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
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a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 970 gm . for a kg . his gain is … % . | "his percentage gain is 100 * 30 / 970 as he is gaining 30 units for his purchase of 970 units . so 3.09 % . answer : a" | a ) 3.09 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(970, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
if 2 ^ ( 2 w ) = 8 ^ ( w − 6 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w − 6 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w − 6 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 18 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 18 w = 18 the answer is c ." | a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | c | divide(multiply(const_12, log(2)), log(2)) | log(n0)|multiply(#0,const_12)|divide(#1,#0)| | general |
tom traveled the entire 20 miles trip . if he did the first 10 miles of at a constant rate 12 miles per hour and the remaining trip of at a constant rate 10 miles per hour , what is the his average speed , in miles per hour ? | avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 10 + 10 ) / ( ( 10 / 12 ) + ( 10 / 10 ) = 120 / 11 = 10.91 mph d | a ) 12.45 mph , b ) 11.25 mph , c ) 10.95 mph , d ) 10.91 mph , e ) 10.56 mph | d | divide(20, add(divide(10, subtract(20, 10)), divide(10, 12))) | divide(n1,n2)|subtract(n0,n1)|divide(n3,#1)|add(#2,#0)|divide(n0,#3)| | physics |
a , b are positive integers . the remainder of a to be divided by 8 is 3 and the remainder of b to be divided by 6 is 5 . which is possible to be the remainder of a * b to be divided by 48 | two ways to do it . . . a = 8 x + 3 . . b = 6 y + 5 . . 1 ) convenient way . . take x and y as 0 , and you will get a * b as 3 * 5 = 15 answer : d | a ) a ) 2 , b ) b ) 6 , c ) c ) 8 , d ) d ) 15 , e ) e ) 20 | d | multiply(3, 5) | multiply(n1,n3) | general |
one pipe can fill a pool 1.25 times faster than a second pipe . when both pipes are opened , they fill the pool in five hours . how long would it take to fill the pool if only the faster pipe is used ? | say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.25 r = 5 r / 4 . since when both pipes are opened , they fill the pool in five hours , then their combined rate is 1 / 5 pool / hour . thus we have that r + 5 r / 4 = 1 / 5 - - > r = 4 / 45 pool / hour , faster pipe fills at ... | a ) 11.25 , b ) 11.52 , c ) 1.25 , d ) 7.2 , e ) 9 | e | divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.25))), 1.25) | add(const_2,const_3)|add(n0,const_1)|inverse(#0)|divide(#2,#1)|inverse(#3)|divide(#4,n0) | physics |
the revenue from sales items in 1996 increases by x percent , compared in 1995 and the revenue in 1997 decreases by x percent , compared in 1996 . if the revenue from sales items in 1997 decreases by 4 percent , compared in 1995 , what is the value of x ? | 1995 - let the value be z 1996 - x % increase - z ( 1 + x / 100 ) 1997 - x % decrease - z ( 1 + x / 100 ) ( 1 - x / 100 ) from 95 to 97 decrease is 4 % hence , [ z - { z ( 1 + x / 100 ) ( 1 - x / 100 ) } ] / z = 4 / 100 solving x = 20 % c is the answer | a ) 10 , b ) 15 , c ) 20 , d ) 30 , e ) 35 | c | sqrt(multiply(4, const_100)) | multiply(n5,const_100)|sqrt(#0) | general |
peter has $ 642986 in his savings account . what is the least amount of money ( in whole number of dollars ) that he must add to his account if he wants to split this money evenly among his 9 children ? | to find the least amount the man should add to his saving account to split the money evenly among his 9 children , he needs to make the total divisible by 9 simply add the individual digits of the total = 6 + 4 + 2 + 9 + 8 + 6 = 35 if you add 1 , the number is divisible by 9 ( 35 + 1 ) correct option : b | a ) $ 642986 , b ) $ 642987 , c ) $ 642988 , d ) $ 642989 , e ) $ 642990 | b | add(642986, divide(9, 9)) | divide(n1,n1)|add(n0,#0) | general |
the measurement of a rectangular box with lid is 25 cmx 4 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 4 , r = 2 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 2 * 2 * 2 = 10.66 π cm 3 answer : d" | a ) 11.12 , b ) 10.11 , c ) 72 , d ) 10.66 , e ) 9.2 | d | multiply(divide(const_4, 3), power(3, 3)) | divide(const_4,n3)|power(n3,n3)|multiply(#0,#1)| | geometry |
a car travels at a speed of 60 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 60 + 60 + 60 + 60 + 60 = 5 * 65 = 300 miles correct answer c" | a ) 425 miles , b ) 625 miles , c ) 300 miles , d ) 225 miles , e ) 625 miles | c | multiply(60, 5) | multiply(n0,n1)| | physics |
find the distance covered by a man walking for 72 min at a speed of 10 km / hr ? | "distance = 10 * 72 / 60 = 12 km answer is a" | a ) 12 km , b ) 3 km , c ) 4 km , d ) 5 km , e ) 6 km | a | divide(multiply(72, divide(multiply(10, const_1000), const_60)), const_1000) | multiply(n1,const_1000)|divide(#0,const_60)|multiply(n0,#1)|divide(#2,const_1000)| | physics |
the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is | "solution clearly , the numbers are ( 23 x 13 ) and ( 23 x 14 ) . larger number = ( 23 x 14 ) = 322 . answer d" | a ) 276 , b ) 299 , c ) 312 , d ) 322 , e ) none | d | multiply(23, 14) | multiply(n0,n2)| | other |
what percent is 2 gm of 1 kg ? | "1 kg = 1000 gm 2 / 1000 × 100 = 200 / 1000 = 1 / 5 = 0.2 % b )" | a ) 0.5 % , b ) 0.2 % , c ) 1.5 % , d ) 2 % , e ) 3 % | b | multiply(divide(2, 1), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a car travels first 160 km at 75 km / hr and the next 160 km at 70 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 70 km / hr time taken to travel next 160 km = distancespeed = 160 / 70 total distance traveled = 160 + 160 = 2 × 160 total time taken = 160 / 75 + 160 / 70 average speed = total distance trave... | a ) 71.11 , b ) 71.12 , c ) 72.4 , d ) 71.17 , e ) 71.13 | c | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 70)))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(const_1,#2)|multiply(n4,#3)| | general |
a train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . what is the length of the train ? | explanation : let x is the length of the train in meter and v is its speed in kmph x / 9 = ( v - 2 ) ( 10 / 36 ) - - - ( 1 ) x / 10 = ( v - 4 ) ( 10 / 36 ) - - - ( 2 ) dividing equation 1 with equation 2 10 / 9 = ( v - 2 ) / ( v - 4 ) = > 10 v - 40 = 9 v - 18 = > v = 22 substituting in equation 1 , x / 9 = 200 / 36 = >... | a ) 62 m , b ) 54 m , c ) 50 m , d ) 55 m , e ) 56 m | c | multiply(9, subtract(subtract(multiply(divide(multiply(4, const_1000), const_3600), 10), multiply(divide(multiply(2, const_1000), const_3600), 9)), divide(multiply(2, const_1000), const_3600))) | multiply(n1,const_1000)|multiply(n0,const_1000)|divide(#0,const_3600)|divide(#1,const_3600)|multiply(n3,#2)|multiply(n2,#3)|subtract(#4,#5)|subtract(#6,#3)|multiply(n2,#7) | physics |
what is 15 percent of 68 ? | "( 15 / 100 ) * 68 = 10.2 the answer is d ." | a ) 7.6 , b ) 8.7 , c ) 9.1 , d ) 10.2 , e ) 11.6 | d | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
if a - b = 2 and a 2 + b 2 = 25 , find the value of ab . | "explanation : 2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 25 - 4 = 21 ab = 10.5 answer : a" | a ) a ) 10.5 , b ) b ) 12 , c ) c ) 15 , d ) d ) 18 , e ) e ) 20 | a | divide(subtract(25, power(2, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
the area of a parallelogram is 98 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 98 = > x = 7 answer : c" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | c | sqrt(divide(98, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry |
tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 2 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | "a is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 2 t ) = 8 t = = > t = 1 / 3 = = > 20 minutes to travel double distance , 2 ( 2 + 2 t ) = 8 t = = > 1 = = > 60 minutes difference , 40 minutes a" | a ) 40 , b ) 72 , c ) 84 , d ) 90 , e ) 108 | a | multiply(subtract(divide(multiply(const_2, const_2), subtract(8, multiply(const_2, 2))), divide(const_2, subtract(8, 2))), const_60) | multiply(const_2,const_2)|multiply(n0,const_2)|subtract(n1,n0)|divide(const_2,#2)|subtract(n1,#1)|divide(#0,#4)|subtract(#5,#3)|multiply(#6,const_60)| | physics |
a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 840 find the share of a . | "explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 / 21 * 840 = 320 answer : e" | a ) 240 , b ) 388 , c ) 379 , d ) 277 , e ) 320 | e | multiply(divide(840, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain |
set # 1 = { a , b , o , d , e } set # 2 = { k , l , m , n , u , p } there are these two sets of letters , and you are going to pick exactly one letter from each set . what is the probability of picking at least one vowel ? | at least questions are best solved by taking the opposite scenario and subtracting it from 1 . probability of choosing no vowel from set 1 is 2 / 5 and set 2 is 5 / 6 . multiply these to get 1 / 3 . therefore , probability of picking at least one vowel = 1 - 1 / 3 = 2 / 3 . answer = e | a ) 1 / 6 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 / 3 , e ) 2 / 3 | e | subtract(1, multiply(divide(const_2, add(const_2, const_3)), divide(add(const_2, const_3), add(add(const_2, const_3), const_1)))) | add(const_2,const_3)|add(#0,const_1)|divide(const_2,#0)|divide(#0,#1)|multiply(#2,#3)|subtract(n0,#4) | general |
the value of x + ( xx ) when x = 3 is : | x + ( xx ) put the value of x = 2 in the above expression we get , 3 + ( 33 ) = 3 + ( 3 × 3 ) = 3 + ( 9 ) = 3 + 9 = 12 b | a ) a ) 36 , b ) b ) 12 , c ) c ) 18 , d ) d ) 64 , e ) e ) 10 | b | add(multiply(3, 3), 3) | multiply(n0,n0)|add(n0,#0) | general |
roy is now 6 years older than julia and half of that amount older than kelly . if in 2 years , roy will be twice as old as julia , then in 2 years what would be roy ’ s age multiplied by kelly ’ s age ? | "r = j + 6 = k + 3 r + 2 = 2 ( j + 2 ) ( j + 6 ) + 2 = 2 j + 4 j = 4 r = 10 k = 7 in 2 years ( r + 2 ) ( k + 2 ) = 12 * 9 = 108 the answer is d ." | a ) 84 , b ) 96 , c ) 100 , d ) 108 , e ) 120 | d | multiply(add(add(6, subtract(6, 2)), 2), add(subtract(add(6, subtract(6, 2)), divide(6, const_2)), 2)) | divide(n0,const_2)|subtract(n0,n1)|add(n0,#1)|add(n1,#2)|subtract(#2,#0)|add(n1,#4)|multiply(#3,#5)| | general |
if x is equal to the sum of the integers from 30 to 40 , inclusive , and y is the number of even integers from 30 to 40 , inclusive , what is the value of x + y ? | "sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 30 + ( 11 - 1 ) * 1 } = 11 * 35 = 385 = x number of even number = ( 40 - 30 ) / 2 + 1 = 6 = y x + y = 385 + 6 = 391 c" | a ) 171 , b ) 281 , c ) 391 , d ) 591 , e ) 601 | c | add(multiply(divide(add(30, 40), const_2), add(subtract(40, 30), const_1)), add(divide(subtract(40, 30), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)| | general |
a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent co... | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 850 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 850 / 70 ~ 12.14 c" | a ) $ 9.40 , b ) $ 11.30 , c ) $ 12.14 , d ) $ 17.80 , e ) $ 22.10 | c | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | add(const_3,const_4)|divide(n0,const_100)|divide(n1,const_4)|divide(#2,const_100)|add(#1,#3)|add(#0,#4)|add(#5,#1)| | general |
a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 100 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for s... | a ) 4,800 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(100, add(const_2, const_3)), const_3), multiply(divide(100, add(const_2, const_3)), const_4)) | add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)| | geometry |
19 balls are numbered 1 to 19 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 9 / 19 p ( 2 nd ball is also even ) = 8 / 18 p ( both balls are even ) = 9 / 19 * 8 / 18 = 4 / 19 the answer is b ." | a ) 2 / 9 , b ) 4 / 19 , c ) 6 / 29 , d ) 8 / 39 , e ) 10 / 49 | b | multiply(divide(add(const_4, const_4), 19), divide(subtract(add(const_4, const_4), 1), subtract(19, 1))) | add(const_4,const_4)|subtract(n0,n1)|divide(#0,n0)|subtract(#0,n1)|divide(#3,#1)|multiply(#2,#4)| | other |
in the first 12 overs of a cricket game , the run rate was only 4.5 . what should be the run rate in the remaining 38 overs to reach the target of 360 runs ? | required run rate = 360 - ( 4.5 x 12 ) / 38 = 306 / 38 = 8.05 option d | a ) 9.55 , b ) 6.25 , c ) 6.75 , d ) 8.05 , e ) 7.25 | d | divide(subtract(360, multiply(12, 4.5)), 38) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
a car during its journey travels 1 hour at a speed of 40 kmph , another 30 minutes at a speed of 60 kmph , and 2 hours at a speed of 60 kmph . the average speed of the car is | first car travels 1 hrs at speed of 40 kmph distance = 40 x 1 = 40 m then car travels 30 min at a speed of 60 kmph distance = 30 min at speed of 60 kmph distance = 60 x 1 / 2 = 30 m at last it travels 2 hours at speed of 60 kmph distance = 60 x 2 = 120 m total distance = 40 + 30 + 120 = 190 total time = 1 + 1 / 2 + 2 =... | a ) 63.07 kmph , b ) 54.28 kmph , c ) 62.02 kmph , d ) 64.02 kmph , e ) none of these | b | divide(add(add(40, 30), add(60, 60)), add(add(1, divide(30, const_60)), 2)) | add(n1,n2)|add(n3,n3)|divide(n2,const_60)|add(#0,#1)|add(n0,#2)|add(n4,#4)|divide(#3,#5) | physics |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 26 seconds , what is the ratio of their speeds ? | "let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 26 s = ( 27 x + 17 y ) / ( x + y ) = 26 = ( 27 x + 17 y ) / = 26 ( x + y ) = x = 9 y = x / y = 9 / 1 answer : b" | a ) 1 : 3 , b ) 9 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | b | divide(subtract(27, 26), subtract(26, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
5358 x 56 = ? | "5358 x 51 = 5358 x ( 50 + 6 ) = 5358 x 50 + 5358 x 6 = 267900 + 32148 = 300048 . c )" | a ) 272258 , b ) 272358 , c ) 300048 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 56), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
excluding stoppages , the speed of a bus is 80 kmph and including stoppages , it is 60 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 20 km less . time taken to cover 20 km = ( 20 / 80 x 60 ) min = 15 min answer : e" | a ) 12 min , b ) 20 min , c ) 25 min , d ) 30 min , e ) 15 min | e | multiply(const_60, divide(subtract(80, 60), 80)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
the average age of a , b and c is 27 years . if the average age of a and c is 29 years , what is the age of b in years ? | age of b = age of ( a + b + c ) â € “ age of ( a + c ) = 27 ã — 3 â € “ 29 ã — 2 = 81 â € “ 58 = 23 years e | a ) 19 , b ) 35 , c ) 20 , d ) 32 , e ) 23 | e | subtract(multiply(27, const_3), multiply(29, const_2)) | multiply(n0,const_3)|multiply(n1,const_2)|subtract(#0,#1) | general |
a man buys a cycle for rs . 1400 and sells it at loss of 15 % . what is the selling price of the cycle ? | "explanation : s . p = 85 % of rs . 1400 ⇒ rs . ( 8510085100 × 1400 ) = rs . 1190 . answer : d" | a ) 1192 , b ) 1191 , c ) 1118 , d ) 1190 , e ) 1112 | d | divide(multiply(subtract(const_100, 15), 1400), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
a strain of bacteria reproduces @ 25 % every 12 min . in how much time will it triple itself ? ? | the original question is : a strain of bacteria reproduces at the rate of 25 % every 12 min . in how much time will it triple itself ? 1.25 ^ x = 3 - - > x = ~ 5 - - > five 12 minute periods = 60 minutes . answer : c . | a ) 96 min , b ) 72 min , c ) 60 min , d ) 48 min , e ) 40 min | c | multiply(divide(log(const_3), log(add(const_1, divide(25, const_100)))), 12) | divide(n0,const_100)|log(const_3)|add(#0,const_1)|log(#2)|divide(#1,#3)|multiply(n1,#4) | general |
a person want to give his money of $ 1000 to his 3 children a , b , c in the ratio 6 : 4 : 5 what is the a + b share ? | "a ' s share = 1000 * 6 / 15 = $ 400 b ' s share = 1000 * 4 / 15 = $ 266.67 a + b = $ 666.67 answer is c" | a ) $ 660.67 , b ) $ 646.67 , c ) $ 666.67 , d ) $ 500.56 , e ) $ 600.24 | c | multiply(divide(3, add(add(6, 3), 5)), 1000) | add(n1,n2)|add(n4,#0)|divide(n1,#1)|multiply(n0,#2)| | other |
in a group of cows and hens , the number of legs are 10 more than twice the number of heads . the number of cows is : | "let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 10 2 x = 10 x = 5 . answer : a" | a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12 | a | subtract(10, const_4) | subtract(n0,const_4)| | general |
a survey was sent to 60 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 60 ) = x / 100 x = 12 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 12 ) % = 2 % answer is a" | a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 4 % | a | floor(multiply(subtract(divide(9, 63), divide(7, 60)), const_100)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)| | gain |
12 persons can complete the work in 18 days . after working for 6 days , 4 more persons added to complete the work fast . in how many days they will complete the work ? | "total work = 12 * 18 = 216 units . after 6 days , work finished = 6 * 12 = 72 units . remaining units = 216 - 72 = 144 units . remaining days = 144 / ( 2 + 4 ) = 9 days the option is e" | a ) 10 , b ) 15 , c ) 17 , d ) 5 , e ) 9 | e | divide(multiply(subtract(18, 6), 12), add(12, 4)) | add(n0,n3)|subtract(n1,n2)|multiply(n0,#1)|divide(#2,#0)| | physics |
if 2 / 3 of the air in a tank is removed with each stroke of a vacuum pump , how many strokes does it take before less than 1 % of the original amount of air in the tank remains ? | "my approach is to find more or less good number and plug in . lets take 90 as the total volume of an air , and the question asks after how many stokes there will be less than 0,9 air in a tank , if each stroke takes 2 / 3 of an air . so after he 1 strokethe volume will be 30 ( 90 - 60 = 30 ) , and so on 2 stroke - 30 ... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(add(add(add(divide(2, 3), divide(divide(2, 3), 3)), divide(divide(divide(2, 3), 3), 3)), divide(divide(divide(divide(2, 3), 3), 3), 3)), divide(divide(divide(divide(divide(2, 3), 3), 3), 3), 3)) | divide(n0,n1)|divide(#0,n1)|add(#0,#1)|divide(#1,n1)|add(#2,#3)|divide(#3,n1)|add(#4,#5)|divide(#5,n1)|add(#6,#7)| | general |
mahesh can do a piece of work in 35 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 35 days = 20 * 1 / 35 = 4 / 7 remaining work = 1 - 4 / 7 = 3 / 7 3 / 7 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 7 / 3 = 70 days answer is e" | a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 70 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 35))), 30)) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)| | physics |
for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.0 ? | "[ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.0 = 6 * 0 + 2 * 7.2 + 8 - 6.0 = 0 + 14.4 + 2 16.4 answer e" | a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 . | e | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.0) | divide(const_4,const_10)|multiply(n1,n4)|subtract(n5,#0)|add(#1,#2)|subtract(#3,n6)| | general |
a parking garage rents parking spaces for $ 10 per week or $ 42 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 42 = 504 $ 520 - 504 = 16 ans b" | a ) $ 14 , b ) $ 16 , c ) $ 22 , d ) $ 24 , e ) $ 26 | b | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(42, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general |
what is the least number which when divided by 8 , 12 , 15 and 20 leaves in each case a remainder of 5 | "lcm of 8 , 12 , 15 and 20 = 120 required number = 120 + 5 = 125 answer : option b" | a ) 110 , b ) 125 , c ) 75 , d ) 170 , e ) 195 | b | add(lcm(lcm(8, 12), lcm(15, 20)), 5) | lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)| | general |
if xy = 4 , x / y = 81 , for positive numbers x and y , y = ? | "very easy question . 2 variables and 2 easy equations . xy = 4 - - - > x = 4 / y - ( i ) x / y = 81 - - - > replacing ( i ) here - - - > 4 / ( y ^ 2 ) = 81 - - - > y ^ 2 = 4 / 81 - - - > y = 2 / 9 or - 2 / 9 the question states that x and y are positive integers . therefore , y = 2 / 9 is the answer . answer c ." | a ) 1 / 2 , b ) 2 , c ) 2 / 9 , d ) 3 , e ) 1 / 6 | c | sqrt(divide(4, 81)) | divide(n0,n1)|sqrt(#0)| | general |
a fellow borrowed a certain sum of money at 6 % per annum at simple interest and in 6 years the interest amounted to rs . 672 less than the sum lent . what was the sum lent ? | "p - 672 = ( p * 6 * 6 ) / 100 p = 1050 answer : a" | a ) 1050 , b ) 1220 , c ) 1250 , d ) 1060 , e ) 1110 | a | divide(672, subtract(const_1, divide(multiply(6, 6), const_100))) | multiply(n0,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)| | gain |
the 7 th grade french and spanish classes each have 21 students in them , and there are 6 students in the 7 th grade taking both languages . if everyone in the 7 th grade is in at least one of the two language classes , how many students are taking just one of the two classes ? | french class + spanish class = 42 students . six students are in both classes so they were counted twice . the number of students taking just one class is 42 - 2 ( 6 ) = 30 . the answer is c . | a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | c | add(subtract(21, 6), subtract(21, 6)) | subtract(n1,n2)|add(#0,#0) | other |
a company that ships boxes to a total of 14 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors... | "back - solving is the best way to solve this problem . you basically need 14 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - c" | a ) 7 , b ) 8 , c ) 5 , d ) 9 , e ) 6 | c | subtract(divide(factorial(subtract(divide(14, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(14, const_2), const_1)) | divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)| | general |
a , b , c hired a car for rs . 520 and used it for 7,8 and 11 hours respectively . hire charges paid by b were : | sol . a : b : c = 7 : 8 : 11 . hire charges paid by b = rs . [ 520 * 8 / 26 ] = rs . 160 answer a | a ) 160 , b ) 180 , c ) 220 , d ) 225 , e ) none | a | multiply(520, divide(multiply(const_4, const_2), add(add(add(const_4, const_3), multiply(const_4, const_2)), 11))) | add(const_3,const_4)|multiply(const_2,const_4)|add(#0,#1)|add(n2,#2)|divide(#1,#3)|multiply(n0,#4) | physics |
maximun number of identical pieces ( of same size ) of a cake by making only 3 cuts ? | since cake is a 3 d object , it will have length , breadth , height . . . first cut across length now we will get 2 pieces then cut across breadth we will get 4 parts in total now . . . now cut across height , finally we will get 8 parts . . . answer : c | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | c | multiply(add(add(power(3, const_3), multiply(add(const_1, const_4), 3)), multiply(3, const_2)), inverse(multiply(3, const_2))) | add(const_1,const_4)|multiply(n0,const_2)|power(n0,const_3)|inverse(#1)|multiply(n0,#0)|add(#4,#2)|add(#5,#1)|multiply(#6,#3) | general |
how many prime numbers exist in 67 x 353 x 1110 ? | "solution : = ( 2 x 3 ) 7 x ( 5 x 7 ) 3 x 1110 = 27 x 37 x 53 x 73 x 1110 thus , there are ( 7 + 7 + 3 + 3 + 10 ) = 30 prime number answer : a" | a ) 30 prime number , b ) 29 prime number , c ) 27 prime number , d ) 31 prime number , e ) none of these | a | multiply(const_4, 67) | multiply(n0,const_4)| | general |
a car travels at a speed of 12 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 12 + 12 + 12 + 12 + 12 = 5 * 12 = 60 miles correct answer b" | a ) 50 miles , b ) 60 miles , c ) 72 miles , d ) 22 miles , e ) 62 miles | b | multiply(12, 5) | multiply(n0,n1)| | physics |
a hostel had provisions for 250 men for 36 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 36 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 36 250 × 36 = 200 x 5 × 36 = 4... | a ) 30 , b ) 45 , c ) 50 , d ) 60 , e ) 65 | b | divide(multiply(250, 36), subtract(250, 50)) | multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)| | gain |
3 , 4 , 6 , 10 , 18 , ( . . . ) | "explanation : 3 3 × 2 - 2 = 4 4 × 2 - 2 = 6 6 × 2 - 2 = 10 10 × 2 - 2 = 18 18 × 2 - 2 = 34 answer : option b" | a ) 22 , b ) 34 , c ) 27 , d ) 32 , e ) 25 | b | subtract(negate(10), multiply(subtract(4, 6), divide(subtract(4, 6), subtract(3, 4)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
the average of runs of a cricket player of 10 innings was 15 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | average after 11 innings = 19 required number of runs = ( 19 * 11 ) - ( 15 * 10 ) = 209 - 150 = 59 . answer : e | a ) 87 , b ) 16 , c ) 10 , d ) 76 , e ) 59 | e | subtract(multiply(add(10, const_1), add(4, 15)), multiply(10, 15)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2) | general |
16 men can complete a piece of work in 25 days . in how many days can 15 men complete that piece of work ? | "16 * 25 = 15 * x = > x = 26.7 days answer : a" | a ) 26.7 days , b ) 77 days , c ) 20 days , d ) 88 days , e ) 44 days | a | divide(multiply(25, 16), 15) | multiply(n0,n1)|divide(#0,n2)| | physics |
a bus trip of 280 miles would have taken 1 hour less if the average speed v for the trip had been greater by 5 miles per hour . what was the average speed v , in miles per hour , for the trip ? | "the time is the distance / speed . the time difference is 1 hour . 280 / v - 280 / ( v + 5 ) = 1 280 ( v + 5 ) - 280 v = ( v ) ( v + 5 ) 1400 = ( v ) ( v + 5 ) 35 * 40 = ( v ) ( v + 5 ) v = 35 mph the answer is a ." | a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | a | divide(subtract(sqrt(add(multiply(multiply(280, 5), const_4), power(5, const_2))), 5), const_2) | multiply(n0,n2)|power(n2,const_2)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n2)|divide(#5,const_2)| | physics |
the total age of a and b is 15 years more than the total age of b and c . c is how many years younger than a ? | "solution [ ( a + b ) - ( b + c ) ] = 15 â € ¹ = â € º a - c = 15 . answer b" | a ) 12 , b ) 15 , c ) c is elder than a , d ) data inadequate , e ) none | b | multiply(15, const_1) | multiply(n0,const_1)| | general |
if the diagonal of a rectangle is 17 cm long and its perimeter is 46 cm , find the area of the rectangle . | let length = x and breadth = y . then , 2 ( x + y ) = 46 or x + y = 23 and x ^ 2 + y ^ 2 = ( 17 ) ^ 2 = 289 . now , ( x + y ) ^ 2 = ( 23 ) ^ 2 < = > ( x ^ 2 + y ^ 2 ) + 2 xy = 529 < = > 289 + 2 xy = 529 = xy = 120 area = xy = 120 cm ^ 2 answer is c . | ['a ) 127 cm ^ 2', 'b ) 125 cm ^ 2', 'c ) 120 cm ^ 2', 'd ) 102 cm ^ 2', 'e ) none of them'] | c | divide(subtract(power(divide(46, const_2), const_2), power(17, const_2)), const_2) | divide(n1,const_2)|power(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|divide(#3,const_2) | geometry |
the toll for crossing a certain bridge is $ 0.75 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 11.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the b... | option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than ... | a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | e | add(multiply(divide(multiply(divide(11, multiply(subtract(0.75, 0.3), const_2)), const_2), const_10), const_2), multiply(divide(11, multiply(subtract(0.75, 0.3), const_2)), const_2)) | subtract(n0,n2)|multiply(#0,const_2)|divide(n1,#1)|multiply(#2,const_2)|divide(#3,const_10)|multiply(#4,const_2)|add(#5,#3) | general |
mother , her daughter and her grand child weighs 160 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | "mother + daughter + child = 160 kg daughter + child = 60 kg mother = 160 - 60 = 100 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 100 = 20 kg so now daughter = 120 - ( mother + child ) = 160 - ( 100 + 20 ) = 40 kg answer : e" | a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 40 | e | subtract(60, divide(subtract(160, 60), 5)) | subtract(n0,n1)|divide(#0,n3)|subtract(n1,#1)| | general |
a batsman makes a score of 100 runs in the 17 th inning and thus increases his averages by 5 . what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 100 then average run after 17 th inning = ( 16 x + 100 ) / 17 ( 16 x + 100 ) / 17 = x + 5 = > 16 x + 100 = 17 x + 85 = > x = 15 x = 15 ; average after 17 th inning = 15 + 5 = 20 answer : c" | a ) 39 , b ) 35 , c ) 20 , d ) 40.5 , e ) 41.5 | c | add(subtract(100, multiply(17, 5)), 5) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
a fair coin is tossed 5 times . what is the probability of getting more heads than tails in 5 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails... | a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(5, const_2), choose(5, const_3)), choose(5, const_4)), choose(5, 5)), power(const_2, 5)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
the difference between the squares of two consecutive numbers is 35 . the numbers are | "explanation : let the numbers be a and ( a + 1 ) ( a + 1 ) 2 − a 2 = 35 ⇒ a 2 + 2 a + 1 − a 2 = 35 ⇒ 2 a = 34 ⇒ 2 a = 34 or a = 17 the numbers are 17 & 18 . correct option : c" | a ) 14,15 , b ) 15,16 , c ) 17,18 , d ) 18,19 , e ) none | c | divide(add(35, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics |
john want to buy a $ 100 trouser at the store , but he think it ’ s too expensive . finally , it goes on sale for $ 60 . what is the percent decrease ? | the is always the difference between our starting and ending points . in this case , it ’ s 100 – 60 = 40 . the “ original ” is our starting point ; in this case , it ’ s 100 . ( 40 / 100 ) * 100 = ( 0.4 ) * 100 = 40 % . c | a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 80 % | c | subtract(100, 60) | subtract(n0,n1) | general |
if rs . 561 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 561 * 2 / 17 = rs . 66 b ' s share = 561 * 3 / 17 = rs . 99 c ' s share = 561 * 12 / 17 = rs . 396 . answer : c" | a ) s . 300 , b ) s . 360 , c ) s . 396 , d ) s . 368 , e ) s . 323 | c | divide(561, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)| | general |
a certain car can travel 56 kilometers on a liter of fuel . if the fuel tank ’ s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers ) | fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 / 1.6 mile speed ( km / hour ) = d / t = 56 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 56 x 3.9 x 3.8 / 5.7 x 1.6 = 78... | a ) 52 , b ) 65 , c ) 78 , d ) 91 , e ) 104 | d | divide(multiply(divide(multiply(56, 3.8), 1.6), 3.9), 5.7) | multiply(n0,n4)|divide(#0,n6)|multiply(n1,#1)|divide(#2,n2) | physics |
if the length of the longest chord of a certain circle is 14 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 14 = 2 * 7 so radius of the circle = 7 correct answer - d" | a ) 2.5 , b ) 5 , c ) 10 , d ) 7 , e ) 9 | d | divide(14, const_2) | divide(n0,const_2)| | geometry |
at a certain organisation , the number of male members went up by 13 % in the year 2001 from year 2000 , and the number of females members went down by 6 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female mem... | men increase by 13 % = = > 1.13 m = males in 2001 women decrease by 6 % = = > 0.94 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.13 m + 0.94 f = 1.012 * ( m + f ) 1.1... | a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 5 , e ) 2 : 1 | d | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(6, const_100)), const_1000)), subtract(multiply(add(const_1, divide(13, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | divide(n4,const_100)|divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#2,const_1)|subtract(const_1,#1)|multiply(#3,const_1000)|multiply(#5,const_1000)|multiply(#4,const_1000)|subtract(#6,#7)|subtract(#8,#6)|divide(#9,#10) | other |
in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the p... | "this is a rather straight forward ratio problem . 1 . 40 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 40 / x thus , x = 1000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total... | a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 1,000 | e | divide(40, divide(2, 50)) | divide(n2,n1)|divide(n0,#0)| | gain |
set a of 8 positive integers may have the same element and have 35 . and set b of 8 positive integers must have different elements and have 35 . when m and n are the greatest possible differences between 35 and other elements ’ sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 35 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 28 and 35 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 7 . so , 28 - 7 = 21 . the correct answer is e . | a ) 20 , b ) 26 , c ) 23 , d ) 25 , e ) 21 | e | subtract(35, add(add(8, const_2), 8)) | add(n0,const_2)|add(n0,#0)|subtract(n1,#1)| | general |
a vessel of capacity 50 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 50 , y = 9 and n = 2 . hence , quantit... | a ) 23.89 , b ) 72.9 , c ) 33.62 , d ) 78.3 , e ) 79.3 | c | subtract(subtract(50, multiply(const_3, const_3)), multiply(divide(subtract(50, multiply(const_3, const_3)), 50), multiply(const_3, const_3))) | multiply(const_3,const_3)|subtract(n0,#0)|divide(#1,n0)|multiply(#2,#0)|subtract(#1,#3)| | physics |
on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 20 percent . what was the total rent the motel actually charged ... | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 20 % . 2 x = 200 = > x = 1000 answer c" | a ) $ 600 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | c | divide(multiply(10, subtract(60, 40)), divide(20, const_100)) | divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0)| | gain |
total of the ages of a , b ahd c at present is 90 years . 10 years ago , the ratio of their ages was 1 : 2 : 3 . what is the age of b at present | explanation : let their ages 10 years ago is x , 2 x and 3 x years . 10 + 2 x + 10 + 3 x + 10 = 90 hence x = 10 b � s present age = ( 2 x + 10 ) = 30 years answer : option b | a ) 25 , b ) 30 , c ) 45 , d ) 60 , e ) 65 | b | add(multiply(divide(subtract(90, add(add(10, 10), 10)), add(add(3, 2), 1)), 2), divide(subtract(90, add(add(10, 10), 10)), add(add(3, 2), 1))) | add(n1,n1)|add(n3,n4)|add(n1,#0)|add(n2,#1)|subtract(n0,#2)|divide(#4,#3)|multiply(n3,#5)|add(#5,#6) | general |
( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) / 8.0 = ? | "dove straight into calculation ( ( - 1.9 ) ( 0.6 ) – ( 2.6 ) ( 1.2 ) ) / 7.0 = - 0.53 answer a" | a ) - 0.53 , b ) 1.0 , c ) 1.07 , d ) 1.71 , e ) 2.71 | a | divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 8.0) | multiply(n0,n1)|multiply(n2,n3)|negate(#0)|subtract(#2,#1)|divide(#3,n4)| | general |
two pipes a and b can separately fill a tank in 12 and 20 minutes respectively . a third pipe c can drain off 45 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ? | 1 / 12 + 1 / 20 - 1 / x = 1 / 15 x = 15 15 * 45 = 675 answer : c | a ) 542 , b ) 540 , c ) 675 , d ) 829 , e ) 279 | c | multiply(multiply(inverse(subtract(add(add(divide(const_1, 12), divide(const_1, 20)), divide(const_1, 45)), divide(const_1, 15))), const_3), 20) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#4,#2)|subtract(#5,#3)|inverse(#6)|multiply(#7,const_3)|multiply(n1,#8) | physics |
a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability e that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . to... | a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | add(const_12,const_3)|divide(n0,n1)|power(#1,const_2)|multiply(#0,#2)| | general |
if 100 < x < 199 and 20 < y < 100 , then the product xy can not be equal to : | "correct answer : ( a ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 100 * 20 . the largest value must be less than 199 * 100 . this means that 2000 < xy < 19,900 . ( a ) is outside of this range , so it is not a possible pro... | a ) 1900 , b ) 19,303 , c ) 19 , 356.732 , d ) 19,502 , e ) 19,909 | a | add(100, const_1) | add(n3,const_1)| | general |
if d | | d ' , d : 2 x - y = 1 and d ' : ( a - 1 ) x + 2 y = x - 2 , then find the value of a . | d : 2 x − y = 1 , d ’ : ( a − 1 ) x + 2 y = x − 2 d : 2 x − y = 1 , d ′ : ( a − 1 ) x + 2 y = x − 2 d ∥ d ′ ⇒ md = m ′ dd ∥ d ′ ⇒ md = md ′ 2 x − y = 1 → y = 2 x − 1 → md = 22 x − y = 1 → y = 2 x − 1 → md = 2 ( a − 1 ) x + 2 y = x − 2 → 2 y = x ( 1 − a + 1 ) − 2 → 2 y = x ( 2 − a ) − 2 → ( a − 1 ) x + 2 y = x − 2 → 2 y... | a ) - 3 , b ) - 2 , c ) - 1 , d ) 0 , e ) 1 | b | subtract(2, multiply(2, 2)) | multiply(n0,n0)|subtract(n0,#0) | general |
money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 2500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | solution : money compounded annually at x % triples in 112 / x years . we need to find the final amount of 2500 at the end of the 28 years . . compounded annually at 8 % by putting the value in formula it will give us = 2500 ( 1.08 ) ^ 28 there must be a relation between these two conditions ? x = 8 % so money will tri... | a ) $ 3,750 , b ) $ 5,600 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 22,500 | e | multiply(2500, power(const_3, divide(28, divide(112, 8)))) | divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2) | gain |
let a be the event that a randomly selected two digit number is divisible by 3 and let b be the event that a randomly selected two digit number is divisible by 2 . what is p ( a and b ) ? | "p ( a and b ) = 1 / 3 * 1 / 2 = 1 / 6 the answer is b ." | a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) 1 / 2 | b | multiply(divide(subtract(3, const_1), multiply(subtract(3, const_1), 3)), divide(multiply(subtract(3, const_1), const_2), multiply(subtract(3, const_1), 3))) | subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)| | physics |
replace x with the appropriate number in 4 , 5 , 14 , 15 , 24 , 25 , x | c list of consecutive numbers that has an ' f ' in the spelling | a ) 30 , b ) 29 , c ) 34 , d ) 21 , e ) 20 | c | add(25, subtract(14, 5)) | subtract(n2,n1)|add(n5,#0) | general |
the radius of a semi circle is 3.5 cm then its perimeter is ? | "diameter = 7 cm 1 / 2 * 22 / 7 * 7 + 7 = 18 answer : d" | a ) 32.8 , b ) 32.4 , c ) 32.1 , d ) 18 , e ) 32.9 | d | add(divide(circumface(3.5), const_2), multiply(3.5, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
if bill can buy 3 pairs of jeans and 2 shirts for $ 69 or 2 pairs of jeans and 3 shirts for $ 71 , how much does one shirt cost ? | "3 j + 2 s = 69 2 j + 3 s = 71 - - - - - - - - - - - - - - - - 5 j + 5 s = 140 - - - - ( divide by 5 ) - - - > j + s = 28 3 j + 2 s = j + 2 ( j + s ) = j + 56 = 69 - - - > j = 13 3 * 13 + 2 s = 69 39 + 2 s = 69 2 s = 30 s = 15 answer : d" | a ) $ 10 , b ) $ 12 , c ) $ 13.20 , d ) $ 15 , e ) $ 16.80 | d | divide(subtract(multiply(71, 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 |
3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 165 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 165 / 7 = 94.2 kgs answer : a" | a ) 94.2 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 165), 7) | add(const_1,n0)|multiply(n2,#0)|divide(#1,n1)| | general |
the area of a rectangular field is equal to 800 square meters . its perimeter is equal to 120 meters . find the width of this rectangle . | "l * w = 800 : area , l is the length and w is the width . 2 l + 2 w = 120 : perimeter l = 60 - w : solve for l ( 60 - w ) * w = 800 : substitute in the area equation w = 20 and l = 40 correct answer d" | a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | divide(subtract(divide(120, const_2), sqrt(subtract(multiply(divide(120, const_2), divide(120, const_2)), multiply(const_4, 800)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|multiply(#0,#0)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry |
lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 20 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 20 3 c = c + 20 c = 10 the answer is c ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | divide(20, subtract(3, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
in a company , 48 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | "the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 52 % of all employees are women , so non - union women are 52 % - 18 % = 34 % 40 % of all employees are non - union . the percent of non - union employees who are women is 34 % ... | a ) 85 % , b ) 80 % , c ) 75 % , d ) 70 % , e ) 65 % | a | multiply(const_100, divide(subtract(subtract(const_100, 48), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | divide(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(n1,#0)|subtract(n1,#3)|subtract(#1,#4)|divide(#5,#2)|multiply(#6,const_100)| | gain |
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 500 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 500 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 500 ) / 100 + ( 1.8 * 700 ) / 100 = 16.6 gms % of liquid x in resultant... | a ) 1.66 % , b ) 1.96 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(500, 0.8), multiply(700, 1.8)), const_1000) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)| | gain |
what number has a 5 : 1 ratio to the number 9 ? | "5 : 1 = x : 9 x = 45 answer : a" | a ) 45 , b ) 50 , c ) 88 , d ) 52 , e ) 12 | a | multiply(9, 5) | multiply(n0,n2)| | other |
there are 100 employees in a room . 99 % are president . how many managers must leave the room to bring down the percentage of president to 98 % ? | "we have 99 presidents and 1 director . that 1 director to compose 2 % of the total number of people , there must be 50 people in the room , hence 50 presidents must leave . answer : b ." | a ) 1 , b ) 2 , c ) 46 , d ) 50 , e ) 97 | b | divide(subtract(multiply(100, divide(99, const_100)), multiply(100, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#1)|subtract(#2,#3)|divide(#5,#4)| | gain |
what is the compound interest paid on a sum of rs . 3000 for the period of 3 years at 10 % per annum . | "solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 / 100 = 11 interest % for 3 rd year = 10 + 10 % ( 10 + 11 ) = 10 + 2.1 = 12.1 total % of interest = 10 + 11 + 12.1 = 33.1 total interest = 33.1 % 3000 = 3000 * ( 33.1 / 100 ) = 993 answer d" | a ) 900 , b ) 930 , c ) 990 , d ) 993 , e ) none of these | d | subtract(multiply(3000, power(add(const_1, divide(10, const_100)), 3)), 3000) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 10 - percent solution ? | "3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 10 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 10 x = 15 since you need a 15 l solution , you must evaporate 35 of the original 50 l solution to get a 10 % sol... | a ) 35 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 15 | a | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
a person borrows rs . 6000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 p . a for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 6000 * 6 * 2 ) / 100 ] - [ ( 6000 * 4 * 2 ) / 100 ] 720 - 480 = 240 gain in 1 year = ( 240 / 2 ) = 120 rs answer : a" | a ) 120 rs , b ) 150 rs , c ) 160 rs , d ) 180 rs , e ) 200 rs | a | divide(subtract(divide(multiply(multiply(6000, 6), 2), const_100), divide(multiply(multiply(6000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain |
a person crosses a 1440 m long street in 12 minutes . what is his speed in km per hour ? | "speed = 1440 / ( 12 x 60 ) m / sec = 2 m / sec . converting m / sec to km / hr = 2 x ( 18 / 5 ) km / hr = 7.2 km / hr . answer : e" | a ) 4.1 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 7.2 | e | divide(divide(1440, const_1000), divide(multiply(12, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics |
an object thrown directly upward is at a height of h feet after t seconds , where h = - 14 ( t - 3 ) ^ 2 + 140 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ? | "we see that h will be a maximum h = 140 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 14 ( 5 - 3 ) ^ 2 + 140 = - 14 ( 4 ) + 140 = 84 the answer is d ." | a ) 62 , b ) 68 , c ) 76 , d ) 84 , e ) 96 | d | add(multiply(negate(14), power(subtract(add(3, 2), 3), 2)), 140) | add(n1,n2)|negate(n0)|subtract(#0,n1)|power(#2,n2)|multiply(#1,#3)|add(n3,#4)| | general |
rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage x . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the val... | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(1... | add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100)| | general |
the amount of time that three people worked on a special project was in the ratio of 2 to 3 to 4 . if the project took 90 hours , how many more hours did the hardest working person work than the person who worked the least ? | "let the persons be a , b , c . hours worked : a = 2 * 90 / 9 = 20 hours b = 3 * 90 / 9 = 30 hours c = 4 * 90 / 9 = 40 hours c is the hardest worker and a worked for the least number of hours . so the difference is 40 - 20 = 20 hours . answer : e" | a ) 18 hours , b ) 22 hours , c ) 25 hours , d ) 26 hours , e ) 20 hours | e | subtract(multiply(divide(90, add(add(2, 3), 4)), 4), multiply(divide(90, add(add(2, 3), 4)), 2)) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n2,#2)|multiply(n0,#2)|subtract(#3,#4)| | physics |
a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 9 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 9 ) x solve for r , r = 33 33 - 9 = 24 mpg d" | a ) 14 , b ) 16 , c ) 21 , d ) 24 , e ) 27 | d | divide(336, divide(subtract(462, 336), 9)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
the ratio of the radius of two circles is 2 : 7 , and then the ratio of their areas is ? | "r 1 : r 2 = 2 : 7 î r 1 ^ 2 : î r 2 ^ 2 r 1 ^ 2 : r 2 ^ 2 = 4 : 49 answer : d" | a ) 1 : 7 , b ) 2 : 9 , c ) 1 : 9 , d ) 4 : 49 , e ) 3 : 4 | d | divide(circle_area(2), circle_area(7)) | circle_area(n0)|circle_area(n1)|divide(#0,#1)| | geometry |
karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many q miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance trav... | a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | divide(n3,n2)|divide(n2,n2)|multiply(n0,#0)|subtract(#1,#0)|add(n0,#2)|divide(#4,#3)|subtract(#5,n0)| | physics |
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