Question

Each packet of SOAP costs Rs. 10. Inside each packet is a gift coupon labelled with one of the letters S, O, A and P. If a customer submits four such coupons that make up the word SOAP, the customer gets a free SOAP packets. Ms. X kept buying packet after packet of SOAP till she could get one set of coupons that formed the word SOAP. How many coupons with label P did she get in the above process?

A. The last label obtained by her was S and the total amount spent was Rs. 210.

B. The total number of vowels obtained was 18.

• A. The questions can be answered by one of the statements, but not by the other
• B. The question can be answered using either of the two statements alone.
• C. The questions can be answered using both the statements together, but cannot be answered using either statement alone.
• D. The question cannot be answered even by using both the statements together.

Hint:

Ms. X stops the moment the last missing letter shows up, so that letter appears exactly once. Count the total coupons from the money spent, then see what is left for P.

Answer 1

The Correct Option is C

Step 1: Understand the buying process.
Ms. X buys packets one after another. Each packet has one coupon, and the coupon is S, O, A or P.
She stops the moment she holds at least one S, one O, one A and one P. So the coupon that completes the set is the last coupon she gets, and that coupon is the only one of its letter in her whole collection.
We are asked for the number of P coupons she collected.

Step 2: Read statement A alone.
Each packet costs Rs. 10, so Rs. 210 means \( 210 \div 10 = 21 \) packets and 21 coupons in all.
The last label was S. The set was completed at that moment, so S appeared only once (the last coupon) and the other 20 coupons are O, A or P, each present at least once.
The 20 coupons can be split in many ways, for example 10 O, 9 A, 1 P or 8 O, 6 A, 6 P. The number of P coupons is not fixed. So A alone is not enough.

Step 3: Read statement B alone.
B tells us only that the number of vowels (O and A together) is 18. It gives no total number of coupons and nothing about which letter came last.
P can be any positive number, because the number of packets bought is unknown. So B alone is not enough.

Step 4: Combine A and B.
From A, the total is 21 coupons and S is the single last coupon, so \( n(S) = 1 \).
From B, \( n(O) + n(A) = 18 \).
The four counts add up to 21:
\[ n(S) + n(O) + n(A) + n(P) = 21 \]
\[ 1 + 18 + n(P) = 21 \Rightarrow n(P) = 2 \]
This is a valid case, since O and A can share 18 coupons with each at least 1, and P = 2 is at least 1.

Step 5: Match with the options.
Option (1) is wrong because neither statement works alone. Option (2) is wrong for the same reason. Option (4) is wrong because the two statements together do give P = 2. Only option (3) fits.

Final Answer:
Both statements are needed together, and neither works alone. P was obtained exactly 2 times, so the answer is option 3. \[ \boxed{3} \]