1. 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 differs by 30 years. Find the ages of Ani and Biju.
Solution:
The difference between the ages of Biju and Ani is 3 years.
Either Biju is 3 years older than Ani or Ani is 3 years older than Biju.
However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.
Let the age of Ani and Biju be x and y years respectively.
Therefore,
age of Ani’s father, Dharam = 2 × x = 2x years and the age of Biju’s sister Cathy y/2 years
By using the information given in the question,
Case (I):
When Ani is older than Biju by 3 years, x − y = 3 ………………….(1)
2x – y/2 = 30
4x – y = 60 ————(2)
Subtracting (1) from (2)
3x = 60 – 3 = 57
x = 57/3 = 19
Therefore, age of Ani = 19 years
And age of Biju = 19 − 3 = 16 years
Case (II):
When Biju is older than Ani, y − x = 3 ……………….(i)
2x – y/2 = 30
4x –y = 60 ———-(2)
Adding (1) and (2)
3x = 63
x = 21
Therefore, age of Ani = 21 years and
age of Bijju = 21 + 3 = 24 years
- 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:
Let those friends were having Rs x and y with them.
Using the information given in the question,
x + 100 = 2(y − 100) x + 100 = 2y – 200
or x − 2y = −300 ……………….(1)
6(x − 10) = (y + 10)
Or 6x − 60 = y + 10
Or 6x − y = 70 ………………………….(2)
Multiplying equation (2) by 2,
12x − 2y = 140———— (3)
Subtracting equation (1) from equation (3), we obtain
11x = 140 + 300
11x = 440
x = 40
Using this in equation (1),
40 − 2y = −300
40 + 300 = 2y
2y = 340
y = 170
Therefore, those friends had Rs 40 and Rs 170 with them respectively.
- 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:
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 was d km. We know that,
Speed = Distance travelled/time taken to travel that distance
x = d/t
Or, d = xt ………………………………… (1)
Using the information given in the question,
(x+10) = d/(t-2)
(x+10)(t-2) = d
xt – 2x + 10t – 20 = d
By using equation (1), we get
-2x + 10t = 20 ———-(2)
(x-10) = d/(t+3)
(x-10)(t+3) = d
xt – 10t + 3x – 30 = d
By using equation (1), we obtain
3x − 10t = 30 ————(3)
Adding equations (2) and (3), we obtain
x = 50
Using equation (2), we obtain
(−2) × (50) + 10t = 20
−100 + 10t = 20 10t = 120 t = 12 hours
From equation (1), we obtain
Distance to travel = d = xt
= 50 × 12
= 600 km
Hence, the distance covered by the train is 600 km.
- 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:
Let the number of rows be x and number of students in a row be y.
Total students of the class = Number of rows × Number of students in a row
= xy
Data given:
Condition 1:
Total number of students = (x − 1) (y + 3)
xy = (x − 1) (y + 3) = xy − y + 3x − 3
3x − y − 3 = 0——–(1)
Condition 2:
Total number of students = (x + 2) (y − 3)
xy = xy + 2y − 3x − 6
3x − 2y = −6 ……………………………….(2)
Subtracting equation (2) from (1),
(3x − y) − (3x − 2y) = 3 − (−6)
− y + 2y = 3 + 6 y = 9
By using equation (i), we obtain
3x − 9 = 3
3x = 9 + 3 = 12
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Number of total students in a class = xy = 4 × 9 = 36
- In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.
Solution:
Given that,
∠C = 3∠B = 2(∠A + ∠B)
3∠B = 2(∠A + ∠B)
3∠B = 2∠A + 2∠B
∠B = 2∠A
2 ∠A − ∠B = 0 ………………………………… (1)
We know that the sum of the measures of all angles of a triangle is 180°. Therefore,
∠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 × 40° = 120°
Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.
- 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:
5x − y = 5
Or, y = 5x − 5
The solution table will be as follows.
x | 0 | 1 | 2 |
y | -5 | 0 | 5 |
3x − y = 3
Or, y = 3x − 3
The solution table will be as follows.
x | 0 | 1 | 2 |
y | -3 | 0 | 3 |
The graphical representation is,
It can be observed that the required triangle is ΔABC formed by these lines and y-axis.
The coordinates of vertices are A (1, 0), B (0, − 3), C (0, − 5).
- 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). x/a – y/b = 0, ax + by = a2 + b2
(iv). (a − b) x + (a + b) y = a2− 2ab – b2
(a + b) (x + y) = a2 + b2
(v) 152x − 378y = − 74
− 378x + 152y = − 604
Solution:
(i) px + qy = p − q ……………………..… (1)
qx − py = p + q ……………………..… (2)
Multiplying equation (1) by p and equation (2) by q,
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
x = (p2+ q2)/( p2+ q2) = 1
From 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,
a2x + aby = ac ………………… (3)
b2x + aby = b + bc …………… (4)
Subtracting equation (4) from equation (3),
(a2 − b2) x = ac − bc − b
x = [c(a-b)-b]/(a2-b2)
From equation (1),
ax + by = c
a{[c(a-b)-b]/(a2-b2)}+by = c
[ac(a-b)-b]/(a2-b2)]+by = c
by = c – [ac(a-b)-b]/(a2-b2)]
by = (a2c-b2c-a2c+abc+ab)/(a2-b2)
by = [abc-b2c+ab]/(a2-b2)
y= [c(a-b)+a]/(a2-b2)
(iii) x/a – y/b = 0
Or, bx − ay = 0 ……………. (1)
ax + by = a2 + b2 ………….. (2)
Multiplying equation (1) and (2) by b and a respectively, we obtain
b2 x − aby = 0 …………….…… (3)
a2 x + aby = a3 + ab2 ……………… (4)
Adding equations (3) and (4), we obtain
b2x + a2 x = a3 + ab2
x (b2 + a2) = a (a2 + b2) x = a
By using (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 = a2 + b2 ………………….… (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
Using 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 = -2ab/(a+b)
(v) 152x – 378y = -74
76x – 189y = -37
x = (189y-37)/76————(1)
-378x + 152y = -604
-189x + 76y = -302 ————(2)
Substituting the value of x in equation (2),
-189[(189y-37)/76] + 76y = -302
− (189)2 y + 189 × 37 + (76)2 y = − 302 × 76
189 × 37 + 302 × 76 = (189)2 y − (76)2 y
6993 + 22952 = (189 − 76) (189 + 76) y
29945 = (113) (265) y
y = 1
From equation (1),
x = 189(1)-37/76
x = 189-37/76 = 152/76
x = 2
- ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.
Solution:
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Therefore, ∠A + ∠C = 180
4y + 20 − 4x = 180
− 4x + 4y = 160
x − y = − 40 ………………………(i)
Also, ∠B + ∠D = 180
3y − 5 − 7x + 5 = 180
− 7x + 3y = 180 …………………..(ii)
Multiplying equation (i) by 3,
3x − 3y = − 120 ……………………(iii)
Adding equations (ii) and (iii),
− 7x + 3x = 180 − 120
− 4x = 60
x = −15
By using equation (i),
x − y = − 40
−15 − y = − 40
y = −15 + 40 = 25
∠A = 4y + 20 = 4(25) + 20 = 120°
∠B = 3y − 5 = 3(25) − 5 = 70°
∠C = − 4x = − 4(− 15) = 60°
∠D = − 7x + 5 = − 7(−15) + 5 = 110°
OMG math. When I was high school math was tough but I passed. When I got to college my grades in math, college algebra, were so low terrifyingly low, I dropped the class a least twice. I sought help from tutors on campus but nothing worked. Please don’t make me take math.
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Oh Ok. Whew!!! Mini Heart attack suddenly vanished.
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Well 😀 that made me lighter 😀 Live long 😀
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Will Do
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