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Learn Exercise 3.7 with Free Lessons & Tips

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Question 1:

ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution 1:

Oppsite angles of cyclic quadrilateral are supplementary.

-7x+5+3y-5=180

-7x+3y=180 eq1

-4x+4y+20=180

-4x+4y=160 eq2

Divide by 4

-x+y=40

Multiply by 3

-3x+3y=120 eq3

-7x+3y=180 eq1

Subtracting eq1 from eq3

4x=-60

x=-15

C= -4x=-4*-15=60

A=180-60=120

D=-7x+5=-7*-15+5=105+5=110

B=180-110=70

Ans

A=120,B=70,C=60,D=110

Comments

Venkatesan Iyer

Question 2:

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)]

Solution 2:

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = - 300        ... (1)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 70            ... (2)

Multiplying equation (2) by 2, we obtain:
12x - 2y = 140        ... (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 - 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Comments

Tasneem

Question 3:

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution 3:

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 * x = 2x years

age of cathy $= \frac{y}{2}$ years.

Case I: Ani is older than Biju by 3 years
x - y = 3 ... (1) $2x-\frac{y}{2}=30$

4x - y = 60 ... (2)

Subtracting (1) from (2), we obtain:
3x = 60 - 3 = 57

$x=\frac{57}{3}=19$

Age of Ani = 19 years
Age of Biju = 19 - 3 = 16 years

Case II: Biju is older than Ani by 3 years
y - x = 3 ... (3)

$2x-\frac{y}{2}=30$

4x - y = 60 ... (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Comments

Tasneem

Question 4:

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution 4:

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Speed= $\frac{disstance}{time}$ , $x=\frac{d}{t}$

$d=xt................(i)$

According to the question, $(x+10)= \frac{d}{t-2}$

By using (i), we get, $-2x+10t=20..............(ii)$

$(x-10)=\frac{d}{t+3}$

By using equation (1), we obtain:
3x - 10t = 30 ... (3)

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

Comments

Tasneem

Question 5:

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution 5:

Let no of rows be x

No of students per row be y

Total no of students is xy.

Condition 1:

(y+3)(x-1)= xy

xy+3x-y-3=xy

3x-y=3 eq1

Second condition

(y-3)(x+2)=xy

xy-3x+2y-6=xy

-3x+2y= 6 eq 2

3x-y =3. 7eq 1

Add them

y=9

substitute y=9

3x-9=3

x=12/3=4

Answer

No of rows 4

No of students per row 9

Total number of students in a class = xy = 4 x 9 = 36

Comments

Venkatesan Iyer

Question 6:

In a triangle ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

Solution 6:

C = 3B = 2(A + B)
3B = 2(A + B)
B = 2A
2A - B = 0 ... (1)

We know that the sum of the measures of all angles of a triangle is 180°.

A +

B +

C = 180°

A +

B + 3

B = 180°

A + 4

B = 180° ... (2)

Multiplying equation (1) by 4, we obtain:
8

A - 4

B = 0 ... (3)

Adding equations (2) and (3), we obtain:

9

A = 180°

A = 20°

From equation (2), we obtain:
20° + 4

B = 180° 4

B = 160°

B = 40°

C = 3

B = 3 x 40° = 120°
Thus, the measure of

B and

C are 20°, 40°, and 120° respectively.

Comments

Tasneem

Question 7:

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Solution 7:

$5x-y=5.............(i)$

For the graph,

x	0	1	2
y	-5	0	5

$3x-y=3............(ii)$

x	0	1	2
y	-3	0	3

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Comments

Tasneem

Question 8:

Solve the following pair of linear equations:

$(i) px+qy=p-q, qx-py=p+q$

$(ii)ax+by=c, bx+ay=1+c$

$(iii)\frac{x}{a}-\frac{y}{b}=0, ax+by=a^{2}+b^{2}$

$(iv)(a-b)x+(a+b)y=a^{2}-2ab-b^{2}, (a+b)(x+y)=a^{2}+b^{2}$

Solution 8:

(i)

px + qy = p - q        ... (1)
qx - py = p + q        ... (2)
Multiplying equation (1) by p and equation (2) by q, we obtain:
p²x + pqy = p² - pq        ... (3)
q²x - pqy = pq + q²        ... (4)
Adding equations (3) and (4), we obtain:
p²x + q²x = p² + q²
(p² + q²)x = p² + q²

Substituting the value of x in equation (1), we obtain:
p(1) + qy = p - q
qy = -q
y = -1

(ii)

ax + by = c ... (1)
bx + ay = 1 + c ... (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
a²x + aby = ac ... (3)
b²x + aby = b + bc ... (4)
Subtracting equation (4) from equation (3),
(a² - b²)x = ac - bc - b

$x=\frac{c(a-b)-b}{a^{2}-b^{2}}$

Substituting the value of x in equation (1), we obtain:

$y=\frac{c(a-b)+a}{a^{2}-b^{2}}$

$(iii) \frac{x}{a}-\frac{y}{b}=0$

bx - ay = 0 ... (1)
ax + by = a² + b² ... (2)
Multiplying equation (1) and (2) by b and a respectively, we obtain:
b²x - aby = 0 ... (3)
a²x + aby = a³ + ab² ... (4)
Adding equations (3) and (4), we obtain:
b²x + a²x = a³ + ab²
x(b² + a²) = a(a² + b²)
x = a
Substituting the value of x in equation (1), we obtain:
b(a) - ay = 0
ab - ay = 0
ay = ab
y = b

(iv)

(a - b)x + (a + b)y = a² - 2ab - b² ... (1)
(a + b)(x + y) = a² + b²

$(a+b)x-(a+b)y= (a^{2}+b^{2})..............(2)$

Subtracting equation (2) from (1), we obtain:
(a - b)x - (a + b)x = (a² - 2ab - b²) - (a² + b²)
(a - b - a - b)x = -2ab - 2b²
-2bx = -2b(a + b)
x = a + b
Substituting the value of x in equation (1), we obtain:
(a - b)(a + b) + (a + b)y = a² - 2ab - b²
a² - b² + (a + b)y = a² - 2ab - b²
(a + b)y = -2ab

$y=\frac{-2ab}{a+b}$

Comments

Tasneem

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

Exercise 3.7

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Learn Exercise 3.7 with Free Lessons & Tips

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

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