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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of 70 players on a football team , 34 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 34 rest = 70 - 34 = 36 left handed = 36 / 3 = 12 right handed = 24 if all thrower are right handed then total right handed is 34 + 24 = 58 so b . 58 is the right answer" | a ) 54 , b ) 58 , c ) 63 , d ) 71 , e ) 92 | b | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 34)), 34) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general |
if the digits 31 in the decimal 0.00031 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00031 ) ? | "99 * 0.31 = 30.69 approx . 31 answer : a" | a ) 31 , b ) 0.31 , c ) 0.0031 , d ) 0.00031 , e ) 3.1 e - 05 | a | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(31, subtract(multiply(10, 10), const_1))), const_1000), 10) | multiply(n2,n2)|subtract(#0,const_1)|divide(n0,#1)|multiply(#0,#1)|multiply(#2,#3)|divide(#4,const_1000)|multiply(n2,#5)| | general |
the perimeter of a triangle is 48 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 48 / 2 = 60 cm 2 answer : e" | a ) 87 cm 2 , b ) 28 cm 2 , c ) 98 cm 2 , d ) 26 cm 2 , e ) 60 cm 2 | e | triangle_area(2.5, 48) | triangle_area(n0,n1)| | geometry |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 24 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 ... | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 16 / 21 | e | divide(subtract(40, 24), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
if the sum of two numbers is 30 and the sum of their squares is 840 , then the product of the numbers is | sol . let the numbers be x and y . then , ( x + y ) = 30 and x 2 + y 2 = 840 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 30 ) 2 - 840 = 900 - 840 = 60 xy = 30 . answer e | a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 30 | e | divide(subtract(power(30, const_2), 840), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2) | general |
pavan travelled for 11 hours . he covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph . find the distance travelled by pavan . | "let the distance travelled be x km . total time = ( x / 2 ) / 30 + ( x / 2 ) / 25 = 11 = > x / 60 + x / 50 = 11 = > ( 5 x + 6 x ) / 300 = 11 = > x = 300 km answer : d" | a ) 298 km , b ) 278 km , c ) 278 km , d ) 300 km , e ) 267 km | d | multiply(inverse(add(inverse(multiply(const_2, 30)), inverse(multiply(25, const_2)))), 11) | multiply(n1,const_2)|multiply(n2,const_2)|inverse(#0)|inverse(#1)|add(#2,#3)|inverse(#4)|multiply(n0,#5)| | physics |
if 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) , then | a | + | b | + | c | + | d | | this is the hard one , definitely a 700 + level question . we need numbers a , b , c , and d such that 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) this means that ac = 6 , bd = β 12 , and ad + bc = 1 . the a & c pair could be ( 1 , 6 ) or ( 2 , 3 ) , in some order . the absolute values of the b & d pair could be ( 1 , 12 ... | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(2, const_3), add(const_3, const_4)) | add(n1,const_3)|add(const_3,const_4)|add(#0,#1) | general |
the speed of a boat in still water in 20 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 4 ) = 24 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 24 = 9.6 km answer is d ." | a ) 9.2 , b ) 10.2 , c ) 9.8 , d ) 9.6 , e ) 10.0 | d | multiply(add(20, 4), divide(24, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
a bag contains 3 red , 5 yellow and 7 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ? | "total number of balls = 3 + 5 + 7 = 15 n ( s ) = 15 c 3 = 455 n ( e ) = 3 c 1 * 5 c 1 * 7 c 1 = 105 probability = 105 / 455 = 3 / 13 answer is c" | a ) 2 / 7 , b ) 3 / 5 , c ) 3 / 13 , d ) 1 / 4 , e ) 7 / 16 | c | divide(multiply(choose(const_4.0, const_2), choose(add(const_3.0, 5), const_1)), choose(add(add(3, 5), 7), 3)) | add(n0,n1)|choose(const_4.0,const_2)|add(n2,#0)|choose(#0,const_1)|choose(#2,n0)|multiply(#1,#3)|divide(#5,#4)| | probability |
two cars are driving toward each other . the first car is traveling at a speed of 100 km / h , which is 25 % faster than the second car ' s speed . if the distance between the cars is 720 km , how many hours will it take until the two cars meet ? | the speed of the first car is 100 km / h . the speed of the second car is 100 / 1.25 = 80 km / h . the two cars complete a total of 180 km each hour . the time it takes the cars to meet is 720 / 180 = 4 hours . the answer is c . | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | c | divide(add(multiply(100, const_4), multiply(divide(multiply(const_4, 100), add(const_1, const_4)), const_4)), add(100, divide(multiply(const_4, 100), add(const_1, const_4)))) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|add(n0,#2)|multiply(#2,const_4)|add(#1,#4)|divide(#5,#3) | physics |
a tiger is chasing its prey at a constant speed . it entire body passes a blade of grass in 1 second . it then runs above a fallen tree trunk that is 20 meters long in 5 seconds . what is the length of the tiger ? | sol . let the length of the tiger be x metres and its speed be y m / sec . then , x / y = 1 β y = x β΄ ( x + 20 ) / 5 = x β x = 5 m . answer b | a ) 6 m , b ) 5 m , c ) 4 m , d ) 3 m , e ) none of the above | b | divide(20, subtract(5, 1)) | subtract(n2,n0)|divide(n1,#0) | physics |
a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 25 % . their gross income from sales for one month increases by what percent ? | "let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.25 x = 1.125 * ( p * x ) , an increase of 12.5 % . the answer is d ." | a ) 5 % , b ) 7.5 % , c ) 10 % , d ) 12.5 % , e ) 15 % | d | subtract(subtract(25, 10), divide(25, 10)) | divide(n1,n0)|subtract(n1,n0)|subtract(#1,#0)| | gain |
5 / 8 th of a job is completed in 10 days . if a person works at the same pace , how many days will he take to complete the job ? | explanation : solution : it is given that 5 / 8 th of the work is completed in 10 days . = > remaining work = 3 / 8 th of total applying unitary method : total work will be completed in 10 * 8 / 5 days = > it takes 16 days to complete total work = > hence , remaining work days = 16 - 10 = 6 days answer : c | a ) 8 , b ) 9 , c ) 6 , d ) 2 , e ) 1 | c | multiply(multiply(subtract(const_1, divide(5, 8)), 10), divide(8, 5)) | divide(n1,n0)|divide(n0,n1)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3) | physics |
find the number of different prime factors of 12800 | "explanation : l . c . m of 12800 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 2 , 5 number of different prime factors is 2 . answer : option b" | a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | b | add(const_2, const_2) | add(const_2,const_2)| | other |
the sum of two numbers is 30 and their difference is 10 . find their product . | let the numbers be x and y . then x + y = 30 and x - y = 10 x = 20 ; y = 10 xy = 20 * 10 = 200 answer : d | a ) 104 , b ) 114 , c ) 315 , d ) 200 , e ) 335 | d | multiply(divide(add(30, 10), add(const_1, const_1)), subtract(divide(add(30, 10), add(const_1, const_1)), 10)) | add(n0,n1)|add(const_1,const_1)|divide(#0,#1)|subtract(#2,n1)|multiply(#2,#3) | general |
difference of two numbers is 1660 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1660 5 * y / 3 β y = 1660 2 * y / 3 = 1660 y = [ ( 1660 * 3 ) / 2 ] = 2490 . one number = 2490 , second number = 5 * y / 3 = 4150 answer is a . | a ) 2490 and 4150 , b ) 249 and 415 , c ) 2400 and 4100 , d ) 2290 and 4350 , e ) 229 and 435 | a | multiply(divide(7.5, 12.5), divide(1660, subtract(const_1, divide(7.5, 12.5)))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2) | gain |
a car is running at a speed of 120 kmph . what distance will it cover in 12 sec ? | "speed = 120 kmph = 120 * 5 / 18 = 33 m / s distance covered in 12 sec = 33 * 12 = 396 m answer is a" | a ) 396 m , b ) 267 m , c ) 180 m , d ) 200 m , e ) 250 m | a | multiply(divide(120, const_3_6), 12) | divide(n0,const_3_6)|multiply(n1,#0)| | physics |
jack and christina are standing 360 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 5 feet per second and christina walks in a straight line toward jack ... | "the relative speed of jack and christina is 5 + 7 = 12 feet per second . the distance between them is 210 feet , hence they will meet in ( time ) = ( distance ) / ( relative speed ) = 360 / 12 = 30 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 12 * 3... | a ) 270 , b ) 300 , c ) 330 , d ) 360 , e ) 390 | d | multiply(divide(360, add(5, 7)), 12) | add(n1,n2)|divide(n0,#0)|multiply(n3,#1)| | physics |
rs . 6000 is lent out in two parts . one part is lent at 6 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 6 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 6 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 4 x / 100 = > 600 - 1 / 25 x = 450 = > x = 3750 amount lent at 10 % = 2250 required ratio = 3750 : 2250 = 75 : 45 = 15 : 9 = 5 : 3 answer : e" | a ) 5 : 1 , b ) 5 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 5 : 3 | e | divide(divide(subtract(multiply(450, const_100), multiply(6000, 6)), subtract(10, 6)), divide(subtract(multiply(450, const_100), multiply(6000, 6)), subtract(10, 6))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)| | gain |
average age of students of an adult school is 52 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | "explanation : let the original no . of students be x . according to situation , 52 x + 120 * 32 = ( x + 120 ) 36 β x = 30 so , required no . of students after joining the new students = x + 120 = 150 answer : b" | a ) 1200 , b ) 150 , c ) 360 , d ) 240 , e ) none of these | b | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(52, add(32, 4))), 4) | add(n2,n3)|multiply(n1,n2)|multiply(n1,#0)|subtract(n0,#0)|subtract(#2,#1)|divide(#4,#3)|multiply(n3,#5)| | general |
the h . c . f and l . c . m of two numbers are 11 and 385 respectively . if one number lies between 75 and 125 , then that number is | "explanation : product of numbers = 11 x 385 = 4235 let the numbers be 11 a and 11 b . then , 11 a x 11 b = 4235 inline fn _ jvn rightarrow ab = 35 now , co - primes with product 35 are ( 1,35 ) and ( 5,7 ) so , the numbers are ( 11 x 1 , 11 x 35 ) and ( 11 x 5 , 11 x 7 ) since one number lies 75 and 125 , the suitable... | a ) 77 , b ) 88 , c ) 99 , d ) 110 , e ) 120 | a | multiply(sqrt(divide(multiply(11, 385), 125)), 125) | multiply(n0,n1)|divide(#0,n3)|sqrt(#1)|multiply(n3,#2)| | physics |
a train running at the speed of 54 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 54 * ( 5 / 18 ) m / sec = 15 m / sec length of train ( distance ) = speed * time ( 15 ) * 9 = 135 meter answer : c" | a ) 150 meter , b ) 299 meter , c ) 135 meter , d ) 155 meter , e ) 144 meter | c | multiply(divide(multiply(54, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
simplify 3 / 4 of 6 0 β 8 / 5 of 60 + ? = 12 | 60 Γ 3 / 4 = 45 , 60 Γ 8 / 5 = 96 45 β 96 + ? = 12 = > 96 + 12 = 108 β 45 = 63 [ 63 + 45 = 108 β 96 = 12 ] option d | a ) 48 , b ) 24 , c ) 56 , d ) 63 , e ) 14 | d | add(negate(subtract(multiply(divide(3, 4), 60), multiply(divide(8, 5), 60))), 12) | divide(n0,n1)|divide(n4,n5)|multiply(n6,#0)|multiply(n6,#1)|subtract(#2,#3)|negate(#4)|add(n7,#5) | general |
a certain junior class has 1000 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "there are 60 siblings in junior class and 60 their pair siblings in the senior class . we want to determine probability of choosing one sibling from junior class and its pair from senior . what is the probability of choosing any sibling from junior class ? 60 / 1000 ( as there are 60 of them ) . what is the probabilit... | a ) 3 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | divide(n3,const_3)| | probability |
what is the average ( arithmetic mean ) of all the multiples of 10 from 10 to 160 inclusive ? | 10 and 190 inclusive would mean there are 15 terms . no need to add up all the terms and further divide by 15 since the terms are evenly spaced in ascending order from 10 , 20 , 30 . . . . . . . . 160 the middle term is the average which is the tenth term = 70 c is the answer . | a ) 90 , b ) 95 , c ) 70 , d ) 105 , e ) 110 | c | divide(subtract(160, multiply(10, const_2)), const_2) | multiply(n0,const_2)|subtract(n2,#0)|divide(#1,const_2) | general |
michael earns $ 7.00 per hour for the first 40 hours he works per week , and twice this rate for overtime . if michael earned $ 320 last week , how many hours did he work ? | $ 7 * 40 + $ 12 * x = $ 320 - - > x = 3 hours . total working hours = 40 + 3 = 43 . answer : a . | a ) 43 , b ) 44 , c ) 45 , d ) 46 , e ) 47 | a | add(divide(40, multiply(7, const_2)), 40) | multiply(n0,const_2)|divide(n1,#0)|add(n1,#1) | physics |
if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 2 years the amount to be paid is ? | "explanation : a = 7500 ( 26 / 25 ) 2 = 8112 answer is d" | a ) rs . 8082 , b ) rs . 7800 , c ) rs . 8100 , d ) rs . 8112 , e ) rs . 9000 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | add(n1,const_100)|divide(#0,const_100)|multiply(#1,#1)|multiply(n0,#2)| | gain |
steve traveled the first 2 hours of his journey at 50 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ? | "distance traveled in 2 hours = 2 * 50 = 100 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 100 = 340 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 340 / 5 = 68 mph answer : a" | a ) 68 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(80, 3), multiply(50, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | add(n0,n2)|divide(const_100,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#2,#3)|subtract(#1,const_0_33)|divide(#4,#0)|add(#6,#5)| | physics |
pat , kate , and mark charged a total of 153 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much time as mark , how many more hours did mark charge to the project than kate ? | "85 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 153 = > x = 17 34 : 17 : 102 102 - 17 = 85 answer a" | a ) 85 , b ) 36 , c ) 72 , d ) 90 , e ) 108 | a | subtract(divide(153, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(153, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2))) | divide(n1,n2)|multiply(n2,const_2)|add(n1,#0)|divide(n1,#1)|add(#2,#3)|divide(n0,#4)|divide(#5,#1)|subtract(#5,#6)| | general |
if d = 1 / ( 2 ^ 3 * 5 ^ 8 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | "another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 8 ) = multiply and divide by 2 ^ 5 = 2 ^ 5 / ( 2 ^ 3 * 2 ^ 5 * 5 ^ 8 ) = 2 ^ 5 / 10 ^ 8 = > non zero digits are 32 = > ans b" | a ) one , b ) two , c ) three , d ) seven , e ) ten | b | add(1, 2) | add(n0,n1)| | general |
the arithmetic mean of 5 negative integers is - 100 . if the average of 3 of these integers is - 90 , what is the least possible value that one of the other 2 integers can have ? | the arithmetic mean of 5 negative integers is ( a + b + c + d + e ) / 5 = - 100 so , the sum of 5 negative integers would be a + b + c + d + e = - 100 * 5 = - 500 similarly , the sum of 3 negative integers would be a + b + c = - 90 * 3 = - 270 thus , the sum of the other 2 integers is - 500 + 270 = - 230 . since we kno... | a ) 229 , b ) - 1 , c ) - 229 , d ) 90 , e ) - 90 | c | add(add(subtract(90, subtract(multiply(5, 90), 100)), divide(90, 3)), const_1) | divide(n3,n2)|multiply(n0,n3)|subtract(#1,n1)|subtract(n3,#2)|add(#0,#3)|add(#4,const_1) | general |
rs . 3150 is divided so that 5 times the first share , thrice the 2 nd share and six times the third share amount to the same . what is the value of the third share ? | "a + b + c = 3150 5 a = 2 b = 6 c = x a : b : c = 1 / 5 : 1 / 2 : 1 / 6 = 6 : 15 : 5 5 / 21 * 3150 = rs 750 answer : d" | a ) s 529 , b ) s 527 , c ) s 570 , d ) s 750 , e ) s 507 | d | multiply(5, divide(3150, add(add(5, 2), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general |
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 94 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 = 94 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 = 90 kms relative speed ( opposite side ) ( as they are moving towards each other speed w... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10 | e | divide(add(94, 6), add(4, 6)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | physics |
excluding stoppages , the speed of a train is 48 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ? | "t = 8 / 48 * 60 = 10 answer : d" | a ) 16 , b ) 17 , c ) 15 , d ) 10 , e ) 12 | d | subtract(const_60, multiply(const_60, divide(40, 48))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
if a child walks at the rate of 5 m / min from his home , he is 6 minutes late for school , if he walks at the rate of 7 m / min he reaches half an hour earlier . how far is his school from his home ? | let the distance between home and school is x . let actual time to reach be t . thus , x / 5 = t + 6 - - - - ( 1 ) and x / 7 = t - 30 - - - - - ( 2 ) solving equation 1 and 2 x = 630 min answer : c | a ) 450 min , b ) 540 min , c ) 630 min , d ) 360 min , e ) none of these | c | divide(add(multiply(const_60, divide(const_1, const_2)), 6), subtract(divide(const_1, 5), divide(const_1, 7))) | divide(const_1,const_2)|divide(const_1,n0)|divide(const_1,n2)|multiply(#0,const_60)|subtract(#1,#2)|add(n1,#3)|divide(#5,#4) | physics |
evaluate : | 9 - 8 ( 3 - 12 ) | - | 5 - 11 | = ? | according to order of operations , inner brackets first . hence | 9 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 9 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 9 + 72 | - | 5 ... | a ) 75 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | subtract(subtract(9, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | subtract(n2,n3)|subtract(n4,n5)|multiply(n1,#0)|negate(#1)|subtract(n0,#2)|subtract(#4,#3) | general |
in town x , 64 percent of the population are employed , and 50 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 50 are employed males , hence 14 % are employed females . ( employed females ) / ( total employed people ) = 14 / 64 = 22 % answer : b ." | a ) 16 % , b ) 22 % , c ) 32 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(64, 50), 64), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a train passes a station platform in 35 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 35 = 15 = > x = 225 m . answer : c" | a ) 228 , b ) 240 , c ) 225 , d ) 166 , e ) 1811 | c | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
the cash difference between the selling prices of an book at a profit of 5 % and 10 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 105 % of x / 110 % of x = 105 / 110 = 21 / 22 = 21 : 22 a" | a ) 21 : 22 , b ) 52 : 33 , c ) 52 : 53 , d ) 45 : 53 , e ) 51 : 34 | a | divide(add(const_100, 5), add(const_100, 10)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
what is the greatest prime factor of 3 ^ 6 - 1 ? | "3 ^ 6 - 1 = ( 3 ^ 3 ) ^ 2 - 1 = ( 27 ^ 2 ) - 1 dividing ( 27 ^ 2 ) - 1 by 13 will give us a reminder of 0 ( hint : ( 2 * 13 + 1 ) ^ 2 - 1 / 13 = ( reminder 1 ) - 1 = 0 hence the greatest prime factor must be 13 . answer : d" | a ) 2 , b ) 3 , c ) 7 , d ) 13 , e ) 17 | d | floor(divide(3, divide(6, const_2))) | divide(n1,const_2)|divide(n0,#0)|floor(#1)| | general |
village p β s population is 1200 greater than village q ' s population . if village q β s population were reduced by 800 people , then village p β s population would be 6 times as large as village q ' s population . what is village q ' s current population ? | "p = q + 1200 . p = 6 ( q - 800 ) . 6 ( q - 800 ) = q + 1200 . 5 q = 6000 . q = 1200 . the answer is d ." | a ) 1050 , b ) 1100 , c ) 1150 , d ) 1200 , e ) 1250 | d | divide(add(1200, multiply(800, 6)), const_3) | multiply(n1,n2)|add(n0,#0)|divide(#1,const_3)| | general |
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 |
how many liters of water must be added to 18 liters of milk and water containing 10 % water to make it 40 % water ? | "by rule of alligation : 40 % - 10 % = 30 % 100 % - 40 % = 60 % quantity of pure water : quantity of the mixture = 1 : 2 there are 18 liters of mixture , so we need to add 9 liters of pure water . the answer is d ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | multiply(divide(18, 10), divide(40, 10)) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)| | gain |
dacid obtained 96 , 95 , 82 , 97 and 95 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 96 + 95 + 82 + 97 + 95 ) / 5 = 465 / 5 = 93 . answer : b" | a ) 79 , b ) 93 , c ) 88 , d ) 88 , e ) 75 | b | divide(add(add(add(add(96, 95), 82), 97), 95), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
the number 0.5 is how much greater than 1 / 6 ? | "let x be the difference then . 5 - 1 / 3 = x 5 / 10 - 1 / 3 = x x = 1 / 3 ans a" | a ) 1 / 3 , b ) 2 / 10 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | a | subtract(0.5, divide(1, 6)) | divide(n1,n2)|subtract(n0,#0)| | general |
there are 22 students in a class . in how many different ways can a committee of 3 students be formed ? | "22 c 3 = 22 * 21 * 20 / 6 = 1540 the answer is c ." | a ) 1180 , b ) 1320 , c ) 1540 , d ) 1760 , e ) 1920 | c | multiply(subtract(const_1, divide(3, 22)), 22) | divide(n1,n0)|subtract(const_1,#0)|multiply(#1,n0)| | probability |
a 70 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 70 = 20 answer : b" | a ) 12 , b ) 20 , c ) 88 , d ) 77 , e ) 14 | b | subtract(70, divide(70, add(divide(2, 5), const_1))) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | physics |
the slant height of a cone is 20 cm and radius of the base is 10 cm , find the curved surface of the cone . | "Ο * 10 * 20 = 628 answer : e" | a ) 330 , b ) 159 , c ) 150 , d ) 200 , e ) 628 | e | multiply(multiply(const_pi, 10), 20) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
evaluate : 6602.5 + 660.25 + 66.025 + 6.6025 + 0.66025 | "6602.5 660.25 66.025 6.6025 + 0.66025 - - - - - - - - - - - - - - - 7336.03775 answer is b ." | a ) 6819.59775 , b ) 7336.03775 , c ) 6918.59775 , d ) 6198.59775 , e ) 6891.59775 | b | subtract(6602.5, multiply(multiply(660.25, 66.025), 6.6025)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
the least number which when divided by 3 , 4 , 5 and 6 leaves a remainder 6 , but when divided by 9 leaves no remainder , is | "explanation : l . c . m of 3 , 4 , 5 , 6 = 60 therefore required number is of the form 60 k + 6 . least value of k for which ( 60 k + 6 ) is divisible by 9 is k = 2 therefore required number = ( 60 x 2 + 6 ) = 126 . answer : b" | a ) 117 , b ) 126 , c ) 252 , d ) 336 , e ) none of these | b | multiply(lcm(lcm(lcm(3, 4), 5), 6), const_2) | lcm(n0,n1)|lcm(n2,#0)|lcm(n3,#1)|multiply(#2,const_2)| | general |
the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 2... | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 20 minutes after it was poured is : f = 120 ( 2 ^ - 20 / 10 ) + 60 f = 120 * 1 / 4 + 60 f = 30 + 60 f = 90 e" | a ) 65 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | e | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general |
the ratio 10 : 20 expressed as percent equals to | "explanation : actually it means 10 is what percent of 20 , which can be calculated as , ( 10 / 20 ) * 100 = 10 * 5 = 50 answer : option a" | a ) 50 % , b ) 80 % , c ) 40 % , d ) 90 % , e ) 100 % | a | multiply(divide(10, 20), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the speed of a train is 120 kmph . what is the distance covered by it in 40 minutes ? | "120 * 40 / 60 = 80 kmph answer : d" | a ) 50 kmph , b ) 60 kmph , c ) 70 kmph , d ) 80 kmph , e ) 85 kmph | d | multiply(divide(40, const_60), 120) | divide(n1,const_60)|multiply(n0,#0)| | physics |
a right triangle aec has to be constructed in the xy - plane so that the right angle is at a and ae is parallel to x axis . the coordinates of a , e and c are integers and satisfy the inequalities - 1 β€ x β€ 7 and 1 β€ y β€ 7 . how many different triangles can be constructed with these properties ? | all the cordinates are integer hence , possible ae values are { - 10 } , { - 11 } . . . . { - 17 } : 8 ways . . . . { 7 , - 1 } , { 70 } . . . . . . { 76 } : 8 ways 9 * 8 ways = 72 for ac values can be { 12 } , { 13 } . . . . { 17 } : 6 ways . . . . . . { 71 } , { 72 } . . . . . { 76 } : 6 ways 7 * 6 = 42 total = 72 * ... | a ) 63 , b ) 336 , c ) 567 , d ) 3024 , e ) 5040 | d | subtract(factorial(7), multiply(multiply(multiply(subtract(7, 1), const_4), subtract(7, 1)), multiply(7, const_2))) | factorial(n1)|multiply(n1,const_2)|subtract(n1,n0)|multiply(#2,const_4)|multiply(#3,#2)|multiply(#4,#1)|subtract(#0,#5) | geometry |
if a ^ 2 + b ^ 2 = 25 and ab = 10 , what is the value of the expression ( a - b ) ^ 2 + ( a + b ) ^ 2 ? | "( a - b ) ^ 2 = a ^ 2 + b ^ 2 - 2 ab = 25 - 20 = 5 ( a + b ) ^ 2 = a ^ 2 + b ^ 2 + 2 ab = 25 + 20 = 45 so ( a + b ) ^ 2 + ( a - b ) ^ 2 = 45 + 5 = 50 d" | a ) 10 , b ) 20 , c ) 30 , d ) 50 , e ) 70 | d | add(multiply(sqrt(25), const_10), const_3) | sqrt(n2)|multiply(#0,const_10)|add(#1,const_3)| | general |
if 25 typists can type 60 letters in 20 minutes , then how many letters will 75 typists working at the same rate complete in 1 hour ? | no . of letters typing by 25 typists in 20 minutes = 60 no . of letters typing by 25 typists in 60 minutes = 60 * 3 = 180 no . of letters typing by 75 typists in 60 minutes = 180 / 25 * 75 = 540 answer : b | a ) 270 , b ) 540 , c ) 610 , d ) 510 , e ) 500 | b | multiply(divide(multiply(60, const_3), 25), 75) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1) | physics |
a shopkeeper labeled the price of his articles so as to earn a profit of 20 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ? | "explanation : let the cp of the article = rs . 100 . then labeled price = rs . 120 . sp = rs . 120 - 10 % of 120 = rs . 120 - 13 = rs . 118 . gain = rs . 118 Γ’ β¬ β rs . 100 = rs . 18 therefore , gain / profit percent = 18 % . answer : option d" | a ) 18 % , b ) 20 % , c ) 17 % , d ) 19 % , e ) none of these | d | subtract(subtract(add(const_100, 20), multiply(add(const_100, 20), divide(10, const_100))), const_100) | add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(#0,#2)|subtract(#3,const_100)| | gain |
a person can row at 9 kmph and still water . he takes 6 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 6.5 = > x / 10 + x / 8 = 13 / 2 = > ( 4 x + 5 x ) / 40 = 13 / 2 = > x = 29 km . answer : c" | a ) 32 , b ) 28 , c ) 29 , d ) 54 , e ) 20 | c | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 6), add(add(9, 1), subtract(9, 1))) | add(n0,n2)|subtract(n0,n2)|add(#0,#1)|multiply(#0,#1)|multiply(n1,#3)|divide(#4,#2)| | physics |
in the junior basketball league there are 18 teams , 2 / 3 of them are bad and Β½ are rich . what ca n ' t be the number of teams that are rich and bad | "total teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . ans d ." | a ) 4 . , b ) 6 . , c ) 7 . , d ) 10 . , e ) 8 . | d | add(multiply(18, divide(const_1, 2)), const_1) | divide(const_1,n1)|multiply(n0,#0)|add(#1,const_1)| | general |
the sum of the present age of henry and jill is 43 . what is their present ages if 5 years ago henry was twice the age of jill ? | "let the age of jill 5 years ago be x , age of henry be 2 x x + 5 + 2 x + 5 = 43 x = 11 present ages will be 16 and 27 answer : a" | a ) and 27 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | a | subtract(43, divide(add(43, 5), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 84 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to Β΅ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 84 % is one st . dev , then on either side of m we have 84 / 2 = 42 % . so , 84 % are to the right and left of m ( = 50 % ) . in other words , our value ... | a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | e | subtract(const_100, divide(subtract(const_100, 84), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general |
the compound ratio of 5 : 8 , 3 : 2 and 4 : 5 ? | "5 / 8 * 3 / 2 * 4 / 5 = 3 / 4 1 : 1 answer : c" | a ) 1 : 1 , b ) 1 : 87 , c ) 3 : 4 , d ) 1 : 9 , e ) 1 : 2 | c | divide(divide(multiply(5, 3), multiply(8, 2)), divide(multiply(3, 4), multiply(2, 5))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)| | other |
by selling 14 pencils for a rupee a man loses 22 % . how many for a rupee should he sell in order to gain 22 % ? | "88 % - - - 14 122 % - - - ? 88 / 122 * 14 = 10 answer : a" | a ) 10 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | a | multiply(divide(const_1, multiply(add(const_100, 22), divide(const_1, subtract(const_100, 22)))), 14) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
i travel the first part of my journey at 40 kmph and the second part at 60 kmph and cover the total distance of 240 km to my destination in 5 hours . how long did the first part of my journey last | the total time of journey = 5 hours . let ' x ' hours be the time that i travelled at 40 kmph therefore , 5 β x hours would be time that i travelled at 60 kmph . hence , i would have covered x Γ 40 + ( 5 β x ) 60 kms in the 5 hours = 240 kms . solving , for x in the equation 40 x + ( 5 β x ) Γ 60 = 240 , we get 40 x + ... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(multiply(60, 5), 240), subtract(60, 40)) | multiply(n1,n3)|subtract(n1,n0)|subtract(#0,n2)|divide(#2,#1) | physics |
if 4 cats can kill 4 rats in 4 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 4 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 4 minutes , so 100 cats can kill 100 rats in 4 minutes answer d" | a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 5 minutes | d | multiply(4, const_1) | multiply(n0,const_1)| | physics |
how many seconds will a 500 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 75 km / hr ? | "speed of train relative to man = 75 - 3 = 72 km / hr . = 72 * 5 / 18 = 20 m / sec . time taken to pass the man = 500 * 1 / 20 = 25 sec . answer : a" | a ) 25 , b ) 30 , c ) 99 , d ) 88 , e ) 61 | a | divide(500, multiply(subtract(75, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if the volume of the cube is 729 , then the surface area of the cube will be | explanation : \ inline \ fn _ jvn a ^ 3 = 729 ; a = 9 surface area = ( 6 x 9 x 9 ) = 486 \ inline \ fn _ jvn cm ^ 2 answer : a ) 486 sq . cm | ['a ) 486', 'b ) 255', 'c ) 2866', 'd ) 265', 'e ) 872'] | a | multiply(multiply(const_3, const_2), power(cube_edge_by_volume(729), const_2)) | cube_edge_by_volume(n0)|multiply(const_2,const_3)|power(#0,const_2)|multiply(#1,#2) | geometry |
when the circumference and area of a circle are numerically equal , then the diameter is numerically equal to | according to question , circumference of circle = area of circle or Ο d = Ο ( d β 2 ) 2 [ where d = diameter ] β΄ d = 4 answer c | ['a ) area', 'b ) circumference', 'c ) 4', 'd ) 2 Ο', 'e ) none of these'] | c | multiply(divide(multiply(const_2, const_pi), const_pi), const_2) | multiply(const_2,const_pi)|divide(#0,const_pi)|multiply(#1,const_2) | geometry |
if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 49 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | ( x + 1 ) ^ 2 < = 49 x < = 6 x > = - 8 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 8 Γ - 6 = 48 minimum possible value of xy is - 8 Γ 8 = - 64 - 64 + 48 = - 16 answer : a | a ) - 16 , b ) - 14 , c ) 0 , d ) 14 , e ) 16 | a | add(sqrt(49), sqrt(64)) | sqrt(n2)|sqrt(n5)|add(#0,#1) | general |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , the number of valid votes that the other candidate got , was | solution number of valid votes = 80 % of 7500 = 6000 . valid votes polled by other candidates = 45 % of 6000 ( 45 / 100 Γ 6000 ) = 2700 . answer b | a ) 2500 , b ) 2700 , c ) 3000 , d ) 3100 , e ) nobe | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4) | gain |
find the circumference and area of radius 6 cm . | area of circle = Ο r Β² = 22 / 7 Γ 6 Γ 6 cm Β² = 113 cm Β² answer : a | ['a ) 113 cm Β²', 'b ) 144 cm Β²', 'c ) 154 cm Β²', 'd ) 184 cm Β²', 'e ) 194 cm Β²'] | a | circle_area(6) | circle_area(n0) | geometry |
cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost r is what percent of the original cost ? | original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600 | a ) r = 200 , b ) r = 600 , c ) r = 800 , d ) r = 1600 , e ) r = 50 | d | multiply(power(const_2, 4), const_100) | power(const_2,n0)|multiply(#0,const_100) | general |
the ratio of three numbers is 5 : 3 : 4 and their sum is 108 . the second number of the three numbers is ? | "5 : 3 : 4 total parts = 12 12 parts - - > 108 1 part - - - - > 9 the second number of the three numbers is = 3 3 parts - - - - > 27 b )" | a ) 23 , b ) 27 , c ) 30 , d ) 32 , e ) 34 | b | sqrt(divide(108, add(power(4, 3), add(power(5, 3), power(3, 3))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
there are two concentric circles with radii 5 and 4 . if the radius of the outer circle is increased by 20 % and the radius of the inner circle decreased by 50 % , by what percent does the area between the circles increase ? | the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 25 pi . the area of the small circle is 16 pi . the area a 1 between the circles is 9 pi . when the big circle ' s radius increases , the new area is 36 pi . when the small circle ' s radius decreases , the new area is 4 pi . the ar... | a ) 56 , b ) 156 , c ) 256 , d ) 356 , e ) 456 | c | multiply(divide(subtract(subtract(power(multiply(5, divide(add(const_100, 20), const_100)), const_2), power(multiply(4, divide(subtract(const_100, 50), const_100)), const_2)), subtract(power(5, const_2), power(4, const_2))), subtract(power(5, const_2), power(4, const_2))), const_100) | add(n2,const_100)|power(n0,const_2)|power(n1,const_2)|subtract(const_100,n3)|divide(#0,const_100)|divide(#3,const_100)|subtract(#1,#2)|multiply(n0,#4)|multiply(n1,#5)|power(#7,const_2)|power(#8,const_2)|subtract(#9,#10)|subtract(#11,#6)|divide(#12,#6)|multiply(#13,const_100) | gain |
a sum of money at simple interest amounts to rs . 815 in 2 years and to rs . 854 in 3 years . the sum is : | "sol . s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 2 years = rs . ( 39 * 2 ) = rs . 78 . Γ’ Λ Β΄ principal = rs . ( 815 - 78 ) = rs . 737 . answer d" | a ) 500 , b ) 698 , c ) 780 , d ) 737 , e ) none | d | subtract(815, divide(multiply(subtract(854, 815), 2), 3)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
the ratio of the area of a square to that of the square drawn on its diagonal is ? | explanation : a 2 : ( a Γ’ Λ Ε‘ 2 ) 2 a 2 : 2 a 2 - > 1 : 2 answer is d | ['a ) 2 : 5', 'b ) 3 : 4', 'c ) 3 : 5', 'd ) 1 : 2', 'e ) 2 : 3'] | d | power(divide(const_1, sqrt(const_2)), const_2) | sqrt(const_2)|divide(const_1,#0)|power(#1,const_2) | geometry |
a rectangular floor that measures 8 meters by 10 meters is to be covered with carpet squares that each measure 2 meters by 2 meters . if the carpet squares cost $ 10 apiece , what is the total cost for the number of carpet squares needed to cover the floor ? | "the width of the rectangular floor ( 8 m ) is a multiple of one side of the square ( 2 m ) , and the length of the floor ( 10 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 8 / 2 ) * ( 10 / 2 ) = 20 . the total cost is 20 * 10 = $ 200 . the answer is , therefore , a... | a ) $ 200 , b ) $ 240 , c ) $ 480 , d ) $ 960 , e ) $ 1,920 | a | multiply(8, 8) | multiply(n0,n0)| | geometry |
there are 6 positive and 8 negative numbers . 4 numbers are choosen at random and multiplied . the probability that the product is positive is : | 6 c 4 / 14 c 4 + ( 6 c 2 * 8 c 2 ) / 14 c 4 + 8 c 4 / 14 c 4 = 505 / 1001 case 1 : only 4 positive no . case 2 : 2 positive and 2 negative no . case 3 : 4 negative no . answer : e | a ) 500 / 1001 , b ) 503 / 1001 , c ) 303 / 1001 , d ) 301 / 1001 , e ) 505 / 1001 | e | divide(add(add(divide(factorial(6), multiply(factorial(4), factorial(const_2))), divide(factorial(8), multiply(factorial(4), factorial(4)))), multiply(divide(factorial(6), multiply(factorial(4), factorial(const_2))), divide(factorial(8), multiply(factorial(6), factorial(const_2))))), divide(factorial(add(6, 8)), multip... | add(n0,n1)|factorial(n0)|factorial(n2)|factorial(const_2)|factorial(n1)|factorial(#0)|multiply(#2,#3)|multiply(#2,#2)|multiply(#1,#3)|subtract(#0,n2)|divide(#1,#6)|divide(#4,#7)|divide(#4,#8)|factorial(#9)|add(#10,#11)|multiply(#10,#12)|multiply(#13,#2)|add(#14,#15)|divide(#5,#16)|divide(#17,#18) | general |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 35 % , then find the population of city atthe end of the year 2007 | required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 35 / 100 ) = 1009125 c | a ) 976374 , b ) 979923 , c ) 1009125 , d ) 2356677 , e ) 1083875 | c | multiply(1000000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6) | gain |
f ( x ) is a function such that f ( x ) + 3 f ( 8 - x ) = x for all real numbers x . find the value of f ( 2 ) . | f ( x ) + 3 f ( 8 - x ) = x : given f ( 2 ) + 3 f ( 6 ) = 2 : x = 2 above f ( 6 ) + 3 f ( 2 ) = 6 : x = 6 above f ( 6 ) = 6 - 3 f ( 2 ) : solve equation c for f ( 6 ) f ( 2 ) + 3 ( 6 - 3 f ( 2 ) ) = 2 : substitute f ( 2 ) = 2 : solve above equation . correct answer is d ) 2 | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | divide(subtract(2, multiply(multiply(3, 3), const_2)), subtract(const_1, multiply(3, 3))) | multiply(n0,n0)|multiply(#0,const_2)|subtract(const_1,#0)|subtract(n2,#1)|divide(#3,#2) | general |
a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 105 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 Γ x + . 50 Γ x + . 25 x = 105 β 1.75 x = 105 β x = 60 coins answer d" | a ) 20 coins , b ) 30 coins , c ) 40 coins , d ) 60 coins , e ) none of these | d | divide(105, add(add(inverse(const_4), inverse(const_2)), const_1)) | inverse(const_4)|inverse(const_2)|add(#0,#1)|add(#2,const_1)|divide(n2,#3)| | general |
john can complete a given task in 20 days . jane will take only 10 days to complete the same task . john and jane set out to complete the task by beginning to work together . however , jane was indisposed 5 days before the work got over . in how many days did the work get over from the time john and jane started to wor... | in such questions , you need to start from the end . last 5 days john works alone and completes 5 * ( 1 / 20 ) = 1 / 4 of the work . so 3 / 4 of the work should have been completed by the two of them together before jane left . their combined rate of work is 1 / 20 + 1 / 10 = 3 / 20 time taken to complete 3 / 4 of the ... | a ) 8 , b ) 10 , c ) 12 , d ) 16 , e ) 20 | b | add(5, divide(subtract(multiply(20, 10), multiply(10, 5)), add(10, 20))) | add(n0,n1)|multiply(n0,n1)|multiply(n1,n2)|subtract(#1,#2)|divide(#3,#0)|add(n2,#4) | physics |
a man sold 20 articles for $ 60 and gained 10 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 10 % ) / 20 = $ 2.70 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 / $ 2.70 = 40 thus , solution c is correct ." | a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | c | divide(original_price_before_loss(10, 90), divide(original_price_before_gain(20, 60), 20)) | original_price_before_gain(n2,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2)| | gain |
list k consists of 10 consecutive integers . if - 4 is the least integer in list k , what is the range of the positive integers in list k ? | "answer = e = 4 if least = - 4 , then largest = 5 range = 5 - 1 = 4" | a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 4 | e | subtract(subtract(subtract(10, 4), const_1), const_1) | subtract(n0,n1)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
in a school of 400 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "44 + 28 + 10 = 82 % 100 β 82 = 18 % 400 * 18 / 100 = 72 answer : c" | a ) 41 , b ) 63 , c ) 72 , d ) 63 , e ) 50 | c | divide(multiply(400, subtract(const_100, add(add(44, 28), 10))), const_100) | add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)| | gain |
a train 120 m in length crosses a telegraph post in 6 seconds . the speed of the train is ? | "s = 120 / 6 * 18 / 5 = 72 kmph answer : a" | a ) 72 kmph , b ) 88 kmph , c ) 54 kmph , d ) 18 kmph , e ) 19 kmph | a | multiply(const_3_6, divide(120, 6)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
two boats are heading towards each other at constant speeds of 3 miles / hr and 21 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 3 + 21 = 24 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 24 * 1 / 60 = 3 / 15 miles apart . answer : e ." | a ) 1 / 12 , b ) 5 / 12 , c ) 1 / 6 , d ) 1 / 3 , e ) 3 / 15 | e | divide(add(21, 3), const_60) | add(n0,n1)|divide(#0,const_60)| | physics |
two trains of equal lengths take 12 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 10 + 6 = 16 m / sec . required time = ( 120 + 120 ) / 16 = 15 sec . answer : d" | a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 15 sec , e ) 23 sec | d | divide(multiply(120, const_2), add(speed(120, 20), speed(120, 12))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 20 % . their gross income from sales for one month increases by what percent ? | "let list price of an item = 100 discount on each item = 10 % discounted price of an item = . 9 * 100 = 90 if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 20 % originally if 10 items were sold in a month , with the new discount 12 ... | a ) 2 % , b ) 4 % , c ) 5 % , d ) 8 % , e ) 12 % | d | subtract(subtract(20, 10), divide(20, 10)) | divide(n1,n0)|subtract(n1,n0)|subtract(#1,#0)| | gain |
a reduction of 48 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "explanation : 40 * ( 48 / 100 ) = 19.2 - - - 64 ? - - - 12 = > rs . 3.6 answer : d" | a ) 2.6 , b ) 8.6 , c ) 7.6 , d ) 3.6 , e ) 1.6 | d | multiply(const_12, divide(multiply(48, divide(48, const_100)), 64)) | divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)| | gain |
what is the remainder when 1271 * 1275 * 1279 * 1285 is divided by 12 ? | "1271 * 1275 * 1277 * 1285 = ( 1272 - 1 ) ( 1272 + 3 ) ( 1284 - 5 ) ( 1284 + 1 ) = ( multiple of 12 - 1 ) ( multiple of 12 + 3 ) ( multiple of 12 - 7 ) ( multiple of 12 + 1 ) now , if we expand this expression all terms but the last one will have a multiple of 12 in them , thus will leave no remainder upon division by ... | a ) 0 , b ) 1 , c ) 11 , d ) 3 , e ) 7 | d | reminder(multiply(1275, 1271), 1279) | multiply(n0,n1)|reminder(#0,n2)| | general |
the s . i . on a certain sum of money for 3 years at 8 % per annum is half the c . i . on rs . 4000 for 2 years at 10 % per annum . the sum placed on s . i . is ? | "c . i . = [ 4000 * ( 1 + 10 / 100 ) 2 - 4000 ] = ( 4000 * 11 / 10 * 11 / 10 - 4000 ) = rs . 840 . sum = ( 420 * 100 ) / ( 3 * 8 ) = rs . 1750 answer : c" | a ) 3000 , b ) 2802 , c ) 1750 , d ) 2990 , e ) 2982 | c | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), 2)), 4000), 2), multiply(3, divide(8, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)| | gain |
in plutarch enterprises , 70 % of the employees are marketers , 20 % are engineers , and the rest are managers . marketers make an average salary of $ 60,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | "for sake of ease , let ' s say there are 10 employees : 7 marketers , 2 engineers , and 1 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 7 * $ 60,000 ) and the engineers ( 2 * $ 80,000 ) > > > $ 800,000 - ... | a ) $ 80,000 , b ) $ 130,000 , c ) $ 240,000 , d ) $ 290,000 , e ) $ 220,000 | e | divide(subtract(subtract(multiply(multiply(add(70, 20), const_1000), const_100), multiply(20, multiply(add(70, 20), const_1000))), multiply(multiply(multiply(divide(20, const_2), 20), const_1000), 70)), 20) | add(n0,n1)|divide(n1,const_2)|multiply(#0,const_1000)|multiply(#1,n1)|multiply(#2,const_100)|multiply(#2,n1)|multiply(#3,const_1000)|multiply(n0,#6)|subtract(#4,#5)|subtract(#8,#7)|divide(#9,n1)| | general |
a person buys an article at rs . 460 . at what price should he sell the article so as to make a profit of 18 % ? | "cost price = rs . 460 profit = 18 % of 460 = rs . 82 selling price = cost price + profit = 460 + 82 = 542 answer : a" | a ) 542 , b ) 882 , c ) 772 , d ) 662 , e ) 521 | a | add(460, multiply(460, divide(18, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
if an integer n is to be chosen at random from the integers 1 to 86 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? | "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 86 , there are 43 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( ... | a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 55 / 86 , e ) 3 / 4 | d | divide(add(divide(86, 2), divide(86, 8)), 86) | divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)| | general |
a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 600 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to c... | let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - 1... | a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | d | divide(multiply(divide(multiply(divide(40, const_100), 600), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1) | general |
the β length of integer x β refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 Γ 2 Γ 3 Γ 5 . ) what is the greatest possible length of integer z if z < 1,000 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 9 = 512 < 1,000 , then the greatest possible length of integer z is 9 . the answer is c ." | a ) 13 , b ) 10 , c ) 9 , d ) 7 , e ) 6 | c | log(power(2, const_10)) | power(n3,const_10)|log(#0)| | general |
in a group of 6 boys and 4 girls are to be selected . in how many different ways can they be selected such that at least one boy should be there ? | no of boys = 6 no of girls = 4 therefore , no of ways at least 1 boy can be selected = 6 c 4 * 4 c 0 + 6 c 3 * 4 c 1 + 6 c 2 * 4 c 2 + 6 c 1 * 4 c 3 = 15 + 80 + 90 + 24 = 209 ans - a | a ) 209 , b ) 200 , c ) 210 , d ) 225 , e ) 221 | a | divide(factorial(add(6, 4)), multiply(factorial(4), factorial(6))) | add(n0,n1)|factorial(n1)|factorial(n0)|factorial(#0)|multiply(#1,#2)|divide(#3,#4) | other |
the youngest of 4 children has siblings who are 1 , 4 , and 7 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 17 , what is the age of the youngest sibling ? | "x + ( x + 1 ) + ( x + 4 ) + ( x + 8 ) = 68 4 x + 12 = 68 4 x = 56 x = 14 the answer is e ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | divide(subtract(multiply(4, 17), add(add(4, 4), 7)), 4) | add(n0,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
a , b , c weighted separately 1 st like a , b , c , then a & b , then b & c , then c & a at last a & b & c , the last weight was 167 , then what will be the average weight of the 7 reading ? | the sum of 7 weights = weights of [ a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] = weights of [ 4 * ( a + b + c ) = 4 * 167 = 668 therefore the average weight of 7 readings is 668 / 7 = 95.428 answer : b | a ) 95 , b ) 95.42 , c ) 95.45 , d ) 94 , e ) 93 | b | divide(multiply(const_4, 167), 7) | multiply(n1,const_4)|divide(#0,n2) | general |
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