| Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
Exercise 3.5
Question 1
The sum of the two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.
Let the two numbers be x and y.
According to the question,
x + y = 18 …(i)
$\frac{1}{x}+\frac{1}{y}=\frac{1}{4}$ …(ii)
Eq. (ii) can be re - written as
$\frac{y+x}{x y}=\frac{1}{4}$ …(iii)
$\frac{y+x}{x y}=\frac{1}{4}$ …(iii)
On putting the value of x + y = 18 in Eq. (iii), we get
$\frac{18}{x y}=\frac{1}{4}$
$\frac{18}{x y}=\frac{1}{4}$
⇒ xy = 72
$\Rightarrow x=\frac{72}{y}$
$\Rightarrow x=\frac{72}{y}$
On putting the value of $x=\frac{72}{y}$ in Eq. (i), we get
$\frac{72}{y}+y=18$
⇒ 72 + y2 = 18y
⇒ y2 – 18y + 72 = 0
⇒ y2 – 12y – 6y + 72 = 0
⇒ y(y – 12) – 6(y – 12) = 0
⇒ (y – 6)(y – 12) = 0
⇒ y = 6 and 12
$\frac{72}{y}+y=18$
⇒ 72 + y2 = 18y
⇒ y2 – 18y + 72 = 0
⇒ y2 – 12y – 6y + 72 = 0
⇒ y(y – 12) – 6(y – 12) = 0
⇒ (y – 6)(y – 12) = 0
⇒ y = 6 and 12
If y=6, then $x=\frac{72}{6}=12$
If y=12, then $x=\frac{72}{12}=6$
Hence, the two numbers are 6 and 12.
Let the two numbers be x and y.
According to the question,
x + y = 15 …(i)
$\frac{1}{x}+\frac{1}{y}=\frac{3}{10}$ …(ii)
Hence, the two numbers are 6 and 12.
Question 2
The sum of two numbers is 15 and sum of their reciprocals is $\frac{3}{10}$. Find the numbers.
Sol :Let the two numbers be x and y.
According to the question,
x + y = 15 …(i)
$\frac{1}{x}+\frac{1}{y}=\frac{3}{10}$ …(ii)
Eq. (ii) can be re - written as
$\frac{y+x}{x y}=\frac{3}{10}$ …(iii)
On putting the value of x+y=18 in Eq. (iii), we get
$\frac{15}{x y}=\frac{3}{10}$
⇒ xy = 50
$\Rightarrow x=\frac{50}{y}$
$\frac{15}{x y}=\frac{3}{10}$
⇒ xy = 50
$\Rightarrow x=\frac{50}{y}$
On putting the value of $x=\frac{50}{y}$ in Eq. (i), we get
$\frac{50}{y}+y=18$
⇒ 50 + y2 = 15y
⇒ y2 – 15y + 50 = 0
⇒ y2 – 10y – 5y + 50 = 0
⇒ y(y – 10) – 5(y – 10) = 0
⇒ (y – 5)(y – 10) = 0
$\frac{50}{y}+y=18$
⇒ 50 + y2 = 15y
⇒ y2 – 15y + 50 = 0
⇒ y2 – 10y – 5y + 50 = 0
⇒ y(y – 10) – 5(y – 10) = 0
⇒ (y – 5)(y – 10) = 0
⇒y=5 and 10
If y=5, then $x=\frac{50}{5}=10$
If y=5, then $x=\frac{50}{5}=10$
If y=10, then $x=\frac{50}{10}=5$
Hence, the two numbers are 5 and 10.
Let the two numbers be x and y.
Hence, the two numbers are 5 and 10.
Question 3
Two numbers are in the ratio of 5 : 6. If 8 is subtracted from each of the numb, they become in the ratio of 4 : 5. Find the numbers.
Sol :Let the two numbers be x and y.
According to the question,
$\frac{x}{y}=\frac{5}{6}$
$\Rightarrow \mathrm{y}=\frac{6 \mathrm{x}}{5}$ …(i)
Also, $\frac{x-8}{y-8}=\frac{4}{5}$
⇒ 5(x – 8) = 4(y – 8)
⇒ 5x – 40 = 4y – 32
⇒ 5x – 4y = 8 …(ii)
⇒ 5(x – 8) = 4(y – 8)
⇒ 5x – 40 = 4y – 32
⇒ 5x – 4y = 8 …(ii)
On putting the value of $y=\frac{6 x}{5}$ in Eq. (ii), we get
$5 x-4\left(\frac{6 x}{5}\right)=8$
$\Rightarrow \frac{25 x-24 x}{5}=8$
⇒ x = 40
$5 x-4\left(\frac{6 x}{5}\right)=8$
$\Rightarrow \frac{25 x-24 x}{5}=8$
⇒ x = 40
On putting the value of x = 40 in Eq. (i), we get
$y=\frac{6 \times 40}{5}=48$
$y=\frac{6 \times 40}{5}=48$
Hence, the two numbers are 40 and 48.
Sol :
Let the two numbers be x and y.
According to the question,
x + y = 16 …(i)
$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$ …(ii)
Question 4
The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Let the two numbers be x and y.
According to the question,
x + y = 16 …(i)
$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$ …(ii)
Eq. (ii) can be re - written as
$\frac{y+x}{x y}=\frac{1}{3}$ …(iii)
On putting the value of x + y = 16 in Eq. (iii), we get
$\frac{16}{x y}=\frac{1}{3}$
⇒xy = 48
⇒xy = 48
$\Rightarrow x=\frac{48}{y}$
On putting the value of $x=\frac{48}{y}$ in Eq. (i), we get
$\frac{48}{y}+y=16$
⇒ 48 + y2 = 16y
⇒ y2 – 16y + 48 = 0
⇒ y2 – 12y – 4y + 48 = 0
⇒ y(y – 12) – 4(y – 12) = 0
⇒ (y – 4)(y – 12) = 0
⇒ y = 4 and 12
$\frac{48}{y}+y=16$
⇒ 48 + y2 = 16y
⇒ y2 – 16y + 48 = 0
⇒ y2 – 12y – 4y + 48 = 0
⇒ y(y – 12) – 4(y – 12) = 0
⇒ (y – 4)(y – 12) = 0
⇒ y = 4 and 12
If y=4 then $x=\frac{48}{4}=12$
If y=12 then $x=\frac{48}{12}=4$
Hence, the two numbers are 4 and 12.
Sol :
Let the two numbers be x and y.
According to the question,
x – y = 3 …(i)
Also, x×y = 54
$\Rightarrow \mathrm{x}=\frac{54}{\mathrm{y}}$ …(ii)
Hence, the two numbers are 4 and 12.
Question 5
Two positive numbers differ by 3 and their product is 54. Find the numbers.
Let the two numbers be x and y.
According to the question,
x – y = 3 …(i)
Also, x×y = 54
$\Rightarrow \mathrm{x}=\frac{54}{\mathrm{y}}$ …(ii)
On putting the value of $x=\frac{54}{y}$ in Eq. (i), we get
$\frac{54}{y}-y=3$
⇒ 54 – y2 = 3y
⇒ y2 + 3y – 54 = 0
⇒ y2 + 9y – 6y – 54 = 0
⇒ y(y + 9) – 6(y + 9) = 0
⇒ (y – 6)(y + 9) = 0
⇒ y = – 9 and 6
But y = – 9 can’t be the one number as it is given that the numbers are positive.
$\frac{54}{y}-y=3$
⇒ 54 – y2 = 3y
⇒ y2 + 3y – 54 = 0
⇒ y2 + 9y – 6y – 54 = 0
⇒ y(y + 9) – 6(y + 9) = 0
⇒ (y – 6)(y + 9) = 0
⇒ y = – 9 and 6
But y = – 9 can’t be the one number as it is given that the numbers are positive.
⇒ y =6, then $x=\frac{54}{6}=9$
Hence, the two numbers are 9 and 6.
Sol :
Let the two numbers be x and y.
Hence, the two numbers are 9 and 6.
Question 6
Two numbers are in the ratio of 3 : 5. If 5 is subtracted from each of the number they become in the ratio of 1 : 2. Find the numbers.
Let the two numbers be x and y.
According to the question,
$\frac{x}{y}=\frac{3}{5}$
$\Rightarrow y=\frac{5 x}{3}$ …(i)
Also, $\frac{x-5}{y-5}=\frac{1}{2}$
⇒ 2(x – 5) = (y – 5)
⇒ 2x – 10 = y – 5
⇒ 2x – y = 5 …(ii)
On putting the value of $y=\frac{5 x}{3}$ in Eq. (ii), we get
$2 x-\left(\frac{5 x}{3}\right)=5$
$\Rightarrow \frac{6 x-5 x}{3}=5$
⇒ x = 15
$\frac{x}{y}=\frac{3}{5}$
$\Rightarrow y=\frac{5 x}{3}$ …(i)
Also, $\frac{x-5}{y-5}=\frac{1}{2}$
⇒ 2(x – 5) = (y – 5)
⇒ 2x – 10 = y – 5
⇒ 2x – y = 5 …(ii)
On putting the value of $y=\frac{5 x}{3}$ in Eq. (ii), we get
$2 x-\left(\frac{5 x}{3}\right)=5$
$\Rightarrow \frac{6 x-5 x}{3}=5$
⇒ x = 15
On putting the value of x = 15 in Eq. (i), we get
$y=\frac{5 \times 15}{3}=25$
Hence, the two numbers are 15 and 25.
Question 7
Two numbers are in the ratio of 3 : 4. If 8 is added to each number, they become in the ratio of 4 : 5. Find the numbers.
Let the two numbers be x and y.
According to the question,
$\frac{x}{y}=\frac{3}{4}$
$\Rightarrow \mathrm{y}=\frac{4 \mathrm{x}}{3}$ …(i)
Also, $\frac{x+8}{y+8}=\frac{4}{5}$
⇒ 5(x + 8) = 4(y + 8)
⇒ 5x + 40 = 4y + 32
⇒ 5x – 4y = – 8 …(ii)
⇒ 5(x + 8) = 4(y + 8)
⇒ 5x + 40 = 4y + 32
⇒ 5x – 4y = – 8 …(ii)
On putting the value of $y=\frac{6 x}{5}$ in Eq. (ii), we get
$5 x-4\left(\frac{4 x}{3}\right)=-8$
$\Rightarrow \frac{15 x-16 x}{3}=-8$
⇒ x = 24
$5 x-4\left(\frac{4 x}{3}\right)=-8$
$\Rightarrow \frac{15 x-16 x}{3}=-8$
⇒ x = 24
On putting the value of x = 24 in Eq. (i), we get
$y=\frac{4 \times 24}{3}=32$
$y=\frac{4 \times 24}{3}=32$
Hence, the two numbers are 24 and 32.
Sol :
Let the two numbers be x and y.
According to the question,
x – y = 2 …(i)
Also, x×y = 360
$\Rightarrow x=\frac{360}{y}$ …(ii)
Question 8
Two numbers differ by 2 and their product is 360. Find the numbers.
Let the two numbers be x and y.
According to the question,
x – y = 2 …(i)
Also, x×y = 360
$\Rightarrow x=\frac{360}{y}$ …(ii)
On putting the value of $x=\frac{360}{y}$ in Eq. (i), we get
$\frac{360}{y}-y=2$
⇒ 360 – y2 = 2y
⇒ y2 + 2y – 360 = 0
⇒ y2 + 20y – 18y – 360 = 0
⇒ y(y + 20) – 18(y + 20) = 0
⇒ (y – 18)(y + 20) = 0
⇒ y = – 20 and 18
But y = – 20 can’t be the one number as it is given that the numbers are positive.
$\frac{360}{y}-y=2$
⇒ 360 – y2 = 2y
⇒ y2 + 2y – 360 = 0
⇒ y2 + 20y – 18y – 360 = 0
⇒ y(y + 20) – 18(y + 20) = 0
⇒ (y – 18)(y + 20) = 0
⇒ y = – 20 and 18
But y = – 20 can’t be the one number as it is given that the numbers are positive.
⇒ y = 18, then $x=\frac{360}{18}=20$
Hence, the two numbers are 20 and 18.
Sol :
Let the two numbers be x and y.
According to the question,
x – y = 4 …(i)
Hence, the two numbers are 20 and 18.
Question 9
Two numbers differ by 4 and their product is 192. Find the numbers.
Let the two numbers be x and y.
According to the question,
x – y = 4 …(i)
Also, x×y = 192
$\Rightarrow x=\frac{192}{y}$ …(ii)
$\Rightarrow x=\frac{192}{y}$ …(ii)
On putting the value of $x=\frac{192}{y}$ in Eq. (i), we get
$\frac{192}{\mathrm{y}}-\mathrm{y}=4$
⇒ 192 – y2 = 4y
⇒ y2 + 4y – 192 = 0
⇒ y2 + 16y – 12y – 192 = 0
⇒ y(y + 16) – 12(y + 16) = 0
⇒ (y – 12)(y + 16) = 0
⇒ y = – 16 and 12
$\frac{192}{\mathrm{y}}-\mathrm{y}=4$
⇒ 192 – y2 = 4y
⇒ y2 + 4y – 192 = 0
⇒ y2 + 16y – 12y – 192 = 0
⇒ y(y + 16) – 12(y + 16) = 0
⇒ (y – 12)(y + 16) = 0
⇒ y = – 16 and 12
But y = – 16 can’t be the one number as it is given that the numbers are positive.
⇒ y = 12, then $x=\frac{192}{12}=16$
Hence, the two numbers are 16 and 12.
Sol :
Let the two numbers be x and y.
According to the question,
x – y = 4 …(i)
Hence, the two numbers are 16 and 12.
Question 10
Two numbers differ by 4 and their product is 96. Find the numbers.
Let the two numbers be x and y.
According to the question,
x – y = 4 …(i)
Also, x×y = 96
$\Rightarrow \mathrm{x}=\frac{96}{\mathrm{y}}$ …(ii)
$\Rightarrow \mathrm{x}=\frac{96}{\mathrm{y}}$ …(ii)
On putting the value of $x=\frac{96}{y}$ in Eq. (i), we get
$\frac{96}{y}-y=4$
⇒ 96 – y2 = 4y
⇒ y2 + 4y – 96 = 0
⇒ y2 + 12y – 8y – 96 = 0
⇒ y(y + 12) – 8(y + 12) = 0
⇒ (y – 8)(y + 12) = 0
⇒ y = – 8 and 12
$\frac{96}{y}-y=4$
⇒ 96 – y2 = 4y
⇒ y2 + 4y – 96 = 0
⇒ y2 + 12y – 8y – 96 = 0
⇒ y(y + 12) – 8(y + 12) = 0
⇒ (y – 8)(y + 12) = 0
⇒ y = – 8 and 12
But y = – 8 can’t be the one number as it is given that the numbers are positive.
⇒ y = 12,then $x=\frac{96}{12}=8$
Hence, the two numbers are 8 and 12.
Given ratio of incomes = 5:4
And the ratio of their expenditures = 7:5
Saving of each person = Rs. 3000
Let incomes of two persons = 5x and 4x
And their expenditures = 7y and 5y
According to the question,
5x – 7y = 3000 …(i)
4x – 5y = 3000 …(ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 5 to make the coefficients of x equal, we get
20x – 28y = 12000 …(iii)
20x – 25y = 15000 …(iv)
On subtracting Eq. (iii) from (iv), we get
20x – 25y – 20x + 28y = 15000 – 12000
⇒ 3y = 3000
⇒ y = 1000
On putting the y = 1000 in Eq. (i), we get
5x – 7y = 3000
⇒ 5x – 7(1000) = 3000
⇒ 5x = 10000
⇒ x = 2000
Thus, monthly income of both the persons are 5(2000) and 4(2000), i.e. Rs. 10000 and Rs. 8000
Let fixed charge = Rs x
and charge per kilometer = Rs y
According to the question,
x + 12y = 45 …(i)
and x + 20y = 73 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 20y – x – 12y = 73 – 45
⇒ 8y = 28
$\Rightarrow \mathrm{y}=\frac{28}{8}=3.5$
Question 11
The monthly incomes of A and B are in the ratio of 5 : 4 and their monthly expenditures are in the ratio of 7 : 5. If each saves Rs. 3000 per month, find the monthly income of each.
Sol :Given ratio of incomes = 5:4
And the ratio of their expenditures = 7:5
Saving of each person = Rs. 3000
Let incomes of two persons = 5x and 4x
And their expenditures = 7y and 5y
According to the question,
5x – 7y = 3000 …(i)
4x – 5y = 3000 …(ii)
On multiplying Eq. (i) by 4 and Eq. (ii) by 5 to make the coefficients of x equal, we get
20x – 28y = 12000 …(iii)
20x – 25y = 15000 …(iv)
On subtracting Eq. (iii) from (iv), we get
20x – 25y – 20x + 28y = 15000 – 12000
⇒ 3y = 3000
⇒ y = 1000
On putting the y = 1000 in Eq. (i), we get
5x – 7y = 3000
⇒ 5x – 7(1000) = 3000
⇒ 5x = 10000
⇒ x = 2000
Thus, monthly income of both the persons are 5(2000) and 4(2000), i.e. Rs. 10000 and Rs. 8000
Question 12
Scooter charges consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a person travels 12 km, he pays Rs. 45 and for travelling 20 km, he pays Rs. 73. Express the above statements in the form of simultaneous equations and hence, find the fixed charges and the rate per km.
Sol :Let fixed charge = Rs x
and charge per kilometer = Rs y
According to the question,
x + 12y = 45 …(i)
and x + 20y = 73 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 20y – x – 12y = 73 – 45
⇒ 8y = 28
$\Rightarrow \mathrm{y}=\frac{28}{8}=3.5$
On putting the value of y = 3.5 in Eq. (i), we get
x + 12(3.5) = 45
⇒ x + 42 = 45
⇒ x = 45 – 42 = 3
Hence, monthly fixed charges is Rs. 3 and charge per kilometer is Rs. 3.5
Sol :
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 22y = 1380 …(i)
In case of student B,
x + 28y = 1680 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y – x – 22y = 1680 – 1380
⇒ 6y = 300
⇒ y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 22(50) = 1380
⇒ x + 1100 = 1380
⇒ x = 1380 – 1100 = 280
Hence, monthly fixed charges is Rs. 280 and cost of food per day is Rs. 50
Sol :
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 110y = 690 …(i)
and x + 200y = 1050 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 200y – x – 110y = 1050 – 690
⇒ 90y = 360
⇒ y = 40
On putting the value of y = 40 in Eq. (i), we get
x + 110(40) = 690
⇒ x + 440 = 690
⇒ x = 690 – 440 = 250
Hence, monthly fixed charges is Rs. 250 and charge per kilometer is Rs. 40
Sol :
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 25y = 1750 …(i)
In case of student B,
x + 28y = 1900 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y – x – 25y = 1900 – 1750
⇒ 3y = 150
⇒ y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 25(50) = 1750
⇒ x + 1250 = 1750
⇒ x = 1750 – 1250 = 500
Hence, monthly fixed charges is Rs. 500 and cost of food per day is Rs. 50
Let fixed rent of the house = Rs. x
And the mess charges per hesd per month = Rs. y
According to the question,
x + 2y = 3900 …(i)
and x + 5y = 7500 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 5y – x – 2y = 7500 – 3900
⇒ 3y = 3600
⇒ y = 1200
On putting the value of y = 1200 in Eq. (i), we get
x + 2 (1200) = 3900
⇒ x + 2400 = 3900
⇒ x = 1500
Hence, fixed rent of the house is Rs. 1500 and the mess charges per head per month is Rs. 1200.
Sol :
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 13y = 96 …(i)
and x + 18y = 131 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 18y – x – 13y = 131 – 96
⇒ 5y = 35
⇒ y = 7
On putting the value of y = 7 in Eq. (i), we get
x + 13 (7) = 96
⇒ x + 91 = 96
⇒ x = 5
Hence, monthly fixed charges is Rs. 5 and charge per kilometer is Rs. 7
Now, amount to be paid for travelling 25 km
= Fixed charge + Rs 7 ×25
= 5 + 175
= Rs. 180
Hence, the amount paid by a person for travelling 25km is Rs. 180
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
After interchanging the digits, New number = x + 10y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) + (x + 10y) = 132
⇒ 11x + 11y = 132
⇒ 11(x + y) = 132
⇒ x + y = 12 …(i)
and 10x + y + 12 = 5(x + y)
⇒ 10x + y + 12 = 5x + 5y
⇒ 10x – 5x + y – 5y = – 12
⇒ 5x – 4y = – 12 …(ii)
From Eq. (i), we get
x = 12 – y …(iii)
On substituting the value of x = 12 – y in Eq. (ii), we get
5(12 – y) – 4y = – 12
⇒ 60 – 5y – 4y = – 12
⇒ – 9y = – 12 – 60
⇒ – 9y = – 72
⇒ y = 8
On putting the value of y = 8 in Eq. (iii), we get
x = 12 – 8 = 4
So, the Original number = 10x + y
= 10×4 + 8
= 48
Hence, the two digit number is 48.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the two digit number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) = 4(x + y)
⇒ 10x + y = 4x + 4y
⇒ 10x – 4x + y – 4y = 0
⇒ 6x – 3y = 0
⇒ 2x – y = 0
⇒ y = 2x …(i)
After interchanging the digits, New number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On substituting the value of y = 2x in Eq. (ii), we get
x – y = – 18
⇒ x – 2x = – 2
⇒ – x = – 2
⇒ x = 2
On putting the value of x = 2 in Eq. (i), we get
y = 2×2 = 4
So, the Original number = 10x + y
= 10×2 + 4
= 20 + 4
= 24
Hence, the two digit number is 24.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
x + 12(3.5) = 45
⇒ x + 42 = 45
⇒ x = 45 – 42 = 3
Hence, monthly fixed charges is Rs. 3 and charge per kilometer is Rs. 3.5
Question 13
A part of monthly hostel charges in a college is fixed and the remaining depend on the number of days one has taken food in the mess. When a student A, takes food for 22 days, he has to pay Rs. 1380 as hostel charges, whereas a student B, who takes food for 28 days, pays Rs. 1680 as hostel charges. Find the fixed charge and the cost of food per day.
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 22y = 1380 …(i)
In case of student B,
x + 28y = 1680 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y – x – 22y = 1680 – 1380
⇒ 6y = 300
⇒ y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 22(50) = 1380
⇒ x + 1100 = 1380
⇒ x = 1380 – 1100 = 280
Hence, monthly fixed charges is Rs. 280 and cost of food per day is Rs. 50
Question 14
Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels 110 km, he pays Rs. 690, and for travelling 200 km, he pays Rs. 1050. Find the fixed charges per day and the rate per km.
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 110y = 690 …(i)
and x + 200y = 1050 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 200y – x – 110y = 1050 – 690
⇒ 90y = 360
⇒ y = 40
On putting the value of y = 40 in Eq. (i), we get
x + 110(40) = 690
⇒ x + 440 = 690
⇒ x = 690 – 440 = 250
Hence, monthly fixed charges is Rs. 250 and charge per kilometer is Rs. 40
Question 15
A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs. 1750 as hostel charges whereas a student a d who takes food for 28 days, pays Rs. 1900 as hostel charges. Find the fixed charges and the cost of the food per day.
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 25y = 1750 …(i)
In case of student B,
x + 28y = 1900 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 28y – x – 25y = 1900 – 1750
⇒ 3y = 150
⇒ y = 50
On putting the value of y = 50 in Eq. (i), we get
x + 25(50) = 1750
⇒ x + 1250 = 1750
⇒ x = 1750 – 1250 = 500
Hence, monthly fixed charges is Rs. 500 and cost of food per day is Rs. 50
Question 16
The total expenditure per month of a household consists of a fixed rent of the house and the mess charges, depending upon the number of people sharing the house. The total monthly expenditure is Rs. 3,900 for 2 people and Rs. 7,500 for 5 people. Find the rent of the house and the mess charges per head per month.
Sol :Let fixed rent of the house = Rs. x
And the mess charges per hesd per month = Rs. y
According to the question,
x + 2y = 3900 …(i)
and x + 5y = 7500 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 5y – x – 2y = 7500 – 3900
⇒ 3y = 3600
⇒ y = 1200
On putting the value of y = 1200 in Eq. (i), we get
x + 2 (1200) = 3900
⇒ x + 2400 = 3900
⇒ x = 1500
Hence, fixed rent of the house is Rs. 1500 and the mess charges per head per month is Rs. 1200.
Question 17
The car rental charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 13 km, the charge paid is Rs. 96 and for a journey of 18 km, the charge paid is Rs. 131. What will a person have to pay for travelling a distance of 25 km?
Let fixed charge = Rs. x
and charge per kilometer = Rs. y
According to the question,
x + 13y = 96 …(i)
and x + 18y = 131 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
x + 18y – x – 13y = 131 – 96
⇒ 5y = 35
⇒ y = 7
On putting the value of y = 7 in Eq. (i), we get
x + 13 (7) = 96
⇒ x + 91 = 96
⇒ x = 5
Hence, monthly fixed charges is Rs. 5 and charge per kilometer is Rs. 7
Now, amount to be paid for travelling 25 km
= Fixed charge + Rs 7 ×25
= 5 + 175
= Rs. 180
Hence, the amount paid by a person for travelling 25km is Rs. 180
Question 18
The sum of a two - digit number and the number formed by interchanging the digits is 132. If 12 is added to the number, the new number becomes 5 times the sum of the digits. Find the number.
Sol :Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
After interchanging the digits, New number = x + 10y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) + (x + 10y) = 132
⇒ 11x + 11y = 132
⇒ 11(x + y) = 132
⇒ x + y = 12 …(i)
and 10x + y + 12 = 5(x + y)
⇒ 10x + y + 12 = 5x + 5y
⇒ 10x – 5x + y – 5y = – 12
⇒ 5x – 4y = – 12 …(ii)
From Eq. (i), we get
x = 12 – y …(iii)
On substituting the value of x = 12 – y in Eq. (ii), we get
5(12 – y) – 4y = – 12
⇒ 60 – 5y – 4y = – 12
⇒ – 9y = – 12 – 60
⇒ – 9y = – 72
⇒ y = 8
On putting the value of y = 8 in Eq. (iii), we get
x = 12 – 8 = 4
So, the Original number = 10x + y
= 10×4 + 8
= 48
Hence, the two digit number is 48.
Question 19
A two - digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the two digit number = 10x + y
The sum of the digit = x + y
According to the question,
(10x + y) = 4(x + y)
⇒ 10x + y = 4x + 4y
⇒ 10x – 4x + y – 4y = 0
⇒ 6x – 3y = 0
⇒ 2x – y = 0
⇒ y = 2x …(i)
After interchanging the digits, New number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On substituting the value of y = 2x in Eq. (ii), we get
x – y = – 18
⇒ x – 2x = – 2
⇒ – x = – 2
⇒ x = 2
On putting the value of x = 2 in Eq. (i), we get
y = 2×2 = 4
So, the Original number = 10x + y
= 10×2 + 4
= 20 + 4
= 24
Hence, the two digit number is 24.
Question 20
A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
$\frac{(10 x+y)}{x+y}=6$
⇒ 10x + y = 6(x + y)
⇒ 10x + y = 6x + 6y
⇒ 10x + y – 6x – 6y
⇒ 4x – 5y = 0 …(i)
The reverse of the number = x + 10y
and 10x + y – 9 = x + 10y
⇒ 10x + y – 9 = x + 10y
⇒ 10x – x + y – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1
⇒ x = y + 1 …(ii)
$\frac{(10 x+y)}{x+y}=6$
⇒ 10x + y = 6(x + y)
⇒ 10x + y = 6x + 6y
⇒ 10x + y – 6x – 6y
⇒ 4x – 5y = 0 …(i)
The reverse of the number = x + 10y
and 10x + y – 9 = x + 10y
⇒ 10x + y – 9 = x + 10y
⇒ 10x – x + y – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1
⇒ x = y + 1 …(ii)
On substituting the value of x = y + 1 in Eq. (i), we get
4x – 5y = 0
⇒ 4(y + 1) – 5y = 0
⇒ 4y + 4 – 5y = 0
⇒ 4 – y = 0
⇒ y = 4
On substituting the value of y = 4 in Eq. (ii), we get
x = y + 1
⇒ x = 4 + 1
⇒ x = 5
So, the Original number = 10x + y
= 10×5 + 4
= 50 + 4
= 54
Hence, the two digit number is 54.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 12 …(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On adding Eq. (i) and (ii) , we get
x + y + x – y = 12 – 2
⇒ 2x = 10
⇒ x = 5
On substituting the value of x = 5 in Eq. (i), we get
x + y = 12
⇒ 5 + y = 12
⇒ y = 7
So, the Original number = 10x + y
= 10×5 + 7
= 50 + 7
= 57
Hence, the two digit number is 57.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 3 + 4(x + y)
⇒ 10x + y = 3 + 4x + 4y
⇒ 10x + y – 4x – 4y = 3
⇒ 6x – 3y = 3
⇒ 2x – y = 1 …(i)
The reverse number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On subtracting Eq. (i) from Eq. (ii) , we get
x – y – 2x + y = – 2 – 1
⇒ – x = – 3
⇒ x = 3
On substituting the value of x = 3 in Eq. (i), we get
2(3) – y = 1
⇒ 6 – y = 1
⇒ – y = 1 – 6
⇒ – y = – 5
⇒ y = 5
So, the Original number = 10x + y
= 10×3 + 5
= 30 + 5
= 35
Hence, the two digit number is 35.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 7(x + y)
⇒ 10x + y = 7x + 7y
⇒ 10x + y – 7x – 7y = 0
⇒ 3x – 6y = 0
⇒ x – 2y = 0
⇒ x = 2y …(i)
The reverse number = x + 10y
and 10x + y – 27 = x + 10y
⇒ 10x + y – 27 = x + 10y
⇒ 10x – x + y – 10y = 27
⇒ 9x – 9y = 27
⇒ x – y = 3 …(ii)
On substituting the value of x = 2y in Eq. (ii), we get
x – y = 3
⇒ 2y – y = 3
⇒ y = 3
On putting the value of y = 3 in Eq. (i), we get
x = 2(3) = 6
So, the Original number = 10x + y
= 10×6 + 3
= 60 + 3
= 63
Hence, the two digit number is 63.
Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 15 …(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 9 = x + 10y
⇒ 10x + y + 9 = x + 10y
⇒ 10x – x + y – 10y = – 9
⇒ 9x – 9y = – 9
⇒ x – y = – 1 …(ii)
On adding Eq. (i) and (ii) , we get
x + y + x – y = 15 – 1
⇒ 2x = 14
⇒ x = 7
On substituting the value of x = 5 in Eq. (i), we get
x + y = 15
⇒ 7 + y = 15
⇒ y = 8
So, the Original number = 10x + y
= 10×7 + 8
= 70 + 8
= 78
Hence, the two digit number is 78.
Sol :
Sol :
Let the numerator = x
and the denominator = y
4x – 5y = 0
⇒ 4(y + 1) – 5y = 0
⇒ 4y + 4 – 5y = 0
⇒ 4 – y = 0
⇒ y = 4
On substituting the value of y = 4 in Eq. (ii), we get
x = y + 1
⇒ x = 4 + 1
⇒ x = 5
So, the Original number = 10x + y
= 10×5 + 4
= 50 + 4
= 54
Hence, the two digit number is 54.
Question 21
The sum of the digits of a two - digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 12 …(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On adding Eq. (i) and (ii) , we get
x + y + x – y = 12 – 2
⇒ 2x = 10
⇒ x = 5
On substituting the value of x = 5 in Eq. (i), we get
x + y = 12
⇒ 5 + y = 12
⇒ y = 7
So, the Original number = 10x + y
= 10×5 + 7
= 50 + 7
= 57
Hence, the two digit number is 57.
Question 22
A two - digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 3 + 4(x + y)
⇒ 10x + y = 3 + 4x + 4y
⇒ 10x + y – 4x – 4y = 3
⇒ 6x – 3y = 3
⇒ 2x – y = 1 …(i)
The reverse number = x + 10y
and 10x + y + 18 = x + 10y
⇒ 10x + y + 18 = x + 10y
⇒ 10x – x + y – 10y = – 18
⇒ 9x – 9y = – 18
⇒ x – y = – 2 …(ii)
On subtracting Eq. (i) from Eq. (ii) , we get
x – y – 2x + y = – 2 – 1
⇒ – x = – 3
⇒ x = 3
On substituting the value of x = 3 in Eq. (i), we get
2(3) – y = 1
⇒ 6 – y = 1
⇒ – y = 1 – 6
⇒ – y = – 5
⇒ y = 5
So, the Original number = 10x + y
= 10×3 + 5
= 30 + 5
= 35
Hence, the two digit number is 35.
Question 23
A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
10x + y = 7(x + y)
⇒ 10x + y = 7x + 7y
⇒ 10x + y – 7x – 7y = 0
⇒ 3x – 6y = 0
⇒ x – 2y = 0
⇒ x = 2y …(i)
The reverse number = x + 10y
and 10x + y – 27 = x + 10y
⇒ 10x + y – 27 = x + 10y
⇒ 10x – x + y – 10y = 27
⇒ 9x – 9y = 27
⇒ x – y = 3 …(ii)
On substituting the value of x = 2y in Eq. (ii), we get
x – y = 3
⇒ 2y – y = 3
⇒ y = 3
On putting the value of y = 3 in Eq. (i), we get
x = 2(3) = 6
So, the Original number = 10x + y
= 10×6 + 3
= 60 + 3
= 63
Hence, the two digit number is 63.
Question 24
The sum of the digits of a two - digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
According to the question,
x + y = 15 …(i)
After interchanging the digits, the number = x + 10y
and 10x + y + 9 = x + 10y
⇒ 10x + y + 9 = x + 10y
⇒ 10x – x + y – 10y = – 9
⇒ 9x – 9y = – 9
⇒ x – y = – 1 …(ii)
On adding Eq. (i) and (ii) , we get
x + y + x – y = 15 – 1
⇒ 2x = 14
⇒ x = 7
On substituting the value of x = 5 in Eq. (i), we get
x + y = 15
⇒ 7 + y = 15
⇒ y = 8
So, the Original number = 10x + y
= 10×7 + 8
= 70 + 8
= 78
Hence, the two digit number is 78.
Question 25
The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Let the numerator = x
and the denominator = y
and the denominator = y
So, the fraction $=\frac{x}{y}$
According to the question,
According to the question,
Condition I:
x + y = 2y – 3
⇒ x + y – 2y = – 3
⇒ x – y = – 3 …(i)
x + y = 2y – 3
⇒ x + y – 2y = – 3
⇒ x – y = – 3 …(i)
Condition II:
$\frac{x-1}{y-1}=\frac{1}{2}$
⇒ 2(x – 1) = y – 1
⇒ 2x – 2 = y – 1
⇒ 2x – y = 1 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2x – y – x + y = 1 + 3
⇒ x = 4
On putting the value of x in Eq. (i), we get
4 – y = – 3
⇒ y = 7
So, the numerator is 4 and the denominator is 7
$\frac{x-1}{y-1}=\frac{1}{2}$
⇒ 2(x – 1) = y – 1
⇒ 2x – 2 = y – 1
⇒ 2x – y = 1 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
2x – y – x + y = 1 + 3
⇒ x = 4
On putting the value of x in Eq. (i), we get
4 – y = – 3
⇒ y = 7
So, the numerator is 4 and the denominator is 7
Hence, the fraction is $\frac{4}{7}$
Question 26
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, then are in the ratio 2 : 3. Determine the fraction.
Let the numerator = x
and the denominator = y
So, the fraction $=\frac{x}{y}$
According to the question,
Condition I:
x + y = 2x + 4
⇒ x + y – 2x = 4
⇒ – x + y = 4
⇒ y = 4 + x …(i)
x + y = 2x + 4
⇒ x + y – 2x = 4
⇒ – x + y = 4
⇒ y = 4 + x …(i)
Condition II:
$\frac{x+3}{y+3}=\frac{2}{3}$
⇒ 3(x + 3) = 2( y + 3)
⇒ 3x + 9 = 2y + 6
⇒ 3x – 2y = – 3 …(ii)
$\frac{x+3}{y+3}=\frac{2}{3}$
⇒ 3(x + 3) = 2( y + 3)
⇒ 3x + 9 = 2y + 6
⇒ 3x – 2y = – 3 …(ii)
On putting the value of y in Eq.(ii) , we get
3x – 2(4 + x) = – 3
⇒ 3x – 8 – 2x = – 3
⇒ x = 5
3x – 2(4 + x) = – 3
⇒ 3x – 8 – 2x = – 3
⇒ x = 5
On putting the value of x in Eq. (i), we get
y = 4 + 5
⇒ y = 9
So, the numerator is 5 and the denominator is 9
Hence, the fraction is $\frac{5}{9}$
Sol :
Let the numerator = x
and the denominator = y
So, the fraction $=\frac{x}{y}$
y = 4 + 5
⇒ y = 9
So, the numerator is 5 and the denominator is 9
Hence, the fraction is $\frac{5}{9}$
Question 27
The sum of the numerator and denominator of a fraction is 8. If 3 is added to both 3 the numerator and the denominator, the fraction becomes 3/4. Find the fraction.Let the numerator = x
and the denominator = y
So, the fraction $=\frac{x}{y}$
According to the question,
Condition I:
x + y = 8
⇒ y = 8 – x …(i)
Condition II:
$\frac{x+3}{y+3}=\frac{3}{4}$
⇒ 4(x + 3) = 3( y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x – 3y = – 3 …(ii)
On putting the value of y in Eq.(ii) , we get
4x – 3(8 – x) = – 3
⇒ 4x – 24 + 3x = – 3
⇒ 7x = 21
⇒ x = 3
On putting the value of x in Eq. (i), we get
y = 8 – 3
⇒ y = 5
Condition I:
x + y = 8
⇒ y = 8 – x …(i)
Condition II:
$\frac{x+3}{y+3}=\frac{3}{4}$
⇒ 4(x + 3) = 3( y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x – 3y = – 3 …(ii)
On putting the value of y in Eq.(ii) , we get
4x – 3(8 – x) = – 3
⇒ 4x – 24 + 3x = – 3
⇒ 7x = 21
⇒ x = 3
On putting the value of x in Eq. (i), we get
y = 8 – 3
⇒ y = 5
So, the numerator is 3 and the denominator is 5
Hence, the fraction is $\frac{3}{5}$
Let the denominator = x
Given that numerator is one less than the denominator
⇒ numerator = x – 1
Question 28
The numerator of a fraction is one less than its denominator. If 3 is added to each of the numerator and denominator, the fraction is increased by 3/28, find the fraction. 28
Sol :Let the denominator = x
Given that numerator is one less than the denominator
⇒ numerator = x – 1
So, the fraction $=\frac{x-1}{x}$
According to the question,
$\frac{x-1+3}{x+3}=\frac{x-1}{x}+\frac{3}{28}$
$\Rightarrow \frac{x+2}{x+3}=\frac{x-1}{x}+\frac{3}{28}$
$\Rightarrow \frac{x+2}{x+3}-\frac{x-1}{x}=\frac{3}{28}$
$\Rightarrow \frac{(x+2) x-(x+3)(x-1)}{(x+3)(x)}=\frac{3}{28}$
⇒ 28{(x2 + 2x) – (x2 –x + 3x – 3)} = 3 (x2 + 3x)
⇒ 28x2 + 56x – 28x2 – 56x + 84 = 3x2 + 9x
⇒ 3x2 + 9x – 84 = 0
⇒ x2 + 3x – 28 = 0
⇒ x2 + 7x – 4x – 28 = 0
⇒ x(x + 7) – 4(x + 7) = 0
⇒ (x – 4) (x + 7) = 0
⇒ x = 4 and – 7
But x is a natural number
Hence, x = 4
So, the fraction is $\frac{x-1}{x}=\frac{4-1}{4}=\frac{3}{4}$
Let the age of father = x years
And the age of his son = y years
According to the question,
x = 3 + 3y ...(i)
Three year here,
Father’s age = (x + 3) years
Son’s age = (y + 3) years
According to the question,
(x + 3) = 10 + 2(y + 3)
⇒ x + 3 = 10 + 2y + 6
⇒ x = 2y + 13 …(ii)
From Eq. (i) and (ii), we get
3 + 3y = 13 + 2y
⇒ 3y – 2y = 13 – 3
⇒ y = 10
On putting the value of y = 7 in Eq. (i), we get
x = 3 + 3(10)
⇒ x = 3 + 30
⇒ x = 33
Hence, the age of father is 33 years and the age of his son is 10 years.
Let the age of a man = x years
And the age of his son = y years
Two years ago,
Man’s age = (x – 2) years
Son’s age = (y – 2) years
According to the question,
(x – 2) = 5(y – 2)
⇒ x – 2 = 5y – 10
⇒ x = 5y – 10 + 2
⇒ x = 5y – 8 ...(i)
Two years later,
Father’s age = (x + 2) years
Son’s age = (y + 2) years
According to the question,
(x + 2) = 8 + 3(y + 2)
⇒ x + 2 = 8 + 3y + 6
⇒ x = 3y + 12 …(ii)
From Eq. (i) and (ii), we get
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
On putting the value of y = 11 in Eq. (i), we get
x = 5(10) – 8
⇒ x = 50 – 8
⇒ x = 42
Hence, the age of man is 42 years and the age of his son is 10 years.
Let the age of two children be x and y
So, the father’s present age = 3(x + y)
After five years,
Age of two children = (x + 5) + (y + 5) years
= ( x + y + 10) years
So, the age of father after five years = 3(x + y) + 5
= 3x + 3y + 5
According to the question,
3x + 3y + 5 = 2(x + y + 10)
⇒ 3x + 3y + 5 = 2x + 2y + 20
⇒ 3x – 2x + 3y – 2y = 20 – 5
⇒ x + y = 15
So, the age of two children = 15 years
And the age of father = 3(15) = 45years
Hence, the age of father is 45 years and the age of his two children is 15 years.
Let the age of A = x years
And the age of B = y years
Five years ago,
A’s age = (x – 5) years
B’s age = (y – 5) years
According to the question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x = 3y – 10 …(i)
Ten years later,
A’s age = (x + 10)
B’s age = (y + 10)
According to the question,
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x = 2y + 10 …(ii)
From Eq. (i) and (ii), we get
3y – 10 = 2y + 10
⇒ 3y – 2y = 10 + 10
⇒ y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 3(20) – 10
⇒ x = 50
Hence, the age of person A is 50years and Age of B is 20years.
Let the age of a man = x years
And the age of his son = y years
Ten years hence,
Man’s age = (x + 10) years
Son’s age = (y + 10) years
According to the question,
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x = 2y + 20 – 10
⇒ x = 2y + 10 ...(i)
Ten years ago,
Father’s age = (x – 10) years
Son’s age = (y – 10) years
According to the question,
(x – 10) = 4(y – 10)
⇒ x – 10 = 4y – 40
⇒ x = 4y – 30 …(ii)
From Eq. (i) and (ii), we get
2y + 10 = 4y – 30
⇒ 2y – 4y = – 30 – 10
⇒ – 2y = – 40
⇒ y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 2y + 10
⇒ x = 2(20) + 10
⇒ x = 50
Hence, the age of man is 50 years and the age of his son is 20 years.
We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°
∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°.
Sol :
We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°
$\frac{x-1+3}{x+3}=\frac{x-1}{x}+\frac{3}{28}$
$\Rightarrow \frac{x+2}{x+3}=\frac{x-1}{x}+\frac{3}{28}$
$\Rightarrow \frac{x+2}{x+3}-\frac{x-1}{x}=\frac{3}{28}$
$\Rightarrow \frac{(x+2) x-(x+3)(x-1)}{(x+3)(x)}=\frac{3}{28}$
⇒ 28{(x2 + 2x) – (x2 –x + 3x – 3)} = 3 (x2 + 3x)
⇒ 28x2 + 56x – 28x2 – 56x + 84 = 3x2 + 9x
⇒ 3x2 + 9x – 84 = 0
⇒ x2 + 3x – 28 = 0
⇒ x2 + 7x – 4x – 28 = 0
⇒ x(x + 7) – 4(x + 7) = 0
⇒ (x – 4) (x + 7) = 0
⇒ x = 4 and – 7
But x is a natural number
Hence, x = 4
So, the fraction is $\frac{x-1}{x}=\frac{4-1}{4}=\frac{3}{4}$
Question 29
The age of the father is 3 years more than 3 times the son's age. 3 years here, the age of the father will be 10 years more than twice the age of the son. Find their present ages.
Sol :Let the age of father = x years
And the age of his son = y years
According to the question,
x = 3 + 3y ...(i)
Three year here,
Father’s age = (x + 3) years
Son’s age = (y + 3) years
According to the question,
(x + 3) = 10 + 2(y + 3)
⇒ x + 3 = 10 + 2y + 6
⇒ x = 2y + 13 …(ii)
From Eq. (i) and (ii), we get
3 + 3y = 13 + 2y
⇒ 3y – 2y = 13 – 3
⇒ y = 10
On putting the value of y = 7 in Eq. (i), we get
x = 3 + 3(10)
⇒ x = 3 + 30
⇒ x = 33
Hence, the age of father is 33 years and the age of his son is 10 years.
Question 30
Two years ago, a man was five times as old as his son. Two years later his age will be 8 more than three times the age of the son. Find the present ages of man and his son.
Sol :Let the age of a man = x years
And the age of his son = y years
Two years ago,
Man’s age = (x – 2) years
Son’s age = (y – 2) years
According to the question,
(x – 2) = 5(y – 2)
⇒ x – 2 = 5y – 10
⇒ x = 5y – 10 + 2
⇒ x = 5y – 8 ...(i)
Two years later,
Father’s age = (x + 2) years
Son’s age = (y + 2) years
According to the question,
(x + 2) = 8 + 3(y + 2)
⇒ x + 2 = 8 + 3y + 6
⇒ x = 3y + 12 …(ii)
From Eq. (i) and (ii), we get
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
On putting the value of y = 11 in Eq. (i), we get
x = 5(10) – 8
⇒ x = 50 – 8
⇒ x = 42
Hence, the age of man is 42 years and the age of his son is 10 years.
Question 31
Father's age is three times the sum of ages of his two children. After 5 years, his age will be twice the sum of ages of two children. Find the age of father.
Sol :Let the age of two children be x and y
So, the father’s present age = 3(x + y)
After five years,
Age of two children = (x + 5) + (y + 5) years
= ( x + y + 10) years
So, the age of father after five years = 3(x + y) + 5
= 3x + 3y + 5
According to the question,
3x + 3y + 5 = 2(x + y + 10)
⇒ 3x + 3y + 5 = 2x + 2y + 20
⇒ 3x – 2x + 3y – 2y = 20 – 5
⇒ x + y = 15
So, the age of two children = 15 years
And the age of father = 3(15) = 45years
Hence, the age of father is 45 years and the age of his two children is 15 years.
Question 32
Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?
Sol :Let the age of A = x years
And the age of B = y years
Five years ago,
A’s age = (x – 5) years
B’s age = (y – 5) years
According to the question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x = 3y – 10 …(i)
Ten years later,
A’s age = (x + 10)
B’s age = (y + 10)
According to the question,
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x = 2y + 10 …(ii)
From Eq. (i) and (ii), we get
3y – 10 = 2y + 10
⇒ 3y – 2y = 10 + 10
⇒ y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 3(20) – 10
⇒ x = 50
Hence, the age of person A is 50years and Age of B is 20years.
Question 33
Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages.
Sol :Let the age of a man = x years
And the age of his son = y years
Ten years hence,
Man’s age = (x + 10) years
Son’s age = (y + 10) years
According to the question,
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x = 2y + 20 – 10
⇒ x = 2y + 10 ...(i)
Ten years ago,
Father’s age = (x – 10) years
Son’s age = (y – 10) years
According to the question,
(x – 10) = 4(y – 10)
⇒ x – 10 = 4y – 40
⇒ x = 4y – 30 …(ii)
From Eq. (i) and (ii), we get
2y + 10 = 4y – 30
⇒ 2y – 4y = – 30 – 10
⇒ – 2y = – 40
⇒ y = 20
On putting the value of y = 20 in Eq. (i), we get
x = 2y + 10
⇒ x = 2(20) + 10
⇒ x = 50
Hence, the age of man is 50 years and the age of his son is 20 years.
Question 34
Find a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10) and ∠D = (4x-5)°. Find the four angles.
Sol :We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°
∴∠A + ∠C = 180° and ∠B + ∠D = 180°
⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180
⇒ 2x + 2y = 180 – 14 and 4x + y – 2 = 180
⇒ x + y = 83 and 4x + y = 182
⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180
⇒ 2x + 2y = 180 – 14 and 4x + y – 2 = 180
⇒ x + y = 83 and 4x + y = 182
So, we get pair of linear equation i.e.
x + y = 83 …(i)
4x + y = 182 …(ii)
x + y = 83 …(i)
4x + y = 182 …(ii)
On subtracting Eq.(i) from (ii), we get
4x + y – x – y = 182 – 83
⇒ 3x = 99
⇒ x = 33
4x + y – x – y = 182 – 83
⇒ 3x = 99
⇒ x = 33
On putting the value of x = 33 in Eq. (i) we get,
33 + y = 83
⇒ y = 83 – 33 = 50
33 + y = 83
⇒ y = 83 – 33 = 50
On putting the values of x and y, we calculate the angles as
∠A = (2x + 43)° = 2(33) + 4 = 66 + 4 = 70°
∠B = (y + 3)° = 50 + 3 = 53°
∠C = (2y + 10)° = 2(50) + 10 = 100 + 10 = 110°
and ∠D = (4x — 5)° = 4(33) – 5 = 132 – 5 = 127°
∠A = (2x + 43)° = 2(33) + 4 = 66 + 4 = 70°
∠B = (y + 3)° = 50 + 3 = 53°
∠C = (2y + 10)° = 2(50) + 10 = 100 + 10 = 110°
and ∠D = (4x — 5)° = 4(33) – 5 = 132 – 5 = 127°
Hence, the angles are ∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°
Question 35
Find the four angles of a cyclic quadrilateral ABCD in which ∠A = (2x — 3)°,
∠B = (y + 7)°, ∠C = (2y + 17)° and ∠D = (4x — 9)°.∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°.
Sol :
We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°
∴∠A + ∠C = 180° and ∠B + ∠D = 180°
⇒ 2x – 3 + 2y + 17 = 180 and y + 7 + 4x – 9 = 180
⇒ 2x + 2y + 14 = 180 and 4x + y – 2 = 180
⇒ 2x + 2y = 180 – 14 and 4x + y = 182
⇒ x + y = 83 and 4x + y = 182
So, we get pair of linear equation i.e.
x + y = 83 …(i)
4x + y = 182 …(ii)
⇒ 2x – 3 + 2y + 17 = 180 and y + 7 + 4x – 9 = 180
⇒ 2x + 2y + 14 = 180 and 4x + y – 2 = 180
⇒ 2x + 2y = 180 – 14 and 4x + y = 182
⇒ x + y = 83 and 4x + y = 182
So, we get pair of linear equation i.e.
x + y = 83 …(i)
4x + y = 182 …(ii)
On subtracting Eq.(i) from (ii), we get
4x + y – x – y = 182 – 83
⇒ 3x = 99
⇒ x = 33
4x + y – x – y = 182 – 83
⇒ 3x = 99
⇒ x = 33
On putting the value of x = 33 in Eq. (i) we get,
33 + y = 83
⇒ y = 83 – 33 = 50
33 + y = 83
⇒ y = 83 – 33 = 50
On putting the values of x and y, we calculate the angles as
∠A = (2x — 3)° = 2(33) – 3 = 66 – 3 = 63°
∠B = (y + 7)° = 50 + 7 = 57°
∠C = (2y + 17)° = 2(50) + 17 = 100 + 17 = 117°
and ∠D = (4x — 9)° = 4(33) – 9 = 132 – 9 = 123°
Hence, the angles are ∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°
Sol :
Sol :
We know that, in a triangle , the sum of three angles is 180°
∴ ∠A + ∠B + ∠C = 180°
According to the question,
x + 3x + y = 180
⇒ 4x + y = 180
⇒ y = 180 – 4x …(i)
Given that 3y — 5x = 30 …(ii)
On substituting the value of y in Eq. (ii), we get
3(180 – 4x) – 5x = 30
⇒ 540 – 12x – 5x = 30
⇒ 540 – 17x = 30
⇒ – 17x = 30 – 540
⇒ – 17x = – 510
⇒ x = 30
Now, we substitute the value of x in Eq.(i), we get
⇒ y = 180 – 4(30)
⇒ y = 60
On putting the value of x and y, we calculate the angles
∠A = x° = 30°
∠B = (3x)° = 3(30) = 90°
and ∠C = y° = 60°
Here, we can see that ∠B = 90° , so triangle is a right angled.
Sol :
Let the length of a rectangle = x m
and the breadth of a rectangle = y m
Then, Area of rectangle = xy m2
Condition I :
Area is reduced by 8m2, when length = (x – 5) m and breadth = (y + 3) m
Then, area of rectangle = (x – 5)×(y + 3) m2
According to the question,
xy – (x – 5)×(y + 3) = 8
⇒ xy – (xy + 3x – 5y – 15) = 8
⇒ xy – xy – 3x + 5y + 15 = 8
⇒ – 3x + 5y = 8 – 15
⇒ 3x – 5y = 7 …(i)
Condition II:
Area is increased by 74m2, when length = (x + 3) m and breadth = (y + 2) m
Then, area of rectangle = (x + 3)×(y + 2) m2
According to the question,
(x + 3)×(y + 2) – xy = 74
⇒ (xy + 3y + 2x + 6) – xy = 74
⇒ xy + 2x + 3y + 6 – xy = 74
⇒ 2x + 3y = 74 – 6
⇒ 2x + 3y = 68 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3, we get
6x – 10y = 14 …(iii)
6x + 9y = 204 …(iv)
On subtracting Eq. (i) from Eq. (ii), we get
6x + 9y – 6x + 10y = 204 – 14
⇒ 19y = 190
⇒ y = 10
On putting the value of y = 10 in Eq. (i), we get
3x – 5 (10) = 7
⇒ 3x – 50 = 7
⇒ 3x = 57
⇒ x = 19
Hence, the length of the rectangle is 19m and the breadth of a rectangle is 10m
Let the breadth of a room = x m
According to the question,
Length of the room = x + 3
Then, Area of room = (x + 3)× (x) m2
= x2 + 3x
Condition II:
Area remains same,
when length = (x + 3 + 3) m = (x + 6) m
and breadth = (x – 2) m
According to the question,
x2 + 3x = (x + 6)(x – 2)
⇒ x2 + 3x = x2 – 2x + 6x – 12
⇒ 3x = 4x – 12
⇒ 3x – 4x = – 12
⇒ x = 12
So, length of the room = (x + 3) = 12 + 3 = 15m
Hence, the length of the room is 15m and the breadth of a room is 12m
Sol :
Let the speed of car I = x km/hr
And the speed of car II = y km/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 6h.
Distance travelled by car I in 6h = 6x km
Distance travelled by car II in 6h = 6y km
Since, they are travelling in same direction, sign should be negative
6x – 6y = 120
⇒ x –y = 20 …(i)
Now, Let two cars meet after 1hr 12min
∠A = (2x — 3)° = 2(33) – 3 = 66 – 3 = 63°
∠B = (y + 7)° = 50 + 7 = 57°
∠C = (2y + 17)° = 2(50) + 17 = 100 + 17 = 117°
and ∠D = (4x — 9)° = 4(33) – 9 = 132 – 9 = 123°
Hence, the angles are ∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°
Question 36
In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.
∠A = 20°, ∠B = 40°, ∠C = 120°.Sol :
We know that, in a triangle , the sum of three angles is 180°
∴ ∠A + ∠B + ∠C = 180° …(a)
According to the question,
$\begin{matrix}\angle \mathrm{C}&=3 \angle \mathrm{B}&=2(\angle \mathrm{A}+\angle \mathrm{B})\\ \text{I}&\text{II}&\text{III}\end{matrix}$
∴ ∠A + ∠B + ∠C = 180° …(a)
According to the question,
$\begin{matrix}\angle \mathrm{C}&=3 \angle \mathrm{B}&=2(\angle \mathrm{A}+\angle \mathrm{B})\\ \text{I}&\text{II}&\text{III}\end{matrix}$
On taking II and III, we get
⇒ 3∠B = 2 (∠A + ∠B)
⇒ 3∠B = 2 ∠A + 2 ∠B
⇒ ∠B = 2 ∠A …(i)
⇒ 3∠B = 2 (∠A + ∠B)
⇒ 3∠B = 2 ∠A + 2 ∠B
⇒ ∠B = 2 ∠A …(i)
Now, on taking I and II, we get
∠C = 3 ∠B
∠C = 3 ∠B
⇒ ∠C = 3(2 ∠A) (from eq. (i))
⇒ ∠C = 6 ∠A …(ii)
⇒ ∠C = 6 ∠A …(ii)
On substituting the value of ∠B and ∠C in Eq. (a), we get
∠A + 2∠A + 6∠A = 180°
⇒ 9∠A = 180°
⇒ ∠A = 20°
On puuting the value of ∠A = 20° in Eq. (i) and (ii), we get
∠B = 2 ∠A = 2(20) = 40°
∠C = 6 ∠A = 6(20) = 120°
Hence, the angles are∠A = 20°, ∠B = 40°, ∠C = 120°
∠A + 2∠A + 6∠A = 180°
⇒ 9∠A = 180°
⇒ ∠A = 20°
On puuting the value of ∠A = 20° in Eq. (i) and (ii), we get
∠B = 2 ∠A = 2(20) = 40°
∠C = 6 ∠A = 6(20) = 120°
Hence, the angles are∠A = 20°, ∠B = 40°, ∠C = 120°
Question 37
In a ΔABC, ∠A = x°, ∠B = (3x)° and ∠C = y°.
If 3y — 5x = 30, show that the triangle is right – angled.We know that, in a triangle , the sum of three angles is 180°
∴ ∠A + ∠B + ∠C = 180°
According to the question,
x + 3x + y = 180
⇒ 4x + y = 180
⇒ y = 180 – 4x …(i)
Given that 3y — 5x = 30 …(ii)
On substituting the value of y in Eq. (ii), we get
3(180 – 4x) – 5x = 30
⇒ 540 – 12x – 5x = 30
⇒ 540 – 17x = 30
⇒ – 17x = 30 – 540
⇒ – 17x = – 510
⇒ x = 30
Now, we substitute the value of x in Eq.(i), we get
⇒ y = 180 – 4(30)
⇒ y = 60
On putting the value of x and y, we calculate the angles
∠A = x° = 30°
∠B = (3x)° = 3(30) = 90°
and ∠C = y° = 60°
Here, we can see that ∠B = 90° , so triangle is a right angled.
Question 38
The area of a rectangle gets reduced by 8 m2, when its length is reduced by 5m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m2. Find the length and the breadth of the rectangle.
Let the length of a rectangle = x m
and the breadth of a rectangle = y m
Then, Area of rectangle = xy m2
Condition I :
Area is reduced by 8m2, when length = (x – 5) m and breadth = (y + 3) m
Then, area of rectangle = (x – 5)×(y + 3) m2
According to the question,
xy – (x – 5)×(y + 3) = 8
⇒ xy – (xy + 3x – 5y – 15) = 8
⇒ xy – xy – 3x + 5y + 15 = 8
⇒ – 3x + 5y = 8 – 15
⇒ 3x – 5y = 7 …(i)
Condition II:
Area is increased by 74m2, when length = (x + 3) m and breadth = (y + 2) m
Then, area of rectangle = (x + 3)×(y + 2) m2
According to the question,
(x + 3)×(y + 2) – xy = 74
⇒ (xy + 3y + 2x + 6) – xy = 74
⇒ xy + 2x + 3y + 6 – xy = 74
⇒ 2x + 3y = 74 – 6
⇒ 2x + 3y = 68 …(ii)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3, we get
6x – 10y = 14 …(iii)
6x + 9y = 204 …(iv)
On subtracting Eq. (i) from Eq. (ii), we get
6x + 9y – 6x + 10y = 204 – 14
⇒ 19y = 190
⇒ y = 10
On putting the value of y = 10 in Eq. (i), we get
3x – 5 (10) = 7
⇒ 3x – 50 = 7
⇒ 3x = 57
⇒ x = 19
Hence, the length of the rectangle is 19m and the breadth of a rectangle is 10m
Question 39
The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.
Sol :Let the breadth of a room = x m
According to the question,
Length of the room = x + 3
Then, Area of room = (x + 3)× (x) m2
= x2 + 3x
Condition II:
Area remains same,
when length = (x + 3 + 3) m = (x + 6) m
and breadth = (x – 2) m
According to the question,
x2 + 3x = (x + 6)(x – 2)
⇒ x2 + 3x = x2 – 2x + 6x – 12
⇒ 3x = 4x – 12
⇒ 3x – 4x = – 12
⇒ x = 12
So, length of the room = (x + 3) = 12 + 3 = 15m
Hence, the length of the room is 15m and the breadth of a room is 12m
Question 40
Two places A and B are 120 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 6 hours, and if they move in opposite directions, they meet in 1 hour 12 minutes. Find the speed of each car.
Let the speed of car I = x km/hr
And the speed of car II = y km/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 6h.
Distance travelled by car I in 6h = 6x km
Distance travelled by car II in 6h = 6y km
Since, they are travelling in same direction, sign should be negative
6x – 6y = 120
⇒ x –y = 20 …(i)
Now, Let two cars meet after 1hr 12min
1hr 12min $=1+\frac{12}{60}=\frac{6}{5} \mathrm{hr}$
Since they are travelling in opposite directions, sign should be positive.
$\frac{6}{5} x+\frac{6}{5} y=120$
⇒ 6x + 6y = 120 × 5
⇒ x + y = 100 …(ii)
$\frac{6}{5} x+\frac{6}{5} y=120$
⇒ 6x + 6y = 120 × 5
⇒ x + y = 100 …(ii)
On adding (i) and (ii) , we get
x – y + x + y = 20 + 100
⇒ 2x = 120
⇒ x = 60
Putting the value of x = 60 in Eq. (i), we get
60 – y = 20
⇒ y = 40
So, the speed of the two cars are 60km/h and 40 km/hr respectively.
Let the speed of a train = x km/hr
And the speed of a car = y km/hr
Total distance travelled = 600km
According to the question,
If he covers 400km by train and rest by car i.e. (600 – 400) = 200km
x – y + x + y = 20 + 100
⇒ 2x = 120
⇒ x = 60
Putting the value of x = 60 in Eq. (i), we get
60 – y = 20
⇒ y = 40
So, the speed of the two cars are 60km/h and 40 km/hr respectively.
Question 41
A train travels a distance of 300 km at a constant speed. If the speed of SE the 2O train is increased by 5 km an hour, the journey would have taken 2 hours less. Find the original speed of the train.
Sol :
Total distance travelled = 300km
Let the speed of train = x km/hr
Let the speed of train = x km/hr
We know that,
time $=\frac{\text { distance }}{\text { speed }}$
time $=\frac{\text { distance }}{\text { speed }}$
Hence, time taken by train $=\frac{300}{x}$
According to the question,
Speed of the train is increased by 5km an hour
∴ the new speed of the train = (x + 5)km/hr
Speed of the train is increased by 5km an hour
∴ the new speed of the train = (x + 5)km/hr
Time taken to cover 300km $=\frac{300}{x+5}$
Given that time taken is 2hrs less from the previous time
$\Rightarrow \frac{300}{x}-\frac{300}{x+5}=2$
$\Rightarrow \frac{300(\mathrm{x}+5)-300(\mathrm{x})}{\mathrm{x}(\mathrm{x}+5)}=2$
⇒ 300x + 1500 – 300x = 2x (x + 5)
⇒ 1500 = 2x2 + 10x
⇒ 750 = x2 + 5x
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x – 25) (x + 30) = 0
⇒ (x – 25) = 0 or (x + 30) = 0
∴ x = 25 or x = – 30
Since, speed can’t be negative.
Hence, the speed of the train is 25km/hr
$\Rightarrow \frac{300}{x}-\frac{300}{x+5}=2$
$\Rightarrow \frac{300(\mathrm{x}+5)-300(\mathrm{x})}{\mathrm{x}(\mathrm{x}+5)}=2$
⇒ 300x + 1500 – 300x = 2x (x + 5)
⇒ 1500 = 2x2 + 10x
⇒ 750 = x2 + 5x
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x – 25) (x + 30) = 0
⇒ (x – 25) = 0 or (x + 30) = 0
∴ x = 25 or x = – 30
Since, speed can’t be negative.
Hence, the speed of the train is 25km/hr
Question 42
A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it has to increase the speed by 250 km/hr from the usual speed. Find its usual speed.
Sol :
Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500km
We know that,
speed $=\frac{\text { distance }}{\text { time }}$
Distance to the destination = 1500km
We know that,
speed $=\frac{\text { distance }}{\text { time }}$
Hence, speed $=\frac{1500}{\mathrm{x}} \mathrm{hrs}$
According to the question,
Plane left 30min later than the scheduled time
Plane left 30min later than the scheduled time
30min $=\frac{30}{60}=\frac{1}{2} \mathrm{hr}$
Time taken by the aeroplane $=x-\frac{1}{2} h r s$
∴ the speed of the plane $=\frac{1500}{x-\frac{1}{2}}$
Given that speed has to increase by 250 km/hr
$\Rightarrow \frac{1500}{x-\frac{1}{2}}-\frac{1500}{x}=250$
$\Rightarrow \frac{1500}{\frac{2 \mathrm{x}-1}{2}}-\frac{1500}{\mathrm{x}}=250$
$\Rightarrow \frac{2}{2 x-1}-\frac{1}{x}=\frac{250}{1500}$
$\Rightarrow \frac{2 x-(2 x-1)}{(2 x-1) x}=\frac{1}{6}$
⇒ 6(2x – 2x + 1) = 2x2 – x
⇒ 6 = 2x2 – x
⇒ 2x2 –x – 6 = 0
⇒ 2x2 – 4x + 3x – 6 = 0
⇒ 2x (x – 2) + 3 (x – 2) = 0
⇒ (2x + 3) (x – 2) = 0
⇒ (2x + 3) = 0 or (x – 2) = 0
$\Rightarrow \frac{1500}{x-\frac{1}{2}}-\frac{1500}{x}=250$
$\Rightarrow \frac{1500}{\frac{2 \mathrm{x}-1}{2}}-\frac{1500}{\mathrm{x}}=250$
$\Rightarrow \frac{2}{2 x-1}-\frac{1}{x}=\frac{250}{1500}$
$\Rightarrow \frac{2 x-(2 x-1)}{(2 x-1) x}=\frac{1}{6}$
⇒ 6(2x – 2x + 1) = 2x2 – x
⇒ 6 = 2x2 – x
⇒ 2x2 –x – 6 = 0
⇒ 2x2 – 4x + 3x – 6 = 0
⇒ 2x (x – 2) + 3 (x – 2) = 0
⇒ (2x + 3) (x – 2) = 0
⇒ (2x + 3) = 0 or (x – 2) = 0
∴ $x=\frac{3}{2}$ or x = 2
Since, time can’t be negative.
Hence, the time taken by the aeroplane is 2hrs and the speed is 750km/hr
Hence, the time taken by the aeroplane is 2hrs and the speed is 750km/hr
Question 43
A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Sol :Let the speed of a train = x km/hr
And the speed of a car = y km/hr
Total distance travelled = 600km
According to the question,
If he covers 400km by train and rest by car i.e. (600 – 400) = 200km
Time take = 6hrs 30min
$=6+\frac{30}{60}=6.5 \mathrm{hrs}$
If he travels 200km by train and rest by car i.e. (600 – 200) = 400km
He takes half hour longer i.e. 7 hours
So, total time = train time + car time
He takes half hour longer i.e. 7 hours
So, total time = train time + car time
We know that,
time $=\frac{\text { distance }}{\text { speed }}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=6.5$ …(i)
$\Rightarrow \frac{200}{x}+\frac{400}{y}=7$ …(ii)
time $=\frac{\text { distance }}{\text { speed }}$
$\Rightarrow \frac{400}{x}+\frac{200}{y}=6.5$ …(i)
$\Rightarrow \frac{200}{x}+\frac{400}{y}=7$ …(ii)
Let take $\frac{1}{\mathrm{x}}=\mathrm{u}$ and $\frac{1}{y}=v$
400u + 200v = 6.5 …(iii)
and 200u + 400v = 7 …(iv)
400u + 200v = 6.5 …(iii)
and 200u + 400v = 7 …(iv)
On multiplying Eq. (iii) by 2 and Eq. (iv) by 4, we get
800u + 400v = 13 …(a)
800u + 1600v = 28 …(b)
On subtracting Eq. (a) from Eq. (b), we get
800u + 1600v – 800u – 400v = 28 – 13
⇒ 1200v = 15
$\Rightarrow \mathrm{v}=\frac{15}{1200}$
$\Rightarrow \mathrm{v}=\frac{1}{80}$
800u + 400v = 13 …(a)
800u + 1600v = 28 …(b)
On subtracting Eq. (a) from Eq. (b), we get
800u + 1600v – 800u – 400v = 28 – 13
⇒ 1200v = 15
$\Rightarrow \mathrm{v}=\frac{15}{1200}$
$\Rightarrow \mathrm{v}=\frac{1}{80}$
On putting the value of v in Eq. (iv), we get
$200 u+400\left(\frac{1}{80}\right)=7$
⇒ 200u + 5 = 7
⇒ 200u = 2
$\Rightarrow \mathrm{u}=\frac{1}{100}$
$200 u+400\left(\frac{1}{80}\right)=7$
⇒ 200u + 5 = 7
⇒ 200u = 2
$\Rightarrow \mathrm{u}=\frac{1}{100}$
So, we get $\mathrm{u}=\frac{1}{100}$ and $v=\frac{1}{80}$
⇒ x = 100 and y = 80
⇒ x = 100 and y = 80
Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.
Sol :
Let the speed of car I = x km/hr
And the speed of car II = y km/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 8h.
Distance travelled by car I in 8h = 8x km
Distance travelled by car II in 8h = 8y km
Since, they are travelling in same direction, sign should be negative
8x – 8y = 80
⇒ x –y = 10 …(i)
Question 44
Places A and B are 80 km apart from each other on a highway. One car starts from A and another from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find speed of the cars.Let the speed of car I = x km/hr
And the speed of car II = y km/hr
Car I starts from point A and Car II starts from point B.
Let two cars meet at C after 8h.
Distance travelled by car I in 8h = 8x km
Distance travelled by car II in 8h = 8y km
Since, they are travelling in same direction, sign should be negative
8x – 8y = 80
⇒ x –y = 10 …(i)
Now, Let two cars meet after 1hr 20 min
1hr 20min $=1+\frac{20}{60}=\frac{4}{3} \mathrm{hr}$
1hr 20min $=1+\frac{20}{60}=\frac{4}{3} \mathrm{hr}$
Since they are travelling in opposite directions, sign should be positive.
$\frac{4}{3} x+\frac{4}{3} y=80$
⇒ 4x + 4y = 240
⇒ x + y = 60 …(ii)
On adding (i) and (ii) , we get
x – y + x + y = 10 + 60
⇒ 2x = 70
⇒ x = 35
Putting the value of x = 25 in Eq. (i), we get
35 – y = 10
⇒ y = 25
So, the speed of the two cars are 35km/h and 25 km/hr respectively.
Sol :
Let speed of the boat in still water = x km/hr
and speed of the stream = y km/hr
Then, the speed of the boat downstream = (x + y)km/hr
And speed of the boat upstream = (x – y)km/hr
⇒ 4x + 4y = 240
⇒ x + y = 60 …(ii)
On adding (i) and (ii) , we get
x – y + x + y = 10 + 60
⇒ 2x = 70
⇒ x = 35
Putting the value of x = 25 in Eq. (i), we get
35 – y = 10
⇒ y = 25
So, the speed of the two cars are 35km/h and 25 km/hr respectively.
Question 45
A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream in the same time. Find the speed of the boat in still water and that of the stream.
Let speed of the boat in still water = x km/hr
and speed of the stream = y km/hr
Then, the speed of the boat downstream = (x + y)km/hr
And speed of the boat upstream = (x – y)km/hr
According to the question
Condition I: When boat goes 16 km upstream, let the time taken be t1.
Then,
Then,
$\mathrm{t}_{1}=\frac{16}{\mathrm{x}-\mathrm{y}} \mathrm{h}$ $\left[\because\right.$ time $\left.=\frac{\text { distance }}{\text { speed }}\right]$
When boat goes 24 km downstream, let the time taken be t2.
When boat goes 24 km downstream, let the time taken be t2.
Then,
$\mathrm{t}_{2}=\frac{24}{\mathrm{x}+\mathrm{y}} \mathrm{h}$
$\mathrm{t}_{2}=\frac{24}{\mathrm{x}+\mathrm{y}} \mathrm{h}$
But total time taken (t1 + t2) = 6 hours
$\therefore \frac{16}{x-y}+\frac{24}{x+y}=6$ …(a)
Condition II: When boat goes 12 km upstream, let the time taken be T1.
Then,
Then,
$\mathrm{T}_{1}=\frac{12}{\mathrm{x}-\mathrm{y}} \mathrm{h}$ $\left[\because\right.$ time $\left.=\frac{\text { distance }}{\text { speed }}\right]$
When boat goes 36 km downstream, let the time taken be T2.
Then,
$\mathrm{T}_{2}=\frac{36}{\mathrm{x}+\mathrm{y}} \mathrm{h}$
$\mathrm{T}_{2}=\frac{36}{\mathrm{x}+\mathrm{y}} \mathrm{h}$
But total time taken (T1 + T2) = 6 hours
$\therefore \frac{12}{x-y}+\frac{36}{x+y}=6$ …(b)
$\therefore \frac{12}{x-y}+\frac{36}{x+y}=6$ …(b)
Now, we solve tis pair of linear equations by elimination method
$\frac{16}{x+y}+\frac{24}{x-y}=6$ …(i)
And $\frac{12}{x+y}+\frac{36}{x-y}=6$ …(ii)
$\frac{16}{x+y}+\frac{24}{x-y}=6$ …(i)
And $\frac{12}{x+y}+\frac{36}{x-y}=6$ …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients equal of first term, we get the equation as
$\frac{48}{x+y}+\frac{72}{x-y}=18$ …(iii)
$\frac{48}{x+y}+\frac{144}{x-y}=24$ …(iv)
$\frac{48}{x+y}+\frac{72}{x-y}=18$ …(iii)
$\frac{48}{x+y}+\frac{144}{x-y}=24$ …(iv)
On substracting Eq. (iii) from Eq. (iv), we get
$\frac{48}{x+y}+\frac{144}{x-y}-\frac{48}{x+y}-\frac{72}{x-y}=24-18$
$\Rightarrow \frac{144-72}{x-y}=6$
$\Rightarrow \frac{72}{x-y}=6$
⇒ x – y = 12 …(a)
$\frac{48}{x+y}+\frac{144}{x-y}-\frac{48}{x+y}-\frac{72}{x-y}=24-18$
$\Rightarrow \frac{144-72}{x-y}=6$
$\Rightarrow \frac{72}{x-y}=6$
⇒ x – y = 12 …(a)
On putting the value of x – y = 12 in Eq. (i), we get
$\Rightarrow \frac{16}{x+y}+\frac{24}{12}=6$
$\Rightarrow \frac{16}{x+y}=6-2$
$\Rightarrow \frac{16}{x+y}=4$
⇒ x + y = 4 …(b)
$\Rightarrow \frac{16}{x+y}+\frac{24}{12}=6$
$\Rightarrow \frac{16}{x+y}=6-2$
$\Rightarrow \frac{16}{x+y}=4$
⇒ x + y = 4 …(b)
Adding Eq. (a) and (b), we get
⇒ 2x = 16
⇒ x = 8
⇒ 2x = 16
⇒ x = 8
On putting value of x = 8 in eq. (a), we get
8 – y = 12
⇒ y = – 4 but speed can’t be negative
⇒ y = 4
8 – y = 12
⇒ y = – 4 but speed can’t be negative
⇒ y = 4
Hence, x = 8 and y = 4 , which is the required solution.
Hence, the speed of the boat in still water is 8km/hr and speed of the stream is 4km/hr
Sol :
Let the speed of a train = x km/hr
And the speed of a car = y km/hr
Total distance travelled = 370km
Question 46
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.
Let the speed of a train = x km/hr
And the speed of a car = y km/hr
Total distance travelled = 370km
According to the question,
If he covers 250km by train and rest by car i.e. (370 – 250) = 120km
Time take = 4hrs
Time take = 4hrs
If he travels 130km by train and rest by car i.e. (370 – 130) = 240km
He takes 18min longer i.e. $4+\frac{18}{60}=4.3$hours
So, total time = train time + car time
We know that,
time $=\frac{\text { distance }}{\text { speed }}$
time $=\frac{\text { distance }}{\text { speed }}$
$\Rightarrow \frac{250}{x}+\frac{120}{y}=4$ …(i)
$\Rightarrow \frac{130}{\mathrm{x}}+\frac{240}{\mathrm{y}}=4.3$ …(ii)
Let take $\frac{1}{x}=u$ and $\frac{1}{\mathrm{y}}=\mathrm{v}$
250u + 120v = 4 …(iii)
and 130u + 240v = 4.3 …(iv)
and 130u + 240v = 4.3 …(iv)
On multiplying Eq. (iii) by 2
500u + 240v = 8 …(v)
500u + 240v = 8 …(v)
On subtracting Eq. (iv) from Eq. (v), we get
500u + 240v – 130u – 240v = 8 – 4.3
⇒ 370u = 3.7
$\Rightarrow \mathrm{u}=\frac{3.7}{370}$
$\Rightarrow \mathrm{u}=\frac{1}{100}$
500u + 240v – 130u – 240v = 8 – 4.3
⇒ 370u = 3.7
$\Rightarrow \mathrm{u}=\frac{3.7}{370}$
$\Rightarrow \mathrm{u}=\frac{1}{100}$
On putting the value of v in Eq. (iv), we get
$130\left(\frac{1}{100}\right)+240 \mathrm{v}=4.3$
⇒ 1.3 + 240v = 4.3
⇒ 240v = 3
$130\left(\frac{1}{100}\right)+240 \mathrm{v}=4.3$
⇒ 1.3 + 240v = 4.3
⇒ 240v = 3
$\Rightarrow \mathrm{v}=\frac{1}{80}$
So, we get $u=\frac{1}{100}$ and $v=\frac{1}{80}$
⇒ x = 100 and y = 80
Hence, the speed of the train is 100km/hr and the speed of the car is 80km/hr.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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