The age difference between Ani and Biju is 3 yrs.
Either Biju is 3 years older than Ani or Ani is 3 years older than Biju.
From both cases, we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.
Let the ages of Ani and Biju be A and B, respectively.
Therefore, the age of Dharam = 2 x A = 2A yrs.
And the age of Biju’s sister Cathy is $\dfrac{B}{2}$ yrs.
By using the information that is given,
Case (i)
When Ani is older than Biju by 3 yrs, then A – B = 3 …..(1)
2A − $\dfrac{B}{2}$ = 30
4A – B = 60 ….(2)
By subtracting the equation (1) from (2), we get;
3A = 60 – 3 = 57
A = $\dfrac{ 57}{3}$ = 19
Therefore, the age of Ani = 19 yrs
And the age of Biju is 19 – 3 = 16 yrs.
Case (ii)
When Biju is older than Ani,
B – A = 3 ….(1)
2A − $\dfrac{B}{2}$ = 30
4A – B = 60 ….(2)
Adding the equations (1) and (2), we get;
3A = 63
A = 21
Therefore, the age of Ani is 21 yrs
And the age of Biju is 21 + 3 = 24 yrs.
Let the capital amount with two friends be Rs. x and Rs. y, respectively.
As per the given,
x + 100 = 2(y − 100)…..(i)
And
6(x − 10) = (y + 10)….(ii)
Consider the equation (i),
x + 100 = 2(y − 100)
x + 100 = 2y − 200
x − 2y = −300…..(iii)
From equation (ii),
6x − 60 = y + 10
6x − y = 70…..(iv)
(iv) × 2 – (iii)
12x – 2y – (x – 2y) = 140 – (-300)
11x = 440
x = 40
Substituting x = 40 in equation (iii), we get;
40 – 2y = -300
2y = 340
y = 170
Therefore, the two friends had Rs. 40 and Rs. 170 with them.
Let the speed of the train be x km/hr and the time taken by the train to travel a distance be t hours, and the d km be the distance.
Speed of the train = $\dfrac{ Distance\:\: travelled \:\:by \:\:train }{ Time\:\: taken\:\: to \:\:travel\:\: that\:\: distance}$
x = $\dfrac{d}{t}$
d = xt …..(i)
Case 1:
When the speed of the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time.
(x + 10) = $\dfrac{d}{(t – 2)}$
(x + 10)(t – 2) = d
xt + 10t – 2x – 20 = d
d + 10t – 2x = 20 + d [From (i)]
10t – 2x = 20…..(ii)
Case 2:
When the train was slower by 10 km/h, it would have taken 3 hours more than the scheduled time.
So, (x – 10) = $\dfrac{d}{(t + 3)}$
(x – 10)(t + 3) = d
xt – 10t + 3x – 30 = d
d – 10t + 3x = 30 + d [From (i)]
-10t + 3x = 30…..(iii)
Adding (ii) and (iii), we get;
x = 50
Thus, the speed of the train is 50 km/h.
Substituting x = 50 in equation (ii), we get;
10t – 100 = 20
10t = 120
t = 12 hours
Distance travelled by train, d = xt
= 50 x 12
= 600 km
Hence, the distance covered by the train is 600 km.
Let x be the number of rows and y be the number of students in a row.
Total students in the class = Number of rows × Number of students in a row
= xy
Case 1:
Total number of students = (x − 1) (y + 3)
xy = (x − 1) (y + 3)
xy = xy − y + 3x − 3
3x − y − 3 = 0
3x − y = 3…..(i)
Case 2:
Total number of students = (x + 2) (y − 3)
xy = xy + 2y − 3x − 6
3x − 2y = −6…..(ii)
Subtracting equation (ii) from (i), we get;
(3x − y) − (3x − 2y) = 3 − (−6)
− y + 2y = 9
y = 9
Substituting y = 9 in equation (i), we get;
3x − 9 = 3
3x = 12
x = 4
Therefore, the total number of students in a class = xy = 4 × 9 = 36
Given,
$\angle$ C = 3 $\angle$ B = 2($\angle$ B + $\angle$ A)
3$\angle$ B = 2 $\angle$ A + 2 $\angle$ B
$\angle$ B = 2 $\angle$ A
2$\angle$ A – $\angle$ B= 0- – – – – – – – – – – – (i)
We know that the sum of a triangle’s interior angles is 180$^{°}$.
Thus, $\angle$ A +$\angle$ B + $\angle$ C = 180$^{°}$
$\angle$ A + $\angle$ B +3 $\angle$ B = 180$^{°}$
$\angle$ A + 4 $\angle$ B = 180$^{°}$– – – – – – – – – – – – – – -(ii)
Multiplying equation (i) by 4, we get;
8 $\angle$ A – 4$\angle$ B = 0- – – – – – – – – – – – (iii)
Adding equations (iii) and (ii), we get;
9 $\angle$ A = 180$^{°}$
$\angle$ A = 20$^{°}$
Using this in equation (ii), we get;
20$^{°}$ + 4$\angle$ B = 180$^{°}$
$\angle$ B = 40$^{°}$
And
$\angle$ C = 3$\angle$ B = 3 x 40 = 120$^{°}$
Therefore, $\angle$ A = 20$^{°}$, $\angle$ B = 40$^{°}$, and $\angle$ C = 120$^{°}$.
Given,
5x – y = 5
⇒ y = 5x – 5
Its solution table will be.
| X | 2 | 1 | 0 |
| Y | 5 | 0 | -5 |
Also given,3x – y = 3
y = 3x – 3
| X | 2 | 1 | 0 |
| Y | 3 | 0 | -3 |
The graphical representation of these lines will be as follows:
From the above graph, we can see that the coordinates of the vertices of the triangle formed by the lines and the y-axis are (1, 0), (0, -5) and (0, -3).
(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 = a$^{2}$ + b$^{2}$
(iv) (a – b)x + (a + b) y = a$^{2}$ – 2ab – b$^{2}$
(a + b)(x + y) = a$^{2}$+ b$^{2}$
(v) 152x – 378y = – 74
–378x + 152y = – 604
(i) px + qy = p – q……………(i)
qx – py = p + q……………….(ii)
Multiplying equation (i) by p and equation (ii) by q, we get;
p$^{2}$x + pqy = p$^{2}$ − pq ………… (iii)
q$^{2}$x − pqy = pq + q$^{2}$ ………… (iv)
Adding equations (iii) and (iv), we get;
p$^{2}$x + q$^{2}$ x = p$^{2}$ + q$^{2}$
(p$^{2}$ + q$^{2}$) x = p$^{2}$ + q$^{2}$
x =$\dfrac{ (p^{2} + q^{2})}{ (p^{2} + q^{2})}$ = 1
Substituting x = 1 in equation (i), we have;
p(1) + qy = p – q
qy = p – q – p
qy = -q
y = -1
(ii) ax + by= c…………………(i)
bx + ay = 1+ c…………..(ii)
Multiplying equation (i) by a and equation (ii) by b, we get;
a$^{2}$x + aby = ac ………………… (iii)
b$^{2}$x + aby = b + bc…………… (iv)
Subtracting equation (iv) from equation (iii),
(a$^{2}$ – b$^{2}$) x = ac − bc– b
x = $\dfrac{(ac − bc – b)}{ (a^{2} – b^{2})}$
x = $\dfrac{ c(a – b) –b }{ (a^{2} – b^{2})}$
From equation (i), we obtain
ax + by = c
$\dfrac{a[c(a − b) − b]}{ (a^{2} – b^{2}) }$+ by = c
$\dfrac{[ac(a−b)−ab]}{(a^{2}– b^{2}) }$+ by = c
by = c – $\dfrac{[ac(a − b) − ab]}{(a^{2}– b^{2})}$
by = $\dfrac{(a^{2}c – b^{2}c – a^{2}c + abc + ab)}{(a^{2} – b^{2})}$
by = $\dfrac{ [abc – b^{2}c + ab]}{ (a^{2} – b^{2})}$
by = $\dfrac{b(ac – bc + a)}{(a^{2}– b^{2})}$
y = $\dfrac{[c(a – b) + a]}{(a^{2} – b^{2})}$
(iii) $\dfrac{x}{a}$ – $\dfrac{y}{b}$ = 0
ax + by = a$^{2}$ + b$^{2}$
$\dfrac{x}{a}$ – $\dfrac{y}{b}$ = 0
⇒ bx − ay = 0 ……. (i)
And
ax + by = a$^{2}$ + b$^{2}$…….. (ii)
Multiplying equations (i) and (ii) by b and a, respectively, we get;
b$^{2}$x − aby = 0 …………… (iii)
a$^{2}$x + aby = a$^{3}$ + ab$^{2}$ …… (iv)
Adding equations (iii) and (iv), we get;
b$^{2}$x + a$^{2}$x = a$^{3}$ + ab$^{2}$
x(b$^{2}$ + a$^{2}$) = a(a$^{2}$ + b$^{2}$)
⇒ x = a
Substituting x = 1 in equation (i), we get;
b(a) − ay = 0
ab − ay = 0
ay = ab
⇒ y = b
(iv) (a – b)x + (a + b) y = a$^{2}$ – 2ab – b$^{2}$
(a + b)(x + y) = a$^{2}$ + b$^{2}$
(a + b) y + (a – b) x = a$^{2}$ − 2ab − b$^{2}$ …………… (i)
(x + y)(a + b) = a$^{2}$ + b$^{2}$
(a + b) y + (a + b) x = a$^{2}$ + b$^{2}$ ………………… (ii)
Subtracting equation (ii) from equation (i), we get;(a − b) x − (a + b) x = (a$^{2}$ − 2ab − b$^{2}$) − (a$^{2}$+ b$^{2}$)
x(a − b − a − b) = − 2ab − 2b$^{2}$
− 2bx = − 2b (a + b)
x = a + b
Substituting x = a + b in equation (i), we get;
y (a + b) + (a + b)(a − b) = a$^{2}$ − 2ab – b$^{2}$
a$^{2}$ − b$^{2}$ + y(a + b) = a$^{2}$− 2ab – b$^{2}$
(a + b)y = −2ab
y = $\dfrac{ -2ab}{(a + b)}$
(v) 152x – 378y = – 74
–378x + 152y = – 604
152x – 378y = – 74 ….(i)
–378x + 152y = – 604….(ii)
From equation (i),
152x + 74 = 378y
y = $\dfrac{(152x + 74)}{378}$
Or
y = $\dfrac{(76x + 37)}{189}$…..(iii)
Substituting the value of y in equation (ii), we get;
-378x + 152 $\dfrac{(76x + 37)}{189}$ = -604
(-378x)189 + [152(76x) + 152(37)] = (-604)(189)
-71442x + 11552x + 5624 = -114156
-59890x = -114156 – 5624 = -119780
x = $\dfrac{-119780}{-59890}$
x = 2
Substituting x = 2 in equation (iii), we get;
y = $\dfrac{76(2) + 37}{189}$
= $\dfrac{(152 + 37)}{189}$
= $\dfrac{189}{189}$
= 1
Therefore, x = 2 and y = 1
Given that ABCD is a cyclic quadrilateral.
As we know, the opposite angles of a cyclic quadrilateral are supplementary.
So,
$\angle$A + $\angle$C = 180
4y + 20 + (-4x) = 180
-4x + 4y = 160
⇒ -x + y = 40….(i)
And
$\angle$B + $\angle$D = 180
3y – 5 + (-7x + 5) = 180
⇒ -7x + 3y = 180…..(ii)
Equation (ii) – 3 × (i),
-7x + 3y – (-3x + 3y) = 180 – 120
-4x = 60
x = -15
Substituting x = -15 in equation (i), we get;
-(-15) + y = 40
y = 40 – 15 = 25
Therefore, x = -15 and y = 25.
Therefore,
$\angle$A = 4$\times$25+20=120$^{°}$
$\angle$B=3$\times$25-5=70$^{°}$
$\angle$C=-4$\times$(-15)=60$^{°}$
$\angle$D=-7$\times$(-15)+5=110$^{°}$