Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
how much 60 % of 50 is greater than 40 % of 30 ? | "( 60 / 100 ) * 50 β ( 40 / 100 ) * 30 30 - 12 = 18 answer : a" | a ) 18 , b ) 28 , c ) 26 , d ) 27 , e ) 23 | a | subtract(divide(multiply(60, 50), const_100), divide(multiply(40, 30), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
you and your friend spent a total of $ 19 for lunch . your friend spent $ 3 more than you . how much did your friend spend on their lunch ? | my lunch = l , my friends lunch = l + 3 ( l ) + ( l + 3 ) = 19 l + l + 3 - 3 = 19 - 3 2 l = 16 l = 8 my friends lunch l + 3 = 8 + 3 = $ 11 , the answer is e | a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 6 , e ) $ 11 | e | add(divide(subtract(19, 3), const_2), 3) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1) | general |
by selling 90 pens , a trader gains the cost of 30 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 90 pens = rs . 90 profit = cost of 30 pens = rs . 30 profit % = 30 / 90 * 100 = 33.3 % answer : a" | a ) 33.3 % , b ) 32.3 % , c ) 31.3 % , d ) 30.3 % , e ) 29.3 % | a | multiply(divide(30, 90), const_100) | divide(n1,n0)|multiply(#0,const_100)| | gain |
a cistern is normally filled in 2 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 2 - 1 / x = 1 / 4 , solving 1 / x = 1 / 4 so in 4 hours full cistern will be empty . answer : a" | a ) 4 hours , b ) 20 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours | a | inverse(subtract(divide(const_1, 2), divide(const_1, add(2, 2)))) | add(n0,n1)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)| | physics |
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 72 kmph . in how much time will the trains cross each other ? | "b relative speed = ( 54 + 72 ) * 5 / 18 = 7 * 5 = 35 mps . the time required = d / s = ( 100 + 100 + 200 ) / 35 = 400 / 35 = 80 / 7 sec ." | a ) 70 / 7 sec , b ) 80 / 7 sec , c ) 40 / 7 sec , d ) 60 / 7 sec , e ) 90 / 7 sec | b | divide(100, multiply(add(54, 72), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
effective annual rate of interest corresponding to nominal rate of 6 % per annum compounded half yearly will be | explanation : let the amount rs 100 for 1 year when compounded half yearly , n = 2 , rate = 6 / 2 = 3 % amount = 100 ( 1 + 3 / 100 ) 2 = 106.09 effective rate = ( 106.09 - 100 ) % = 6.09 % option a | a ) 6.09 % , b ) 6.10 % , c ) 6.12 % , d ) 6.14 % , e ) none of these | a | subtract(multiply(power(add(divide(divide(6, const_2), const_100), const_1), const_2), const_100), const_100) | divide(n0,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(#3,const_100)|subtract(#4,const_100) | gain |
in a certain business school class , p students are accounting majors , q students are finance majors , r students are marketing majors , and s students are strategy majors . if pqrs = 1365 , and if 1 < p < q < r < s , how many students in the class are finance majors ? | "pqrs = 1365 = 3 * 5 * 7 * 13 since 1 < p < q < r < s , the number of students who are finance majors is q = 5 . the answer is b ." | a ) 3 , b ) 5 , c ) 8 , d ) 11 , e ) 17 | b | divide(1365, multiply(multiply(add(const_2, const_3), add(add(const_2, const_3), const_2)), add(const_10, const_3))) | add(const_10,const_3)|add(const_2,const_3)|add(#1,const_2)|multiply(#1,#2)|multiply(#0,#3)|divide(n0,#4)| | gain |
a train 500 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 500 + x ) / 60 = 65 / 3 x = 800 m . answer : c" | a ) 298 m , b ) 300 m , c ) 800 m , d ) 967 m , e ) 1181 m | c | divide(500, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the average age of 3 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 3 * 2 ) years = 6 years . sum of ages of two new men = ( 21 + 23 + 6 ) years = 50 years average age of two new men = ( 50 / 2 ) years = 25 years . answer : c" | a ) 22 , b ) 30 , c ) 25 , d ) 38 , e ) 27 | c | add(divide(add(21, 23), const_2), multiply(const_1, 3)) | add(n1,n2)|multiply(n0,const_1)|divide(#0,const_2)|add(#2,#1)| | general |
points a , b , and , c have xy - coordinates ( 2,0 ) , ( 8,12 ) , and ( 14,0 ) , respectively . points x , y , and z have xy - coordinates ( 6,0 ) , ( 8,4 ) , and ( 10,0 ) , respectively . what fraction s of the area of triangle abc is the area of triangle xyz ? | "if you notice , both triangles abc and xyz have a side on x axis . we can take these sides as bases for each triangle , therefore area of abc is 1 / 2 * 12 * 12 ( height of abc is the y coordinate of the third point ( 8,12 ) ) similarly area of xyz is 1 / 2 * 4 * 4 dividing area of xyz with that of abc gives s = 1 / 9... | a ) 1 / 9 , b ) 1 / 8 , c ) 1 / 6 , d ) 1 / 5 , e ) 1 / 3 | a | divide(divide(power(const_4, const_2), const_2), divide(power(add(const_10, const_2), const_2), const_2)) | add(const_10,const_2)|power(const_4,const_2)|divide(#1,const_2)|power(#0,const_2)|divide(#3,const_2)|divide(#2,#4)| | geometry |
a van takes 6 hours to cover a distance of 270 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distence = 270 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 270 / 9 = 30 kmph b" | a ) 50 kmph , b ) 30 kmph , c ) 70 kmph , d ) 80 kmph , e ) 90 kmph | b | divide(270, multiply(divide(3, 2), 6)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
working together , printer a and printer b would finish the task in 40 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 4 pages a minute more than printer a ? | 40 * a + 40 * b = x pages in 40 mins printer a will print = 40 / 60 * x pages = 2 / 3 * x pages thus in 40 mins printer printer b will print x - 2 / 3 * x = 1 / 3 * x pages also it is given that printer b prints 4 more pages per min that printer a . in 40 mins printer b will print 160 more pages than printer a thus 2 /... | a ) 250 , b ) 375 , c ) 450 , d ) 480 , e ) 500 | d | multiply(divide(4, subtract(divide(60, 40), const_1)), const_60) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,const_60) | physics |
if 36 men do a work in 70 days , in how many days will 40 men do it ? | "36 * 70 = 40 * x x = 63 days answer : b" | a ) 64 , b ) 63 , c ) 65 , d ) 66 , e ) 67 | b | divide(multiply(36, 70), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
last year , for every 100 million vehicles that traveled on a certain highway , 75 vehicles were involved in accidents . if 6 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 ) | "to solve we will set up a proportion . we know that β 100 million vehicles is to 75 accidents as 6 billion vehicles is to x accidents β . to express everything in terms of β millions β , we can use 6,000 million rather than 6 billion . creating a proportion we have : 100 / 75 = 6,000 / x cross multiplying gives us : 1... | a ) 2500 , b ) 4500 , c ) 3500 , d ) 4000 , e ) 1300 | b | multiply(75, multiply(6, const_10)) | multiply(n2,const_10)|multiply(n1,#0)| | general |
on a certain road , 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 60 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 10 % of them exceeded the speed limit and received the ticket , i . e . x / 10 . again , suppose total no . of motorists who exceeded the speed limit are y . 60 % of y exceeded the speed limit but did n ' t received the ticket , i . e . 3 y / 5 . it means 2 y / 5 received the ticket . h... | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 25 % , e ) 30 % | d | multiply(divide(10, subtract(const_100, 60)), const_100) | subtract(const_100,n1)|divide(n0,#0)|multiply(#1,const_100)| | gain |
a and b can do a piece of work in 48 days , a and c can do it in 60 days , b and c can do it 72 days . in what time can c alone do it ? | explanation : ( a + b ) 1 day work = 1 / 48 ( b + c ) 1 day work = 1 / 72 ( a + c ) 1 day work = 1 / 60 by adding 2 ( a + b + c ) = 1 / 48 + 1 / 72 + 1 / 60 = 37 / 720 ( a + b + c ) 1 day work = 37 / 1440 a 1 day work = 37 / 1440 β 1 / 72 = 17 / 1440 = 85 days ( approximately ) answer : option c | a ) 84 days , b ) 84 Β½ days , c ) 85 days , d ) 87 days , e ) 89 days | c | add(subtract(72, 48), 60) | subtract(n2,n0)|add(n1,#0) | physics |
if x is an integer such that 1 < x < 9 , 2 < x < 15 , 7 > x > β 1 , 4 > x > 0 , and x + 1 < 5 , then x is | "1 < x < 9 , 2 < x < 15 , - 1 < x < 7 0 < x < 4 x + 1 < 5 from above : 2 < x < 4 - - > x = 3 . answer : a ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(subtract(5, 1), 1) | subtract(n9,n8)|subtract(#0,n8)| | general |
1 = 1,2 = 2,3 = 9,4 = 16,5 = 25 , then 9 = ? | "1 = 1,2 = 2,3 = 9,4 = 16,5 = 25 , then 9 = ? 9 = 3 check the third eqn . answer : c" | a ) 1 , b ) 255 , c ) 3 , d ) 445 , e ) 235 | c | divide(subtract(subtract(25, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2) | add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)| | general |
a train 300 m long can cross an electric pole in 10 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 300 / 10 s = 30 m / sec speed = 30 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 108 kmph answer : e" | a ) 88 kmph , b ) 89 kmph , c ) 72 kmph , d ) 16 kmph , e ) 108 kmph | e | divide(divide(300, const_1000), divide(10, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
if a book is sold at 10 % profit instead of 10 % loss , it would have brought rs 14 more . find out the cost price of the book | "let c . p . of the book be rs . β x β given , 1.10 x - 0.90 x = 14 = > 0.2 x = 14 = 14 / 0.2 = rs 70 answer : d" | a ) 75 , b ) 72 , c ) 60 , d ) 70 , e ) 80 | d | divide(multiply(const_100, divide(10, const_2)), 10) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0)| | gain |
in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 55 , what is the greatest possible value of b Γ a ? | "you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 25 ( next number wo n ' t have 110 as the first 3 digit... | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 25 | e | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 55), const_1), 55), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 55), const_1), 55), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000),... | power(n0,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,n0)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)| | general |
how many liters of water must be evaporated from 50 liters of a 3 percent sugar solution to get a 4 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.03 ( 50 ) = 0.04 ( 50 - x ) 0.04 x = 2 - 1.5 x = 0.5 / 0.04 = 12.5 liters the answer is a ." | a ) 12.5 , b ) 15.5 , c ) 17.5 , d ) 21.5 , e ) 24.5 | a | subtract(50, multiply(divide(50, const_100), 4)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 & that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidateswho passed the examination is ? | "let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 β 24 y = 4200 - 1800 β΄ y = 2400 / 24 = 100 c" | a ) 50 , b ) 90 , c ) 100 , d ) 120 , e ) 140 | c | divide(subtract(multiply(120, 35), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n4)|subtract(n2,n4)|subtract(#0,#1)|divide(#3,#2)| | general |
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 4 feet . how many feet does the squirrel travels if the post is 16 feet tall and 2 feet in circumference ? | "total circuit = 16 / 4 = 4 total feet squirrel travels = 4 * 2 = 8 feet answer : b" | a ) 10 feet , b ) 8 feet , c ) 13 feet , d ) 15 feet , e ) 18 feet | b | multiply(divide(16, 4), 2) | divide(n1,n0)|multiply(n2,#0)| | geometry |
if ( t - 8 ) is a factor of t ^ 2 - kt - 46 , then k = | "t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 14 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) e" | a ) 16 , b ) 12 , c ) 2 , d ) 6 , e ) 14 | e | add(const_10, 2) | add(n1,const_10)| | general |
what will be the cost of building a fence around a rectangular plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 50 ? | let the side of the rectangle plot be a ft . a * b = 289 = > 1.7 b * b = 289 b = 13.03 a = 22.17 length of the fence = perimeter of the plot = 2 ( a + b ) = 70.39 ft . cost of building the fence = 70.39 * 50 = rs . 3519.5 answer : c | ['a ) a ) rs . 3944', 'b ) b ) rs . 3948', 'c ) c ) rs . 3519.5', 'd ) d ) rs . 3949', 'e ) e ) rs . 3923'] | c | multiply(square_perimeter(sqrt(289)), 50) | sqrt(n0)|square_perimeter(#0)|multiply(n1,#1) | geometry |
how many liters of water must be evaporated from 50 liters of a 8 - percent sugar solution to get a 10 - percent solution ? | "8 % of a 50 liter solution is 4 l which is 10 % of the solution at the end . the solution at the end must be 40 l . we need to evaporate 10 liters . the answer is b ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | b | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
. in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 70 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are... | registered voters = 100 d = 60 r = 40 70 % of d ( 60 ) = 42 20 % of r ( 40 ) = 8 total voter % of registered voters = 50 / 100 50 % ans a | a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 57 % | a | add(multiply(60, divide(70, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4) | gain |
a box contains 12 apples , 11 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ? | "p ( choosing a red apple 7 times in a row ) = 11 / 12 * 10 / 11 * 9 / 10 * 8 / 9 * 7 / 8 * 6 / 7 * 5 / 6 = 5 / 12 < 0.5 the answer is e ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(multiply(0.5, 11), const_1) | multiply(n2,n1)|add(#0,const_1)| | general |
a train 110 meters long completely crosses a 280 meters long bridge in 38 seconds . what is the speed of the train is ? | "s = ( 110 + 280 / 38 = 390 / 38 * 18 / 5 = 40 answer : c" | a ) 32 , b ) 45 , c ) 40 , d ) 66 , e ) 09 | c | divide(divide(add(110, 280), const_1000), divide(38, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics |
in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many f cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer f = 44 . d" | a ) 0.33 , b ) 0.44 , c ) 0.55 , d ) 44 , e ) 55 | d | multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)| | general |
paulson spends 75 % of his income . his income is increased by 20 % and he increased his expenditure by 10 % . find the percentage increase in his savings | "let the original income = rs . 100 . then , expenditure = rs . 75 and savings = rs . 25 new income = rs . 120 , new expenditure = rs . ( ( 110 / 100 ) * 75 ) = rs . 165 / 2 new savings = rs . ( 120 - ( 165 / 2 ) ) = rs . 75 / 2 increase in savings = rs . ( ( 75 / 2 ) - 25 ) = rs . 25 / 2 increase % = ( ( 25 / 2 ) * ( ... | a ) 30 % , b ) 35 % , c ) 40 % , d ) 45 % , e ) 50 % | e | multiply(divide(subtract(subtract(add(const_1, divide(20, const_100)), multiply(add(const_1, divide(10, const_100)), divide(75, const_100))), subtract(const_1, divide(75, const_100))), subtract(const_1, divide(75, const_100))), const_100) | divide(n1,const_100)|divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#2)|subtract(#3,#6)|subtract(#7,#5)|divide(#8,#5)|multiply(#9,const_100)| | general |
if ( 1 β 1.5 ) n = 1 , then n = | "( 1 β 1.5 ) n = 1 simplify to get : - 0.5 n = 1 rewrite as ( - 1 / 2 ) n = 1 multiply both sides by - 2 to get : n = - 2 answer : c" | a ) β 400 , b ) β 140 , c ) β 2 , d ) 4 , e ) 400 | c | negate(multiply(1, const_4)) | multiply(n2,const_4)|negate(#0)| | general |
in a college , the ratio of the number of boys to girls is 8 : 5 . if there are 160 girls , the total number of students in the college is | "explanation : let the number of boys and girls be 8 x and 5 x . total number of students = 13 x = 13 * 32 = 416 . answer : d" | a ) 2447 , b ) 377 , c ) 667 , d ) 416 , e ) 871 | d | add(160, multiply(divide(160, 5), 8)) | divide(n2,n1)|multiply(n0,#0)|add(n2,#1)| | other |
the average weight of 5 students decreases by 12 kg when one of them weighing 72 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 12 kg = > ( 72 - x ) / 5 = 12 = > x = 12 answer : a" | a ) 12 kg , b ) 10 kg , c ) 20 kg , d ) 72 kg , e ) none of these | a | subtract(72, multiply(5, 12)) | multiply(n0,n1)|subtract(n2,#0)| | general |
find the least multiple of 23 , which when divided by 1821 and 24 leaves remainders 710 and 13 respectively . | solution here ( 18 - 7 ) = 11 , ( 21 - 10 ) = 11 and ( 24 - 13 ) = 11 . l . c . m . of 1821 and 24 is 504 . let required number be 504 k - 11 . least value of k for which ( 504 k - 11 ) is divisible 23 is k = 6 . so , required number = 504 Γ 6 - 11 = 3024 - 11 = 3013 . answer b | a ) 3002 , b ) 3013 , c ) 3024 , d ) 3036 , e ) none of these | b | multiply(add(add(multiply(24, const_10), multiply(24, const_10)), multiply(add(const_4, const_4), const_3)), add(subtract(13, const_10), const_3)) | add(const_4,const_4)|multiply(n2,const_10)|subtract(n4,const_10)|add(#1,#1)|add(#2,const_3)|multiply(#0,const_3)|add(#3,#5)|multiply(#6,#4) | general |
a boy is travelling from his home to school at 5 km / hr and reached 7 min late . next day he traveled at 10 km / hr and reached 8 min early . distance between home and school ? | let the distance be x t 1 = x / 5 hr t 2 = x / 10 hr difference in time = 7 + 8 = 15 = 1 / 4 hr x / 5 - x / 10 = 1 / 4 x / 10 = 1 / 4 x = 2.5 km answer is a | a ) 2.5 km , b ) 3.5 km , c ) 4.5 km , d ) 5.5 km , e ) 6.5 km | a | divide(add(divide(7, const_60), divide(8, const_60)), divide(const_1, 10)) | divide(n1,const_60)|divide(n3,const_60)|divide(const_1,n2)|add(#0,#1)|divide(#3,#2) | physics |
how many of the positive factors of 16 are not factors of 20 | "factors of 16 - 1 , 2 , 4 , 8 , 16 factors of 20 - 1 , 2 , 4 , 5 , 10 , 20 . comparing both , we have 2 factors of 16 which are not factors of 20 - 8,16 answer : a" | a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | a | divide(20, 16) | divide(n1,n0)| | other |
an investor deposited $ 10,000 to open a new savings account that earned 8 percent annual interest , compounded quarterly . if there were no other transactions in the account , what was the amount of money in the account 6 months after the account was opened ? | "the amount in the account after 6 months is 1.02 * 1.02 ( $ 10,000 ) = $ 10,404 . the answer is d ." | a ) $ 10,200 , b ) $ 10,202 , c ) $ 10,400 , d ) $ 10,404 , e ) $ 10,800 | d | multiply(multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2)), power(add(divide(divide(8, 8), const_100), const_1), const_2)) | add(const_2,const_3)|divide(n1,n1)|divide(#1,const_100)|multiply(#0,const_100)|multiply(#0,const_2)|add(#2,const_1)|multiply(#3,#4)|power(#5,const_2)|multiply(#6,#7)| | gain |
a pipe can fill a cistern in 77 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 77 minutes 1 / 11 part filled in = 77 * 1 / 11 = 7 minutes answer is e" | a ) 5 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 7 min | e | multiply(77, divide(1, 11)) | divide(n1,n2)|multiply(n0,#0)| | physics |
a can do a job in 10 days and b in 20 days . if they work on it together for 4 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 10 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 10 + 1 / 20 = 3 / 20 a + b 4 days work = 3 / 20 * 4 = 3 / 5 remaining work = 1 - 3 / 5 = 2 / 5 answer is a" | a ) 2 / 5 , b ) 8 / 15 , c ) 3 / 11 , d ) 1 / 12 , e ) 6 / 13 | a | subtract(const_1, multiply(4, add(divide(const_1, 10), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
a trader sells 20 meters of cloth for rs . 4500 at the profit of rs . 10 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4500 / 20 = rs . 225 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 225 - rs . 10 = rs . 215 . answer : c" | a ) rs . 380 , b ) rs . 215 , c ) rs . 190 , d ) rs . 395 , e ) none of these | c | subtract(divide(4500, 20), 10) | divide(n1,n0)|subtract(#0,n2)| | physics |
sahil purchased a machine at rs 10000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it . | "explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 10000 + 5000 + 1000 = 16000 150 % of 16000 = 150 / 100 * 16000 = 24000 option b" | a ) rs . 22000 , b ) rs . 24000 , c ) rs . 26000 , d ) rs . 28000 , e ) none of these | b | add(add(add(10000, 5000), 1000), multiply(divide(add(add(10000, 5000), 1000), const_100), 50)) | add(n0,n1)|add(n2,#0)|divide(#1,const_100)|multiply(n3,#2)|add(#1,#3)| | gain |
the ages of 2 persons differ by 35 years . if 10 years ago the elder one be 6 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 35 6 ( x - 10 ) = x + 35 - 10 x = 15 age of elder person = 15 + 35 = 50 answer is d" | a ) 30 , b ) 48.5 , c ) 52 , d ) 50 , e ) 55 | d | subtract(add(divide(multiply(2, 35), subtract(35, const_1)), 35), 2) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 38 pages per hour . if tom starts reading a novel at 4 : 30 , and jan begins reading an identical copy of the same book at 5 : 18 , at what time will they be reading the same page ? | "since tom reads an average of 1 page every 2 minutes , tom will read 24 pages in the first 48 minutes . jan can catch tom at a rate of 8 pages per hour , so it will take 3 hours to catch tom . the answer is b ." | a ) 7 : 48 , b ) 8 : 18 , c ) 8 : 48 , d ) 9 : 18 , e ) 9 : 48 | b | divide(add(5, divide(multiply(add(subtract(5, 4), divide(subtract(18, 30), const_60)), 30), subtract(38, 30))), 18) | subtract(n5,n3)|subtract(n4,n2)|subtract(n1,n0)|divide(#0,const_60)|add(#3,#1)|multiply(n0,#4)|divide(#5,#2)|add(n4,#6)|divide(#7,n5)| | general |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the integers between 1 and 50 ? | "sum = 50 * 51 / 2 = 25 * 51 = 1275 answer is a" | a ) 1275 , b ) 1542 , c ) 985 , d ) 1024 , e ) 1125 | a | add(divide(subtract(subtract(50, 1), add(1, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
simplify : 896 x 896 - 204 x 204 | "( 896 ) ^ 2 - ( 204 ) ^ 2 = ( 896 + 204 ) ( 896 - 204 ) = 1100 x 692 = 761200 . answer is a ." | a ) 761200 , b ) 761400 , c ) 761800 , d ) 761500 , e ) none of them | a | add(multiply(896, 896), multiply(204, 204)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general |
a searchlight on top of the watch - tower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | 2 revolutions per minute = 1 revolution every 30 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than 30 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least 5 seconds ) = 1 - p ( man in dark for max of 5 seconds ) = 1 -... | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | e | subtract(const_1, divide(5, divide(multiply(const_60, const_1), 2))) | multiply(const_1,const_60)|divide(#0,n0)|divide(n1,#1)|subtract(const_1,#2) | physics |
{ - 10 , - 6 , - 5 , - 4 , - 2.5 , - 1 , 0 , 2.5 , 4 , 6 , 7 , 10 } a number is to be selected at random from the set above . what is the probability that the number selected will be a solution of the equation ( x - 5 ) ( x + 10 ) ( 2 x - 5 ) = 0 ? | roots of the equation ( x - 5 ) ( x + 10 ) ( 2 x - 5 ) = 0 are x = 5 , x = - 10 and x = 5 / 2 = 2.5 . so , two solutions present in our set of 12 distinct numbers , therefore p = 2 / 12 = 1 / 6 . answer : b . | a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | b | divide(2, multiply(6, 2)) | multiply(n1,n14)|divide(n14,#0) | general |
a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability y that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total o... | a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2)| | probability |
how many positive factors of 42 are not factors of 56 ? | "42 = 2 * 3 * 7 . so , total number of factors for 42 = 2 * 2 * 2 = 8 factors for 42 = 1 , 2,3 , 6,7 , 14,21 and 42 56 = 2 * 2 * 2 * 7 so , total number of factors for 56 = 4 * 2 = 8 factors of 56 = 1 , 2,4 , 7,8 , 14,28 and 56 . factors same in 42 and 56 are 1 , 2,7 and 14 . hence , answer will be d ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(subtract(56, const_10), const_10) | subtract(n1,const_10)|divide(#0,const_10)| | other |
when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 40 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 60 % ( 100 - 40 ) . therefore 0.5 x amount of flower - nectar = 0.60 x amount of honey = 0.60 x 1 kg therefore amount of flower - nectar needed = ( 0.60 / 0.51 ) kg = 1.2 kgs answer : a" | a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | a | divide(divide(subtract(const_100, 40), const_100), divide(50, const_100)) | divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0)| | general |
sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 7000 ? | "total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 42000 - 34009 ) = rs . 7991 answer : d" | a ) 4966 , b ) 2477 , c ) 2877 , d ) 7991 , e ) 8881 | d | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 7000), 5) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5)| | general |
the number of sailors on a ship is 50 % more than the number of officers . the ratio of sailors to officers would be | "sailor = 1.50 * officer sailor / officer = 1.50 / 1 = 150 / 100 = 3 / 2 answer will be c" | a ) 33 : 25 , b ) 47 : 20 , c ) 3 : 2 , d ) 22 : 25 , e ) 97 : 50 | c | divide(add(50, const_100), const_100) | add(n0,const_100)|divide(#0,const_100)| | other |
in the x - y plane , there are 4 points ( 0,0 ) , ( 0,4 ) , ( 5,4 ) , and ( 5,0 ) . if these 4 points makes a rectangle , what is the probability that x + y < 4 ? | the line y = - x + 4 intersects the rectangle and these three points of intersection ( 0,0 ) , ( 0,4 ) and ( 4,0 ) form a triangle . the points below the line y = - x + 4 satisfy x + y < 4 . the area of this triangle is ( 1 / 2 ) ( 4 ) ( 4 ) = 8 the area of the rectangle is 20 . p ( x + y < 4 ) = 8 / 20 = 2 / 5 the ans... | a ) 2 / 3 , b ) 2 / 5 , c ) 3 / 4 , d ) 3 / 7 , e ) 5 / 8 | b | divide(divide(multiply(4, 4), const_2), multiply(4, add(const_1, const_4))) | add(const_1,const_4)|multiply(n0,n0)|divide(#1,const_2)|multiply(n0,#0)|divide(#2,#3) | general |
we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value q of the ratio a / b ? | q - > a / b = 20 * 21 * β¦ β¦ * 39 / 21 * 22 * β¦ . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a . | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | divide(n6,n6)|add(#0,#0)|divide(#0,#1) | general |
3 numbers are in the ratio of 2 : 3 : 4 . if the sum of the squares of the extremes is 180 , then the middle number is : | number be = 2 x , 3 x , 4 x , ( 2 x ) 2 + ( 4 x ) 2 = 180 20 x = 180 β x = 9 β x = 3 , midline number = 3 Γ 3 = 9 answer : d | a ) 6 , b ) 12 , c ) 15 , d ) 9 , e ) 7 | d | multiply(3, sqrt(divide(180, add(power(2, const_2), power(4, const_2))))) | power(n1,const_2)|power(n3,const_2)|add(#0,#1)|divide(n4,#2)|sqrt(#3)|multiply(n0,#4) | general |
20 men do a work in 20 days . how many men are needed to finish the work in 5 days ? | "men required to finish the work in 5 days = 20 * 20 / 5 = 80 answer is c" | a ) 50 , b ) 20 , c ) 80 , d ) 10 , e ) 15 | c | divide(multiply(20, 20), 5) | multiply(n0,n1)|divide(#0,n2)| | physics |
a man can row his boat with the stream at 22 km / h and against the stream in 10 km / h . the man ' s rate is ? | "explanation : ds = 22 us = 10 s = ? s = ( 22 - 10 ) / 2 = 6 kmph answer : b" | a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 4 kmph , e ) 9 kmph | b | divide(subtract(22, 10), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
suganya and suriya are partners in a business . suganya invests rs . 35,000 for 8 months and suriya invests rs . 42,000 for 10 months . out of a profit of rs . 31,570 . suganya ' s share is | "solution ratio of their shares = ( 35000 Γ 8 ) : ( 42000 Γ 10 ) = 2 : 3 . suganya ' s share = rs . ( 31570 Γ 2 / 5 ) = rs . 12628 . answer c" | a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 18,040 , d ) rs . 18,942 , e ) none | c | multiply(multiply(const_0_25, const_100), 10) | multiply(const_0_25,const_100)|multiply(n3,#0)| | gain |
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 7 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ? | "distance = relative speed * time = ( 7 + 3 ) * 4 = 40 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : b" | a ) 22 km , b ) 40 km , c ) 65 km , d ) 18 km , e ) 16 km | b | multiply(add(7, 3), 4) | add(n0,n1)|multiply(n2,#0)| | physics |
if xy denotes x is raised to the power y , find the last two digits of 19413846 + 19614181 | 1941 ^ 3846 unit digit = 1 ^ 4 = 1 10 th digit = 4 * 6 = 24 = 2 means last two digit = 41 and 1972 ^ 4181 unit digit = 1 10 th digit = 7 * 2 = 14 mean last two digit = 41 hence , 41 + 41 = 82 | a ) 67 , b ) 74 , c ) 82 , d ) 98 , e ) 34 | c | subtract(subtract(add(subtract(subtract(reminder(reminder(19614181, power(const_100, const_2)), const_100), const_10), const_10), reminder(floor(divide(19413846, power(const_100, const_2))), const_100)), const_10), const_10) | power(const_100,const_2)|divide(n0,#0)|reminder(n1,#0)|floor(#1)|reminder(#2,const_100)|reminder(#3,const_100)|subtract(#4,const_10)|subtract(#6,const_10)|add(#5,#7)|subtract(#8,const_10)|subtract(#9,const_10) | general |
all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 156 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial qua... | "a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 156 liters was transferred from cont... | a ) 1248 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(156, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | divide(n0,const_100)|inverse(const_2)|subtract(const_1,#0)|subtract(#1,#2)|divide(n1,#3)| | general |
the chance of rain on any given day in tel - aviv is 50 % . what is the probability a that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ? | "chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days a = 15 / 64 ans a it is ." | a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64 | a | divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), const_2)))) | choose(n2,n1)|divide(const_1,const_2)|power(#1,n1)|power(#1,const_2)|multiply(#2,#3)|inverse(#4)|divide(#0,#5)| | probability |
a 300 m long train crosses a platform in 27 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 27 = 50 / 3 x = 150 m . answer : e" | a ) 286 m , b ) 350 m , c ) 277 m , d ) 278 m , e ) 150 m | e | subtract(multiply(speed(300, 18), 27), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics |
of 67 players on a cricket team , 37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 67 thrower = 37 rest = 67 - 37 = 30 left handed = 30 / 3 = 10 right handed = 20 if all thrower are right handed then total right handed is 30 + 20 = 57 so c . 57 is the right answer" | a ) 54 , b ) 55 , c ) 57 , d ) 71 , e ) 92 | c | add(37, multiply(subtract(const_1, divide(const_1, const_3)), subtract(67, 37))) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general |
a can do a half of certain work in 70 days and b one third of the same in 35 days . they together will do the whole work in . | "a = 140 days b = 105 days 1 / 140 + 1 / 105 = 7 / 420 = 1 / 60 = > 60 days answer : d" | a ) 88 days , b ) 27 days , c ) 78 days , d ) 60 days , e ) 17 days | d | add(inverse(multiply(70, const_2)), inverse(multiply(35, const_3))) | multiply(n0,const_2)|multiply(n1,const_3)|inverse(#0)|inverse(#1)|add(#2,#3)| | physics |
when 1 / 10 percent of 4,000 is subtracted from 1 / 10 of 4,000 , the difference is | "we can break this problem into two parts : 1 ) what is 1 / 10 percent of 4,000 ? 2 ) what is 1 / 10 of 4,000 ? to calculate 1 / 10 percent of 4,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 10... | a ) 0 , b ) 50 , c ) 396 , d ) 495 , e ) 500 | c | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | add(const_3,const_4)|divide(n0,n1)|divide(#1,const_100)|multiply(#0,const_1000)|multiply(#1,#3)|multiply(#2,#3)|subtract(#4,#5)| | general |
if n = 2.0823 and n * is the decimal obtained by rounding n to the nearest hundredth , what is the value of n * β n ? | "n * = 2.08 n * - n = 2.08 - 2.0823 - 0.0023 answer : b" | a ) - 0.0053 , b ) - 0.0023 , c ) 0.0007 , d ) 0.0047 , e ) 0.0153 | b | subtract(subtract(2.0823, divide(divide(add(multiply(const_2, const_10), const_3), const_100), const_100)), 2.0823) | multiply(const_10,const_2)|add(#0,const_3)|divide(#1,const_100)|divide(#2,const_100)|subtract(n0,#3)|subtract(#4,n0)| | general |
if jack walked 7 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ? | "distance walked in 1 hour and 15 mins = 7 miles speed per hour = distance / time = 7 / ( 5 / 4 ) = 5.6 miles per hour answer c" | a ) 4 , b ) 4.5 , c ) 5.6 , d ) 6.25 , e ) 15 | c | divide(multiply(7, const_60), add(15, const_60)) | add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0)| | physics |
a grocery shop has a sale of rs . 6635 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | let the sale in the sixth month = x then ( 6635 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6635 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ 6500 = > 34209 + x = 39000 = > x = 39000 β 34209 = 4791 answer : d | a ) 4857 , b ) 4184 , c ) 4012 , d ) 4791 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6635, 6927)), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5) | general |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 150 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 150 kilometers ? 150 Γ£ Β· 40 = 3.75 Γ£ β 1 gallon = 3.75 gallons correct answer is d ) 3.75 gallons" | a ) 3.5 gallons , b ) 2.7 gallons , c ) 5.75 gallons , d ) 3.75 gallons , e ) 7.50 gallons | d | divide(150, 40) | divide(n1,n0)| | physics |
a fellow borrowed a certain sum of money at 5 % per annum at simple interest and in 5 years the interest amounted to rs . 2250 less than the sum lent . what was the sum lent ? | "p - 2250 = ( p * 5 * 5 ) / 100 p = 3000 answer : e" | a ) 1050 , b ) 1220 , c ) 1250 , d ) 1060 , e ) 3000 | e | divide(2250, subtract(const_1, divide(multiply(5, 5), const_100))) | multiply(n0,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)| | gain |
in a certain country 1 / 3 of 4 = 8 . assuming the same proportion , what would be the value of 1 / 8 of 4 ? | "e 3" | a ) 2 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | multiply(divide(3, 4), 4) | divide(n1,n2)|multiply(n6,#0)| | general |
the average age of a group of 12 students is 20 years . if 4 more students join the group , the average age increases by 1 year . the average age of the new students is | "total age of 12 std = 20 * 12 total age of 16 std = 21 * 16 total age of 4 std = ( 21 * 16 ) - 20 * 12 = 96 avg of 4 std = ( 96 ) / 4 = 24 answer : c" | a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | c | divide(subtract(multiply(add(20, 1), add(20, 1)), multiply(20, 12)), 4) | add(n1,n3)|multiply(n0,n1)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,n2)| | general |
golu started from his house towards north . after covering a distance of 8 km . he turned towards left and covered a distance of 6 km . what is the shortest distance now from his house ? | c - - - - - - - - - - - - - b ! ! ! ! ! a ab = 8 km bc = 6 km ac = sqrtof ( 8 ^ 2 + 6 ^ 2 ) ac = 10 km answer : a | a ) 10 km . , b ) 16 km . , c ) 14 km , d ) 2 km . , e ) 4 km . | a | sqrt(add(power(8, const_2), power(6, const_2))) | power(n0,const_2)|power(n1,const_2)|add(#0,#1)|sqrt(#2) | physics |
a tank can be filled by a tap in 20 minutes and by another tap in 60 minutes . both the taps are kept open for 10 minutes and then the first tap is shut off . after this , the tank will be completely filled in what time ? | explanation : how we can solve this question ? first we will calculate the work done for 10 mins , then we will get the remaining work , then we will find answer with one tap work , as part filled by tap a in 1 min = 1 / 20 part filled by tap b in 1 min = 1 / 60 ( a + b ) ' s 10 mins work = 10 β ( 1 / 20 + 1 / 60 ) = 1... | a ) 10 mins , b ) 15 mins , c ) 20 mins , d ) 25 mins , e ) 30 mins | c | divide(subtract(const_1, multiply(10, add(divide(const_1, 20), divide(const_1, 60)))), divide(const_1, 60)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|divide(#4,#1) | physics |
renu can do a piece of work in 6 days , but with the help of her friend suma , she can do it in 3 days . in what time suma can do it alone ? | "renu Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 6 suma Γ’ β¬ β’ s one day Γ’ β¬ β’ s work = 1 / 3 - 1 / 6 = 1 / 6 suma can do it alone in 6 days . answer : a" | a ) 6 , b ) 12 , c ) 14 , d ) 15 , e ) 17 | a | inverse(subtract(divide(const_1, 3), divide(const_1, 6))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)| | physics |
an engineer undertakes a project to build a road 15 km long in 300 days and employs 50 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "50 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proport... | a ) a . 43 , b ) b . 45 , c ) c . 75 , d ) d . 68 , e ) e . 60 | c | subtract(divide(multiply(multiply(50, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 50) | subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)| | physics |
a train running at the speed of 60 km / hr crosses a pole in 12 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 12 ) m = 200 m . answer : d" | a ) 187 m , b ) 278 m , c ) 876 m , d ) 200 m , e ) 267 m | d | multiply(divide(multiply(60, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a is twice as fast as b . if b alone can do a piece of work in 24 days , in what time can a and b together complete the work ? | a can do the work in 24 / 2 i . e . , 12 days . a and b ' s one day ' s work = 1 / 12 + 1 / 24 = ( 2 + 1 ) / 24 = 1 / 8 so a and b together can do the work in 8 days . answer : e | a ) 10 , b ) 12 , c ) 6 , d ) 2 , e ) 8 | e | inverse(add(divide(const_1, 24), multiply(divide(const_1, 24), const_2))) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2) | physics |
the owner of a furniture shop charges his customer 32 % more than the cost price . if a customer paid rs . 5400 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 5400 ( 100 / 132 ) = rs . 4090 . answer : a" | a ) rs . 4090 , b ) rs . 4067 , c ) rs . 6290 , d ) rs . 6725 , e ) rs . 6708 | a | divide(5400, add(const_1, divide(32, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
in the coordinate plane , a circle centered on point ( - 3 , - 4 ) passes through point ( - 1 , 1 ) . what is the area of the circle ? | r ^ 2 = ( - 3 + 1 ) ^ 2 + ( - 4 - 1 ) ^ 2 = 4 + 25 = 29 area of circle = Ο r ^ 2 = 29 Ο answer : c | ['a ) 9 Ο', 'b ) 18 Ο', 'c ) 29 Ο', 'd ) 37 Ο', 'e ) 41 Ο'] | c | circle_area(sqrt(add(power(subtract(3, 1), const_2), power(add(1, 4), const_2)))) | add(n1,n2)|subtract(n0,n2)|power(#1,const_2)|power(#0,const_2)|add(#2,#3)|sqrt(#4)|circle_area(#5) | geometry |
city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 5 : 00 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? | "train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 5 : 00 pm + 2 hrs = 7 : 00 pm answer : e . 7 : 00" | a ) 5 : 00 , b ) 5 : 30 , c ) 6 : 00 , d ) 6 : 30 , e ) 7 : 00 | e | divide(add(5, const_2), 00) | add(n4,const_2)|divide(#0,n5)| | physics |
a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price b per t - shirt that the vendor paid for these purchases ? | correct answer : a explanation : the relevant formula for this problem is average b = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ... | a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00 | a | add(divide(add(10, 14), const_2), add(const_0_25, const_0_25)) | add(const_0_25,const_0_25)|add(n0,n1)|divide(#1,const_2)|add(#0,#2) | general |
a cylinder is 6 cms in diameter and 6 cms in height . if spheres of the same size are made from the material obtained , what is the diameter of each sphere ? | since the sphere is made out of the cylinder material their volume will be same volume of cylinder = pi * 3 ^ 2 * 6 volume of one sphere = 4 / 3 * pi * r ^ 3 ( where r is radius of sphere ) so if there are n such sphere ' s equating the volumes , pi * r ^ 2 * h = n * ( 4 / 3 ) * pi * r ^ 3 n is the number of spheres , ... | ['a ) 5 cms', 'b ) 2 cms', 'c ) 3 cms', 'd ) 4 cms', 'e ) 6 cms'] | c | floor(divide(multiply(multiply(const_3, const_3), 6), multiply(divide(const_4, const_3), multiply(6, const_2)))) | divide(const_4,const_3)|multiply(const_3,const_3)|multiply(n0,const_2)|multiply(n0,#1)|multiply(#0,#2)|divide(#3,#4)|floor(#5) | physics |
a number x is 8 times another number y . the percentage that y is less than x is | "say y = 1 and x = 8 . then y = 1 is less than x = 8 by ( 8 - 1 ) / 8 * 100 = 7 / 8 * 100 = 87.5 % . answer : b ." | a ) 12.5 % , b ) 87.5 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(8, const_1), 8), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general |
if 20 typists can type 44 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 44 letters , so 30 typists can type = 44 * 30 / 20 44 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 44 * 30 * 60 / 20 * 20 = 198 c is the answer" | a ) 63 , b ) 72 , c ) 198 , d ) 216 , e ) 400 | c | multiply(divide(multiply(44, const_3), 20), 30) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)| | physics |
if ( 1 / 5 ) ^ m * ( 1 / 2 ) ^ 5 = 1 / ( ( 10 ) ^ 5 ) , then m = | "( 1 / 5 ) ^ m ( 1 / 2 ) ^ 5 = 1 / { ( 10 ) ^ 5 } ( 10 ) ^ 5 / 2 ^ 5 = 5 ^ m [ ( 2 ^ 5 ) ( 5 ^ 5 ) ] / [ 2 ^ 5 ] = 5 ^ m 5 ^ 5 = 5 ^ m m = 5 e" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(add(multiply(divide(1, 2), 10), multiply(divide(1, 5), 10)), divide(5, 10)) | divide(n0,n3)|divide(n0,n1)|divide(n1,n6)|multiply(n6,#0)|multiply(n6,#1)|add(#3,#4)|subtract(#5,#2)| | general |
amithab ' s average expenditure for the january to june is rs . 4200 and he spends rs . 1200 in january and rs . 1500 in july . the average expenditure for the months of febraury to july is : | "explanation : amithab ' s total expenditure for jan - june = 4200 x 6 = 25200 expenditure for feb - june = 25200 - 1200 = 24000 expenditure for the months of feb - july = 24000 + 1500 = 25500 the average expenditure = { 25500 } { 6 } = 4250 answer : a" | a ) rs . 4250 , b ) rs . 4288 , c ) rs . 4227 , d ) rs . 4218 , e ) rs . 4219 | a | add(4200, divide(subtract(1500, 1200), multiply(const_2, const_3))) | multiply(const_2,const_3)|subtract(n2,n1)|divide(#1,#0)|add(n0,#2)| | general |
a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 105 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 105 ( multiplying factor is 7 : ( 105 / 15 ) ) then typical dosage would be 2 * 7 = 14 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 14 ) * 100 = 14.28 % e is the answer ." | a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 14.28 % | e | multiply(divide(subtract(multiply(divide(2, 15), 105), 18), multiply(divide(2, 15), 105)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100)| | gain |
a discount electronics store normally sells all merchandise at a discount of 5 percent to 45 percent off the suggested retail price . if , during a special sale , an additional 2 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 400 before any discount ? | "original price : 400 $ max first discount = - 45 % thus : 400 Γ’ Λ β ( 1 Γ’ Λ β 45 / 100 ) = 220 second discount on the discounted price = - 2 % thus : 220 Γ’ Λ β ( 1 Γ’ Λ β 2 / 100 ) = 215.60 answer e" | a ) $ 215.00 , b ) $ 200.00 , c ) $ 215.80 , d ) $ 182.00 , e ) $ 215.60 | e | multiply(400, divide(add(45, 2), const_100)) | add(n1,n2)|divide(#0,const_100)|multiply(n3,#1)| | gain |
a man sold 20 articles for $ 40 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 40 * ( 100 % - 20 % ) / 20 = $ 1.6 required production costs for a loss of 20 % : $ 90 * ( 100 % + 20 % ) = $ 108 number of articles to be sold for $ 108 to incur a 20 % loss : $ 108 / $ 1.6 = 67.5 thus , solution e is correct ." | a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 67.5 | e | divide(original_price_before_loss(20, 90), divide(original_price_before_gain(20, 40), 20)) | original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2)| | gain |
two trains are moving in the same direction at 162 kmph and 18 kmph . the faster train crosses a man in the slower train in 33 seconds . find the length of the faster train ? | "relative speed = ( 162 - 18 ) * 5 / 18 = 8 * 5 = 40 mps . distance covered in 33 sec = 33 * 40 = 1320 m . the length of the faster train = 1320 m . answer : c" | a ) 1300 , b ) 1310 , c ) 1320 , d ) 1330 , e ) 1340 | c | multiply(divide(subtract(162, 18), const_3_6), 33) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served only for 9 months and receiveduniform and rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a ) rs . 90 , b ) rs . 100 , c ) rs . 130 , d ) rs . 170 , e ) rs . 190 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general |
500 x 223 + 500 x 77 = ? | "= 8796 x ( 223 + 77 ) ( by distributive law ) = 500 x 300 = 150000 answer is c" | a ) 2736900 , b ) 2738800 , c ) 150000 , d ) 2716740 , e ) none of them | c | multiply(500, power(223, 500)) | power(n1,n2)|multiply(n0,#0)| | general |
how many factors of 220 are odd numbers greater than 1 ? | when factorized , 220 has 4 prime factors . of these prime factors 2 are odd and 2 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 1 . the total number of odd factors greater than 1 are 3 . ( option a ) | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(const_60.0, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(const_60.0,#0)| | other |
in xyz business school , 88 % are taking their specialization as finance at the same time 76 % taking their specialization as marketing . if 90 % are taking their specialization as finance then , what is % of marketing specialization students ? | percentage of finance specialization = 88 ; percentage of marketing specialization = 75 % ; total = 90 * 100 ; percentage of finance specialization = 90 - 88 = 2 ; percentage of marketing specialization = 100 - 75 = 25 ; then percentage of marketing specialization = 0.75 * 24 = 18 . then total percentage of marketing s... | a ) 55 , b ) 75 , c ) 43 , d ) 52 , e ) 45 | c | subtract(subtract(add(divide(subtract(88, subtract(90, 76)), const_3), divide(subtract(88, subtract(90, 76)), const_3)), const_3), subtract(add(divide(subtract(88, subtract(90, 76)), const_3), divide(subtract(88, subtract(90, 76)), const_3)), subtract(add(divide(subtract(88, subtract(90, 76)), const_3), divide(subtract... | subtract(n2,n1)|subtract(n0,#0)|divide(#1,const_3)|add(#2,#2)|subtract(#3,const_3)|subtract(#3,#4)|subtract(#4,#5) | gain |
the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least Β½ foot when they are inside the traffic lane , what should be the limit y on th... | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of... | a ) 5 Β½ ft , b ) 7 Β½ ft , c ) 8 Β½ ft , d ) 9 Β½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | divide(n0,const_2)|divide(const_1,const_2)|subtract(#0,#1)| | physics |
in the rectangular coordinate system , points ( 4 , 0 ) and ( β 4 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c | "it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 4 b" | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above | b | sqrt(power(4, const_2)) | power(n0,const_2)|sqrt(#0)| | geometry |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.