options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) 79 kmph , b ) 80 kmph , c ) 34 kmph , d ) 23 kmph , e ) 14 kmph | b | divide(add(70, 90), const_2) | the speed of a car is 70 km in the first hour and 90 km in the second hour . what is the average speed of the car ? | "s = ( 70 + 90 ) / 2 = 80 kmph answer : b" | a = 70 + 90
b = a / 2
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed z ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this z we get x = 8 . b ." | a = 96 / 16
b = 96 / 16
c = 50 / 100
d = 1 + c
e = b / d
f = e * 4
g = f * 4
h = 96 / 16
i = h ** 2
j = g + i
k = math.sqrt(j)
l = a + k
m = l / 2
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a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 16.7 % | e | multiply(divide(subtract(multiply(divide(2, 15), 90), 18), multiply(divide(2, 15), 90)), const_100) | a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 90 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 90 ( multiplying factor is 6 : ( 90 / 15 ) ) then typical dosage would be 2 * 6 = 12 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 12 ) * 100 = 16.67 % e is the answer ." | a = 2 / 15
b = a * 90
c = b - 18
d = 2 / 15
e = d * 90
f = c / e
g = f * 100
|
a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | d | floor(add(power(1152, divide(const_1, const_3)), const_2)) | what is the smallest positive integer x , such that 1152 x is a perfect cube ? | we need to make 1152 x a perfect cube , hence we need to have the powers a multiple of 3 1152 = 2 ^ 7 * 3 ^ 2 the minimum value of x for which 1152 x is a perfect cube = 2 ^ 2 * 3 = 12 correct option : d | a = 1 / 3
b = 1152 ** a
c = b + 2
d = math.floor(c)
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a ) 75 % , b ) 58 % , c ) 42 % , d ) 14.5 % , e ) 19.6 % | e | multiply(divide(multiply(divide(60, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | an association of mathematics teachers has 1,600 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? | total umber of members = 1600 number of members that cast votes = 525 since , winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60 / 100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315 / 1600 ) * 100 = 19.6 % answer e | a = 60 / 100
b = a * 525
c = 4 ** 2
d = 100 * c
e = b / d
f = e * 100
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | b | divide(add(125, 40), 15) | a no . when divided by 125 gives a remainder 40 , what remainder will be obtainedby dividingthe same no . 15 ? | "125 + 40 = 165 / 15 = 11 ( remainder ) b" | a = 125 + 40
b = a / 15
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a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 1 | b | divide(add(add(3, 4), 5), 4) | the area of a triangle will be when a = 3 m , b = 4 m , c = 5 m , a , b , c being lengths of respective sides ? | s = ( 3 + 4 + 5 ) / 2 = 6 answer : b | a = 3 + 4
b = a + 5
c = b / 4
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a ) a . 150 , b ) b . 300 , c ) c . 600 , d ) d . 1,200 , e ) e . 2,400 | b | multiply(0.5, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100))) | a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.5 per box per month for the record storage , what is the total amount that the company pays each month for record storage ? | "volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.5 = 300 answer : b" | a = 15 * 12
b = a * 10
c = b / 100
d = c / 100
e = d / 100
f = 1 / 8
g = 0 * 5
|
a ) 588 , b ) 263 , c ) 120 , d ) 160 , e ) 1260 | a | multiply(divide(multiply(7, subtract(7, const_1)), const_2), divide(multiply(8, subtract(8, const_1)), const_2)) | if 7 parallel lines in a plane is intersected by a family of another 8 parallel lines , how many parallelograms are there in the network thus formed ? | "parallelogram can formed by 2 horizontal and 2 vertical lines for horizontal 7 c 2 for vertical 8 c 2 total parallelogram is 7 c 2 * 8 c 2 = 21 * 28 = 588 answer : a" | a = 7 - 1
b = 7 * a
c = b / 2
d = 8 - 1
e = 8 * d
f = e / 2
g = c * f
|
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) none of them | b | multiply(divide(subtract(490, 465.5), 490), const_100) | if the a radio is sold for rs 490 and sold for rs 465.50 . find loss percentage . | cost price = rs 490 , selling price = 465.50 . loss = rs ( 490 - 465.50 ) = rs 24.50 . loss % = [ ( 24.50 / 490 ) * 100 ] % = 5 % answer is b | a = 490 - 465
b = a / 490
c = b * 100
|
a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 16 , d ) 1 / 32 , e ) 1 / 64 | d | power(divide(1, 2), 5) | if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first flip but not on the last 4 flips ? | "p ( htttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is d ." | a = 1 / 2
b = a ** 5
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a ) a ) 452000 , b ) b ) 562000 , c ) c ) 800000 , d ) d ) 1000000 , e ) e ) 652000 | d | multiply(divide(50000, subtract(const_100, add(add(multiply(20, 3), 30), 5))), const_100) | a person distributed 20 % of his income to his 3 children each . he deposited 30 % of his income to his wife ' s account . he donated 5 % of remaining amount to an orphan house . finally he has $ 50000 . find his total income ? | "3 children got = 3 * 20 % = 60 % wife got = 30 % orphan house = 5 % total = 60 + 30 + 5 = 95 % remaining = 100 - 95 = 5 % 5 % = 50000 100 % = 50000 * 100 / 5 = $ 1000000 answer is d" | a = 20 * 3
b = a + 30
c = b + 5
d = 100 - c
e = 50000 / d
f = e * 100
|
a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | b | subtract(divide(add(30, 0), const_2), divide(add(60, 0), const_2)) | the average ( arithmetic mean ) of the even integers from 0 to 60 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 30 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 60 is 60 / 2 = 30 the average of the even numbers from 0 to 30 is 30 / 2 = 15 the answer is b ." | a = 30 + 0
b = a / 2
c = 60 + 0
d = c / 2
e = b - d
|
a ) 5 , b ) 6 , c ) 10 , d ) 9 , e ) 5 | c | divide(200, multiply(add(60, 10), const_0_2778)) | a train 200 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 10 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 10 = 70 km / hr . = 70 * 5 / 18 = 175 / 9 m / sec . time taken to pass the men = 200 * 9 / 175 = 10 sec . answer : c" | a = 60 + 10
b = a * const_0_2778
c = 200 / b
|
a ) 3 , b ) 11 , c ) 19 , d ) 27 , e ) 35 | a | add(3, add(4, add(6, add(5, multiply(5, const_2))))) | what is the least number which should be added to 2697 so that the sum is exactly divisible by 5 , 6 , 4 , and 3 ? | "l . c . m . of 5 , 6 , 4 and 3 = 60 . when dividing 2697 by 60 , the remainder is 57 . the number to be added = 60 - 57 = 3 . the answer is a ." | a = 5 * 2
b = 5 + a
c = 6 + b
d = 4 + c
e = 3 + d
|
a ) 15 , b ) 30 , c ) 31 , d ) 33 , e ) 47 | e | subtract(divide(subtract(subtract(150, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(150, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 10 to 150 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 150 using ap formula , we can say 150 = 10 + ( n - 1 ) * 2 or n = 71 . for 6 , 12,18 , . . . 96 using ap formula , we can say 150 = 12 + ( n - 1 ) * 6 or n = 24 . hence , only divisible by 2 but not 3 = 71 - 24 = 47 . hence , answer e" | a = 150 - 10
b = a - 2
c = b / 2
d = 150 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 30 , b ) 25 , c ) 12 , d ) 10 , e ) 32 | a | inverse(multiply(divide(inverse(20), add(const_2, const_1)), const_2)) | a is twice as good as b . and together they finish a piece of work in 20 days . in how many days will a alone finish the work | "( a ' s 1 day work ) : ( b ' s 1 day work ) = 2 : 1 a + b 1 day work = 1 / 20 a ' s 1 day work = ( 1 / 20 ) * ( 2 / 3 ) = 1 / 30 a alone can finish the work in 30 days answer is a" | a = 1/(20)
b = 2 + 1
c = a / b
d = c * 2
e = 1/(d)
|
a ) $ 100 , b ) $ 200 , c ) $ 300 , d ) $ 400 , e ) $ 500 | d | divide(100, divide(add(20, 5), const_100)) | a tradesman sold an article at a loss of 20 % . if the selling price had been increased by $ 100 , there would have been a gain of 5 % . what was the cost price of the article ? | "let c . p . be $ x then , ( 105 % of x ) - ( 80 % of x ) = 100 or 25 % of x = 100 x / 4 = 100 x = 400 c . p . = $ 400 correct option is d" | a = 20 + 5
b = a / 100
c = 100 / b
|
a ) 2 % , b ) 5 % , c ) 8 % , d ) 10 % , e ) 15 % | a | subtract(subtract(add(const_100, 20), divide(multiply(add(const_100, 20), 15), const_100)), const_100) | a shop keeper marked 20 % above the cost price and offered 15 % discount then find it ' s net profit ? | net profit = 20 - 15 + ( 20 * ( - 15 ) / 100 ) = 2 % answer is a | a = 100 + 20
b = 100 + 20
c = b * 15
d = c / 100
e = a - d
f = e - 100
|
a ) rs . 130 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | a | divide(65, 0.5) | if 0.5 % of a = 65 paise , then the value of a is ? | answer ∵ 0.5 / 100 of a = 65 / 100 ∴ a = rs . ( 65 / 0.5 ) = rs . 130 correct option : a | a = 65 / 0
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a ) 1 , b ) 4 , c ) 9 , d ) 11 , e ) no such integers exist | d | add(multiply(add(add(1, 1), 1), add(1, 1)), add(add(1, 1), add(add(1, 1), 1))) | suppose a , b , and c are positive integers with a < b < c such that 1 / a + 1 / b + 1 / c = 1 . what is a + b + c ? | first note that we must have 1 / a < 1 , so a > 1 . since 1 / a > 1 / b > 1 / c , we must also have 1 / a > 1 / 3 ; so a < 3 . thus , a / 2 . now 1 / b + 1 / c = 1 / 2 where 2 < b < c . similar to before , 1 / b > 1 / 4 , so b < 4 . thus , b = 3 . with a = 2 and b = 3 we have 1 / 2 + 1 / 3 + 1 / c = 1 , which is satisfied when c = 6 . to conclude , a + b + c = 2 + 3 + 6 = 11 . correct answer d | a = 1 + 1
b = a + 1
c = 1 + 1
d = b * c
e = 1 + 1
f = 1 + 1
g = f + 1
h = e + g
i = d + h
|
a ) 276 , b ) 84 , c ) 88 , d ) 90 , e ) 342 | b | add(lcm(lcm(5, 10), lcm(15, 20)), 24) | what is the smallest number which when diminished by 24 , is divisible 5 , 10 , 15 and 20 ? | "required number = ( lcm of 5 , 10 , 15 and 20 ) + 24 = 60 + 24 = 84 option b" | a = math.lcm(5, 10)
b = math.lcm(15, 20)
c = math.lcm(a, b)
d = c + 24
|
['a ) 2 / 3', 'b ) 3 / 4', 'c ) 4 / 5', 'd ) 5 / 6', 'e ) 6 / 7'] | a | divide(square_perimeter(1), 6) | a 6 - meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point ? | the longer wire will form a square with an area more than 1 if the wire is cut at a point within two meters of either end . the probability of this is 4 / 6 = 2 / 3 . the answer is a . | a = square_perimeter / (
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a ) 70 , b ) 245 , c ) 150 , d ) 300 , e ) 350 | d | divide(42, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100))) | in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 42 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ? | "since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 42 , so 0.14 * { total } = 42 - - > { total } = 300 . answer : d ." | a = 100 - 30
b = a / 100
c = 20 / 100
d = b * c
e = 42 / d
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a ) 382 , b ) 782 , c ) 278 , d ) 270 , e ) 220 | e | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 300) | a goods train runs at the speed of 72 km / hr and crosses a 300 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 300 ) / 26 = 20 x = 220 m . answer : e" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 300
|
a ) 4 , b ) 7 , c ) 6 , d ) 12 , e ) it can not be determined from the information given . | c | subtract(multiply(3, const_10), multiply(const_4, 3)) | two sets of 3 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? | "a = ( 2 , 3,4 ) , sum of this = 9 b = ( 4 , 5,6 ) , sum of this = 15 , the differenct between 15 - 9 = 6 hence , 6 is the answer i . e . c" | a = 3 * 10
b = 4 * 3
c = a - b
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a ) 12 , b ) 10 , c ) 7 , d ) 9 , e ) 8 | d | divide(subtract(multiply(80, 5), multiply(5, 44)), subtract(80, 60)) | the average mark of the students of a class in a particular exam is 60 . if 5 students whose average mark in that exam is 44 are excluded , the average mark of the remaining will be 80 . find the number of students who wrote the exam ? | "let the number of students who wrote the exam be x . total marks of students = 60 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 60 x - ( 5 * 44 ) = 80 ( x - 5 ) 180 = 20 x = > x = 9 answer : d" | a = 80 * 5
b = 5 * 44
c = a - b
d = 80 - 60
e = c / d
|
a ) 5.9 % , b ) 5.5 % , c ) 6 % , d ) 6.4 % , e ) 7.5 % | b | divide(const_100, 18) | in how many years will a sum of money doubles itself at 18 % per annum on simple interest ? | "p = ( p * 18 * r ) / 100 r = 5.5 % answer : b" | a = 100 / 18
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a ) 30 , b ) 40 , c ) 60 , d ) none , e ) 74 | e | divide(55.5, 75) | how many pieces of 75 cm can be cut from a rope 55.5 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 55.5 meters long is = ( 55.5 meters ) / ( 75 cm ) = ( 55.5 meters ) / ( 0.75 meters ) = 74 answer : e" | a = 55 / 5
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a ) 80 , b ) 60 , c ) - 60 , d ) - 70 , e ) none | c | subtract(250, subtract(380, 70)) | a merchant gains or loses , in a bargain , a certain sum . in a second bargain , he gains 380 dollars , and , in a third , loses 70 . in the end he finds he has gained 250 dollars , by the 3 together . how much did he gain or lose bv the first ? | in this sum , as the profit and loss are opposite in their nature , they must be distinguished by contrary signs . if the profit is marked + , the loss must be - . let x = the sum required . then according to the statement x + 380 - 70 = 250 and x = - 60 . answer c | a = 380 - 70
b = 250 - a
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a ) 26630 , b ) 26640 , c ) 36620 , d ) 26620 , e ) 26820 | d | add(add(add(20000, multiply(20000, divide(10, const_100))), multiply(add(20000, multiply(20000, divide(10, const_100))), divide(10, const_100))), multiply(add(add(20000, multiply(20000, divide(10, const_100))), multiply(add(20000, multiply(20000, divide(10, const_100))), divide(10, const_100))), divide(10, const_100))) | population is 20000 . pop increases by 10 % every year , then the pop after 3 years is ? | population after 1 st year = 20000 * 10 / 100 = 2000 = = = > 20000 + 2000 = 22000 population after 2 nd year = 22000 * 10 / 100 = 2200 = = = > 22000 + 2200 = 24200 population after 3 rd year = 24200 * 10 / 100 = 2420 = = = > 24200 + 2420 = 26620 answer : d | a = 10 / 100
b = 20000 * a
c = 20000 + b
d = 10 / 100
e = 20000 * d
f = 20000 + e
g = 10 / 100
h = f * g
i = c + h
j = 10 / 100
k = 20000 * j
l = 20000 + k
m = 10 / 100
n = 20000 * m
o = 20000 + n
p = 10 / 100
q = o * p
r = l + q
s = 10 / 100
t = r * s
u = i + t
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a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | a | add(const_2, const_2) | find the number of different prime factors of 6440 | "explanation : l . c . m of 6440 = 2 x 2 x 2 x 5 x 7 x 23 2 , 5 , 7,23 number of different prime factors is 4 . answer : option a" | a = 2 + 2
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a ) 240 meters , b ) 260 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | b | subtract(multiply(divide(multiply(72, const_1000), const_3600), 31), multiply(divide(multiply(72, const_1000), const_3600), 18)) | a train traveling at 72 kmph crosses a platform in 31 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ? | "speed of train = 72 * ( 5 / 18 ) = 20 m / s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 31 = 620 m . so , length of platform = 620 - 360 = 260 m imo , answer b" | a = 72 * 1000
b = a / 3600
c = b * 31
d = 72 * 1000
e = d / 3600
f = e * 18
g = c - f
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a ) 1 / 4 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22 | a | multiply(divide(5, 8), subtract(const_1, divide(3, 5))) | wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 5 / 8 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ? | p ( 1 and not 2 ) = 5 / 8 * ( 1 - 3 / 5 ) = 1 / 4 answer : a . | a = 5 / 8
b = 3 / 5
c = 1 - b
d = a * c
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a ) 20 , b ) 25 , c ) 55 , d ) 65 , e ) 80 | a | subtract(multiply(35, 4), multiply(40, const_3)) | joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 35 . he was allowed to drop his lowest score . after doing so , his average test score improved to 40 . what is the lowest test score that was dropped ? | the arithmetic mean of 4 equally weighted tests was 35 . so what we can assume is that we have 4 test scores , each 35 . he dropped his lowest score and the avg went to 40 . this means that the lowest score was not 35 and other three scores had given the lowest score 5 each to make it up to 35 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 35 . so the lowest score = 35 - 15 = 20 answer ( a ) | a = 35 * 4
b = 40 * 3
c = a - b
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a ) 220 , b ) 760 , c ) 990 , d ) 880 , e ) 660 | b | subtract(divide(multiply(760, subtract(22, const_2)), 10), 760) | there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 10 days more ? | "760 - - - - 22 760 - - - - 20 x - - - - - 10 x * 10 = 760 * 20 x = 1520 760 - - - - - - - 760 answer : b" | a = 22 - 2
b = 760 * a
c = b / 10
d = c - 760
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a ) 5 , b ) 10 , c ) 3 , d ) 12 , e ) 6 | c | divide(15, multiply(add(15, 3), const_0_2778)) | the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 15 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 15 / 5 = 3 seconds . answer : c" | a = 15 + 3
b = a * const_0_2778
c = 15 / b
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a ) 38.3 % , b ) 25 % , c ) 77 % , d ) 99 % , e ) 12 % | a | subtract(multiply(divide(const_100, 723), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 723 grams per kg , what is his percent ? | "723 - - - 277 100 - - - ? = > 38.3 % answer : a" | a = 100 / 723
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
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a ) rs . 1600 , b ) rs . 1500 , c ) rs . 2000 , d ) rs . 2500 , e ) none of the above | a | multiply(multiply(subtract(4, 3), 800), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 800 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 800 x = 800 . b ' s share = rs . 2 x = rs . ( 2 x 800 ) = rs . 1600 . answer = a" | a = 4 - 3
b = a * 800
c = b * 3
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a ) 4 : 15 , b ) 1 : 3 , c ) 2 : 5 , d ) 4 : 5 , e ) 7 : 6 | a | divide(divide(4, 3), 5) | if the ratio of apples to bananas is 4 to 3 and the ratio of bananas to cucumbers is 1 to 5 , what is the ratio of apples to cucumbers ? | "apples : bananas = 4 : 3 bananas : cucumbers = 1 : 5 = 3 : 15 so if i have bananas , i ' d have 4 apples and 15 cucumbers apples : cucumbers = 4 : 15 answer : a" | a = 4 / 3
b = a / 5
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a ) 16 , b ) 1 , c ) 2 , d ) 6 , e ) 14 | b | add(const_10, 2) | if ( t - 8 ) is a factor of t ^ 2 - kt - 40 , then k = | "t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 1 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ) ( t + m ) b" | a = 10 + 2
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a ) 1.5 , b ) 2.0 , c ) 2.5 , d ) 3.0 , e ) 3.5 | d | divide(const_1, add(divide(const_1, 4), divide(divide(const_1, 4), const_3))) | p alone can complete a job in 4 days . the work done by q alone in one day is equal to one - third of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | p ' s rate is 1 / 4 q ' s rate is 1 / 12 the combined rate is 1 / 4 + 1 / 12 = 1 / 3 if they work together , the job will take 3 days . the answer is d . | a = 1 / 4
b = 1 / 4
c = b / 3
d = a + c
e = 1 / d
|
a ) 8 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | subtract(subtract(12, 2), 2) | a man ' s speed with the current is 12 km / hr and the speed of the current is 2 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 12 - 2 ) km / hr = 10 km / hr . man ' s rate against the current = ( 10 - 2 ) km / hr = 8 km / hr . answer : a" | a = 12 - 2
b = a - 2
|
a ) 6.25 , b ) 5.25 , c ) 6.29 , d ) 6.39 , e ) 6.13 | b | divide(subtract(242, multiply(10, 3.2)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 242 runs ? | "required run rate = [ 242 - ( 3.2 * 10 ) ] / 40 = 210 / 40 = 5.25 answer : b" | a = 10 * 3
b = 242 - a
c = b / 40
|
a ) 512 , b ) 768 , c ) 1024 , d ) 131072 , e ) 409600 | d | subtract(power(4, add(8, const_1)), const_1) | the population of a bacteria colony doubles every day . if it was started 8 days ago with 4 bacteria and each bacteria lives for 12 days , how large is the colony today ? | "4 ^ 8 ( 2 ) = 131072 the answer is d ." | a = 8 + 1
b = 4 ** a
c = b - 1
|
a ) 63 / 64 , b ) 31 / 37 , c ) 31 / 18 , d ) 31 / 10 , e ) 31 / 19 | a | multiply(multiply(divide(const_1, const_2), multiply(multiply(multiply(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)), divide(const_1, const_2))), const_1) | a coin is tossed six times . what is the probability that there is at the least one tail ? | let p ( t ) be the probability of getting least one tail when the coin is tossed six times . = there is not even a single tail . i . e . all the outcomes are heads . = 1 / 64 ; p ( t ) = 1 - 1 / 64 = 63 / 64 answer : a | a = 1 / 2
b = 1 / 2
c = 1 / 2
d = b * c
e = 1 / 2
f = d * e
g = 1 / 2
h = f * g
i = a * h
j = i * 1
|
a ) 64 , b ) 62 , c ) 63 , d ) 61 , e ) 56 | e | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 5,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . e" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 9 % , b ) 8 % , c ) 10 % , d ) 10.5 % , e ) 7.5 % | c | divide(multiply(const_100, 2500), multiply(5000, 5)) | what is the rate percent when the simple interest on rs . 5000 amount to rs . 2500 in 5 years ? | "interest for 1 year = 2500 / 5 = 500 interest on rs 5000 p / a = 500 interest rate = 500 / 5000 * 100 = 10 % answer : c" | a = 100 * 2500
b = 5000 * 5
c = a / b
|
a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction q of the total distance did he cover while the sun was shining on him ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is q = 2 / 14 = 1 / 7 . answer : e | a = math.lcm(2, 3)
b = a / 2
c = b - 2
|
a ) 3 , b ) 4 , c ) 5 , d ) 1 , e ) 2 | e | divide(multiply(16, 2), 16) | given the two equations 3 x + 2 y = 16 and 5 x + 3 y = 26 , by how much does r exceed s ? | solve by simultaneous equations . my answer is e . x = 4 y = 2 | a = 16 * 2
b = a / 16
|
a ) 24 , b ) 25 , c ) 26.6 m , d ) 26 , e ) 27 | c | multiply(sqrt(divide(divide(423, 3), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 300 % . if rs . 423 is required to paint the floor at the rate of rs . 3 per sq m , then 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 + 400 % of b = l + 4 b = 5 b area of the floor = 423 / 3 = 141 sq m l b = 141 i . e . , l * l / 5 = 141 l 2 = 705 = > l = 26.6 m . answer : c" | a = 423 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
|
a ) 1 \ 5 , b ) 1 \ 6 , c ) 1 \ 7 , d ) 1 \ 8 , e ) 1 \ 9 | b | add(divide(const_1, 18), multiply(divide(const_1, 18), const_2)) | a can finish a work in 18 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day | "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 18 b ' s 1 day work = 1 / 9 [ because b take half time than a ] ( a + b ) ' s one day work = ( 1 / 18 + 1 / 9 ) = ( 1 + 2 / 18 ) = 1 / 6 so in one day 1 / 6 work will be done . option b" | a = 1 / 18
b = 1 / 18
c = b * 2
d = a + c
|
a ) 22.51 , b ) 22.0 , c ) 22.15 , d ) 22.17 , e ) 22.63 | b | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(18, 10)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 18 cm and breadth 10 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 18 + 10 ) = 56 cm parameter of the square = 56 cm i . e . 4 a = 56 a = 14 diameter of the semicircle = 15 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 14 ) = 1 / 2 ( 22 / 7 ) ( 14 ) = 308 / 14 = 22 cm to two decimal places answer : b" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) 65 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | a | multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100) | the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 198,000 . what was the percent increase in the cost of the home ? | "increase = 198000 - 120000 = 78000 % increase = 78000 * 100 / 120000 = 65 % answer : option a" | a = 4 * 4
b = 12 * a
c = b * 1000
d = 12 * 1000
e = d * 10
f = c - e
g = 12 * 1000
h = g * 10
i = f / h
j = i * 100
|
a ) 42000 , b ) 46000 , c ) 54000 , d ) 40000 , e ) none of these | d | add(divide(multiply(subtract(14000, divide(multiply(25, 14000), const_100)), add(const_2, const_3)), add(add(const_2, const_3), add(const_2, const_4))), divide(multiply(25, 14000), const_100)) | a is a working partner and b is a sleeping partner in a business . a puts in 80,000 and b 70,000 . a gets 25 % of the profit for managing the business , and the rest is divided in proportion to their capitals . find the profit if b get 14000 . | "let amount distributed between a and b is x ratio of capital between a and b is 8 : 7 if b get 14000 then x * 7 / 15 = 14000 x * 7 = 210000 x = 30000 let total profit = y y * 75 / 100 = 30000 y = 40000 answer d" | a = 25 * 14000
b = a / 100
c = 14000 - b
d = 2 + 3
e = c * d
f = 2 + 3
g = 2 + 4
h = f + g
i = e / h
j = 25 * 14000
k = j / 100
l = i + k
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | add(divide(20, 5), const_1) | how many integers are divisible by 5 between 10 ! and 10 ! + 20 inclusive ? | a - 7 10 ! is divisible by 5 there are 4 numbers between 10 ! and 10 ! + 20 that are divisible by 5 . hence 5 | a = 20 / 5
b = a + 1
|
a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 4.5 , e ) 5.5 | d | divide(divide(multiply(150, multiply(15, const_2)), const_1000), divide(15, const_10)) | a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 150 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | "the distance between the nest and the ditch is 150 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 150 = 4500 . d = st 4500 / 1.5 = t , i think we can take 4500 meters as 4.5 km , then only we get t = 4.5 . ( 1000 meters = 1 km ) d )" | a = 15 * 2
b = 150 * a
c = b / 1000
d = 15 / 10
e = c / d
|
a ) 22.8 kg , b ) 34.2 kg , c ) 28 kg , d ) 26.5 kg , e ) none of these | b | divide(multiply(9, 42.75), 11.25) | if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 9 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 9 : : 42.75 : x = > 11.25 x x = 9 x 42.75 = > x = ( 9 x 42.75 ) / 11.25 = > x = 34.2 answer : b" | a = 9 * 42
b = a / 11
|
a ) $ 1.60 , b ) $ 16.00 , c ) $ 96.00 , d ) $ 108.00 , e ) $ 196.00 | b | multiply(divide(3.20, 120), surface_cube(10)) | if paint costs $ 3.20 per quart , and a quart covers 120 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | "total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 120 = 5 cost will be 5 * 3.2 = $ 16 ans : b" | a = 3 / 20
b = a * surface_cube
|
a ) 5 / 24 , b ) 6 / 24 , c ) 7 / 24 , d ) 8 / 24 , e ) 1 / 3 | e | subtract(divide(3, 6), divide(3, 18)) | if 3 / p = 6 & 3 / q = 18 then p - q = ? | "p = 3 / 6 , q = 3 / 18 = > q = 1 / 6 therefore p - q = ( 1 / 2 ) - ( 1 / 6 ) = 1 / 3 answer : e" | a = 3 / 6
b = 3 / 18
c = a - b
|
a ) 10.16 sec , b ) 18.8 sec , c ) 14.8 sec , d ) 10.8 sec , e ) 11.16 sec | e | divide(add(150, 160), multiply(add(60, 40), const_0_2778)) | two trains 150 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 150 + 160 = 310 m . required time = 310 * 9 / 250 = 279 / 25 = 11.16 sec . answer : e" | a = 150 + 160
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) 24 | b | subtract(divide(factorial(subtract(divide(12, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(12, const_2), const_1)) | a company that ships boxes to a total of 12 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 needed for the coding ? ( assume that the order of colors in a pair does not matter ) | let # of colors needed be nn , then it must be true that n + c 2 n ≥ 12 n + cn 2 - # of ways to choose the pair of different colors from nn colors when order does n ' t matter ) - - > n + n ( n − 1 ) / 2 ≥ 12 - - > 2 n + n ( n − 1 ) ≥ 24 - - > n is an integer ( it represents # of colors ) n ≥ 5 - - > nmin = 5 answer : b . | a = 12 / 2
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = math.factorial(2)
f = d * e
g = c / f
h = 12 / 2
i = h - 1
j = g - i
|
a ) 25 , b ) 27 , c ) 30 , d ) 35 , e ) 37 | d | subtract(multiply(5, 27), multiply(subtract(5, const_1), 25)) | the average of 5 numbers is 27 . if one number is excluded the average become 25 . the excluded number is : | solution excluded number = ( 27 × 5 ) - ( 25 × 4 ) = 135 - 100 = 35 . answer d | a = 5 * 27
b = 5 - 1
c = b * 25
d = a - c
|
a ) 11 , b ) 10 , c ) 5 , d ) 9 , e ) 8 | a | multiply(const_3600, divide(divide(200, const_1000), add(60, 6))) | a train 200 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 200 * 3 / 55 ] sec = 11 sec answer : a" | a = 200 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 3888 , b ) 2988 , c ) 2777 , d ) 8925 , e ) 288 | d | divide(divide(multiply(4016.25, const_100), 9), 5) | a sum fetched a total simple interest of rs . 4016.25 at the rate of 9 % p . a . in 5 years . what is the sum ? | principal = ( 100 * 4016.25 ) / ( 9 * 5 ) = rs . 8925 . answer : d | a = 4016 * 25
b = a / 9
c = b / 5
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | c | divide(subtract(add(24, add(24, 7)), multiply(const_3.0, const_3.0)), const_2) | the captain of a cricket team of 11 members is 24 years old and the wicket keeper is 7 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 24 + 31 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is c ." | a = 24 + 7
b = 24 + a
c = 3 * 0
d = b - c
e = d / 2
|
a ) 1 / 6 , b ) 2 / 9 , c ) 6 / 20 , d ) 13 / 20 , e ) 3 / 4 | d | multiply(divide(subtract(16, 3), 16), divide(subtract(subtract(16, 3), const_1), subtract(16, const_1))) | in a box of 16 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 3 defective in a lot of 16 pens i . e . probability of first pen not being defective = ( 13 / 16 ) i . e . probability of second pen not being defective = ( 12 / 15 ) [ 15 pen remaining with 12 defective remaining considering that first was defective ] probability of both pen being non - defective = ( 12 / 15 ) * ( 13 / 16 ) = 13 / 20 answer : option d" | a = 16 - 3
b = a / 16
c = 16 - 3
d = c - 1
e = 16 - 1
f = d / e
g = b * f
|
a ) 250 , b ) 260 , c ) 270 , d ) 280 , e ) 285 | b | lcm(divide(multiply(2.6, const_10), const_2), divide(multiply(4, const_10), const_2)) | during a pizza buffet where a eats more times 2.6 than b , and b eats 4 times less than c . find the least number of times all the 3 has to eat | a eats more than b if b eats 1 times than the ratio of a and b is a : b is 2.6 : 1 or 13 : 5 and as b eat 4 times less the c the the ratio of b : c is 5 : 20 the the least number of times all three has eat is the lcm of a , b , c that is 260 . . answer : b | a = 2 * 6
b = a / 2
c = 4 * 10
d = c / 2
e = math.lcm(b, d)
|
a ) 78 , b ) 58 , c ) 390 , d ) 310 , e ) 66 | a | divide(add(add(add(add(subtract(multiply(62, add(const_4, const_1)), add(add(add(28, 42), 78), 104)), 48), 62), 98), 124), add(const_4, const_1)) | if the mean of numbers 28 , x , 42 , 78 and 104 is 62 , what is the mean of 48 , 62 , 98 , 124 and x ? | "x is common to both the series . so , x is not going to make a difference to the average . only the remaining 4 numbers will contribute to the difference in average between the two series . sum of the 4 numbers , excluding x , of the first series is 28 + 42 + 78 + 104 = 252 sum of the 4 numbers , excluding x , of the second series is 48 + 62 + 98 + 124 = 332 the difference between the sum of the two sets of numbers = 332 - 252 = 80 the sum of the second series is 80 more than the sum of the first series . if the sum of the second series is 80 more , the average of the second series will be 80 / 5 = 16 more than the first series . therefore , the average of the second series = 62 + 16 = 78 . answer a" | a = 4 + 1
b = 62 * a
c = 28 + 42
d = c + 78
e = d + 104
f = b - e
g = f + 48
h = g + 62
i = h + 98
j = i + 124
k = 4 + 1
l = j / k
|
a ) 391 , b ) 392 , c ) 410 , d ) 423 , e ) 445 | b | divide(450, add(const_1, divide(15, const_100))) | company p had 15 percent more employees in december than it had in january . if company p had 450 employees in december , how many employees did it have in january ? | "d = number of employees in december j = number of employees in january j x 1.15 = d j x 1.15 = 450 j = 450 / 1.15 j = 45,000 / 115 = 392 thus b is the correct answer ." | a = 15 / 100
b = 1 + a
c = 450 / b
|
a ) 15 , b ) 24 , c ) 25 , d ) 26 , e ) 13 | c | divide(add(7, 5), 0.5) | salley ' s internet provider offers unlimited traffic which costs $ 0.5 per day charged off at 12 a . m . the service is discontinued when client ' s debt exceeds $ 5 . if today in the morning salley makes a payment of $ 7 , for how many days , including today , will she be connected to the internet without any additional payments , given her initial balance was $ 0 ? | day 0 - payment of $ 7 is made in the morning ( after 12 am midnight ) day 1 onwards ( after 12 am midnight ) internet service will begin and will run for 7 * 2 = 14 days i . e . at the end of day 15 balance will be zero . now salley can still use the internet till her debt reaches $ 5 which means total of 5 * 2 = 10 days . now the question states for how many days , including today , will she be connected to the internet without any additional paymentsi . e . we will have to include day 0 also in our final answer . therefore total no . of days = 1 + 14 + 10 = 25 = c | a = 7 + 5
b = a / 0
|
a ) s . 8,000 , b ) s . 9,000 , c ) s . 20,000 , d ) s . 10,000 , e ) s . 50,000 | a | subtract(multiply(5, const_4), const_12) | a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 700 more than that by the previous schema , his sales were worth ? | [ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 700 x = 8000 answer a | a = 5 * 4
b = a - 12
|
a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 10 rs | e | divide(150, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 150 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 150 rs . 5 m + 5 m + 5 m - - - - - 150 rs . 15 m - - - - - - 150 rs . = > 1 m = 10 rs . answer : e" | a = 3 * 5
b = 150 / a
|
a ) 57 : 11 , b ) - 57 : 11 , c ) 11 : 10 , d ) - 39 : 5 , e ) - 1 : 10 | d | divide(add(multiply(5, 3), multiply(4, 6)), subtract(3, multiply(2, 4))) | if a : b : : 3 : 4 , then what is ( 5 a + 6 b ) : ( a - 2 b ) ? | "a / b = 3 / 4 dividing numerator & denominator of ' ( 5 a + 6 b ) / ( a - 2 b ) ' by b , [ 5 ( a / b ) + 6 ] / [ ( a / b ) - 2 ] = [ 5 * ( 3 / 4 ) + 6 ] / [ ( 3 / 4 ) - 2 ] = - 39 / 5 answer : d" | a = 5 * 3
b = 4 * 6
c = a + b
d = 2 * 4
e = 3 - d
f = c / e
|
a ) 1.5 , b ) 6 , c ) 7.5 , d ) 8.5 , e ) 2 | c | subtract(multiply(const_4, const_3), divide(divide(multiply(45000, multiply(const_4, const_3)), 36000), 2)) | a began business with rs . 45000 and was joined afterwards by b with rs . 36000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | "45 * 12 : 36 * x = 2 : 1 x = 7.5 12 - 7.5 = 4.5 answer : c" | a = 4 * 3
b = 4 * 3
c = 45000 * b
d = c / 36000
e = d / 2
f = a - e
|
a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | d | divide(divide(add(4, 3), 3), divide(add(const_4.0, 4), const_2)) | in the xy - coordinate system , what is the slope of the line that goes through point ( 2 , 4 ) and is equidistant from the two points p = ( 9 , 3 ) and q = ( 5 , 9 ) ? | "first , get the middle coordinate between ( 9,3 ) and ( 5,9 ) . x = 9 + ( 5 - 9 ) / 2 = 7 y = 3 + ( 9 - 3 ) / 2 = 6 second , get the slope of ( 7,6 ) and ( 2,4 ) . m = 6 - 4 / 7 - 2 = 2 / 5 = 0.4 answer : d" | a = 4 + 3
b = a / 3
c = 4 + 0
d = c / 2
e = b / d
|
a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 22 | e | divide(add(160, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 160 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 160 + 280 = 440 m . the time required = d / s = 440 / 20 = 22 sec . answer : e" | a = 160 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 77.14 cm , b ) 47.14 cm , c ) 26.71 cm , d ) 94.94 cm , e ) 23.57 cm | c | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(20, 14)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 20 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 20 + 14 ) = 68 cm parameter of the square = 68 cm i . e . 4 a = 68 a = 17 diameter of the semicircle = 17 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 17 ) = 1 / 2 ( 22 / 7 ) ( 17 ) = 374 / 14 = 26.71 cm to two decimal places answer : c" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) 6 % , b ) 14 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | b | multiply(add(divide(const_1, 10), divide(const_1, 25)), const_100) | if x > 0 , x / 10 + x / 25 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 10 + 100 / 25 = 10 + 4 = 14 14 is 14 % of 100 = b" | a = 1 / 10
b = 1 / 25
c = a + b
d = c * 100
|
a ) 62 , b ) 52 , c ) 32 , d ) 12 , e ) 22 | e | subtract(30, multiply(multiply(12, const_2.0), 2)) | evaluate : 30 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 30 - 12 * 3 * 2 = 30 - 8 = 22 correct answer e" | a = 12 * 2
b = a * 2
c = 30 - b
|
a ) 50 , b ) 30 , c ) 40 , d ) 60 , e ) 80 | b | add(20, 10) | one day , raviraj left home and cycled 20 km southwards , turned right and cycled 10 km and turned right and cycled 20 km and turned left and cycled 20 km . how many kilometres will he have to cycle to reach his home straight ? | raviraj starts from home at a , moves 20 km in south upto b . then he turns right and moves 10 km upto c , then he turns right and moves 20 km upto d , then he turns lefts and moves 20 km upto e . so from image it is clear that , if he moves straight then he will have to move ad + de , ad = bc = 10 km so , he will have to move 10 + 20 = 30 km answer b | a = 20 + 10
|
a ) 45 , b ) 47 , c ) 49 , d ) 51 , e ) 53 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 370), multiply(36, const_0_2778)) | a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 370 - meter long platform ? | let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 370 ) / 10 = 51 seconds the answer is d . | a = 36 * const_0_2778
b = a * 14
c = b + 370
d = 36 * const_0_2778
e = c / d
|
a ) 26 seconds , b ) 36 seconds , c ) 46 seconds , d ) 56 seconds , e ) none of these | b | divide(add(360, 140), divide(multiply(50, const_1000), const_3600)) | a train is 360 meter long is running at a speed of 50 km / hour . in what time will it pass a bridge of 140 meter length . | "explanation : speed = 50 km / hr = 50 * ( 5 / 18 ) m / sec = 125 / 9 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 / 125 / 9 = 36 seconds answer : b" | a = 360 + 140
b = 50 * 1000
c = b / 3600
d = a / c
|
a ) 690 , b ) 860 , c ) 560 , d ) 875 , e ) 880 | a | divide(subtract(multiply(1500, divide(25, const_100)), 30), divide(50, const_100)) | if 50 % of x is 30 less than 25 % of 1500 , then x is ? | "50 % of x = x / 2 ; 25 % of 1500 = 25 / 100 * 1500 = 375 given that , x / 2 = 375 - 30 = > x / 2 = 345 = > x = 690 . answer : a" | a = 25 / 100
b = 1500 * a
c = b - 30
d = 50 / 100
e = c / d
|
a ) 52 , b ) 75 , c ) 92 , d ) 136 , e ) 150 | e | divide(multiply(300, divide(40, subtract(44, 36))), divide(add(36, 44), subtract(44, 36))) | an investor can sell her microtron stock for 36 $ per share and her dynaco stock for 44 $ per share , if she sells 300 shares altogether , some of each stock , at an average price per share of 40 $ , how many shares of dynaco stock has she sold ? | w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) = ( 44 - 40 ) / ( 40 - 36 ) = 4 / 4 = 1 / 1 = number of microtron stocks / number of dynaco stocks so for every 1 microtron stock , she sold 1 dynaco stock . so out of 300 total stocks , ( 1 / 1 ) th i . e . 300 / 2 = 150 must be dynaco stock . answer ( e ) | a = 44 - 36
b = 40 / a
c = 300 * b
d = 36 + 44
e = 44 - 36
f = d / e
g = c / f
|
a ) 78 years , b ) 22 years , c ) 33 years , d ) 66 years , e ) 65 years | c | divide(subtract(35, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 35 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "let ' s son age is x , then father age is x + 35 . = > 2 ( x + 2 ) = ( x + 35 + 2 ) = > 2 x + 4 = x + 37 = > x = 33 years answer : c" | a = 2 * 2
b = a - 2
c = 35 - b
d = 2 - 1
e = c / d
|
a ) 89 , b ) 20 , c ) 36 , d ) 24 , e ) 34 | d | divide(add(120, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 120 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 120 + 120 = 360 m . time taken = 240 / 10 = 24 sec . answer : option d" | a = 120 + 120
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
['a ) 8 m', 'b ) 10 m', 'c ) 12 m', 'd ) 15 m', 'e ) 17 m'] | b | divide(sqrt(divide(200, divide(const_1, const_2))), const_2) | the width of a rectangular hall is ½ of its length . if the area of the hall is 200 sq . m , what is the difference between its length and breadth ? | let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 200 = x * 1 x / 2 x ² = 400 x = 20 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 20 / 2 = 10 m answer : b | a = 1 / 2
b = 200 / a
c = math.sqrt(b)
d = c / 2
|
a ) 15 , b ) 20 , c ) 88 , d ) 71 , e ) 11 | a | divide(multiply(multiply(3, const_2), 10), subtract(10, multiply(3, const_2))) | a work which could be finished in 10 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | "x - - - - - - - 10 ( x + 10 ) - - - - 6 x * 10 = ( x + 10 ) 6 x = 15 \ answer : a" | a = 3 * 2
b = a * 10
c = 3 * 2
d = 10 - c
e = b / d
|
a ) 1 , b ) 40 , c ) 20 , d ) 26 , e ) 30 | e | multiply(divide(30, 30), 30) | in a dairy farm , 30 cows eat 30 bags of husk in 30 days . in how many days one cow will eat one bag of husk ? | "explanation : one bag of husk = 30 cows per day ⇒ 30 × 1 × 30 = 1 × 30 × x for one cow = 30 days answer : e" | a = 30 / 30
b = a * 30
|
a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | d | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 56 hours . if harry and annie were paid the same amount last week , how many hours did harry work last week ? | "annie earned 40 x + 16 ( 2 x ) = 72 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 72 x ( 1.5 x ) ( h ) = 87 x h = 58 hours the answer is d ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
|
a ) 18 % , b ) 19.78 % , c ) 22 % , d ) 23 % , e ) 25 % | b | multiply(divide(subtract(divide(add(const_100, 9), subtract(const_10, const_3)), divide(const_100, const_10)), divide(const_100, const_10)), const_100) | car dealership x , the total profit from sales increased by 9 percent over the previous years , while the number of cars sold decreased by 9 percent over the previous years . approximately what was the average percent increase in profit per car over the previous year . | "let profit be x and cars sold be y originally profit / car = x / y now it is 1.09 x / 0.91 y = 109 / 91 ( x / y ) increase in profit per car = ( ( 18 / 91 ) ( x / y ) ) / ( x / y ) * 100 = 19.78 % answer = b" | a = 100 + 9
b = 10 - 3
c = a / b
d = 100 / 10
e = c - d
f = 100 / 10
g = e / f
h = g * 100
|
a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(100, reminder(5, 7)) | when positive integer n is divided by 2 , the remainder is 2 . when n is divided by 7 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and next multiple of 7 will be 84 + 21 > 100 . . ans 5 . . c" | a = 100 - reminder
|
a ) 24 , b ) 45 , c ) 48 , d ) 51 , e ) 44 | a | add(45, 30) | two goods trains each 250 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 250 + 250 = 500 m . required time = 500 * 6 / 125 = 24 sec . answer : option a" | a = 45 + 30
|
a ) 5 , b ) 1 / 5 , c ) - 1 / 6 , d ) - 9 / 5 , e ) - 5 | d | divide(add(2, 7), subtract(2, 7)) | if x / y = 2 / 7 , then ( x + y ) / ( x - y ) = ? | "any x and y satisfying x / y = 2 / 7 should give the same value for ( x + y ) / ( x - y ) . say x = 2 and y = 7 , then ( x + y ) / ( x - y ) = ( 2 + 7 ) / ( 2 - 7 ) = - 9 / 5 . answer : d ." | a = 2 + 7
b = 2 - 7
c = a / b
|
a ) 22 , b ) 36 , c ) 48 , d ) 99 , e ) 23 | b | add(70, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 70 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 70 + 30 = 100 km / hr . 100 * 5 / 18 = 250 / 9 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 9 / 250 = 36 sec answer : b" | a = 70 + 30
|
a ) 50 days , b ) 102.7 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(140, const_100), 60)) | p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 140 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250 days and hence r alone in 390 days p works for 60 days - - > 60 / 200 work done = > 30 % q works for 140 days - - > 140 / 250 work done = > 56 % 260 % work left . . . r alone will take 26 % * 395 = 102.7 days answer is ( b )" | a = 4 * 50
b = 140 / 100
c = b * 60
d = a - c
|
a ) 126 kmph , b ) 77 kmph , c ) 54 kmph , d ) 71 kmph , e ) 88 kmph | a | multiply(const_3_6, divide(560, 16)) | a train 560 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 560 / 16 * 18 / 5 = 126 kmph answer : a" | a = 560 / 16
b = const_3_6 * a
|
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 2000 , e ) none of these | c | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 3,000 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 . 3000 at 5 % per annum = ( 3000 × 5 × 1 ) / 100 = rs . 150 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 3000 + x ) × 6 × 1 / 100 . = > 150 + 2 x / 25 = ( 3000 + x ) × 6 × 1 / 100 . = > 15000 + 8 x = ( 3000 + x ) × 6 . = > 15000 + 8 x = 18000 + 6 x . = > 2 x = 3000 . = > x = 1500 . answer : c" | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
|
a ) 12 pm , b ) 22 pm , c ) 72 pm , d ) 92 pm , e ) 23 pm | a | add(divide(subtract(300, 50), add(multiply(add(const_1, divide(50, const_100)), 50), 50)), 10) | k is 50 % faster than l . if l starts at 9 a . m . and k starts at 10 a . m . l travels at a speed of 50 km / hr . if l and k are 300 kms apart , the time when they meet when they travel in opposite direction is ? | explanation : let ' t ' be the time after which they met since l starts . given k is 50 % faster than l 50 t + 1.5 * 50 ( t - 1 ) = 300 50 t + 75 t = 300 + 75 t = 375 / 125 = 3 hrs past the time that l starts so they meet at ( 9 + 3 ) hrs = 12 : 00 noon . answer : a | a = 300 - 50
b = 50 / 100
c = 1 + b
d = c * 50
e = d + 50
f = a / e
g = f + 10
|
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