Question 15 Marks
Solve: $4-\frac{2(\text{z}-4)}{3}=\frac{1}{2}(2\text{z}+5)$
Answer$4-\frac{2(\text{z}-4)}{3}=\frac{1}{2}(2\text{z}+5)$
$\Rightarrow\frac{12-2(\text{z}-4)}{3}=\frac{1(2\text{z}+5)}{2}$
$\text{(L.C.M of 1 and 3 is 3)}$
$\Rightarrow\frac{12-2\text{z}+8}{3}=\frac{2\text{z}+5}{2}$
$\Rightarrow\frac{20-2\text{z}}{3}=\frac{2\text{z}+5}{2}$
$\Rightarrow2(20-2\text{z})=3(2\text{z}+5)$
$\text{(by cross multiplication)}$
$\Rightarrow40-4\text{z}=6\text{z}+15$
$\Rightarrow40-15=6\text{z}+4\text{z}$
$\Rightarrow25=10\text{z}$
$\Rightarrow10\text{z}=5$
$(\text{by transposition})$
$\Rightarrow\text{z}=\frac{25}{10}=\frac{5}{2}$
$\therefore\text{z}=\frac{5}{2}$
View full question & answer→Question 25 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.
AnswerLet 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)
$\Rightarrow72 = 8\text{x} - 7\text{x}\Rightarrow 72 = \text{x}\Rightarrow \text{x} = 72$ Total number.
View full question & answer→Question 35 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 $3):$
$\Rightarrow3(3\text{t}-2)-4(2\text{t}+3)=4(2-3\text{t})$
$\Rightarrow9\text{t} - 6 - 8\text{t} - 12 = 8 - 12\text{t}$
$\Rightarrow 9\text{t} - 8\text{t} - 6 - 12 = 8 - 12\text{t}$
$\Rightarrow\text{t} - 18 = 8 - 12\text{t}$
$\Rightarrow\text{t} + 12\text{t} = 18 + 8$
$\Rightarrow13\text{t} = 26$
$\Rightarrow\text{t} = \frac{26}{13} = 2$
$\therefore \text{t} = 2$
View full question & answer→Question 45 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?
AnswerLet 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 $= ( \text{x}-29 ) = 46 - 29 = 17 \text{ years}$
View full question & answer→Question 55 Marks
The distance between two stations is $300\ km$. Two motorcyclists start simultaneously from these stations and move towards each other. The speed of one of them is $7\ km/ h$ more than that of the other. If the distance between them after $2$ hours of their start is $34\ km$, find the speed of each motorcyclist. Check your solution.
AnswerLet 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 $
$\Rightarrow 300 - (2x + 2x + 14) = 34 $
$\Rightarrow 300 - 4x - 14 = 34 286 - 4x = 34$
$\Rightarrow 286 - 34 = 4x$
$ \Rightarrow 252 = 4x$
$\Rightarrow\text{x} = \frac{252}{4} = 63$
Therefore, the speed of the first motorcyclist is $63\ km/ h.$
The speed of the second motorcyclist is $(x + 7) = (63 + 7) = 70\ km/ h$.
Check:
The distance covered by the first motorcyclist in $2$ hours $= 63 \times 2 = 126\ km$
The distance covered by the second motorcyclist in $2$ hours $= 70 \times 2 = 140\ km$
The distance between the motorcyclists after $2$ hours $= 300 - (126 + 140) = 34\ km$ (which is the same as given)
Therefore, the speeds of the motorcyclists are $63\ km/h$ and $70\ km/h$, respectively.
View full question & answer→Question 65 Marks
Solve: $\frac{\text{5x}-4}{6}=\text{4x}+1-\frac{3\text{z}+10}{2}$
Answer$\frac{\text{5x}-4}{6}=\text{4x}+1-\frac{3\text{z}+10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{2(4\text{x}+1)-3\text{x}-10}{2}$
$(\text{L.C.M}\text{ of }1\text{ and }2\text{ is }2)$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{8\text{x}+2-3\text{x}-10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{8\text{x}-3\text{x}+2-10}{2}$
$\Rightarrow\frac{5\text{x}-4}{6}=\frac{5\text{x}-8}{2}$
$\Rightarrow2(5\text{x}-4)=6(5\text{x}-8)$
$\Rightarrow10\text{x}-8=30\text{x}-48$
$\Rightarrow10\text{x}-30\text{x}=-48+8$
$\Rightarrow10\text{x}-30\text{x}=-48+8$
$\Rightarrow-20\text{x}=-40$
$\Rightarrow\text{x}=\frac{-40}{-20}=2$
$\therefore\text{x}=2$
View full question & answer→Question 75 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.
AnswerLet 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 $=\frac{15}{22}$
View full question & answer→Question 85 Marks
An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by $4\ cm$ and the base decreased by $2\ cm$, the area of the triangle remains the same. Find the base and the altitude of the triangle.
AnswerLet 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 $=\frac{5}{3} \text{x}=\frac{5}{3}(12)=20 \text{m}$
View full question & answer→Question 95 Marks
The length of a rectangle exceeds its breadth by $7\ cm$. If the length is decreased by $4\ cm$ and the breadth is increased by $3\ cm$, 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.
AnswerLet 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 $= (x + 7) = (9 + 7) = 16\ cm$
View full question & answer→Question 105 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) $
$\Rightarrow12\text{x} + 42 -45\text{x} - 165 = 20\text{x} - 70$
$-33\text{x}-123=20\text{x}$
View full question & answer→Question 115 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)$
$(\text{L.C.M}\text{ of }1\text{ and }30\text{ is }30)$
$\Rightarrow5\text{x}-\frac{(\text{x}+1)}{3}=\frac{30\text{x}+1}{5}$
$\Rightarrow\frac{15\text{x}-\text{x}-1}{3}=\frac{30\text{x}+1}{5}$
$(\text{L.C.M}\text{ of }1\text{ and }3\text{ is }3)$
$\Rightarrow\frac{14\text{x}-1}{3}=\frac{30\text{x}+1}{5}$
$\Rightarrow5(14\text{x}-1)=3(30\text{x}+1)$
$(\text{by cross multiplication})$
$\Rightarrow70\text{x}-5=90\text{x}+3$
$\Rightarrow70\text{x}-90\text{x}=3+5$
$\Rightarrow-20\text{x}=8$
$\Rightarrow\text{x}=\frac{8}{-20}=\frac{-2}{5}$
$\therefore\text{x}=-\frac{2}{5}$
View full question & answer→Question 125 Marks
The width of a rectangle is two-thirds its length. If the perimeter is $180$ metres, find the dimensions of the rectangle.
AnswerLet 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 $=\frac{3}{2}\text{x}=\frac{3}{2}(36)=54\text{m}$
View full question & answer→Question 135 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.
AnswerDenominator, $d = x$ 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 $=\frac{7}{12}$
View full question & answer→Question 145 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.
AnswerLet 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 $62.$
View full question & answer→Question 155 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.
AnswerLet 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 $39.$
View full question & answer→Question 165 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)
$\Rightarrow\text{y} = -\frac{66}{22 }= -3$
$\therefore\text{y} = -3$
View full question & answer→Question 175 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.
AnswerLet 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
View full question & answer→