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the speed of a boat in still water is 30 kmph . what is the speed of the stream if the boat can cover 80 km downstream or 40 km upstream in the same time ? | x = the speed of the stream ( 30 + x ) / ( 30 - x ) = 2 / 1 30 + x = 60 - 2 x 3 x = 30 x = 10 km / hour if the speed of the stream is 10 km / hour , then the ' downstream ' speed of the boat is 30 + 10 = 40 km / hour and the ' upstream ' speed of the boat is 30 - 10 = 20 km / hour . in that way , if the boat traveled for 2 hours , it would travel 2 x 40 = 80 km downstream and 2 x 20 = 40 km / hour upstream . answer : b | a ) 11 kmph , b ) 10 kmph , c ) 12 kmph , d ) 16 kmph , e ) 15 kmph | b | divide(30, add(const_1, const_2)) | add(const_1,const_2)|divide(n0,#0) | physics | B |
it costs $ 6 for the first 1 / 4 hour to use the copier at kinkos . after the first ¼ hour it costs $ 8 per hour . if a certain customer uses the copier for 4 hours and 25 minutes , how much will it cost him ? | 4 hrs 25 min = 265 min first 15 min - - - - - - > $ 6 time left is 250 min . . . now , 60 min costs $ 8 1 min costs $ 8 / 60 250 min costs $ 8 / 60 * 250 = > $ 33.33 so , total cost will be $ 33.33 + $ 6 = > $ 39.33 the answer will be ( e ) $ 39.33 | a ) $ 23.45 , b ) $ 65.33 , c ) $ 40 , d ) $ 38.27 , e ) $ 39.33 | e | add(add(add(25, 8), 6), divide(add(25, 8), const_100)) | add(n3,n5)|add(n0,#0)|divide(#0,const_100)|add(#1,#2) | physics | E |
convert 2.0 hectares in ares | "2.0 hectares in ares 1 hectare = 100 ares therefore , 2.0 hectares = 2.0 × 100 ares = 200 ares . answer - d" | a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 200 ares . , e ) 250 ares . | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 2.0), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics | D |
a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 77.44 , the number of the member is the group is : | "money collected = ( 77.44 x 100 ) paise = 7744 paise . number of members = square root of 7744 = 88 . answer : option d" | a ) 57 , b ) 67 , c ) 77 , d ) 88 , e ) 97 | d | sqrt(multiply(77.44, const_100)) | multiply(n0,const_100)|sqrt(#0)| | general | D |
q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 20 , what is the greatest possible value of q minus r ? | "a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 20 . the greatest multiple of 9 which is less than 20 is 18 . the answer is d ." | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | multiply(reminder(20, subtract(const_10, const_1)), subtract(const_10, const_1)) | subtract(const_10,const_1)|reminder(n0,#0)|multiply(#1,#0)| | general | D |
if - 3 x + 2 y = 28 and 3 x + 6 y = 84 , what is the product of x and y ? | "given - 3 x + 2 y = 28 - - - eq 1 3 x + 6 y = 84 - - eq 2 sum both eqns we get 8 y = 112 = > y = 14 sum 2 y in eq 1 = > - 3 x - 28 = 28 . = > x = 0 now xy = 0 * 14 = 0 . option c is correct answer ." | a ) 264 . , b ) 428 , c ) 0 , d ) 462 , e ) 642 | c | multiply(divide(add(84, multiply(divide(add(28, 84), 6), 6)), 3), divide(add(28, 84), 6)) | add(n2,n5)|divide(#0,n4)|multiply(n4,#1)|add(n5,#2)|divide(#3,n0)|multiply(#4,#1)| | general | C |
if jamie ' s income is rs 20000 , calculate his savings given that his income and expenditure are in the ratio 7 : 5 . ? | let the income and the expenditure of the person be rs . 7 x and rs . 7 x respectively . income , 7 x = 28000 = > x = 4000 savings = income - expenditure = 7 x - 5 x = 2 x so , savings = rs . 8000 answer : d | a ) rs 7500 , b ) rs 4000 , c ) rs 2000 , d ) rs 8000 , e ) rs 7000 | d | multiply(divide(5, add(7, 5)), 20000) | add(n1,n2)|divide(n2,#0)|multiply(n0,#1) | other | D |
( x ) + 1315 + 9211 - 1569 = 11901 . calculate the value of x | x + 1315 + 9211 - 1569 = 11901 = x + 1315 + 9211 = 11901 + 1569 = x + 10526 = 13470 = x = 98329 - 10526 = 87803 answer is e | a ) 87801 , b ) 87811 , c ) 87862 , d ) 11803 , e ) 87803 | e | multiply(subtract(subtract(add(11901, 1569), 9211), 1315), divide(const_60, const_2)) | add(n2,n3)|divide(const_60,const_2)|subtract(#0,n1)|subtract(#2,n0)|multiply(#1,#3) | general | E |
for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 7.2 – 7.2 * ? | since y * is defined as the greatest positive even integer less than or equal to y , then 7.2 * = 6 ( the greatest positive even integer less than or equal to 7.2 is 6 ) . hence , 7.2 – 7.2 * = 7.2 - 6 = 1.2 answer : a . | a ) 1.2 , b ) 0.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | a | subtract(7.2, subtract(floor(7.2), const_1)) | floor(n0)|subtract(#0,const_1)|subtract(n0,#1) | general | A |
how many zeros does 1000 ! end with ? | "according to above 1000 ! has 1000 / 5 + 1000 / 25 + 1000 / 125 + 1000 / 625 = 200 + 40 + 8 + 1 = 249 trailing zeros . answer : c" | a ) 20 o , b ) 24 o , c ) 249 , d ) 30 o , e ) 325 | c | add(add(divide(1000, add(const_4, const_1)), divide(subtract(1000, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(1000, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|subtract(n0,#0)|divide(#4,#2)|divide(#4,#3)|add(#1,#5)|add(#7,#6)| | other | C |
a mixture contains alcohol and water in the ratio 4 : 3 . if 6 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 6 ) = 4 / 5 20 x = 4 ( 3 x + 6 ) 8 x = 24 x = 3 quantity of alcohol = ( 4 x 3 ) litres = 12 litres . answer is d" | a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 12 litres , e ) 8 litres | d | multiply(6, const_1) | multiply(n2,const_1)| | general | D |
a runs twice as fast as b and gives b a start of 64 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 64 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 64 m is 64 / 1 = 64 m so the total time taken by a and b to reach = 2 * 64 = 128 m answer : b" | a ) 75 m . , b ) 128 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | b | multiply(64, const_2) | multiply(n0,const_2)| | physics | B |
a & b started a partnership business . a ' s investment was thrice the investment of b and the period of his investment was two times the period of investments of b . if b received rs 5000 as profit , what is their total profit ? | "explanation : suppose b ' s investment = x . then a ' s investment = 3 x suppose bs period of investment = y , then a ' s period of investment = 2 y a : b = 3 x * 2 y : xy = 6 : 1 total profit * 1 / 7 = 5000 = > total profit = 5000 * 7 = 35000 . answer : option d" | a ) 28000 , b ) 30000 , c ) 32000 , d ) 35000 , e ) none of these | d | divide(5000, divide(multiply(const_1, const_1), add(multiply(const_3, const_2), multiply(const_1, const_1)))) | multiply(const_1,const_1)|multiply(const_2,const_3)|add(#1,#0)|divide(#0,#2)|divide(n0,#3)| | general | D |
convert 0.38 in to a vulgar fraction ? | "answer 0.38 = 38 / 100 = 19 / 50 correct option : d" | a ) 18 / 50 , b ) 16 / 50 , c ) 17 / 50 , d ) 19 / 50 , e ) none | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.38), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics | D |
of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 1 / 3 times the fraction of first - year students who have declared a major , what fraction of all the students in the dormitory are second - year students who have not declared a major ? | tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 1 / 3 ( 1 / 5 ( x / 2 ) ) = 1 / 30 ( x ) - - - > non major = x / 2 - 1 / 30 ( x ) = 7 / 15 ( x ) hence 7 / 15 a | a ) 7 / 15 , b ) 1 / 5 , c ) 4 / 15 , d ) 1 / 3 , e ) 2 / 5 | a | subtract(divide(1, 2), divide(const_1, multiply(3, const_10))) | divide(n0,n1)|multiply(n5,const_10)|divide(const_1,#1)|subtract(#0,#2) | general | A |
122 * 252 - 12234 = ? | "d ? = 122 * 252 - 12234 = 30744 - 12234 = 18510" | a ) 15310 , b ) 18870 , c ) 24510 , d ) 18510 , e ) 42510 | d | multiply(122, 252) | multiply(n0,n1)| | general | D |
in a colony of 70 residents , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of educated to uneducated persons is 8 : 27 , then find the ratio of the number of educated to uneducated men in the colony ? | number of men in the colony = 4 / 7 * 70 = 40 . number of women in the colony = 3 / 7 * 70 = 40 . number educated women in the colony = 1 / 5 * 30 = 6 . number of uneducated women in the colony = 4 / 5 * 50 = 24 . number of educated persons in the colony = 8 / 35 * 70 = 16 . as 6 females are educated , remaining 10 educated persons must be men . number of uneducated men in the colony = 40 - 10 = 30 . number of educated men and uneducated men are in the ratio 10 : 30 i . e . , 1 : 3 . answer : e | a ) 1 : 7 , b ) 2 : 3 , c ) 1 : 8 , d ) 2 : 3 , e ) 1 : 3 | e | divide(subtract(multiply(divide(70, add(8, 27)), 8), divide(multiply(divide(70, add(4, 3)), 3), add(1, 4))), subtract(multiply(divide(70, add(8, 27)), 27), multiply(divide(multiply(divide(70, add(4, 3)), 3), add(1, 4)), 4))) | add(n5,n6)|add(n1,n2)|add(n1,n3)|divide(n0,#0)|divide(n0,#1)|multiply(n5,#3)|multiply(n2,#4)|multiply(n6,#3)|divide(#6,#2)|multiply(n1,#8)|subtract(#5,#8)|subtract(#7,#9)|divide(#10,#11) | other | E |
the sale price sarees listed for rs . 750 after successive discount is 20 % and 15 % is ? | "750 * ( 80 / 100 ) * ( 85 / 100 ) = 570 answer : b" | a ) 227 , b ) 570 , c ) 342 , d ) 680 , e ) 230 | b | subtract(subtract(750, divide(multiply(750, 20), const_100)), divide(multiply(subtract(750, divide(multiply(750, 20), const_100)), 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain | B |
find the compound ratio of ( 1 : 2 ) , ( 1 : 3 ) and ( 3 : 5 ) is | "required ratio = 1 / 2 * 1 / 3 * 3 / 5 = 1 / 10 = 1 : 10 answer is a" | a ) 1 : 10 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 3 : 2 | a | multiply(divide(1, 2), multiply(divide(1, 2), divide(1, 2))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other | A |
a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 104 million in sales ? | "( 9 / 104 ) / ( 3 / 20 ) = 30 / 54 = 57,6 % it means that 9 / 108 represents only 57,6 % . therefore a decrease of 42 % . answer d" | a ) 8 % , b ) 15 % , c ) 45 % , d ) 42 % , e ) 56 % | d | multiply(divide(3, 20), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | D |
running at the same constant rate , 15 identical machines can produce a total of 45 bags per minute . at this rate , how many bags could 150 such machines produce in 8 minutes ? | "let ' s take the approach that uses the answer choices to eliminate wasted time . 45 / 15 = 3 bags per minute per machine . 150 machines = 450 per minute . 8 minutes worth = 3600 bags . looking at the answers it is clear . . . we can only choose ( d ) . the correct answer is d ." | a ) 9,00 , b ) 1,800 , c ) 2,700 , d ) 3,600 , e ) 4,800 | d | multiply(multiply(divide(45, 15), 8), 150) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain | D |
think of a number , divide it by 5 and add 23 to it . the result is 42 . what is the number thought of ? | "explanation : 42 - 23 = 19 19 x 5 = 95 answer : e" | a ) 24 , b ) 77 , c ) 297 , d ) 267 , e ) 95 | e | multiply(subtract(42, 23), 5) | subtract(n2,n1)|multiply(n0,#0)| | general | E |
300 × ? + ( 12 + 4 ) × 1 / 8 = 602 | explanation : = > 300 × ? + ( 12 + 4 ) × 1 / 8 = 602 = > 300 × ? = 602 - ( 12 + 4 ) × 1 / 8 = > 300 × ? = 602 - 2 = 600 = > ? = 600 / 300 = 2 answer : option d | a ) a ) 4 , b ) b ) 3 , c ) c ) 5 , d ) d ) 2 , e ) e ) 8 | d | divide(subtract(602, multiply(add(12, 4), divide(1, 8))), 300) | add(n1,n2)|divide(n3,n4)|multiply(#0,#1)|subtract(n5,#2)|divide(#3,n0) | general | D |
a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . how many hours are required for b to copy 26 papers ? | a can copy 50 papers in 10 hrs , while a & b can copy 70 papers in 10 hrs . it means b can copy 20 papers in 10 hrs . then time taken by b to copy 26 papers = 26 * 10 / 20 = 13 hours answer : c | a ) 11 hours , b ) 12 hours , c ) 13 hours , d ) 14 hours , e ) 18 hours | c | divide(multiply(26, 10), subtract(70, 50)) | multiply(n1,n4)|subtract(n2,n0)|divide(#0,#1) | physics | C |
the average salary of the employees in a office is rs . 120 / month . the avg salary of officers is rs . 420 and of non officers is rs 110 . if the no . of officers is 15 , then find the no of nonofficers in the office . | "let no . of non - officers be x 15 * 420 + x * 110 = ( x + 15 ) 120 x = 450 d" | a ) 400 , b ) 420 , c ) 430 , d ) 450 , e ) 510 | d | divide(subtract(multiply(15, 420), multiply(15, 120)), subtract(120, 110)) | multiply(n1,n3)|multiply(n0,n3)|subtract(n0,n2)|subtract(#0,#1)|divide(#3,#2)| | general | D |
tough and tricky questions : arithmetic . ( 56 ^ 2 + 56 ^ 2 ) / 28 ^ 2 = | ans is 8 my approach was : ( 56 ^ 2 + 56 ^ 2 ) / 28 ^ 2 = 56 ( 56 + 56 ) / 28 * 28 = 56 * 112 / 28 * 28 = 2 * 4 = 8 d | a ) 4 , b ) 18 , c ) 29 , d ) 8 , e ) 116 | d | divide(add(power(56, 2), power(56, 2)), power(28, 2)) | power(n0,n1)|power(n4,n1)|add(#0,#0)|divide(#2,#1) | general | D |
for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 920 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 1000 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 920 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 920 so , 3 y < 408 = = > y < 136 so , y can take any value which is less than 136 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . d" | a ) 14 , b ) 12 , c ) 10 , d ) 16 , e ) 18 | d | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general | D |
a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 12 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s ) | "the distance the sound traveled to the man is 12 * 330 = 3960 meters the answer is e ." | a ) 3120 , b ) 3330 , c ) 3540 , d ) 3750 , e ) 3960 | e | multiply(330, 12) | multiply(n1,n3)| | physics | E |
a circular ground whose diameter is 40 metres , has a garden of area 1100 m ^ 2 around it . what is the wide of the path of the garden ? | "req . area = ï € [ ( 20 ) 2 â € “ ( r ) 2 ] = 22 â „ 7 ã — ( 400 - r ^ 2 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] ie ) 22 / 7 ( 400 - r ^ 2 ) = 1100 , ie ) r ^ 2 = 50 , r = 7.07 m answer b" | a ) 8.07 , b ) 7.07 , c ) 6.07 , d ) 7.0 , e ) 8.5 | b | subtract(circle_area(add(divide(40, 1100), 1100)), circle_area(divide(40, 1100))) | divide(n0,n1)|add(n1,#0)|circle_area(#0)|circle_area(#1)|subtract(#3,#2)| | physics | B |
x is the product of each integer from 1 to 75 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | "the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k ≤ n x = 1 * 2 * 3 . . . . * 75 = 50 ! no . of trailing zeros in 75 ! = 75 / 5 + 75 / 5 ^ 2 = 15 + 3 = 18 100 ^ k = 10 ^ 2 k → k = 18 / 2 = 9 d" | a ) 8 , b ) 7 , c ) 6 , d ) 9 , e ) 10 | d | add(divide(75, const_10), 1) | divide(n1,const_10)|add(#0,n0)| | general | D |
find the value of a from ( 15 ) ^ 2 x 8 ^ 3 ã · 256 = a . | given exp . = ( 15 ) ^ 2 x 8 ^ 3 ã · 256 = a = 225 x 512 ã · 256 = 450 e | a ) 250 , b ) 420 , c ) 440 , d ) 650 , e ) 450 | e | divide(multiply(power(15, 2), power(8, 3)), 256) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4) | general | E |
a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates q apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates q / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | add(const_2,const_3)|add(const_3,const_4)|divide(n2,const_2)|multiply(n2,const_2)|multiply(#0,#2)|multiply(n2,#1)|add(#4,#3)|divide(#6,#5)| | physics | C |
a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 6.30 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 6.30 when c . p = ( 100 * 6.30 ) / 21 = 30 answer : a" | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | a | divide(multiply(6.30, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | add(n0,const_100)|add(n3,const_100)|multiply(n2,const_100)|subtract(const_100,n1)|multiply(#1,#3)|divide(#4,const_100)|subtract(#0,#5)|divide(#2,#6)| | gain | A |
a man can row 5 kmph in still water . when the river is running at 2.3 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 5 s = 2.3 ds = 6.3 us = 2.7 x / 6.3 + x / 2.7 = 1 x = 1.89 d = 1.89 * 2 = 3.78 answer : d" | a ) 2.91 , b ) 3.48 , c ) 2.98 , d ) 3.78 , e ) 4.21 | d | multiply(divide(multiply(add(5, 2.3), subtract(5, 2.3)), add(add(5, 2.3), subtract(5, 2.3))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics | D |
the number of boxes in a warehouse can be divided evenly into 12 equal shipments by boat or 32 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 12 and 32 = 96 answer c" | a ) 27 , b ) 33 , c ) 96 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 12) | multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_3)|multiply(n0,#2)| | general | C |
how many cubes of 4 cm edge can be put in a cubical box of 1 m edge . | "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 4 * 4 * 4 = 15625 note : 1 m = 100 cm answer : b" | a ) 17725 cm , b ) 15625 cm , c ) 12786 cm , d ) 12617 cm , e ) 12187 cm | b | divide(volume_cube(1), volume_cube(divide(4, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | physics | B |
kathleen can paint a room in 3 hours , and anthony can paint an identical room in 4 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "( 1 / 3 + 1 / 4 ) t = 2 t = 24 / 7 answer : d" | a ) 8 / 15 , b ) 4 / 3 , c ) 15 / 8 , d ) 24 / 7 , e ) 15 / 4 | d | multiply(divide(const_1, add(divide(const_1, 3), divide(const_1, 4))), const_2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_2)| | physics | D |
how many multiples of 8 are less than 7600 , and also multiples of 19 ? | "lcm of 8 & 19 = 152 tried dividing 7600 by 152 got quotient 50 ' so c is answer" | a ) 104 , b ) 100 , c ) 50 , d ) 89 , e ) 90 | c | add(divide(subtract(19, 7600), 8), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | C |
an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 4 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 4 y and y . 2 ( 3 x + 2 x ) = 2 ( 4 y + y ) 5 x = 5 y x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 4 y * y = 6 x ^ 2 : 4 y ^ 2 = 6 x ^ 2 : 4 x ^ 2 = 6 : 4 = 3 : 2 the answer is b . | ['a ) 5 : 8', 'b ) 3 : 2', 'c ) 6 : 1', 'd ) 8 : 7', 'e ) 5 : 6'] | b | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), const_2), add(4, 1)), 4))) | add(n2,n3)|rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|divide(#2,const_2)|divide(#3,#0)|multiply(n2,#4)|rectangle_area(#4,#5)|divide(#1,#6) | geometry | B |
a part of certain sum of money is invested at 16 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 16 = 3 : 4 answer : e" | a ) 5 : 4 , b ) 1 : 4 , c ) 3 : 1 , d ) 3 : 5 , e ) 3 : 4 | e | multiply(divide(12, const_100), 16) | divide(n1,const_100)|multiply(n0,#0)| | gain | E |
if n divided by 7 has a remainder of 2 , what is the remainder when 2 times n is divided by 7 ? | "as per question = > n = 7 p + 2 for some integer p hence 2 n = > 14 q + 4 = > remainder = > 4 for some integer q alternatively = > n = 2 > 2 n = > 4 = > 4 divided by 7 will leave a remainder 4 hence d" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6 | d | multiply(2, 2) | multiply(n1,n2)| | general | D |
if the average ( arithmetic mean ) of 2 a + 16 , 3 a - 8 is 94 , what is the value of a ? | "am of 2 a + 16 , 3 a - 8 = 2 a + 16 + 3 a - 8 / 2 = 5 a + 8 / 2 given that 5 a + 8 / 2 = 94 a = 36 answer is d" | a ) 25 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | d | subtract(multiply(16, const_2), multiply(2, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general | D |
total 60 cows 20 cow gives each 2 liter milk 20 cow gives each 3 / 4 liter milk 20 cow gives each 1 / 4 liter milk this is split into 3 son per each 20 cows & 20 liter milk how ? | 20 cow 2 liter each = 40 liter 20 cow 3 / 4 liter each = 3 / 4 = 0.75 * 20 = 15 20 cow 1 / 4 liter each = 1 / 4 = 0.25 * 20 = 5 add 40 + 15 + 5 = 60 milk split into 3 son each 20 liter then 60 / 3 = 20 answer : a | a ) 20 , b ) 10 , c ) 15 , d ) 16 , e ) 18 | a | add(add(divide(3, 4), multiply(divide(3, 4), 20)), multiply(const_0_25, 20)) | divide(n4,n5)|multiply(n1,const_0_25)|multiply(n1,#0)|add(#0,#2)|add(#3,#1) | general | A |
the total age of a and b is 10 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 10 â € ¹ = â € º a - c = 10 . answer b" | a ) 12 , b ) 10 , c ) c is elder than a , d ) data inadequate , e ) none | b | multiply(10, const_1) | multiply(n0,const_1)| | general | B |
a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 80 points . to the nearest tenth of a point , what was the student ’ s average grade for the entire two - year period ? | let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 80 * 5 = 400 student ' s average = ( 600 + 400 ) / 11 = 90.91 answer d | a ) 79 , b ) 89 , c ) 95 , d ) 90.91 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 80)), add(6, 5))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|floor(#4) | general | D |
find the area of trapezium whose parallel sides are 12 cm and 16 cm long , and the distance between them is 14 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 12 + 16 ) * ( 14 ) = 196 cm 2 answer : d" | a ) 288 cm 2 , b ) 277 cm 2 , c ) 285 cm 2 , d ) 196 cm 2 , e ) 287 cm 2 | d | quadrilateral_area(14, 16, 12) | quadrilateral_area(n2,n1,n0)| | physics | D |
sachin is younger than rahul by 8 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 8 , so ( x - 8 ) / x = 7 / 9 = > 9 x - 72 = 7 x = > 2 x = 72 = > x = 36 so sachin age is 36 - 8 = 28 answer : e" | a ) 24.58 , b ) 24.5 , c ) 26 , d ) 27 , e ) 28 | e | multiply(divide(8, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other | E |
what is the area of square field whose side of length 13 m ? | "13 * 13 = 169 sq m answer : a" | a ) 169 , b ) 196 , c ) 266 , d ) 288 , e ) 261 | a | square_area(13) | square_area(n0)| | geometry | A |
there are two concentric circles with radii 8 and 4 . if the radius of the outer circle is increased by 25 % and the radius of the inner circle decreased by 25 % , by what percent does the area between the circles increase ? | "the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 64 pi . the area of the small circle is 16 pi . the area a 1 between the circles is 48 pi . when the big circle ' s radius increases , the new area is 100 pi . when the small circle ' s radius decreases , the new area is 9 pi . the area a 2 between the circles is 91 pi . the ratio of a 2 / a 1 is 91 / 48 = 1.9 which is an increase of 90 % . the answer is d ." | a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | d | multiply(divide(subtract(subtract(power(multiply(8, divide(add(const_100, 25), const_100)), const_2), power(multiply(4, divide(subtract(const_100, 25), const_100)), const_2)), subtract(power(8, const_2), power(4, const_2))), subtract(power(8, const_2), power(4, const_2))), const_100) | add(n2,const_100)|power(n0,const_2)|power(n1,const_2)|subtract(const_100,n3)|divide(#0,const_100)|divide(#3,const_100)|subtract(#1,#2)|multiply(n0,#4)|multiply(n1,#5)|power(#7,const_2)|power(#8,const_2)|subtract(#9,#10)|subtract(#11,#6)|divide(#12,#6)|multiply(#13,const_100)| | gain | D |
what is the square root of 239,121 ? | "according to divisibility rules , 239,121 is divisible by 9 . 2 + 3 + 9 + 1 + 2 + 1 = 18 this means , that the answer choice must be divisible by 3 a ) 4 + 7 + 6 = 17 no b ) 4 + 8 + 9 = 21 yes c ) 4 + 9 + 7 = 20 no d ) 5 + 1 + 1 = 7 no e ) 5 + 2 + 4 = 11 no b is the only answer" | a ) 476 , b ) 489 , c ) 497 , d ) 511 , e ) 524 | b | circle_area(divide(239,121, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other | B |
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent r should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked r will be 25 % greater than the original number . the answer is c ." | a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | multiply(const_10,const_4)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,const_10)|divide(#0,#3)|subtract(#4,const_4)|divide(#5,const_4)|multiply(#6,const_100)| | general | C |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 15 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 15 ) = 285 cm 2 answer : c" | a ) 288 , b ) 276 , c ) 285 , d ) 299 , e ) 261 | c | quadrilateral_area(15, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics | C |
a 300 m long train crosses a platform in 54 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 54 = 50 / 3 3 x + 900 = 2700 = > x = 600 m . answer : c" | a ) 287 , b ) 350 , c ) 600 , d ) 277 , e ) 122 | c | subtract(multiply(speed(300, 18), 54), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | C |
the area of a square field is 28808 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ? | "area of a square field = 28808 sq m let the side of square = a a ^ 2 = 28808 = > a = 169.73 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 169.73 = 240 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 240 / 120 = 2 min answer c" | a ) 2.5 min , b ) 3 min , c ) 2 min , d ) 4 min , e ) 4.5 min | c | divide(28808, multiply(7.2, const_1000)) | multiply(n1,const_1000)|divide(n0,#0)| | geometry | C |
if a lends rs . 3500 to b at 10 % p . a . and b lends the same sum to c at 11.5 % p . a . , then the gain of b ( in rs . ) in a period of 3 years is | explanation : we need to calculate the profit of b . it will be , si on the rate b lends - si on the rate b gets gain of b = 3500 × 11.5 × 3 / 100 − 3500 × 10 × 3 / 100 = 157.50 option d | a ) rs . 154.50 , b ) rs . 155.50 , c ) rs . 156.50 , d ) rs . 157.50 , e ) none of these | d | subtract(divide(multiply(multiply(3500, 11.5), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(n3,#0)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5) | gain | D |
in what time will a train 90 meters long cross an electric pole , if its speed is 124 km / hr | "explanation : first convert speed into m / sec speed = 124 * ( 5 / 18 ) = 34 m / sec time = distance / speed = 90 / 34 = 2.6 seconds option d" | a ) 5 seconds , b ) 4.5 seconds , c ) 3 seconds , d ) 2.6 seconds , e ) none of these | d | divide(90, multiply(124, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | D |
if jack walked 7 miles in 1 hour and 15 minutes , what was his rate of walking in miles per hour ? | "distance walked in 1 hour and 15 mins = 7 miles speed per hour = distance / time = 7 / ( 5 / 4 ) = 5.6 miles per hour answer c" | a ) 4 , b ) 4.5 , c ) 5.6 , d ) 6.25 , e ) 15 | c | divide(multiply(7, const_60), add(15, const_60)) | add(n2,const_60)|multiply(n0,const_60)|divide(#1,#0)| | physics | C |
running at the same constant rate , 16 identical machines can produce a total of 32 shirt per minute . at this rate , how many shirt could 8 such machines produce in 10 minutes ? | "let ' s take the approach that uses the answer choices to eliminate wasted time . 32 / 16 = 2 shirt per minute per machine . 8 machines = 16 per minute . 10 minutes worth = 160 shirt . looking at the answers it is clear . . . we can only choose ( d ) the correct answer is d ." | a ) 911 , b ) 100 , c ) 101 , d ) 160 , e ) 785 | d | multiply(multiply(divide(32, 16), 10), 8) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain | D |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 39 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 39 = > 4 / x = 1 / 39 x = 39 ∗ 4 = 156 mins option d" | a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 156 minw , e ) none of these | d | multiply(add(const_1, const_4), 39) | add(const_1,const_4)|multiply(n0,#0)| | physics | D |
a person distributed 20 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 40000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 30 % orphan house = 5 % total = 60 + 30 + 5 = 95 % remaining = 100 - 95 = 5 % 5 % = 40000 100 % = 40000 * 100 / 5 = $ 800000 answer is c" | a ) 452000 , b ) 562000 , c ) 800000 , d ) 500000 , e ) 652000 | c | multiply(divide(40000, subtract(const_100, add(add(multiply(20, 3), 30), 5))), const_100) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|subtract(const_100,#2)|divide(n4,#3)|multiply(#4,const_100)| | gain | C |
today joelle opened an interest - bearing savings account and deposited $ 6000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d | a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315 | d | multiply(6000, power(add(const_1, divide(5, const_100)), const_2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2) | gain | D |
a number when divided by 45 , gives 30 as quotient and 0 as remainder . what will be the remainder when dividing the same number by 15 | explanation : p ÷ 45 = 30 = > p = 30 * 45 = 1350 p / 15 = 1350 / 15 = 90 , remainder = 0 answer : option a | a ) a ) 0 , b ) b ) 3 , c ) c ) 4 , d ) d ) 6 , e ) e ) 7 | a | divide(multiply(45, 30), 15) | multiply(n0,n1)|divide(#0,n3) | general | A |
when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of q ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . q * 23 + 39 = 25 * ( q + 1 ) q = 7 ans . ( a )" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,n3)| | general | A |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 4 : 3 . the ratio of economy cars to sport utility vehicles is 6 : 5 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 4 : 3 - - > e : l = 4 : 3 = 24 : 18 . the ratio of economy cars to sport utility vehicles is 6 : 5 - - > e : s = 6 : 5 = 24 : 20 . thus , l : s = 18 : 20 = 9 : 10 . answer : a ." | a ) 9 : 10 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | a | divide(divide(multiply(const_4, const_3.0), multiply(5, 5)), divide(multiply(5, const_4), multiply(3, const_4))) | multiply(const_3.0,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other | A |
find the simple interest on rs . 64,000 at 16 2 / 3 % per annum for 9 months . | "p = rs . 64000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 64,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8000 answer is b ." | a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | b | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain | B |
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 884 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 884 - 815 ) = rs . 69 . s . i . for 3 years = rs . ( 69 x 3 ) = rs . 207 . principal = rs . ( 815 - 207 ) = rs . 608 . answer : option a" | a ) s . 608 , b ) s . 690 , c ) s . 698 , d ) s . 700 , e ) s . 760 | a | subtract(815, divide(multiply(subtract(884, 815), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | A |
when 1 + 2 = 23 , 2 + 3 = 65 , 3 + 4 = 127 , then 4 + 5 = ? | "1 + 2 = > 1 x 2 = 2 & 1 + 2 = 3 = > 2 & 3 = > 23 2 + 3 = > 2 ã — 3 = 6 & 2 + 3 = 5 = > 6 & 6 = > 65 3 + 4 = > 3 ã — 4 = 12 & 3 + 4 = 7 = > 12 & 7 = > 127 then 4 + 5 = > 4 ã — 5 = 20 & 4 + 5 = 9 = > 20 & 9 = > 209 answer : a" | a ) 209 , b ) 250 , c ) 265 , d ) 280 , e ) 225 | a | add(multiply(multiply(4, 5), const_10), 1) | multiply(n7,n10)|multiply(#0,const_10)|add(n0,#1)| | general | A |
if m and n are positive integers and m ^ 2 + n ^ 2 = 10 , what is the value of m ^ 3 + n ^ 3 ? | "you need to integers which squared are equal 10 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 stop . the integers ca n ' t be greater than 3 or we will score above 10 . the second integer need to be picked up the same way . 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 the only pair that matches is 3 ^ 2 + 1 ^ 2 = 10 . so 3 ^ 3 + 1 ^ 3 = 28 . answer a . )" | a ) 28 , b ) 224 , c ) 320 , d ) 512 , e ) 1,600 | a | gcd(10, const_4) | gcd(n2,const_4)| | general | A |
two boats are heading towards each other at constant speeds of 3 miles / hr and 21 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 3 + 21 = 24 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 24 * 1 / 60 = 3 / 15 miles apart . answer : e ." | a ) 1 / 12 , b ) 5 / 12 , c ) 1 / 6 , d ) 1 / 3 , e ) 3 / 15 | e | divide(add(21, 3), const_60) | add(n0,n1)|divide(#0,const_60)| | physics | E |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 59 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 59 ) = 481.20 hence answer will be ( e )" | a ) 320.20 $ , b ) 380.20 $ , c ) 421.20 $ , d ) 451.20 $ , e ) 481.20 $ | e | multiply(multiply(0.65, 59), 12) | multiply(n1,n3)|multiply(n2,#0)| | general | E |
the length of the rectangular field is double its width . inside the field there is square shaped pond 8 m long . if the area of the pond is 1 / 50 of the area of the field . what is the length of the field ? | "explanation : a / 50 = 8 * 8 = > a = 8 * 8 * 50 x * 2 x = 8 * 8 * 50 x = 40 = > 2 x = 80 answer : option e" | a ) 73 , b ) 32 , c ) 34 , d ) 43 , e ) 80 | e | sqrt(divide(multiply(square_area(8), 50), inverse(const_2))) | inverse(const_2)|square_area(n0)|multiply(n2,#1)|divide(#2,#0)|sqrt(#3)| | geometry | E |
xy = 1 then what is ( 7 ^ ( x + y ) ^ 2 ) / ( 7 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 7 ^ 4 = 2401 answer a" | a ) 2401 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | a | power(7, multiply(const_4, 1)) | multiply(n0,const_4)|power(n1,#0)| | general | A |
if 50 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ? | "50 % of ( x - y ) = 20 % of ( x + y ) 50 / 100 ( x - y ) = 20 / 100 ( x + y ) 3 x = 7 y required percentage = y / x * 100 = 3 y / 7 y * 100 = 42.85 % answer is b" | a ) 50.5 % , b ) 42.8 % , c ) 22.2 % , d ) 33.3 % , e ) 25 % | b | multiply(divide(subtract(50, 20), add(50, 20)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general | B |
a small company reduced its faculty by approximately 13 percent to 263 employees . what was the original number of employees ? | if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 263 x = 302 so the original number of employees is 302 correct answer - e | a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | e | divide(263, divide(subtract(const_100, 13), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1) | gain | E |
what is the difference between the largest number and the least number written with the digits 9 , 3 , 5 , 7 ? | "explanation : 3579 9753 - - - - - - - - - - - - 6174 answer : e" | a ) 6084 , b ) 3788 , c ) 2077 , d ) 2721 , e ) 6174 | e | subtract(add(add(add(multiply(multiply(9, const_100), const_10), multiply(7, const_100)), multiply(3, const_10)), 5), add(add(add(const_1000, multiply(3, const_100)), multiply(7, const_10)), 9)) | multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)| | general | E |
a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 360 - meter long platform ? | "let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 360 ) / 10 = 50 seconds the answer is d ." | a ) 44 , b ) 46 , c ) 48 , d ) 50 , e ) 52 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 360), multiply(36, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)| | physics | D |
the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 π · r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many q grams ? | weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b ) | ['a ) q = 16', 'b ) q = 32', 'c ) 64', 'd ) 128', 'e ) 512'] | b | multiply(8, 4) | multiply(n0,n3) | geometry | B |
if bill can buy 3 pairs of jeans and 2 shirts for $ 69 or 2 pairs of jeans and 3 shirts for $ 86 , how much does one shirt cost ? | "3 j + 2 s = 69 2 j + 3 s = 86 - - - - - - - - - - - - - - - - 5 j + 5 s = 155 - - - - ( divide by 5 ) - - - > j + s = 31 3 j + 2 s = j + 2 ( j + s ) = j + 62 = 69 - - - > j = 7 3 * 7 + 2 s = 69 21 + 2 s = 69 2 s = 48 s = 24 answer : e" | a ) $ 10 , b ) $ 12 , c ) $ 13.20 , d ) $ 15 , e ) $ 24 | e | divide(subtract(multiply(86, 3), multiply(69, 2)), subtract(multiply(3, 3), multiply(2, 2))) | multiply(n0,n5)|multiply(n1,n2)|multiply(n0,n0)|multiply(n1,n1)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)| | general | E |
the roof of an apartment building is rectangular and its length is 5 times longer than its width . if the area of the roof is 720 feet squared , what is the difference between the length and the width of the roof ? | "answer is e : 48 let w be the width , so length is 5 w . therefore : w * 5 w = 720 , solving for , w = 12 , so 5 w - w = 4 w = 4 * 12 = 48" | a ) 38 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 48 . | e | subtract(multiply(sqrt(divide(720, 5)), 5), sqrt(divide(720, 5))) | divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)| | geometry | E |
the sum of two numbers is 528 and their h . c . f . is 33 . the number of pairs of numbers satisfying the above conditions is : | "explanation : let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 . = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . â ˆ ´ required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 x 7 , 33 x 9 ) . the number of such pairs is 4 . answer is a" | a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15 | a | multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33)) | add(n0,n1)|add(const_1,const_1)|divide(#0,#1)|subtract(#2,n1)|multiply(#2,#3)| | general | A |
in a hostel , the number of students decreased by 15 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previous year ? | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 15 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.85 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.85 * 1.2 ) = > f ( new ) = 0.9804 f therefore the new cost of food must be 98.04 % of the old cost , or the cost of food must decrease by 1.96 % ( option e )" | a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 1.96 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 15), add(const_100, 20)))), const_100) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general | E |
1 / 4 of all married couples have more than one child . 1 / 5 of all married couples have more than 3 children . what fraction of all married couples have 2 or 3 children ? | "plug in simple numbers . take 100 couples for example . 1 / 4 of 100 couples have more than one child = 25 couples . 1 / 5 of 100 couples have more than 3 kids = 20 couples . this implies that 20 couples are a subset of 25 couples and the complement of 25 couples within those 100 couples , which equals 75 couples have either one or no kids at all . we need to find couples that have 2 or 3 kids , so essentially , it is 25 - 20 = 5 . fraction will be 5 / 100 = 1 / 20 . option c" | a ) 1 / 5 , b ) b . 1 / 4 , c ) 1 / 20 , d ) 3 / 5 , e ) it can not be determined from the given information . | c | subtract(divide(1, 4), divide(1, 4)) | divide(n0,n1)|divide(n2,n1)|subtract(#0,#1)| | general | C |
the average age of 28 students in a group is 11 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "age of the teacher = ( 29 × 12 – 28 × 11 ) years = 40 years . answer c" | a ) 36 , b ) 38 , c ) 40 , d ) can not be determined , e ) none of these | c | add(28, const_1) | add(n0,const_1)| | general | C |
60 kg of an alloy a is mixed with 100 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 2 and alloy b has tin and copper in the ratio 1 : 4 , then the amount of tin in the new alloy is ? | quantity of tin in 60 kg of a = 60 * 2 / 5 = 24 kg quantity of tin in 100 kg of b = 100 * 1 / 5 = 20 kg quantity of tin in the new alloy = 24 + 20 = 44 kg answer is b | a ) 24 kg , b ) 44 kg , c ) 20 kg , d ) 30 kg , e ) 52 kg | b | add(multiply(divide(60, add(3, 2)), 2), multiply(divide(100, add(1, 4)), 1)) | add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5) | other | B |
a and b started a business in partnership investing rs . 20000 and rs . 15000 respectively . after 6 months , c joined them with rs . 20000 . whatwill be b ' s share in total profit of rs . 20000 earned at the end of 2 years from the startingof the business ? | a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 20000 x 3 / 10 = rs . 6,000 . d | a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,000 , e ) s . 7,500 | d | divide(multiply(20000, divide(multiply(15000, multiply(const_12, const_2)), add(add(multiply(20000, multiply(const_12, const_2)), multiply(15000, multiply(const_12, const_2))), multiply(subtract(multiply(const_12, const_2), 6), 20000)))), const_12) | multiply(const_12,const_2)|multiply(n1,#0)|multiply(n0,#0)|subtract(#0,n2)|add(#2,#1)|multiply(n0,#3)|add(#4,#5)|divide(#1,#6)|multiply(n0,#7)|divide(#8,const_12) | gain | D |
the mean of 40 observations was 36 . it was found later that an observation 34 was wrongly taken as 20 . the corrected new mean is | "explanation : correct sum = ( 36 * 40 + 34 - 20 ) = 1454 correct mean = = 1454 / 40 = 36.35 answer : b" | a ) 76.55 , b ) 36.35 , c ) 46.15 , d ) 16.05 , e ) 20 | b | divide(add(multiply(36, 40), subtract(subtract(40, const_2), 20)), 40) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general | B |
when leo imported a certain item , he paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1000 . if the amount of the import tax that leo paid was $ 87.50 , what was the total value of the item ? | let the value of the item be $ x , then 0.07 ( x − 1000 ) = 87.5 - - > x − 1000 = 1250 x − 1000 = 1250 - - > x = 2250 . answer : c . | a ) $ 1600 , b ) $ 1850 , c ) $ 2250 , d ) $ 2400 , e ) $ 2750 | c | add(divide(87.5, divide(7, const_100)), 1000) | divide(n0,const_100)|divide(n2,#0)|add(n1,#1) | general | C |
in an election between two candidates - lange and sobel - 70 % of the voters voted for sobel . of the election ` s voters , 60 % were male . if 35 % of the female voters voted for lange , what percentage of the male voters voted for sobel ? | m f total l 16 14 30 s 44 70 tot 60 40 100 35 % of 40 = 14 - - - - - - female voters that means 16 male voted for l so 60 - 16 = 44 d is correct | a ) 14 , b ) 16 , c ) 26 , d ) 44 , e ) 65 | d | subtract(70, subtract(subtract(const_100, 60), divide(multiply(subtract(const_100, 60), 35), const_100))) | subtract(const_100,n1)|multiply(n2,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(n0,#3) | gain | D |
working together , printer a and printer b would finish the task in 10 minutes . printer a alone would finish the task in 30 minutes . how many pages does the task contain if printer b prints 5 pages a minute more than printer a ? | "10 * a + 10 * b = x pages in 10 mins printer a will print = 10 / 30 * x pages = 1 / 3 * x pages thus in 10 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 5 more pages per min that printer a . in 10 mins printer b will print 50 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 50 = > x = 150 pages answer : b" | a ) 125 , b ) 150 , c ) 175 , d ) 200 , e ) 225 | b | multiply(divide(5, subtract(divide(30, 10), const_1)), 30) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)| | physics | B |
this year , mbb consulting fired 8 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ? | "100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 8 % reduction in employees lead to 92 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 92 employees is 92 * 1100 = 101200 now the new salary is more than previous salary by x % . x = ( 101200 / 100000 ) * 100 = 101.2 % so the answer is b" | a ) 98.5 % , b ) 101.2 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | b | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 8), const_4)), multiply(const_100, 10)) | multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)| | general | B |
1 + 2 + 3 + . . . + 12 = ? | 1 + 2 + 3 + ⋯ + n = ∑ n = n ( n + 1 ) / 2 1 + 2 + 3 + ⋯ + 12 = n ( n + 1 ) / 2 = ( 12 ( 12 + 1 ) ) / 2 = ( 12 × 13 ) / 2 = 6 × 13 = 78 answer is c | a ) 76 , b ) 68 , c ) 78 , d ) 66 , e ) 67 | c | divide(multiply(12, add(12, 1)), 2) | add(n0,n3)|multiply(n3,#0)|divide(#1,n1) | general | C |
what is the units digit of 2222 ^ ( 333 ) * 3333 ^ ( 222 ) ? | "( 2222 ^ 333 ) * ( 3333 ^ 222 ) = 2222 ^ 111 * ( 2222 ^ 222 * 3333 ^ 222 ) here please pay attention to the fact that the unit digit of multiplication of 2222 and 3333 is 6 ( 2222 ^ 222 * 3333 ^ 222 ) . since 6 powered in any number more than 0 results in 6 as a units digit , as a result we have - 6 * 2222 ^ 111 2 has a cycle of 4 . 111 = 27 * 4 + 3 . 2 ^ 3 = 8 6 * 8 = 48 so the units digit is 8 , and the answer is e" | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general | E |
the distance from city a to city b is 30 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | "the time it takes bob to drive to city b is 30 / 40 = 0.75 hours . alice needs to take less than 0.25 hours for the trip . alice needs to exceed a constant speed of 30 / 0.25 = 120 miles per hour . the answer is c ." | a ) 45 , b ) 88 , c ) 120 , d ) 152 , e ) 154 | c | divide(30, subtract(divide(30, 40), divide(30, const_60))) | divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2)| | physics | C |
the tax on a commodity is diminished by 20 % and its consumption increased by 25 % . the effect on revenue is ? | "100 * 100 = 10000 80 * 125 = 10000 - - - - - - - - - - - 10000 - - - - - - - - - - - 0 100 - - - - - - - - - - - ? = > 0 % no change answer : d" | a ) 2 % decrease , b ) 8 % decrease , c ) 9 % decrease , d ) 0 % no change , e ) 2 % decrease | d | subtract(const_100, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general | D |
for a race a distance of 224 meters can be covered by p in 8 seconds and q in 32 seconds . by what distance does p defeat q eventually ? | "explanation : this is a simple speed time problem . given conditions : = > speed of p = 224 / 8 = 28 m / s = > speed of q = 224 / 32 = 7 m / s = > difference in time taken = 24 seconds therefore , distance covered by p in that time = 28 m / s x 24 seconds = 672 metres answer : a" | a ) 672 m , b ) 6738 m , c ) 634 m , d ) 671 m , e ) 636 m | a | subtract(224, multiply(8, speed(224, 32))) | speed(n0,n2)|multiply(n1,#0)|subtract(n0,#1)| | physics | A |
pr is tangent to a circle at point p . q is another point on circle such that pq is diameter of circle and rq cuts circle at m . if radius of circle is 4 units and pr = 6 units . find ratio of triangle pmr to pqr . | let pm = y qm = x mr = 10 - x in triangle pqm , pm ^ 2 + qm ^ 2 = pq ^ 2 = > x ^ 2 + y ^ 2 = 64 . . . . . ( 1 ) in triangle pmr , pm ^ 2 + mr ^ 2 = pr ^ 2 y ^ 2 + ( 10 - x ) ^ 2 = 36 . . . . . . ( 2 ) solving for x and y x = 6.4 y = 4.8 taking permineter of triangle pqr = 24 perimeter of triangle pmr = 14.4 take ratio 14.4 / 24 = 0.6 answer : c | ['a ) 11 / 20', 'b ) 3 / 5', 'c ) 13 / 20', 'd ) 18 / 25', 'e ) 18 / 26'] | c | divide(triangle_perimeter(6, divide(multiply(multiply(4, const_2), 6), sqrt(add(power(multiply(4, const_2), const_2), power(6, const_2)))), divide(multiply(6, 6), multiply(4, const_2))), triangle_perimeter(multiply(4, const_2), 6, sqrt(add(power(multiply(4, const_2), const_2), power(6, const_2))))) | multiply(n0,const_2)|multiply(n1,n1)|power(n1,const_2)|divide(#1,#0)|multiply(n1,#0)|power(#0,const_2)|add(#5,#2)|sqrt(#6)|divide(#4,#7)|triangle_perimeter(n1,#0,#7)|triangle_perimeter(n1,#8,#3)|divide(#10,#9) | geometry | C |
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 37 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 37 sec = 37 * 10 = 370 m . the length of the faster train = 370 m . answer : b" | a ) 270 , b ) 370 , c ) 266 , d ) 299 , e ) 126 | b | multiply(divide(subtract(72, 36), const_3_6), 37) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics | B |
the current of a stream runs at the rate of 5 kmph . a boat goes 6 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 5 m = x ds = x + 5 us = x - 5 6 / ( x + 5 ) + 6 / ( x - 5 ) = 2 x = 8.83 answer : c" | a ) a ) 7.63 , b ) b ) 2.6 , c ) c ) 8.83 , d ) d ) 6.69 , e ) e ) 3 | c | divide(power(5, 2), 2) | power(n0,n2)|divide(#0,n2)| | physics | C |
what is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between the two sides is 150 o ? | explanatory answer if two sides of a triangle and the included angle ' y ' is known , then the area of the triangle = 1 / 2 * ( product of sides ) * sin y substituting the values in the formula , we get 1 / 2 * 8 * 12 * sin 150 o = 1 / 2 * 8 * 12 * 1 / 2 . note : sin 150 o = 1 / 2 area = 24 sq units . choice a | ['a ) 24 sq units', 'b ) 48 sq units', 'c ) 24 under root 3', 'd ) 48 under root 3', 'e ) such a triangle does not exist'] | a | subtract(triangle_area(add(multiply(8, divide(sqrt(const_3), const_2)), 12), divide(8, const_2)), triangle_area(divide(8, const_2), multiply(8, divide(sqrt(const_3), const_2)))) | divide(n0,const_2)|sqrt(const_3)|divide(#1,const_2)|multiply(n0,#2)|add(n1,#3)|triangle_area(#0,#3)|triangle_area(#4,#0)|subtract(#6,#5) | geometry | A |
the sum of three consecutive integers is 102 . find the lowest of the three ? | "explanation : three consecutive numbers can be taken as ( p - 1 ) , p , ( p + 1 ) . so , ( p - 1 ) + p + ( p + 1 ) = 102 3 p = 102 = > p = 34 . the lowest of the three = ( p - 1 ) = 34 - 1 = 33 . answer b" | a ) 15 , b ) 33 , c ) 56 , d ) 96 , e ) 88 | b | add(add(power(add(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(102, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | B |
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