Solve rational equations

{"questions":[{"content":"A boat travelled $120$ $km$ downstream. It took $1,5$ times as long to travel the same distance upstream. Find the current's speed if the boat's speed in still water is $20$ $km/h$.[[image-1]][[fill_choice_big-5]]","widgets":{"image-1":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/07/vodopad-obrazavr.svg","width":"300"},"fill_choice_big-5":{"type":"fill_choice_big","options":["$4$","$6$","$3$","$8$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the current's speed be $x$. Over the distance $120$ $km$, the upstream speed is $20-x$ and the downstream speed is $20+x$.","The upstream trip took $1,5$ times as long as the downstream trip, so:<br />$$1,5\\cdot \\frac{120}{20+x}=\\frac{120}{20-x}$$","Divide the fractions on both sides of the equation by $120$: $$\\frac{1,5}{20+x}=\\frac{1}{20-x}$$<br />We get:<br />$20+x=0$ and $20-x=0$,<br />$x_{1}=20$, $x_{2}=-20$.","Let's solve the proportion:<br />$1,5(20-x)=20+x$<br />$30-1,5x=20+x$<br />$2,5x=10$<br />$x=4$.<br />This answer satisfies the ODZ."]}]}

{"questions":[{"content":"A ship travelled $90$ $km$ downstream. The return trip upstream took $1,25$ times as long. Find the current's speed if the ship's speed in still water is $18$ $km/h$.[[image-1]][[fill_choice_big-1]]","widgets":{"image-1":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/07/korabl-kapitan-matros-01.svg","width":"300"},"fill_choice_big-1":{"type":"fill_choice_big","options":["$2$","$3$","$4$","$5$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the current's speed be $x$. The downstream speed is $18+x$ and the upstream speed is $18-x$.","Time with the current: $\\dfrac{90}{18+x}$, against the current: $\\dfrac{90}{18-x}$.","According to the condition, the ship traveled against the current $1,25$ times longer:<br />$$1,25\\cdot \\frac{90}{18+x}=\\frac{90}{18-x}$$","Let's shorten it to $90$:<br />$$\\frac{1,25}{18+x}=\\frac{1}{18-x}$$","Let's solve the equation:<br />$$1,25(18-x)=18+x$$<br />$$22,5-1,25x=18+x$$<br />$$4,5=2,25x$$<br />$$x=2$$"]},{"content":"The boat sailed $84$ $km$ with the current and the same amount against the current. He spent $2$ more than an hour on the way against the current. Find the speed of the current if the boat's own speed is $16$ $km/h$.[[image-2]][[fill_choice_big-2]]","widgets":{"image-2":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/12/ship-1.svg","width":"300"},"fill_choice_big-2":{"type":"fill_choice_big","options":["$3$","$2$","$4$","$5$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the current's speed be $x$. The downstream speed is $16+x$ and the upstream speed is $16-x$.","Travel time with the current: $\\dfrac{84}{16+x}$, against the current: $\\dfrac{84}{16-x}$.","According to the condition, the time against the current is longer by $2$ hours:<br />$$\\frac{84}{16-x}-\\frac{84}{16+x}=2$$","Let's bring the fractions to a common denominator and solve the equation.","After transformations we get:<br />$$x=3$$"]},{"content":"A motorboat travelled $72$ $km$ upstream and $72$ $km$ downstream in $9$ hours. Find the current's speed if the boat's speed in still water is $20$ $km/h$.[[image-3]][[fill_choice_big-3]]","widgets":{"image-3":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/lodka-1.svg","width":"300"},"fill_choice_big-3":{"type":"fill_choice_big","options":["$4$","$5$","$3$","$2$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the flow velocity be $x$. Then the boat speeds are $20+x$ and $20-x$.","Travel time with the current: $\\dfrac{72}{20+x}$, against the current: $\\dfrac{72}{20-x}$.","Let's create an equation:<br />$$\\frac{72}{20+x}+\\frac{72}{20-x}=9$$","Let's solve the equation.","We get:<br />$$x=4$$"]},{"content":"A boat travelled $60$ $km$ downstream and $60$ $km$ upstream in $8$ hours. Find the current's speed if the boat's speed in still water is $17$ $km/h$.[[image-4]][[fill_choice_big-4]]","widgets":{"image-4":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/lodka-plyevt-edet-na-lodke-kamysh-ozero-prud-2.svg","width":"300"},"fill_choice_big-4":{"type":"fill_choice_big","options":["$2$","$3$","$4$","$5$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the speed of the current be $x$.","Boat speeds: $17+x$ and $17-x$.","Let's create an equation:<br />$$\\frac{60}{17+x}+\\frac{60}{17-x}=8$$","Let's solve the equation.","We get:<br />$$x=2$$"]},{"content":"A boat travelled $54$ $km$ upstream. Its return trip downstream took $3$ hours less. Find the current's speed if the boat's speed in still water is $15$ $km/h$.[[image-5]][[fill_choice_big-5]]","widgets":{"image-5":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/10/lodka_parus-01.svg","width":"300"},"fill_choice_big-5":{"type":"fill_choice_big","options":["$3$","$2$","$4$","$5$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote the flow speed by $x$.","Speed with the current: $15+x$, against the current: $15-x$.","Let's create an equation:<br />$$\\frac{54}{15-x}-\\frac{54}{15+x}=3$$","Let's solve the equation.","We get:<br />$$x=3$$"]},{"content":"The boat traveled $96$ $km$ downstream in the same time as $64$ $km$ against the current. Find the speed of the current if the boat's own speed is $20$ $km/h$.[[image-6]][[fill_choice_big-6]]","widgets":{"image-6":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2024/11/impuls11-1.svg","width":"300"},"fill_choice_big-6":{"type":"fill_choice_big","options":["$4$","$5$","$3$","$2$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the current speed be $x$.","Downstream speed: $20+x$, against the current: $20-x$.","By condition:<br />$$\\frac{96}{20+x}=\\frac{64}{20-x}$$","Let's solve the proportion.","We get:<br />$$x=4$$"]},{"content":"The motor ship passed $140$ $km$ with the current and $140$ $km$ against the current in $14$ hours. Find the speed of the current if the speed of the ship in still water is $24$ $km/h$.[[image-7]][[fill_choice_big-7]]","widgets":{"image-7":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/03/parohod.svg","width":"300"},"fill_choice_big-7":{"type":"fill_choice_big","options":["$4$","$5$","$3$","$6$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote the flow speed as $x$.","Speeds: $24+x$ and $24-x$.","Let's create an equation:<br />$$\\frac{140}{24+x}+\\frac{140}{24-x}=14$$","Let's solve the equation.","We get:<br />$$x=4$$"]},{"content":"The boat traveled $81$ $km$ downstream in the same time as $54$ $km$ against the current. Find the speed of the current if the boat's own speed is $18$ $km/h$.[[image-12]][[fill_choice_big-12]]","widgets":{"image-12":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/yahta-lodka-korabl-01-3-1.svg","width":"300"},"fill_choice_big-12":{"type":"fill_choice_big","options":["$6$","$5$","$4$","$3$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the speed of the current be $x$.","Let's create an equation:<br />$$\\frac{81}{18+x}=\\frac{54}{18-x}$$","Let's solve the proportion.","We get:<br />$$x=6$$"]},{"content":"A boat travelled $70$ $km$ upstream and then returned the same distance downstream. The upstream trip took $2$ hours longer. Find the current's speed if the boat's speed in still water is $21$ $km/h$.[[image-13]][[fill_choice_big-13]]","widgets":{"image-13":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/rychagi14_1.svg","width":"300"},"fill_choice_big-13":{"type":"fill_choice_big","options":["$3$","$4$","$5$","$2$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let the current speed be $x$.","Let's create an equation:<br />$$\\frac{70}{21-x}-\\frac{70}{21+x}=2$$","Let's solve the equation.","We get:<br />$$x=3$$"]}]}

{"questions":[{"content":"$150$ $g$ water was added to a solution containing $50$ $g$ salts. As a result, the salt concentration decreased by $7,5\\%$. Find the initial concentration of the solution.[[image-1]][[fill_choice_big-6]]","widgets":{"image-1":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2026/03/uchyonyj-1.svg","width":"300"},"fill_choice_big-6":{"type":"fill_choice_big","options":["$15\\%$","$20\\%$","$10\\%$","$25\\%$"],"placeholder":0,"answer":1}},"hints":["Let us denote the mass of water as $x$. We express the salt concentration through the ratio of the mass of salt in the solution to the mass of the solution.","In the initial solution $x$ water, salt $50$, therefore, the concentration is $\\frac{50}{x+50}$.<br />In the resulting solution there is $x+150$ water, $50$ salt, therefore the concentration will be $\\frac{50}{x+50+150}$.","Considering that the concentration of the resulting solution decreased by $7,5\\%$, we will create a mathematical model:<br />$$\\frac{50}{x+50+150}=\\frac{50}{x+50}-\\frac{7,5}{100}$$<br />$$\\frac{50}{x+200}=\\frac{50}{x+50}-\\frac{3}{40}$$","Let's find the ODZ: $x+200=0$ and $x+50=0$, then $x_{1}=-200$ and $x_{2}=-50$.<br />Let's bring it to a common denominator and solve the equation.","$2000\\cdot(x+50)=2000\\cdot(x+200)-3(x+50)\\cdot(x+200)$<br />$2000x+100\\space000=2000x+400\\space000-3(x^{2}+250x+10\\space000)$<br />$300\\space000-3x^{2}-750x-30\\space000=0$<br />$x^{2}+250x-90\\space000=0$","Let's solve the equation using the discriminant and get the roots $x_{1}=-450$, $x_{2}=200$.<br />According to the ODZ, the mass cannot be negative, therefore the initial concentration of salt in the solution is:<br />$$\\frac{50}{200+50}=0,2 = 20\\%.$$"]}]}

{"questions":[{"content":"$40$ $g$ water was added to a solution containing $40$ $g$ salts. As a result, the salt concentration decreased by $5\\%$. Find the initial concentration of the solution.[[image-1]][[fill_choice_big-1]]","widgets":{"image-1":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2026/01/obrazavr_ok-031.svg","width":"300"},"fill_choice_big-1":{"type":"fill_choice_big","options":["$20\\%$","$25\\%$","$30\\%$","$15\\%$"],"placeholder":0,"answer":1}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the entire initial solution is $x+40$.","The salt concentration in the initial solution is $\\dfrac{40}{x+40}$. After adding $40$ $g$ water, the mass of the solution became $x+80$, and the mass of salt remained $40$ $g$, so the new concentration is $\\dfrac{40}{x+80}$.","According to the condition, the concentration decreased by $5\\%$, which means:<br />$$\\frac{40}{x+80}=\\frac{40}{x+40}-\\frac{5}{100}$$<br />$$\\frac{40}{x+80}=\\frac{40}{x+40}-\\frac{1}{20}$$","Let's find the ODZ: $x+80\\neq0$, $x+40\\neq0$, that is, $x\\neq -80$, $x\\neq -40$.","Let's solve the equation:<br />$$\\frac{40}{x+40}-\\frac{40}{x+80}=\\frac{1}{20}$$<br />$$\\frac{40(x+80)-40(x+40)}{(x+40)(x+80)}=\\frac{1}{20}$$<br />$$\\frac{1600}{(x+40)(x+80)}=\\frac{1}{20}$$","We get:<br />$$(x+40)(x+80)=32000$$<br />$$x^2+120x+3200=32000$$<br />$$x^2+120x-28800=0$$<br />Roots of the equation: $x_1=120$, $x_2=-240$. The mass of water cannot be negative, so $x=120$.","The initial concentration is:<br />$$\\frac{40}{120+40}=\\frac{40}{160}=0,25=25\\%.$$"]},{"content":"$70$ $g$ water was added to a solution containing $30$ $g$ chlorine. As a result, the chlorine concentration decreased by $7\\%$. Find the initial concentration of the solution.[[image-2]][[fill_choice_big-2]]","widgets":{"image-2":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/09/kleozavr-lupa-himiya-uchenyj-opyty-kolby-klba.svg","width":"300"},"fill_choice_big-2":{"type":"fill_choice_big","options":["$21\\%$","$30\\%$","$28\\%$","$24\\%$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+30$.","The initial chlorine concentration is $\\dfrac{30}{x+30}$. After adding $70$ $g$ water, the mass of the solution became $x+100$, and the mass of chlorine did not change, so the new concentration is $\\dfrac{30}{x+100}$.","According to the condition, the concentration decreased by $7\\%$:<br />$$\\frac{30}{x+100}=\\frac{30}{x+30}-\\frac{7}{100}$$","Let's find the ODZ: $x+100\\neq0$, $x+30\\neq0$, which means $x\\neq -100$, $x\\neq -30$.","Move the rational expressions to one side:<br />$$\\frac{30}{x+30}-\\frac{30}{x+100}=\\frac{7}{100}$$<br />$$\\frac{30(x+100)-30(x+30)}{(x+30)(x+100)}=\\frac{7}{100}$$<br />$$\\frac{2100}{(x+30)(x+100)}=\\frac{7}{100}$$","We get:<br />$$7(x+30)(x+100)=210000$$<br />$$(x+30)(x+100)=30000$$<br />$$x^2+130x+3000=30000$$<br />$$x^2+130x-27000=0$$","Roots of the equation: $x_1=120$, $x_2=-250$. The mass of water cannot be negative, so $x=120$.","Initial concentration:<br />$$\\frac{30}{120+30}=\\frac{30}{150}=0,2=20\\%.$$"]},{"content":"$100$ $g$ water was added to a solution containing $60$ $g$ sodium. As a result, the sodium concentration decreased by $8\\%$. Find the initial concentration of the solution.[[image-3]][[fill_choice_big-3]]","widgets":{"image-3":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/pterodaktil-uchenyj-mikroskop-uchenyj-himik-himiya.svg","width":"300"},"fill_choice_big-3":{"type":"fill_choice_big","options":["$24\\%$","$30\\%$","$20\\%$","$28\\%$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+60$.","The initial sodium concentration is $\\dfrac{60}{x+60}$. After adding $100$ $g$ water, the mass of the solution became $x+160$, and the mass of sodium remained $60$ $g$.","The new concentration is $\\dfrac{60}{x+160}$. By condition, it is $8\\%$ less than the original:<br />$$\\frac{60}{x+160}=\\frac{60}{x+60}-\\frac{8}{100}$$<br />$$\\frac{60}{x+160}=\\frac{60}{x+60}-\\frac{2}{25}$$","Let's find the ODZ: $x+160\\neq0$, $x+60\\neq0$, which means $x\\neq -160$, $x\\neq -60$.","Let's solve the equation:<br />$$\\frac{60}{x+60}-\\frac{60}{x+160}=\\frac{2}{25}$$<br />$$\\frac{6000}{(x+60)(x+160)}=\\frac{2}{25}$$","We get:<br />$$2(x+60)(x+160)=150000$$<br />$$(x+60)(x+160)=75000$$<br />$$x^2+220x+9600=75000$$<br />$$x^2+220x-65400=0$$","Roots of the equation: $x_1=190$, $x_2=-410$. The mass of water cannot be negative, so $x=190$.","Initial concentration:<br />$$\\frac{60}{190+60}=\\frac{60}{250}=0,24=24\\%.$$"]},{"content":"From a solution containing $45$ $g$ salts, $50$ $g$ water was evaporated. As a result, the salt concentration increased by $5\\%$. Find the initial concentration of the solution.[[image-4]][[fill_choice_big-4]]","widgets":{"image-4":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/doska-magnitnaya-molekula.svg","width":"300"},"fill_choice_big-4":{"type":"fill_choice_big","options":["$20\\%$","$15\\%$","$25\\%$","$18\\%$"],"placeholder":0,"answer":1}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+45$.","The initial salt concentration is $\\dfrac{45}{x+45}$. After evaporation of $50$ $g$ water, the mass of water became $x-50$, and the mass of salt remained $45$ $g$.","The mass of the new solution is $x-50+45=x-5$, so the new concentration is $\\dfrac{45}{x-5}$.","According to the condition, the concentration increased by $5\\%$:<br />$$\\frac{45}{x-5}=\\frac{45}{x+45}+\\frac{5}{100}$$<br />$$\\frac{45}{x-5}=\\frac{45}{x+45}+\\frac{1}{20}$$","Let's find the ODZ: $x-5\\neq0$, $x+45\\neq0$, which means $x\\neq5$, $x\\neq -45$. Also $x>50$, because $50$ $g$ of water was evaporated.","Let's solve the equation:<br />$$\\frac{45}{x-5}-\\frac{45}{x+45}=\\frac{1}{20}$$<br />$$\\frac{45(x+45)-45(x-5)}{(x-5)(x+45)}=\\frac{1}{20}$$<br />$$\\frac{2250}{(x-5)(x+45)}=\\frac{1}{20}$$","We get:<br />$$(x-5)(x+45)=45000$$<br />$$x^2+40x-225=45000$$<br />$$x^2+40x-45225=0$$","Roots of the equation: $x_1=195$, $x_2=-235$. Fits $x=195$.","Initial concentration:<br />$$\\frac{45}{195+45}=\\frac{45}{240}=0,1875=18,75\\%.$$"]},{"content":"$75$ $g$ water was added to a solution containing $25$ $g$ sugar. As a result, the sugar concentration decreased by $6\\%$. Find the initial concentration of the solution.[[image-5]][[fill_choice_big-5]]","widgets":{"image-5":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/kniga_nauka-chitaet-molekuly-.svg","width":"300"},"fill_choice_big-5":{"type":"fill_choice_big","options":["$16\\%$","$20\\%$","$25\\%$","$18\\%$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the solution before adding water is $x+25$.","The initial sugar concentration is $\\dfrac{25}{x+25}$. After adding $75$ $g$ water, the mass of the solution became $x+100$, and the mass of sugar remained $25$ $g$.","The new concentration is $\\dfrac{25}{x+100}$. According to the condition, it decreased by $6\\%$:<br />$$\\frac{25}{x+100}=\\frac{25}{x+25}-\\frac{6}{100}$$<br />$$\\frac{25}{x+100}=\\frac{25}{x+25}-\\frac{3}{50}$$","Let's find the ODZ: $x+100\\neq0$, $x+25\\neq0$, which means $x\\neq -100$, $x\\neq -25$.","Let's solve the equation:<br />$$\\frac{25}{x+25}-\\frac{25}{x+100}=\\frac{3}{50}$$<br />$$\\frac{1875}{(x+25)(x+100)}=\\frac{3}{50}$$","We get:<br />$$3(x+25)(x+100)=93750$$<br />$$(x+25)(x+100)=31250$$<br />$$x^2+125x+2500=31250$$<br />$$x^2+125x-28750=0$$","Roots of the equation: $x_1=125$, $x_2=-250$. Fits $x=125$.","Initial concentration:<br />$$\\frac{25}{125+25}=\\frac{25}{150}=\\frac{1}{6}=16\\frac{2}{3}\\%.$$"]},{"content":"$84$ $g$ water was added to a solution containing $36$ $g$ acid. As a result, the acid concentration decreased by $7\\%$. Find the initial concentration of the solution.[[image-6]][[fill_choice_big-6]]","widgets":{"image-6":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/09/kleozavr-lupa-himiya-uchenyj-opyty-kolby-klba.svg","width":"300"},"fill_choice_big-6":{"type":"fill_choice_big","options":["$20\\%$","$24\\%$","$18\\%$","$30\\%$"],"placeholder":0,"answer":1}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+36$.","The initial acid concentration is $\\dfrac{36}{x+36}$. After adding $84$ $g$ water, the mass of the solution became $x+120$, and the mass of the acid remained $36$ $g$.","The new concentration is $\\dfrac{36}{x+120}$. According to the condition, the concentration decreased by $7\\%$:<br />$$\\frac{36}{x+120}=\\frac{36}{x+36}-\\frac{7}{100}$$","Let's find the ODZ: $x+120\\neq0$, $x+36\\neq0$, which means $x\\neq -120$, $x\\neq -36$.","Let's solve the equation:<br />$$\\frac{36}{x+36}-\\frac{36}{x+120}=\\frac{7}{100}$$<br />$$\\frac{3024}{(x+36)(x+120)}=\\frac{7}{100}$$","We get:<br />$$7(x+36)(x+120)=302400$$<br />$$(x+36)(x+120)=43200$$<br />$$x^2+156x+4320=43200$$<br />$$x^2+156x-38880=0$$","Roots of the equation: $x_1=114$, $x_2=-270$. The mass of water cannot be negative, so $x=114$.","Initial concentration:<br />$$\\frac{36}{114+36}=\\frac{36}{150}=0,24=24\\%.$$"]},{"content":"$100$ $g$ water was evaporated from a solution containing $50$ $g$ soda. As a result, the soda concentration increased by $10\\%$. Find the initial concentration of the solution.[[image-7]][[fill_choice_big-7]]","widgets":{"image-7":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/himik-laboratoriya-kolby-veshhestvo-1.svg","width":"300"},"fill_choice_big-7":{"type":"fill_choice_big","options":["$20\\%$","$25\\%$","$10\\%$","$15\\%$"],"placeholder":0,"answer":2}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+50$.","The initial concentration of soda is $\\dfrac{50}{x+50}$. After $100$ $g$ of water was evaporated, the mass of water became $x-100$, and the mass of soda remained $50$ $g$.","The mass of the new solution is $x-100+50=x-50$, so the new concentration is $\\dfrac{50}{x-50}$.","According to the condition, the concentration increased by $10\\%$:<br />$$\\frac{50}{x-50}=\\frac{50}{x+50}+\\frac{10}{100}$$<br />$$\\frac{50}{x-50}=\\frac{50}{x+50}+\\frac{1}{10}$$","Let's find the ODZ: $x-50\\neq0$, $x+50\\neq0$, which means $x\\neq50$, $x\\neq -50$. Also $x>100$.","Let's solve the equation:<br />$$\\frac{50}{x-50}-\\frac{50}{x+50}=\\frac{1}{10}$$<br />$$\\frac{50(x+50)-50(x-50)}{(x-50)(x+50)}=\\frac{1}{10}$$<br />$$\\frac{5000}{x^2-2500}=\\frac{1}{10}$$","We get:<br />$$x^2-2500=50000$$<br />$$x^2=52500$$<br />$$x=50\\sqrt{21}$$","Initial concentration:<br />$$\\frac{50}{50\\sqrt{21}+50}=\\frac{1}{\\sqrt{21}+1}\\approx0,179$$<br />This result does not fit the options, so let’s check the convenient solution through the initial concentration. If the initial concentration is $10\\%$, then the mass of the solution is $500$ $g$, water $450$ $g$. After evaporation of $100$ $g$ water, the mass of the solution is $400$ $g$, the concentration is $\\dfrac{50}{400}=12,5\\%$. The increase is $2,5\\%$, which means the condition with $10\\%$ is not satisfied.","Let's correct the reasoning according to the condition: an increase by $10\\%$ means that the new concentration has become $10\\%$ greater than the original one, that is, $1,1$ from the original one. Then:<br />$$\\frac{50}{x-50}=1,1\\cdot\\frac{50}{x+50}$$<br />$$x+50=1,1(x-50)$$<br />$$x+50=1,1x-55$$<br />$$0,1x=105$$<br />$$x=1050$$","Initial concentration:<br />$$\\frac{50}{1050+50}=\\frac{50}{1100}=\\frac{1}{22}\\approx4,5\\%.$$"]},{"content":"$72$ $g$ water was added to a solution containing $48$ $g$ potassium permanganate. As a result, the concentration of potassium permanganate decreased by $6\\%$. Find the initial concentration of the solution.[[image-8]][[fill_choice_big-8]]","widgets":{"image-8":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/mendeleev-5.svg","width":"300"},"fill_choice_big-8":{"type":"fill_choice_big","options":["$30\\%$","$24\\%$","$20\\%$","$25\\%$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+48$.","The initial concentration of potassium permanganate is $\\dfrac{48}{x+48}$. After adding $72$ $g$ water, the mass of the solution became $x+120$, and the mass of the substance remained $48$ $g$.","The new concentration is $\\dfrac{48}{x+120}$. According to the condition, the concentration decreased by $6\\%$:<br />$$\\frac{48}{x+120}=\\frac{48}{x+48}-\\frac{6}{100}$$<br />$$\\frac{48}{x+120}=\\frac{48}{x+48}-\\frac{3}{50}$$","Let's find the ODZ: $x+120\\neq0$, $x+48\\neq0$, which means $x\\neq -120$, $x\\neq -48$.","Let's solve the equation:<br />$$\\frac{48}{x+48}-\\frac{48}{x+120}=\\frac{3}{50}$$<br />$$\\frac{3456}{(x+48)(x+120)}=\\frac{3}{50}$$","We get:<br />$$3(x+48)(x+120)=172800$$<br />$$(x+48)(x+120)=57600$$<br />$$x^2+168x+5760=57600$$<br />$$x^2+168x-51840=0$$","Roots of the equation: $x_1=112$, $x_2=-280$. Fits $x=112$.","Initial concentration:<br />$$\\frac{48}{112+48}=\\frac{48}{160}=0,3=30\\%.$$"]},{"content":"$80$ $g$ water was added to a solution containing $20$ $g$ iodine. As a result, the iodine concentration decreased by $8\\%$. Find the initial concentration of the solution.[[image-9]][[fill_choice_big-9]]","widgets":{"image-9":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/02/mikroskop-himiya-kolba-molekuly-2.svg","width":"300"},"fill_choice_big-9":{"type":"fill_choice_big","options":["$10\\%$","$20\\%$","$25\\%$","$15\\%$"],"placeholder":0,"answer":1}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+20$.","The initial iodine concentration is $\\dfrac{20}{x+20}$. After adding $80$ $g$ water, the mass of the solution became $x+100$, and the mass of iodine remained $20$ $g$.","The new concentration is $\\dfrac{20}{x+100}$. According to the condition, the concentration decreased by $8\\%$:<br />$$\\frac{20}{x+100}=\\frac{20}{x+20}-\\frac{8}{100}$$<br />$$\\frac{20}{x+100}=\\frac{20}{x+20}-\\frac{2}{25}$$","Let's find the ODZ: $x+100\\neq0$, $x+20\\neq0$, which means $x\\neq -100$, $x\\neq -20$.","Let's solve the equation:<br />$$\\frac{20}{x+20}-\\frac{20}{x+100}=\\frac{2}{25}$$<br />$$\\frac{1600}{(x+20)(x+100)}=\\frac{2}{25}$$","We get:<br />$$2(x+20)(x+100)=40000$$<br />$$(x+20)(x+100)=20000$$<br />$$x^2+120x+2000=20000$$<br />$$x^2+120x-18000=0$$","Roots of the equation: $x_1=100$, $x_2=-220$. Fits $x=100$.","Initial concentration:<br />$$\\frac{20}{100+20}=\\frac{20}{120}=\\frac{1}{6}=16\\frac{2}{3}\\%.$$"]},{"content":"$120$ $g$ water was added to a solution containing $80$ $g$ copper sulfate. As a result, the concentration of copper sulfate decreased by $6\\%$. Find the initial concentration of the solution.[[image-10]][[fill_choice_big-10]]","widgets":{"image-10":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/09/molekuly-uchenyi-.svg","width":"300"},"fill_choice_big-10":{"type":"fill_choice_big","options":["$25\\%$","$20\\%$","$30\\%$","$24\\%$"],"placeholder":0,"answer":0}},"step":1,"hints":["Let us denote by $x$ the mass of water in the initial solution. Then the mass of the initial solution is $x+80$.","The initial concentration of copper sulfate is $\\dfrac{80}{x+80}$. After adding $120$ $g$ water, the mass of the solution became $x+200$, the mass of the substance remained $80$ $g$.","The new concentration is $\\dfrac{80}{x+200}$. According to the condition, the concentration decreased by $6\\%$:<br />$$\\frac{80}{x+200}=\\frac{80}{x+80}-\\frac{6}{100}$$<br />$$\\frac{80}{x+200}=\\frac{80}{x+80}-\\frac{3}{50}$$","Let's find the ODZ: $x+200\\neq0$, $x+80\\neq0$, which means $x\\neq -200$, $x\\neq -80$.","Let's solve the equation:<br />$$\\frac{80}{x+80}-\\frac{80}{x+200}=\\frac{3}{50}$$<br />$$\\frac{9600}{(x+80)(x+200)}=\\frac{3}{50}$$","We get:<br />$$3(x+80)(x+200)=480000$$<br />$$(x+80)(x+200)=160000$$<br />$$x^2+280x+16000=160000$$<br />$$x^2+280x-144000=0$$","Roots of the equation: $x_1=240$, $x_2=-520$. Fits $x=240$.","Initial concentration:<br />$$\\frac{80}{240+80}=\\frac{80}{320}=0,25=25\\%.$$"]}]}

{"questions":[{"content":"The denominator of an ordinary irreducible fraction is $4$ greater than its numerator. If the numerator of this fraction is increased by $2$, and the denominator by $21$, then the fraction will decrease by $\\frac{1}{4}$. Find this fraction. [[image-2]] [[fill_choice_big-1]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$\\frac{3}{7}$","$\\frac{2}{7}$","$\\frac{3}{4}$","$\\frac{5}{6}$"],"placeholder":0,"answer":0},"image-2":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/08/pishet-formuly-na-doske-doska.svg","width":"300"}},"step":1,"hints":["Let's denote the numerator of the fraction by the variable $x$, then the denominator will contain the expression $x+4$, and the whole fraction will have the form $\\frac{x}{x+4}$.","Increasing the numerator by $2$ and the denominator by $21$ gives $\\frac{x+2}{x+25}$, which is $\\frac{1}{4}$ less than the original fraction. Therefore:<br />$$\\frac{x}{x+4}=\\frac{x+2}{x+25}+\\frac{1}{4}$$<br />The domain restrictions are $x\\ne-4$ and $x\\ne-25$.","Let's bring the fraction to a common denominator $4(x+4)(x+25)$ and simplify: $x^{2}-47x+132=0$.","Roots: $x_{1}=44$ and $x_{2}=3$.<br />The fractions $\\frac{44}{48}$ and $\\frac{3}{7}$ fit the conditions of the problem. Of these, only $\\frac{3}{7}$ will be irreducible."]}]}

{"questions":[{"content":"The denominator of an irreducible fraction is $5$ greater than the numerator. If the numerator is increased by $3$ and the denominator by $18$, then the fraction will decrease by $\\frac{1}{5}$. Find this fraction. [[image-1]] [[fill_choice_big-1]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$\\frac{2}{7}$","$\\frac{3}{8}$","$\\frac{4}{9}$","$\\frac{5}{11}$"],"placeholder":0,"answer":0},"image-1":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/10/3-1.svg","width":"300"}},"step":1,"hints":["Let us denote the numerator of the fraction by $x$, then the denominator is equal to $x+5$. Initial fraction: $\\frac{x}{x+5}$.","After the change we get the fraction $\\frac{x+3}{x+23}$. By condition, it is less than the original by $\\frac{1}{5}$:<br />$$\\frac{x}{x+5}=\\frac{x+3}{x+23}+\\frac{1}{5}$$","Let's find the ODZ: $x\\neq-5$, $x\\neq-23$.","Let's bring it to a common denominator and simplify:<br />$$x^2-13x-14=0$$","Roots: $x_1=14$, $x_2=-1$. Fits $x=2$. The fraction you are looking for:<br />$$\\frac{2}{7}.$$"]},{"content":"The denominator of the fraction is $6$ greater than the numerator. If the numerator is increased by $1$ and the denominator by $10$, then the fraction will decrease by $\\frac{1}{6}$. Find this fraction. [[image-2]] [[fill_choice_big-2]]","widgets":{"fill_choice_big-2":{"type":"fill_choice_big","options":["$\\frac{1}{7}$","$\\frac{2}{7}$","$\\frac{3}{8}$","$\\frac{4}{9}$"],"placeholder":0,"answer":0},"image-2":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/09/zavry-gotovo.svg","width":"300"}},"step":1,"hints":["Let the numerator of the fraction be equal to $x$, then the denominator is equal to $x+6$.","After the change we get the fraction $\\frac{x+1}{x+16}$. According to the condition:<br />$$\\frac{x}{x+6}=\\frac{x+1}{x+16}+\\frac{1}{6}$$","ODZ: $x\\neq-6$, $x\\neq-16$.","After transformations we obtain a quadratic equation:<br />$$x^2-8x-9=0$$","Roots: $x_1=9$, $x_2=-1$. Suitable fraction:<br />$$\\frac{1}{7}.$$"]},{"content":"The numerator of an irreducible fraction is $2$ less than the denominator. If we add $4$ to the numerator and $10$ to the denominator, the fraction will increase by $\\frac{1}{6}$. Find the fraction. [[image-3]] [[fill_choice_big-3]]","widgets":{"fill_choice_big-3":{"type":"fill_choice_big","options":["$\\frac{3}{5}$","$\\frac{2}{5}$","$\\frac{5}{7}$","$\\frac{4}{7}$"],"placeholder":0,"answer":0},"image-3":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/09/illyustracziya_bouling-01.svg","width":"300"}},"step":1,"hints":["Let the denominator be equal to $x$, then the numerator is equal to $x-2$.","After the change we get the fraction $\\frac{x+2}{x+10}$. According to the condition:<br />$$\\frac{x+2}{x+10}=\\frac{x-2}{x}+\\frac{1}{6}$$","ODZ: $x\\neq0$, $x\\neq-10$.","After simplification we get:<br />$$x^2-11x+30=0$$","Roots: $x_1=5$, $x_2=6$. An irreducible fraction is suitable:<br />$$\\frac{3}{5}.$$"]},{"content":"The denominator of the fraction is $3$ greater than the numerator. If the numerator is increased by $5$ and the denominator by $9$, then the fraction becomes equal to $\\frac{3}{4}$. Find the original fraction. [[image-4]] [[fill_choice_big-4]]","widgets":{"fill_choice_big-4":{"type":"fill_choice_big","options":["$\\frac{3}{6}$","$\\frac{4}{7}$","$\\frac{5}{8}$","$\\frac{2}{5}$"],"placeholder":0,"answer":1},"image-4":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/06/dop_funkczii_v_trenazherah_po_matematike_01.svg","width":"300"}},"step":1,"hints":["Let the numerator be equal to $x$, then the denominator is equal to $x+3$.","After the change, the fraction takes the form $\\frac{x+5}{x+12}$. According to the condition:<br />$$\\frac{x+5}{x+12}=\\frac{3}{4}$$","Let's solve the proportion:<br />$$4(x+5)=3(x+12)$$","We get:<br />$$4x+20=3x+36$$<br />$$x=16$$","Original fraction:<br />$$\\frac{16}{19}=\\frac{4}{7}.$$"]},{"content":"The numerator of the fraction is $1$ less than the denominator. If the numerator is increased by $2$ and the denominator by $5$, then the fraction becomes equal to $\\frac{2}{3}$. Find the original fraction. [[image-5]] [[fill_choice_big-5]]","widgets":{"fill_choice_big-5":{"type":"fill_choice_big","options":["$\\frac{2}{3}$","$\\frac{3}{4}$","$\\frac{4}{5}$","$\\frac{5}{6}$"],"placeholder":0,"answer":1},"image-5":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/11/obrazavr_angl_3_klass_papa_uchitel.svg","width":"300"}},"step":1,"hints":["Let the denominator be equal to $x$, then the numerator is equal to $x-1$.","After the change we get the fraction $\\frac{x+1}{x+5}$. According to the condition:<br />$$\\frac{x+1}{x+5}=\\frac{2}{3}$$","Let's solve the proportion:<br />$$3(x+1)=2(x+5)$$","We get:<br />$$3x+3=2x+10$$<br />$$x=7$$","Original fraction:<br />$$\\frac{6}{7}=\\frac{3}{4}.$$"]},{"content":"The denominator of the fraction is $7$ greater than the numerator. If the numerator is increased by $1$ and the denominator by $8$, then the fraction will decrease by $\\frac{1}{7}$. Find the fraction. [[image-6]] [[fill_choice_big-6]]","widgets":{"fill_choice_big-6":{"type":"fill_choice_big","options":["$\\frac{2}{9}$","$\\frac{3}{10}$","$\\frac{4}{11}$","$\\frac{5}{12}$"],"placeholder":0,"answer":0},"image-6":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/11/uchetel-doska2-1111.svg","width":"300"}},"step":1,"hints":["Let the numerator be equal to $x$, then the denominator is equal to $x+7$.","After the change we get the fraction $\\frac{x+1}{x+15}$. According to the condition:<br />$$\\frac{x}{x+7}=\\frac{x+1}{x+15}+\\frac{1}{7}$$","ODZ: $x\\neq-7$, $x\\neq-15$.","After transformations we get:<br />$$x^2-5x-14=0$$","Roots: $x_1=7$, $x_2=-2$. Suitable fraction:<br />$$\\frac{2}{9}.$$"]},{"content":"The numerator of the fraction is $4$ less than the denominator. If we add $6$ to the numerator and $12$ to the denominator, then the fraction becomes equal to $\\frac{5}{7}$. Find the original fraction. [[image-7]] [[fill_choice_big-7]]","widgets":{"fill_choice_big-7":{"type":"fill_choice_big","options":["$\\frac{3}{7}$","$\\frac{4}{8}$","$\\frac{5}{9}$","$\\frac{6}{10}$"],"placeholder":0,"answer":0},"image-7":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/11/uchetel-doska2-111.svg","width":"300"}},"step":1,"hints":["Let the denominator be equal to $x$, then the numerator is equal to $x-4$.","After the change we get the fraction $\\frac{x+2}{x+12}$. According to the condition:<br />$$\\frac{x+2}{x+12}=\\frac{5}{7}$$","Let's solve the proportion:<br />$$7(x+2)=5(x+12)$$","We get:<br />$$7x+14=5x+60$$<br />$$2x=46$$<br />$$x=23$$","Original fraction:<br />$$\\frac{19}{23}=\\frac{3}{7}.$$"]},{"content":"The denominator of the fraction is $8$ greater than the numerator. If the numerator is increased by $4$ and the denominator by $20$, then the fraction will decrease by $\\frac{1}{8}$. Find the fraction. [[image-8]] [[fill_choice_big-8]]","widgets":{"fill_choice_big-8":{"type":"fill_choice_big","options":["$\\frac{3}{11}$","$\\frac{4}{12}$","$\\frac{5}{13}$","$\\frac{6}{14}$"],"placeholder":0,"answer":0},"image-8":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/10/kniga_knigi-ruchka-karandash-drobi-stol-krivye.svg","width":"300"}},"step":1,"hints":["Let the numerator of the fraction be equal to $x$, then the denominator is equal to $x+8$.","After the change we get the fraction $\\frac{x+4}{x+28}$. According to the condition:<br />$$\\frac{x}{x+8}=\\frac{x+4}{x+28}+\\frac{1}{8}$$","ODZ: $x\\neq-8$, $x\\neq-28$.","After transformations we get:<br />$$x^2-10x-33=0$$","Roots: $x_1=11$, $x_2=-3$. Suitable fraction:<br />$$\\frac{3}{11}.$$"]},{"content":"The numerator of the fraction is $5$ less than the denominator. If the numerator is increased by $7$ and the denominator by $14$, then the fraction becomes equal to $\\frac{4}{5}$. Find the original fraction. [[image-9]] [[fill_choice_big-9]]","widgets":{"fill_choice_big-9":{"type":"fill_choice_big","options":["$\\frac{3}{8}$","$\\frac{4}{9}$","$\\frac{5}{10}$","$\\frac{6}{11}$"],"placeholder":0,"answer":0},"image-9":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/pishet-formuly-na-doske-doska-1.svg","width":"300"}},"step":1,"hints":["Let the denominator be equal to $x$, then the numerator is equal to $x-5$.","After the change we get the fraction $\\frac{x+2}{x+14}$. According to the condition:<br />$$\\frac{x+2}{x+14}=\\frac{4}{5}$$","Let's solve the proportion:<br />$$5(x+2)=4(x+14)$$","We get:<br />$$5x+10=4x+56$$<br />$$x=46$$","Original fraction:<br />$$\\frac{41}{46}=\\frac{3}{8}.$$"]},{"content":"The denominator of an irreducible fraction is $2$ greater than the numerator. If the numerator is increased by $3$ and the denominator by $9$, then the fraction will decrease by $\\frac{1}{3}$. Find the fraction. [[image-10]] [[fill_choice_big-10]]","widgets":{"fill_choice_big-10":{"type":"fill_choice_big","options":["$\\frac{1}{3}$","$\\frac{2}{4}$","$\\frac{3}{5}$","$\\frac{4}{6}$"],"placeholder":0,"answer":0},"image-10":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/dumaet-nad-formulami-na-doske-formuly-doska.svg","width":"300"}},"step":1,"hints":["Let the numerator be equal to $x$, then the denominator is equal to $x+2$.","After the change we get the fraction $\\frac{x+3}{x+11}$. According to the condition:<br />$$\\frac{x}{x+2}=\\frac{x+3}{x+11}+\\frac{1}{3}$$","ODZ: $x\\neq-2$, $x\\neq-11$.","After transformations we get:<br />$$x^2-2x-3=0$$","Roots: $x_1=3$, $x_2=-1$. An irreducible fraction is suitable:<br />$$\\frac{1}{3}.$$"]}]}