General Aptitude Test 4

Consider only one term (x₁ + y₁)

The number of items which have only single x = 1

Consider two terms (x₁ + y₁) (x₂ + y₂)

= x₁x₂ + x₁y₂ + y₁x₂ + y₁y₂

The number of items which have only single x = 2

Consider three terms (x₁ + y₁) (x₂ + y₂) (x₃ + y₃)

= (x₁x₂ + x₁y₂ + y₁x₂ + y₁y₂) (x₃ + y₃)

= x₁x₂x₃ + x₁x₂y₃ + x₁y₂x₃ + x₁y₂y₃ + y₁x₂x₃ + y₁x₂y₃ + y₁y₂x₃ + y₁y₂y₃

The number of items which have only single x = 3

In the same way,

Concept:

A common geometric progression with ‘r’ as common ratio has the form.

a, ar, ar²

Application:

Since a, b, c are in GP

a = a

b = ar

c = ar²

\(\frac{a-b}{b-c}=\frac{a-ar}{ar-a{{r}^{2}}}=\frac{a\left( 1-r \right)}{ar\left( 1-r \right)}=\frac{1}{r}\)

\(\frac{a-b}{b-c}=\frac{1}{r}\)

Given r = 2

\(\frac{a-b}{b-c}=\frac{1}{2}\)

According to the given information,

log(P) = (1/3)log(Q) = (1/7)log(R)

log(P) = (1/3)log(Q)

⇒ Q = P³

It is also given that,

log(P) = (1/7)log(R)

⇒ P⁷ = R

Substituting the value of Q in all options,

1) P² = Q³R² = P⁹R²

⇒ P^-7 = R²

2) P⁴ = Q³R⁴ = P⁹R⁴

⇒ P^-5 = R⁴

3) Q⁷ = P³R⁴

⇒ P²¹ = P³ R⁴

⇒ P⁹ = R²

4) Q³ = P²R

⇒ P⁹ = P²R

⇒ P⁷ = R

|4x - 7| = 5

4x – 7 = 5             and       – (4x - 7) = 5

⇒ 4x = 12                          4x – 7 = -5

X = 3                                 4x = 2

                                          X = 0.5

Now when x = 3

2|x| - |-x| = 2 |3| - |-3| = 3

When x = 0.5

2 |0.5| - |-0.5| = 0.5

∴ Ans is 3, 0.5.

For numbers from 700 to 1000, the digit at hundred's place can be 7, 8 or 9. So, three possibilities are there.

Now, for the digit at ten's place to be greater than the digit at unit's place:

If ten’s digit is 2, the possible numbers at unit’s place = 0, 1 (total numbers = 2)

If ten’s digit is 3, the possible numbers at unit’s place = 0, 1, 2 (total numbers = 3)

Similarly,

If ten’s digit is 7, the possible numbers at unit’s place = 0, 1, 2, 3, 4, 5, 6 (total numbers = 7)

Now, for the digit at ten's place to be greater than the digit at unit's place, the total numbers is equal to the value of ten's digit.

When the hundred’s place is 7, total possible numbers are = 1 + 2 + 3 + 4 + 5 + 6 = 21

When the hundred’s place is 8, total possible numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

When the hundred’s place is 8, total possible numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Given equations are

x² + y² = 1 … (1)

u² + v² = 1 … (2)

xu + yv = 0 … (3)

From equation (3), x = -yv/u

From equation (2), v² = 1 – u²

Substitute these values in equation (1),

\({\left( { - \frac{{yv}}{u}} \right)^2} + {y^2} = 1\)

⇒ y²v² + u²y² = u²

⇒ y² (1 - u²) + y² u² = u²

⇒ y = u … (4)

Similarly, x = v

Now, from equation (1)

x² + y² = 1 ⇒ x² + u² = 1

⇒ 1 – y² + 1 – v² = 1 ⇒ y² + v² = 1

Now, from equations (3) and (4)

xu + yv = 0

⇒ xy + uv = 0

Now, xv + yu = v (v) + (u) u = v² + u² = 1

The probability that a component manufactured by machine 1 is, P(M1) = 0.6

The probability that a component manufactured by machine 2 is, P(M2) = 0.4

The probability that a component is defective and manufactured by machine 1 is,

P(d/M1) = 0.02

The probability that a component is defective and manufactured by machine 1 is,

P(d/M2) = 0.03

Now, a randomly drawn autocomponent from the combined lot is found defective, the probability that it was manufactured by M2 is,

\(P\left( {M2/d} \right) = \frac{{P\left( {d/M2} \right)P\left( {M2} \right)}}{{P\left( {d/M1} \right)P\left( {M1} \right) + P\left( {d/M2} \right)P\left( {M2} \right)}}\)

\( = \frac{{0.03\left( {0.4} \right)}}{{0.02\left( {0.6} \right) + 0.03\left( {0.4} \right)}} = 0.5\)

To get maximum possible number, assume that no students have taken two subjects i.e. every student either taken one subject or three subjects.

Let x is the maximum number of students that have taken three subjects.

Total number students = 300

Total number of students that have taken only on subject = 300 – x

Number of students taken M600 = 100

Number of students taken only M600 = 100 – x

Number of students taken C600 = 200

Number of students taken only C600 = 200 – x

Number of students taken E600 = 60

Number of students taken only E600 = 60 – x

Now, 300 – x = 100 – x + 200 – x + 60 – x

According to the given information:

ax² + bx + c = 0

Let’s assume α and β are the roots of the equations.

α + β = -b/a       ---(1)

αβ = c/a             ---(2)

In the given equation,

𝑎𝑥² + 𝑏|𝑥| + 𝑐 = 0

If x > 0

The equation will be:

𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0

But, this equation has negative roots.

Which is not possible.

No roots exists.

If x < 0

The equation will be,

𝑎𝑥² – 𝑏𝑥 + 𝑐 = 0

And, α’ + β’ = b/a

α’β’ = c/a

On comparing with Equation (1) and (2),

α’ = -α

β’ = -β

It is given that α and β are negative. And, we assumed α’ and β’ to be negative also.

Again, this is not possible.