**1- Wire cutting problem questions;**

*** If cut from one side = Slip Amount: Initial Length of Wire X Cut Ratio / 2

*** If cut from both sides at the same time = Slip Amount: Initial Length of Wire X | Cut difference | / 2 *** If cut from both sides (cut from the remaining side)=

Slip Amount: First Wire Length X [first cut ratio – (remaining ratio X second cut ratio)] / 2

**2- Queue problem questions;**

*** Maximum number of people in the queue (from head to front, end to back) = Queue max: End to End + End to End + Person in Between

*** At least how many people are in the queue (from head to back, end to front)= Queue min: From Top to Line + Last Line – Person in Between – 2 People *** How many people are in line? rank (to know a person’s rank)=

Rank: Upper Row + Last Row – 1 (Him)

**3- Candle problem questions;**

*** The ratio of the length of the candles after a certain hour (burning) =

1.Candle length – (decreasing length per hour X hours) / 2.Candle length – (decreasing length per hour X hours):

Note: If the ratio is asked, burning times are found, it is considered as the size of the direct candles.

Note: If the ratio is a simple fraction, the 1st candle in the equation; It is a candle with a long flashing time.

*** The difference in the size of the candles after a certain hour (burning) =

1.Candle aspect ratio – (height decreasing per hour X hours) 2.Candle height ratio – (decreasing height per hour X hours): Note: If height difference is asked, it is the smallest common multiple of the burning times. found, k is next to the smallest of the middle layer, not known as n. Note: The 1st candle in the equation; It is a candle with a long flashing time. Otherwise, the result will be negative.

**4- Problem questions for numbering book pages;**

*** In books with a 2-digit page number; *** In books with a 3-digit page number; *** In books with a 4-digit page number;

**5- Forward-backward step problem questions;**

Number of digits used = 2n-9 (n: number of pages) Number of digits used = 3n-108 (n: number of pages) Number of digits used = 4n-1107 (n: number of pages)

** Number of steps away from where it started = Output from TotalSteps/(Number of forward steps+number of backward steps); Let quotient=b, let remainder=k. [ b X (Number of forward steps – number of backward steps) ] + [ number of forward steps – | k – number of forward steps | ]

*** Let the number of steps it moves away from / (Number of forward steps – number of backward steps) be =b and remainder=k.

Maximum total number of steps taken by the number of steps away =

[ b X (Number of forward steps + number of backward steps) ] – k

*** Let the number of steps away / (Number of forward steps – number of backward steps) be =b, remainder=k

never mind.

The minimum total number of steps taken by the number of steps he moves away =

[ (b-1) X (Number of forward steps + number of backward steps) ] + [k + (Number of forward steps – number of backward steps)]

**6- True-false-according to the net, the problem of points;**

*** Number of correct= D, Number of wrong=Y, Number of blanks=B Total Number of Questions= D+Y+(B if any) Net Number=D–Y/4

**7- Chicken-rabbit/ piggy bank-penny-money problem questions;**

*** number of rabbits=r, number of chickens=c,

Since the rabbit has 4 legs and the chicken has 2 legs; Total number of feet = r.4 + c.2;

*** Out of 40 coins; 50 cents piece=a, 25 cents piece= 40-a

All coins in the piggy bank = a.50 + (40-a).25

**8- Bacteria / Lotus-Bamboo plant problem questions;**

*** Let m be the number of days, and k be the number of days that have passed.

k. number of bacteria per day = m. (n to the k)

*** The lotus plant that grows every day as much as the previous day k. if it completely covers the pond where it is located during the day;

(k-1). half of the day, (k-2). occupies 1⁄4 of the day.

*** However, if each day grows twice as much as the previous day (3 times); (k-1). 1/3 of the day, (k-2). It covers 1/9 of the day.

Therefore, if it grows n times (n+1 times) = (k-1). day; 1/(n+1), (k-2).day; 1/(n+1) over 2

**9- Bottle-water problem questions;**

*** Let Bottle = b, Water = w. b+w= X grams, If half of the water is poured, b + w/2 = Y Grams; It is destroyed by addition/subtraction to either b or w party. The question is found.

**10- Cow-feed, scout-eat problem questions**

*** Let there be enough feed for x days for a number of cows. If 5 cows are slaughtered after 5 days;

if — (x-5) days are enough for a cow

(a-5 cows) — ? day enough? With inverse proportion a.(x-5)=(a-5).?

11- Pencil-notebook-eraser / Trouser-shirt-tie problem questions

*** If pencil:k, notebook:d, eraser:s, 2 pencils + 3 notebooks + 1 eraser = a dollar, 6 pencils + 8 notebooks + 4 erasers = b dollars; 1k+1s+1d=a and b types?

(3 k + 4 d + 2 s = b/2) – (2k + 3d + 1s = a) = 1k + 1d + 1s = b/2 – a, multiplying or dividing by the same numbers and adding or subtracting 1k from side to side ,1d,1s price is available.

**12- Ball bouncing problem questions**

*** If a ball dropped from a height h bounces a/b of its previous height after each hit, then falls again and rises after hitting the ground; n. The height it reaches after hitting the ground = h.(a/b)

*** A ball dropped from a height h bounces a/b of its previous height after each hitting the ground, then falls again and hits the ground and rises, and n. If the total distance traveled vertically after hitting the ground once is asked;

Total distance taken vertically = h + 2. h. (a/b) to the 1 + 2nd h. (a/b) to the 2 + …. + 2. h. (a/b) to the (n-1)+ h. (a/b) to the n

**13- Civil servant salary problem questions**

*** Let the civil servant salary: M, kitchen expenditure: MH, bill expenditure: FH, rent expenditure: KH.

He devotes 1/6 of his salary to his MH, 1/5 of the remainder to his FH, and 1/4 of the remainder to his KH. If you have 1,500 TL left;

M. (1-1/6). (1-1/5). (1-1/4) = 1.500 TL salary can be found here.

M. (1-1/6). (1-1/5). (1/4)= HR ; M. (1-1/6). (1/5)=FH ; M. (1/6)= MH

**14- Reading books / number of days problem questions***** Number of days the whole book is read = Page (rate) of the first part of the book read / pages per day +

Page (rate) of the rest of the book / pages per day

**15- Rent/meal money problem questions divided by the group**

*** Let the rental price be k, the number of people in the group: g, the number of people leaving or joining the group: a, the money per person: p.

k = g. p k= (g-a).px k= g. p k= (g+a).py The k values are equalized. Here x and y: give the change in money per capita.

**16- Rank-increasing student problem questions**

*** Let class size be m, number of rows be s.

If a number of students sit on the benches and b students stand, m= (s.a) + b

If y rows remain empty when x number of students sit on the rows, m= (s.x) – (y.x) ; The letters m are equated to each other. The number of rows is found and the current is reached.

When 3 students sit on the benches, 4 students stand. When 4 students sit on the benches, 1 row remains empty, and 1 person sits in 1 row.

m= 3s + 4 = 4s – 4(one row is completely empty) – 3(there are no 3 people in one row)

**17- Linear function/graph problem questions**

*** If f(x)=ax+b then y=ax+b. Write Y when X is 0 and X when Y is 0 into the equation y=ax+b to create 2 equations. Then find a and b using the side-by-side elimination method. After putting it in the Y=AX+B equation, put the value in x according to the situation and condition, and reach y, that is, the result.

**18- Problem questions for male/female students **with glasses, without glasses *** Gl: With glasses, Gs: Without glasses, M: Male students. , B: Ms. never mind

gl. gs. Exy Bmn

The table is drawn, the size of the class (x+y+m+n) is given at the beginning, equations are created (since the student without glasses is 17, y+n=17) according to the next data, and the desired one is reached by the method of elimination side by side.

**19- The male student’s male and female student’s girlfriends problem questions**

*** Let e be the number of male students, k the number of female students. Class size is e+k. The number of male student friends of a male student is e-1 (except himself), and the number of female student friends is k. The number of female student friends of a female student is k-1 (except herself), and the number of male student friends is e.

(If the number of male student friends of a man is 7 less than twice the number of his female friends, and the number of female student friends of a female student in that class is 3 more than half of his male friends; e-1=2k-7, k-1=e/2 + 3 equations form)

**20- Reading as many books as the previous day, problem questions**

*** If he reads twice (as many) books as before, he reads x+2x+4x+8x=15x books on the 1st day x, 2x on the 2nd day, 4x on the 3rd day, 8x on the 4th day. In this case, he will have read 4x books on the 3rd day, and x+2x+4x=7x books at the end of the 3rd day. (The day and the end of the day give different equation results)

*** If he reads 2 times more than before (more, expensive words mean 3 times as much) every day, 1st day x, 2nd day 3x, 3rd day 9x, 4th day 27x in total x+3x+9x+ He reads 27x=40x books. In this case, he will have read 9x books on the 3rd day, and x+3x+9x=13x books at the end of the 3rd day. (The day and the end of the day give different equation results)

*** If he reads n times (as many) books as the previous day and the number of days is given more, 1st day x, 2nd day nx, 3rd day nkarex, 4th day nküpx, so 9th day n is 8 (9-1). He has read x books.

**21- Stair-step problem questions**

*** Let the number of stair steps be m, the number of steps be a. If 2 each, m/2=a (the number of steps taken while exiting),

m/3=a (number of steps taken while descending)

**22- Soda gift to the soda cap problem questions**

*** Total number of soda caps: k, get 1 soda as a gift for 5 caps.

The quotient and the remainder from k/5 (how many caps were gifted with 1 soda) are added up and divided by 5 again; The quotient and remainder are summed and divided by 5 again, the operations are repeated until the quotient+remainder is not divisible by 5. All partitions are summed up and the last remaining is added. The result found is how many soda can you can buy.

**23- Taximeter / Telephone top up problem questions**

*** Total cost: T, opening fee: a, cost per km (every minute): ü, let km (minute): k. (Two

different km.li travel is given, opening fee or another km. The total fee is asked in

**24- Adding numbers to the numerator and denominator of a fraction problem questions**

*** Whether the numerator/denominator of the fraction is a/b, the value of the fraction will be ax/bx.

Let m be the number added to the numerator, n added to the denominator, and the final value of the fraction be e/f. It is solved as ax + m / bx + n = e/f.

T= a + k.u

**25- Age problem questions**

*** Age problems mostly consist of 2 equations that are solved by the elimination or substitution method side by side. The equations are independent of each other. The first equation is the ratio between the ages of the people, the excess, etc. It introduces by giving features. The second equation is years later, years ago, when each other’s age, etc. created with features.

*** If the father’s age is 1 less than 3 times the age of his daughter: B= 3K – 1,

After 6 years, if the father’s age is 8 more than twice the age of his daughter: B+6= 2(F+6)+8 / (Both grow by 6 years)

After 6 years, if the father’s age is 10 less than 4 times the current age of the sheath: B+6= 4K – 10 / (Only the father grows by 6 years)

When the daughter comes to the age of the father; K—-B, B—-B+(B-K)father, the age difference grows. When the father was at the age of his daughter; B—-K, K—-K-(B-K) girl gets smaller as the age difference.

*** Let Kemal’s age be K, Mustafa’s age M, let K>M

If Kemal was born 2 years ago: K+2; If Mustafa was born 3 years later: M-3 Kemal was at K-Mage when Mustafa was born

If there are 4 years before Mustafa was born when Kemal was Mustafa’s age: K—-M, M—-M- (K-

M)=-4 (When Kemal is Mustafa’s age, Mustafa gets smaller as well as the age difference.

if it is 4 years, it is -4 years old)

Let the difference of ages be F and the sum of their ages be T. Sum of their ages 3 years ago T=M+F+3+3, age difference F

constant.

*** If the age of the mother is twice the sum of the ages of her 3 sons: A=2T,

4 years ago, if mother’s age is 2 times more than 5 times the sum of her sons’ ages: A-4=5(T-12)+2/ (As the mother gets 4 years younger, the sons total age decreases by 12)

If the mother’s age 4 years ago is 14 more than the sum of her sons’ present ages: A-4= T+14 / (As the mother gets younger by 4 years, the sum of her sons’ ages does not change)

When the sum of the sons’ ages reaches their mother’s age; since mother’s age=2T,sum of sons’ ages=T

When T— is 2T, if the mother is 2T — 2T+ (2T-T)/3 (if the sum of the three children’s ages increases by 3k, the mother will increase by 1k, so 1/3 of the age difference between the mother total(T) grows up.)

If children are born 3 years apart, their ages can be accepted as n, n+3, n+6. If the eldest child is asked how old is the minimum, the sum of the ages of the children is divided by the number of 3 children and it is tried to be sequential (close to each other).

*** For questions including date of birth; Let Dt: date of birth (19ab), current year (year of speech): By, age: Y.

Y=By-Dt. If a person’s age in 2000 is equal to the digits of his date of birth; 2000-19ab=1+9+a+b; 2000-(1900+10a+b)=10+a+b is solved.

**26- Movement problem questions**

*** General Equation: X=V.t ; X=distance (km-m); V=speed (km/h – m/min – m/sec) , t=time (h – min – sec)

*** 1 m/s=3.6 km/h equation is very useful in unit flip questions.

*** if two vehicles move in opposite directions (meaning towards each other) from different cities; X= (V1+V2).t

*** if two vehicles move in the same direction (faster catches slower) from different cities; X= (V1-V2).t (V1>V2)

*** if two movements move in opposite directions (meaning towards each other) from the same place in a circular track; X= (V1+V2).t where X=Periphery=2.pi.r. In this type of questions n. match times may be asked.

*** if two movements move in the same direction from the same place in a circular track (the fast one overlaps the slow one); X= (V1-V2).t here is X= Perimeter=2.pi.r. In this type of questions n. You may be asked about the times of coming together. Each time they come side by side, the fast vehicle with speed V1 puts the other one in one lap.

*** If a moving at speed V traveling between A and B had moved faster in km/h, the distance between A and B would be x m. if he would take the distance b hours earlier; X=V.t=(V+a).(t-b) the two equations are equalized and the result is reached.

*** X=V.t ; While X and V are directly proportional to X and t; V is inversely proportional to t. The faster you get to the destination, the less time it takes to reach it. The faster you travel in a fixed time, the further you travel. The more time you travel at a constant speed, the more you make.

*** Let the speed of the wind be Vr and the speed of the plane be Vu. X= (Vu-Vr).t if traveling against the wind, if traveling in the same direction as the wind

X= (Vu+Vr).t

*** Let the velocity of your current be Va and the velocity of the swimmer be Vy. The swimmer who goes to the opposite shore in the same direction with the current and comes back to the first shore; X=(Vy+Va).t goes with the equation, X=(Vy-Va).t goes with the equation.(Your current velocity is constant and always in the same direction). Let the travel time be t1 and the return time be t2. t1 < t2.

*** In questions of train-tunnel-tunnel distance, m/min, m/sec, km/h unit conversions should be considered. Let the length of the train be b, the length of the tunnel: n, the distance to the tunnel: m and the speed of the train be Vt.

If the train is at the entrance of the tunnel (not yet): m=Vt. T

If the train has completely entered the tunnel (if the back of the train is at the entrance of the tunnel): m+b=Vt.t

If the train is completely out of the tunnel (if the back of the train is at the exit of the train): m+b+n=Vt.t

*** In athlete-race questions, the units are usually m/min or m/sec, it should be noted. The paths taken by the 1st, 2nd and 3rd runners at a given moment of the race are directly proportional to their own speed. Since they have reached the point they are at in the same time and have not changed their speed in the race, their speeds are 6v, 4v and v, respectively, if the 1st 600m, 2nd 400m, and 3rd competitor have traveled 100m. 1. He has 300m to finish the race and 300m at 6v speed. 2nd Racer is 200m at 4v speed, 3rd Racer is 50m at v speed. takes the lead.

*** Average velocity (Vort) = Total distance (x) / Total time (t)

Let the speed on one part of the road (going) be Va, and let my other wife be the speed on the road (turn) Vb. If the going and returning paths are equal, if the part paths traveled are equal: Vort= 2. Va. Vb / Va+Vb (harmonic mean)

If the times are equal: Vort= Va+Vb/2 (arithmetic mean)

*** In order for two vehicles moving towards each other (in the opposite direction) to meet and pass each other,

In total, they must travel as much as the sum of their heights.

*** In order for the faster vehicle moving in the same direction to catch and pass the slower vehicle, the fast vehicle must travel as much as the sum of its own length and the distance traveled by the slower vehicle.

*** If a vehicle takes the distance between A-B with a speed of Va on the way, and between B-A with a speed of V on the turn, and the total journey takes as long as T; If Va and Vb can be simplified, they are simplified, put k next to Va and the duration of this turn, k next to Vb, which gives the duration of the departure (inverse proportion). Two k expressions are added together and equalized to T (total time). K is found and thus the distance between A and B is found.

The other way is found with Vort= 2.Va.Vb / Va+Vb. Then, X is found from X=Vort.t and divided by 2 to reach between A and B.

*** At Va and Vb speeds, the distance between the two vehicles is x m. They meet at point C t hours after their movement in the opposite direction (towards each other) from different points (A and B cities) that are at a distance of each other and continue on their way. The fast vehicle coming from A goes to B right after the encounter, and the slow moving vehicle departing from B goes to A. if it catches the distance;

The total distance traveled by the fast mover is x + (x-y); The total distance (x-y) taken by the slow one is calculated using Va and Vb in direct proportion accordingly.

*** If a moving road from A to B moves with V speed in 1/4 of it, 2V in 1/3 of the rest, and 2 times increasing its final speed in the remaining part;

If the path is 4x, 1/4 of it is x, 1/3 of the remaining 3x is x, and the last remaining is 2x. In the first x part (with X=V.t) there is time t), in the second x part

(with X=2V.t) the time t/2 is found. In the last remaining 2x part, the speed is 2V+4V=6v and (with 2X=6V.t) the time t/3 is found.

If the speed is increased by 1/5;

After finding [1+(1/5)]=6/5, flip it over, multiply by the time. The elapsed time becomes t.5/6. (Inverse proportion)

If the speed is reduced by 2/7;

After finding [1-(2/7)]=5/7, turn it over, multiply by the time. The elapsed time becomes t.7/5. (Inverse proportion)

**27- Worker-pool problem questions**

*** General equation: t/A + t/B + t/C=1

t: the time they worked together (spent on the same job), A: the time (capacity) of the first worker to finish alone,

B: Second worker alone time to finish (capacity), C: Third worker alone finish time (capacity),

On the right of the equation 1: The work is finished (while working together, if 1/3 of the work is finished, 1/3, if 2/5 of the work remains, 3/5 is finished and 3/5 is written)

The smaller the value of A, B, or C in the general equation, the more capable (fast) the worker is.

If A, B and C are three workers of equal capacity, they can all be given the same value as x.

If the capacity of A is three times that of B and half of that of C, then C has the fastest (most capacity). Therefore, the value starts from C.

C is given x, second capacity A is given 2x, and the slowest B is given 6x. (inverse proportion)

According to the data in the previous item, if worker A increases his capacity by 20% (100% is added by 20%=120% is reached and reversed, multiplied by 2x, which is the capacity of A), the new capacity is found with 2x.100/120. (inverse proportion)

According to the data in the previous item, if worker B reduces his capacity by 20% (subtracted from 100% by 20%=80% is reached and reversed, multiplied by 6x the capacity of B), the new capacity is found with 6x.100/80. (inverse proportion)

If workers A, B and C started working together and completed 3 times the work; t/A + t/B + t/C = 3 solves the equation.

If workers A, B and C started and finished with a one-day break; (t+2)/A + (t+1)/B + t/C = 1 takes the equation to the solution.

If workers A, B and C started with an interval of two days and finished half of them in 8 days; 8/A + 6/B + 4/C = 1/2 takes the equation to the solution. The time spent by the first person who started the job or never left the job is the completion time of the job.

If workers A, B and C work together 2 hours after B, 3 hours after that, C quits and A has finished the remaining work; (2+3+x)/A + (2)/B + (2+3)/C = 1 takes the equation to the solution. In this equation, x is the time to finish the remaining work, and (2+3+x) is the time to finish all the work.

If there are n workers with equal capacities, they can be converted into a single worker with high capacity. Instead of t/x + t/x + …. + t/x=1

t/ (x/n) =1 can be written. (inverse proportion: if 1 person does 1 job in 10 hours, 5 people (with the same power as the first person) will do that job in 10/5=2 hours)

In master-journalist questions; The number of masters is multiplied by the day and the day of 1 master finishing that work, the number of apprentices multiplied by the day and the day of 1 apprentice finishing that work is found. Then the days found and the work done are equated with the right proportion. Thus, the total work done by 1 master and 1 journeyman in an equalized day and the total work asked in the question are directly proportional to the total number of days.

*** General equation: t/A + t/B + t/C=1

In pool problems;

t: the time that the taps stay open together, A: The time (capacity) of the first tap to fill the empty pool alone,

B: Time to fill the empty pool alone (capacity) of the second tap, C: Time to empty the empty pool alone (capacity) of the third tap to its own level,

On the right of the equation 1: The pool is full (1/3 if 1/3 of the work is finished after opening together, if 2/5 of the pool remains empty, 3/5 is filled and 3/5 is written)

**28- Mixing problem questions**

*** General Equation: Amount of solute (gr) / Total Mixture (gr) = x / 100

** Let the amount of solute (usually salt, sugar or alcohol) be: ç, total mixture: k. The mixture is %xituzise: k.(x/100)/k begins to solve the question.

If y% of the mixture is water: The question begins to be solved with the equation k.(100-y)/100 / k.

Note: Even if water is given in the question, we always write the amount of salt, sugar or alcohol in the share part.

If there is a gr water and b gr salt in the mixture, the equation b / (a+b) = x / 100 gives the percentage of salt. If there is a gr water and b gr salt in the mixture, the equation a / (a+b) = x / 100 gives the percentage of water.

Note: Even if the water percentage is asked in the question, the equation is solved according to salt, sugar or alcohol (what is written in the numerator part), the result found is subtracted from 100% and the result is reached.

** 300

20% of a gram mixture is sugar. If x g sugar is added to the mixture; A. (20/100) + x / A+x sugar amount is added to both the numerator and the denominator.

20% of a gram mixture is sugar. If x gr water is added to the mixture; A. (20/100) / A+x water amount is only added to the denominator.

20% of a gram mixture is sugar. If x g of water is evaporated from the mixture; A. (20/100) / A-x is only subtracted from the denominator since only water evaporates.

20% of a gram mixture is sugar. If x g sugar and y g water are added to the mixture; A. (20/100) +x / A + x + y sugar amount is added to both the numerator and the denominator, water is added to the denominator only.

There is x gr salt in a gram mixture. If 1⁄4 of this mixture is poured; (x – (x/4)) / (A- (A/4)) 1⁄4 of both the salt and the total mixture is reduced.

There is 90 g of salt in a 1 g mixture. 1/3 of this mixture is poured and instead of the poured amount;

If salt is added; 90-30+100 / 300-100+100 (the amount poured is 100 gr. 100 gr of salt is added to both the numerator and the denominator.

If water is added; 90-30 / 300-100+100 (the amount poured is 100 gr. 100 gr of water is only added to the denominator.)

If the mixture consisting of 20% salt is added; 90-30+20 / 300-100+100 (the amount poured is 100 gr. 20% of the 100 gr mixture, that is 20 grams of salt, is added to the denominator and salt to the fraction.)

** A gr. x g salt in the mixture, B g. There is y grams of salt in the mixture. If two mixtures are mixed in a bowl; x+y / A+B = ? / one hundred

** A gr. x g salt in the mixture, B g. There is y grams of salt in the mixture. If 1/3 of A gr of mixture and 1/4 of B gr of mixture are mixed in another bowl; (x.1/3+y.1/4) / (A/3+B/4) = ? / one hundred

** 200 gr. 40gr salt, 300 gr. There is 30 g of salt in the mixture. A gr. If half of the mixture is poured into B and mixed, 1/5 of the B mixture is poured into A;

A= 40-20 / 200-100 ie 20/100 —- B= 30+20 / 300+100 ie 1/5 of 50 / 400 will be poured into 10/80 A 20+10 / 100+80= 30 / 180 = ? / one hundred

** While going over % in general mixing problems, it is calculated over 24 for gold carat questions and over 1000 for silver carat questions.

If there is x gr of pure gold in the A gr jewelry; x/A = ? The setting can be found with the equation / 24.

If the Y carat jewelry is B gram; y/24 = ? Pure gold gram in jewelery with equation B

can be found.

It contains x gr. pure gold, A gr. if y g of pure gold is added to the jewelry; x+y / A+y = ? / 24 last

settings can be found.

** Selling mixed products at different prices and weights; The prices and kilos of each different product are multiplied, and after all are added together, the average price can be found by dividing by the total kilo.

1- Percent-Profit/Loss problems

20% of a number; x.20/100

40% of 30% of which number; x.(30./100).(40/100)

Sum of 25% and 35% of a number; x.(25/100) + x.(35/100)

The difference between 45% and 15% of which number; x.(45/100) – x(15/100)

20% more than a number; x + x.(20/100) briefly x.(120/100)

Which number is 10% less; x – x(10/100) x in short.(90/100)

15% more than 25% less than a number; x.(75/100).(115/100)

The difference between which number is 30% more and 20% less; x.(130/100) – x.(80/100)

30% less than 10% more than 20% of a number; x.(110/100).(20/100).(70/100)

Let the money in the pocket be p. 20% of the pocket money (salary) is spent, 25% of the rest is spent,

if 15% of the remainder is spent and x TL remains; p.(80/100).(75/100).(85/100)=x where p can be found.

If the number of A increases and becomes B; B-A / A = ? / 100 (give a percentage increase)

If the number of A decreases and becomes B; A-B / A= ? / 100 (give percent decrease)

By how many % the number A is more than the number B: A – B / B (Always large number – small number;

in the denominator, the word is taken from -subjector relative; in comparison, the basenumber is written)

How many % the number B is more than the number A: B – A / A (The numerator part is always large number – small number; denominator part is taken from -subjector relative; in comparison, basenumber is written)

What is x% of A in B: A.(x/100)=B.(?/100)

A number is what % of B: A=B.(?/100)

A number is how many percent more than B: A=B.[(100+x)/100] : x gives the direct answer.

A number is how many % less than B: A=B.[(100-x)/100] : x gives a direct answer.

x % of A number: multiply A by x then divide by 100.

Number with x% of A: Divide A by x then multiply by 100.

%xyA%ys of the numbers: (let’s say the numberB) B.(x/100)=the numberB is found by Abudenkle and

The result is obtained with the equation B.(y/100).

** Wheat-flour-dough problem questions (Wheat: B); If flour is obtained from x% of wheat and dough is obtained from y% of flour, in order to obtain A gram dough: B.(x/100).(y/100)=A can reach the weight of wheat.

** Let the number of boys be: e, number of girls: k in male-female student problem questions. If x% of the class is male

Ratio of girls to class: (100-x) / 100, ratio of girls to boys: (100-x) / x

if a number of girls come from outside; (e+k): means class size. (e+k). (x/100) / (e+k) + a (male)

If we go through the percentage, the number of girls coming from outside is only in the denominator, that is, in the total number.

added)

If b numbers of men come from outside: (e+k). (x/100) +b / (e+k) + b (in terms of male percentage)

the number of men coming from outside is added to both the numerator, that is, the number of men, and the denominator, that is, the whole existing)

** Let the number of boys be: e, the number of girls: k in male-female student problem questions. If more than a % of the class is girls; Minimum number of girls; (e+k).(a/100) + 1.

Max number of men; (e+k).[(100-a)/100]-1.

** Let the number of boys be: e, the number of girls: k in male-female student problem questions. If 45% of the class is girls;

Female ratio: 45%, male ratio: 55%. Therefore, k=9m, e:11m comes out from the proportioning and simplification of 45 and 55. Accordingly, the number of female students will definitely be a multiple of 9, and the number of male students will be a multiple of 11. In total available (9m+11m=20m) it will be a multiple of 20.

According to the above data (e+k)=20m, m is found from here and k is reached if it is multiplied by 9.

According to the above data (e+k), that is, if the total is more than 600; More than 600, which is a multiple of 20

the first number is reached. The number 620 found is divided by 20 and the number m, that is, 31, is found. Multiply by 11 to get the minimum number of men 341.

** In tire stretching (length extension) problem questions; If the tire of x cm length stretches 130% when pulled and becomes A cm:

x. (100+130) / 100 = A with this equation, the initial length of the tire can be reached.

** Capacity (power) and time, waste (grape-soap drying / egg breaking) and cost etc. in inverse proportional problem questions;

If the capacity (or number of people) increases by x% [(100+x)/100] for the newly created time (first time: t):

t. [100 / (100+x)) … (invert multiply technique)

If the capacity (or number of people) decreases by x% [(100-x)/100] for the newly formed time (initial time: t):

t. [100 / (100-x)] … (invert multiplication technique)

If x% of the egg is broken (if x% of the grape dries up), the initial cost of the egg (let it be m) varies according to the sales situation (even though the cost is not out of pocket plus money, the cost has increased depending on the sales situation) In other words, the cost incurred:

m. [100 / (100-x)] … (invert multiplication technique)

The question of adding water to milk, which comes out very rarely, is the opposite of wastage. If x% water is added to milk, the initial cost of milk

(Let it be m) varies according to the sales situation (although the cost does not come out as minus money because there is an addition, the cost has decreased according to the sales situation) That is, the cost incurred: m. [100 / (100+x)] … (invert multiply technique)

** Trade; Let Purchase price + Expense = Cost price + Profit = Label price – Discount = Selling price Cost: m, label price: e, sale price: s.

If x%profit is made on cost: m.(100+x)/100=s(ore)

If x% loss is incurred: m. (100-x) / 100 = s (or e

If x% profit is made on the label(sales): e.(100+x)/100=s

If x% loss is made on the label (over sales): e.(100-x) / 100 = s

If x% profit is made on sales: p. (100-x) / 100 = m (subtracted from % when snow)

If x% loss is made by sales: p. (100+x) / 100 = m (added to % when damaged)

If a y% discount is made after x% profit is made on the cost (over that price): m. [(100+x)/100] . [(100-y)/100] = h

If the label is increased by y% after an increase of x%: e. [(100+x)/100] . [(100+y)/100] = h

If an increase of x% is made on the label, y% profit is made on the cost: e. [(100+x)/100] = m.

[(100+y)/100]

If there is a loss of y% on the cost when x% discount is made from the sale (over the sale): p. [(100-x)/100] = m. [(100-y)/100]

** Bought for X and sold to Y;

Profit is earned if X < Y. % Profit rate = (Y-X) / X = (? / 100) … cost (buying) in the denominator. If X > Y, it is damaged. % Loss ratio = (X-Y) / X = (? /100) … cost (buy) in the denominator.

**In inflation/salary problem questions;

Inflation: 20%, if salaries are increased by 32% (inflation: let the price of bread rise. bread price: 100); (salary:100)

100.(120/100)=120(increased price of bread); 100.(132/100)=132(increase salary)

If the increased salary > the increased price of the bread, the purchasing power of the civil servant increases.

Ratio = 132-120/120 (the denominator will always be the price of bread, that is, the price with inflation)

Note: If the inflation rate is greater than the salary increase rate, the purchasing power of the civil servant decreases. In this case, ” (inflationary price – increased salary) / inflationary price (again) ”’

** In the problem questions where the customer increases when the prices are reduced;

Let’s assume the unit price is 100 and the number of customers is 100. If the number of customers increases by 40% when the prices are reduced by 25%; 100.100—-75.140 = 10.000TL—-10.500TL will be.5%profit