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 ) 5 / 2 , b ) 10 / 3 , c ) 15 / 4 , d ) 25 / 3 , e ) 25 / 6 | e | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 5))), const_2)) | if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 5 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 2 d / 5 j = n / 5 = 2 d / 25 the average of november and january is ( n + j ) / 2 = 12 d / 25 / 2 = 6 d / 25 d is 25 / 6 times the average of november and january . the answer is e ." | a = 2 / 5
b = 2 / 5
c = 1 / 5
d = b * c
e = a + d
f = e / 2
g = 1 / f
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | multiply(6, divide(3, 2)) | john was thrice as old as tom 6 years ago . john will be 3 / 2 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 3 / 2 * ( t + 6 ) 2 j + 12 = 3 t + 18 2 ( 3 t - 12 ) + 12 = 3 t + 18 3 t = 30 t = 10 the answer is c ." | a = 3 / 2
b = 6 * a
|
a ) 92 , b ) 85 , c ) 136.67 , d ) 140.67 , e ) 150 | c | divide(divide(multiply(160, const_100), 60), const_2) | a rainstorm increased the amount of water stored in state j reservoirs from 130 billion gallons to 160 billion gallons . if the storm increased the amount of water in the reservoirs to 60 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | "let total capacity be x we know 160 = 0.60 x x = 160 / 0.60 = 266.67 prior to storm , we had 130 bn gallons 266.67 - 130 = 136.67 answer : c" | a = 160 * 100
b = a / 60
c = b / 2
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | b | divide(73, 11) | what is the 8 th digit to the right of the decimal point in the decimal equivalent of 73 / 11 ? | "73 / 11 = 6.6363 . . . . 63 is non - terminating repeating decimal . the 8 th digit to the right of decimal point will be 3 . answer b" | a = 73 / 11
|
a ) rs . 300 , b ) rs . 360 , c ) rs . 389 , d ) rs . 368 , e ) rs . 323 | b | divide(510, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | if rs . 510 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 510 * 2 / 17 = rs . 60 b ' s share = 510 * 3 / 17 = rs . 90 c ' s share = 510 * 12 / 17 = rs . 360 . answer : b" | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 510 / f
|
a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | if p # q denotes the least common multiple of p and q , then p = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , ... | a = 18 * 24
b = 12 * 16
c = a - b
d = c / 2
e = d + 24
|
a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 3 | d | multiply(const_3.0, 1) | if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 6 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 6 = d" | a = 3 * 0
|
a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 64 | d | divide(multiply(8, subtract(8, const_1)), const_2) | there are 8 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 7 teams . . . so total no of matches = 8 x 7 = 56 . now , each match is played twice = > 56 x 2 but 2 teams play a match = > 56 x 2 / 2 = 56 . ans : d" | a = 8 - 1
b = 8 * a
c = b / 2
|
a ) 13.2 % , b ) 15.2 % , c ) 14.28 % , d ) 32 1 / 3 % , e ) 33 1 / 2 % | c | divide(multiply(3.5, const_100), 4) | a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3.5 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3.5 he selling at 3.5 a rupee , his sp = 4 profit = sp - cp = 4 - 3.5 = 0.5 profit percent = 0.5 / 3.5 * 100 = 14.28 % answer : c" | a = 3 * 5
b = a / 4
|
a ) 32.5 % , b ) 40 % , c ) 60 % , d ) 65 % , e ) can not be determined | c | multiply(add(multiply(divide(21, add(21, 4)), divide(5, add(5, 6))), multiply(divide(2, add(2, 3)), divide(6, add(5, 6)))), const_100) | solution a is made up of alcohol and water mixed in the ratio of 21 : 4 by volume ; solution b is made up of alcohol and water mixed in the ratio of 2 : 3 by volume . if solution a and solution b are mixed in the ratio of 5 : 6 by volume , what percent of the resultant mixture is alcohol ? | let mixture after mixing a and b is 110 ml ( number assumed for calculation because of 5 : 6 ) now solution a is 50 ml and solu b is 60 ml further in a 100 ml a contains 84 ml alch so 50 ml contain 42 ml in mix b 100 ml solution contain 40 ml alc , 60 ml of b , it will be 24 ml alcohol . so 42 + 24 = 66 in 110 ml solut... | a = 21 + 4
b = 21 / a
c = 5 + 6
d = 5 / c
e = b * d
f = 2 + 3
g = 2 / f
h = 5 + 6
i = 6 / h
j = g * i
k = e + j
l = k * 100
|
a ) 95 , b ) 20 , c ) 85 , d ) 75 , e ) 35 | c | multiply(40, const_1) | at veridux corporation , there are 180 employees . of these , 90 are female , and the rest are males . there are a total of 40 managers , and the rest of the employees are associates . if there are a total of 135 male associates , how many female managers are there ? | "well , first let â € ™ s take care of the â € œ totals â € . the numbers in the â € œ totals â € row must add up . if 90 are females , the other 180 â € “ 90 = 90 must be males . similarly , the numbers in the â € œ totals â € column must add up . if 40 are managers , then the other 180 â € “ 40 = 140 must be as... | a = 40 * 1
|
a ) 270 % , b ) 310 % , c ) 350 % , d ) 390 % , e ) 430 % | d | multiply(divide(divide(subtract(multiply(55, const_100), multiply(20, const_100)), subtract(const_100, 55)), 20), const_100) | keats library purchases a number of new books , all in the category of biography , and the library does not acquire any other books . with the addition of the new biographies , the biography collection of the library amounts to 55 % of the total number of books in the library . if prior to the purchase , only 20 % of t... | "let x be the number of new biographies added to the library . let b be the original number of biographies , so the original number of books was 5 b . 0.55 ( 5 b + x ) = b + x 1.75 b = 0.45 x x = 3.9 b the answer is d ." | a = 55 * 100
b = 20 * 100
c = a - b
d = 100 - 55
e = c / d
f = e / 20
g = f * 100
|
a ) 1667 , b ) 6789 , c ) 1200 , d ) 1100 , e ) 1421 | d | divide(836, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 836 . find the actual cost of an article ? | cp * ( 76 / 100 ) = 836 cp = 11 * 100 = > cp = 1100 answer : d | a = 24 / 100
b = 1 - a
c = 836 / b
|
a ) 4 / 6 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 2 / 3 | a | divide(sqrt(16), sqrt(36)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 36 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 16 / 36 - - > height / height = 4 / 6 . answer : ... | a = math.sqrt(16)
b = math.sqrt(36)
c = a / b
|
a ) 225 , b ) 275 , c ) 325 , d ) 350 , e ) 600 | e | divide(add(1500, 300), const_3) | in a weight - lifting competition , the total weight of joe ' s two lifts was 1500 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | "this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 1500 pounds . we sum the two variables to obtain : f ... | a = 1500 + 300
b = a / 3
|
a ) 123 , b ) 109 , c ) 100 , d ) 156 , e ) 240 | b | subtract(109.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 109.25 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 109.25 - 0.25 ) = rs . 109 . answer : b" | a = 1 / 4
b = 109 - 25
|
a ) 110 , b ) 330 , c ) 550 , d ) 462 , e ) 880 | d | multiply(divide(880, const_100), subtract(const_100, 47.5)) | if 47.5 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "students enrolled in biology are 47.5 % and therefore not enrolled are 52.5 % . so of 880 is 880 * . 525 = 462 answer is d 462" | a = 880 / 100
b = 100 - 47
c = a * b
|
a ) 173 m , b ) 200 m , c ) 273 m , d ) 300 m , e ) 373 m | c | multiply(add(sqrt(const_3), const_1), 100) | two ships are sailing in the sea on the two sides of a lighthouse . the angle of elevation of the top of the lighthouse is observed from the ships are 30 ° and 45 ° respectively . if the lighthouse is 100 m high , the distance between the two ships is : | let ab be the lighthouse and c and d be the positions of the ships . then , ab = 100 m , acb = 30 ° and adb = 45 ° . ab = tan 30 ° = 1 ac = ab x 3 = 1003 m . ac 3 ab = tan 45 ° = 1 ad = ab = 100 m . ad cd = ( ac + ad ) = ( 1003 + 100 ) m = 100 ( 3 + 1 ) = ( 100 x 2.73 ) m = 273 m . answer = c | a = math.sqrt(3)
b = a + 1
c = b * 100
|
a ) 42 , b ) 45 , c ) 48 , d ) 50 , e ) 55 | c | divide(multiply(divide(multiply(10, 60), subtract(80, 60)), 80), add(20, divide(multiply(10, 60), subtract(80, 60)))) | a tank can supply water to a village for 80 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 60 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 10 liters per day results in a loss of 600 liters in 60 days . so , those 600 liters were for 20 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 80 = 2400 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last... | a = 10 * 60
b = 80 - 60
c = a / b
d = c * 80
e = 10 * 60
f = 80 - 60
g = e / f
h = 20 + g
i = d / h
|
a ) 20 days , b ) 10 days , c ) 6 8 / 7 days , d ) 8 days , e ) 7 days | c | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 25))) | a can do a work in 25 days and b can do it in 10 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 25 b ' s 1 day ' s work = 1 / 10 they work together = 1 / 25 + 1 / 10 = 7 / 50 = 50 / 7 = 6 8 / 7 days answer : option c" | a = 1 / 10
b = 1 / 25
c = a - b
d = 1 / c
|
a ) 28 , b ) 29 , c ) 34 , d ) 38 , e ) 36 | d | subtract(add(20, 20), add(1, 1)) | a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the difference between the maximum and minimum possible values of ' a - b ' ? | possible values of a and b can be 54 ; 45 and 120 ; 201 maximum possible value for a - b is 20 - 1 = 19 minimum possible value for a - b is 1 - 20 = - 19 19 - ( - 19 ) = 38 ans is d | a = 20 + 20
b = 1 + 1
c = a - b
|
['a ) 2', 'b ) 4', 'c ) 34', 'd ) 38', 'e ) 40'] | c | divide(subtract(power(18, const_2), power(16, const_2)), const_2) | the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 18 - inch screen would be how many square inches greater than the area of a square 16 - inch screen ? | pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 18 ^ 2 / 2 - 16 ^ 2 / 2 = 68 / 2 = 34 ans c . | a = 18 ** 2
b = 16 ** 2
c = a - b
d = c / 2
|
a ) 6 , b ) 9 , c ) 12 , d ) 24 , e ) 48 | c | divide(multiply(divide(80, const_100), 24), divide(70, const_100)) | an alloy weighing 24 ounces is 70 percent gold . how many ounces of pure gold must be added to create an alloy that is 80 percent gold ? | "in 24 ounces , gold is 24 * ( 70 / 100 ) = 16.8 ounces . now we add x ounces of pure gold to make it 90 % gold . so 16.8 + x = ( 24 + x ) * 80 / 100 = > x = 12 . answer is c ." | a = 80 / 100
b = a * 24
c = 70 / 100
d = b / c
|
a ) 1000 , b ) 2217 , c ) 2889 , d ) 2777 , e ) 1300 | e | divide(1456, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1456 ? | "1456 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1300 answer : e" | a = 2 * 5
b = a + 2
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1456 / f
|
a ) 1123 , b ) 1234 , c ) 1349 , d ) 1455 , e ) 1567 | c | add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(add(4, 3), add(4, 3))) | what is the 4 digit no . in which the 1 st digit is 1 / 3 of the second , the 3 rd is the sum of the 1 st and 2 nd , and the last is 3 times the second ? | first digit is 1 / 3 second digit = > the numbers can be 1 & 3 , 2 & 6 , 3 & 9 . first + second = third = > we can eliminate 3 & 9 since 3 + 9 = 12 . last is 3 times the second = > we can eliminate option 2 & 6 since 3 * 6 = 18 . hence the number is 1349 c | a = 100 * 10
b = 3 * 100
c = a + b
d = 4 + 3
e = 4 + 3
f = d * e
g = c + f
|
a ) 101 , b ) 225 , c ) 304 , d ) 324 , e ) 336 | a | divide(76, subtract(const_1, divide(const_1, const_4))) | what number is 76 more than one - fourth of itself ? | "1 / 4 x + 76 = x that means 76 = 3 / 4 x x = ( 76 * 4 ) / 3 = 304 / 3 = 101 a is the answer" | a = 1 / 4
b = 1 - a
c = 76 / b
|
a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint w , in gallons , that must be purchased in order to measure... | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , l... | a = const_0_25 * 2
b = 2 * 5
c = a + b
|
a ) 600 ft , b ) 660 ft , c ) 670 ft , d ) 680 ft , e ) 700 ft | b | multiply(30, divide(multiply(15, 5280), const_3600)) | someone on a skateboard is traveling 15 miles per hour . how many feet does she travel in 30 seconds ? ( 1 mile = 5280 feet ) | "per second = > 15 * 5280 ft / 60 * 60 = 22 ft 30 seconds = > 22 * 30 = 660 ft answer : b" | a = 15 * 5280
b = a / 3600
c = 30 * b
|
a ) 2000 , b ) 4500 , c ) 5000 , d ) 7200 , e ) 9000 | d | divide(72000, 10) | a company produces 72000 bottles of water everyday . if a case can hold 10 bottles of water . how many cases are required by the company to hold its one day production | "number of bottles that can be held in a case = 10 number of cases required to hold 72000 bottles = 72000 / 10 = 7200 cases . so the answer is d = 7200" | a = 72000 / 10
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a ) 100 , b ) 120 , c ) 149 , d ) 150 , e ) 151 | d | add(add(power(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(99, const_10), const_2), const_4), const_2), power(add... | the sum of 99 consecutive integers is 9999 . what is the greatest integer in the set ? | "sum of n numbers = ( n ( 2 a + ( n - 1 ) d ) ) / 2 where n = no of terms a = 1 st term , d - difference between two terms ( ie d = a 2 - a 1 ) here sum = 9999 , n = 99 d = 1 ( since consecutive numbers ) 9999 = ( 99 / 2 ) * ( 2 a + ( 99 - 1 ) 1 ) from this ' a ' ( ie the 1 st term ) = 52 nth term in a sequence : nth t... | a = 99 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 99 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 99 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 99 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 4 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | add(divide(multiply(add(24, const_1), 60), const_1000), 3.5) | a train of 24 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 3.5 km long ? | "d = 25 * 60 + 3500 = 5000 m t = 5000 / 60 * 18 / 5 = 300 sec = 5 mins answer : c" | a = 24 + 1
b = a * 60
c = b / 1000
d = c + 3
|
a ) 9 % decrease , b ) 8 % decrease , c ) 6 % decrease , d ) 1 % decrease , e ) 2 % decrease | b | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | the tax on a commodity is diminished by 20 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 80 * 115 = 9200 - - - - - - - - - - - 10000 - - - - - - - - - - - 800 100 - - - - - - - - - - - ? = > 8 % decrease answer : b" | a = 15 / 100
b = 1 + a
c = 20 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | multiply(sqrt(divide(divide(484, 3), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 484 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 484 / 3 = 161.33 sq m l b = 161.33 i . e . , l * l / 3 = 161.33 l ^ 2 = 484 = > l = 22 . d" | a = 484 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) 125 , b ) 272 , c ) 278 , d ) 277 , e ) 112 | a | multiply(divide(multiply(50, const_1000), const_3600), 9) | a train running at the speed of 50 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec length of train ( distance ) = speed * time ( 125 / 9 ) * 9 = 125 meter answer : a" | a = 50 * 1000
b = a / 3600
c = b * 9
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 33 ? | "first , let ' s translate the expression 21 # 33 , using the definition given : 21 # 33 = 21 × 24 × 27 × 30 × 33 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 33 = 3 ^ 6 ( 7 × 8 × 9 × 10 × 11 ) now let ' s replace each consecutive integer with its prime factorization ... | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
|
a ) 775 , b ) 620 , c ) 750 , d ) 800 , e ) 720 | a | divide(620, subtract(const_1, divide(20, const_100))) | after decreasing 20 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 80 / 100 ) = 620 cp = 7.75 * 100 = > cp = 775 answer : a" | a = 20 / 100
b = 1 - a
c = 620 / b
|
a ) 32256 , b ) 24000 , c ) 24936 , d ) 25600 , e ) none | d | multiply(divide(const_100, 90), 23040) | 90 % of the population of a village is 23040 . the total population of the village is ? | "answer ∵ 90 % of p = 23040 ∴ p = ( 23040 x 100 ) / 90 = 25600 correct option : d" | a = 100 / 90
b = a * 23040
|
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 8,820 multiplied by y is the square of an integer , then y must be | "i just tried plugging in the numbers and found out that 5 * 8820 = 44 , 100 , which is a square of 210 b" | a = 2 + 3
b = a * 2
|
a ) 36.21 , b ) 77.11 , c ) 54.12 , d ) 33.0 , e ) 20.25 | e | subtract(multiply(40, multiply(const_60.0, const_0_2778)), 100) | a train 100 m long crosses a platform 125 m long in 40 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 40 s = 225 / 40 * 18 / 5 = 20.25 kmph answer : e" | a = const_60 * 0
b = 40 * a
c = b - 100
|
a ) 15 , b ) 48 , c ) 54 , d ) 33 , e ) 45 | b | divide(multiply(8, 18), subtract(18, 15)) | a group of men decided to do a work in 15 days , but 8 of them became absent . if the rest of the group did the work in 18 days , find the original number of men ? | "original number of men = 8 * 18 / ( 18 - 15 ) = 48 answer is b" | a = 8 * 18
b = 18 - 15
c = a / b
|
a ) 1 / 5 , b ) 3 / 20 , c ) 1 / 7 , d ) 2 / 15 , e ) 3 / 28 | e | multiply(divide(const_2, 5), divide(const_1, const_4)) | x , y , and z are all unique numbers . if x is chosen randomly from the set { 5 , 6 , 7 , 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "p ( x is prime ) = 3 / 7 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 3 / 7 * 1 / 4 = 3 / 28 the answer is e ." | a = 2 / 5
b = 1 / 4
c = a * b
|
a ) 14 days , b ) 20 days , c ) 22 days , d ) 24 days , e ) 26 days | d | divide(multiply(2, const_3), subtract(divide(add(divide(multiply(2, const_3), 4), add(divide(multiply(2, const_3), 2), divide(multiply(2, const_3), 3))), const_2), divide(multiply(2, const_3), 2))) | a and b can do a piece of work in 2 days , b and c in 3 days , c and a in 4 days . how long will c take to do it ? | "2 c = 1 / 3 + 1 / 4 – 1 / 2 = 1 / 12 c = 1 / 24 = > 24 days answer : d" | a = 2 * 3
b = 2 * 3
c = b / 4
d = 2 * 3
e = d / 2
f = 2 * 3
g = f / 3
h = e + g
i = c + h
j = i / 2
k = 2 * 3
l = k / 2
m = j - l
n = a / m
|
a ) $ 54.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 134.40 | e | divide(multiply(subtract(1720, 600), 12), const_100) | a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 12 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : e the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.12 ( $ 1120 ) = $ 134.40 . the correct answer is e ." | a = 1720 - 600
b = a * 12
c = b / 100
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | a | divide(subtract(36, subtract(multiply(4, divide(10, 2)), 10)), 10) | the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 36 , what is the value of x ? | "the areas are ( x + 5 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 5 and 2 x - 5 . if we add the two perimeters : 4 ( x + 5 ) + 4 ( 2 x - 5 ) = 36 12 x = 36 x = 3 the answer is a ." | a = 10 / 2
b = 4 * a
c = b - 10
d = 36 - c
e = d / 10
|
a ) 0.005 , b ) 0.002 , c ) 0.001 , d ) 0.0003 , e ) 0.0002 | d | divide(0.3, 1000) | when magnified 1000 times by an electron microscope , the image of a certain circular piece of tissue has a diameter of 0.3 centimeter . the actual diameter of the tissue , in centimeters , is | it is very easy if x is the diameter , then the magnified length is 1000 x . ince 1000 x = 0.3 then x = 0.3 / 1000 = 0.0003 . the answer is d | a = 0 / 3
|
a ) 66 , b ) 46 , c ) 49 , d ) 53 , e ) 86 | a | add(add(multiply(5, const_2), 3), add(multiply(5, multiply(const_2, 3)), 3)) | a group of n students can be divided into equal groups of 4 with 3 student left over or equal groups of 5 with 3 students left over . what is the sum of the two smallest possible values of n ? | "4 x + 3 = 5 y + 3 . . . . . . . . . . . ie : 4 x - 5 y = 0 x , y must be > 1 and y is even ie ( 2 , 4,6 , . . etc ) if y = 2 thus x = fraction ( not possible ) if y = 4 thus x = 5 n = 23 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 43 23 + 43 = 66 . . . . . a" | a = 5 * 2
b = a + 3
c = 2 * 3
d = 5 * c
e = d + 3
f = b + e
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(add(add(add(add(1, 1), 1), 3), 4), 1) | q - 1 : how many two digit numbers of distinct digits can be formed by using digits 1 , 2 , 3 , 4 , 5 , 6 and 7 such that the numbers are divisible by 5 ? | "6 two digit numbers divisible by 5 must end in either 0 or 5 . since there is no 0 , all correct answers are a combination of the other digits and 5 . there are six other numbers plus the 5 so there are 6 correct answers . correct answer is d ." | a = 1 + 1
b = a + 1
c = b + 3
d = c + 4
e = d + 1
|
a ) 125 miles , b ) 225 miles , c ) 585 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 9) | a car travels at a speed of 65 miles per hour . how far will it travel in 9 hours ? | "during each hour , the car travels 65 miles . for 9 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 = 9 × 65 = 585 miles correct answer is c ) 585 miles" | a = 65 * 9
|
a ) 21 , b ) 30 , c ) 11 , d ) 10 , e ) 9 | b | divide(subtract(105, 45), const_2) | how many of the integers between 45 and 105 are even ? | "number start between 45 to 105 is 60 numbers half of them is even . . which is 30 answer : b" | a = 105 - 45
b = a / 2
|
a ) 654 , b ) 655 , c ) 656 , d ) 657 , e ) 470 | e | multiply(divide(subtract(const_100, 50), const_100), 940.00) | yearly subscription to professional magazines cost a company $ 940.00 . to make a 50 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 50 / 100 = 470 so the cut in amount is 470 the less amount to be spend is 940 - 470 = 470 answer : e" | a = 100 - 50
b = a / 100
c = b * 940
|
a ) 32.25 % , b ) 23.34 % , c ) 36 % , d ) 39 % , e ) 29 % | a | multiply(subtract(multiply(divide(add(const_100, 15), const_100), divide(add(const_100, 15), const_100)), const_1), const_100) | price of a book increases 15 % successively ( 2 times ) what is the new price of the book more compared to that of the old price : | new price is 1.15 * 1.15 * old price = 1.3225 * old price increase in price = 0.3225 * old price % increase = 100 * 0.3225 * old price / old price = 32.25 % answer : a | a = 100 + 15
b = a / 100
c = 100 + 15
d = c / 100
e = b * d
f = e - 1
g = f * 100
|
a ) 90 , b ) 120 , c ) 315 , d ) 180 , e ) 200 | c | divide(multiply(multiply(7, const_4), multiply(6, 7)), power(factorial(2), 2)) | there are 7 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 7 c 2 number of ways of selecting 2 non fiction books = 6 c 2 7 c 2 * 6 c 2 = 21 * 15 = 315 answer : c" | a = 7 * 4
b = 6 * 7
c = a * b
d = math.factorial(2)
e = d ** 2
f = c / e
|
a ) 1000 , b ) 975 , c ) 980 , d ) 1020 , e ) 1080 | b | add(multiply(6, 74), multiply(9, 59)) | andrew purchased 6 kg of grapes at the rate of 74 per kg and 9 kg of mangoes at the rate of 59 per kg . how much amount did he pay to the shopkeeper ? | cost of 6 kg grapes = 74 × 6 = 444 . cost of 9 kg of mangoes = 59 × 9 = 531 . total cost he has to pay = 444 + 531 = 975 b | a = 6 * 74
b = 9 * 59
c = a + b
|
a ) $ 24000 , b ) $ 23000 , c ) $ 22000 , d ) $ 21000 , e ) $ 20000 | e | multiply(4, 5000) | the ratio of the incomes of uma and bala is 4 : 3 and the ratio of their expenditure is 3 : 2 . if at the end of the year , each saves $ 5000 then the income of uma is ? | "let the income of uma and bala be $ 4 x and $ 3 x let their expenditures be $ 3 y and $ 2 y 4 x - 3 y = 5000 - - - - - - - 1 ) 3 x - 2 y = 5000 - - - - - - - 2 ) from 1 ) and 2 ) x = 5000 uma ' s income = 4 x = 4 * 5000 = $ 20000 answer is e" | a = 4 * 5000
|
a ) 392 , b ) 229 , c ) 756 , d ) 493 , e ) 540 | c | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5040) | 3 / 4 of 1 / 2 of 2 / 5 of 5040 = ? | c 756 ? = 5040 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 756 | a = 3 / 4
b = 1 / 2
c = a * b
d = 2 / 5
e = c * d
f = e * 5040
|
a ) 156 , b ) 167 , c ) 157 , d ) 342 , e ) 380 | d | multiply(add(17, const_1), add(add(17, const_1), const_1)) | there are 17 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 19 from 19 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 19 p ₂ ways . ² ⁰ p ₂ = 19 * 18 = 342 . answer : d" | a = 17 + 1
b = 17 + 1
c = b + 1
d = a * c
|
a ) 800 , b ) 380 , c ) 360 , d ) 478 , e ) 566 | b | multiply(add(18, const_2), subtract(add(18, const_2), const_1)) | there are 18 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 = 20 from 20 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 20 p 2 ways . 20 p 2 = 20 * 19 = 380 answer : b" | a = 18 + 2
b = 18 + 2
c = b - 1
d = a * c
|
a ) 2 / 15 , b ) 7 / 15 , c ) 10 / 30 , d ) 7 / 30 , e ) 8 / 30 | b | divide(3, const_10) | tickets numbered 1 to 30 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ? | "here , the sample space s = ( 1 , 2 , 3 , 4 , 5 , . . . , 29 , 30 ) . let e = the event of getting a multiple of 3 or 5 . e = ( 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 20 , 21 , 24 , 25 , 27 , 30 ) p ( e ) = n ( e ) / n ( s ) = 14 / 30 = 7 / 15 answer : b" | a = 3 / 10
|
a ) 96 kmph , b ) 94 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 6), const_3_6) | a 160 meter long train crosses a man standing on the platform in 6 sec . what is the speed of the train ? | s = 160 / 6 * 18 / 5 = 96 kmph answer : a | a = 160 / 6
b = a * const_3_6
|
a ) 12 : 1 , b ) 1 : 2 , c ) 4 : 1 , d ) 2 : 5 , e ) 2 : 6 | b | divide(subtract(divide(6, add(3, 7)), divide(3, 7)), subtract(divide(3, 7), divide(3, add(3, 7)))) | an alloy of copper and zinc contains copper and zinc in the ratio 6 : 4 . another alloy of copper and zinc contains copper and zinc in the ratio 3 : 7 . in what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc ? | "let alloy _ 1 be x units , and let alloy _ 2 be y units . so , fraction of copper in alloy _ 1 = 6 x / 10 , and fraction of zinc in alloy _ 1 = 4 x / 10 similarly , fraction of copper in alloy _ 2 = 3 y / 10 , and fraction of zinc in alloy _ 2 = 7 y / 10 . mixing them , we get copper = 6 x / 10 + 3 y / 10 ; zinc = 4 x... | a = 3 + 7
b = 6 / a
c = 3 / 7
d = b - c
e = 3 / 7
f = 3 + 7
g = 3 / f
h = e - g
i = d / h
|
a ) 5.18 % , b ) 6.18 % , c ) 7.18 % , d ) 8.18 % , e ) . 7 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(6, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 7,500 gallons of a solution that is 6 percent sodium chloride by volume . if 2,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 7,500 gallons of a solution that is 6 % sodium chloride by volume . this means that there are 0.06 x 7,500 = 450 gallons of sodium chloride . when 2,000 gallons of water evaporate we are left with 5,500 gallons of solution . from here we can determine what percent of the 5,500 gallon solution is sodium c... | a = 100 * 100
b = 6 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | b | divide(subtract(add(multiply(40.2, 10), add(13, 16)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 16 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 – 13 + 31 – 16 = 404 hence , new average = 404 ⁄ 10 = 40.4 answer b" | a = 40 * 2
b = 13 + 16
c = a + b
d = c - 31
e = d / 10
|
a ) 177 , b ) 150 , c ) 180 , d ) 716 , e ) 616 | c | divide(705, add(add(divide(4, 3), divide(4, 5)), const_1)) | rs . 705 is divided amongst a , b , c so that 3 times a ' s share , 5 times b ' s share and 4 times c ' s share are all equal . find b ' s share ? | a + b + c = 705 3 a = 5 b = 4 c = x a : b : c = 1 / 3 : 1 / 5 : 1 / 4 = 20 : 12 : 15 12 / 47 * 705 = rs . 180 answer : c | a = 4 / 3
b = 4 / 5
c = a + b
d = c + 1
e = 705 / d
|
a ) 187 , b ) 133 , c ) 142 , d ) 178 , e ) 175 | b | subtract(subtract(175, divide(multiply(175, 20), const_100)), divide(multiply(subtract(175, divide(multiply(175, 20), const_100)), 5), const_100)) | the sale price sarees listed for rs . 175 after successive discount is 20 % and 5 % is ? | "175 * ( 80 / 100 ) * ( 95 / 100 ) = 133 answer : b" | a = 175 * 20
b = a / 100
c = 175 - b
d = 175 * 20
e = d / 100
f = 175 - e
g = f * 5
h = g / 100
i = c - h
|
a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 12 | e | add(7, divide(multiply(7, subtract(12000, 9500)), subtract(9500, 6000))) | the average salary of all the workers in a workshop is rs . 9500 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | let the total number of workers be x . then , 9500 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = > 3500 x = 42000 = x = 12 . answer : e | a = 12000 - 9500
b = 7 * a
c = 9500 - 6000
d = b / c
e = 7 + d
|
a ) 1,4 , b ) - 1,4 , c ) 2,3 , d ) 2 , - 3 , e ) 3,4 | b | multiply(4, 2) | solve x 2 – 3 x – 4 = 0 . | "this one factors easily : x 2 – 3 x – 4 = 0 ( x + 1 ) ( x – 4 ) = 0 x + 1 = 0 or x – 4 = 0 x = – 1 or x = 4 the solution is x = – 1 , 4 answer b - 1,4" | a = 4 * 2
|
a ) 223.78 , b ) 37.78 , c ) 246.0 , d ) 278.88 , e ) 102.5 | e | divide(61.50, divide(const_4, 5)) | a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 2 / 5 % his yearly income diminishes by rs . 61.50 . his capital is | "explanation : capital = rs . x , then 3 / 5 x = 61.5 x = 102.50 answer : e ) rs . 102.50" | a = 4 / 5
b = 61 / 50
|
a ) 200000 , b ) 20000 , c ) 200029 , d ) 288778 , e ) 277789 | b | divide(482, subtract(power(add(divide(divide(20, const_2), const_100), const_1), const_4), power(add(divide(20, const_100), const_1), const_2))) | a sum of money lent at compound interest for 2 years at 20 % per annum would fetch rs . 482 more , if the interest was payable half yearly than if it was payable annually . the sum is | explanation : let sum = rs . x c . i . when compounded half yearly = c . i . when compounded annually = = > x = 20000 answer : b ) 20000 | a = 20 / 2
b = a / 100
c = b + 1
d = c ** 4
e = 20 / 100
f = e + 1
g = f ** 2
h = d - g
i = 482 / h
|
a ) 15 hrs , b ) 18 hrs , c ) 19 hrs , d ) 17 hrs , e ) 16 hrs | b | subtract(divide(20, subtract(3, 2)), 2) | a monkey start climbing up a tree 20 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 17 hrs it wil be 17 ft . bt since the height of the tree is 20 ft , so if the monkey hops up the tree in the next hr i . e 18 th hr then it reaches at the top of the tree . hence it takes 18 hrs for m... | a = 3 - 2
b = 20 / a
c = b - 2
|
a ) 25.5 , b ) 16.7 , c ) 25 , d ) . 25 , e ) none of these | b | divide(1, 0.06) | 1 / 0.06 is equal to | "explanation : 1 / 0.06 = ( 1 * 100 ) / 6 = 100 / 6 = 16.7 option b" | a = 1 / 0
|
a ) s 4000 , b ) s 2800 , c ) s 1800 , d ) s 3500 , e ) s 3100 | c | divide(subtract(multiply(3600, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | rs 3600 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | "explanation : average rate = ( 144 / 3600 ) * 100 = 4 ratio = 1 : 1 so , first part = ( 1 / 2 ) * 3600 = rs 1800 . answer : c" | a = 5 / 100
b = 3600 * a
c = b - 144
d = 5 / 100
e = 3 / 100
f = d - e
g = c / f
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(const_60, multiply(divide(160, 200), const_60)) | without stoppages , a train travels certain distance with an average speed of 200 km / h , and with stoppages , it covers the same distance with an average speed of 160 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 40 km less . time taken to cover 40 km = 40 â „ 200 h = 1 â „ 5 h = 1 â „ 5 ã — 60 min = 12 min answer e" | a = 160 / 200
b = a * const_60
c = const_60 - b
|
a ) 320.8 $ , b ) 387.8 $ , c ) 420.8 $ , d ) 457.8 $ , e ) 480.8 $ | d | multiply(multiply(0.65, 56), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 56 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 56 ) = 457.8 hence answer will be ( d )" | a = 0 * 65
b = a * 12
|
a ) 66 % , b ) 67.5 % , c ) 69 % , d ) 70 % , e ) 72 % | a | add(multiply(divide(divide(25, const_100), subtract(1, divide(1, 2))), const_100), 2) | the price of an item is discounted 2 percent on day 1 of a sale . on day 2 , the item is discounted another 12 percent , and on day 3 , it is discounted an additional 25 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 2 % discount = 980 price in day 2 after 12 % discount = 862.4 price in day 3 after 25 % discount = 646.8 so , price in day 3 as percentage of the sale price on day 1 will be = 646.8 / 980 * 100 = > 66 % answer will definitely be ( a )" | a = 25 / 100
b = 1 / 2
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 1150 , b ) 1160 , c ) 1140 , d ) 1152 , e ) 1145 | b | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 5,7) | find the smallest number of four digits exactly divisible by 4 , 5,7 and 8 . | "smallest number of four digits is 1000 . required number must be divisible by l . c . m . of 4,5 , 7,8 i . e 280 , on dividing 1000 by 280 , we get 120 as remainder . therefore , required number = 1000 + ( 280 â € “ 120 ) = 1160 . answer is b ." | a = 100 * 100
b = 10 * a
c = b - 100
d = c + 5
|
a ) 25000 , b ) 20000 , c ) 18000 , d ) 14000 , e ) 15000 | e | divide(divide(multiply(2700, const_100), 6), 3) | a sum fetched a total simple interest of rs . 2700 at the rate of 6 p . c . p . a . in 3 years . what is the sum ? | "sol . principal = rs . [ 100 * 2700 / 6 * 3 ] = rs . [ 270000 / 18 ] = rs . 15000 . answer e" | a = 2700 * 100
b = a / 6
c = b / 3
|
a ) 10 m , b ) 14 m , c ) 17 m , d ) 19 m , e ) 20 m | b | multiply(divide(56, multiply(20, 21)), multiply(35, 3)) | if 20 men can build a water fountain 56 metres long in 21 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 21 : 3 : : 56 : x therefore ( 20 x 21 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 420 = 14 hence , the required length is 14 m . answer : b" | a = 20 * 21
b = 56 / a
c = 35 * 3
d = b * c
|
a ) 1 , b ) 2 , c ) 10 , d ) 8 , e ) 16 | c | multiply(const_2, sqrt(power(5, const_2))) | a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 5 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 5 ( pir ^ 2 ) pir ^ 2 = 10 pir r = 10 answer : c" | a = 5 ** 2
b = math.sqrt(a)
c = 2 * b
|
a ) 4 , b ) 2 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(multiply(3, 6), add(subtract(13, 6), 3)), 6) | the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 3 . if the sum of the first and the last number is 13 . what is the last numbers ? | a + b + c = 18 b + c + d = 9 a + d = 13 a – d = 9 a + d = 13 2 d = 4 d = 2 answer : b | a = 3 * 6
b = 13 - 6
c = b + 3
d = a - c
e = d - 6
|
a ) 16 , b ) 24 , c ) 32 , d ) 48 , e ) 54 | c | multiply(multiply(const_2, 8), const_2) | a classroom has equal number of boys and girls . 8 girls left to play kho - kho , leaving twice as many boys as girls in the classroom . what was the total number of girls and boys present initially ? | after 8 girls left remaining 8 girls now boys 16 are twice as many as remaining girls . initially boys = 16 and girls = 16 . answer : c | a = 2 * 8
b = a * 2
|
a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | d | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . what is the present age of the mother ? | "let present age of the mother = 5 x then , present age of the person = 2 x 5 x + 8 = 2 ( 2 x + 8 ) 5 x + 8 = 4 x + 16 x = 8 present age of the mother = 5 x = 40 answer is d ." | a = 2 + 3
b = 8 - a
c = 1 / 2
d = 2 + 3
e = 2 / d
f = c - e
g = b / f
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | divide(add(const_4, const_2), const_1) | a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube ? | "given : positive integer n is a perfect square and a perfect cube - - > n is of a form of n = x 6 for some positive integer x - - > 0 < x ^ 6 < 10 ^ 3 - - > 0 < x ^ 2 < 10 - - > x can be 1 , 2 or 3 hence nn can be 1 ^ 6 , 2 ^ 6 or 3 ^ 6 . answer : b ." | a = 4 + 2
b = a / 1
|
a ) 30 kmph , b ) 40 kmph , c ) 78 kmph , d ) 15 kmph , e ) 23 kmph | a | subtract(50, 20) | the speed of a boat in still water is 50 kmph and the speed of the current is 20 kmph . find the speed and upstream ? | speed downstream = 50 + 20 = 70 kmph speed upstream = 50 - 20 = 30 kmph answer : a | a = 50 - 20
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a ) 48 / 25 , b ) 36 / 25 , c ) 6 / 5 , d ) 5 / 6 , e ) 25 / 36 | a | multiply(divide(divide(1, const_3.0), divide(1, 5)), divide(6, 5)) | if 5 x = 6 y and xy ≠ 0 , what is the ratio of 1 / 5 * x to 1 / 8 * y ? | "5 x = 6 y = > x / y = 6 / 5 1 / 5 * x to 1 / 8 * y = x / y * 8 / 5 = ( 6 / 5 ) * ( 8 / 5 ) = 48 / 25 ans : a" | a = 1 / 3
b = 1 / 5
c = a / b
d = 6 / 5
e = c * d
|
a ) 4.5 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 6.5 | b | multiply(divide(10, 12), 6) | when a number is divided by 6 & then multiply by 12 the answer is 10 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 10 = > 2 x = 10 = > x = 5 b" | a = 10 / 12
b = a * 6
|
a ) 6 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | b | subtract(divide(factorial(subtract(divide(13, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(13, const_2), const_1)) | a company that ships boxes to a total of 13 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors... | "back - solving is the best way to solve this problem . you basically need 13 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - b" | a = 13 / 2
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = math.factorial(2)
f = d * e
g = c / f
h = 13 / 2
i = h - 1
j = g - i
|
a ) 160 km , b ) 170 km , c ) 180 km , d ) 190 km , e ) 150 km | a | add(multiply(20, subtract(12, 10)), multiply(40, const_3)) | a and b start from house at 10 am . they travel on the mg road at 20 kmph and 40 kmph . there is a junction t on their path . a turns left at t junction at 12 : 00 noon , b reaches t earlier , and turns right . both of them continue to travel till 2 pm . what is the distance between a and b at 2 pm | at 12 a will travel = 20 * 2 = 40 km b will travel this 40 km in 40 / 40 = 1 hr i . e . by 11 am . after t junction for a - distance travelled = 2 * 20 = 40 km for b distance travelled = 40 * 3 = 120 so distance between a & b is = 120 + 40 = 160 km answer : a | a = 12 - 10
b = 20 * a
c = 40 * 3
d = b + c
|
a ) 4 / 7 , b ) 5 / 7 , c ) 3 / 7 , d ) 2 / 7 , e ) 1 / 7 | d | divide(4, add(add(add(subtract(1980, 1972), const_1), 4), const_1)) | the circulation for magazine p in 1971 was 4 times the average ( arithmetic mean ) yearly circulation for magazine p for the years 1972 - 1980 . what is the ratio of the circulation in 1971 to the total circulation during 1971 - 1980 for magazine p ? | there are 9 years from 1972 - 1980 , inclusive . let ' s say the average circulation every year between 1972 - 1980 inclusive is x . so the total circulation is 9 x from 1972 - 1980 , inclusive . in 1971 , the circulation is 4 x . so total circulation for 1971 - 1980 is 4 x + 9 x = 13 x . ratio of circulation in 1971 t... | a = 1980 - 1972
b = a + 1
c = b + 4
d = c + 1
e = 4 / d
|
a ) $ 16.32 , b ) $ 18.36 , c ) $ 21.60 , d ) $ 24.48 , e ) $ 28.80 | b | multiply(divide(subtract(const_100, 15), const_100), multiply(0.30, 72)) | the regular price per can of a certain brand of soda is $ 0.30 . if the regular price per can is discounted 15 percent when the soda is purchased in 24 - can cases , what is the price of 72 cans of this brand of soda purchased in 24 - can cases ? | "the discounted price of one can of soda is ( 0.85 ) ( $ 0.30 ) , or $ 0.255 . therefore , the price of 72 cans of soda at the discounted price would be ( 72 ) ( $ 0.255 ) = 18.36 answer : b ." | a = 100 - 15
b = a / 100
c = 0 * 30
d = b * c
|
a ) 48 , b ) 24 , c ) 60 , d ) 72 , e ) 84 | a | divide(12, subtract(const_1, sqrt(divide(9, 16)))) | a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 9 / 16 . how many marbles are in the bag ? | "ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) / x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probab... | a = 9 / 16
b = math.sqrt(a)
c = 1 - b
d = 12 / c
|
a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 53.1 | e | divide(add(multiply(26, 40), multiply(50, 60)), add(26, 50)) | the average marks of a class of 26 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | sum of the marks for the class of 26 students = 26 * 40 = 1040 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 76 students = 1040 + 3000 = 4040 average marks of all the students = 4040 / 76 = 53.15 answer : e | a = 26 * 40
b = 50 * 60
c = a + b
d = 26 + 50
e = c / d
|
a ) 51 , b ) 50 , c ) 88 , d ) 65 , e ) 22 | a | divide(divide(subtract(125, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | a train 125 m long passes a man , running at 6 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 6 ) km / hr . x - 6 = 45 = = > x = 51 km / hr . answer : a" | a = 6 * const_0_2778
b = a * 6
c = 125 - b
d = c / 6
e = d / const_0_2778
|
a ) 5 min , b ) 6 min , c ) 7 and 1 / 2 min , d ) 8 min , e ) 10 min | c | multiply(subtract(divide(divide(multiply(multiply(subtract(15, 10), divide(15, const_60)), const_60), 10), const_2), const_0_25), 10) | tom and john traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour , respectively . after 15 minutes tom passed john , tom reaches a certain gas station , how many l minutes it takes john to reach the station ? | since the question states “ after 15 minutes ” , we can say tom traveled 15 / 4 km for 15 minutes as he can travel 15 km per hour . hence , using the same logic , we can say john traveled 10 / 4 km as he travels 10 km per hour . so , john has to travel ( 15 / 4 ) - ( 10 / 4 ) km = 5 / 4 km more . since john ’ s speed i... | a = 15 - 10
b = 15 / const_60
c = a * b
d = c * const_60
e = d / 10
f = e / 2
g = f - const_0_25
h = g * 10
|
a ) 5 , b ) 24 , c ) 78 , d ) 90 , e ) 54 | b | divide(multiply(12, 8), subtract(12, 8)) | pipe a can fill a tank in 8 hours . due to a leak at the bottom , it takes 12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 8 - 1 / x = 1 / 12 = > 1 / x = 1 / 8 - 1 / 12 = ( 3 - 2 ) / 24 = 1 / 24 = > x = 24 . answer : b" | a = 12 * 8
b = 12 - 8
c = a / b
|
a ) 1 / 38 , b ) 1 / 48 , c ) 1 / 56 , d ) 1 / 64 , e ) 1 / 68 | d | multiply(multiply(multiply(divide(const_1, 8), divide(const_1, 8)), divide(const_1, 8)), divide(const_1, 8)) | three 8 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 , 666 , 777 and 888 . those are eight in number . further the three dice can fall in 8 * 8 * 8 = 512 ways . hence the probability is 8 / 512 = 1 / 64 answer : d" | a = 1 / 8
b = 1 / 8
c = a * b
d = 1 / 8
e = c * d
f = 1 / 8
g = e * f
|
a ) 80 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | c | add(50, multiply(50, divide(subtract(90, 50), subtract(90, 50)))) | an escalator is descending at constant speed . a walks down and takes 50 steps to reach the bottom . b runs down and takes 90 steps in the same time as a takes 10 steps . how many steps are visible when the escalator is not operating ? | lets suppose that a walks down 1 step / min and escalator moves n steps / min it is given that a takes 50 steps to reach the bottom in the same time escalator would have covered 50 n steps so total steps on escalator is 50 + 50 n . again it is given that b takes 90 steps to reach the bottom and time taken by him for th... | a = 90 - 50
b = 90 - 50
c = a / b
d = 50 * c
e = 50 + d
|
a ) 2.6 kg , b ) 4 kg , c ) 3.4 kg , d ) 1.8 kg , e ) none of these | a | divide(subtract(multiply(30, 10), multiply(multiply(divide(2, const_100), 30), const_100)), multiply(30, const_3)) | how many kg of sugar must be added to 30 kg of 2 % solution of sugar and water to increase it to a 10 % solution ? | explanation : solution : amount of sugar in 30 kg solution = ( 2 / 100 * 30 ) kg = 0.6 kg . let x kg of sugar be added . then , ( 0.6 + x ) / ( 30 + x ) = 10 / 100 60 + 100 x = 300 + 10 x x = 240 / 90 = 8 / 3 = 2.6 answer : a | a = 30 * 10
b = 2 / 100
c = b * 30
d = c * 100
e = a - d
f = 30 * 3
g = e / f
|
a ) 19 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | d | add(divide(18, const_2), 1) | each of the integers from 1 to 18 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even | out of the 18 integers : 9 are odd and 9 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 9 odd numbers first 2 . 10 th n... | a = 18 / 2
b = a + 1
|
a ) - 1 , b ) - 1 / 2 , c ) - 1 / 4 , d ) 1 / 4 , e ) 1 / 2 | d | divide(1, add(2, 2)) | suppose f ( x ) is an odd function for which f ( x + 2 ) = f ( x ) for all x , and f ( x ) = x 2 for x 2 ( 0 , 1 ) . compute f ( - 3 / 2 ) + f ( 1 ) . | because f is periodic , we know that f ( - 3 / 2 ) = f ( 1 / 2 ) = ( 1 / 2 ) 2 = 1 = 4 . because f is odd , we know that f ( 1 ) = - f ( / 1 ) , but because f is periodic , f ( 1 ) = f ( - 1 ) . therefore , f ( 1 ) = 0 and the answer is 1 = 4 . correct answer d | a = 2 + 2
b = 1 / a
|
a ) 16 % , b ) 12 % , c ) 9 % , d ) 10 % , e ) 45 % | c | multiply(divide(1980, add(multiply(5000, 2), multiply(3000, 4))), const_100) | a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1980 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 1980 100 r + 120 r = 1980 r = 9 % answer : c" | a = 5000 * 2
b = 3000 * 4
c = a + b
d = 1980 / c
e = d * 100
|
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