Directions: Below question is followed by two statements labelled I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below: What is last two non-zero digits of nCr I. n = r + 2 II. r = 25% of (30

Directions: Below question is followed by two statements labelled I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below:

What is last two non-zero digits of ⁿC_r

I. n = r + 2

II. r = 25% of (30 - 40% of 25)

A Statement I alone is sufficient to answer the question.

B Statement II alone is sufficient to answer the question.

C Statement I and Statement II together are sufficient, but neither of the two alone is sufficient to answer the question.

D Either Statement I or Statement I alone is sufficient to answer the question.

E Neither Statement I nor Statement II is sufficient to answer the question.

Solution

Correct Answer: Option C

Given, n = r + 2

⇒ⁿC_r= ^(r+2)C_r = \(\frac{{\left( {r + 2} \right)!}}{{\left( {r!} \right) \times \left( {2!} \right)}} = \frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2}\)

No comprehensive answer can be derived from this expression.

Hence, statement I is insufficient to solve the question.

From statement II:

r = 25% of (30 - 40% of 25)

We use BODMAS to solve certain parts of the question to get a simplified form.

BODMAS stands for:

B – Brackets

O – Of (this simply stands for multiplication)

D – Division

M – Multiplication

A – Addition

S – Subtraction

The above is the standard order in which a given question is simplified.

On the RHS, we have

25% of (30 - 40% of 25)

As per the BODMAS, we solve the expression in the brackets first.

⇒30 - 40% of 25

Even within the bracket, we found an “OF”. Since there are no other brackets inside the bracket, we solve this part first.

40% of 25 = \(\frac{{40}}{{100}} \times 25 = 10\)

⇒ 30 – 10 = 20

Hence,

25% of 20 = \(\frac{{25}}{{100}} \times 20 = 5\;\)

Hence, r = 5.

Although the value of r has been found, the last two non-zero digits have not been found.

Hence, statement II is insufficient to solve this question.

Combining the two by substituting the value of r in \(\frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2},\)

We have \(\frac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2} = \frac{{6 \times 7}}{2} = 21\)

Hence, the two non-zero digits are 1 and 2.

Hence, both the statements combined can give you the solution.

Solution

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