options stringlengths 37 300 | correct stringclasses 5 values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 198 , b ) 288 , c ) 363 , d ) 396 , e ) 468 | e | multiply(multiply(power(2, 2), power(3, 2)), 13) | if 2 ^ 4 , 3 ^ 3 , and 13 ^ 3 are factors of the product of 1452 and w , where w is a positive integer , what is the smallest possible value of w ? | i will go with e ( pending elements to match is 2 ^ 2 * 3 ^ 2 * 13 ^ 1 = 468 | a = 2 ** 2
b = 3 ** 2
c = a * b
d = c * 13
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a ) 10 / 20 , b ) 1 / 20 , c ) 8 / 20 , d ) 3 / 20 , e ) 7 / 20 | c | divide(multiply(3, const_5), 20) | find the probability that a number selected from numbers 1 , 2 , 3 , . . . , 20 is a prime number , when each of the given numbers is equally likely to be selected ? | "let x be the event of selecting a prime number . x = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } n ( x ) = 8 , n ( s ) = 20 hence , the required probability is 8 / 20 . answer : c" | a = 3 * 5
b = a / 20
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a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many w eggs does the plant process per day ? | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there w = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ." | a = 99 - 96
b = 12 / a
c = b * 100
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a ) 35 , b ) 25 , c ) 20 , d ) 30 , e ) 15 | d | divide(multiply(multiply(6, const_3), const_3), multiply(const_3, const_3)) | if two painters can complete two rooms in two hours , how many rooms painted if 10 painters work for 6 hours ? | "explanation : two painters can complete two rooms in two hours . so 30 rooms can be painted in 6 hrs by 10 painters answer : d ) 10 painters" | a = 6 * 3
b = a * 3
c = 3 * 3
d = b / c
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a ) 1 / 4 , b ) 3 / 8 , c ) 3 / 16 , d ) 5 / 16 , e ) 11 / 32 | d | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting exactly 2 heads in a single throw of five fair coins ? | "one possible case is hhttt . p ( hhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 there are 5 c 2 = 10 possible cases . p ( 2 heads ) = 10 * 1 / 32 = 5 / 16 the answer is d ." | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
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a ) 1 : 2 , b ) 2 : 5 , c ) 3 : 4 , d ) 3 : 5 , e ) 3 : 9 | b | divide(divide(multiply(2, 3), multiply(3, 4)), divide(multiply(3, 4), multiply(const_2.0, 5))) | the compound ratio of 2 : 3 , 3 : 4 and 4 : 5 ? | 2 / 3 * 3 / 4 * 4 / 5 = 2 / 5 2 : 5 answer : b | a = 2 * 3
b = 3 * 4
c = a / b
d = 3 * 4
e = 2 * 0
f = d / e
g = c / f
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a ) 42 , b ) 70 , c ) 150 , d ) 165 , e ) 315 | c | multiply(multiply(5, 3), 6) | a certain university will select 2 of 6 candidates eligible to fill a position in the mathematics department and 3 of 5 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "6 c 2 * 5 c 3 = 15 * 10 = 150 the answer is ( c )" | a = 5 * 3
b = a * 6
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a ) 0.05 , b ) 0.45 , c ) 0.15 , d ) 0.5 , e ) 0.55 | b | subtract(add(0.25, 0.6), 0.4) | the probability that event b occurs is 0.4 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.6 , what is the probability that event a will occur ? | "p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.6 = 0.4 + p ( a ) - 0.25 p ( a ) = 0.45 ans : b" | a = 0 + 25
b = a - 0
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a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 9 | e | divide(divide(1, const_1), multiply(add(multiply(1, const_1), 2), const_3)) | when anunfaircoin is tossed twice , the probability of getting one tails and one heads is 1 / 2 . what is the probability of getting two heads and two tails if the coin is tossed 4 times ? | let the probability of heads be x and the probability of tails be 1 - x . so , we are given that p ( one tails and one heads ) = 2 * x * ( 1 - x ) = 1 / 2 . p ( two heads and two tails ) = 4 ! / ( 2 ! 2 ! ) * x ^ 2 * ( 1 - x ) ^ 2 = 6 * ( x * ( 1 - x ) ) ^ 2 . we are multiplying by 4 ! / ( 2 ! 2 ! ) because hhtt scenario can occur in several ways : hhtt , htht , thht , . . . ( permutation of 4 letters hhtt where 2 t ' s and 2 h ' s are the same ) . since from 2 * x * ( 1 - x ) = 1 / 2 it follows that x * ( 1 - x ) = 1 / 4 , then 4 * ( x * ( 1 - x ) ) ^ 2 = 4 * ( 1 / 6 ) ^ 2 = 1 / 9 . answer : e . | a = 1 / 1
b = 1 * 1
c = b + 2
d = c * 3
e = a / d
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a ) 700 m , b ) 720 m , c ) 712 m , d ) 600 m , e ) 640 m | c | divide(multiply(15, multiply(1.9, const_1000)), 40) | amar takes as much time in running 15 meters as a car takes in covering 40 meters . what will be the distance covered by amar during the time the car covers 1.9 km ? | distance covered by amar = 15 / 40 ( 1.9 km ) = 3 / 8 ( 1900 ) = 712 m answer : c | a = 1 * 9
b = 15 * a
c = b / 40
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a ) 24 % , b ) 25 % , c ) 30 % , d ) 33 % , e ) 40 % | a | multiply(subtract(divide(add(const_100, 55), add(const_100, 25)), const_1), const_100) | at the end of the first quarter , the share price of a certain mutual fund was 25 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 55 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of the first quarter to the end of the second quarter ? | "another method is to use the formula for 2 successive percentage changes : total = a + b + ab / 100 55 = 25 + b + 25 b / 100 b = 24 answer ( a )" | a = 100 + 55
b = 100 + 25
c = a / b
d = c - 1
e = d * 100
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a ) 15 , b ) 20 , c ) 30 , d ) 40 , e ) 45 | d | divide(subtract(200, 80), 3) | a marketing firm determined that , of 200 households surveyed , 80 used neither brand e nor brand b soap , 60 used only brand e soap , and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 households surveyed used both brands of soap ? | solution for soap e and soap b ( d ) 40 | a = 200 - 80
b = a / 3
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a ) s . 259 , b ) s . 289 , c ) s . 299 , d ) s . 300 , e ) s . 389 | a | divide(multiply(570, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 570 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 570 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 570 β y [ 12 / 10 + 1 ] = 570 β 22 y / 10 = 570 β 22 y = 5700 β y = 5700 / 22 = rs . 259 a ) | a = 1 + 4
b = a * 2
c = 570 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
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a ) 30 sec , b ) 32 sec , c ) 36 sec , d ) 38 sec , e ) 52 sec | e | multiply(const_3600, divide(divide(390, const_1000), add(25, 2))) | a train 390 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 Γ 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 390 Γ 2 / 15 ) sec = 52 sec answer : e" | a = 390 / 1000
b = 25 + 2
c = a / b
d = 3600 * c
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a ) 800 , b ) 300 , c ) 700 , d ) 400 , e ) none of these | b | divide(30, 90) | if 90 % of a = 30 % of b and b = x % of a , then the value of x is ? | "0.9 a = 0.3 b so b = 3 a means 300 % of a answer : b" | a = 30 / 90
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 3 | a | divide(add(multiply(multiply(const_0_25, 20), 14), 20), add(multiply(const_0_25, 20), const_1)) | data sufficiency question : if solution ( x ) with alcohol ( 14 % ) and solution ( y ) with alcohol ( 20 % ) mixed together to form 24 liters of mixture . | explanation : let a liters of first solution , and ' b ' liters of second solution has been taken . so using weighted average formula = n 1 Γ a 1 + n 2 Γ a 2 n 1 + n 2 n 1 Γ a 1 + n 2 Γ a 2 n 1 + n 2 = n 1 Γ 14 + n 2 Γ 20 n 1 + n 2 = 15 n 1 Γ 14 + n 2 Γ 20 n 1 + n 2 = 15 β n 1 = 5 n 2 β n 1 n 2 = 51 β n 1 = 5 n 2 β n 1 n 2 = 51 as total volume was given as 24 liters , first solution = 16 Γ 24 = 416 Γ 24 = 4 so statement 1 is sufficient . answer : a | a = const_0_25 * 20
b = a * 14
c = b + 20
d = const_0_25 * 20
e = d + 1
f = c / e
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a ) 4.5 , b ) 4.2 , c ) 4.7 , d ) 4.8 , e ) none of these | b | divide(subtract(multiply(3.95, 6), add(multiply(3.8, const_2), multiply(3.85, const_2))), const_2) | the average of 6 number is 3.95 . the average of two of them is 3.8 , while the average of the other two is 3.85 . what is the average of the remaining two number ? | solution : sum of the remaining two numbers = ( 3.95 Γ 6 ) - [ ( 3.8 Γ 2 ) + ( 3.85 Γ 2 ) ] = 23.70 - ( 7.6 + 7.7 ) = 23.70 - 15.3 = 8.4 . β΄ required average = 8.4 / 2 = 4.2 answer b | a = 3 * 95
b = 3 * 8
c = 3 * 85
d = b + c
e = a - d
f = e / 2
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a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23 | e | add(subtract(100, multiply(12, 7)), 7) | a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points q that an individual player might have scored ? | general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points q for 12 th player is 100 - 77 = 23 . answer : e . | a = 12 * 7
b = 100 - a
c = b + 7
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a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 14 | d | multiply(divide(7, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3) | jill has 7 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ? | let the number of each size of jar = w then 1 / 4 w + 1 / 2 w + w = 7 1 3 / 4 w = 7 w = 4 the total number of jars = 3 w = 12 answer : d | a = 1 / 2
b = const_0_25 + a
c = 1 + b
d = 7 / c
e = d * 3
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a ) 8 , b ) 14 , c ) 16 , d ) 17 , e ) 18 | b | divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 50), 5) | the sum of ages of 5 children born 2 years different each is 50 yrs . what is the age of the elder child ? | let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) and ( x + 8 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 50 5 x = 30 x = 6 x + 8 = 6 + 8 = 14 answer : b | a = 2 + 4
b = 2 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 50
g = f / 5
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a ) 7458 , b ) 7456 , c ) 7450 , d ) 7454 , e ) 7453 | c | divide(2086, divide(subtract(add(const_100, 14), subtract(const_100, 14)), const_100)) | vijay sells a cupboard at 14 % below cost price . had he got rs . 2086 more , he would have made a profit of 14 % . what is the cost price of the cupboard ? | "explanation : cost price = 2086 / ( 0.14 + 0.14 ) = 2086 / 0.28 = rs . 7450 answer : c" | a = 100 + 14
b = 100 - 14
c = a - b
d = c / 100
e = 2086 / d
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | subtract(80, subtract(390, multiply(80, 4))) | we bought a total of 80 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 390 . how many math books did we buy ? | "m + h = 80 h = 80 - m 4 m + 5 h = 390 4 m + 5 * ( 80 - m ) = 390 m = 10 the answer is a ." | a = 80 * 4
b = 390 - a
c = 80 - b
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a ) 144 mins , b ) 140 mins , c ) 136 mins , d ) 132 mins , e ) 130 mins | a | multiply(add(const_1, const_4), 36) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 36 minutes , then the slower pipe alone will be able to fill the tank in | "explanation : let the slower pipe alone fill the tank in x minutes then faster will fill in x / 3 minutes . part filled by slower pipe in 1 minute = 1 / x part filled by faster pipe in 1 minute = 3 / x part filled by both in 1 minute = 1 / x + 3 / x = 1 / 36 = > 4 / x = 1 / 36 x = 36 Γ’ Λ β 4 = 144 mins answer is a" | a = 1 + 4
b = a * 36
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a ) 180 , b ) 1800 , c ) 18 , d ) 18000 , e ) 1.8 | b | divide(multiply(36, 15), divide(multiply(6, 5), const_100)) | a hall 36 m long and 15 m broad is to be paved with stones , each measuring 6 dm by 5 dm . the number of stones required is : | "area of the hall = 3600 * 1500 area of each stone = ( 60 * 50 ) therefore , number of stones = ( 3600 * 1500 / 60 * 50 ) = 1800 answer : b" | a = 36 * 15
b = 6 * 5
c = b / 100
d = a / c
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a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48 | c | divide(10, subtract(divide(5, 4), const_1)) | walking with 4 / 5 of my usual speed , i arrive at the bus stop 10 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 10 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 40 the answer is c ." | a = 5 / 4
b = a - 1
c = 10 / b
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a ) 32 / 15 , b ) 3 / 8 , c ) 8 / 5 , d ) 5 / 8 , e ) 5 / 3 | a | divide(divide(480, 3), divide(300, 4)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 480 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 480 km time taken by eddy = 3 hours average speed of eddy = 480 / 3 = 160 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 160 / 75 = 32 / 15 answer a" | a = 480 / 3
b = 300 / 4
c = a / b
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a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | b | multiply(divide(2, 20), const_100) | a pharmaceutical company received $ 2 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "( 9 / 108 ) / ( 2 / 20 ) = 30 / 54 = 55,5 % it means that 9 / 108 represents only 83,3 % . therefore a decrease of 15 % . answer b" | a = 2 / 20
b = a * 100
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a ) 16 days , b ) 12 days , c ) 8 days , d ) 6 days , e ) 18 days | c | inverse(add(divide(const_1, 24), divide(const_1, divide(24, const_2)))) | ram , who is half as efficient as krish , will take 24 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ? | "answer ram takes 24 days to complete the task , if he works alone . krish is twice as efficient as ram is . so , working alone , krish will take half the time to complete the task . i . e . , 12 days . ram will complete 1 / 24 th of the task in a day . krish will complete 1 / 12 th of the task in a day . when they work together , they will complete 1 / 24 + 1 / 12 = 1 / 8 th of the task in a day . therefore , when they work together they will complete the task in 8 days . choice c" | a = 1 / 24
b = 24 / 2
c = 1 / b
d = a + c
e = 1/(d)
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a ) 197 , b ) 280 , c ) 189 , d ) 278 , e ) 315 | e | divide(multiply(540, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | i bought two books ; for rs . 540 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 540 - x ) 119 / 100 x = 315 answer : e" | a = 100 + 19
b = 540 * a
c = 100 - 15
d = 100 + 19
e = c + d
f = b / e
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a ) 30 % , b ) 10 % , c ) 25 % , d ) 19 % , e ) 50 % | c | multiply(divide(subtract(2492, 1980), 1980), const_100) | a sum of money deposited at c . i . amounts to rs . 1980 in 3 years and to rs . 2492 in 4 years . find the rate percent ? | "1980 - - - 512 100 - - - ? = > 25 % answer : c" | a = 2492 - 1980
b = a / 1980
c = b * 100
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a ) 20 cm , b ) 10 cm , c ) 25 cm , d ) 50 cm , e ) 28 cm | b | multiply(20, const_4) | what is the greatest possible length which can be used to measure exactly the lengths 10 m , 14 m 50 cm and 20 m 40 cm cm ? | "required length = hcf of 1000 cm , 1450 cm , 2040 cm = 10 cm answer is b" | a = 20 * 4
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a ) 6 , b ) 90 , c ) 12 , d ) 24 , e ) 48 | b | divide(multiply(divide(90, const_100), 18), divide(70, const_100)) | an alloy weighing 18 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 90 percent gold ? | "in 24 ounces , gold is 18 * ( 70 / 100 ) = 12.6 ounces . now we add x ounces of pure gold to make it 90 % gold . so 12.6 + x = ( 24 + x ) * 90 / 100 = > x = 90 . answer is b ." | a = 90 / 100
b = a * 18
c = 70 / 100
d = b / c
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a ) 30 , b ) 40 , c ) 50 , d ) 55 , e ) 70 | c | multiply(subtract(const_100, 80), subtract(divide(const_100, 55), divide(80, 80))) | john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 55 miles per hour , what is v in miles per hour ? | "hibunuel the question seems incorrect . it should not be 80 % at the speed of 80 . however if it ' s 20 % at the speed of 80 , answer comes out 55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it , which comes as : 0.8 d / 80 + 0.2 d / v = d / 55 = > v = 50 ( option c ) ." | a = 100 - 80
b = 100 / 55
c = 80 / 80
d = b - c
e = a * d
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a ) 36 kmph , b ) 40 kmph , c ) 42 kmph , d ) 45 kmph , e ) 38 kmph | a | divide(divide(400, const_1000), divide(40, const_3600)) | a train 400 m long can cross an electric pole in 40 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 400 / 40 s = 10 m / sec speed = 10 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 36 kmph answer : a" | a = 400 / 1000
b = 40 / 3600
c = a / b
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a ) 41 , b ) 40.6 , c ) 40.8 , d ) 42 , e ) 42.6 | b | divide(divide(subtract(220, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 220 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 20 seconds . the speed of the train is : | "speed of the train relative to man = 220 / 20 = 11 m / sec . = 11 x 18 / 5 km / hr = 198 / 5 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 198 / 540.6 km / hr . answer : b" | a = 5 * const_0_2778
b = a * 5
c = 220 - b
d = c / 5
e = d / const_0_2778
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a ) 102 % , b ) 105 % , c ) 95 % , d ) 93.5 % , e ) 90 % | d | add(subtract(subtract(10, 15), divide(multiply(15, 10), const_100)), const_100) | the price of a certain painting increased by 10 % during the first year and decreased by 15 % during the second year . the price of the painting at the end of the 2 - year period was what percent of the original price ? | "easiest thing to do : assume that price is 100 price at the end of yr 1 : 100 + 10 = 110 price at the end of year 2 = 110 - 110 * 0.15 = 110 * 0.85 = 93.5 hence required answer = ( 93.5 / 100 ) * 100 % = 93.5 % answer is d ." | a = 10 - 15
b = 15 * 10
c = b / 100
d = a - c
e = d + 100
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a ) 80 m , b ) 100 m , c ) 120 m , d ) 180 m , e ) none | e | multiply(multiply(add(36, 45), const_0_2778), 16) | two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 16 seconds . the length of the faster train is | "solution relative speed = ( 36 + 45 ) km / hr = ( 81 x 5 / 18 ) m / sec = ( 45 / 2 ) m / sec length of the train = ( 45 / 2 x 16 ) m = 360 m . answer e" | a = 36 + 45
b = a * const_0_2778
c = b * 16
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a ) 1100 , 1750 , b ) 1140 , 1800 , c ) 940 , 2000 , d ) 1190 , 1750 , e ) 1000 : 1690 | d | multiply(divide(2940, add(add(multiply(7000, 6), multiply(subtract(7000, 3000), subtract(const_12, 6))), add(multiply(15000, 6), multiply(add(15000, 3000), subtract(const_12, 6))))), add(multiply(7000, 6), multiply(subtract(7000, 3000), subtract(const_12, 6)))) | a and b began business with rs . 7000 and rs . 15000 after 6 months , a advances rs . 3000 and b withdraws rs . 5000 . at the end of the year , their profits amounted to rs . 2940 find the share of a and b respectively . | "7 * 6 + 10 * 6 ) : ( 15 * 6 + 10 * 6 ) = 102 : 150 = 51 : 75 = 17 : 25 17 : 25 a ' s share = 17 / 42 * 2940 = 1190 . b ' s share = 25 / 42 * 2940 = 1750 . a and b shares respectively 1190 and 1750 . answer : d" | a = 7000 * 6
b = 7000 - 3000
c = 12 - 6
d = b * c
e = a + d
f = 15000 * 6
g = 15000 + 3000
h = 12 - 6
i = g * h
j = f + i
k = e + j
l = 2940 / k
m = 7000 * 6
n = 7000 - 3000
o = 12 - 6
p = n * o
q = m + p
r = l * q
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a ) 39 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 % | a | add(divide(multiply(15, subtract(const_100, 40)), const_100), subtract(40, 10)) | in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 15 % of the people who do not have the job of their choice have a university diploma . if 40 % of the people have the job of their choice , what percent of the people have a university diploma ? | "setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = 10 % job of choice total = 40 % not job of choice with diploma = . 15 x not job of choice w / o diploma = . 85 x total not job of choice = x total with diploma total without diploma total citizen = 100 if 40 % of people have their job of choice , then 60 % of people do not have their job of choice . 15 % of 60 % = 9 % . we can also see that 30 % of the people have their job of choice and a diploma ( 40 % - 10 % = 30 % ) . 30 % + 9 % = 39 % . therefore 39 % of the people in country z have a diploma . ans a" | a = 100 - 40
b = 15 * a
c = b / 100
d = 40 - 10
e = c + d
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a ) rs . 150.50 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 625 , e ) none | d | divide(676, power(add(divide(4, const_100), const_1), 2)) | the present worth of rs . 676 due in 2 years at 4 % per annum compound interest is | solution present worth = rs . [ 676 / ( 1 + 4 / 100 ) Β² ] = rs . ( 676 x 25 / 26 x 25 / 26 ) = rs . 625 answer d | a = 4 / 100
b = a + 1
c = b ** 2
d = 676 / c
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a ) 4 % , b ) 5 % , c ) 6 % , d ) 8 % , e ) 10 % | d | floor(multiply(const_100, divide(subtract(2591.36, 2400), 2400))) | a shopkeeper sold an article for rs 2591.36 . approximately what was his profit percent if the cost price of the article was rs 2400 | "explanation : gain % = ( 191.36 * 100 / 2400 ) = 7.97 % = 8 % approx option d" | a = 2591 - 36
b = a / 2400
c = 100 * b
d = math.floor(c)
|
a ) 79 , b ) 99 , c ) 88 , d ) 85 , e ) 75 | d | divide(add(add(add(add(86, 89), 82), 87), 81), divide(const_10, const_2)) | dacid obtained 86 , 89 , 82 , 87 and 81 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 86 + 89 + 82 + 87 + 81 ) / 5 = 425 / 5 = 85 . answer : d" | a = 86 + 89
b = a + 82
c = b + 87
d = c + 81
e = 10 / 2
f = d / e
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a ) 23 , b ) 25 , c ) 27 , d ) 29 , e ) 31 | d | subtract(add(multiply(divide(783, divide(783, 2)), const_4), const_10), const_1) | the product of two positive integers is 783 and their difference is 2 . what is the bigger number ? | "let ' s use trial and error to find the two numbers . 30 * 28 = 840 ( too high ) 29 * 27 = 783 the answer is d ." | a = 783 / 2
b = 783 / a
c = b * 4
d = c + 10
e = d - 1
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(1,000, const_100), subtract(700, subtract(700, 150))) | matt gets a $ 1,000 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 700 , how many sales has matt made ? | "let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1000 new average = ( xy + 1000 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1000 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1000 ) = ( xy + x + 150 y + 150 ) i . e . ( 850 ) = ( x + 150 y ) new commission = 700 = 150 + x i . e . x = 550 i . e . y = 2 new sales = y + 1 = 3 answer : option a" | a = 1 - 0
b = 700 - 150
c = 700 - b
d = a / c
|
['a ) β 10 / 2', 'b ) β 10', 'c ) 2 β 10', 'd ) 5', 'e ) 10'] | c | multiply(sqrt(divide(divide(22, const_pi), 0.7)), const_2) | water is poured into a cylindrical well at the rate of 22 cubic meters per hour , the level of water in the tank rises at a rate of 0.7 meters per hour . which of the following best approximates the diameter of the tank in meters ? | here ' s my re - worked solution : volume = 22 = pi ( r ^ 2 ) ( 0.7 ) 22 = 22 / 7 ( 7 / 10 ) ( r ^ 2 ) 10 = r ^ 2 r = sqrt ( 10 ) diameter = 2 * sqrt ( 10 ) ans : c | a = 22 / math.pi
b = a / 0
c = math.sqrt(b)
d = c * 2
|
a ) 3 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | c | subtract(multiply(4, 3), 4) | how many different 3 - digit multiples of 5 can be composed of digits 1 , 2 , 3 , 4 , and 5 if none of the digits is repeated ? | can you explain your method . i solved it differently xyz , where z must be 5 . therefore 1 variation of digit in z . y can be any of the 4 possible choices . x can be any of the 3 possible choices . 4 + 3 + 1 = 8 c | a = 4 * 3
b = a - 4
|
a ) 16 , b ) 18 , c ) 14 , d ) 20 , e ) 22 | d | subtract(divide(multiply(add(floor(divide(subtract(sqrt(add(multiply(multiply(575, 2), const_4), 1)), 1), const_2)), 1), add(add(floor(divide(subtract(sqrt(add(multiply(multiply(575, 2), const_4), 1)), 1), const_2)), 1), 1)), 2), 575) | a student finds the sum 1 + 2 + 3 + . . . as his patience runs out . he found the sum as 575 . when the teacher declared the result wrong , the student realized that he missed a number . what was the number the student missed ? | explanation : sum of 1 st 10 natural numbers ( 1 β 10 ) = 1 + 2 + β¦ 10 = 55 . sum of next 10 natural numbers ( 11 β 20 ) = 10 + 11 + 12 + β¦ 20 = 100 + 55 = 155 . sum of next 10 natural numbers ( 31 β 40 ) = 255 . sum of numbers from 31 β 40 = 355 and so on and on . so , we have here 55 + 155 + 255 = 465 . if we add 355 to 465 , it will be way more than 575 . so , we can add one by one . 465 + 31 + 32 + 33 = 561 . 465 + 31 + 32 + 33 + 34 = 595 . thus , the sum should be 595 instead of 575 , and the missing number is 20 . answer : d | a = 575 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d - 1
f = e / 2
g = math.floor(f)
h = g + 1
i = 575 * 2
j = i * 4
k = j + 1
l = math.sqrt(k)
m = l - 1
n = m / 2
o = math.floor(n)
p = o + 1
q = p + 1
r = h * q
s = r / 2
t = s - 575
|
a ) $ 6.5 , b ) $ 6 , c ) $ 7 , d ) $ 8 , e ) $ 9 | b | divide(divide(300, 25), 2) | dhoni spends $ 300 buying his favorite dolls . if he buys only small monkey dolls , which are $ 2 cheaper than the large monkey dolls , he could buy 25 more dolls than if he were to buy only large monkey dolls . how much does a large monkey doll cost ? | a is not an integer . so we start with b . if large doll costs $ 6 , then he can buy 300 / 6 = 50 large dolls and 300 / 4 = 75 small dolls . difference is 25 , which is we wanted . answer b . | a = 300 / 25
b = a / 2
|
a ) 20 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50 | a | add(20, 20) | set x consists of 5 integers and has median of 20 and a range of 20 . what is the value of the greatest possible integer that can be present in the set ? | "note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between and hence can not be the median . ( d ) 43 range of 20 means the smallest integer will be 23 . so 20 can not lie in between and hence can not be the median . ( c ) 40 range of 20 means the smallest integer will be 20 . 20 can lie in between such as : 20 , 20 , 20 , 20 , 20 this is possible . hence it is the greatest such number . answer ( a )" | a = 20 + 20
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a ) 939.6 , b ) 928.8 , c ) 927.78 , d ) 926.82 , e ) 902.1 | a | divide(1120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 8), const_100), const_1)) | find the principle on a certain sum of money at 8 % per annum for 2 2 / 5 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 8 * 12 / 5 ) / 100 ] p = 939.60 answer : a" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 8
e = d / 100
f = e + 1
g = 1120 / f
|
a ) 60 % , b ) 28 % , c ) 30 % , d ) 56 % , e ) 73 % | b | subtract(divide(multiply(add(const_100, 21.6), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 21.6 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 121.60 let marked price be rs . x . then , 95 / 100 x = 121.60 x = 12160 / 95 = rs . 128 now , s . p . = rs . 128 , c . p . = rs . 100 profit % = 28 % . answer : b" | a = 100 + 21
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 22 , b ) 21 , c ) 88 , d ) 30 , e ) 29 | d | add(10, divide(multiply(10, subtract(12000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 10 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "explanation : lot the total number of workers be v then , 8 ooov = ( 12000 * 10 ) + 6000 ( v - 10 ) < = > 2000 v = 60000 < = > v = 30 answer : d ) 30" | a = 12000 - 8000
b = 10 * a
c = 8000 - 6000
d = b / c
e = 10 + d
|
a ) 90 sec , b ) 190 sec , c ) 290 sec , d ) 490 sec , e ) 390 sec | a | subtract(divide(multiply(const_1, const_1000), divide(100, 10)), 10) | in a kilometer race , a beats b by 100 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 100 = 100 sec . time taken by a = 100 - 10 = 90 sec . answer : a" | a = 1 * 1000
b = 100 / 10
c = a / b
d = c - 10
|
a ) 480 , b ) 14,400 , c ) 24,200 , d ) 28,800 , e ) 48,000 | b | multiply(4, const_3600) | a space shuttle orbits the earth at about 4 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 4 * 3600 = 14,400 answer : b" | a = 4 * 3600
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a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | c | divide(multiply(choose(3, 2), choose(3, 1)), choose(add(add(3, 4), 3), 3)) | a jar contains only red , yellow , and orange marbles . if there are 3 red , 3 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 3 / 22 . this helped me realise that for the 3 possible orderings the probabbility is the same . so , it should be ( 1 / 45 ) * ( 2 ) , which indeed is 2 / 45 . c" | a = math.comb(3, 2)
b = math.comb(3, 1)
c = a * b
d = 3 + 4
e = d + 3
f = math.comb(e, 3)
g = c / f
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | multiply(divide(35, add(const_4, const_3)), const_4) | there are 2 friends peter and paul . peter age is twice as old as paul when peter was as old as paul is now . sum of the present ages of peter and paul is 35 . what is the present age of peter ? | let pauls age be ' a ' and peters be ' e ' e + a = 35 ( given ) e = 35 - a - - - - - - eq 1 when peter was as old as paul that is when peter ' s age was ' a ' pauls age was a - ( e - a ) = 2 a - e now peters age is e = 2 ( 2 a - e ) e = 4 a - 2 e 3 e = 4 a 3 ( 35 - a ) = 4 a 105 - 3 a = 4 a a = 15 hence e = 20 answer : c | a = 4 + 3
b = 35 / a
c = b * 4
|
a ) 8 , b ) 9 , c ) 7 , d ) 62 , e ) 2 | b | sqrt(divide(162, const_2)) | the area of a parallelogram is 162 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 162 = > x = 9 answer : b" | a = 162 / 2
b = math.sqrt(a)
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a ) 24 , b ) 34 , c ) 28 , d ) 40 , e ) 14 | d | sqrt(subtract(power(4, const_2), power(10, const_2))) | 34 . the side surface of a cylinder is rolled with a rectangular plate . if the height of a cylinder tank is 10 feet and the perimeter of the circular base is 4 feet , what is the area of the rectangular plate ? | think of a pringles can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both you can multiply them together to find the area , 40 . ( correct answer d ) | a = 4 ** 2
b = 10 ** 2
c = a - b
d = math.sqrt(c)
|
a ) 2988 , b ) 4543 , c ) 4400 , d ) 2871 , e ) 881 | b | floor(divide(3553, multiply(divide(subtract(const_100, 8), const_100), divide(subtract(const_100, 15), const_100)))) | 8 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3553 , how much was it in the beginning ? | "x * ( 92 / 100 ) * ( 85 / 100 ) = 3553 x = 4543 answer : b" | a = 100 - 8
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 3553 / e
g = math.floor(f)
|
a ) 10 , b ) 13 , c ) 14 , d ) 15 , e ) 19 | e | divide(multiply(multiply(57, 12), 5), multiply(30, 6)) | 57 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "according to the chain rule , m 1 x t 1 = m 2 x t 2 therefore , 57 x 12 x 5 = 30 x 6 x x x = 19 hence , the number of days = 19 . answer : e" | a = 57 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 18 , b ) 750 , c ) 24 , d ) 120 , e ) none of these | a | divide(circle_area(divide(30, multiply(2, const_pi))), 4) | how many plants will be there in a circular bed whose outer edge measure 30 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 30 cm area of circular bed = ( 30 ) 2 β 4 Ο space for each plant = 4 cm 2 β΄ required number of plants = ( 30 ) 2 β 4 Ο Γ· 4 = 17.89 = 18 ( approx ) answer a" | a = 2 * math.pi
b = 30 / a
c = circle_area / (
|
a ) 21 , b ) 35 , c ) 39 , d ) 51 , e ) 57 | e | multiply(subtract(20, const_4), const_3) | the product z of two prime numbers is between 20 and 60 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 18 but less than 36 then what is z ? | "the smallest possible product is 57 which is 3 * 19 . all other products are too big . the answer is e ." | a = 20 - 4
b = a * 3
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a ) 3 , b ) 22 1 / 4 , c ) 22 1 / 2 , d ) 22 1 / 1 , e ) 22 1 / 3 | a | divide(subtract(divide(100, 10), divide(60, 15)), const_2) | a boat goes 100 km downstream in 10 hours , and 60 m upstream in 15 hours . the speed of the stream is ? | "100 - - - 10 ds = 10 ? - - - - 1 60 - - - - 15 us = 4 ? - - - - - 1 s = ( 10 - 4 ) / 2 = 3 kmp answer : a" | a = 100 / 10
b = 60 / 15
c = a - b
d = c / 2
|
a ) 16 ^ 4 , b ) ( 4 ! ) ^ 4 , c ) 16 ! / ( 2 ! ) ^ 4 , d ) 16 ! / 2 ! , e ) 4 ^ 16 | c | divide(divide(factorial(16), power(const_2, 4)), divide(factorial(16), power(const_2, 4))) | in how many ways can 16 different gifts be divided among 4 children such that each child receives exactly two gifts ? | total 16 different gifts , and 4 children . thus any one child gets 16 c 2 gifts , then the other child gets 14 c 2 gifts ( 16 total - 2 already given ) , then the third one gets 12 c 2 gifts , and the last child gets 10 c 2 gifts . since order in which each child gets the gift is not imp , thus , ans : 16 c 2 * 14 c 2 * 12 c 2 * 10 c 2 = 16 ! / ( 2 ! ) ^ 4 c | a = math.factorial(16)
b = 2 ** 4
c = a / b
d = math.factorial(16)
e = 2 ** 4
f = d / e
g = c / f
|
a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16 | d | add(4, const_4) | how many different values of positive integer x , for which | x + 4 | < x , are there ? | answer d i opted to put the random value option . i used 0 , 4 , - 4 and the the extreme of 60 and - 60 . . i was able to solve it in 1 : 09 d | a = 4 + 4
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 1 | e | divide(subtract(subtract(add(add(15, 18), 12), 4), 39), 2) | at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 15 employees are trained to work in the family buffet , 18 are trained to work in the dining room , and 12 are trained to work in the snack bar . if 4 employees are trained to work in exactly 2 restaurants , how many employees are trained to work in all 3 restaurants ? | "39 = 15 + 18 + 12 - 4 - 2 x 2 x = 15 + 18 + 12 - 4 - 39 = 41 - 39 = 2 x = 1 e" | a = 15 + 18
b = a + 12
c = b - 4
d = c - 39
e = d / 2
|
a ) $ 28 , b ) $ 82 , c ) $ 172 , d ) $ 138 , e ) $ 192 | c | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(2, const_100)), multiply(add(60, 2), 150)), 150) | a computer manufacturer produces a certain electronic component at a cost of $ 60 per component . shipping costs for delivering the components are $ 2 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "this question can be solved easily using a equation model . 150 * p = 16500 ( fixed cost ) + 150 * 60 + 150 * 2 . = 16500 + 9000 + 300 = p = $ 172 . c" | a = 4 * 4
b = a * 1000
c = 2 * 100
d = b + c
e = 60 + 2
f = e * 150
g = d + f
h = g / 150
|
a ) 53.3 , b ) 53.8 , c ) 53.7 , d ) 53.5 , e ) 353.3 | a | multiply(180, power(subtract(1, divide(60, 180)), 3)) | a beaker contains 180 liters of alcohol . on 1 st day , 60 l of alcohol is taken out and replaced by water . 2 nd day , 60 l of mixture iss taken out and replaced by water and the process continues day after day . what will be the quantity of alcohol in beaker after 3 days | use the formula , final alcohol = initial alcohol Γ ( 1 β replacementquantityfinalvolume ) n Γ ( 1 β replacementquantityfinalvolume ) n final alcohol = 180 ( 1 β 60180 ) 3180 ( 1 β 60180 ) 3 = 180 Γ ( 23 ) 3 = 53.3 ans : a | a = 60 / 180
b = 1 - a
c = b ** 3
d = 180 * c
|
a ) none , b ) one , c ) two , d ) three , e ) four | d | add(add(const_1, const_1), 1) | how many two - digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6 ? | the possible number n can be written as follow : n = multiple of lcm ( 6,10 ) + 1 st such number n = 30 x + 1 possible values = 1 , 31 , 61 , 91 answer : 3 such 2 digit number . answer : d | a = 1 + 1
b = a + 1
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a ) sec , b ) sec , c ) sec , d ) sec , e ) sec | d | divide(add(110, 150), multiply(60, const_0_2778)) | how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 150 m in length ? | "d = 110 + 150 = 260 m s = 60 * 5 / 18 = 50 / 3 t = 260 * 3 / 50 = 15.6 sec answer : d" | a = 110 + 150
b = 60 * const_0_2778
c = a / b
|
a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 37 kmph | e | divide(add(100, 50), add(divide(100, 30), divide(50, 70))) | a car travels uphill at 30 km / hr and downhill at 70 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 60 = 5 / 7 ; avg speed = 150 / ( 10 / 3 + 5 / 7 ) = 37 kmph answer : e | a = 100 + 50
b = 100 / 30
c = 50 / 70
d = b + c
e = a / d
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a ) 515 . , b ) 545 . , c ) 618 . , d ) 687 . , e ) 666 . | d | add(226, 423) | in the faculty of reverse - engineering , 226 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 75 % of the total ? | "answer is d : 687 solution : total number of students studying both are 423 + 226 - 134 = 515 ( subtracting the 134 since they were included in the both the other numbers already ) . so 75 % of total is 515 , so 100 % is approx . 687 ." | a = 226 + 423
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a ) 550 , b ) 245 , c ) 150 , d ) 35 , e ) 350 | a | divide(77, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 77 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 77 , so 0.14 * { total } = 77 - - > { total } = 550 . answer : a ." | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 77 / d
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a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | d | add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4) | if x = 1 + β 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 1 ? | "answer x = 1 + β 2 β΄ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 1 = x 2 ( x - 2 ) 2 + 1 = ( 1 + β 2 ) 2 ( 1 + β 2 - 2 ) 2 + 1 = ( β 2 + 1 ) 2 ( β 2 - 1 ) 2 + 1 = [ ( β 2 ) 2 - ( 1 ) 2 ] 2 + 1 = ( 2 - 1 ) 2 = 1 + 1 = 2 correct option : d" | a = math.sqrt(2)
b = 1 + a
c = b ** 2
d = math.sqrt(2)
e = 1 + d
f = e - 2
g = f ** 2
h = c * g
i = h + 4
|
a ) 54 , b ) 72 , c ) 96 , d ) 144 , e ) 432 | a | multiply(divide(factorial(9), factorial(8)), divide(factorial(4), multiply(const_2, const_2))) | vlaudia can choose any two of 4 different candles and any 8 of 9 different flowers for a centerpiece arrangement . given these choices , how many candle + flower groupings can she select ? | vlaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement . 2 candles from 4 candles : 4 c 2 : 6 8 flowers from 9 flowers : 9 c 8 : 9 so total grouping 9 * 6 = 54 . answer is a . | a = math.factorial(9)
b = math.factorial(8)
c = a / b
d = math.factorial(4)
e = 2 * 2
f = d / e
g = c * f
|
a ) 13 , b ) 14 , c ) 5 , d ) 6 , e ) 7 | c | subtract(subtract(divide(divide(100, const_2), const_2), const_10), const_10) | how many odd integers from 1 to 100 ( both inclusive ) have odd number of factors ? | integers having odd number of factors will be perfect squares . odd numbers will have odd perfect squares . thus , the possible values for the perfect squares are : 1 , 9,25 , 49,81 and the corresponding integers are 1 , 3,5 , 7,9 ( more than 4 ) . thus c is the correct answer . | a = 100 / 2
b = a / 2
c = b - 10
d = c - 10
|
a ) 1.5 , b ) 1.6 , c ) 2.5 , d ) 1.8 , e ) 1.9 | c | multiply(divide(25, const_100), 10) | how many litres of pure acid are there in 10 litres of a 25 % solution | explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 10 * 25 / 100 = 2.5 answer : option c | a = 25 / 100
b = a * 10
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a ) 105 , b ) 110 , c ) 115 , d ) 120 , e ) 125 | a | divide(factorial(15), multiply(factorial(subtract(15, const_2)), factorial(const_2))) | there are 15 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "15 c 2 = 105 the answer is a ." | a = math.factorial(15)
b = 15 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 80 , b ) 138 , c ) 100 , d ) 129 , e ) 85 | e | divide(add(6500, 2000), add(65, 35)) | rahim bought 65 books for rs . 6500 from one shop and 35 books for rs . 2000 from another . what is the average price he paid per book ? | "average price per book = ( 6500 + 2000 ) / ( 65 + 35 ) = 8500 / 100 = rs . 85 answer : e" | a = 6500 + 2000
b = 65 + 35
c = a / b
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a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | multiply(divide(12, 13), divide(12, 13)) | at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 10 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 3 to 13 . how many cooks does the restaurant have ? | "originally there were 3 k cooks and 10 k waiters . 13 k = 10 k + 12 k = 4 there are 12 cooks . the answer is d ." | a = 12 / 13
b = 12 / 13
c = a * b
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a ) s . 800 , b ) s . 85 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | multiply(divide(subtract(multiply(9, 800), multiply(multiply(const_3, const_4), 400)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 800 and a uniform after one year service . he served only for 9 months and received uniform and rs . 400 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 800 = 600 400 - - - - - - - - - - - - - 200 1 / 4 - - - - - - - - 200 1 - - - - - - - - - ? = > rs . 800 answer : a" | a = 9 * 800
b = 3 * 4
c = b * 400
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
|
a ) $ 4.65 , b ) $ 4.10 , c ) $ 3.60 , d ) $ 3.20 , e ) $ 2.80 | e | divide(1.96, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | a customer purchased a package of ground beef at a cost of $ 1.96 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 30 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.96 . let x be the price per pound for the steak . then 0.7 x = 196 x = 196 / 0.7 = $ 2.80 the answer is e ." | a = const_0_25 * 2
b = const_0_33 * 1
c = a + b
d = 1 / 96
|
a ) $ 28 , b ) $ 82 , c ) $ 110 , d ) $ 138 , e ) $ 197 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(7, const_100)), multiply(add(80, 7), 150)), 150) | a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 7 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "by the question , the equation would be 150 p - 87 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 197 . answer e for me ." | a = 4 * 4
b = a * 1000
c = 7 * 100
d = b + c
e = 80 + 7
f = e * 150
g = d + f
h = g / 150
|
a ) 92 , b ) 84 , c ) 24 , d ) 60 , e ) 184 | b | divide(multiply(24, 36), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 24 and 36 ? | "prime factorization of 24 = 2 x 2 x 2 x 3 prime factorization of 36 = 2 x 2 x 3 x 3 gcf = 12 lcm = 72 sum = 84 answer b ." | a = 24 * 36
b = a / 4
|
a ) 125 miles , b ) 225 miles , c ) 325 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 5) | a car travels at a speed of 65 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 65 + 65 + 65 + 65 + 65 = 5 Γ 65 = 325 miles correct answer is c ) 325 miles" | a = 65 * 5
|
a ) 1 / 140 , b ) 1 / 180 , c ) 27 / 410 , d ) 47 / 120 , e ) 57 / 120 | d | add(add(divide(1, 6), divide(1, 10)), divide(1, 8)) | in a race where 12 cars are running , the chance that car x will win is 1 / 6 , that y will win is 1 / 10 and that z will win is 1 / 8 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 6 + 1 / 10 + 1 / 8 = 47 / 120 answer : d" | a = 1 / 6
b = 1 / 10
c = a + b
d = 1 / 8
e = c + d
|
a ) 1 / 6 , b ) 2 / 9 , c ) 6 / 11 , d ) 9 / 16 , e ) 3 / 4 | a | multiply(divide(subtract(12, 3), 12), divide(subtract(subtract(12, 3), const_1), subtract(12, const_1))) | in a box of 12 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that both pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 3 defective in a lot of 12 pens i . e . probability of first pen being defective = ( 1 / 3 ) i . e . probability of second pen being defective = ( 1 / 2 ) probability of both pen being defective = ( 1 / 2 ) * ( 1 / 3 ) = 1 / 6 answer : option a" | a = 12 - 3
b = a / 12
c = 12 - 3
d = c - 1
e = 12 - 1
f = d / e
g = b * f
|
a ) 52 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | d | divide(multiply(4.2, const_60), 5) | the timing of a college is from 12 p . m to 4.20 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | explanation : total time a student spends in college = 4 hours 20 minutes = 260 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 260 β 20 ) / 5 = 48 minutes answer d | a = 4 * 2
b = a / 5
|
a ) 1200 , b ) 50 , c ) 2400 , d ) 100 , e ) 600 | c | divide(1200, subtract(const_1, divide(50, const_100))) | after decreasing 50 % in the price of an article costs rs . 1200 . find the actual cost of an article ? | "cp * ( 50 / 100 ) = 1200 cp = 24 * 100 = > cp = 2400 answer : c" | a = 50 / 100
b = 1 - a
c = 1200 / b
|
a ) 112 % , b ) b . 100 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(60, const_100))) | mary ' s income is 60 % more than tim ' s income and tim ' s income is 30 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.7 = 70 m = 70 * 1.6 = 112 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 112 * 100 / 100 = 112 ans : a" | a = 100 - 30
b = 100 - 30
c = 60 / 100
d = b * c
e = a + d
|
a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 30 | d | subtract(power(3, const_3), const_1) | a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 5 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | we have 3 places ( - - - ) to fill & we can use only digits 2 , 4 , 5 first place can be filled in 3 ways 2 nd place can be filled in 3 ways because we can repeat the digits 3 rd place can be filled in 3 ways because we can repeat the digits total no of codes = 3 x 3 x 3 = 27 but one combination ( 555 ) will not be even so total no of even codes = 27 - 1 = 26 answer d | a = 3 ** 3
b = a - 1
|
a ) 345652 , b ) 547678 , c ) 356789 , d ) 778990 , e ) 859622 | e | subtract(859622, reminder(859622, 456)) | which no . need to add to 859622 to get a number exactly divisible by 456 ? | "dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 β 62 = 394 should be added to 859622 . e" | a = 859622 - reminder
|
a ) 228 , b ) 108 , c ) 1266 , d ) 75 , e ) 211 | d | multiply(divide(250, 12), const_3_6) | a 250 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 250 / 12 * 18 / 5 = 75 kmph answer : d" | a = 250 / 12
b = a * const_3_6
|
a ) 2.3 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | a | divide(120, multiply(185, const_0_2778)) | in what time will a train 120 m long cross an electric pole , it its speed be 185 km / hr ? | "speed = 185 * 5 / 18 = 51 m / sec time taken = 120 / 51 = 2.3 sec . answer : a" | a = 185 * const_0_2778
b = 120 / a
|
a ) 25 % , b ) 33 % , c ) 29 % , d ) 55 % , e ) 45 % | b | subtract(multiply(divide(const_100, 750), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 750 grams per kg , what is his percent ? | "750 - - - 250 100 - - - ? = > 33.33 % answer : b" | a = 100 / 750
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
|
a ) 10 : 7 , b ) 9 : 8 , c ) 4 : 5 , d ) 13 : 11 , e ) 14 : 8 | c | divide(subtract(25, 21), subtract(21, 16)) | what ratio must a shopkeeper mix peas and soybean of rs . 16 and rs . 25 / kg , as to obtain a mixture of rs . 21 ? | "correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 4 : 5" | a = 25 - 21
b = 21 - 16
c = a / b
|
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent e should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked e will be 25 % greater than the original number . the answer is c ." | a = 10 * 4
b = 100 - 20
c = b / 100
d = c * 10
e = a / d
f = e - 4
g = f / 4
h = g * 100
|
['a ) 3 : 2', 'b ) 3 : 4', 'c ) 4 : 5', 'd ) 2 : 3', 'e ) 2 : 5'] | b | divide(rectangle_area(15, 20), square_area(20)) | a rectangle has 20 meter length and 15 meter breadth . find ratio between area of rectangle and square whose side is same with rectangle ' s length . | area of rectangle = 20 * 15 = 300 area of square = 20 ^ 2 = 400 ratio of rectangle ' area and square ' s area = 300 : 400 = 3 : 4 answer : b | a = rectangle_area / (
|
a ) 9 , b ) 11 , c ) 15 , d ) 5 , e ) 20 | a | subtract(43, multiply(17, const_2)) | a number when divided by 221 gives a remainder 43 , what remainder will be obtained by dividing the same number 17 ? | 221 + 43 = 264 / 17 = 9 answer a | a = 17 * 2
b = 43 - a
|
a ) 21 , b ) 30 , c ) 20 , d ) 16 , e ) 12 | a | multiply(subtract(subtract(const_1, divide(8, const_100)), divide(20, 50)), 50) | george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 8 % of the money for taxi fare and buys 20 mangoes . how many oranges can he buy ? | "let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 8 % of 200 = 16 for taxi fare , so he is left with 184 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 84 ( 184 - 100 ) no of oranges he can buy = 84 / 4 = > 21 so , george can buy 21 oranges . a" | a = 8 / 100
b = 1 - a
c = 20 / 50
d = b - c
e = d * 50
|
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