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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x , y , and z are consecutive numbers and x > y > z . also , 2 x + 3 y + 3 z = 5 y + 8 . what is the value of z ? | if x , y , and z are consecutive numbers and x > y > z , then y = z + 1 and x = z + 2 . 2 x + 3 y + 3 z = 5 y + 8 2 z + 4 + 3 z + 3 + 3 z = 5 z + 5 + 8 3 z = 6 z = 2 the answer is a . | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(subtract(8, subtract(multiply(const_4, const_2), multiply(const_2, const_3))), const_3) | multiply(const_2,const_4)|multiply(const_2,const_3)|subtract(#0,#1)|subtract(n4,#2)|divide(#3,const_3) | general |
in the coordinate plane , points ( x , 6 ) and ( 10 , y ) are on line k . if line k passes through the origin and has slope 1 / 2 , then x * y = | "line k passes through the origin and has slope 1 / 2 means that its equation is y = 1 / 2 * x . thus : ( x , 6 ) = ( 12 , 6 ) and ( 10 , y ) = ( 10,5 ) - - > x * y = 12 * 5 = 60 . answer : c" | a ) 52 , b ) 46 , c ) 60 , d ) 50 , e ) 35 | c | multiply(multiply(10, 2), divide(6, 2)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general |
a vessel of capacity 3 litre has 25 % of alcohol and another vessel of capacity 5 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "25 % of 3 litres = 0.75 litres 40 % of 5 litres = 2 litres therefore , total quantity of alcohol is 2.75 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 27.5 % c" | a ) 31 % . , b ) 71 % . , c ) 27.5 % . , d ) 29 % . , e ) 51 % . | c | multiply(divide(add(multiply(divide(25, const_100), 3), multiply(divide(40, const_100), 5)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general |
a jar contains 18 balls . 3 blue balls are removed from the jar and not replaced . now the probability of getting a blue ball is 1 / 5 then how many blue balls jar contains initially ? | x / 15 = 1 / 5 x = 3 3 + 3 ( removed 3 blue balls ) = 6 answer : e | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(3, multiply(subtract(18, 3), divide(1, 5))) | divide(n2,n3)|subtract(n0,n1)|multiply(#0,#1)|add(n1,#2) | probability |
a rectangular grassy plot 110 m by 65 cm has a gravel path . 5 cm wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . mt ? | area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = ( 110 - 5 ) * ( 65 - 5 ) = 6300 sq m area of the path = 7150 - 6300 = 850 sq m cost of gravelling the path = 850 * 80 / 100 = 680 rs answer : a | a ) 680 rs , b ) 540 rs , c ) 480 rs , d ) 640 rs , e ) 280 rs | a | multiply(divide(80, const_100), subtract(multiply(110, 65), multiply(subtract(110, 5), subtract(65, 5)))) | divide(n3,const_100)|multiply(n0,n1)|subtract(n0,n2)|subtract(n1,n2)|multiply(#2,#3)|subtract(#1,#4)|multiply(#0,#5) | physics |
a motorcyclist goes from nagpur to delhi , a distance of 600 kms at an average of 60 kmph speed . another man starts from nagpur by car 5 Γ’ Β½ hours after the first , and reaches delhi Γ’ Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | t = 600 / 60 = 10 h t = 10 - 6 = 4 time ratio = 10 : 4 = 5 : 2 speed ratio = 2 : 5 answer : e | a ) 1 : 2 , b ) 1 : 7 , c ) 1 : 9 , d ) 1 : 5 , e ) 2 : 5 | e | divide(60, divide(600, subtract(subtract(subtract(divide(600, 60), 5), inverse(const_2)), inverse(const_2)))) | divide(n0,n1)|inverse(const_2)|subtract(#0,n2)|subtract(#2,#1)|subtract(#3,#1)|divide(n0,#4)|divide(n1,#5) | physics |
according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 12 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 120 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounc... | a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | d | add(120, multiply(3, const_10)) | multiply(n1,const_10)|add(n4,#0)| | general |
there are 20 balls in a jar . you take out 2 blue balls without putting them back inside , and now the probability of pulling out a blue ball is 1 / 5 . how many blue balls were there in the beginning ? | "12 = 3 blue balls + 18 / 2 answer : a" | a ) 12 . , b ) 9 . , c ) 8 . , d ) 7 . , e ) 6 . | a | add(divide(subtract(20, 2), 5), 2) | subtract(n0,n1)|divide(#0,n3)|add(n1,#1)| | probability |
the volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is : | sol . required volume = volume of a sphere of radius 1 cm = [ 4 / 3 β * 1 * 1 * 1 ] cm Β³ = 4 / 3 β cm Β³ answer b | ['a ) 3 / 4 β', 'b ) 4 / 3 β', 'c ) 5 β', 'd ) 5 / 9 β', 'e ) none'] | b | volume_sphere(1) | volume_sphere(n0) | geometry |
a = { 20 , 1 , - 3 , 6 , - 8 } b = { - 1 , 2 , - 4 , 7 , - 620 } if a is a number that is randomly selected from set a , and b is a number that is randomly selected from set b , what is the probability that ab > 0 ? | for the product of 2 numbers to be positive either both of them must be positive or both of them must be negative : p ( positive , positive ) = 3 / 6 * 3 / 6 = 9 / 36 ; p ( negative , negative ) = 2 / 6 * 3 / 6 = 6 / 36 . p = 9 / 36 + 6 / 36 = 15 / 36 = 5 / 12 . answer : c . | a ) 1 / 4 , b ) 1 / 3 , c ) 5 / 12 , d ) 4 / 9 , e ) 1 / 2 | c | subtract(divide(add(multiply(const_2, const_3), multiply(const_3, const_2)), power(add(1, 4), const_2)), divide(const_2, power(add(1, 4), const_2))) | add(n1,n7)|multiply(const_2,const_3)|add(#1,#1)|power(#0,const_2)|divide(#2,#3)|divide(const_2,#3)|subtract(#4,#5) | general |
if x is 40 percent greater than 88 , then x = | "x = 88 * 1.4 = 123.2 so the answer is c ." | a ) 68 , b ) 70.4 , c ) 123.2 , d ) 105.6 , e ) 108 | c | add(88, multiply(divide(40, const_100), 88)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
the perimeter of an isosceles right triangle is 4 + 4 sq rt 2 . what is the length of the hypotenuse of the triangle ? | "side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 4 + 4 sqrt 2 or , a . ( 2 + sqrt 2 ) = 4 ( 1 + sqrt 2 ) a = 4 . ( 1 + sqrt 2 ) / 2 + sqrt 2 = 4 * 2.414 / 3.414 = then hypotenuse = 2.82 a" | a ) 2.82 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(multiply(4, sqrt(2)), 2) | sqrt(n2)|multiply(n0,#0)|divide(#1,n2)| | geometry |
an 11 - meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | the longer wire will form a square with an area more than 4 if the wire is cut at a point within three meters of either end . the probability of this is 6 / 11 . the answer is e . | ['a ) 3 / 7', 'b ) 5 / 8', 'c ) 3 / 10', 'd ) 5 / 11', 'e ) 6 / 11'] | e | multiply(const_2, divide(subtract(11, square_perimeter(sqrt(4))), 11)) | sqrt(n1)|square_perimeter(#0)|subtract(n0,#1)|divide(#2,n0)|multiply(#3,const_2) | geometry |
how many different positive integers are factors of 36 ? | "6 Γ 6 = 3 ^ 2 Γ 2 ^ 2 so total factors = ( 2 + 1 ) ( 2 + 1 ) = 9 answer : d" | a ) 4 , b ) 6 , c ) 7 , d ) 9 , e ) 11 | d | add(power(const_2, const_2), const_2) | power(const_2,const_2)|add(#0,const_2)| | other |
the list price of an article is rs . 65 . a customer pays rs . 57.33 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "option e explanation : 65 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 57.33 x = 2 %" | a ) 8 % , b ) 7 % , c ) 10 % , d ) 12 % , e ) 2 % | e | multiply(divide(subtract(subtract(65, multiply(65, divide(10, const_100))), 57.33), subtract(65, multiply(65, divide(10, const_100)))), const_100) | divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)| | gain |
the supplementary of an angle is thrice its complimentary . find the angle . | let angle = x ; by question - - 180 - x = 3 ( 90 - x ) x = 45 answer : b | a ) 60 , b ) 45 , c ) 90 , d ) 180 , e ) 120 | b | divide(subtract(multiply(const_3, subtract(const_100, const_10)), multiply(subtract(const_100, const_10), const_2)), const_2) | subtract(const_100,const_10)|multiply(#0,const_3)|multiply(#0,const_2)|subtract(#1,#2)|divide(#3,const_2) | geometry |
9 log 9 ( 2 ) = ? | exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 2 ) = 2 correct answer b | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(log(multiply(9, 9)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0) | other |
there are 2 vessels a & b containing 25 liters each of pure milk and pure water respectively . 5 liters of milk from a is taken and poured into b , then 6 liters of mixture from b is taken and poured in a . what is ratio of water a and b respectively . | at first a cotain milk = 25 water = 0 b cotain milk = 00 water = 25 5 litres of milk taken from a and poured in b then a cotain milk = 20 water = 0 b cotain milk = 5 water = 25 6 litres ( 1 lit milk & 5 lit water ) of mixture from b and poured in a . so , finally a cotain milk = 21 water = 5 b cotain milk = 4 water = 2... | a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 1 : 6 , e ) 1 : 7 | b | divide(multiply(divide(5, add(25, 5)), 6), divide(subtract(25, 5), 5)) | add(n1,n2)|subtract(n1,n2)|divide(n2,#0)|divide(#1,n2)|multiply(n3,#2)|divide(#4,#3) | other |
tim and Γ© lan are 150 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and Γ© lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets Γ© lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice e meet the 2 : 1 ( tim : elan = 100 : 50 ) ratio of maintaining total distance traveled 150 miles socorrect answer e" | a ) 30 miles . , b ) 35 miles . , c ) 45 miles . , d ) 90 miles . , e ) 100 miles . | e | multiply(divide(10, add(5, 10)), 150) | add(n1,n2)|divide(n1,#0)|multiply(n0,#1)| | physics |
16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 4 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 4 girls = 8 Γ£ β 6 / 4 + 12 = 12 + 4 = 16 girls 9 days to complete the work answer : option e" | a ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | e | divide(const_1, add(multiply(4, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16)))) | divide(const_1,n2)|divide(#0,n1)|divide(#0,n0)|multiply(n4,#1)|multiply(n3,#2)|add(#3,#4)|divide(const_1,#5)| | physics |
30 pens and 75 pencils were purchased for 630 . if the average price of a pencil was 2.00 , find the average price of a pen . | "since average price of a pencil = 2 β΄ price of 75 pencils = 150 β΄ price of 30 pens = ( 630 β 150 ) = 480 β΄ average price of a pen = 480 β 60 = 16 answer e" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 16 | e | divide(subtract(630, multiply(75, 2.00)), 30) | multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0)| | general |
a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 1,600 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 8400 = 5 / 84 which is about 5.95 % the answer is e ." | a ) 5.55 % , b ) 5.65 % , c ) 5.75 % , d ) 5.85 % , e ) 5.95 % | e | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain |
find the cost of fencing around a circular field of diameter 34 m at the rate of rs . 2 a meter ? | "2 * 22 / 7 * 17 = 106 106 * 2 = rs . 212 answer : b" | a ) 215 , b ) 212 , c ) 278 , d ) 279 , e ) 222 | b | multiply(circumface(divide(34, const_2)), 2) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
a man covered a certain distance at some speed . if he had moved 3 kmph faster , he would have taken 40 minutes less . if he had moved 2 kmph slower , he would have taken 40 minutes more . what is the the distance in km ? | "let the distance be x km , the speed in which he moved = v kmph time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes β xv β xv + 3 = 4060 β x [ 1 v β 1 v + 3 ] = 23 β x [ v + 3 β vv ( v + 3 ) ] = 23 β 2 v ( v + 3 ) = 9 x . . . . . . . . . . . . . . . . ( equation 1 ) time taken wh... | a ) 33 , b ) 37 , c ) 40 , d ) 38 , e ) 27 | c | multiply(divide(add(multiply(2, divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))), multiply(2, divide(40, const_60))), divide(40, const_60)), divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))) | divide(n1,const_60)|subtract(n0,n2)|multiply(n2,#0)|multiply(n0,#0)|add(#2,#3)|divide(#4,#1)|multiply(n2,#5)|add(#6,#2)|divide(#7,#0)|multiply(#8,#5)| | physics |
if x + y = 280 , x - y = 200 , for integers of x and y , y = ? | "x + y = 280 x - y = 200 2 x = 80 x = 40 y = 240 answer is b" | a ) 200 , b ) 240 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(280, 200), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
there are 10 students named alphabetically from a to j . what is the probability that a and d do not sit together if all 10 sit around a circular table ? | number of students = 10 number of ways 10 students can sit around a circular table = ( 10 - 1 ) ! = 9 ! number of ways a and d sit together ( consider a and d as one entity ) = ( 9 - 1 ) ! = 8 ! * 2 number of ways a and d do not sit together = 9 ! - ( 8 ! * 2 ) probability = ( 9 ! - ( 8 ! * 2 ) ) / 9 ! = 1 - 2 / 9 = 7 ... | a ) 2 / 9 , b ) 2 / 5 , c ) 7 / 9 , d ) 4 / 5 , e ) 8 / 9 | c | subtract(const_1, divide(const_2, 10)) | divide(const_2,n0)|subtract(const_1,#0) | probability |
a pipe can empty 2 / 3 rd of a cistern in 10 mins . in 8 mins , what part of the cistern will be empty ? | "2 / 3 - - - - 10 ? - - - - - 8 = = > 8 / 15 c" | a ) 2 / 3 , b ) 3 / 7 , c ) 8 / 15 , d ) 3 / 8 , e ) 4 / 7 | c | divide(multiply(divide(2, 3), 8), 10) | divide(n0,n1)|multiply(n3,#0)|divide(#1,n2)| | physics |
if $ 10 be allowed as true discount on a bill of $ 110 due at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is : | s . i . on $ ( 110 - 10 ) for a certain time = $ 10 . s . i . on $ 100 for double the time = $ 20 . t . d . on $ 120 = $ ( 120 - 100 ) = $ 20 . t . d . on $ 110 = $ ( 20 / 120 * 100 ) = $ 18.33 answer : a | a ) $ 18.33 , b ) $ 22.33 , c ) $ 28.33 , d ) $ 26.23 , e ) $ 16.23 | a | multiply(divide(multiply(const_2, 10), add(110, 10)), 110) | add(n0,n1)|multiply(n0,const_2)|divide(#1,#0)|multiply(n1,#2) | general |
if you multiply two integers together and then add 4 , the result is 40 . which of the following could not be the sum of the two numbers ? | let the two integers equal x and y , and then create the following equation and simplify : xy + 4 = 40 xy = 36 so x and y are a pair of integers that equal 36 . try adding all possible combinations of two integers that multiply out to 36 : 1 Γ 36 = 36 1 + 36 = 37 2 Γ 18 = 36 2 + 18 = 20 3 Γ 12 = 36 3 + 12 = 15 4 Γ 9 = ... | a ) 12 , b ) 13 , c ) 15 , d ) 18 , e ) 20 | d | divide(subtract(40, 4), const_2) | subtract(n1,n0)|divide(#0,const_2) | general |
a train 460 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 460 + 140 = 600 m required time = 600 * 2 / 25 = 48 sec answer : c" | a ) 40 , b ) 87 , c ) 48 , d ) 21 , e ) 14 | c | divide(460, multiply(subtract(45, 140), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 45 kg . what might be the weight of the new person ? | "a 65 kg total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 64 + 20 ) kg = 65 kg ." | a ) 65 kg , b ) 90 kg , c ) 85 kg , d ) data inadequate , e ) none of these | a | add(multiply(8, 2.5), 45) | multiply(n0,n1)|add(n2,#0)| | general |
a shopkeeper loses 15 % , if an article is sold for rs . 187 . what should be the selling price of the article to gain 20 % ? | "given that sp = rs . 187 and loss = 15 % cp = [ 100 ( sp ) ] / ( 100 - l % ) = ( 100 * 187 ) / 85 = 20 * 6 = rs . 220 . to get 20 % profit , new sp = [ ( 100 + p % ) cp ] / 100 = ( 220 * 120 ) / 100 = rs . 264 answer : c" | a ) s . 247 , b ) s . 248 , c ) s . 264 , d ) s . 329 , e ) s . 412 | c | add(divide(187, subtract(const_1, divide(15, const_100))), multiply(divide(187, subtract(const_1, divide(15, const_100))), divide(20, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n1,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
if 9 a - b = 10 b + 65 = - 12 b - 2 a , what is the value of 11 a + 11 b ? | "( i ) 9 a - 11 b = 65 ( ii ) 2 a + 22 b = - 65 adding ( i ) and ( ii ) : 11 a + 11 b = 0 the answer is c ." | a ) - 11 , b ) - 5 , c ) 0 , d ) 5 , e ) 11 | c | multiply(negate(multiply(divide(65, 2), 2)), 11) | divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)| | general |
find the average of all numbers between 1 and 76 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 + 70 + 75 ) / 15 = 600 / 15 = 40 answer : option a" | a ) 40 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
a cricketer makes a score of 134 runs in the 22 nd inning and thus increases his average by 3.5 . find his average after 22 nd inning . | explanation : let the average after 22 nd innings = x then average after 21 th innings = ( x - 3.5 ) therefore 21 ( x - 3.5 ) + 134 = 22 x therefore x = 60.5 answer : b | a ) 40 , b ) 60.5 , c ) 52 , d ) 55 , e ) 36 | b | add(subtract(134, multiply(3.5, 22)), 3.5) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1) | general |
for each of her sales , a saleswoman receives a commission equal to 20 percent of the first $ 500 of the total amount of the sale , plus 30 percent of the total amount in excess of $ 500 . if the total amount of one of her sales was $ 800 , the saleswoman β s commission was approximately what percent of the total amoun... | "total sales = 800 commission = ( 20 / 100 ) * 500 + ( 30 / 100 ) * 300 = 100 + 90 = 190 % commission = ( 190 / 800 ) * 100 = 23.7 ~ 24 % answer is b" | a ) 22 % , b ) 24 percent , c ) 25 % , d ) 28 % , e ) 27 % | b | multiply(divide(add(multiply(divide(20, const_100), 500), multiply(divide(30, const_100), subtract(800, 500))), 800), const_100) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n1)|multiply(n1,#0)|multiply(#1,#2)|add(#3,#4)|divide(#5,n4)|multiply(#6,const_100)| | general |
in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 6 rows in all and a total of 135 bricks in the wall , how many bricks does the bottom row contain ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 + x - 5 = 135 6 x - 15 = 135 6 x = 120 x = 20 answer : b" | a ) 21 , b ) 20 , c ) 22 , d ) 19 , e ) 24 | b | divide(subtract(subtract(subtract(subtract(135, const_1), const_2), const_3), const_4), 6) | subtract(n1,const_1)|subtract(#0,const_2)|subtract(#1,const_3)|subtract(#2,const_4)|divide(#3,n0)| | general |
the average marks of a class of 25 students is 50 and that of another class of 40 students is 65 . find the average marks of all the students ? | "sum of the marks for the class of 25 students = 25 * 50 = 1250 sum of the marks for the class of 40 students = 40 * 65 = 2600 sum of the marks for the class of 65 students = 1250 + 2600 = 3850 average marks of all the students = 4850 / 65 = 59 . answer : d" | a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 59 , e ) 52.9 | d | divide(add(multiply(25, 50), multiply(40, 65)), add(25, 40)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
how many plants will be there in a circular bed whose outer edge measure 34 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 34 cm area of circular bed = ( 34 ) 2 Γ’ Β β 4 Γ― β¬ space for each plant = 4 cm 2 Γ’ Λ Β΄ required number of plants = ( 34 ) 2 Γ’ Β β 4 Γ― β¬ Γ£ Β· 4 = 22.98 = 23 ( approx ) answer c" | a ) 18 , b ) 750 , c ) 23 , d ) 120 , e ) none of these | c | divide(circle_area(divide(34, multiply(2, const_pi))), 4) | multiply(n2,const_pi)|divide(n0,#0)|circle_area(#1)|divide(#2,n1)| | physics |
a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 10 % ? | "cost price = rs . 500 profit = 10 % of 500 = rs . 50 selling price = cost price + profit = 500 + 50 = 550 answer : a" | a ) 550 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(10, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
stacy and helon are 30 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour faster than helon ' s constant rate of 5 miles / hour . if helon starts her journey 24 minutes after stacy , how far from the original destination has helon walked when the two meet... | original distance between s and h = 30 miles . speed of s = 5 + 1 = 6 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 24 / 60 = t + 2 / 5 hours . now , the total distances traveled by s and h = 20 miles - - - > 6 * ( t + 2 / 5 ) + 5 * t = 30 - - - > t = 138 / 55 hours . thus h h... | a ) 5 miles , b ) 8 miles , c ) 6 miles , d ) 13 miles , e ) 12 miles | d | multiply(divide(subtract(30, divide(multiply(add(5, 1), 24), const_60)), add(5, add(5, 1))), 5) | add(n1,n2)|add(n2,#0)|multiply(n3,#0)|divide(#2,const_60)|subtract(n0,#3)|divide(#4,#1)|multiply(n2,#5) | physics |
if log 2 = 0.30103 and log 3 = 0.4771 , find the number of digits in ( 648 ) 5 | "log ( 648 ) 5 = 5 log ( 648 ) = 5 log ( 81 Γ 8 ) = 5 [ log ( 81 ) + log ( 8 ) ] = 5 [ log ( 34 ) + log ( 23 ) ] = 5 [ 4 log ( 3 ) + 3 log ( 2 ) ] = 5 [ 4 Γ 0.4771 + 3 Γ 0.30103 ] = 5 ( 1.9084 + 0.90309 ) = 5 Γ 2.81149 β 14.05 answer : a" | a ) 14.05 , b ) 14.02 , c ) 277 , d ) 288 , e ) 222 | a | multiply(0.30103, divide(0.30103, 0.4771)) | divide(n1,n3)|multiply(n1,#0)| | other |
if the ratio of the areas of two squares is 225 : 256 , then the ratio of their perimeters is : | explanation : a 2 / b 2 = 225 / 256 = 15 / 16 < = > 4 a / 4 b = 4 β 15 / 4 β 16 = 15 / 16 = 15 : 16 option c | ['a ) 15 : 12', 'b ) 15 : 14', 'c ) 15 : 16', 'd ) 15 : 22', 'e ) none of these'] | c | sqrt(divide(225, 256)) | divide(n0,n1)|sqrt(#0) | geometry |
what least number must be subtracted from 12702 to get number exactly 99 ? | "explanation : divide the given number by 99 and find the remainder . if you subtract the remainder from the given number then it is exactly divisible by 99 . 99 ) 12702 ( 128 99 280 198 822 792 30 required number is 30 . answer is b" | a ) 49 , b ) 30 , c ) 29 , d ) 31 , e ) 32 | b | subtract(12702, multiply(floor(divide(12702, 99)), 99)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
if x is 20 percent greater than 40 , then x = | "x is 20 % greater than 40 means x is 1.2 times 40 ( in other words 40 + 20 / 100 * 40 = 1.2 * 40 ) therefore , x = 1.2 * 40 = 48 answer : a" | a ) 48 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 108 | a | add(40, multiply(divide(20, const_100), 40)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
in a theater , the first row has 14 seats and each row has 2 more seats than previous row . if the last row has 56 seats , what is the total number of seats in the theater ? | "the number of seats in the theater is 14 + ( 14 + 2 ) + . . . + ( 14 + 42 ) = 22 ( 14 ) + 2 ( 1 + 2 + . . . + 21 ) = 22 ( 14 ) + 2 ( 21 ) ( 22 ) / 2 = 22 ( 14 + 21 ) = 22 ( 35 ) = 770 the answer is a ." | a ) 770 , b ) 780 , c ) 790 , d ) 800 , e ) 810 | a | multiply(divide(add(14, 56), const_2), divide(add(subtract(56, 14), 2), 2)) | add(n0,n2)|subtract(n2,n0)|add(n1,#1)|divide(#0,const_2)|divide(#2,n1)|multiply(#3,#4)| | general |
a certain automobile company β s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 2 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 60 vehicles that are not speedsters , how many speedster convert... | "total vehicle = 2 / 3 of speedster + 1 / 3 of others . speedster convertibles = 2 / 3 total vehicle * 4 / 5 given : 1 / 3 constitutes 60 vehicles . hence 2 / 3 constitutes 120 speedster convertibls = 120 * 4 / 5 = 96 b" | a ) 36 , b ) 96 , c ) 100 , d ) 76 , e ) 72 | b | multiply(multiply(divide(60, subtract(const_1, divide(2, 3))), divide(2, 3)), divide(4, 5)) | divide(n2,n3)|divide(n0,n1)|subtract(const_1,#1)|divide(n4,#2)|multiply(#3,#1)|multiply(#0,#4)| | physics |
a bag of potatoes weighs 12 lbs divided by half of its weight . how much does the bag of potatoes weight ? | "sol . 12 Γ· 1 = 12 . answer : c" | a ) 20 lb , b ) 18 lb , c ) 12 lb , d ) 15 lb , e ) 5 lb | c | divide(12, const_1) | divide(n0,const_1)| | general |
the price of 80 apples is equal to that of 120 oranges . the price of 60 apples and 75 oranges together is rs . 1320 . the total price of 25 apples and 40 oranges is | "let the price of one apple = a and price of one orange = b the price of 80 apples is equal to that of 120 oranges 80 a = 120 b = > 2 a = 3 b β b = 2 a / 3 - - - - - ( equation 1 ) price of 60 apples and 75 oranges together is rs . 1320 = > 60 a + 75 b = 1320 = > 4 a + 5 b = 88 β 4 a + 5 ( 2 a ) / 3 = 88 ( β΅ substitute... | a ) 620 , b ) 610 , c ) 630 , d ) 625 , e ) 635 | a | add(multiply(divide(1320, add(60, multiply(75, divide(80, 120)))), 25), multiply(multiply(divide(1320, add(60, multiply(75, divide(80, 120)))), divide(80, 120)), 40)) | divide(n0,n1)|multiply(n3,#0)|add(n2,#1)|divide(n4,#2)|multiply(n5,#3)|multiply(#3,#0)|multiply(n6,#5)|add(#4,#6)| | general |
two vessels a and b contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively . find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ration 8 : 5 ? | let the c . p . of spirit be re . 1 litre . spirit in 1 litre mix . of a = 5 / 7 litre , c . p . of 1 litre mix . in a = re . 5 / 7 spirit in 1 litre mix . of b = 7 / 13 litre , c . p . of 1 litre mix . in b = re . 7 / 13 spirit in 1 litre mix . of c = 8 / 13 litre , mean price = re . 8 / 13 . by the rule of alligation... | a ) 4 : 3 , b ) 3 : 4 , c ) 5 : 6 , d ) 7 : 9 , e ) none | d | divide(7, add(2, 7)) | add(n1,n2)|divide(n2,#0) | other |
a scale 7 ft . 6 inches long is divided into 5 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 7 * 12 ) + 6 = 90 inches length of each of the 5 parts = 90 / 5 = 18 inches answer : d" | a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 18 inches , e ) 66 inches | d | divide(add(multiply(7, const_12), 6), 5) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general |
a cheese factory sells its cheese in rectangular blocks . a normal block has a volume of three cubic feet . if a large block has twice the width , twice the depth , and three times the length of a normal block , what is the volume of cheese in a large block in cubic feet ? | volume of cube = lbh = 3 new cube l , b , h are increases of 3 l , 2 b , 2 h new volume of cube = 3 l * 2 b * 2 h = 12 lbh = 12 * 3 = 36 answer : e | ['a ) 18', 'b ) 6', 'c ) 27', 'd ) 48', 'e ) 36'] | e | multiply(multiply(multiply(const_2, const_2), const_3), const_3) | multiply(const_2,const_2)|multiply(#0,const_3)|multiply(#1,const_3) | geometry |
having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.60 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 60 cents ( 4 / 15 ) x = 60 x = $ 2.25 the answer is a ." | a ) $ 2.25 , b ) $ 2.75 , c ) $ 3.00 , d ) $ 3.50 , e ) $ 3.75 | a | divide(multiply(multiply(3, 5), 0.60), const_4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_4)| | general |
there are 14 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "10 players are there . two players play one game with one another . so 14 c 2 = 14 * 13 / 2 = 91 so option a is correct" | a ) 91 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | a | divide(multiply(14, subtract(14, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 60 percent of its games that season , what was the total number of games that the team played ? | "60 % is 20 % - points above 40 % and 20 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 1 : 1 . so the team played a total of 30 + 30 = 60 games . the answer is b ." | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | divide(multiply(30, divide(40, const_100)), subtract(divide(80, const_100), divide(60, const_100))) | divide(n0,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n1,#0)|subtract(#1,#2)|divide(#3,#4)| | gain |
cereal a is 11 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "2 % is 2 % - points below 4 % and 11 % is 7 % - points above 4 % . the ratio of a : b should be 2 : 7 . the answer is c ." | a ) 2 : 5 , b ) 1 : 3 , c ) 2 : 7 , d ) 3 : 4 , e ) 1 : 5 | c | divide(subtract(4, 2), subtract(11, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
10 years ago a was half of b in age . if the ratio of their present ages is 3 : 4 , what will be the total of their present ages | explanation : let a ' s age 10 years ago = x years . then , b ' s age 10 years ago = 2 x years . ( x + 10 ) / ( 2 x + lo ) = 3 / 4 = > x = 5 . so , the total of their present ages = ( x + 10 + 2 x + 10 ) = ( 3 x + 20 ) = 35 years . answer : option b | a ) 30 , b ) 35 , c ) 37 , d ) 41 , e ) 43 | b | add(multiply(3, divide(subtract(10, divide(10, const_2)), subtract(3, divide(4, const_2)))), multiply(4, divide(subtract(10, divide(10, const_2)), subtract(3, divide(4, const_2))))) | divide(n0,const_2)|divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#1)|divide(#2,#3)|multiply(n1,#4)|multiply(n2,#4)|add(#5,#6) | general |
a rectangular tiled patio is composed of 126 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 126 tiles , and it will still be rectangular . how many rows are in the tile patio before ... | suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 126 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , n... | ['a ) 5', 'b ) 6', 'c ) 9', 'd ) 13', 'e ) 28'] | c | divide(126, divide(add(negate(4), sqrt(add(power(4, const_2), multiply(4, multiply(126, 2))))), const_2)) | multiply(n0,n1)|negate(n2)|power(n2,const_2)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,const_2)|divide(n0,#7) | geometry |
in a class of 39 students 26 play football and play 20 long tennis , if 17 play above , many play neither ? | "26 + 20 - 17 = 29 39 - 29 = 10 play neither answer is c" | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | subtract(39, subtract(add(26, 20), 17)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
a rectangular floor is covered by a rug except for a strip 2 meters wide along each of the four edge . if the floor is 10 meters by 8 meters , what is the area of the rug in square meters ? | "a strip of 2 meters is covering the inner rectangular rug for all 4 sides . length of inner rug = 10 - ( 2 * 2 ) breadth of inner rug = 8 - ( 2 * 2 ) area of rug = 6 * 4 = 24 sq . mt Γ’ nswer : a" | a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | a | rectangle_area(subtract(10, multiply(2, const_2)), subtract(8, multiply(2, const_2))) | multiply(n0,const_2)|subtract(n1,#0)|subtract(n2,#0)|rectangle_area(#1,#2)| | geometry |
if a * b denotes the greatest common divisor of a and b , then ( ( 12 * 16 ) * ( 18 * 12 ) ) = ? | "the greatest common divisor of 12 and 16 is 4 . hence 12 * 16 = 4 ( note that * here denotes the function not multiplication ) . the greatest common divisor of 18 and 12 is 6 . hence 18 * 12 = 6 . hence ( ( 12 * 16 ) * ( 18 * 12 ) ) = 4 * 6 . the greatest common divisor of 4 and 6 is 2 . answer ; e ." | a ) 24 , b ) 12 , c ) 6 , d ) 4 , e ) 2 | e | divide(divide(18, const_3), const_3) | divide(n2,const_3)|divide(#0,const_3)| | general |
if 24 men take 15 days to to complete a job , in how many days can 18 men finish that work ? | ans . 20 days | a ) 20 , b ) 19 , c ) 18 , d ) 17 , e ) 16 | a | divide(multiply(24, 15), 18) | multiply(n0,n1)|divide(#0,n2)| | physics |
the circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long . what is the distance travelled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | "total distance - x x / 40 - x / 48 = 5 x = 1200 ft answer a" | a ) 1200 ft , b ) 800 ft , c ) 900 ft , d ) 1000 ft , e ) 1500 ft | a | multiply(40, add(divide(multiply(40, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | divide(const_10,const_2)|multiply(n0,#0)|divide(#1,const_3)|add(#2,#0)|multiply(n0,#3)| | physics |
a train requires 10 seconds to pass a pole while it requires 30 seconds to cross a stationary train which is 600 mtrs long . find the speed of the train . | "in 10 s the train crosses the pole and in 30 sec the train crosses one more stationary train in 20 sec the train travels a distance of 600 mtrs speed = 600 / 20 = 30 m / s = 30 ( 3600 / 1000 ) = 30 * 18 / 5 = 108 kmph answer : c" | a ) 110 kmph , b ) 120 kmph , c ) 108 kmph , d ) 100 kmph , e ) 98 kmph | c | multiply(divide(600, subtract(30, 10)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6)| | physics |
the average of runs of a cricket player of 10 innings was 36 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 36 so , total = average x no . of innings = 36 x 10 = 360 . now increase in avg = 4 runs . so , new avg = 36 + 4 = 40 runs total runs = new avg x new no . of innings = 40 x 11 = 440 runs made in the 11 th inning = 440 - 360 = 80 answer : b" | a ) 76 , b ) 80 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 36)), multiply(10, 36)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
john had a stock of 800 books in his bookshop . he sold 62 on monday , 62 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 62 + 62 + 60 + 48 + 40 = 272 let m be the books not sold m = 800 - n = 1400 - 272 = 528 percentage books not sold / total number of books = 528 / 800 = 0.66 = 66 % correct answer b" | a ) 78 % , b ) 66 % , c ) 62 % , d ) 64 % , e ) 60 % | b | multiply(divide(subtract(800, add(add(add(62, 62), add(60, 48)), 40)), 800), const_100) | add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | gain |
a certain city with a population of 180,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 180 k x = 15,000 answer : d" | a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 15,000 , e ) e ) 11,100 | d | multiply(multiply(const_4, const_2), const_100) | multiply(const_2,const_4)|multiply(#0,const_100)| | general |
a palindrome is a number that reads the same front - to - back as it does back - to - front ( e . g . 202 , 575 , 1991 , etc . ) p is the smallest integer greater than 100 that is both a prime and a palindrome . what is the sum of the digits of p ? | "given that p is smallest integer greater than 200 - assume there is a 3 - digit that satisfies the above conditions . let the number be xyx ; question asks us the values of 2 x + y we can straight away cross out options a ) and d ) - sum of digits 3 or 6 implies it is divisible by 3 - - - > we know that p is a prime n... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(const_3, const_3) | add(const_3,const_3)| | general |
circular gears l and r start to rotate at the same time at the same rate . gear l makes 20 complete revolutions per minute and gear r makes 40 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 6 more revolutions than gear l ? | gear l - - 20 rotations per 60 seconds - - 2 rotation per 6 seconds . gear r - - 40 rotations per 60 seconds - - 4 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 2 rotations hence every 6 seconds the difference between the number of rotations ... | a ) a ) 6 , b ) b ) 8 , c ) c ) 10 , d ) d ) 12 , e ) e ) 15 | a | divide(divide(6, subtract(divide(40, const_60), divide(20, const_60))), const_3) | divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|divide(#3,const_3) | physics |
two pipes a and b can fill a tank in 4 and 12 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min . = 14 part filled by b in 1 min . = 1 / 12 part filled by ( a + b ) in 1 min . = 1 / 4 + 1 / 12 = 1 / 3 . both the pipes can fill the tank in 3 minutes . answer : b" | a ) 12 , b ) 3 , c ) 6 , d ) 9 , e ) 10 | b | add(multiply(4, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 4), divide(const_1, 12)), const_2)), 4), const_60)) | divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)| | physics |
a girl was asked to multiply a certain number by 43 . she multiplied it by 34 and got his answer less than the correct one by 1206 . find the number to be multiplied . | "let the required number be x . then , 43 x β 34 x = 1206 or 9 x = 1206 or x = 134 . required number = 134 . answer : c" | a ) 130 , b ) 132 , c ) 134 , d ) 136 , e ) 138 | c | divide(1206, subtract(43, 34)) | subtract(n0,n1)|divide(n2,#0)| | general |
a company pays 18.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ? | "explanation : dividend on 1 share = ( 18.5 * 50 ) / 100 = rs . 9.25 rs . 25 is income on an investment of rs . 100 rs . 9.25 is income on an investment of rs . ( 9.25 * 100 ) / 25 = rs . 37 answer : d" | a ) 25 , b ) 66 , c ) 18 , d ) 37 , e ) 01 | d | divide(multiply(divide(multiply(18.5, 50), const_100), const_100), 25) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,const_100)|divide(#2,n2)| | gain |
1000 men have provisions for 20 days . if 650 more men join them , for how many days will the provisions last now ? | "1000 * 20 = 1650 * x x = 12.1 answer : e" | a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 12.2 , e ) 12.1 | e | divide(multiply(20, 1000), add(1000, 650)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
david ' s bank ' s saving amount is decreased 10 % due to loan payment and current balance is rs . 90000 . find the actual balance before deduction ? | 10 % decreased 90 % balance = 90000 100 % = 90000 / 90 * 100 = 100000 answer : e | a ) 8000 , b ) 8500 , c ) 9000 , d ) 9500 , e ) 100000 | e | divide(multiply(90000, const_100), subtract(const_100, 10)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1) | gain |
the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 2200 , then what is the income of p 1 ? | "let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x β 3 y = 2200 β¦ ( i ) and 4 x β 2 y = 2200 β¦ β¦ . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 2200 - > x = 1100 p 1 β s income = rs 5 * 1100 =... | a ) s . 800 , b ) s . 2400 , c ) s . 4000 , d ) s . 5500 , e ) s . 4200 | d | multiply(subtract(multiply(divide(2200, 2), 3), 2200), 5) | divide(n6,n1)|multiply(n4,#0)|subtract(#1,n6)|multiply(n2,#2)| | other |
3 math classes : x , y , and z , take an algebra test . the average score in class x is 83 . the average score in class y is 76 . the average score in class z is 85 . the average score of all students in classes x and y together is 79 . the average score of all students in classes y and z together is 81 . what is the a... | explanation : let the number of students in classes x , y and z be a , b and c respectively . then total of x = 83 a total of y = 76 b total of z = 85 c and , ( 83 a + 76 b ) / ( a + b ) = 79 . i . e 4 a = 3 b . also , ( 76 b + 85 c ) / ( b + c ) = 81 . i . e 4 c = 5 b . hence , b = ( 4 / 3 ) a , c = ( 5 / 4 ) b = ( 5 ... | a ) 81 , b ) 81.5 , c ) 82 , d ) 84.5 , e ) none of these | b | divide(add(multiply(83, subtract(79, 76)), add(multiply(85, subtract(81, 76)), multiply(76, subtract(83, 79)))), add(subtract(81, 76), add(subtract(83, 79), subtract(79, 76)))) | subtract(n5,n2)|subtract(n1,n4)|subtract(n4,n2)|add(#1,#2)|multiply(n3,#0)|multiply(n2,#1)|multiply(n1,#2)|add(#4,#5)|add(#3,#0)|add(#7,#6)|divide(#9,#8) | general |
a can finish a piece of work in 4 days . b can do it in 14 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 14 = 1 = > x = 5 days answer : b" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 14), const_2))), divide(const_1, 14)) | divide(const_1,n0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics |
the population of a town increased from 1 , 34,000 to 2 , 32,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 232500 - 134000 ) = 98500 . increase % = ( 98500 / 134000 x 100 ) % = 73 % . required average = ( 73 / 10 ) % = 7.3 % . answer : option b" | a ) 4.37 % , b ) 7.3 % , c ) 7.6 % , d ) 8.75 % , e ) none | b | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
if r is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of r ? | answer is c . numbers in format of 3 ^ k in the series from 1 to 30 inclusive r are : 3 * 1 , 3 * 2 , 3 * 3 , 3 * 4 , 3 * 5 , 3 * 3 * 2 , 3 * 7 , 3 * 8 , 3 * 3 * 3 , 3 * 10 . total number of 3 = 13 . so k = 14 . | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | subtract(divide(30, const_2), 1) | divide(n1,const_2)|subtract(#0,n0) | general |
rs . 460 was divided among 41 boys and girls such that each boy rs . 12 and each girl got rs . 8 . what is the number of boys ? | explanation : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - solution 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - assume that the number of b... | a ) 33 , b ) 30 , c ) 36 , d ) 28 , e ) 26 | a | divide(subtract(460, multiply(8, 41)), const_4) | multiply(n1,n3)|subtract(n0,#0)|divide(#1,const_4) | general |
a shopkeeper sells 200 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 200 = rs . 90 loss per metre = rs . 5 cp per metre = 90 + 5 = rs . 95 . answer : e | a ) 12 , b ) 27 , c ) 29 , d ) 55 , e ) 95 | e | add(divide(18000, 200), 5) | divide(n1,n0)|add(n2,#0)| | gain |
find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 . | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 2 = 2 ^ 22 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer is a ." | a ) 29 , b ) 27 , c ) 28 , d ) 24 , e ) 21 | a | add(add(multiply(2, 11), 5), 2) | multiply(n1,n5)|add(n3,#0)|add(n5,#1)| | general |
1235 1253 1325 . . . . . . . . . + 5321 the addition problem above shows 4 of the 24 different integers that can be formed by using each of the digits 12 , 35 exactly once in each integer . what is the sum of these 24 integers ? | in each of the units , tens , hundreds , and thousands place , each digit will appear 6 times . then the sum equals 6 ( 1111 ) + 6 ( 2222 ) + 6 ( 3333 ) + 6 ( 5555 ) = 73,326 the answer is a . | a ) 73,326 , b ) 78,524 , c ) 84,820 , d ) 89,918 , e ) 95,560 | a | divide(add(add(add(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), multiply(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), ... | divide(n6,const_10)|divide(n7,const_10)|divide(n5,n4)|reminder(n6,const_10)|reminder(n7,const_10)|floor(#0)|floor(#1)|add(#5,#3)|add(#7,#6)|add(#8,#4)|multiply(#9,#2)|multiply(#10,const_1000)|multiply(#10,const_100)|multiply(#10,const_10)|add(#10,#11)|add(#14,#12)|add(#15,#13)|divide(#16,const_1000) | general |
a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . if 45 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | "construct an equation : 15 / 100 = x / 45 - > x = 6.75 answer : e ." | a ) 3.0 , b ) 5.0 , c ) 5.5 , d ) 6.5 , e ) 6.75 | e | multiply(45, divide(15, 100)) | divide(n0,n1)|multiply(n2,#0)| | physics |
two trains of equal lengths take 5 sec and 15 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 / 5 = 24 m / sec . speed of the second train = 120 / 15 = 8 m / sec . relative speed = 24 + 8 = 32 m / sec . required time = ( 120 + 120 ) / 32 = 7.5 sec . answer : c" | a ) 22 , b ) 12 , c ) 7.5 , d ) 99 , e ) 21 | c | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 5))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
machine a and machine b are each used to manufacture 880 sprockets . it takes machine a 10 hours longer to produce 880 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 880 qt = 800 for a : ( t + 10 ) ( q ) = 880 qt + 10 q = 880 800 + 10 q = 880 q = 8 so a can produce 8 / hour . then b can produce = 8 ( 1.1 ) = 8.8 / hour . c" | a ) 6 , b ) 6.6 , c ) 8.8 , d ) 100 , e ) 110 | c | divide(880, divide(multiply(multiply(10, 880), divide(add(const_100, 10), const_100)), subtract(multiply(880, divide(add(const_100, 10), const_100)), 880))) | add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)| | gain |
an empty fuel tank with a capacity of 204 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 204 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 204 - a gallons of fuel b is 0.16 ( 204 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 204 - a ) = 30 - - > a =... | a ) 160 , b ) 150 , c ) 100 , d ) 66 , e ) 50 | d | divide(subtract(multiply(204, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
what is the largest 4 digit number exactly divisible by 50 ? | "largest 4 digit number = 9999 9999 Γ· 50 = 199 , remainder = 49 hence largest 4 digit number exactly divisible by 50 = 9999 - 49 = 9950 answer : c" | a ) 4676 , b ) 4678 , c ) 9950 , d ) 9504 , e ) 9936 | c | multiply(floor(divide(power(const_10, 4), 50)), 50) | power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)| | general |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind poe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of poe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . poe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena ca... | a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1512 respectively . what is the rate of interest ? | "1512 - 1400 = 112 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 112 ) / ( 1400 ) = 8 % the answer is d ." | a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | d | divide(multiply(subtract(1512, 1400), const_100), 1400) | subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)| | gain |
in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder . the dividend is | "divisor = ( 6 * 3 ) + 2 = 20 5 * quotient = 20 quotient = 4 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 4 ) + 6 = 86 . a" | a ) 86 , b ) 90 , c ) 92 , d ) 94 , e ) 96 | a | add(multiply(add(multiply(6, const_3), 2), divide(add(multiply(6, const_3), 2), 5)), 6) | multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3)| | general |
a number when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 31 | "explanation : p Γ· 44 = 432 = > p = 432 * 44 = 19008 p / 31 = 19008 / 31 = 613 , remainder = 5 option a" | a ) 5 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | a | divide(multiply(44, 432), 31) | multiply(n0,n1)|divide(#0,n3)| | general |
if $ 120 invested at a certain rate of simple interest amounts to $ 160 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ? | "120 amounts to 160 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 160 120 + 120 * ( r / 100 ) * ( 3 ) = 160 = > r = 100 / 9 150 in 6 years = principal + interest = 150 + 150 * ( r / 100 ) * ( 6 ) 250 answer is e ." | a ) $ 190 , b ) $ 180 , c ) $ 200 , d ) $ 240 , e ) $ 250 | e | add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(160, 120), 120), 120), 3)), 120)) | multiply(n3,n4)|subtract(n1,n0)|multiply(#1,n0)|divide(#2,n0)|divide(#3,n2)|multiply(#4,#0)|divide(#5,n0)|add(n3,#6)| | gain |
the population of a town increased from 1 , 33,400 to 1 , 93,500 in a decade . the average percent increase of population per year is : | "explanation : increase in 10 years = ( 193500 - 133400 ) = 60100 . increase % = ( 60100 / 133400 x 100 ) % = 45 % . required average = ( 45 / 10 ) % = 4.5 % . answer : option a" | a ) 4.5 % , b ) 5.5 % , c ) 6.5 % , d ) 8.75 % , e ) none | a | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(1, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(1, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
a person distributed 20 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 60000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 30 % orphan house = 5 % total = 60 + 30 + 5 = 95 % remaining = 100 - 95 = 5 % 5 % = 60000 100 % = 60000 * 100 / 5 = $ 1200000 answer is a" | a ) a ) 1200000 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 500000 , e ) e ) 652000 | a | multiply(divide(60000, subtract(const_100, add(add(multiply(20, 3), 30), 5))), const_100) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|subtract(const_100,#2)|divide(n4,#3)|multiply(#4,const_100)| | gain |
a = 5 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "a = 5 ^ 15 - 625 ^ 3 = > 5 ^ 15 - ( 5 ^ 4 ) ^ 3 = > 5 ^ 15 - 5 ^ 12 = 5 ^ 12 ( 5 ^ 3 - 1 ) = 5 ^ 12 * 124 124 = 31 * 4 a / x is integer for condition of 2 < p < x only 5 and 31 satisfies this hence answer is c" | a ) none , b ) one , c ) two , d ) three , e ) four | c | subtract(15, multiply(3, const_4)) | multiply(n3,const_4)|subtract(n1,#0)| | general |
a car takes 6 hours to cover a distance of 360 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 360 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 360 / 9 = 40 kmph answer c ." | a ) 60 , b ) 50 , c ) 40 , d ) 70 , e ) 65 | c | divide(360, divide(multiply(6, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics |
if 4 people are selected from a group of 7 married couples , what is the probability that none of them would be married to each other ? | "if we are to select 4 people from 7 couples without any restriction , how many ways can we make the selection ? 14 ! / 4 ! 10 ! = 1001 if we are to select 4 people from 7 couples with restriction that no married couple can both make it to the group , only a representative ? 7 ! / 4 ! 3 ! = 35 but we know that to selec... | a ) 1 / 33 , b ) 2 / 33 , c ) 1 / 3 , d ) 16 / 33 , e ) 47 / 84 | e | multiply(multiply(multiply(divide(multiply(7, const_2), multiply(7, const_2)), divide(multiply(4, 4), subtract(multiply(7, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | multiply(n0,n0)|multiply(n1,const_2)|divide(#1,#1)|subtract(#0,const_2)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#0)|divide(#0,#5)|subtract(#3,const_2)|divide(#8,#4)|multiply(#2,#7)|multiply(#6,#10)|multiply(#9,#11)| | probability |
a shopkeeper has 100 kg of apples . he sells 50 % of these at 25 % profit and remaining 50 % at 30 % profit . find his % profit on total . | "total number of apples = 100 let the cost price be x selling price at 25 % profit = 1.25 x selling price at 30 % profit = 1.3 x profit % = ( ( sp - cp ) / cp ) * 100 profit % = ( ( 1 / 2 ) * 100 * 1.25 x + ( 1 / 2 ) * 100 * 1.3 x - 100 x ) / 100 x * 100 = ( 255 - 200 ) / 2 = 27.5 % answer is a" | a ) 27.5 % , b ) 25.6 % , c ) 31.5 % , d ) 35.9 % , e ) 29.5 % | a | divide(multiply(subtract(add(multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 25), const_100)), multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 30), const_100))), 100), const_100), 100) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)| | gain |
a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways... | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . whe... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(multiply(multiply(multiply(const_4, 4), 2), 1), multiply(multiply(4, 2), 1)) | multiply(n5,const_4)|multiply(n1,n5)|multiply(n1,#0)|multiply(n3,#1)|multiply(n3,#2)|divide(#4,#3)| | general |
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