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 ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | e | subtract(75, divide(75, add(divide(add(8, const_1), multiply(8, const_2)), const_1))) | ` ` i am 8 times as old as you were when i was as old as you are ` ` , said a man to his son . find out their present ages if the sum of the their ages is 75 years . | present age of boy x and previous age of boy y so present age of dad : 8 y and previous age is x so the gap between duration must be same so 8 y - x = x - y - - > x = 9 y / 2 and x + 8 y = 75 ( given ) - - > 9 y / 2 + y = 75 so y = 6 and and x = 27 therefore fathers age is 8 y = 8 * 6 = 48 and child ' s age = x = 27 answer : e | a = 8 + 1
b = 8 * 2
c = a / b
d = c + 1
e = 75 / d
f = 75 - e
|
a ) 9 , b ) 14 , c ) 17 , d ) 23 , e ) 26 | e | add(add(multiply(4, 3), 10), 4) | ada and paul received their scores on 3 tests . on the first test , ada ' s score was 10 points higher than paul ' s score . on the second test , ada ' s score was 4 points higher than paul ' s score . if paul ' s average ( arithmetic mean ) score on the 3 tests was 4 points higher than ada ' s average score on the 3 tests , then paul ' s score on the third test was how many points higher than ada ' s score ? | my take is option d ( 23 ) i followed a simple approach ( explained below ) : test 1 : ada ' s score = paul ' s score + 10 test 2 : ada ' s score = paul ' s score + 4 avg . of paul ' s score = 4 points higher than avg . of ada ' s score this implies that : sum of paul ' s score [ 3 tests ] = 12 points higher than sum of ada ' s score [ 3 tests ] ( 12 points higher since 3 points were given in terms of average of 3 scores ) so , paul needs to score 23 points higher than ada in test 3 , since paul needs to compensate for the lower score in test 1 and test 2 ( 26 = 10 + 4 + 12 ) e | a = 4 * 3
b = a + 10
c = b + 4
|
a ) $ 1250 , b ) $ 1300 , c ) $ 1750 , d ) $ 2015 , e ) $ 1452 | c | divide(divide(subtract(multiply(4000, power(add(const_1, divide(10, const_100)), const_2)), 4000), 2), multiply(3, divide(8, const_100))) | the simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on $ 4000 for 2 years at 10 % per annum . the sum placed on simple interest is ? | c . i . = 4000 * ( 1 + 10 / 100 ) ^ 2 - 4000 = 4000 * 11 / 10 * 11 / 10 - 4000 = $ 840 sum = 420 * 100 / 3 * 8 = $ 1750 answer is c | a = 10 / 100
b = 1 + a
c = b ** 2
d = 4000 * c
e = d - 4000
f = e / 2
g = 8 / 100
h = 3 * g
i = f / h
|
a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 178.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(156, add(add(1, 1), 2)), add(126, 135)), 2) | teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 156 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 156 ) / \ x Γ’ Λ β 156 22.50 = > x Γ’ Λ β ( 156 / 22.50 ) = 1 . = > x Γ’ Λ β 156 = 22.50 . = > x = 178.50 rs . answer : c" | a = 1 + 1
b = a + 2
c = 156 * b
d = 126 + 135
e = c - d
f = e / 2
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number of employees who take the subway to work ? | "let part time emp = x let full time emp = y then , 48 = x + y . . . . . . . . . ( 1 ) ( 1 / 3 ) x + ( 1 / 4 ) y = no . of ppl taking the subway 4 x + 3 y / 12 = no . of ppl taking the subway using 1 x / 12 + 3 * 48 / 12 = no . of ppl taking the subway so , the minimum value of x has to be 12 . hence to maximize the no . put 36 for x 36 / 12 + 12 / 4 = 15 answer : d" | a = 48 * 3
b = 3 * 4
c = a / b
d = 3 * 4
e = 48 - d
f = 48 * 3
g = 3 * 4
h = f / g
i = e / h
j = c + i
|
a ) 5 , b ) 7 , c ) 49 , d ) 20 , e ) 25 | c | divide(multiply(21, 21), multiply(3, 3)) | what is the maximum number of pieces of birthday cake of size 3 β by 3 β that can be cut from a cake 21 β by 21 β ? | "the prompt is essentially asking for the maximum number of 3 x 3 squares that can be cut from a larger 21 by 21 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 7 ' pieces ' each , we have ( 7 ) ( 7 ) = 49 total smaller squares ( at maximum ) . c" | a = 21 * 21
b = 3 * 3
c = a / b
|
a ) 44 , b ) 45 , c ) 46 , d ) 47 , e ) 48 | d | floor(divide(multiply(37, 9), 7)) | in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 37 . in the company , what is the maximum number of non - managers in a department that has 9 managers ? | "9 / 7 * 37 = 47.6 the answer is d ." | a = 37 * 9
b = a / 7
c = math.floor(b)
|
a ) 177 m , b ) 189 m , c ) 140 m , d ) 178 m , e ) 188 m | c | divide(12, subtract(divide(12, 10), 7)) | a train covers a distance of 12 km in 10 min . if it takes 7 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 7 = 140 m . answer : c" | a = 12 / 10
b = a - 7
c = 12 / b
|
a ) 26 , b ) 27 , c ) 24 , d ) 18 , e ) 11 | c | subtract(78, subtract(add(41, 22), 9)) | in a class of 78 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | "explanation : you could solve this by drawing a venn diagram . a simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking french to find the number taking only french . likewise find those taking only german . then we have : total = only french + only german + both + neither 78 = ( 41 - 9 ) + ( 22 - 9 ) + 9 + neither . not enrolled students = 24 answer : c" | a = 41 + 22
b = a - 9
c = 78 - b
|
a ) 2 / 131 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | subtract(add(const_4, const_4), const_1) | if ( n + 2 ) ! / n ! = 182 , n = ? | "( n + 2 ) ! / n ! = 182 rewrite as : [ ( n + 2 ) ( n + 1 ) ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] / [ ( n ) ( n - 1 ) ( n - 2 ) . . . . ( 3 ) ( 2 ) ( 1 ) ] = 132 cancel out terms : ( n + 2 ) ( n + 1 ) = 132 from here , we might just test the answer choices . since ( 14 ) ( 13 ) = 182 , we can see that n = 12 e" | a = 4 + 4
b = a - 1
|
a ) 1 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 3 , e ) none | a | divide(subtract(74, 69), subtract(69, 64)) | in what ratio must tea at rs . 64 per kg be mixed with tea at rs . 74 per kg so that the mixture must be worth rs . 69 per kg ? | "required ratio = 500 : 500 = 1 : 1 answer a" | a = 74 - 69
b = 69 - 64
c = a / b
|
a ) 48 , b ) 60 , c ) 56 , d ) 64 , e ) 68 | d | multiply(divide(multiply(const_2, 240), 27), divide(const_3600, const_1000)) | 36 . a 240 m long train crosses a platform of equal length in 27 s . what is the speed of the train in km / h ? | by the formula , speed = distance / time = ( 240 + 240 ) / 27 x ( 18 / 5 ) = 480 / 27 x 18 / 5 = 64 km / h answer : d | a = 2 * 240
b = a / 27
c = 3600 / 1000
d = b * c
|
['a ) 100 m', 'b ) 125 m', 'c ) 150 m', 'd ) 278 m', 'e ) 300 m'] | e | divide(divide(divide(multiply(divide(333.18, 24.68), const_1000), const_3), const_10), divide(const_3, const_2)) | the base of a triangular field is three times its altitude . if the cost of cultivating the field at rs . 24.68 per hectare be rs . 333.18 , find its base and height . | sol . area of the field = total cost / rate = ( 333.18 / 25.6 ) hectares = 13.5 hectares ( 13.5 x 10000 ) m ^ 2 = 135000 m ^ 2 . let altitude = x metres and base = 3 x metres . then , ( 1 / 2 ) * 3 x * x = 135000 < = > x ^ 2 = 90000 < = > x = 300 . base = 900 m and altitude = 300 m . ans : e | a = 333 / 18
b = a * 1000
c = b / 3
d = c / 10
e = 3 / 2
f = d / e
|
a ) $ 811 , b ) $ 922 , c ) $ 1033 , d ) $ 1144 , e ) $ 1255 | d | divide(multiply(multiply(11000, divide(1540, divide(multiply(15000, 8), const_100))), 8), const_100) | a , b and c enter into a partnership by investing $ 11000 , $ 15000 and $ 23000 respectively . at the end of 8 months , b receives $ 1540 as his share . find the share of a . | "the ratio of capital of a , b and c = 11000 : 15000 : 23000 = 11 : 15 : 23 a ' s share = ( 11 / 15 ) * 1560 = $ 1144 the answer is d ." | a = 15000 * 8
b = a / 100
c = 1540 / b
d = 11000 * c
e = d * 8
f = e / 100
|
a ) 2 , b ) 4 , c ) 5 , d ) 3 , e ) 7 | d | add(const_4, const_4) | if each year the population of the country grows by 30 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 130 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 169 . . . . . . . . . . . . . 2 year elapsed year 2004 : population > 200 . . . . . . . . . . . . . . . . . . 3 year elapsed answer : d" | a = 4 + 4
|
a ) 33 , b ) 28 , c ) 27 , d ) 24 , e ) 23 | b | divide(add(24, 26), const_2) | if the median of a list of numbers is m , the first quartile of the list is the median of the numbers in the list that are less than m . what is the first quartile of the list of numbers 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 ? | "it is given that a quartile is the middle number of all numbers less than median . . so lets arrange the number in ascending order - 42 , 24 , 30 , 34 , 26 , 36 , 33 and 35 24 , 26 , 30 , 33 , 34 , 35 , 36 , 42 . . . numbers less than median are 24 , 26 , 30 , 33 . . the median of these numbers = center of 26 and 30 = 28 b" | a = 24 + 26
b = a / 2
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(divide(20, 3), const_1) | how many integers are divisible by 3 between 20 ! and 20 ! + 20 inclusive ? | "b - 7 20 ! is divisible by 3 there are 6 numbers between 10 ! and 10 ! + 20 that are divisible by 3 . hence 7" | a = 20 / 3
b = a + 1
|
a ) 9.25 , b ) 5.25 , c ) 7.25 , d ) 6.25 , e ) 11.25 | e | multiply(divide(27, const_60), add(20, 5)) | the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 27 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 27 / 60 ) ) km = 11.25 km . answer : e" | a = 27 / const_60
b = 20 + 5
c = a * b
|
a ) 1 / 33 , b ) 2 / 63 , c ) 1 / 3 , d ) 43 / 62 , e ) 11 / 12 | d | multiply(multiply(multiply(divide(multiply(10, const_2), multiply(10, const_2)), divide(multiply(4, 4), subtract(multiply(10, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | if 4 people are selected from a group of 10 married couples , what is the probability that none of them would be married to each other ? | "if we are to select 4 people from 10 couples without any restriction , how many ways can we make the selection ? 20 ! / 4 ! 16 ! = 4845 if we are to select 4 people from 10 couples with restriction that no married couple can both make it to the group , only a representative ? 10 ! / 4 ! 6 ! = 210 but we know that to select a person from each couple , take 2 possibilities 210 * 2 * 2 * 2 * 2 = 3360 probability = desired / all possibilities = 3360 / 4845 = 43 / 62 answer : d" | a = 10 * 2
b = 10 * 2
c = a / b
d = 4 * 4
e = 10 * 2
f = e - 1
g = d / f
h = c * g
i = 4 * 4
j = i - 2
k = 4 * 4
l = j / k
m = h * l
n = 4 * 4
o = n - 2
p = o - 2
q = 4 * 4
r = q - 1
s = p / r
t = m * s
|
a ) 724 , b ) 804 , c ) 814 , d ) 9000 , e ) none | d | divide(multiply(add(multiply(11, const_100), 25), multiply(50, const_100)), power(25, const_2)) | what is the least number of square tiles required to pave the floor of a room 50 m 00 cm long and 11 m 25 cm broad ? | solution length of largest tile = h . c . f . of 5000 cm & 1125 cm = 25 cm . area of each tile = ( 25 x 25 ) cm 2 β΄ required number of tiles = [ 5000 x 1125 / 25 x 25 ] = 9000 . answer d | a = 11 * 100
b = a + 25
c = 50 * 100
d = b * c
e = 25 ** 2
f = d / e
|
a ) 320 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 528 $ | e | multiply(multiply(0.65, 65), 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 65 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 * 65 ) = 21 + 507 = > 528 hence answer will be ( e ) 528" | a = 0 * 65
b = a * 12
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a ) 37 , b ) 41 , c ) 50 , d ) 59 , e ) 33 | e | add(multiply(37, const_2), multiply(subtract(69, multiply(37, const_2)), 37)) | if the average of 37 , 69 , 47 and x is 46.5 , what is x ? | "x = 33 because : ( 37 + 69 + 47 + x ) / 4 = 46.5 ( 37 + 69 + 47 + x ) / 4 * 4 = 46.5 * 4 ( 37 + 69 + 47 + x ) = 186 ( 37 + 69 + 47 + x ) - 153 = 186 - 153 x = 33 therefore , the answer is e , 33 ." | a = 37 * 2
b = 37 * 2
c = 69 - b
d = c * 37
e = a + d
|
a ) 20 , b ) 45 , c ) 55 , d ) 65 , e ) 80 | b | subtract(multiply(60, 4), multiply(65, const_3)) | joe β s average ( arithmetic mean ) test score across 4 equally weighted tests was 60 . he was allowed to drop his lowest score . after doing so , his average test score improved to 65 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 60 . so what we can assume is that we have 4 test scores , each 60 . he dropped his lowest score and the avg went to 65 . this means that the lowest score was not 60 and other three scores had given the lowest score 5 each to make it up to 60 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 60 . so the lowest score = 60 - 15 = 45 answer ( b )" | a = 60 * 4
b = 65 * 3
c = a - b
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 0 | e | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | if the number 992 , 132,24 x is divisible by 11 , what must be the value of x ? | multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 992 , 132,24 x sum of digits at odd places = 9 + 2 + 3 + 2 + x = 16 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 4 = 16 ( ii ) ( i ) - ( ii ) = 16 + x - 16 = x - 0 hence x should be = 0 to make this a multiple of 11 ( 0 ) option e | a = 3 * 10
b = a + 1
c = b * 2
d = c * 4
e = d * 11
|
a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | b | add(add(9, 8), 5) | in kaya ' s teacher ' s desk there are 9 pink highlighters , 8 yellow highlighters , and 5 blue highlighters . how many highlighters are there in all ? | add the numbers of highlighters . 9 + 8 + 5 = 22 . answer is b . | a = 9 + 8
b = a + 5
|
a ) 36 , b ) 72 , c ) 120 , d ) 144 , e ) 180 | c | divide(500, add(divide(100, const_60), divide(150, const_60))) | a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 500 bolts ? | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 500 bolts 500 * 6 / 25 = 120 mins ans c" | a = 100 / const_60
b = 150 / const_60
c = a + b
d = 500 / c
|
a ) 71 , b ) 74 , c ) 78 , d ) 70 , e ) 80 | c | divide(subtract(subtract(multiply(75, subtract(34, const_2)), 28), 34), subtract(subtract(34, const_2), 3)) | the average mark of a class of thirty two students is 75 . if 3 students whose marks are 28 and 34 are removed , then find the approximate average mark of the remaining students of the class . | exp . total mark of 32 students = 75 * 32 = 2400 , total mark after the removal of 2 students = 2400 β ( 28 + 34 ) = 2400 β 62 = 2338 approximate average mark = 2338 / ( 32 - 2 ) = 2338 / 30 = 78 answer : c | a = 34 - 2
b = 75 * a
c = b - 28
d = c - 34
e = 34 - 2
f = e - 3
g = d / f
|
a ) 2 days , b ) 3.5 days , c ) 2.2 days , d ) 4 days , e ) 5.7 days | c | divide(const_1, add(divide(const_1, 12), add(divide(const_1, 6), divide(const_1, 5)))) | a can do a piece of work in 6 days , and b can do it in 5 days . if c , who can do the work in 12 days , joins them , how long will they take to complete the work ? | a , b , and c do the work in = 6 * 5 * 12 / 6 * 5 + 5 * 12 + 6 * 12 = 360 / 162 = 2.2 days answer is c | a = 1 / 12
b = 1 / 6
c = 1 / 5
d = b + c
e = a + d
f = 1 / e
|
a ) 186 , b ) 120 , c ) 152 , d ) 220 , e ) 220 | a | subtract(multiply(20, 20), add(multiply(5, 14), multiply(9, 16))) | the average age of 20 students of a class is 20 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 20 th student is | "explanation : age of the 20 th student = [ 20 * 20 - ( 14 * 5 + 16 * 9 ) ] = ( 400 - 214 ) = 186 years . answer : a" | a = 20 * 20
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
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a ) 19 / 35 , b ) 23 / 42 , c ) 27 / 55 , d ) 31 / 63 , e ) 35 / 74 | d | divide(const_4, add(multiply(const_4, 5), const_1)) | tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 5 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ? | "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 21 / 30 ) = 10 / 21 hours peter worked for 31 / 21 hours so he painted 31 / 21 * 1 / 3 = 31 / 63 of the room . the answer is d ." | a = 4 * 5
b = a + 1
c = 4 / b
|
a ) $ 4.70 , b ) $ 4.80 , c ) $ 3.85 , d ) $ 3.80 , e ) $ 3.70 | e | subtract(10.5, multiply(const_2, 3.4)) | little roshini had $ 10.50 . he spent some money on sweets and gave to his two friends $ 3.40 each . how much money was spent for sweets ? | roshini spent and gave to his two friends a total of x + 3.40 + 3.40 = $ 3.65 money spent for sweet = 10.50 - 6.80 = $ 3.70 correct answer is e ) $ 3.70 | a = 2 * 3
b = 10 - 5
|
a ) 3 , b ) 4 , c ) 18 / 5 , d ) 9 , e ) 12 | c | divide(const_1, multiply(9, add(divide(const_1, 36), divide(const_1, const_2.0)))) | a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used the same number of each type of machine to do job in 12 hours , how many machine r were used ? | "yes there is a typo in the question , i got the same ques on my gmat prep last week , and the questions goes as : a company has two types of machines , type r and type s . operating at a constant rate a machine of r does a certain job in 36 hours and a machine of type s does the job in 9 hours . if the company used the same number of each type of machine to do job in 2 hours , how many machine r were used ? so for a job to be done in 2 hours r = 1 / 2 r _ a ( rate of machine r ) = 1 / 36 r _ s ( rate of machine s ) = 1 / 9 lets say x machines are used to attain the desired rate , thus x / 36 + x / 9 = 1 / 2 ( desired r = 1 / 2 i . e . to complete the job in 2 hours ) ( x + 4 x ) / 36 = 1 / 2 5 x / 36 = 1 / 2 x = 18 / 5 . qa = 18 / 5 ( answer c )" | a = 1 / 36
b = 1 / 2
c = a + b
d = 9 * c
e = 1 / d
|
a ) 40 , b ) 50 , c ) 60 , d ) 45 , e ) 56 | c | add(subtract(110, multiply(11, 5)), 5) | a batsman makes a score of 110 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 110 = 11 x x = 110 - 50 = 60 answer is c" | a = 11 * 5
b = 110 - a
c = b + 5
|
a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) 240 | e | multiply(divide(const_60, 15), 1) | if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 1 hour ? | "answer = 4 * 60 = 240 answer is e" | a = const_60 / 15
b = a * 1
|
a ) 16,8 , b ) 10,2 , c ) 17,6 , d ) 14,10 , e ) 12,4 | a | add(subtract(multiply(divide(const_10, const_2), 6), divide(add(8, multiply(divide(const_10, const_2), 6)), const_2)), divide(const_10, const_2)) | the difference of two numbers is 8 and one - fourth of their sum is 6 . find the numbers . | "let numbers be x and y . equation ( i ) : x - y = 8 equation ( ii ) : ( x + y ) / 4 = 6 solve system of equations : x = 16 , y = 8 a is the correct answer ." | a = 10 / 2
b = a * 6
c = 10 / 2
d = c * 6
e = 8 + d
f = e / 2
g = b - f
h = 10 / 2
i = g + h
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(multiply(divide(45, add(const_1, divide(80, 100))), const_2), 45) | a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 45 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | "x is the buy price . x + 80 % x = 1,8 x is the sale price = 45 so x , the buy price becomes 45 / 1,8 = 25 % 100 mark up means x + 100 % x = 2 x so 50 . 50 - 45 = 5 answer : e" | a = 80 / 100
b = 1 + a
c = 45 / b
d = c * 2
e = d - 45
|
a ) 1 , b ) 2 , c ) 8 , d ) 4 , e ) 5 | c | subtract(subtract(multiply(1250, power(add(const_1, divide(8, const_100)), 2)), 1250), multiply(multiply(1250, divide(8, const_100)), 2)) | indu gave bindu rs . 1250 on compound interest for 2 years at 8 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "1250 = d ( 100 / 4 ) 8 d = 8 answer : c" | a = 8 / 100
b = 1 + a
c = b ** 2
d = 1250 * c
e = d - 1250
f = 8 / 100
g = 1250 * f
h = g * 2
i = e - h
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a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | c | add(multiply(0.25, subtract(5, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 7.50 + 0.25 ( s β 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ? | oa is definitely wrong . the answer should be c . | a = 5 - 1
b = 0 * 25
c = b + 0
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a ) 540 , b ) 240 , c ) 440 , d ) 140 , e ) 340 | b | power(sqrt(divide(12, const_4)), const_3) | the lateral surface area of cuboid length 12 m , breadth 8 m and height 6 m . | "cuboid lateral surface = 2 h ( l + b ) = 2 Γ 6 ( 20 ) = 240 m ( power 2 ) answer is b" | a = 12 / 4
b = math.sqrt(a)
c = b ** 3
|
a ) 648 , b ) 1800 , c ) 2700 , d ) 2160 , e ) none of these | d | multiply(multiply(divide(270, 6), 4), 12) | running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 12 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) Γ’ Λ Β΄ 6 Γ£ β 1 Γ£ β x = 12 Γ£ β 4 Γ£ β 270 Γ’ β‘ β x = 12 x 4 x 270 / 6 = 2160 . answer d" | a = 270 / 6
b = a * 4
c = b * 12
|
a ) 65 , b ) 100 , c ) 115 , d ) 130 , e ) 260 | e | add(multiply(40, const_2), 60) | of the 400 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 40 percent of the employees prefer city y and 60 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ? | "240 prefer x ( group 1 ) ; 160 prefer y ( group 2 ) . city y needs 300 people : letall 160 who prefer y ( entire group 2 ) be relocated there , the rest 140 will be those who prefer x from group 1 ; city x needs 100 people : 240 - 140 = 100 from group 1 will be relocated to x , which they prefer . so , the highest possible number of employees who will be relocated to the city they prefer is 160 + 100 = 260 . answer : e ." | a = 40 * 2
b = a + 60
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(divide(22, add(add(divide(20, const_100), const_1), const_1)), add(divide(20, const_100), const_1)) | two friends plan to walk along a 22 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 20 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ? | "if q complete x kilometers , then p completes 1.2 x kilometers . x + 1.2 x = 22 2.2 x = 22 x = 10 then p will have have walked 1.2 * 10 = 12 km . the answer is c ." | a = 20 / 100
b = a + 1
c = b + 1
d = 22 / c
e = 20 / 100
f = e + 1
g = d * f
|
a ) 50 , b ) 45 , c ) 55 , d ) 60 , e ) 65 | d | divide(subtract(const_100, 40), divide(40, 40)) | a car traveled 40 % of the way from town x to town y at an average speed of 40 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ? | "assume total distance = 100 miles time taken for 40 miles = 40 / 40 = 1 hour time taken for the rest of the 60 miles = 60 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 60 / v hence v = 60 mph answer : d" | a = 100 - 40
b = 40 / 40
c = a / b
|
a ) 28 , b ) 30 , c ) 32 , d ) 34 , e ) 31 | e | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , what is the weight of b ? | "let the weight of a , b and c are a , b and c respectively . average weight of a , b and c = 45 a + b + c = 45 Γ 3 = 135 - - - equation ( 1 ) average weight of a and b = 40 a + b = 40 Γ 2 = 80 - - - equation ( 2 ) average weight of b and c = 43 b + c = 43 Γ 2 = 86 - - - equation ( 3 ) equation ( 2 ) + equation ( 3 ) - equation ( 1 ) = > a + b + b + c - ( a + b + c ) = 80 + 86 - 135 = > b = 80 + 86 - 135 = 166 - 135 = 31 weight of b = 31 kg answer is e ." | a = 40 * 2
b = 43 * 2
c = a + b
d = 45 * 3
e = c - d
|
a ) 1 : 4 , b ) 4 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | c | divide(divide(const_100, add(const_100, 80)), divide(const_100, add(const_100, 20))) | bert and rebecca were looking at the price of a condominium . the price of the condominium was 80 % more than bert had in savings , and separately , the same price was also 20 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings . | "suppose bert had 100 so price becomes 180 , this 180 = 1.2 times r ' s saving . . so r ' s saving becomes 150 so required ratio is 100 : 150 = 2 : 3 answer : c" | a = 100 + 80
b = 100 / a
c = 100 + 20
d = 100 / c
e = b / d
|
a ) 32.5 % , b ) 32.9 % , c ) 92.5 % , d ) 38.5 % , e ) 37.5 % | a | multiply(divide(subtract(multiply(divide(25, const_100), 1000), multiply(divide(120, const_100), multiply(1000, divide(60, const_100)))), subtract(1000, multiply(1000, divide(60, const_100)))), const_100) | of the 1000 inhabitants of a town , 60 % are males of whom 120 % are literate . if , of all the inhabitants , 25 % are literate , then what percent of the females of the town are literate ? | "explanation : number of males = 60 % of 1000 = 600 . number of females = ( 1000 - 600 ) = 400 . number of literates = 25 % of 1000 = 250 . number of literate males = 20 % of 600 = 120 . number of literate females = ( 250 - 120 ) = 130 . required pecentage = ( 130 / 400 * 100 ) % = 32.5 % . answer : a ) 32.5 %" | a = 25 / 100
b = a * 1000
c = 120 / 100
d = 60 / 100
e = 1000 * d
f = c * e
g = b - f
h = 60 / 100
i = 1000 * h
j = 1000 - i
k = g / j
l = k * 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | subtract(subtract(10, 4), const_1) | if 4 < x < 6 < y < 10 , then what is the greatest possible positive integer difference of x and y ? | "4 < x < 6 < y < 10 ; 4 < x y < 10 4 + y < x + 10 y - x < 6 . positive integer difference is 5 ( for example y = 9.5 and x = 4.5 ) answer : c ." | a = 10 - 4
b = a - 1
|
a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | b | divide(1, divide(add(multiply(const_3600, divide(1, 48)), 15), const_3600)) | a car traveling at a certain constant speed takes 15 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 48 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "48 * t = 1 km = > t = 1 / 48 km / h v * ( t + 15 / 3600 ) = 1 v ( 1 / 48 + 15 / 3600 ) = 1 v ( 90 / 3600 ) = 1 v = 40 km / h the answer is b ." | a = 1 / 48
b = 3600 * a
c = b + 15
d = c / 3600
e = 1 / d
|
a ) 320 $ , b ) 420 $ , c ) 490 $ , d ) 516 $ , e ) 680 $ | d | multiply(multiply(0.75, 55), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.75 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.75 + 0.75 * 12 * 55 = 516 hence - d | a = 0 * 75
b = a * 12
|
a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 1,476 | c | divide(divide(multiply(multiply(15, add(15, const_1)), add(multiply(15, 2), const_1)), 6), divide(multiply(multiply(15, add(15, const_1)), add(multiply(15, 2), const_1)), 6)) | the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n 3 / 3 + cn 2 + n / 6 , where c is a constant . what is the sum of the first 15 positive perfect squares ? | "for n = 1 , 1 = 1 / 3 + c + 1 / 6 1 = 1 / 2 + c = > c = 1 / 2 15 * 15 * 15 / 3 + 1 / 2 * 15 * 15 + 15 / 6 = 1240 answer : c" | a = 15 + 1
b = 15 * a
c = 15 * 2
d = c + 1
e = b * d
f = e / 6
g = 15 + 1
h = 15 * g
i = 15 * 2
j = i + 1
k = h * j
l = k / 6
m = f / l
|
['a ) 1276', 'b ) 1200', 'c ) 2832', 'd ) 1800', 'e ) 1236'] | d | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, const_2), power(divide(1200, 30), const_2)))), 15) | a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 15 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | explanation : l * 30 = 1200 Γ¨ l = 40 40 + 30 + 50 = 120 120 * 15 = 1800 answer : option d | a = 1200 / 30
b = 30 + a
c = 30 ** 2
d = 1200 / 30
e = d ** 2
f = c + e
g = math.sqrt(f)
h = b + g
i = h * 15
|
a ) 21 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | a | divide(multiply(7, subtract(7, const_1)), const_2) | there are 7 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "7 players are there . two players play one game with one another . so 7 c 2 = 7 * 6 / 2 = 21 so option a is correct" | a = 7 - 1
b = 7 * a
c = b / 2
|
a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | b | subtract(119, subtract(add(144, 119), 238)) | in a graduating class of 238 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 238 = 25 greatest - smallest is 144 - 25 = 119 so answer must be b . 119" | a = 144 + 119
b = a - 238
c = 119 - b
|
a ) 43 , b ) 36 , c ) 28 , d ) 45 , e ) 11 | d | multiply(subtract(const_1, divide(10, const_100.0)), 50) | there are 50 students in a class . if 10 % are absent on a particular day , find the number of students present in the class . | "number of students absent on a particular day = 10 % of 50 i . e . , 10 / 100 Γ 50 = 5 therefore , the number of students present = 50 - 5 = 45 students . answer : d" | a = 10 / 100
b = 1 - a
c = b * 50
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(190, multiply(add(300, 42), const_0_2778)) | the speed at which a woman can row a boat in still water is 300 kmph . if he rows downstream , where the speed of current is 42 kmph , what time will he take to cover 190 metres ? | "speed of the boat downstream = 300 + 42 = 342 kmph = 342 * 5 / 18 = 95 m / s hence time taken to cover 190 m = 190 / 95 = 2 seconds . answer : b" | a = 300 + 42
b = a * const_0_2778
c = 190 / b
|
['a ) 12', 'b ) 96', 'c ) 50', 'd ) 72', 'e ) 144'] | d | power(sqrt(divide(power(multiply(6, const_2), const_2), const_2)), const_2) | in a circle with a radius of 6 , what is the area of the biggest rectangle that can be cut out of this circle ? | the biggest rectangle would be a square , whose diagonal is 12 ( 6 + 6 ) if side of square is a , 144 = a ^ 2 + a ^ 2 = 2 a ^ 2 so area of square = a ^ 2 = 144 / 2 = 72 answer : d | a = 6 * 2
b = a ** 2
c = b / 2
d = math.sqrt(c)
e = d ** 2
|
a ) 86.6 km , b ) 46.6 km , c ) 33 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 3), divide(44, const_60)) | the speed of a boat in still water in 42 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 3 ) = 45 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time Γ speed = 11 / 15 Γ 45 = 33 km answer : c" | a = 42 + 3
b = 44 / const_60
c = a * b
|
a ) 55.8 , b ) 51.8 , c ) 53.8 , d ) 56.8 , e ) 52.8 | d | subtract(add(subtract(subtract(add(add(multiply(45, const_1), const_4), const_12), const_4), const_0_25), const_1), const_1) | find the total average marks of all the students in 2 separate classes , if the average marks of students in the first class of 39 students is 45 and that of another class of 35 students is 70 . | sum of the marks for the class of 39 students = 39 * 45 = 1755 sum of the marks for the class of 35 students = 35 * 70 = 2450 sum of the marks for the class of 74 students = 1755 + 2450 = 4205 average marks of all the students = 4205 / 74 = 56.8 answer : d | a = 45 * 1
b = a + 4
c = b + 12
d = c - 4
e = d - const_0_25
f = e + 1
g = f - 1
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(divide(const_3.0, const_4), const_100)) | on an order of 4 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 5 dozens free . equivalent discount = 1 / 5 * 100 = 20 % . answer b ) 20 %" | a = 3 / 0
b = a * 100
c = 100 - b
|
a ) 33 , b ) 57 , c ) 54 , d ) 99 , e ) 01 | b | divide(multiply(300, subtract(const_100, add(27, 54))), const_100) | in an examination , 300 students appeared . out of these students ; 27 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed . | "the number of students with first division = 27 % of 300 = 27 / 100 Γ 300 = 8100 / 100 = 81 and , the number of students with second division = 54 % of 300 = 54 / 100 Γ 300 = 16200 / 100 = 162 therefore , the number of students who just passed = 300 β ( 81 + 162 ) = 57 answer : b" | a = 27 + 54
b = 100 - a
c = 300 * b
d = c / 100
|
a ) 21 : 23 , b ) 23 : 45 , c ) 22 : 7 , d ) 25 : 29 , e ) none of these | c | divide(multiply(77000, const_12), multiply(42000, add(const_4, const_3))) | x starts a business with rs . 77000 . y joins in the business after 5 months with rs . 42000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 77000 * 12 : 42000 * 7 = 77 * 12 : 42 * 7 = 11 * 2 : 7 = 22 : 7 . answer : option c" | a = 77000 * 12
b = 4 + 3
c = 42000 * b
d = a / c
|
a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 1 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 10)), const_100)) | what is the % change in the area of a rectangle when its length increases by 10 % and its width decreases by 10 % ? | "explanatory answer whenever you encounter problems like this , use a numerical example and then proceed . for ease of computation , it is safe in most cases , to assume the length to be 100 units and the width to be 100 units . ( remember , a square is a rectangle too and the problem works the same way when you assume different values for length and width . computation becomes a bit tedious with different values for length and width ) area of a rectangle = length * width = 100 * 100 = 10,000 sq units . when the length increases by 10 % , the new length becomes 110 units . and as the width decreases by 10 % , new width becomes 90 units . therefore , new area = 110 * 90 = 9900 sq units . new area is 100 sq units lesser than the original area . % change in area = ( ( change in area ) / ( original area ) ) * 100 = ( 100 / 10,000 ) * 100 = 1 % decrease in area the correct choice is ( d )" | a = 100 + 10
b = 100 - 10
c = a * b
d = c / 100
e = 100 - d
|
a ) 9 km , b ) 72.5 km , c ) 190.75 km , d ) 885.5 km , e ) none of these | d | divide(multiply(80.5, 6.6), 0.6) | on a scale of map , 0.6 cm represent 6.6 km . if the distance between the points on the map is 80.5 cm , the actual distance between these points is : | solution let the actual distance be x km . then , more distance on the map , more is the actual distance ( direct proportion ) β΄ 0.6 : 80.5 : : 6.6 : x β 0.6 x = 80.5 Γ 6.6 β x = β x = 80.5 x 6.6 / 0.6 = x = 885.5 . answer d | a = 80 * 5
b = a / 0
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | subtract(9, const_1) | there are 9 cities numbered 1 to 9 . from how many cities the flight can start so as to reach the city 8 either directly or indirectly such that the path formed is divisible by 3 . | start from 1 : 1368 , start from 2 : 2358 , start from 3 : 378 , start from 4 : 48 , start from 5 : 528 , start from 6 : 678 , start from 7 : 78 , why start from 8 ? ? ? , start from 9 : 918 . so flights can start from all 8 other cities . answer : b | a = 9 - 1
|
a ) 221 , b ) 287 , c ) 400 , d ) 288 , e ) 671 | e | divide(740, power(add(const_1, divide(5, const_100)), 2)) | find the sum lend at c . i . at 5 p . c per annum will amount to rs . 740 in 2 years ? | "explanation : 740 = p ( 21 / 20 ) 2 p = 671 answer : e" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 740 / c
|
a ) rs . 10123.77 , b ) rs . 10123.21 , c ) rs . 10123.20 , d ) rs . 10123.28 , e ) rs . 10123.21 | c | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 25,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 25000 * ( 1 + 12 / 100 ) 3 ] = 25000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 35123.20 - 25000 ) = rs . 10123.20 answer : c" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) β 4.72 , b ) β 3.82 , c ) β 3.58 , d ) β 2.68 , e ) 0.57 | d | subtract(multiply(divide(divide(subtract(power(5, 2), power(1.9, 3.7)), const_1000), const_1000), 5), divide(divide(subtract(power(5, 2), power(1.9, 3.7)), const_1000), const_1000)) | what is the value of 5 x ^ 2 β 1.9 x β 3.7 for x = β 0.3 ? | "5 x ^ 2 - 1.9 x - 3.7 , x = - 0.3 firstly , 5 x ^ 2 can be written as 5 * x * x - substituting one of the x with - 0.3 we have - 1.5 x - 1.9 x - 3.7 - - - > - 3.4 x - 3.7 - - - > - 3.4 ( - 0.3 ) - 3.7 - - - > 1.02 - 3.7 = - 2.68 answer : d" | a = 5 ** 2
b = 1 ** 9
c = a - b
d = c / 1000
e = d / 1000
f = e * 5
g = 5 ** 2
h = 1 ** 9
i = g - h
j = i / 1000
k = j / 1000
l = f - k
|
a ) 12040 , b ) 12140 , c ) 12240 , d ) 12340 , e ) 12440 | c | add(multiply(divide(2, const_100), 12000), 12000) | the income of a man increase consistently by 2 % per year . if his present income is rs . 12000 then what will his income be after 1 year ? | explanation : income = 12000 x 1.02 = 12240 answer : option c | a = 2 / 100
b = a * 12000
c = b + 12000
|
a ) 88 , b ) 89 , c ) 99 , d ) 54 , e ) 90 | a | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(2, const_10)), const_3), 88) | find the largest 2 digit number which is exactly divisible by 88 ? | "largest 2 digit number is 99 after doing 99 Γ· 88 we get remainder 11 hence largest 2 digit number exactly divisible by 88 = 99 - 11 = 88 a" | a = 100 * 100
b = 100 * 10
c = a + b
d = 100 * 3
e = c + d
f = 2 * 10
g = e + f
h = g + 3
i = h * 88
|
a ) $ 60 , b ) $ 64 , c ) $ 80 , d ) $ 96 , e ) can not be determined | c | divide(add(80, 80), 4) | if greg buys 3 shirts , 6 trousers and 4 ties , the total cost is $ 80 . if greg buys 7 shirts , 2 trousers and 4 ties , the total cost is $ 80 . how much will it cost him to buy 5 trousers , 4 shirts and 4 ties ? | "solution : 3 x + 6 y + 4 z = 80 7 x + 2 y + 4 z = 80 adding both the equations = 10 x + 8 y + 8 z = 160 5 x + 4 y + 4 z = 80 ans c" | a = 80 + 80
b = a / 4
|
a ) 4.5 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 9 | e | multiply(divide(18, 12), 6) | when a number is divided by 6 & then multiply by 12 the answer is 18 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 18 = > 2 x = 18 = > x = 9 e" | a = 18 / 12
b = a * 6
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | e | divide(divide(400, subtract(divide(400, 40), divide(400, 48))), 40) | donovan and michael are racing around a circular 400 - meter track . if donovan runs each lap in 48 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 400 / 48 = > 50 / 6 2 . speed / rate of michael = distance / time = > 400 / 40 = > 10 relative speed between them = 10 - 50 / 6 = > 10 / 6 ( we subtract the rates if moving in the same direction and add the rates if moving in the opposite direction ) in order to pass donovan - distance to be covered = 400 , relative rate = 10 / 6 total time taken by micheal to surpass donovan = distance / rate = > 400 * 6 / 10 = > 2400 / 10 = > 240 no . of laps taken by michael = total time / michael ' s rate = > 240 / 40 = > 6 hence correct answer is 6 laps . e" | a = 400 / 40
b = 400 / 48
c = a - b
d = 400 / c
e = d / 40
|
a ) 66 m square , b ) 49 m square , c ) 77 m square , d ) 44 m square , e ) 33 m square | b | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 6))), multiply(4, 6)) | a cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm . find the total area of the wet surface . | "area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 4 * 1.25 + 6 * 1.25 ) ] + 6 * 4 = 49 m square answer : b" | a = 25 / 100
b = a + 1
c = b * 4
d = 25 / 100
e = d + 1
f = e * 6
g = c + f
h = 2 * g
i = 4 * 6
j = h + i
|
a ) 4 % , b ) 10 % , c ) 97.14 % , d ) 90.14 % , e ) 20 % | c | multiply(divide(subtract(divide(subtract(const_100, 65), const_100), subtract(divide(40, const_100), multiply(divide(65, const_100), divide(60, const_100)))), divide(subtract(const_100, 65), const_100)), const_100) | 65 % of the employees of a company are men . 60 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ? | "no of employees = 100 ( say ) men = 65 women = 35 men speaking french = 0.60 * 65 = 39 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 39 = 1 and women not speaking french = 35 - 1 = 34 % of women not speaking french = 34 / 35 * 100 = 97.14 % answer c" | a = 100 - 65
b = a / 100
c = 40 / 100
d = 65 / 100
e = 60 / 100
f = d * e
g = c - f
h = b - g
i = 100 - 65
j = i / 100
k = h / j
l = k * 100
|
a ) 8 days , b ) 2 days , c ) 6 days , d ) 7 days , e ) 4 days | c | divide(multiply(6, 6), divide(subtract(multiply(6, 6), multiply(add(divide(multiply(6, 6), 6), divide(multiply(6, 6), 6)), 2)), 2)) | a can do a piece of work in 6 days . b can do it in 6 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 6 - 1 / 6 = 1 / 6 = > 6 days answer : c" | a = 6 * 6
b = 6 * 6
c = 6 * 6
d = c / 6
e = 6 * 6
f = e / 6
g = d + f
h = g * 2
i = b - h
j = i / 2
k = a / j
|
a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | e | divide(200, const_2) | what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 200 ? | "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 200 / 2 = 100 answer is e" | a = 200 / 2
|
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | divide(1500, 5) | 57 + 58 = 115 . how many such 2 consecutive numbers are there less than 1500 when added gives a sum which is divisible by 5 ? | "since 2 + 3 = 5 & 7 + 8 = 15 any combination with these no be will give u desirable result . . . so total no in 100 will be 20 & that ' s why in 1000 , it will be 20 x 15 = 300 . answer : c" | a = 1500 / 5
|
a ) 6800 , b ) 5800 , c ) 4800 , d ) 6500 , e ) none of these | a | add(divide(multiply(subtract(10000, divide(multiply(20, 10000), const_100)), add(const_2, const_3)), add(add(const_2, const_3), add(const_2, const_4))), divide(multiply(20, 10000), const_100)) | a is a working partner and b is a sleeping partner in a business . a puts in 60,000 and b 40,000 . a gets 20 % of the profit for managing the business , and the rest is divided in proportion to their capitals . find the share of a in profit of 10000 . | "the amount a gets for managing = 20 % of rs . 10000 = 2000 remaining profit = 10000 β 2000 = 8000 this is to be divided in the ratio 3 : 2 . share of a = 3 / 5 of 8000 = 4800 β total share of a = 4800 + 2000 = 6800 . answer a" | a = 20 * 10000
b = a / 100
c = 10000 - b
d = 2 + 3
e = c * d
f = 2 + 3
g = 2 + 4
h = f + g
i = e / h
j = 20 * 10000
k = j / 100
l = i + k
|
a ) $ 4152 , b ) $ 1951 , c ) $ 2258 , d ) $ 8978 , e ) $ 8875 | b | subtract(multiply(15,624, power(add(const_1, divide(16, const_100)), 9)), 15,624) | find the compound interest on $ 15,624 for 9 months at 16 % per annum compounded quarterly . | "p = $ 15625 , n = 9 months = 3 quarters , r = 16 % p . a = 4 % per quarter . amount = $ [ 15625 * ( 1 + ( 4 / 100 ) ^ 3 ) ] = $ ( 15625 * ( 26 / 25 ) * ( 26 / 25 ) * ( 26 / 25 ) ) = $ 17576 . c . i = $ ( 17576 - 15625 ) = $ 1951 . answer ( b )" | a = 16 / 100
b = 1 + a
c = b ** 9
d = 15 * 624
e = d - 15
|
a ) 5 / 28 , b ) 3 / 28 , c ) 1 / 28 , d ) 3 / 28 , e ) 3 / 27 | a | divide(const_2, add(const_3, const_4)) | find the probability that a year chosen at random has 53 mondays . | "explanation : there are 2 kinds of year leap year - probability = 1 / 4 ( out of 4 years one is a leap year ) non leap year - probability = 3 / 4 so it ' s either this or that now in non leap year , there are 365 days meaning 52 weeks with one day extra that one day can be monday out of 7 possible days . hence the probability = 1 / 7 similarly in leap year , there are 366 days and so 366 / 7 = 52 weeks with 2 days extra now probability of occurrence of monday out of possible 7 days = 2 / 7 so what we essentially want is ( a non leap year and monday ) or ( a leap year and monday ) so it ' ll be [ 3 / 4 * 1 / 7 ] + [ 1 / 4 * 2 / 7 ] = 5 / 28 answer : a" | a = 3 + 4
b = 2 / a
|
a ) $ 200 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600 | a | divide(multiply(subtract(const_100, 10), divide(360, const_2)), const_100) | a pair of articles was bought for $ 360 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 360 / 2 = $ 180 let m . p = $ x 90 % of x = 360 x = 180 * 100 / 90 = $ 200 answer is a" | a = 100 - 10
b = 360 / 2
c = a * b
d = c / 100
|
a ) 19828.88 , b ) 19828.8 , c ) 19828.87 , d ) 19828.81 , e ) 19828.82 | b | add(divide(2828.80, subtract(power(add(const_1, divide(8, const_100)), const_2), const_1)), 2828.80) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 8 % p . a . was rs . 2828.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned | "let the sum be rs . p p { [ 1 + 8 / 100 ] 2 - 1 } = 2828.80 p ( 8 / 100 ) ( 2 + 8 / 100 ) = 2828.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 2828.80 / ( 0.08 ) ( 2.08 ) = 1360 / 0.08 = 17000 principal + interest = rs . 19828.80 answer : b" | a = 8 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 2828 / 80
f = e + 2828
|
a ) 8 , b ) 11 , c ) 14 , d ) 12 , e ) 20 | d | power(3, 2) | if x ^ 2 + y ^ 2 = 18 and xy = 3 , then ( x β y ) ^ 2 = | "but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 18 - 2 * 3 = 12 . . answer d" | a = 3 ** 2
|
a ) 40 m 2 , b ) 44 m 2 , c ) 48 m 2 , d ) 36 m 2 , e ) none of these | b | multiply(5, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 5 revolutions ? ( use Ο = 22 β 7 ) | "required area covered in 5 revolutions = 5 Γ 2 Ο rh = 5 Γ 2 Γ 22 β 7 Γ 0.7 Γ 2 = 44 m 2 answer b" | a = 22 / 7
b = 2 * a
c = 1 / 4
d = b * c
e = d * 2
f = 5 * e
|
a ) 127 , b ) 128 , c ) 130 , d ) 131 , e ) 132 | e | add(multiply(16, 8), 4) | what is the dividend . divisor 16 , the quotient is 8 and the remainder is 4 | "e = d * q + r e = 16 * 8 + 4 e = 128 + 4 e = 132" | a = 16 * 8
b = a + 4
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 12 | e | subtract(1, multiply(divide(1, 3), divide(1, 2))) | harold and millicent are getting married and need to combine their already - full libraries . if harold , who has 1 / 2 as many books as millicent , brings 1 / 3 of his books to their new home , then millicent will have enough room to bring 1 / 4 of her books to their new home . what fraction of millicent ' s old library capacity is the new home ' s library capacity ? | "because we see h willbring 1 / 3 of his booksto the new home - - > try to pick a number that isdivisible by 3 . before : assume h = 30 books h = 1 / 2 m - - > m = 60 books after : h ' = 1 / 3 h = 10 books m ' = 1 / 4 m = 15 books total = 25 books m ' = 25 = 5 / 12 * 60 ratio : 5 / 12 ans : e" | a = 1 / 3
b = 1 / 2
c = a * b
d = 1 - c
|
a ) 1 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | e | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 36)), divide(const_1, 18))) | if a , b and c together can finish a piece of work in 4 days . a alone in 36 days and b in 18 days , then c alone can do it in ? | "c = 1 / 4 - 1 / 36 β 1 / 18 = 1 / 6 = > 6 days ' answer : e" | a = 1 / 36
b = const_0_25 - a
c = 1 / 18
d = b - c
e = 1 / d
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) none of the above | a | divide(subtract(multiply(180, 1), multiply(60, 1)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 1 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 1 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 1 ) Γ 60 + x Γ 120 + ( x β 1 ) Γ 180 / ( x + 1 ) + x + ( x β 1 ) = 100 β 360 x β 120 / 3 x = 100 β 60 x = 120 β x = 2 answer a" | a = 180 * 1
b = 60 * 1
c = a - b
d = 60 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
|
a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 2400 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 2400 x = 40 the numbers are 3 * 40 , 4 * 40 , 5 * 40 hence required h . c . f . = 40 answer is c" | a = 3 * 5
b = a * 100
c = 4 * 5
d = b + c
|
['a ) 4457', 'b ) 4567', 'c ) 4235', 'd ) 4547', 'e ) 4675'] | a | multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), const_3) | the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 3 per metre approximately | explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . Ο r 2 = 175600 β ( r ) 2 = ( 175600 x ( 7 / 22 ) ) β r = 236.37 m . circumference = 2 Ο r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . answer : option a | a = 17 / 56
b = math.sqrt(a)
c = b * 100
d = circumface * (
|
a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 9 | b | divide(add(35, 25), 15) | a certain no . whendivided by 35 leaves a remainder 25 , what is the remainder if the same no . be dividedby 15 ? | explanation : 35 + 25 = 60 / 15 = 4 ( remainder ) b | a = 35 + 25
b = a / 15
|
a ) 0 , b ) 3 , c ) 5 , d ) 8 , e ) 2 | e | factorial(divide(36, 17)) | . on dividing a number by 357 , we get 36 as remainder . on dividing the same number 17 , what will be the remainder ? | let x be the number and y be the quotient . then , x = 357 x y + 36 = ( 17 x 21 x y ) + ( 17 x 2 ) + 2 = 17 x ( 21 y + 2 ) + 2 ) required remainder = 2 . answer : option e | a = 36 / 17
b = math.factorial(a)
|
a ) 700 , b ) 800 , c ) 900 , d ) 1000 , e ) 1100 | a | divide(add(multiply(multiply(const_1, const_2), 500), 400), const_2) | there is a point p on the circle . a and b started running in two constant different speeds . a in clockwise and b in anti - clockwise . first time 500 m in clockwise from p then 400 anti - clockwise . if b is yet to complete one round , what is the circumference of the circle ? | let the speeds of a & b be u and v resp . and let circumference be c . let their first meeting be after time t then ut + vt = c and ut = 500 β΄ vt = c β 500 and u / v = 500 / ( c β 500 ) for their second meeting after time t , ut + vt = 2 c and vt = 400 β΄ ut = 2 c β 400 and u / v = ( 2 c β 400 ) / 400 hence 500 / ( c β 500 ) = ( 2 c β 400 ) / 400 simplifying , 100000 = ( c β 200 ) ( c β 500 ) = c Β² β 700 c + 100000 β c = 700 answer : a | a = 1 * 2
b = a * 500
c = b + 400
d = c / 2
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | divide(subtract(55, multiply(5, 2)), add(10, 5)) | carina has 55 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 55 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 55 - > y = 45 / 15 . = 3 e | a = 5 * 2
b = 55 - a
c = 10 + 5
d = b / c
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(add(2, multiply(16, 5)), multiply(16, 5)) | how many two - digit numbers yield a remainder of 2 when divided by both 5 and 16 ? | "easier to start with numbers that are of the form 16 p + 2 - - - > 18,34 , 50,66 , 82,96 . out of these , there is only one number ( 82 ) is also of the form 5 q + 2 . thus 1 is the answer . b is the correct answer ." | a = 16 * 5
b = 2 + a
c = 16 * 5
d = b - c
|
a ) 135 , b ) 129 , c ) 131 , d ) 138 , e ) 141 | d | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(8, const_100)), multiply(add(50, 8), 200)), 200) | a computer manufacturer produces a certain electronic component at a cost of $ 50 per component . shipping costs for delivering the components are $ 8 per unit . further , the manufacturer has costs of $ 16,000 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 200 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 16000 is a fixed cost each component is $ 58 ( $ 50 to produce , $ 8 to ship ) manufacturer will be producing and selling 200 components so therefore the equation to find price would be 200 * p = 16000 + ( 200 * 50 ) + ( 200 * 8 ) p = ( 16000 + 10000 + 1600 ) / 200 p = 138 answer : d" | a = 4 * 4
b = a * 1000
c = 8 * 100
d = b + c
e = 50 + 8
f = e * 200
g = d + f
h = g / 200
|
a ) 1218 , b ) 1212 , c ) 1210 , d ) 3 . 11.14 , e ) 291.1 | d | multiply(subtract(power(18, const_2), power(15, const_2)), divide(add(multiply(15, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 15 m to 18 m , how much additional grassy ground shall it graze ? | "Ο ( 182 β 152 ) = 311.14 answer : d" | a = 18 ** 2
b = 15 ** 2
c = a - b
d = 15 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
['a ) 2.21 %', 'b ) 2.07 %', 'c ) 2.08 %', 'd ) 2.01 %', 'e ) 2.11 %'] | d | divide(add(multiply(const_2, const_100), 1), const_100) | the radius of a circle is increased by 1 % . find how much % does its area increases ? | r = 100 r = 101 r 2 = 10000 r 2 = 10201 10000 - - - - 201 100 - - - - ? = > 2.01 % answer : d | a = 2 * 100
b = a + 1
c = b / 100
|
a ) 10 , b ) 20 , c ) 84 , d ) 94 , e ) 178 | a | subtract(94, 84) | in gabriel ' s sheep herd , every sheep has either fleas or lice or both . half of the sheep have lice , and 84 sheep have both pests . if 94 sheep have lice , how many sheep have only fleas ? | n ( lice ) = n ( only lice ) + n ( both lice and fleas ) n ( both lice and fleas ) = 84 n ( lice ) = 94 n ( only lice ) = 94 - 84 = 10 since , it is also said that exactly half the sheep have lice , it can be inferred that half the number of sheep is 94 . therefore , the total number of sheep having only fleas is 10 ans : ( option a ) | a = 94 - 84
|
a ) 56 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(7, 8)), const_100) | tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 7 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 7 = 112 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 112 pi / 200 pi = 0.56 a" | a = 8 / 10
b = a ** 2
c = 7 / 8
d = b * c
e = d * 100
|
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