**Exercise 3.11**

**If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however, the length is reduced by 1 unit and the breadth is increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.**

**Solution:**

Let the length of the rectangle be x units.

Let the breadth of the rectangle be y units.

Hence, area of rectangle =xy units

According to the question,

Case 1;

Length is increased by 2 unit=x+2

Breadth is reduced by 2 units=y-2

And, the area is reduced by 28 square units=xy-28

Then, the equation becomes,

\((x+2)(y-2)=xy-28\)

\(\Rightarrow xy-2x+2y-4=xy-28\) \(\Rightarrow -2x+2y-4+28=0\) \(\Rightarrow -2x+2y+24=0\) \(\Rightarrow 2x-2y-24=0\)Therefore, 2x – 2y – 24 = 0 ——- (i)

Case 2;

Length is reduced by 1 unit=x-1

Breadth is increased by 2 units=y+2

And, the area is increased by 33 square units.

Then, the equation becomes,

\((x-1)(y+2)=xy+33\) \(\Rightarrow xy+2x-y-2=x+33\) \(\Rightarrow 2x-y-2-33=0\) \(\Rightarrow 2x-y-35=0\)Therefore, 2x – y – 35 = 0 ——- (ii)

Thus the system of linear equations obtained is:

2x – 2y – 24 = 0

2x – y – 35 = 0

Using cross multiplication, we get

\(\frac{x}{(-2*-35)-(-1*-24)}=\frac{y}{(2*-35)-(2*-24)}=\frac{1}{(2*-1)-(2*-2)}\) \(\frac{x}{70-24}=\frac{-y}{-70+48}=\frac{1}{-2+4}\) \(\frac{x}{46}=\frac{-y}{-22}=\frac{1}{2}\) \(x=\frac{46}{2}\) \(x=23\)And

\(y=\frac{22}{2}\) \(y=11\)The length of actual rectangle is 23 units.

The breadth of actual rectangle is 11 units.

Thus, the area of the actual rectangle = length x breadth,

\(=x\times y\)=23 x 11

= 253 square units

Hence, the area of rectangle is 253 square units.

**The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and the breadth is increased by 5 metres. Find the dimensions of the rectangle.**

**Solution:**

Let the length of the rectangle be x units.

Let the breadth of the rectangle be y units.

Hence, area of rectangle =xy units

According to the question,

Case 1;

Length is increased by 7 metres=x+7

Breadth is decreased by 3 metres=y-3

And, the area of the rectangle remains same=xy.

Then, the equation becomes,

\(xy=(x+7)(y-3)\) \(xy=xy+7y-3x-21\)3x – 7y + 21 = 0 ——– (i)

Case 2;

Length is decreased by 7 metres=x-7

Breadth is increased by 5 metres=y+5

And, the area of the rectangle remains same=xy.

Then, the equation becomes,

\(xy=(x-7)(y+5)\) \(xy=xy-7y+5x-35\)5x – 7y – 35 = 0 ——– (ii)

Thus the system of linear equations obtained is:

3x – 7y + 21 = 0

5x – 7y – 35 = 0

Using cross-multiplication, we get,

\(\frac{x}{(-7\times -5)-(-7\times 21)}=\frac{y}{(3\times -35)-(5\times 21)}=\frac{1}{(3\times -7)-(5\times -7)}\) \(\frac{x}{245+147}=\frac{-y}{-105-105}=\frac{1}{-21+35}\) \(\frac{x}{392}=\frac{-y}{-210}=\frac{1}{14}\) \(x=\frac{392}{14}\) \(x=28\)And

\(y=\frac{210}{14}\) \(y=15\)Therefore, the length of the actual rectangle is 28 meters.

The breadth of the actual rectangle is 15 metres.

**In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the triangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres. Find the dimension of the rectangle.**

**Solution:**

Let the length of the rectangle be x units.

Let the breadth of the rectangle be y units.

Hence, area of rectangle =xy units

According to the question,

Case 1;

Length is increased by 3 metres=x+3

Breadth is reduced by 4 metres =y-4

And, the area of the rectangle is reduced by 67 sq. metres =xy-67.

Then, the equation becomes,

xy – 67 = (x + 3)(y – 4)

xy – 67 = xy + 3y – 4x – 12

4xy – 3y – 67 + 12 = 0

4x – 3y – 55 = 0 —— (i)

Case 1;

Length is reduced by 1 metre =x-1

Breadth is increased by 4 metre =y+4

And, the area of the rectangle is increased by 89 sq. metres =xy+89.

Then, the equation becomes,

xy +89 = (x -1)(y + 4)

4x – y – 93 = 0 ——- (ii)

Thus the system of linear equations obtained is:

4x – 3y – 55 = 0

4x – y – 93 = 0

Using cross multiplication, we get,

\(\frac{x}{(-3\times -93)-(-1\times -55)}=\frac{-y}{(4\times -93)-(4\times -55)}=\frac{1}{(4\times -1)-(4\times -3)}\) \(\frac{x}{279-55}=\frac{-y}{-372+220}=\frac{1}{-4+12}\) \(\frac{x}{224}=\frac{-y}{-152}=\frac{1}{8}\) \(x=\frac{224}{8}\) \(x=28\)And

\(y=\frac{152}{8}\) \(y=19\)Therefore, the length of rectangle is 28 metre

The breadth of rectangle is 19 metre.

**The income of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs. 1250, find their incomes.**

**Solution:**

Let the income be denoted by x.

Let the expenditure be denoted by y.

Then, according to the question,

The income of X is Rs. 8x and the expenditure of X is 19y.

The income of Y is Rs. 7x and the expenditure of Y is 16y.

Then, we get,

Saving of X = 8x – 19y = 1250

Saving of Y = 7x – 16y = 1250

Hence, the qiven equations are

8x – 19y – 1250 = 0 —– (i)

7x – 16y – 1250 = 0 —– (ii)

Using cross-multiplication method, we have

\(\frac{x}{(-19\times -1250)-(-16\times -1250)}=\frac{-y}{(8\times -1250)-(7\times -1250)}=\frac{1}{(8\times -16)-(7\times -19)}\) \(\frac{x}{23750-20000}=\frac{-y}{-10000+8750}=\frac{1}{-128+133}\) \(\frac{x}{3750}=\frac{y}{1250}=\frac{1}{5}\) \(x=\frac{3750}{5}\) \(x=750\)Since x=750,

The income of X = 8x

= 8 x 750

= 6000

The income of Y = 7x

= 7 x 750

= 5250

Therefore, the income of X is Rs. 6000

The income of Y is Rs. 5250

**A and B each has some money. If A gives Rs. 30 to B, then B will have twice the money left with A. But, if B gives Rs. 10 to A, then A will have thrice as much as is left with B. How much money does each have?**

**Solution:**

Let the money with A be Rs.x

Let the money with B be Rs.y.

According to the question,

Case 1: if A gives Rs. 30 to B, then B will have twice the money left with A.

Then the equation becomes,

y + 30 = 2(x – 30)

y + 30 = 2x – 60

2x – y – 60 – 30 = 0

2x – y – 90 = 0 —— (i)

Case 2: If B gives Rs. 10 to A, then A will have thrice as much as is left with B,

x + 10 = 3(y – 10)

x + 10 = 3y – 10

x – 3y + 10 + 30 = 0

x – 3y + 40 = 0 —— (ii)

To equate equation (i) and (ii), we multiply equation (ii) with 2

On multiplying equation (ii) with 2, we get,

2x – 6y + 80 = 0

Subtract equation (ii) from (i), we get,

y = 34

Now, on substituting y = 34 in equation (i), we get,

x = 62

Therefore, the money with A be Rs. 62

The money with B be Rs. 34

** **

**2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?**

**Solution:**

Let the time required for a man alone to finish the work be x days.

Let the time required for a boy alone to finish the work be y days.

Then,

The work done by a man in one day =

\(\frac{1}{x}\)The work done by a boy in one day =

\(\frac{1}{y}\)The work done by 2 men in one day =

\(\frac{2}{x}\)The work done by 7 boys in one day =

\(\frac{7}{y}\)From the question, we know that,

2 men and 7 boys together can finish the work in 4 days

\(4\left ( \frac{2}{x}+\frac{7}{y} \right )=1\) \(\frac{8}{x}+\frac{28}{y}=1\)——–(i)

Again 4 men and 4 boys can finish the work in 3 days

\(3\left ( \frac{4}{x}+\frac{4}{y} \right )=1\) \(\frac{12}{x}+\frac{12}{y}=1\)——–(ii)

Substituting

\(\frac{1}{x}=u\)and

\(\frac{1}{y}=v\)in equation (i) and (ii), we get

8u + 28v = 1

12u + 12v = 1

8u + 28v – 1 = 0 —— (iii)

12u + 12v – 1 = 0 —— (iv)

Using cross multiplication, we get,

\(u=\frac{1}{15}\) \(\frac{1}{x}=\frac{1}{15}\)x = 15

and

\(v=\frac{1}{60}\) \(\frac{1}{y}=\frac{1}{60}\)y = 60

Therefore,

The time required for a man alone to finish the work is 15 days while the time required for a boy alone to finish the work is 60 days.

**In a \(\Delta\)**ABC,

\(\angle\)A = x

\(\angle\)^{o},B = (3x – 2)

\(\angle\)^{o},C = y

\(\angle\)^{o}, Also,C –

\(\angle\)**B = 9**^{o}. Find the three angles.

**Solution:**

According to the question,

\(\angle\)A = x^{o},

B = (3x – 2)^{o},

C = y^{o}

It is also given that,

\(\angle\)C –

\(\angle\)B = 9^{o}

Substituting the value for

\(\angle\)C = y^{o} in above equation we get,

y^{o} = 7^{o} + 3x^{o}

A +

\(\angle\)B +

\(\angle\)C = 180^{o}

A = x^{o} = 25^{o}

B = (3x – 2)^{o} = 73^{o}

C = (7 + 3x)^{o} = 82^{o}

Therefore,

\(\angle\)A = 25^{o},

B = 73^{o},

C = 82^{o}

**In a cyclic quadrilateral ABCD, \(\angle\)**A = (2x + 4)

\(\angle\)^{o},B = (y + 3)

\(\angle\)^{o},C = (2y + 10)

\(\angle\)^{o},**D = (4x – 5)**^{o}. Find the four angles.

**Solution:**

The sum of the opposite angles of cyclic quadrilateral = 180^{o}.

We know that, in the cyclic quadrilateral ABCD,

angles

\(\angle\)A and

\(\angle\)C and angles

\(\angle\)B and

\(\angle\)D pairs of opposite angles

Hence,

\(\angle\)A +

\(\angle\)C = 180^{o} and

B +

\(\angle\)D = 180^{o}

Taking the value of

\(\angle\)A +

\(\angle\)C = 180^{o}

Substituting

\(\angle\)A = (2x + 4)^{o} and

C = (2y + 10)^{o}

We get,

2x + 4 + 2y + 10 = 180^{o}

2x + 2y + 14 = 180^{o}

2x + 2y = 180^{o }– 14^{o}

2x + 2y = 166 —— (i)

Taking the value of

\(\angle\)B +

\(\angle\)D = 180^{o}

Substituting

\(\angle\)B = (y+3)^{o} and

D = (4x – 5)^{o}

We get,

y + 3 + 4x – 5 = 180^{o}

4x + y – 5 + 3 = 180^{o}

4x + y – 2 = 180^{o}

4x + y = 180^{o} + 2^{o}

4x + y = 182^{o} ——- (ii)

To equate equations (i) and (ii), we multiply equation (ii) by 2.

On multiplying equation (ii) by 2, we get,

8x + 2y = 364 —— (iii)

Now, subtract equation (iii) from (i)

-6x = -198

\(x=\frac{-198}{-6}\)x = 33^{o}

On substituting the value of x = 33^{o} in equation (ii) we get

4x + y = 182

132 + y = 182

y = 182 – 132

y = 50

Hence, we can find out the angles of a cyclic quadrilateral by the following method:

\(\angle\)A = 2x + 4

= 66 + 4

= 70^{o}

B = y + 3

= 50 + 3

= 53^{o}

C = 2y + 10

= 100 + 10

= 110^{o}

D = 4x – 5

= 132 – 5

= 127^{o}

Therefore, the angles of cyclic quadrilateral ABCD are

**\(\angle\)**

A = 70^{o},

B = 53^{o},

C = 110^{o},

**D = 127 ^{o}**

** **

**In a \(\Delta\)**ABC,

\(\angle\)A = x

\(\angle\)^{o},B = 3x

\(\angle\)^{o},**C = y**^{o}. If 3y – 5x = 30, prove that the triangle is right angled.

**Solution:**

To prove:

\(\Delta\)ABC is right angled.

Given:

\(\angle\)A = x^{o},

B = 3x^{o} and

C = y^{o}

Sum of the three angles in a triangle is 180^{o}

i.e.,

\(\angle\)A +

\(\angle\)B +

\(\angle\)C = 180^{o}

x + 3x + y = 180^{o}

4x + y = 180 —— (i)

Given in the question, 3y – 5x = 30 —– (ii)

To equating 4x + y = 180 and 3y – 5x = 30, we multiply equation (i) by 3,

On multiplying equation (i) by 3, we get,

12x + 36y = 540 —– (iii)

Subtracting equation (ii) from equation (iii)

17x = 510

\(x=\frac{510}{17}\)x = 30^{o}

Substituting the value of x = 30^{o} in equation (i) we get

4x + y = 180

120 + y = 180

y = 180 – 120

y = 60^{o}

Angles

\(\angle\)A,

\(\angle\)B and

\(\angle\)C are

\(\angle\)A = x^{o}

= 30^{o}

B = 3x^{o}

= 90^{o}

C = y^{o}

= 60^{o}

A right angled triangle is a triangle with any one side right angled to other, i.e., 90^{o} to other.

Therefore,

\(\angle\)B = 90^{o}.

The triangle ABC is right angled. Hence proved.

**The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs. 89 and for a journey of 20 km, the charge paid is Rs. 145. What will a person have to pay for travelling a distance of 30 km?**

**Solution:**

Let the fixed charges of car be Rs. x per km

Let the running charges of car be Rs. y per km

According to the question,

x + 12y = 89 —— (i)

x + 20y = 145 —— (ii)

On subtraction of (i) from (ii), we get,

-8y = -56

\(y=\frac{-56}{-8}\)y = 7

Substituting the value of y = 7 in equation (i) we get

x + 12y = 89

x + 84 = 89

x = 89 – 84

x = 5

Hence, the total charges for travelling a distance of 30 km can be calculated as:

= x + 30y

= 5 + 210

= Rs. 215

Therefore, a person has to pay Rs. 215 for travelling a distance of 30 km.

**A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day.**

**Solution:**

Let the fixed charges of hostel = Rs. x

Let the cost of food charges = Rs. y

According to the question,

x + 20y = 1000 —— (i)

x + 26y = 1180 ——- (ii)

On subtracting equation (ii) from (i) we get

-6y = -180

\(y=\frac{-180}{-6}\)y = 30

Substituting the value for y =30 in equation (i) we get

x + 20y = 1000

x + 600 = 1000

x = 1000 – 600

x = 400

Therefore, the fixed charges of hostel is Rs. 400

The cost of food per day is Rs. 30

**Half the perimeter of a garden, whose length is 4 more than its width is 36 m. Find the dimension of the garden.**

**Solution:**

Let perimeter of rectangular garden = 2(l + b).

Then, according to the question, half the perimeter of a garden will be 36 m

½ x 2(l + b) = 36

l + b = 36 ——- (i)

When the length is four more than its width then l=(b + 4)

On substituting the value for l = b + 4 in equation (i), we get,

l + b = 36

b + 4 + b = 36

2b = 36 – 4

2b = 32

b = 32/2

b = 16

Substituting the value for b = 16 in equation (i), we get,

(l + b) = 36

l + 16 = 36

l = 36 -16

l = 20

Therefore, the width of rectangular garden = 16 m

And the length of rectangular garden = 20 m

**The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.**

**Solution:**

The sum of supplementary angles = 180^{o}

Let the larger supplementary angle be x.

Then, we know that,

x + y = 180^{o} —– (i)

According to the question,

Suppose the larger of supplementary angles exceeds the smaller by 18 degrees,

Then we get the equation,

x = y + 18 —— (ii)

Substituting the value of x = y + 18 in equation (i), we get,

x + y = 180^{o}

y + 18 + y = 180^{o}

2y + 18 = 180^{o}

2y = 180^{o} – 18^{o}

2y = 162^{o}

y = (162/2)^{o}

y = 81^{o}

Substituting the value of y = 81^{o} in equation (ii), we get

x = y+ 18

x = 81 + 18

x = 90^{o}

Therefore, the larger supplementary angle is 99^{o}

While the smaller supplementary angle is 81^{o}

^{ }

**2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.**

**Solution:**

Let the time required for a woman alone to finish the work be x days.

Let the time required for a man alone to finish the work be y days.

Then,

The work done by a woman in one day =

\(\frac{1}{x}\)The work done by a man in one day =

\(\frac{1}{y}\)The work done by 2 women in one day =

\(\frac{2}{x}\)The work done by 5 men in one day=

\(\frac{5}{y}\)From the question, we know that,

Case 1:

2 women and 5 men can finish the work in 4 days

\(4\left ( \frac{2}{x}+\frac{5}{y} \right )=1\) \(\frac{8}{x}+\frac{20}{y}=1\)——–(i)

Case 2:

3 women and 6 men can finish the work in 3 days

\(3\left ( \frac{3}{x}+\frac{6}{y} \right )=1\) \(\frac{9}{x}+\frac{18}{y}=1\)——–(ii)

Substituting

\(\frac{1}{x}=u\)and

\(\frac{1}{y}=v\)in equation (i) and (ii), we get

8u + 20v – 1 = 0 —— (iii)

9u + 18v – 1 = 0 —— (iv)

Using cross multiplication, we get,

\(\frac{u}{(20\times -1)-(18\times -1)}=\frac{-v}{(8\times -1)-(9\times -1)}=\frac{1}{(8\times 18)-(9\times 20)}\) \(\frac{u}{-20+18}=\frac{-v}{-8+9}=\frac{1}{144-180}\) \(\frac{u}{-2}=\frac{-v}{1}=\frac{1}{-36}\) \(u=\frac{-2}{-36}\) \(u=\frac{1}{18}\) \(v=\frac{-1}{-36}\) \(v=\frac{1}{36}\)Now,

\(u=\frac{1}{18}\) \(\frac{1}{x}=\frac{1}{18}\)x = 18

\(v=\frac{1}{36}\) \(\frac{1}{y}=\frac{1}{36}\)y = 36

Therefore,

The time required for a woman alone to finish the work be 18 days.

The time required for a man alone to finish the work be 36 days.

**Meena went to a bank to withdraw Rs. 2000. Shw asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes Rs. 50 and Rs. 100 she received.**

**Solution:**

Let the number of notes of Rs. 50 = x

Let the number of notes of Rs. 100 = y

According to the question,

Case 1:

Suppose Meena asks for Rs. 50 and Rs. 100 notes only, we get the equation,

50x + 100y = 2000

x + 2y = 40 ——- (i)

Case 2:

Suppose Meena got 25 notes in all, we get the equation,

x + y = 25 ——- (ii)

Subtracting the equation (ii) from (i) we get

y = 15

Substituting the value of y = 15 in equation (ii), we get

x + y = 25

x + 15 = 25

x = 25 – 15

x = 10

Hence, x = 10 and y = 15

Therefore, Meena has 10 notes of Rs. 50 and 15 notes of Rs. 100 each.

**A wizard having powers of mystic in candations and magical medicines seeing a cock fight going on, spoke privately to both the owners of cocks. To one he said; if your bird wins, then you give me your stake-money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourth. From both of them his gain would be only 12 gold coins. Find the stake of money each of the cock-owners have.**

**Solution:**

Let the strike money of first cock-owner be Rs. x

Let the strike money of second cock-owner be Rs. y.

According to the question,

We have,

For first cock-owner:

\(x-\frac{3}{4}y=12\) \(\frac{4x-3y}{4}=12\)4x – 3y = 48 —– (i)

For second cock-owner:

\(y-\frac{2}{3}x=12\) \(\frac{3y-2x}{3}=12\)3y – 2x = 36 ——- (ii)

Subtracting equation (ii) from (i), we get

2x = 84

x = 84/2 = 42

Substituting the value of x = 42 in equation (ii) we fetch,

3y – 2x = 36

3y – 84 = 36

3y = 36 + 84

3y = 120

y = 120/3 = 40

Therefore, the stake of money of first cock-owner is Rs. 42 and of second cock-owner is Rs. 40.

**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 arrow there would be 2 rows more. Find the number of students in the class.**

**Solution:**

Let the number of students be x

Let the number of rows be y.

Then the number of students in each row =

\(\frac{x}{y}\)According to the question,

If three students are extra in each row, then there is one row less that is when each row has

\(\left ( \frac{x}{y}+3 \right )\)students.

The number of students in each row=

\(\left ( \frac{x}{y}+3 \right )\)The number of rows = (y – 1)

We know that, Total number of students = number of rows x number of student in each row

\(x=\left ( \frac{x}{y}+3 \right )(y-1)\) \(x=(x+3y-\frac{x}{y}-3)\) \(0=\frac{-x}{y}+x-x+3y-3\) \(0=\frac{-x}{y}+3y-3\)—— (i)

If three students are less in each row then there are 2 rows more that is when each row has

\(\left ( \frac{x}{y}-3 \right )(y+2)\)Since, total number of students = number of rows x number of students in each row

\(x=\left ( \frac{x}{y}-3 \right )(y+2)\) \(x=x-3y+\frac{2y}{x}-6\) \(0=\frac{2x}{y}+x-x-3y-6\) \(0=\frac{2x}{y}-3y-6\)——- (ii)

Substituting,

\(\frac{x}{y}=u\)in (i) and (ii) we get

-u + 3y – 3 = 0 ——- (iii)

2u – 3y – 6 = 0 —— (iv)

Now, on adding (iii) and (iv) we get

u – 9 = 0

u = 9

Substituting u = 9 in equation (iii) we get

-u + 3y – 3 = 0

-9 + 3y – 3 = 0

3y – 12 = 0

3y = 12

y = 12/3 = 4

Now,

\(\frac{x}{y}=9\) \(\frac{x}{4}=9\)x = 36

Therefore, the number of students in the class is 36.

**One says, “Give me 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?**

**Solution:**

**Solution:**

Le one friend have Rs *x* and the other friend have Rs *y* with them.

According to the question,

*x* + 100 = 2(*y* − 100)

*x* + 100 = 2*y* − 200

*x* − 2*y* = −300 …(i)

And, 6(*x* − 10) = (*y* + 10)

6*x* − 60 = *y* + 10

6*x* − *y* = 70 …(ii)

Multiplying equation (ii) by 2,

We get,

12*x* − 2*y* = 140 …(iii)

Subtracting equation (i) from (iii),

We get,

11*x* = 140 + 300

11*x* = 440

*x* = 40

Substituting x=40 in equation (i),

We get,

40 − 2*y* = −300

40 + 300 = 2*y*

2*y* = 340

*y* = 170

Therefore, one of the friends had Rs 40 and the other had Rs 170 with them respectively.