Coupon B: A 15% discount on a minimum spend of Rs. 500 in one transaction, up to a maximum discount of Rs. 300.
Coupon C: A at discount of Rs. 100 on a minimum spend of Rs. 600 in one transaction.
Coupon D: A 10% discount on a minimum spend of Rs. 250 in one transaction, up to a maximum discount of Rs. 100.
Coupon E: A at discount of Rs. 50 on a minimum spend of Rs. 200 in one transaction.
The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction. Four customers used four different discount coupons for their respective transactions in such a way that they obtained a total discount of Rs. 710. Which discount coupon was not used?
E
Step-by-Step Solution
Key idea: This is a combinatorial optimization problem involving discount caps. The key is to determine the maximum possible discount each coupon can provide and use elimination to find which combination can reach the target sum.
Step 1: Analyze the maximum discount for each coupon.
- Coupon A: Flat Rs. 250 (Min spend 1200). Max = 250.
- Coupon B: 15% off, max Rs. 300 (Min spend 500). Max = 300.
- Coupon C: Flat Rs. 100 (Min spend 600). Max = 100.
- Coupon D: 10% off, max Rs. 100 (Min spend 250). Max = 100.
- Coupon E: Flat Rs. 50 (Min spend 200). Max = 50.
Step 2: Calculate the maximum possible total discount if one coupon is left out.
The total discount obtained is Rs. 710. We need to find which coupon was NOT used. This means the sum of the discounts of the 4 used coupons must be exactly 710.
Let's check the maximum possible sum for each scenario (leaving one out):
- Leave out A: Max sum = B(300) + C(100) + D(100) + E(50) = 550. (Cannot reach 710).
- Leave out B: Max sum = A(250) + C(100) + D(100) + E(50) = 500. (Cannot reach 710).
- Leave out C: Max sum = A(250) + B(300) + D(100) + E(50) = 700. (Cannot reach 710).
- Leave out D: Max sum = A(250) + B(300) + C(100) + E(50) = 700. (Cannot reach 710).
- Leave out E: Max sum = A(250) + B(300) + C(100) + D(100) = 750. (Can reach 710, since 710 <= 750).
Step 3: Verify the valid scenario.
If E is left out, the used coupons are A, B, C, D.
Their discounts must sum to 710.
A gives 250, C gives 100. Sum so far = 350.
Remaining needed from B and D = 710 - 350 = 360.
Since B can give up to 300 and D can give up to 100, their maximum combined is 400.
We need 360, which is perfectly possible (e.g., B gives 300, D gives 60; or B gives 260, D gives 100).
Therefore, the coupon not used is Coupon E.
Answer: Coupon E