4\text{z}$
$\Rightarrow25=10\text{z}$
$\Rightarrow10\text{z}=5$ $(\text{by transposition})$
$\Rightarrow\text{z}=\frac{25}{10}=\frac{5}{2}$
$\therefore\text{z}=\frac{5}{2}$

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Question 24 Marks

Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find thenumber of deer in the herd.

Answer

Let the number of deer in the herd be x.
The number of deer grazing in the field is $\Big(\frac{1}{2}\Big)\text{x}$
Remaining deer $=\text{x}-\frac{\text{x}}{2}=\frac{\text{x}}{2}$
Number of deer playing nearby $=\frac{3}{4}\Big(\frac{\text{x}}{2}\Big)=\frac{3}{8}\text{x}$
The number of deer drinking water from the pond is 9.
$\therefore9+\frac{3}{8}\text{x}+\frac{1}{2}\text{x}=\text{x}$
$\Rightarrow\frac{72+3\text{x}+4\text{x}}{8}=\text{x}$
(multiplying the L.H.S. by 8, which is the L.C.M. of 1, 8 and 2)
$\Rightarrow72+7\text{x}=8\text{x}$ (by cross multiplication)
$\Rightarrow72 = 8\text{x} - 7\text{x}⇒ 72 = \text{x}⇒ \text{x} = 72$ Total number.

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Question 34 Marks

Solve:
$\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2}{3}-\text{t}$

Answer

$\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2}{3}-\text{t}$$\Rightarrow\frac{\text{3t}-2}{4}-\frac{\text{2t}+3}{3}=\frac{2-3\text{t}}{3}$ (3 is the L.C.M. of 1 and 3)
$\Rightarrow12\Big(\frac{\text{3t}-2}{4}\Big)-12\Big(\frac{\text{2t}+3}{3}\Big)=12\Big(\frac{2-3\text{t}}{3}\Big)$
(multiplying throughout by 12, which is the L.C.M. of 4, 3 and 3):
$\Rightarrow3(3\text{t}-2)-4(2\text{t}+3)=4(2-3\text{t})$
$\Rightarrow9\text{t} - 6 - 8\text{t} - 12 = 8 - 12\text{t}$
$\Rightarrow 9\text{t} - 8\text{t} - 6 - 12 = 8 - 12\text{t}$
$\Rightarrow\text{t} - 18 = 8 - 12\text{t}$
$\Rightarrow\text{t} + 12\text{t} = 18 + 8$
$\Rightarrow13\text{t} = 26$
$\Rightarrow\text{t} = \frac{26}{13} = 2$
$\therefore \text{t} = 2$

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Question 44 Marks

Monu's father is 26 years younger than Monu's grandfather and 29 years older than Monu. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Answer

Let the age of Monu's father be x years.
The age of Monu's grandfather will be $(\text{x}+26)$
Then, the age of Monu will be $( \text{x}-29 ).$
$\therefore\text{x} + ( \text{x}+26 ) + (\text{x}-29 ) = 135$
$\Rightarrow \text{x} + \text{x} + 26 +\text{ x} - 29 = 135$
$.\Rightarrow 3\text{x} -3 = 135$
$\Rightarrow 3\text{x} = 135 + 3$
$\Rightarrow 3\text{x} = 138$
$\Rightarrow \text{x} = \frac{138}{3} =46$
$\therefore$ Age of Monu's father $= 46 \text{ years}$
Age of Monu's grandfather $= (\text{x}+26 ) = ( 46+26 ) =72 \text{ years}$
Age of Monu $= ( \text{x}-29 ) = 46 - 29 = 17 \text{ years}$

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Question 54 Marks

The distance between two stations is 300km. Two motorcyclists start simultaneously from these stations and move towards each other. The speed of one of them is 7km/ h more than that of the other. If the distance between them after 2 hours of their start is 34km, find the speed of each motorcyclist. Check your solution.

Answer

Let the speed of one motorcyclist be x km/ h. So, the speed of the other motorcyclist will be (x + 7)km/ h. Distance travelled by the first motorcyclist in 2 hours = 2x km Distance travelled by the second motorcyclist in 2 hours = 2(x + 7)km Therefore, 300 - (2x + (2x + 14)) = 34 ⇒ 300 - (2x + 2x + 14) = 34 ⇒ 300 - 4x - 14 = 34 286 - 4x = 34 ⇒ 286 - 34 = 4x ⇒ 252 = 4x $\Rightarrow\text{x} = \frac{252}{4} = 63$ Therefore, the speed of the first motorcyclist is 63km/ h.The speed of the second motorcyclist is (x + 7) = (63 + 7) = 70km/ h.Check:
The distance covered by the first motorcyclist in 2 hours = 63 × 2 = 126km The distance covered by the second motorcyclist in 2 hours = 70 × 2 = 140km The distance between the motorcyclists after 2 hours = 300 - (126 + 140) = 34km (which is the same as given) Therefore, the speeds of the motorcyclists are 63km/h and 70km/h, respectively.

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Question 64 Marks

Solve:
$\frac{\text{5x}-4}{6}=\text{4x}+1-\frac{3\text{z}+10}{2}$

Answer

$\frac{\text{5x}-4}{6}=\text{4x}+1-\frac{3\text{z}+10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{2(4\text{x}+1)-3\text{x}-10}{2}$ $(\text{L.C.M}\text{ of }1\text{ and }2\text{ is }2)$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{8\text{x}+2-3\text{x}-10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{8\text{x}-3\text{x}+2-10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{5\text{x}-8}{2}$
$\Rightarrow2(5\text{x}-4)=6(5\text{x}-8)$
$\Rightarrow10\text{x}-8=30\text{x}-48$
$\Rightarrow10\text{x}-30\text{x}=-48+8$
$\Rightarrow10\text{x}-30\text{x}=-48+8$
$\Rightarrow-20\text{x}=-40$
$\Rightarrow\text{x}=\frac{-40}{-20}=2$
$\therefore\text{x}=2$

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Question 74 Marks

The denominator of a rational number is greater than its numerator by 7. If the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2. Find the original number.

Answer

Let the numerator be x.
The denominator is greater than the numerator by 7.
$\therefore(\text{x}+7)$
$\therefore\frac{\text{x}+17}{(\text{x}+7)-6}=2$
$\Rightarrow\frac{\text{x}+17}{\text{x}+1}=2$
$\Rightarrow\text{x}+17=2(\text{x}+1)$ (by cross multiplication)
$\Rightarrow\text{x}+17=2\text{x}+2$
$\Rightarrow\text{x}-2\text{x}=2-17$
$\Rightarrow-\text{x}=-15$
$\Rightarrow\text{x}=15$
Therefore, the numerator is 15.
Denominator $= (\text{ x}+7 ) = ( 15+7 ) = 22$
$\therefore$ Original number $=\frac{15}{22}$

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Question 84 Marks

An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by 4cm and the base decreased by 2cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Answer

Let the length of the base of the triangle be x cm.
Then, its altitude will be $\frac{5}{3}\text{x}\text{ cm}.$
Area of the triangle $=\frac{1}{2}( \text{x})\Big(\frac{5}{3} \text{x}\Big)=\frac{5}{6} \text{x}^2$
$\Rightarrow\frac{1}{2}(\text{x}-2)\Big(\frac{5}{3}\text{x}+4\Big)=\frac{5}{6}\text{x}^2$
$\Rightarrow\Big(\frac{ \text{x}-2}{2}\Big)\Big(\frac{5 \text{x}+12}{3}\Big)=\frac{5 \text{x}^2}{6}$
$\Rightarrow\frac{5 \text{x}^2+12 \text{x}-10-24}{6}=\frac{5 \text{x}^2}{6}$
$\Rightarrow 5 \text{x}^2 + 2 \text{x} - 24 = 5 \text{x}^2$
(cancelling the denominators from both the sides since they are same)
$\Rightarrow5 \text{x}^2 - 5 \text{x}^2 +2 \text{x} = 24$
$\Rightarrow2\text{x} = 24$
$\Rightarrow \text{x} = \frac{24}{2} =12\text{m}$
Therefore, the base of the triangle is 12m.
Altitude of the triangle $=\frac{5}{3} \text{x}=\frac{5}{3}(12)=20 \text{m}$

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Question 94 Marks

The length of a rectangle exceeds its breadth by 7cm. If the length is decreased by 4cm and the breadth is increased by 3cm, the area of the new rectangle is the same as the area of the original rectangle. Find the length and the breadth of the original rectangle.

Answer

Let the breadth of the original rectangle be x cm .
Then, its length will be $(x+7) cm$.
The area of the rectangle will be $(x)(x+7) cm ^2$.
$\therefore(x+3)(x+7-4)=(x)(x+7)$
$\Rightarrow(x+3)(x+3)=x^2+7 x$
$\Rightarrow x^2+3 x+3 x+9=x^2+7 x$
$\Rightarrow x^2+6 x+9=x^2+7 x$
$\Rightarrow 9=x^2-x^2+7 x-6 x$
$\Rightarrow 9=x$
$\Rightarrow x=9 \text { (by transposition) }$
Breadth of the original rectangle $=9 cm$
Length of the original rectangle $=(x+7)=(9+7)=16 cm$

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Question 104 Marks

Solve:
$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$

Answer

$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
$\Rightarrow\frac{2\text{x} + 7}{5} - \frac{3\text{x} + 11}{2} = \frac{2\text{x} + 8 - 15}{3} $ (L.C.M of 3 and 1 is 3)
$\Rightarrow30\Big(\frac{2\text{x} + 7}{5}\Big) -30\Big( \frac{3\text{x} + 11}{2}\Big) =30\Big( \frac{2\text{x} + 8 - 15}{3} \Big)$
(multiplying throughout by 30, which is the L.C.M. of 5, 2 and 3)
$\Rightarrow6 (2\text{x} + 7) - 15 (3\text{x} + 11) = 10 (2\text{x} + 8 - 15) $
$\Rightarrow12\text{x} + 42 -45\text{x} - 165 = 20\text{x} - 70$
$-33\text{x}-123=20\text{x}$

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Question 114 Marks

Solve:
$5\text{x}-\frac{1}{3}(\text{x}+1)=6(\text{x}+\frac{1}{30})$

Answer

$5\text{x}-\frac{1}{3}(\text{x}+1)=6(\text{x}+\frac{1}{30})$$\Rightarrow\text{5x}-\frac{1(\text{x}+1)}{3}=6\Big(\frac{30\text{x}+1}{30}\Big)$ $(\text{L.C.M}\text{ of }1\text{ and }30\text{ is }30)$
$\Rightarrow5\text{x}-\frac{(\text{x}+1)}{3}=\frac{30\text{x}+1}{5}$
$\Rightarrow\frac{15\text{x}-\text{x}-1}{3}=\frac{30\text{x}+1}{5}$ $(\text{L.C.M}\text{ of }1\text{ and }3\text{ is }3)$
$\Rightarrow\frac{14\text{x}-1}{3}=\frac{30\text{x}+1}{5}$
$\Rightarrow5(14\text{x}-1)=3(30\text{x}+1)$ $(\text{by cross multiplication})$
$\Rightarrow70\text{x}-5=90\text{x}+3$
$\Rightarrow70\text{x}-90\text{x}=3+5$
$\Rightarrow-20\text{x}=8$
$\Rightarrow\text{x}=\frac{8}{-20}=\frac{-2}{5}$
$\therefore\text{x}=-\frac{2}{5}$

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Question 124 Marks

The width of a rectangle is two-thirds its length. If the perimeter is 180 metres, find the dimensions of the rectangle.

Answer

Let the width of the rectangle be x cm.
It is $\frac{2}{3}$ of the length of the rectangle.
This means that the length of the rectangle will be $\frac{3}{2}\text{x}.$
Perimeter of the rectangle $=2(\text{x})+2\Big(\frac{3}{2}\Big)\text{x}=180\text{m}$
$\therefore2\text{x}+\frac{6\text{x}}{2}=180$
$\Rightarrow\frac{4\text{x}+6\text{x}}{2}=180$ (taking the L.C.M. of 1 on the L.H.S. of the equation)
$\Rightarrow10\text{x}=2\times180$ (by cross multiplication)
$\Rightarrow10\text{x}=360$
$\Rightarrow\text{x}=\frac{360}{10}=36$
Therefore, the width of the rectangle is 36 m.
Length of the rectangle will be $=\frac{3}{2}\text{x}=\frac{3}{2}(36)=54\text{m}$

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Question 134 Marks

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

Answer

Denominator, d = x
It is given that twice the numerator is equal to two more than the denominator.
$\therefore$ Twice of numerator, $2\text{n}=\text{x}+2$
$\therefore$ Numerator, $\text{n}=\frac{\text{x}+2}{2}$
$\therefore\frac{\text{n}+3}{\text{d}+3}=\frac{2}{3}$
$\Rightarrow3(\text{n}+3)=2(\text{d}+3)$ (by cross multiplication)
$\Rightarrow3\text{n}+9=2\text{d}+6$
$\Rightarrow3\text{n}-2\text{d}=6-9$
$\Rightarrow3\text{n}-2\text{d}=-3$
On replace d by x and n by $\frac{\text{x}+2}{2}$
$\Rightarrow3\Big(\frac{\text{x}+2}{2}\Big)-2\text{x}=-3$
$\Rightarrow\frac{3\text{x}+6-4\text{x}}{2}=-3$ (taking the L.C.M. of 2 and 1 as 2)
$\Rightarrow6-\text{x}=-6$ (by cross multipliction)
$\Rightarrow-\text{x}=-6-6$
$\Rightarrow\text{x}=12$
The denominator is 12.
$\therefore$ Numerator $=\frac{\text{x}+2}{2}=\frac{12+2}{2}=\frac{14}{2}=7$
$\therefore$ Original fraction $=\frac{7}{12}$

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Question 144 Marks

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.

Answer

Let the digit in the units place be x.
Digit in the tens place =3x
Original number = 10(3x) + x = 30x + x
On reversing the digits, we have x at the tens place and (3x) at the units place.
$\therefore$ New number =10(3x) + 3x = 10x + 3x
New number = Original number - 36
$\Rightarrow 10\text{x} + 3\text{x} = 30\text{x} + \text{x} - 36$
$\Rightarrow13\text{x} = 31\text{x} - 36$
$\Rightarrow36 = 31\text{x} - 13\text{x}$
$\Rightarrow36 = 18\text{x}$
$\Rightarrow18\text{x} = 36$
$\Rightarrow\text{x}=\frac{36}{18}=2$
Therefore, the digit in the units place is 2.
Digit in the tens place $=(3\text{x})=3\times2=6$
Therefore, the original number is 62.
Check:
New number + 36 = Original Number
$26+26=62$
Hence, both the conditions are satisfied.
Therefore, the original number is 62.

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Question 154 Marks

The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.

Answer

Let the digit in the units place be x.
Digit in the tens place $=(12-\text{x})$
$\therefore$ Original number $=10(12-\text{x})+\text{x}=120-9\text{x}$
On reversing the digits, we have x at the tens place and $(12-\text{x})$at the units place.
$\therefore$ New number $=10\text{x}+12-\text{x}=9\text{x}+12$
New number - Original number = 54
$\Rightarrow9\text{x}+12-(120-9\text{x})=54$
$\Rightarrow9\text{x}+12-120+9\text{x}=54$
$\Rightarrow18\text{x}-108=54$
$\Rightarrow18\text{x}=54+108$
$\Rightarrow18\text{x}=162$
$\Rightarrow\text{x}=\frac{162}{18}=9$
Therefore, the digit in the units place is 9.
Digit in tens place $=(12-\text{x})=(12-9)=3$
Therefore, the original number is 39.
Check:
The original number is 39.
Sum of the digits in the original number $=(3+9)=12$
New number obtained on reversing the digits $=93$
New number - Original number $=(93-39)=54$
Thus, both the given conditions are satisfied by 39.
Hence, the original number is 39.

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Question 164 Marks

Solve:
$\frac{3(\text{y}−5)}{4}−4\text{y}=3-\frac{(\text{y}-3)}{2}$

Answer

$\frac{3(\text{y}−5)}{4}−4\text{y}=3-\frac{(\text{y}-3)}{2}$
$\Rightarrow \frac{3\text{y} - 15}{4} - 4\text{y} = 3 - \frac{\text{y} - 3}{2}$
$\Rightarrow \frac{3\text{y} - 15 - 16\text{y}}{4} = 3 -\frac{ \text{y} - 3}{2}$ (L.C.M. of 4 and 1 is 4)
$\Rightarrow \frac{-13\text{y} − 15}{4} = \frac{6 - \text{y}+ 3}{2} $
$\Rightarrow \frac{-13\text{y} - 15}{4} = \frac{9 - \text{y}}{2}$
$\Rightarrow 2(-13\text{y}-15) = 4(9 - \text{y})$
$\Rightarrow -26\text{y} - 30 = 36 - 4\text{y}$
$\Rightarrow -26\text{y} + 4\text{y} = 36 + 30$
$\Rightarrow-22\text{y}=66$
$\Rightarrow-22\text{y}=66$ (multiplying both the sides with a - ve sign)
$\Rightarrow\text{y} = -\frac{66}{22 }= -3$
$\therefore\text{y} = -3$

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Question 174 Marks

5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.

Answer

Let the present age of the son be x years and that of the father be f years.5 years back, the father was 7 times as old as his son. $\therefore$ (f - 5) = 7(x - 5) f = 7x - 35 + 5 f = 7x - 30.... (1) After 5 years, ages of the father and son will be (f + 5) and (x + 5), respectively.After 5 years, the father will be three times older than his son. $\therefore$ (f + 5) = 3(x + 5) 7x - 30 + 5 = 3x + 15 [inserting the value of f from equation (1)] 7x - 25 = 3x + 15 7x - 3x = 25 + 15 4x = 40 $\text{x} = \frac{40}{4}= 10$ Therefore, the present age of the son is 10 years. Father's present age = ( 7x - 30) = 7(10) - 30 = 70 - 30 = 40 years

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