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 ) 40.3 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | a | divide(subtract(add(multiply(40.2, 10), add(13, 17)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 17 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 17 = 403 hence , new average = 403 β 10 = 40.3 answer a" | a = 40 * 2
b = 13 + 17
c = a + b
d = c - 31
e = d / 10
|
a ) 4 , b ) 5 / 2 , c ) 2 , d ) 3 / 2 , e ) 5 / 4 | e | divide(add(multiply(3, const_4), 3), const_10) | nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 4 screws and 2 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 4 s and 2 n = 4 s + 2 * 4 s = 12 s . . ratio = 15 s / 12 s = 15 / 12 = 5 / 4 answer : e" | a = 3 * 4
b = a + 3
c = b / 10
|
a ) rs . 15000 , b ) rs . 15550 , c ) rs . 24750 , d ) rs . 16500 , e ) none of these | c | multiply(1200, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1200 per sq . metre . | "solution area of the floor = ( 5.5 Γ 3.75 ) m 2 = 20.625 m 2 β΄ cost of paving = rs . ( 1200 Γ 20.625 ) = 24750 . answer c" | a = 5 * 5
b = 1200 * a
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a ) 30 , b ) 20 , c ) 19 , d ) 29 , e ) 39 | a | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 6), divide(2, 3))) | 15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 6 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 6 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 6 * ( 2 / 3 ) ) days = > 30 days answer : a | a = 15 * 21
b = a * 8
c = 21 * 6
d = 2 / 3
e = c * d
f = b / e
|
a ) $ 540 , b ) $ 580 , c ) $ 600 , d ) $ 650 , e ) $ 680 | d | subtract(9750, add(add(divide(9750, add(const_1, const_2)), divide(9750, add(4, const_1))), divide(9750, add(divide(3, const_2), const_1)))) | 4 brothers adam , bill , charles and david together contributed certain amount of money and purchased a car . the contribution made by adam is half that of the total contribution made by the other 3 brothers , contribution made by bill is one - fourth of the total contribution made by the other 3 brothers and the contribution made by charles is two - third of the total contribution made by the other 3 brothers . if the total price of the car is $ 9750 , find the contribution made by david . | given that a : ( b + c + d ) = 1 : 2 . hence a contributed 1 / 3 rd of the total price . given b : ( a + c + d ) = 1 : 4 hence b contributed 1 / 5 th of the total price . given c : ( a + b + d ) = 2 : 3 hence c contributed 2 / 5 th of the total price . thus the contribution made by a , b and c = 1 / 3 + 1 / 5 + 2 / 5 = 14 / 15 so , the remaining 1 / 15 th of the price is contributed by david . hence david ' s contribution = 1 / 15 Γ 9750 = $ 650 . the answer is d . | a = 1 + 2
b = 9750 / a
c = 4 + 1
d = 9750 / c
e = b + d
f = 3 / 2
g = f + 1
h = 9750 / g
i = e + h
j = 9750 - i
|
a ) 900 , b ) 980 , c ) 1200 , d ) 1240 , e ) 2000 | e | subtract(divide(6000, 2), divide(6000, 6)) | share rs . 6000 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ? | "amount received by sanjay . 4 / 12 x 6000 = 2000 = ( related ratio / sum of ratio ) x total amount so , the amount received by sanjay is 2000 . e" | a = 6000 / 2
b = 6000 / 6
c = a - b
|
a ) 40 , b ) 60 , c ) 45 , d ) 39 , e ) none | b | divide(120, 2) | a worker makes a toy in every 2 h . if he works for 120 h , then how many toys will he make ? | "no . of toys = 120 / 2 = 60 answer : b" | a = 120 / 2
|
a ) 2500 , b ) 2400 , c ) 2300 , d ) 2200 , e ) 2100 | a | add(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of all odd numbers up to 100 . | "the given numbers are 1 , 3 , 5 , 7 , . . . , 99 . this is an a . p . with a = 1 and d = 2 . let it contain n terms . then , 1 + ( n - 1 ) x 2 = 99 or n = 50 . required sum = n / 2 ( first term + last term ) = 50 / 2 ( 1 + 99 ) = 2500 . answer is a ." | a = 100 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 100 / e
g = f / 2
h = 100 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 100 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 450 , b ) 760 , c ) 110 , d ) 180 , e ) 400 | b | multiply(multiply(subtract(20, const_1), 4), divide(20, const_2)) | there are , in a certain league , 20 teams , and each team face another team for a total of 4 times . how many games are played in the season ? | "by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 760 option b ." | a = 20 - 1
b = a * 4
c = 20 / 2
d = b * c
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 2 , d ) 4 / 5 , e ) 6 | e | divide(divide(12, 15), divide(2, 15)) | if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a Γ’ Λ Βͺ b ) = 12 / 15 find p ( b | a ) | "p ( b | a ) = p ( a Γ’ Λ Βͺ b ) / p ( a ) p ( b | a ) = ( 12 / 15 ) / ( 2 / 15 ) = 6 . e" | a = 12 / 15
b = 2 / 15
c = a / b
|
['a ) 100', 'b ) 250', 'c ) 1875', 'd ) 1200', 'e ) 5625'] | c | multiply(volume_rectangular_prism(20, 25, divide(6, add(const_10, const_2))), 7.5) | the water level in a rectangular swimming pool measuring 20 feet by 25 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons ) | 6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 20 * 25 * 1 / 2 = 250 feet ^ 3 of water must be removed , which equals to 250 * 7.5 = 1875 gallons . answer : c . | a = 10 + 2
b = 6 / a
c = volume_rectangular_prism * (
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 18 | d | multiply(divide(const_100, 12), multiply(divide(9, const_100), 20)) | a man buys rs . 20 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | dividend on rs 20 = 9 / 100 * 20 = 9 / 5 is 12 income 100 9 / 5 is an income on rs ( 100 / 12 * 9 / 5 ) = rs 15 answer d | a = 100 / 12
b = 9 / 100
c = b * 20
d = a * c
|
a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 71.17 , e ) 81.81 | e | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 90)))) | a car travels first 160 km at 75 km / hr and the next 160 km at 90 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 90 km / hr time taken to travel next 160 km = distancespeed = 160 / 90 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 160 / 75 + 160 / 90 average speed = total distance traveled / total time taken = 320 / ( 160 / 75 + 160 / 90 ) = 81.81 km / hr answer : e" | a = 160 / 75
b = 160 / 90
c = a + b
d = 1 / c
e = 320 * d
|
a ) 60 , b ) 78 , c ) 267 , d ) 29 , e ) 27 | a | subtract(add(200, 360), 500) | a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 360 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 360 - - - - - - - - - - - - - - a + b + 2 c = 560 a + b + c = 500 - - - - - - - - - - - - - - - - c = 60 answer : a" | a = 200 + 360
b = a - 500
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(add(sqrt(add(multiply(multiply(190, const_2), const_4), const_1)), const_1), const_2) | if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 190 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 190 n * ( n - 1 ) = 380 = 20 * 19 ( trial and error ) the answer is d ." | a = 190 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
a ) 10 men , b ) 30 men , c ) 40 men , d ) 20 men , e ) none of these | d | divide(10, subtract(const_1, divide(10, add(10, 10)))) | a certain number of men can finish a piece of work in 10 days . however , if there were 10 less men it will take 10 days more for the work to be finished . how many men were there originally ? | "let x be the number of men originally thus , 10 x = ( 10 + 10 ) ( x - 10 ) 10 x = 20 ( x - 10 ) x = 2 ( x - 10 ) x = 20 answer : d" | a = 10 + 10
b = 10 / a
c = 1 - b
d = 10 / c
|
a ) 20 , b ) 21 , c ) 25 , d ) 30 , e ) 55 | b | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 4 and 34 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 5 = 105 / 5 = 21 option b" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
['a ) 21', 'b ) 787', 'c ) 37', 'd ) 26', 'e ) 01'] | a | multiply(divide(8, 3), 8) | how can a cake ( circular ) be cut into 8 pieces by making just 3 cuts ? | explanation : cut the cake using three mutually perpendicular planes . it leaves 8 pieces , one each in each octant . answera | a = 8 / 3
b = a * 8
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a ) rs . 360 , b ) rs . 432 , c ) rs . 540 , d ) rs . 1080 , e ) rs . 1280 | a | divide(multiply(60, 72), subtract(72, 60)) | the bankers discount of a certain sum of money is rs . 72 and the true discount on the same sum for the same time is rs . 60 . the sum due is : | "sum = ( b . d * t . d ) / ( b . d - t . d ) ( 72 * 60 ) / 72 - 60 ; 360 answer : a" | a = 60 * 72
b = 72 - 60
c = a / b
|
a ) rs . 15,000 , b ) rs . 15,500 , c ) rs . 15,600 , d ) rs . 20,625 , e ) none | d | multiply(1000, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 1000 per sq . metre . | "solution area of the floor = ( 5.5 x 3.75 ) m Β² = 20.635 m Β² cost of paying = rs . ( 1000 x 20.625 ) = rs . 20625 . answer d" | a = 5 * 5
b = 1000 * a
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(17.28, multiply(multiply(3.6, 0.2), 2)) | find the value of x in [ ( 17.28 / x ) / ( 3.6 * 0.2 ) ] = 2 | ( 17.28 / x ) = 2 * 3.6 * 0.2 x = ( 17.28 / 1.44 ) = ( 1728 / 14 ) = 12 . option b | a = 3 * 6
b = a * 2
c = 17 / 28
|
a ) 600 , b ) 800 , c ) 1000 , d ) 1200 , e ) 1400 | c | multiply(subtract(62, 22), add(divide(subtract(70, 22), const_2), const_1)) | set a contains all the even numbers between 22 and 70 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 40 more than the corresponding term in set a . the difference of the sums = 25 * 40 = 1000 . the answer is c ." | a = 62 - 22
b = 70 - 22
c = b / 2
d = c + 1
e = a * d
|
a ) 27 , b ) 87 , c ) 48 , d ) 77 , e ) 51 | c | divide(add(multiply(40, add(const_1, divide(50, const_100))), 40), const_2) | a person travels from p to q a speed of 40 km / hr and returns by increasing his speed by 50 % . what is his average speed for both the trips ? | "speed on return trip = 150 % of 40 = 60 km / hr . average speed = ( 2 * 40 * 60 ) / ( 40 + 60 ) = 48 km / hr . answer : c" | a = 50 / 100
b = 1 + a
c = 40 * b
d = c + 40
e = d / 2
|
a ) 12.0 , b ) 30.0 , c ) 30.9 , d ) 40.0 , e ) 60.0 | a | subtract(multiply(9.2, 5), multiply(6.8, 5)) | the average of 5 numbers is 6.8 . if one of the numbers is multiplied by a factor of 2 , the average of the numbers increases to 9.2 . what number is multiplied by 2 ? | the average of 5 numbers is 6.8 the sum of 5 numbers will be 6.8 x 5 = 34 the average of 5 number after one of the number is multiplied by 2 is 9.2 the sum of the numbers will now be 9.2 x 5 = 46 so the sum has increased by 46 - 34 = 12 let the number multiplied by 2 be n then , 2 n = n + 12 or n = 12 answer : - a | a = 9 * 2
b = 6 * 8
c = a - b
|
a ) 321 , b ) 276 , c ) 342 , d ) 228 , e ) 245 | d | subtract(subtract(345, divide(multiply(345, 12), const_100)), divide(multiply(subtract(345, divide(multiply(345, 12), const_100)), 25), const_100)) | the sale price sarees listed for rs . 345 after successive discount is 12 % and 25 % is ? | "explanation : 345 * ( 88 / 100 ) * ( 75 / 100 ) = 228 answer : d" | a = 345 * 12
b = a / 100
c = 345 - b
d = 345 * 12
e = d / 100
f = 345 - e
g = f * 25
h = g / 100
i = c - h
|
['a ) 16', 'b ) 24', 'c ) 32', 'd ) 48', 'e ) none of these'] | a | divide(subtract(subtract(multiply(20, 10), multiply(add(add(add(20, 20), add(const_3, const_3)), add(const_3, const_3)), const_2)), divide(subtract(multiply(20, 10), multiply(add(add(add(20, 20), add(const_3, const_3)), add(const_3, const_3)), const_2)), const_3)), const_4) | the length and breadth of the floor of the room are 20 feet and 10 feet respectively . square tiles of 2 feet length of different colours are to be laid on the floor . black tiles are laid in the first row on all sides . if white tiles are laid in the one - third of the remaining and blue tiles in the rest , how many blue tiles will be there ? | area left after laying black tiles = [ ( 20 β 4 ) Γ ( 10 β 4 ) ] sq . ft . = 96 sq . ft . area under white tiles = ( 1 β 3 Γ 96 ) sq . ft = 32 sq . ft . area under blue tiles = ( 96 β 32 ) sq . ft = 64 sq . ft . number of blue tiles = 64 / 2 Γ 2 = 16 . answer a | a = 20 * 10
b = 20 + 20
c = 3 + 3
d = b + c
e = 3 + 3
f = d + e
g = f * 2
h = a - g
i = 20 * 10
j = 20 + 20
k = 3 + 3
l = j + k
m = 3 + 3
n = l + m
o = n * 2
p = i - o
q = p / 3
r = h - q
s = r / 4
|
a ) 9 , b ) 19 , c ) 27 , d ) 26 , e ) 31 | c | subtract(36, subtract(add(36, 32), 59)) | in a group of 89 students , 36 are taking history , and 32 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ? | "number of students taking history = h = 36 number of students taking statistics = s = 32 total number of students = t = 89 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 89 - 59 = 30 letxbe the number of students taking both history and statistics . then t = h + s + n - x or 89 = 36 + 32 + 30 - x or x = 9 now , number of students taking only history will be h - x or 36 - 9 = 27 answer : - c" | a = 36 + 32
b = a - 59
c = 36 - b
|
a ) 160 , b ) 220 , c ) 264 , d ) 360 , e ) 420 | c | add(divide(multiply(divide(55, const_100), 22), subtract(divide(60, const_100), divide(55, const_100))), 22) | the workforce of company x is 60 % female . the company hired 22 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 22 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 22 ) 0.05 x = 12.1 x = 242 - this is quantity of employees before adding 22 men so after adding it will be 264 answer is c" | a = 55 / 100
b = a * 22
c = 60 / 100
d = 55 / 100
e = c - d
f = b / e
g = f + 22
|
a ) rs 8.5 , b ) rs 9.5 , c ) rs 10.5 , d ) rs 11.5 , e ) none of these | c | divide(multiply(8, add(const_100, 5)), subtract(const_100, 20)) | a fruit seller sells mangoes at the rate of rs . 8 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % | explanation : 80 : 8 = 105 : x x = ( 8 Γ 105 / 80 ) = rs 10.5 option c | a = 100 + 5
b = 8 * a
c = 100 - 20
d = b / c
|
a ) 3 / 6 , b ) 3 / 2 , c ) 2 / 3 , d ) 3 / 5 , e ) 5 / 2 | c | divide(subtract(27, 21), subtract(21, 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 21 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 ) = 21 = = > 27 x + 17 y = 21 x + 21 y = = > 6 x = 4 y = = > x / y = 2 / 3 . answer : c" | a = 27 - 21
b = 21 - 17
c = a / b
|
a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(9), log(power(9, 11))) | if n = 9 ^ 11 β 9 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 9 ^ 3 - 9 . now 9 ^ 3 has last digit 9 . , we subtract 9 from 9 , thus 9 - 9 = 0 is the answer . option a | a = math.log(9)
b = 9 ** 11
c = math.log(b)
d = a / c
|
a ) 3 sec , b ) 2.6 sec , c ) 3.6 sec , d ) 8.6 sec , e ) 7.6 sec | d | multiply(const_3600, divide(divide(add(210, 210), const_1000), add(90, 85))) | two trains each 210 m in length each , are running on two parallel lines in opposite directions . if one goes at the speed of 90 km / h while the other travels at 85 km / h . how long will it take for them to pass each other completely . | "explanation : d = 210 m + 210 m = 420 m rs = 90 + 85 = 175 * 5 / 18 = 243 / 5 t = 420 * 5 / 243 = 8.6 sec answer : option d" | a = 210 + 210
b = a / 1000
c = 90 + 85
d = b / c
e = 3600 * d
|
a ) 3.75 days , b ) 4.44 days , c ) 5 days , d ) 6.44 days , e ) 7 days | b | divide(multiply(10, 8), add(8, 10)) | b completes a work in 8 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | "1 / 8 + 1 / 10 = 9 / 40 40 / 9 = 4.44 days answer : b" | a = 10 * 8
b = 8 + 10
c = a / b
|
a ) 2500 , b ) 2800 , c ) 3500 , d ) 1900 , e ) 1500 | a | divide(divide(multiply(1000, const_100), 10), const_4) | a sum fetched a total simple interest of rs . 1000 at the rate of 10 p . c . p . a . in 4 years . what is the sum ? | sol . principal = rs . [ 100 * 1000 / 10 * 4 ] = rs . [ 100000 / 40 ] = rs . 2500 . answer a | a = 1000 * 100
b = a / 10
c = b / 4
|
a ) $ 3 , b ) $ 6 , c ) $ 9 , d ) $ 12 , e ) $ 15 | b | subtract(divide(360, divide(10, subtract(divide(const_3, const_2), const_1))), divide(360, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | p and q are the only two applicants qualified for a short - term research project that pays 360 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job . candidate p β s hourly wage is how many dollars greater than candidate q β s hourly wage ? | "let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 360 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 360 / 30 = $ 12 p ' s hourly wage is 360 / ( t - 10 ) = $ 18 which is $ 6 per hour more . the answer is b ." | a = 3 / 2
b = a - 1
c = 10 / b
d = 360 / c
e = 3 / 2
f = e - 1
g = 10 / f
h = g + 10
i = 360 / h
j = d - i
|
a ) 25 % , b ) 50 % , c ) 69 % , d ) 80 % , e ) 100 % | c | multiply(divide(divide(3, 5), divide(7, 8)), const_100) | a cylinder of height h is 7 / 8 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 7 / 8 new cylinder 3 / 5 so 3 / 5 / 7 / 8 = 3 / 5 * 8 / 7 = 0.69 or 69 % . answer c" | a = 3 / 5
b = 7 / 8
c = a / b
d = c * 100
|
a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 1000 | e | divide(250, subtract(const_1, divide(3, 4))) | linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 250 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 250 . so 1 / 4 of her savings is $ 250 . so her original savings are 4 times $ 250 = $ 1000 correct answer e" | a = 3 / 4
b = 1 - a
c = 250 / b
|
a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 39.5 % , e ) 40 % | d | subtract(multiply(divide(subtract(const_100, 7), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 7 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.93 * 150 x = 139.5 x hence total profit = 39.5 % option d" | a = 100 - 7
b = a / 100
c = 100 + 20
d = 100 + 25
e = d / 100
f = c * e
g = b * f
h = g - 100
|
a ) $ 880 , b ) $ 950 , c ) $ 1100 , d ) $ 1240 , e ) $ 1360 | a | multiply(divide(56, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 56 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 56 0.07 = 56 p = 800 the original selling price was 800 * 1.1 = 880 . the correct answer is a ." | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 + f
h = e - g
i = 56 / h
j = 10 / 100
k = 1 + j
l = i * k
|
a ) 188 m , b ) 876 m , c ) 251 m , d ) 50 m , e ) 45 m | d | multiply(100, subtract(const_2, const_1)) | a train speeds past a pole in 10 seconds and a platform 100 m long in 30 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 100 / 30 = x / 10 x = 50 m . answer : d" | a = 2 - 1
b = 100 * a
|
a ) 18 , b ) 25 , c ) 16 , d ) 72 , e ) 27 | c | divide(subtract(112, multiply(16, 6)), subtract(17, 16)) | suraj has a certain average of runs for 16 innings . in the 17 th innings he scores 112 runs thereby increasing his average by 6 runs . what is his average after the 17 th innings ? | "to improve his average by 6 runs per innings he has to contribute 16 x 6 = 96 runs for the previous 16 innings . thus , the average after the 17 th innings = 112 - 96 = 16 . answer : c" | a = 16 * 6
b = 112 - a
c = 17 - 16
d = b / c
|
a ) 12,4 , b ) 6,3 , c ) 9,3 , d ) 6,6 , e ) none of these | a | divide(subtract(16, 8), const_2) | a man can row downstream at the rate of 16 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 16 + 8 ) = 12 km / hr rate of current = 1 / 2 ( 16 - 8 ) = 4 km / hr answer : option a" | a = 16 - 8
b = a / 2
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a ) 164 , b ) 174 , c ) 184 , d ) 194 , e ) 204 | c | divide(multiply(multiply(const_2, add(11, const_1)), add(multiply(11, const_2), const_1)), const_3) | the average of squares of first 11 consecutive even numbers is : | 2 ( n + 1 ) ( 2 n + 1 ) / 3 ( 2 * 12 * 23 ) / 3 = 2 * 4 * 23 = 184 answer : c | a = 11 + 1
b = 2 * a
c = 11 * 2
d = c + 1
e = b * d
f = e / 3
|
a ) 126 , b ) 156 , c ) 190 , d ) 321 , e ) 220 | e | multiply(add(divide(subtract(divide(divide(1430, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(1430, 6.5), const_2), 10), const_2), 10)), const_2) | the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1430 . the perimeter of the plot is ? | "sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1430 x = 50 required perimeter = 2 ( 50 + 60 ) = 220 e" | a = 1430 / 6
b = a / 2
c = b - 10
d = c / 2
e = 1430 / 6
f = e / 2
g = f - 10
h = g / 2
i = h + 10
j = d + i
k = j * 2
|
a ) 8 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | e | divide(subtract(69, 54), subtract(54, 51)) | rahul played weel in this season . his current batting averagge is 51 . if he score 69 runs in today match . his batting average will become 54 . how many matches had he played in this season . | 51 x + 69 = 54 ( x + 1 ) = > 3 x = 15 = > x = 5 answer : e | a = 69 - 54
b = 54 - 51
c = a / b
|
a ) 62.2 , b ) 2000 , c ) 1700 , d ) 1729 , e ) none | d | subtract(multiply(divide(1852, const_100), 1230), multiply(divide(const_1, const_3), multiply(divide(1852, const_100), 1230))) | 1852 - 1230 Γ· 10.00 = ? | "answer given expression = 1852 - 1230 Γ· 10.00 = 1852 - 123 = 1729 correct option : d" | a = 1852 / 100
b = a * 1230
c = 1 / 3
d = 1852 / 100
e = d * 1230
f = c * e
g = b - f
|
a ) 4063 , b ) 5325 , c ) 5351 , d ) 6000 , e ) 6154 | e | divide(multiply(5000, divide(8, const_100)), divide(6.5, const_100)) | last year a certain bond price with a face value of 5000 yielded 8 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ? | "interest = 0.08 * 5000 = 0.065 * selling price - - > selling price = 0.08 * 5000 / 0.065 - - > selling price = ~ 6,154 answer : e ." | a = 8 / 100
b = 5000 * a
c = 6 / 5
d = b / c
|
a ) 10 , b ) 99 , c ) 27 , d ) 4 , e ) 29 | d | multiply(divide(multiply(4, 5), subtract(multiply(8, 4), multiply(3, 4))), 4) | mixture contains alcohol and water in the ratio 4 : 3 . if 8 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 32 x = 4 ( 3 x + 5 ) 20 x = 20 x = 1 quantity of alcohol = ( 4 x 1 ) litres = 4 litres . answer : d" | a = 4 * 5
b = 8 * 4
c = 3 * 4
d = b - c
e = a / d
f = e * 4
|
a ) 124 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | d | add(subtract(const_100, 50), multiply(subtract(const_100, 50), divide(60, const_100))) | mary ' s income is 60 % more than tim ' s income and tim ' s income is 50 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.5 = 50 m = 50 * 1.6 = 80 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 80 * 100 / 100 = 80 ans : d" | a = 100 - 50
b = 100 - 50
c = 60 / 100
d = b * c
e = a + d
|
a ) 11 , b ) 50 , c ) 78 , d ) 27 , e ) 18 | c | add(40, 38) | laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 40 kmph and prasanna moves southward at a speed of 38 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 40 kmph x 1 hr = 40 km total southward prasanna ' s distance = 38 kmph x 1 hr = 38 km total distance between prasanna and laxmi is = 40 + 38 = 78 km . answer : c" | a = 40 + 38
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a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 36 % | b | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(70, 100))), add(400, 100)), 100) | a grocer has 400 pounds of coffee in stock , 20 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 70 percent is decaffeinated , what percent , by weight , of the grocer β s stock of coffee is decaffeinated ? | "1 . 20 % of 400 = 80 pounds of decaffeinated coffee 2 . 70 % of 100 = 70 pounds of decaffeinated coffee 3 . wt have 150 pounds of decaffeinated out of 500 pounds , that means 150 / 500 * 100 % = 30 % . the correct answer is b ." | a = 20 / 100
b = a * 400
c = 70 / 100
d = 100 * c
e = b + d
f = 400 + 100
g = e / f
h = g * 100
|
a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.27 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.5), 1.567), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | the value of ( 34.31 * 0.5 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.5 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 26.882 / 11.845 = 2.27 answer : d" | a = 34 * 31
b = a * 1
c = 7 * 57
d = c * 0
e = b / d
f = e / 10
|
a ) 9 years , b ) 11 years , c ) 14 years , d ) 21 years , e ) 25 years | b | subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16))) | the average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 5 + 16 x 9 ) ] = ( 225 - 214 ) = 11 years . answer b" | a = 15 * 15
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
|
a ) 14300 , b ) 14400 , c ) 14500 , d ) 14600 , e ) 15400 | b | add(10000, multiply(divide(multiply(10000, 20), const_100), 2)) | the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ 120 / 100 Γ 120 / 100 = 14400 b )" | a = 10000 * 20
b = a / 100
c = b * 2
d = 10000 + c
|
a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.25 , e ) 0.0156 | e | power(divide(1, 4), const_3) | what is the decimal equivalent of ( 1 / 4 ) ^ 3 ? | ( 1 / 4 ) ^ 3 = 1 / 64 = 0.0156 answer : e | a = 1 / 4
b = a ** 3
|
a ) 79 , b ) 89 , c ) 95 , d ) 72.72 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 40)), add(6, 5))) | a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 40 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 40 * 5 = 200 student ' s average = ( 600 + 200 ) / 11 = 72.72 answer d" | a = 6 * 100
b = 5 * 40
c = a + b
d = 6 + 5
e = c / d
f = math.floor(e)
|
a ) 87.5 , b ) 89.5 , c ) 82.3 , d ) 82.1 , e ) 92.4 | a | divide(multiply(add(66.5, divide(multiply(66.5, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 66.5 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 66.5 sp = 66.5 * ( 125 / 100 ) = 83.125 mp * ( 95 / 100 ) = 59.375 mp = 87.5 answer : a" | a = 66 * 5
b = a / 100
c = 66 + 5
d = c * 100
e = 100 - 5
f = d / e
|
a ) 1 / 36 , b ) 1 / 216 , c ) 1 / 48 , d ) 1 / 55 , e ) 1 / 90 | a | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | three 6 faced dice are thrown together . the probability that all the three show the same number on them is - . | "it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 and 666 . those are six in number . further the three dice can fall in 6 * 6 * 6 = 216 ways . hence the probability is 6 / 216 = 1 / 36 answer a" | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 1008 , b ) 1015 , c ) 1022 , d ) 1032 , e ) none of these | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 7) | the smallest number which when diminished by 7 , is divisible 12 , 16 , 18 , 21 and 28 is : | "explanation required number = ( l . c . m . of 12,16 , 18 , 21 , 28 ) + 7 = 1008 + 7 = 1015 answer b" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 7
|
a ) 10 , b ) 11 , c ) 12 , d ) 16 , e ) 14 | d | add(15, const_1) | the average of first 15 even numbers is ? | "sum of 15 even numbers = 15 * 16 = 240 average = 240 / 15 = 16 answer : d" | a = 15 + 1
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a ) 48 , b ) 50 , c ) 62 , d ) 78 , e ) 100 | a | multiply(subtract(divide(multiply(divide(3, add(3, 2)), 100), divide(3, add(3, 4))), 100), divide(add(const_10, const_2), const_10)) | in 100 litres mixture milk and water are in the ratio 3 : 2 . after adding how many liters of water its ratio will become 3 : 4 | milk quantity = 3 / 5 * 60 = 36 water quantity = 100 - 36 = 64 new ratio of m : w = 36 : 64 + x = 3 : 4 192 + 3 x = 144 x = 48 answer is a | a = 3 + 2
b = 3 / a
c = b * 100
d = 3 + 4
e = 3 / d
f = c / e
g = f - 100
h = 10 + 2
i = h / 10
j = g * i
|
a ) 5768 , b ) 8925 , c ) 8965 , d ) 6474 , e ) 8723 | c | divide(divide(multiply(4034.25, const_100), 9), 5) | a sum fetched a total simple interest of 4034.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4034.25 ) / ( 9 x 5 ) = 403425 / 45 = 8965 . answer c" | a = 4034 * 25
b = a / 9
c = b / 5
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a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100) | rani bought more apples than oranges . she sells apples at βΉ 23 apiece and makes 15 % profit . she sells oranges at βΉ 10 apiece and makes 25 % profit . if she gets βΉ 653 after selling all the apples and oranges , find her profit percentage w . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % w = ( 97 / 556 ) * 100 = 17.4 % answer : b" | a = 2 * 10
b = 2 * 10
c = b + 1
d = a * c
e = 25 / 100
f = e + 1
g = 10 / f
h = 15 + 2
i = g * h
j = d + i
k = 653 - j
l = 2 * 10
m = 2 * 10
n = m + 1
o = l * n
p = 25 / 100
q = p + 1
r = 10 / q
s = 15 + 2
t = r * s
u = o + t
v = k / u
w = v * 100
|
a ) 2 : 9 , b ) 2 : 5 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | b | divide(subtract(4, 2), subtract(9, 4)) | cereal a is 9 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "( 9 / 100 ) a + ( 2 / 100 ) b = ( 4 / 100 ) ( a + b ) 5 a = 2 b = > a / b = 2 / 5 answer is b ." | a = 4 - 2
b = 9 - 4
c = a / b
|
a ) 33 , b ) 56 , c ) 55 , d ) 82 , e ) 62 | b | divide(add(180, 660), multiply(54, const_0_2778)) | how long does a train 180 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | "explanation : t = ( 660 + 180 ) / 54 * 18 / 5 t = 56 answer : option b" | a = 180 + 660
b = 54 * const_0_2778
c = a / b
|
a ) rs . 200 , b ) rs . 100 , c ) rs . 300 , d ) rs . 50 , e ) rs . 90 | b | subtract(divide(divide(500, add(divide(1, 2), divide(1, 3))), 2), divide(divide(500, add(divide(1, 2), divide(1, 3))), 3)) | a profit of rs . 500 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 500 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 300 and 200 . difference in profit share = 300 - 200 = 100 answer : b" | a = 1 / 2
b = 1 / 3
c = a + b
d = 500 / c
e = d / 2
f = 1 / 2
g = 1 / 3
h = f + g
i = 500 / h
j = i / 3
k = e - j
|
a ) s . 12 , b ) s . 15 , c ) s . 18 , d ) s . 21 , e ) s . 42 | e | multiply(divide(const_100, 12), multiply(divide(9, const_100), 56)) | a man buys rs . 56 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | "dividend on rs . 56 = rs . 9 / 100 x 56 = rs . 5.04 . rs . 12 is an income on rs . 100 . rs . 5.04 is an income on rs . 100 / 12 x 5.04 = rs . 42 . answer : option e" | a = 100 / 12
b = 9 / 100
c = b * 56
d = a * c
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 55 | e | subtract(multiply(multiply(divide(440, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(440, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), subtract(const_1, divide(25, const_100)))) | of 440 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 440 = 1.25 b + b - 0.25 b - - > b = 220 , a = 1.25 b = 275 and { both } = 0.25 b = 55 . the number of students who read only book a is { a } - { both } = 125 - 25 = 100 ; the number of students who read only book b is { b } - { both } = 275 - 55 - 220 ; the difference is 220 - 55 = 165 . answer : e ." | a = 25 / 100
b = 20 / 100
c = a / b
d = c + 1
e = 25 / 100
f = d - e
g = 440 / f
h = 25 / 100
i = 20 / 100
j = h / i
k = g * j
l = 20 / 100
m = 1 - l
n = k * m
o = 25 / 100
p = 20 / 100
q = o / p
r = q + 1
s = 25 / 100
t = r - s
u = 440 / t
v = 25 / 100
w = 1 - v
x = u * w
y = n - x
|
a ) 10.5 % , b ) 12.5 % , c ) 14.3 % , d ) 22 % , e ) 30 % | c | multiply(divide(10, subtract(const_100, 30)), const_100) | on a certain road , 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 30 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 10 % of them exceeded the speed limit and received the ticket , i . e . x / 10 . again , suppose total no . of motorists who exceeded the speed limit are y . 30 % of y exceeded the speed limit but did n ' t received the ticket , i . e . 3 y / 10 . it means 7 y / 10 received the ticket . hence , 7 y / 10 = x / 10 or y / x = 1 / 7 or y / x * 100 = 1 / 7 * 100 = 14.3 % c" | a = 100 - 30
b = 10 / a
c = b * 100
|
a ) 10 hrs , b ) 15 hrs , c ) 20 hrs , d ) 25 hrs , e ) 18 hrs | b | divide(750, 50) | ajay can ride 50 km in 1 hour . in how many hours he can ride 750 km ? | "1 hour he ride 50 km he ride 750 km in = 750 / 50 * 1 = 15 hours answer is b" | a = 750 / 50
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a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 17 | c | divide(47, 3) | a number of 47 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 14 * 3 + 1 * 5 = 47 - - > 14 + 1 = 15 . answer : c ." | a = 47 / 3
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a ) 200 , b ) 240 , c ) 198 , d ) 190 , e ) 322 | a | divide(add(12, 178), divide(95, const_100)) | 12 is subtracted from 95 % of a number , the result is 178 . find the number ? | "( 95 / 100 ) * x Γ’ β¬ β 12 = 178 9.5 x = 1900 x = 200 answer : a" | a = 12 + 178
b = 95 / 100
c = a / b
|
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these | d | divide(subtract(multiply(8000, 6), multiply(8000, 5)), subtract(8, 6)) | a man invests rs . 8000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | explanation : interest on rs . 8000 at 5 % per annum = ( 8000 Γ 5 Γ 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ 8 Γ 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) Γ 6 Γ 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) Γ 6 Γ 1 / 100 . = > 40000 + 8 x = ( 8000 + x ) Γ 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d | a = 8000 * 6
b = 8000 * 5
c = a - b
d = 8 - 6
e = c / d
|
a ) $ 4.31 , b ) $ 4.51 , c ) $ 4.21 , d ) $ 5.21 , e ) $ 5.61 | c | divide(4, add(add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5)), divide(const_1, 6))) | the cost per unit of electricity during a 4 year period is as follows : year 1 , $ 3 per unit ; year 2 , $ 4 per unit ; year 3 , $ 5 per unit ; year 4 , $ 6 per unit . if a factory spent equal dollars on electricity each year , what was the average cost of a unit of electricity during the 4 year period ? | let money spent each year = $ 60 units bought first year = 20 units bought second year = 15 units bought third year = 12 units bought fourth year = 10 total number of units = 57 total money spent = 60 * 4 = 240 average = 240 / 57 = 4.21 answer : c | a = 1 / 3
b = 1 / 4
c = a + b
d = 1 / 5
e = c + d
f = 1 / 6
g = e + f
h = 4 / g
|
a ) 0.5 , b ) 1.25 , c ) 1.5 , d ) 2.5 , e ) 3.5 | c | divide(45, multiply(108, const_0_2778)) | in what time will a train 45 m long cross an electric pole , it its speed be 108 km / hr ? | "speed = 108 * 5 / 18 = 30 m / sec time taken = 45 / 30 = 1.5 sec . answer : c" | a = 108 * const_0_2778
b = 45 / a
|
a ) 3 / 4 , b ) 11 / 15 , c ) 7 / 10 , d ) 1 / 6 , e ) 2 / 15 | a | divide(add(10, 5), multiply(10, const_2)) | in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 10 of the housing units , including 1 / 4 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | 1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 4 ( equiped with cable tv ) i . e . 1 / 4 ( 1 / 5 ) = 1 / 20 therefore only video cassette recorder = 1 / 10 - 1 / 20 = 1 / 20 total = 1 / 5 + 1 / 20 + neither cable tv nor videocassette recorders 1 = 1 / 4 + neither cable tv nor videocassette recorders therefore neither cable tv nor videocassette recorders = 1 - 1 / 4 = 3 / 4 hence a . | a = 10 + 5
b = 10 * 2
c = a / b
|
a ) 50 , b ) 26 , c ) 27 , d ) 62.5 , e ) 21 | d | divide(multiply(45, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 45 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 45 = 25 / 9 = > x = 65.5 . answer : d" | a = 46 - 36
b = a * 1000
c = b / 3600
d = 45 * c
e = d / 2
|
a ) 19 , b ) 29 , c ) 39 , d ) 49 , e ) 59 | c | add(subtract(87, multiply(17, 3)), 3) | a batsman makes a score of 87 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "solution let the average after 17 th inning = x . then , average after 16 th inning = ( x - 3 ) average = 16 ( x - 3 ) + 87 = 17 x or x = ( 87 - 48 ) = 39 . answer c" | a = 17 * 3
b = 87 - a
c = b + 3
|
a ) 1 second , b ) 4.5 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these | a | divide(100, multiply(360, const_0_2778)) | in what time will a train 100 meters long cross an electric pole , if its speed is 360 km / hr | "explanation : first convert speed into m / sec speed = 360 * ( 5 / 18 ) = 100 m / sec time = distance / speed = 100 / 100 = 1 second answer : a" | a = 360 * const_0_2778
b = 100 / a
|
a ) 6 % . , b ) 7.5 % . , c ) 9.2 % . , d ) 10.5 % . , e ) 26.5 % . | e | multiply(divide(add(subtract(subtract(340, multiply(340, divide(64, const_100))), multiply(340, divide(9, const_100))), 3.2), add(add(add(340, 3.2), 8), 6.8)), const_100) | a 340 - liter solution of kola is made from 64 % water , 9 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 8 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ? | "denominator : 340 + 8 + 3.2 + 6.8 = 358 numerator : 340 ( 1 - . 64 - . 09 ) + 3.2 340 ( 0.27 ) + 3.2 91.8 + 3.2 95 ratio : 95 / 358 = 0.265 answer : e" | a = 64 / 100
b = 340 * a
c = 340 - b
d = 9 / 100
e = 340 * d
f = c - e
g = f + 3
h = 340 + 3
i = h + 8
j = i + 6
k = g / j
l = k * 100
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | divide(log(divide(multiply(power(5, 11), power(9, 6)), 3)), log(15)) | if ( 5 ^ 11 ) ( 9 ^ 6 ) = 3 ( 15 ^ x ) , what is the value of x ? | "( 5 ^ 11 ) ( 9 ^ 6 ) = 3 ( 15 ^ x ) = > 5 ^ 11 * 3 ^ 12 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 11 * 3 ^ 12 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 11 answer c" | a = 5 ** 11
b = 9 ** 6
c = a * b
d = c / 3
e = math.log(d)
f = math.log(15)
g = e / f
|
a ) 26 , b ) 32 , c ) 25 , d ) 27 , e ) 29 | b | add(add(multiply(5, 11), 5), 5) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 5 = 2 ^ 22 x 7 ^ 5 x 11 ^ 5 total number of prime factors = ( 22 + 5 + 5 ) = 32 . answer is b ." | a = 5 * 11
b = a + 5
c = b + 5
|
a ) 21 , b ) 36 , c ) 38 , d ) 40 , e ) 42 | a | divide(subtract(180, subtract(power(multiply(5, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 5 inches wide on each side , the total area would have been ( m + 180 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 2 ) ( b + 2 ) = m - - - - ( 2 ) and ( l + 10 ) ( b + 10 ) = m + 180 - - - - - - - - - > ( l + 10 ) ( b + 10 ) - 180 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 2 ) ( b + 2 ) = ( l + 10 ) ( b + 10 ) - 180 lb + 2 l + 2 b + 4 = lb + 10 l + 10 b + 100 - 180 simplify 8 l + 8 b = 84 - - - - - - > 2 ( l + b ) = 21 ( check eq ( 1 ) ) answer is a" | a = 5 * 2
b = a ** 2
c = 1 * 2
d = c ** 2
e = b - d
f = 180 - e
g = f / 2
|
a ) 79 , b ) 89 , c ) 95 , d ) 77.27 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 50)), add(6, 5))) | a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 50 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 50 * 5 = 250 student ' s average = ( 600 + 250 ) / 11 = 77.27 answer d" | a = 6 * 100
b = 5 * 50
c = a + b
d = 6 + 5
e = c / d
f = math.floor(e)
|
a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.5 , e ) $ 5.40 | d | add(2.35, multiply(0.35, divide(3.6, divide(2, 5)))) | jim β s taxi service charges an initial fee of $ 2.35 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim β s taxi service = 2.35 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.35 + ( 3.6 / . 4 ) * . 35 = 2.35 + 9 * . 35 = 5.5 $ answer d" | a = 2 / 5
b = 3 / 6
c = 0 * 35
d = 2 + 35
|
a ) 50 % , b ) 67 % , c ) 83 % , d ) 90 % , e ) 96 % | c | multiply(subtract(1, power(divide(divide(9, const_2), 9), const_2)), const_100) | a miniature roulette wheel is divided into 9 equal sectors , each bearing a distinct integer from 1 to 9 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun three times , approximately what is the probability that the product of the three winning sectors β integers will be even ? | "the only way to have an odd product is if all 3 integers are odd . p ( odd product ) = 5 / 9 * 5 / 9 * 5 / 9 = 125 / 729 p ( even product ) = 1 - 125 / 729 = 604 / 729 which is about 83 % the answer is c ." | a = 9 / 2
b = a / 9
c = b ** 2
d = 1 - c
e = d * 100
|
a ) 24 , b ) 18 , c ) 30 , d ) 15 , e ) 12 | b | multiply(6, 3) | local kennel has cats and dogs in the ratio of 2 : 3 . if there are 6 fewer cats than dogs , how many dogs are in the kennel ? | lets work with the data given to us . we know that there ratio of cats to dogs is 2 : 3 or cats 2 dogs 3 we can write number of cats as 2 x and number of dogs as 3 x and we know that 3 x - 2 x = 6 ( therefore x = 6 ) then # of dogs = 3 x 6 = 18 answer is b | a = 6 * 3
|
a ) - 2 , b ) 2 / 3 , c ) - 2 / 3 , d ) 1 / 4 , e ) 3 / 2 | b | divide(const_1, 3) | in the coordinate plane a slope of the line k is 3 / 2 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 3 / 2 mx + m , from 0 = 3 / 2 mx + m we get x = 2 / 3 . hence , the correct answer choice is b . | a = 1 / 3
|
a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 6800 , e ) s . 5600 | e | divide(multiply(multiply(multiply(24, 4), 4), 140000), multiply(multiply(multiply(10, 16), 6), 10)) | 10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 140000 , find the cost of a camel ? | cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 140000 p = rs . [ ( 24 * 4 * 4 * 140000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 53760000 / 9600 ) = > p = rs . 5600 answer : e | a = 24 * 4
b = a * 4
c = b * 140000
d = 10 * 16
e = d * 6
f = e * 10
g = c / f
|
a ) s . 933.33 , b ) s . 1009 , c ) s . 1007 , d ) s . 1006 , e ) s . 1002 | a | divide(multiply(140, const_100), subtract(add(const_100, 5), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 5 % . what is the cost price ? | "explanation : 90 % 105 % - - - - - - - - 15 % - - - - 140 100 % - - - - ? = > rs . 933.33 answer : a" | a = 140 * 100
b = 100 + 5
c = 100 - 10
d = b - c
e = a / d
|
a ) 14 , b ) 13 , c ) 12 , d ) 8 , e ) 10 | d | subtract(subtract(subtract(15, const_2), const_4), const_1) | how many positive integers less than 15 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | the number = 2 a + 3 b < 20 when a = 1 , b = 1 , 2 , 3 , 4 , 5 - > 2 a = 2 ; 3 b = 3 , 6 , 9 - > the number = 5 , 8 , 11 - - > 3 numbers when a = 2 , b = 1 , 2,3 - > . . . . - - > 3 numbers when a = 3 , b = 1,2 , 3,4 - - > . . . . - - > 2 numbers total number is already 8 . look at the answer there is no number greater than 8 - - > we dont need to try any more answer must be d | a = 15 - 2
b = a - 4
c = b - 1
|
a ) 110 , b ) 135 , c ) 150 , d ) 165 , e ) 185 | e | add(add(add(add(31, const_2), add(add(31, const_2), const_2)), add(add(add(31, const_2), const_2), const_2)), 41) | the sum of all consecutive odd integers from β 31 to 41 , inclusive , is | "the sum of the odd numbers from - 31 to + 31 is 0 . let ' s add the remaining numbers . 33 + 35 + 37 + 39 + 41 = 5 ( 37 ) = 185 the answer is e ." | a = 31 + 2
b = 31 + 2
c = b + 2
d = a + c
e = 31 + 2
f = e + 2
g = f + 2
h = d + g
i = h + 41
|
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 6 : 5 , e ) 2 : 3 | d | multiply(divide(2, 5), multiply(divide(2, 5), divide(6, 5))) | find the compound ratio of ( 2 : 5 ) , ( 6 : 11 ) and ( 11 : 2 ) is | "required ratio = 2 / 5 * 6 / 11 * 11 / 2 = 2 / 1 = 6 : 5 answer is d" | a = 2 / 5
b = 2 / 5
c = 6 / 5
d = b * c
e = a * d
|
a ) 16 , b ) 8 , c ) 9 , d ) 4 , e ) 3 | a | divide(multiply(subtract(const_100, 4), 24), add(const_100, 44)) | if a man lost 4 % by selling oranges at the rate of 24 a rupee at how many a rupee must he sell them to gain 44 % ? | "96 % - - - - 24 144 % - - - - ? 96 / 144 * 24 = 16 answer : a" | a = 100 - 4
b = a * 24
c = 100 + 44
d = b / c
|
a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40 | d | add(divide(subtract(multiply(multiply(6, 2), 3), multiply(6, 2)), 2), subtract(multiply(multiply(6, 2), 3), multiply(6, 2))) | the ratio of the number of females to males at a party was 1 : 2 but when 6 females and 6 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 6 ) = 2 x - 6 x = 12 there were 3 x = 36 people at the party originally . the answer is d ." | a = 6 * 2
b = a * 3
c = 6 * 2
d = b - c
e = d / 2
f = 6 * 2
g = f * 3
h = 6 * 2
i = g - h
j = e + i
|
a ) 70 min , b ) 94 min , c ) 109 min , d ) 114 min , e ) 120 min | d | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 6 robots altogether , if the robots build only one robot at a time ? | "first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 = 5 / 2 so they complete one robot in 2 / 5 th of an hour , i . e . 24 minutes - ( 3 ) now the 4 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 3 so they complete one robot in 1 / 3 th of an hour , i . e . 20 minutes - ( 4 ) and now we have 5 robots so total = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) = 114 minutes answer - d" | a = 1/(1)
b = 1/(2)
c = a + b
d = 1/(2)
e = c + d
f = 1/(e)
g = 1/(1)
h = 1/(2)
i = g + h
j = 1/(i)
k = f + j
|
a ) 50.4 % , b ) 23.8 % , c ) 66.5 % , d ) 34.9 % , e ) 43.5 % | c | multiply(divide(subtract(1200, add(add(add(75, 50), add(64, 78)), 135)), 1200), const_100) | john had a stock of 1200 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ? | let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 1200 - n = 1200 - 402 = 798 percentage books not sold / total number of books = 798 / 1200 = 0.665 = 66.5 % correct answer c | a = 75 + 50
b = 64 + 78
c = a + b
d = c + 135
e = 1200 - d
f = e / 1200
g = f * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | floor(divide(16, subtract(5, const_1))) | if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 16 ? | "@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 16 . then 4 n < 16 and n < 4 . the largest possible integer is n = 3 . the answer is c ." | a = 5 - 1
b = 16 / a
c = math.floor(b)
|
a ) 15 , b ) 20 , c ) 25 , d ) 32.5 , e ) 35 | d | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 2 and 63 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 ) / 12 = 390 / 12 = 32.5 answer : option d" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
a ) 15 litres , b ) 14 litres , c ) 30 litres , d ) 22 litres , e ) 8 litres | b | multiply(7, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 7 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 7 ) = 4 / 5 20 x = 4 ( 3 x + 7 ) 8 x = 28 x = 3.5 quantity of alcohol = ( 4 x 3.5 ) litres = 14 litres . answer is b ." | a = 7 * 1
|
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