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Official TSA Date & Time Questions (2008-2019) |
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Speed as well as accuracy is important in this section. Work quickly, or you might not finish the paper. There are no penalties for incorrect responses, only marks for correct answers, so you should attempt all questions. Each question is worth one mark.
Calculators are NOT permitted.
Here you will find all TSA Date & Time questions that have been written 2008-2019.
A long-running play in London’s West End is performed every evening except Sunday. In addition, there are matinee performances every Wednesday, Thursday and Saturday.
The correct answer is D.
The play performed every evening except Sunday, which is 6 days per week. There are matinee performances every Wednesday, Thursday and Saturday as well, which is an additional 3 days per week. To find the maximum number of performances in 31 consecutive days, we need to include as many Wednesdays, Thursdays and or Saturdays as possible, because these are the days on which the play is performed twice. There will be 4 full weeks in these 31 days, which is 4×7=28 days. In these 4 weeks, there will be 9 plays performed per week: 9×4=36 plays performed. We have 31-28=3 days left. If we started the month on a Wednesday (so Wednesday=1st), the final 3 days after our 4 weeks would be Wednesday, Thursday and Friday. As we would not be able to include all of Wednesday, Thursday and Saturday (the days when there are matinee performances) in the 3 days extra to our 4 full weeks, including 2 of them is the best we can get. From these final 3 days, we would have 2+2+1=5 performances. So in total over the 31 days, the maximum number of performances we could have would be 36+5=41.
Below is part of an itinerary for flying from London to Kampala. It involves changing flights at Dubai with a wait of 5 hours 15 minutes. All times are local times. Dubai is 4 hours ahead of London and Kampala is 1 hour behind Dubai.
DEPART LONDON (HEATHROW) 22:30 FRI 30.01.04
DEPART DUBAI 14:30 SAT 31.01.04
ARRIVE KAMPALA (ENTEBE) 20:45 SAT 31.01.04
The correct answer is B.
Kampala is 1 hour behind Dubai, which is 4 hours ahead of London, so Kampala is 3 hours ahead of London. So the arrival time at Kampala 20:45 SAT 31.01.04 in Kampala local time would be 17:45 SAT 31.01.04 in London local time. The depart from London was at 22:30 FRI 30.01.04 (London local time). The time difference between the depart from London and the arrival at Kampala would be the difference in time between 22:30 FRI 30.01.04 and 17:45 SAT 31.01.04, which is 19hrs and 15minutes in total. But, in the question it is given that there is a wait of 5 hours 15 minutes at Dubai, so subtract this from 19hrs and 15minutes to calculate the time spent flying, which would be 14hrs, answer B. The depart time at Dubai does not need to be used to find the answer to this question.
Hockey is an 11-a-side game, but a team may consist of up to 16 players, with unlimited substitutions allowed throughout a match of 70 minutes duration.
Roger captains the Buccaneers hockey team. He has 16 players for today’s match. He will play the whole match himself, as will the goalkeeper. He intends to rotate all the others in such a way that each of them spends the same total amount of time on the pitch.
The correct answer is B.
Because neither Roger nor the goalkeeper will leave the pitch, there are 16-2=14 player remaining who need to be rotated and 11-2=9 available spaces for the remaining players on the pitch. As the duration of the match is 70 minutes, the calculation to obtain the number of minutes each of the remaining players should spend on the pitch would be: 9/14×70=45 minutes.
Julie lives in London. She has three daughters; April, who lives in New York (where the local time is 5 hours behind London), May, who lives in Vancouver (8 hours behind London), and June, who lives in Tokyo ( 9 hours ahead of London). They all communicate with each other regularly.
Last week April received a text message from May that said “call me on Wednesday at 23.30 your time.” Unfortunately, the message had been sent to April by mistake and was intended for June.
The correct answer is C.
April lives in New York which is 5 hours behind London and June lives in Tokyo which is 9 hours ahead of London. Therefore June’s time zone is 5+9=14hrs ahead of April’s. Therefore April would call May 14hrs later than June would, and since the message was intended for June, may would receive the call 14hrs late.
Acme Haulage pays dividends to its shareholders every six months. Better Ball Bearings pays its shareholders every eight months and Koffmore Chemists pays every eighteen months.
In January 1990 all three companies paid me a dividend.
The correct answer is D.
Find the lowest common multiple of 6, 8 and 18 (the number of months each company pays its shareholders) which is 72 months. Convert this to years: 72/12=6 years. 6 years from January 1990 is January 1996.
A woman has to get up for work at seven o’clock in the morning. She wakes up and looks over at her digital clock. Unfortunately, the batteries are low and the figures on the display, showing hours and minutes, aren’t showing up properly. All she can see is the display below:
It is light outside, so she cannot be more than twelve hours late.
The correct answer is E.
Work backwards from the answers given, starting with E as it is the greatest amount of time. 2hrs and 59min after 7am is 09:59am. This can be shown on the image of the clock given by adding more lines but not omitting any of the existing lines, therefore E is possible and must be the answer as it is the greatest amount of time out of all of the answer options given.
The ferry from Harport to Selmer is at sea for 1 hour 15 minutes and at its berth at either end for 30 minutes. The hovercraft service between the same ports takes only 45 minutes and is berthed for 15 minutes before returning.
The correct answer is B.
Start from answer A as this is earliest time given. At 9:45 (1hr and 45mins after first leaving Harport), the ferry just leaving Selmer as it has spent 1hr and 15 minutes at sea and 30 minutes at Selmer. The hovercraft will be just returning to Harport as it has spent 45 minutes at sea, 15 minutes at Selmer and another 45 minutes at sea. Therefore at 9:45 they are at different ports, so A is not the answer. Now check answer B. At 12:45 (4hrs and 45mins after first leaving Harport), the ferry is just arriving at Selmer as it has spent 1hr 15mins at sea, 30mins at Selmer, a second 1hr and 15mins at sea, 30mins at Harport and a third 1hr 45mins at sea. The hovercraft will also be just arriving at Selmer as it has spent 45mins at sea, 15mins at Selmer, a second 45mins at sea, 15mins at Harport, a third 45mins at sea, a second 15mins at Selmer, a fourth 45mins at sea, a second 15mins at Harport and a fifth 45mins at sea. Therefore, as both are arriving at Selmer, the answer is B, 12:45.
Dates may be written in an eight digit form. For instance, 19 January 2005 may be written 19-01-2005.
The correct answer is D.
An airplane flight crew starts its day in Rome and does two round trips to London in the day. On each arrival at an airport they take the next scheduled flight The timetable is shown below (all times are local):
Rome – London |
London – Rome | ||
Depart | Arrive | Depart |
Arrive |
09:05 |
09:55 | 10:30 | 13:45 |
12:05 | 12:55 | 13:30 |
16:45 |
15:05 |
15:55 | 16:30 | 19:45 |
18:05 | 18:55 | 19:30 |
22:45 |
The correct answer is D.
To find the time of take-off for their first flight, we need to look for the earliest time in the depart column of the Rome-London section of the table (since the flight crew starts its day in Rome). This is 09:05. They arrive in London before the 10:30 take-off back to Rome, where they arrive at 13:45. They then must take the 15:05 flight to London (as they missed the 12:05 flight but want the next earliest flight), where they arrive at 15:55. This means they make it in time to catch the 16:30 flight back to Rome, where they arrive at 19:45, having now made 2 round trips. The total time this took was: 19:45-09:05=10hrs and 40min.
The Tickton town hall clock is rather a It shows the correct time every hour on the hour, but the minute hand travels three times as fast down to 6 as it does back up to 12, though within each half of its revolution its speed is constant.
The correct answer is D.
We’ve been told that the clock is able to show the correct time every hour on the hour, hence the minute hand must return back to 12 within an hour. The minute hand travels down to 6 3x faster than it travels up to 12. In other words, this means that the minute hand travels down to 6 in 1/3 the time that it travels up to 12. If we were to look at this in terms of a ratio this would mean 1/3:1 which can be simplified to 1:3. Splitting an hour into this ratio gives us 15:45 minutes. This means that it takes the minute hand 15 mins to travel down to 6 but 45 mins to travel back up to 12 subsequently. As the speed within each half turn is constant, this means that at quarter to the hour the minute hand is halfway through the last 45 mins of the hour. 45/2=22.5 so 22.5 mins remain until the next hour and the correct answer is D.
On only 360 occasions during the twentieth century was it possible to write the date in the form DD/MM/YYYY using eight different One example was 28th July 1956 (28/07/1956).
Eleven of Arthur’s birthdays have been such dates (though not his date of birth), and in 1974 and 1983 the two digits that did not form part of the date made up his age at the time.
The correct answer is D.
Crossing out the digits included within the years 1974 and 1983 gives us the remaining four digits 1,2,5 and 6 which would go towards forming the day and month of Arthur’s birthday, which will stay constant. This means that the two digits that are not part of the date and make up his age are 3 and 8 in 1974 and 4 and 7 in 1983. He must be 38 in 1974 and 47 in 1983 as this is the only combination that means that the increases in age by 9 between those two years. Therefore 1974-38 will give us his birth year which is 1936. The correct answer is therefore D.
The correct answer is B.
Performances nightly Sun-Fri= Sun, Mon, Tue, Wed, Thurs, Fri=6 days per week. Saturday 7th to 28th of December=3 weeks, so during this time period there would be 6×3=18 nightly performances . However, the theatre is closed on the 25th of December so there are actually 17 nightly performances. There are 2 saturday performances each week, so there would be 2×3=6 saturday performances during the given time period, however both the start and end date are on a saturday, which means there are an extra 2 saturday performances= 8 saturday performances in total. Additionally, theres are afternoon performances on weds, thurs and sat. For weds and thurs during this period this would mean 2×3=6 extra performances, including the afternoon saturday performances this would be 6+4=10 afternoon performances in total. However, the theatre is closed on the 25th which is a Wednesday, so subtract 1 performance away from this number, giving 9 afternoon performances during this period. Altogether, there would be: 17 nightly performances+8 saturday performances+9 afternoon performances=34 performances in total.
I have a bird automaton that, when activated, flaps its wings and whistles three different tunes successively, selected at random from its repertoire of It performs one of its ten tunes for 10 seconds, six of them for 15 seconds each and three of them for 20 seconds each.
When my granddaughter visits she is fascinated, activating it again and again as soon as it stops each time. On her last visit she kept it going for a whole hour, and I lost count of how many times she activated it.
The correct answer is C.
For the bird to be activated the maximum number of times, we want it to play it’s shortest possible combination of 3 songs each repetition. Since there is only one 10 second song, we will use this, and there are six 15 second songs so we can use two of these. So each repeition of 3 songs played would last: 10+15+15=40 seconds. In each minute there are 60 seconds, and there are 60 of these in each hour, so there are 60×60=3600 seconds in 1 hour. So the maximum possible number of times the bird could have been activated is: 3600/40=90 times.
There is a half-hourly train service that operates along the Rowbina Valley, between Erd and Leaving Erd at 10 and 40 minutes past each hour, trains stop at Aregon, Lowley, Ergen, Uble and Gidion before arriving at Teovil 51 minutes after departure from Erd. In the opposite direction, trains leave Teovil at 20 and 50 minutes past each hour, arriving at Erd 51 minutes later.
The trains normally average a speed of 60 km/h, but this rises to 80 km/h if they are running late. They stop for three minutes at each station en route, but this is reduced to two minutes when running late.
Yesterday the 10.20 am train from Teovil arrived at Uble on time, but because of a technical fault, remained at Uble for 22 minutes. There were no further delays.
The correct answer is C.
Until the train reaches Uble, it is still running on time, and therefore it will be travelling the 12km distance between Teovil and Uble at 60km/h. This means that it will take 12 mins to travel between these stations. It will also stop for 3 minutes at Gidion. Therefore, the journey up to Uble will take 12+3 =15mins. It then waits at Uble for 22mins. During the remainder of the journey the train will be running late so it will travel the 24km from Uble to Erd at 80km/h. This means that the train will cover the distance in 18 minutes. It will also wait for 2 minutes at Ergen, Lowly and Aregon, which is an additional 6 minutes. So overall it will take 15mins+22mins+18mins+6mins = 61 minutes. Since the train would usually travel from Teovil to Erd in 51 mins, this means that it will be 10 minutes late once it arrives at Erd. The right answer is therefore C.
Two neighbours work for the same company and share the journey to work, driving alternately in strict rotation. They work Monday to Friday each week and every other Saturday. (They always work the same Saturdays as each other.)
The correct answer is C.
The maximum number of days in a calendar month is 31. Assume the 1st day is a Monday. Person 1 drives Mon, Wed and Fri for this week. They work this Sat but person 2 must drive. Therefore for the 2nd week of the month Person 1 drives Mon, Wed and Fri again. This Sat they do no work, so person 2 must drive on the Mon of the 3rd week. So in the 3rd week, person 1 drives Tue, Thurs and Sat and in the 4th week Tue and Thurs (not Sat because they are not working this Sat). These for weeks include 28 days, so there are 3 days remaining for the 5th week, which are Mon, Tue and Wed. Person 1 works Mon and Wed in the 5th week. In total, person 1 has worked: 3 days in week 1, 3 in week 2, 3 days in week 3, 2 days in week 4 and 2 days in week 5. In total person 1 drives: 3+3+3+2+2=13 days, which is the maximum possible (as the month both starts and ends with person 1 driving and the working Saturdays are arranged in order to maximise the number of times person 1 drives).
A main road crosses a side road and the junction is controlled by a set of traffic lights. The normal sequence is for the lights to be green for 25 seconds on the main road followed by 15 seconds of green for the side road. There is also a pedestrian phase in the lights when all the road lights are If a pedestrian presses a button to cross any of the roads then, after the side road lights have been green, the pedestrian lights turn green for 10 seconds. Following a pedestrian phase, the main road green period is increased to 35 seconds to clear the build-up. Whenever any of the lights change there is a two-second period when all lights are red. The time on amber is included in the green times given above.
The correct answer is D.
Assume the side road lights are initially green. The pedestrian lights then turn green fro 10 seconds, followed by the main road lights for 35 seconds, which takes 45 seconds altogether before the side road lights can turn green. However, there is a 2 second period between each light change, and in this case there would be three changes (from side road green to pedestrian green, from pedestrian green to main road green and from main road green back to side road green), which would add another 2×3=6 seconds. So in total the side road would have to wait: 45+6=51 seconds.
Lucy allowed an hour to travel to a job interview, intending to arrive exactly on time. The train which she took arrived ten minutes late by her The taxi which she then took got caught in a traffic jam for an unexpected 15 minutes and walking to the appropriate room took five minutes longer than anticipated. The clock in the interview room was running 30 minutes slower than Lucy’s watch which was running ten minutes fast.
The correct answer is E.
Lucy’s watch was running 10 minutes fast and the train was 10 minutes late according to her watch, therefore the train actually arrived on time. There were 20 additional minutes added to Lucy’s planned 1 hour journey (15 from traffic jam and 5 minutes from walking), so the total journey time was 1hr 20mins. This means in actual time she would have arrived 20mins late, however the clock in the interview room was running 20 minutes slower than the actual time (30mins slower than Lucy’s 10mins fast watch). This means that according to the clock in the interview room, Lucy arrived on time.
Allan wishes to visit a client on floor 8 of a building with 10 floors (including the ground floor). He enters the lift on the ground floor. His client has explained by email that the floors are indicated by an illuminated matrix but there is a notice stating that two of the lights are not working.
Allan enters the lift and pushes the button. Just then he starts to sneeze, distracting his attention from the display. The lift leaves the ground floor, travels, then stops. He glances at the display and concludes he could be on floor 8 so he gets out. In fact he is on the wrong floor.
The correct answer is A.
Look for the numbers that could make an 8 when 2 lights are added to them in any positions. These numbers would be 0, 2, 3, 5, 6 and 9. However, we know that the lift has left the ground floor (floor 0) so Allan could have gotten out on any of floors 2, 3, 5, 6, or 9 instead of floor 8, which means he could have got out at one of five different floors by mistake.
William wants to cook roast beef and potatoes for himself and four friends. From his cookbook he gets the following information:
After removing from the oven, the meat should be allowed to stand for 10 minutes.
William buys a joint of the correct weight and decides to cook it medium rare in a hot oven.
The correct answer is D.
William is cooking for 5 people (himself and 4 friends) and is cooking the joint medium rare in a hot oven. 300g of beef per person: 300×5=1500g. From the 1st table, to cook the beef medium rare in a hot oven you need 15 min per 500g plus 15 extra min; so to cook 1500g you need (15×3)+15=60min. Working backwards from when the beef is ready to serve, he needs to put the beef in the oven: 10min(for the meat to stand)+60min=70min before he intends to serve it. Potatoes take 30min in the hot oven and do not need to stand before serving, so the potatoes need to be put in the oven 70-30=40min after putting the beef in the oven.
I stayed in a hotel last night. My room was at the front of the building. At one point I looked out of the window and saw:
on an illuminated digital clock on the other side of the road. I took the opportunity to reset my watch, which had stopped sometime earlier.
This morning, when I looked out, I realised that what I could see across the road was a reflection in the building opposite of a clock outside the hotel. According to my watch (which I now knew I had reset incorrectly last night) the time was supposedly nine minutes past ten.
The correct answer is E.
I have booked a flight from London to Auckland, which is in a time zone 12 hours ahead of GMT. I am due to leave London at 17:30 GMT on 19 August and will arrive in Auckland at 06:15 on 21 August (local time). My journey includes a 1-hour stop in Los Angeles and a 1.5-hour stop in Hawaii.
The correct answer is B.
They leave London at 17:30 GMT on 19th August, which will be 05:30 on 20th August in Auckland (as it is 12 hours ahead). That means that the journey lasts from 05:30 (20th) to 06:15 (21st) which is 24 hours and 45 mins. Take away 2.5 hours for the two stops, giving us 22 hours and 15 mins in the air which is 22 ¼ hours so the answer is B.
The planet Melpomene has three moons, Othello, Hamlet and Romeo, each moving at a constant speed in the same direction in circular orbits.
Othello takes 20 days to complete one orbit of Melpomene, whereas Hamlet takes 45 days and Romeo takes 120 days.
The correct answer is B.
Sun, 02 Oct 2022 11:06:56
can u explain this please, I dont know how you get the answer
Sat, 15 Oct 2022 11:03:39
You could either calculate the LCM of all 3 numbers (360) and divide this by the HCF which is 5 to get 72 OR you could simplify the numbers by dividing all 3 by their HCF of 5 and then calculate the LCM of 4, 9 and 24 which gives a value of 72 days as well. I am not sure as to the reasoning behind this method working however it provides the correct answer and requires the position of the orbit at which all 3 moons align rather than the number of orbits at which all 3 align which would lead to 360 days. I hope this helps :)
Nathan lives in London. His older brother Mark lives in San Diego, his younger brother Ben lives in Barcelona and his sister Isabel lives in Nairobi. Nathan is trying to arrange an online call between all four siblings. Each of the siblings has agreed to be available between 08:00 and 20:00 local time.
San Diego is 8 hours behind London time.
Barcelona is 1 hour ahead of London time.
Nairobi is 2 hours ahead of London time.
The correct answer is D.
This question becomes much each to approach if you create a timeline and mark in the times that each sibling is able to be online in terms of London time. It then becomes clear that the first time point at which 3 siblings are available at the same time is 8.00am (London time) and this continues until 7.00pm, after which only Nathan and Mark are online. Therefore, the correct answer is D, 11 hours.
Most railway stations have digital clocks that display the time in 24-hour format. I recently undertook a train journey from Exeter to Aberdeen. As the train left Exeter, the clock on the platform showed:
I lost track of time as I read a book and then dozed. I eventually became aware of the train coming to rest and the guard announcing ‘Carlisle’.
I opened my eyes, and found that I could only see the top of the platform clock (because of the hoarding in the way) as follows:
The correct answer is C.
This is a relatively straightforward question, all you have to do is recognise the numbers within the platform clock, having been given only the top half. Quickly running through each of the digits (remembering how they are presented on a 24-hour clock – which is particularly different for the digit 4 for example) tells you that the only possible time on the clock can be 14:57 i.e. the time is 3 o’clock to the nearest hour, so the correct answer option is C.
I have a digital clock that works in 24-hour format (that is, after 23:59, it goes to 00:00). The patterns for each number are made up of segments, as shown below:
The correct answer is D.
First check which numbers have the highest number of segments. We can see that 8 has the highest (7 segments) while 6 and 0 have the next highest (6 segments). We can then think about the possible combinations present on a clock displaying a valid time. The first number can only ever be 0, 1 or 2. From these we can see that 0 has the highest number of segments, and it also doesn’t restrict the numbers for the second number (like 2 would). For the second number we can easily select the number with the highest number of segments, 8, as it gives a valid hour. For the minutes section we can’t have 88, which would give the highest number of segments but an invalid time, nor can we have 68, but we can have 08. As a result, the time with the highest number of segments is 08:08, with 26 segments and the correct answer is D
There is an automatic photograph booth in my local shopping centre. It takes exactly 2 minutes from the time that the money is inserted for the photographs to be taken, then the developed photographs appear 4 minutes later.
Money may only be inserted when the green light is on. This light goes out immediately after money is inserted, and comes back on again exactly 3 minutes later.
This morning, I arrived at the booth just as it was being switched on, and I was sixth in the queue. The first person inserted their money immediately, and the rest of us all inserted our money as soon as we were allowed to.
The correct answer is A.
As the first person inserts their money immediately, their picture will be taken within 2 minutes, however it will take 3 minutes from the time that the money was inserted for the green light to come back on and for the next person to insert their money. As a result, for each subsequent person in the queue you need to add 3 minutes. As you are the 6th person, that means you have to do 5×3= 15 minutes, so you can insert your money 15 minutes after the booth was switched on and the correct answer is A. It can be tempting to do 6 x 3 =18 mins however remember that the first person did not have to wait and inserted their money as soon as the booth was switched on, so it’s only the subsequent individuals that had to wait 3 minutes. Also remember that the 4 minutes to receive photographs is just an additional piece of information to distract you, as the text states that each person inserts their money as soon as they can, i.e. as soon as the green light turns back on.
When riding my bike in the dark, I use two rear bike lamps. Both of the lamps are in ‘flashing mode’, which means that they repeatedly emit light for a set period of time at set intervals. The reasons for using two lights are that it reduces the amount of time when neither of the lamps emit light, and that if one of the lamps stops working, other road users can still see me on the road. One of the lamps shines for one second and goes dark for one second repeatedly; the other lamp shines for two seconds and goes dark for two seconds repeatedly. I always turn both lamps on simultaneously at the start of my ride.
The correct answer is C.
Lamp 1 is lit up for 1 second, then off for one second. Lamp 2 is lit up for two seconds, then off for two seconds. They are both switched on at the same time. Consider the first four seconds. If o represents on and x represents off, lamp 1 will be oxox, lamp 2 will be ooxx. Therefore in second three, they are both off. We conclude every four seconds, they are both off for 1 second. We want how many seconds this occurs in a minute: 4 x 15 = 60 so it will happen 15 times a minute, so the answer is C.
The single ferry to Pepper Island starts each day from mainland Seatown and crosses to the island port of Kaysville. It runs back and forth from 9:30 am to 4:10 pm on Saturdays, Sundays and Bank Holidays, and from 10:30 am to 2:20 pm from Monday to Friday.
Each crossing takes 50 minutes. Between crossings, there is a 10-minute stopover for cleaning and refuelling. This is extended to 20 minutes on Saturdays, Sundays and Bank Holidays to allow for the increased cleaning requirements.
The correct answer is B.
Thu, 29 Jun 2023 17:01:21
Can you provide the explanation please?
Wed, 11 Oct 2023 12:08:57
I am not sure if this is right but do we have to round numbers in this question? 4:10 pm - 9:30 am gives 400 mins, which can be divided by 70 mins because 50 mins is the travel time and 20 mins is the time for cleaning. This gives 5 and 5/7 crossings- almost 6. And then 2.20 - 10.30 gives 230 mins that we then can divide by 60 cuz 50 for crossing and 10 for cleaning, so we get 3 and 5/6 which is almost 4 crossings. If we take 6 as the number of crossings for holidays and weekends and 4 for weekdays we can have 6*10= 60 (10 is the sum of weekends and holidays) and 8*21= 84. And the sum of those gives 144. Does it mean that we have to assume that the ferry will leave the port even if it will not come back by the time it is supposed to stop running?
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