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 |
|---|---|---|---|---|---|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | divide(70, divide(subtract(const_100, 25), subtract(25, 10))) | in 70 liters of a mixture of milk and water , the quantity of water is 10 % . how much water should be added so that new mixture may 25 % water . | there are 10 % of water in 70 l - - > there are 7 l of waters ; after x liters of water is added amount of water becomes ( 7 + x ) liters , which on the other hand is 1 / 4 th ( 25 % ) of new ( 70 + x ) liters of mixture : 7 + x = 1 / 4 * ( 70 + x ) - - > 28 + 4 x = 70 + x - - > x = 14 answer : a | a = 100 - 25
b = 25 - 10
c = a / b
d = 70 / c
|
a ) 27 , b ) 38 , c ) 37 , d ) 30 , e ) 28 | a | multiply(divide(150, 24), const_3_6) | an athlete runs 150 metres race in 24 seconds . what is his speed ? | "speed = distance / time = 150 / 20 = 7.5 m / s = 7.5 * 18 / 5 = 27 km / hr answer : a" | a = 150 / 24
b = a * const_3_6
|
a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | b | multiply(7, divide(multiply(add(7, 15), subtract(15, multiply(divide(5, add(7, 5)), 15))), subtract(multiply(15, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 15 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | "ci * vi = cf * vf ( 7 / 12 ) * ( v 1 - 15 ) = ( 7 / 16 ) * v 1 ( v 1 - 15 ) / v 1 = 3 / 4 15 accounts for the difference of 1 on ratio scale so initial volume = v 1 = 4 * 15 = 60 litres . 7 / 12 of the initial mixture was liquid a so liquid a was ( 7 / 12 ) * 60 = 35 litres . answer : b" | a = 7 + 15
b = 7 + 5
c = 5 / b
d = c * 15
e = 15 - d
f = a * e
g = 15 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
a ) 17 , b ) 42 , c ) 69 , d ) 47 , e ) 52 | c | multiply(divide(divide(divide(divide(1780, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(1780, const_1000), const_3), const_3), const_3)) | find the smallest number that must be added to 1780 to make it a perfect square . | "4 1780 ( 42 16 82 180 164 16 number to be added = ( 43 ) 2 - 1780 = 1849 - 1780 = 69 . answer c" | a = 1780 / 1000
b = a / 3
c = b / 3
d = c / 3
e = 1780 / 1000
f = e / 3
g = f / 3
h = g / 3
i = d * h
|
a ) 4 : 5 , b ) 5 : 3 , c ) 4 : 4 , d ) 4 : 8 , e ) 4 : 1 | b | divide(sqrt(25), sqrt(9)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 25 : β 9 = 5 : 3 answer : b" | a = math.sqrt(25)
b = math.sqrt(9)
c = a / b
|
a ) a ) 1.33 , b ) b ) 5 , c ) c ) 7 , d ) d ) 5 , e ) e ) 8 | a | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(55, 40)), divide(const_1, const_2)) | a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 55 kmph when will the owner over take the thief from the start ? | "explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 55 40 d = 20 rs = 55 β 40 = 15 t = 20 / 15 = 1.33 hours answer : option a" | a = 1 / 2
b = a * 40
c = 55 - 40
d = b / c
e = 1 / 2
f = d - e
|
a ) 115 , b ) 520 , c ) 550 , d ) 750 , e ) 796 | d | add(50, divide(80000, 800)) | a newspaper manufacturer produces 800 units of a certain model each month at a cost to the manufacturer of Β£ 50 per unit and all of the produced units are sold each month . what is the minimum selling price per unit that will ensure that the monthly profit ( revenue from sales minus production costs ) on the sales of these units will be at least Β£ 80000 ? | 800 ( x - 50 ) β₯ 80000 x - 50 β₯ 800 x β₯ 750 answer : option d | a = 80000 / 800
b = 50 + a
|
a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 2 , e ) 17 : 3 | c | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 85,000 and rs . 20,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 20000 = 17 : 4 . answer : c" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 10 , b ) 15 , c ) 3 , d ) 6 , e ) 12 | d | inverse(add(divide(const_1, 10), divide(const_1, 15))) | x does a work in 10 days . y does the same work in 15 days . in how many days they together will do the same work ? | "x ' s 1 day ' s work = 1 / 10 y ' s 1 day ' s work = 1 / 15 ( x + y ) ' s 1 day ' s work = ( 1 / 10 + 1 / 15 ) = 1 / 6 both together will finish the work in 6 days . correct option is d" | a = 1 / 10
b = 1 / 15
c = a + b
d = 1/(c)
|
a ) 400 , b ) 1550 , c ) 1575 , d ) 1600 , e ) 1625 | c | add(rectangle_area(200, 5), rectangle_area(120, 5)) | a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area y covered by the two roads ? | "area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads y = 1000 + 600 - 25 = 1575 answer : option c" | a = rectangle_area + (
|
a ) 1977452 / 1977455 , b ) 1982466 / 1982469 , c ) 1982487 / 1982490 , d ) 1982456 / 1982459 , e ) 1982465 / 1982468 | c | divide(subtract(const_3600, multiply(const_3, const_3)), subtract(const_3600, const_4)) | which expression has the greatest value ? | "the difference between each numerator / denominator pair is three . the largest expression ( which in this case will also be the closest to one ) is the expression where ' 3 ' is the smallest fraction of the numerator ( or the denominator ) . the largest numerator is 1982487 , or c , and this will give an expression with a value closest to one . we can also demonstrate that this approach works with some simple logic . taking numerator / denominator pairs that differ by three . . . 1 / 4 = 0.25 2 / 5 = 0.4 3 / 6 = 0.5 . . . 27 / 30 = 0.9 . . . 1982487 / 1982490 = 0.999998486 . . . answer : c" | a = 3 * 3
b = 3600 - a
c = 3600 - 4
d = b / c
|
a ) 48 kmph , b ) 54 kmph , c ) 66 kmph , d ) 82 kmph , e ) 69 kmph | d | subtract(multiply(const_3_6, divide(add(108, 112), 6)), 50) | a train 108 m long moving at a speed of 50 kmph crosses a train 112 m long coming from opp direction in 6 seconds . the speed of the second train is | both trains are moving in opposite direction means we need to add their speeds . i . e : 1 st train speed is = 50 kmph let the 2 nd train speed = x kmph relative speed = ( 50 + x ) kmph = ( 50 + x ) * 5 / 18 = ( 250 + 5 x ) / 18 m / sec distance covered by two trains = 108 + 112 = 220 m therefore , 220 / ( ( 250 + 5 x ) / 18 ) = 6 = > 250 + 5 x = 660 = > x = 82 km / hr . answer : d | a = 108 + 112
b = a / 6
c = const_3_6 * b
d = c - 50
|
a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 31 | e | add(2, 29) | what is the greatest prime factor of 4 ^ 17 - 2 ^ 29 ? | i ' m happy to help with this . we know 4 = 2 ^ 2 , so 4 ^ 17 = ( 2 ^ 2 ) ^ 17 = 2 ^ ( 2 * 17 ) = 2 ^ 34 that takes advantage of a law of exponents that says ( a ^ n ) ^ m = a ^ ( n * m ) so , 4 ^ 17 - 2 ^ 29 = 2 ^ 34 - 2 ^ 29 = 2 ^ ( 29 + 5 ) - 2 ^ 29 = ( 2 ^ 29 ) * ( 2 * 6 ) - 2 ^ 29 = ( 2 ^ 6 - 1 ) * ( 2 ^ 29 ) = ( 32 - 1 ) * ( 2 ^ 29 ) = 31 * ( 2 ^ 29 ) the prime factors of 31 are 31 , so the largest prime factor is 31 , answer choice d . here ' s a blog you may find helpful . http : / / magoosh . com / gmat / 2012 / gmat - math - factors / does all that make sense ? please let me know if you have any further questions . mike wow . i am floored by how great of an explanation you provided . posts like that make me really think that doing thousands of practice problems with good explanations beats out reading books on math every day of the week . e | a = 2 + 29
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a ) 5 : 1 , b ) 10 : 5 , c ) 15 : 2 , d ) 20 : 2 , e ) 25 : 2 | d | divide(multiply(5, 4), multiply(1, 2)) | jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 5 to 2 and the ratio of indian to british stamps is 4 to 1 , what is the ratio of us to british stamps ? | "u / i = 5 / 2 i / b = 4 / 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 20 : 8 : 2 i . e . u : b = 20 : 2 answer : option d" | a = 5 * 4
b = 1 * 2
c = a / b
|
a ) 2288 , b ) 2877 , c ) 5760 , d ) 7636 , e ) 5728 | c | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1500, multiply(const_4, const_100))) | by investing in 16 % stock at 64 , one earns rs . 1500 . the investment made is ? | "to earn rs . , investment = rs . 64 . to earn rs . 1500 , investment = = rs . 5760 . answer : c" | a = 4 * 2
b = 4 * 2
c = a * b
d = c * 3
e = 100 / 2
f = d / e
g = 4 * 100
h = 1500 - g
i = f * h
|
a ) 344 , b ) 251 , c ) 411 , d ) 212 , e ) 120 | e | divide(add(138, 102), 2) | a student , after his teacher ' s question , chose a number , multiplied it by 2 , then subtracted 138 from the result and got 102 . what was the number he chose ? | let x be the number he chose , then 2 β
x β 138 = 102 2 x = 240 x = 120 correct answer is e ) 120 | a = 138 + 102
b = a / 2
|
a ) 148 , b ) 210 , c ) 314 , d ) 320 , e ) 344 | a | divide(multiply(36, 16), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 36 and 16 ? | prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 16 = 2 x 2 x 2 x 2 gcf = 4 lcm = 144 sum = 148 . answer a . | a = 36 * 16
b = a / 4
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | e | divide(28, subtract(3, const_1)) | lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 28 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 28 3 c = c + 28 c = 14 the answer is e ." | a = 3 - 1
b = 28 / a
|
a ) rs 6.56 , b ) rs 9.56 , c ) rs 10.56 , d ) rs 11.56 , e ) none of these | a | divide(multiply(5, add(const_100, 5)), subtract(const_100, 20)) | a fruit seller sells mangoes at the rate of rs . 5 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 80 : 5 = 105 : x x = ( 5 Γ 105 / 80 ) = rs 6.56 option a" | a = 100 + 5
b = 5 * a
c = 100 - 20
d = b / c
|
a ) m , b ) m , c ) m , d ) m , e ) m | c | divide(15, subtract(divide(15, 10), 6)) | a train covers a distance of 15 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 15 / 10 * 60 ) km / hr = ( 90 * 5 / 18 ) m / sec = 25 m / sec . length of the train = 25 * 6 = 150 m . answer : c" | a = 15 / 10
b = a - 6
c = 15 / b
|
a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 400,000 , e ) $ 600,000 | e | multiply(multiply(400, const_100), const_10) | a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.1 % on a company β s payroll above $ 200,000 . if belfried industries paid $ 400 in this special municipal payroll tax , then they must have had a payroll of ? | "answer : e , ( with different approach ) : the 400 paid is 0.1 % of the additional amount above 200,000 . let it be x now 0.1 % of x = 400 therefore x = 400,000 total = 200,000 + x = 600,000" | a = 400 * 100
b = a * 10
|
a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 15 , d ) 7 / 15 , e ) 7 / 16 | d | divide(choose(7, const_2), choose(add(7, 3), const_2)) | there are 7 red shoes & 3 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probablity is 7 c 2 from 10 shoes probablity of taking 2 red shoe is 7 c 2 / 10 c 2 = 7 / 15 answer : d" | a = math.comb(7, 2)
b = 7 + 3
c = math.comb(b, 2)
d = a / c
|
a ) 87 days , b ) 20 days , c ) 26 days , d ) 19 days , e ) 16 days | e | divide(multiply(8, 16), divide(subtract(multiply(8, 16), multiply(add(divide(multiply(8, 16), 8), divide(multiply(8, 16), 16)), 4)), 4)) | a can do a piece of work in 8 days . b can do it in 16 days . with the assistance of c they completed the work in 4 days . find in how many days can c alone do it ? | "c = 1 / 4 - 1 / 8 - 1 / 16 = 1 / 16 = > 16 days answer : e" | a = 8 * 16
b = 8 * 16
c = 8 * 16
d = c / 8
e = 8 * 16
f = e / 16
g = d + f
h = g * 4
i = b - h
j = i / 4
k = a / j
|
a ) 2200 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4714.3 | e | multiply(circumface(10), 15) | the radius of a cylinder is 10 m , height 15 m . the volume of the cylinder is : | "cylinder volume = Γ― β¬ r ( power 2 ) h = 22 / 7 Γ£ β 10 Γ£ β 10 Γ£ β 15 = 4714.3 m ( power 3 ) answer is e ." | a = circumface * (
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a ) 10 % , b ) 6 % , c ) 15 % , d ) 17 % , e ) 20 % | b | multiply(divide(subtract(divide(50, const_100), divide(47, const_100)), subtract(const_1, divide(47, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 47 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "let me try a simpler one . lets assume that candidate got 47 % votes and total votes is 100 . candidate won = 47 remaining = 53 to get 50 % , candidate requires 3 votes from 100 which is 3 % and 3 votes from 53 . 3 / 53 = . 056 = 5.6 % which is approx 6 % . hence the answer is b ." | a = 50 / 100
b = 47 / 100
c = a - b
d = 47 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) 12 , b ) 77 , c ) 66 , d ) 99 , e ) 11 | c | subtract(divide(multiply(1.50, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.25, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.50 km and 1.25 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.50 + 1.25 = 2.75 km = 2750 m . required time = 2750 * 3 / 125 = 66 sec . answer : c" | a = 1 * 50
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 25
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) s . 6000 , b ) s . 6300 , c ) s . 4736 , d ) s . 6480 , e ) s . 7000 | c | multiply(multiply(64, 74), divide(30, 10)) | what is the cost of leveling the field in the form of parallelogram at the rate of rs . 30 / 10 sq . metre , whose base & perpendicular distance from the other side being 64 m & 74 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 64 * 74 = 4736 m . total cost of levelling = rs . 4736 c" | a = 64 * 74
b = 30 / 10
c = a * b
|
a ) 50 , b ) 198 , c ) 380 , d ) 398 , e ) 400 | b | multiply(inverse(20), multiply(multiply(const_100, 20), add(const_4, const_4))) | when 1 / 20 % of 4,000 is subtracted from 1 / 20 of 4,000 , the difference is | "1 / 20 % of 4000 = 2 1 / 20 of 4000 = 200 200 - 2 = 198 ans : b" | a = 1/(20)
b = 100 * 20
c = 4 + 4
d = b * c
e = a * d
|
a ) 6 % , b ) 8 % , c ) 4 % , d ) 2 % , e ) 9 % | c | multiply(divide(2, 24), const_100) | what percent is 2 minutes 24 seconds of an hour ? | "2 mt 24 s = 144 s ( 144 * 100 ) / ( 60 * 60 ) = 4 % answer : c" | a = 2 / 24
b = a * 100
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 7 | b | subtract(divide(8, const_2), multiply(43, 43)) | what is the remainder when 43 * 49 is divided by 8 ? | "we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 43 * 49 ) / 8 } } = remainder of ( 43 / 8 ) * remainder of ( 49 / 8 ) = 3 * 1 = 3 answer : b" | a = 8 / 2
b = 43 * 43
c = a - b
|
a ) 512 , b ) 510 , c ) 514 , d ) 516 , e ) 518 | b | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | the difference between a number and its two - fifth is 510 . what is 60 % of that number ? | "let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 60 % of 850 = 510 . answer : b" | a = 4 + 1
b = 510 * a
c = 1 + 2
d = b / c
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a ) 9 % , b ) 11 % , c ) 13 % , d ) 15 % , e ) 17 % | d | multiply(divide(add(multiply(40, divide(5, const_100)), 5.5), add(40, add(5.5, 4.5))), const_100) | a 40 - liter solution of alcohol and water is 5 percent alcohol . if 5.5 liters of alcohol and 4.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 5.5 ) / 50 = 7.5 / 50 = 15 % the answer is d ." | a = 5 / 100
b = 40 * a
c = b + 5
d = 5 + 5
e = 40 + d
f = c / e
g = f * 100
|
a ) 15,000 , b ) 25,000 , c ) 40,000 , d ) 48,000 , e ) 60,000 | d | multiply(divide(add(multiply(5, 16000), 16000), const_4), const_2) | in the first round of the elections , the only two candidates got exactly the same number of votes . during the second round , 16000 votes switched from the first candidate to the second one . the total number of votes remained the same in both rounds , and no other votes switched sides . if , in the second round , the winning candidate got 5 times as many votes as the other candidate , how many people have voted in each round ? | let a be the first candidate and b be the second one . given x + 16000 = 5 ( x - 16000 ) = > x = 24000 2 x = 48000 imo option d . | a = 5 * 16000
b = a + 16000
c = b / 4
d = c * 2
|
a ) 20 , b ) 10 , c ) 30 , d ) 50 , e ) 40 | e | divide(add(120, 80), add(4, 1)) | in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 80 questions and secures 120 marks , how many questions does he answer correctly ? | "let the number of correct answers be x then , number of wrong answers = ( 80 - x ) 4 x β ( 80 - x ) = 120 = > 4 x β 80 + x = 120 = > 5 x = 200 = > x = 200 / 5 = 40 i . e . , he does 40 questions correctly answer is e ." | a = 120 + 80
b = 4 + 1
c = a / b
|
a ) 540 m , b ) 960 m , c ) 1080 m , d ) 1020 m , e ) 1120 m | c | multiply(120, add(4, 5)) | after striking the floor , a rubber ball rebounds to 4 / 5 th of the height from which it has fallen . find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres . | solution : the first drop is 120 metres . after this the ball will rise by 96 metres and fall by 96 metres . this process will continue in the form of infinite gp with common ratio 0.8 and first term 96 . the required answer is given by the formula : a / ( 1 - r ) now , [ { 120 / ( 1 / 5 ) } + { 96 / ( 1 / 5 ) } ] = 1080 m . answer c | a = 4 + 5
b = 120 * a
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a ) s . 2000 , b ) s . 2500 , c ) s . 3000 , d ) s . 3500 , e ) s . 4000 | b | multiply(divide(divide(250, 5), 2), const_100) | a sum was put a simple interest at a certain rate for 5 years . had it been put at 2 % higher rate , it would have fetched rs . 250 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x Γ£ β ( r + 2 ) Γ£ β 5 / 100 Γ’ Λ β x Γ£ β r Γ£ β 5 / 100 = 250 Γ’ β‘ β x Γ£ β 2 Γ£ β 5 / 100 = 250 Γ’ β‘ β x Γ£ β 5 / 100 = 125 Γ’ β‘ β 5 x = 12500 Γ’ β‘ β x = 2500 answer : option b" | a = 250 / 5
b = a / 2
c = b * 100
|
a ) 4676 , b ) 4678 , c ) 9975 , d ) 9504 , e ) 9936 | c | multiply(floor(divide(power(const_10, 4), 35)), 35) | what is the largest 4 digit number exactly divisible by 35 ? | "largest 4 digit number = 9999 9999 Γ· 35 = 285 , remainder = 24 hence largest 4 digit number exactly divisible by 88 = 9999 - 24 = 9975 answer : c" | a = 10 ** 4
b = a / 35
c = math.floor(b)
d = c * 35
|
a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | b | divide(105, add(16, 5)) | a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 105 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 105 / 21 = 5 hours option b" | a = 16 + 5
b = 105 / a
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a ) 15 , b ) 125 , c ) 25 , d ) 75 , e ) 50 | d | add(divide(100, const_2), divide(50, const_2)) | the sum of the average ( arithmetic mean ) of the even integers from 0 to 100 inclusive and the average ( arithmetic mean ) of the even integers from 0 to 50 inclusive ? | so , according to a mean of a set of even numbers from 0 to 100 = ( 0 + 100 ) / 2 = 50 and mean of a set of even numbers from 0 to 50 = ( 0 + 50 ) / 2 = 25 sum = 50 + 25 = 75 answer : d . | a = 100 / 2
b = 50 / 2
c = a + b
|
['a ) 20', 'b ) 30', 'c ) 40', 'd ) 50', 'e ) 60'] | b | divide(divide(multiply(150, const_4), const_10), const_2) | mr . loyd wants to fence his square shaped land of 150 sqft each side . if a pole is laid every 30 ft how many poles do he need ? | if each side is 120 feet . . then total perimeter is 150 * 4 = 600 poles every 20 feet hence no of poles = 600 / 20 = 30 answer : b | a = 150 * 4
b = a / 10
c = b / 2
|
a ) $ 150 , b ) $ 200 , c ) $ 320 , d ) $ 650 , e ) $ 810 | b | multiply(divide(multiply(1.92, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.50) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 1.92 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 1 , 920,000 / 4800 = 400 boxes total cost = 400 Γ $ 0.5 = $ 200 the answer is b ." | a = 1000 * 1000
b = 1 * 92
c = 20 * 20
d = c * 12
e = b / d
f = e * 0
|
a ) s . 8600 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | a | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | find the simple interest on rs . 68,800 at 16 2 / 3 % per annum for 9 months . | "p = rs . 68800 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 68,800 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8600 answer is a ." | a = 2 * 3
b = a * 100
c = b * 100
d = 3 * 3
e = d * 100
f = 3 + 2
g = f * 2
h = e * g
i = c + h
j = 16 * 3
k = j + 2
l = k / 3
m = i * l
n = 3 * 3
o = 2 * 3
p = 2 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
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['a ) 18 β 9 / 4 β Ο', 'b ) 22 β 11 / 2 β Ο', 'c ) 9 + 9 / 4 β Ο', 'd ) 9 + 9 / 2 β Ο', 'e ) 9 / 2 β 9 / 4 β Ο'] | b | subtract(22, multiply(divide(22, const_4), const_pi)) | the area of square abcd above is 22 . the shaded region is the intersection of the square and a circular region centered at c . what is the area of the shaded region ? | = square - quarter circle = 22 - pi * 22 / 4 = 22 - pi * 11 / 2 answer : b | a = 22 / 4
b = a * math.pi
c = 22 - b
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a ) 2 / 9 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 11 , e ) 7 / 9 | d | divide(const_4.0, 11) | in the list 1 , 2 , 3 , 4 , 5 , 5 , 5 , 5 , 7 , 11 , 21 , what fraction of the data is less than the mode ? | "mode : the mode of any set is the term which has the highest frequency ( occurrence ) highest frequent term in the set is 5 ( with frequency 4 ) hence mode = 5 four terms ( 1 , 2 , 3 , 4 ) out of a total of 11 terms are less than mode of the set . fraction of set that are less than mode of set = 4 / 11 answer : option d" | a = 4 / 0
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a ) 15360 , b ) 15720 , c ) 15060 , d ) 14930 , e ) 16075 | b | subtract(add(6000, 14000), add(multiply(6000, divide(20, const_100)), multiply(divide(23, const_100), 14000))) | a soft drink company had 6000 small and 14000 big bottles in storage . if 20 % of small 23 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 14000 - ( 0.2 * 6000 + 0.22 * 14000 ) = 15720 . answer : b ." | a = 6000 + 14000
b = 20 / 100
c = 6000 * b
d = 23 / 100
e = d * 14000
f = c + e
g = a - f
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a ) 10 mps , b ) 76 mps , c ) 20 mps , d ) 97 mps , e ) 16 mps | c | multiply(const_0_2778, 72) | express a speed of 72 kmph in meters per second ? | "72 * 5 / 18 = 20 mps answer : c" | a = const_0_2778 * 72
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a ) 33 , b ) 36 , c ) 39 , d ) 42 , e ) 45 | e | divide(factorial(10), multiply(factorial(subtract(10, const_2)), factorial(const_2))) | there are 10 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "10 c 2 = 45 the answer is e ." | a = math.factorial(10)
b = 10 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | e | divide(subtract(5, 4), add(4, 2)) | if two of the 4 expressions x + y , x + 5 y , x - y , 5 x + y are chosen at random , what is the probability that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | there are total 4 c 2 = 64 c 2 = 6 ways to choose 2 equations out of 4 at random . we want the product of 2 equations in format x 2 β ( by ) 2 x 2 β ( by ) 2 , where b is a constant coefficient of term y 2 y 2 . also note equation does not contain any xyxy term . there are only 2 equations , ( x + y ) & ( x - y ) which gives the product x 2 β y 2 x 2 β y 2 . product of all other equations will contain term xyxy so answer = 1 / 6 ans : e | a = 5 - 4
b = 4 + 2
c = a / b
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a ) $ 36 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | a | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(7, const_2), 19)), const_2) | bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 7 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 19 left in his pocket . what was the value of n ? | "started to test answer d if he had 60 , then he spent 15 at hardware store now he was left with 45 $ he spent 7 dollars on cleaning , thus he remained with 38 $ he then spent 1 / 2 of 38 , or 19 , and was left with 19 . hence , the only option that can be right is a ." | a = 1 / 4
b = 1 - a
c = 1/(b)
d = 7 / 2
e = d + 19
f = c * e
g = f * 2
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(add(divide(32, 8), divide(48, 8)), const_2) | a man swims downstream 48 km and upstream 32 km taking 8 hours each time ; what is the speed of the current ? | "48 - - - 8 ds = 6 ? - - - - 1 32 - - - - 8 us = 4 ? - - - - 1 s = ? s = ( 6 - 4 ) / 2 = 1 answer : a" | a = 32 / 8
b = 48 / 8
c = a + b
d = c / 2
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a ) 11 sec , b ) 6 sec , c ) 77 sec , d ) 14 sec , e ) 12 | b | divide(100, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 100 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 100 * 3 / 50 = 6 sec . answer : b" | a = 63 - 3
b = a * const_0_2778
c = 100 / b
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a ) 1 : 7 , b ) 2 : 3 , c ) 1 : 8 , d ) 2 : 3 , e ) 1 : 3 | e | divide(subtract(multiply(divide(70, add(8, 27)), 8), divide(multiply(divide(70, add(4, 3)), 3), add(1, 4))), subtract(multiply(divide(70, add(8, 27)), 27), multiply(divide(multiply(divide(70, add(4, 3)), 3), add(1, 4)), 4))) | in a colony of 70 residents , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of educated to uneducated persons is 8 : 27 , then find the ratio of the number of educated to uneducated men in the colony ? | number of men in the colony = 4 / 7 * 70 = 40 . number of women in the colony = 3 / 7 * 70 = 40 . number educated women in the colony = 1 / 5 * 30 = 6 . number of uneducated women in the colony = 4 / 5 * 50 = 24 . number of educated persons in the colony = 8 / 35 * 70 = 16 . as 6 females are educated , remaining 10 educated persons must be men . number of uneducated men in the colony = 40 - 10 = 30 . number of educated men and uneducated men are in the ratio 10 : 30 i . e . , 1 : 3 . answer : e | a = 8 + 27
b = 70 / a
c = b * 8
d = 4 + 3
e = 70 / d
f = e * 3
g = 1 + 4
h = f / g
i = c - h
j = 8 + 27
k = 70 / j
l = k * 27
m = 4 + 3
n = 70 / m
o = n * 3
p = 1 + 4
q = o / p
r = q * 4
s = l - r
t = i / s
|
['a ) 1 : 5', 'b ) 1 : 25', 'c ) 1 : 125', 'd ) 1 : 625', 'e ) none'] | b | divide(surface_cube(1), surface_cube(5)) | a cube of edge 5 cm is cut into cubes of each edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to | solution required ratio = ( 6 x 1 x 1 / 6 x 5 x 5 ) βΉ = βΊ 1 / 25 βΉ = βΊ 1 : 125 . answer b | a = surface_cube / (
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a ) 1235 , b ) 1346 , c ) 1590 , d ) 1635 , e ) 1489 | c | multiply(divide(subtract(1365, 15), subtract(7, const_1)), 7) | find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 7 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 7 x + 15 6 x = 1350 x = 225 large number = 225 + 1365 = 1590 c" | a = 1365 - 15
b = 7 - 1
c = a / b
d = c * 7
|
a ) 15 , b ) 13 , c ) 20 , d ) 21 , e ) 26 | b | divide(add(multiply(const_10, const_10), 13), subtract(const_12, const_3)) | alex age after 13 years will be 10 times his age 10 years back . what is the present age of alex ? | let alex ' s present age be x alex ' s age after 17 years = x + 17 alex ' s age 10 years back = x - 10 x + 17 = 10 ( x - 10 ) x = 13 rajeev ' s present age = 13 years answer is b | a = 10 * 10
b = a + 13
c = 12 - 3
d = b / c
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of q ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . q * 23 + 39 = 25 * ( q + 1 ) q = 7 ans . ( a )" | a = 39 - 15
b = 2 + 1
c = a / b
d = c - 1
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a ) 99 , b ) 18 , c ) 16 , d ) 198 , e ) 15 | d | divide(multiply(60, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 60 m long and 16 1 / 2 m board ? | "60 * 33 / 2 = 5 / 2 * 2 * x = > x = 198 answer : d" | a = 1 / 2
b = 16 + a
c = 60 * b
d = 1 / 2
e = 2 + d
f = e * 2
g = c / f
|
a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | c | divide(subtract(add(26, add(26, 7)), multiply(7, 7)), const_2) | the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 7 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x Γ’ β¬ β ( 26 + 33 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 9 x = 50 2 x = 50 x = 25 . so , average age of the team is 25 years . answer c" | a = 26 + 7
b = 26 + a
c = 7 * 7
d = b - c
e = d / 2
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a ) 11 / 3 , b ) 11 / 2 , c ) 242 , d ) 6 ^ 11 - 1 , e ) 3 ^ 11 - 1 | d | subtract(power(6, add(9, const_2)), const_1) | if 6 ^ k + 6 ^ k = ( 6 ^ 9 ) ^ ( 6 ^ 9 ) - 6 ^ k , then k = ? | "6 ^ k + 6 ^ k = ( 6 ^ 9 ) ^ 6 ^ 9 - 6 ^ k 6 * ( 6 ^ k ) = 6 ^ ( 36 * 6 ^ 9 ) = 6 ^ ( 6 ^ 2 * 6 ^ 9 ) = 6 ^ ( 6 ^ 11 ) 6 ^ k + 1 = 6 ^ ( 6 ^ 11 ) so k + 1 = 6 ^ 11 so k = 6 ^ 11 - 1 answer is d" | a = 9 + 2
b = 6 ** a
c = b - 1
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a ) 132 , b ) 145 , c ) 165 ` , d ) 400 , e ) 300 | a | multiply(add(12, const_2), subtract(add(12, const_2), const_1)) | there are 12 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 12 from 12 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 12 p 2 ways . 12 p 2 = 12 * 11 = 870 answer : a" | a = 12 + 2
b = 12 + 2
c = b - 1
d = a * c
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a ) 24 , b ) 26 , c ) 27 , d ) 29 , e ) 45 | e | sqrt(divide(150, add(power(5, 3), add(power(2, 3), power(3, 3))))) | the ratio of three numbers is 2 : 3 : 5 and their sum is 150 . the second number of the three numbers is ? | "2 : 3 : 5 total parts = 10 10 parts - - > 150 1 part - - - - > 15 the second number of the three numbers is = 3 * 15 = 45 answer : e" | a = 5 ** 3
b = 2 ** 3
c = 3 ** 3
d = b + c
e = a + d
f = 150 / e
g = math.sqrt(f)
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a ) a ) 3800 , b ) b ) 4200 , c ) c ) 4400 , d ) d ) 4500 , e ) e ) 4600 | b | floor(divide(3213, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 15), const_100)))) | 10 % 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 3213 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 85 / 100 ) = 3213 x = 4200 answer : b" | a = 100 - 10
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 3213 / e
g = math.floor(f)
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a ) 89 , b ) 20 , c ) 36 , d ) 88 , e ) 30 | e | divide(add(180, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 180 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 180 + 120 = 300 m . time taken = 300 / 10 = 30 sec . answer : option e" | a = 180 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 25 , b ) 9 , c ) 39 , d ) 61 , e ) 16 | b | multiply(multiply(4, divide(6, 4)), divide(6, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 6 mat - weavers in 6 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 6 : : 4 : x days 4 : 6 4 * 4 * x = 6 * 6 * 4 x = ( 6 * 6 * 4 ) / ( 4 x 4 ) x = 9 . answer is b ." | a = 6 / 4
b = 4 * a
c = 6 / 4
d = b * c
|
a ) 33 , b ) 38 , c ) 45 , d ) 28 , e ) 19 | c | subtract(56, 12) | nitin ranks 12 th in a class of 56 students . what is rank from the last ? | "explanation : number students behind the nitin in rank = ( 56 - 12 ) = 44 nitin is 45 th from the last answer : c ) 45" | a = 56 - 12
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a ) 400 , b ) 370.4 , c ) 376.98 , d ) 340.9 , e ) 350.03 | c | multiply(circumface(divide(40, const_2)), 3) | find the cost of fencing around a circular field of diameter 40 m at the rate of rs . 3 a meter ? | "2 * 22 / 7 * 20 = 125.66 125.66 * 3 = rs . 376.98 answer : c" | a = 40 / 2
b = circumface * (
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a ) $ 600 , b ) $ 740 , c ) $ 860 , d ) $ 980 , e ) $ 1,140 | c | multiply(add(add(subtract(subtract(50, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20) | tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 50 people bought tickets to the concert , what was the total revenue from ticket sales ? | "price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 20 tickets = 20 * 20 = 400 revenue generated from sales of 50 tickets = 120 + 340 + 400 = 860 $ answer c" | a = 50 - 20
b = a - 10
c = 40 / 100
d = 1 - c
e = d * 10
f = b + e
g = 15 / 100
h = 1 - g
i = h * 20
j = f + i
k = j * 20
|
a ) 2.25 , b ) 3.25 , c ) 4.25 , d ) 5.25 , e ) 0.25 | e | subtract(power(1.5, 2), 2) | x + ( 1 / x ) = 1.5 find x ^ 2 + ( 1 / x ^ 2 ) | "squaring on both sides ( x + 1 / x ) ^ 2 = 1.5 ^ 2 x ^ 2 + 1 / x ^ 2 = 2.25 - 2 x ^ 2 + 1 / x ^ 2 = 0.25 answer : e" | a = 1 ** 5
b = a - 2
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a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 10,000 | b | divide(const_180, const_1000) | the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 36 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 36 pi r = 3 so area = 4 . pi . r ^ 2 = 36 . pi = 36 x 3.14 = 100 ( approx ) b" | a = const_180 / 1000
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a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | c | divide(80, const_2) | a soccer store typically sells replica jerseys at a discount of 30 percent to 40 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 = 30 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , c is the answer" | a = 80 / 2
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a ) 1 / 3 , b ) 2 / 3 , c ) 4 / 3 , d ) 35 / 3 , e ) 10 / 3 | d | subtract(multiply(divide(4, 3), 4), 3) | at a certain paint store forest green is made by mixing 4 parts blue paint with 3 parts yellow paint . verdant green is made by mixing 4 parts yellow paint with 3 parts blue paint . how many liters of yellow paint must be added to 35 liters of forest green to change it to verdant green ? | "35 liter of forset green have 20 liter of blue and 15 liter of yellow suppose we add x liter of yellow to make it a verdant green so the ratio of blue to yellow in verdant green is ΒΎ so the equation is blue / yellow = 20 / ( 15 + x ) = ΒΎ 45 + 3 x = 80 = > x = 35 / 3 answer : d" | a = 4 / 3
b = a * 4
c = b - 3
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a ) 6 , b ) 7 , c ) 23 , d ) 17 , e ) 42 | e | subtract(divide(power(negate(12), 2), 3), 2) | find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 12 x + 3 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 12 / 1 = - 12 ab = 3 / 1 = 3 hence a / b + b / a = [ ( - 12 ) 2 - 2 ( 3 ) ] / 3 = 126 / 3 = 42 . e )" | a = negate ** (
b = a / 2
c = b - 3
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a ) 20 , b ) 24 , c ) 25 , d ) 27 , e ) 30 | b | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 60)), const_2), divide(const_1, 60))) | p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 60 days to complete the same work . then q alone can do it in | "work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 60 work done by p , q and r in 1 day = 1 / 10 + 1 / 60 = 7 / 60 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day Γ£ β 2 = 7 / 60 = > work done by p in 1 day = 7 / 120 = > work done by q and r in 1 day = 7 / 120 hence work done by q in 1 day = 7 / 120 Γ’ β¬ β 1 / 60 = 1 / 24 so q alone can do the work in 24 days answer is b ." | a = 1 / 10
b = 1 / 60
c = a + b
d = c / 2
e = 1 / 60
f = d - e
g = 1 / f
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15 | c | multiply(12, const_1) | the total age of a and b is 12 years more than the total age of b and c . c is how many year younger than a ? | "given that , a + b = 12 + b + c β a - c = 12 therefore , c is younger than a by 12 years answer is c ." | a = 12 * 1
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a ) 1 , b ) 10 , c ) 125 , d ) 50 , e ) 20 | c | divide(multiply(divide(multiply(25, 1000), const_100), 50), const_100) | there are 1000 students in a school and among them 25 % of them attends chess class . 50 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | "25 % of 1000 gives 250 . so 250 attends chess and 50 % of 250 gives 125 so 125 enrolled for swimming answer : c" | a = 25 * 1000
b = a / 100
c = b * 50
d = c / 100
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a ) 125 , b ) 95 , c ) 80 , d ) 61.8 , e ) 50 | d | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, const_2))), 120), 60) | the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 60 minutes after it was poured was how many degrees fahrenheit ? | answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 60 minutes after it was poured is : f = 120 ( 2 ^ - 60 / 10 ) + 60 f = 120 * 1 / 64 + 60 f = 15 / 8 + 60 f = 495 / 8 = 61.8 d | a = 60 / 10
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
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['a ) 17', 'b ) 19', 'c ) 16', 'd ) 12', 'e ) 14'] | d | divide(add(add(add(const_100, const_3), add(multiply(8, const_10), const_3)), add(multiply(10, 6), const_3)), subtract(add(const_1, multiply(const_2, 10)), const_0_25)) | the edges of three iron cubes are 6 cm ; 8 cm ; 10 cm respectively . a new cube was made by melting them . find the edge of the new cube ? | 63 + 83 + 103 = a 3 = > a = 12 answer : d | a = 100 + 3
b = 8 * 10
c = b + 3
d = a + c
e = 10 * 6
f = e + 3
g = d + f
h = 2 * 10
i = 1 + h
j = i - const_0_25
k = g / j
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a ) 76 , b ) 76.5 , c ) 85 , d ) 80 , e ) none of these | c | add(multiply(8, 2.5), 65) | the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 8 x 2.5 ) kg = 20 kg . weight of new person = ( 65 + 20 ) kg = 85 kg . answer : c" | a = 8 * 2
b = a + 65
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a ) 4 : 1 , b ) 1 : 4 , c ) 5 : 2 , d ) 2 : 3 , e ) 2 : 5 | c | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 5,000 and rs . 2,000 respectively at the end of year then ratio of their investments are | "ratio = 5000 / 2000 = 5 : 2 answer : c" | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
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a ) 380 ft , b ) 400 ft , c ) 410.05 ft , d ) 329.85 ft , e ) 420.85 ft | d | multiply(45, divide(multiply(5, 5280), const_3600)) | someone on a skateboard is traveling 5 miles per hour . how many feet does she travel in 45 seconds ? ( 1 mile = 5280 feet ) | "per second = > 5 * 5280 ft / 60 * 60 = 7.33 ft 45 seconds = > 7.33 * 45 = 329.85 ft answer : d" | a = 5 * 5280
b = a / 3600
c = 45 * b
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a ) 132 , b ) 138 , c ) 144 , d ) 128 , e ) 116 | a | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(4, const_100)), multiply(add(70, 4), 250)), 250) | a computer manufacturer produces a certain electronic component at a cost of $ 70 per component . shipping costs for delivering the components are $ 4 per unit . further , the manufacturer has costs of $ 14,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 250 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 14500 is a fixed cost each component is $ 74 ( $ 70 to produce , $ 4 to ship ) manufacturer will be producing and selling 250 components so therefore the equation to find price would be 250 * p = 14500 + ( 250 * 70 ) + ( 250 * 4 ) p = ( 14500 + 17500 + 1000 ) / 250 p = 132 answer : a" | a = 4 * 4
b = a * 1000
c = 4 * 100
d = b + c
e = 70 + 4
f = e * 250
g = d + f
h = g / 250
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a ) 3 / 7 , b ) 3 / 2 , c ) 3 / 5 , d ) 3 / 1 , e ) 3 / 3 | a | divide(subtract(27, 20), subtract(20, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 20 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 20 = = > 27 x + 17 y = 20 x + 20 y = = > 7 x = 3 y = = > x / y = 3 / 7 . answer : a" | a = 27 - 20
b = 20 - 17
c = a / b
|
a ) 15 liters , b ) 21 liters , c ) 12 liters , d ) 18 liters , e ) 25 liters | b | multiply(divide(add(multiply(9, multiply(divide(7, add(7, 5)), 9)), multiply(7, subtract(9, multiply(divide(5, add(7, 5)), 9)))), subtract(multiply(9, 7), multiply(5, 7))), 7) | a bucket contains 2 mixtures a and b in the ratio 7 : 5 . if 9 liters of the mixture is replaced by 9 liters of liquid b , then the ratio of the two liquids becomes 7 : 9 . how much of the liquid a was there in the bucket ? | 1 st ratio = 7 : 5 2 nd ratio = 7 : 9 difference of cross products ratio = 7 * 9 - 7 * 5 = 28 common factor of the 1 st ratio = ( quantity replaced / sum of terms in 1 st ratio ) + ( quantity replaced * term a in 2 nd ratio / difference ) = ( 9 / 7 + 5 ) + ( 9 * 7 / 28 ) = 3 quantity of a = 7 * 3 = 21 liters answer is b | a = 7 + 5
b = 7 / a
c = b * 9
d = 9 * c
e = 7 + 5
f = 5 / e
g = f * 9
h = 9 - g
i = 7 * h
j = d + i
k = 9 * 7
l = 5 * 7
m = k - l
n = j / m
o = n * 7
|
a ) 20 % , b ) 70 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 53), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 53 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 61.2 bears per 36 hours a week or 1.7 bears / hr ( [ 40 bears * 1.53 ] / [ 40 hrs * . 90 ] ) . [ ( 1.7 - 1 ) / 1 ] * 100 % = 70 % answer : b" | a = 100 - 53
b = a - 10
c = 10 / b
d = c * 100
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | a | multiply(1, 5) | two numbers are in the ratio of 1 : 2 . if 5 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 5 ) / ( y + 5 ) = 3 / 5 = > x = 10 and y = 20 answer : a" | a = 1 * 5
|
a ) 80 , b ) 100 , c ) 75 , d ) 90 , e ) none of these | e | divide(72, multiply(divide(60, const_100), divide(3, 5))) | if 60 % of 3 / 5 of a number is 72 , then the number is ? | "let the number be x . then 60 % of 3 / 5 of x = 72 60 / 100 * 3 / 5 * x = 72 x = ( 72 * 25 / 9 ) = 200 required number = 200 . correct option : e" | a = 60 / 100
b = 3 / 5
c = a * b
d = 72 / c
|
a ) $ 1.09 , b ) $ 2.69 , c ) $ 2.25 , d ) $ 2.37 , e ) $ 2.50 | b | divide(add(12.00, multiply(1.75, subtract(9, 2))), 9) | the cost to park a car in a certain parking garage is $ 12.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 12 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 12 + 7 * 1.75 = 24.25 thus the average parking price = 24.25 / 9 = 2.69 $ b is the correct answer ." | a = 9 - 2
b = 1 * 75
c = 12 + 0
d = c / 9
|
a ) 1 / 11 , b ) 1 / 12 , c ) 1 / 13 , d ) 3 / 19 , e ) 3 / 43 | d | divide(divide(divide(1, 8), divide(2, 3)), add(divide(divide(1, 8), divide(2, 3)), 1)) | in a certain large company , the ratio of college graduates with a graduate degree to non - college graduates is 1 : 8 , and ratio of college graduates without a graduate degree to non - college graduates is 2 : 3 . if one picks a random college graduate at this large company , what is the probability t this college graduate has a graduate degree ? | "in believe the answer is d . please see below for explanation . 0 ) we are told the following ratios cgd - college graduate with degree ncg - non college graduate cgn - college graduate no degree cgd ncg cgn 1 8 3 2 in order to make cgd and cgn comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get cgd ncg cgn 3 24 16 if one picks a random college graduate at this large company , what is the probability this college graduate has a graduate degree ? nr of cgd = 3 nr of cg = 3 + 16 = 19 probability t of cgd / ( cg ) - > 3 / 19 answer d" | a = 1 / 8
b = 2 / 3
c = a / b
d = 1 / 8
e = 2 / 3
f = d / e
g = f + 1
h = c / g
|
a ) 20 min , b ) 22 min , c ) 15 min , d ) 18 min , e ) 24 min | e | divide(6, subtract(divide(5, 4), const_1)) | walking with 4 / 5 of my usual speed , i arrive at the bus stop 6 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 + 6 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 24 the answer is e ." | a = 5 / 4
b = a - 1
c = 6 / b
|
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 t 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 ) 600 km , b ) 700 km , c ) 800 km , d ) 900 km , e ) 1000 km | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 500), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 500 miles to km ? | "800 km 500 * 1.6 = 800 answer : c" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 500
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 5 , b ) 7 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(multiply(2, multiply(4, const_10)), multiply(divide(3, 2), multiply(4, const_10))), 4) | audrey 4 hours to complete a certain job . ferris can do the same job in 3 hours . audrey and ferris decided to collaborate on the job , working at their respective rates . while audrey worked continuously , ferris took 4 breaks of equal length . if the two completed the job together in 2 hours , how many minutes long was each of ferris β breaks ? | "audery and ferris collective work rate : 1 / 4 + 1 / 3 = 7 / 12 collective work time = 12 / 7 = 1.7 hrs job was actually done in = 2 ( includes breaks ) breaks = actual time taken - collective work time = 2 - 1.7 = . 3 hrs = 1 / 2 so ferrais took 4 breaks = . 3 / 4 = . 07 hrs = 7 m so answer is b ) 10 mins" | a = 4 * 10
b = 2 * a
c = 3 / 2
d = 4 * 10
e = c * d
f = b - e
g = f / 4
|
a ) 60 , b ) 70 , c ) 75 , d ) 100 , e ) 120 | c | add(15, 30) | each week a restaurant serving mexican food uses the same volume of chili paste , which comes in either 25 - ounce cans or 15 - ounce cans of chili paste . if the restaurant must order 30 more of the smaller cans than the larger cans to fulfill its weekly needs , then how manysmallercans are required to fulfill its weekly needs ? | "let x be the number of 25 ounce cans . therefore ( x + 30 ) is the number of 15 ounce cans . total volume is same , therefore 25 x = 15 ( x + 30 ) 10 x = 450 x = 45 therefore , number of 15 ounce cans = 45 + 30 = 75 ans - c" | a = 15 + 30
|
a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | b | subtract(multiply(const_10, 160), add(multiply(3, 100), multiply(5, 150))) | a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 160 . find the unknown rate of two blankets ? | "10 * 160 = 1600 3 * 100 + 5 * 150 = 1050 1600 β 1050 = 550 answer : b" | a = 10 * 160
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 227 , b ) 266 , c ) 230 , d ) 288 , e ) 200 | e | subtract(multiply(divide(300, 18), 30), 300) | a 300 meter long train crosses a platform in 30 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 30 = 50 / 3 3 ( x + 300 ) = 1500 Γ¨ x = 200 m . answer : e" | a = 300 / 18
b = a * 30
c = b - 300
|
a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16 | d | add(6, 6) | how many different values of positive integer x , for which | x + 6 | < x , are there ? | "answer d i opted to put the random value option . i used 0 , 6 , - 6 and the the extreme of 20 and - 20 . . i was able to solve it in 1 : 09 d" | a = 6 + 6
|
a ) 25 , b ) 30 , c ) 50 , d ) 75 , e ) 40 | c | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 36), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . if the faster train passes the slower train in 36 seconds , what is the length of each train ? | "assume the length of each train = x total distance covered for overtaking the slower train = x + x = 2 x relative speed = 46 - 36 = 10 km / hr = ( 10 Γ 10 ) / 36 = 100 / 36 m / s time = 36 seconds 2 x / ( 100 / 36 ) = 36 = > ( 2 x Γ 36 ) / 100 = 36 = > x = 50 meter answer is c ." | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 36
e = d / 2
|
a ) 0.9 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | a | multiply(const_3_6, divide(9, 36)) | convert the 9 / 36 m / s into kilometers per hour ? | "9 / 36 m / s = 9 / 36 * 18 / 5 = 9 / 10 = 0.9 kmph . answer : a" | a = 9 / 36
b = const_3_6 * a
|
a ) 2.5 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | c | divide(120, multiply(160, const_0_2778)) | in what time will a train 120 m long cross an electric pole , it its speed be 160 km / hr ? | "speed = 160 * 5 / 18 = 44 m / sec time taken = 120 / 44 = 2.7 sec . answer : c" | a = 160 * const_0_2778
b = 120 / a
|
a ) 0 , b ) 8 , c ) 16 , d ) 24 , e ) 4 | d | divide(divide(64, const_4), const_2) | all the faces of cubes are painted with red colour . it is then cut into 64 equal small cubes . find how many small cubes are there whose two adjacent faces are coloured red ? | "there are 64 small cubes , hence one side side of the big cube is 3 β 64 = 4 cm number of small cubes having two adjacent faces coloured red = ( x - 2 ) x no . of edges = ( 4 - 2 ) x 12 = 24 answer : d" | a = 64 / 4
b = a / 2
|
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