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Answer the following using a truth table and a Karnaugh map.
A food delivery app offers free home delivery to its customers who meet any of the following…
A food delivery app offers free home delivery to its customers who meet any of the following criteria.
• The order is above ₹ 1000 and payment is made through UPI
OR
• Food is ordered from a partner restaurant and payment is made through UPI
OR
• The customer uses the app for the first time and places order above ₹ 1000
The inputs are:
(In all the above cases, 1 indicates YES, 0 indicates NO)
Output: D - Denotes free home delivery [1 indicates YES and 0 indicates NO in all cases]
| INPUTS | |
|---|---|
| A | Order is above ₹ 1000 |
| U | Payment is done through UPI |
| P | Food is ordered from a partner restaurant |
| F | Customer uses the app for the first time |
(i)[5.0]
Draw a truth table for the inputs and the outputs given above. Write the SOP expression for D(A, U, P, F).
(ii)(a)[2.5]
Reduce the above expression D(A, U, P, F) by using 4-variable Karnaugh map, showing the various groups (i.e., octal, quads and pairs).
(ii)(b)[2.5]
Draw the logic gate diagram for the reduced expression using NAND gates only. Assume that the variables and their complements are available as inputs.
Draw: logic gate diagram using NAND gates only for the reduced expression
Answer
Answer
AI3(i): | A | U | P | F | D |
D(A,U,P,F) = Σm(6,7,9,11,12,13,14,15)
SOP expression:
D(A,U,P,F) = A'UPF' + A'UPF + AU'P'F + AU'PF + AUP'F' + AUP'F + AUPF' + AUPF
3(ii): K-map (variables A, U, P, F) grouping the eight 1s (minterms 6, 7, 9, 11, 12, 13, 14, 15) into three quads (verified with the boolean tool):
Quad 1: cells 12, 13, 14, 15 (A=1, U=1) → term A.U
Quad 2: cells 9, 11, 13, 15 (A=1, F=1) → term A.F
Quad 3: cells 6, 7, 14, 15 (U=1, P=1) → term U.P
Reduced (minimal) SOP: D(A,U,P,F) = A.U + A.F + U.P
With NAND gates only (NAND-NAND): one 2-input NAND gate for each term and a 3-input NAND gate combining them, since $[(AU)' \cdot (AF)' \cdot (UP)']' = AU + AF + UP$ by De Morgan's law.
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
From ISC 2025 Improvement Computer Science Paper 1, question 3.
Check your working with the Boolean algebra solver: the steps law by law, the K-map and the logic circuit for any expression.