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
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if 15 % of 30 % of 50 % of a number is 126 , then what is the number ? | "let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 126 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 126 = > a = 10 * 20 * 10 * 2 = 5600 . answer : b" | a ) 5000 , b ) 5600 , c ) 5400 , d ) 4500 , e ) none of these | b | divide(126, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100))) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,const_100)|multiply(#1,#2)|multiply(#0,#3)|divide(n3,#4)| | gain |
a full stationary oil tank that is a right circular cylinder has a radius of 100 feet and a height of 25 feet . oil is pumped from the stationary tank to an oil truck that has a tank that is a right circular cylinder until the truck ' s tank is completely filled . if the truck ' s tank has a radius of 5 feet and a heig... | "the volume of oil pumped to the tank = the volume of oil taken away from stationary cylinder . pi * 25 * 12 = pi * h * 100 * 100 ( h is distance that the oil level dropped ) h = 300 / 10,000 = 3 / 100 = 0.03 ft the answer is d ." | a ) 2.5 , b ) 0.025 , c ) 0.05 , d ) 0.03 , e ) 3 | d | divide(volume_cylinder(5, 12), circle_area(100)) | circle_area(n0)|volume_cylinder(n2,n3)|divide(#1,#0)| | geometry |
company x provides bottled water to its employees in 5 liter bottles , $ 2 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 100 fixed monthly rate and $ 1 for 10 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 10 l bottles will be x / 2 ( to equate the vol . ) since the total cost will be equal , 2 x = 100 + x / 2 so x = 133.33 or 134 . answer is ( d ) . | a ) 121 , b ) 125 , c ) 132 , d ) 134 , e ) 138 | d | divide(100, subtract(5, divide(const_1, 1))) | divide(const_1,n4)|subtract(n1,#0)|divide(n3,#1)| | general |
when the price of an article was reduced by 35 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 35 % , the new price is 0.65 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.65 p ) ( 1.8 n ) = 1.17 np , which is 1.17 times the old revenue , so is 17 % greater . answer : a" | a ) 17 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 35)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | gain |
two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 10 seconds . the length of the faster train is | "solution relative speed = ( 36 + 45 ) km / hr = ( 81 x 5 / 18 ) m / sec = ( 45 / 2 ) m / sec length of the train = ( 45 / 2 x 10 ) m = 225 m . answer a" | a ) 225 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) none | a | multiply(multiply(add(36, 45), const_0_2778), 10) | add(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)| | physics |
s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits r are in s ( 99 ) ? | "focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square .... | a ) r = 350 , b ) r = 353 , c ) r = 354 , d ) r = 356 , e ) 357 | b | add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2))) | add(n6,const_60)|multiply(n0,n4)|multiply(n2,n4)|add(n6,#0)|multiply(n2,#2)|add(#1,#4)|multiply(n6,#3)|add(#3,#6)|add(#7,#5)| | general |
with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 6 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 6 2 x / 15 - x / 8 = 6 = > x = 720 km . answer : c" | a ) 187 km , b ) 480 km , c ) 720 km , d ) 297 km , e ) 671 km | c | divide(1, 2) | divide(n3,n4)| | physics |
if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 2 , what is the value of 2 ^ a * 2 ^ b ? | "2 ^ ab = 2 ^ 2 therefore ab = 2 either a = 1 or 2 or b = 2 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 3 = 8 a" | a ) 8 , b ) 4 , c ) 16 , d ) 32 , e ) 64 | a | multiply(power(2, 2), 2) | power(n0,n0)|multiply(n0,#0)| | general |
when 242 is divided by a certain divisor the remainder obtained is 4 . when 698 is divided by the same divisor the remainder obtained is 8 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 7 . what is the value of the divisor ? | "let that divisor be x since remainder is 4 or 8 it means divisor is greater than 8 . now 242 - 4 = 238 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 8 = 690 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 238 + 690 ) + 4 + 8 = x ( k + l ) + 12 when we divide this numb... | a ) 5 , b ) 17 , c ) 13 , d ) 23 , e ) none of these | a | subtract(add(4, 8), 7) | add(n1,n3)|subtract(#0,n6)| | general |
in a college , the ratio of the number of boys to girls is 6 : 5 . if there are 200 girls , the total number of students in the college is | "explanation : let the boy are 6 x and girls are 5 x = > 5 x = 200 = > x = 40 total students = 6 x + 5 x = 11 x = 11 ( 40 ) = 440 option a" | a ) 440 , b ) 520 , c ) 620 , d ) 720 , e ) 820 | a | add(200, multiply(divide(200, 5), 6)) | divide(n2,n1)|multiply(n0,#0)|add(n2,#1)| | other |
what is the greater of the two numbers whose product is 1280 , given that the sum of the two numbers exceeds their difference by 32 ? | "let the greater and the smaller number be g and s respectively . g * s = 1288 g + s exceeds g - s by 32 i . e . , g + s - ( g - s ) = 32 i . e . , 2 s = 32 = > s = 16 . g = 1280 / s = 80 . answer : e" | a ) a ) 90 , b ) b ) 100 , c ) c ) 110 , d ) d ) 70 , e ) 80 | e | divide(1280, multiply(power(const_2, const_4), const_2)) | power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)| | general |
p , q and r have rs . 8000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 8000 - r ) = > 3 r = 16000 - 2 r = > 5 r = 16000 = > r = 3200 . answer : a" | a ) rs . 3200 , b ) rs . 3600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | a | multiply(divide(8000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | divide(const_2,const_3)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general |
an error 2 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is | "100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 102 ) 2 - ( 100 ) 2 ] = ( 102 + 100 ) x ( 102 - 100 ) = 404 cm 2 . percentage error = 404 x 100 % = 4.04 % 100 x 100 a" | a ) 4.04 % , b ) 4.00 % , c ) 4.14 % , d ) 5.04 % , e ) none | a | divide(multiply(subtract(square_area(add(const_100, 2)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 1000 m ? | when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 1000 m , the number of meters by which a beats c = ( 1000 * 212.5 ) / 1000 = 21... | a ) 122.9 m , b ) 127.5 m . , c ) 122.2 m , d ) 222.9 m , e ) 212.5 m | e | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | subtract(n0,n1)|subtract(n2,n1)|multiply(#0,#1)|divide(#2,n2)|subtract(n0,#3) | physics |
how many positive integers between 1 and 300 are there such that they are multiples of 9 ? | "multiples of 9 = 9 , 18,27 , - - - - - 297 number of multiples of 15 = > 9 * 33 = 297 answer is b" | a ) 31 , b ) 33 , c ) 35 , d ) 36 , e ) 23 | b | divide(subtract(300, 1), 9) | subtract(n1,n0)|divide(#0,n2)| | general |
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 5 feet . how many feet does the squirrel travels if the post is 25 feet tall and 3 feet in circumference ? | total circuit = 25 / 5 = 5 total feet squirrel travels = 5 * 3 = 15 feet answer : d | ['a ) 10 feet', 'b ) 12 feet', 'c ) 13 feet', 'd ) 15 feet', 'e ) 18 feet'] | d | multiply(divide(25, 5), 3) | divide(n1,n0)|multiply(n2,#0) | geometry |
a man swims downstream 48 km and upstream 42 km taking 6 hours each time ; what is the speed of the current ? | "explanation : 48 - - - 6 ds = 8 ? - - - - 1 42 - - - - 6 us = 7 ? - - - - 1 s = ? s = ( 8 - 7 ) / 2 = 0.5 answer : option b" | a ) 3 kmph , b ) 0.5 kmph , c ) 13 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph | b | divide(add(divide(42, 6), divide(48, 6)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
the average weight of 16 girls in a class is 50.25 kg and that of the remaining 8 girls is 45.15 kg . find the average weights of all the girls in the class . | required average = ( 50.25 x 16 + 45.15 x 8 ) / ( 16 + 8 ) = ( 804 + 361.20 ) / 24 = 1165.20 / 24 = 48.55 kg answer is a | a ) a ) 48.55 , b ) b ) 49 , c ) c ) 51 , d ) d ) 61 , e ) e ) 62 | a | divide(add(multiply(16, 50.25), multiply(8, 45.15)), add(16, 8)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0) | general |
if rs . 782 is divided into 3 parts , proportional to 1 / 2 : 2 / 3 : 3 / 4 find the first part . | explanation : 1 / 2 : 2 / 3 : 3 / 4 = 6 : 8 : 9 first part = 782 ∗ 6 / 23 = 204 option b | a ) 190 , b ) 204 , c ) 220 , d ) 230 , e ) none of these | b | multiply(782, divide(divide(const_12, 2), add(add(divide(const_12, 2), multiply(divide(const_12, 3), const_2)), multiply(divide(const_12, 4), const_3)))) | divide(const_12,n3)|divide(const_12,n1)|divide(const_12,n7)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(#0,#6)|multiply(n0,#7) | general |
the hcf of two numbers is 50 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 50 hcf will be always a factor of lcm 50 is factor of lcm other two factors are 11 & 12 then the numbers are ( 50 * 11 ) and ( 50 x 12 ) = 550 and 600 answer : option e" | a ) 462 , b ) 450 , c ) 480 , d ) 504 , e ) 600 | e | multiply(50, 12) | multiply(n0,n2)| | other |
find the value for m ? 21 ( m + n ) + 21 = 21 ( - m + n ) + 21 | 21 m + 21 n - 21 = - 21 m + 21 n + 21 42 m = 42 = > m = 1 c | a ) 0 , b ) - 1 , c ) 1 , d ) 2 , e ) 3 | c | subtract(21, multiply(const_2, const_10)) | multiply(const_10,const_2)|subtract(n0,#0) | general |
find 40 % of 320 | "we know that r % of m is equal to r / 100 × m . so , we have 40 % of 320 40 / 100 × 320 = 128 answer : c" | a ) 96 , b ) 94 , c ) 128 , d ) 74 , e ) 110 | c | divide(40, 320) | divide(n0,n1)| | gain |
at 1 : 00 pm , a truck left city p and headed toward city q at a constant speed of 50 km / h . one hour later , a car left city q and headed toward city p along the same road at a constant speed of 40 km / h . if the distance between city p and city q is 230 km , at what time will the truck and the car meet each other ... | "at 2 : 00 pm , the truck and the car are 180 km apart . the truck and the car complete a distance of 90 km each hour . the time it takes to meet is 180 / 90 = 2 hours . they will meet at 4 : 00 pm . the answer is a ." | a ) 4 : 00 , b ) 4 : 30 , c ) 5 : 00 , d ) 5 : 30 , e ) 6 : 00 | a | divide(subtract(divide(230, 50), 1), add(1, divide(40, 50))) | divide(n4,n2)|divide(n3,n2)|add(n0,#1)|subtract(#0,n0)|divide(#3,#2)| | physics |
a certain telescope increases the visual range at a particular location from 50 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | original visual range = 50 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 50 ) / 50 * 100 % = 2 * 100 % = 200 % answer e | a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 200 % | e | multiply(divide(subtract(150, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100) | general |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 minutes , then the area of the park ( in sq . m ) is | "perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1600 or x = 160 . length = 480 m and breadth = 320 m . area = ( 480 x 320 ) m 2 = 153600 m 2 . c" | a ) 153200 , b ) 153400 , c ) 153600 , d ) 154000 , e ) 154200 | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
a contractor undertakes to built a walls in 50 days . he employs 60 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | "60 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 60 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 60 × 25 × 0.6 / 25 × 0.4 = 90 men answerb" | a ) 25 , b ) 90 , c ) 35 , d ) 20 , e ) none of these | b | divide(multiply(60, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0)| | physics |
excluding stoppages , the average speed of a bus is 50 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 50 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 50 - 40 ) km i . e 10 km at 50 km / hr . stoppage time = 10 / 50 hrs = 12 min answer : a" | a ) 12 min , b ) 18 min , c ) 16 min , d ) 20 min , e ) 26 min | a | subtract(multiply(const_1, const_60), multiply(divide(40, 50), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general |
the largest prime number is ? | "the largest prime number is 13 . b )" | a ) 11 , b ) 13 , c ) 5 , d ) 7 , e ) 9 | b | square_area(const_pi) | square_area(const_pi)| | general |
the total marks obtained by a student in mathematics and physics is 20 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 20 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 20 + 20 ) / 2 = 20 . answer : c" | a ) 40 , b ) 26 , c ) 20 , d ) 28 , e ) 22 | c | divide(add(20, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
how many numbers from 10 to 10000 are exactly divisible by 9 ? | "10 / 9 = 1 and 10000 / 9 = 1111 = = > 1111 - 1 = 1110 . therefore 1110 answer : c" | a ) 900 , b ) 1000 , c ) 1110 , d ) 1200 , e ) 1400 | c | add(divide(subtract(multiply(floor(divide(10000, 9)), 9), multiply(add(floor(divide(10, 9)), const_1), 9)), 9), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
the average age of 36 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "explanation : age of the teacher = ( 37 * 15 - 36 * 14 ) years = 51 years . answer : c" | a ) 21 years , b ) 71 years , c ) 51 years , d ) 91 years , e ) 11 years | c | add(36, const_1) | add(n0,const_1)| | general |
find the principle on a certain sum of money at 5 % per annum for 4 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 4 ) / 100 ] p = 933 answer : a" | a ) 933 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 4), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n2,#2)| | gain |
out of 10 students in a class , 4 are from maharashtra , 3 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ? | "total possible ways of selecting 4 students out of 10 = 10 c 4 = ( 10 * 9 * 8 * 7 ) / ( 4 * 3 * 2 * 1 ) = 210 the number of ways of selecting a 4 students in which no student belongs to karnataka = 7 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 7 c 4 = 175 probability = 175 / 210 = 5 /... | a ) 1 / 3 , b ) 4 / 9 , c ) 5 / 6 , d ) 3 / 7 , e ) 1 / 4 | c | divide(subtract(divide(divide(factorial(10), factorial(subtract(10, const_4))), factorial(const_4)), divide(divide(factorial(subtract(10, 3)), factorial(subtract(subtract(10, 3), const_4))), factorial(const_4))), divide(divide(factorial(10), factorial(subtract(10, const_4))), factorial(const_4))) | factorial(n0)|factorial(const_4)|subtract(n0,const_4)|subtract(n0,n2)|factorial(#2)|factorial(#3)|subtract(#3,const_4)|divide(#0,#4)|factorial(#6)|divide(#7,#1)|divide(#5,#8)|divide(#10,#1)|subtract(#9,#11)|divide(#12,#9)| | probability |
a sum was put at simple interest at a certain rate for 3 years had it been put at 3 % higher rate , it would have fetched 81 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 81 = p × 3 x 3 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 81 × 100 / 3 x 3 = 900 answer e" | a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900 | e | divide(81, multiply(divide(3, const_100), 3)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain |
the least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ratio less than 16 : 21 , is | "explanation : let x is subtracted . then , ( 6 − x ) / ( 7 − x ) < 16 / 21 21 ( 6 — x ) < 16 ( 7 — x ) = > 5 x > 14 = x > 2.8 least such number is 3 option a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(divide(subtract(multiply(21, 6), multiply(7, 16)), subtract(21, 16)), divide(const_2, const_10)) | divide(const_2,const_10)|multiply(n0,n3)|multiply(n1,n2)|subtract(n3,n2)|subtract(#1,#2)|divide(#4,#3)|add(#5,#0)| | other |
the total cost of 100 paper plates and 200 paper cups is $ 6.00 at the same rates what is the total cost of 20 of the plates and 40 of the cups ? | u dont need to go through all this what u have with u is 100 p + 200 c = $ 6.00 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 1.20 therefore oa is c | a ) $ . 90 , b ) $ 1.00 , c ) $ 1.20 , d ) $ 1.50 , e ) $ 1.60 | c | multiply(divide(20, 100), 6) | divide(n3,n0)|multiply(n2,#0) | gain |
if the price of sugar rises from rs . 3 per kg to rs . 5 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 3 = x × 5 = x = 60 kg . ∴ reduction in consumption = 40 % . answer e | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 40 % | e | multiply(subtract(const_1, divide(multiply(const_1, 3), 5)), const_100) | multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100) | general |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 300 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1500 m t = d / s t = 1500 / 10 t = 150 sec answer : d" | a ) 176 sec , b ) 190 sec , c ) 178 sec , d ) 150 sec , e ) 276 sec | d | divide(add(1200, 300), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics |
find the average of all prime numbers between 30 and 50 . | "sol . there are five prime numbers between 30 and 50 . they are 31 , 37 , 41 , 43 and 47 ∴ required average = [ 31 + 37 + 41 + 43 + 47 / 5 ] = 199 / 5 = 39.8 answer c" | a ) 30 , b ) 33.6 , c ) 39.8 , d ) 44.9 , e ) none | c | divide(add(add(add(30, const_1), add(add(const_4.0, const_1), const_2)), add(subtract(50, const_4.0), subtract(50, const_2))), 30) | add(n0,const_1)|subtract(n1,const_4.0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,const_4)| | general |
there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability w that the both positions will be taken by members of just one party ? | "w probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total can... | a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2 | d | multiply(multiply(divide(subtract(divide(50, 2), const_1), subtract(50, const_1)), divide(divide(50, 2), 50)), 2) | divide(n1,n0)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#4,#2)|multiply(n0,#5)| | other |
a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 30 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 30 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could ha... | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 30 chips lost and minimize # of $ 100 chips lost , which means that # of $ 30 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 30 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that... | a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 2,030 | e | subtract(multiply(const_3, const_1000), add(multiply(divide(add(16, 2), 2), 30), multiply(subtract(divide(add(16, 2), 2), 2), 100))) | add(n3,n5)|multiply(const_1000,const_3)|divide(#0,n5)|multiply(n1,#2)|subtract(#2,n5)|multiply(n2,#4)|add(#3,#5)|subtract(#1,#6)| | general |
an association of mathematics teachers has 1,260 members . only 445 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? | "total umber of members = 1260 number of members that cast votes = 445 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 445 = 267 percent of total membership that voted for winning candidate = ( 267 / 1260 ) * 100 = 21.19 % answer e" | a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 22 % | e | multiply(divide(multiply(divide(60, const_100), 445), multiply(const_100, power(const_4, const_2))), const_100) | divide(n2,const_100)|power(const_4,const_2)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
what is the smallest positive integer x , such that 5000 x is a perfect cube ? | "take out the factors of 5000 that will come 10 ^ 3 * 5 . for perfect cube you need every no . raise to the power 3 . for 5000 x to be a perfect cube , you need two 5 that means 25 . d is the answer ." | a ) 4 , b ) 6 , c ) 8 , d ) 25 , e ) 18 | d | add(const_3, const_4) | add(const_3,const_4)| | geometry |
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 6 - 1 / 12 ) = 1 / 12 . work done by the inlet in 1 min . = ( 1 / 12 × 1 / 60 ) = 0.001389 volume of 0.001389 part = 6 litres . therefore , volume of whole = 1 / 0.001389 x 6 ‹ = › 4320 litres . answer d" | a ) 7580 , b ) 7960 , c ) 8290 , d ) 4320 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(6, const_60)), const_1)) | multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)| | physics |
maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 24 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he me... | "total distance = 24 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 20 kms relative speed ( opposite side ) ( as they are moving towards each other speed w... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | divide(add(24, 6), add(4, 6)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | physics |
one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 3 robots altogether , if the robots build only one robot at a t... | "first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) and now we have 3 robots in 40 minutes answer - a" | a ) 40 min , b ) 94 min , c ) 110 min , d ) 112 min , e ) 120 min | a | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)| | physics |
if 20 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | "let x be the number of students in the school . 0.2 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.2 x / 0.25 = 0.8 x students went to the camping which is 80 % . the answer is d" | a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 75 | d | divide(20, divide(subtract(100, 75), 100)) | subtract(n1,n2)|divide(#0,n1)|divide(n0,#1)| | general |
in a group of 400 readers who read science fiction or literacy works or both , 250 read science fiction and 230 read literacy works . how many read both science fiction and literacy works ? | "consider total number of reader n ( s u l ) = 400 people who read science fiction n ( s ) = 250 people who read literacy works n ( l ) = 230 both science fiction and literacy n ( s ∩ l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s ∩ l ) 400 = 250 + 230 - n ( s ∩ l ) n ( s ∩ l ) = 480 - 400 n ( s ∩ l ) = 80 so people r... | a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | b | subtract(add(250, 230), 400) | add(n1,n2)|subtract(#0,n0)| | other |
if 12 men and 16 boys can do a piece of work in 5 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy ? | 12 m + 16 b - - - - - 5 days 13 m + 24 b - - - - - - - 4 days 60 m + 80 b = 52 m + 96 b 8 m = 16 b = > 1 m = 2 b m : b = 2 : 1 answer : d | a ) 2 : 5 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 2 | d | divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13))) | multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5) | physics |
the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 10 miles per gallon , will not get him there . by how many miles per gallon must the aircraft ’ s fuel efficiency be improved to make the fli... | "actual miles / gallon is = 480 / 40 = 12 miles / gallon . current engine miles / gallon is 10 miles / gallon . additional 2 miles / gallon is required to match the actual mileage . imo option a ." | a ) 2 , b ) 4 , c ) 12 , d ) 40 , e ) 160 | a | subtract(divide(480, 40), 10) | divide(n1,n0)|subtract(#0,n2)| | physics |
a certain bus driver is paid a regular rate of $ 12 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compens... | "for 40 hrs = 40 * 12 = 480 excess = 976 - 480 = 496 for extra hours = . 75 ( 12 ) = 9 + 12 = 21 number of extra hrs = 496 / 21 = 23.62 total hrs = 40 + 23.62 = 63.62 = 64 approx . answer d 64" | a ) 60 , b ) 68 , c ) 65 , d ) 64 , e ) 61 | d | add(40, divide(subtract(976, multiply(12, 40)), divide(multiply(12, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
three numbers are in the ratio 3 : 4 : 5 and their average is 36 . the largest number is : | "explanation : let the numbers be 3 x , 4 x and 5 x , then , ( 3 x + 4 x + 5 x ) / 3 = 36 = > 12 x = 36 * 3 = > x = 9 largest number 5 x = 5 * 9 = 45 answer : e" | a ) 30 , b ) 98 , c ) 27 , d ) 21 , e ) 45 | e | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
5 cats can catch 5 mice in 5 minutes . how many cats does it take to catch 100 mice in 100 minutes ? | let x be no . of cat 5 cat catch 5 mice in 5 min 5 cat catch 1 mice in ( 5 / 5 = 1 min ) 1 cat catch 1 mice in 1 * 5 = 5 min 1 cat catch 100 mice in 5 * 100 min x cat catch 100 mice in ( 5 * 100 ) / x min a / q ( 5 * 100 ) / x = 100 = > 5 * 100 = x * 100 = > x = 5 answer : e | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(100, multiply(5, const_4)) | multiply(n0,const_4)|divide(n3,#0) | physics |
the tax on a commodity is diminished by 19 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 81 * 115 = 9315 - - - - - - - - - - - 10000 - - - - - - - - - - - 685 100 - - - - - - - - - - - ? = > 6.85 % decrease answer : d" | a ) 2 % , b ) 8.85 % , c ) 5.85 % , d ) 6.85 % , e ) 1 % | d | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
when a train travels at a speed of 60 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 60 km / h 50 km / h - - - - > x + 15 distance is equal so 60 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 60 x = 50 x + 750 so the would be in km and x = 75 so 75 km answer : a" | a ) 75 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | divide(n2,const_60)|multiply(n1,#0)|divide(#1,n1)|multiply(#2,const_60)| | physics |
if x is less than y by 40 % then y exceed x by : | "using formula ( x / ( 100 - x ) * 100 ) where x is percentage decrease ( here it is 40 % ) = > 40 / ( 100 - 40 ) * 100 = 66.66 % answer : d" | a ) 33.33 % , b ) 25 % , c ) 75 % , d ) 66.66 % , e ) none of these | d | multiply(subtract(divide(const_100, subtract(const_100, 40)), const_1), const_100) | subtract(const_100,n0)|divide(const_100,#0)|subtract(#1,const_1)|multiply(#2,const_100)| | general |
p is 20 % more efficient than q . p can complete a work in 22 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 / 22 let work done by q in 1 day = q q × ( 120 / 100 ) = 1 / 22 q = 100 / ( 22 × 120 ) = 10 / ( 22 × 12 ) the work done by p and q in 1 day = 1 / 22 + 10 / ( 22 × 12 ) = 22 / ( 22 × 12 ) = 1 / 12 p and q together can do the work in 12 days . the answer is c ." | a ) 8 , b ) 11 , c ) 12 , d ) 15 , e ) 16 | c | divide(multiply(add(divide(20, const_100), const_1), 22), add(add(divide(20, 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 |
a school has 4 section of chemistry in class x having 55 , 35 , 45 and 42 students . the mean marks obtained in chemistry test are 50 , 60 , 55 and 45 respectively for the 4 sections . determine the overall average of marks per student . | "required average marks = 55 ã — 50 + 35 ã — 60 + 45 ã — 55 + 42 ã — 45 / 55 + 35 + 45 + 42 = 2750 + 2100 + 2475 + 1890 / 162 = 9215 â „ 177 = 52.06 answer b" | a ) 50.15 , b ) 52.06 , c ) 51.25 , d ) 53.06 , e ) none of the above | b | divide(add(add(add(multiply(55, 50), multiply(35, 60)), multiply(45, 55)), multiply(42, 45)), add(add(add(55, 35), 45), 42)) | add(n1,n2)|multiply(n1,n5)|multiply(n2,n6)|multiply(n3,n7)|multiply(n3,n4)|add(#1,#2)|add(n3,#0)|add(#5,#3)|add(n4,#6)|add(#7,#4)|divide(#9,#8)| | general |
rs . 2600 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 2600 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 2600 = rs . 1200 answer : a" | a ) s . 1200 , b ) s . 1542 , c ) s . 1528 , d ) s . 1540 , e ) s . 1549 | a | multiply(4, divide(2600, add(add(4, 2), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general |
the difference of two numbers is 1355 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1355 ) . x + 1355 = 6 x + 15 5 x = 1340 x = 268 smaller number = 270 . answer d" | a ) 240 , b ) 270 , c ) 295 , d ) 268 , e ) 252 | d | divide(add(1355, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general |
if a radio is purchased for rs 490 and sold for rs 465.50 find the loss % ? | c . p = rs 490 s . p = rs 465.50 loss = c . p - s . p = 490 - 465.50 = rs 24.50 loss % = ( loss * 100 ) / c . p % = ( 24.50 * 100 ) / 490 % = 5 % answer a 5 % | a ) 5 % , b ) 4 % , c ) 6 % , d ) 8 % , e ) 7 % | a | divide(multiply(subtract(490, 465.5), const_100), 490) | subtract(n0,n1)|multiply(#0,const_100)|divide(#1,n0) | gain |
if a person cover 3 / 5 part of his journey by train , 7 / 20 part of his journey by bus and rest 6.5 km . by walk . what is total distance of his journey ? | let total distance be = x km . then , 3 x / 5 + 7 x / 20 + 6.5 = x ; 12 x + 7 x + 130 = 20 x ; x = 130 hence total distance = 130 km . answer d | a ) 65 km , b ) 100 km , c ) 120 km , d ) 130 km , e ) 75 km | d | divide(6.5, subtract(const_1, add(divide(3, 5), divide(7, 20)))) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(const_1,#2)|divide(n4,#3) | general |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 144 grams of water ? | "see h / o = 2 / 16 add 1 to both sides : h / o + 1 = 2 / 16 + 1 take lcm ( h + o ) / o = ( 2 + 16 ) / 16 water / o = 18 / 16 now that we have the water to oxygen ratio , put in the amount of water i . e . 144 gm = > o = 144 * 16 / 18 = 128 answer : d" | a ) 16 , b ) 72 , c ) 112 , d ) 128 , e ) 142 | d | multiply(2, divide(144, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
the h . c . f . of two numbers is 16 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is : | "clearly , the numbers are ( 16 x 13 ) and ( 16 x 14 ) . larger number = ( 23 x 14 ) = 224 . answer : option c" | a ) 276 , b ) 299 , c ) 224 , d ) 345 , e ) 365 | c | multiply(16, 14) | multiply(n0,n2)| | other |
at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 21 employees are trained to work in the family buffet , 18 are trained to work in the dinin... | "39 = 21 + 18 + 12 - 4 - 2 x 2 x = 21 + 18 + 12 - 4 - 39 = 47 - 39 = 8 x = 4 c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(subtract(subtract(add(add(21, 18), 12), 4), 39), 2) | add(n4,n5)|add(n6,#0)|subtract(#1,n7)|subtract(#2,n0)|divide(#3,n8)| | physics |
how many 4 digit number contain number 2 ? | "total 4 digit no . = 9 * 10 * 10 * 10 = 9000 not containing 2 = 8 * 9 * 9 * 9 = 5832 total 4 digit number contain 2 = 9000 - 5832 = 3168 answer : a" | a ) 3168 , b ) 3268 , c ) 3368 , d ) 3468 , e ) 3568 | a | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(4, 4)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
a bookstore has a shelf that contains biographies which normally sell for $ 20 each and mysteries that normally sell for $ 12 each . during a sale , the biographies and mysteries are discounted at different rates so that a customer saves a total of $ 19 from the normal price by buying 5 discounted biographies and 3 dis... | let b be the discount on biographies and m be the discount on mysteries so . , b + m = 0.435 - - - - - ( 1 ) and ( 20 * 5 + 12 * 3 ) - ( 20 * 5 * ( 1 - b ) + 12 * 3 * ( 1 - m ) ) = 19 - - > 100 ( 1 - ( 1 - b ) ) + 36 ( 1 - ( 1 - m ) = 19 100 b + 36 m = 19 - - - - - - ( 2 ) solving 12 . , we get m = 0.375 = 37.5 % d | a ) 24 % , b ) 25 % , c ) 30 % , d ) 37.5 % , e ) 40 % | d | multiply(subtract(const_1, divide(multiply(20, const_2), add(add(20, 43), const_1))), const_100) | add(n0,n5)|multiply(n0,const_2)|add(#0,const_1)|divide(#1,#2)|subtract(const_1,#3)|multiply(#4,const_100) | gain |
calculate the average of all the numbers between 15 and 55 which are divisible by 4 . | explanation : numbers divisible by 4 are 16,20 , 24,28 , 32,36 , 40,44 , 48,52 , , , average = ( 16 + 20 + 24 + 28 + 32 + 36 + 40 + 44 + 48 + 52 ) / 10 = 340 / 10 = 34 answer : c | a ) 39 , b ) 33 , c ) 34 , d ) 30 , e ) 36 | c | multiply(divide(add(add(floor(divide(15, 4)), const_1), floor(divide(55, 4))), const_2), 4) | divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6) | general |
what is the remainder when 9 ^ 1 + 9 ^ 2 + 9 ^ 3 + . . . + 9 ^ 9 is divided by 9 ? | "notice that in the brackets we have the sum of 9 odd multiples of 9 , which yields remainder of 0 upon division by 9 . answer : c" | a ) 2 , b ) 3 , c ) 0 , d ) 5 , e ) none of the above | c | divide(subtract(power(9, 1), power(9, 1)), 9) | power(n0,n1)|power(n2,n1)|subtract(#0,#1)|divide(#2,n4)| | general |
by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 25 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 160 sp = cp - loss = 640 - 160 = rs . 480 answer : b" | a ) rs . 488 , b ) rs . 480 , c ) rs . 482 , d ) rs . 487 , e ) rs . 481 | b | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 25), const_100)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
if ( a + b ) = 16 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 16 = > a = 16 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 16 - b + 3 - c = > 19 - ( b + c ) = > 19 - 9 = 10 . option a . . ." | a ) 10 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | a | subtract(add(16, 3), 9) | add(n0,n2)|subtract(#0,n1)| | general |
rs . 880 is divided amongst a , b , c so that 4 times a ' s share , 5 times b ' s share and 10 times c ' s share are all equal . find c ' s share ? | a + b + c = 880 4 a = 5 b = 10 c = x a : b : c = 1 / 4 : 1 / 5 : 1 / 10 = 5 : 4 : 2 2 / 11 * 880 = rs . 160 answer : d | a ) 177 , b ) 150 , c ) 817 , d ) 160 , e ) 616 | d | divide(880, add(add(divide(10, 4), divide(10, 5)), const_1)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3) | general |
divide $ 900 among a , b in the ratio 1 : 2 . how many $ that a get ? | "sum of ratio terms = 1 + 2 = 3 a = 900 * 1 / 3 = $ 300 answer is b" | a ) $ 50 , b ) $ 300 , c ) $ 150 , d ) $ 250 , e ) $ 600 | b | divide(900, 1) | divide(n0,n1)| | other |
what is the square root of 169 ? | "13 x 13 = 169 answer c" | a ) 8 , b ) 16 , c ) 13 , d ) 42 , e ) 25 | c | circle_area(divide(169, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
a car is running at a speed of 96 kmph . what distance will it cover in 14 sec ? | "speed = 96 kmph = 96 * 5 / 18 = 27 m / s distance covered in 14 sec = 27 * 14 = 378 m answer is a" | a ) 378 m , b ) 350 m , c ) 380 m , d ) 200 m , e ) 250 m | a | multiply(divide(96, const_3_6), 14) | divide(n0,const_3_6)|multiply(n1,#0)| | physics |
in a certain aquarium , the number of red fish is 3 times the number of the green fish , and the number of blue fish is half the number of the red fish . if the ratio of green fish to red fish were to be doubled , and the ratio of red fish to blue fish were to be doubled , then the ratio of blue fish to green fish in t... | lots of little places to make a mistake or panic here . = ) r = 3 g b = r / 2 g / r = 1 / 3 - - - - > doubled g / r = 2 / 3 r / b = 2 - - - - > doubled r / b = 4 b / g = ( b / r ) * ( r / g ) = ( 1 / 4 ) * ( 3 / 2 ) = 3 / 8 answer : e | a ) 3 : 2 , b ) 1 : 1 , c ) 1 : 2 , d ) 3 : 4 , e ) 3 : 8 | e | multiply(inverse(multiply(const_2, const_2)), inverse(multiply(const_2, divide(const_1, 3)))) | divide(const_1,n0)|multiply(const_2,const_2)|inverse(#1)|multiply(#0,const_2)|inverse(#3)|multiply(#2,#4) | general |
in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.845 per pound ? | "using weighted average method : let x be the proportion in which the $ 0.8 per pound flour is mixed with $ 0.9 per pound flour . thus 0.8 * x + 0.9 * ( 1 - x ) = 0.845 0.9 – 0.1 x = 0.845 x = 0.55 thus ratio of both flours is 5 : 4 e" | a ) 1 : 3 , b ) 1 : 2 , c ) 1 : 1 , d ) 2 : 1 , e ) 5 : 4 | e | divide(0.845, add(0.8, 0.9)) | add(n0,n1)|divide(n2,#0)| | general |
a car drives 90 miles on local roads at 30 mph , and 75 miles on the highway at 60 mph , what is the average speed of the entire trip ? | so the answer is plainly c . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 90 miles , and v = 30 mph so we can obtain time : 30 * t = 90 or t = 90 / 30 = 3 hours the needed time to cover 90 miles in the same way we should divide ... | a ) 36 mph , b ) 40 mph , c ) 39 mph , d ) 42 mph , e ) 38 mph | c | divide(add(90, 75), add(divide(90, 30), divide(75, 60))) | add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3) | physics |
in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 18 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 18 ) = 36 / 27 = 4 / 3 . answer : b . | a ) 2 / 1 , b ) 4 / 3 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 2 | b | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 18)) | add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2) | general |
a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 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 / $ 2.40 = 45 thus , solution a is correct ." | a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | a | divide(original_price_before_loss(20, 90), divide(original_price_before_gain(20, 60), 20)) | original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2)| | gain |
if x > 3000 , then the value of ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) is closest to ? | "assume x = 3002 ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) = 9006 / ( 2 * ( 3002 ) ^ 1.21 - 1111 ^ 1.3 ) = 9006 / 23154 = = 4 / 10 d" | a ) 1 / 6 , b ) 1 / 3 , c ) 10 / 21 , d ) 4 / 10 , e ) 3 / 2 | d | multiply(subtract(const_2, 2), const_2) | subtract(const_2,n2)|multiply(#0,const_2)| | general |
john left home and drove at the rate of 55 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ? | "the total distance d traveled by john is given by d = 55 * 2 + 3 * 55 = 275 miles . answer e" | a ) 235 miles . , b ) 245 miles . , c ) 255 miles . , d ) 265 miles . , e ) 275 miles . | e | add(multiply(55, 2), multiply(3, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | physics |
a circular well with a diameter of 4 metres , is dug to a depth of 14 metres . what is the volume of the earth dug out ? | "solution volume = π r 2 h ‹ = › ( 22 / 7 × 2 × 2 × 14 ) m 3 ‹ = › 176 m 3 . answer e" | a ) 32 m 3 , b ) 36 m 3 , c ) 40 m 3 , d ) 44 m 3 , e ) 176 m 3 | e | volume_cylinder(divide(4, const_2), 14) | divide(n0,const_2)|volume_cylinder(#0,n1)| | geometry |
six bells commence tolling together and toll at intervals of 2 , 4,6 , 8,10 and 12 seconds respectively . in 30 m 9 nutes , how many times do they toll together ? | "solution l . c . m . of 2,4 , 6,8 , 10,12 is 120 . so , the bells will toll together after every 120 seconds , i . e . , 2 minutes in 30 minutes , they will toll together left [ ( 30 / 2 ) + 1 ] = 16 times . answer d" | a ) 4 , b ) 10 , c ) 15 , d ) 16 , e ) 17 | d | divide(30, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 30)) | add(n0,const_3)|multiply(n0,#0)|multiply(#1,const_3)|multiply(n0,#2)|multiply(n0,#3)|divide(#4,n4)|divide(n4,#5)| | physics |
a man cycling along the road noticed that every 6 minutes a bus overtakes him and every 2 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ? | "let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 6 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 6 , d = 6 b - 6 c ; every 2 minutes cyclist meets an oncoming bus : \ frac { d } { b + ... | a ) 3 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 10 minutes | a | divide(subtract(6, divide(6, divide(add(2, 6), subtract(6, 2)))), const_1) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|divide(n0,#2)|subtract(n0,#3)|divide(#4,const_1)| | physics |
what is the area of a square field whose diagonal of length 30 m ? | "d 2 / 2 = ( 30 * 30 ) / 2 = 450 answer : c" | a ) 287 , b ) 269 , c ) 450 , d ) 200 , e ) 230 | c | divide(square_area(30), const_2) | square_area(n0)|divide(#0,const_2)| | geometry |
what is the least number should be added to 1056 , so the sum of the number is completely divisible by 25 ? | "( 1056 / 25 ) gives remainder 6 19 + 6 = 23 , so we need to add 19 a" | a ) 19 , b ) 20 , c ) 32 , d ) 41 , e ) 53 | a | subtract(multiply(add(floor(divide(1056, 25)), const_1), 25), 1056) | divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)| | general |
a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 50 paise per sq . metre | "area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 50 = 42500 p in rs = 42500 / 100 = rs 425 answer : a" | a ) s 425 , b ) s 780 , c ) s 880 , d ) s 480 , e ) s 980 | a | divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 50), const_100) | multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics |
a man bought 40 shares of rs . 60 at 7 discount , the rate of dividend being 1212 % the rate of interest obtained is | "explanation : face value of a share = rs . 60 he bought each share at rs . 60 - rs . 7 = rs . 53 number of shares = 40 dividend = 12 1 / 2 % = 25 / 2 % dividend per share = 60 × 25 / 2 × 100 = rs . 7.5 total dividend = ( 40 × 7.5 ) ie , he got a dividend of ( 40 × 7.5 ) for an investment of rs . ( 40 × 53 ) interest o... | a ) 13.64 % , b ) 15.5 % , c ) 14 % , d ) 14.15 % , e ) 14.95 % | d | divide(multiply(multiply(40, 60), divide(1212, const_100)), multiply(40, subtract(60, 7))) | divide(n3,const_100)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#1)|multiply(n0,#2)|divide(#3,#4)| | gain |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m ^ 3 ) is : | "clearly , l = ( 48 - 16 ) m = 32 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 32 x 20 x 8 ) m 3 = 5120 m ^ 3 . answer e" | a ) 2130 , b ) 3620 , c ) 4512 , d ) 5620 , e ) 5120 | e | volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
if 853845 bars of chocolate need to be placed in little boxes and each box can contain at most 9 bars of chocolate . how many bars of chocolate will be in the last unfilled box ? | the number of bars of chocolate that can be placed in the little boxes of 9 bars is the first number that is divisible by 9 that occurs before 853845 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 853845 ... | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | d | subtract(853845, multiply(floor(divide(853845, 9)), 9)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2) | general |
the length of a bridge in meters , which a train 110 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 110 = 375 meters x = 265 meters the answer is b ." | a ) 245 , b ) 265 , c ) 295 , d ) 325 , e ) 375 | b | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 110) | multiply(n1,const_1000)|divide(#0,const_3600)|multiply(n2,#1)|subtract(#2,n0)| | physics |
income and expenditure of a person are in the ratio 9 : 8 . if the income of the person is rs . 36000 , then find his savings ? | "let the income and the expenditure of the person be rs . 9 x and rs . 7 x respectively . income , 9 x = 36000 = > x = 4000 savings = income - expenditure = 9 x - 8 x = x so , savings = rs . 4000 . answer : c" | a ) 2000 , b ) 3000 , c ) 4000 , d ) 5000 , e ) 6000 | c | subtract(36000, multiply(divide(8, 9), 36000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
light glows for every 16 seconds . how many times did it between 1 : 57 : 58 and 3 : 20 : 47 am | "the diff in sec between 1 : 57 : 58 and 3 : 20 : 47 is 4969 sec , 4969 / 16 = 310 . so total 311 times light ll glow answer : a" | a ) 311 , b ) 322 , c ) 313 , d ) 314 , e ) 385 | a | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 16) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)| | physics |
the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 294 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 294 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 294 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 294 r 2 = ( 294 * 2 ) / 3 = 196 = > r = 14 rate = 14 % answer : option d" | a ) 18 , b ) 12 , c ) 9 , d ) 14 , e ) 16 | d | sqrt(294) | sqrt(n2)| | gain |
a thief goes away with a santro car at a speed of 30 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 60 kmph when will the owner over take the thief from the start ? | "- - - - - - - - - - - 30 - - - - - - - - - - - - - - - - - - - - | 60 30 d = 30 rs = 60 â € “ 30 = 30 t = 30 / 30 = 1 hours answer : a" | a ) 1 hours , b ) 2 hours , c ) 2 / 3 hours , d ) 1 / 3 hours , e ) 2 / 5 hours | a | subtract(divide(multiply(divide(const_1, const_2), 30), subtract(60, 30)), divide(const_1, const_2)) | divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)| | physics |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 10000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 1000 = > x = 10000 . b ' s share = rs . 2 x = 2 * 10000 = rs . 20000 . answer : e" | a ) a ) 82390 , b ) b ) 29000 , c ) c ) 20005 , d ) d ) 23930 , e ) e ) 20000 | e | multiply(multiply(subtract(4, 3), 10000), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general |
one night a certain hotel rented 1 / 2 of its rooms , including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | "the rooms which were not rented is 1 / 2 the ac rooms which were not rented is ( 1 / 3 ) * ( 3 / 5 ) = 1 / 5 the percentage of unrented rooms which were ac rooms is ( 1 / 5 ) / ( 1 / 2 ) = 2 / 5 = 40 % the answer is d ." | a ) 25 % , b ) 30 % , c ) 35 % , d ) 40 % , e ) 50 % | d | multiply(divide(multiply(subtract(const_1, divide(const_2.0, 3)), multiply(divide(1, 2), const_100)), subtract(const_100, multiply(divide(1, 2), const_100))), const_100) | divide(n3,n5)|divide(n2,n0)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_100,#2)|divide(#4,#5)|multiply(#6,const_100)| | gain |
a class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the class room ? | explanation : length = 6 m 24 cm = 624 cm width = 4 m 32 cm = 432 cm hcf of 624 and 432 = 48 number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 . answer is b | a ) 115 , b ) 117 , c ) 116 , d ) 114 , e ) 112 | b | divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(add(24, 24), add(24, 24))) | add(n1,n1)|multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#1)|add(n3,#2)|multiply(#0,#0)|multiply(#3,#4)|divide(#6,#5) | general |
128 % of 5 / 8 = | "should be simple . 1.28 * 5 / 8 = 6.4 / 8 = 0.8 correct option : d" | a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.8 , e ) 1.0 | d | divide(multiply(divide(multiply(8, 5), const_100), 128), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | general |
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