Quantitative Ability Test - 5

The number of ways to select r things out of n given things wherein r ≤ n is given by: 

Given: There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Here, we have to find in how many ways a student can fail in the examination.

In order to fail in the examination the student can fail in 1 or 2 or 3 or 4 or 5 subjects out of the 5 subjects in each case.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects = ⁵C₁

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects = ⁵C₂

Case 3 : Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects = ⁵C₃

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects = ⁵C₄

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. = ⁵C₅

∴ Total number of ways in which a student can fail in an examination

= ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = 31

Hence, the correct option is (A).

Given

A certain sum is given on Compound Interest at some rate. The sum amounts to Rs. 7350 and Rs. 8575 at the end of 2 years and 3 years respectively.

Interest for 1 year is the same whether it's Simple Interest or the compound Interest.

Now, interest of third year

= 8575 − 7350 = 1225

Principal for this interest is 7350 if Compound Interest is taken.

If 7350 is the principal then interest = 1225

if 100 is the principal interest

Let the unit digit of the number be y and the tens digit of this number be x.

So, the number is 10x+y

Then, according to the question,

⇒ xy = 12⋯ (1)

Also, the number after interchanging the digits = 10y + x

⇒10x + y + 36 =10y + x

⇒ 9x − 9y = −36

⇒ x − y = −4

⇒ x = (y − 4) ⋯(2)

On substituting the value of x in equation (1), we get,

y⋅(y − 4) = 12

⇒ y² − 4y − 12 = 0

⇒ y² − 6y + 2y − 12 = 0

⇒ y(y − 6) + 2(y − 6) = 0

⇒ (y − 6)(y + 2) = 0

⇒ y = 6 or −2

But the unit digit of the two-digit number can't be negative.

⇒ y = 6

On substituting this value in equation (2), we get,

x = 6 − 4

⇒ x = 2

⇒ 10x + y = 10 × 2 + 6 = 26

∴ The number is 26.

Hence, the correct option is (C).

It is given that, the coordinates on which the line joins are (1, 2, −3) and (−2, 3, 1).

Let the direction cosines be l, m, n.

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two point which joins the line.

∴ x₁ = 1, y₁ = 2, z₁ = −3

and x₂ = −2, y₂ = 3, z₂ = 1

Now, we know that,

Probability of getting a queen of clubs = (1 / 52).

Probability of getting a king of hearts = (1 / 52).

The probability of getting either a queen of clubs or a king of hearts

= (1 / 52) + (1 / 52) = (2 / 52) = (1 / 26).