Question 14 Marks
In a shooting competition a marks man receives $50$ paise if he hits the mark and pays $20$ paise if he misses it. He tried $100$ shots and was paid $Rs. 29$. How many times did he hit the mark?
AnswerLet the marksman hit $x$ times and miss $y$ times
Therefore, $x - y = 100z$
$\Rightarrow y = x - 100\dots......(i)$
Also given, Amount earned
$= Rs.29$
$= (29 \times 100)$ paise
$= 2900$ paise
Therefore,
$50x - 20y = 2900$
$5x - 2y = 290 \dots.......(ii)$
Substituting $(i)$ in $(ii)$
$5x - 2(x - 100) = 29$
$5x - 2x + 200 = 290$
$3x = 90$
$x = 30$
Therefore, marks man hits $30$ times and misses $70$ times.
View full question & answer→Question 24 Marks
A boy played $100$ games, gaining $Rs. 50 $ on each game that he won, and losing $Rs. 20$ for each game that he lost. If on the whole he gained $Rs. 800$, find the number of games that he won.
AnswerLet number of games won be $x$ and number of games lost be $y.$
$x + y = 100$
$y = 100 -x........(i)$
Also given,
$Rs.$ gained $= Rs.800$
$50x - 20y = 800$
$\Rightarrow 5x - 2y = 80 \dots.....(ii)$
Substituting $(i) 9n (ii)$
$5x - 2(100 - x) = 80$
$\Rightarrow 5x - 200 + 2x = 80$
$\Rightarrow 7x = 280$
$\Rightarrow x = 40$
Number of games won $= 40.$
View full question & answer→Question 34 Marks
A worker is employed on the condition that he will be paid $Rs. 60$ for each day that he works and fined $Rs. 20$ for each day that he remains absent. If he is paid $Rs. 1000$ for the month of September, find the number of days that he worked.
AnswerLet working days be $x$ and non$-$working days be $y.$
$x + y = 30$
$y = 30 - x........(i)$
Also given, amount paid to him $= Rs.1000$
$60x - 20y = 1000$
$\Rightarrow 3x - y = 50\dots.....(ii)$
Substituting $(i)$ in $(ii)$
$3x - (30 - x) = 50$
$\Rightarrow 3x - 30 + x = 50$
$\Rightarrow 4x = 80$
$\Rightarrow x = 20$
Working days $= 20.$
View full question & answer→Question 44 Marks
In a factory a worker is paid $Rs. 20$ per hour for normal work and double the rate for overtime work. If he worked for $56$ hours in a week, find the number of hours of his normal work if he receives $Rs. 1440$ in all.
AnswerLet normal working hours be $x$ and overtime working hours be $y.$
$x + y = 56$
$y = 56 - x .......(i)$
Also given, total amount received $= Rs.1440$
$20x + 40y = 1440$
$\Rightarrow x + 2y = 72 .....(ii)$
Substituting $(i)$ and $(ii)$
$x + 2(56 - x) - 72$
$\Rightarrow x + 112 - 2x = 72$
$\Rightarrow x - 2x = 72 - 112$
$\Rightarrow x = 40$
Normal working hours $= 40$ hrs.
View full question & answer→Question 54 Marks
A man leaves half his property to his wife, one$-$third to his son and the remaining to his daughter. If the daughter's share is $Rs. 15000$, how much money did the man leave? How much money did his wife get?
AnswerLet $x$ be the total money left by the man.
Share of wife $=\frac{1}{2} x$
Share of son $=\frac{1}{3} x$
Share of daughter $=x-\frac{1}{2} x-\frac{1}{3} x=0.166667 x$
But daughter's share $= Rs. 15000$
$\Rightarrow 0.166667 \times \operatorname{Rs} .15000$
$\Rightarrow x=R s .90000$
Share of wife $=\frac{1}{2} x=\frac{1}{2} \times Rs. 90000= Rs. 45000$
Therefore, Man left $Rs. 90000$ and wife's share is $Rs. 45000 .$
View full question & answer→Question 64 Marks
How many kilograms of tea at $Rs.50$ per $kg$ should be mixed with $35 \ kg$ of tea costing $Rs.60$ per $kg$ so as to sell the mixture at $Rs.57$ per $kg$ without gaining or losing anything in transaction?
AnswerLet $x \ Kg$ of $Rs.50$ per $kg$ tea should be added with $35 \ kg$ of tea costing $Rs. \frac{60}{ \ kg }$.
Total weight of tea after mixing $=(x+35) g$
Selling price of $(x+35) \ Kg =(x+35) 57 R s$.
But here $S.P. = C.P.$
$C.P.$ of the tea
$=\operatorname{Rs} \cdot(50 x +35 \times 60) $
$\Rightarrow( x +35) 57=50 x +2100 $
$\Rightarrow 57 x +1995=50 x +2100 $
$\Rightarrow 7 x =105 $
$\Rightarrow x =15 \ kg$
Therefore, $15 \ kg$ of $Rs. 50$ per $kg$ should be added.
View full question & answer→Question 74 Marks
A $90 \ kg$ solution has $10\%$ salt. How much water must be evaporated to have the solution with $20\%$ salt?
AnswerQuantity of solution $=90 \ kg$
Quantity of salt
$=90 \ kg \times \frac{10}{100} $
$=9 \ kg$
Let $x \ kg$ of water be ecaporated.
Therefore,
$\frac{9}{90-x}=\frac{20}{100} $
$\Rightarrow 900-1800-20 x $
$\Rightarrow 20 x=900 $
$\Rightarrow x=45$
Therefore, $45 \ kg$ of water must be evaporated.
View full question & answer→Question 84 Marks
A $12\ liter$ solution is $33 \frac{1}{3} \%$ acid. How much water must be added to get the solution having $20 \%$ acid?
AnswerLet $x$ be the water added
then,
$12(33.33\%) = 20\%(x + 12)$
$\Rightarrow 12 \times 0.3333 = 0.2(x + 12)$
$\Rightarrow 3.999 = 0.2x + 2.4$
$\Rightarrow 0.2x = 1.596$
$\Rightarrow x = 7.95\ It = 8\ It$
Approximately, $8$ litres of water nust be added.
View full question & answer→Question 94 Marks
A man invested $Rs. 35000$, a part of it at $12\%$ and the rest at $14\%$. If he received a total annual interest of $Rs. 4460$, how much did he invest at each rate?
AnswerLet investment at $12\%$ be $x$ and at $14\%$ be $y.$
Then, $x + y = 35000$
$\Rightarrow y = 35000 - x \dots......(i)$
And $0.12x + 0.14y = 4460 \dots......(ii)$
Substituting $(i)$ in $(ii)$
$0.12x + 0.14(35000 - x) = 4460$
$\Rightarrow 0.12x + 4900 - 0.14x = 4460$
$\Rightarrow 0.02x = 440$
$\Rightarrow x = 22000$
Substitutng value of $x$ in $(i)$
$\Rightarrow y = 35000 - 22000$
$\Rightarrow y = 13000$
Therefore, he invested $Rs.22000$ at $12\%$ and $Rs.13000$ at $14\%.$
View full question & answer→Question 104 Marks
There are certain benches in a classroom. If $4$ students sit on each bench, three benches are left vacant and if $3$ students sit on each bench, $3$ students are left standing. What is the total number of students in the class?
AnswerLet the number of students be $x$ and number of benches be $y.$
$(i)$ When $4$ students sit on one bench:
Number of benches occupied $= y - 3$
$\Rightarrow x = 4(y - 3) \dots......(i)$
$(ii)$When $3$ students sit on one bench:
$3$ student are left standing
$\Rightarrow x = 3 = 3y$
$\Rightarrow x = 3y + 3 \dots......(ii)$
From $(i)$ and $(ii)$
$4(y - 3) = 3y + 3$
$4y - 12 = 3y + 3$
$y = 15$
But $x = 4(y - 3) \dots...($from $(i))$
$\Rightarrow x = 4(15 - 3)$
$\Rightarrow x = 48$
Therefore, Number of students in the class $= 48.$
View full question & answer→Question 114 Marks
The length of a rectangle is $30$ more than its breadth. The perimeter of the rectangle is $180 \ cm$. Find the length and breadth of the rectangle.
AnswerLet the breadth of the rectangle be $x \ cm.$
Then, length is $(x + 30)cm.$
Given, The perimeter of the rectangle is $180\ cm.$
$\Rightarrow 2($length $\times$ breadth$) = 180\ cm$
$\Rightarrow 2(x + x + 30)cm = 180\ cm$
$\Rightarrow 2x + 30 = 90\ cm$
$\Rightarrow 2x = 60\ cm$
$\Rightarrow x = 30\ cm$
Thus, the breadth of the rectangle is $30\ cm.$
Then, length is $60\ cm.$
View full question & answer→Question 124 Marks
A man is double his son's age. Twenty years ago, he was six times his son's age. Find the present age of the father and the son.
AnswerLet the present age of the son be $x$ years.
Then, the man's age $=2 x$ years.
$20$ years ago, their ages will be $( x-20 )$years and $(2 x-20)$ years respectively.
Given, Twenty years ago, son was six times his son's age.
As per given conditions,
$\frac{(x-20)}{(2 x-20)}=\frac{1}{6} $
$\Rightarrow 6 x-120=2 x-20 $
$\Rightarrow 4 x=100 $
$\Rightarrow x=25$ years.
$\Rightarrow y=50$ years.
Thus, the present age of the son is $25$ years.
Then, the man's age $=50$ years.
View full question & answer→Question 134 Marks
A boy is now one$-$third as old as his father. Twelve years hence he will be half as old as his father. Determine the present ages of the boy and that of his father.
AnswerGiven, present age of boy is one$-$third as that of his father.
So, let the present age of the son be $x$ years.
Then, present age of father is $3 x$ years.
Twelve years hence, their ages becomes, $(x+12)$ and $(3 x+12)$ respectively.
As per given conditions,
$\frac{(x+12)}{(3 x+12)}=\frac{1}{2}$
$\Rightarrow 2 x +24=3 x +12$
$\Rightarrow x =12$ years
Thus, the present age of the son is $12$ years.
Then, present age of father is $36$ years.
View full question & answer→Question 144 Marks
The age of a man is three times the age of his son. After $10$ years, the age of the man will be double that of his son. Find their present ages.
AnswerLet the age of the son be $x$ years.
Then, the man's age $=3 x$ years.
After $10$ years, their ages becomes
$(x+10)$ and $(3 x+10)$ respectively.
As per given conditions,
$\frac{(x+10)}{(3 x+10)}=\frac{1}{2}$
$\Rightarrow 2 x+20=3 x+10$
$\Rightarrow x=10$ years
Thus, the present age of the son is $10$ years.
Then, the main's age $=30$ years.
View full question & answer→Question 154 Marks
The present age of a man is double the age of his son. After $8$ years, the ratio of their ages will be $7:4.$ Find the present ages of the man and his son.
AnswerLet the present age of the son be $x$ years.
Then, the father's age $=2 x$ years.
After $8$ years, their ages will be $(x+8)$ years and $(2 x+8)$ years respectively.
Given, After $8$ years, the ratio of their ages will be $7: 4$.
As per given conditions,
$ \frac{(x+8)}{(2 x+8)}=\frac{4}{7}$
$\Rightarrow 7 x +56=8 x +32$
$\Rightarrow x =24$ years.
Thus, the present age of the son is $24$ years.
Then, the father's age $=48$ years.
View full question & answer→Question 164 Marks
The ages of $A$ and $B$ are in the ratio $7:5.$ Ten years hence, the ratio of their ages will be $9:7.$ Find their ages.
AnswerGiven, the present ages of $A$ and $B$ are in the ratio $7: 5$.
$\therefore$ Their present ages are $7 x: 5 x$.
After $10$ years, their ages will be :
$(7 x+10)$ and $(5 x+10)$ respectively.
New ratio of their ages $=9: 7$
As per given conditions,
$ \frac{(7 x+10)}{(5 x+10)}=\frac{9}{7}$
$\Rightarrow 7(7 x+10)=9(5 x+10)$
$\Rightarrow 49 x+70=45 x+90$
$\Rightarrow 4 x=20$
$\Rightarrow x=5$ years.
Thus, the present ages of $A$ and $B$ are $35$ and $25$ respectively.
View full question & answer→Question 174 Marks
The length of a room exceeds its breadth by $3 \ m$. If both the length and breadth, are increased by $1\ m$, then the area of the room is increased by $18 \ cm^2$. Find the length and breadth of the room.
AnswerLet the breadth of the rectangle be $x m.$
Then, the length $= (x + 3)m.$
Then, Area of the room $= x(x + 3)m^2$
If both the length and breadth, are increased by $1\ m$, then:
Increased length $= (x + 4)m.$
Increased Breadth $= (x + 1)m.$
Given, If both the length and breadth, are increased by $1\ m$, then the area of the room is increased by $18\ cm^2$
$\Rightarrow (x + 1)(x + 4) = x(x + 3) + 18$
$\Rightarrow (x^2 + x + 4x + 4) = x^2+ 3x + 18$
$\Rightarrow 5x - 3x = 18 - 4 = 14$
$\Rightarrow 2x = 14$
$\Rightarrow x = 7\ cm$
Thus, the breadth of the rectangle is $7\ m$.
Then, the length $= 10\ m.$
View full question & answer→Question 184 Marks
$A's$ age is six times that of $B's$ age. $15$ years hence $A$ will be three times as old as $B$; find their ages.
AnswerLet the Age of $B$ be $x$ years.
Then, the age of $A$ becomes $6x$ years.
After $15$ years,
Age of $A$ after $15$ years $= 6x + 15$
Age of $B$ after $15$ years $= x + 15$
Given, $15$ years hence $A$ will be there times as old as $B.$
$\Rightarrow 6x + 15 = 3(x + 15)$
$\Rightarrow 6x + 15 = 3x + 45$
$\Rightarrow 3x = 30$
$\Rightarrow x = 10$ years.
Thus, the Age of $B$ be $10$ years.
Then, the age of $A$ becomes $60$ years.
View full question & answer→Question 194 Marks
In an isosceles triangle, each of the two equal sides is $4 \ cm$ more than the base. If the perimeter of the triangle is $29 \ cm$, find the sides of the triangle.
AnswerLet the base of the triangle be $x \ cm.$
So, each of its equal sides $= (x + 4)cm$
As per the given condition,
perimeter of the triangle $= 29\ cm$
$\Rightarrow x + x + 4 + x + 4 = 29$
$\Rightarrow 3x = 21$
$\Rightarrow x = 7\ cm$
So,
$x + 4$
$= 7 + 4$
$= 11\ cm$
Hence, the sides are $7\ cm,11\ cm$ and $11\ cm.$
View full question & answer→Question 204 Marks
A man leaves half his property to his wife, one$-$third to his son, and the remaining to his daughter. If the daughter's share is $Rs. 15000,$ how much money did the man leave? How much money did his wife get?
AnswerLet the amount of property be $x$.
Daughter' share $=x=\frac{1}{2} x-\frac{1}{3} x$
As per the given condition,
$x-\frac{1}{2} x-\frac{1}{3} x=15000 $
$\Rightarrow \frac{6 x-3 x-2 x}{6}=15000 $
$\Rightarrow \frac{x}{6}=15000 $
$\Rightarrow x=90000 $
$\frac{1}{2} x=\frac{90000}{2}=45000$
Hence, the amount of property he left was $Rs. 90000$ and
the amount his wife got was $Rs. 45000 .$
View full question & answer→Question 214 Marks
In a shooting competition a marksman receives $Rs. 50$ if he hits the mark and pays $Rs. 20$ if he misses it. He tried $Rs. 100$ shots and was paid $Rs. 100.$ How many times did he hit the mark?
AnswerLet the number of times he hit the mark be $x$.
So, the number of times he misses it be $100-x$.
As per the given condition,
$50 x-20(100-x)=100 $
$\Rightarrow 50 x+2000+20 x=100 $
$\Rightarrow 50 x+20 x=2100 $
$\Rightarrow 70 x=2100 $
$\Rightarrow x=\frac{2100}{70} $
$\Rightarrow x=30$
Hence, he hit the mark $30$ times.
View full question & answer→Question 224 Marks
A boy played $100$ games, gaining $Rs. 50$ on each game that he won, and losing $Rs. 20$ for each game that he lost. If on the whole he gained $Rs. 800,$ find the number of games that he won.
AnswerLet the number of games that he won be $x$.
So, the number of games that he loss is $100-x$.
As per the given condition,
$50 x -2000+20 x =800 $
$\Rightarrow 50 x -2000+20 x =800 $
$\Rightarrow 50 x +20 x =2800 $
$\Rightarrow 70 x =2800 $
$\Rightarrow x =\frac{2800}{70} $
$\Rightarrow x =40$
Hence he won $40$ games.
View full question & answer→Question 234 Marks
The speed of a boat in still water is $8\ km/h.$ It takes the same time in going $20\ km$ in downstream as it takes in going $12 \ km$ upstream. Find the speed of the stream.
AnswerDistance travelled downstream $=20 \ km$.
Distance travelled upstream $=12 \ km$.
Given, The speed of a boat in still water is $8 \ km / h$.
Let the speed of a boat in still water is $8 \ km / h$.
Let the speed of the stream $=x \ km / hr$.
Relative speed upstream $=(8-x) \ km / hr$
Relative speed downstream $=(8+x) \ km / hr$
Time taken to go upstream $=\frac{12}{8-x_0}$ hrs
Time taken to go downstream $=\frac{20}{8-x}$ hrs
As per given condition, boat takes the same time in going $20 \ km$ in downstream as it takes in going $12 \ km$ upstream.
$ \Rightarrow \frac{12}{8-x}=\frac{20}{8+x}$
$12(8+x)=20(8-x)$
$\Rightarrow 12 \times 8+12 x=20 \times 8-20 x$
$\Rightarrow 8(20-12)=32 x$
$\Rightarrow 64=32 x$
$\Rightarrow x=2 \ km / hr $
Thus, the speed of the stream $=2 \ km / hr$.
View full question & answer→Question 244 Marks
The distance between two places Bangalore and Hyderabad is $660 \ km.$ Anand starts from Bangalore with a certain uniform speed while Sonu starts from Hyderabad at the same time with a uniform speed that is $4\ km$ more than that of Anand. If they meet each other after $5$ hours, find the speed of each.
AnswerLet the speed of Anand be $x \ km/hr.$
Then Speed of Sonu $= (x + 4)km/hr.$
In 5 hours, Anand travels $= 5x \ km$
In 5 hours, Sonu travels $= 5(x + 4)km$
Distance between two places $= 660\ km$
The total distance travelled by Anand and Sonu will be same as the distance between two places
$\text{ATQ}, 5x + 5(x + 4) = 660$
$\Rightarrow 10x + 20 = 660$
$\Rightarrow 10x = 640$
$\Rightarrow x = 64 \ km/hr$
Speed of Anand $= 64\ km/hr$
Speed of Sonu $= 64 + 4 = 68\ km/hr.$
View full question & answer→Question 254 Marks
The present age of a man is double the age of his son. After $8$ years, the ratio of their ages will be $7 : 4.$ Find the present ages of the man and his son.
AnswerLet the son's age be $x$ years,
so the man's age is $2 x$ years.
As per the given condition,
$\frac{2 x+8}{x+8}=\frac{7}{4} $
$\Rightarrow 4(2 x+8)=7(x+8) $
$\Rightarrow 8 x+32=7 x+56 $
$\Rightarrow 8 x-7 x=56-32 $
$\Rightarrow x=24$
Son's age $=x$ years $=24$ years
Man's age $=2 x$ years $=2(24)=48$ years
Hence, son's age is $24$ years and man's age is $48$ years.
View full question & answer→Question 264 Marks
The ages of $P$ and $Q$ are in the ratio $7 : 5.$ Ten years hence, the ratio of their ages will be $9 : 7.$ Find their ages.
AnswerLet the common multiple be $x$.
So, $P 's$ age be $7 x$ years and $Q 's$ age be $5 x$ years.
$10$ years hence, means after $10$ years.
As per the given condition,
$ \frac{7 x+10}{5 x+10}=\frac{9}{7}$
$\Rightarrow 7(7 x+10)=9(5 x+10)$
$\Rightarrow 49 x+70=45 x+90$
$\Rightarrow 49 x-45 x=90-70$
$\Rightarrow 4 x=20$
$\Rightarrow x=5 $
$P's$ age $=7$ years $=7(5)=35$ years
$Q's$ age $=5 x=5(5)=25$ years
Hence, $P^{\prime} s$ age is $35$ years and $Q^{\prime} s$ age is $25$ years.
View full question & answer→Question 274 Marks
A police car is ordered to chase a speeding car which is $5 \ km$ ahead and is travelling at an average speed of $80 \ km/h$. The police car is running at an average speed of $100 \ km/h$. How long it take for the police car to overtake the speeding car?
AnswerThe average speed of speeding car is $80 \ km / h$.
The average speed of police car is $100 \ km / h$.
As the police car and the speeding car is moving in the same direction so
Therefore there average speed would be subtracted to get the total speed
$ =100 lm / hr -80 \ km / hr$
$=20 \ km / hr $
Therefore the car will travel $20 \ km$ in $1$ hour
i.e. $20 \ km$ in $60$ minutes
So, the time taken by car to travel $1 \ km =\frac{60}{20}=3$ minutes
So. time taken by car to cover a distance of $5 \ km$ as the speeding car is $5 \ km$ ahead
$ =\frac{60}{20} \times 5$
$=15$ minutes.
View full question & answer→Question 284 Marks
The measures of the angles of a quadrilateral are: $(2x - 4)^\circ , (5x - 10)^\circ , (4x - 8)^\circ $ and $(7x - 14)^\circ $. find $ x.$
AnswerWe know that the sum of all angles of a quadrilateral are $360^{\circ} \ldots ($By angle sum property of a quadrilateral$)$
Thus,
$(2 x-4)^{\circ}+(5 x-10)^{\circ}+(4 x-8)^{\circ}+(7 x-14)^{\circ}=360^{\circ}$
$\Rightarrow$ Collecting like terms, we get:
$(2 x+5 x+4 x+7 x)+(-4-10-8-14)^{\circ}=360^{\circ} $
$\Rightarrow 18 x-36^{\circ}-360^{\circ}=0 $
$\Rightarrow x=\frac{396^{\circ}}{18} $
$=22^{\circ}$
View full question & answer→Question 294 Marks
The measures of the angles of a triangle are $(3x - 5)^\circ , (3++ 5)^\circ $ and $6x^\circ $ Find $x$ and state that the type of triangle formed.
AnswerWe know that the sum of all angles of a triangle are $180^{\circ} $(By angle sum property$)$
Thus, we get: $(3 x-5)^{\circ}+(3 x+5)^{\circ}+6 x^{\circ}=180^{\circ}$
$\Rightarrow$ Collecting like terms, we get:
$(2 x)^{\circ}+(5-5)^{\circ}=180^{\circ} $
$\Rightarrow x=\frac{180}{}^{\circ}=15^{\circ} .$
Thus, the angles of the triangle are respectively:
$(3 x-5)^{\circ}=(3 \times 15-5)^{\circ}=40^{\circ} $
$(3 x-5)^{\circ}=(3 \times 15+5)^{\circ}=50^{\circ}$
And $6 x ^{\circ}=6 \times 15^{\circ}=90^{\circ}$
Since on e of the angles is $90^{\circ}$, the type of triangle formed is right angled.
View full question & answer→Question 304 Marks
Two numbers are in the ratio $11:13.$ If the smaller number is $286,$ find the bigger one.
AnswerLet the two numbers be $x$ and $y$ Then, $x: y=11: 13$
The smaller number $=$
$\frac{11}{11+13} \text { of }(x+y)=286 \ldots ($Given$)$
$\Rightarrow \frac{11}{24} \times(x+y)=286 $
$\Rightarrow(x+y) $
$=\frac{286 \times 24}{11} $
$=624$
The bigger number $=$
$\frac{13}{11+13}$ of $(624) $
$\Rightarrow \frac{13}{24} \times(624)$
$=338$
View full question & answer→Question 314 Marks
Find the value of $x$ if the difference of one$-$third of $(x + 7)$ and one$-$fifth of $(3x - 2)$ is $3.$
AnswerGiven that:
$\frac{1}{3}(x+7)-\frac{1}{5}(3 x-2)=3$
Taking $\text{LCM}$,
$\frac{5(x+7)-3(3 x-2)}{15}=3$
$\Rightarrow$ On simplying and Cross $-$ multiplying, we get:
$5 x-9 x+35+6=15 \times 3 $
$\Rightarrow-4 x+41=45 $
$\Rightarrow-4 x=4 $
$\Rightarrow x=-1 .$
View full question & answer→Question 324 Marks
The sum of three consecutive odd natural numbers is $99$. Find the numbers.
AnswerLet the three consecutive odd natural numbers be $2n + 1, 2n + 3, 2n + 5$
Then, $\sum = 2n + 1 + 2n + 3 + 2n + 5 = 99$
$⇒ 6n + 9 = 99$
$⇒ 6n = 90$
$n = 15$
$⇒ 2n + 1 = 31, 2n + 3 = 33, 2n + 5 = 55$
Thus, the $3$ natural numbers are: $31, 33, 35.$
View full question & answer→Question 334 Marks
Find the value of $x$ for which the expression $\frac{x}{5}+2$ and $\frac{x}{3}-4$ are equal.
AnswerWe are to find the value of $x$ for which the two expressions are equal:
$\text { i.e. } \frac{x}{5}+2=\frac{x}{3}-4 $
$\Rightarrow \frac{x}{5}-\frac{x}{3}=-2-4 $
$\Rightarrow$ Taking $\text{LCM},$
$\frac{3 x-5 x}{15}=-6 $
$\Rightarrow \frac{-2 x}{15}=-6$
On Cross $-$ multiplying, we get:
$2 x =15 \times 6$
$\Rightarrow x =45$
View full question & answer→Question 344 Marks
Solve the following equations: $a(x - 2a) +b (x - 2b) = 4ab$
Answer$a(x-2 a)+b(x-2 b)=4 a b$
On simplifying, we get:
$\Rightarrow x a-2 a^2+b x-2 b^2=4 a b$
$\Rightarrow$ On Collecting like terms, we get:
$\Rightarrow x(a+b)=4 a b+2\left(a^2+b^2\right) $
$\Rightarrow x=\frac{2\left(a^2+b^2+2 a b\right)}{a+b} $
$=\frac{2(a+b)^2}{a+b} $
$=2(a+b)$
View full question & answer→Question 354 Marks
Solve the following equations for the unknown: $\frac{x+13}{x^2-1}+\frac{5}{x+1}=\frac{7}{x+1}$
Answer$\frac{x+13}{x^2-1}+\frac{5}{x+1}=\frac{7}{x+1} $
$\Rightarrow \frac{x+13}{(x-1)(x+1)}+\frac{5}{x+1}=\frac{7}{x+1}$
$\text{LCM}$ of all the denominators in the equation is $(x-1)(x+1)$.
Multiply throughout by the $\text{LCM}.$
$x+13+5(x-1)=7(x-1) $
$\Rightarrow x+13+5 x-5=7 x-7 $
$\Rightarrow 13-5+7=7 x-5 x-x $
$\Rightarrow x=15$
View full question & answer→Question 364 Marks
Solve the following equations for the unknown: $\frac{3}{x+1}-\frac{x-6}{x^2-1}=\frac{12}{x-1}$
Answer$\frac{3}{x+1}-\frac{x-6}{x^2-1}=\frac{12}{x-1} $
$\frac{3}{x+1}-\frac{x-6}{(x-1)(x+1)}=\frac{12}{x-1}$
$\text{LCM}$ of all the denominators in the equation is $(x-1)(x+1)$.
Multiply throughout by the $\text{LCM}.$
$3(x-1)-(x-6)=12(x+1) $
$\Rightarrow 3 x-3-x+6=12 x+12 $
$\Rightarrow 3 x-12 x-x=12-6+3 $
$\Rightarrow-10 x=9 $
$\Rightarrow x=-\frac{9}{10}$
View full question & answer→Question 374 Marks
Solve the following equations for the unknown: $11-\frac{3}{x}=\frac{5}{x}+3$
Answer$11-\frac{3}{x}=\frac{5}{x}+3$
$ \Rightarrow 11-3=\frac{5}{x}+\frac{3}{x} $
$\Rightarrow \frac{5}{x}+\frac{3}{x}=11-3 $
$\Rightarrow \frac{5+3}{x}=8$
$ \Rightarrow \frac{8}{x}=8 $
$ \Rightarrow x=\frac{8}{8} $
$ \Rightarrow x=1 .$
View full question & answer→Question 384 Marks
Solve the following equation for the unknown: $\frac{a-1}{2}-\frac{a+1}{3}=5-a$
Answer$\frac{a-1}{2}-\frac{a+1}{3}=5-a $
$\Rightarrow \frac{a-1}{2}-\frac{a+1}{3}+a=5$
Take the $\text{LCM}$ of all the denominators, that is, $2$ and $3$ which is $6 .$
Multiply throughout by the $\text{LCM}$
$\Rightarrow 3(a-1)-2(a+1)+6 a=30 $
$\Rightarrow 3 a-3-2 a-2+6 a=30 $
$\Rightarrow 7 a=35 $
$\Rightarrow a=5 .$
View full question & answer→Question 394 Marks
Solve the following equation for the unknown: $\frac{4}{5} x-21=\frac{3}{4} x-20$
Answer$\frac{4}{5} x-21=\frac{3}{4} x-20 $
$ \Rightarrow \frac{4}{5} x-\frac{3}{4} x=21-20 $
$ \Rightarrow \frac{16 x}{20}-\frac{15 x}{20}=21-20 $
$ \Rightarrow \frac{16 x-15 x}{20}=1 $
$ \Rightarrow 16 x-15 x=20 $
$ \Rightarrow x=20$
View full question & answer→Question 404 Marks
Solve the following equation for the unknown: $\frac{2(x-1)}{9}-\frac{x-1}{2}=0$
Answer$\frac{2(x-1)}{9}-\frac{x-1}{2}=0 $
$ \Rightarrow \frac{4 x-2}{9}-\frac{x-1}{2}=0 $
$ \Rightarrow \frac{4 x-2}{9}=\frac{x-1}{2}$
$ \Rightarrow 2(4 x-2)=9(x-1) $
$ \Rightarrow 8 x-4=9 x-9$
$\Rightarrow x=5 .$
View full question & answer→Question 414 Marks
Solve the following equation for the unknown: $\frac{x}{2}+\frac{x}{4}+\frac{x}{8}=7$
Answer$\frac{x}{2}+\frac{x}{4}+\frac{x}{8}=7$
$ \Rightarrow \frac{4 x+2 x+x}{8}=7$
$ \Rightarrow \frac{7 x}{8}=7$
$ \Rightarrow 7 x=(7)(8)$
$ \Rightarrow x=\frac{7 \times 8}{7} $
$\Rightarrow x=8 .$
View full question & answer→Question 424 Marks
Solve the following equation for the unknown: $\frac{9 y}{4}-\frac{5 y}{3}=\frac{1}{5}$
Answer$\frac{9 y}{4}-\frac{5 y}{3}=\frac{1}{5}$
$ \Rightarrow \frac{27 y-20 y}{12}=\frac{1}{5}$
$ \Rightarrow \frac{7 y}{12}=\frac{1}{5} $
$ \Rightarrow(7 y)(5)=12 $
$\Rightarrow y=\frac{12}{7 \times 5} $
$\Rightarrow y=\frac{12}{35} .$
View full question & answer→Question 434 Marks
Solve the following equation for the unknown: $\frac{1}{2} p+\frac{3}{4} p=p-3$
Answer$\frac{1}{2} p +\frac{3}{4} p = p -3$
$ \Rightarrow \frac{1}{2} p +\frac{3}{4} p - p =-3$
$ \Rightarrow \frac{ p }{2}+\frac{3 p }{4}- p =-3 $
$\Rightarrow \frac{2 p +3 p -4 p }{4}=-3$
$\Rightarrow \frac{ p }{4}=-3$
$ \Rightarrow p =(-3)(4) $
$ \Rightarrow p =-12 .$
View full question & answer→Question 444 Marks
Solve the following equations for the unknown: $(6p + 9)^2 + (8p - 7)^2 = (10p + 3)^2 - 71$
Answer$(6p + 9)^2 + (8p - 7)^2 = (10p + 3)^2 - 71$
$[36p^2 + 81 + 2(6p)(9)] + [64p^2 + 49 - 2(8p)(7)] = 100p^2 + 9 + 2(10p)(3) - 71$
$\Rightarrow 100p^2 + 130 + 2 \times 54p - 2 \times 54p - 2 \times 56p$
$= 100p^2 + 9 - 71 + 60p$
Collecting like terms,
$\Rightarrow 130 + 71 - 9 = 60p + 4p$
$\Rightarrow 192 = 64p$
$\Rightarrow p = 3.$
View full question & answer→Question 454 Marks
Solve the following equations for the unknown: $(a + 2)(2a +5) = 2(a + 1)^2+ 13$
Answer$(a + 2)(2a +5) = 2(a + 1)^2+ 13$
Collecting like terms,
$\Rightarrow a(2a + 5) + 2 (2a + 5) = 2a^2 + 2 + 4a + 13$
$\Rightarrow 2a^2+ 5a + 4a + 10 - (2a^2+ 4a + 15) = 0$
$\Rightarrow 5a + 10 - 15 = 0$
$\Rightarrow a = 1$ is the unknown.
View full question & answer→Question 464 Marks
In the following equations, verify if the given value is a solution of the equation: $\frac{3 x-1}{4}+\frac{3}{4}=2 ; x= 2$
Answer$\frac{3 x-1}{4}+\frac{3}{4}=2 ; x=2$
Simplifying, we get:
$\Rightarrow \frac{3 x-1+3}{4}=2 $
$\Rightarrow 3 x+2=8 $
$\Rightarrow 3 x=6 $
$\Rightarrow x=2$
Thus, $x=2$ is a solution of the equation $\frac{3 x-1}{4}+\frac{3}{4}=2 ;$
$x=2$.
View full question & answer→Question 474 Marks
If $a = 2x - 5, 3b = 3x + 1$ and if $a : b = 3 : 2,$ find the value of $x.$
Answer$\frac{ a }{ b }=\frac{3}{2} \ldots($Given$)$
$\Rightarrow$ On Cross $-$ multiplying, we get:
$2 a =3 b \cdots(1)$
Also, $a=2 x-5,3 b=3 x+1 \cdots (2)$
From $(1)$ and $(2),$
$2(2 x-5)=3 x+1 $
$\Rightarrow 4 x-10-3 x-1=0 $
$\Rightarrow x=11 .$
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