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hello everybody.welcome to edwin media gh a channel.dedicated to reliable and useful.educational.information today we're going to look at.some common pass questions and their.solutions in numeracy exam.[Music].welcome back now before we start if you.are new to the channel.please consider subscribing also hit the.notification icon to be notified.anytime we upload new content.the gtl examination is due in few weeks.from now.and i'm not gonna bore you with any.unnecessary content.i'll go straight to the point as i said.earlier in this video we're going to.look at some recurring past questions in.their solutions.all right let's get in now to our first.question.an exam was conducted for 25 people.12 of them packed exams what percentage.of the people.filled the exam now this is a very.simple question.we have been given some values so.all that we need to do is to start with.those values all right.so from the question we know the rule.that number of people's.data in the class the total number of.peoples in the class that is 25 peoples.and also we also know the number of.people that pass the examination.that is 12 people so with these two.values.we should be able to get the number of.peoples that fill the examination.so we have 25 as a total number of.peoples in the classroom.and then 12 as the number of people that.pass the examination.now with this the number of people's.whole field is now going to be.x minus y now note that i'm using x to.denote the total number of peoples in.the class.and also i'm using y to determine the.twitter total i'm using y to denote.the number of people who pass the.examination so.x which is number of people who uh now.the total number of people in the class.and then y number of people who pass the.examination so x.minus y and that should give us 25 minus.12.all right so then 25 minus 12.and that is 13. so our total number of.people so total number of people that.actually i feel the examination.as certain right now the question said.the percentage of people's that feel the.exam.not the number of people who feel the.exam so we are looking for the.percentage.so at this point we need to convert this.value.into a percentage aspect.so we are converting this into a.percentage.now to do that we're taking the third.thing that we've got in the third and.the number of people that failed.we're taking that over 25 okay.so 13 of 25 times 100.and that gives us 52 percent all right.so it means that.the number of people that failed or the.percentage of people.that fuel the exam is 20 52.all right so this is the answer this is.the answer so we worked it.after getting the number then we now.convert it into a percentage.aspect all right now there's also an.alternative way of doing this.without coming down to convert it here.so the alternative method is here.now you know overall that 25 peoples.represent hundred percent.okay we can use this to represent that.because the 25 peoples that's the two.third number of people in the classroom.so.using 100 or using it in percentage-wise.we can say that 25 percent 25 peoples.represent.100 if one of them is out assuming 25.people.and once people is out of the 25.it's no more going to be hundred percent.it's going to be less than 100.so since the 25 that the total number of.people we have there.that can be represented as 100 right.and also the number of peoples that.passed here.is 12 we know that 12 so.taking this to off we are now we in this.aspect.you want to convert the number of people.that passed.into the percentage wise so we take that.the more people uh that passed.that's 12 divided by 25 times 100.and that gives us 48 so 48.of the people's past the exam so.since we know that the total number of.people's percentage of the number of.people in the class is hundred.and then the percentage of the peoples.that are passed all that we need to do.is to take.that hundred hundred percent minus the.forty eight percent.and that gives us fifty two percent so.therefore.um percentage of people who feel the.examination is 52.so any of the method that you use you.will still arrive at the same answer.so when you meet a question like this is.um.you have to decide if you feel.comfortable going.with the first alternative that i gave.off the first option that i gave.you can go that way after you finish.everything that you the last it is for.you to convert it.or you can decide to convert it and then.put all of them in percentage right and.do that.subtraction when you have already.converted them into.um percentage so this is the question.and this a few of the approaches that.you can use to solve.this particular question now to the next.question.the competition grant for a school.increased from.four thousand eight hundred cities to.six thousand.cities in one year what was the.percentage.increase in the grant quickly to the.solution.now in this is this question is also.similar to the.previous question we just looked at all.right we know our percentage and we know.our initial.grant that was given to us and then we.know our.um our.value the new value that we have which.is the six thousand so.as at this moment we are receiving six.thousand.as our computation grant previously.we're receiving 4.800 but it has now been increased to 6.000.so our initial value here and our.current value here.is so the initial value or the initial.grant is.thousand eight hundred and the current.value is.six thousand so with this knowing our.current value and the initial grant.we we should be able to get them.increase.okay the increase in amount so the.increase in amount.is going to be six thousand which is our.current grant.minus our initial grant which is four.thousand eight hundred.so therefore additional grant is.thousand.two hundred cities all right so also.you're going to convert it because the.question is that percentage increase.in the grant so we're going to convert.this also into a percentage.so that's what we did here 1200 cities.divided by.4800 cities times hundred and that.that gives us twenty five percent so um.twenty five percent i think i made me i.repeated the people's year.so this shouldn't be people's.now to our next question a district.assembly.plans to build a public parking lot the.diagram below represents.the proposed shape of the parking lot.so we have the diagram here you can look.at it here.and then also we have the questions here.so the first question or question is.what is the value of y in the diagram so.the value of this here.the y here the diagram what is the value.and then also.uh b the district assembly so the b part.of the question says the district.assembly wants to fence the parking lot.all right they want to fence this whole.parking lot so this is the shape of the.parking lot.now how many meters of fencing materials.will be required.to fence this particular parking lot so.this is the question and this is the.diagram you have here.let's quickly look at the solution and.how we can approach a question like this.now we have it so the first aspect is.what is the value of.y in the in the diagram so you know this.is y here they're asking us of this.value.so we know 100 meters 50 meters y.120 meters and then this is the 50.meters or so you don't know the value.of this aspect as one as well.now trying too much or trying to use.comparison wise.we can say that this line here is the.same as this line here.all right the hundred meters here is the.same as from here.y all the way to this point so.we can do that dotted line here to.denote that.okay so therefore we can also say that.we can name this as x and then say that.y well i said that 100 meters.is the same as y plus x because this to.from this point here to the point.at the bottom here is x and then from.here upwards is y.so y plus x is the same as 100 meters.all right so using comparison we can.write that down.so 100 meters across y plus x.all right but also using the same.comparison method.we can also say that xcm then.the measurement of x here is the same as.the measurement here.so s can also stand in for 50.50 meters so x here the measurement of x.here.is also the same as the measurement of.uh 50 here so this line and this line.are also the same.so x can act as 50 meters looking at.this here.the same comparison so 50 meters across.x here so that's what we have so.therefore it means we can have.x to be substituted here since x equals.50 meters.so we can have uh therefore 100 meters.across.y plus 50 meters 100 meters it was.y plus 50 meters now with this when we.take.change of subject we should be able to.get 100 meters across.y um plus 50 meters which will give us.y equals 100 meters minus 50 meters.so it means that y equals 50 meters.so the value of y here is.50 meters.now to the next question here the second.part of this question.it says the district assembly want to.fix the parking lot how many meters.of fence and fencing materials will be.required.and this question what we.what the question is requiring from us.what the question is requiring from us.here is to calculate the perimeter of.the whole.parking lot all right so we're going to.add this side 100 meters.plus the 50 meters like this 50 meters.that's the y that we just.um obtained in the previous in the.previous.question the 120 meters and then 50.meters here as well and then this.portion also will be added because.that's the perimeter so the perimeter.here.the perimeter is the length of the outer.edge of a shape.so that is perimeter so this is from the.cambridge advanced uh linux dictionary.third addition so there's a.perimeter that's the length of the outer.edge.of a shape so we need to make sure we.add up.everything with the other the whole.distance.around this this particular parking lot.okay so with that we can now say p.which is the perimeter is 100 meters.this hundred meters cm.plus 50 meters the 50 meters at the top.here plus 50 meters that's this 50.meters as well.plus 120 meters this particular.dimension here.plus 50 meters here that's also here.and then plus so now we need to add this.side.and we don't know it so we need to find.out this.measurement here so using the same.comparison method that we use in the.previous part of the question.of this particular question we can also.equate it here saying that extending.this line.you know that this is the same.measurement as this side.so this line should be the same.measurement as this line so this.should be 50 meters so this is 50 meters.just like this because they all have the.same measurement right.now since they have the same measurement.this is 50 and this is 120.meters because this line comprised of.this aspect and this aspect.so we can now add the 50 meters.and the 120 meters to get the.measurement of this particular.line here so that is 170 meters.here all right so therefore we can now.put.170 meters here and then continue.our calculation so adding up everything.all the figures around this particular.um shape we end up having 540 meters so.the perimeter.of this particular shape.is 540 meters.now to our next question a farmer with.7 by 8 centimeters land decided to.construct a triangular house.with dimensions 3 centimeters.4 centimeters and five centimeters at.the middle of his farm.what is the area of the farmed portion.quickly to the solution now this.question.has given us two different aspects okay.so the farmer.has a portion of land and the land is.seven by eight.centimeters okay so with this value.seven by eight centimeters you know.overall that.we know very well that one of the side.will be longer than the other okay so.with this and then with our previous.knowledge and shapes.you know we wrote a triangle we know the.word that.uh sorry a rectangle fits into this.dimension.because in a red for a rectangle one.side is longer than the other.for a square all the sides are equal.but for a rectangle one side is longer.than the other.so seven by eight centimeters would.definitely mean that.the shape of the farm is in a.rectangular form.all right so we can now put that down.knowing the order that is the.rectangular form.and know that this is not drawn to scale.okay.so this is the shape of the farm.now we have to figure out whose side is.seven centimeters and then which side is.also eight centimeters.so the longer side is the eight.centimeters definitely that's the.highest figure.and then the shorter side is the seven.centimeters.so that is also here all right so we've.gotten.this aspect that's a fiber with seven by.eight centimeters land.so we have gotten that aspect sorted out.seven by eight centimeters that's in a.rectangular form.that we've also known that this side the.longer side is.eight centimeters the shorter side is.there.seven centimeters now to the second part.of the question all right he said.he want to construct a triangular house.with.dimensions three centimeters four.centimeters and five centimeters.at the middle of his farm so we know.also.in fact the question has even given us a.clue that a triangular house a.triangular house meaning is going to be.in a triangle.all right and some questions might even.be silenced about this.they will just give you the dimensions.three centimeters four centimeters of.five centimeters.you should be able to know that these.three values should be able to.form a triangle for you but the question.was very.explicit and gave us triangular house.giving us a cheap day triangular house.so you know we need a triangle at the.middle.of his farm so we have our triangle here.drawn at the middle of the farm and then.we need to label it.so we have 3 centimeters one side.another side also.four centimeters and also five.centimeters.so this is the side of the house is.building so this is the.dimensions of the house is building at.the middle.of his farm okay now the question is.what is the area of the farmed portion.so the portion that he has formed what.is that what is the area of that portion.okay now let's look at the area of this.particular triangle this whole farm that.is having let's look at the area of that.woofer.the area of the triangle of the.rectangle.now the area of the farm is length times.breadth.that's the formula for finding the area.of a rectangle.length times breadth now the length here.is in centimeters.and then the breadth is seven.centimeters.so this gives us 56 centimeters squared.right this is an area so it should be.squared 56 centimeters squared.we have gotten the area of this.particular rectangle.the next we need to do is to find the.area of the house's building all right.so the house is building is in a.triangular form.and then we need to find the area of.that also.so to find the area of a triangle is.half times breadth times height.that's the area of a triangle half.of breadth and height that is what we.are doing.so we take our height.and then we take we we take our half.multiply it by our breath which is the.four centimeters here.and then we also multiply by our height.which is three centimeters.here right and then the answer here is.six.centimeters squared so we've now gotten.that.the area of the house and then the area.of the farm.now remember that the area of the.of the house is in the area of the farm.because he's devoting in the farm he's.building.on that same piece of land so he's.he can no more farm on the whole portion.of the land.portion of the land is going to be used.for the is going to be used for the.building.so since we know the area of that.particular.of the whole farm and then we also know.the area.of the house that we.should be able to get the portion that.he's able to use for farming.solely for farming so we'll move on to.do that now.and then we find the area so area of the.farm minus.area of the house taking out this area.from it.from the whole rectangle or from the.whole farm.we are able to get um.an answer here that is 56 centimeters.that's the area.of the whole farm minus 6 centimeter.square that's the area of the house.and then the answer is 50 centimeters.squared so therefore the area of the.free.of the farmed portion is 50 centimeters.squared all right there is it so.50 centimeters squared the circle graph.below represents classes a.and b so we have class a.boy is 46 percent and then girls x we.don't know the value of girls.and then total is 50. b also boys.x and that because you don't know the.value of boys and then girls 40 percent.so the total here is 60. so the total.number of people's.in class b is 60..the question is how many peoples in the.two classes.are girls to approach this question.there are several ways.of approaching this question i'm going.to take one of the steps.you can try to work it the other way and.then.also see if you arrive at the same.answer i'm going to approach it.by converting everything into percentage.okay so i'm taking it.from the percentage wise using.percentage.i know very well that the total number.of peoples in the class is 50.right now it means the total number of.people we have in the class is 50 and.that will mean that 100 because these 50.people that you have in the class if you.mark register.it means that and they are all present.then it means 100.of the class is present simple 100.denoting the total number of peoples in.the class apart from this hundred.percent.we have also been given the number of.the percentage of boys in the class.so we can take a hundred percent and.then subtract.the percentage of boys from the hundred.percent that the total number of people.the total percentage of the class and.they will that will give us the.percentage of the girls.we take 100 minus 46 percent.then we get 54 that is the percentage of.girls.in the class therefore 50 percent.of 50 across 27 now the 50.so we are now converting it back to the.actual number.okay so i want to know the number of.people the number of girls.in class a when we take 54 percent the.54 percent.is the percentage of girls in the class.and the 50.is the total number of people or.people's in class a.so 54 percent times 50 equals 27.right 27 so we have 27 girls.in class a now when we get to class b.in class b also we have the total number.of peoples to be 60.and then we can also use hundred percent.to denote that.we have been given the percentage of.girls which is 40.so straight straight away we're going to.take 40 percent.times 60 okay 40 percent.is the percentage of girls and then.60 is the total number of girls in the.class so we are converted to the actual.number of.girls in the class so 40 percent times.60.that's 24 all right so in class b.we have 24 girls in class a 27 girls.now the question is how many people's.how many people's in the two classes.are girls to get that we take the two.numbers.add them together and then we have our.two term number of peoples who who are.girls in the two classes.that is 51 here.all right this is all that i have for.you in this particular.lesson before i leave i will leave you.with one question.for you to also look at it this is the.question find the.area of the seeded portion portion.please take this question try your hands.on it.ntc examination is getting closer let's.try to practice as.much as we can and then solve it.find the area of the shaded portion this.particular portion.excluding this this side all right.that's all that i have for you in this.video.i hope you like it if you do please.please and please hit the like button.please share this video to anyone that.will benefit from it and if you are new.to the channel.please consider subscribing and hit the.bell notification icon for you to be.notified every time we upload new video.now in our next video we'll look at.prime numbers.whole numbers real composite numbers.factorials and how their questions.and their questions are structured on.the ntc.so for all of that and more i'll see you.in the next video.

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