Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5
values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6
values | answer stringclasses 5
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a man buys rs . 56 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | "dividend on rs . 56 = rs . 9 / 100 x 56 = rs . 5.04 . rs . 12 is an income on rs . 100 . rs . 5.04 is an income on rs . 100 / 12 x 5.04 = rs . 42 . answer : option e" | a ) s . 12 , b ) s . 15 , c ) s . 18 , d ) s . 21 , e ) s . 42 | e | multiply(divide(const_100, 12), multiply(divide(9, const_100), 56)) | divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2)| | gain | E |
the number which exceeds 12 % of it by 52.8 is : | "explanation : let the number be x . then , x – 12 % of x = 52.8 x – ( 12 / 100 ) x = 52.8 x ( 1 – 12 / 100 ) = 52.8 ( 88 / 100 ) x = 52.8 x = ( 100 x 52.8 ) / 88 = 60 answer : a" | a ) 60 , b ) 77 , c ) 269 , d ) 26 , e ) 91 | a | divide(multiply(52.8, const_100), subtract(const_100, 12)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | A |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 9 ) = 5 / 36 the cistern will be filled in 36 / 5 hrs i . e . , 7.2 hrs . answer : d" | a ) 2.9 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 7.2 hrs , e ) 8.6 hrs | d | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | D |
if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 3 = 2 d / 15 the average of november and january is ( n + j ) / 2 = 8 d / 15 / 2 = 4 d / 15 d is 15 / 4 times the average of november and january . the answer is d ." | a ) 5 / 3 , b ) 5 / 4 , c ) 10 / 3 , d ) 15 / 4 , e ) 15 / 2 | d | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 3))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general | D |
think of a number , divide it by 5 and add 8 to it . the result is 61 . what is the number thought of ? | "explanation : 61 - 6 = 53 53 x 5 = 265 answer : d" | a ) 24 , b ) 77 , c ) 297 , d ) 265 , e ) 29 | d | multiply(subtract(61, 8), 5) | subtract(n2,n1)|multiply(n0,#0)| | general | D |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 470 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 -... | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 470 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 770 a + b + c = 70 per hour so for eight hrs = 70 * 8 = 560 =... | a ) 400 , b ) 475 , c ) 550 , d ) 560 , e ) 700 | d | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0)| | physics | D |
danny obtained 76 , 65 , 82 , 67 and 75 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | average = ( 76 + 65 + 82 + 67 + 75 ) / 5 = 365 / 5 = 73 . answer : e | a ) a ) 87 , b ) b ) 99 , c ) c ) 68 , d ) d ) 82 , e ) e ) 73 | e | divide(add(add(add(add(76, 65), 82), 67), 75), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general | E |
what number times ( 1 ⁄ 4 ) ^ 2 will give the value of 4 ^ 3 ? | "x * ( 1 / 4 ) ^ 2 = 4 ^ 3 x = 4 ^ 2 * 4 ^ 3 = 4 ^ 5 = 1024 the answer is e ." | a ) 4 , b ) 16 , c ) 64 , d ) 256 , e ) 1024 | e | multiply(power(const_2.0, 4), power(4, 3)) | power(n2,n1)|power(n1,n4)|multiply(#0,#1)| | general | E |
what is log 3 ( 4 ) log 4 ( 5 ) . . . log 80 ( 81 ) ? | recall the change of base formula logb ( a ) = ln ( a ) ln ( b ) : ( alternatively , we can substitute logc for ln on the right , as long as the base is the same on the top and on the bottom . ) using this , we can rewrite the entire expression using natural logarithms , as ln ( 4 ) ln ( 3 ) ln ( 5 ) ln ( 4 ) ln ( 81 )... | a ) 1 , b ) 4 , c ) 81 , d ) log 80 243 , e ) e 81 | b | divide(log(81), log(3)) | log(n5)|log(n0)|divide(#0,#1) | other | B |
( 132 ) ^ 7 × ( 132 ) ^ ? = ( 132 ) ^ 11.5 . | "( 132 ) ^ 7 × ( 132 ) ^ x = ( 132 ) ^ 11.5 = > 7 + x = 11.5 = > x = 11.5 - 7 = 4.5 answer is a" | a ) 4.5 , b ) 3 , c ) 4 , d ) 3.5 , e ) 8.5 | a | divide(multiply(power(132, 7), power(132, 132)), 11.5) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4)| | general | A |
a certain bacteria colony doubles in size every day for 16 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ? | if there is one bacteria colony , then it will reach the limit of its habitat in 16 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 15 times . answer : d . similar questions to pra... | a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19 | d | subtract(16, divide(16, 16)) | divide(n0,n0)|subtract(n0,#0) | physics | D |
find the area of a rhombus one side of which measures 20 cm and one diagonal is 23 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 23 x 32 ) cm 2 = 368 ... | a ) 100 cm 2 , b ) 150 cm 2 , c ) 300 cm 2 , d ) 368 cm 2 , e ) 400 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 23), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(23, 23)))), 23) | divide(const_1,const_2)|multiply(n0,n0)|multiply(n1,n1)|multiply(n1,#0)|multiply(#1,const_4)|subtract(#4,#2)|sqrt(#5)|multiply(#3,#6)|add(n1,#7)| | geometry | D |
the total price of a basic computer and printer are $ 2,500 . if the same printer had been purchased with an enhanced computer whose price was $ 500 more than the price of the basic computer , then the price of the printer would have been 1 / 3 of that total . what was the price of the basic computer ? | "let the price of basic computer be c and the price of the printer be p : c + p = $ 2,500 . the price of the enhanced computer will be c + 500 and total price for that computer and the printer will be 2,500 + 500 = $ 3,000 . now , we are told that the price of the printer is 1 / 3 of that new total price : p = 1 / 3 * ... | a ) 1500 , b ) 1600 , c ) 1750 , d ) 1900 , e ) 2000 | a | subtract(multiply(multiply(const_0_25, const_1000), const_10), divide(add(500, multiply(multiply(const_0_25, const_1000), const_10)), 3)) | multiply(const_0_25,const_1000)|multiply(#0,const_10)|add(n1,#1)|divide(#2,n3)|subtract(#1,#3)| | general | A |
the height of the wall is 6 times its width and lenght of the wall is 7 times its height . if the volume of the wall be 16128 cu . m . its width is | explanation : let width = x then , height = 6 x and length = 42 x 42 x × 6 x × x = 16128 x = 4 answer : a | ['a ) 4 m', 'b ) 5 m', 'c ) 6 m', 'd ) 7 m', 'e ) 8 m'] | a | power(divide(16128, multiply(multiply(6, 7), 6)), divide(const_1, const_3)) | divide(const_1,const_3)|multiply(n0,n1)|multiply(n0,#1)|divide(n2,#2)|power(#3,#0) | physics | A |
angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "let the weekly sales of both = x 550 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 23500 answer : a" | a ) 23,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(1,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 1,000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,n2)|floor(#6)| | general | A |
speed of a boat in standing water is 16 kmph and the speed of the stream is 2 kmph . a man rows to a place at a distance of 7380 km and comes back to the starting point . the total time taken by him is : | "explanation : speed downstream = ( 16 + 2 ) = 18 kmph speed upstream = ( 16 - 2 ) = 14 kmph total time taken = 7380 / 18 + 7380 / 14 = 410 + 527.1 = 937.1 hours answer : option b" | a ) 914.2 hours , b ) 937.1 hours , c ) 915 hours , d ) 905 hours , e ) 915 hours | b | add(divide(7380, add(16, 2)), divide(7380, subtract(16, 2))) | add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3)| | physics | B |
a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 6 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | "ci * vi = cf * vf ( 7 / 12 ) * ( v 1 - 6 ) = ( 7 / 16 ) * v 1 ( v 1 - 6 ) / v 1 = 3 / 4 6 accounts for the difference of 1 on ratio scale so initial volume = v 1 = 4 * 6 = 24 litres . 7 / 12 of the initial mixture was liquid a so liquid a was ( 7 / 12 ) * 24 = 14 litres . answer : c" | a ) 12 , b ) 18 , c ) 14 , d ) 20 , e ) 22 | c | multiply(7, divide(multiply(add(7, 6), subtract(6, multiply(divide(5, add(7, 5)), 6))), subtract(multiply(6, 7), multiply(7, 5)))) | add(n0,n4)|add(n0,n1)|multiply(n0,n4)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | other | C |
( 1000 ) 7 ÷ ( 10 ) 15 = ? | "explanation : = ( 103 ) 7 / ( 10 ) 15 = ( 10 ) 21 / ( 10 ) 15 = 10 ( 6 ) = 1000000 option c" | a ) 10 , b ) 100 , c ) 1000000 , d ) 10000 , e ) none of these | c | multiply(1000, 10) | multiply(n0,n2)| | general | C |
what is the remainder when 8 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 3 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? d | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(8, 8)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general | D |
15.06 * 0.0001 = ? | "explanation : clearly after decimal 6 digits should be there . option b" | a ) 15060000 , b ) 0.001506 , c ) 0.01506 , d ) 0.1506 , e ) none of these | b | multiply(divide(15.06, 0.0001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | B |
0.999991 + 0.111119 = ? | "0.999991 + 0.111119 = 0.999991 + 0.11111 + 0.00009 = ( 0.999991 + 0.00009 ) + 0.11111 = 1 + 0.11111 = 1.11111 d" | a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.11111 , e ) 1.1111 | d | multiply(divide(0.999991, 0.111119), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | D |
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 2 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 2 w = 120 = > w = 60 answer d" | a ) 65 , b ) 63 , c ) 52 , d ) 60 , e ) 68 | d | divide(rectangle_area(5, 24), 2) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry | D |
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 . 10 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 10 = rs . 6750 answer : d" | a ) 6350 , b ) 7357 , c ) 6328 , d ) 6750 , e ) 7560 | d | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 10))), rectangle_area(75, 55)), 10) | multiply(n2,const_2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry | D |
two trains 170 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 170 + 160 = 330 m . required time = 330 * 9 / 250 = 297 / 25 = 11.88 sec . answer : e" | a ) 10.6 sec , b ) 18.8 sec , c ) 14.8 sec , d ) 10.88 sec , e ) 11.88 sec | e | divide(add(170, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | E |
on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 20 percent . what was the total rent the motel actually charged ... | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 20 % . 2 x = 200 = > x = 1000 answer c" | a ) $ 600 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | c | divide(multiply(10, subtract(60, 40)), divide(20, const_100)) | divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0)| | gain | C |
how many cubes of 4 cm edge can be cut out of a cube of 28 cm edge | "explanation : number of cubes = ( 28 x 28 x 28 ) / ( 4 x 4 x 4 ) = 343 answer : a" | a ) 343 , b ) 232 , c ) 216 , d ) 484 , e ) none of these | a | divide(volume_cube(28), volume_cube(divide(4, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | probability | A |
in a class of 78 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "explanation : you could solve this by drawing a venn diagram . a simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking french to find the number taking only french . likewise find those taking only german . then we have : total = only french + only german ... | a ) 26 , b ) 27 , c ) 24 , d ) 18 , e ) 11 | c | subtract(78, subtract(add(41, 22), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other | C |
how many even 4 - digit numbers can be formed , so that the numbers are divisible by 4 and no two digits are repeated ? | number is divisible by 4 if the last two digits form a number divisible by 4 . therefore last two digits can be : 00 ; 04 ; 08 ; 12 16 ; . . . 96 . basically multiples of 4 in the range 0 - 96 , inclusive . multiples of 4 in a range 0 - 96 , inclusive are last multiple in the range − first multiple in the range 4 + 1 =... | a ) 336 , b ) 784 , c ) 1120 , d ) 1804 , e ) 1936 | c | add(multiply(multiply(subtract(subtract(const_10, const_2), const_1), subtract(subtract(const_10, const_2), const_1)), subtract(subtract(divide(const_100, 4), const_3), multiply(const_2, const_3))), multiply(multiply(subtract(const_10, const_2), subtract(subtract(const_10, const_2), const_1)), multiply(const_2, const_3... | divide(const_100,n0)|multiply(const_2,const_3)|subtract(const_10,const_2)|subtract(#2,const_1)|subtract(#0,const_3)|multiply(#3,#3)|multiply(#2,#3)|subtract(#4,#1)|multiply(#5,#7)|multiply(#6,#1)|add(#8,#9) | general | C |
what is the units digit of ( 147 ^ 25 ) ^ 50 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 25 has the form 4 n + 1 so the units digit is 7 inside the bracket . the exponent 50 has the form 4 n + 2 , so the units digit is 9 . the answer is e . | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2) | general | E |
an inspector rejects 0.08 % of the meters as defective . how many will be examine to project ? | let the number of meters to be examined be x . then , 0.08 % of x = 2 [ ( 8 / 100 ) * ( 1 / 100 ) * x ] = 2 x = [ ( 2 * 100 * 100 ) / 8 ] = 2500 answer is d . | a ) 2000 , b ) 2300 , c ) 2700 , d ) 2500 , e ) 250 | d | divide(multiply(multiply(const_2, const_100), const_100), multiply(0.08, const_100)) | multiply(const_100,const_2)|multiply(n0,const_100)|multiply(#0,const_100)|divide(#2,#1) | gain | D |
n is the greatest number which divides 1305 , 4665 and 6905 and gives the same remainder in each case . what is the sum of the digits in n ? | if the remainder is same in each case and remainder is not given , hcf of the differences of the numbers is the required greatest number 6905 - 1305 = 5600 6905 - 4665 = 2240 4665 - 1305 = 3360 hence , the greatest number which divides 1305 , 4665 and 6905 and gives the same remainder , n = hcf of 5600 , 2240 , 3360 = ... | a ) 23 , b ) 28 , c ) 4 , d ) 1 , e ) 25 | c | add(add(divide(subtract(divide(subtract(divide(gcd(gcd(subtract(4665, 1305), subtract(6905, 4665)), subtract(6905, 1305)), const_10), reminder(divide(gcd(gcd(subtract(4665, 1305), subtract(6905, 4665)), subtract(6905, 1305)), const_10), const_10)), const_10), reminder(divide(subtract(divide(gcd(gcd(subtract(4665, 1305)... | subtract(n1,n0)|subtract(n2,n1)|subtract(n2,n0)|gcd(#0,#1)|gcd(#3,#2)|divide(#4,const_10)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|reminder(#8,const_10)|subtract(#8,#9)|divide(#10,const_10)|add(#11,#9)|add(#12,#6) | general | C |
a goods train runs at the speed of 72 km / hr and crosses a 250 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 250 ) / 26 = 20 x = 270 m . answer : d" | a ) 228 , b ) 2000 , c ) 267 , d ) 270 , e ) 274 | d | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 250) | divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)| | physics | D |
what least number should be subtracted from 13603 such that the remainder is divisible by 87 ? | "13603 ÷ 87 = 156 , remainder = 31 hence 31 is the least number which can be subtracted from 13603 such that the remainder is divisible by 87 answer is e" | a ) 27 , b ) 29 , c ) 28 , d ) 30 , e ) 31 | e | reminder(13603, 87) | reminder(n0,n1)| | general | E |
tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 56 people bought tickets to the concert , what was the total revenue from ticket sales ? | "price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 26 tickets = 20 * 26 = 520 revenue generated from sales of 56 tickets = 120 + ... | a ) $ 600 , b ) $ 740 , c ) $ 850 , d ) $ 980 , e ) $ 1,140 | d | multiply(add(add(subtract(subtract(56, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n0)|subtract(const_1,#0)|subtract(#2,n1)|subtract(const_1,#1)|multiply(n1,#3)|multiply(n0,#5)|add(#6,#4)|add(#8,#7)|multiply(n0,#9)| | gain | D |
how many numbers between 100 and 600 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 600 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 600 : 600 / 42 = 14.28 hence , 42 * 14 = 588 is the last number divisible by 42 within 60... | a ) 11 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | subtract(divide(600, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)| | general | B |
a 35 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 35 = 10 answer : a" | a ) 10 , b ) 20 , c ) 88 , d ) 77 , e ) 14 | a | subtract(35, divide(35, add(divide(2, 5), const_1))) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | physics | A |
how many seconds will it take for a car that is traveling at a constant rate of 14 miles per hour to travel a distance of 96 yards ? ( 1 mile = 1,160 yards ) | "speed = 14 miles / hr = 6.84 yard / s distance = 96 yards time = distance / speed = 96 / 6.84 = 14 sec ans - d" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 17 | d | divide(96, multiply(divide(14, const_3600), multiply(subtract(add(14, 14), const_4), const_10))) | add(n0,n0)|divide(n0,const_3600)|subtract(#0,const_4)|multiply(#2,const_10)|multiply(#1,#3)|divide(n1,#4)| | physics | D |
in how many years will a sum of money doubles itself at 12 % per annum on simple interest ? | "p = ( p * 12 * r ) / 100 r = 8 % answer : d" | a ) 15 % , b ) 20 % , c ) 10 % , d ) 8 % , e ) 22 % | d | divide(const_100, 12) | divide(const_100,n0)| | gain | D |
the average weight of a group of boys is 30 kg . after a boy of weight 38 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 38 kg joins the group , total weight of boys = 30 x + 38 so 30 x + 38 = 31 ( x + 1 ) = > x = 7 . answer : e" | a ) a ) 4 , b ) b ) 8 , c ) c ) 6 , d ) d ) 2 , e ) e ) 7 | e | add(subtract(38, add(30, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general | E |
x , a , z , and b are single digit positive integers . x = 1 / 5 a . z = 1 / 5 b . ( 10 a + b ) – ( 10 x + z ) could not equal | a = 5 x , b = 5 z therefore ( 5 x * 10 + 5 z ) - ( 10 x + z ) = ( 5 - 1 ) ( 10 x + z ) = . ( 10 x + z ) number should be divisible by 4 b | a ) 36 , b ) 33 , c ) 44 , d ) 64 , e ) 56 | b | add(add(subtract(add(multiply(5, 5), multiply(5, 10)), add(multiply(5, 10), 5)), 10), const_3) | multiply(n1,n1)|multiply(n1,n4)|add(#0,#1)|add(n1,#1)|subtract(#2,#3)|add(n4,#4)|add(#5,const_3) | general | B |
p and q invested in a business . the profit earned was divided in the ratio 3 : 5 . if p invested rs 12000 , the amount invested by q is | "let the amount invested by q = q 12000 : q = 3 : 5 ⇒ 12000 × 5 = 3 q ⇒ q = ( 12000 × 5 ) / 3 = 30000 answer is a" | a ) 30000 , b ) 50000 , c ) 40000 , d ) 20000 , e ) 60000 | a | multiply(divide(12000, const_2.0), 5) | divide(n2,const_2.0)|multiply(n1,#0)| | gain | A |
find the slope of the line perpendicular to the line y = ( 1 / 6 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 6 . if m is the slope of the line perpendicular to the given line , then m × ( 1 / 6 ) = - 1 solve for m m = - 6 correct answer c ) - 6" | a ) 1 , b ) 2 , c ) - 6 , d ) 4 , e ) 5 | c | divide(1, 6) | divide(n0,n1)| | general | C |
the total age of a and b is 16 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 16 â € ¹ = â € º a - c = 16 . answer a" | a ) 16 , b ) 24 , c ) c is elder than a , d ) data inadequate , e ) none | a | multiply(16, const_1) | multiply(n0,const_1)| | general | A |
40 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ? | "explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 40 = 53.33 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capac... | a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these | a | multiply(1200, multiply(800, divide(40, multiply(800, 600)))) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n4,#2)| | physics | A |
after a storm deposits 110 billion gallons of water into the city reservoir , the reservoir is 60 % full . if the original contents of the reservoir totaled 220 billion gallons , the reservoir was approximately what percentage full before the storm ? | when the storm deposited 110 billion gallons , volume of water in the reservoir = 220 + 110 = 360 billion gallons if this is only 60 % of the capacity of the reservoir , the total capacity of the reservoir = 330 / 0.6 = 550 billion gallons therefore percentage of reservoir that was full before the storm = ( 220 / 550 )... | a ) 40 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 % | a | multiply(divide(220, divide(add(110, 220), divide(60, const_100))), const_100) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|divide(n2,#2)|multiply(#3,const_100) | general | A |
a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 20 miles per hour . if the ride from marker a to marker b lasted 3 t... | a - b = 120 miles b - c = 60 miles avg speed = 20 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 20 = 180 / 4 t t = 135 b - c = 135 mph answer b | a ) 40 , b ) 135 , c ) 50 , d ) 55 , e ) 60 | b | multiply(divide(divide(add(divide(120, const_2), 120), 20), const_4), divide(120, const_2)) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|divide(#2,const_4)|multiply(#3,#0) | physics | B |
population is 21000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 21000 * 10 / 100 = 2100 = = = > 21000 + 2100 = 23100 population after 2 nd year = 23100 * 10 / 100 = 2310 = = = > 23100 + 2310 = 25410 population after 3 rd year = 254100 * 10 / 100 = 2541 = = = > 25410 + 2541 = 27951 answer : b" | a ) 26630 , b ) 27951 , c ) 36621 , d ) 26621 , e ) 26821 | b | multiply(21000, add(const_1, divide(multiply(3, 10), const_100))) | multiply(n1,n2)|divide(#0,const_100)|add(#1,const_1)|multiply(n0,#2)| | gain | B |
tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 3 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | "d is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 3 t ) = 8 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 3 t ) = 8 t = = > 2 = = > 120 minutes difference , 96 minutes d" | a ) 60 , b ) 72 , c ) 84 , d ) 96 , e ) 108 | d | multiply(subtract(divide(multiply(const_2, const_2), subtract(8, multiply(const_2, 3))), divide(const_2, subtract(8, 3))), const_60) | multiply(const_2,const_2)|multiply(n0,const_2)|subtract(n1,n0)|divide(const_2,#2)|subtract(n1,#1)|divide(#0,#4)|subtract(#5,#3)|multiply(#6,const_60)| | physics | D |
a company produces 65000 bottles of water everyday . if a case can hold 13 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 13 number of cases required to hold 65000 bottles = 65000 / 13 = 5000 cases . so the answer is c = 5000" | a ) 2000 , b ) 4500 , c ) 5000 , d ) 8000 , e ) 9000 | c | divide(65000, 13) | divide(n0,n1)| | physics | C |
if x = 6 ^ 36 and x ^ x = 6 ^ k , what is k ? | "solution : we know that x = 6 ^ 36 which implies x ^ x = ( 6 ^ 36 ) ^ ( 6 ^ 36 ) = 6 ^ ( 36 * 6 ^ 36 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 6 ^ ( 6 ^ 2 * 6 ^ 36 ) = 6 ^ ( 6 ^ ( 2 + 36 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 6 ^ ( 6 ^ 38 ) given that x ^ x = 6 ^ k so 6 ^ ( 6 ^ 38 ) = 6... | a ) 6 ^ 36 , b ) 6 ^ 38 , c ) 6 ^ 72 , d ) 6 ^ 216 , e ) 6 ^ 432 | b | multiply(power(6, 36), 36) | power(n0,n1)|multiply(n1,#0)| | general | B |
evaluate : 22 + sqrt ( - 4 + 6 * 4 * 3 ) = ? | "according to order of operations , inner brackets first where 6 ? 4 ? 3 is first calculated since it has a multiplication and a division . 6 * 4 * 3 = 24 * 3 = 8 hence 22 + sqrt ( - 4 + 6 * 4 * 3 ) = 22 + sqrt ( - 4 + 8 ) = 22 + sqrt ( 4 ) = 22 + 2 = 24 correct answer c" | a ) 4 , b ) 14 , c ) 24 , d ) 34 , e ) 44 | c | add(22, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3)| | general | C |
a man ’ s current age is ( 2 / 5 ) of the age of his father . after 8 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | a 40 let , father ’ s current age is a years . then , man ’ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 8 ] = ( 1 / 2 ) ( a + 8 ) 2 ( 2 a + 40 ) = 5 ( a + 8 ) a = 40 | a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 39 | a | divide(subtract(multiply(8, divide(2, 1)), 8), subtract(const_1, multiply(divide(2, 5), divide(2, 1)))) | divide(n0,n3)|divide(n0,n1)|multiply(n2,#0)|multiply(#1,#0)|subtract(#2,n2)|subtract(const_1,#3)|divide(#4,#5) | general | A |
what is the difference between the c . i . on rs . 6000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 6000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 6000 * 26 / 25 * 51 / 50 = rs . 6364.8 c . i . when interest is compounded half - yearly = [ 6000 * ( 1 + 2 / 100 ) 2 ] = ( 6000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 6367.25 difference = ( 6367.25 - 6364.8 ) = rs . ... | a ) s . 2.04 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.83 , e ) s . 2.45 | e | subtract(multiply(6000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(6000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | divide(n3,const_100)|divide(n4,const_100)|add(#0,n1)|add(#1,n1)|multiply(#2,#2)|multiply(#2,#3)|multiply(#2,#4)|multiply(n0,#5)|multiply(n0,#6)|subtract(#8,#7)| | general | E |
of 70 players on a football team , 31 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 31 rest = 70 - 31 = 39 left handed = 39 / 3 = 13 right handed = 26 if all thrower are right handed then total right handed is 31 + 26 = 57 so a . 57 is the right answer" | a ) 57 , b ) 59 , c ) 63 , d ) 71 , e ) 92 | a | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 31)), 31) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general | A |
a parking garage rents parking spaces for $ 15 per week or $ 30 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 15 = 780 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 30 = 360 $ 780 - 360 = 420 ans d" | a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 420 , e ) $ 260 | d | subtract(multiply(add(multiply(15, add(const_3, const_2)), const_2), 15), multiply(30, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general | D |
a cistern is filled by pipe a in 10 hours and the full cistern can be leaked out by an exhaust pipe b in 12 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 10 - 1 / 12 ) hrs = 1 / 60 = 60 hrs answer : b" | a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | b | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | B |
a , b and c started a business with capitals of rs . 6000 , rs . 8000 , and rs . 10000 respectively . at the end of the year , the profit share of b is rs 1000 . the difference between the profit shares of a and c is | ratio of investments of a , b and c is 6000 : 8000 : 10000 = 3 : 4 : 5 . and also given that , profit share of ‘ b ’ is rs . 1000 . now required difference is 5 – 3 = 2 parts . therefore , required difference = 2 / 4 ( 1000 ) = rs . 500 . answer : c | a ) rs . 400 , b ) rs . 450 , c ) rs . 500 , d ) rs . 550 , e ) rs . 650 | c | multiply(divide(subtract(10000, 6000), 8000), 1000) | subtract(n2,n0)|divide(#0,n1)|multiply(n3,#1) | gain | C |
a will states that the estate would be divided among 6 sons with the remainder donated to charity . one stipulation in the will is that no one among the sons can receive an amount within 12 % of another ' s amount . if one of the beneficiaries received $ 30000 , what is the smallest possible range between the highest a... | 1 st - 30000 . 2 nd - 0.88 * 30000 = 26400 3 rd - 0.88 * 26400 = 23232 4 th - 0.88 * 23232 = 20 , 444.16 5 th - 0.88 * 20 , 444.16 = ~ 17 , 990.86 6 th - 0.88 * 17 , 990.86 = ~ 15 , 831.96 range = 30000 - 15831 = 14168 answer : b . | a ) $ 4096 , b ) $ 14043 , c ) $ 7892 , d ) $ 17736 , e ) $ 15336 | b | multiply(multiply(30000, divide(12, const_100)), const_4) | divide(n1,const_100)|multiply(n2,#0)|multiply(#1,const_4) | general | B |
an article is bought for rs . 600 and sold for rs . 450 , find the loss percent ? | "600 - - - - 150 100 - - - - ? = > 16 2 / 3 % answer : e" | a ) 16 % , b ) 18 % , c ) 19 % , d ) 20 % , e ) 25 % | e | subtract(const_100, divide(multiply(450, const_100), 600)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain | E |
a person was asked to state his age in years . his reply was , ` ` take my age 3 years hence , multiply it by 3 and subtract 3 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 , 3 ( x + 3 ) - 3 ( x - 3 ) = x < = > ( 3 x + 9 ) - ( 3 x - 9 ) = x < = > x = 18 . . answer : a ) 18" | a ) 18 , b ) 92 , c ) 27 , d ) 26 , e ) 19 | a | add(multiply(3, 3), multiply(3, 3)) | multiply(n0,n1)|add(#0,#0)| | general | A |
if x + ( 1 / x ) = 5 , what is the value of r = x ^ 2 + ( 1 / x ) ^ 2 ? | "squaring on both sides , x ^ 2 + ( 1 / x ) ^ 2 + 2 ( x ) ( 1 / x ) = 5 ^ 2 x ^ 2 + ( 1 / x ) ^ 2 = 23 answer : c" | a ) r = 21 , b ) r = 22 , c ) r = 23 , d ) 24 , e ) 27 | c | subtract(power(5, 2), 2) | power(n1,n2)|subtract(#0,n2)| | general | C |
a rectangular floor that measures 6 meters by 10 meters is to be covered with carpet squares that each measure 2 meters by 2 meters . if the carpet squares cost $ 15 apiece , what is the total cost for the number of carpet squares needed to cover the floor ? | "the width of the rectangular floor ( 6 m ) is a multiple of one side of the square ( 2 m ) , and the length of the floor ( 10 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 6 / 2 ) * ( 10 / 2 ) = 15 . the total cost is 15 * 15 = $ 225 . the answer is , therefore , d... | a ) $ 200 , b ) $ 240 , c ) $ 480 , d ) $ 225 , e ) $ 1,920 | d | multiply(15, 15) | multiply(n4,n4)| | geometry | D |
a man invests rs . 8000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | explanation : interest on rs . 8000 at 5 % per annum = ( 8000 × 5 × 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) × 6 × 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) × 6 × 1... | a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these | d | divide(subtract(multiply(8000, 6), multiply(8000, 5)), subtract(8, 6)) | multiply(n0,n3)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2) | gain | D |
for the positive integers x , x + 2 , x + 4 , x + 7 , and x + 32 , the mean is how much greater than the median ? | "mean = ( x + x + 2 + x + 4 + x + 7 + x + 32 ) / 5 = ( 5 x + 45 ) / 5 = x + 9 median = x + 4 thus mean - median = x + 9 - ( x + 4 ) = 5 answer = e" | a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | e | subtract(divide(add(add(add(2, 4), 7), 32), add(4, const_1)), 4) | add(n0,n1)|add(const_1,n1)|add(n2,#0)|add(n3,#2)|divide(#3,#1)|subtract(#4,n1)| | general | E |
a merchant purchased a jacket for $ 48 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 40 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant ’ s gross profit on this ... | "actual cost = $ 48 sp = actual cost + mark up = actual cost + 40 % sp = 48 * 100 / 60 on sale sp = 80 / 100 ( 48 * 100 / 60 ) = 64 gross profit = $ 16 answer is e" | a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 16 | e | subtract(multiply(divide(48, subtract(const_1, divide(40, const_100))), subtract(const_1, divide(20, const_100))), 48) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|divide(n0,#2)|multiply(#4,#3)|subtract(#5,n0)| | gain | E |
the ratio of male to female in a class is 2 : 3 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle sho... | let the common ratio be x . . so m = 2 x and f = 3 x and total = 5 x 1 of m = 2 x and 1 of f = 3 x . . total preferring that carrer = 5 x now 5 x = 360 , so x = 360 / 5 = 72 . . so x * 8 / 3 = 72 * 8 / 3 = 192 d | ['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 72 degree', 'e ) e ) 204 degree'] | d | multiply(divide(const_1, 3), multiply(divide(3, add(2, 3)), const_360)) | add(n0,n1)|divide(const_1,n1)|divide(n1,#0)|multiply(#2,const_360)|multiply(#1,#3) | geometry | D |
what is 15 percent of 54 ? | "( 15 / 100 ) * 54 = 8.1 the answer is b ." | a ) 7.3 , b ) 8.1 , c ) 9.5 , d ) 10.2 , e ) 11.6 | b | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain | B |
in a sports club with 30 members , 17 play badminton and 17 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 17 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 17 - x ) + x + 2 = 30 36 - 2 x + x = 30 36 - x = 30 x = 6 so 6 members play both badminton and tennis . b" | a ) 7 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | subtract(add(add(17, 17), 2), 30) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | other | B |
a boat running downstream covers a distance of 20 km in 2 hours while for covering the same distance upstream , it takes 5 hours . what is the speed of the boat in still water ? | explanation : rate downstream = ( 20 / 2 ) kmph = 10 kmph ; rate upstream = ( 20 / 5 ) kmph = 4 kmph speed in still water = 1 / 2 ( 10 + 4 ) kmph = 7 kmph answer : b | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 1 | b | divide(add(divide(20, 2), divide(20, 5)), const_2) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2) | physics | B |
a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.08 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the f... | "1 st day fine - 0.08 2 nd day fine - 0.08 * 2 = 0.16 ( as doubling gives lower value ) 3 rd day fine - 0.16 * 2 = 0.32 ( as doubling gives lower value ) 4 th day fine - 0.32 + 0.3 = 0.62 ( as doubling gives higher value we add 0.3 this time ) answer : a ." | a ) $ 0.62 , b ) $ 0.70 , c ) $ 0.80 , d ) $ 0.90 , e ) $ 1.00 | a | add(multiply(multiply(multiply(0.08, const_2), const_2), const_2), 0.30) | multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|add(n1,#2)| | general | A |
if w is the set of all the integers between 49 and 76 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 75 , the smallest is 51 ) : 75 - 51 = 24 step 2 . divide by 3 : 24 / 3 = 8 step 3 . add 1 : 8 + 1 = 9 . so there are 9 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . nu... | a ) 19 , b ) 91 , c ) 41 , d ) 18 , e ) 14 | a | subtract(add(floor(divide(subtract(76, 49), 3)), divide(subtract(76, 49), 2)), floor(divide(subtract(76, 49), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other | A |
solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 72 % percent carbonated water , what percent of the volume of the mixture is p ? | 72 % is 8 % - points below 80 % and 17 % - points above 55 % . so the ratio of solution p to solution q is 17 : 8 . mixture p is 17 / 25 = 68 % of the volume of mixture pq . the answer is e . | a ) 48 % , b ) 54 % , c ) 60 % , d ) 64 % , e ) 68 % | e | multiply(divide(subtract(divide(72, const_100), divide(55, const_100)), add(subtract(divide(72, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(72, const_100)))), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#0)|add(#3,#4)|divide(#3,#5)|multiply(#6,const_100) | gain | E |
a candidate got 35 % of the votes polled and he lost to his rival by 2460 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2460 100 % - - - - - - - - - ? = > 8200 answer : d" | a ) 7500 , b ) 3388 , c ) 2665 , d ) 8200 , e ) 2661 | d | divide(2460, 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 | D |
david obtained 76 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "solution : average = ( 76 + 65 + 82 + 67 + 85 ) / 5 = 375 / 5 = 75 . hence , average = 75 . answer : option c" | a ) 65 , b ) 69 , c ) 75 , d ) none of these , e ) can not be determined | c | divide(add(add(add(add(76, 65), 82), 67), 85), add(const_2, const_3)) | add(n0,n1)|add(const_2,const_3)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | C |
if 10 spiders make 5 webs in 5 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 10 : 1 ( webs ) 1 : 5 } : : x : 5 â ‡ ’ 10 ã — 1 ã — 5 = 1 ã — 5 ã — x â ‡ ’ x = 10 answer : option a" | a ) 10 , b ) 9 , c ) 5 , d ) 15 , e ) 20 | a | multiply(1, 10) | multiply(n0,n3)| | physics | A |
while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 3 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to p... | "both computers together process files at a rate of 240 / 12 + 240 / 3 = 20 + 80 = 100 files per hour . the time required to process 240 files is 240 / 100 = 2.4 hours the answer is e ." | a ) 1.6 , b ) 1.8 , c ) 2 , d ) 2.2 , e ) 2.4 | e | divide(240, add(divide(240, 12), divide(240, 3))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(n0,#2)| | physics | E |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 10000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 10000 x 7 ) + 6000 ( x - 7 ) x = 14 . answer e" | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 14 | e | add(7, divide(multiply(7, subtract(10000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general | E |
the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 23 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ? | let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 23 = 2 r a = 12 / 5 r and r = 69 a = 165.6 required total cost = 4 * 165.6 + 3 * 69 + 5 * 23 = 662.4 + 207 + 115 = $ 984.40 b | a ) 347.4 , b ) 984.4 , c ) 877.4 , d ) 637.4 , e ) 667.4 | b | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(23, 6), 2))), multiply(3, divide(multiply(23, 6), 2))), multiply(5, 23)) | divide(n1,n0)|multiply(n2,n4)|multiply(n4,n7)|divide(#1,n3)|multiply(#0,#3)|multiply(n6,#3)|multiply(n5,#4)|add(#6,#5)|add(#7,#2) | general | B |
how many integers are divisible by 3 between 20 ! and 20 ! + 20 inclusive ? | "b - 7 20 ! is divisible by 3 there are 6 numbers between 10 ! and 10 ! + 20 that are divisible by 3 . hence 7" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(divide(20, 3), const_1) | divide(n3,n0)|add(#0,const_1)| | general | B |
the dimensions of a field are 10 m by 10 m . a pit 10 m long , 5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 10 * 5 * 4 = 200 m ^ 3 . the remaining area of the field is 10 * 10 - 10 * 5 = 50 m ^ 2 . 200 m ^ 3 of the earth evenly spread over the area of 50 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 200 / 50 = 4 m . answer : c" | a ) 2 m , b ) 3 m , c ) 4 m , d ) 5 m , e ) 1.5 m | c | divide(multiply(10, 10), subtract(rectangle_area(10, 10), rectangle_area(5, 10))) | multiply(n1,n2)|rectangle_area(n0,n1)|rectangle_area(n2,n3)|subtract(#1,#2)|divide(#0,#3)| | other | C |
from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 5 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 5 x = 45 - - > 21 / 20 * x = 45 - - > x = 45 * 20 / 21 = 300 / 7 = ~ 43 . answer : e ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 43 | e | multiply(divide(const_100, add(const_100, 5)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain | E |
a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 126 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | "we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 126 / 25 = 5 . . . . so the max is 5 126 / 20 = 6 . . . so the max is 6 126 - 125 = 1 ( 5 big box or 5 small box + 1 big box ) answer : c" | a ) 2 , b ) 4 , c ) 1 , d ) 5 , e ) 3 | c | subtract(25, 20) | subtract(n0,n1)| | general | C |
a , b , c hired a car for rs . 520 and used it for 7 , 8 and 11 hours respectively . hire charges paid by b were ? | a : b : c = 7 : 8 : 11 . hire charges paid by b = rs . ( 520 * 8 / 26 ) = rs . 160 . answer : b | a ) 127 , b ) 160 , c ) 287 , d ) 237 , e ) 111 | b | multiply(520, divide(8, add(add(7, 8), 11))) | add(n1,n2)|add(n3,#0)|divide(n2,#1)|multiply(n0,#2) | physics | B |
a set s = { x , - 8 , - 5 , - 3 , 3 , 6 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ? | "median of the set = ( - 3 + 3 ) / 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 18 - 16 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer a" | a ) 2 , b ) 0 , c ) - 1 , d ) can not be determined , e ) 1 | a | subtract(subtract(subtract(add(add(9, 8), 3), 3), 5), 8) | add(n0,n5)|add(n2,#0)|subtract(#1,n2)|subtract(#2,n1)|subtract(#3,n0)| | general | A |
the diagonals of two squares are in the ratio of 2 : 5 . find the ratio of their areas . | explanation : let the diagonals of the squares be 2 x and 5 x . then ratio of their areas will be area of square = 1 / 2 ∗ diagonal 2 1 / 2 ∗ 2 x 2 : 1 / 2 ∗ 5 x 2 4 x 2 : 25 x 2 = 4 : 25 option a | ['a ) 4 : 25', 'b ) 4 : 15', 'c ) 3 : 25', 'd ) 3 : 15', 'e ) none of these'] | a | divide(const_4, power(5, const_2)) | power(n1,const_2)|divide(const_4,#0) | geometry | A |
on dividing a n by 9 , remainder is 8 . the quotient obtained when divided by 11 , leaves remainder 9 . now the quotient when divided by 13 , leaves remainder 8 . find the remainder when when the n is divided by 1287 | take from the last step division suppose no . is n which is divided by 13 and remainder 8 = 13 n + 8 now 13 n + 8 will be the no . which is divided by 11 and remainder is 9 = [ 11 * ( 13 n + 8 ) ] + 9 now this no . is used for the very first step which is divider is 9 and remainder is 8 = { 9 * [ 11 * ( 13 n + 8 ) ] + ... | a ) 871 , b ) 881 , c ) 891 , d ) 904 , e ) 987 | b | multiply(multiply(9, 9), 11) | multiply(n0,n0)|multiply(n2,#0) | general | B |
what is the ratio of 3 / 7 to the product 2 * ( 7 / 3 ) ? | "3 / 7 / 14 / 3 = 9 / 98 . . . imo option b ." | a ) 3 : 7 , b ) 9 : 98 , c ) 3 : 21 , d ) 1 : 7 , e ) 3 : 49 | b | divide(3, const_60) | divide(n0,const_60)| | general | B |
what is the square root of 9 ? | "3 x 3 = 9 answer b" | a ) 1 , b ) 3 , c ) 2 , d ) 4 , e ) 9 | b | circle_area(divide(9, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other | B |
what percent of 75 is 45 ? | "75 * x / 100 = 45 x = 4 * 45 / 3 x = 60 ans : e" | a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 40 % , e ) 60 % | e | multiply(divide(75, 45), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain | E |
six bells commence tolling together and toll at intervals of 2 , 4,6 , 8,10 and 12 seconds respectively . in 30 m 9 nutes , how many times do they toll together ? | "solution l . c . m . of 2,4 , 6,8 , 10,12 is 120 . so , the bells will toll together after every 120 seconds , i . e . , 2 minutes in 30 minutes , they will toll together left [ ( 30 / 2 ) + 1 ] = 16 times . answer d" | a ) 4 , b ) 10 , c ) 15 , d ) 16 , e ) 17 | d | divide(30, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 30)) | add(n0,const_3)|multiply(n0,#0)|multiply(#1,const_3)|multiply(n0,#2)|multiply(n0,#3)|divide(#4,n4)|divide(n4,#5)| | physics | D |
a student traveled 30 percent of the distance of the trip alone , continued another 20 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.3 x + 20 miles + 0.5 x = x 20 miles = 0.2 x x = 100 miles the answer is d ." | a ) 240 , b ) 200 , c ) 160 , d ) 100 , e ) 50 | d | divide(20, subtract(subtract(const_1, inverse(30)), divide(const_1, const_2))) | divide(const_1,const_2)|inverse(n0)|subtract(const_1,#1)|subtract(#2,#0)|divide(n1,#3)| | physics | D |
a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 65 % . staff are allowed a further 60 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 65 - 60 + ( - 65 ) ( - 60 ) / 100 = - 86 sale price = d * ( 1 - 86 / 100 ) sale price = . 14 * d answer ( d )" | a ) 0.15 d , b ) 0.16 d , c ) 0.65 d , d ) 0.14 d , e ) 0.05 d | d | subtract(divide(subtract(const_100, 65), const_100), multiply(divide(subtract(const_100, 65), const_100), divide(60, const_100))) | divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)| | gain | D |
two good train each 650 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 650 + 650 ) m = 1300 m . required time = ( 1300 x 6 / 125 ) sec = 62.4 sec . answer c" | a ) 12.4 sec , b ) 24.3 sec , c ) 62.4 sec , d ) 60.1 sec , e ) none | c | multiply(multiply(650, inverse(multiply(add(45, 30), const_0_2778))), const_2) | add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2)| | physics | C |
for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 7 of whom are females , by how much does the index for the females exceed the index for the males in the gr... | "index for females = ( 20 - 7 ) / 20 = 13 / 20 = 0.65 index for males = ( 20 - 13 / 20 = 7 / 20 = 0.35 index for females exceeds males by 0.65 - 0.35 = 0.3 answer : d" | a ) 0.05 , b ) 0.0625 , c ) 0.2 , d ) 0.3 , e ) 0.6 | d | subtract(divide(subtract(20, 7), 20), divide(7, 20)) | divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)| | general | D |
a and b walk around a circular track . they start at 9 a . m . from the same point in the opposite directions . a and b walk at a speed of 2 rounds per hour and 3 rounds per hour respectively . how many times shall they cross each other before 11 : 00 a . m . ? | "sol . relative speed = ( 2 + 3 ) = 5 rounds per hour . so , they cross each other 5 times in an hour . hence , they cross each other 15 times before 11 : 00 a . m . answer d" | a ) 8 , b ) 7 , c ) 6 , d ) 15 , e ) 3 | d | add(add(2, 3), add(2, 3)) | add(n1,n2)|add(#0,#0)| | physics | D |
a certain roller coaster has 5 cars , and a passenger is equally likely to ride in any 1 of the 5 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 5 times , what is the probability that the passenger will ride in each of the 5 cars ? | "if he is to ride 5 times and since he can choose any of the 5 cars each time , total number of ways is = 5 * 5 * 5 * 5 * 5 = 3125 now the number of ways if he is to choose a different car each time is = 5 * 4 * 3 * 2 * 1 = 120 so the probability is = 120 / 3125 = 24 / 625 answer : b" | a ) 0 , b ) 24 / 625 , c ) 2 / 9 , d ) 1 / 3 , e ) 1 | b | multiply(factorial(5), power(divide(1, 5), 5)) | divide(n1,n0)|factorial(n0)|power(#0,n0)|multiply(#1,#2)| | general | B |
sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 0.5 % more . she would get how much ? | "( 8000 * 3 * 0.5 ) / 100 = 120 9200 - - - - - - - - 9320 answer : a" | a ) 9320 , b ) 96288 , c ) 26667 , d ) 1662 , e ) 2882 | a | add(multiply(multiply(add(divide(0.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)| | gain | A |
of the diplomats who attended a summit conference : 20 spoke french , 32 did not speak hindi and 20 % of the diplomats spoke neither french nor hindi . if 10 % of the diplomats spoke both french and hindi , then how many diplomats attended the conference ? | "2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer" | a ) 70 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | d | divide(subtract(32, 20), divide(10, const_100)) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,#0)| | other | D |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 950 grams per kg , what is his percent ? | 950 - - - 50 100 - - - ? = > 5.26 % answer : d | a ) 22 % , b ) 25 % , c ) 77 % , d ) 5.26 % , e ) 12 % | d | subtract(multiply(divide(const_100, 950), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100) | gain | D |
when 6 + 2 = 85 , 8 + 3 = 115 , 7 + 5 = 125 , then 5 + 8 = ? | "6 + 2 = > 6 + 2 = 8 = > 8 ã — 10 + 5 = 85 8 + 3 = > 8 + 3 = 11 = > 11 ã — 10 + 5 = 115 7 + 5 = > 7 + 5 = 12 = > 12 ã — 10 + 5 = 125 then 5 + 8 = > 5 + 8 = 13 = > 13 ã — 10 + 5 = 135 answer : d" | a ) 145 , b ) 185 , c ) 245 , d ) 135 , e ) 140 | d | add(multiply(multiply(5, 8), const_10), 6) | multiply(n7,n10)|multiply(#0,const_10)|add(n0,#1)| | general | D |
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