Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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what is the smallest positive integer nn such that √ 6,480 ∗ n is a perfect cube ? | "sol : let ' s factorize 6480 and we get 6480 = 3 ^ 4 * 2 ^ 4 * 5 now we need to see for what minimum value of n √ n * 6480 = a ^ 3 where a is an integer so from 6480 we already have 2 ^ 4 * 3 ^ 4 * 5 * n √ n = ( 2 ^ 2 ) ^ 3 * ( 3 ^ 2 ) ^ 3 * ( 5 ) ^ 3 why cause a is an integer which will need to be have the same factors which are in lhs solving for n √ n = ( 2 ^ 6 * 3 ^ 6 * 5 ^ 3 ) / 2 ^ 4 * 3 ^ 4 * 5 and we get n √ n = 2 ^ 2 * 3 ^ 2 * 5 ^ 2 or n = 2 ^ 4 * 3 ^ 4 * 5 ^ 4 or 30 ^ 4 answer is e" | a ) 5 , b ) 5 ^ 2 , c ) 30 , d ) 30 ^ 2 , e ) 30 ^ 4 | e | add(const_3, const_4) | add(const_3,const_4)| | geometry | E |
the ratio of buses to cars on river road is 1 to 17 . if there are 80 fewer buses than cars on river road , how many cars are on river road ? | b / c = 1 / 17 c - b = 80 . . . . . . . . . > b = c - 80 ( c - 80 ) / c = 1 / 17 testing answers . clearly eliminate acde put c = 85 . . . . . . . . . > ( 85 - 80 ) / 85 = 5 / 85 = 1 / 17 answer : b | a ) 40 , b ) 85 , c ) 60 , d ) 30 , e ) 20 | b | multiply(divide(80, subtract(17, 1)), 17) | subtract(n1,n0)|divide(n2,#0)|multiply(n1,#1) | other | B |
caleb and kyle built completed the construction of a shed in 10 and half days . if they were to work separately , how long will it take each for each of them to build the shed , if it will take caleb 2 day earlier than kyle ? | work = ( a ) ( b ) / ( a + b ) where a and b are the individual times of each entity . here , we ' re told that ( working together ) the two workers would complete a job in 12 days . this means that ( individually ) each of them would take more than 10 days to do the job . answers e , a and c are illogical , since the individual times must both be greater than 10 days . so we can test the values for answers b and d . using the values for answers b and d . . . answer b : ( 20 ) ( 22 ) / ( 20 + 22 ) = 440 / 42 = 10.5 this is a match final answer : c | a ) 10 and 12 , b ) 9 and 11 , c ) 20 and 22 , d ) 8 and 10 , e ) 19 and 21 | c | add(add(multiply(10, 2), 2), multiply(multiply(10, 10), multiply(10, 2))) | multiply(n0,n1)|multiply(n0,n0)|add(n1,#0)|multiply(#1,#0)|add(#2,#3) | physics | C |
if jack walked 4 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 = 4 miles speed per hour = distance / time = 4 / ( 5 / 4 ) = 3.2 miles per hour answer a | a ) 3.2 , b ) 4.5 , c ) 6 , d ) 6.25 , e ) 15 | a | divide(multiply(4, const_60), add(15, const_60)) | add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0) | physics | A |
there are 750 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 750 - x x / 2 + 750 - x / 4 = 1 / 3 * ( 750 ) = 250 x = 250 x / 2 = 125 is supposed to be the answer answer : c" | a ) 75 , b ) 100 , c ) 125 , d ) 175 , e ) 225 | c | divide(subtract(multiply(divide(750, const_3), const_4), 750), const_2) | divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)| | general | C |
if x , y , and z are positive integers , and 2 x = 5 y = 6 z , then the least possible value of x + y + z is | "take lcm of 2,5 and 6 = 30 now 2 x = 30 = > x = 15 5 y = 30 = > y = 6 6 z = 30 = > z = 5 15 + 6 + 5 = 26 . option b ." | a ) 15 , b ) 26 , c ) 37 , d ) 42 , e ) 60 | b | add(add(divide(divide(multiply(multiply(2, 5), 6), const_2), 2), divide(divide(multiply(multiply(2, 5), 6), const_2), 5)), divide(divide(multiply(multiply(2, 5), 6), const_2), 6)) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|divide(#2,n0)|divide(#2,n1)|divide(#2,n2)|add(#3,#4)|add(#6,#5)| | general | B |
if the numerator of a fraction be increased by 20 % and its denominator be diminished by 10 % , the value of the fraction is 20 / 21 . find the original fraction ? | "let the original fraction be x / y then , 120 % of x / 90 % of y = 20 / 21 120 x / 90 y = 20 / 21 x / y = 5 / 7 answer is d" | a ) 1 / 2 , b ) 3 / 5 , c ) 4 / 7 , d ) 5 / 7 , e ) 7 / 9 | d | divide(multiply(subtract(const_100, 10), 20), multiply(add(const_100, 20), 21)) | add(n0,const_100)|subtract(const_100,n1)|multiply(n2,#1)|multiply(n3,#0)|divide(#2,#3)| | general | D |
what is 10 - 8 + 6 - 4 + . . . + ( - 14 ) ? | the expression considers all even numbers between 10 and - 14 with alternate addition and subtraction of the numbers . the numbers to be used are : 10 , 8 , 6 , 4 , 2 , 0 , - 2 , - 4 , - 6 , - 8 , - 10 , - 12 , and - 14 now , the first term is positive and the next term is subtracted . so , the required expression becomes , 10 - 8 + 6 - 4 + 2 - 0 + ( - 2 ) - ( - 4 ) + ( - 6 ) - ( - 8 ) + ( - 10 ) - ( - 12 ) + ( - 14 ) = 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 - 14 = 42 - 44 = - 2 hence the correct answer choice is e . | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) - 2 | e | subtract(6, 8) | subtract(n2,n1) | general | E |
a rock is dropped into a well and the distance traveled is 16 t 2 feet , where t is the time . if the water splash is heard 3 seconds after the rock was dropped , and that the speed of sound is 1100 ft / sec , approximate the height of the well | let t 1 be the time it takes the rock to reach the bottom of the well . if h is the height of the well , we can write h = 16 t 1 2 let t 2 be the time it takes sound wave to reach the top of the well . we can write h = 1100 t 2 the relationship between t 1 and t 2 is t 1 + t 2 = 3 eliminate h and combine the equations h = 16 t 1 2 and h = 1100 t 2 to obtain 16 t 1 2 = 1100 t 2 we now substitute t 2 by 3 - t 1 in the above equation 16 t 1 2 = 1100 ( 3 - t 1 ) the above is a quadratic equation that may be written as 16 t 1 2 + 1100 t 1 - 3300 = 0 the above equations has two solutions and only one of them is positive and is given by t 1 = 2.88 seconds ( 2 decimal places ) we now calculate the height h of the well h = 16 t 1 2 = 16 ( 2.88 ) 2 = 132.7 feet answer c | a ) 134.8 feet , b ) 136.2 feet , c ) 132.7 feet , d ) 132.2 feet , e ) 115.3 feet | c | multiply(16, multiply(divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)), divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)))) | multiply(n3,n3)|multiply(n2,n3)|multiply(n0,const_4)|multiply(n0,const_2)|multiply(#1,#2)|add(#0,#4)|sqrt(#5)|subtract(#6,n3)|divide(#7,#3)|multiply(#8,#8)|multiply(n0,#9) | physics | C |
600 men have provisions for 20 days . if 200 more men join them , for how many days will the provisions last now ? | "600 * 20 = 800 * x x = 15 . answer : a" | a ) 15 , b ) 11 , c ) 10 , d ) 8 , e ) 7 | a | divide(multiply(20, 600), add(600, 200)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics | A |
line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 8 ) and ( 6 , - 4 ) . what is the slope of line m ? | "the midpoint of ( 2,8 ) and ( 6 , - 4 ) is ( 4,2 ) . the slope of a line through ( 0 , - 2 ) and ( 4,2 ) is ( 2 - ( - 2 ) ) / ( 4 - 0 ) = 4 / 4 = 1 the answer is e ." | a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) 1 | e | divide(add(divide(subtract(8, 4), 2), 2), divide(add(2, 6), 2)) | add(n1,n3)|subtract(n2,n4)|divide(#1,n1)|divide(#0,n1)|add(#2,n0)|divide(#4,#3)| | general | E |
if the volume of two cubes are in the ratio 64 : 8 , the ratio of their edges is : | "explanation : let the edges be a and b of two cubes , then a 3 / b 3 = 64 / 8 = > ( a / b ) 3 = ( 4 / 2 ) 3 a / b = 2 / 1 = > a : b = 2 : 1 option d" | a ) 3 : 1 , b ) 3 : 2 , c ) 3 : 5 , d ) 2 : 1 , e ) none of these | d | cube_edge_by_volume(64) | cube_edge_by_volume(n0)| | geometry | D |
if a truck is traveling at a constant rate of 72 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 72 km / hr = > 72,000 m / hr in one minute = > 72000 / 60 = 1200 meters in one sec = > 1200 / 60 = 20 meters time = total distance need to be covered / avg . speed = > 600 / 20 = 30 and hence the answer : c" | a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 48 | c | multiply(divide(divide(600, 1000), 72), const_3600) | divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)| | physics | C |
the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 and that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidates who passed the examination is ? | let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 ⇒ 24 y = 4200 - 1800 ∴ y = 2400 / 24 = 100 c | a ) 80 , b ) 90 , c ) 100 , d ) 120 , e ) 130 | 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 | C |
p is 40 % more efficient than q . p can complete a work in 24 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 24 let work done by q in 1 day = q q × ( 140 / 100 ) = 1 / 24 q = 100 / ( 24 × 140 ) = 10 / ( 24 × 14 ) the work done by p and q in 1 day = 1 / 24 + 10 / ( 24 × 14 ) = 24 / ( 24 × 14 ) = 1 / 14 p and q together can do the work in 14 days . the answer is d ." | a ) 8 , b ) 9 , c ) 11 , d ) 14 , e ) 15 | d | divide(multiply(add(divide(40, const_100), const_1), 24), add(add(divide(40, const_100), const_1), const_1)) | divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n1,#1)|divide(#3,#2)| | physics | D |
in an election , candidate a got 70 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 70 % therefore , the number of valid votes polled in favour of candidate a = 70 % of 476000 = 70 / 100 × 476000 = 33320000 / 100 = 333200 d )" | a ) 330000 , b ) 340000 , c ) 347000 , d ) 333200 , e ) 357000 | d | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(70, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain | D |
a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many j cubes are painted on only 2 faces ? | "1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all on the edge of the cube , in themiddleof the edge . there are 4 in front , 4 in back and 4 more on thestripthat runs around the left / top / right / bottom of the cube . j = 4 + 4 + 4 = 12 . answer a" | a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16 | a | multiply(const_4, power(27, divide(const_1, const_3))) | divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)| | geometry | A |
how many positive integers d between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e . | a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(const_100, const_1), subtract(add(add(add(add(divide(const_100, const_2), const_1), add(divide(subtract(const_100, const_1), const_3), const_1)), add(divide(const_100, 5), const_1)), add(divide(subtract(const_100, const_10), multiply(multiply(3, 5), 2)), const_1)), add(add(add(divide(subtract(const_100, multiply(const_3, const_2)), multiply(const_3, const_2)), const_1), add(divide(subtract(const_100, const_10), multiply(3, 5)), const_1)), add(divide(const_100, const_10), const_1)))) | add(const_1,const_100)|divide(const_100,const_2)|divide(const_100,n4)|divide(const_100,const_10)|multiply(n3,n4)|multiply(const_2,const_3)|subtract(const_100,const_1)|subtract(const_100,const_10)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|divide(#6,const_3)|divide(#7,#4)|multiply(n2,#4)|subtract(const_100,#5)|add(#11,const_1)|add(#12,const_1)|divide(#7,#13)|divide(#14,#5)|add(#8,#15)|add(#17,const_1)|add(#18,const_1)|add(#19,#9)|add(#21,#16)|add(#22,#20)|add(#23,#10)|subtract(#24,#25)|subtract(#0,#26) | other | E |
each day a man meets his wife at the train station after work , and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived 16 minutes earlier than usual . how much time did the man spend walking ? | "as they arrived 16 minutes earlier than usual , they saved 16 minutes on round trip from home to station ( home - station - home ) - - > 8 minutes in each direction ( home - station ) - - > wife meets husband 8 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time , so he must have spent waking the rest of the time before their meeting , which is hour - 8 minutes = 52 minutes . answer : e" | a ) 45 minutes , b ) 50 minutes , c ) 40 minutes , d ) 55 minutes , e ) 52 minutes | e | subtract(const_60, divide(16, const_2)) | divide(n0,const_2)|subtract(const_60,#0)| | physics | E |
a can run 2 km distance in 2 min 20 seconds , while b can run this distance in 3 min 5 sec . . by how much distance can a beat b ? | a takes time 2.20 minutes = 140 sec b takes time 3 minutes = 185 sec difference = 185 - 140 = 45 sec now we are to find distance covered in 40 sec by b 180 sec = 2000 m 1 sec = 100 / 9 m 45 sec = 45 x 100 / 9 = 500 m answer : b | a ) 900 m , b ) 500 m , c ) 120 m , d ) 180 m , e ) 190 m | b | subtract(multiply(2, const_1000), multiply(divide(multiply(2, const_1000), add(multiply(3, const_60), 5)), add(multiply(2, const_60), 20))) | multiply(n0,const_1000)|multiply(n0,const_60)|multiply(n3,const_60)|add(n2,#1)|add(n4,#2)|divide(#0,#4)|multiply(#3,#5)|subtract(#0,#6) | physics | B |
the ratio of the adjacent angles of a parallelogram is 5 : 10 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 5 x + 10 x = 180 ° or , 15 x = 180 ° or , x = 12 ° hence the angles of the parallelogram are 60 ° and 120 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y + 6 y + 7 y + 12 y = 360 ° or , 30 y = 360 ° or , y = 12 ° hence the angles of the quadrilateral are 60 ° , 72 , 84 ° and 144 ° will be 60 ° + 84 ° = 144 ° answer : c" | a ) 168 ° , b ) 228 ° , c ) 144 ° , d ) 224 ° , e ) none of these | c | multiply(divide(multiply(add(multiply(multiply(5, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 10) | add(n2,n3)|multiply(const_2,n0)|add(n4,#0)|multiply(#1,const_10)|add(#3,const_100)|add(n5,#2)|multiply(#4,const_2)|divide(#6,#5)|multiply(n1,#7)| | other | C |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 14 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 14 ) + ( 0.65 * 14 * 55 ) = 24.5 + 500.5 = > 525 hence answer will be ( c ) 525" | a ) 320 $ , b ) 380 $ , c ) 525 $ , d ) 456 $ , e ) 480 $ | c | multiply(multiply(0.65, 55), 14) | multiply(n1,n3)|multiply(n2,#0)| | general | C |
what is the least number . which should be added to 0.0355 to make it a perfect square ? | "0.0355 + 0.0006 = 0.0361 ( 0.19 ) ^ 2 answer : e" | a ) 0.0005 , b ) 0.0016 , c ) 0.0056 , d ) 0.0066 , e ) 0.0006 | e | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0355) | add(const_12,const_4)|add(#0,const_2)|divide(#1,const_100)|multiply(#2,#2)|subtract(#3,n0)| | general | E |
a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 5 volts , what was the smallest possible recording in volts ? | "if average of 3 is 5 so sum of 3 should be 15 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 6 and 6 so a has to be 3 b" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | subtract(multiply(3, 5), 10) | multiply(n2,n3)|subtract(#0,n1)| | general | B |
a textile manufacturing firm employees 50 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 50 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | "explanation : profit = 5 , 00,000 − ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 × ( 49 / 50 ) − 150000 × ( 49 / 50 ) − 75000 . = > rs 2 , 68,000 . decrease in profit = > 2 , 75,000 − 2 , 68,000 = > rs . 7,000 . answer : b" | a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | b | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(50, 1), 50), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(50, 1), 50))), 75000)) | multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10)| | general | B |
3 candidates in an election and received 5136 , 7636 and 11628 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 5136 + 7636 + 11628 ) = 24400 so , required percentage = 11628 / 24400 * 100 = 48 % b" | a ) 45 % , b ) 48 % , c ) 57 % , d ) 59 % , e ) 61 % | b | multiply(divide(11628, add(add(5136, 7636), 11628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)| | gain | B |
when an amount was distributed among 14 boys , each of them got $ 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys . what was the amount ? | "let the total amount be rs . x the , x / 14 - x / 18 = 80 = = > 2 x / 126 = 80 = = > x / 63 = 63 x 80 = 5040 . hence the total amount is 5040 . answer c ." | a ) 5050 , b ) 5020 , c ) 5040 , d ) 5030 , e ) 5075 | c | divide(80, subtract(divide(const_1, 14), divide(const_1, 18))) | divide(const_1,n0)|divide(const_1,n2)|subtract(#0,#1)|divide(n1,#2)| | general | C |
veena ranks 65 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | "total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 65 + 1 = 96 answer : a" | a ) 96 , b ) 108 , c ) 110 , d ) 90 , e ) 93 | a | subtract(subtract(subtract(182, 22), subtract(65, const_1)), const_4) | subtract(n1,n2)|subtract(n0,const_1)|subtract(#0,#1)|subtract(#2,const_4)| | other | A |
a number is doubled and 9 is added . if the resultant is trebled , it becomes 57 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 57 2 x = 10 = > x = 5 answer : a" | a ) 5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | a | divide(subtract(57, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general | A |
if 2 log ( 4 * 5 ^ 2 ) = x , find x | "2 ( log 2 ^ 2 * 5 ^ 2 ) = x 2 log ( 5 * 2 ) ^ 2 = x 2 * 2 log ( 5 * 2 ) = x 4 log 10 = x log 10 base 10 = 1 so 4 * 1 = x x = 4 answer : d" | a ) 5 , b ) 6 , c ) 7 , d ) 4 , e ) 8 | d | power(multiply(4, power(5, 2)), 2) | power(n2,n3)|multiply(n1,#0)|power(#1,n0)| | general | D |
the sum of three consecutive numbers is 87 . the greatest among these three numbers is : | "sol . let the numbers be x , x + 1 and x + 2 . then , x + ( x + 1 ) + ( x + 2 ) = 87 ⇔ 3 x = 84 ⇔ x = 28 . greatest number = ( x + 2 ) = 30 . answer d" | a ) 22 , b ) 24 , c ) 26 , d ) 30 , e ) 44 | d | divide(add(87, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics | D |
if ( n + 2 ) ! / n ! = 182 , n = ? | "( n + 2 ) ! / n ! = 182 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 132 from here , we might just test the answer choices . since ( 14 ) ( 13 ) = 182 , we can see that n = 12 e" | a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1)| | general | E |
the difference between the squares of two consecutive numbers is 35 . the numbers are | "explanation : let the numbers be a and ( a + 1 ) ( a + 1 ) 2 − a 2 = 35 ⇒ a 2 + 2 a + 1 − a 2 = 35 ⇒ 2 a = 34 ⇒ 2 a = 34 or a = 17 the numbers are 17 & 18 . correct option : c" | a ) 14,15 , b ) 15,16 , c ) 17,18 , d ) 18,19 , e ) none | c | divide(add(35, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics | C |
a palindrome is a word or a number that reads the same forward and backward . for example , 2442 and 111 are palindromes . if 5 - digit palindromes are formed using one or more of the digits 1 , 2 , 3 , 4 , 5 , 6 , and 7 , how many palindromes are possible ? | "there are 7 choices for each of the first three digits . the number of possible palindromes is 7 ^ 3 = 343 . the answer is e ." | a ) 35 , b ) 49 , c ) 105 , d ) 255 , e ) 343 | e | power(5, 3) | power(n2,n5)| | general | E |
how many of the positive factors of 26 , 16 and how many common factors are there in numbers ? | "factors of 26 - 1 , 2 , 13 and 26 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 2 answer : b" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(16, 26) | divide(n1,n0)| | other | B |
at the faculty of aerospace engineering , 312 students study random - processing methods , 234 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "312 + 234 - 112 ( since 112 is counted twice ) = 434 b is the answer" | a ) 430 , b ) 434 , c ) 438 , d ) 444 , e ) 446 | b | add(subtract(312, divide(112, const_2)), subtract(234, divide(112, const_2))) | divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|add(#1,#2)| | other | B |
if 3 men or 4 women can reap a field in 43 days how long will 7 men and 5 women take to reap it ? | "explanation : 3 men reap 1 / 43 field in 1 day 1 man reap 1 / ( 3 x 43 ) 4 women reap 1 / 43 field in 1 day 1 woman reap 1 / ( 43 x 4 ) 7 men and 5 women reap ( 7 / ( 3 x 43 ) + 5 / ( 4 x 43 ) ) = 1 / 12 in 1 day 7 men and 5 women will reap the field in 12 days answer : option b" | a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | b | divide(43, add(multiply(7, divide(43, 3)), multiply(5, divide(43, 4)))) | divide(n2,n0)|divide(n2,n1)|multiply(n3,#0)|multiply(n4,#1)|add(#2,#3)|divide(n2,#4)| | physics | B |
on the coordinate plane , points p and u are defined by the coordinates ( - 1,0 ) and ( 3,3 ) , respectively , and are connected to form a chord of a circle which also lies on the plane . if the area of the circle is ( 25 / 4 ) π , what are the coordinates of the center of the circle ? | although it took me 3 mins to solve this question using all those equations , later i thought this question can be solved easily using options . one property to keep in mind - a line passing through the centre of the circle bisects the chord ( or passes from the mid point of the chord ) . now mid point of chord here is ( - 1 + 3 ) / 2 , ( 3 + 0 ) / 2 i . e . ( 1 , 1.5 ) now luckily we have this in our ans . choice . so definitely this is the ans . it also indictaes that pu is the diameter of the circle . there can be a case when pu is not a diameter but in that case also the y - coordinate will remain same as it is the midpoint of the chord and we are moving up in the st . line to locate the centre of the circle . if ans choices are all distinct ( y cordinates ) only check for y cordinate and mark the ans = d | ['a ) ( 1.5 , 1 )', 'b ) ( 2 , - 5 )', 'c ) ( 0,0 )', 'd ) ( 1 , 1.5 )', 'e ) ( 2,2 )'] | d | add(multiply(const_3, const_3), sqrt(divide(25, 4))) | divide(n2,n3)|multiply(const_3,const_3)|sqrt(#0)|add(#1,#2) | geometry | D |
if a = 105 and a ^ 3 = 21 * 25 * 35 * b , what is the value of b ? | "first step will be to break down all the numbers into their prime factors . 105 = 3 * 5 * 7 21 = 7 * 3 25 = 5 * 5 35 = 7 * 5 so , ( 105 ) ^ 3 = 3 * 7 * 5 * 5 * 7 * 5 * b therefore ( 3 * 5 * 7 ) ^ 3 = 3 * 5 ^ 3 * 7 ^ 2 * b therefore , b = 3 ^ 3 * 5 ^ 3 * 7 ^ 3 / 3 * 5 ^ 3 * 7 ^ 2 b = 3 ^ 2 * 7 = 9 * 7 = 63 correct answer e ." | a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 63 | e | divide(power(105, 3), multiply(multiply(21, 25), 35)) | multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)| | general | E |
a train 360 m long runs with a speed of 45 km / hr . what time will it take to pass a platform of 180 m long ? | "speed = 45 km / hr = 45 ã — ( 5 / 18 ) m / s = 150 / 12 = 50 / 4 = 25 / 2 m / s total distance = length of the train + length of the platform = 360 + 180 = 540 meter time taken to cross the platform = 540 / ( 25 / 2 ) = 540 ã — 2 / 25 = 43.2 seconds answer : c" | a ) 38 sec , b ) 35 sec , c ) 43.2 sec , d ) 40 sec , e ) none of these | c | multiply(divide(add(divide(180, const_1000), divide(360, const_1000)), 45), const_3600) | divide(n2,const_1000)|divide(n0,const_1000)|add(#0,#1)|divide(#2,n1)|multiply(#3,const_3600)| | physics | C |
10 percent of the women in a college class are science majors , and the non - science majors make up 60 % of the class . what percentage of the men are science majors if 40 % of the class are men ? | science majors make up 0.4 of the class . 60 % of the class are women and 0.1 * 0.6 = 0.06 of the class are female science majors . then 0.34 of the class are male science majors . 0.4 x = 0.34 x = 0.85 = 85 % the answer is e . | a ) 45 % , b ) 55 % , c ) 65 % , d ) 75 % , e ) 85 % | e | multiply(add(divide(subtract(add(60, 40), multiply(10, const_2)), const_100), divide(divide(10, const_2), const_100)), const_100) | add(n1,n2)|divide(n0,const_2)|multiply(n0,const_2)|divide(#1,const_100)|subtract(#0,#2)|divide(#4,const_100)|add(#5,#3)|multiply(#6,const_100) | gain | E |
pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed o ? | let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this o we get x = 8 . b . | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), const_4), power(divide(96, 16), const_2)))), const_2) | divide(n0,n2)|divide(n3,const_100)|add(#1,const_1)|power(#0,const_2)|divide(#0,#2)|multiply(n1,#4)|multiply(#5,const_4)|add(#6,#3)|sqrt(#7)|add(#0,#8)|divide(#9,const_2) | physics | B |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 55 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 55 = rs . 150 cp on 40 metres = 150 x 40 = rs . 6000 profit earned on 40 metres cloth = rs . 8200 – rs . 6000 = rs . 2200 . answer : option d | a ) rs . 950 , b ) rs . 1500 , c ) rs . 1000 , d ) rs . 2200 , e ) none of these | d | multiply(55, 40) | multiply(n0,n2) | gain | D |
the tax on a commodity is diminished by 32 % but its consumption is increased by 12 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 68 * 112 = 7616 10000 - - - - - - - 2384 100 - - - - - - - ? = 24 % answer : e" | a ) 20 % , b ) 18 % , c ) 15 % , d ) 12 % , e ) 24 % | e | subtract(const_100, divide(multiply(add(const_100, 12), subtract(const_100, 32)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | general | E |
one side of a rectangular field is 16 m and one of its diagonal is 17 m . find the area of the field . | "solution other side = √ ( 17 ) 2 - ( 16 ) 2 = √ 289 - 256 = √ 33 = 5.7 m . ∴ area = ( 16 x 5.7 ) m 2 = 91.2 m 2 . answer c" | a ) 100 , b ) 120 , c ) 91.2 , d ) 180 , e ) none | c | rectangle_area(16, sqrt(subtract(power(17, const_2), power(16, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry | C |
the difference between the local value and the face value of 7 in the numeral 32675149 is | "explanation : ( local value of 7 ) - ( face value of 7 ) = ( 70000 - 7 ) = 69993 b )" | a ) 69990 , b ) 69993 , c ) 70000 , d ) 77000 , e ) 78000 | b | subtract(multiply(7, const_1000), 7) | multiply(n0,const_1000)|subtract(#0,n0)| | general | B |
a reduction of 20 % in the price of salt enables a lady to obtain 20 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 20 ? - - - 1 = > rs . 1 100 - - - 80 ? - - - 1 = > rs . 1.25 answer : c" | a ) 12.6 , b ) 1.21 , c ) 1.25 , d ) 12.4 , e ) 12.7 | c | multiply(divide(divide(multiply(divide(20, const_100), 100), 20), multiply(divide(20, const_100), 100)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain | C |
two employees m and n are paid a total of $ 572 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 572 2.2 n = 572 n = 260 the answer is c ." | a ) $ 220 , b ) $ 240 , c ) $ 260 , d ) $ 300 , e ) $ 320 | c | divide(572, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general | C |
a student chose a number , multiplied it by 2 , then subtracted 148 from the result and got 110 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 148 = 110 2 x = 258 x = 129 correct answer b" | a ) 90 , b ) 129 , c ) 120 , d ) 160 , e ) 200 | b | divide(add(110, 148), 2) | add(n1,n2)|divide(#0,n0)| | general | B |
he total marks obtained by a student in physics , chemistry and mathematics is 180 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 180 + p c + m = 180 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 180 / 2 = 90 . answer : e | a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 90 | e | divide(180, const_2) | divide(n0,const_2) | general | E |
a can do a job in 30 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "c 1 / 30 + 1 / 30 = 1 / 15 15 * 2 = 30 days" | a ) 14 days , b ) 26 days , c ) 30 days , d ) 31 days , e ) 39 days | c | add(divide(const_1, 30), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics | C |
how many cubes of 5 cm edge can be cut out of a cube of 15 cm edge | "explanation : number of cubes = ( 15 x 15 x 15 ) / ( 5 x 5 x 5 ) = 27 answer : c" | a ) 36 , b ) 232 , c ) 27 , d ) 48 , e ) none of these | c | divide(volume_cube(15), volume_cube(divide(5, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | probability | C |
the average runs of a cricket player of 15 innings was 25 . how many runs must he make in his next innings so as to increase his average of runs by 6 ? | "explanation : average after 16 innings = 31 required number of runs = ( 31 * 16 ) – ( 25 * 15 ) = 496 – 375 = 121 answer a" | a ) 121 , b ) 421 , c ) 143 , d ) 176 , e ) 84 | a | subtract(multiply(add(15, const_1), add(25, 6)), multiply(25, 15)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general | A |
having scored 95 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ? | "explanation : let the average score of the first 18 innings be n 18 n + 95 = 19 ( n + 4 ) = > n = 19 so , average score after 19 th innings = x + 4 = 23 . answer : c" | a ) 28 , b ) 27 , c ) 23 , d ) 22 , e ) 24 | c | add(subtract(95, multiply(19, 4)), 4) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general | C |
two ships are sailing in the sea on the two sides of a lighthouse . the angle of elevation of the top of the lighthouse is observed from the ships are 30 ° and 45 ° respectively . if the lighthouse is 100 m high , the distance between the two ships is : | let ab be the lighthouse and c and d be the positions of the ships . then , ab = 100 m , acb = 30 ° and adb = 45 ° . ab = tan 30 ° = 1 ac = ab x 3 = 1003 m . ac 3 ab = tan 45 ° = 1 ad = ab = 100 m . ad cd = ( ac + ad ) = ( 1003 + 100 ) m = 100 ( 3 + 1 ) = ( 100 x 2.73 ) m = 273 m . answer = c | a ) 173 m , b ) 200 m , c ) 273 m , d ) 300 m , e ) 373 m | c | multiply(add(sqrt(const_3), const_1), 100) | sqrt(const_3)|add(#0,const_1)|multiply(n2,#1) | physics | C |
the length of a rectangle is reduced by 25 % . by what % would the width have to be increased to maintainthe original area ? | "sol . required change = ( 25 * 100 ) / ( 100 - 25 ) = 25 % d" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 33 % , e ) 40 % | d | multiply(divide(subtract(const_1, divide(subtract(const_100, 25), const_100)), divide(subtract(const_100, 25), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry | D |
a person walks from one end to the other of a 100 - meter long moving walkway at a constant rate in 50 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 150 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | let v be the speed of the person and let x be the speed of the walkway . 50 ( v + x ) = 100 then 150 ( v + x ) = 300 150 ( v - x ) = 100 when we add the two equations : 300 v = 400 v = 4 / 3 time = 100 / ( 4 / 3 ) = 75 seconds the answer is e . | a ) 63 , b ) 66 , c ) 69 , d ) 72 , e ) 75 | e | divide(100, divide(add(divide(100, 50), divide(100, 150)), const_2)) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|divide(n0,#3) | physics | E |
2 friends a and b running up hill and then to get down length of road - 440 yads a on his return journey met b going up at 20 yards from top if a has finished race 5 minute earlier than b then how much time a had taken to complete the race ? | total journey = 440 * 2 = 880 a meet b 20 yards from top in getting down it means he has covered 440 + 20 = 460 yards while b is 420 yards . so he is 40 yards ahead of b which is equals to 5 minute . so 40 yards in 5 min 880 yards will be in 5 * 880 / ( 40 ) = 110 minute answer : b | a ) 100 minute , b ) 110 minute , c ) 120 minute , d ) 130 minute , e ) 140 minute | b | multiply(divide(5, subtract(divide(440, const_10), const_4)), multiply(440, 2)) | divide(n1,const_10)|multiply(n0,n1)|subtract(#0,const_4)|divide(n3,#2)|multiply(#3,#1) | general | B |
find the perimeter of a triangle with sides measuring 5 centimeters , 20 centimeters and 30 centimeters . | "perimeter of triangle = a + b + c perimeter = 5 cm + 20 cm + 30 cm = 55 cm answer : d ." | a ) 30 cm , b ) 40 cm , c ) 50 cm , d ) 55 cm , e ) 60 cm | d | triangle_perimeter(5, 20, 30) | triangle_perimeter(n0,n1,n2)| | geometry | D |
the inner circumference of a circular race track , 14 m wide is 440 m . find the radius of the outer circle . | "explanation : let inner radius be r meters . { \ color { black } then , 2 \ pi r = 440 \ rightarrow r = 440 \ times \ frac { 7 } { 22 } \ times \ frac { 1 } { 2 } = 70 m } radius of outer circle = 70 + 14 = 84 m answer : a ) 84 m" | a ) 84 , b ) 12 , c ) 67 , d ) 28 , e ) 21 | a | add(14, divide(divide(440, const_pi), const_2)) | divide(n1,const_pi)|divide(#0,const_2)|add(n0,#1)| | physics | A |
if log 2 = 0.3010 and log 3 = 0.4771 , the values of log 5 512 is | "ans : log 5512 = { \ color { black } \ frac { \ log 512 } { \ log 5 } } = { \ color { black } \ frac { \ log 2 ^ { 9 } } { \ log ( \ frac { 10 } { 2 } ) } } = { \ color { black } \ frac { 9 \ log 2 } { \ log 10 - \ log 2 } } = { \ color { black } \ frac { ( 9 \ times 0.3010 ) } { 1 - 0.3010 } } = { \ color { black } \ frac { 2.709 } { 0.699 } } = { \ color { black } \ frac { 2709 } { 699 } } = 3.876 answer : b ) 3.875" | a ) 3.879 , b ) 3.875 , c ) 3.872 , d ) 3.828 , e ) 3.192 | b | multiply(0.3010, divide(0.3010, 0.4771)) | divide(n1,n3)|multiply(n1,#0)| | other | B |
when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum e of all possible 2 - digit remainders for x / y ? | "ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . e = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616 . b" | a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784 | b | divide(59.32, subtract(2, floor(2))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general | B |
the volume v of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate , in cubic meters per minutes , at which the water is flowing into the swimming pool ? | the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e | ['a ) 0.125', 'b ) 0.25', 'c ) 0.5', 'd ) 0.75', 'e ) not enough information to determine the rate'] | e | multiply(square_area(cube_edge_by_volume(840)), 0.5) | cube_edge_by_volume(n0)|square_area(#0)|multiply(n1,#1) | geometry | E |
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 60 / 5 = 12 seconds . answer : d" | a ) 22 seconds , b ) 65 seconds , c ) 78 seconds , d ) 12 seconds , e ) 21 seconds | d | divide(60, multiply(add(15, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | D |
a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 60 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 60 hours = 3,600 / 60 seconds = 60 seconds ; time to cover 1 kilometer at regular speed is 60 + 5 = 65 seconds = 65 / 3,600 hours = 1 / 55 hours ; so , we get that to cover 1 kilometer 1 / 55 hours is needed - - > regular speed 55 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : e" | a ) 70 , b ) 72 , c ) 74 , d ) 75 , e ) 55 | e | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 5), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics | E |
vijay lent out an amount rs . 10000 into two parts , one at 8 % p . a . and the remaining at 10 % p . a . both on simple interest . at the end of the year he received rs . 850 as total interest . what was the amount he lent out at 8 % pa . a ? | let the amount lent out at 8 % p . a . be rs . a = > ( a * 8 ) / 100 + [ ( 10000 - a ) * 10 ] / 100 = 850 = > a = rs . 15000 . answer : a | a ) rs . 15000 , b ) rs . 6000 , c ) rs . 25000 , d ) rs . 10000 , e ) rs . 18000 | a | multiply(subtract(add(multiply(divide(10, const_100), 10000), divide(8, const_100)), 850), const_100) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#1)|add(#0,#2)|subtract(#3,n3)|multiply(#4,const_100) | gain | A |
for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 1000 , what is the maximum possible sum of the length of x and the length of y ? | basically the length of an integer is the sum of the powers of its prime factors . for example the length of 24 is 4 because 24 = 2 ^ 3 * 3 ^ 1 - - > 3 + 1 = 4 . given : x + 3 y < 1,000 . now , to maximize the length of x or y ( to maximize the sum of the powers of their primes ) we should minimize their prime bases . minimum prime base is 2 : so if x = 2 ^ 9 = 512 then its length is 9 - - > 512 + 3 y < 1,000 - - > y < 162.7 - - > maximum length of y can be 7 as 2 ^ 7 = 128 - - > 9 + 7 = 16 . answer : d . | a ) 5 , b ) 6 , c ) 15 , d ) 16 , e ) 18 | d | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2) | general | D |
dean winchester has got a long wooden stock of size 60 feet , he need to cut small stock of size 1 feet long using his axe . he takes 5 minutes to cut one small stock ( 1 feet ) , how long will he take to make 60 such small stocks ? | solution : 295 minutes . when he cut 59 stocks , the 60 th stock will remain . 59 * 5 = 295 minutes . answer c | a ) 296 minutes . , b ) 297 minutes . , c ) 295 minutes . , d ) 294 minutes . , e ) none | c | multiply(60, 5) | multiply(n0,n2) | physics | C |
if 15 % of a is the same as 20 % of b , then a : b is : | "expl : 15 % of a i = 20 % of b = 15 a / 100 = 20 b / 100 = 4 / 3 = 4 : 3 answer : b" | a ) 3 : 4 , b ) 4 : 3 , c ) 17 : 16 , d ) 16 : 17 , e ) none of these | b | divide(divide(20, const_100), divide(15, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)| | gain | B |
find the value for m ? 19 ( m + n ) + 17 = 19 ( - m + n ) - 135 | 19 m + 19 n + 17 = - 19 m + 19 n - 135 38 m = - 152 = > m = - 4 e | a ) 0 , b ) - 1 , c ) 1 , d ) 2 , e ) - 4 | e | add(135, 17) | add(n1,n3) | general | E |
find the largest 2 digit number which is exactly divisible by 88 ? | "largest 2 digit number is 99 after doing 99 ÷ 88 we get remainder 11 hence largest 2 digit number exactly divisible by 88 = 99 - 11 = 88 a" | a ) 88 , b ) 89 , c ) 99 , d ) 54 , e ) 90 | a | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(2, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7)| | general | A |
how many 0 ' s are there in the binary form of 6 * 1024 + 4 * 64 + 2 | 6 * 1024 + 4 * 64 + 2 = 6144 + 256 + 2 = 6402 in binary code 6402 base 10 = 11001 000000 10 in base 2 . so 9 zeros are there . answer : e | a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 9 | e | subtract(add(6, 4), const_1) | add(n1,n3)|subtract(#0,const_1) | general | E |
in a school 10 % of the boys are same in number as 1 / 2 th of the girls . what is the ratio of boys to the girls in the school ? | "10 % of b = 1 / 2 g 10 b / 100 = g / 2 b = 5 g / 1 b / g = 5 / 1 b : g = 5 : 1 answer is a" | a ) 5 : 1 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(10, const_100), divide(1, 2)) | divide(n0,const_100)|divide(n1,n2)|divide(#0,#1)| | other | A |
if y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer , then y must be | "3150 * y = a ^ 2 3 * 1050 * y = a ^ 2 3 * 5 * 210 * y = a ^ 2 3 * 5 * 7 * 3 * 2 * 5 * y = a ^ 2 2 * 3 ^ 2 * 5 ^ 2 * 7 * y = a ^ 2 concept : the factors of squared integer should occur in pair . so , the only numbers left are 2 * 7 hence 14 ; answer : e" | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry | E |
two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 4 litres from vessel p is mixed with 8 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 4 = 5 / 2 litres quantity of alcohol in vessel q = 87.5 / 100 * 8 = 7 / 1 litres quantity of alcohol in the mixture formed = 5 / 2 + 7 / 1 = 19 / 2 = 9.50 litres as 12 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 9.50 : 2.50 = 38 : 10 . answer : d" | a ) 19 : 1 , b ) 19 : 4 , c ) 19 : 8 , d ) 38 : 10 , e ) 38 : 2 | d | divide(add(divide(multiply(62.5, 4), const_100), divide(multiply(87.5, 8), const_100)), add(subtract(4, divide(multiply(62.5, 4), const_100)), subtract(8, divide(multiply(87.5, 8), const_100)))) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|subtract(n2,#2)|subtract(n3,#3)|add(#5,#6)|divide(#4,#7)| | other | D |
tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 65 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 × 8 = 560 . cost of 9 kg of mangoes = 65 × 9 = 585 . total cost he has to pay = 560 + 585 = 1145 . b )" | a ) a ) 1040 , b ) b ) 1145 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1075 | b | add(multiply(8, 70), multiply(9, 65)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | B |
what is the greatest 6 - digit number when divided by 8 , 9 , and 10 leaves a remainder of 6 , 7 , and 8 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 8 when we divide by 10 , so b is the only possible answer . | a ) 456780 , b ) 678918 , c ) 997479 , d ) 997916 , e ) 997920 | b | subtract(subtract(multiply(const_1000, const_1000), const_1), multiply(multiply(lcm(lcm(8, 9), lcm(10, 9)), lcm(10, 9)), 10)) | lcm(n1,n2)|lcm(n2,n3)|multiply(const_1000,const_1000)|lcm(#0,#1)|subtract(#2,const_1)|multiply(#3,#1)|multiply(n3,#5)|subtract(#4,#6) | general | B |
the average marks of a class of 22 students is 40 and that of another class of 28 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 22 students = 22 * 40 = 880 sum of the marks for the class of 28 students = 28 * 60 = 1680 sum of the marks for the class of 50 students = 880 + 1680 = 2560 average marks of all the students = 2560 / 50 = 51.2 answer : a" | a ) 51.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 52.1 | a | divide(add(multiply(22, 40), multiply(28, 60)), add(22, 28)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general | A |
a train running at the speed of 50 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 50 * 5 / 18 = 125 / 9 m / sec length of the train = speed * time = 125 / 9 * 9 = 125 m answer : a" | a ) 125 m , b ) 150 m , c ) 187 m , d ) 167 m , e ) 197 m | a | multiply(divide(multiply(50, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | A |
a 300 meter long train crosses a platform in 27 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 27 = 50 / 3 3 ( x + 300 ) = 1350 è x = 150 m . answer : a" | a ) 150 m , b ) 278 m , c ) 350 m , d ) 228 m , e ) 282 m | a | subtract(multiply(divide(300, 18), 27), 300) | divide(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | A |
the regular price per can of a certain brand of soda is $ 0.40 . if the regular price per can is discounted 15 percent when the soda is purchased in 24 - can cases , what is the price of 100 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.85 ) ( $ 0.40 ) , or $ 0.34 . therefore , the price of 72 cans of soda at the discounted price would be ( 100 ) ( $ 0.34 ) = 34 answer : d ." | a ) $ 16.32 , b ) $ 18.00 , c ) $ 21.60 , d ) $ 34 , e ) $ 28.80 | d | multiply(divide(subtract(const_100, 15), const_100), multiply(0.40, 100)) | multiply(n0,n3)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)| | gain | D |
the sub - duplicate ratio of 25 : 16 is | "root ( 25 ) : root ( 16 ) = 5 : 4 answer : d" | a ) 4 : 3 , b ) 1 : 2 , c ) 1 : 3 , d ) 5 : 4 , e ) 2 : 3 | d | divide(sqrt(25), sqrt(16)) | sqrt(n0)|sqrt(n1)|divide(#0,#1)| | other | D |
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 72 . what is the difference between the two digits of that number ? | "sol . let the ten ’ s digit be x and unit ’ s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 72 ⇔ 9 ( x - y ) = 72 ⇔ x - y = 8 answer d" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | d | divide(72, subtract(const_10, const_1)) | subtract(const_10,const_1)|divide(n0,#0)| | general | D |
a man can row downstream at the rate of 26 kmph and upstream at 12 kmph . find the man ’ s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 26 + 12 ) = 19 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 26 - 12 ) = 1 / 2 ( 14 ) = 7 kmph answer is d ." | a ) 8.5 , b ) 8.0 , c ) 9.5 , d ) 7.0 , e ) 8.25 | d | divide(subtract(26, 12), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain | D |
a man ' s speed with the current is 15 km / hr and the speed of the current is 2.8 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.8 km / hr hence , speed of the man = 15 - 2.8 = 12.2 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.2 - 2.8 = 9.4 km / hr answer is a ." | a ) 9.4 , b ) 20 , c ) 50 , d ) 30 , e ) 40 | a | subtract(subtract(15, 2.8), 2.8) | subtract(n0,n1)|subtract(#0,n1)| | gain | A |
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 |
1000 men have provisions for 17 days . if 500 more men join them , for how many days will the provisions last now ? | "1000 * 17 = 1500 * x x = 11.3 answer : c" | a ) 12.9 , b ) 12.5 , c ) 11.3 , d ) 12.2 , e ) 12.1 | c | divide(multiply(17, 1000), add(1000, 500)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics | C |
excluding stoppages , the speed of a bus is 70 km / hr and including stoppages , it is 40 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 30 km less . time taken to cover 30 km = 30 / 70 * 60 = 25 min . answer : e" | a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 25 min | e | multiply(const_60, divide(subtract(70, 40), 70)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics | E |
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.40 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 â € “ 6 = 666 * 1.4 = 932.4 answer : c" | a ) 399 , b ) 272 , c ) 932.4 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.40), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics | C |
carl can wash all the windows of his house in 8 hours . his wife maggie can wash all the windows in 4 hours . how many hours will it take for both of them working together to wash all the windows ? | "work hrs = ab / ( a + b ) = 32 / 12 = 3 2 / 2 answer is c" | a ) 2 , b ) 2 1 / 4 , c ) 3 2 / 2 , d ) 4 1 / 2 , e ) 5 | c | multiply(divide(const_60, add(divide(const_1, 8), divide(const_1, 4))), add(divide(const_1, 8), divide(const_1, 4))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_60,#2)|multiply(#2,#3)| | physics | C |
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 18 - meter platform and caught it after it had traveled 53.4 meters . how many times did the ball bounce ? | "ans : 6 division of total diatance travelled will be 18 + 18 + 9 + 4.5 + 2.25 + 1.125 + 0.5 ans b" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(divide(18, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | general | B |
a cargo ship carrying 4 kinds of items , doohickies , geegaws , widgets , and yamyams , arrives at the port . each item weighs 2 , 11 , 5 , and 7 pounds , respectively , and each item is weighed as it is unloaded . if , in the middle of the unloading process , the product of the individual weights of the unloaded items equals 104 , 350400 pounds , how many widgets have been unloaded ? | we need to know the number of widgets ( which weigh 5 pounds each ) . the number of times that 5 divides the number is related to the number of times that 10 divides the number . when we divide 104 , 350,400 by 100 , we get 104 , 350,400 = 1 , 043,504 * 5 ^ 2 * 2 ^ 2 . 1 , 043,504 is not divisible by 5 , thus there are 2 widgets . the answer is a . | a ) 2 , b ) 3 , c ) 4 , d ) 25 , e ) 20 , 870,080 | a | add(const_1, const_1) | add(const_1,const_1) | general | A |
i sold a book at a profit of 10 % . had i sold it for $ 100 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 100 5 % of cost = $ 100 cost = 100 * 100 / 5 = $ 2000 answer is a" | a ) $ 2000 , b ) $ 2500 , c ) $ 3000 , d ) $ 3120 , e ) $ 1540 | a | divide(multiply(divide(multiply(100, 10), subtract(15, 10)), const_100), 10) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain | A |
the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 6.6 . the true discount is | solution t . d = [ b . g x 100 / r x t ] = rs . ( 6.6 x 100 / 12 x 1 ) = rs . 55 . answer c | a ) rs . 72 , b ) rs . 36 , c ) rs . 55 , d ) rs . 50 , e ) none | c | divide(6.6, divide(12, const_100)) | divide(n1,const_100)|divide(n2,#0) | gain | C |
a person bought 118 glass bowls at a rate of rs . 12 per bowl . he sold 102 of them at rs . 15 and the remaining broke . what is the percentage gain for a ? | cp = 118 * 12 = 1416 and sp = 102 * 15 = 1530 gain % = 100 * ( 1530 - 1416 ) / 1100 = 11400 / 1416 = 475 / 59 answer : d | a ) 40 , b ) 300 / 11 , c ) 243 / 7 , d ) 475 / 59 , e ) 279 / 8 | d | multiply(divide(subtract(multiply(102, 15), multiply(118, 12)), multiply(118, 12)), const_100) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100) | gain | D |
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 | E |
find the compound interest on $ 100000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 100000 * ( 1 + 2 / 100 ) ^ 4 = 100000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 108243.21 c . i . = 108243.21 - 100000 = $ 8243.21 answer is e" | a ) $ 645.56 , b ) $ 824.32 , c ) $ 954.26 , d ) $ 745.69 , e ) $ 8243.21 | e | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 100000), 100000) | divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)| | gain | E |
someone on a skateboard is traveling 5 miles per hour . how many feet does she travel in 45 seconds ? ( 1 mile = 5280 feet ) | "per second = > 5 * 5280 ft / 60 * 60 = 7.33 ft 45 seconds = > 7.33 * 45 = 329.85 ft answer : d" | a ) 380 ft , b ) 400 ft , c ) 410.05 ft , d ) 329.85 ft , e ) 420.85 ft | d | multiply(45, divide(multiply(5, 5280), const_3600)) | multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)| | physics | D |
if w , x , y and z are distinct odd positive integers , then the maximum value of the expression ( w ^ 2 + x ^ 2 ) * ( y ^ 2 + z ^ 2 ) can be divisible by ? | any odd number square will give odd number only . . similarly wen we add 2 odd numbers we will get even number . . wen we multiply 2 even owe will get even nuber . . so the result of the above expression is an even number only . . hence it is divisible by 2 . . . answer : a | a ) 2 , b ) 3 , c ) 5 , d ) 8 , e ) 10 | a | divide(multiply(add(2, 2), 2), multiply(2, 2)) | add(n0,n0)|multiply(n0,n0)|multiply(n0,#0)|divide(#2,#1) | general | A |
a person buys an article at rs . 800 . at what price should he sell the article so as to make a profit of 35 % ? | "cost price = rs . 800 profit = 35 % of 800 = rs . 280 selling price = cost price + profit = 800 + 280 = 1080 answer : e" | a ) 600 , b ) 277 , c ) 269 , d ) 261 , e ) 1080 | e | add(800, multiply(800, divide(35, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | E |
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