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 ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) 65 sec | b | multiply(divide(500, divide(multiply(const_1000, add(45, 30)), const_3600)), const_2) | two goods train each 500 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 . | "explanation : relative speed = = ( 45 + 30 ) km / hr = 75 x 5 / 18 m / sec = 125 / 6 m / sec we have to find the time taken by the slower train to pass the driver of the faster train and not the complete train . so , distance covered = length of the slower train . therefore , distance covered = 500 m . required time =... | a = 45 + 30
b = 1000 * a
c = b / 3600
d = 500 / c
e = d * 2
|
a ) 11 , b ) 13 , c ) 14 , d ) 12 , e ) 10 | e | divide(subtract(multiply(add(const_4, const_1), 5), 5), subtract(add(const_4, const_1), const_3)) | the age of somu is one - third his father ' s . 5 years back he was one - fifth of his father ' s age . what is his persent age ? | "explanation : let somu ' s age be x and that of his father be 3 x . so , x - 5 = 3 x - 5 / 5 = x = 10 answer : option e" | a = 4 + 1
b = a * 5
c = b - 5
d = 4 + 1
e = d - 3
f = c / e
|
a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number w of tickets sold by the male members of the association | "given that , total average w sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ." | a = 1 + 2
b = 66 * a
c = 70 * 2
d = b - c
|
a ) 68 days , b ) 70 days , c ) 73 days , d ) 75 days , e ) 77 days | d | add(divide(subtract(const_1, multiply(inverse(15), 3)), inverse(90)), 3) | amit and ananthu can do a work in 15 days and 90 days respectively . amit started the work and left after 3 days . ananthu took over and completed the work . in how many days was the total work completed ? | "amit β s one day β s work = 1 / 15 amit β s 3 day β s work = 1 / 15 * 3 = 1 / 5 work left = 1 - 1 / 5 = 4 / 5 ananthu β s one day β s work = 1 / 90 ananthu can do work in = 4 / 5 * 90 = 72 days so total days = 72 + 3 = 75 days answer : d" | a = 1/(15)
b = a * 3
c = 1 - b
d = 1/(90)
e = c / d
f = e + 3
|
a ) 1 : 8 , b ) 1 : 4 , c ) 2 : 1 , d ) 4 : 1 , e ) 12 : 1 | e | multiply(6, const_2) | city x has a population 6 times as great as the population of city y , which has a population twice as great as the population of city z . what is the ratio of the population of city x to the population of city z ? | "x = 6 y , y = 2 * z x : y , y : z 6 : 1 , 2 : 1 12 : 2 , 2 : 1 so , x : z = 12 : 1 ( e )" | a = 6 * 2
|
a ) 10 / 12 , b ) 16 / 28 , c ) 24 / 11 , d ) 42 / 16 , e ) 18 / 12 | c | divide(const_1, divide(add(add(inverse(2), inverse(4)), inverse(6)), const_2)) | a and b can do a work in 2 days , b and c in 4 days and c and a in 6 days . in how many days will the work be completed , if all three of them work together ? | one day work of a and b = 1 / 2 one day work of b and c = 1 / 4 one day work of c and a = 1 / 6 2 ( a + b + c ) = 1 / 2 + 1 / 4 + 1 / 6 . 2 ( a + b + c ) = 11 / 12 ( a + b + c ) = 11 / 24 number of days required = 24 / 11 days . answer : c | a = 1/(2)
b = 1/(4)
c = a + b
d = 1/(6)
e = c + d
f = e / 2
g = 1 / f
|
a ) 692 , b ) 700 , c ) 677 , d ) 654 , e ) 665 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 7), add(const_2, const_4)) | what is the sum of all the multiples of 7 between 30 and 100 ? | "you first have to know all the multiples of 7 between 30 and 100 . they are 35 , 42 , 49 , 56 , 63 , 70 , 77 , 84 , 91 , and 98 . if you add all these numbers together , you get 665 . final answer : e" | a = 12 + 2
b = a + 1
c = 12 + 2
d = 2 + 4
e = d + 4
f = 10 - 1
g = e + f
h = 2 + 4
i = h + 4
j = g + i
k = 10 + 2
l = j + k
m = c + l
n = b + m
o = 12 + 2
p = o + 1
q = p + 1
r = n + q
s = r + 7
t = 2 + 4
u = s + t
|
a ) 23 , b ) 25 , c ) 11 , d ) 5 , e ) 2 | b | add(divide(add(20, 24), const_2), multiply(1, const_2)) | if { x } is the product of all even integers from 1 to x inclusive , what is the greatest prime factor of { 20 } + { 24 } ? | "soln : { 24 } + { 20 } = 24 * { 20 } + { 20 } = 25 * { 20 } answer : b" | a = 20 + 24
b = a / 2
c = 1 * 2
d = b + c
|
a ) 87 , b ) 91 , c ) 100 , d ) 96 , e ) 101 | b | gcd(1001, 910) | the maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 1001 number of pencils = 910 required number of students = h . c . f . of 1001 and 910 = 91 answer is b" | a = math.gcd(1001, 910)
|
a ) 33 , b ) 36 , c ) 39 , d ) 42 , e ) 45 | c | add(divide(subtract(subtract(250, const_2), add(20, const_2)), add(const_2, 4)), const_1) | how many even integers n , such that 20 < = n < = 250 are of the form 3 k + 4 , where k is any natural number ? | "the first number is 22 = 16 + 6 ( 1 ) . we can continue adding 6 to make a list : 22 , 28 , 34 , . . . the last number is 250 = 16 + 6 ( 39 ) there are 39 numbers in the list . the answer is c ." | a = 250 - 2
b = 20 + 2
c = a - b
d = 2 + 4
e = c / d
f = e + 1
|
a ) $ 69 , b ) $ 78 , c ) $ 90 , d ) $ 86 , e ) $ 77 | c | subtract(divide(8925, 85), 15) | a trader sells 85 meters of cloth for $ 8925 at the profit of $ 15 per metre of cloth . what is the cost price of one metre of cloth ? | c $ 90 sp of 1 m of cloth = 8925 / 85 = $ 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = $ 105 - $ 15 = $ 90 . | a = 8925 / 85
b = a - 15
|
a ) 15 , b ) 10 , c ) 9 , d ) 6 , e ) 4 | b | subtract(multiply(12, const_2), multiply(7, const_2)) | a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 12 : 8 : 7 , respectively , by weight . if 54 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ? | "we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 12 x : 8 x : 7 x we are given that 54 pounds of the mixture is prepared... | a = 12 * 2
b = 7 * 2
c = a - b
|
a ) 2443200 , b ) 3643200 , c ) 5443200 , d ) 1443200 , e ) 2163200 | c | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | of the six - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "of the six - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for p - 9 ways ; choosing the digit for q - 8 ways ; choosing the digit for r - 7... | a = 10 - 1
b = 10 - 2
c = a * b
d = c * 1
|
a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 400 | a | divide(multiply(330, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 330 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 330 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 330 Γ’ β‘ β y [ 12 / 10 + 1 ] = 330 Γ’ β‘ β 22 y / 10 = 330 Γ’ β‘ β 22 y = 3300 Γ’ β‘ β y = 3300 / 22 = 300 / 2 = rs . 150 a )" | a = 1 + 4
b = a * 2
c = 330 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 540 , b ) 520 , c ) 619 , d ) 649 , e ) 700 | b | multiply(divide(338, divide(add(const_100, 30), const_100)), 2) | if the price of a certain computer increased 30 percent from d dollars to 338 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 338 ( given ) i . e . d = 338 / 1.3 = $ 260 i . e . 2 d = 2 * 260 = 520 answer : option b" | a = 100 + 30
b = a / 100
c = 338 / b
d = c * 2
|
a ) rs . 4000 , b ) rs . 665.6 , c ) rs . 520 , d ) rs . 166.4 , e ) none | c | divide(multiply(divide(divide(add(divide(multiply(4000, 8), const_100), divide(multiply(add(4000, divide(multiply(4000, 8), const_100)), 8), const_100)), 2), 2), const_100), 16) | simple interest on a certain sum of money for 2 years at 16 % per annum is one fourth the compound interest on rs . 4000 for 2 years at 8 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 4000 x ( 1 + 8 / 100 ) Γ’ Β² - 4000 ] rs . ( 4000 x 108 / 100 x 108 / 100 - 4000 ) = rs . 665.6 . sum = rs . [ 166.4 x 100 / 2 x 16 ] = rs . 520 . answer c" | a = 4000 * 8
b = a / 100
c = 4000 * 8
d = c / 100
e = 4000 + d
f = e * 8
g = f / 100
h = b + g
i = h / 2
j = i / 2
k = j * 100
l = k / 16
|
a ) 48 , b ) 66 , c ) 76 , d ) 78 , e ) none of these | e | subtract(multiply(multiply(multiply(5, 5), 1), const_2), multiply(multiply(multiply(5, 1), const_2), const_2)) | each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 1 and the 5 can not be adjacent digits in the integer , how many 5 - digit integers are possible ? | "number of arrangements using 5 distinct digits = 5 ! number of arrangements in which 1 and 5 are adjacent - consider 1 and 5 together as one group . now you have 4 numbers / groups to arrange which can be done in 4 ! ways . in each of these arrangements , 1 and 5 can be arranged as 15 or 51 . number of arrangements in... | a = 5 * 5
b = a * 1
c = b * 2
d = 5 * 1
e = d * 2
f = e * 2
g = c - f
|
a ) 30 , b ) 80 , c ) 20 , d ) 68 , e ) 26 | d | multiply(subtract(79, 45), const_2) | a pupil ' s marks were wrongly entered as 79 instead of 45 . due to that the average marks for the class got increased by half . the number of pupils in the class is : | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 . x / 2 = ( 79 - 45 ) = > x / 2 = 34 = > x = 68 . answer : d" | a = 79 - 45
b = a * 2
|
a ) 40 , b ) 44 , c ) 48 , d ) 50 , e ) 8 | e | add(add(multiply(5, divide(8, subtract(multiply(divide(4, 5), 5), 2))), divide(8, subtract(multiply(divide(4, 5), 5), 2))), 8) | in a can , there is a mixture of milk and water in the ratio 2 : 5 . if it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 4 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 2 / 7 ( t - 8 ) after adding milk , quantity of milk in the mixture = 4 / 9 t . 4 t / 9 - 8 = 2 / 7 ( t - 8 ) 2 t = 72 - 56 = > t = 8 . answer : e" | a = 4 / 5
b = a * 5
c = b - 2
d = 8 / c
e = 5 * d
f = 4 / 5
g = f * 5
h = g - 2
i = 8 / h
j = e + i
k = j + 8
|
a ) 26799 , b ) 24000 , c ) 50000 , d ) 29973 , e ) 12312 | c | multiply(divide(const_100, 90), 45000) | 90 % of the population of a village is 45000 . the total population of the village is ? | "x * ( 90 / 100 ) = 45000 x = 500 * 100 x = 50000 answer : c" | a = 100 / 90
b = a * 45000
|
a ) 60 , b ) 51 , c ) 53 , d ) 56 , e ) 50 | b | add(52, const_1) | total 52 matches are conducted in knockout match type . how many players will be participated in that tournament ? | "51 players answer : b" | a = 52 + 1
|
a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 8 | e | multiply(divide(divide(add(add(multiply(multiply(const_4, const_2), const_10), multiply(multiply(const_4, const_2), const_100)), multiply(const_12, const_1000)), 230), 7), 1) | the ratio between the number of sheep and the number of horses at the stewart farm is 1 to 7 , if each horse is fed 230 ounces of horse food per day and the farm needs a total 12,880 ounces of horse food per day , what is the number of sheep in the farm ? | "let the number of sheeps and horses be 1 x and 7 x . now total number of horses = total consumption of horse food / consumption per horse = 12880 / 230 = 56 , which is equal to 7 x . = > x = 8 sheeps = 1 x = 1 * 8 = 8 . hence e" | a = 4 * 2
b = a * 10
c = 4 * 2
d = c * 100
e = b + d
f = 12 * 1000
g = e + f
h = g / 230
i = h / 7
j = i * 1
|
a ) 58 , b ) 62 , c ) 66 , d ) 70 , e ) 74 | c | divide(factorial(12), multiply(factorial(subtract(12, const_2)), factorial(const_2))) | there are 12 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | 12 c 2 = 66 the answer is c . | a = math.factorial(12)
b = 12 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 1288 , b ) 1299 , c ) 1000 , d ) 10000 , e ) 11600 | e | multiply(add(45, const_100), subtract(const_100, 20)) | the length of a rectangle is increased by 45 % and its breadth is decreased by 20 % . what is the effect on its area ? | "100 * 100 = 10000 145 * 80 = 11600 answer : e" | a = 45 + 100
b = 100 - 20
c = a * b
|
a ) $ 830 , b ) $ 800 , c ) $ 770 , d ) $ 730 , e ) $ 700 | b | subtract(1270, add(multiply(50, subtract(15, floor(divide(15, const_2)))), multiply(floor(divide(15, const_2)), 10))) | michael cashed a check for $ 1270 and received only $ 10 and $ 50 bills in return . during the course of a day , he used 15 bills and then lost the rest of the money . if the number of $ 10 bills used was either one more or one less than the number of $ 50 bills used , what is the minimum possible amount of money that ... | 1 . 15 bills spent 2 . number of $ 10 bills is one more or one less than number of $ 50 bills . so , 2 choices 1 . 7 * 10 + 8 * 50 = $ 470 2 . 8 * 10 + 7 * 50 = $ 430 least money lost implies most money spent . i . e $ 470 therefore , lost money = 1270 - 470 = 800 . answer b | a = 15 / 2
b = math.floor(a)
c = 15 - b
d = 50 * c
e = 15 / 2
f = math.floor(e)
g = f * 10
h = d + g
i = 1270 - h
|
a ) 8 , b ) 5 , c ) 9 , d ) 6 , e ) 10 | e | divide(subtract(180, multiply(3, 40)), multiply(3, const_2)) | a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 180 , how many hours overtime did he work ? | "at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 180 - $ 120 = $ 60 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 60 / $ 6 = 10 ans is e" | a = 3 * 40
b = 180 - a
c = 3 * 2
d = b / c
|
a ) 12 , b ) 20 , c ) 48 , d ) 55 , e ) 60 | b | multiply(subtract(multiply(3, 4), 10), 10) | if a farmer sells 5 of his goats , his stock of feed will last for 4 more days than planned , but if he buys 10 more goats , he will run out of feed 3 days earlier than planned . if no goats are sold or bought , the farmer will be exactly on schedule . how many goats does the farmer have ? | say farmer has n goat and he is good for d days . : - we have 3 equations given in question : - ( n - 5 ) * d + 4 = ( n + 10 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 10 d - 3 n = 30 4 n - 5 d = 20 = > n = 20 ans b it is ! | a = 3 * 4
b = a - 10
c = b * 10
|
a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 90 | e | multiply(divide(multiply(90, 2), add(30, 90)), const_60) | cole drove from home to work at an average speed of 30 kmh . he then returned home at an average speed of 90 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 30 time from work to home = x / 90 total time = 2 hrs ( x / 30 ) + ( x / 90 ) = 2 solving for x , we get x = 45 time from home to work in minutes = ( 45 ) * 60 / 30 = 90 minutes ans = e" | a = 90 * 2
b = 30 + 90
c = a / b
d = c * const_60
|
a ) 44 % , b ) 100 % , c ) 15 % , d ) 125 % , e ) 400 % | d | multiply(divide(subtract(divide(2, 3), divide(const_2.0, 2)), divide(3, 2)), const_100) | when tom works alone he chops 3 lb . salad in 2 minutes , and when tammy works alone she chops 2 lb . salad in 3 minutes . they start working together , and after some time finish chopping 65 lb . of salad . of those 80 lb . , the salad quantity chopped by tammy is what percent lesser than the quantifying chopped by to... | "tom chops 4 lbs in 6 minutes tammy chops 9 lbs in 6 minutes so in the same amount of time , tammy chops 125 % more than tom , since 9 is 125 % greater than 4 . so 125 % is the answer . note that the actual time does n ' t matter . if you multiply the time each work by x , you ' ll multiply the work each does by x , an... | a = 2 / 3
b = 2 / 0
c = a - b
d = 3 / 2
e = c / d
f = e * 100
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | what is the smallest integer y for which 27 ^ y > 3 ^ 24 ? | "27 ^ y > 3 ^ 24 converting into the same bases : 27 ^ y > 27 ^ 8 therefore for the equation to hold true , y > 8 or y = 9 option c" | a = 24 / 3
b = math.floor(a)
c = b + 1
|
a ) 0.35 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.2 days | e | inverse(subtract(inverse(4), inverse(20))) | a and b together can do a piece of work in 4 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "b = 1 / 4 β 1 / 20 = 0.2 days answer : e" | a = 1/(4)
b = 1/(20)
c = a - b
d = 1/(c)
|
a ) 0 , b ) 1 , c ) 3 , d ) 6 , e ) 7 | b | subtract(1, 1) | a bell curve ( normal distribution ) has a mean of β 1 and a standard deviation of 1 / 8 . how many integer values z are within 3 standard deviations of the mean ? | got the question correctly - - the second item - - [ highlight ] bthe list of elements in the set is required . [ / highlight ] is not required . with the new information , there is only one integer value ( - 1 ) that is between ( - 1.375 , - 0.625 ) i . e . , falls within the three 3 sd range . b | a = 1 - 1
|
a ) 9.4 , b ) 20 , c ) 50 , d ) 30 , e ) 40 | a | subtract(subtract(15, 2.8), 2.8) | a man ' s speed with the current is 15 km / hr and the speed of the current is 2.8 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.8 km / hr hence , speed of the man = 15 - 2.8 = 12.2 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.2 - 2.8 = 9.4 km / hr answer is a ." | a = 15 - 2
b = a - 2
|
a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6 | b | divide(2, 15) | 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 2 / 15 = | "we need to determine the result of 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 let β s add the given fractions in two groups . in the group of the first three fractions , notice that 1 / 3 and 1 / 2 share a common denominator of 6 with 5 / 6 . 1 / 2 + 1 / 3 = 3 / 6 + 2 / 6 = 5 / 6 thus , 5 / 6 β 5 / 6 = 0 looking a... | a = 2 / 15
|
a ) 4500 , b ) 2678 , c ) 2689 , d ) 2761 , e ) 2882 | a | divide(multiply(multiply(2, 45), multiply(3, const_1000)), multiply(const_1, const_60)) | a river 2 m deep and 45 m wide is flowing at the rate of 3 kmph the amount of water that runs into the sea per minute is ? | "( 3000 * 2 * 5 ) / 60 = 4500 m 3 answer : a" | a = 2 * 45
b = 3 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(multiply(180, 2), multiply(60, 2)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 2 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) Γ 60 + x Γ 120 + ( x β 2 ) Γ 180 / ( x + 2 ) + x + ( x β 2 ) = 100 β 360 x β 240 / 3 x = 100 β 60 x = 240 β x = 4 answer e" | a = 180 * 2
b = 60 * 2
c = a - b
d = 60 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
|
a ) 4 , b ) 12 , c ) 15 , d ) 20 , e ) 24 | c | divide(subtract(multiply(70, 30), 1800), subtract(70, 50)) | tourist purchased a total of 30 travelers checks in $ 50 and $ 100 denominations . the total worth of the travelers checks is $ 1800 . how many checks of $ 50 denominations can he spend so that average amount ( arithmetic mean ) of the remaining travelers checks is $ 70 ? | "you could set - up a quick table and brute force the answer . a 4 * 50 200 1800 - 200 1600 26 61.54 b 12 * 50 600 1800 - 600 1200 18 66.67 c 15 * 50 750 1800 - 750 1050 15 70.00 d 20 * 50 1000 1800 - 1000 800 10 80.00 e 24 * 50 1200 1800 - 1200 600 6 100.00 answer is c" | a = 70 * 30
b = a - 1800
c = 70 - 50
d = b / c
|
a ) 0.44 , b ) 0.5 , c ) 0.56 , d ) 0.62 , e ) 0.68 | c | subtract(const_1, add(multiply(divide(40, const_100), divide(50, const_100)), multiply(divide(40, const_100), subtract(const_1, divide(40, const_100))))) | in a graduating class , 40 percent of the students are male . in this class , 50 percent of the male students and 40 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "let x be the total number of students . the number students who are younger than 25 is 0.5 * 0.4 x + 0.6 * 0.6 x = 0.56 x the answer is c ." | a = 40 / 100
b = 50 / 100
c = a * b
d = 40 / 100
e = 40 / 100
f = 1 - e
g = d * f
h = c + g
i = 1 - h
|
a ) 9 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | add(divide(20, const_4), const_2) | up to 20 , how many no . are odd and prime both ? | we have to calculate it manually . odd 1,3 , 5,7 , 9,11 , 13,15 , 17,19 prime 2,3 , 5,7 , 11,13 , 17,19 common 3 , 5,7 , 11,13 , 17,19 total no . 7 answer d | a = 20 / 4
b = a + 2
|
a ) 52.6 , b ) 52.9 , c ) 52.8 , d ) 52.1 , e ) 82.5 | e | multiply(multiply(power(15, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 15 metro and whose angle at the center is 42 Γ’ Β° is ? | "42 / 360 * 22 / 7 * 15 * 15 = 82.5 m 2 answer : e" | a = 15 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 42 / g
i = f * h
|
a ) 6 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | a | sqrt(add(power(multiply(8, divide(1, 42)), 42), power(multiply(subtract(8, 42), divide(1, 42)), 42))) | 9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 42 miles per hour faster than the first runner ' s rate . how far apart , to the neare... | "if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 8 m / h - - > 4 m in 1 / 2 hr side 2 = 10 m / h - - > 5 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 5 ^ 2 d ^ 2 = 41 = ~ 6 answer option a" | a = 1 / 42
b = 8 * a
c = b ** 42
d = 8 - 42
e = 1 / 42
f = d * e
g = f ** 42
h = c + g
i = math.sqrt(h)
|
a ) 40 , b ) 50 , c ) 55 , d ) 60 , e ) 19 | e | add(subtract(divide(multiply(multiply(2, 4), 5), 2), divide(multiply(multiply(2, 4), 5), 4)), divide(multiply(multiply(2, 4), 5), 5)) | if x , y , and z are positive integers and 2 x = 4 y = 5 z , then the least possible value of x + y + z is | "given 2 x = 4 y = 5 z x + y + z in terms of x = x + ( 2 x / 4 ) + ( 2 x / 5 ) = 38 x / 20 = 19 x / 10 now checking with each of the answers and see which value gives a minimum integer value . a x = 10 / 19 * 40 , not an integer b , c , e can be ruled out similarly . d is minimum value as x = 19 * 10 / 19 = 10 answer i... | a = 2 * 4
b = a * 5
c = b / 2
d = 2 * 4
e = d * 5
f = e / 4
g = c - f
h = 2 * 4
i = h * 5
j = i / 5
k = g + j
|
a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 3 | b | divide(68, add(13, 4)) | a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 68 km downstream . | "speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 68 km downstream = 68 / 17 = 4 hrs ans - b" | a = 13 + 4
b = 68 / a
|
a ) 1 / 45 , b ) 1 / 2 , c ) 1 / 8 , d ) 2 / 3 , e ) 2 / 11 | b | multiply(divide(add(const_2, const_1), add(multiply(5, const_2), const_2)), const_2) | the events a and b are independent , the probability that event a occurs is greater than 0 , and the probability that event a occurs is twice the probability that event b occurs . the probability that at least one of events a and b occurs is 5 times the probability that both events a and b occur . what is the probabili... | let us say probability of a occuring is a . let us say probability of b occuring is b . a = 2 b probability ( either a or b or both ) = 5 times probability ( a and b ) a * ( 1 - b ) + b * ( 1 - a ) + ab = 5 * ab substituting a = 2 b in the second equation : 2 b * ( 1 - b ) + b * ( 1 - 2 b ) + 2 b * b = 5 * 2 b * b 3 b ... | a = 2 + 1
b = 5 * 2
c = b + 2
d = a / c
e = d * 2
|
a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 23 / 32 , e ) 3 / 4 | d | divide(add(const_1, divide(subtract(multiply(divide(7.5, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 7.5 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the med... | "the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ... | a = 7 / 5
b = a * 2
c = b - 1
d = 3 / 4
e = d * 2
f = e - 1
g = 1 / f
h = c / g
i = 1 + h
j = i / 2
|
a ) 700 , b ) 300 , c ) 1000 , d ) 360 , e ) 2400 | a | divide(subtract(multiply(divide(6, const_100), 1400), multiply(1400, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | barbata invests $ 1400 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | let the additional invested amount for 8 % interest be x ; equation will be ; 1400 + 0.05 * 2400 + x + 0.08 x = 1400 + x + 0.06 ( 2400 + x ) 0.05 * 1400 + 0.08 x = 0.06 x + 0.06 * 1400 0.02 x = 1400 ( 0.06 - 0.05 ) x = 1400 * 0.01 / 0.02 = 700 ans : a | a = 6 / 100
b = a * 1400
c = 5 / 100
d = 1400 * c
e = b - d
f = 8 / 100
g = 6 / 100
h = f - g
i = e / h
|
a ) 1 coin , b ) 2 coins , c ) 3 coins , d ) 4 coins , e ) coins | a | subtract(multiply(add(floor(divide(12004, 7)), const_1), 7), 12004) | a person has 12004 coins of silver . how much would he need to add so that he can distribute equally among his 7 children in whole numbers ? | he would need to add one coin more so 12005 is fully divisible by 7 and each would get 1715 coins . so the correct answer is a | a = 12004 / 7
b = math.floor(a)
c = b + 1
d = c * 7
e = d - 12004
|
a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 55.5 m 2 , e ) 57 m 2 | d | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 7))), multiply(4, 7)) | a cistern 7 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 7 x 1.25 ) + 7 x 4 ] m 2 = 55.5 m 2 . answer : option d" | a = 25 / 100
b = a + 1
c = b * 4
d = 25 / 100
e = d + 1
f = e * 7
g = c + f
h = 2 * g
i = 4 * 7
j = h + i
|
a ) 420 , b ) 550 , c ) 490 , d ) 570 , e ) 457 | d | subtract(multiply(const_10, 162), add(multiply(3, 100), multiply(5, 150))) | a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 162 . find the unknown rate of two blankets ? | "10 * 162 = 1620 3 * 100 + 5 * 150 = 1050 1620 β 1050 = 570 answer : d" | a = 10 * 162
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 10 , b ) 25 , c ) 60 , d ) 16 , e ) 20 | d | divide(multiply(12, 40), subtract(40, 10)) | a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 12 days , a finished the remaining work . in how many days a alone can finish the job ? | "a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 12 days whole work will be done by a in 12 * 4 / 3 = 16 days answer is d" | a = 12 * 40
b = 40 - 10
c = a / b
|
a ) 500 , b ) 277 , c ) 466.7 , d ) 297 , e ) 111 | c | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 20)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 20 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 120 = rs . 466.7 answer : c" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 20
g = e / f
|
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | e | divide(multiply(1, 48), 4) | the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 4 . the sum of numbers is : | "let the numbers be 1 x and 4 x . then , their l . c . m = 4 x . so , 4 x = 48 or x = 12 . the numbers are 12 and 48 . hence , required sum = ( 12 + 48 ) = 60 . answer : e" | a = 1 * 48
b = a / 4
|
a ) 6.06 % , b ) 6.07 % , c ) 6.08 % , d ) 6.09 % , e ) none | d | add(add(divide(6, const_2), divide(6, const_2)), divide(multiply(divide(6, const_2), divide(6, const_2)), const_100)) | the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half yearly is | "solution amount of rs . 100 for 1 year when compounded half yearly = rs . [ 100 x ( 1 + 3 / 100 ) 2 ] = rs . 106.09 % . effective rate = ( 106.09 - 100 ) % = rs . 6.09 % . answer d" | a = 6 / 2
b = 6 / 2
c = a + b
d = 6 / 2
e = 6 / 2
f = d * e
g = f / 100
h = c + g
|
a ) $ 2.5 , b ) $ 15 , c ) $ 50 , d ) $ 100 , e ) $ 105 | a | subtract(subtract(multiply(500, power(add(const_1, divide(10, const_100)), 2)), 500), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 1,000), 1,000)) | on a certain day , tim invested $ 500 at 10 percent annual interest , compounded annually , and lana invested 1,000 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim β s investment in the first 2 years was how much greater than the total amount of interest earned by lana β ... | "compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 500 ( 1.10 ) ^... | a = 10 / 100
b = 1 + a
c = b ** 2
d = 500 * c
e = d - 500
f = 5 / 100
g = 1 + f
h = g ** 2
i = h * 1
j = i - 1
k = e - j
|
a ) 50 , b ) 88 , c ) 65 , d ) 55 , e ) 22 | c | divide(multiply(36, divide(multiply(subtract(49, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 49 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 49 - 36 = 13 km / hr . = 13 * 5 / 18 = 65 / 18 m / sec . 2 x / 36 = 65 / 18 = > x = 65 . answer : c" | a = 49 - 36
b = a * 1000
c = b / 3600
d = 36 * c
e = d / 2
|
a ) 122336 , b ) 145456 , c ) 122347 , d ) 126666 , e ) 383838 | a | subtract(1222343, multiply(multiply(12, 3), 2)) | evaluate : 1222343 - 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 1222343 - 12 * 3 * 2 = 122343 - 8 = 122336 correct answer a" | a = 12 * 3
b = a * 2
c = 1222343 - b
|
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | b | subtract(divide(multiply(divide(60, 2), 5), 2), 60) | the ratio of boarders to day students at a school was originally 2 to 5 . however , after a number of new boarders join the initial 60 boarders , the ratio changed to 1 to 2 . if no boarders became day students and vice versa , and no students left the school , how many new boarders joined the school ? | "let x be the number of new boarders . the ratio changed from 2 : 5 = 4 : 10 up to 1 : 2 = 5 : 10 . 60 / ( 60 + x ) = 4 / 5 x = 15 the answer is b ." | a = 60 / 2
b = a * 5
c = b / 2
d = c - 60
|
a ) 2250 , b ) 2450 , c ) 2650 , d ) 2850 , e ) 3050 | b | divide(833, subtract(subtract(const_1, divide(33, const_100)), divide(33, const_100))) | a candidate got 33 % of the votes polled and he lost to his rival by 833 votes . how many votes were cast ? | "let x be the total number of votes . 0.33 x + 833 = 0.67 x 0.34 x = 833 x = 833 / 0.34 = 2450 the answer is b ." | a = 33 / 100
b = 1 - a
c = 33 / 100
d = b - c
e = 833 / d
|
a ) 5.88 % decrease , b ) 0.88 % decrease , c ) 0.88 % increase , d ) 1.88 % decrease , e ) 2.88 % decrease | b | multiply(divide(subtract(multiply(add(const_100, 18), subtract(const_100, 16)), multiply(const_100, const_100)), multiply(const_100, const_100)), const_100) | calculate the effect changes in dimension of a rectangle will have on its area , if length is increased by 18 % and its breadth is decreased by 16 % ? | "let l and b be 100 each 100 * 100 = 10000 l increase by 18 % = 118 b decrease by 16 % = 84 118 * 84 = 9912 0.88 % decrease answer : b" | a = 100 + 18
b = 100 - 16
c = a * b
d = 100 * 100
e = c - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) 40 minutes , b ) 1 hour , c ) 1 hour 15 min , d ) 1 hour 30 min , e ) 1 hour 10 min | d | divide(6, divide(add(multiply(divide(1, 10), const_60), divide(2, 2)), const_2)) | a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes . how long will it take to go 6 km in stationary water ? | "speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 10 / 60 ) = 6 kmhr speed in still water = 1 / 2 ( 2 + 6 ) = 4 kmhr time taken in stationary = 6 / 4 = 1 hrs 30 min answer : d" | a = 1 / 10
b = a * const_60
c = 2 / 2
d = b + c
e = d / 2
f = 6 / e
|
a ) 20 % , b ) 80 % , c ) 90 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 71), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 71 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent ? | "let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 68.4 bears per 36 hours a week or 1.9 bears / hr ( [ 40 bears * 1.71 ] / [ 40 hrs * . 90 ] ) . [ ( 1.9 - 1 ) / 1 ] * 100 % = 90 % answer : c" | a = 100 - 71
b = a - 10
c = 10 / b
d = c * 100
|
['a ) 10', 'b ) 12', 'c ) 8', 'd ) 14', 'e ) 9'] | a | sqrt(add(power(8, const_2), power(6, const_2))) | 34 . the side surface of a cylinder can is rolled with a rectangular plate . if the height of a cylinder tank is 8 feet and the perimeter of the circular base is 6 feet , what is the diagonal of the rectangular plate ? | think of a can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both , use pythagoreans theorem or properties of 3 - 4 - 5 triangles to solve for the hypo... | a = 8 ** 2
b = 6 ** 2
c = a + b
d = math.sqrt(c)
|
a ) 32 , b ) 545 , c ) 39 , d ) 40 , e ) 09 | c | divide(divide(add(120, 480), const_1000), divide(55, const_3600)) | a train 120 meters long completely crosses a 480 meters long bridge in 55 seconds . what is the speed of the train is ? | s = ( 120 + 480 ) / 45 = 600 / 55 * 18 / 5 = 39 answer : c | a = 120 + 480
b = a / 1000
c = 55 / 3600
d = b / c
|
a ) 48 , b ) 56 , c ) 64 , d ) 72 , e ) 80 | a | multiply(multiply(2, 3), 2) | employees of a certain company are each to receive a unique 6 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , and 5 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ? | "there are 4 ! ways to make codes starting with 12 . there are 4 ! ways to make codes starting with 24 . the number of codes is 2 * 4 ! = 48 . the answer is a ." | a = 2 * 3
b = a * 2
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | divide(add(20, 2), subtract(65, 63)) | in an examination , a pupil β s average marks were 63 per paper . if he had obtained 20 more marks for his geography paper and 2 more marks for his history paper , his average per paper would have been 65 . how many papers were there in the examination ? | solution let the number pf papers be x . then , 63 x + 20 + 2 = 65 x 65 x - 63 x = 22 2 x = 22 x = 11 . answer d | a = 20 + 2
b = 65 - 63
c = a / b
|
a ) 20 , b ) 15 , c ) 16 , d ) 25 , e ) 30 | c | divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100)) | if 15 % of 40 is greater than 25 % of a number by 2 , then find the number is ? | "15 / 100 * 40 - 25 / 100 * x = 2 or x / 4 = 4 so x = 16 answer c" | a = 15 / 100
b = a * 40
c = b - 2
d = 25 / 100
e = c / d
|
a ) 5 % , b ) 8 % , c ) 3 % , d ) 4.5 % , e ) 1 % | d | divide(multiply(const_100, 144), multiply(800, 4)) | what is the rate percent when the simple interest on rs . 800 amount to rs . 144 in 4 years ? | "144 = ( 800 * 4 * r ) / 100 r = 4.5 % answer : d" | a = 100 * 144
b = 800 * 4
c = a / b
|
a ) 11 , b ) 14 , c ) 13 , d ) 15 , e ) 16 | a | add(multiply(2, 2), 7) | if p / q = 2 / 7 , then 2 p + q = ? | "let p = 2 , q = 7 then 2 * 2 + 7 = 11 so 2 p + q = 11 . answer : a" | a = 2 * 2
b = a + 7
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(multiply(10, 2.00), multiply(1.60, 10)) | martin bought 10 concert tickets , some at the full price of $ 2.00 per ticket , and some at a discounted price of $ 1.60 per ticket . if he spent a total of $ 17.60 , how many discounted tickets did he buy ? | "let x be the number of tickets he bought at $ 2 per ticket . then 2 x + ( 10 - x ) 1.6 = 17.6 0.4 x = 1.6 = > x = 4 discounted tickets = 10 - x = 6 ans : d" | a = 10 * 2
b = 1 * 60
c = a - b
|
a ) 263 , b ) 293 , c ) 323 , d ) 353 , e ) 383 | e | add(add(add(add(add(add(add(50, const_1), add(add(50, const_1), const_1)), add(add(add(50, const_1), const_1), const_2)), add(add(add(add(50, const_1), const_1), const_2), const_1)), add(add(add(add(add(50, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(50, const_1), const_1), const_2), cons... | the sum of the non - prime numbers between 50 and 60 , non - inclusive , is | sum of consecutive integers from 51 to 59 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 51 + 59 ) / 2 * 9 = 55 * 9 = 495 sum of non - prime numbers b / w 50 and 60 , non inclusive = = = > 495 - 112 ( i . e . , 53 + 59 , being the prime # s in the range ) = 383 answer : e | a = 50 + 1
b = 50 + 1
c = b + 1
d = a + c
e = 50 + 1
f = e + 1
g = f + 2
h = d + g
i = 50 + 1
j = i + 1
k = j + 2
l = k + 1
m = h + l
n = 50 + 1
o = n + 1
p = o + 2
q = p + 1
r = q + 1
s = m + r
t = 50 + 1
u = t + 1
v = u + 2
w = v + 1
x = w + 1
y = x + 1
z = s + y
A = 50 + 1
B = A + 1
C = B... |
a ) 14 , b ) 47 , c ) 70 , d ) 180 , e ) 240 | c | add(add(divide(500, 40), divide(500, 40)), 40) | frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 500 square feet . how many feet offence does frank use ? | "area = length x breadth 500 = 40 x breadth so , breadth = 15 units fencing required is - breadth + breadth + length 15 + 15 + 40 = > 70 feet answer must be ( c ) 70" | a = 500 / 40
b = 500 / 40
c = a + b
d = c + 40
|
a ) 30 , b ) 31 , c ) 50 , d ) 90 , e ) 89 | c | multiply(2, 15) | the class mean score on a test was 80 , and the standard deviation was 15 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ? | "1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 95 to 65 , where 95 is within sd above the mean and 65 within 1 sd below the mean 2 sd = 15 twice = 30 from the the mean , which is 9... | a = 2 * 15
|
a ) a ) 10 , b ) b ) 12 , c ) c ) 24 , d ) d ) 60 , e ) e ) 200 | e | multiply(factorial(const_4.0), factorial(3)) | there are 3 red chips and 3 blue ones . when arranged in a row , they form a certain color pattern , for example rbrrb . how many color patterns ? | "using anagram method : @nl 6 _ 5 _ 4 _ 3 _ 2 _ 1 r _ r _ r _ b _ b _ b so . . 6 ! / number of repeated letters ( 3 ! ) ( 3 ! ) = 20 ans : e" | a = math.factorial(4)
b = math.factorial(3)
c = a * b
|
a ) 8 , b ) 77 , c ) 7 , d ) 18 , e ) 34 | c | add(subtract(add(32, 8), subtract(32, 8)), const_1) | the average age of applicants for a new job is 32 , with a standard deviation of 8 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the max... | "minimum age = average - 1 standard deviation = 32 - 8 = 24 maximum age = average + 1 standard deviation = 32 + 8 = 40 maximum number of different ages of the applicants = 30 - 24 + 1 = 7 answer c" | a = 32 + 8
b = 32 - 8
c = a - b
d = c + 1
|
a ) 36 , b ) 77 , c ) 88 , d ) 55 , e ) 221 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number is : | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a = 3 * 5
b = 9 / a
c = 3 * b
d = 3 * 5
e = 9 / d
f = 5 * e
g = c + f
|
a ) 22 , b ) 119 , c ) 110 , d ) 109 , e ) 12 | b | subtract(divide(multiply(const_1, const_1000), divide(48, 6)), 6) | in a kilometer race , a beats b by 48 meters or 6 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 6 ) / 48 = 125 sec . time taken by a = 125 - 6 = 119 sec . answer : b" | a = 1 * 1000
b = 48 / 6
c = a / b
d = c - 6
|
a ) 20029 , b ) 20000 , c ) 10000 , d ) 20027 , e ) 20026 | c | divide(200, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 200 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 200 = > x = 10000 answer : c" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 200 / h
|
a ) 85.5 , b ) 86.5 , c ) 87.5 , d ) 110 , e ) 89.5 | d | add(multiply(10, 5), 60) | the average weight of 10 person ' s increases by 5 kg when a new person comes in place of one of them weighing 60 kg . what is the weight of the new person ? | "total increase in weight = 10 Γ 5 = 50 if x is the weight of the new person , total increase in weight = x β 60 = > 50 = x - 60 = > x = 50 + 60 = 110 answer : d" | a = 10 * 5
b = a + 60
|
a ) 22 seconds , b ) 65 seconds , c ) 78 seconds , d ) 12 seconds , e ) 21 seconds | d | divide(60, 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 60 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 60 m = 60 / 5 = 12 seconds . answer : d" | a = 15 + 3
b = a * const_0_2778
c = 60 / b
|
a ) 6 : 4 , b ) 6 : 14 , c ) 3 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(10, const_100))))) | solution a is 10 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 -... | a = 80 / 100
b = 50 * a
c = 50 / 100
d = 50 * c
e = b - d
f = 80 / 100
g = 10 / 100
h = f - g
i = e / h
j = 80 / 100
k = 50 * j
l = 50 / 100
m = 50 * l
n = k - m
o = 80 / 100
p = 10 / 100
q = o - p
r = n / q
s = 50 - r
t = i / s
|
a ) 62.5 , b ) 67.5 , c ) 62.7 , d ) 62.2 , e ) 62.9 | b | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 12)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 12 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 88 / 100 ) = 59.375 mp = 67.5 answer : b" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 12
f = d / e
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 , d ) 4 / 5 , e ) 4 / 7 | c | divide(divide(6, 15), divide(2, 15)) | if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a Γ’ Λ Βͺ b ) = 6 / 15 find p ( b | a ) | p ( b | a ) = p ( a Γ’ Λ Βͺ b ) / p ( a ) p ( b | a ) = ( 6 / 15 ) / ( 2 / 15 ) = 3 . c | a = 6 / 15
b = 2 / 15
c = a / b
|
a ) - 4 , b ) 4 , c ) 10 , d ) 27 , e ) 40 | d | subtract(power(subtract(9, 3), 2), power(const_3, const_2)) | if x + y = 9 and x β y = 3 , then x ^ 2 - y ^ 2 = | the fastest approach has already been shown . here ' s one more option . given : x + y = 9 x β y = 3 add the two equations to get : 2 x = 12 , which means x = 6 if x = 6 , we can plug that value into either equation , to conclude that y = 3 if x = 6 and y = 3 , then x Β² - y Β² = 6 Β² - 3 Β² = 27 answer : d | a = 9 - 3
b = a ** 2
c = 3 ** 2
d = b - c
|
a ) 100 , b ) 250 , c ) 260 , d ) 300 , e ) 350 | d | add(multiply(5, divide(160, 8)), add(160, multiply(5, 8))) | in a primary school , the ratio of the number of girls to boys is 5 : 8 . if there are 160 girls , the total number of students in the college is : | let the number of girls and boys be 5 x and 8 x . total number of students = 13 x = 13 x 32 = 416 . answer : d | a = 160 / 8
b = 5 * a
c = 5 * 8
d = 160 + c
e = b + d
|
a ) 525 . , b ) 412 . , c ) 432 . , d ) 502 . , e ) 522 . | a | multiply(divide(multiply(49, 6), subtract(49, 35)), 35) | a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 35 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "pastry is divided in 35 customers equally . so , total number of pastry must be a multiple of 35 only option a satisfies the condition , and hence is the answer" | a = 49 * 6
b = 49 - 35
c = a / b
d = c * 35
|
a ) 8624 , b ) 8634 , c ) 8654 , d ) 8644 , e ) 8677 | a | subtract(divide(multiply(77, add(77, const_1)), const_2), divide(multiply(subtract(74, const_1), 74), const_2)) | ( 74 + 75 + 76 + 77 + . . . . . + 149 + 150 ) = ? | "explanation : sum of first n natural numbers = ( 1 + 2 + 3 + . . . . . + n ) = n ( n + 1 ) / 2 ( 74 + 75 + 76 + . . . . . + 150 ) = ( 1 + 2 + . . . . + 73 + 74 + 75 + . . . . . + 150 ) - ( 1 + 2 + . . . . + 73 ) = ( 1 / 2 x 150 x 151 ) - ( 1 / 2 x 73 x 74 ) = 11325 - 2701 = 8624 answer : option a" | a = 77 + 1
b = 77 * a
c = b / 2
d = 74 - 1
e = d * 74
f = e / 2
g = c - f
|
a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 75.00 | e | multiply(subtract(10, divide(multiply(17, 8), const_100)), 10) | an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 17 % . also , if a customer will β buy at least 10 β items they will deduct an additional 8 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 27 % , thus he must pay slightly more than $ 73 . only answer choice e fits . answer : e" | a = 17 * 8
b = a / 100
c = 10 - b
d = c * 10
|
a ) 5729 , b ) 5760 , c ) 2889 , d ) 2870 , e ) 4320 | e | divide(multiply(3, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 3 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 3 = 4320 . answer : e" | a = 8 * const_60
b = 3 * a
c = 8 * const_60
d = 6 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 55 % | a | multiply(divide(const_3, add(const_3, const_2)), const_100) | a feed store sells two varieties of birdseed : brand a , which is 65 % millet and 35 % sunflower , and brand b , which is 40 % millet and 60 % safflower . if a customer purchases a mix of the two types of birdseed that is 50 % millet , what percent of the mix is brand a ? | "yes there is a simple method : consider the following method brand a : 65 % millet and 35 % sunflower brand b : 40 % millet and 60 % safflower mix : 50 % millet here the weighted average is 50 % , now brand a has 65 % millet , which is 15 % more than the weighted average of mix = + 0.15 a - - - - - - - - - - - - - - -... | a = 3 + 2
b = 3 / a
c = b * 100
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | subtract(subtract(16, subtract(10, 4)), 6) | of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 16 boys and 44 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you w... | a = 10 - 4
b = 16 - a
c = b - 6
|
a ) 9 / 29 , b ) 3 / 23 , c ) 3 / 8 , d ) 17 / 29 , e ) 3 / 4 | b | divide(multiply(1, 1), add(multiply(multiply(4, 1), const_2), multiply(1, 1))) | pipe p can drain the liquid from a tank in 1 / 4 the time that it takes pipe q to drain it and in 2 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | "suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 1 / 4 ) rq ] = 4 also , rp = rr / ( 2 / 3 ) = > 4 = rr / ( 2 / 3 ) = > rr = 8 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 4 + 8 / 3 = 23 / 3 = 1 / ( 3 / 23 ) thus running simultaneo... | a = 1 * 1
b = 4 * 1
c = b * 2
d = 1 * 1
e = c + d
f = a / e
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | add(multiply(0.5, const_10), const_1) | a box contains 10 apples , 9 of which are red . an apple is drawn from the box and its color is noted before it is eaten . this is done a total of n times , and the probability that a red apple is drawn each time is less than 0.5 . what is the smallest possible value of n ? | "when you choose ( and then eat ) the first apple , the probability of that apple being red is 9 / 10 . so if we do the activity 1 times , the probability of it being red is 9 / 10 . for 2 times , it is ( 9 / 10 ) * ( 8 / 9 ) for 3 times , it is ( 9 / 10 ) * ( 8 / 9 ) * ( 7 / 8 ) you can notice that the numerator of th... | a = 0 * 5
b = a + 1
|
a ) 68 , b ) 42 , c ) 55 , d ) 18 , e ) 44 | c | divide(subtract(210, power(10, 2)), 2) | if a - b = 10 and a 2 + b 2 = 210 , find the value of ab . | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 210 - 100 = 110 ab = 55 . answer : c" | a = 10 ** 2
b = 210 - a
c = b / 2
|
a ) 22 , b ) 21 , c ) 20 , d ) 19 , e ) 18 | e | add(multiply(multiply(divide(20, const_100), 10), multiply(divide(20, const_100), 10)), divide(subtract(150, 10), 10)) | in a 150 member association consisting of men and women , exactly 10 % of men and exactly 20 % women are homeowners . what is the least number of members who are homeowners ? | solution simple out of 150 10 % are male i . e 15 and 20 % are female i . e 30 , so total homeowner is 45 . now min number homeowner is 15 and max is 30 so question ask us to find least and 18 has least value among all option . so ans is 18 . ans : e | a = 20 / 100
b = a * 10
c = 20 / 100
d = c * 10
e = b * d
f = 150 - 10
g = f / 10
h = e + g
|
['a ) 6.978 m', 'b ) 6.928 m', 'c ) 6.829 m', 'd ) 6.248 m', 'e ) 6.982 m'] | b | sqrt(add(power(multiply(sqrt(const_2), 4), const_2), power(4, const_2))) | the length of the longest rod that can fit in a cubical room of 4 m side is ? | the length of the longest rod = 4 sq . root of 3 = 6.928 m answer : b | a = math.sqrt(2)
b = a * 4
c = b ** 2
d = 4 ** 2
e = c + d
f = math.sqrt(e)
|
a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 432 | d | multiply(16, const_1) | if $ 5,000 is invested in an account that earns 16 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | "solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.16 / 2 ) ^ 2 = 5416 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 16 / 100 ) = 5400 difference = 5416 - 5400 = 16 $ d" | a = 16 * 1
|
['a ) 140 cm', 'b ) 240 cm 2', 'c ) 560 cm 2', 'd ) none of these', 'e ) can not be determined'] | a | multiply(divide(subtract(multiply(38, const_2), 34), const_3), subtract(38, multiply(divide(subtract(multiply(38, const_2), 34), const_3), const_2))) | a rectangular paper , when folded into two congruent parts had a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides . what is the area of the paper ? | solution when folded along breadth , we have : 2 ( l / 2 + b ) = 34 or l + 2 b = 34 when folded along length , we have : 2 ( l / 2 + b ) = 38 or 2 l + b = 38 solving ( i ) and ( ii ) , we get : l = 14 and b = 10 . β΄ area of the paper = ( 1410 ) cm 2 = 140 cm 2 answer a | a = 38 * 2
b = a - 34
c = b / 3
d = 38 * 2
e = d - 34
f = e / 3
g = f * 2
h = 38 - g
i = c * h
|
a ) 0.5 , b ) 0.6 , c ) 0.48 , d ) 0.65 , e ) 0.45 | c | divide(subtract(add(add(180, divide(180, divide(const_100, 20))), 50), 180), 180) | after giving a discount of rs . 50 the shopkeeper still gets a profit of 20 % , if the cost price is rs . 180 . find the markup % ? | "cost price = 180 s . p = 180 * 120 / 100 = 216 disc = 50 so . . . mark price = 216 + 50 = 266 . . . . . . mark up % = 266 - 180 / 180 = 86 / 180 = . 48 or 48 % answer : c" | a = 100 / 20
b = 180 / a
c = 180 + b
d = c + 50
e = d - 180
f = e / 180
|
a ) 100 , b ) 2250 , c ) 750 , d ) 1200 , e ) 5625 | b | multiply(volume_rectangular_prism(60, 10, divide(6, add(const_10, const_2))), 7.5) | the water level in a rectangular swimming pool measuring 60 feet by 10 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 60 * 10 * 1 / 2 = 300 feet ^ 3 of water must be removed , which equals to 300 * 7.5 = 2250 gallons . answer : b" | a = 10 + 2
b = 6 / a
c = volume_rectangular_prism * (
|
a ) 4670 , b ) 4546 , c ) 4556 , d ) 4675 , e ) 5760 | e | multiply(multiply(multiply(6, 4), const_60), 4) | a leak in the bottom of a tank can empty the tank in 6 hrs . an pipe fills water atthe rate of 4 ltrs / min . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hrs . the capacity of the tank is ? | 1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 e | a = 6 * 4
b = a * const_60
c = b * 4
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.