Questions NCERT Solutions Class 11 Computer Science Chapter 4

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Start by reading the dividend and the divisor from the user.
2. Check the divisor before division because division by zero is not valid.
3. If the divisor is not zero, compute quotient = dividend / divisor.
4. Print the quotient as the final output.

INPUT num1
INPUT num2
IF num2 != 0 THEN
  quotient = num1 / num2
  PRINT quotient
ELSE
  PRINT "Division not possible"
END IF

Final answer: INPUT num1, INPUT num2, IF num2 is not 0 THEN quotient = num1 / num2 and PRINT quotient ELSE PRINT division not possible.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Set p1 = 0, p2 = 0 and flip_count = 0.
2. Repeat while both players have less than three wins and fewer than five flips have been read.
3. Read the flip result. If it is 1, increment p1. If it is 2, increment p2.
4. After every valid flip, check whether p1 or p2 has reached 3 and print the winner.

SET p1 = 0, p2 = 0, count = 0
WHILE p1 < 3 AND p2 < 3 AND count < 5
  INPUT win
  IF win == 1 THEN p1 = p1 + 1
  ELSE IF win == 2 THEN p2 = p2 + 1
  count = count + 1
END WHILE
IF p1 == 3 THEN PRINT "Player 1"
ELSE PRINT "Player 2"

Final answer: Maintain two counters. Read flip results until p1 = 3 or p2 = 3, then print Player 1 or Player 2 as the cake winner.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Start the number at 10 because the lower limit is included.
2. Print the current number.
3. Add 5 to move to the next multiple of 5.
4. Continue while the number is less than or equal to 25.

SET num = 10
WHILE num <= 25
  PRINT num
  num = num + 5
END WHILE

Final answer: The pseudocode prints 10, 15, 20 and 25.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Choose an activity with a fixed count.
2. Set a counter to track how many times the action has happened.
3. Repeat the action until the counter reaches the fixed count.
4. Stop after the required number of repetitions.

SET count = 1
WHILE count <= 10
  PRINT count
  count = count + 1
END WHILE

Final answer: Printing the first 10 natural numbers is a loop executed a certain number of times.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Set total = 0 before collecting money.
2. While total is less than 200, ask the next person for a contribution.
3. Read the amount and add it to total.
4. When total becomes at least 200, stop and print that the target has been collected.

SET total = 0
WHILE total < 200
  INPUT amount
  total = total + amount
END WHILE
PRINT "Target collected"
PRINT total

Final answer: Keep adding each contribution to total until total >= 200, then stop collecting.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read price and quantity.
2. Compute bill = price * quantity.
3. Compute gst = bill * 5 / 100.
4. Compute bill_gst = bill + gst and print both bill values.

INPUT price
INPUT quantity
bill = price * quantity
gst = bill * 5 / 100
bill_gst = bill + gst
PRINT bill
PRINT bill_gst

Final answer: bill = price * quantity and Bill GST = bill + 0.05 * bill.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read marks in Computer Science, Mathematics and Physics.
2. Add the three marks to get aggregate.
3. Since each subject is out of 100, the maximum total is 300.
4. Compute percentage = aggregate / 300 * 100, which is also aggregate / 3.

INPUT cs
INPUT maths
INPUT physics
aggregate = cs + maths + physics
percentage = aggregate / 3
PRINT aggregate
PRINT percentage

Final answer: aggregate = cs + maths + physics and percentage = aggregate / 3.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read the two numbers.
2. Compare the first number with the second number.
3. If the first is greater, print it as the greatest.
4. Otherwise print the second number as the greatest.

INPUT num1
INPUT num2
IF num1 > num2 THEN
  PRINT num1
ELSE
  PRINT num2
END IF

Final answer: If num1 > num2, print num1 as greatest, otherwise print num2 as greatest.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read the number.
2. Test whether it lies from 5 up to but not including 15 and print GREEN.
3. Else test whether it lies from 15 up to but not including 25 and print BLUE.
4. Else test whether it lies from 25 up to and including 35 and print ORANGE, otherwise print the default message.

INPUT n
IF n >= 5 AND n < 15 THEN PRINT "GREEN"
ELSE IF n >= 15 AND n < 25 THEN PRINT "BLUE"
ELSE IF n >= 25 AND n <= 35 THEN PRINT "ORANGE"
ELSE PRINT "ALL COLOURS ARE BEAUTIFUL"
END IF

Final answer: Use an if, else if chain for 5 <= n < 15, 15 <= n < 25, 25 <= n <= 35 and an else case.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read four numbers a, b, c and d.
2. Set largest = a and smallest = a.
3. Compare b, c and d one by one with largest and smallest.
4. Update largest when a checked number is bigger, and update smallest when it is smaller.

INPUT a, b, c, d
largest = a
smallest = a
FOR each value in b, c, d
  IF value > largest THEN largest = value
  IF value < smallest THEN smallest = value
END FOR
PRINT largest, smallest

Final answer: After all comparisons, print the final values stored in largest and smallest.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read units consumed.
2. If units are up to 100, charge units * 5.
3. If units are between 101 and 250, charge 100 * 5 plus the remaining units at 10.
4. If units are above 250, charge first 100 at 5, next 150 at 10 and remaining units at 20.
5. Add Rs 75 meter charge to get the total bill.

INPUT units
IF units <= 100 THEN bill = units * 5
ELSE IF units <= 250 THEN bill = 100 * 5 + (units - 100) * 10
ELSE bill = 100 * 5 + 150 * 10 + (units - 250) * 20
total = bill + 75
PRINT total

Final answer: Total bill is the slab charge plus Rs 75 meter charge.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. A conditional checks a condition.
2. If the condition is true, one set of steps is executed.
3. If the condition is false, another set may be executed.
4. They are required when a program must make a decision based on input or state.

IF condition THEN
  steps for true case
ELSE
  steps for false case
END IF

Final answer: Conditionals are decision-making statements such as IF and ELSE. They are required when a program has to choose actions based on a condition.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. An arrow shows the flow of control.
2. A rectangle represents a process step.
3. An oval or terminator represents start or stop.
4. A parallelogram represents data input or output.
5. A diamond represents decision making.

Arrow -> Flow of Control
Rectangle -> Process Step
Oval -> Start/Stop
Parallelogram -> Data
Diamond -> Decision Making

Final answer: Arrow: flow of control; rectangle: process step; oval: start or stop; parallelogram: data; diamond: decision making.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Add a check for the day and school timing before leaving.
2. Add options for breakfast, books, identity card and homework check.
3. Add alternate transport choices such as walking, bicycle, auto, car or bus.
4. Add decision steps for missing the bus, traffic or bad weather.

Wake up
Check school day and time
Get ready
Pack bag and lunch box
IF bus is available THEN take bus
ELSE IF bicycle is available THEN ride bicycle
ELSE arrange another transport
Reach school or college

Final answer: Improve the algorithm by adding readiness checks and transport alternatives before reaching school or college.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read n from the user.
2. Set fact = 1 and counter = 1.
3. Repeat while counter is less than or equal to n.
4. Multiply fact by counter, then increment counter.
5. Print fact after the loop ends.

INPUT n
fact = 1
i = 1
WHILE i <= n
  fact = fact * i
  i = i + 1
END WHILE
PRINT fact

Final answer: Initialize fact = 1, multiply it by every integer from 1 to n, and print fact.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. Read the number and store a copy as original.
2. Extract hundreds, tens and ones digits using division and remainder operations.
3. Compute sum = hundreds cube + tens cube + ones cube.
4. If sum equals original, display Armstrong number, otherwise display not Armstrong.

INPUT n
original = n
ones = n MOD 10
n = n DIV 10
tens = n MOD 10
hundreds = n DIV 10
sum = hundreds^3 + tens^3 + ones^3
IF sum == original THEN PRINT "Armstrong"
ELSE PRINT "Not Armstrong"

Final answer: Compare the sum of cubes of the three digits with the original number.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. The original condition Number < 9 wrongly classifies 9 as Double Digit.
2. The original condition Number < 99 wrongly classifies 99 as Big.
3. Single digit positive numbers are less than 10.
4. Double digit positive numbers are from 10 to 99, so the second condition should be Number < 100.

INPUT Number
IF Number < 10 THEN PRINT "Single Digit"
ELSE IF Number < 100 THEN PRINT "Double Digit"
ELSE PRINT "Big"
END IF

Final answer: Correct algorithm: IF Number < 10 print Single Digit; ELSE IF Number < 100 print Double Digit; ELSE print Big.

Expert view: Choose the control structure first, then test the boundary or stopping condition.

1. The phrase positive integers up to 100 means 1, 2, 3, ..., 100.
2. The given condition accepts 0, but 0 is not positive.
3. It also does not explicitly check whether the input is an integer.
4. Improve it by accepting only integer values with 1 <= Number <= 100.

INPUT Number
IF Number is integer AND Number >= 1 AND Number <= 100 THEN
  ACCEPT
ELSE
  REJECT
END IF

Final answer: It fails on 0 because 0 is accepted even though it is not positive. Improved condition: IF Number is an integer AND Number >= 1 AND Number <= 100 THEN ACCEPT ELSE REJECT.