8th Grade > Mathematics
LINEAR EQUATIONS IN ONE VARIABLE MCQs
:
x+55x+8 = 1138
38x + 190 = 55x + 88
55x -38x = 190 - 88
17x = 102
x = 10217 = 6
:
A
53x = 2527
By cross multiplication, we get
5 × 27 = 3x×25
⇒3x=5×2725
⇒x=5×273×25
⇒x = 95
= 1.8, which lies between 1 and 2.
:
C and D
Let the age of Sunil be 3x.
Then, the age of Sunil's father would be 7x.
Four years later,
Sunil's age = 3x+4 and
Sunil's father's age = 7x+4.
Then, the sum of their ages four years later =3x+4+7x+4=10x+8.
Given that, 10x+8=78.
⇒10x=70 ⇒x=7
∴ Present age of Sunil = 3 × 7
= 21 years
Present age of Sunil's father = 7 × 7
= 49 years
:
B
Given,
x+56−x+19=x+34
Multiplying both sides by 36 (LCM of 4, 6 and 9), we get,
(x+56×36)−(x+19×36)=(x+34×36)
⇒6(x+5)–4(x+1)=9(x+3)
⇒6x+30–4x−4=9x+27
⇒2x+26=9x+27
Transposing 9x from RHS to LHS, and 26 from LHS to RHS, we get,
2x−9x=27−26
⇒−7x=1
Dividing both sides by -7, we get,
x=−17
:
A
Linear dependency signifies that "a","b"and "c" are linearly related, which means that if one of them can be found, the other two can be found.
Hence, the given equation can be reduced to a linear equation if the exact relation between "a","b"and "c" is given.
For a better understanding, let us express the dependencies as equations.
Since b is linearly dependent on a, we can write b = ka + p, where k and p are some constants. Note that in this equation, b and a are the only variables, whereas k and p are known.
Similarly, a can be written as a = gc + u, where g and u are constants (known values).
Substituting these values of a and b in the given equation, we get,
3a + 2b + 4c = 5
3(gc + u) + 2(ka + p) + 4c = 5
3gc + 3u + 2p + 2k(gc + u) + 4c = 5
(3g + 2kg + 4)c + (3u + 2p + 2ku - 5) = 0.
Note that in the final equation, the only c is unknown, the others are all constants and hence is a linear equation in one variable.
:
A
x+94+2x+35=10 is the given equation.
We can check if x=11 is a solution by substituting this value in the LHS to check whether it is equal to RHS or not
(11+9)4+(22+3)5
= 5 + 5
= 10
Thus we see that x=11 makes LHS = RHS and is the solution of the given equation.
:
B
A variable is an unknown quantity whose value is not constant.
In the given equation 3a + b + 6ab = 0, "a" and "b" are the variables. "ab" is not considered as a separate variable because, if "a" and "b" are individually found, "ab" can be found. Hence, there are 2 variables in the equation and not 3.
:
A
3x+1 = 65x−1
On cross multiplication, we get
⇒3(5x−1)=6(x+1)⇒ 15x−3=6x+6⇒ 9x=9⇒ x=1
:
A
Given that the denomination notes of 1000, 500, 100 are in the ratio 2:3:5.
Number of 1000 denomination notes with Pawan = 2x
Therefore, amount = ₹ 2000x
Number of 500 denomination notes with Pawan = 3x
Therefore, amount = ₹ 1500x
Number of 100 denomination notes with Pawan = 5x
Therefore, amount = ₹ 500x
Total amount = ₹ 5,00,000 (Given)
2000x+1500x+500x=₹5,00,000
4000x=5,00,000
x=5,00,0004,000=125
Therefore, number of 1000 denomination notes = 2x=2×125
= 250 notes.
:
C
Let the number be x.
'Subract ¼ from the number' is mathematically written as 'x - ¼'.
Multiplying the above obtained result by ½ is mathematically shown as:
'(x - ¼) × ½'.
According to the question,
(x−14)×12=14
⇒ x−14=2×14
(by multiplying both the sides by 2)
⇒ x−14=12
⇒ x = 12+14
(by transposing ¼ from LHS to the RHS)
⇒ x = 34
So, the required number is 34.