AS Computer Science 9618 Paper 1 Solution - Fastpapers Mock Championship 2026

追加: 2026-05-09

[00:00:00]All right guys, today we are going to do the paper of Cambridge International S and A level competitions paper one. So this paper was made for the fast paper UK's mock championship 2026. So overall this paper was maybe a bit harder than the normal paper but it's not that too hard. All right, so let's start.

[00:00:23]A smart home has an has an automatic air conditioning system. The system Y operates according the following criteria. We have THDM the parameters.

[00:00:31]This is also written here. Then we have a description here. We have the binary value and the conditions. Right? The air conditioner system turns on Y if the temperature is too high. Right? So the temperature value will be this and the humidity level is acceptable.

[00:00:49]Humidity level is this. Right? So we have t dot humidity level is acceptable. So we have h here and acceptable means we have a not right. So this thing that I'm doing is called boolean algebra. You can also write this in word like t and not h then we have a and between all of them. The door is closed or the manual override is off. The door is closed means what? D not right or the manual over it is off. So M not and we have this. All right. So now we need to draw this. So we have a T normally then I have H.

[00:01:37]This will have a N gate.

[00:01:45]Then I will have E which is having an end gate with the not H that we have.

[00:01:58]So we now need to do the end gate for the whole operation which is being and with expression we are going to draw now. So we have not D and not M. These gates are having a or get between them. So let's draw this and finally we will draw another branch to the end G. Let's try this question.

[00:02:37]So consider the following logic expression. We have expression given here. A truth table for the logic expression is given. So we need to find out the value of this part right.

[00:02:47]And we can just inverse that. All right.

[00:02:49]So for that we can find the value of this part at first then we find the value of this part then we use the end gate between them.

[00:02:58]So let's start with line number one. We have P is 0. Q is 0. P q 0 nor 0 is equal to 1. Right? This part is equal to 1. Now R is zero. P is zero. So not P is going to not P is going to be 1. And this is a R. So that means if this value is one, the value of R doesn't exactly matter. So this will evalute to one. One and one is going to be zero. Sorry one and one is going to be one and not of that is going to be zero. So this should evaluate to zero.

[00:03:37]All right. Again since not P is going to be one here.

[00:03:43]So this R doesn't matter and this is unchanged. So one and one is one and not that is zero.

[00:03:53]So this is an error that we identified.

[00:03:55]Next line number three. P is 0, Q is zero. So p q is going to be equal to zero and we don't have to evaluate this part because anything ended with zero is going to be zero and not of that is going to be one again same scenario here this part actually evaluates to what zero because we have a norate and zero and anything is going to be zero and not of that is going to be one. So here we have a similar trend. All of this is going to output a zero here. So that means all of this will evalate to one by default.

[00:04:35]So let's erase this. So the mistakes are in this line 1 3 and the final line eight. So we have 1 3 and 8 here.

[00:04:50]Now the next question state one difference between k byte and terabyte.

[00:04:55]So we know k by is 1024 gab byte.

[00:05:07]So GB is going to be 1024 GB while terabyte is going to be 10 1,000 GB, right? Because debate is binary prefix and terabyte is a denary prefix.

[00:05:27]So let's input that one too by is binary prefix while terabyte is dary prefix.

[00:05:52]Let's keep it aside because this is an extra point for the first point is enough to score full marks in this question.

[00:06:01]Now the next question give the number of buys in give. So you don't actually have to write the exact value you can keep this in multiples. For example here let's try to convert this.

[00:06:19]So we have GB byte and then we have after GB we have me maybe.

[00:06:30]So we multiply with 1 024 and we get baby and we multiply this by 1024 to get k.

[00:06:51]So we can just write 1024 into 1024 into 1024.

[00:06:58]And this is enough for the answer. We don't have to evaluate fully.

[00:07:05]All right. Let's try the next addition now.

[00:07:12]So 0 + 1 is 1. 1 + 1 is 0. We have a carry of 1. So 1 + 1 is going to be zero. We have a carry of 1. 1 + 1 is going to be zero. We have a carry of 1.

[00:07:23]1 + 1 + 1 is 1. Carry of 1. 1 + 1 is 0.

[00:07:27]Carry of 1. 1 + 1 is 0 and we have a carry of 1. So 1 + 1 + 1 is 1 and the carry of one. So we denote carry using the bracket. So you need to show your working. That means you have to show that there are carries.

[00:07:44]All right. Next question. and describe the error that occurred when you added the binary number in part C1.

[00:07:51]So let's type it out.

[00:07:56]There was overflow, right? Because the result of the operation cannot be represented in the number of bits available.

[00:08:19]All right. In this question, we have a binary subression. So the value above is higher than the value given below. All right. So let's try one method. So in the first method, we will convert this part of the number into the two's complement form to get a negative number and then we can add this with the first number. So let's try to convert this into two's complement. So this becomes 1 1 0 1 1 0 0 1 by flipping all the bits.

[00:08:47]This is called the first one complement, right? So then we add the value of one to this number. So this becomes 1 1 0 1 1 0 1 0. Then let's add it this with the first number 0 1 1 0 0 1 1. Let's do the addition.

[00:09:17]So after doing addition we have 0 + 1 is 1. 1 + 1 is 0. 0 + 0 is 0.

[00:09:29]And then let's continue the addition using the carries.

[00:09:33]So we have one carry and 1 + 1 is going to be zero. We have another one carry. 1 + 1 plus is 1. Another carry 1 + 1 is zero and we have a one here. Since we are you are subtracting a smaller number from a larger number. Your answer should be positive. That means the number of bits that we have have available is eight.

[00:09:58]Right? So this is the answer and this one will not be inputed.

[00:10:08]Now for the next part let's try the gener normal method that we use generary by using borrows and all.

[00:10:24]So 1 - 0 is 1. 1 - 1 is 0. And here we have to perform a carry borrowing right.

[00:10:33]So we remove this one and the next one becomes 1 zero. So we remove this one zero again it becomes one and we have 1 0 in the next part. So 1 0 - 1 meaning 2 - 1 is going to be equal to 1. So we write a one here. Then we have 1 - 0 and this is equal to 1 and we have 0 - 0 is 0. 1 - 1 is 0. 1 - 0 is 1 and 0 - 0 is 0.

[00:11:00]So this is the method two borrowing.

[00:11:24]Now let's go to the next question.

[00:11:26]The next question you are The audio message is recorded with a sampling rate of 40 kHz and a sampling rate of 32. The recording is the recording is stereo meaning it has two channels. The is 32 bits. Calculate the file size of the recording. Giving your answer in megabytes and show your working.

[00:11:46]All right. Now let's try the calculation.

[00:11:52]So we are going to multiply 40 kilohz meaning 40 into 1,00 multiplied by 32. This is the sampling resistance and then we multiply this by the duration which is 10 into 60 because we need to convert this into seconds. We multiply this 2 by 2 because it has two channels.

[00:12:21]And finally, we divide this by at first 1,000 because the answer that we got was in bits. We need to convert this in megabytes. So we divide this by 1,000 at first. Then we divide this by 1,000 once again.

[00:12:42]And then we divide this by eight to convert it into bite. So since we have a large calculation here, we can easily simplify this. All right. So let's use a different color and 1,000 gets removed with 1,000. Then we remove the zeros.

[00:13:02]We have 32 and 8. We can just simplify this making 32 8. So let's multiply now four.

[00:13:14]This will be different color. of 4 into 4 into 6 into 2.

[00:13:24]So let's 2 and into 6.

[00:13:34]So after we do the calculation we get 192.

[00:13:38]Answer is 192.

[00:13:47]All right, coming to the next question.

[00:13:49]Provides um pro the following incomplete table contains four network devices and their description.

[00:13:57]Complete the trace table by writing the missing devices and missing description.

[00:14:06]Provide a MAC address to the device to identify it on the network. So this is a W NIC wireless because since this is providing wireless network. Next we have router.

[00:14:21]It will receive and send data between two networks which are operating on the same protocol.

[00:14:41]hardware component that provides radio communication from the central US to nodes on the network and vice versa.

[00:14:47]This is going to be a WAP wireless access point. You can write it in short form if you want.

[00:14:57]Next we have a repeater.

[00:15:01]So repeater does what? It restores restores digital signal so that it can be transmitted over large distance.

[00:15:24]Now to describe model provides access to internet using telephone lines.

[00:15:43]All right. Next question. Local area network lands can be made up of several sub networks. give two benefits of dividing the network into subarrow by submitting the land. Improve security as well as it will reduce congestion.

[00:16:03]So let's write that reduce congestion and let's write some extra points. For example, this will allow an extension to the network.

[00:16:18]Then performance will be improved.

[00:16:34]All right, let's go to the next question. A subnet pack is used when subnetting a LAN. Two devices on the LAN are located in different subs. The IP address and corresponding subnet masks are shown.

[00:16:48]So we have a table where the values are given.

[00:17:03]So the first part is the network ID right and the second part is the host ID. So we have 50.1 which is the host ID the second part and here we have 161.52.24 which is the network ID and three is the host ID. So 161.52.24 24 is the network ID. We do not need to write the final point here.

[00:17:32]Next question. A factory makes toy cars.

[00:17:36]A the factory uses a convertible system to move cars between different stages of production. System counts the number of cars at the end of the production. A sensor is positioned above the conveyorable. Identify one appropriate type of sensor that can be used.

[00:17:50]So we can write infrared sensor here.

[00:18:00]So imagine this a situation where there is a sensor above the conveyor belt and the below part is conveyor. So imagine this circle as a car. So whenever this passes by the sensor it is being detected. So this can be done using an infrared sensor.

[00:18:30]The next question, a second automated system removes cars that do not meet the required weight from the production line. Describe the role of an actuator in the second system.

[00:18:40]So, let's type it down.

[00:18:43]actuator will generate a signal and causes an action.

[00:18:52]It will push and push an arm to remove the car with incorrect weight.

[00:19:10]All right, next question. The factory uses many machines that contain embedded system. Identify two features of the embedded system. So this is very easy.

[00:19:18]The first feature is dedicated to a single task and then we can write built into a larger system.

[00:19:34]So identify one drawback of embedded system.

[00:19:41]So one drawback is going to be difficult to change or upgrade.

[00:19:52]So one point is enough. So we don't write maybe you can write they cannot be easily adapted for another task or troubleshooting false is a specialization task or devices are thrown away.

[00:20:07]The system includes a central central unit that uses static RAM to store temporary data during operation. The system is connected to a robotic arm and a control panel used by workers. Explain why static RAM is used in the control unit instead of dynamic RAM.

[00:20:23]So let's type out the answer.

[00:20:29]Static RAM has faster access time because it does not need to be refreshed.

[00:20:46]Since this is a continuous operation, static ramps will be more reliable.

[00:21:02]The robotic arm uses EROM to store it control program. Explain two reasons why EROM is suitable for storing this program. We have to explain in this question.

[00:21:13]So writing the points very briefly is not enough. EROM is not volatile. So the control program is not lost when power is gone.

[00:21:24]We have to explain this a bit right. The program is not volatile.

[00:21:30]So the control program is lost when power is gone.

[00:21:39]It's not lost when the power is gone.

[00:21:43]Next part.

[00:21:47]Let's write about the program.

[00:21:56]EROM can be electronically erased and reprogrammed. Meaning we can update the instruction.

[00:22:15]This is the word update. The instruction of the robotic arm fast paper storage data of students who signed up for the mock templation. The owner of the company wants to ensure the security and integrity of data. So why the data needs to be kept secure? to prevent the data from being lost or corrupted or updated.

[00:22:56]One way the data stored in a computer can be kept secure is by using a backup software. Give two other ways the data stored in a computer can be kept secure.

[00:23:04]So the first point is we can install a firewall install a firewall.

[00:23:21]Next part, we use encryption to make the read data unreadable.

[00:23:34]All right, next question.

[00:23:38]Next part of the question. The data about the students is currently stored on paper. The data needs to be transferred into a computer system. Data validation verification are used to help maintain the individual data. Identify and describe one method of data verification that can be used when transferring the data from paper to the computer. So we can write double entry in this point.

[00:24:00]And for description, let's write enter data twice and compare both to check if these are the same.

[00:24:18]We can also include the point for visual check here.

[00:24:22]But you need to make sure it is verification when we are transferring the data from written paper to the computer. Right?

[00:24:31]Meaning we are manually entering the data or something. All right. Next question. The company needs to transfer the phone number of each candidate into the computer system. The phone number is a Bangladeshi number exactly 11 digits long and starting with 01. Give one example of how each of the following data validation rules can be used to validate the phone number. It is phone number it is entered into the system.

[00:24:56]All right. We have a range check here.

[00:24:58]So let's describe this.

[00:25:05]The number must be within 01 then we have nine zeros.

[00:25:16]and 0 1 then we have nine 9ines.

[00:25:25]All right. Now for the presence check make sure a phone number has been been entered.

[00:25:35]For length check we have to make sure this 11 digit right. So make sure the number is exactly 11 digits long.

[00:25:46]Next question. Explain why the data in the system may not be correct even after validating and verifying the data.

[00:25:55]So maybe we enter the data using the region that we have. But we cannot guarantee that the data was correct in the first place. Right? Validation will check that the data is reasonable.

[00:26:09]But it cannot check if the accuracy accurate data is entered.

[00:26:19]Verification will check if data matches the source but it cannot check if original data is accurate or not.

[00:26:44]All right, next question.

[00:26:47]All right, next question. A programmer uses both a compiler and a interpreter to translate a battle game written in a highle language. Describe the advantages of using the compiler compared to the interpreter to translate the game. So we need to write the advantages of the interpreter here.

[00:27:08]So trans the game is translated at once.

[00:27:16]A complete machine code version will be produced.

[00:27:27]Why which can be run after compilation?

[00:27:39]Multiple errors are detected at once.

[00:27:51]The game can be distributed independently of the source code.

[00:28:09]So we can include more points for the compiler for the compiler if you want here.

[00:28:18]Next question. State one. Reason why some highle languages are partially compiled and partially interpreted.

[00:28:28]That's right. Partially compiled programs may be used in different platforms as these are interpreted when run.

[00:28:55]Next question. identify two features that support the visual presentation of the code in a typical ID integrated development service s environment.

[00:29:04]So pretty print and then we have auto indentation two debugging of the code. So at first we had at first we had visualization. Now we need to give debugging.

[00:29:35]Here we debugging right.

[00:29:40]So we have single stepping and we have break points.

[00:29:50]So we have a table here. The following table shows part of the instruction set for a process. The processor has one, the accumulator and the index register. So this is the question.

[00:30:04]So this is immediate addressing. We load the number exactly in. So we load zero in.

[00:30:15]Then we have this is line number 77.

[00:30:22]Now we have line number 78 LDX 210. So we can find out index addressing from from the address from address plus the content of the ex. Copy the content of this calculator address to the accumulator. So we will add 210 value with the ex. So this becomes 98.

[00:30:44]Then we will go to the next line 79 which is to compare with 67. So these are not equal. We won't do anything. Now line number 80 jump if equal again we won't do anything because these are not equal. Then we have line number 81.

[00:30:59]So this is to add 200. 200 is the address. All right. This is not the value. This is the address. So we will add zero which is the value from 200. We will add this to 98. So this becomes 98.

[00:31:13]Next line number 82 we have store st 201. So we will store the value that we have to 201.

[00:31:25]Then we have line number 83.

[00:31:28]So let's check this immediate addressing load the number to accumulator.

[00:31:35]So we will load one to the accumulator.

[00:31:37]Now we have line 84 which is to add 200.

[00:31:40]So the value of 200 the value is zero right? So we will add zero. Now line number 85 we will store this in 200. So store the value of one inside 200.

[00:31:51]Now line number 86 increment.

[00:31:55]So we will increase the value of in index register by one. Now line number 87 we will have ldx 210.

[00:32:05]So we will add we will add one to the value 200 10. So we will have 211. So we will load the value from 211 into line 87. Now line number 88 is to compare with the value 67. And these are not equal. So go to line number 89 now. So this is jump if not equal. So we'll jump to line number 81.

[00:32:32]So now line number 81 line number 81 is to add 200. So we will add 97 with the value stored in 200 which is one. So this becomes 98. Now line number 82 we will store this in 201. So we will write this value under the column of 201.

[00:32:52]Now line number 83. This is LDM1.

[00:32:56]Right?

[00:33:01]So we will load the value of one into LDM in the calculator. Line number 84 is add with 200.

[00:33:09]So this becomes two. Line number 85 is store this in 200.

[00:33:18]So we will write the value of two under the 200's column. Now line number 86 is to increment index register.

[00:33:31]So that column will become two. Now line number 87 is ldx 210.

[00:33:39]So we will add the value 210 + 2 is equals to 212. So we will the value from 212 into this column and this become 67.

[00:33:47]Now line number 88 we will compare with 67 and these are equal. So line number 89 we won't run this because these are equal. Now we have line number 90 which is LD 211.

[00:34:02]So we load the value of 97 into accumulator.

[00:34:14]Now line number 92 we will output. So the corresponding value of 97 in S is a.

[00:34:28]So now we have to output 52 which is four. Then we go to line number 19 which is the end.

[00:34:39]All right let's do this now. The part state the purpose of this part of an assembly language program.

[00:34:52]So let's try it out. We have address 305, address 2110, we have the address 211. So let's write the value. Let's be a x y and z. We have accumulator.

[00:35:08]So we will load accum 110 which is a x into accumulator. Then we'll store this in 3005. Then we will load y in the accumulator. Right? So we will store this in 210. So now we will load the value in 305 which is x again and store this in 211. So you notice that we have completely swapped 210 and 211 and 305 acted like a temporary variable for us. So the purpose of this assembly language program is to swap contents of 210 and 211.

[00:35:53]All right. Next question is for bit manipulation. We are given with the B wise operation. We now have to evaluate some values. The current corner of the ACC are 0 1 0 1 0 1 0 1. So the result of the execution of the following instruction. So we need to run the end operation with one one one one 0 0 0 right. So zero and one.

[00:36:16]So right part will become zeros and for the first part we have 0 1 0 1 and the right part is all zeros.

[00:36:26]Next part the result of the execution of the following instruction.

[00:36:31]So we will have a or between this. So 1 one 1 0 0 0 0 right. So we have a or argate. So the first four values becomes one by default.

[00:36:44]So one one one then 0 1 0 0.

[00:36:58]All right. So the result of after the execution of the following instructions.

[00:37:01]So this is zorg exclusive r meaning only one values can be one. So let's write the value.

[00:37:08]So the values where the first value is one one second value is another one one. Third value is going to be output of give output of one. Then the fifth value will give output of one and the last value will give a output of one because this is a zorgate.

[00:37:27]Then we write the remaining values.

[00:37:31]Next part the current card of the ACC are this show the result of the ex of the following instruction.

[00:37:38]So if I have left shift left I by three places I can easily remove the first three and keep on writing and add last three elements. All right. So this becomes 1 1 0 0 0 1 0 0 0 next part. Take one or more boxes in each row to indicate whether task is performing first pass, second pass or two pass assembing instructions. This is first first pass generates the object code. Second pass resolve comments is first pass assembly language with assembly language program one line at a time. This is both.

[00:38:21]Then we have the question of a database.

[00:38:29]A company rents custom cars to customer and stores data about the rentals using relational. The company creates a relational database called car rentals.

[00:38:38]The database stores data about the customer cars and rentals.

[00:38:44]The database car rentals is designed to include the following tables. So let's draw the year diagram. You will notice that some fields are common, right? So customer ID is a field from customer.

[00:39:00]Then I have the field car ID in service ID which is from car.

[00:39:07]And we also have another field which is inside rental the car ID.

[00:39:24]So let's complete the ER diagram. So one car can have be in many rentals, right?

[00:39:30]This is one. And now we can see one customer can be in many rentals because rental has customer ID and car ID. So many ids can be used in different different rows.

[00:39:41]Then for cars and service we will have a connection with car.

[00:39:47]So many car can be of one service.

[00:39:51]Right. The table shows the data for the table car. So we given write a structure query language to define the table car include constraints or restriction on the data can be entered into the field where appropriate. So we have to mention the constraint. This can include primary key or not null or the length.

[00:40:10]All right, let's start writing create table car. Then we have some fields here.

[00:40:20]Let's put a semicolon at the end. We have the field of car ID, manufacturer, model and year.

[00:40:26]So in after some indentation we have car ID which is of type character and six characters.

[00:40:40]This is the primary key.

[00:40:44]All right. Next we have manufacturer which is we can include this as vure of 20 characters a very high number and we set this to not null.

[00:41:01]Then after manufacturer we have model. So in model again we use virture of 20 and not null.

[00:41:19]The last part here is going to be integer and this is also going to be not null.

[00:41:27]So then we end the SQL code.

[00:41:37]Now coming to the next question.

[00:41:41]The company wants to find the total number of rentals for each car manufacturer.

[00:41:48]Write the SQL to return the car manufacturer and total number of rentals for that manufacturer. We use select.

[00:41:55]Then we need that manufacturer name right. So we will write manufacturer.

[00:42:02]We can write car dot manufacturer or just manufacturer. Since this is a unique field car manufacturer we also need to have return the total number of rentals.

[00:42:17]Right? So we use count.

[00:42:27]So we use count and inside we use rental ID the number of rentals.

[00:42:34]The next time we will write the name of the table from car rental.

[00:42:52]These are the two tables.

[00:42:55]We can write this as rental dot rental ID where car dot car ID is going to be equal to rental dot car ID. Right?

[00:43:11]Because this is the common field between these two.

[00:43:22]In the next line we have to use group by because the data needs to return for each manufacturer. So we will use this line called group by because this is a group data. We are grouping similar datas right? Group by manufacturer. We are grouping by the manufacturer name.

[00:43:38]So we write this.

[00:43:41]All right. This solution has an alternative which is when we use inner joint to solve this. Right? So let's now see the syntax for that.

[00:43:58]So let's write alternate solution.

[00:44:02]Select again manufacturer and count rental ID.

[00:44:12]Now from the first table which is car.

[00:44:16]Now we in the second table in rental on the name of the field car dot car ID equal to rental dot rental ID. The rental car ID we will group by manufacturer and this is the answer.

[00:44:44]All right guys, thank you for watching this video and here we have solved the paper one for fast of championship. If you like this video do give it a like and consider for registering for the next championship.

[00:45:06]Thank you.