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 ) 19.6 , b ) 20.6 , c ) 21.6 , d ) 22.6 , e ) 23.6 | b | subtract(divide(multiply(multiply(1.5, const_0_2778), const_100), const_2), divide(multiply(1.5, const_0_2778), const_2)) | jack and paul are running 1000 m . jack runs 1.5 km / hr faster than paul but also finishes the race 1.5 sec earlier . what is the speed of jack ? ? | distance ( given ) 1000 m paul speed ( let ) = x m / s time ( paul take ) = 1000 / x sec jack speed = x + 1.5 * ( 5 / 18 ) m / s tym ( jack ) = 1000 / ( x + 7.5 / 18 ) s tym ( paul ) - tym ( jack ) = 1.5 solving for x we get 20.20 speed of jack is 20.20 + 7.5 / 18 = 20.61 answer : b | a = 1 * 5
b = a * 100
c = b / 2
d = 1 * 5
e = d / 2
f = c - e
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 2 | a | divide(subtract(multiply(20, 26), multiply(20, 14)), add(26, 14)) | the speed of boat in still water is 20 km / hr . if it travels 26 km downstream and 14 km upstream in same time , what is the speed of the stream ? | explanation : let x be speed of stream speed of boat downstream = ( 20 + x ) speed of boat upstream = ( 20 – x ) it is given that boat takes same time to travel 26 km downstream and 14 km upstream [ 26 / ( 20 + x ) ] = [ 14 / ( 20 – x ) ] 520 – 26 x = 280 + 14 x 40 x = 240 x = 6 km / hr answer : a | a = 20 * 26
b = 20 * 14
c = a - b
d = 26 + 14
e = c / d
|
a ) 145 cm , b ) 140 cm , c ) 142 cm , d ) 138 cm , e ) 135 cm | b | subtract(147, divide(multiply(147, 5), const_100)) | on my sister ' s birthday , she was 147 cm in height , having grown 5 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.05 x = 147 x = 140 the answer is b ." | a = 147 * 5
b = a / 100
c = 147 - b
|
a ) 65 , b ) 75 , c ) 60 , d ) 55 , e ) 50 | b | multiply(divide(15, const_2), 10) | if the sum of the 4 th term and the 12 th term of an arithmetic progression is 10 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 10 i . e . , ( a + 3 d ) + ( a + 11 d ) = 10 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 10 - - - - - - - - - - - - - - - from ( 1 ) = 75 answer : b" | a = 15 / 2
b = a * 10
|
a ) 80 , b ) 75 , c ) 56 , d ) 50 , e ) 105 | b | add(100, divide(subtract(65, 70), subtract(divide(70, 100), divide(50, 100)))) | during a certain season , a team won 65 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ? | "we are first given that a team won 65 percent of its first 100 games . this means the team won 0.65 x 100 = 65 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 equals the number of remaining games in the season . thus we can say : 0.5 ( t – 100 ) = number of wins for remaining games 0.5 t – 50 = number of wins for remaining games lastly , we are given that team won 70 percent of all games played in the season . that is , they won 0.7 t games in the entire season . with this we can set up the equation : number of first 100 games won + number of games won for remaining games = total number of games won in the entire season 65 + 0.5 t – 50 = 0.7 t 15 = 0.2 t 150 = 2 t 75 = t answer is b ." | a = 65 - 70
b = 70 / 100
c = 50 / 100
d = b - c
e = a / d
f = 100 + e
|
a ) 4 / 9 , b ) 2 4 / 9 , c ) 3 4 / 9 , d ) 4 4 / 9 , e ) none of these | d | divide(const_1, add(divide(const_1, 8), divide(const_1, 10))) | worker a takes 8 hours to do a job . worker b takes 10 hours to do a job . how long should it take both a and b , working together to do same job . | "explanation : in this type of questions , first we need to calculate 1 hours work , then their collective work as , a ' s 1 hour work is 1 / 8 b ' s 1 hour work is 1 / 10 ( a + b ) ' s 1 hour work = 1 / 8 + 1 / 10 = 9 / 40 so both will finish the work in 40 / 9 hours = 4 4 / 9 option d" | a = 1 / 8
b = 1 / 10
c = a + b
d = 1 / c
|
a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9 | a | divide(divide(add(18, 12), const_2), const_2) | a man can row downstream at 18 kmph and upstream at 12 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 12 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 30 = > x = 15 , y = 3 . answer : a" | a = 18 + 12
b = a / 2
c = b / 2
|
a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 7 / 15 | e | divide(subtract(60, 25), subtract(const_100, 25)) | the weight of a glass of jar is 25 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 25 g weight of coffee beans = 75 g weight of jar and remaining beans = 60 g weight of remaining beans = 35 g fraction remaining = 35 / 75 = 7 / 15 answer is e ." | a = 60 - 25
b = 100 - 25
c = a / b
|
a ) 2000 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2500), 16), add(multiply(16, 16), 4)) | one ton has 2500 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2500 pound = 13 * 2500 * 16 ounces . hence the answer is ( 13 * 2500 * 16 ) / 260 = 2000 . answer : a ." | a = 13 * 2500
b = a * 16
c = 16 * 16
d = c + 4
e = b / d
|
a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 70.20 % , e ) none of these | d | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(28, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 28 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | girls = 900 , boys = 1100 passed = ( 28 % of 1100 ) + ( 32 % of 900 ) = 308 + 288 = 596 failed = 2000 - 596 = 1404 failed % = [ ( 1404 / 2000 ) x 100 ] % = 70.2 % . answer : d | a = 100 * 10
b = 2 * a
c = 28 / 100
d = 100 * 10
e = 2 * d
f = e - 900
g = c * f
h = b - g
i = 32 / 100
j = i * 900
k = h - j
l = 100 * 10
m = 2 * l
n = k / m
o = n * 100
|
a ) s 1200 , b ) s 1500 , c ) s 1600 , d ) s 2000 , e ) s 1700 | d | multiply(divide(1500, 3), subtract(7, 3)) | an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 3 : 7 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | "explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 3 x and r = 7 x . then , 5 x = 1500 ? x = 500 . p = 1500 , q = 1500 and r = 3500 . hence , ( r - p ) = ( 3500 - 1500 ) = 2000 answer : d" | a = 1500 / 3
b = 7 - 3
c = a * b
|
a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 16 | c | divide(add(subtract(529, multiply(divide(19, 23), 529)), const_4), add(const_4, const_4)) | a , b and c are employed to do apiece of work for rs . 529 . a and c are supposed to finish 19 / 23 of the work together . how much shall be paid to b ? | c 12 12 men can complete a work in 18 days . six days after they started working , 4 more men joined them . how many days will all of them together complete the remaining work ? | a = 19 / 23
b = a * 529
c = 529 - b
d = c + 4
e = 4 + 4
f = d / e
|
a ) 2898 , b ) 277 , c ) 310 , d ) 297 , e ) 435 | c | divide(990, multiply(subtract(78, 1), const_0_2778)) | a train 990 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 990 + x ) / 60 = 65 / 3 x = 310 m . answer : c" | a = 78 - 1
b = a * const_0_2778
c = 990 / b
|
a ) 6.80 days , b ) 4.80 days , c ) 3.80 days , d ) 2 days , e ) 1.80 days | e | divide(multiply(multiply(2, 3), 30), multiply(20, 5)) | calculate how many days it will take for 5 boys to paint a 30 m long fence if 2 boys can paint a 20 m long fence in 3 days , | "the length of wall painted by one boy in one day = 20 / 2 * 1 / 3 = 3.33 m no . of days required to paint 50 m cloth by 8 boys = 30 / 5 * 1 / 3.33 = 1.80 days . e" | a = 2 * 3
b = a * 30
c = 20 * 5
d = b / c
|
a ) 1980 , b ) 1982 , c ) 1930 , d ) 1988 , e ) 1680 | a | multiply(add(divide(48, 69), 28), 69) | find the value of ( 28 + 48 / 69 ) × 69 | "= ( 28 + 48 / 69 ) × 69 = ( 1932 + 48 ) / 69 × 69 = 1980 / 69 × 69 = 1980 answer is a ." | a = 48 / 69
b = a + 28
c = b * 69
|
a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | e | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 20 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the lowest possible sale price ? | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 50 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , e is the answer" | a = 80 / 2
|
a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 31 / 45 , e ) 11 / 15 | d | multiply(add(multiply(4, 3), 1), multiply(divide(1, 3), divide(1, 5))) | a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 4 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the purposes of renovating the hospital ' s original wing , all of the patients in the hospital ' s original beds must be transferred to beds in the new tower . if patients are neither admitted nor discharged during the transfer , what fraction of the beds in the new tower will be unoccupied once the transfer is complete ? | "i think d - 31 / 45 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 4 x so old = x , new = 3 x now 1 / 3 of x are occupied and 1 / 5 of 3 x are occupied which simplifies to ( 3 / 5 ) x we are shifting 1 / 3 of x to the new ward so there will now be : 1 / 3 of x plus 3 / 5 of x occupied in the new ward . add them up to get 14 / 15 of x there are 3 x beds in new tower so ratio is : ( 14 / 15 ) x / 3 x = 14 / 45 of x subtract that from 45 / 45 of x and you get the number of un - occupied beds to total capacity of new tower = 31 / 45 d" | a = 4 * 3
b = a + 1
c = 1 / 3
d = 1 / 5
e = c * d
f = b * e
|
a ) 131 , b ) 135 , c ) 139 , d ) 152 , e ) 188 | d | add(multiply(divide(subtract(212, 32), const_3), const_2), 32) | if jake loses 32 pounds , he will weigh twice as much as his sister . together they now weigh 212 pounds . what is jake ’ s present weight , in pounds ? | lets say j is the weight of jack and s is the wt of his sister . if he loses 32 pounds , he s twice as heavy as his sister . j - 32 = 2 * s also , together they weight 212 pounds j + s = 212 solvong the 2 equation , we get j = 152 pounds ! d | a = 212 - 32
b = a / 3
c = b * 2
d = c + 32
|
a ) 225 km / hr , b ) 50 km / hr , c ) 99 km / hr , d ) 88 km / hr , e ) 66 km / hr | b | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 sec . the speed of the train is ? | "speed of the train relative to man = 125 / 10 = 25 / 2 m / sec . = 25 / 2 * 18 / 5 = 45 km / hr let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 45 = > x = 50 km / hr . answer : b" | a = 5 * const_0_2778
b = a * 5
c = 125 - b
d = c / 5
e = d / const_0_2778
|
a ) 2.6 , b ) 5.8 , c ) 6.2 , d ) 7.5 , e ) 1.2 | d | divide(multiply(multiply(18, 15), 2), multiply(multiply(6, 4), 3)) | how many cuboids of length 6 m , width 4 m and height 3 m can be farmed from a cuboid of 18 m length , 15 m width and 2 m height . | "( 18 ã — 15 ã — 2 ) / ( 6 ã — 4 ã — 3 ) = 7.5 answer is d ." | a = 18 * 15
b = a * 2
c = 6 * 4
d = c * 3
e = b / d
|
a ) 26 , b ) 72 , c ) 28 , d ) 82 , e ) 27 | c | divide(add(180, 100), multiply(36, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 180 meters long if the speed of the train is 36 kmph ? | "explanation : d = 100 + 180 = 280 s = 36 * 5 / 18 = 10 mps t = 280 / 10 = 28 sec answer : option c" | a = 180 + 100
b = 36 * const_0_2778
c = a / b
|
a ) 3 , b ) 12 , c ) 18 , d ) 24 , e ) 27 | a | divide(divide(36, const_3), const_4) | there are 36 students in a certain geometry class . if one third of the students are boys and one fourth of the boys are under 6 feet tall , how many boys in the class are under 6 feet tall ? | total students = 36 one third of the students are boys = 1 / 3 * 36 = 12 one fourth of the boys are under six feet tall = 1 / 4 * 12 = 3 . . . therefore number of boys in the class under six feet tall = 3 . . . answer a . . . . alternatively . . . number of boys in the class under six feet tall = 1 / 4 of 1 / 3 of total students = 1 / 4 * 1 / 3 * 36 = 3 . . . answer a | a = 36 / 3
b = a / 4
|
a ) 225 cm 2 , b ) 150 cm 2 , c ) 127 cm 2 , d ) 177 cm 2 , e ) 187 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(6, 9)), 30) | find the area of the quadrilateral of one of its diagonals is 30 cm and its off sets 9 cm and 6 cm ? | "1 / 2 * 30 ( 9 + 6 ) = 225 cm 2 answer : a" | a = 1 / 2
b = 6 + 9
c = a * b
d = c * 30
|
a ) 46.2 % , b ) 54.54 % , c ) 63 % , d ) 56.52 % , e ) none of these | d | multiply(divide(subtract(138, add(multiply(12, const_4), multiply(2, multiply(const_2, const_3)))), 138), const_100) | a cricketer scored 138 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | explanation : number of runs made by running = 138 - ( 12 x 4 + 2 x 6 ) = 138 - ( 60 ) = 78 now , we need to calculate 72 is what percent of 138 . = > 78 / 138 * 100 = 56.52 % answer : d | a = 12 * 4
b = 2 * 3
c = 2 * b
d = a + c
e = 138 - d
f = e / 138
g = f * 100
|
['a ) no change', 'b ) 16 % increase', 'c ) 8 % decrease', 'd ) 16 % decrease', 'e ) none of thes'] | d | subtract(const_100, divide(multiply(subtract(const_100, 40), add(const_100, 40)), const_100)) | if the height of a triangle is decreased by 40 % , land its base is increased by 40 % , what will be the effect on its area ? | regd effect = ∣ ∣ 40 − 40 − 40 × 40100 ∣ ∣ % = − 16 % | 40 − 40 − 40 × 40100 | % = − 16 % i . e . , the area will decrease by 16 % answer d | a = 100 - 40
b = 100 + 40
c = a * b
d = c / 100
e = 100 - d
|
a ) 35000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) none | d | multiply(divide(40,000, 2), 3) | p and q invested in a shop . the profits were divided in the ratio of 2 : 3 respectively . if p invested rs . 40,000 , the amount invested by q is : | "suppose q invested rs . y . then , 40000 / y = 2 / 3 or y = [ 40000 x 3 / 2 ] = 60000 . answer d" | a = 40 / 0
b = a * 3
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a ) $ 39.00 , b ) $ 39.10 , c ) $ 39.25 , d ) $ 39.30 , e ) $ 39.20 | e | add(8, multiply(subtract(divide(8, divide(1, 5)), const_1), 0.8)) | if taxi fares were $ 8.00 for the first 1 / 5 mile and $ 0.8 for each 1 / 5 mile there after , then the taxi fare for a 8 - mile ride was | in 8 miles , initial 1 / 5 mile charge is $ 8 rest of the distance = 8 - ( 1 / 5 ) = 39 / 5 rest of the distance charge = 39 ( 0.8 ) = $ 31.2 ( as the charge is 0.8 for every 1 / 5 mile ) = > total charge for 4 miles = 8 + 31.2 = 39.2 answer is e | a = 1 / 5
b = 8 / a
c = b - 1
d = c * 0
e = 8 + d
|
a ) 9000 cubic meters , b ) 10500 cubic meters , c ) 11750 cubic meters , d ) 12000 cubic meters , e ) 12500 cubic meters | d | multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1)))) | with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity q of the pool ? | "d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) / ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or , 120 x = 80 ( x + 50 ) . or x = 100 . hence , the capacity of pool = 120 x = 12000 cubic meters ." | a = 48 / const_60
b = 4 + 1
c = a * b
d = const_60 * c
e = 50 * d
|
a ) 1890 , b ) 1953 , c ) 3780 , d ) 3843 , e ) 3906 | a | divide(factorial(63), multiply(factorial(subtract(63, const_2)), factorial(const_2))) | how many diagonals does a 63 - sided convex polygon have ? | "a 63 - sided convex polygon has 63 vertices . if we examine a single vertex , we can see that we can connect it with 60 other vertices to create a diagonal . note : there are 60 options because we ca n ' t connect the vertex to itself , and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 63 vertices can be connected with 60 vertices to create a diagonal then . . . . . . the total number of diagonals = ( 63 ) ( 60 ) = 3780 however , before we select answer choice c , we must recognize that we have counted every diagonal twice . for example , we might connect vertex a with vertex f and count that as 1 diagonal , and at the same time we connect vertex f with vertex a and count that as 1 diagonal . of course these diagonals are the same . to account for counting each diagonal twice , we must divide 3780 by 2 to get : 1890 answer : a" | a = math.factorial(63)
b = 63 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 20 , e ) e ) 25 | c | add(multiply(divide(add(50000, 20000), subtract(subtract(100000, add(50000, 20000)), 20000)), const_2), const_1) | in a graduating class , the difference between the highest and lowest salaries is $ 100000 . the median salary is $ 50000 higher than the lowest salary and the average salary is $ 20000 higher than the median . what is the minimum number of students t in the class ? | the difference between the highest and lowest salaries is $ 100000 . so there are at least 2 people - say one with salary 0 and the other with 100 k . no salary will be outside this range . median = 50 k more than lowest . so median is right in the center of lowest and highest since lowest and highest differ by 100 k . in our example , median = 50 k . since there are more than 2 people , there would probably be a person at 50 k . mean = 20 k more than median so in our example , mean salary = 70 k on the number line , 0 . . . . . . . . 50 k ( median ) . . . . . . . . 100 k mean = 70 k so there must be people more toward 100 k to bring the mean up to 70 k . since we want to add minimum people , we will add people at 100 k to quickly make up the right side deficit . 0 and 50 k are ( 70 k + 20 k ) = 90 k away from 70 k . 100 k is 30 k away from 70 k . to bring the mean to 70 k , we will add two people at 100 k each to get : 0 . . . . 50 k . . . . . 100 k , 100 k , 100 k but when we add more people to the right of 70 k , the median will shift to the right . we need to keep the median at 50 k . so every time we add people to the right of 70 k , we need to add people at 50 k too to balance the median . 50 k is 20 k less than 70 k while 100 k is 30 k more than 70 k . to keep the mean same , we need to add 2 people at 100 k for every 3 people we add at 50 k . so if we add 3 people at 50 k and 2 people at 100 k , we get : 0 , . . . 50 k , 50 k , 50 k , 50 k , . . . 100 k , 100 k , 100 k , 100 k , 100 k the median is not at 50 k yet . add another 3 people at 50 k and another 2 at 100 k to get 0 , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k now the median is 50 k and mean is 70 k . total number of people is 15 . answer ( c ) | a = 50000 + 20000
b = 50000 + 20000
c = 100000 - b
d = c - 20000
e = a / d
f = e * 2
g = f + 1
|
a ) 18 , b ) 20 , c ) 16 , d ) 22 , e ) 24 | c | multiply(divide(10, const_100), 10) | uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 20 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 10 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 90 % * total = 36 = > ( 9 / 10 ) total = 36 = > total = 360 / 9 = > total = 40 oz , from there , you must find 10 % of the total 40 oz of the mixture . 10 % * total = > ( 1 / 10 ) ( 40 ) = 4 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 20 - 4 = 16 oz chocolate left over . answer : c" | a = 10 / 100
b = a * 10
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a ) 0 , b ) 2 , c ) 4.8 , d ) 4.67 , e ) 10 | c | divide(subtract(64, subtract(multiply(4, divide(10, 2)), 10)), 10) | the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 12 x + 9 . if the sum of the perimeters of both squares is 64 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 5 ) ^ 2 a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 l 2 = 2 x - 3 p 1 = 4 ( x + 5 ) p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 64 4 ( x + 5 ) + 4 ( 2 x - 3 ) = 64 . . . . . . . . . . . . > x = 4.8 answer : c" | a = 10 / 2
b = 4 * a
c = b - 10
d = 64 - c
e = d / 10
|
a ) 70 sec , b ) 60 sec , c ) 82 sec , d ) 75 sec , e ) 62 sec | d | divide(add(100, 1400), multiply(72, const_0_2778)) | how long does a train 100 m long traveling at 72 kmph takes to cross a tunnel of 1400 m in length ? | d = 1400 + 100 = 1500 m s = 72 * 5 / 18 = 20 t = 1500 * 1 / 20 = 75 sec answer : d | a = 100 + 1400
b = 72 * const_0_2778
c = a / b
|
a ) 48 , b ) 60 , c ) 72 , d ) 84 , e ) 96 | a | divide(60, divide(add(divide(60, 30), divide(60, 120)), const_2)) | a person walks from one end to the other of a 60 - meter long moving walkway at a constant rate in 30 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 120 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | "let v be the speed of the person and let x be the speed of the walkway . 30 ( v + x ) = 60 then 120 ( v + x ) = 240 120 ( v - x ) = 60 when we add the two equations : 240 v = 300 v = 5 / 4 time = 60 / ( 5 / 4 ) = 48 seconds the answer is a ." | a = 60 / 30
b = 60 / 120
c = a + b
d = c / 2
e = 60 / d
|
a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | c | divide(subtract(add(add(2.50, 5.00), multiply(0.25, 22)), 2.50), 0.25) | mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 22 miles , how many miles was mike ' s ride ? | "the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 22 ) = $ 13 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 13 0.25 * x = 10.5 x = 42 miles the answer is c ." | a = 2 + 50
b = 0 * 25
c = a + b
d = c - 2
e = d / 0
|
a ) 4 , b ) 7.6 , c ) 6.6 , d ) 7.6 , e ) 9.6 | c | divide(multiply(250, 85), 3200) | what is the dividend from tony ' s stock , if he invested rs . 3200 at 85 to obtain an income of rs . 250 . | by investing rs . 3200 , income = rs . 250 by investing rs . 85 , income = 250 ã — 85 / 3200 = 6.6 ie , dividend = 6.6 % answer is c . | a = 250 * 85
b = a / 3200
|
a ) 11 , b ) 21 , c ) 28 , d ) 42 , e ) none of these | b | multiply(add(const_1, const_2), 14) | a is twice as good a workman as b and together they finish a piece of work in 14 days . the number of days taken by a alone to finish the work is : | "( a ' s 1 day ' s work ) : ( b ' s 1 day ' s work ) = 2 : 1 ( a + b ) ' s 1 day ' s work = 1 / 14 divide 1 / 14 in the ratio 2 : 1 a ' s 1 day ' s work = ( 1 / 14 x 2 / 3 ) = 1 / 21 hence , a alone can finish the work in 21 days . correct option : b" | a = 1 + 2
b = a * 14
|
a ) 23 , b ) 25 , c ) 27 , d ) 29 , e ) 31 | b | add(add(divide(subtract(160, 40), 5), const_1), const_1) | a test has 160 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain group of test - takers attempted different numbers of questions , but each test - taker still received the same net score of 40 . what is the maximum possible number of such test - takers ? | a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 points , a test taker must have at least 40 correct answers . then c = > 40 . for every correct question above 40 , the test taker has 4 incorrect answers . then , the i = 4 * ( c - 40 ) . also , i + c < = 160 . thus 5 c < = 320 and so c < = 64 . then 40 < = c < = 64 and c can have 25 possible values . the answer is b . | a = 160 - 40
b = a / 5
c = b + 1
d = c + 1
|
['a ) 7.50', 'b ) 8.75', 'c ) 7.75', 'd ) 9.50', 'e ) 9.75'] | e | multiply(65, divide(15, 100)) | a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . if 65 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | we are given that a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of glucose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of glucose would be in the container when we have 45 cubic centimeters of solution , we can set up a proportion . we can say : “ 15 grams of glucose is to 100 cubic centimeters of solution as x grams of glucose is to 65 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 15 / 100 = x / 65 when we cross multiply we obtain : ( 15 ) ( 65 ) = 100 x 975 = 100 x 9.75 = x there are 9.75 grams of glucose in the solution in the container . the answer is e . | a = 15 / 100
b = 65 * a
|
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | e | multiply(divide(150, 500), const_100) | the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 150 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 150 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 150 ) * 100 / 500 = 30 % answer : option e" | a = 150 / 500
b = a * 100
|
a ) 11 , b ) 28 , c ) 39 , d ) 40 , e ) 43 | d | divide(multiply(subtract(const_100, 4), 60), add(const_100, 44)) | if a man lost 4 % by selling oranges at the rate of 60 a rupee at how many a rupee must he sell them to gain 44 % ? | "96 % - - - - 60 144 % - - - - ? 96 / 144 * 60 = 40 answer : d" | a = 100 - 4
b = a * 60
c = 100 + 44
d = b / c
|
a ) 6.24 km , b ) 6 km , c ) 8.84 km , d ) 5.66 km , e ) 10 km | c | multiply(divide(multiply(add(9, 1.2), subtract(9, 1.2)), add(add(9, 1.2), subtract(9, 1.2))), const_2) | a man can row 9 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 9 s = 1.2 ds = 10.2 us = 7.8 x / 10.2 + x / 7.8 = 1 x = 4.42 d = 4.42 * 2 = 8.84 answer : c" | a = 9 + 1
b = 9 - 1
c = a * b
d = 9 + 1
e = 9 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 158 m , b ) 150 m , c ) 110 m , d ) 130 m , e ) 157 m | b | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 15 sec and a platform 100 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 100 ) / 25 = x / 15 = > x = 150 m . answer : b" | a = 2 - 1
b = 100 * a
|
a ) 10.6 , b ) 27 , c ) 10.4 , d ) 10.8 , e ) 10.1 | b | divide(add(300, 450), multiply(add(60, 40), const_0_2778)) | two trains 300 m and 450 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 300 + 450 = 750 m . required time = 700 * 9 / 250 = 27 sec . answer : b" | a = 300 + 450
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 33 4 / 3 % , b ) 33 1 / 7 % , c ) 50 % , d ) 32 1 / 3 % , e ) 60 % | c | divide(multiply(3, const_100), 6) | a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 3 a rupee . find his net profit or loss percent ? | the total number of mangoes bought by the shopkeeper be 18 . if he buys 6 a rupee , his cp = 3 he selling at 3 a rupee , his sp = 6 profit = sp - cp = 6 - 3 = 3 profit percent = 3 / 6 * 100 = 50 % answer : c | a = 3 * 100
b = a / 6
|
a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 40 % , e ) 250 % | c | multiply(divide(180, 45), const_100) | what percent of 180 is 45 ? | "here ' s an algebraic approach . let p = the required percent . this means p / 100 = the percent as a fraction . so , ` ` what percent of 180 is 45 ? ' ' becomes ` ` ( p / 100 ) of 180 = 45 or : ( p / 100 ) ( 180 ) = 45 divide both sides by 180 to get : p / 100 = 45 / 180 simplify right side : p / 100 = 1 / 4 multiply both sides by 100 to get : p = 25 answer : c" | a = 180 / 45
b = a * 100
|
a ) 65 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 42 mtr . | e | multiply(112, divide(multiply(15, 3), multiply(20, 6))) | if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 15 men in 3 days ? | "20 men is 6 days can build 112 metres 15 men in 3 days can build = 112 * ( 15 / 20 ) x ( 3 / 6 ) = 42 meters answer : e ." | a = 15 * 3
b = 20 * 6
c = a / b
d = 112 * c
|
a ) 2.31 days , b ) 4.31 days , c ) 5.31 days , d ) 6.31 days , e ) 7.31 days | a | divide(multiply(10, 3), add(3, 10)) | b completes a work in 3 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | 1 / 3 + 1 / 10 = 13 / 30 30 / 13 = 2.31 days answer : a | a = 10 * 3
b = 3 + 10
c = a / b
|
a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,000 , e ) s . 7,500 | d | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 20,000 earned at the end of 2 years from the startingof the business ? | a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 20000 x 3 / 10 = rs . 6,000 . d | a = 4 + 1
b = a * 4
c = 3 * 4
d = 2 * c
e = b * d
|
a ) 49.5 kmph , b ) 49.0 kmph , c ) 48.5 kmph , d ) 48.0 kmph , e ) none of these | d | inverse(add(divide(inverse(60), const_2), inverse(80))) | nick took a bus from home to market , that travels at 80 kmph . while walking back at 8 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 60 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 80 , by walking = x / 8 , by cycling = x / 60 hours : . average speed = total distance / total time = 8 x / x / 40 + x / 8 + x / 60 = 8 * 60 / 1.5 + 7.5 + 1 = 12.0 answer : d | a = 1/(60)
b = a / 2
c = 1/(80)
d = b + c
e = 1/(d)
|
a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 180 kg , e ) 110 kg | d | multiply(multiply(multiply(6, 3), divide(1, const_100)), const_1000) | a boat having a length 6 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "volume of water displaced = ( 6 x 3 x 0.01 ) m 3 = 0.18 m 3 . mass of man = volume of water displaced x density of water = ( 0.18 x 1000 ) kg = 180 kg . answer : d" | a = 6 * 3
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 22 , b ) 38 , c ) 62 , d ) 94 , e ) 672 | d | multiply(2, divide(divide(1551, add(const_10, const_1)), const_3)) | a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 1551 , what is the value of n ? | "factorise 1551 . . 3 * 11 * 47 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 47 = 94 . . ans : d" | a = 10 + 1
b = 1551 / a
c = b / 3
d = 2 * c
|
a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.11111 | e | multiply(divide(0.999999, 0.111111), const_100) | 0.999999 + 0.111111 = ? | "0.999999 + 0.111111 = 0.999999 + 0.11111 + 0.000001 = ( 0.999999 + 0.000001 ) + 0.11111 = 1 + 0.11111 = 1.11111 e" | a = 0 / 999999
b = a * 100
|
a ) 30 % gain , b ) 30 % loss , c ) 40 % gain , d ) 60 % loss , e ) 50 % loss | d | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 10)), divide(multiply(const_100, 25), 10))) | if the cost price of 10 articles is same as the selling price of 25 articles . find the gain or loss percentage ? | "explanation : 10 cp = 25 sp cost price cp = 25 selling price sp = 10 formula = ( sp - cp ) / cp * 100 = ( 10 - 25 ) / 25 * 100 = 60 % loss answer : option d" | a = 100 * 25
b = a / 10
c = 100 - b
d = 100 * 25
e = d / 10
f = c / e
g = 100 * f
|
a ) 600 , b ) 784 , c ) 1080 , d ) 1200 , e ) 1440 | b | divide(112, divide(multiply(const_2, subtract(120, 112)), 112)) | julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 112 and $ 120 from the simple interest account and the compound interest account respectively . if the interest rates for both accounts were the same , what was the amount of julie ' s initial savings ? | $ 112 for 2 years = $ 56 per year . extra $ 8 yearned with the compound interest is the percent yearned on percent . so , $ 8 is yearned on $ 56 , which means that the interest = 7 % . this on the other hand means that half of the savings = 56 * 7 = $ 392 . twice of that = $ 784 . answer : b . | a = 120 - 112
b = 2 * a
c = b / 112
d = 112 / c
|
a ) 2.5 % , b ) 40 % , c ) 5 % , d ) 15 % , e ) 25 % | b | multiply(divide(subtract(70, 30), add(70, 30)), const_100) | if 70 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ? | "70 % of ( x - y ) = 30 % of ( x + y ) ( 70 / 100 ) ( x - y ) = ( 30 / 100 ) ( x + y ) 7 ( x - y ) = 3 ( x + y ) 4 x = 10 y x = 5 / 2 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / ( 5 / 2 ) y ) x 100 ) = 40 % answer is b ." | a = 70 - 30
b = 70 + 30
c = a / b
d = c * 100
|
a ) 8120 , b ) 8123 , c ) 8222 , d ) 8322 , e ) 8422 | b | divide(1760, subtract(1, add(add(divide(1, 3), divide(1, 4)), divide(1, 5)))) | a man spends 1 / 3 rd of his salary on food . he spends 1 / 4 th on rent and 1 / 5 th on cloths . if he is left with 1760 , then who much salary does he earn . | "let salary be x total spent = ( x / 3 + x / 4 + x / 5 ) = ( 47 / 60 ) * x left = ( x - 47 x / 60 ) = ( 13 / 60 ) * x given , ( 13 / 60 ) * x = 1760 x = ( 1760 * 60 ) / 13 = 8123.077 = 8123 ( approax ) answer : b" | a = 1 / 3
b = 1 / 4
c = a + b
d = 1 / 5
e = c + d
f = 1 - e
g = 1760 / f
|
a ) 7 , b ) 8 , c ) 9 , d ) 3 , e ) 2 | d | divide(divide(divide(lcm(2, 5476), 5476), const_4), const_4) | what is the least value of x , so that 2 x 5476 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 6 = 24 + x least value of x may be ' 3 ' , so that the total 24 + 3 = 27 is divisible by 9 . answer : option d" | a = math.lcm(2, 5476)
b = a / 5476
c = b / 4
d = c / 4
|
a ) 6 cm , b ) 12 cm , c ) 35 cm , d ) 44 cm , e ) none of these | c | multiply(55, divide(7, 11)) | the length of the ribbon was originally 55 cm . it was reduced in the ratio 11 : 7 . what is its length now ? | explanation : length of ribbon originally = 55 cm let the original length be 11 x and reduced length be 7 x . but 11 x = 55 cm x = 55 / 11 cm = 5 cm therefore , reduced length = 7 x cm = 7 5 cm = 35 cm answer : c | a = 7 / 11
b = 55 * a
|
a ) 19.39 , b ) 19.3 , c ) 19.32 , d ) 19.33 , e ) 19.36 | b | add(13.3, divide(add(multiply(6, divide(const_3, const_2)), 3), subtract(8, 6))) | at 12.00 hours , j starts to walk from his house at 6 kmph . at 13.30 , p follows him from j ' s house on his bicycle at 8 kmph . when will j be 3 km behind p ? | by the time p starts j is 1.5 hr x 6 = 9 km away from his house . j is 3 km behind when p is 3 km ahead of him . ie . , p has to cover 12 km . so he takes 12 / ( 8 - 6 ) = 6 hrs after 13.30 . so the required time is 19.30 hrs answer : b | a = 3 / 2
b = 6 * a
c = b + 3
d = 8 - 6
e = c / d
f = 13 + 3
|
a ) 30 , b ) 50 , c ) 66 , d ) 44 , e ) 48 | e | divide(800, multiply(subtract(65, 5), const_0_2778)) | how many seconds will a 800 m long train take to cross a man walking with a speed of 5 km / hr in the direction of the moving train if the speed of the train is 65 km / hr ? | "speed of train relative to man = 65 - 5 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 800 * 3 / 50 = 48 sec . answer : e" | a = 65 - 5
b = a * const_0_2778
c = 800 / b
|
a ) 140 , b ) 66 , c ) 100 , d ) 99 , e ) 120 | c | add(add(power(add(add(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), const_2))) | the sum of five consecutive even numbers is 500 . find the middle number of the five ? | "middle number = 500 / 5 = 100 ans c" | a = 500 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 500 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 500 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 500 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 12 days , b ) 15 days , c ) 21 / 2 days , d ) 21 days , e ) 22 days | c | inverse(subtract(inverse(6), inverse(14))) | a and b together can do a piece of work in 6 days and a alone can do it in 14 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 14 work in 1 day b alone can do ( 1 / 6 - 1 / 14 ) = 2 / 21 work in 1 day = > complete work can be done in 21 / 2 days by b answer : option c" | a = 1/(6)
b = 1/(14)
c = a - b
d = 1/(c)
|
a ) 1 / 9 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(2, 2), add(multiply(divide(2, 6), const_60), multiply(divide(2, 8), const_60))) | triathlete dan runs along a 2 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 8 miles per hour and swims at a rate of 6 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 4 miles round trip . running part : ( 2 / 8 = 1 / 4 * 60 = 15 minutes ) swimming part : ( 2 / 6 = 1 / 3 * 60 = 20 minutes ) 4 miles in ( 15 + 20 ) minutes 4 / 35 = 1 / 9 mile per minute answer : 1 / 9 mile per minute" | a = 2 + 2
b = 2 / 6
c = b * const_60
d = 2 / 8
e = d * const_60
f = c + e
g = a / f
|
a ) s . 80 , b ) s . 85 , c ) s . 90 , d ) s . 100 , e ) s . 200 | e | multiply(divide(subtract(multiply(9, 600), 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 . 600 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 * 600 = 450 400 - - - - - - - - - - - - - 50 1 / 4 - - - - - - - - 50 1 - - - - - - - - - ? = > rs . 200 answer : e | a = 9 * 600
b = 3 * 4
c = b * 400
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
|
a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 36 | e | divide(multiply(power(12, 1), power(6, 4)), 432) | find the value of y from ( 12 ) ^ 1 x 6 ^ 4 ÷ 432 = y ? | "36 e" | a = 12 ** 1
b = 6 ** 4
c = a * b
d = c / 432
|
a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them | a | add(40, lcm(20, 25,35)) | find the least number which when divided by 20 , 25,35 and 40 leaves remainders 14 , 19,29 and 34 respectively . | "here , ( 20 - 14 ) = 6 , ( 25 – 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 20,25 , 35,40 ) – 6 = 1394 . answer is a ." | a = math.lcm(20, 25)
b = 40 + a
|
a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 5 | c | multiply(8, 1) | if 1 / 8 of a pencil is black , 1 / 2 of the remaining is white and the remaining 3 1 / 2 cm is blue , find the total length of the pencil . | let the total length of the pencil be x cm . black part is 1 / 8 of the length so x / 8 . so the remaining part is x - x / 8 = = > 8 x / 8 - x / 8 = = > 7 x / 8 cm . white part = 1 / 2 * 7 x / 8 = 7 x / 16 cm . so the remaining part is 7 x / 8 - 7 x / 16 = = > 14 x / 16 - 7 x / 16 = 7 x / 16 which is equal to 3 1 / 2 . 7 x / 16 = 7 / 2 cross multiply and we get x = 16 / 2 = 8 cm . so the total length of the pencil is 8 cm . the correct answer is option c ) 8 . | a = 8 * 1
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a ) 21 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | b | subtract(25, add(floor(divide(9, const_2)), const_1)) | marcella has 25 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "greatest number of paired shoes means greatest number of lost pairs 9 shoes have 4 pairs and 1 extra shoes of a pair . so a otal of 5 pairs she loses leftover pair = 20 answer : b" | a = 9 / 2
b = math.floor(a)
c = b + 1
d = 25 - c
|
a ) 724 m , b ) 704 m , c ) 287 m , d ) 278 m , e ) 253.4 m | e | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 180), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 180 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 180 resolutions . = 180 * 2 * 22 / 7 * 22.4 = 25344 cm = 253.4 m answer : e" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 180
i = h / 100
|
a ) 100 , b ) 120 , c ) 140 , d ) 150 , e ) 160 | e | add(multiply(subtract(80, 5), 2), multiply(2, 10)) | if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 80 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day ? | "case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 80 ) = ( r + 5 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 2 ) x = 160 ans e" | a = 80 - 5
b = a * 2
c = 2 * 10
d = b + c
|
a ) rs . 20 , b ) rs . 25 , c ) rs . 30 , d ) rs . 35 , e ) rs . 40 | c | divide(divide(multiply(1500, 20), const_100), 10) | a reduction of 20 % in the price of oil enables a house wife to obtain 10 kgs more for rs . 1500 , what is the reduced price for kg ? | "explanation : 1500 * ( 20 / 100 ) = 300 - - - - 10 ? - - - - 1 = > rs . 30 answer : c" | a = 1500 * 20
b = a / 100
c = b / 10
|
a ) 440 , b ) 488 , c ) 477 , d ) 426 , e ) 422 | a | multiply(20, multiply(99, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 99 km / hr . what is the length of the platform ? | "speed = 99 * 5 / 18 = 27.5 m / sec . length of the train = 27.5 * 20 = 550 m . let the length of the platform be x m . then , ( x + 550 ) / 36 = 27.5 = > x = 440 m . answer : a" | a = 99 * const_0_2778
b = 20 * a
|
a ) 75 , b ) 70 , c ) 85 , d ) 90 , e ) 95 | b | subtract(100, 20) | 100 white and black tiles will be used to form a 20 x 20 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | "answer = b please refer diagram below 90 - 20 = 70" | a = 100 - 20
|
a ) 2 : 3 , b ) 1 : 2 , c ) 4 : 3 , d ) 1 : 1 , e ) 3 : 2 | a | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(const_2.0, const_100)), subtract(divide(6, const_100), divide(multiply(divide(6, const_100), const_2), const_3))) | two numbers a and b are such that the sum of 6 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "explanation : 6 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 6 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) ⇒ 6 a + 4 b = 2 / 3 ( 6 a + 8 b ) ⇒ 18 a + 12 b = 12 a + 16 b ⇒ 6 a = 4 b ⇒ ab = 4 / 6 = 2 / 3 ⇒ a : b = 2 : 3 answer : option a" | a = 8 / 100
b = a * 4
c = b / 3
d = 2 / 0
e = c - d
f = 6 / 100
g = 6 / 100
h = g * 2
i = h / 3
j = f - i
k = e / j
|
a ) 45 , b ) 28 , c ) 65 , d ) 36 , e ) 42 | d | multiply(divide(add(const_1, subtract(const_10, const_1)), const_2), subtract(const_10, const_1)) | if you add all the numbers on your mobile phone except 9 , what is the answer ? | "we have to add 0 to 8 to find the answer . therefore 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 answer is d" | a = 10 - 1
b = 1 + a
c = b / 2
d = 10 - 1
e = c * d
|
a ) 125 , b ) 160 , c ) 200 , d ) 250 , e ) 320 | c | add(add(70, divide(multiply(70, 25), subtract(const_100, add(40, 25)))), divide(multiply(70, 40), subtract(const_100, add(40, 25)))) | a fruit drink is made of orange , watermelon , and grape juice , where 25 percent of the drink is orange juice and 40 percent is watermelon juice . if the drink is made with 70 ounces of grape juice , how many ounces is the drink total ? | "let the total number of ounces in the drink be x . % of orange = 25 % % of watermelon = 40 % % of grape = 100 % - 65 % = 35 % 0.35 x = 70 x = 200 therefore there are a total of 200 ounces in the drink . the answer is c ." | a = 70 * 25
b = 40 + 25
c = 100 - b
d = a / c
e = 70 + d
f = 70 * 40
g = 40 + 25
h = 100 - g
i = f / h
j = e + i
|
a ) 18.75 % , b ) 23 % , c ) 30 % , d ) 50 % , e ) 100 % | b | multiply(divide(subtract(add(const_100, 60), add(const_100, 30)), add(const_100, 30)), const_100) | the wages earned by robin is 30 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ? | let wage of erica = 10 wage of robin = 1.3 * 10 = 13 wage of charles = 1.6 * 10 = 16 percentage by which wage earned by charles is more than that earned by robin = ( 16 - 13 ) / 13 * 100 % = 3 / 13 * 100 % = 23 % answer b | a = 100 + 60
b = 100 + 30
c = a - b
d = 100 + 30
e = c / d
f = e * 100
|
['a ) 868 sq . cm', 'b ) 600 sq . cm', 'c ) 700 sq . cm', 'd ) 900 sq . cm', 'e ) none of these'] | c | multiply(const_2, add(add(multiply(12, 14), multiply(12, 7)), multiply(14, 7))) | find the surface area of a cuboid 12 m long , 14 m broad and 7 m high | explanation : surface area = [ 2 ( 12 x 14 + 14 x 7 + 12 x 7 ) ] cm 2 = ( 2 x 350 ) cm 2 = 700 cm 2 . answer : c | a = 12 * 14
b = 12 * 7
c = a + b
d = 14 * 7
e = c + d
f = 2 * e
|
a ) 86.6 km , b ) 46.6 km , c ) 35.9 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 7), divide(44, const_60)) | the speed of a boat in still water in 42 km / hr and the rate of current is 7 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 7 ) = 49 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 49 = 35.9 km answer : c" | a = 42 + 7
b = 44 / const_60
c = a * b
|
a ) 3 years , b ) 4 years , c ) 6 years , d ) 7 years , e ) 8 years | c | subtract(subtract(divide(60, 5), 3), 3) | the sum of ages of 5 children born at the intervals of 3 years each is 60 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 60 5 x = 30 x = 6 . age of the youngest child = x = 6 years . c )" | a = 60 / 5
b = a - 3
c = b - 3
|
a ) 57.5 litres , b ) 56.67 litres , c ) 57.67 litres , d ) 56.5 litres , e ) 50 litres | e | multiply(100, subtract(const_1, sqrt(divide(9, 36)))) | a 100 - litre mixture of milk and water contains 36 litres of milk . ' x ' litres of this mixture is removed and replaced with an equal quantum of water . if the process is repeated once , then the concentration of the milk stands reduced at 9 % . what is the value of x ? | "working formula . . . initial concentration * initial volume = final concentration * final volume . let x is the part removed from 100 lts . 36 % ( 1 - x / 100 ) ^ 2 = 9 % * 100 % ( 1 - x / 100 ) ^ 2 = 9 / 36 - - - - - - > ( 1 - x / 100 ) ^ 2 = ( 3 / 6 ) ^ 2 100 - x = 300 / 6 x = 50 ans e" | a = 9 / 36
b = math.sqrt(a)
c = 1 - b
d = 100 * c
|
['a ) 3 : 4', 'b ) 3 : 9', 'c ) 3 : 7', 'd ) 3 : 1', 'e ) 3 : 2'] | a | divide(3, 4) | the length of each side of an equilateral triangle having an area of 4 √ 3 cm 2 is ? | let the speed of the train be x km / hr and that of the car be y km / hr . then , 120 / x + 480 / y = 8 or 1 / x + 4 / y = 1 / 15 - - - ( i ) and , 200 / x + 400 / y = 25 / 3 or 1 / x + 2 / y = 1 / 24 - - - ( ii ) solving ( i ) and ( ii ) , we get x = 60 and y = 80 ratio of speeds = 60 : 80 = 3 : 4 answer : a | a = 3 / 4
|
a ) 52 m , b ) 62 m , c ) 72 m , d ) 82 m , e ) 92 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 24)), const_4) | length of a rectangular plot is 24 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 24 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 24 ) + x ] = 200 2 x + 24 = 100 2 x = 76 x = 38 . hence , length = x + 24 = 62 m b" | a = 5300 / 26
b = 2 * 24
c = a + b
d = c / 4
|
a ) 6.125 , b ) 9.125 , c ) 12.125 , d ) 15.125 , e ) 18.125 | d | subtract(add(multiply(7, divide(multiply(2, 2), 7)), 9), multiply(2, power(divide(multiply(2, 2), 7), 2))) | if x is real , find the maximum value of the expression - 2 x ^ 2 + 7 x + 9 . | "this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 7 x + 9 = ( - 2 x + 9 ) ( x + 1 ) the roots are 9 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 1.75 the maximum value is - 2 ( 1.75 ) ^ 2 + 7 ( 1.75 ) + 9 = 15.125 the answer is d ." | a = 2 * 2
b = a / 7
c = 7 * b
d = c + 9
e = 2 * 2
f = e / 7
g = f ** 2
h = 2 * g
i = d - h
|
a ) rs . 6003 , b ) rs . 6029 , c ) rs . 4650 , d ) rs . 6108 , e ) rs . 6011 | c | subtract(multiply(add(1500, 150), add(20, const_1)), multiply(1500, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1500 . if the manager ' s salary is added , then the average salary increases by rs . 150 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1650 * 21 - 1500 * 20 ) = rs . 4650 . answer : c" | a = 1500 + 150
b = 20 + 1
c = a * b
d = 1500 * 20
e = c - d
|
a ) 0.45 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6 | e | subtract(const_1, subtract(add(0.25, 0.3), 0.15)) | the probability of two events a and b are 0.25 and 0.30 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________ | we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 30 - . 15 = . 40 but the probability of neither a nor b = 1 - . 40 = 0.60 answer : e | a = 0 + 25
b = a - 0
c = 1 - b
|
['a ) 5.0', 'b ) 6.0', 'c ) 6.8', 'd ) 5.8', 'e ) 5.5'] | e | subtract(divide(power(6, const_2), const_2), divide(power(5, const_2), const_2)) | the size of a flat - screen tablet is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 6 - inch flat - screen tablet than a square 5 - inch flat - screen tablet ? | if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 6 - inch flat - screen television the diagonal ( hypotenuse ) = 6 so , we can apply the pythagorean theorem to get x ² + x ² = 6 ² simplify : 2 x ² = 6 ² divide both sides by 2 to get : x ² = 6 ² / 2 since the area of the square = x ² , we can see that the area of this square is 6 ² / 2 square 5 - inch flat - screen television the diagonal ( hypotenuse ) = 5 so , we can apply the pythagorean theorem to get x ² + x ² = 5 ² simplify : 2 x ² = 5 ² divide both sides by 2 to get : x ² = 5 ² / 2 since the area of the square = x ² , we can see that the area of this square is 5 ² / 2 difference in areas = 6 ² / 2 - 5 ² / 2 = ( 6 ² - 5 ² ) / 2 = ( 36 - 25 ) / 2 = 11 / 2 = 5.5 e | a = 6 ** 2
b = a / 2
c = 5 ** 2
d = c / 2
e = b - d
|
a ) 33 , b ) 38 , c ) 70 , d ) 123 , e ) 12 | b | add(add(add(divide(subtract(140, const_12), 4), const_2), const_2), const_2) | the sum of 4 consecutive even numbers is 140 . what would be the largest number ? | let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x + 1 ) = 38 . answer : b | a = 140 - 12
b = a / 4
c = b + 2
d = c + 2
e = d + 2
|
a ) 2 / 3 , b ) 1 / 4 , c ) 2 / 5 , d ) 3 / 7 , e ) 2 / 9 | a | divide(subtract(subtract(const_100, 50), multiply(const_3, const_10)), multiply(const_3, const_10)) | in a sample of college students , 50 percent are third year students and 70 percent are not second - year students . what fraction of those students who are not third - year students are second - year students ? | required answer = 40 / 60 . = 2 / 3 answer is a | a = 100 - 50
b = 3 * 10
c = a - b
d = 3 * 10
e = c / d
|
a ) 5.35 % , b ) 6.35 % , c ) 7.35 % , d ) 8.35 % , e ) 9.35 % | b | divide(multiply(subtract(75.50, add(12.50, add(multiply(3, 7.35), multiply(3, 12.05)))), const_100), 75.50) | a camera lens filter kit containing 5 filters sells for $ 75.50 . if the filters are purchased individually , 3 of them are priced at $ 7.35 each , 3 at $ 12.05 each , 1 at $ 12.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ? | "cost of kit = $ 75.50 if filters are purchased individually - $ 7.35 * 3 + $ 12.05 * 3 + $ 12.50 = $ 70.70 amount saved = $ 75.50 - $ 70.70 = $ 4.80 required % age = ( $ 4.80 / $ 75.50 ) * 100 = 6.35 % so , the correct answer is b ." | a = 3 * 7
b = 3 * 12
c = a + b
d = 12 + 50
e = 75 - 50
f = e * 100
g = f / 75
|
a ) 1 / 32 , b ) 1 / 28 , c ) 1 / 24 , d ) 1 / 16 , e ) 1 / 14 | c | divide(const_1, add(20, const_4)) | if a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20 , what is the probability e that this product will be a multiple of 45 ? | "there are 3 single digit multiple of 3 , that is , 3 , 6,9 . there are 8 prime nos less than 20 - 2,3 , 5,7 , 11,13 , 17,19 total outcome - 8 * 3 = 24 favourable outcome = 1 ( 9 * 5 ) hence required probability e = 1 / 24 . answer c ." | a = 20 + 4
b = 1 / a
|
a ) 100 , b ) 120 , c ) 400 , d ) 260 , e ) 380 | c | divide(16, subtract(204.04, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 16 . if n / j = 204.04 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 204.04 here 204 is the quotient . given that remainder = 16 so , 204.04 = 204 + 16 / j so , j = 400 answer : c" | a = 4 * 10
b = a + 2
c = 100 + b
d = 204 - 4
e = 16 / d
|
a ) 360 , b ) 420 , c ) 510 , d ) 320 , e ) 280 | a | lcm(lcm(24, 36), 40) | find the lowest common multiple of 24 , 36 and 40 . | "lcm = 2 * 2 * 2 * 3 * 3 * 5 = 360 . answer is a" | a = math.lcm(24, 36)
b = math.lcm(a, 40)
|
a ) 0.2 % , b ) 2 % , c ) 5 % , d ) 20 % , e ) 500 % | d | multiply(divide(14, 70), const_100) | 14 is what percent of 70 ? | "14 = x * 70 / 100 x = 20 % ans ; d" | a = 14 / 70
b = a * 100
|
a ) 27 , b ) 28 , c ) 19 , d ) 11 , e ) 12 | a | divide(140, add(const_2, const_pi)) | the perimeter of a semi circle is 140 cm then the radius is ? | "36 / 7 r = 140 = > r = 27 answer : a" | a = 2 + math.pi
b = 140 / a
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 35 | b | multiply(divide(divide(3, 5), 4), const_100) | among the members of the malmo family , there are 3 times as many members who do not wear glasses as members who do . half of the malmo family members have blue eyes , and 4 / 5 of the family members who have blue eyes do not wear glasses . what percent of the malmo family members wear glasses and do not have blue eyes ? | number in black are given and in red are calculated . now , since x + 3 x = 100 , then x = 25 and therefore = 25 − 10 = 15 answer : b . | a = 3 / 5
b = a / 4
c = b * 100
|
a ) 227 , b ) 140 , c ) 160 , d ) 120 , e ) 171 | c | multiply(12, multiply(72, const_0_2778)) | a train passes a station platform in 20 sec and a man standing on the platform in 12 sec . if the speed of the train is 72 km / hr . what is the length of the platform ? | "speed = 72 * 5 / 18 = 20 m / sec . length of the train = 20 * 12 = 240 m . let the length of the platform be x m . then , ( x + 240 ) / 20 = 20 = > x = 160 m . answer : c" | a = 72 * const_0_2778
b = 12 * a
|
a ) 09 pm , b ) 07 pm , c ) 03 pm , d ) 05 am , e ) 03 am | c | add(divide(add(870, 75), add(60, 75)), 8) | the distance between two cities a and b is 870 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 870 . therefore , 60 x + 75 ( x - 1 ) = 870 . = > x = 7 . so , they meet at ( 8 + 7 ) i . e , 15 = > 3 pm answer : c )" | a = 870 + 75
b = 60 + 75
c = a / b
d = c + 8
|
a ) - 8 , b ) - 9 , c ) - 10 , d ) - 12 , e ) - 16 | d | subtract(10, subtract(38, 12)) | apple , orange , mango . mango + apple = 12 . mango + orange = 10 . mango + apple + orange = 38 . calculate the no of mango ? | mango + apple = 12 . . . . . . . . . . . . . ( a ) mango + orange = 10 . . . . . . . . . . ( b ) adding a and b 2 mango + apple + orange = 22 but mango + apple + orange = 38 . . . . given which is not possible . in this case , no of mangoes is negative ( - 16 ) answer : d | a = 38 - 12
b = 10 - a
|
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