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Maths 4 Marks Questions

P-1 Linear Equations in Two Variables · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 62 questions.

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Question 34 Marks
A three digit number is equal to 17 times the sum of its digits; If the digits are reversed, the new number is 198 more than the old number ; also the sum of extreme digits is less than the middle digit by unity. Find the original number.
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Question 64 Marks
The perimeter of a rectangle is 40 cm. The length of the rectangle is
more than double its breadth by 2. Find length and breadth.
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Question 74 Marks
Solve : $(a-b) x+(a+b) y=a^2-2 a b-b^2$ and $(a+b)(x+y)=a^2+b^2$
Answer
$(a-b) x+(a+b) y=a^2-2 a b-b^2$...(i)
$(a+b)(x+y) \quad=a^2+b^2$
$\therefore \quad(a+b) x+(a+b) y=a^2+b^2$...(ii)
Subtracting (ii) from (i),
Image
$\therefore \quad(a-b-a-b) x=-2 b(a+b)$
$\therefore \quad-2 b x=-2 b(a+b)$
$\therefore \quad x=\frac{-2 b(a+b)}{-2 b}$
$\therefore \quad x=(a+b)$
Substituting value of x in (i), we get,
$(a-b)(a+b)+(a+b) y=a^2-2 a b-b^2$
$\therefore \quad a^2-b^2+(a+b) y=a^2-2 a b-b^2$
$\therefore \quad(a+b) y=-2 a b$
$\therefore \quad y=\frac{-2 a b}{a+b}$
$\therefore$$x=(a+b)$ and $y=\frac{-2 a b}{a+b}$ is the solution of the given equations.
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Question 84 Marks
A train covered a certain distance at a uniform speed. If the train would have been $6 \ km/hr$ faster, it would have taken $4$ hours less than the schedule time. And, if train were slower by $6 \ km/hr,$ it would have taken $6$ hours more than the schedule time. Find the length of the journey.
Answer
Let the original speed of train be $x \ km/hr$ and the actual time taken be $y$ hours.
Then, distance $=$ speed $\times$ time
$=(x y) \ km$
According to the first condition,
$(x+6)(y-4)=x y$
$\therefore x y-4 x+6 y-24=x y$
$\therefore-4 x+6 y=24$
$\therefore-2 x+3 y=12 ...(i)$
According to the second condition,
$(x-6)(y+6)=x y$
$\therefore x y+6 x-6 y-36=x y$
$\therefore 6 x-6 y=36$
$\therefore x-y=6 ...(ii)$
Multiplying $(ii)$ by $2,$ we get,
$2 x-2 y=12 ...(iii)$
Adding equations $(i)$ and $(iii)$ we get,
Image
Substituting the value of $y$ in equation $(i)$ we get,
$x-24=6$
$\therefore x=6+24$
$\therefore x=30$
$\therefore$ The original speed $= 30 \ km/hr$ and time $= 24$ hours.
Hence, length of the journey $=$ speed $\times$ time
$=30 \times 24$
$=720 \ km$
$\therefore$ Length of the journey is $720 \ km.$
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Question 94 Marks
The sum of the digits of a number consisting of three digits is 12. The middle digit is equal to half of the sum of the other two. If the order of the digit be reversed, the number diminished by 198. Find the number.
Answer
DigitsHTU
Original
number		x$\frac{1}{2}(x+y)$y
Reversed
number		y$\frac{1}{2}(y+x)$x
Let the digit in the hundredth place be x and the digit in the units place be y.
$\therefore$ The middle digit $=\frac{1}{2}(x+y)$
The sum of the digits is 12. (Given)
$\therefore x+\frac{1}{2}(x+y)+y=12$
$\therefore \quad 2 x+x+y+2 y=24$
$\therefore \quad 3 x+3 y=24$
$\therefore \quad x+y=8$... (i)
The original number $=100 x+10 \times \frac{1}{2}(x+y)+y$
$=100 x+5(x+y)+y$
$=100 x+5 x+5 y+y$
$=105 x+6 y$
The reversed number $=100 y+10 \times \frac{1}{2}(y+x)+x$
$=100 y+5(y+x)+x$
$=100 y+5 y+5 x+x$
$=6 x+105 y$
Now,
Reversed number = Original number – 198 (Given)
$\therefore \quad 6 x+105 y=105 x+6 y-198$
$\therefore 6 x-105 x+105 y-6 y=-198$
$\therefore \quad-99 x+99 y=-198$
$\therefore \quad x-y=2$ ....(ii)
Adding (i) and (ii), we get,
Image
$\therefore$ $x=5$
Substituting $x=5$ in (i),
$5+y=8$
$\therefore \quad y=8-5$
$\therefore \quad y=3$
$\therefore$ The required number $=105 x+6 y$
$=105 \times 5+6 \times 3$
= 525 + 18
= 543
$\therefore$ The required number is 543
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Question 104 Marks
A person deposits ₹ $x$ in savings bank account at the rate of $5 \%$ per annum and ₹ y in fixed deposit at $10 \%$ per annum. At the end of one year, he gets ₹ 400 as total interest. If he deposits ₹ y in savings bank account and ₹ $x$ in fixed deposit, he would get ₹ 350 as total interest. Find the total amount he deposited.
Answer
Amount deposited in savings account =₹ $x$
Amount deposited in fixed deposit account =₹ y
$\therefore \quad$ Total amount deposited =₹ $(x+y)$
Interest received on Savings account ₹ $\frac{5 x}{100}$
Interest received on fixed deposit account ₹ $\frac{10 y}{100}$
According to the first condition,
$\frac{5 x}{100}+\frac{10 y}{100}=400$
$\therefore \quad 5 x+10 y=40,000$ ...(i)
According to the second condition,
$\frac{10 x}{100}+\frac{5 y}{100}=350$
$\therefore 5 y+10 x=35,000$
$\therefore 10 x+5 y=35,000$...(ii)
Adding equation (i) and (ii), we get,
Image
$\therefore x+y=5,000$ [Dividing both sides by 15]
$\therefore$ Total amount deposited is ₹ 5000 .
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Question 114 Marks
Solve: $2^x+3^y=17,2^{x+2}-3^{y+1}=5$
Answer
$2^x+3^y=17...(i)$
$2^{x+2}-3^{y+1}=5$
$\therefore \quad 2^x \cdot 2^2-3^y \cdot 3^1=5$
$\therefore \quad 4.2^x-3.3^y=5 ...(ii)$
Substituting $2^x=m$ and $3^y=n$, we get,
$m+n=17...(iii)$
$4 m-3 n=5...(iv)$
Multiplying eq. $(iii)$ by $3,$
$ 3m + 3n = 51...(v)$
Adding $(iv)$ and $(v)$ we get,
Image
$\therefore \quad m=\frac{56}{7}$
$\therefore \quad m=8$
Substituting $m=8$ in $(iii),$
$8+n=17$
$\therefore \quad n=17-8$
$\therefore \quad n=9$
Resubstituting the value of $m$ and $n$, we get,
$ m=2^x$
$\therefore 8=2^x$
$\therefore 2^3=2^x$
$\therefore x=3$
$n=3^y$
$9=3^y$
$3^2=3^y$
$y=2$
$\therefore x=3$ and $y=2$ is the solution.
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Question 124 Marks
Find the area of the triangle formed by the following lines and X axis.
$4 x-3 y+4=0$ and $4 x+3 y-20=0$
Answer
$4 x-3 y+4=0 \quad \therefore y=\frac{4 x+4}{3}$
x25-4
y48-4
(x, y)(2, 4)(5, 8)(-4, -4)
$4 x+3 y-20=0 \quad \therefore y=\frac{20-4 x}{3}$
x2-15
y480
(x, y)(2, 4)(-1, 8)(5, 0)
Image
For $\triangle ABC , A (2,4), B (-1,0)$ and $C(5,0)$
$\therefore A (\triangle ABC )=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 6 \times 4$
$\therefore$ $A(\triangle A B C)=12$ sq. units
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Question 134 Marks
The forewheel of a carriage makes $6$ revolutions more than the rearwheel in going $120 m$ . If the diameter of the forewheel be increased by $\frac{1}{4}$ its present diameter and the diameter of the rearwheel be increased by $\frac{1}{5}$ of its present diameter, then the forewheel makes $4$ revolutions more than the rearwheel in going the same distance. Find the circumference of each wheel of the carriage.
Answer
Let the circumference of the forewheel be $x m$ and circumference of rearwheel be $y m$.
Number of revolutions $=\frac{\text { Distance covered }}{\text { circumference }}$
$\therefore $ Distance covered by each wheel $=120 m$
$\therefore $ Number of revolutions made by forewheel $=\frac{120}{x}$
$\therefore $ Number of revolutions made by rearwheel $=\frac{120}{y}$
According to the first condition,
$\frac{120}{x}=\frac{120}{y}+6$
$\therefore \frac{120}{x}-\frac{120}{y}=6$
$\therefore \frac{20}{x}-\frac{20}{y}=1 ...(i)$
Now, $c=\pi d$
$\therefore $ As diameter increases, circumference also increases.
Increase in circumference of forewheel $=\frac{x}{4}$
New circumference of forewheel $=x+\frac{x}{4}$
$=\frac{4 x+x}{4}$
$=\left(\frac{5 x}{4}\right) m $
New circumference of rearwheel $=y+\frac{y}{5}$
$=\frac{5 y+y}{5}$
$=\left(\frac{6 y}{5}\right) m $
According to the second condition,
$\frac{120}{5 x / 4}=\frac{120}{6 y / 5}+4$
$\therefore \frac{120 \times 4}{5 x}=\frac{120 \times 5}{6 y}+4$
$\therefore \frac{96}{x}=\frac{100}{y}+4$
$\therefore \frac{96}{x}-\frac{100}{y}=4 ...(ii)$
Substituting $\frac{1}{x}=a$ and $\frac{1}{y}=b$, we get,
$20a – 20b = 1 ...(iii)$
$96a – 100b = 4...(iv)$
Multiplying equation $(iii)$ by $5,$ we get
$100a – 100b = 5 ...(v)$
Subtracting $(v)$ from $(iv),$
Image
$\therefore a=\frac{1}{4}$
Substituting value of $a=\frac{1}{4}$ in $(iii),$ we get
$20 \times \frac{1}{4}-20 b=1$
$\therefore 5-20 b=1$
$\therefore 5-1=20 b$
$\therefore 4=20 b$
$\therefore b=\frac{4}{20}$
$\therefore b=\frac{1}{5}$
Resubstituting the values of $a$ and $b,$
$a=\frac{1}{x}=\frac{1}{4}$ and $b=\frac{1}{y}=\frac{1}{5}$
$\therefore x=4$ and $y=5$
$\therefore$Circumference of forewheel is $4 m$ and circumference of rearwheel is $5 m$.
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Question 144 Marks
When the son will be as old as his father today, the sum of their ages then will be 126; when the father was as old as his son is today, the sum of their ages then was 38. Find their present ages.
Answer
Let the present age of father be x years and son be
y years.
Son’s age when he will be as old as his father
today = x years.
$\therefore \quad$ Father's age at that time $=[x+(x-y)]$ years
According to the first condition,
$x+x+(x-y)=126$
$\therefore \quad 3 x-y=126$ ...(i)
Father’s age when he was as old as son today = y years
$\therefore \quad$ Son's age at that time $=[y-(x-y)]$ years
According to the second condition,
$y+y-(x-y)=38$
$\therefore \quad y+y-x+y=38$
$\therefore \quad-x+3 y=38$ ...(ii)
Multiplying (ii) by 3,
$-3 x+9 y=114$ ...(iii)
Adding (i) and (iii), we get
Image
$\therefore \quad y=\frac{240}{8}$
y = 30
Substituting the value of y in equation (i) we get,
$3 x-30=126$
$\therefore \quad 3 x=126+30$
$\therefore \quad 3 x=156$
$\therefore \quad x=\frac{156}{3}$
$\therefore \quad x= 5 2$
$\therefore$ The present age of father and son are 52 years and 30 years respectively.
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Question 154 Marks
Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.
Answer
Let the speed of the train be x km/hr and speed of the taxi be y km/hr.
Time $=\frac{\text { Distance }}{\text { Speed }}$
Time taken to cover 300 km by train $=\frac{300}{x}$ hrs.
Time taken to cover 200 km by taxi $=\frac{200}{y}$ hrs.
According to the first condition,
$\frac{300}{x}+\frac{200}{y}=5 \frac{30}{60}$ [ 1 hour $=60 min$.]
$\therefore \quad \frac{300}{x}+\frac{200}{y}=\frac{11}{2}$ ...(i)
Similarly,
Time taken to cover 260 km by train $=\frac{260}{x} hrs$.
Time taken to cover 240 km by taxi $=\frac{240}{y}$ hrs.
According to the second condition,
$\frac{260}{x}+\frac{240}{y}=5 \frac{36}{60}$
$\therefore \quad \frac{260}{x}+\frac{240}{y}=\frac{28}{5}$ ...(ii)
Substituting $\frac{1}{x}=m$ and $\frac{1}{y}=n$ in equations (i) and (ii),
$300 m+200 n=\frac{11}{2}$
i.e. $600 m+400 n=11$ ...(iii)
Similarly, $260 m+240 n=\frac{28}{5}$
i.e. $1300 m+1200 n=28$...(iv)
Multiplying eq. (iii) by 3 ,
$1800 m+1200 n=33$ ...(v)
Subtracting (iv) from (v),
Image
$\therefore \quad m=\frac{5}{500}$
$\therefore \quad m=\frac{1}{100}$
Substituting value of m in equation (iii) we get,
$600 \times \frac{1}{100}+400 n=11$
$\therefore \quad 6+400 n=11$
$\therefore \quad 400 n=11-6$
$\therefore \quad 400 n=5$
$\therefore \quad n=\frac{5}{400}$
$\therefore \quad n=\frac{1}{80}$
Resubstituting the values of $m$ and $n$,
$\frac{1}{x}=\frac{1}{100}$ and $\frac{1}{y}=\frac{1}{80}$
$\therefore x=100$ and $y=80$
The speed of the train is $100 km / hr$ and the speed of the taxi is $80 km / hr$.
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Question 164 Marks
The weight of a bucket is 15 kg , when it is filled with water upto $\frac{3}{5}$ of its capacity. And the weight is 19 kg , if it is filled with water upto $\frac{4}{5}$ of its capacity. Find the weight of bucket, if it is completely filled with water.
Answer
Let the weight of empty bucket be x kg and the weight of water the bucket can hold to its full capacity be y kg.
According to the first condition,
$x+\frac{3}{5} y=15$
$\therefore \quad 5 x+3 y=75$ ...(i)
According to the second condition,
$x+\frac{4}{5} y=19$
$\therefore \quad 5 x+4 y=95$...(ii)
Subtracting (ii) from (i),
Image
$\therefore y=20$
Substituting y = 20 in (i), we get,
$5 x+3(20)=75$
$\therefore \quad 5 x+60=75$
$\therefore \quad 5 x=75-60$
$\therefore \quad 5 x=15$
$\therefore \quad x=3$
$\therefore$The weight of bucket when it is completely filled with water = $x+y$
= 3 + 20
= 23 kg.
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Question 224 Marks
A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, If he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer
$100 km / hr$ and $80 km / hr$
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Question 344 Marks
A three digit number is equal to 17 times the sum of its digits; If the digits are reversed, the new number is 198 more than the old number ; also the sum of extreme digits is less than the middle digit by unity. Find the original number.
Answer
Let the digit in hundreds place be x and that in unit place be y.

HTunit
xx+y+1y

$\therefore$ the three digit number is $100 x+10(x+y+1)+y$
$=100 x+10 x+10 y+10+y=110 x+11 y+10$
the sum of the digits in the given number $=x+(x+y+1)+y=2 x+2 y+1$
$\therefore$ From first condition
$\begin{aligned}
& \text { Given number }=17 \times(\text { sum of the digits }) \\
& \therefore 110 x+11 y+10=17 \times(2 x+2 y+1) \\
& \therefore 110 x+11 y+10=34 x+34 y+17 \\
& \therefore 76 x-23 y=7 \ldots \text { (I) }
\end{aligned}$
The number obtained by reversing the digits
$\begin{aligned}
& =100 y+10(x+y+1)+x=110 y+11 x+10 \\
& \text { Given number }=110 x+11 y+10
\end{aligned}$
From $2^{\text {nd }}$ condition, Given number $+198=$ new number.
$\begin{aligned}
& 110 x+11 y+10+198=110 y+11 x+10 \\
& 99 x-99 y=-198 \\
& \quad x-y=-2 \\
& \therefore x=y-2 \ldots \text { (II) }
\end{aligned}$
Substitute this value of $x$ in equation (I).
$\begin{gathered}
\therefore 76(y-2)-23 y=7 \\
\therefore 76 y-152-23 y=7 \\
53 y=159
\end{gathered}$
$\therefore \quad y=3 \quad \therefore$ the digit in units place is $=3$
Substitute this value in equation (II)
$\begin{aligned}
& \quad x=y-2 \\
& \therefore \quad x=3-2=1 \\
& \therefore \quad x=1
\end{aligned}$
$\therefore \quad x=1 \quad \therefore \text { The digit in hundred's place is } 1$
the digit in ten's place is $3+1+1=5$
$\therefore$ the number is 153 .

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Question 354 Marks

Image
Answer
Let the speed of the boat in still water be $x km / hr$ and the speed of water current be $y km / hr$
$\therefore$ speed of boat in downstream $=(x+y) km / hr$. and that in upstream $=(x-y) km / hr$.
Now distance $=$ speed $\times$ time
$\therefore \text { time }=\frac{\text { distance }}{\text { speed }}$
Time taken by the boat to travel $16 km$ upstream $=\frac{16}{x-y}$ hours and it takes $\frac{24}{x+y}$ hours to travel $24 km$ downstream.
from first condition -
$\frac{16}{x-y}+\frac{24}{x+y}=6 \ldots (I)$
from $2^{\text {nd }}$ condition
$\frac{36}{x-y}+\frac{48}{x+y}=13 \ldots (II)$
By replacing $\frac{1}{x-y}$ by $m$ and $\frac{1}{x+y}$ by $n$ we get
$\begin{aligned}
& 16 m+24 n=6 \ldots \text { (III) } \\
& 36 m+48 n=13 \ldots \text { (IV) }
\end{aligned}$
Solving equations (III) and (IV) $m=\frac{1}{4}, n=\frac{1}{12}$
Repalcing $m, n$ by their original values we get
$x-y=4 \ldots \text { (V) } x+y=12 \ldots \text { (VI) }$
Solving equations (V), (VI) we get $x=8, y=4$
$\therefore$ speed of the boat in still water is $8 km / hr$. and speed of water current is $4 km / hr$.
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Question 364 Marks

Image
Answer
Let selling price of each analogue watch be $₹ x$ \
Selling price of each digital watch be ₹ $y$ 
From first condition -
$11 x+6 y=4330 \ldots (2)$
from $2^{\text {nd }}$ condition -
$22 x+5 y=7330 \ldots (3)$
multiplying equation (I) by 2 we get,
$22 x+12 y=8660 \ldots \text { (III) }$
subtract equation (III) from equation (II).
$\begin{gathered}
22 x+5 y=7330 \ldots \\
-\quad 22 x \pm 12 y=\_8660 \ldots \\
\hline-7 y=-1330 \\
\therefore \quad y=190
\end{gathered}$
Substitute this value of $y$ in equation (I)
$\begin{aligned}
11 x+6 y & =4330 \\
\therefore 11 x+6(190) & =4330 \\
\therefore \quad 11 x+1140 & =4330 \\
\therefore \quad 11 x & =3190 \\
\therefore \quad x & =290
\end{aligned}$
$\therefore$ selling price of each analogue watch is ₹ 290 and of each digital watch is ₹ 190 .
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Question 374 Marks
The perimeter of a rectangle is 40 cm. The length of the rectangle is
more than double its breadth by 2. Find length and breadth.
Answer
Let length of rectangle be $x cm$ and breadth be $y cm$.
From first condition -
$\begin{aligned}
2(x+y) & =40 \\
x+y & =20 \ldots(1)
\end{aligned}$
From $2^{\text {nd }}$ condition -
$\begin{gathered}
x=2 y+2 \\
\therefore x-2 y=2 \ldots(2)
\end{gathered}$
Let's solve eq. (I), (II) by determinant method
$\begin{aligned}
& \quad \begin{aligned}
x+y=20 \\
x-2 y=2
\end{aligned} \\
& D =\left|\begin{array}{rr}
1 & 1 \\
1 & -2
\end{array}\right|=[1 \times(-2)]-(1 \times 1)=-2-1=-3 \\
& D _x=\left|\begin{array}{rr}
20 & 1 \\
2 & -2
\end{array}\right|=[20 \times(-2)]-(1 \times 2)=-40-2=-42 \\
& D _y=\left|\begin{array}{rr}
1 & 20 \\
1 & 2
\end{array}\right|=(1 \times 2)-(20 \times 1)=2-20=-18 \\
& x=\frac{ D _x}{ D } \text { and } y=\frac{ D _y}{ D } \\
& \therefore x=\frac{-42}{-3} \text { and } y=\frac{-18}{-3} \\
& \therefore x=14, \quad y=6
\end{aligned}$
$\therefore$ Length of the rectangle is $14 cm$ and breadth is $6 cm$.
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Question 424 Marks
When the son will be as old as his father today, the sum of their ages then will be 126; when the father was as old as his son is today, the sum of their ages then was 38. Find their present ages.
Answer
Let the present age of father be x years and son be
y years.
Son’s age when he will be as old as his father
today = x years.
$\therefore \quad$ Father's age at that time $=[x+(x-y)]$ years
According to the first condition,
$x+x+(x-y)=126$
$\therefore \quad 3 x-y=126$ ...(i)
Father’s age when he was as old as son today = y years
$\therefore \quad$ Son's age at that time $=[y-(x-y)]$ years
According to the second condition,
$y+y-(x-y)=38$
$\therefore \quad y+y-x+y=38$
$\therefore \quad-x+3 y=38$ ...(ii)
Multiplying (ii) by 3,
$-3 x+9 y=114$ ...(iii)
Adding (i) and (iii), we get
Image
$\therefore \quad y=\frac{240}{8}$
y = 30
Substituting the value of y in equation (i) we get,
$3 x-30=126$
$\therefore \quad 3 x=126+30$
$\therefore \quad 3 x=156$
$\therefore \quad x=\frac{156}{3}$
$\therefore \quad x= 5 2$
$\therefore$ The present age of father and son are 52 years and 30 years respectively.
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Question 434 Marks
The weight of a bucket is 15 kg , when it is filled with water upto $\frac{3}{5}$ of its capacity. And the weight is 19 kg , if it is filled with water upto $\frac{4}{5}$ of its capacity. Find the weight of bucket, if it is completely filled with water.
Answer
Let the weight of empty bucket be x kg and the weight of water the bucket can hold to its full capacity be y kg.
According to the first condition,
$x+\frac{3}{5} y=15$
$\therefore \quad 5 x+3 y=75$ ...(i)
According to the second condition,
$x+\frac{4}{5} y=19$
$\therefore \quad 5 x+4 y=95$...(ii)
Subtracting (ii) from (i),
Image
$\therefore y=20$
Substituting y = 20 in (i), we get,
$5 x+3(20)=75$
$\therefore \quad 5 x+60=75$
$\therefore \quad 5 x=75-60$
$\therefore \quad 5 x=15$
$\therefore \quad x=3$
$\therefore$The weight of bucket when it is completely filled with water = $x+y$
= 3 + 20
= 23 kg.
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Question 444 Marks
The sum of the digits of a number consisting of three digits is 12. The middle digit is equal to half of the sum of the other two. If the order of the digit be reversed, the number diminished by 198. Find the number.
Answer
DigitsHTU
Original
number		x$\frac{1}{2}(x+y)$y
Reversed
number		y$\frac{1}{2}(y+x)$x
Let the digit in the hundredth place be x and the digit in the units place be y.
$\therefore$ The middle digit $=\frac{1}{2}(x+y)$
The sum of the digits is 12. (Given)
$\therefore x+\frac{1}{2}(x+y)+y=12$
$\therefore \quad 2 x+x+y+2 y=24$
$\therefore \quad 3 x+3 y=24$
$\therefore \quad x+y=8$... (i)
The original number $=100 x+10 \times \frac{1}{2}(x+y)+y$
$=100 x+5(x+y)+y$
$=100 x+5 x+5 y+y$
$=105 x+6 y$
The reversed number $=100 y+10 \times \frac{1}{2}(y+x)+x$
$=100 y+5(y+x)+x$
$=100 y+5 y+5 x+x$
$=6 x+105 y$
Now,
Reversed number = Original number – 198 (Given)
$\therefore \quad 6 x+105 y=105 x+6 y-198$
$\therefore 6 x-105 x+105 y-6 y=-198$
$\therefore \quad-99 x+99 y=-198$
$\therefore \quad x-y=2$ ....(ii)
Adding (i) and (ii), we get,
Image
$\therefore$ $x=5$
Substituting $x=5$ in (i),
$5+y=8$
$\therefore \quad y=8-5$
$\therefore \quad y=3$
$\therefore$ The required number $=105 x+6 y$
$=105 \times 5+6 \times 3$
= 525 + 18
= 543
$\therefore$ The required number is 543
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Question 454 Marks
The forewheel of a carriage makes 6 revolutions more than the rearwheel in going 120 m . If the diameter of the forewheel be increased by $\frac{1}{4}$ its present diameter and the diameter of the rearwheel be increased by $\frac{1}{5}$ of its present diameter, then the forewheel makes 4 revolutions more than the rearwheel in going the same distance. Find the circumference of each wheel of the carriage.
Answer
Let the circumference of the forewheel be x m and circumference of rearwheel be y m.
Number of revolutions $=\frac{\text { Distance covered }}{\text { circumference }}$
$\therefore \quad$ Distance covered by each wheel $=120 m$
$\therefore \quad$ Number of revolutions made by forewheel $=\frac{120}{x}$
$\therefore \quad$ Number of revolutions made by rearwheel $=\frac{120}{y}$
According to the first condition,
$\frac{120}{x}=\frac{120}{y}+6$
$\therefore \quad \frac{120}{x}-\frac{120}{y}=6$
$\therefore \quad \frac{20}{x}-\frac{20}{y}=1$ ...(i)
Now, $c=\pi d$
$\therefore \quad$ As diameter increases, circumference also increases.
Increase in circumference of forewheel $=\frac{x}{4}$
New circumference of forewheel $=x+\frac{x}{4}$
$\begin{array}{l}=\frac{4 x+x}{4} \\ =\left(\frac{5 x}{4}\right) m \end{array}$
New circumference of rearwheel $=y+\frac{y}{5}$
$\begin{array}{l}=\frac{5 y+y}{5} \\ =\left(\frac{6 y}{5}\right) m \end{array}$
According to the second condition,
$\frac{120}{5 x / 4}=\frac{120}{6 y / 5}+4$
$\therefore \quad \frac{120 \times 4}{5 x}=\frac{120 \times 5}{6 y}+4$
$\therefore \quad \frac{96}{x}=\frac{100}{y}+4$
$\therefore \quad \frac{96}{x}-\frac{100}{y}=4$ ...(ii)
Substituting $\frac{1}{x}=a$ and $\frac{1}{y}=b$, we get,
20a – 20b = 1 ...(iii)
96a – 100b = 4...(iv)
Multiplying equation (iii) by 5, we get
100a – 100b = 5 ...(v)
Subtracting (v) from (iv),
Image
$\therefore \quad a=\frac{1}{4}$
Substituting value of $a=\frac{1}{4}$ in (iii), we get
$20 \times \frac{1}{4}-20 b=1$
$\therefore 5-20 b=1$
$\therefore \quad 5-1=20 b$
$\therefore \quad 4=20 b$
$\therefore \quad b=\frac{4}{20}$
$\therefore \quad b=\frac{1}{5}$
Resubstituting the values of a and b,
$a=\frac{1}{x}=\frac{1}{4}$ and $b=\frac{1}{y}=\frac{1}{5}$
$\therefore \quad x=4$ and $y=5$
$\therefore$Circumference of forewheel is 4 m and circumference of rearwheel is 5 m.
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Question 464 Marks
Solve : $(a-b) x+(a+b) y=a^2-2 a b-b^2$ and $(a+b)(x+y)=a^2+b^2$
Answer
$(a-b) x+(a+b) y=a^2-2 a b-b^2$...(i)
$(a+b)(x+y) \quad=a^2+b^2$
$\therefore \quad(a+b) x+(a+b) y=a^2+b^2$...(ii)
Subtracting (ii) from (i),
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$\therefore \quad(a-b-a-b) x=-2 b(a+b)$
$\therefore \quad-2 b x=-2 b(a+b)$
$\therefore \quad x=\frac{-2 b(a+b)}{-2 b}$
$\therefore \quad x=(a+b)$
Substituting value of x in (i), we get,
$(a-b)(a+b)+(a+b) y=a^2-2 a b-b^2$
$\therefore \quad a^2-b^2+(a+b) y=a^2-2 a b-b^2$
$\therefore \quad(a+b) y=-2 a b$
$\therefore \quad y=\frac{-2 a b}{a+b}$
$\therefore$$x=(a+b)$ and $y=\frac{-2 a b}{a+b}$ is the solution of the given equations.
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Question 474 Marks
Solve: $2^x+3^y=17,2^{x+2}-3^{y+1}=5$
Answer
$2^x+3^y=17$...(i)
$2^{x+2}-3^{y+1}=5$
$\therefore \quad 2^x \cdot 2^2-3^y \cdot 3^1=5$
$\therefore \quad 4.2^x-3.3^y=5$ ...(ii)
Substituting $2^x=m$ and $3^y=n$, we get,
$m+n=17$...(iii)
$4 m-3 n=5$...(iv)
Multiplying eq. (iii) by 3,
 3m + 3n = 51...(v)
Adding (iv) and (v) we get,
Image
$\therefore \quad m=\frac{56}{7}$
$\therefore \quad m=8$
Substituting $m=8$ in (iii),
$8+n=17$
$\therefore \quad n=17-8$
$\therefore \quad n=9$
Resubstituting the value of m and n, we get,
$\begin{aligned} & m=2^x \\ \therefore & 8=2^x \\ \therefore & 2^3=2^x \\ \therefore & x=3\end{aligned}$
$\begin{array}{l}n=3^y \\ 9=3^y \\ 3^2=3^y \\ y=2\end{array}$
$\therefore$ $x=3$ and $y=2$ is the solution.
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Question 484 Marks
Solve : $\frac{7}{2 x+1}+\frac{13}{y+2}=27 ; \frac{13}{2 x+1}+\frac{7}{y+2}=33$
Answer
$x=\frac{-1}{4}, y=-1$
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Question 524 Marks
Find the area of the triangle formed by the following lines and X axis.
$4 x-3 y+4=0$ and $4 x+3 y-20=0$
Answer
$4 x-3 y+4=0 \quad \therefore y=\frac{4 x+4}{3}$
x25-4
y48-4
(x, y)(2, 4)(5, 8)(-4, -4)
$4 x+3 y-20=0 \quad \therefore y=\frac{20-4 x}{3}$
x2-15
y480
(x, y)(2, 4)(-1, 8)(5, 0)
Image
For $\triangle ABC , A (2,4), B (-1,0)$ and $C(5,0)$
$\therefore A (\triangle ABC )=\frac{1}{2} \times b \times h$
$=\frac{1}{2} \times 6 \times 4$
$\therefore$ $A(\triangle A B C)=12$ sq. units
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Question 534 Marks
A train covered a certain distance at a uniform speed. If the train would have been 6 km/hr faster, it would have taken 4 hours less than the schedule time. And, if train were slower by 6 km/hr, it would have taken 6 hours more than the schedule time. Find the length of the journey.
Answer
Let the original speed of train be $x$ km/hr and the actual time taken be y hours.
Then, distance = speed × time
$=(x y) km$
According to the first condition,
$(x+6)(y-4)=x y$
$\therefore \quad x y-4 x+6 y-24=x y$
$\therefore-4 x+6 y=24$
$\therefore-2 x+3 y=12$ ...(i)
According to the second condition,
$(x-6)(y+6)=x y$
$\therefore \quad x y+6 x-6 y-36=x y$
$\therefore \quad 6 x-6 y=36$
$\therefore \quad x-y=6$ ...(ii)
Multiplying (ii) by 2, we get,
$2 x-2 y=12$ ...(iii)
Adding equations (i) and (iii) we get,
Image
Substituting the value of y in equation (i) we get,
$x-24=6$
$\therefore \quad x=6+24$
$\therefore \quad x=30$
$\therefore$The original speed = 30 km/hr and time = 24 hours.
Hence, length of the journey = speed × time
$\begin{array}{l}=30 \times 24 \\ =720 km\end{array}$
$\therefore$ Length of the journey is 720 km.
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Question 544 Marks
A person deposits ₹ $x$ in savings bank account at the rate of $5 \%$ per annum and ₹ y in fixed deposit at $10 \%$ per annum. At the end of one year, he gets ₹ 400 as total interest. If he deposits ₹ y in savings bank account and ₹ $x$ in fixed deposit, he would get ₹ 350 as total interest. Find the total amount he deposited.
Answer
Amount deposited in savings account =₹ $x$
Amount deposited in fixed deposit account =₹ y
$\therefore \quad$ Total amount deposited =₹ $(x+y)$
Interest received on Savings account ₹ $\frac{5 x}{100}$
Interest received on fixed deposit account ₹ $\frac{10 y}{100}$
According to the first condition,
$\frac{5 x}{100}+\frac{10 y}{100}=400$
$\therefore \quad 5 x+10 y=40,000$ ...(i)
According to the second condition,
$\frac{10 x}{100}+\frac{5 y}{100}=350$
$\therefore 5 y+10 x=35,000$
$\therefore 10 x+5 y=35,000$...(ii)
Adding equation (i) and (ii), we get,
Image
$\therefore x+y=5,000$ [Dividing both sides by 15]
$\therefore$ Total amount deposited is ₹ 5000 .
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Question 554 Marks
A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, If he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer
$100 km / hr$ and $80 km / hr$
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Question 564 Marks
Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.
Answer
Let the speed of the train be x km/hr and speed of the taxi be y km/hr.
Time $=\frac{\text { Distance }}{\text { Speed }}$
Time taken to cover 300 km by train $=\frac{300}{x}$ hrs.
Time taken to cover 200 km by taxi $=\frac{200}{y}$ hrs.
According to the first condition,
$\frac{300}{x}+\frac{200}{y}=5 \frac{30}{60}$ [ 1 hour $=60 min$.]
$\therefore \quad \frac{300}{x}+\frac{200}{y}=\frac{11}{2}$ ...(i)
Similarly,
Time taken to cover 260 km by train $=\frac{260}{x} hrs$.
Time taken to cover 240 km by taxi $=\frac{240}{y}$ hrs.
According to the second condition,
$\frac{260}{x}+\frac{240}{y}=5 \frac{36}{60}$
$\therefore \quad \frac{260}{x}+\frac{240}{y}=\frac{28}{5}$ ...(ii)
Substituting $\frac{1}{x}=m$ and $\frac{1}{y}=n$ in equations (i) and (ii),
$300 m+200 n=\frac{11}{2}$
i.e. $600 m+400 n=11$ ...(iii)
Similarly, $260 m+240 n=\frac{28}{5}$
i.e. $1300 m+1200 n=28$...(iv)
Multiplying eq. (iii) by 3 ,
$1800 m+1200 n=33$ ...(v)
Subtracting (iv) from (v),
Image
$\therefore \quad m=\frac{5}{500}$
$\therefore \quad m=\frac{1}{100}$
Substituting value of m in equation (iii) we get,
$600 \times \frac{1}{100}+400 n=11$
$\therefore \quad 6+400 n=11$
$\therefore \quad 400 n=11-6$
$\therefore \quad 400 n=5$
$\therefore \quad n=\frac{5}{400}$
$\therefore \quad n=\frac{1}{80}$
Resubstituting the values of $m$ and $n$,
$\frac{1}{x}=\frac{1}{100}$ and $\frac{1}{y}=\frac{1}{80}$
$\therefore x=100$ and $y=80$
The speed of the train is $100 km / hr$ and the speed of the taxi is $80 km / hr$.
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