Page No 3.111
Q.3. In a rectangle, if the length is increased by 3 meters and breadth is decreased by 4 meters, the area of the rectangle is reduced by 67 square meters. If length is reduced by 1 meter and breadth is increased by 4 meters, the area is increased by 89 Sq. meters. Find the dimensions of the rectangle.
Ans. Let the length and breadth of the rectangle be x and y units respectively
Then, area of rectangle = xy square units
If the length is increased by 3 meters and breath is reduced each by 4 square meters the area is reduced by 67 square units
Therefore,
xy - 67 = (x + 3) (y - 4)
xy - 67 = xy + 3y - 4x - 12
4x - 3y - 67 + 12 = 0
4x - 3y - 55 = 0 ...(i)
Then the length is reduced by 1 meter and breadth is increased by 4 meter then the area is increased by 89 square units
Therefore, 0 = 4x - y - 93 ...(ii)
Thus, we get the following system of linear equation
4x - 3y - 55 = 0
4x - y - 93 = 0
By using cross multiplication we have
x = 28
and
y = 19
Hence, the length of rectangle is 28 meter,
The breath of rectangle is 19 meter.
Q.4. The incomes 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.
Ans. Let the income of X be Rs 8x and the income of Y be Rs 7x further let the expenditure of X be 19y and the expenditure of Y be 16y respectively then,
Saving of x = 8x - 19y
Saving of Y = 7x - 16y
8x - 19y = 1250
7x - 16y = 1250
8x - 19y - 1250 = 0 ...(i)
7x - 16y - 1250 = 0 ...(ii)
Solving equation (i) and (ii) by cross- multiplication, we have
The monthly income of X = 8x
= 8 x 750
= 6000
The monthly income of Y = 7x
= 7 x 750
= 5250
Hence the monthly income of X is Rs. 6000
The monthly income of Y is Rs. 5250
Q.5. 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?
Ans. Let the money with A be Rs x and the money with B be Rs y.
If A gives Rs 30 to B, Then B will have twice the money left with A, According to the condition we have,
y + 30 = 2(x - 30)
y + 30 = 2x - 60
0 = 2x - y - 60 - 30
0 = 2x - y - 90 ...(i)
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 - 30
x - 3y + 10 + 30 = 0
x - 3y + 40 = 0 ...(ii)
By multiplying equation (ii) with 2 we get, 2x - 6y + 80 = 0
By subtracting (ii) from (i) we get,
By substituting y = 34 in equation (i) we get
x = 62
Hence the money with A be Rs. 62 and the money with B be Rs. 34
Q.6. ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y − 5)°, ∠C = (−4x)° and ∠D = (7x + 5)°. Find the four angles.
Ans. We know that the sum of the opposite angles of cyclic quadrilateral is 180° in the cyclic quadrilateral ABCD angles A and C and angles B and D pairs of opposite angles
Therefore
∠A + C = 180° and ∠B + ∠D = 180°
∠A + C = 180°
By substituting ∠A = (4y + 20)° and ∠C = (- 4x)° we get
4y + 20 - 4x = 180°
- 4x + 4y + 20 = 180°
- 4x + 4y = 180° - 20
- 4x + 4y = 160°
4x - 4y = - 160°
Divide both sides of equation by 4 we get
x - y = - 40°
x - y + 40° = 0 ...(i)
∠B + ∠D = 180°
By substituting ∠B = (3y - 5)° and ∠D = (7x + 5)° we get
3y - 5 + 7x + 5 = 180°
7x + 3y = 180
7x + 3y - 180 = 0 ...(ii)
By multiplying equation (i) by 3 we get
3x - 3y + 120° = 0 ...(iii)
By subtracting equation (iii) from (ii) we get
By substituting x = 6° in equation (i) we get
x - y + 40° = 0
6 - y + 40 = 0
- 1y = - 40 - 6
- 1y = - 46
The angles of a cyclic quadrilateral are
∠A = 4y + 20
= 4 x 46 + 20
= 184 + 20
= 204°
∠B = 3y - 5
= 3 x 46 - 5
= 138 - 5
= 133°
∠C = - 4x°
= -4(6)
= - 24°
∠D = 7x + 5
= 7 x 6 + 5
= 42 + 5
= 47°
Hence the angles of quadrilateral are ∠A = 204°,∠B = 133°, ∠C = - 24°, ∠D = 47°
Q.7. 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?
Ans. A man can alone finish the work in days and one boy alone can finish it in y days then
One mans one days work = 1/x
One boys one days work = 1/y
2 men one day work = 2/x
7 boys one day work = 7/y
Since 2 men and 7 boys can finish the work in 4 days
Again 4 men and 4 boys can finish the work in 3 days
Putting in equation (i) and (ii) we get
8u + 28v = 1
12u + 12v = 1
8u + 28v - 1 ... (iii)
12u + 12v - 1 ...(iv)
By using cross multiplication we have
Now,
and
Hence, one man alone can finish the work in 15 days and one boy alone can finish the work in 60 days.
Q.8. In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°. Find the three angles.
Ans. Let ∠A = x°, ∠B = (3x − 2)°, ∠C = y° and
∠C−∠B = 9°
⇒ ∠C = 9°+∠B
⇒∠C = 9°+3x°− 2°
⇒∠C = 7°+3x°
Substitute ∠C = y° in above equation we get ,
y° = 7° + 3x°
∠A + ∠B + ∠C = 180°
⇒ x°+ (3x°−2°)+(7°+3x°) = 180°
⇒ 7x°+5°=180°
⇒ 7x°=180°−5°=175°
∠A = x° = 25°
∠B = (3x−2)° = 3(25°)−2°=73°
∠C = (7°+3x°) = 7°+3(25)°=82°
∠A=25°, ∠B=73°, ∠C = 82°
Hence, the answer.
Q.9. In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x − 5)°. Find the four angles.
Ans. We know that the sum of the opposite angles of cyclic quadrilateral is 180°. In the cyclic quadrilateral ABCD, angles A and C and angles B and D pairs of opposite angles
Therefore ∠A + ∠C = 180° and ∠B + ∠D = 180°
Taking ∠A + ∠C = 180°
By substituting ∠A = (2x + 4)° and ∠C = (2y + 10)°, we get
2x + 4 + 2y + 10 = 180
2x + 2y + 14 = 180°
2x + 2y = 180° - 14°
2x + 2y = 166 ...(i)
Taking ∠B + ∠D = 180°
By substituting ∠B = (y + 3)° and ∠D (4x + 5)° we get
y + 3 + 4x - 5 = 180°
4x + y - 5 + 3 = 180°
4x + y - 2 = 180°
4x + y = 180° + 2°
4x + y = 182° ...(ii)
By multiplying equation (ii) by 2 we get 8x + 2y = 364 ...(iii)
By subtracting equation (iii) from (i) we get
By substituting x = 33° in equation (ii) we get
4x + y = 182
4 x 33 + y = 182
132 + y = 182
y = 182 - 132
y = 50
The angles of a cyclic quadrilateral are
∠A = 2x + 4
= 2x 33 + 4
= 66 + 4
= 70°
∠B = y + 3
= 50 + 3
= 53°
∠C = 2y + 10°
= 2 x 50 + 10
= 100 + 10
= 110°
∠D = 4x - 5
= 4 x 33 - 5
= 132 - 5
= 127°
Hence, the angles of cyclic quadrilateral ABCD are ∠A = 70°, ∠B = 53°, ∠C = 110°, ∠D = 127°
Q.10. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awanded for each correct answer and 2 marks been deducted for each incorrect answer, the Yash would have scored 50 marks. How many question were there in the test?
Ans. Let take right answer will beand wrong answer will be y.
Hence total number of questions will be x + y ...(i)
If yash scored 40 marks in atleast getting 3 marks for each right answer and losing 1 mark for each wrong answer then
3x - 1y = 40 ...(ii)
If 4 marks awarded for each right answer and 2 marks deduced for each wrong answer the he scored 50 marks
4x - 2y = 50 ...(iii)
By multiplying equation (i) by 2 we get
6x - 2y = 80 ...(iv)
By subtracting (iii) from (iv) we get
Putting x = 15 in equation (ii) we have
3x - 1y = 40
3 x 15 - 1y = 40
45 - 1y = 40
- 1y = 40 - 45
Total number question will be
= x + y
= 15 + 5
= 20
Hence, the total number of question is 20
Q.11. In a ∆ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y − 5x = 30, prove that the triangle is right angled.
Ans. We have to prove that the triangle is right
Given ∠A = x°, ∠B = 3x° and ∠C = y°
Sum of three angles in triangle are ∠A + ∠B + ∠C = 180°
∠A + ∠B + ∠C = 180°
x + 3x + y = 180
4x + y = 180 ...(i)
By solving 4x + y = 180 with 3y - 5x = 30 we get,
4x + y = 180
- 5x + 3y = 30 ...(ii)
Multiplying equation (i) by 3 we get
12x + 3y = 540 ...(iii)
Subtracting equation (ii) from (iii)
x = 30°
Substituting x = 30° in equation (i) we get
4x + y = 180
4 x 30 + y = 180
120 + y = 180
y = 180 - 120
y = 60°
Angles ∠A, ∠B and ∠C are
∠A = x°
= 30°
∠B = 3x°
= 3 x 30°
= 90°
∠C = y°
= 60°
A right angled triangle is a triangle in which one side should has a right angle that is 90° in it.
Hence, ∠B = 90° The triangle ABC is right angled
Q.12. 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 the journey of 20 km, the charge paid is Rs 145. What will a person have to pay for travelling a distance of 30 km?
Ans. Let the fixed charges of car be Rs. x per km and the running charges be Rs. y km/hr
According to the given condition we have
x + 12y = 89 ...(i)
x + 20y = 145 ....(ii)
Putting y = 7 in equation (i) we get
x + 12y = 89
x + 12 x 7 = 89
x + 84 = 89
x = 89 - 84
x = 5
Therefore, Total charges for travelling distance of 30 km
= x + 30
= 5 + 30 x 7
= 5 + 210
= Rs. 215
Hence, A person have to pay Rs. 215 for travelling a distance of 30 km.
Page No 3.112
Q.13. 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 foods for 20 days, he has to pay Rs 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.
Ans. Let the fixed charges of hostel be Rs. x and the cost of food charges be Rs. y per day
According to the given condition we have,
x + 20y = 1000 ...(i)
x + 26 y = 1180 ...(ii)
Subtracting equation (ii) from equation (i) we get
Putting y = 30 in equation (i) we get
x + 20y = 1000
x + 20 x 30 = 1000
x + 600 = 1000
x = 1000 - 600
x = 400
Hence, the fixed charges of hostel is Rs 400.
The cost of food per day is Rs 30.
in equation (i) and (ii) we get
