Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
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john found that the average of 15 numbers is 40 . if 12 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 52 option a" | a ) 52 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 12) | add(n1,n2)| | general |
a and b together can do a piece of work in 2 days . if a alone can do the same work in 22 days , then b alone can do the same work in ? | "b = 1 / 2 β 1 / 22 = 0.45 days answer : b" | a ) 0.35 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | b | inverse(subtract(inverse(2), inverse(22))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
if 1 / ( x + 3 ) + 1 / ( x - 3 ) = 1 / ( x + 3 ) , what is the value of x ? | "if we solve the question , we get x = - 3 . option : e" | a ) 2 , b ) 3 , c ) - 1 , d ) - 2 , e ) - 3 | e | subtract(add(1, 1), 1) | add(n0,n0)|subtract(#0,n0)| | general |
the area of a square field is 1600 km 2 . how long will it take for a horse to run around at the speed of 16 km / h ? | explanation area of field = 1600 km 2 . then , each side of field = β 1600 = 40 km distance covered by the horse = perimeter of square field = 40 Γ 4 = 160 km β΄ time taken by horse = distances / peed = 160 / 16 = 10 h answer b | a ) 12 h , b ) 10 h , c ) 8 h , d ) 6 h , e ) none of these | b | divide(multiply(const_4, sqrt(1600)), 16) | sqrt(n0)|multiply(#0,const_4)|divide(#1,n2)| | geometry |
how many 4 - digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ? . | the number of ways u can chose two digits from 9 digits is 9 c 2 = ( 9 * 8 ) / 2 . and the no of ways u can arrange two digits like asked is 4 c 2 = ( 4 * 3 ) / 2 . total ways is 36 * 6 = 216 . answer : d | a ) 400 , b ) 1728 , c ) 108 , d ) 216 , e ) 432 | d | multiply(divide(multiply(9, subtract(9, 1)), const_2), divide(multiply(4, const_3), const_2)) | multiply(n0,const_3)|subtract(n2,n1)|divide(#0,const_2)|multiply(n2,#1)|divide(#3,const_2)|multiply(#4,#2) | general |
the charge for a single room at hotel p is 50 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | p = 0.5 r = 0.9 g r = 0.9 g / 0.5 = 1.8 * g thus r is 80 % greater than g . the answer is d . | a ) 45 % , b ) 50 % , c ) 60 % , d ) 80 % , e ) 90 % | d | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 50), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 50), subtract(const_100, 10)), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100) | gain |
0.009 / x = 0.03 . find the value of x | "x = 0.009 / 0.03 = 0.3 answer : c" | a ) 0.0003 , b ) 0.03 , c ) 0.3 , d ) 3 , e ) 30 | c | divide(0.009, 0.03) | divide(n0,n1)| | general |
a man rows his boat 90 km downstream and 70 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 70 / ( 3 ) = 23 kmph the speed of the stream = ( 30 - 23 ) / 2 = 3.5 kmph answer : c" | a ) 76 kmph , b ) 6 kmph , c ) 3.5 kmph , d ) 4.5 kmph , e ) 4 kmph | c | divide(subtract(divide(90, 3), divide(70, 3)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
one night 18 percent of the female officers on a police force were on duty . if 180 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 180 number of female officers on duty on that night = 180 / 2 = 90 ( 18 / 100 ) * f = 90 = > f = 500 answer d" | a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | d | divide(divide(180, const_2), divide(18, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
there are 24 students in a seventh grade class . they decided to plant birches and roses at the school ' s backyard . while each girl planted 3 roses , every three boys planted 1 birch . by the end of the day they planted 24 plants . how many birches and roses were planted ? | "let x be the number of roses . then the number of birches is 24 β x , and the number of boys is 3 Γ ( 24 β x ) . if each girl planted 3 roses , there are x 3 girls in the class . we know that there are 24 students in the class . therefore x 3 + 3 ( 24 β x ) = 24 x + 9 ( 24 β x ) = 3 β
24 x + 216 β 9 x = 72 216 β 72 = ... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(multiply(3, 24), 24), subtract(multiply(3, 3), 1)) | multiply(n0,n1)|multiply(n1,n1)|subtract(#0,n0)|subtract(#1,n2)|divide(#2,#3)| | physics |
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,900 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ? | "median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.9 so , answer is d ." | a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,900 , e ) 48,500 | d | multiply(multiply(const_100, const_100), const_4) | multiply(const_100,const_100)|multiply(#0,const_4)| | general |
if shares of two persons in profits are rs . 400 and rs . 1200 then ratio of their capitals is | profit = time * capital so 400 : 1200 = 1 : 3 answer : d | a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 2 | d | divide(400, 1200) | divide(n0,n1) | other |
rectangular tile each of size 60 cm by 56 cm must be laid horizontally on a rectangular floor of size 560 cm by 240 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 60 * 56 = 3360 area of floor = 560 * 240 = 134400 no of tiles = 134400 / 3360 = 40 so , the no of tile = 40 answer : d" | a ) 50 , b ) 54 , c ) 47 , d ) 40 , e ) 45 | d | divide(multiply(560, 240), multiply(60, 56)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
the l . c . m of two numbers is 495 and their h . c . f is 5 . if the sum of the numbers is 10 , then their difference is | "solution let the numbers be x and ( 100 - x ) . then , x ( 100 - x ) = 5 Γ 495 βΉ = βΊ x 2 - 100 x + 2475 = 0 . βΉ = βΊ ( x - 55 ) ( x - 45 ) = 0 . x = 55 or x = 45 . therefore , the numbers are 45 and 55 . required difference = ( 55 - 45 ) = 10 . answer a" | a ) 10 , b ) 46 , c ) 70 , d ) 90 , e ) none | a | divide(multiply(5, 495), 10) | multiply(n0,n1)|divide(#0,n2)| | general |
how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 9 can be used only once ? | 4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with two 9 - s , 9 x 9 x... | a ) 20 , b ) 150 , c ) 225 , d ) 304 , e ) 320 | d | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), const_4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,const_4)|subtract(#3,#4) | physics |
there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 45 , then find the smaller number ? | "let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 45 3 x = 45 = > x = 15 = > smaller number = 5 x = 75 . answer : e" | a ) 25 , b ) 66 , c ) 77 , d ) 88 , e ) 75 | e | divide(multiply(45, 5), const_4) | multiply(n0,n2)|divide(#0,const_4)| | other |
the sum of ages of 4 children born 2 years different each is 36 years . what is the age of the elder child ? | "let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) = 36 4 x = 24 x = 6 . x + 6 = 6 + 6 = 12 answer : d" | a ) 8 , b ) 9 , c ) 10 , d ) 12 , e ) 17 | d | divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 36), 4) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
if ajay completes a certain work in 8 days and vijay can complete the same work in 24 days , find the time in which the work will be completed if both ajay and vijay work together . | explanation : on working together ajay and vijay can complete the work in ( 8 * 24 ) / ( 8 + 24 ) = 192 / 32 = 6 days answer : a | a ) 6 days , b ) 9 days , c ) 2 days , d ) 7 days , e ) 1 days | a | inverse(add(inverse(8), inverse(24))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2) | physics |
the length of the diagonal of a square is 12 cm . find its area and perimeter . | diagonal of a square = 12 cm area of square = 1 / 2 ( d ) Β² = 1 / 2 ( 12 ) Β² = 1 / 2 Γ 12 Γ 12 = 72 side of a square = β area = β 72 = β ( 2 Γ 2 Γ 2 Γ 3 Γ 3 ) = 2 Γ 3 β 2 = 6 Γ 1.41 = 8.46 cm perimeter of square = 4 Γ 8.46 = 33.84 cm answer option : b | ['a ) 31.84 cm', 'b ) 33.84 cm', 'c ) 35.84 cm', 'd ) 38.84 cm', 'e ) 38.92 cm'] | b | square_perimeter(divide(12, sqrt(const_2))) | sqrt(const_2)|divide(n0,#0)|square_perimeter(#1) | geometry |
a cube of edge 15 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 20 cm * 14 cm , find the rise in water level ? | "increase in volume = volume of the cube = 15 * 15 * 15 cm ^ 3 rise in water level = volume / area = 15 * 15 * 15 / 20 * 14 = 12.05 cm answer is b" | a ) 6 cm , b ) 12.05 cm , c ) 11.25 cm , d ) 15.12 cm , e ) 20.62 cm | b | divide(volume_cube(15), multiply(20, 14)) | multiply(n1,n2)|volume_cube(n0)|divide(#1,#0)| | geometry |
joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 9 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 9 ) * 360 = 40 gallons . remaning paint = 320 gallons paint used in the second week = ( 1 / 5 ) * 320 = 64 gallons total paint used = 104 gallons . option c" | a ) 18 , b ) 144 , c ) 104 , d ) 216 , e ) 250 | c | add(multiply(divide(360, 9), 1), divide(subtract(360, multiply(divide(360, 9), 1)), 5)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)| | physics |
what is the sum of 59 consecutive integers from - 29 inclusive , in a increasing order ? | "from - 29 to - 1 - - > 29 nos . zero - - > 1 number from + 1 to + 29 - - > 29 nos . when we add up nos . from - 29 to + 29 sum will be zero for total 59 nos . sum of these 59 nos . = 0 . d is the answer ." | a ) - 29 , b ) 29 , c ) - 30 , d ) 0 , e ) 60 | d | add(29, const_1) | add(n1,const_1)| | general |
if an integer n is to be selected at random from 1 to 400 , inclusive , what is probability n ( n + 1 ) will be divisible by 8 ? | "for n ( n + 1 ) to be a multiple of 8 , either n or n + 1 has to be a multiple of 8 . thus n must be of the form 8 k or 8 k - 1 . the probability is 2 / 8 = 1 / 4 . the answer is a ." | a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | a | divide(const_2, 8) | divide(const_2,n3)| | general |
students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 185 cms . what was the ratio of men to women ? | "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 185 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 185 m ) / ( w + m ) = 180 170 w + 185 m = 180 w + 1... | a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 2 : 1 , e ) 3 : 1 | d | divide(subtract(180, 170), subtract(185, 180)) | subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)| | general |
a and b start walking towards each other at 6 pm at speed of 4 kmph and 6 kmph . they were initially 18 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 18 / 6 + 3 = 18 / 9 = 2 hrs after 6 pm = 8 pm answer is a" | a ) 8 pm , b ) 6 pm , c ) 7 pm , d ) 10 pm , e ) 5 pm | a | add(6, divide(18, add(4, 6))) | add(n1,n2)|divide(n3,#0)|add(n0,#1)| | physics |
calculate the l . c . m of 5 / 6 , 4 / 7 , 9 / 10 , 8 / 9 is : | "required l . c . m = l . c . m . of 5 , 4 , 9 , 8 / h . c . f . of 6 , 7 , 10 , 9 = 360 / 1 = 360 answer is e" | a ) 560 , b ) 364 , c ) 320 , d ) 330 , e ) 360 | e | multiply(multiply(5, 4), multiply(9, 8)) | multiply(n0,n2)|multiply(n4,n6)|multiply(#0,#1)| | physics |
find the least positive integer divisible by each of the integers between 5 through 10 ( including 10 ) . | the integer should be divisible by : 5 , 6 , 7 , 89 , and 10 . the least common multiple of these integers is lcm = 420 . answer : 5 * 3 ^ 2 * 7 * 2 ^ 3 the answer is e ) | a ) 560 , b ) 380 , c ) 900 , d ) 2560 , e ) 2520 | e | multiply(multiply(multiply(power(const_2, const_3), power(const_3, const_2)), 5), add(const_3, const_4)) | add(const_3,const_4)|power(const_2,const_3)|power(const_3,const_2)|multiply(#1,#2)|multiply(n0,#3)|multiply(#0,#4) | general |
a number when divided by a divisor leaves a remainder of 25 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 25 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 38 as d is also integer and 38 is a prime number , the d should be 38 to satisfy the above equation . hence answer is ' d '" | a ) 12 , b ) 13 , c ) 35 , d ) 38 , e ) 59 | d | add(25, 11) | add(n0,n1)| | general |
dexter and prexter are competing with each other in a friendly community competition in a pool of 50 m length and the race is for 1000 m . dexter crosses 50 m in 2 min and prexter in 3 min 15 sec . each time they meet / cross each other , they do handshake ' s . how many such handshake ' s will happen if they start fro... | explanation : when dexter completes second round , they do handshake once . now for every round which dexter completes , there will be one hand shake as the ratio of speed is 13 : 8 . d and p will meet at the pool end only after d completes 26 rounds . in the 20 th round , d finish the race and the total handshake ' s ... | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 22 | c | subtract(divide(1000, 50), const_1) | divide(n1,n0)|subtract(#0,const_1) | physics |
find the average of all numbers between 6 and 34 which are divisible by 5 | "explanation : average = ( 10 + 15 + 20 + 25 + 305 ) = 1005 = 20 answer : option b" | a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | b | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
two numbers are less than third number by 35 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 35 ) % of x = 65 % of x = 65 x / 100 second number is ( 63 x / 100 ) difference = 65 x / 100 - 63 x / 100 = x / 50 so required percentage is , difference is what percent of first number ( x / 50 * 100 / 65 x * 100 ) % = 33 % answer : c" | a ) 8 % , b ) 10 % , c ) 33 % , d ) 41 % , e ) 42 % | c | subtract(multiply(divide(subtract(37, 35), subtract(const_100, 35)), const_100), const_10) | subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|subtract(#3,const_10)| | gain |
sam β s car was fined when he gave joe and peter a ride , so they decided to help sam pay the fine . joe paid $ 8 more than 1 / 4 of the fine and peter paid $ 8 less than 1 / 3 of the fine , leaving pay $ 15 less than 1 / 2 the fine to complete the payment . what fraction of the fine did sam pay ? | "call the fine f . joe paid ( 1 / 4 ) f + 4 and peter paid ( 1 / 3 ) f β 4 , leaving ( 1 / 2 ) f β 5 left . if we add those three up , they should add up to f . f = [ ( 1 / 4 ) f + 8 ] + [ ( 1 / 3 ) f β 8 ] + [ ( 1 / 2 ) f β 15 ] f = ( 1 / 4 ) f + ( 1 / 3 ) f + ( 1 / 2 ) f β 15 multiply all terms by 12 to clear the fra... | a ) $ 70 , b ) $ 72 , c ) $ 73 , d ) $ 74 , e ) $ 75 | e | add(multiply(add(multiply(add(divide(1, 2), add(divide(1, 4), divide(1, 3))), 15), 15), 2), const_10) | divide(n1,n2)|divide(n1,n5)|divide(n1,n8)|add(#0,#1)|add(#3,#2)|multiply(n6,#4)|add(n6,#5)|multiply(n8,#6)|add(#7,const_10)| | general |
according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 12 ounces of water to make a smoothie . how many 2 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounce... | a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | c | add(150, multiply(3, const_10)) | multiply(n1,const_10)|add(n4,#0) | general |
a man gains 60 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be . | "explanation : let the c . p . = x , then s . p . = ( 160 / 100 ) x = 8 x / 5 new s . p . = 2 ( 8 x / 5 ) = 16 x / 5 profit = 16 x / 5 - x = 16 x / 5 profit % = ( profit / c . p . ) * 100 = > ( 16 x / 5 ) * ( 1 / x ) * 100 = 320 % option b" | a ) 130 % , b ) 320 % , c ) 150 % , d ) 160 % , e ) 170 % | b | add(multiply(subtract(multiply(add(const_1, divide(60, const_100)), const_2), const_1), const_100), const_100) | divide(n0,const_100)|add(#0,const_1)|multiply(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|add(#4,const_100)| | gain |
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 54 . what is the difference between the two digits of that number ? | sol . let the ten β s digit be x and unit β s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 54 β 9 ( x - y ) = 54 β x - y = 6 answer c | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(54, subtract(const_10, const_1)) | subtract(const_10,const_1)|divide(n0,#0) | general |
the ratio , by volume , of soap to alcohol to water in a certain solution is 2 : 40 : 70 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water... | "soap : alcohol initial ratio soap : alcohol : water - - > 2 : 40 : 70 initial soap : alcohol = 2 / 40 = 2 : 40 after doubled soap : alcohol = 2 * 2 / 40 = 4 : 40 initial soap : water = 2 / 70 = 2 : 70 after halved soap : water : 1 / 2 * 2 / 70 = 1 / 70 = 1 : 70 after soap : alcohol : water - - > 4 : 40 : 280 - - > 1 :... | a ) 500 , b ) 600 , c ) 650 , d ) 700 , e ) 720 | d | divide(divide(divide(divide(divide(volume_rectangular_prism(100, 70, 40), const_3), const_2), const_4), const_4.0), const_4) | volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,const_4)|divide(#3,const_4.0)|divide(#4,const_4)| | geometry |
a grocery shop has a sale of rs . 6435 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6435 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6435 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ 6500 = > 34009 + x = 39000 = > x = 39000 β 34009 = 4991 answer : d" | a ) 4857 , b ) 4184 , c ) 4012 , d ) 4991 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6435, 6927)), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
if rs . 420 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | 80 = ( 420 * 4 * r ) / 100 r = 7.14 % i = ( 420 * 6 * 7.14 ) / 100 = 180 420 + 180 = 600 answer : d | a ) s . 575 , b ) s . 595 , c ) s . 590 , d ) s . 600 , e ) s . 585 | d | subtract(multiply(subtract(540, 420), 6), subtract(540, 420)) | subtract(n1,n0)|multiply(n3,#0)|subtract(#1,#0) | gain |
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 |
if 40 % of a number exceeds 20 % of it by 12 , then find the number ? | "use the elimination method to find the correct option . of all the options only 60 fits 40 % of 60 = 24 20 % of 60 = 12 24 - 12 = 12 required number is 60 . answer : a" | a ) 60 , b ) 50 , c ) 47 , d ) 55 , e ) 80 | a | add(divide(12, divide(subtract(40, 20), const_100)), 20) | subtract(n0,n1)|divide(#0,const_100)|divide(n2,#1)|add(n1,#2)| | gain |
a discount electronics store normally sells all merchandise at a discount of 5 percent to 25 percent off the suggested retail price . if , during a special sale , an additional 25 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 300 before any discount ? | "original price : 300 $ max first discount = - 25 % thus : 300 Γ’ Λ β ( 1 Γ’ Λ β 25 / 100 ) = 225 second discount on the discounted price = - 25 % thus : 225 Γ’ Λ β ( 1 Γ’ Λ β 25 / 100 ) = 168.75 answer c" | a ) $ 130.00 , b ) $ 160.75 , c ) $ 168.75 , d ) $ 182.00 , e ) $ 168.00 | c | multiply(300, divide(add(25, 25), const_100)) | add(n1,n2)|divide(#0,const_100)|multiply(n3,#1)| | gain |
a train covers a distance of 12 km in 10 minutes . if it takes 6 seconds to pass a telegraph post , then the length of the train is : | "sol . speed = [ 12 / 10 * 60 ] km / hr = [ 72 * 5 / 18 ] m / sec = 20 m / sec . length of the train = ( speed * time ) = ( 20 * 6 ) m = 120 m . answer a" | a ) 120 m , b ) 140 m , c ) 240 m , d ) 300 m , e ) none | a | divide(const_100.0, subtract(divide(const_100.0, 10), 6)) | divide(const_100.0,n1)|subtract(#0,n2)|divide(const_100.0,#1)| | physics |
workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.20 lacs . how many workers were they ? | "n * 50 = ( 320000 - 300000 ) = 20000 n = 20000 / 50 = 400 b" | a ) 220 , b ) 400 , c ) 500 , d ) 560 , e ) 590 | b | divide(multiply(multiply(subtract(3.20, 3), const_1000), const_100), 50) | subtract(n2,n0)|multiply(#0,const_1000)|multiply(#1,const_100)|divide(#2,n1)| | general |
each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 1 and the 2 can not be adjacent digits in the integer , how many 5 - digit integers are possible ? | "number of arrangements using 5 distinct digits = 5 ! number of arrangements in which 1 and 2 are adjacent - consider 1 and 2 together as one group . now you have 4 numbers / groups to arrange which can be done in 4 ! ways . in each of these arrangements , 1 and 2 can be arranged as 12 or 21 . number of arrangements in... | a ) 48 , b ) 66 , c ) none of these , d ) 78 , e ) 90 | c | subtract(multiply(multiply(multiply(5, 2), 1), const_2), multiply(multiply(multiply(2, 1), const_2), const_2)) | multiply(n1,n4)|multiply(n3,n4)|multiply(n3,#0)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|subtract(#4,#5)| | general |
if a rectangular billboard has an area of 130 square feet and a perimeter of 46 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 130 and a perimeter of 46 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 130 , the shorter side will almost certainly be ... | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 130)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
a train running at the speed of 60 km / hr crosses a pole in 6 seconds . what is the length of the train ? | "speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 6 m = 100 m . option c" | a ) 120 metres , b ) 180 metres , c ) 100 metres , d ) 228 metres , e ) 150 metres | c | multiply(divide(multiply(60, const_1000), const_3600), 6) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
find the greatest number which on dividing 3815 and 4521 , leaves a reminder of 31 and 33 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 3815 - 31 ) and ( 4521 - 33 ) = hcf ( 3784 , 4488 ) = 88 option a" | a ) 88 , b ) 56 , c ) 89 , d ) 86 , e ) 96 | a | gcd(subtract(3815, 31), subtract(4521, 33)) | subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)| | general |
16 men can complete a piece of work in 28 days . in how many days can 20 men complete that piece of work ? | "16 * 28 = 20 * x = > x = 22.4 days answer : c" | a ) 55 days , b ) 77 days , c ) 22.4 days , d ) 88 days , e ) 44 days | c | divide(multiply(28, 16), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 25 : β 26 = 5 : 4 answer : c" | a ) 4 : 9 , b ) 4 : 3 , c ) 5 : 4 , d ) 4 : 1 , e ) 4 : 2 | c | divide(sqrt(25), sqrt(16)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
two sides of a triangle have lengths x and w and meet at a right angle . if the perimeter of the triangle is 4 x , what is the ratio of x to w ? | ahhhh . . . . my bad . thank you , you ' re correct . 8 x = 6 w - - > x / w = 3 / 4 . actually b is the right answer . | ['a ) a ) 2 : 3', 'b ) b ) 3 : 4', 'c ) c ) 4 : 3', 'd ) d ) 3 : 2', 'e ) e ) 2 : 1'] | b | divide(multiply(subtract(4, const_1), const_2), subtract(power(subtract(4, const_1), const_2), const_1)) | subtract(n0,const_1)|multiply(#0,const_2)|power(#0,const_2)|subtract(#2,const_1)|divide(#1,#3) | geometry |
a man deposits $ 2000 in a bank at 4 % per annum and $ 3000 in uti at 14 % per annum . find the rate of interest for the whole sum . | following the above formula , we have the rate of interest for the whole sum = ( 2000 * 4 ) + ( 3000 * 14 ) / 2000 + 3000 = 50000 / 5000 = 10 % . answer is d | a ) 15 % , b ) 18 % , c ) 11 % , d ) 17 % , e ) 10 % | d | multiply(const_100, divide(subtract(add(p_after_gain(4, 2000), p_after_gain(14, 3000)), add(2000, 3000)), 3000)) | add(n0,n2)|p_after_gain(n1,n0)|p_after_gain(n3,n2)|add(#1,#2)|subtract(#3,#0)|divide(#4,n2)|multiply(#5,const_100) | gain |
the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 55.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 55 = 14 $ ( 14 / 55 ) * 100 = 25.5 % answer : a" | a ) 25.5 , b ) 30.4 , c ) 31.8 , d ) 32.5 , e ) 33.8 | a | multiply(divide(subtract(69.0, 55.0), 69.0), const_100) | subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
convert 30 miles into yards ? | "1 mile = 1760 yards 30 miles = 30 * 1760 = 52800 yards answer is c" | a ) 25630 yards , b ) 35200 yards , c ) 52800 yards , d ) 42560 yards , e ) 41520 yards | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 30), 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 |
a garrison of 2000 men has provisions for 65 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ? | "2000 - - - - 65 2000 - - - - 50 x - - - - - 20 x * 20 = 2000 * 50 x = 5000 2000 - - - - - - - 3000 answer : d" | a ) 1977 , b ) 1893 , c ) 1979 , d ) 3000 , e ) 1278 | d | subtract(divide(subtract(multiply(2000, 65), multiply(2000, 15)), 20), 2000) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
in what time will a railway train 70 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 70 / 36 * 18 / 5 = 7 sec answer : b" | a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 2 sec | b | divide(70, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 20 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ? | "wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 20 emlpoyee paid = . 8 * 240 = 192 $ answer c" | a ) $ 198 , b ) $ 216 , c ) $ 192 , d ) $ 230 , e ) $ 240 | c | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 20)) | divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4)| | gain |
what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 5 ? | find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 5 seeing the options after dividing and finding the reminder of 2 my answer was b | a ) 897 , b ) 98,910 , c ) 64,749 , d ) 49,700 , e ) 56,720 | b | multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 5), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3))) | multiply(n0,n0)|subtract(const_1000,n0)|multiply(#0,const_10)|add(#2,#0)|add(n0,#3)|add(#4,#1)|subtract(#1,#4)|divide(#6,n2)|add(#7,const_1)|divide(#8,n1)|multiply(#5,#9) | general |
the speed of a train is 72 kmph . what is the distance covered by it in 20 minutes ? | "72 * 20 / 60 = 24 kmph answer : e" | a ) 20 kmph , b ) 28 kmph , c ) 25 kmph , d ) 22 kmph , e ) 24 kmph | e | multiply(divide(20, const_60), 72) | divide(n1,const_60)|multiply(n0,#0)| | physics |
a math teacher has 27 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 27 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at le... | ['a ) 9', 'b ) 10', 'c ) 11', 'd ) 12', 'e ) 13'] | b | subtract(subtract(divide(divide(multiply(27, 8), const_10), const_2), const_0_25), const_0_25) | multiply(n0,n1)|divide(#0,const_10)|divide(#1,const_2)|subtract(#2,const_0_25)|subtract(#3,const_0_25) | geometry |
at 6 β² o a clock ticks 6 times . the time between first and last ticks is 45 seconds . how long does it tick at 12 β² o clock | "explanation : for ticking 6 times , there are 5 intervals . each interval has time duration of 45 / 5 = 9 secs at 12 o ' clock , there are 11 intervals , so total time for 11 intervals = 11 Γ 9 = 99 secs . answer : e" | a ) 47 , b ) 76 , c ) 28 , d ) 66 , e ) 99 | e | multiply(divide(45, subtract(6, const_1)), subtract(12, const_1)) | subtract(n0,const_1)|subtract(n3,const_1)|divide(n2,#0)|multiply(#2,#1)| | physics |
the total marks obtained by a student in mathematics and physics is 70 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 70 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 70 + 20 ) / 2 = 45 . answer : b" | a ) 40 , b ) 45 , c ) 27 , d ) 28 , e ) 22 | b | divide(add(70, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
the ratio of the money with ravi and giri is 6 : 7 and that with giri and kiran is 6 : 15 . if ravi has $ 36 , how much money does kiran have ? | rita : sita : kavita 6 : 7 6 : 15 36 : 42 : 105 the ratio of money with ravi , giri and kiran is 36 : 42 : 105 we see that 36 = $ 36 then 105 = $ 105 answer is e | a ) $ 36 , b ) $ 42 , c ) $ 60 , d ) $ 75 , e ) $ 105 | e | multiply(divide(multiply(divide(36, 6), 7), divide(36, 6)), 15) | divide(n4,n0)|multiply(n1,#0)|divide(#1,#0)|multiply(n3,#2) | other |
( 3 x + 2 ) ( 2 x - 7 ) = ax ^ 2 + kx + n . what is the value of a - n + k ? | expanding we have 6 x ^ 2 - 21 x + 4 x - 14 6 x ^ 2 - 17 x - 14 taking coefficients , a = 6 , k = - 17 , n = - 14 therefore a - n + k = 6 - ( - 14 ) - 17 = 20 - 17 = 3 the answer is a . | a ) 3 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | add(add(multiply(3, 2), multiply(7, 2)), subtract(multiply(2, 2), multiply(7, 3))) | multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5) | general |
how many positive integers h between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the inte... | a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(add(subtract(300, 200), const_1), add(subtract(add(add(add(floor(divide(add(subtract(300, 200), const_1), 2)), const_1), add(floor(divide(subtract(300, 200), 3)), const_1)), add(divide(subtract(300, 200), 5), const_1)), add(add(floor(divide(divide(subtract(300, 200), const_2), const_3)), add(floor(divide(divid... | divide(const_100,const_3)|subtract(n1,n0)|add(#1,const_1)|divide(#0,const_10)|divide(#1,n3)|divide(#1,n4)|divide(#1,const_2)|divide(#1,const_10)|add(#5,const_1)|add(#7,const_1)|divide(#2,n2)|divide(#5,n3)|divide(#6,const_3)|floor(#3)|floor(#4)|add(#13,const_1)|add(#14,const_1)|floor(#10)|floor(#11)|floor(#12)|add(#17,c... | other |
if the sides of a triangle are 91 cm , 84 cm and 35 cm , what is its area ? | "the triangle with sides 91 cm , 84 cm and 35 cm is right angled , where the hypotenuse is 91 cm . area of the triangle = 1 / 2 * 84 * 35 = 1470 cm 2 answer : b" | a ) 120 cm 2 , b ) 1470 cm 2 , c ) 216 cm 2 , d ) 197 cm 2 , e ) 275 cm 2 | b | divide(multiply(84, 35), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 28 ) ? | "f ( 28 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 * 28 the greatest prime factor in this list is 13 . the answer is d ." | a ) 2 , b ) 7 , c ) 11 , d ) 13 , e ) 17 | d | subtract(divide(28, 2), const_1) | divide(n8,n0)|subtract(#0,const_1)| | general |
mike needs 30 % to pass . if he scored 212 marks and falls short by 16 marks , what was the maximum marks he could have got ? | if mike had scored 16 marks more , he could have scored 30 % therefore , mike required 212 + 16 = 228 marks let the maximum marks be m . then 30 % of m = 228 ( 30 / 100 ) Γ m = 228 m = ( 228 Γ 100 ) / 30 m = 22800 / 30 m = 760 answer : c | a ) 643 , b ) 677 , c ) 760 , d ) 767 , e ) 832 | c | divide(add(212, 16), divide(30, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1) | gain |
a student completes a 3 part exam as follows . the student completes sections a in 18 minutes and takes an equal time to complete the two remaining sections . what is the ratio of time spent on section c to section a if the total duration is 1 hour ? | "the time spent on section b and section c is 21 minutes each . the ratio of c to a is 21 : 18 = 7 : 6 the answer is d ." | a ) 4 : 3 , b ) 5 : 4 , c ) 6 : 5 , d ) 7 : 6 , e ) 8 : 7 | d | divide(3, const_4) | divide(n0,const_4)| | physics |
. a car covers a distance of 742 km in 7 hours . find its speed ? | "742 / 5 = 148 kmph answer : c" | a ) 104 , b ) 55 , c ) 148 , d ) 150 , e ) 159 | c | divide(742, 7) | divide(n0,n1)| | physics |
how many bricks each measuring 20 cm x 10 cm x 8 cm , will be needed to build a wall 10 m x 8 m x 24.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 1000 x 800 x 24.5 ) / ( 20 x 10 x 8 ) = 12250 answer : a" | a ) 12250 , b ) 13400 , c ) 12500 , d ) 14340 , e ) none of these | a | divide(multiply(multiply(10, 8), 24.5), divide(divide(multiply(multiply(20, 10), 8), const_100), const_100)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#3,const_100)|divide(#4,const_100)|divide(#2,#5)| | physics |
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 $ 2,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo β s and isabella β s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal bas... | a ) 19,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(2,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 2,000)) | 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 |
if 45 % of z is 120 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 120 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 120 / 100 ) * ( 3 / 4 ) x i . e . z = ( 120 * 3 ) x / ( 45 * 4 ) i . e . z = ( 2 ) x = ( 200 / 100 ) x i . e . z is 200 % of x answer : option a" | a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50 | a | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(120, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)| | gain |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 13,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years l... | answer equals d in 48 years . i thought by 50 th year it would reach 100,500 . options should have been separated more widely for clarity . | a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000 | d | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
what is the square root of 81 ? | "9 x 9 = 81 answer d" | a ) 2 , b ) 8 , c ) 12 , d ) 9 , e ) 16 | d | circle_area(divide(81, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
how is 4 % expressed as a decimal fraction ? | "4 / 100 = 0.04 answer : b" | a ) 0.4 , b ) 0.04 , c ) 0.004 , d ) 0.0004 , e ) 4 | b | divide(4, const_100) | divide(n0,const_100)| | gain |
in a certain school , the ratio of boys to girls is 5 to 13 . if there are 128 more girls than boys , how many boys are there ? | "the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 128 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 128 . so 1 part corresponds to 128 / 8 = 16 and so 5 parts correspond to 5 * 16 = 80 . e" | a ) 27 , b ) 36 , c ) 45 , d ) 72 , e ) 80 | e | subtract(divide(128, subtract(const_1, divide(5, 13))), 128) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|subtract(#2,n2)| | other |
i remember during the school days , the teacher asked the class ` ` can you tell me the sum of the first 50 odd numbers ? ` ` . i ran quickly to the teacher and told her ` ` the answer is 2500 ' ' . the teacher replied ` ` lucky guess ' ' . she then asked me ` ` can you tell me the sum of first 75 odd numbers ? ` ` . i... | solution : 5625 n ^ 1 75 * 75 = 5625 ( sum of first 75 odd numbers ) . 50 * 50 = 2500 ( sum of first 50 odd numbers ) . answer b | a ) 5624 , b ) 5625 , c ) 5626 , d ) 5627 , e ) none | b | power(75, const_2) | power(n2,const_2) | physics |
in one hour , a boat goes 11 km along the stream and 7 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 7 ) km / hr = 9 kmph . answer d" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | d | divide(add(11, 7), const_2) | add(n0,n1)|divide(#0,const_2)| | gain |
the average of first five prime numbers greater than 10 is ? | "11 + 13 + 17 + 19 + 23 = 83 / 5 = 16.6 answer : a" | a ) 16.6 , b ) 12.9 , c ) 31.3 , d ) 19.2 , e ) 15.7 | a | add(10, const_1) | add(n0,const_1)| | general |
following an increase in prices , the price of a candy box was 15 pounds and the price of a can of soda was 6 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "price of candy before price increase = 15 / 1.25 = 12 price of soda before price increase = 6 / 1.5 = 4 total price = 12 + 4 = 16 e is the answer" | a ) 11 . , b ) 12 . , c ) 13 . , d ) 14 . , e ) 16 | e | add(divide(multiply(15, const_100), add(const_100, 25)), divide(multiply(6, const_100), add(const_100, 50))) | add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|multiply(n1,const_100)|divide(#2,#0)|divide(#3,#1)|add(#4,#5)| | general |
city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 4 : 20 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? | "train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 4 : 20 pm + 2 hrs = 6 : 20 pm answer : d . 6 : 20" | a ) 5 : 00 , b ) 5 : 30 , c ) 6 : 00 , d ) 6 : 20 , e ) 7 : 00 | d | divide(add(4, const_2), 20) | add(n4,const_2)|divide(#0,n5)| | physics |
a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 46 instead of 56 then find the correct average marks up to two decimal places . | "total marks = 35 x 72 = 2520 corrected total marks = 2520 - 46 + 56 = 2530 correct average = 2530 / 35 = 72.28 answer : b" | a ) 73.41 , b ) 72.28 , c ) 72.43 , d ) 73.43 , e ) can not be determined | b | divide(subtract(multiply(35, 72), subtract(56, 46)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
a number when divided by a divisor leaves a remainder of 21 . when twice the original number is divided by the same divisor , the remainder is 11 . what is the value of the divisor ? | "let the number is n , the divisor = d , i will make the two equations - n = xd + 21 2 n = yd + 11 where x and y are integers solving them : d ( y - 2 x ) = 34 as d is also integer and 34 is a prime number , the d should be 34 to satisfy the above equation . hence answer is ' d '" | a ) 12 , b ) 13 , c ) 34 , d ) 37 , e ) 59 | d | add(21, 11) | add(n0,n1)| | general |
last year a certain bond price with a face value of 5000 yielded 7 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ? | "interest = 0.07 * 5000 = 0.065 * selling price - - > selling price = 0.07 * 5000 / 0.065 - - > selling price = ~ 5,385 answer : c ." | a ) 4063 , b ) 5325 , c ) 5385 , d ) 6000 , e ) 6154 | c | divide(multiply(5000, divide(7, const_100)), divide(6.5, const_100)) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)| | gain |
the length and the breadth of a rectangle are in the ratio of 3 : 2 respectively . if the sides of the rectangle are extended on each side by 1 metre , the ratio of length to breadth becomes 10 : 7 . find the area of the original rectangle in square metres . | let the length and breadth be l and b respectively . l β b = 3 β 2 or l = 3 β 2 b . . . . . ( i ) l + 1 / b + 1 = 10 / 7 or 7 l - 10 b = 3 . . . . . ( ii ) from eq . ( i ) 10.5 b β l 0 b = 6 or , 0.5 b = 3 or , b = 6 and l = 9 area = l Γ b = 6 Γ 9 = 54 m 2 answer e | ['a ) 256', 'b ) 150', 'c ) 280', 'd ) data inadequate', 'e ) none of these'] | e | rectangle_area(subtract(7, 1), subtract(10, 1)) | subtract(n4,n2)|subtract(n3,n2)|rectangle_area(#0,#1) | geometry |
32 meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ? | area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 32 = > x / 360 * 2 * pi * r + 2 r = 32 = > 2 a / r + 2 r = 32 = > a = r 16 - r ^ 2 = r 16 - r ^ 2 we will now max using derivations max value of a will found at a = 0 i . e 16 - 2 r = 0 r = 8 e | ['a ) 2 β 2', 'b ) 2 β 5', 'c ) 5', 'd ) 4 β 2', 'e ) 8'] | e | divide(32, const_4) | divide(n0,const_4) | geometry |
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 |
compute the perimeter of a square with area 36 . | if the length of the square is s , then s 2 = 36 so s = 6 . the perimeter is therefore 4 s = 24 . correct answer a | ['a ) 24', 'b ) 27', 'c ) 25', 'd ) 26', 'e ) 29'] | a | multiply(sqrt(36), const_4) | sqrt(n0)|multiply(#0,const_4) | geometry |
a can run 4 times as fast as b and gives b a start of 72 m . how long should the race course be so that a and b might reach in the same time ? | "speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 72 m he gains 72 * ( 4 / 3 ) i . e 96 m answer : e" | a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 96 m | e | add(multiply(4, divide(divide(72, 4), subtract(4, const_1))), 72) | divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)| | physics |
evaluate : 30 - 12 Γ· 3 Γ 2 = ? | "according to order of operations , 12 Γ· 3 Γ 2 ( division and multiplication ) is done first from left to right 12 Γ· 3 Γ 2 = 4 Γ 2 = 8 hence 30 - 12 Γ· 3 Γ 2 = 30 - 8 = 22 correct answer b ) 22" | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | b | subtract(30, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a certain telephone company offers two plans , a and b . under plan a , the company charges a total of $ 0.60 for the first 5 minutes of each call and $ 0.06 per minute thereafter . under plan b , the company charges $ 0.08 per minute of each call . what is the duration of a call , in minutes , for which the company ch... | "let the duration , in minutes , for which the company charges the same under plan a and plan b be t minutes . then under plan a the cost would be $ 0.6 + 0.06 ( t - 5 ) and under plan b the cost would be $ 0.08 t . we want these amount to be equal : 0.6 + 0.06 ( t - 5 ) = 0.08 t - - > 60 + 6 ( t - 5 ) = 8 t - - > t = ... | a ) 2 , b ) 9 , c ) 15 , d ) 21 , e ) 30 | c | divide(subtract(0.60, multiply(0.06, 5)), subtract(0.08, 0.06)) | multiply(n1,n2)|subtract(n3,n2)|subtract(n0,#0)|divide(#2,#1)| | other |
an employer pays rs . 30 for each day a worker works , and forfeits rs . 5 for each day he is idle . at the end of 60 days , a worker gets rs . 500 . for how many days did the worker remain idle ? | explanation : suppose the worker remained idle for m days . then , he worked for ( 60 - m ) days . 30 ( 60 - m ) β 5 m = 500 1800 β 25 m = 500 25 m = 1300 m = 52 so , the worker remained idle for 52 days . answer : a | a ) 52 , b ) 27 , c ) 99 , d ) 61 , e ) 11 | a | divide(subtract(500, multiply(60, 30)), negate(subtract(30, 5))) | multiply(n0,n2)|subtract(n0,n1)|negate(#1)|subtract(n3,#0)|divide(#3,#2) | physics |
a can finish a piece of work in 5 days . b can do it in 15 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 5 + ( 2 + x ) / 15 = 1 = > x = 7 days answer : e" | a ) 3 days , b ) 5 days , c ) 4 days , d ) 6 days , e ) 7 days | e | divide(subtract(const_1, add(multiply(divide(const_1, 5), const_2), multiply(divide(const_1, 15), const_2))), divide(const_1, 15)) | divide(const_1,n0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 16 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 16 ) = 304 cm 2 answer : a" | a ) 304 cm 2 , b ) 179 cm 2 , c ) 285 cm 2 , d ) 167 cm 2 , e ) 197 cm 2 | a | quadrilateral_area(16, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
a policeman noticed a criminal from a distance of 180 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 5 minutes ? | "explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 5 / 60 ) km = 1 / 12 km = 83.3 m . . ' . distance between the criminal and policeman = ( 180 - 83.3 ) m = 96.7 m . answer : a" | a ) 96.7 m , b ) 120 m , c ) 130 m , d ) 150 m , e ) none of these | a | subtract(180, multiply(divide(5, const_60), const_1000)) | divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)| | physics |
8269 Γ£ β 9998 = ? | "8269 Γ£ β ( 10000 - 2 ) 82690000 - 16538 = 82673462 answer b" | a ) 85432864 , b ) 82673462 , c ) 86985434 , d ) 68954365 , e ) 65486954 | b | multiply(divide(8269, 9998), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
let f ( x ) = x ^ 2 + bx + c . if f ( 1 ) = 0 and f ( - 8 ) = 0 , then f ( x ) crosses the y - axis at what y - coordinate ? | "when x = 1 and when x = - 8 , the expression f ( x ) = x Β² + bx + c equals 0 . then f ( x ) = ( x - 1 ) ( x + 8 ) f ( 0 ) = - 8 the answer is a ." | a ) - 8 , b ) - 1 , c ) 0 , d ) 1 , e ) 8 | a | negate(divide(subtract(power(8, 2), 1), add(8, 1))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general |
two numbers are in the ratio of 5 : 7 . if 12 is subtracted from each number , they are in the ratio of 11 : 17 . what are the two numbers ? | "( 5 x - 12 ) : ( 7 x - 12 ) = 11 y : 17 y 5 x - 12 = 11 y 7 x - 12 = 17 y 84 - 60 = 85 y - 77 y 24 = 8 y y = 3 5 x - 12 = 33 x = 9 the two numbers are 5 x and 7 x which are 45 and 63 . the answer is a ." | a ) 45 , 63 , b ) 50 , 70 , c ) 55 , 77 , d ) 60 , 84 , e ) 65 , 91 | a | subtract(multiply(multiply(subtract(multiply(17, 12), multiply(11, 12)), 5), divide(7, 5)), multiply(subtract(multiply(17, 12), multiply(11, 12)), 5)) | divide(n1,n0)|multiply(n2,n4)|multiply(n2,n3)|subtract(#1,#2)|multiply(n0,#3)|multiply(#0,#4)|subtract(#5,#4)| | other |
the sale price sarees listed for rs . 600 after successive discount is 10 % and 5 % is ? | "600 * ( 90 / 100 ) * ( 95 / 100 ) = 513 answer : b" | a ) 227 , b ) 513 , c ) 342 , d ) 787 , e ) 191 | b | subtract(subtract(600, divide(multiply(600, 10), const_100)), divide(multiply(subtract(600, divide(multiply(600, 10), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 6720 votes were polled , then how many valid votes did b get ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given t... | a ) 1888 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2184 | e | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 6720), divide(multiply(15, 6720), const_100)), const_2) | divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(const_1,#0)|multiply(n2,#3)|subtract(#4,#2)|divide(#5,const_2)| | general |
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