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 values |
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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 - hour day ? | "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 = answer = d" | 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 ) ln ( 80 ) = ln ( 81 ) ln ( 3 ) = log 3 ( 81 ) = log 3 ( 34 ) = 4 : correct answer b | 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 practice : hope it helps . | 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 cm 2 answer : option d" | 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 * $ 3,000 = $ 1000 plug this value in the first equation : c + 1000 = $ 2,500 - - > c = $ 1,500 . answer : a ." | 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 money ? | "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 for that night ? | "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 + both + neither 78 = ( 41 - 9 ) + ( 22 - 9 ) + 9 + neither . not enrolled students = 24 answer : c" | 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 = 25 last multiple in the range − first multiple in the range 4 + 1 = 25 but 3 numbers out of these 25 are not formed with distinct digits : 00 , 44 , and 88 . hence the numbers we are looking for can have only 22 endings . if there is 0 in the ending ( 04 , 08 , 20 , 40 , 60 , 80 - total 6 such numbers ) , then the first and second digit can take 8 and 7 choices each = 56 choices total . as there are 6 numbers with 0 in ending , hence total 6 * 56 = 336 . if there is no 0 in the ending ( total 22 - 6 with zero = 16 such numbers ) , then the first digit can take 7 choices ( 10 - 2 digits in the ending and zero , so total 3 digits = 7 , as 4 - digit number can not start with zero ) and the second digit can take 7 choices too ( 10 digits - 3 digits we ' ve already used ) = 7 * 7 = 49 choices total . as there are 16 numbers without zero in ending , hence total 16 * 49 = 784 . total : 336 + 784 = 1120 answer : c . | 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 = 1120 sum of digits in n = sum of digits in 1120 = 1 + 1 + 2 + 0 = 4 answer is c | 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(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), const_10)), const_10), reminder(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), const_10)), reminder(divide(gcd(gcd(subtract(4665, 1305), subtract(6905, 4665)), subtract(6905, 1305)), const_10), const_10)) | 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 + 340 + 520 = 980 $ answer d" | 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 600 hence , total numbers divisible by 2 , 3 , 7 together are ( 14 – 2 ) = 12 answer : b" | 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 capacity of engine doubles , the diesel requirement will double too . therefore , with a 1200 cc engine , quantity of diesel required = 1200 / 800 * 53.33 = 80 litres . answer a" | 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 ) * 100 = 40 % option a | 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 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | 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 difference , in minutes , between the amount of time it takes tom to cover half of the distance that linda has covered and the amount of time it takes tom to cover twice the distance that linda has covered ? | "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 ^ k since the base is same the exponent will also be same therefore k = 6 ^ 38 answer : b" | 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 . 2.45 . answer : e" | 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 and lowest amounts ( rounded to the nearest dollar ) among the 6 sons ? | 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 / 100 . = > 40000 + 8 x = ( 8000 + x ) × 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d | 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 sale ? | "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 should be used to represent a career that is preferred by one of the males and one of the females in the class ? | 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 fourth day it is overdue ? | "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 . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 76 , the smallest is 50 ) : 76 - 50 = 26 step 2 . divide by 2 : 26 / 2 = 13 step 3 . add 1 : 13 + 1 = 14 . so there are 14 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 9 multiples of 3 and the 14 multiples of 2 : 9 + 14 = 23 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 9 + 14 = 23 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 23 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 72 - 54 = 18 step 2 . divide by 6 : 18 / 6 = 3 step 3 . add 1 : 3 + 1 = 4 . so there are 4 multiples of 6 within the range : we counted 4 numbers twice . subtract the 4 multiples of 6 from the sum of the multiples of 2 and 3 : = 9 + 14 - 4 = 23 - 4 = 19 therefore , the final number of multiples of 2 , 3 or 6 is 19 . hence , this is the correct answer . ( a )" | 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 process a total of 240 files ? | "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 ) ] + 9 } + 8 solve and ans is 881 answer : b | 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 group ? | "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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