Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
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the price of lunch for 20 people was $ 300 including a 20 % gratuity for service . what was the average price per person , excluding the gratuity ? | "clearly b is the answer i used poe here lets consider option ( b ) 20 * 12.5 = 250 now 250 ( 120 / 100 ) = 300 = > possible answer imo b" | a ) 11.73 , b ) 12.5 , c ) 13.8 , d ) 14 , e ) 15.87 | b | divide(divide(multiply(300, subtract(const_100, 20)), const_100), 20) | subtract(const_100,n2)|multiply(n1,#0)|divide(#1,const_100)|divide(#2,n0)| | general |
the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 28 - inch flat - screen television than a square 25 - inch flat - screen television ? | if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 28 - inch flat - screen television the diagonal ( hypotenuse ) = 28 so , we can apply the pythagorean theorem to get x ² + x ² = 28 ² s... | ['a ) 79.5', 'b ) 89.3', 'c ) 85.5', 'd ) 75.4', 'e ) 72.5'] | a | subtract(divide(power(28, const_2), const_2), divide(power(25, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3) | geometry |
the sector of a circle has radius of 14 cm and central angle 144 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 144 / 360 * 2 * 22 / 7 * 14 ) + 2 ( 14 ) = 35.2 + 28 = 63.2 cm answer : b" | a ) 60 cm , b ) 63.2 cm , c ) 70 cm , d ) 80 cm , e ) 90 cm | b | multiply(multiply(const_2, divide(multiply(subtract(14, const_3), const_2), add(const_4, const_3))), 14) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
a number is doubled and 4 is added . if the resultant is divided by 5 , it becomes 25 . what is that number ? | "solution let the number be x . then , ( 2 x + 4 ) / 5 ‹ = › 25 ‹ = › 2 x + 4 = 125 ‹ = › 2 x = 121 x = 60.5 answer c" | a ) 73.5 , b ) 68 , c ) 60.5 , d ) none of these , e ) can not be determined | c | divide(subtract(5, multiply(const_3, 4)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
chris mixed 4 pounds of raisins with 4 pounds of nuts . if a pound of nuts costs 3 times as much as a pound of raisins , then the total cost of the raisins was what fraction of the total cost of the mixture ? | "1 lbs of raisin = $ 1 4 lbs of raisin = $ 4 1 lbs of nuts = $ 3 4 lbs of nuts = $ 12 total value of mixture = 12 + 4 = 16 fraction of the value of raisin = 4 / 16 = 1 / 4 ans : c" | a ) 1 / 7 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 3 / 7 | c | divide(4, add(multiply(4, 3), 4)) | multiply(n1,n2)|add(n0,#0)|divide(n0,#1)| | general |
how many 3 digit numbers are divisible by 6 in all ? | "explanation : required numbers are 102 , 108 , 114 , . . . , 996 this is an a . p . in which a = 102 , d = 6 and l = 996 let the number of terms be n . then , a + ( n - 1 ) d = 996 102 + ( n - 1 ) x 6 = 996 6 x ( n - 1 ) = 894 ( n - 1 ) = 149 n = 150 . answer is b" | a ) 149 , b ) 150 , c ) 151 , d ) 166 , e ) 172 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(3, 3)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
1907 x 1907 = ? | "1907 x 1907 = ( 1907 ) 2 = ( 1900 + 7 ) 2 = ( 1900 ) 2 + ( 7 ) 2 + ( 2 x 1900 x 7 ) = 3610000 + 49 + 26600 . = 3636649 . d )" | a ) a ) 3623216 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3636649 , e ) e ) 3625216 | d | multiply(divide(1907, 1907), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
an amount of rs . 3120 was divided among a , b and c , in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of c ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 / 12 : 4 / 12 : 3 / 12 = 6 : 4 : 3 . share of c = 3 / 13 * 3120 = rs . 720 an... | a ) 720 , b ) 360 , c ) 250 , d ) 280 , e ) 290 | a | multiply(divide(3120, add(add(2, 3), 4)), 4) | add(n2,n4)|add(n6,#0)|divide(n0,#1)|multiply(n6,#2)| | general |
a certain company ’ s profit in 1996 was 10 percent greater than its profit in 1995 , and its profit in 1997 was 20 percent greater than its profit in 1996 . the company ’ s profit in 1997 was what percent greater than its profit in 1995 ? | profit in 1995 - 100 profit in 1996 - 110 % increae profit in 1997 in comparison to 1995 = 10 + 110 * 20 % = 32 correct option : c | a ) 5 % , b ) 18 % , c ) 32 % , d ) 35 % , e ) 38 % | c | multiply(subtract(multiply(add(divide(20, const_100), const_1), add(const_1, divide(10, const_100))), const_1), const_100) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain |
if a * b * c = ( √ ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 11 . | "6 * 15 * 11 = ( √ ( 6 + 2 ) ( 15 + 3 ) ) / ( 11 + 1 ) = ( √ 8 * 18 ) / 12 = ( √ 144 ) / 12 = 12 / 124 = 1 . answer is c" | a ) 8 , b ) 5 , c ) 1 , d ) 3 , e ) 4 | c | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(11, 1)) | add(n0,n3)|add(n1,n4)|add(n2,n5)|multiply(#0,#1)|sqrt(#3)|divide(#4,#2)| | general |
an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 5 : 8 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | "explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 5 x and r = 8 x . then , 5 x = 1500 ? x = 300 . p = 900 , q = 1500 and r = 2400 . hence , ( r - p ) = ( 2400 - 900 ) = 1500 answer : b" | a ) s 1200 , b ) s 1500 , c ) s 1600 , d ) s 1900 , e ) s 1700 | b | multiply(divide(1500, 5), subtract(8, 3)) | divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)| | other |
in a caravan , in addition to 50 hens , there are 45 goats and 8 camels with some keepers . if the total number of feet be 224 more than the number of heads in the caravan , the number of keepers is | explanation : let number of keepers be x . then , total number of feet = 2 x 50 + 4 x 45 + 4 x 8 + 2 x = 2 x + 312 . total number of heads = 50 + 45 + 8 + x = 103 + x . therefore ( 2 x + 312 ) = ( 103 + x ) + 224 or x = 15 . answer : d | a ) 5 , b ) 8 , c ) 10 , d ) 15 , e ) 20 | d | subtract(add(add(add(50, 45), 8), 224), add(add(multiply(const_2, 50), multiply(const_4, 45)), multiply(const_4, 8))) | add(n0,n1)|multiply(n0,const_2)|multiply(n1,const_4)|multiply(n2,const_4)|add(n2,#0)|add(#1,#2)|add(n3,#4)|add(#5,#3)|subtract(#6,#7) | general |
a 1500 m long train crosses a tree in 120 sec , how much time will it take to pass a platform 500 m long ? | "l = s * t s = 1500 / 120 s = 12.5 m / sec . total length ( d ) = 2000 m t = d / s t = 2000 / 12.5 t = 160 sec answer : b" | a ) 288 , b ) 160 , c ) 188 , d ) 188 , e ) 12 | b | divide(add(1500, 500), divide(1500, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics |
n is a positive integer . when n + 1 is divided by 6 , the remainder is 4 . what is the remainder when n is divided by 2 ? | n + 1 = 6 a + 4 i . e . n + 1 = 10 , 16 , 22 , 28 , . . . etc . i . e . n = 9 , 15 , 21 , 27 , . . . etc . when n is divided by 2 remainder is always 1 answer : c | a ) 6 , b ) 5 , c ) 1 , d ) 3 , e ) 2 | c | divide(6, add(4, 2)) | add(n2,n3)|divide(n1,#0) | general |
a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 2 = rs . 1350 answerb" | a ) 2999 , b ) 1350 , c ) 2677 , d ) 2898 , e ) 2787 | b | multiply(subtract(rectangle_area(add(75, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(75, 55)), 2) | multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry |
if 9 / ( 5 + 3 / x ) = 1 , then x = | the expression 9 / ( 5 + 3 / x ) = 1 should have been equal to something . if 9 / ( 5 + 3 / x ) = 1 = > 9 x / ( 5 x + 3 ) = 1 = > 9 x = 5 x + 3 = > 4 x = 3 = > x = 3 / 4 correct option : c | a ) 3 , b ) 1 , c ) 3 / 4 , d ) - 1 / 3 , e ) - 3 | c | divide(3, subtract(9, 5)) | subtract(n0,n1)|divide(n2,#0) | general |
if price of t . v set is reduced by 25 % , then its sale increases by 75 % , find net effect on sale value | "- a + b + ( ( - a ) ( b ) / 100 ) = - 25 + 75 + ( - 25 * 75 ) / 100 = - 25 + 75 - 19 = 31 answer : a" | a ) 31 , b ) 32 , c ) 46 , d ) 47 , e ) 48 | a | multiply(subtract(multiply(divide(subtract(const_100, 25), const_100), divide(add(const_100, 75), const_100)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
25 , 50 , 100 , 175 , 275 , ( . . . . ) | "explanation : the pattern is 25 , 50 , 75 , 100 , etc . hence 125 = 400 answer : a" | a ) 400 , b ) 300 , c ) 425 , d ) 450 , e ) 325 | a | subtract(negate(175), multiply(subtract(50, 100), divide(subtract(50, 100), subtract(25, 50)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 9 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | "let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 9 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 9 * ( 2 / 3 ) ) days = > 20 days answer : b" | a ) 30 , b ) 20 , c ) 19 , d ) 29 , e ) 39 | b | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 9), divide(2, 3))) | divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4)| | physics |
kathleen can paint a room in 3 hours , and anthony can paint an identical room in 4 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "( 1 / 3 + 1 / 4 ) t = 2 t = 24 / 7 answer : d" | a ) 8 / 15 , b ) 4 / 3 , c ) 15 / 8 , d ) 24 / 7 , e ) 15 / 4 | d | multiply(divide(const_1, add(divide(const_1, 3), divide(const_1, 4))), const_2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_2)| | physics |
how many prime numbers are between 25 / 8 and 213 / 6 ? | "25 / 8 = 4 - 213 / 6 = 36 - prime numbers between 4 and 36 are 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , and 31 - sign signifies that the number is marginally less . answer d" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | floor(const_2) | floor(const_2)| | general |
on the number line , the number p is twice as many units from - 2 as - 2 is from 4 . if p is less than – 2 , what is the value of p ? | "distance between - 2 and 4 = 6 since number p is twice as many units from - 2 as - 2 is from 4 , therefore p can be - 14 since , p is less than - 2 , the value of p = - 14 answer a" | a ) – 14 , b ) – 10 , c ) – 6 , d ) 10 , e ) 14 | a | subtract(negate(2), multiply(2, subtract(2, negate(2)))) | negate(n2)|subtract(n0,#0)|multiply(#1,n0)|subtract(#0,#2)| | general |
two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 1 : 2 and 3 : 1 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | "vessel a = 300 gallons - - > milk = 100 , water = 200 ; vessel b = 500 gallons - - > milk = 375 , water = 125 ; vessel a + b = 800 gallons - - > milk = 475 , water 325 . the ratio = 475 / 325 - - > 19 / 13 answer : c" | a ) 18 : 13 , b ) 16 : 13 , c ) 19 : 13 , d ) 14 : 11 , e ) 13 : 11 | c | power(1, 3) | power(n2,n0)| | other |
a bus travels first half distance between two places with a speed of 40 kmph and the rest half distance with a speed of 60 kmph . the average speed of the bus is ? | explanation : we know that speed = distance traveled / time taken let the total distance traveled by the car is 2 x km . then time taken by it to cover first half is hour . and for second half is hour . then average speed = total distance travelled / total time taken . i . e . = > = 48 kmph . answer : a | a ) 48 kmph , b ) 78 kmph , c ) 28 kmph , d ) 18 kmph , e ) 98 kmph | a | divide(add(40, 60), const_2) | add(n0,n1)|divide(#0,const_2) | physics |
in a hostel there were 100 students . to accommodate 20 more students the average is decreased by rupees 5 . but total expenditure increased by rs . 400 . find the total expenditure of the hostel now ? | "100 x + 400 = 12 ( x – 5 ) x = 50 100 * 50 + 400 = 5400 answer : c" | a ) 54888 , b ) 2777 , c ) 5400 , d ) 1278 , e ) 2881 | c | multiply(subtract(divide(add(multiply(add(100, 20), 5), 400), 20), 5), add(100, 20)) | add(n0,n1)|multiply(n2,#0)|add(n3,#1)|divide(#2,n1)|subtract(#3,n2)|multiply(#0,#4)| | general |
the ratio between the number of passengers travelling by i and ii class between the two railway stations is 1 : 50 , whereas the ratio of i and ii class fares between the same stations is 3 : 1 . if on a particular day rs . 1,325 were collected from the passengers travelling between these stations , then what was the a... | let x be the number of passengers and y be the fare taken from passengers . 3 xy + 50 xy = 1325 = > xy = 25 amount collected from ii class passengers = 25 × 50 = rs . 1250 . answer : d | a ) rs . 750 , b ) rs . 1000 , c ) rs . 850 , d ) rs . 1250 , e ) none of these | d | multiply(multiply(50, 1), divide(add(add(add(multiply(const_10, const_100), multiply(const_3, const_100)), multiply(const_2, const_10)), add(const_2, const_3)), add(multiply(1, 3), multiply(50, 1)))) | add(const_2,const_3)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(const_10,const_2)|multiply(n0,n2)|multiply(n0,n1)|add(#1,#2)|add(#4,#5)|add(#6,#3)|add(#8,#0)|divide(#9,#7)|multiply(#10,#5) | other |
a , b , c subscribe rs . 50,000 for a business . if a subscribes rs . 4000 more than b and b rs . 5000 more than c , out of a total profit of rs . 38,000 , what will be the amount a receives ? | "total amount invested = 50000 assume that investment of c = x . then investment of b = 5000 + x , investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000 ⇒ 3 x + 14000 = 50000 ⇒ 3 x = 50000 – 14000 = 36000 ⇒ x = 36000 / 3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 inves... | a ) 14700 , b ) 15960 , c ) 14900 , d ) 14300 , e ) 14000 | b | multiply(add(divide(subtract(multiply(5000, const_10), add(add(4000, 5000), 5000)), const_3), add(4000, 5000)), divide(multiply(multiply(const_3, const_12), const_1000), multiply(5000, const_10))) | add(n1,n2)|multiply(n2,const_10)|multiply(const_12,const_3)|add(n2,#0)|multiply(#2,const_1000)|divide(#4,#1)|subtract(#1,#3)|divide(#6,const_3)|add(#0,#7)|multiply(#8,#5)| | general |
( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) / 4.0 = ? | "dove straight into calculation but quickly realized that the sum of two negatives is a negative so there is only one option . - 1.065 answer c" | a ) 1.07 , b ) 1.0 , c ) - 1.065 , d ) 1.71 , e ) 2.71 | c | divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 4.0) | multiply(n0,n1)|multiply(n2,n3)|negate(#0)|subtract(#2,#1)|divide(#3,n4)| | general |
if â € œ * â € is called â € œ + â € , â € œ / â € is called â € œ * â € , â € œ - â € is called â € œ / â € , â € œ + â € is called â € œ - â € . 120 / 40 â € “ 30 * 20 - 5 = ? | "explanation : given : 120 / 40 â € “ 30 * 20 - 5 = ? substituting the coded symbols for mathematical operations , we get , 120 * 40 / 30 + 20 / 5 = ? 120 * 1.33 + 4 = ? 159.6 + 4 = 163.6 answer : d" | a ) 178.5 , b ) 156.6 , c ) 160.4 , d ) 163.6 , e ) 164.1 | d | add(multiply(divide(30, 20), divide(120, 40)), 5) | divide(n2,n3)|divide(n0,n1)|multiply(#0,#1)|add(n4,#2)| | general |
the hypotenuse of a right triangle is 2 centimeters more than the longer side of the triangle . the shorter side of the triangle is 7 centimeters less than the longer side . find the length of the hypotenuse . | let a and b be the two sides of the triangle such that a is longer than b . the statement ` ` the hypotenuse of a right triangle is 2 centimeters more than the longer side of the triangle ' ' may be fomulated by h = a + 2 or a = h - 2 the statement ` ` the shorter side of the triangle is 7 centimeters less than the lon... | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | a | add(divide(add(multiply(add(7, 2), 2), sqrt(subtract(power(multiply(add(7, 2), 2), const_2), multiply(subtract(power(7, const_2), power(2, const_2)), const_4)))), const_2), 2) | add(n0,n1)|power(n1,const_2)|power(n0,const_2)|multiply(n0,#0)|subtract(#1,#2)|multiply(#4,const_4)|power(#3,const_2)|subtract(#6,#5)|sqrt(#7)|add(#3,#8)|divide(#9,const_2)|add(n0,#10) | geometry |
how long does a train 120 m long running at the speed of 70 km / hr takes to cross a bridge 145 m length ? | "speed = 70 * 5 / 18 = 19 m / sec total distance covered = 120 + 145 = 265 m . required time = 265 / 19 = 13.9 sec . answer : d" | a ) 13.9 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 13.9 sec , e ) 47.98 sec | d | divide(add(120, 145), multiply(70, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 650 , what is the least possible value of n ? | "650 = 2 * 5 * 5 * 13 so the least value forncan be 13 . d" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | divide(divide(divide(divide(650, const_2), const_3), const_4), divide(const_10, const_2)) | divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)| | general |
a rectangular field is to be fenced on three sides leaving a side of 30 feet uncovered . if the area of the field is 600 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 30 * breadth = 600 = > breadth = 20 feet area to be fenced = 2 b + l = 2 * 20 + 30 = 70 feet answer : option a" | a ) 70 feet , b ) 52 feet , c ) 32 feet , d ) 12 feet , e ) 55 feet | a | add(multiply(divide(600, 30), const_2), 30) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
two employees x and y are paid a total of rs . 560 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 560 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 560 ⇒ y [ 12 / 10 + 1 ] = 560 ⇒ 22 y / 10 = 560 ⇒ 22 y = 5600 ⇒ y = 5600 / 22 = rs . 254 ( approx ) a )" | a ) s . 254 , b ) s . 284 , c ) s . 290 , d ) s . 304 , e ) s . 380 | a | divide(multiply(560, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
calculate the sum of first 28 natural numbers . | "solution we know that ( 1 + 2 + 3 + . . . . . + 28 ) = n ( n + 1 ) / 2 therefore ( 1 + 2 + 3 + . . . . + 28 ) = ( 28 × 29 / 2 ) = 406 . answer b" | a ) 426 , b ) 406 , c ) 401 , d ) 409 , e ) 405 | b | multiply(add(28, const_1), divide(28, const_2)) | add(n0,const_1)|divide(n0,const_2)|multiply(#0,#1)| | general |
when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 4 times , what is the probability that it will land on the same side each time ? | "must be either 4 tails or 4 heads 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 + 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 1 / 16 + 1 / 16 = 1 / 8 answer : a" | a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 16 , d ) 1 / 2 , e ) 1 / 32 | a | divide(const_1, power(const_2, 4)) | power(const_2,n0)|divide(const_1,#0)| | probability |
trapezoid jklm in the x - y plane has coordinates j = ( – 2 , – 3 ) , k = ( – 2 , 1 ) , l = ( 6 , 7 ) , and m = ( 6 , – 3 ) . what is its perimeter ? | "jk = 4 lm = 10 kl = using distance formula 10 jm = using distance formula 8 sum of all is 32 a" | a ) 32 , b ) 36 , c ) 38 , d ) 40 , e ) ( f ) 42 | a | add(add(add(const_10, add(1, 3)), add(2, 6)), add(7, 3)) | add(n1,n3)|add(n0,n4)|add(n1,n5)|add(#0,const_10)|add(#3,#1)|add(#4,#2)| | physics |
the c . p of 8 books is equal to the s . p of 16 books . find his gain % or loss % ? | "explanation : 8 cp = 16 sp 16 - - - 8 cp loss 100 - - - ? = > 50 % loss answer : b" | a ) 60 % , b ) 50 % , c ) 70 % , d ) 90 % , e ) 55 % | b | multiply(subtract(const_1, divide(8, 16)), const_100) | divide(n0,n1)|subtract(const_1,#0)|multiply(#1,const_100)| | gain |
a & b started a partnership business . a ' s investment was thrice the investment of b and the period of his investment was two times the period of investments of b . if b received rs 5000 as profit , what is their total profit ? | "explanation : suppose b ' s investment = x . then a ' s investment = 3 x suppose bs period of investment = y , then a ' s period of investment = 2 y a : b = 3 x * 2 y : xy = 6 : 1 total profit * 1 / 7 = 5000 = > total profit = 5000 * 7 = 35000 . answer : option d" | a ) 28000 , b ) 30000 , c ) 32000 , d ) 35000 , e ) none of these | d | divide(5000, divide(multiply(const_1, const_1), add(multiply(const_3, const_2), multiply(const_1, const_1)))) | multiply(const_1,const_1)|multiply(const_2,const_3)|add(#1,#0)|divide(#0,#2)|divide(n0,#3)| | general |
a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 1 day . for how many number of days should a and c should work together to complete the remaining work ? | b work 1 / 8 * 1 = 1 / 8 remaining work = 1 - 1 / 8 = 7 / 8 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 take reciprocal 7 / 8 * remaining work = 8 / 7 * 1 / 4 = 2 / 7 answer : b | a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 7 | b | add(divide(1, 6), divide(1, 8)) | divide(n3,n0)|divide(n3,n1)|add(#0,#1) | physics |
the ratio by weight , measured in pounds , of books to clothes to electronics in a suitcase initially stands at 7 : 4 : 3 . someone removes 6 pounds of clothing from the suitcase , thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? | "the weights of the items in the suitcase are 7 k , 4 k , and 3 k . if removing 6 pounds of clothes doubles the ratio of books to clothes , then 6 pounds represents half the weight of the clothes . 2 k = 6 pounds and then k = 3 pounds . the electronics weigh 3 ( 3 ) = 9 pounds . the answer is e ." | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | multiply(3, divide(negate(multiply(add(add(7, 4), 3), 6)), subtract(multiply(add(add(7, 4), 3), const_2), multiply(add(add(7, 4), 3), 4)))) | add(n0,n1)|add(n2,#0)|multiply(n3,#1)|multiply(#1,const_2)|multiply(n1,#1)|negate(#2)|subtract(#3,#4)|divide(#5,#6)|multiply(n2,#7)| | other |
a bag consists of 50 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | "bag consists of 50 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 50 - 5 - 9 = 36 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 50 + 36 / 50 = ... | a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 9 / 10 | e | divide(add(subtract(50, add(5, 9)), 9), 50) | add(n1,n2)|subtract(n0,#0)|add(n2,#1)|divide(#2,n0)| | probability |
how many integers between 100 and 200 , inclusive , can not be evenly divided by 3 nor 5 ? | "the total numbers between 100 and 200 , inclusive , is 101 . 3 * 34 = 102 and 3 * 66 = 198 so the number of multiples of 3 is 33 . 5 * 20 = 100 and 5 * 40 = 200 so the number of multiples of 5 is 21 . however , the multiples of 15 have been counted twice . 15 * 7 = 105 and 15 * 13 = 195 so the number of multiples of 1... | a ) 46 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | e | subtract(200, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
for the positive integers x , x + 2 , x + 4 , x + 7 , and x + 22 , the mean is how much greater than the median ? | mean = ( x + x + 2 + x + 4 + x + 7 + x + 22 ) / 5 = ( 5 x + 35 ) / 5 = x + 7 median = x + 4 thus mean - median = x + 7 - ( x + 4 ) = 3 answer = d | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 7 | d | subtract(divide(add(add(add(2, 4), 7), 22), add(const_4, const_1)), 4) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|divide(#3,#1)|subtract(#4,n1) | 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 ) 398 km , b ) 892 km , c ) 387 km , d ) 300 km , e ) 230 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 |
3 / 4 of 1 / 2 of 2 / 5 of 5100 = ? | "c 765 ? = 5100 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 765" | a ) 392 , b ) 229 , c ) 765 , d ) 493 , e ) 540 | c | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5100) | divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)| | general |
in order to obtain an income of rs . 900 from 20 % stock at rs . 102 , one must make an investment of | "explanation : market value = rs . 102 . required income = rs . 900 . here face value is not given . take face value as rs . 100 if it is not given in the question to obtain rs . 20 ( ie , 20 % of the face value 100 ) , investment = rs . 102 to obtain rs . 900 , investment = 102 / 20 ã — 900 = rs . 4590 answer : option... | a ) rs . 6000 , b ) rs . 4590 , c ) rs . 4560 , d ) rs . 4500 , e ) rs . 4570 | b | multiply(divide(102, 20), 900) | divide(n2,n1)|multiply(n0,#0)| | gain |
what is the value of ( p + q ) / ( p - q ) if p / q is 10 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 10 + 1 ) / ( 10 - 1 ) = 11 / 9 = 11 / 9 answer : d" | a ) 4 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 11 / 9 , e ) 8 / 7 | d | divide(multiply(add(add(const_100, const_60), const_1), 10), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general |
if the price of a certain bond on may 1 st was 1 / 2 the price of the bond on june 1 st and the price of the bond on july 1 st was 50 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1... | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 1 / 2 * 12 = 6 ; the price on july 1 st = 6 * 1.5 = 9 . the average price of the bond on may 1 st and july 1 st = ( 6 + 9 ) / 2 = 7.5 . the price of the bond on june 1 st ( 12 ) is 8 / 5 times ( 160 % ) the average price of the bond on may 1 st and july ... | a ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 160 % | e | add(multiply(50, const_0_33), add(const_100, const_0_33)) | add(const_0_33,const_100)|multiply(n5,const_0_33)|add(#0,#1)| | general |
an investment compounds annually at an interest rate of 33.34 % what is the smallest investment period by which time the investment will more than triple in value ? | "assume initial amount is x annual interest is 34.1 % so after 1 year the amount will become x * ( 100 + 33.34 ) / 100 = > x * 4 / 3 now we need to find n for x * ( 4 / 3 ) ^ n = 3 x or in other words n = 4 e" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | e | add(floor(divide(log(const_3), log(add(const_1, divide(33.34, const_100))))), const_1) | divide(n0,const_100)|log(const_3)|add(#0,const_1)|log(#2)|divide(#1,#3)|floor(#4)|add(#5,const_1)| | gain |
a no . when divided by 220 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 67 ? | "221 + 43 = 264 / 17 = 9 ( remainder ) c" | a ) 4 , b ) 6 , c ) 9 , d ) 11 , e ) 14 | c | divide(add(220, 43), 67) | add(n0,n1)|divide(#0,n2)| | general |
a and b ’ s salaries together amount to rs . 3,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is b ’ s salary ? | "( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 3000 4 b = 3000 = > b = 750 answer b" | a ) 777 , b ) 750 , c ) 789 , d ) 776 , e ) 881 | b | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6)| | gain |
ritesh and co . generated revenue of rs . 1,700 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 1,700 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1700 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) × 1700 . = > 13,600 . hence , the total revenue by end of 2007 is rs . 13,600 . in 2006 , revenue... | a ) 18.38 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these | a | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | add(const_1,const_4)|add(const_2,const_4)|divide(n2,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#3,const_100)|add(#6,#5)|multiply(#6,const_2)|add(#8,#4)|divide(#7,#2)|divide(#9,#10)|multiply(#11,const_100)| | gain |
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1000 answer : a" | a ) 1000 , b ) 3889 , c ) 8908 , d ) 3780 , e ) 2709 | a | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general |
the radius of a cylindrical vessel is 7 cm and height is 5 cm . find the whole surface of the cylinder ? | "r = 7 h = 5 2 π r ( h + r ) = 2 * 22 / 7 * 7 ( 12 ) = 528 answer : b" | a ) 281 , b ) 528 , c ) 440 , d ) 767 , e ) 1981 | b | surface_cylinder(7, 5) | surface_cylinder(n0,n1)| | geometry |
a man took a loan at rate of 12 % per annum simple interest . after 3 years he had to pay 5400 interest . the principal amount borrowed by him was . | "explanation : s . i . = p ∗ r ∗ t / 100 = > p = s . i . ∗ 100 / r ∗ t = > p = 5400 ∗ 100 / 12 ∗ 3 = rs 15000 option b" | a ) rs 14000 , b ) rs 15000 , c ) rs 16000 , d ) rs 17000 , e ) none of these | b | divide(multiply(5400, const_100), multiply(12, 3)) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)| | gain |
in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 / 2 , and in every year men are reduces 12 % . what is population of after 2 year . | "next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 11 % so total man = 150 * . 88 = 132 women will be = 175 - 132 = 43 so population after two years = [ 135 ^ 2 + 43 ^ 2 ] ^ . 5 = 138.8 = 139 so population after two year = 139 answer : a" | a ) 139 , b ) 141 , c ) 142 , d ) 143 , e ) 144 | a | power(add(power(multiply(subtract(1, divide(1, 12)), 150), 2), power(subtract(power(add(power(150, 2), power(90, 2)), divide(1, 2)), multiply(subtract(1, divide(1, 12)), 150)), 2)), divide(1, 2)) | divide(n4,n6)|divide(n4,n2)|power(n0,n2)|power(n1,n2)|add(#2,#3)|subtract(n4,#0)|multiply(n0,#5)|power(#4,#1)|power(#6,n2)|subtract(#7,#6)|power(#9,n2)|add(#8,#10)|power(#11,#1)| | general |
evaluate : 58 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 58 - 12 * 3 * 2 = 58 - 8 = 50 correct answer b" | a ) 42 , b ) 50 , c ) 62 , d ) 72 , e ) 82 | b | subtract(58, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
jenny has recently moved house and is working out her optimal commute . she can take an indirect route which involves taking a bus for 1 mile and then walk 0.75 miles . or she can take a direct route and walk for 1.5 miles and this takes her 30 minutes . what is the most amount time that the bus journey can take for je... | if jenny can walk 1.5 miles in 30 minutes , then she can talk 0.75 miles in half the time i . e . 15 minutes . for jenny to be indifferent between the two routes , the two journeys must have the same travel time . the direct route has a travel time of 30 minutes ( walking ) . the indirect route has a travel time of 15 ... | a ) 5 minutes , b ) 10 minutes , c ) 15 minutes , d ) 20 minutes , e ) 25 minutes | c | subtract(30, divide(0.75, divide(1.5, 30))) | divide(n2,n3)|divide(n1,#0)|subtract(n3,#1) | physics |
seller selling an apple for rs . 15 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 15 loss = cp 16 loss = cp − sp = cp − 15 ⇒ cp 16 = cp − 15 ⇒ 15 cp 16 = 15 ⇒ cp 16 = 1 ⇒ cp = 16 c" | a ) 10 , b ) 12 , c ) 16 , d ) 18 , e ) 20 | c | add(15, 1) | add(n0,n1)| | general |
if the volume of a cube is x ^ 3 cubic units , what is the number of square units in the surface area of the cube ? | the question basically asks you to calculate the surface area of the cube . volume of the cube = x ^ 3 this means each side of the cube = x hence area of each side = x ^ 2 total surfaces in a cube = 6 total surface area = 6 * x 2 correct option : d | ['a ) x ^ 2', 'b ) x ^ 3', 'c ) x ^ 6', 'd ) 6 x ^ 2', 'e ) 6 x ^ 3'] | d | multiply(multiply(const_2, const_3), multiply(const_1, const_1)) | multiply(const_2,const_3)|multiply(const_1,const_1)|multiply(#0,#1) | geometry |
the area of a square field is 28808 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ? | "area of a square field = 28808 sq m let the side of square = a a ^ 2 = 28808 = > a = 169.73 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 169.73 = 240 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 240 / 120 = 2 min answer c" | a ) 2.5 min , b ) 3 min , c ) 2 min , d ) 4 min , e ) 4.5 min | c | divide(28808, multiply(7.2, const_1000)) | multiply(n1,const_1000)|divide(n0,#0)| | geometry |
if a town of 35,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a ) a . 36000 , b ) b . 37000 , c ) c . 38000 , d ) d . 39000 , e ) e . 40000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | add(const_3,const_4)|add(#0,const_3)|multiply(n2,#0)|multiply(#2,const_100)|multiply(#1,#3) | gain |
there are 12 slate rocks , 17 pumice rocks , and 8 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "12 / 37 * 11 / 36 = 11 / 111 the answer is d ." | a ) 5 / 36 , b ) 7 / 64 , c ) 9 / 83 , d ) 11 / 111 , e ) 13 / 135 | d | multiply(divide(12, add(add(12, 17), 8)), divide(subtract(12, const_1), subtract(add(add(12, 17), 8), const_1))) | add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)| | other |
one day a car rental agency rented 3 / 4 of its cars , including 5 / 6 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 6 ) ( 3 / 5 ) = 1 / 10 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 1 / 10 ) / ( 1 / 4 ) = 2 / 5 the answer is e ." | a ) 13 / 20 , b ) 11 / 15 , c ) 7 / 10 , d ) 4 / 7 , e ) 2 / 5 | e | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 5), 6)), 6), divide(multiply(divide(multiply(4, multiply(multiply(4, 5), 6)), 6), 4), 5)), subtract(multiply(multiply(4, 5), 6), divide(multiply(3, multiply(multiply(4, 5), 6)), 4))), const_100) | multiply(n1,n2)|multiply(n3,#0)|multiply(n1,#1)|multiply(#1,const_2)|divide(#2,n3)|divide(#3,n1)|multiply(n1,#4)|subtract(#1,#5)|divide(#6,n2)|subtract(#4,#8)|divide(#9,#7)|multiply(#10,const_100)| | general |
how many words , with or without meaning , can be formed using all letters of the word red using each letter exactly once ? | the word red has exactly 3 letters which are all different . therefore the number of words that can be formed = number of permutations of 3 letters taken all at a time . = p ( 3 , 3 ) = 3 ! = 3 x 2 × 1 = 6 answer : a | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | factorial(const_3) | factorial(const_3) | general |
of the 65 cars on a car lot , 30 have air - conditioning , 45 have air bag , and 12 have both air - conditioning and air bag . how many of the cars on the lot have neither air - conditioning nor air bag ? | "total - neither = all air conditioning + all air bags - both or 65 - neither = 30 + 45 - 12 = 63 . = > neither = 2 , hence c . answer : c" | a ) 2 , b ) 8 , c ) 10 , d ) 15 , e ) 18 | c | subtract(65, subtract(add(30, 45), 12)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
a and b can do a piece of work in 2 days . with the help of c they finish the work in 1 days . c alone can do that piece of work in ? | c = 1 / 1 – 1 / 2 = 1 / 2 = > 2 days answer : e | a ) 1.5 days , b ) 4 days , c ) 3 days , d ) 1 day , e ) 2 days | e | inverse(subtract(const_1, divide(1, 2))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1) | physics |
four machines , each working at the same constant rate , together can complete a certain job in 20 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 10 days ? | "rate of one machine = 1 job / ( 20 * 4 ) days let x = number of machines needed to complete the job in 10 days 1 / ( 4 * 20 ) * 10 * x = 1 job x = 8 8 - 4 = 4 answer : c" | a ) 5 , b ) 3 , c ) 4 , d ) 6 , e ) 1 | c | subtract(divide(multiply(20, add(const_4, const_1)), 10), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,n1)|subtract(#2,#0)| | physics |
a car gets 32 miles to the gallon . if it is modified to use a solar panel , it will use only 80 percent as much fuel as it does now . if the fuel tank holds 12 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ? | originally , the distance the car could go on a full tank was 12 * 32 = 384 miles . after it has been modified , the car can go 32 / 0.8 = 40 miles per gallon . on a full tank , the car can go 12 * 40 = 480 miles , thus 96 miles more . the answer is d . | a ) 84 , b ) 88 , c ) 92 , d ) 96 , e ) 100 | d | subtract(multiply(multiply(divide(32, 80), const_100), 12), multiply(32, 12)) | divide(n0,n1)|multiply(n0,n2)|multiply(#0,const_100)|multiply(n2,#2)|subtract(#3,#1) | physics |
john bought 9.25 m of cloth for $ 434.75 . find the cost price per metre . | "cloth bought by john = 9.25 m cost of 9.25 m = $ 434.75 cost per metre = 434.75 ÷ 9.25 the cost of the cloth per metre = $ 47 answers : b" | a ) 46 , b ) 47 , c ) 58 , d ) 56 , e ) 54 | b | divide(434.75, 9.25) | divide(n1,n0)| | physics |
the average of 60 results is 40 and the average of other 40 results is 60 . what is the average of all the results ? | "answer sum of 100 result = sum of 60 result + sum of 40 result . = 60 x 40 + 40 x 60 = 4800 / 100 correct option : c" | a ) 24 , b ) 25 , c ) 48 , d ) 50 , e ) none | c | divide(add(multiply(60, 40), multiply(40, 60)), add(60, 40)) | add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)| | general |
3 , 7,12 , 18,25 , . . . . . . . . . . . . . . 6 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 answer : d" | a ) 43 , b ) 53 , c ) 63 , d ) 33 , e ) 78 | d | subtract(negate(6), multiply(subtract(7,12, 18,25), divide(subtract(7,12, 18,25), subtract(3, 7,12)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a person was asked to state his age in years . his reply was , ` ` take my age 3 years hence , multiply it by 4 and subtract 4 times my age 3 years ago and you will know how old i am . ' ' what was the age of the person ? | explanation : let the present age of person be x years . then , 4 ( x + 3 ) - 4 ( x - 3 ) = x < = > ( 4 x + 12 ) - ( 4 x - 12 ) = x < = > x = 24 . . answer : d | a ) 18 , b ) 92 , c ) 27 , d ) 24 , e ) 19 | d | add(multiply(3, 4), multiply(3, 4)) | multiply(n0,n1)|add(#0,#0) | general |
there are cats got together and decided to kill the mice of 999984 . each cat kills equal number of mice and each cat kills more number of mice than cats there were . then what are the number of cats ? | "999984 can be written as 1000000 â € “ 16 = 10002 â € “ 42 ie of the form a 2 - b 2 = ( a + b ) ( a - b ) = ( 1000 + 4 ) * ( 1000 - 4 ) = ( 1004 ) * ( 996 ) given that number of cats is less than number if mice . so number of cats is 996 and number of mice were 1004 answer c" | a ) 941,1009 , b ) 991,1001 , c ) 996,1004 , d ) 791,1009 , e ) 931,1009 | c | divide(999984, add(multiply(const_100, const_10), add(const_3, const_2))) | add(const_2,const_3)|multiply(const_10,const_100)|add(#0,#1)|divide(n0,#2)| | general |
100 kg of an alloy a is mixed with 200 kg of alloy b . if alloy a has lead and tin in the ratio 5 : 3 and alloy b has tin and copper in the ratio 2 : 3 , then the amount of tin in the new alloy is ? | "quantity of tin in 100 kg of a = 100 * 3 / 8 = 37.5 kg quantity of tin in 200 kg of b = 200 * 2 / 5 = 80 kg quantity of tin in the new alloy = 37.5 + 80 = 117.5 kg answer is c" | a ) 100.6 kg , b ) 120.3 kg , c ) 117.5 kg , d ) 110.8 kg , e ) 114 kg | c | add(multiply(divide(100, add(5, 3)), 3), multiply(divide(200, add(2, 3)), 2)) | add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)| | other |
a certain automobile company ’ s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 1 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 30 vehicles that are not speedsters , how many speedster convert... | "total vehicle = 1 / 3 of speedster + 2 / 3 of others . speedster convertibles = 1 / 3 total vehicle * 4 / 5 given : 2 / 3 constitutes 30 vehicles . hence 1 / 3 constitutes 15 speedster convertibls = 15 * 4 / 5 = 12 c" | a ) 13 , b ) 18 , c ) 12 , d ) 15 , e ) 17 | c | multiply(multiply(divide(30, subtract(const_1, divide(1, 3))), divide(1, 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 group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 35 adults have their meal , find the total number of children that can be catered with the remaining food . | "explanation : as there is meal for 70 adults and 35 have their meal , the meal left can be catered to 35 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 35 adults = 45 children hence , the meal can be catered to 45 children . answer : d" | a ) 33 , b ) 54 , c ) 18 , d ) 45 , e ) 01 | d | multiply(subtract(70, 35), divide(90, 70)) | divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1)| | general |
the monthly rent of a shop of dimension 10 feet × 10 feet is rs . 1300 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 1300 / ( 10 * 10 ) = 13 & annual rent per square feet = 12 * 13 = 156 answer : b" | a ) 43 , b ) 156 , c ) 68 , d ) 87 , e ) 92 | b | multiply(add(const_10, const_2), divide(1300, rectangle_area(10, 10))) | add(const_10,const_2)|rectangle_area(n0,n1)|divide(n2,#1)|multiply(#0,#2)| | geometry |
a candidate got 35 % of the votes polled and he lost to his rival by 2280 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2280 100 % - - - - - - - - - ? = > 7600 answer : e" | a ) 7500 , b ) 3388 , c ) 2665 , d ) 2888 , e ) 7600 | e | divide(2280, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
it takes 3 workers a total of 9 hours to build a giant machine with each worker working at the same rate . if 3 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 2 : 00 pm , at what time will the machine be complete ? | "3 workers build 1 / 9 of a machine in one hour . 1 worker builds 1 / 27 of a machine in one hour . in the first 3 hours , 3 workers build 3 * ( 1 / 27 ) * 3 = 9 / 27 of a machine from 2 : 00 to 3 : 00 , 4 workers build another 4 / 27 . the total is 13 / 27 . from 3 : 00 to 4 : 00 , 5 workers build another 5 / 27 . the... | a ) 5 : 15 , b ) 5 : 26 , c ) 5 : 37 , d ) 5 : 48 , e ) 5 : 59 | b | subtract(divide(multiply(3, 9), 9), const_0_33) | multiply(n0,n1)|divide(#0,n1)|subtract(#1,const_0_33)| | general |
the ratio of the volumes of a cube to that of the sphere which will fit inside the cube is ? | 52 * 40 * 29 = 60320 48 * 36 * 27 = 46650 - - - - - - - - - - - - 13664 13664 * 1 / 2 = 6832 = > 6.832 kg answer : d | ['a ) 6.862 kg', 'b ) 6.838 kg', 'c ) 6.632 kg', 'd ) 6.832 kg', 'e ) 6.834 kg'] | d | divide(volume_cube(const_1), volume_sphere(divide(const_1, const_2))) | divide(const_1,const_2)|volume_cube(const_1)|volume_sphere(#0)|divide(#1,#2) | geometry |
a and b start walking towards each other at 1 pm at speed of 5 kmph and 7 kmph . they were initially 24 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 24 / 7 + 5 = 24 / 12 = 2 hrs after 1 pm = 3 pm answer is c" | a ) 4 pm , b ) 6 pm , c ) 3 pm , d ) 10 pm , e ) 5 pm | c | add(1, divide(24, add(5, 7))) | add(n1,n2)|divide(n3,#0)|add(n0,#1)| | physics |
a train running at the speed of 60 km / hr crosses a pole in 27 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 27 ) m = 450 m . answer : c" | a ) 186 m , b ) 176 m , c ) 450 m , d ) 150 m , e ) 765 m | c | multiply(divide(multiply(60, const_1000), const_3600), 27) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
fresh grapes contain 75 % by weight while dried grapes contain 25 % water by weight . what is the weight of dry grapes available from 200 kg of fresh grapes ? | "from the question we know : 200 kg * 75 % = 150 kg of water in the fresh grapes 200 kg - 150 kg of water = 50 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 25 % of the weight of the dry graps is water and we know that 50 kg is non - water mass we can se... | a ) 40.00 kg , b ) 45.50 kg , c ) 66.67 kg , d ) 80.56 kg , e ) 90.00 kg | c | multiply(divide(divide(multiply(subtract(const_100, 75), 200), const_100), subtract(const_100, 25)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain |
the ratio of the volumes of two cubes is 1000 : 3375 . what is the ratio of their total surface areas ? | "ratio of the sides = â ³ â ˆ š 1000 : â ³ â ˆ š 3375 = 10 : 15 ratio of surface areas = 10 ^ 2 : 15 ^ 2 = 100 : 225 = 4 : 9 answer : b" | a ) 81 : 121 , b ) 4 : 9 , c ) 729 : 1331 , d ) 27 : 121 , e ) none of these | b | power(divide(1000, 3375), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry |
a man goes into a barbershop and gets a $ 15 haircut . he pays the barber with a $ 20 bill , but the barber does n ' t have change because it is an especially slow day . the barber goes to the flower shop next to his own and trades the $ 20 bill for a $ 10 dollar bill and two $ 5 bills . then he gives the man his $ 5 i... | he lost $ 5 . the lady from the flower shop gave him $ 20 in change , but eventually got $ 20 back . she broke even . the man who got the haircut gave the barber nothing ( except counterfeit money ) but got $ 5 back from the barber . so the barber lost only $ 5 , and gave a free haircut . | a ) $ 5 , b ) $ 10 , c ) $ 12 , d ) $ 15 , e ) $ 20 | a | subtract(20, 15) | subtract(n1,n0) | general |
if the cost price of 72 articles is equal to the selling price of 60 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 60 y = 72 x y = 1.2 x the answer is c ." | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | c | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 60), 72)), divide(multiply(const_100, 60), 72))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
of all the homes on gotham street , 1 / 3 are termite - ridden , and 5 / 8 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | "the fraction of homes which are termite - infested but not collapsing is 3 / 8 * 1 / 3 = 1 / 8 the answer is e ." | a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 8 | e | divide(add(1, const_2.0), multiply(3, 8)) | add(n0,const_2.0)|multiply(n1,n3)|divide(#0,#1)| | general |
what will be the compound interest on a sum of rs . 22,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 22000 * ( 1 + 12 / 100 ) 3 ] = 22000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 30908.42 c . i . = ( 30908.42 - 22000 ) = rs . 8908.42 answer : a" | a ) rs . 8908.42 , b ) rs . 9720 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
of the votes cast on a certain proposal , 68 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . ) | "let x be the total number of votes cast . 0.6 x = 0.4 x + 68 0.2 x = 68 x = 340 the answer is b ." | a ) 300 , b ) 340 , c ) 380 , d ) 420 , e ) 460 | b | divide(68, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n0,#2)| | gain |
a box is in the shape of a semicircle with a radius of 12 . what is the approximate perimeter of the semicircle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 12 + 2 * 12 = 61.68 approximately 62 answer b | ['a ) 54', 'b ) 62', 'c ) 25', 'd ) 34', 'e ) 60'] | b | add(multiply(const_pi, 12), multiply(const_2, 12)) | multiply(n0,const_pi)|multiply(n0,const_2)|add(#0,#1) | geometry |
a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates x apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates x / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | add(const_2,const_3)|add(const_3,const_4)|divide(n2,const_2)|multiply(n2,const_2)|multiply(#0,#2)|multiply(n2,#1)|add(#4,#3)|divide(#6,#5)| | physics |
in traveling from a dormitory to a certain city , a student went 1 / 5 of the way by foot , 2 / 3 of the way by bus , and the remaining 4 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 5 x + 2 / 3 x + 4 x - 13 / 15 x = 4 2 / 15 x = 4 = > so x = ( 15 / 2 ) * 4 = 30 km answer a" | a ) 30 , b ) 45 , c ) 60 , d ) 90 , e ) 120 | a | multiply(4, inverse(subtract(1, add(divide(1, 5), divide(2, 3))))) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(n0,#2)|inverse(#3)|multiply(n4,#4)| | physics |
a tree of height 36 m is on one edge of a road broke at a certain height . it fell in such a way that the top of the tree touches the other edge of the road . if the breadth of the road is 12 m , then what is the height at which the tree broke ? | let the tree was broken at x meters height from the ground and 36 - x be the length of other part of the tree . from the diagram , ( 36 − x ) 2 = x 2 + 122 ( 36 − x ) 2 = x 2 + 122 ⇒ 1296 − 72 x + x 2 = x 2 + 144 ⇒ 1296 − 72 x + x 2 = x 2 + 144 ⇒ 72 x = 1296 − 144 ⇒ 72 x = 1296 − 144 ⇒ x = 16 answer : a | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | divide(subtract(power(36, const_2), power(12, const_2)), multiply(36, const_2)) | multiply(n0,const_2)|power(n0,const_2)|power(n1,const_2)|subtract(#1,#2)|divide(#3,#0) | physics |
if y is the smallest positive integer such that 3,675 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 12 * 3675 = 44 , 100 , which is a square of 210 e" | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 12 | e | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
if a person walks at 16 km / hr instead of 12 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 12 = ( x + 20 ) / 16 x - 60 = > x = 60 km . answer : e" | a ) 50 , b ) 65 , c ) 70 , d ) 61 , e ) 60 | e | multiply(12, divide(20, subtract(16, 12))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
a man is 29 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 29 ) + 2 = 2 ( x + 2 ) x + 31 = 2 x + 4 x = 27 . answer : d" | a ) 14 years , b ) 18 years , c ) 20 years , d ) 27 years , e ) 16 years | d | divide(subtract(29, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
the average of 25 results is 18 . the average of first 12 of them is 10 and that of last 12 is 20 . find the 13 th result ? | 13 th result = sum of 25 results - sum of 24 results = 18 * 25 - 12 * 10 - 12 * 20 = 450 - 120 - 240 = 90 answer is c | a ) 60 , b ) 75 , c ) 90 , d ) 100 , e ) 110 | c | subtract(subtract(multiply(25, 18), multiply(12, 20)), multiply(12, 10)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2) | general |
the average weight of a group of boys is 30 kg . after a boy of weight 36 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 36 kg joins the group , total weight of boys = 30 x + 36 so 30 x + 36 = 31 ( x + 1 ) = > x = 5 . answer : a" | a ) a ) 5 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 1 | a | add(subtract(36, add(30, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
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