Word problems (applied practice)

{"questions":[{"content":"In the year $2008$, $40\\space000$ people lived in the city block. In the year $2009$, as a result of the construction of new houses, the number of residents increased by $8\\%,$ and in the year $2010$ by $9\\%$ compared to the year $2009$. How many people began to live in the block in the year $2010$?<br />[[fill_choice_big-1]][[image-61]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$47\\space088$","$47\\space788$","$47\\space188$"],"placeholder":0,"answer":0},"image-61":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/10/zagryaznenie-gorod-doma-zdaniya-01.svg","width":"300"}},"hints":["$1.$ Find the number of inhabitants in the year $2009$:<br />$40\\space000 \\cdot 1.08 = 43\\space200$<br /><br />$2.$ Find the number of inhabitants in the year $2010$:<br />$43\\space200 \\cdot 1.09 = 47\\space088$<br /><br />Answer: In the year $2010$, $47\\space088$ people lived in the block."]},{"content":"On Monday, the company's shares rose in price by a certain amount of percent, and on Tuesday they fell in price by the same amount of percent. As a result, they began to cost $4\\%$ cheaper than when trading opened on Monday. By what percentage did the company's shares rise in price on Monday?<br />[[fill_choice_big-4]][[image-106]]","widgets":{"fill_choice_big-4":{"type":"fill_choice_big","options":["$20$","$18$","$15$"],"placeholder":0,"answer":0},"image-106":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/04/zavod-1.svg","width":"300"}},"hints":["$1.$ Let the initial price of shares be equal to $100$ units.<br /><br />$2.$ On Monday, the shares rose in price by $x$ $\\%$, their value became:<br />$$100 \\cdot \\left(1 + \\frac{x}{100}\\right)$$<br />$3.$ On Tuesday, shares fell in price by $x\\%$, their value became:<br />$$100 \\cdot \\left(1 + \\frac{x}{100}\\right) \\cdot \\left(1 — \\frac{x}{100}\\right)$$<br />$4.$ According to the condition, the final price of shares is $4\\%$ less than the initial one:<br />$$100 \\cdot \\left(1 + \\frac{x}{100}\\right) \\cdot \\left(1-\\frac{x}{100}\\right) = 100 \\cdot 0.96$$<br />$5.$ Simplify the equation:<br />$$\\left(1 + \\frac{x}{100}\\right)\\left(1-\\frac{x}{100}\\right) = 0.96$$ $$1-\\left(\\frac{x}{100}\\right)^2 = 0.96$$ $$\\left(\\frac{x}{100}\\right)^2 = 0.04$$ $$\\frac{x}{100} = 0.2$$ $$x = 20$$<br /><br />Answer: on Monday the shares rose in price by $20\\%.$"]},{"content":"Four shirts are cheaper than a jacket on $8\\%.$ What percentage are five shirts more expensive than a jacket?<br />[[fill_choice_big-57]][[image-155]]","widgets":{"fill_choice_big-57":{"type":"fill_choice_big","options":["$15$","$10$","$20$"],"placeholder":0,"answer":0},"image-155":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/11/makintosh.svg","width":"300"}},"hints":["$1.$ The cost of four shirts is $92\\%$ the cost of a jacket.<br /><br />$2.$ This means that the cost of one shirt is:<br />$\\dfrac{92\\%}{4} = 23\\%$ jacket cost.<br /><br />$3.$ Therefore, the cost of five shirts is:<br />$5 \\cdot 23\\% = 115\\%$ jacket cost.<br /><br />$4.$ This exceeds the cost of the jacket by:<br />$115\\%-100\\% = 15\\%.$<br /><br />Answer: $15\\%.$"]},{"content":"The family consists of a husband, wife and their student daughter. If the husband's salary doubled, the total family income would increase by $67\\%.$. If the daughter's scholarship were tripled, the total family income would decrease by $4\\%$. What percentage of the total family income is the wife's salary?<br />[[fill_choice_big-81]][[image-208]]","widgets":{"fill_choice_big-81":{"type":"fill_choice_big","options":["$27$","$33$","$23$"],"placeholder":0,"answer":0},"image-208":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/dengi.svg","width":"300"}},"hints":["$1.$ Let the total family income be equal to $100\\%$.<br /><br />$2.$ If the husband's salary doubles, the total family income will increase by $67\\%$. This means that the husband's salary is $67\\%$ of the total family income.<br /><br />$3.$ If the daughter's scholarship is reduced by three times, the total family income will be reduced by $4\\%$. This means that $\\dfrac{2}{3}$ scholarships equal $4\\%$ family income. Therefore, the daughter's entire scholarship amounts to:<br />Daughter's scholarship = $\\dfrac{4\\%}{\\dfrac{2}{3}} = 6\\% \\, \\text{of the family income}.$<br /><br />$4.$ Now find the wife’s share of income:<br />Wife's income = $100\\%-67\\%-6\\% = 27\\%.$<br /><br />Answer: the wife's salary is approximately $27\\%$ of the total family income.<br />"]},{"content":"The price of a refrigerator in a store decreases annually by the same percentage from the previous price. Determine by what percentage the price of the refrigerator decreased each year if, put up for sale for $20\\space000$ rubles, two years later it was sold for $15\\space842$ rubles.[[fill_choice_big-154]][[image-265]]","widgets":{"fill_choice_big-154":{"type":"fill_choice_big","options":["$11$","$22$","$17$"],"placeholder":0,"answer":0},"image-265":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/fridge.svg","width":"300"}},"hints":["1. Let $x\\%$ be the percentage of price reduction per year. Then every year the price is $(100-x)\\%$ from the previous one.<br />2. After the first year the price will be:<br />$$20\\space000 \\cdot \\dfrac{100-x}{100}$$3. After the second year the price will be: $$20\\space000 \\cdot \\left(\\dfrac{100-x}{100}\\right)^2 = 15\\space842$$ 4. We solve the equation:<br />$$\\left(\\dfrac{100-x}{100}\\right)^2 = \\dfrac{15\\space842}{20\\space000}$$ $$\\left(\\dfrac{100-x}{100}\\right)^2 = 0.7921$$ $$\\dfrac{100-x}{100} = \\sqrt{0.7921}$$ $$\\dfrac{100-x}{100} = 0.89$$ $$100-x = 89$$ $$x = 11$$ Answer: every year the price decreased by $11\\%.$"]},{"content":"Mitya, Anton, Gosha and Boris established a company with an authorized capital of $200\\space000$ rubles. Mitya contributed $14\\%$ of the authorized capital, Anton - $42\\space000$ rubles, Gosha - $12\\%$ of the authorized capital, and Boris contributed the rest of the capital. The founders agreed to divide the annual profit in proportion to the contribution made to the authorized capital. What amount of profit $1\\space000\\space000$ rubles is due to Boris?<br />[[fill_choice_big-91]][[image-326]]","widgets":{"fill_choice_big-91":{"type":"fill_choice_big","options":["$530\\space000$","$500\\space000$","$570\\space000$"],"placeholder":0,"answer":0},"image-326":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/za-stolom-doklad.svg"}},"hints":["$1.$ Calculate the shares of each founder in the authorized capital:<br />Mitya: $14\\%$ from $200\\space000 = 0.14 \\cdot 200\\space000 = 28\\space000$ rubles.<br />Anton: $42\\space000$ rubles (given).<br />Gosha: $12\\%$ from $200\\space000 = 0.12 \\cdot 200\\space000 = 24\\space000$ rubles.<br />Boris: the rest of the capital.<br /><br />$2.$ Find Boris's share:<br />$200\\space000-(28\\space000 + 42\\space000 + 24\\space000) = 106\\space000$ rubles.<br /><br />$3.$ Calculate Boris's share as a percentage:<br />$\\dfrac{106\\space000}{200\\space000} \\cdot 100\\% = 53\\%.$<br /><br />$4.$ Find the amount of profit due to Boris:<br />$1\\space000\\space000 \\cdot 0.53 = 530\\space000$ rubles.<br /><br />Answer: Boris is owed $530\\space000$ rubles."]},{"content":"Four friends - Dima, Katya, Petya and Masha - decided to open a small cafe. Dima contributed $35\\%$ of the total amount to the authorized capital, Katya contributed $60\\space000$ rubles, Petya contributed $25\\%$ of the total amount, and Masha contributed the rest. Based on the results of the first year of operation, the cafe received a profit of $800\\space000$ rubles, which they decided to divide in proportion to the invested funds. What amount will Masha receive?<br />[[fill_choice_big-517]][[image-375]]","widgets":{"fill_choice_big-517":{"type":"fill_choice_big","options":["$80\\space000$","$82\\space000$","$78\\space000$"],"placeholder":0,"answer":0},"image-375":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/06/fastfud-kafe.svg","width":"300"}},"hints":["$1.$ Let us denote the total amount of the authorized capital by $x$ rubles.<br /><br />$2.$ Calculate the shares of each participant:<br />Dima: $35\\%$ from $x = 0.35x$ rubles.<br />Katya: $60\\space000$ rubles.<br />Petya: $25\\%$ from $x = 0.25x$ rubles.<br />Masha: the rest of the capital.<br /><br />$3.$ Create an equation to determine the total amount:<br />$x = 0.35x + 60\\space000 + 0.25x + $ Masha's contribution.<br /><br />$4.$ Find Masha's contribution:<br />$x-(0.35x + 60\\space000 + 0.25x) = 0.4x-60\\space000.$<br /><br />$5.$ Calculate Masha's share as a percentage:<br />$0.4x-60\\space000 = x -0.6x-60\\space000;$<br />$0.4x-60\\space000 = 0.4x-60\\space000.$<br /><br />$6.$ Find the total amount of capital:<br />$0.6x = 60\\space000;$<br />$x = 100\\space000$ rubles.<br /><br />$7.$ Calculate Masha's contribution:<br />$0.4 \\cdot 100\\space000-60\\space000 = 40\\space000-60\\space000 = 10\\space000$ rubles.<br /><br />$8.$ Find Masha's share as a percentage:<br />$\\dfrac{10\\space000}{100\\space000} \\cdot 100\\% = 10\\%.$<br /><br />$9.$ Calculate Masha's profit:<br />$800\\space000 \\cdot 0.1 = 80\\space000$ rubles.<br /><br />Answer: Masha will receive $80\\space000$ rubles."]},{"content":"In the year $2020$, $25\\space000$ inhabitants lived in a small town. In the year $2021$, after the opening of a new industrial enterprise, the population increased by $6\\%.$. In the year $2022$, thanks to the development of infrastructure, the number of residents increased by another $7\\%$ compared to the previous year. How many people lived in the city in the year $2022$?<br />[[fill_choice_big-583]][[image-444]]","widgets":{"fill_choice_big-583":{"type":"fill_choice_big","options":["$28\\space355$","$22\\space455$","$24\\space280$"],"placeholder":0,"answer":0},"image-444":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/idet-po-ulitse-vniz-1-2.svg","width":"300"}},"hints":["$1.$ Let us denote:<br />$a$ — number of inhabitants in $2020$ year.<br />$b$ — number of inhabitants in $2021$ year.<br />$c$ — number of inhabitants in $2022$ year.<br /><br />$2.$ From the problem statement:<br />in $2021$ the population increased by $6\\%.$ This means that $b = a \\cdot 1.06;$<br />in $2022$ the population increased by $7\\%.$ This means that $c = b \\cdot 1.07.$<br /><br />$3.$ Substitute the values:<br />$a = 25\\space000;$<br />$b = 25\\space000 \\cdot 1.06 = 26\\space500;$<br />$c = 26\\space500 \\cdot 1.07 = 28\\space355.$<br />Answer: in the year $2022$ there were $28\\space355$ people living in the city."]},{"content":"On Thursday, the company's shares rose in price by a certain number of percent, and on Friday they fell in price by the same number of percent. As a result, they began to cost $36\\%$ cheaper than at the opening of trading on Thursday. By what percentage did the company's shares rise in price on Thursday?<br />[[fill_choice_big-457]][[image-517]]","widgets":{"fill_choice_big-457":{"type":"fill_choice_big","options":["$60$","$40$","$48$"],"placeholder":0,"answer":0},"image-517":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/11/dengi-koshelek-monety-.svg","width":"300"}},"hints":["$1.$ Let the initial price of shares be $x.$<br />$2$. Let the shares rise in price on Thursday at $p\\%.$<br />$3.$ After the rise in price on Thursday, the price became: $$x\\cdot(1 + \\dfrac{p}{100})$$ $4.$ After the fall on Friday, the price became: $$x \\cdot (1 + \\dfrac{p}{100}) \\cdot (1-\\dfrac{p}{100})$$ $5.$ According to the condition, the final cost is $36\\%$ less than the initial one: $$x \\cdot (1 + \\frac{p}{100}) \\cdot (1-\\frac{p}{100}) = x \\cdot 0.64$$ $6.$ We reduce $x$: $$(1 + \\frac{p}{100}) \\cdot (1 — \\frac{p}{100}) = 0.64$$ $7.$ We use the formula for the difference of squares: $$1-(\\dfrac{p}{100})^2 = 0.64$$ $8.$ We solve the equation: $$(\\dfrac{p}{100})^2 = 0.36$$ $$\\dfrac{p}{100} = 0.6$$ $$p = 60$$<br />Answer: the company's shares rose by $60\\%.$ on Thursday"]},{"content":"Ten shirts are cheaper than a jacket on $10\\%.$ What percentage are twelve shirts more expensive than a jacket?<br />[[fill_choice_big-774]][[image-594]]","widgets":{"fill_choice_big-774":{"type":"fill_choice_big","options":["$8$","$10$","$12$"],"placeholder":0,"answer":0},"image-594":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/11/obrazavr_angl-9-urok_kurtka.svg","width":"300"}},"hints":["$1.$ Take the cost of the jacket as $100\\%$.<br /><br />$2.$ Then the cost of $10$ shirts is $90\\%$ of the cost of the jacket.<br /><br />$3.$ Find the cost of one shirt: $1$ shirt = $\\dfrac{90\\%}{10} = 9\\%$ from the cost of the jacket.<br /><br />$4.$ Find the cost of $12$ shirts: $12$ shirts = $12 \\cdot 9\\% = 108\\%$ from the cost of the jacket.<br /><br />$5.$ Percentage difference between the cost of $12$ shirts and jackets: $108\\%-100\\% = 8\\%.$<br /><br />Answer: twelve shirts are more expensive than a jacket by $8\\%.$"]}]}

{"questions":[{"content":"Two cars left from point $A$ to point $B$ at the same time. The first one drove at a constant speed all the way. The second one drove the first half of the journey at a speed of $24\\space km/h,$ and the second half of the journey at a speed $16\\space km/h$ greater than the speed of the first, as a result of which it arrived at point B at the same time as the first car. Find the speed of the first car. Give your answer in km/h.<br />[[fill_choice_big-1]][[image-51]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$32$","$29$","$33$"],"placeholder":0,"answer":0},"image-51":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/05/obrazavr_deyatelnost_dop-03-2.svg","width":"300"}},"hints":["Let $v$ km/h be the speed of the first car, then the speed of the second car in the second half of the journey is equal to $v + 16$ km/h. Take the distance between points as $1$. The cars were on the road at the same time, so we have:<br /><br />$$\\dfrac{0.5}{24} + \\dfrac{0.5}{v + 16} = \\dfrac{1}{v}$$ Bring it to a common denominator:<br />$$\\dfrac{0.5(v + 16) + 0.5 \\cdot 24}{24(v + 16)} = \\dfrac{1}{v}$$ Simplify the numerator:<br />$$0.5v + 8 + 12 = 0.5v + 20$$Then the equation will take the form:<br />$$\\dfrac{0.5v + 20}{24(v + 16)} = \\dfrac{1}{v}$$ Multiply crosswise:<br />$$v(0.5v + 20) = 24(v + 16)$$ Expand the brackets:<br />$$0.5v^2 + 20v = 24v + 384$$ Move everything to one side:<br />$$0.5v^2-4v-384 = 0$$ Multiply by $2,$ to get rid of the fraction:<br />$$v^2-8v-768 = 0$$ Solve the quadratic equation:<br />$$v = \\dfrac{8 \\pm \\sqrt{64 + 3072}}{2} = \\dfrac{8 \\pm \\sqrt{3136}}{2} = \\dfrac{8 \\pm 56}{2}$$ We get the roots:<br />$$v_1 = \\dfrac{64}{2} = 32$$ $$ v_2 = \\dfrac{-48}{2} = -24$$The negative root has no physical meaning.<br /><br />Answer: the speed of the first car is $32$ km/h."]},{"content":"Two cars left from point $A$ to point $B$ at the same time. The first one drove at a constant speed all the way. The second drove the first half of the journey at a speed lower than the speed of the first at $13\\space km/h,$ and the second half of the journey at a speed of $78,$, as a result of which he arrived at point B at the same time as the first car. Find the speed of the first car if it is known that it is greater than $48\\space km/h.$ Give your answer in km/h.[[fill_choice_big-4]][[image-96]]","widgets":{"fill_choice_big-4":{"type":"fill_choice_big","options":["$52$","$50$","$55$"],"placeholder":0,"answer":0},"image-96":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/05/obrazavr_svetofor-01.svg","width":"300"}},"hints":["Let $v$ km/h be the speed of the first car, then the speed of the second car in the first half of the journey is equal to $(v-13)$ km/h. Take the distance between points as $2S.$<br /><br />Travel time for the first car: $$t_1 = \\frac{2S}{v}$$ Travel time for the second car: $$t_2 = \\frac{S}{v-13} + \\frac{S}{78}$$ According to the conditions of the problem $t_1 = t_2,$ therefore: $$\\frac{2S}{v} = \\frac{S}{v-13} + \\frac{S}{78}$$ Reduce by $S$: $$\\frac{2}{v} = \\frac{1}{v-13} + \\frac{1}{78}$$ Reduce to a common denominator: $$\\frac{2}{v} = \\frac{78 + (v-13)}{78(v-13)} = \\frac{v + 65}{78(v-13)}$$ Multiply crosswise: $$2 \\cdot 78(v-13) = v(v + 65)$$ $$156v-2\\space028 = v^2 + 65v$$ $$v^2-91v + 2\\space028 = 0$$ $$v = \\frac{91 \\pm \\sqrt{91^2-4 \\cdot 2\\space028}}{2}$$ $$v= \\frac{91 \\pm \\sqrt{8\\space281-8\\space112}}{2}$$ $$ v=\\frac{91 \\pm \\sqrt{169}}{2} = \\frac{91 \\pm 13}{2}$$ $$v_1 = \\frac{104}{2} = 52$$ $$ \\quad v_2 = \\frac{78}{2} = 39$$ Since by condition $v > 48$, then $v = 52$.<br />Answer: the speed of the first car is $52$ km/h."]},{"content":"From point $A$ to point $B,$, the distance between which is $75$km, a motorist and a cyclist left at the same time. It is known that in an hour a motorist travels more on a $40\\space km/h$ than a cyclist. Determine the speed of the cyclist if it is known that he arrived at point B $6$hours later than the motorist. Give your answer in km/h.[[fill_choice_big-56]][[image-145]]","widgets":{"fill_choice_big-56":{"type":"fill_choice_big","options":["$10$","$12$","$15$"],"placeholder":0,"answer":0},"image-145":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/na-velike-velosiped-01-2-3.svg","width":"300"}},"hints":["$1.$ Let $v$ km/h be the speed of the cyclist<br />$2.$ Then the speed of the motorist: $(v + 40)$ km/h<br />$3.$ Cyclist travel time: $\\dfrac{75}{v}$ hours<br />$4.$ Travel time for a motorist: $\\dfrac{75}{v+40}$ hours<br />$5.$ By condition: $\\dfrac{75}{v}-\\dfrac{75}{v+40} = 6$<br /><br />Solve the equation:<br /><br />$$\\frac{75}{v}-\\frac{75}{v+40} = 6$$ $$\\frac{75(v+40)-75v}{v(v+40)} = 6$$ $$\\frac{75v + 3\\space000-75v}{v(v+40)} = 6$$ $$\\frac{3\\space000}{v(v+40)} = 6$$ $$3\\space000 = 6v(v+40)$$ $$500 = v(v+40)$$ $$v^2 + 40v-500 = 0$$<br />Solving the quadratic equation:<br /><br />$$v = \\dfrac{-40 \\pm \\sqrt{1\\space600 + 2\\space000}}{2}$$ $$v = \\frac{-40 \\pm \\sqrt{3\\space600}}{2} = \\dfrac{-40 \\pm 60}{2}$$ $$v_1 = 10$$ $$ v_2 = -50$$ Since the speed cannot be negative, then $v = 10.$ Answer: the speed of the cyclist is $10\\ km/h.$"]},{"content":"A cyclist rides at a constant speed from city $A$ to city $B,$ a distance of $70\\space km.$ The next day, he returns to $A$ at a speed $3\\space km/h$ higher and stops for $3$ hours along the way. The return trip takes the same total time as the trip from $A$ to $B.$ Find his speed on the trip from $B$ to $A.$ Give your answer in km/h.[[fill_choice_big-146]][[image-211]]","widgets":{"fill_choice_big-146":{"type":"fill_choice_big","options":["$10$","$12$","$13$"],"placeholder":0,"answer":0},"image-211":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/na-velike-velosiped-01-2-3.svg","width":"300"}},"hints":["$1.$ Travel time from $A$ to $B{:}$ $\\dfrac{70}{v-3}.$<br /><br />$2.$ Travel time from $B$ to $A{:}$ $\\dfrac{70}{v} + 3.$<br /><br />$3.$ By condition: $\\dfrac{70}{v-3} = \\dfrac{70}{v} + 3.$<br /><br />Solve the equation: $$\\dfrac{70}{v-3} = \\dfrac{70}{v} + 3$$ $$\\dfrac{70}{v-3} = \\dfrac{70 + 3v}{v}$$ $$70v = 70(v-3) + 3v(v-3)$$ $$70v = 70v-210 + 3v^2 — 9v$$ $$3v^2-9v-210 = 0$$ $$v^2-3v-70 = 0$$ Solve the quadratic equation: $$v = \\frac{3 \\pm \\sqrt{9 + 280}}{2} = \\frac{3 \\pm \\sqrt{289}}{2} = \\frac{3 \\pm 17}{2}$$ $$v_1 = 10$$ $$\\quad v_2 = -7$$ Since the speed cannot be negative, then $v = 10.$ <br />Answer: the speed of a cyclist on the way from $B$ to $A$ is $10\\ km/h.$"]},{"content":"A cyclist rides at a constant speed from city $A$ to city $B,$ a distance of $98\\space km.$ The next day, his speed on the return trip is $7\\space km/h$ higher. He stops for $7$ hours along the way. The return trip takes the same total time as the trip from $A$ to $B.$ Find his speed on the trip from A to B. Give your answer in km/h.[[fill_choice_big-247]][[image-268]]","widgets":{"fill_choice_big-247":{"type":"fill_choice_big","options":["$7$","$11$","$9$"],"placeholder":0,"answer":0},"image-268":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/10/gorod-lyudi-zanyatoj-mashiny-1-1.svg","width":"300"}},"hints":["$1.$ Travel time from $A$ to $B{:}$ $\\dfrac{98}{v}.$<br /><br />$2.$ Travel time from $B$ to $A{:}$ $\\dfrac{98}{v+7} + 7.$<br /><br />$3.$ By condition: $\\frac{98}{v} = \\dfrac{98}{v+7} + 7.$<br /><br />Solve the equation:<br />$$\\frac{98}{v} = \\frac{98}{v+7} + 7$$ $$\\frac{98}{v} = \\frac{98 + 7(v+7)}{v+7}$$ $$\\frac{98}{v} = \\frac{98 + 7v + 49}{v+7}$$ $$\\frac{98}{v} = \\frac{147 + 7v}{v+7}$$ $$98(v+7) = v(147 + 7v)$$ $$98v + 686 = 147v + 7v^2$$ $$7v^2 + 49v-686 = 0$$ $$v^2 + 7v — 98 = 0$$<br />Solving the quadratic equation:<br /><br />$$v = \\frac{-7 \\pm \\sqrt{49 + 392}}{2} = \\frac{-7 \\pm \\sqrt{441}}{2} = \\frac{-7 \\pm 21}{2}$$ $$v_1 = 7$$ $$ v_2 = -14$$ Since the speed cannot be negative, then $v = 7.$<br /><br />Answer: the speed of a cyclist on the way from $A$ to $B$ is $7\\ km/h.$"]},{"content":"Two cyclists set out on a $240$-kilometer ride at the same time. The first one drove at a speed $1\\space km/h$ higher than the speed of the second one, and arrived at the finish line $1$ an hour earlier than the second one. Find the speed of the cyclist who came to the finish line first. Give your answer in km/h.[[fill_choice_big-764]][[image-329]]","widgets":{"fill_choice_big-764":{"type":"fill_choice_big","options":["$16$","$26$","$19$"],"placeholder":0,"answer":0},"image-329":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/bicycle-1.svg","width":"300"}},"hints":["$1.$ First cyclist time: $\\dfrac{240}{v+1}.$<br /><br />$2.$ Time of the second cyclist: $\\dfrac{240}{v}.$<br /><br />$3.$ By condition: $\\dfrac{240}{v}-\\dfrac{240}{v+1} = 1.$<br /><br />Solve the equation: $$\\frac{240}{v}-\\frac{240}{v+1} = 1$$ $$\\frac{240(v+1)-240v}{v(v+1)} = 1$$ $$\\frac{240v + 240-240v}{v(v+1)} = 1$$ $$\\frac{240}{v(v+1)} = 1$$ $$240 = v(v+1)$$ $$v^2 + v-240 = 0$$ Solve the quadratic equation: $$v = \\frac{-1 \\pm \\sqrt{1 + 960}}{2} = \\frac{-1 \\pm \\sqrt{961}}{2} = \\frac{-1 \\pm 31}{2}$$ $$v_1 = 15$$ $$v_2 = -16$$ Since the speed cannot be negative, then $v = 15.$<br />Speed of the first cyclist: $v + 1 = 15 + 1 = 16.$<br /><br />Answer: the speed of the first cyclist is $16\\ km/h.$"]},{"content":"The distance between the cities $A$ and $B$ is equal to $435\\space km.$ The first car left the city $A$ for the city $B$ at the speed $60\\space km/h$, and an hour after that the second car left the city B towards it at the speed $65\\space km/h$. At what distance from the city $A$ will the cars meet? Give your answer in kilometers.[[fill_choice_big-686]][[image-394]]","widgets":{"fill_choice_big-686":{"type":"fill_choice_big","options":["$240$","$242$","$250$"],"placeholder":0,"answer":0},"image-394":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/mashina-01.svg","width":"300"}},"hints":["$1.$ In the first hour, the first car passed: $$60 \\cdot 1 = 60\\ km$$ $2.$ Remaining distance: $$435-60 = 375\\ km$$ $3.$ Speed of approach of the cars: $$60 + 65 = 125\\ km/h$$ $4.$ Time until the meeting after the departure of the second car: $$\\frac{375}{125} = 3\\ h$$ $5.$ Total time of movement of the first car: $$1 + 3 = 4\\ h$$ $6.$ Distance from city A to the meeting place: $$60 \\cdot 4 = 240\\ km$$ Answer: cars will meet at a distance of $240\\ km$ from city $A.$"]},{"content":"The distance between the cities $A$ and $B$ is equal to $470\\space km.$ The first car left the city $A$ for the city $B$, and $3$ hours after that, the second car left the city $B$ towards it at a speed of $60\\space km/h$. Find the speed of the first car if the cars met at a distance $350\\space km$ from the city $A$. Give your answer in km/h.<br /><br />[[fill_choice_big-617]][[image-463]]","widgets":{"fill_choice_big-617":{"type":"fill_choice_big","options":["$70$","$75$","$77$"],"placeholder":0,"answer":0},"image-463":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/09/probka.svg","width":"300"}},"hints":["$1.$ Distance traveled by the second car before the meeting:<br /> $$470-350 = 120\\ km$$ $2.$ Travel time of the second car: $$t_2 = \\frac{120}{60} = 2\\ h$$ $3.$ Travel time of the first car: $$t_1 = t_2 + 3 = 2 + 3 = 5\\ h$$ $4.$ Speed of the first car: $$v_1 = \\frac{350}{5} = 70\\ km/h$$ Answer: speed of the first car is $70\\ km/h.$"]},{"content":"A motorcyclist and a cyclist rode out from the cities of $A$ and $B$ towards each other. The motorcyclist arrived at $B$ $3$ hours earlier than the cyclist arrived at $A,$, and they met $48$ minutes after leaving. How many hours did the cyclist spend on the journey from point $B$ to point $A$?[[fill_choice_big-557]][[image-536]]","widgets":{"fill_choice_big-557":{"type":"fill_choice_big","options":["$4$","$5$","$6$"],"placeholder":0,"answer":0},"image-536":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/11/italiya-3.svg","width":"300"}},"hints":["$1.$ Let $x$ be the time of the cyclist from $B$ to $A.$<br /><br />$2.$ Then the motorcyclist's time: $x-3.$<br /><br />$3.$ Cyclist speed: $\\dfrac{1}{x}.$<br /><br />$4.$ Rider speed: $\\dfrac{1}{x-3}.$ <br />$5.$ The cyclist's path to the meeting: $$\\dfrac{1}{x} \\cdot \\dfrac{4}{5} = \\dfrac{4}{5x}$$ $6.$ The motorcyclist's path to the meeting: $$\\dfrac{1}{x-3} \\cdot \\dfrac{4}{5} = \\dfrac{4}{5(x-3)}$$ $7.$ The sum of the paths to the meeting is equal to $1$: $$\\dfrac{4}{5x} + \\dfrac{4}{5(x-3)} = 1$$ $8.$ Multiply by $5x(x-3)$: $$4(x-3) + 4x = 5x(x-3)$$ $9.$ Expand brackets: $$4x-12 + 4x = 5x^2-15x$$ $10.$ Reduce to standard form: $$5x^2-23x + 12 = 0$$ $11.$ Solve the quadratic equation:<br />$$D = 23^2-4 \\cdot 5 \\cdot 12 = 529-240 = 289$$ $$x_1 = \\frac{23-17}{10} = \\frac{6}{10} = \\frac{3}{5}$$ - not suitable; $$x_2 = \\frac{23 + 17}{10} = \\frac{40}{10} = 4$$ Answer: the cyclist spent $4$ hours traveling from point $B$ to point $A.$"]},{"content":"Every minute a freight train travels $750$ meters less than an express train, and on the way to $180\\space km$ it takes $2$ hours more time than an express train. Find the speed of the freight train. Give your answer in km/h.<br />[[fill_choice_big-506]][[image-613]]","widgets":{"fill_choice_big-506":{"type":"fill_choice_big","options":["$45$","$43$","$37$"],"placeholder":0,"answer":0},"image-613":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/10/poezd.svg","width":"300"}},"hints":["$1.$ In $1$ minute, a freight train travels $0,75\\ km$ less.<br /><br />$2.$ For $1$ hour $(60$ minutes $)$ a freight train travels less by: $$0,75 \\cdot 60 = 45\\ km$$ $3.$ Let $v_t$ be the speed of the freight train.<br /><br />$4.$ Then the speed of the fast train: $v_t + 45\\ km/h.$<br /><br />$5.$ Freight train time: $t_t = \\dfrac{180}{v_t}.$<br /><br />$6.$ Fast train times: $t_s = \\dfrac{180}{v_t + 45}.$<br /><br />$7.$ By condition: $t_t-t_s = 2.$<br /><br />$8.$ Create an equation:<br /> $$\\frac{180}{v_t}-\\frac{180}{v_t + 45} = 2$$ $9.$ We reduce to a common denominator: $$\\frac{180(v_t + 45)-180v_t}{v_t(v_t + 45)} = 2$$$10.$ We simplify the numerator: $$\\frac{180v_t + 8\\space100-180v_t}{v_t(v_t + 45)} = 2$$ $11.$ We get: $$\\frac{8\\space100}{v_t(v_t + 45)} = 2$$ $12.$ We multiply both parts by the denominator: $$8\\space100 = 2v_t(v_t + 45)$$ $13.$ We expand brackets: $$8\\space100 = 2v_t^2 + 90v_t$$ $14.$ Divide by 2: $$4\\space050 = v_t^2 + 45v_t$$ $15.$ Reduce to standard form: $$v_t^2 + 45v_t-4\\space050 = 0$$ $16.$ Solve the quadratic equation: $$D = 45^2 + 4 \\cdot 4\\space050 = 2\\space025 + 16\\space200 = 18\\space225$$ $$v_t = \\frac{-45 + \\sqrt{18\\space225}}{2} = \\frac{-45 + 135}{2} = \\frac{90}{2} = 45$$ Answer: the speed of a freight train is $45\\ km/h.$"]}]}

{"questions":[{"content":"Two pedestrians set off simultaneously in the same direction from the same place for a walk along a park alley. The speed of the first is $1.5\\space km/h$ greater than the speed of the second. After how many minutes will the distance between pedestrians become equal to $300$ meters?[[fill_choice_big-1]][[image-31]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$12$","$11$","$10$"],"placeholder":0,"answer":0},"image-31":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/10/quiet-tishina-golubi-pticza-park-01.svg","width":"300"}},"hints":["$1.$ Convert the speed difference into meters per minute: $$1,5\\ km/h = 1500\\ m/h$$ $$1500\\ m/h : 60\\ min = 25\\ m/min$$ $2.$ We use the formula to find time: $$t = \\frac{S}{V}$$ $3.$ We substitute the values:<br />$$t = \\frac{300\\ m}{25\\ m/min} = 12\\ min$$ Answer: the distance between pedestrians will become equal to $300\\ m$ in $12$ minutes.<br />"]},{"content":"The first cyclist left the village along the highway at a speed of $15\\space km/h.$ An hour after him, a second cyclist left the same village in the same direction at a speed of $10\\space km/h$, and an hour after that, a third. Find the speed of the third cyclist if he first caught up with the second, and after $2$ hours $20$ minutes after that he caught up with the first. Give your answer in km/h.[[fill_choice_big-7]][[image-76]]","widgets":{"fill_choice_big-7":{"type":"fill_choice_big","options":["$25$","$21$","$22$"],"placeholder":0,"answer":0},"image-76":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/na-velike-velosiped-01-2-3.svg","width":"300"}},"hints":["$1.$ Let $x$ be the speed of the third cyclist.<br /><br />$2.$ The third left through $2\\ h$ after the first.<br /><br />$3.$ During these $2\\ h$ the first passed: $15 \\cdot 2 = 30\\ km.$<br /><br />$4.$ The second one left through $1\\ h$ after the first one, which means it passed: $10 \\cdot 1 = 10\\ km.$<br /><br />$5.$ The third one catches up with the second one:<br /> $\\dfrac{10}{x-10}$ - time until meeting with the second one.<br /><br />$6.$ The third one catches up with the first one:<br /> $\\dfrac{30}{x-15}$ - time until meeting the first one.<br /><br />$7.$ By condition:<br />$$\\frac{30}{x-15} = \\frac{10}{x-10} + \\frac{7}{3}$$ $8.$ We reduce to a common denominator: $$\\frac{30(x-10)-10(x-15)}{(x-15)(x-10)} = \\frac{7}{3}$$ $9.$ We simplify the numerator: $$\\frac{30x-300-10x + 150}{(x-15)(x-10)} = \\frac{7}{3}$$ $10.$ We get: $$\\frac{20x-150}{(x-15)(x-10)} = \\frac{7}{3}$$ $11.$ We multiply both sides by $3(x-15)(x-10){:}$ $$60x-450 = 7(x^2-25x + 150)$$ $12.$ Open the brackets: $$60x-450 = 7x^2-175x + 1\\space050$$ $13.$ Reduce to standard form: $$7x^2-235x + 1\\space500 = 0$$ $14.$ Solve the quadratic equation: $$D = 235^2-4 \\cdot 7 \\cdot 1\\space500 = 55\\space225-42\\space000 = 13\\space225$$ $$x_1 = \\frac{235-\\sqrt{13\\space225}}{14} = \\frac{235-115}{14} = \\frac{120}{14} = \\frac{60}{7}$$ $x_1$ - not suitable; $$x_2 = \\frac{235 + \\sqrt{13\\space225}}{14} = \\frac{235 + 115}{14} = \\frac{350}{14} = 25$$Answer: the speed of the third cyclist is $25\\ km/h.$"]},{"content":"Half the time spent on the road, the car was traveling at a speed of $74\\space km/h,$ and the other half of the time, at a speed of $66\\space km/h.$ Find the average speed of the car throughout the entire journey. Give your answer in km/h.[[fill_choice_big-40]][[image-125]]","widgets":{"fill_choice_big-40":{"type":"fill_choice_big","options":["$70$","$64$","$60$"],"placeholder":0,"answer":0},"image-125":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/mashina_derevnya-1.svg","width":"300"}},"hints":["$1.$ Let half the time be equal to $x$ hours.<br />$2.$ Then all the time: $2x$ hours.<br />$3.$ During the first half of the time the car traveled: $$S_1 = 74x\\ km$$ $4.$ During the second half of the time the car traveled: $$S_2 = 66x\\ km$$ $5.$ Overall path: $$S = S_1 + S_2 = 74x + 66x = 140x\\ km$$ $6.$ The average speed is found by the formula: $$v_{avg} = \\frac{S}{t}$$ $7.$ We substitute the values: $$v_{avg} = \\frac{140x}{2x} = \\frac{140}{2} = 70\\ km/h$$Answer: the average speed of the car is equal to $70\\ km/h.$"]},{"content":"The car drove the first third of the route at a speed of $60\\space km/h,$, the second third at a speed of $120\\space km/h,$, and the last third at a speed of $110\\space km/h.$ Find the average speed of the car throughout the entire journey. Give your answer in km/h.[[fill_choice_big-133]][[image-178]]","widgets":{"fill_choice_big-133":{"type":"fill_choice_big","options":["$88$","$78$","$84$"],"placeholder":0,"answer":0},"image-178":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/06/bolid-01.svg","width":"300"}},"hints":["$1.$ Let the entire path be equal to $x$ $km.$<br /><br />$2.$ Each section of the path is equal to $\\dfrac{x}{3}$ $km.$<br /><br />$3.$ Time in the first section: $t_1 = \\dfrac{x/3}{60} = \\dfrac{x}{180}$ $hours.$<br /><br />$4.$ Time in the second section: $t_2 = \\dfrac{x/3}{120} = \\dfrac{x}{360}$ $hours.$<br /><br />$5.$ Time in the third section: $t_3 = \\dfrac{x/3}{110} = \\dfrac{x}{330}$ $hours.$<br /><br />$6.$ The average speed is found by the formula:<br /> $$v_{avg} = \\dfrac{S}{t} = \\frac{3}{\\frac{1}{60} + \\frac{1}{120} + \\frac{1}{110}}$$ $7.$ Reduce to a common denominator: $$v_{avg} = \\frac{3 \\cdot 120 \\cdot 11}{22 + 11 + 12} = \\frac{3\\space960}{45} = 88\\ km/h$$<br />Answer: the average speed of the car is $88\\ km/h.$"]},{"content":"For the first two hours the car drove at a speed of $50\\space km/h,$, the next hour at a speed of $100\\space km/h,$ and then for two hours at a speed of $75\\space km/h.$ Find the average speed of the car throughout the entire journey. Give your answer in km/h.[[fill_choice_big-213]][[image-235]]","widgets":{"fill_choice_big-213":{"type":"fill_choice_big","options":["$70$","$72$","$77$"],"placeholder":0,"answer":0},"image-235":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/09/bezhit-beg-sport-01-2.svg","width":"300"}},"hints":["$1.$ Find the path on each section:<br /> $$S_1 = 50 \\cdot 2 = 100\\ km$$ $$S_2 = 100 \\cdot 1 = 100\\ km$$ $$S_3 = 75 \\cdot 2 = 150\\ km$$ $2.$ General path:<br /> $$S = S_1 + S_2 + S_3 = 100 + 100 + 150 = 350\\ km$$ $3.$ Total time: $$t = 2 + 1 + 2 = 5\\ h$$ $4.$ The average speed is found by the formula: $$v_{avg} = \\frac{S}{t} = \\frac{350}{5} = 70\\ km/h$$ Answer: the average speed of the car is $70\\ km/h.$"]},{"content":"First $190\\space km$ the car was traveling at a speed of $50\\space km/h,$ the next $180\\space km$—at a speed of $90\\space km/h,$ and then $170\\space km$—at a speed of $100\\space km/h.$ Find the average speed of the car throughout the entire journey. Give your answer in km/h.[[fill_choice_big-300]][[image-296]]","widgets":{"fill_choice_big-300":{"type":"fill_choice_big","options":["$72$","$71$","$73$"],"placeholder":0,"answer":0},"image-296":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/05/obrazavr_deyatelnost_dop-03-2.svg","width":"300"}},"hints":["$1.$ Find the total distance:<br /> $$S = S_1 + S_2 + S_3 = 190 + 180 + 170 = 540\\ km$$ $2.$ Find the time at each section:<br /> $$t_1 = \\frac{190}{50} = 3,8\\ h$$ $$t_2 = \\frac{180}{90} = 2\\ h$$ $$t_3 = \\frac{170}{100} = 1.7\\ h$$<br />$3.$ Total time: $$t = t_1 + t_2 + t_3 = 3.8 + 2 + 1.7 = 7.5\\ h$$ $4.$ Average speed is found by the formula:<br /> $$v_{avg} = \\frac{S}{t} = \\frac{540}{7.5} = 72\\ km/h$$ Answer: the average speed of the car is $72\\ km/h.$"]},{"content":"A train, moving uniformly at a speed of $80\\space km/h,$, passes a roadside pole in $36$ seconds. Find the length of the train in meters.[[fill_choice_big-434]][[image-361]]","widgets":{"fill_choice_big-434":{"type":"fill_choice_big","options":["$800$","$780$","$790$"],"placeholder":0,"answer":0},"image-361":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/station.svg","width":"300"}},"hints":["$1.$ Convert the speed from km/h to m/s:<br /> $$v = 80\\ km/h = \\frac{80 \\cdot 1\\space000}{3\\space600}$$ $$v= \\frac{80\\space000}{3\\space600} = \\frac{800}{36} = \\frac{200}{9}\\ m/s$$ $2.$ The length of the train is equal to the product of speed and time: $$L = v \\cdot t = \\frac{200}{9} \\cdot 36 = 200 \\cdot 4 = 800\\ m$$Answer: the length of the train is $800\\ m.$"]},{"content":"A train, moving uniformly at a speed of $60\\space km/h,$, passes a forest belt, the length of which is $400$ meters, in $1$ a minute. Find the length of the train in meters.[[fill_choice_big-578]][[image-430]]","widgets":{"fill_choice_big-578":{"type":"fill_choice_big","options":["$600$","$520$","$780$"],"placeholder":0,"answer":0},"image-430":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/05/poezd-stancziya-.svg","width":"300"}},"hints":["$1.$ In 1 minute the train covers a distance equal to $1\\space km,$ since:<br /> $v = 60\\ km/h \\Rightarrow$ in $1$ hour $(3\\space600\\space s)$ passes $60\\space km$<br /> $\\Rightarrow$ for $60$ with passes $\\dfrac{60}{3\\space600} \\cdot 60 = 1\\ km = 1\\space000\\ m.$<br /><br />$2.$ During this time, the train travels a distance equal to the sum of the lengths of the forest belt and the train itself: $$S = L_{forest} + L_{train}$$ $3.$ According to the path formula:<br /> $$S = v \\cdot t = 1\\space000\\ m$$ $4.$ Find the length of the train:<br /> $$L_{train} = S-L_{forest} = 1\\space000-400 = 600\\ m$$ Answer: the length of the train is $600\\ m.$"]},{"content":"The road between points $A$ and $B$ consists of ascent and descent, and its length is equal to $8\\space km.$ The tourist traveled from $A$ to $B$ in $5$ hours. His time of movement on the descent was $1$hour. At what speed did the tourist walk on the descent, if his speed on the ascent is less than the speed on the descent by $3\\space km/h?$[[fill_choice_big-727]][[image-503]]","widgets":{"fill_choice_big-727":{"type":"fill_choice_big","options":["$4$","$5$","$6$"],"placeholder":0,"answer":0},"image-503":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/05/obrazavr_poznanie-03.svg","width":"300"}},"hints":["$1.$ Let the speed on the descent be equal to $x\\ km/h$.<br />$2.$ Then the speed on the rise is equal to $(x-3)\\ km/h$.<br />$3.$ Rise time: $5-1=4\\ h$.<br />$4.$ Trigger length: $x \\cdot 1 = x\\ km$.<br />$5.$ Lift length: $$(x-3) \\cdot 4 = 4(x-3)\\ km$$ $6.$ Create the equation: $$x + 4(x-3) = 8$$ $7.$ Solve the equation:<br /> $$x + 4x-12 = 8$$ $$5x = 20$$ $$x = 4$$ Answer: the speed of the tourist on the descent is $4\\ km/h.$"]},{"content":"Ivan and Alexey agreed to meet in Moscow. Ivan calls Alexey and finds out that he is $275\\space km$ from Moscow and is driving at a constant speed $75\\space km/h.$ Ivan at the time of the conversation is $255\\space km$ from Moscow and still has to make a $50$-minute stop on the way. How fast should Ivan drive to arrive in Moscow at the same time as Alexey?<br />[[fill_choice_big-922]][[image-580]]","widgets":{"fill_choice_big-922":{"type":"fill_choice_big","options":["$90$","$85$","$75$"],"placeholder":0,"answer":0},"image-580":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/11/krasnaya-ploshchad.svg","width":"300"}},"hints":["$1.$ Find the time for Alexey to get to Moscow:<br />$$t_A = \\frac{275}{75} = \\frac{11}{3}\\ h = 3\\ h\\ 40\\ min$$ $2.$ Ivan has a stop $50$ minutes, which means he has time to move: $$t_I = 3\\ h\\ 40\\ min-50\\ min = 2\\ h\\ 50\\ min = \\frac{17}{6}\\ h$$ $3.$ Ivan must travel $255$ km in $\\dfrac{17}{6}$ hours, which means his speed must be: $$v = \\frac{255}{\\frac{17}{6}} = 255 \\cdot \\frac{6}{17} = 90\\ km/h$$ Answer: Ivan must travel at a speed $90\\ km/h.$"]}]}

{"questions":[{"content":"The motor boat passed against the current of the river $112\\space km$ and returned to the point of departure, spending $6$ less hours on the return journey. Find the speed of the current if the speed of the boat in still water is $11\\space km/h.$ Give your answer in km/h.<br />[[fill_choice_big-1]][[image-111]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$3$","$5$","$4$"],"placeholder":0,"answer":0},"image-111":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/rychagi14_1.svg","width":"300"}},"hints":["$1.$ Let the time of movement against the current: $(x+6)\\ h.$<br />$2.$ Time of movement with the current: $x\\ h.$<br />$3.$ Speed against current: $11-v\\ km/h.$<br />$4.$ Speed with current: $11+v\\ km/h.$<br />$5.$ Create the equations:<br />$$\\dfrac{112}{x+6} = 11-v$$ $$\\dfrac{112}{x} = 11+v$$ $6.$ Add these equations: $$\\frac{112}{x+6} + \\frac{112}{x} = 22$$ $7.$ Simplify:<br />$$112x + 112(x+6) = 22x(x+6)$$ $$56x + 56(x+6) = 11x(x+6)$$ $$56x + 56x + 336 = 11x^2 + 66x$$ $$11x^2-46x — 336 = 0$$ $8.$ Solve the quadratic equation: $$D = (-46)^2-4 \\cdot 11 \\cdot (-336) = 16\\space900$$ $$\\sqrt{D} = 130$$ $9.$ Find the roots: $$x_1 = \\frac{46 + 130}{22} = 8$$ $$x_2 = \\frac{46-130}{22} = -\\frac{42}{11}$$ $10.$ Take the positive root $x = 8.$ <br />$11.$ Find the flow speed: $$\\frac{112}{8} = 11+v$$ $$14 = 11+v$$ $$v = 3\\ km/h$$ Answer: the flow speed is $3\\ km/h.$"]},{"content":"The motor boat passed against the current of the river $255\\space km$ and returned to the point of departure, spending an hour less on the return journey $2$. Find the speed of the boat in still water if the current speed is $1\\space km/h.$ Give your answer in km/h.[[fill_choice_big-56]][[image-156]]","widgets":{"fill_choice_big-56":{"type":"fill_choice_big","options":["$16$","$11$","$14$"],"placeholder":0,"answer":0},"image-156":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2024/12/dlinavolny12.svg","width":"300"}},"hints":["$1.$ Speed against current: $x-1\\ km/h.$<br /><br />$2.$ Speed with current: $x+1\\ km/h.$<br /><br />$3.$ Time against the current: $\\dfrac{255}{x-1}\\ h.$<br /><br />$4.$ Downstream time: $\\dfrac{255}{x+1}\\ h.$<br /><br />$5.$ Create an equation:<br />$$\\frac{255}{x-1}-\\frac{255}{x+1} = 2$$ $6.$ We reduce to a common denominator: $$\\frac{255(x+1)-255(x-1)}{(x-1)(x+1)} = 2$$ $7.$ We simplify the numerator: $$\\frac{255x + 255-255x + 255}{x^2-1} = 2$$ $8.$ We get: $$\\frac{510}{x^2-1} = 2$$ $9.$ We solve: $$510 = 2(x^2-1)$$ $$255 = x^2-1$$ $$x^2 = 256$$ $$x = 16$$ Answer: the speed of the boat in still water is $16\\ km/h.$"]},{"content":"The motor boat in $10:00$ left the point $A$ to the point $B,$ located in $30\\space km$ from $A.$ Having stayed at the point $B$ $2$ for an hour $30$ minutes, the boat went back and returned to the point $A$ in $18:00.$ Determine (at km/h) the boat's own speed, if it is known that the river current speed is $1\\space km/h.$[[fill_choice_big-41]][[image-205]]","widgets":{"fill_choice_big-41":{"type":"fill_choice_big","options":["$11$","$12$","$10$"],"placeholder":0,"answer":0},"image-205":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/10/angliya-dom-doma-zdaniya-zdanie_15.svg","width":"300"}},"hints":["$1.$ Find the total travel time:<br />$18:00-10:00-2.5\\ h = 5.5\\ h$<br /><br />$2.$ Speed against current: $x-1\\ km/h.$<br />$3.$ Speed with current: $x+1\\ km/h.$<br /><br />$4.$ Create an equation:<br />$$\\frac{30}{x-1} + \\frac{30}{x+1} = 5.5$$ $5.$ We reduce to a common denominator: $$\\frac{30(x+1) + 30(x-1)}{(x-1)(x+1)} = 5.5$$ $6.$ We simplify: $$\\frac{30x + 30 + 30x-30}{x^2-1} = 5.5$$ $$\\frac{60x}{x^2-1} = 5.5$$ $7.$ We solve:<br />$$60x = 5.5(x^2-1)$$ $$60x = 5.5x^2-5.5$$ $$5.5x^2-60x-5.5 = 0$$ $8.$ Multiply by 2 for convenience: $$11x^2-120x-11 = 0$$ $9.$ Solve the quadratic equation: $$x = \\frac{60 \\pm \\sqrt{3\\space600 + 121}}{11}$$ $$x = \\frac{60 \\pm 61}{11}$$ $10.$ Get the roots: $$x_1 = \\frac{121}{11} = 11$$ $$x_2 = \\frac{-1}{11}$$ $x_2$ - not suitable. <br /><br />Answer: the boat's own speed is $11\\ km/h.$"]},{"content":"The motor ship travels along the river to the destination $200\\space km$ and after stopping it returns to the point of departure. Find the speed of the current if the speed of the ship in still water is $15\\space km/h,$, the stay lasts $10$ hours, and the ship returns to the point of departure $40$ hours after departure. Give your answer in km/h.[[fill_choice_big-305]][[image-258]]","widgets":{"fill_choice_big-305":{"type":"fill_choice_big","options":["$5$","$4$","$7$"],"placeholder":0,"answer":0},"image-258":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/yahta-lodka-korabl.svg","width":"300"}},"hints":["$1.$ Speed with current: $15 + u\\ km/h$<br />$2.$ Speed against current: $15-u\\ km/h$<br /><br />$3.$ Travel time excluding parking: $$40-10 = 30\\ h$$ $4.$ Create an equation: $$\\frac{200}{15-u} + \\frac{200}{15+u} = 30$$ $5.$ Reduce to a common denominator: $$\\frac{200(15+u) + 200(15-u)}{(15-u)(15+u)} = 30$$ $6.$ Simplify: $$\\frac{200 \\cdot 15 + 200u + 200 \\cdot 15-200u}{225-u^2} = 30$$ $$\\frac{200 \\cdot 30}{225-u^2} = 30$$ $7.$ Solve:$$\\frac{6\\space000}{225-u^2} = 30$$ $$6\\space000 = 30(225-u^2)$$ $$200 = 225-u^2$$ $$u^2 = 25$$ $$u = 5$$ Answer: the speed of the river is $5\\ km/h.$"]},{"content":"The motor ship travels along the river to the destination $255\\space km$ and after stopping it returns to the point of departure. Find the speed of the ship in still water if the speed of the current is $1\\space km/h,$, the stay lasts $2$ $hours$, and the ship returns to the point of departure through $34$ $hours$ after sailing from it. Give your answer in km/h.[[fill_choice_big-152]][[image-315]]","widgets":{"fill_choice_big-152":{"type":"fill_choice_big","options":["$16$","$14$","$18$"],"placeholder":0,"answer":0},"image-315":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/07/korabl-kapitan-matros-01.svg","width":"300"}},"hints":["$1.$ Speed with current: $u + 1\\ km/h$<br />$2.$ Speed against current: $u-1\\ km/h$<br /><br />$3.$ Travel time excluding parking:<br />$$34-2 = 32\\ h$$ $4.$ Create an equation: $$\\frac{255}{u+1} + \\frac{255}{u-1} = 32$$ $5.$ Reduce to a common denominator: $$\\frac{255(u-1) + 255(u+1)}{(u-1)(u+1)} = 32$$ $6.$ Simplify: $$\\frac{255u-255 + 255u + 255}{u^2-1} = 32$$ $$\\frac{510u}{u^2-1} = 32$$ $7.$ Solve: $$510u = 32(u^2-1)$$ $$510u = 32u^2-32$$ $$32u^2-510u-32 = 0$$ $8.$ We solve the quadratic equation: $$u = \\frac{510 \\pm \\sqrt{510^2 + 4 \\cdot 32^2}}{2 \\cdot 32}$$ $$u = \\frac{510 \\pm \\sqrt{260\\space100 + 4\\space096}}{64}$$ $$u = \\frac{510 \\pm \\sqrt{264\\space196}}{64}$$ $$u = \\frac{510 \\pm 514}{64}$$ $9.$ We get the roots: $$u_1 = \\frac{1\\space024}{64} = 16$$ $$u_2 = \\frac{-4}{64} = -\\frac{1}{16}$$ $u_2$ - not suitable.<br /><br />Answer: the speed of the ship in still water is $16\\ km/h.$"]},{"content":"From the pier $A$ to the pier $B$ the first motor ship departed at a constant speed, and through $1$ $hour$ after that the second one departed after it at a speed $1\\space km/h$ higher. The distance between the piers is $420\\space km.$ Find the speed of the first ship if both ships arrived at point B at the same time. Give your answer in km/h.[[fill_choice_big-347]][[image-391]]","widgets":{"fill_choice_big-347":{"type":"fill_choice_big","options":["$20$","$22$","$23$"],"placeholder":0,"answer":0},"image-391":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/plavaniesudov2.svg","width":"300"}},"hints":["$1.$ Travel time of the first ship: $\\dfrac{420}{v}.$<br /><br />$2.$ Travel time of the second ship: $\\dfrac{420}{v+1}.$<br /><br />$3.$ Since the second ship departed an hour later, but arrived at the same time, the time difference is $1$ $hour:$ $$\\frac{420}{v}-\\frac{420}{v+1} = 1$$ $4.$ We reduce it to a common denominator: $$\\frac{420(v+1)-420v}{v(v+1)} = 1$$ $5.$ We simplify: $$\\frac{420v + 420-420v}{v(v+1)} = 1$$ $$\\frac{420}{v(v+1)} = 1$$ $6.$ We solve: $$420 = v(v+1)$$ $$v^2 + v-420 = 0$$ $7.$ We solve the quadratic equation: $$v = \\frac{-1 \\pm \\sqrt{1 + 4 \\cdot 420}}{2}$$ $$v = \\frac{-1 \\pm \\sqrt{1\\space681}}{2}$$ $$v = \\frac{-1 \\pm 41}{2}$$ $8.$ We get the roots:<br />$$v_1 = \\frac{40}{2} = 20$$ $$v_2 = \\frac{-42}{2} = -21$$ $v_2$ - not suitable.<br /><br />Answer: the speed of the first ship is $20\\ km/h.$"]},{"content":"From the pier $A$ to the pier $B$ the first motor ship departed at a constant speed, and through $1$ $hour$ after that the second one departed after it at a speed $1$$km/h$ higher. The distance between the piers is equal to $110 \\space km.$ Find the speed of the second ship if it arrived at point $B$ at the same time as the first.<br />Give your answer in km/h.[[fill_choice_big-403]][[image-456]]","widgets":{"fill_choice_big-403":{"type":"fill_choice_big","options":["$11$","$10$","$9$"],"placeholder":0,"answer":0},"image-456":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/03/yahta-lodka-korabl.svg","width":"300"}},"hints":["$1.$ Travel time of the first ship: $\\dfrac{110}{v}.$<br />$2.$ Travel time of the second ship: $\\dfrac{110}{v+1}.$<br /><br />$3.$ Since the second ship departed an hour later, but arrived at the same time, the time difference is equal to $1$ hour: $$\\frac{110}{v}-\\frac{110}{v+1} = 1$$ $4.$ We reduce to a common denominator: $$\\frac{110(v+1)-110v}{v(v+1)} = 1$$ $5.$ We simplify: $$\\frac{110v + 110-110v}{v(v+1)} = 1$$ $$\\frac{110}{v(v+1)} = 1$$ $6.$ We solve: $$110 = v(v+1)$$ $$v^2 + v-110 = 0$$ $7.$ We solve the quadratic equation: $$v = \\frac{-1 \\pm \\sqrt{1 + 4 \\cdot 110}}{2}$$ $$v = \\frac{-1 \\pm \\sqrt{441}}{2}$$ $$v = \\frac{-1 \\pm 21}{2}$$ $8.$ We get the roots:<br />$$v_1 = \\frac{20}{2} = 10$$ $$v_2 = \\frac{-22}{2} = -11$$ $v_2$ - not suitable.<br /><br />Answer: the speed of the first motor ship is $10\\ km/h$, therefore, the speed of the second motor ship: $10 + 1 = 11\\ km/h.$"]},{"content":"The barge in $10:00$ left from point $A$ to point $B,$ located in $15\\space km$ from $A.$ Having stayed at point $B$ $1$ for an hour $20$ minutes, the barge went back and returned to point $A$ in $16:00$ on the same day. Determine the speed of the river flow if it is known that the barge’s own speed is $7\\space km/h.$[[fill_choice_big-561]][[image-525]]","widgets":{"fill_choice_big-561":{"type":"fill_choice_big","options":["$2$","$6$","$4$"],"placeholder":0,"answer":0},"image-525":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2023/07/korabl-kapitan-matros-01.svg","width":"300"}},"hints":["$1.$ Speed with current: $7 + u\\ km/h.$<br />$2.$ Speed against current: $7-u\\ km/h.$<br />$3.$ Travel time excluding parking: $$6-\\frac{4}{3} = \\frac{18}{3}-\\frac{4}{3} = \\frac{14}{3}\\ h$$ $4.$ Create an equation: $$\\frac{15}{7-u} + \\frac{15}{7+u} = \\frac{14}{3}$$ $5.$ Reduce to a common denominator: $$\\frac{15(7+u) + 15(7-u)}{(7-u)(7+u)} = \\frac{14}{3}$$ $6.$ Simplify: $$\\frac{15 \\cdot 7 + 15u + 15 \\cdot 7-15u}{49-u^2} = \\frac{14}{3}$$ $$\\frac{15 \\cdot 14}{49-u^2} = \\frac{14}{3}$$ $7.$ Solve: $$\\frac{210}{49-u^2} = \\frac{14}{3}$$ $$210 \\cdot 3 = 14(49-u^2)$$ $$630 = 686-14u^2$$ $$14u^2 = 56$$ $$u^2 = 4$$ $$u = 2$$ Answer: the speed of the river is $2\\ km/h.$"]},{"content":"The piers $A$ and $B$ are located on the lake, the distance between them is $390\\space km.$ The barge set off at a constant speed from $A$ to $B.$ The next day after arrival, it set off back at a speed $3\\space km/h$ more than before, stopping along the way for $9$ hours. As a result, she spent the same amount of time on the way back as on the way from $A$ to $B.$ Find the speed of the barge on the way from $A$ to $B.$ Give your answer in km/h.[[fill_choice_big-902]][[image-598]]","widgets":{"fill_choice_big-902":{"type":"fill_choice_big","options":["$10$","$12$","$11$"],"placeholder":0,"answer":0},"image-598":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/utki-prud-01-2.svg","width":"300"}},"hints":["$1.$ Travel time from $A$ to $B{:}$ $\\dfrac{390}{u}.$<br /><br />$2.$ Travel time from $B$ to $A{:}$ $\\dfrac{390}{u+3}.$<br /><br />$3.$ Create an equation:<br />$$\\frac{390}{u} = \\frac{390}{u+3} + 9$$ $4.$ We reduce to a common denominator: $$\\frac{390}{u} = \\frac{390 + 9(u+3)}{u+3}$$ $5.$ We simplify: $$\\frac{390}{u} = \\frac{390 + 9u + 27}{u+3}$$ $$\\frac{390}{u} = \\frac{417 + 9u}{u+3}$$ $6.$ We solve: $$390(u+3) = u(417 + 9u)$$ $$390u + 1\\space170 = 417u + 9u^2$$ $$9u^2 + 27u-1\\space170 = 0$$ $$u^2 + 3u-130 = 0$$ $7.$ We solve quadratic equation: $$u = \\frac{-3 \\pm \\sqrt{9 + 520}}{2}$$ $$u = \\frac{-3 \\pm \\sqrt{529}}{2}$$ $$u = \\frac{-3 \\pm 23}{2}$$ $8.$ We get the roots: $$u_1 = \\frac{20}{2} = 10$$ $$u_2 = \\frac{-26}{2} = -13$$ $u_2$ - does not fit.<br />Answer: the speed of the barge on the way from $A$ to $B$ is equal to $10\\ km/h.$"]},{"content":"The traveler crossed the sea on a yacht with an average speed of $20\\space km/h.$. He flew back on a sports plane with a speed of $480\\space km/h.$ Find the average speed of the traveler throughout the entire journey. Give your answer in km/h.[[fill_choice_big-980]][[image-675]]","widgets":{"fill_choice_big-980":{"type":"fill_choice_big","options":["$38.4$","$38.6$","$39.2$"],"placeholder":0,"answer":0},"image-675":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2026/03/pilot-1.svg","width":"300"}},"hints":["$1.$ Travel time on the yacht: $\\dfrac{x}{20}$ hours.<br />$2.$ Travel time by plane: $\\dfrac{x}{480}$ hours.<br />$3.$ Total travel time: $$\\frac{x}{20} + \\frac{x}{480} = \\frac{24x}{480} + \\frac{x}{480} = \\frac{25x}{480}\\space h$$<br />$4.$ Average speed is the ratio of the entire path to the entire time of movement:<br />$$v_{avg} = \\frac{2x}{\\frac{25x}{480}} = 2x \\cdot \\frac{480}{25x} = \\frac{960}{25} = 38.4\\ km/h$$<br />Answer: the average speed of the traveler throughout the entire journey is $38.4\\ km/h.$<br /><br />"]}]}

{"questions":[{"content":"The first worker completes an order for $110$ parts $1$ hour faster than the second. How many parts does the second worker make per hour, if it is known that the first worker makes $1$ more parts in an hour?<br />[[fill_choice_big-1]][[image-31]]","widgets":{"fill_choice_big-1":{"type":"fill_choice_big","options":["$10$","$12$","$11$"],"placeholder":0,"answer":0},"image-31":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/obrazavr_konveer-01.svg","width":"300"}},"hints":["$1.$ Let $x$ be the productivity of the second worker (parts/hour).<br /><br />$2.$ Then $(x+1)$ is the productivity of the first worker.<br /><br />$3.$ Working hours of the second worker: $\\dfrac{110}{x}$ hours.<br /><br />$4.$ Working hours of the first worker: $\\dfrac{110}{x+1}$ hours.<br /><br />$5.$ According to the condition: $$\\frac{110}{x}-\\frac{110}{x+1} = 1$$ $6.$ Solve the equation: $$\\frac{110}{x}-\\frac{110}{x+1} = 1$$ $$110(x+1)-110x = x(x+1)$$ $$110x + 110-110x = x^2 + x$$ $$110 = x^2 + x$$ $$x^2 + x-110 = 0$$ $7.$ Solve the quadratic equation:<br />$$D = 1^2 + 4 \\cdot 110 = 441$$ $$x_1 = \\frac{-1 + \\sqrt{441}}{2} = \\frac{-1 + 21}{2} = 10$$ $$x_2 = \\frac{-1-\\sqrt{441}}{2} = \\frac{-1-21}{2} = -11$$ $x_2$ - not suitable.<br /><br />Answer: The second worker makes $10$ parts per hour."]},{"content":"The first worker completes an order for $156$ parts $1$ hour faster than the second. How many parts does the first worker make per hour, if it is known that he makes $1$ more parts per hour?[[fill_choice_big-31]][[image-76]]","widgets":{"fill_choice_big-31":{"type":"fill_choice_big","options":["$13$","$11$","$15$"],"placeholder":0,"answer":0},"image-76":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/05/ponyatie_shestirenki-01.svg","width":"300"}},"hints":["$1.$ Let $n$ be the productivity of the first worker (parts/hour).<br /><br />$2.$ Then $(n-1)$ is the productivity of the second worker.<br /><br />$3.$ Working hours of the first worker: $\\dfrac{156}{n}$ hours.<br /><br />$4.$ Working hours of the second worker: $\\dfrac{156}{n-1}$ hours.<br /><br />$5.$ According to the condition: $$\\frac{156}{n-1}-\\frac{156}{n} = 1$$ $6.$ Solve the equation: $$\\frac{156}{n-1}-\\frac{156}{n} = 1$$ $$\\frac{156n-156(n-1)}{n(n-1)} = 1$$ $$\\frac{156n-156n + 156}{n(n-1)} = 1$$ $$\\frac{156}{n(n-1)} = 1$$ $$156 = n(n-1)$$$$n^2-n-156 = 0$$ $7.$ Solve the quadratic equation: $$D = (-1)^2 + 4 \\cdot 156 = 1 + 624 = 625$$ $$n_1 = \\frac{1 + \\sqrt{625}}{2} = \\frac{1 + 25}{2} = 13$$ $$n_2 = \\frac{1-\\sqrt{625}}{2} = \\frac{1-25}{2} = -12$$ $n_2$ - not suitable.<br /><br />Answer: The first worker makes $13$ parts per hour."]},{"content":"Two workers working together can complete the job in $12$ days. In how many days, working separately, will the first worker complete this work if he completes the same part of the work in two days as the second worker does in three days? [[fill_choice_big-66]][[image-125]]","widgets":{"fill_choice_big-66":{"type":"fill_choice_big","options":["$20$","$22$","$19$"],"placeholder":0,"answer":0},"image-125":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/vnutri-doma-01-1.svg","width":"300"}},"hints":["$1.$ Let $x$ be the number of days in which the first worker will complete all the work.<br /><br />$2.$ Then the second worker will complete all the work in $1.5x$ days (since $2$ of the first day = $3$ of the second day).<br /><br />$3.$ Productivity of the first worker: $\\dfrac{1}{x}$ work per day.<br /><br />$4.$ Productivity of the second worker: $\\dfrac{1}{1.5x}$ work per day.<br /><br />$5.$ Total productivity: $$\\frac{1}{x} + \\frac{1}{1.5x} = \\frac{1.5 + 1}{1.5x} = \\frac{2.5}{1.5x} = \\frac{5}{3x}$$ $6.$ According to the condition, together they perform the work in $12$ days: $$\\frac{5}{3x} \\cdot 12 = 1$$ $7.$ Solve the equation: $$\\frac{60}{3x} = 1$$ $$20 = x$$ Answer: the first worker will complete the work in $20$ days."]},{"content":"The first pipe passes $1$ a liter of water per minute less than the second. How many liters of water per minute does the first pipe pass if it fills a tank with a volume of $110$ liters $1$ a minute longer than the second pipe?[[fill_choice_big-147]][[image-178]]","widgets":{"fill_choice_big-147":{"type":"fill_choice_big","options":["$10$","$11$","$9$"],"placeholder":0,"answer":0},"image-178":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2024/12/rezonans14.svg","width":"300"}},"hints":["$1.$ Let $k$ be the rate of the second pipe (l/min).<br /><br />$2.$ Then $(k-1)$ is the productivity of the first pipe.<br /><br />$3.$ First pipe filling time: $\\dfrac{110}{k-1}$ minutes.<br /><br />$4.$ Second pipe filling time: $\\dfrac{110}{k}$ minutes.<br /><br />$5.$ According to the condition: $$\\frac{110}{k-1}-\\frac{110}{k} = 1$$ $6.$ Solve the equation:<br />$$\\frac{110}{k-1}-\\frac{110}{k} = 1$$ $$\\frac{110k-110(k-1)}{k(k-1)} = 1$$ $$\\frac{110k-110k + 110}{k(k-1)} = 1$$ $$\\frac{110}{k(k-1)} = 1$$ $$110 = k(k-1)$$ $$k^2-k-110 = 0$$ $7.$ Solve the quadratic equation:<br />$$D = (-1)^2 + 4 \\cdot 110 = 1 + 440 = 441$$ $$k_1 = \\frac{1 + \\sqrt{441}}{2} = \\frac{1 + 21}{2} = 11$$ $$k_2 = \\frac{1-\\sqrt{441}}{2} = \\frac{1-21}{2} = -10$$ $k_2$ - not suitable.<br /><br />$8.$ Second pipe capacity: $11$ liters/min.<br /><br />$9.$ First pipe capacity: $11-1 = 10$ liters/min.<br /><br />Answer: the first pipe passes $10$ liters of water per minute."]},{"content":"The first pipe passes $5$ liters of water per minute less than the second. How many liters of water per minute does the second pipe flow if it fills a tank with a volume of $375$ liters $10$ minutes faster than the first pipe fills a tank with a volume of $500$ liters?[[fill_choice_big-228]][[image-235]]","widgets":{"fill_choice_big-228":{"type":"fill_choice_big","options":["$25$","$22$","$21$"],"placeholder":0,"answer":0},"image-235":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/08/personazh-s-lupoi-.svg","width":"300"}},"hints":["$1.$ Let $x$ be the rate of the second pipe (l/min).<br />$2.$ Then $(x-5)$ is the productivity of the first pipe.<br /><br />$3.$ Second pipe filling time: $\\dfrac{375}{x}$ minutes.<br /><br />$4.$ First pipe filling time: $\\dfrac{500}{x-5}$ minutes.<br /><br />$5.$ According to the condition: $$\\frac{500}{x-5}-\\frac{375}{x} = 10$$ $6.$ Solve the equation:<br />$$\\frac{500}{x-5}-\\frac{375}{x} = 10$$ $$\\frac{500x-375(x-5)}{x(x-5)} = 10$$ $$\\frac{500x-375x + 1\\space875}{x(x-5)} = 10$$ $$\\frac{125x + 1\\space875}{x(x-5)} = 10$$ $7.$ Multiply both sides by $x(x-5){:}$ $$125x + 1\\space875 = 10x(x-5)$$ $$125x + 1\\space875 = 10x^2-50x$$ $$10x^2-175x-1\\space875 = 0$$ $8.$ Divide the equation by $5{:}$ $$2x^2-35x-375 = 0$$ $9.$ Solve quadratic equation: $$D = (-35)^2 + 4 \\cdot 2 \\cdot 375 = 1\\space225 + 3\\space000 = 4\\space225$$ $$x_1 = \\frac{35-\\sqrt{4\\space225}}{4} = \\frac{35-65}{4} = -\\frac{30}{4}$$ $x_1$ - does not fit. $$x_2 = \\frac{35 + \\sqrt{4\\space225}}{4} = \\frac{35 + 65}{4} = 25$$ Answer: the second pipe passes $25$ liters of water per minute."]},{"content":"Each of two workers with the same qualifications can complete an order in $15$ hours. $3$ hours after one of them began fulfilling the order, a second worker joined him, and they completed the work on the order together. How many hours did it take to complete the entire order?[[fill_choice_big-395]][[image-296]]","widgets":{"fill_choice_big-395":{"type":"fill_choice_big","options":["$9$","$10$","$11$"],"placeholder":0,"answer":0},"image-296":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/07/rabota-opazdyvaet-chasy-ofis.svg","width":"300"}},"hints":["$1.$ We will accept the entire order for $1.$<br /><br />$2.$ Productivity of each worker: $\\dfrac{1}{15}$ orders per hour.<br /><br />$3.$ In the first $3$ hours the first worker completed:<br />$$\\dfrac{1}{15} \\cdot 3 = \\dfrac{3}{15} = \\dfrac{1}{5} \\space job$$<br />$4.$ When working together, their overall productivity:<br />$$\\dfrac{1}{15} + \\dfrac{1}{15} = \\dfrac{2}{15}\\space job\\space in\\space hour$$<br />$5.$ All that remains is to complete:<br />$$1-\\dfrac{1}{5} = \\dfrac{5}{5}-\\dfrac{1}{5} = \\dfrac{4}{5}\\space job$$<br />$6.$ Collaboration time:<br />$$\\dfrac{4}{5} : \\dfrac{2}{15} = \\dfrac{4}{5} \\cdot \\dfrac{15}{2} = \\dfrac{60}{10} = 6\\space hours$$<br />$7.$ Total order lead time:<br />$$3 + 6 = 9\\space hours$$ Answer: It took $9$ hours to complete the entire order."]},{"content":"One master can complete an order in $12$ hours, and another in $6$ hours. How many hours will it take both masters to complete the order, working together?[[fill_choice_big-512]][[image-361]]","widgets":{"fill_choice_big-512":{"type":"fill_choice_big","options":["$4$","$6$","$8$"],"placeholder":0,"answer":0},"image-361":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/08/stolyar-instrumenty-ugol2.svg","width":"300"}},"hints":["$1.$ We will accept the entire order for $1.$<br /><br />$2.$ Productivity of the first master: $\\dfrac{1}{12}$ orders per hour.<br /><br />$3.$ Productivity of the second master: $\\dfrac{1}{6}$ orders per hour.<br /><br />$4.$ Overall collaboration work rate:<br />$$\\frac{1}{12} + \\frac{1}{6} = \\frac{1}{12} + \\frac{2}{12} = \\frac{3}{12} = \\frac{1}{4}\\space of\\space the\\space job\\space per\\space hour$$<br />$5.$ Lead time when working together:<br />$$1 : \\frac{1}{4} = 1 \\cdot 4 = 4\\space hours$$ Answer: working together, both masters will complete the order in $4$ hours."]},{"content":"The first pump fills the tank in $20$ minutes, the second in $30$ minutes, and the third in $1$ hour. How many minutes will it take three pumps to fill the tank, working simultaneously?[[fill_choice_big-656]][[image-430]]","widgets":{"fill_choice_big-656":{"type":"fill_choice_big","options":["$10$","$12$","$8$"],"placeholder":0,"answer":0},"image-430":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/10/very-well.svg","width":"300"}},"hints":["$1.$ Take the entire tank as $1.$<br /><br />$2.$ Capacity of the first pump: $\\dfrac{1}{20}$ tanks per minute.<br /><br />$3.$ Second pump capacity: $\\dfrac{1}{30}$ tanks per minute.<br /><br />$4.$ Third pump capacity: $\\dfrac{1}{60}$ tank per minute.<br /><br />$5.$ Total collaboration work rate per minute:<br />$$\\frac{1}{20} + \\frac{1}{30} + \\frac{1}{60} = \\frac{3}{60} + \\frac{2}{60} + \\frac{1}{60} = \\frac{6}{60} = \\frac{1}{10}$$ $6.$ Tank filling time when working together:<br />$$1 : \\dfrac{1}{10} = 1 \\cdot 10 = 10\\space minutes$$ Answer: three pumps, working simultaneously, will fill the tank in $10$ minutes.<br />"]},{"content":"Dasha and Masha weed the garden bed in $12$ minutes, and Masha alone - in $20$ minutes. How many minutes does it take Dasha to weed a garden bed alone?[[fill_choice_big-805]][[image-503]]","widgets":{"fill_choice_big-805":{"type":"fill_choice_big","options":["$30$","$32$","$28$"],"placeholder":0,"answer":0},"image-503":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2022/06/repa-morkovka-zelen-01.svg","width":"300"}},"hints":["$1.$ We will take all the work of weeding the beds as $1.$<br /><br />$2.$ Masha's productivity: $\\dfrac{1}{20}$ beds per minute.<br /><br />$3.$ Productivity of Dasha and Masha together: $\\dfrac{1}{12}$ beds per minute.<br />$4.$ Let Dasha’s productivity be equal to $\\dfrac{1}{x}$ beds per minute<br /><br />$5.$ Create the equation: $$\\frac{1}{20} + \\frac{1}{x} = \\frac{1}{12}$$ $6.$ Solve the equation:<br />$$\\frac{1}{20} + \\frac{1}{x} = \\frac{1}{12}$$ $$\\frac{120}{20} + \\frac{120}{x} = \\frac{120}{12}$$ $$6 + \\frac{120}{x} = 10$$ $$\\frac{120}{x} = 4$$ $$x = \\frac{120}{4}$$ $$x = 30$$ Answer: Dasha weeds the bed in $30$ minutes."]},{"content":"Petya and Vanya perform the same test. Petya answers an hour for $8$ test questions, and Vanya answers $9.$ questions. They simultaneously began answering the test questions, and Petya finished his test $20$ minutes later than Vanya. How many questions does the test contain?[[fill_choice_big-1000]][[image-580]]","widgets":{"fill_choice_big-1000":{"type":"fill_choice_big","options":["$24$","$22$","$20$"],"placeholder":0,"answer":0},"image-580":{"type":"image","url":"/wp-content/themes/Main/img/math-port/2025/02/test-01.svg","width":"300"}},"hints":["$1.$ Let $x$ be the number of questions in the test.<br /><br />$2.$ Petya's time to complete the test: $\\dfrac{x}{8}$ hours.<br /><br />$3.$ Vanya's time to complete the test: $\\dfrac{x}{9}$ hours.<br /><br />$4.$ Convert $20$ minutes into hours:<br /> $$\\dfrac{20}{60} = \\dfrac{1}{3}$$ $5.$ Create an equation: $$\\frac{x}{8}-\\frac{x}{9} = \\frac{1}{3}$$ $6.$ Solve the equation: $$\\frac{9x-8x}{72} = \\frac{1}{3}$$ $$\\frac{x}{72} = \\frac{1}{3}$$ $$x = \\frac{72}{3}$$ $$x = 24$$ Answer: the test contains $24$ questions."]}]}