MCQ: Linear Equations - 2 - SSC CGL MCQ

Let us assume the abscissa coordinate be x and Ordinate be y.
According to question,
ordinate four times its abscissa
y = 4x
⇒ y - 4x = 0

According to question,
40 notebooks at the rate of Rs. 18 per notebook and 55 pencils at the rate of Rs. 8 per pencil.
Amount paid = Rs. (40 x 18 + 55 x 8)
Amount paid = Rs. (720 + 440)
Amount paid = Rs. 1160

Let us assume the original fraction be x/y.
According to question,
200x/100 / 200y/100 = 14/5
2x/2y = 14/5
⇒x/y = 14/5

Let the ten's digit be x. Then, number = 10x + 3 and sum of digits = ( x + 3 )
So ,(x+3) = 1/7(10x + 3)
⇔ 7x + 21 = 10x + 3
⇔ 3x = 18
⇔ x=6
Hence, the number = 10x + 3 = (10 x 6) + 3 = 63.

Let the numbers be x and y.
According to question,
The ratio of two numbers is 4:7.
⇒ x/y = 4/7
⇒ 7x = 4y
∴ 7x - 4y = 0 ........................(1)
Again According to question,
If each of those numbers increased by 30, their ratio will become 5:8
(x + 30) / (y + 30) = 5/8
⇒ 8(x + 30) = 5(y + 30)
⇒ 8x + 240 = 5y + 150
⇒ 8x - 5y = -90 ..........................(2)
Multiply 5 with equation (1) , we will get.
35x - 20y = 0 ..................(3)
Multiply 4 with equation (2), we will get.
32x - 20y = -360 ............(4)
Subtracts the Equation (4) from Equation (3). we will get,
35x - 20y - (32x - 20y) = 0 - (-360)
⇒ 35x - 20y - 32x + 20y = 360
⇒ 3x = 360
⇒ x = 120
Put the value of x in equation (1) to get the value of y.
7x - 4y = 0
⇒ 7(120) - 4y = 0
⇒ 840 - 4y = 0
⇒ 4y = 840
⇒ y = 210
∴ Average of the numbers = (x + y)/2
put the vale of x and y.
∴ Average of the numbers = (120 + 210)/2
∴ Average of the numbers = 330/2
∴ Average of the numbers = 165

Suppose that the Ravi present age is A years.
According to the given question,
A/4 + A/5 + A/3 + 13 = A
⇒ (15A + 12A + 20A)/60 = A - 13
⇒ 47A = 60A - 780
⇒60A - 47A = 780
⇒13A = 780
∴ A = 780/13 = 60 years

Let us assume the digits of the original number are unit's digit a and ten's digit b.
The Original Number will be 10a + b.
After interchanging the digits the new number will be 10b + a.
According to question,
The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54.
New Number = Original Number - 54
10b + a = 10a + b - 54
⇒ 10b + a - 10a - b = -54
⇒ 9b - 9a = -54
⇒ a - b = 6....................................(1)
Again according to question,
Sum of the digits of original number = 10
a + b = 10..................................................(2)
Add the equation (1) and (2), we will get
a - b + a + b = 10 + 6
2a = 16
a = 8
Put the value of a in Equation (2) , we will get
8 + b = 10
b = 10 - 8
b = 2
Put the value of a and b for original number, we will get
10a + b = 10 x 8 + 2 = 80 + 2 = 82

Let us assume the present age of father = F year and Son's present age = S years
According to question, 5 years ago,
Father's age = F - 5 and Son's age = S - 5.
According to the question,
The age of the father 5 years ago was 5 times the age of his son.
F - 5 = 5(S - 5)
F - 5 = 5S - 25.....................(1)
At present the father's age is 3 times that of his son.
F = 3S.................................(2)
Put the value of F from equation (2) in equation (1), we will get
⇒ 3S - 5 = 5S - 25
⇒ 25 - 5 = 5S - 3S
⇒ 20 = 2S
⇒ 10 = S
⇒ S = 10.
Put the value of S in Equation (2). we will get,
F = 3S = 3 x 10 = 30
So the present Age of Father = 30.

Let us assume N be the given number.
According to the question,
Sum of third , fourth and fifth part of a number exceeds half of the number by 34.
N/3 + N/4 + N/5 - 34 = N/2
N/3 + N/4 + N/5 - N/2 = 34
(20N + 15N + 12N - 30N)/60 = 34
17N/60 = 34
N = 34 x 60/17
N = 2 x 60
N = 120

For infinite solution
a₁/a₂ = b₁/₂ = c₁/c₂
⇒ K/12 = 3/K = (-K + 3)/ -K
⇒ K/12 = 3/K
⇒ K² = 36
∴ K = √36 = 6

Given, (x/4) + (y/3) = 10/24
⇒ (3x + 4y)/12 = 10/24
∴ 3x + 4y = 5 ...(i)
and (x/2) + y = 1
x + 2y = 2 ...(ii)
On multiplying Eq. (ii) by 2 and subtracting from Eq. (i).
3x + 4y = 5
2x + 4y = 4
- - -
--------------------------
x = 1
On putting the value of x in Eq. (ii), we get
x + 2y = 2
⇒ 1 + 2y = 2
⇒ y = 1/2

∴ x + y = 1 + 1/2 = 3/2

Let the number be x and y.
Then, according to the question,
x + y = 15 ...(1)
x - y = 3 ...(ii)
on adding Eqs. (i) and (ii), we get
2x = 18
⇒ x = 9
On putting the value of in Eq. (i), we get
y = 6
∴ Product = xy = 54

6x + 1 > 7 - 4x
⇒ x > 3/5
∴ 3/5 < x ≤ 2

Let hens = H, goats = G
According to the question,
H + G = 90 ...(i)
2H + 4G = 248 ...(ii)
On multiplying Eq. (i) by 2 and subtracting from Eq. (ii) , we get
2H + 2G = 180
2H + 4G = 248
-------------------
-2G = -68
∴ G = 34

3x + 7y = 75 ..(i)
5x - 5y = 25
∴ x - y = 5 ...(ii)
On multiplying Eq. (ii) by 7 and adding to Eq. (i), we get
3x + 7y = 75
7x - 7y = 35
------------------
10x = 110
x = 11
∴ Putting the value of x in Eq. (ii), we get
11 - y = 5
∴ y = 6
∴ x + y = 6 + 11 = 17