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 ) 21 , b ) 91 , c ) 11 , d ) 17 , e ) 12 | a | subtract(divide(multiply(60, 50), const_100), divide(multiply(30, 30), const_100)) | how much 60 % of 50 is greater than 30 % of 30 ? | "( 60 / 100 ) * 50 β ( 30 / 100 ) * 30 30 - 9 = 21 answer : a" | a = 60 * 50
b = a / 100
c = 30 * 30
d = c / 100
e = b - d
|
a ) 22 , b ) 33 , c ) 22 , d ) 41 , e ) 50 | e | divide(subtract(multiply(50, 50), add(45, 55)), subtract(50, const_2)) | the average of 50 numbers id 50 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | "explanation : total of 50 numbers = ( 50 Γ 50 ) = 2500 total of 48 numbers = ( 2500 - ( 45 + 55 ) ] = 2400 required average = 2400 / 48 = 50 answer : e" | a = 50 * 50
b = 45 + 55
c = a - b
d = 50 - 2
e = c / d
|
a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6 | e | divide(multiply(9, 18), add(9, 18)) | a can do a work in 9 days . b can do the same work in 18 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 9 b rate = 1 / 18 ( a + b ) rate = ( 1 / 9 ) + ( 1 / 18 ) = 1 / 6 a & b finish the work in 6 days correct option is e" | a = 9 * 18
b = 9 + 18
c = a / b
|
a ) 112 , b ) 77 , c ) 267 , d ) 14 , e ) 99 | d | subtract(divide(672, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | how many numbers between 100 and 672 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 672 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 672 : 672 / 42 = 14 hence , 42 * 16 = 672 is the last number divisible by 42 within 672 hence , total numbers divisible by 2 , 3 , 7 together are ( 16 Γ’ β¬ β 2 ) = 14 answer : d" | a = 2 * 3
b = a * 7
c = 672 / b
d = 2 * 3
e = d * 7
f = 100 / e
g = c - f
|
a ) 23,500 , b ) 15,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | b | floor(divide(divide(subtract(550, multiply(3,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 3,000)) | angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 3,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo β s and isabella β s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal base salary plus commission . so angelo β s earnings are 550 + ( 0.08 ) ( x β 3,000 ) , and isabella β s are 0.10 x . set up the equation and solve : 550 + ( 0.08 ) ( x β 3,000 ) = 0.10 x distribute the 0.08 : 550 + 0.08 x β 240 = 0.10 x combine terms and subtract 0.08 x from both sides : 310 = 0.02 x divide both sides by 0.02 : 15,500 = x your answer is b ." | a = 8 / 100
b = 3 * 0
c = 550 - b
d = 10 / 100
e = 8 / 100
f = d - e
g = c / f
h = g / 3
i = math.floor(h)
|
a ) 0 , b ) 233 , c ) 500 , d ) 695 , e ) 789 | e | divide(add(const_1000, 586.75), const_2) | a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 586.75 . how many boxes of plain cookies did she sell ? | "let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1586.75 = > x = 789" | a = 1000 + 586
b = a / 2
|
a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 520 | e | subtract(multiply(const_10, 150), 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 . 157 . find the unknown rate of two blankets ? | "10 * 157 = 1570 3 * 100 + 5 * 150 = 1050 1570 β 1050 = 520 answer : e" | a = 10 * 150
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 0.02 , b ) 0.2 , c ) 2 , d ) 20 , e ) none of these | a | multiply(divide(0.0004, 0.02), const_100) | 0.0004 ? = 0.02 | "explanation : required answer = 0.0004 / 0.02 = 0.04 / 2 = 0.02 . answer : option a" | a = 0 / 4
b = a * 100
|
a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability t that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability t = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b" | a = math.comb(10, 3)
b = 10 + 10
c = math.comb(b, 3)
d = a / c
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 33 % , e ) 40 % | d | multiply(divide(subtract(const_1, divide(subtract(const_100, 25), const_100)), divide(subtract(const_100, 25), const_100)), const_100) | the length of a rectangle is reduced by 25 % . by what % would the width have to be increased to maintainthe original area ? | "sol . required change = ( 25 * 100 ) / ( 100 - 25 ) = 25 % d" | a = 100 - 25
b = a / 100
c = 1 - b
d = 100 - 25
e = d / 100
f = c / e
g = f * 100
|
a ) 2 % , b ) 5 % , c ) 7 % , d ) 28 % , e ) 63 % | c | floor(multiply(subtract(divide(9, 63), divide(7, 95)), const_100)) | a survey was sent to 95 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 95 ) = x / 100 x = 7 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 7 ) % = 7 % answer is c" | a = 9 / 63
b = 7 / 95
c = a - b
d = c * 100
e = math.floor(d)
|
a ) $ 4035 , b ) $ 4036 , c ) $ 9035 , d ) $ 16140 , e ) $ 36140 | c | divide(add(add(add(multiply(50, multiply(const_1000, const_1000)), 200000), multiply(20000, 5000)), add(multiply(30, multiply(const_1000, const_1000)), 500000)), 20000) | tough and tricky questions : word problems . ak car company wants to manufacture a new car known as model x , and it is trying to determine how many cars it needs to sell in order to make an annual profit of $ 30 , 500000 . the annual fixed costs for model x total $ 50 , 200000 . in addition , each model x car has an average cost of $ 5000 per vehicle to manufacture . if the company forecasts it will sell 20000 model x cars this year , at what price must the company sell the car to achieve the desired annual profit ? | the ak company plans to make a profit of $ 30 , 500000 and it needs to earn the fixed costs of $ 50 , 200000 , so it needs to earn $ 80 , 700000 on top of the costs for the production of the 20000 cars . therefore , price of each car needs to be the unit costs plus $ 80 , 700000 / 20000 . after removing the 0 ' s , we end with the unit costs plus $ 8070 / 2 , which is $ 5000 plus $ 4035 . therefore , answer c is correct . | a = 1000 * 1000
b = 50 * a
c = b + 200000
d = 20000 * 5000
e = c + d
f = 1000 * 1000
g = 30 * f
h = g + 500000
i = e + h
j = i / 20000
|
a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | c | add(150, multiply(3, const_10)) | according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 15 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounces of water , that would mean we would need 150 packets of smoothie mix ( along with 12 ( 150 ) ounces of water for a total of 150 packets . however , we know , after a more careful reading of the stimulus , that the 3 ounces are not included in the 12 ounces . as such , the answer has to be less than 150 packets , since 150 would be too much powder considering you already have 150 ( 12 ) ounces of water and need less packets than water to make a smoothie . as such , the only answer less than 150 is 120 , a . does this make sense ? or am i way off base ? c" | a = 3 * 10
b = 150 + a
|
a ) 149 , b ) 150 , c ) 151 , d ) 166 , e ) none | b | subtract(divide(const_1000, 6), divide(const_100, 6)) | how many 3 - digit numbers are divisible by 6 in all ? | solution required numbers are 102 , 018114 , . . . . 996 . this is an a . p with a = 102 , d = 6 . let the number of its terms be n . then a + ( n - 1 ) d βΉ = βΊ 102 + ( n - 1 ) Γ 6 = 996 βΉ = βΊ n = 150 . answer b | a = 1000 / 6
b = 100 / 6
c = a - b
|
a ) 30 minutes , b ) 17 minutes , c ) 7 minutes , d ) 10 minutes , e ) 12 minutes | c | subtract(add(divide(const_1, 10), divide(const_1, 25)), divide(const_1, 45)) | two pipes can fill a tank in 10 minutes and 25 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 10 + 1 / 25 - 1 / 45 = ( 45 + 18 - 10 ) / 450 = 53 / 450 so , the tank becomes full in 7 minutes . answer : c" | a = 1 / 10
b = 1 / 25
c = a + b
d = 1 / 45
e = c - d
|
a ) 729 , b ) 754 , c ) 784 , d ) 783 , e ) 108 | c | multiply(power(2, 4), power(7, 2)) | if 7 ^ k = 2 , then 7 ^ ( 4 k + 2 ) = | "7 ^ k = 2 7 ^ 4 k = 2 ^ 4 7 ^ 4 k = 16 7 ^ ( 4 k + 2 ) = 7 ^ 4 k * 7 ^ 2 = 16 * 49 = 784 answer : c" | a = 2 ** 4
b = 7 ** 2
c = a * b
|
a ) 29 / 30 , b ) 24 / 25 , c ) 19 / 20 , d ) 14 / 15 , e ) 9 / 10 | b | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 3), 5)), 5), divide(multiply(divide(multiply(4, multiply(multiply(4, 3), 5)), 5), 4), 3)), subtract(multiply(multiply(4, 3), 5), divide(multiply(3, multiply(multiply(4, 3), 5)), 4))), const_100) | one day a car rental agency rented 3 / 4 of its cars , including 3 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 2 / 5 ) ( 3 / 5 ) = 6 / 25 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 6 / 25 ) / ( 1 / 4 ) = 24 / 25 the answer is b ." | a = 4 * 3
b = a * 5
c = 4 * b
d = c / 5
e = 4 * 3
f = e * 5
g = 4 * f
h = g / 5
i = h * 4
j = i / 3
k = d - j
l = 4 * 3
m = l * 5
n = 4 * 3
o = n * 5
p = 3 * o
q = p / 4
r = m - q
s = k / r
t = s * 100
|
a ) $ 2.15 , b ) $ 2.05 , c ) $ 2.45 , d ) $ 2.25 , e ) $ 2.85 | b | subtract(5.10, add(1.05, add(1.00, 1.00))) | little john had $ 5.10 . he spent $ 1.05 on sweets and gave to his two friends $ 1.00 each . how much money was left ? | "john spent and gave to his two friends a total of 1.05 + 1.00 + 1.00 = $ 3.05 money left 5.10 - 3.05 = $ 2.05 answer : b" | a = 1 + 0
b = 1 + 5
c = 5 - 10
|
a ) 3 / 4 , b ) 3 / 2 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 5 | d | divide(add(add(add(choose(3, const_2), choose(3, const_3)), choose(3, const_4)), choose(3, 3)), power(const_2, 3)) | a fair coin is tossed 3 times . what is the probability of getting at least 2 heads ? | "let ' s find the probability of 2 heads , 3 heads p ( hhh ) = ( ( 1 / 2 ) ^ 3 = 1 / 8 . p ( hht ) = ( 3 ! / 2 ! ) * ( 1 / 2 ) ^ 3 = 3 / 8 total probablity = 1 / 8 + 3 / 8 = 1 / 2 answer ( d )" | a = math.comb(3, 2)
b = math.comb(3, 3)
c = a + b
d = math.comb(3, 4)
e = c + d
f = math.comb(3, 3)
g = e + f
h = 2 ** 3
i = g / h
|
a ) 6300 , b ) 2765 , c ) 6000 , d ) 1298 , e ) 1261 | a | multiply(multiply(const_1, const_12), divide(18900, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18900 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18900 = 6300 answer : a" | a = 1 * 12
b = 1 * 12
c = 12 - 6
d = c * 2
e = b + d
f = 12 - 8
g = f * 3
h = e + g
i = 18900 / h
j = a * i
|
a ) $ 0.25 , b ) $ 0.01 , c ) $ 0.05 , d ) $ 0.03 , e ) $ 0.002 | b | subtract(divide(2.99, 130), divide(8.25, 250)) | bottle r contains 250 capsules and costs $ 8.25 . bottle t contains 130 capsules and costs $ 2.99 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | "cost per capsule in r is 8.25 / 250 = 0.825 / 25 = 0.033 cost per capsule in t is 2.99 / 130 = 0.023 the difference is 0.01 the answer is b" | a = 2 / 99
b = 8 / 25
c = a - b
|
a ) 621 , b ) 652 , c ) 660 , d ) 690 , e ) 693 | a | multiply(divide(598, add(const_1, divide(30, const_100))), add(const_1, divide(35, const_100))) | if albert β s monthly earnings rise by 30 % , he would earn $ 598 . if , instead , his earnings rise by only 35 % , how much ( in $ ) would he earn this month ? | "= 598 / 1.3 β 1.35 = 621 = 621 answer is a" | a = 30 / 100
b = 1 + a
c = 598 / b
d = 35 / 100
e = 1 + d
f = c * e
|
a ) 0 , b ) 1 , c ) 6 , d ) 3 , e ) 4 | c | subtract(multiply(multiply(multiply(3421, 6721), 4584), 6784), subtract(multiply(multiply(multiply(3421, 6721), 4584), 6784), add(const_4, const_4))) | the unit digit in the product 3421 * 6721 * 4584 * 6784 is ? | "unit digit in the given product = unit digit in 1 * 1 * 4 * 4 = 6 answer is c" | a = 3421 * 6721
b = a * 4584
c = b * 6784
d = 3421 * 6721
e = d * 4584
f = e * 6784
g = 4 + 4
h = f - g
i = c - h
|
a ) 22 , b ) 77 , c ) 70 , d ) 68 , e ) 71 | d | add(add(multiply(divide(60, 4), const_2), 4), add(divide(60, 4), 4)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 4 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 4 years , their ages will be 44 and 24 . sum of their ages = 44 + 24 = 68 . answer : d" | a = 60 / 4
b = a * 2
c = b + 4
d = 60 / 4
e = d + 4
f = c + e
|
a ) 8 kmph , b ) 9 kmph , c ) 12 kmph , d ) 14 kmph , e ) 16 kmph | c | divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), subtract(divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), 10), 1)) | robert is travelling on his cycle andhas calculated to reach point a at 2 pm . if he travels at 10 kmph , he will reach there at 12 pm if he travels at 15 kmph . at what speed must he travel to reach a at 1 pm ? | let the distance travelled by x km . then , x - x = 2 10 15 3 x - 2 x = 60 x = 60 km . time taken to travel 60 km at 10 km / hr = 60 hrs = 6 hrs . 10 so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . required speed = 60 kmph . = 12 kmph . 5 c | a = 1 / 10
b = 1 / 15
c = a - b
d = 2 / c
e = 1 / 10
f = 1 / 15
g = e - f
h = 2 / g
i = h / 10
j = i - 1
k = d / j
|
a ) 9 , b ) 12 , c ) 18 , d ) 20 , e ) 30 | e | subtract(36, multiply(const_2, divide(divide(36, const_2), add(const_1, const_4)))) | let f ( x , y ) be defined as the remainder when ( x β y ) ! is divided by x . if x = 36 , what is the maximum value of y for which f ( x , y ) = 0 ? | "the question is finding y such that ( 36 - y ) ! is a multiple of 36 . that means we need to have 2 ^ 2 * 3 ^ 2 in ( 36 - y ) ! 6 ! is the smallest factorial number with 2 ^ 2 * 3 ^ 2 as a factor . 36 - y = 6 y = 30 the answer is e ." | a = 36 / 2
b = 1 + 4
c = a / b
d = 2 * c
e = 36 - d
|
a ) 2.25 . , b ) 3.00 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | b | divide(divide(4032, 28), multiply(subtract(28, const_4), const_2)) | a computer factory produces 4032 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 4032 / ( 28 * 24 * 2 ) = 3 answer b | a = 4032 / 28
b = 28 - 4
c = b * 2
d = a / c
|
a ) 1 : 16 , b ) 1 : 6 , c ) 1 : 9 , d ) 1 : 3 , e ) 1 : 2 | a | divide(circle_area(1), circle_area(4)) | the ratio of the radius of two circles is 1 : 4 , and then the ratio of their areas is ? | "r 1 : r 2 = 1 : 4 Ο r 12 : Ο r 22 r 12 : r 22 = 1 : 16 answer : a" | a = circle_area / (
|
a ) 25 , b ) 28 , c ) 30 , d ) 35 , e ) 39 | a | subtract(multiply(add(39, 3), add(15, 3)), multiply(39, 15)) | the average age of a class of 39 students is 15 years . if the age of the teacher be included , then the average increased by 3 months . find the age of the teacher ? | "total age of 39 persons = 39 * 15 = 585 years average age of 40 persons = 15 yrs 3 months = 61 / 4 years total age of 40 persons = 61 / 4 * 40 = 610 years age of the teacher = 610 - 585 = 25 years answer is a" | a = 39 + 3
b = 15 + 3
c = a * b
d = 39 * 15
e = c - d
|
a ) $ 1124 million , b ) $ 1120 million , c ) $ 1144 million , d ) $ 1240 million , e ) $ 1488 million | e | subtract(multiply(488, divide(const_12, const_2)), multiply(1.44, const_1000)) | country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 488 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 488 mill / 2 month i . e . $ 244 mill / month difference / month = 244 - 120 = 124 difference / year = $ 124 mill * 12 = $ 1488 mill answer : e | a = 12 / 2
b = 488 * a
c = 1 * 44
d = b - c
|
a ) 252 , b ) 243 , c ) 521 , d ) 361 , e ) 600 | a | sqrt(divide(18, add(power(7, 2), add(power(2, 2), power(2, 2))))) | the ratio of 2 numbers is 2 : 7 and their h . c . f . is 18 . their l . c . m . is ? | "let the numbers be 2 x and 7 x their h . c . f . = 18 so the numbers are 2 * 18 , 7 * 18 = 36,126 l . c . m . = 252 answer is a" | a = 7 ** 2
b = 2 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 18 / e
g = math.sqrt(f)
|
a ) 34 , b ) 32 , c ) 30 , d ) 26 , e ) 28 | d | subtract(110, add(add(38, const_1), 45)) | 110 people are attending a newspaper conference . 45 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ? | "{ total } = { writers } + { editors } - { both } + { neither } . { total } = 110 ; { writers } = 45 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 110 = 45 + { editors } - x + 2 x - - > x = 65 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 39 , thus x = { both } = 65 - 39 = 26 . answer : d ." | a = 38 + 1
b = a + 45
c = 110 - b
|
a ) 64 % , b ) 68 % , c ) 72 % , d ) 74 % , e ) 83 % | d | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | in may mrs lee ' s earnings were 70 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 70 in may rest of the family ' s income = 30 in june mrs lees income = 70 * 120 / 100 = 84 in june total income = 84 + 30 = 114 % of mrs lee ' s income = 84 / 114 = 73.68 ( d )" | a = 100 + 20
b = 100 + 20
c = b + 100
d = a / c
e = d * 100
|
a ) 32 , b ) 33 , c ) 34 , d ) 35 , e ) 36 | e | add(divide(multiply(7, subtract(multiply(8, add(4, const_1)), multiply(7, add(4, const_1)))), subtract(multiply(7, 7), multiply(6, 8))), const_1) | one year ago , the ratio of roonie β s and ronaldo β s age was 6 : 7 respectively . 4 years hence , this ratio would become 7 : 8 . how old is ronaldo ? | 7 ( x - 1 ) = 6 ( y - 1 ) = > 7 x - 6 y - 1 = 0 - - - - eq 1 8 ( x + 4 ) = 7 ( y + 4 ) = > 8 x - 7 y + 4 = 0 - - - - eq 2 on solving eq 1 n eq 2 y = 36 answer : e | a = 4 + 1
b = 8 * a
c = 4 + 1
d = 7 * c
e = b - d
f = 7 * e
g = 7 * 7
h = 6 * 8
i = g - h
j = f / i
k = j + 1
|
a ) rs . 500 , b ) rs . 1800 , c ) rs . 2000 , d ) rs . 2500 , e ) none of the above | b | multiply(multiply(subtract(4, 3), 900), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 900 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 900 x = 900 . b ' s share = rs . 2 x = rs . ( 2 x 900 ) = rs . 1800 . answer = b" | a = 4 - 3
b = a * 900
c = b * 3
|
a ) 16 seconds , b ) 34 seconds , c ) 9 seconds , d ) 12 seconds , e ) 15 seconds | c | divide(40, multiply(add(14, 2), const_0_2778)) | the speed at which a man can row a boat in still water is 14 kmph . if he rows downstream , where the speed of current is 2 kmph , what time will he take to cover 40 metres ? | "speed of the boat downstream = 14 + 2 = 16 kmph = 16 * 5 / 18 = 40 / 9 m / s hence time taken to cover 40 m = 40 * 9 / 40 = 9 seconds . answer : c" | a = 14 + 2
b = a * const_0_2778
c = 40 / b
|
a ) rs . 3240 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | c | multiply(8640, divide(add(divide(subtract(90000, add(6000, 3000)), const_3), 6000), 90000)) | a , b and c started a business with a total investment of rs . 90000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find c ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 90000 = > x + ( x - 3000 ) + ( x + 3000 ) = 90000 = > 3 x = 90000 = > x = 30000 hence , a ' s investment = rs . 33000 b ' s investment = rs . 27000 c ' s investment = rs . 30000 ratio of the capitals of a , b and c = 33000 : 27000 : 30000 = 11 : 9 : 10 a ' s share = rs . [ ( 10 / 30 ) Γ£ β 8640 ] = rs . 2880 answer : option c" | a = 6000 + 3000
b = 90000 - a
c = b / 3
d = c + 6000
e = d / 90000
f = 8640 * e
|
a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 9), add(2, 2)), 6) | the ratio of the present age of sandy to that of molly is 9 : 2 . sixteen years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | "let the present age of sandy be 9 x years and that of molly be 2 x years . ( 9 x + 16 ) / ( 2 x + 16 ) = 5 / 2 8 x = 48 x = 6 six years ago , sandy ' s age was 9 ( 6 ) - 6 = 48 the answer is d ." | a = 10 + 6
b = 5 * a
c = 10 + 6
d = 2 * c
e = b - d
f = e * 9
g = 2 + 2
h = f / g
i = h - 6
|
a ) 90 , b ) 610 , c ) 729 , d ) 900 , e ) 1000 | a | multiply(multiply(multiply(multiply(3, const_10), const_10), const_10), const_10) | how many 3 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 121 . | "take the task of building palindromes and break it intostages . stage 1 : select the hundreds digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the tens digit we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 2 in 10 ways important : at this point , the remaining digits are alreadylocked in . stage 3 : select the units digit this digit must be the same as the thousands digit ( which we already chose in stage 2 ) so , we can complete this stage in 1 way . by thefundamental counting principle ( fcp ) , we can complete all 5 stages ( and thus build a 3 - digit palindrome ) in ( 9 ) ( 10 ) ( 1 ) ways ( = 900 ways ) answer : a" | a = 3 * 10
b = a * 10
c = b * 10
d = c * 10
|
a ) 1.5 , b ) 3.0 , c ) 3.9 , d ) 4.0 , e ) 6.0 | e | divide(subtract(multiply(9.2, 5), multiply(6.8, 5)), const_2) | the average of the 5 numbers is 6.8 . if one of the numbers is multiplied by 3 , the average of the numbers increases to 9.2 . which of the 5 numbers is multiplied by 3 ? | initial avg is 6.8 so the summation of the nos is 6.8 * 5 = 34 now the new avg is 9.2 so summation will be 9.2 * 5 = 46 . so change in summation is 12 . this comes when a number is multiplied by 3 . in the original sum we already have that number added once . to bring additional 12 increment by multiplying by 3 , we get a total increase of 18 . thus , 6 is multiplied by 3 . answer : e | a = 9 * 2
b = 6 * 8
c = a - b
d = c / 2
|
a ) 380 , b ) 350 , c ) 334 , d ) 310 , e ) none of these | c | multiply(3192, power(add(const_4, const_1), const_4)) | ( 3192 + 6160 ) / 28 | "explanation : as per bodmas rule , first we will solve the equation in bracket then we will go for division = ( 9352 ) / 28 = 334 option c" | a = 4 + 1
b = a ** 4
c = 3192 * b
|
a ) 115 , b ) 120 , c ) 110 , d ) 100 , e ) 90 | c | divide(multiply(multiply(100, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | a ferry can transport 100 tons of vehicles . automobiles range in weight from 1,800 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "to get maximum vehicles we must take into consideration the minimum weight i . e 1800 pounds here since , 1 ton = 2000 pounds 78 tons will be 200,000 pounds from the answer choices : let max number of vehicles be 110 total weight will be = 110 * 1800 = 198000 pounds , which is lesser than the maximum weight allowed . ans : c" | a = 100 * 2
b = a * 1000
c = 1000 + 100
d = c + 100
e = d + 100
f = e + 100
g = f + 100
h = g + 100
i = b / h
|
a ) 600 , b ) 500 , c ) 800 , d ) 200 , e ) 900 | b | divide(25, subtract(204.05, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 25 . if n / j = 204.05 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 204.05 here 204 is the quotient . given that remainder = 25 so , 204.05 = 204 + 25 / j so , j = 500 answer : b" | a = 4 * 10
b = a + 2
c = 100 + b
d = 204 - 5
e = 25 / d
|
a ) s . 10,000 , b ) s . 10,200 , c ) s . 10,400 , d ) s . 10,700 , e ) s . 10,800 | b | multiply(30000, divide(divide(add(50000, subtract(5000, 4000)), const_3), 50000)) | a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30000 , b receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 30000 x 17 / 50 ) = rs . 10,200 . b | a = 5000 - 4000
b = 50000 + a
c = b / 3
d = c / 50000
e = 30000 * d
|
a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 14 | d | divide(add(divide(18, 3), divide(48, 3)), const_2) | a man swims downstream 48 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "48 - - - 3 ds = 16 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 16 + 6 ) / 2 = 11 answer : d" | a = 18 / 3
b = 48 / 3
c = a + b
d = c / 2
|
a ) 20 , b ) 30 , c ) 15 , d ) 10 , e ) 80 | a | multiply(divide(const_60, 30), 10) | if the population of a certain country increases at the rate of one person every 30 seconds , by how many persons does the population increase in 10 minutes ? | "answer = 2 * 10 = 20 answer = a" | a = const_60 / 30
b = a * 10
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 15)), add(divide(const_1, 20), divide(const_1, 30))), 15) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 15 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 15 days , a can do 3 / 4 of the project . thus a and b must complete 1 / 4 of the project , which takes 3 days . the total number of days is 3 + 15 = 18 . the answer is d ." | a = 1 / 30
b = a * 15
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 15
|
a ) 25 , b ) 20 , c ) 35 , d ) 36 , e ) 38 | a | subtract(choose(7, 4), choose(subtract(7, 2), 2)) | a meeting has to be conducted with 4 managers . find the number of ways in which the managers be selected from among 7 managers , if 2 managers will not attend the meeting together ? | "we can either choose all 4 people from 5 manager who have no problems or choose 3 from the 5 and 1 from the 2 managers who have a problem sitting together so 5 c 4 + ( 5 c 3 * 2 c 1 ) this is 5 + 20 = 25 answer : a" | a = math.comb(7, 4)
b = 7 - 2
c = math.comb(b, 2)
d = a - c
|
a ) 25 sec , b ) 23 sec , c ) 34 sec , d ) 22 sec , e ) 6 sec | e | multiply(const_3600, divide(divide(add(150, 150), const_1000), add(95, 85))) | two trains each 150 m in length each , are running on two parallel lines in opposite directions . if one goes at the speed of 95 km / h while the other travels at 85 km / h . how long will it take for them to pass each other completely . | "explanation : d = 150 m + 150 m = 300 m rs = 95 + 85 = 180 * 5 / 18 = 50 t = 300 * 1 / 50 = 6 sec answer : option e" | a = 150 + 150
b = a / 1000
c = 95 + 85
d = b / c
e = 3600 * d
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | divide(multiply(add(divide(10, const_100), const_1), 21), add(add(divide(10, const_100), const_1), const_1)) | p is 10 % more efficient than q . p can complete a work in 21 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 21 let work done by q in 1 day = q q Γ ( 110 / 100 ) = 1 / 21 q = 100 / ( 21 Γ 110 ) = 10 / ( 21 Γ 11 ) the work done by p and q in 1 day = 1 / 21 + 10 / ( 21 Γ 11 ) = 21 / ( 21 Γ 11 ) = 1 / 11 p and q together can do the work in 11 days . the answer is c ." | a = 10 / 100
b = a + 1
c = b * 21
d = 10 / 100
e = d + 1
f = e + 1
g = c / f
|
a ) 17 kg , b ) 28 kg , c ) 26 kg , d ) 31 kg , e ) 35 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(44, const_3)) | the average weight of a , b and c is 44 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let d sum of a , b , c is 3 * 44 = 132 and sum of a and b s 2 * 40 = 80 sum of b and c is 2 * 43 = 86 hence 80 + 86 - 132 = 28 ans = 28 answer : b" | a = 40 * 2
b = 43 * 2
c = a + b
d = 44 * 3
e = c - d
|
a ) 4.2 , b ) 4.8 , c ) 5.6 , d ) 6.4 , e ) 7.2 | c | subtract(40, 30) | a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 100 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 2.4 hours to travel 240 km , so c leaves 5.6 hours after a . the answer is c ." | a = 40 - 30
|
a ) 92.5 % , b ) 89.4 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(4, const_100)))), add(600, 400)), const_100) | a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 4 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ 600 = 9000 / 100 = 90 number of rotten bananas = 4 % of 400 = 16 therefore , total number of rotten fruits = 90 + 16 = 106 therefore number of fruits in good condition = 1000 - 106 = 894 therefore percentage of fruits in good condition = ( 894 / 1000 Γ 100 ) % = ( 89400 / 1000 ) % = 89.4 % answer : b" | a = 600 + 400
b = 15 / 100
c = 600 * b
d = 4 / 100
e = 400 * d
f = c + e
g = a - f
h = 600 + 400
i = g / h
j = i * 100
|
a ) 54.0 , b ) 54.9 , c ) 55.5 , d ) 57.0 , e ) 63.0 | c | add(add(divide(multiply(multiply(const_1, const_100), 18), add(multiply(const_1, const_100), 20)), divide(multiply(25, 18), 20)), 18) | james took a 3 - hour bike ride . in the second hour he travlled 18 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | let the distance travelled in the first hour be x . thus , 1.2 x = 18 , x = 15 . now , the distance travelled in the 3 rd hour = 18 + 1 / 4 β 18 = the only option ending with a 0.5 in the decimal place is c . answer : c | a = 1 * 100
b = a * 18
c = 1 * 100
d = c + 20
e = b / d
f = 25 * 18
g = f / 20
h = e + g
i = h + 18
|
a ) 4.5 % , b ) 4.7 % , c ) 4.9 % , d ) 5.1 % , e ) 5.3 % | e | multiply(divide(add(const_1, divide(multiply(7.5, 600), const_100)), const_1000), const_100) | by weight , liquid x makes up 2 percent of solution p and 7.5 percent of solution q . if 400 grams of solution p are mixed with 600 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 2 ( 400 ) / 100 + 7.5 ( 600 ) / 100 = 8 + 45 = 53 grams . 53 / 1000 = 5.3 % the answer is e ." | a = 7 * 5
b = a / 100
c = 1 + b
d = c / 1000
e = d * 100
|
a ) 240 meters , b ) 360 meters , c ) 420 meters , d ) 720 meters , e ) can not be determined | d | subtract(multiply(divide(multiply(216, const_1000), const_3600), 30), multiply(divide(multiply(216, const_1000), const_3600), 12)) | a train traveling at 216 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length of the train plus the length of the platform . the extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover . i . e . , length of the platform . compute length of platform length of the platform = speed of train * extra time taken to cross the platform . length of platform = 216 kmph * 12 seconds convert 216 kmph into m / sec 1 kmph = 5 / 18 m / s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . β΄ 216 kmph = 5 / 18 β 216 = 60 m / sec therefore , length of the platform = 60 m / s * 12 sec = 720 meters . choice d" | a = 216 * 1000
b = a / 3600
c = b * 30
d = 216 * 1000
e = d / 3600
f = e * 12
g = c - f
|
a ) 31 , b ) 28 , c ) 19 , d ) 33 , e ) 25 | c | divide(190, const_10) | for my anniversary , i decided to surprise my wife . since she is a voracious reader , i decided to collect a lot of books for her . on the first day of the month , i bought one book , on the second , i bought two and on the third , i bought three . this process went on till the anniversary and on the day , i had 190 books with me to gift her . can you calculate , on which day is our anniversary ? | c 19 rd day 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 = 190 thus our anniversary falls on the 19 rd day of the month . you can apply other formulas to shorten the process but that is the simplest way to do it . | a = 190 / 10
|
a ) $ 1220 , b ) $ 1120 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 3), 30), multiply(subtract(100, 30), 10)) | at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 3 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ? | "i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 3 ) = $ 120 total revenue : 1000 + 120 = $ 1120 the answer is b ." | a = 10 + 3
b = a * 30
c = 100 - 30
d = c * 10
e = b + d
|
a ) 75 kg . , b ) 80 kg . , c ) 86 kg . , d ) 90 kg . , e ) 92 kg . | e | divide(add(multiply(7, 94), add(110, 60)), add(7, const_2)) | there are 7 players in a bowling team with an average weight of 94 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 94 * 7 + 110 + 60 ) / 9 = 92 kgs e is the answer" | a = 7 * 94
b = 110 + 60
c = a + b
d = 7 + 2
e = c / d
|
a ) 85 , b ) 80 , c ) 95 , d ) 90 , e ) 106 | e | add(multiply(12, 4), 58) | the average weight of 12 person ' s increases by 4 kg when a new person comes in place of one of them weighing 58 kg . what is the weight of the new person ? | "total increase in weight = 12 Γ 4 = 48 if x is the weight of the new person , total increase in weight = x β 58 = > 48 = x - 58 = > x = 48 + 58 = 106 answer : e" | a = 12 * 4
b = a + 58
|
['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | b | floor(cube_edge_by_volume(125)) | a cube 125 mt is colored pink on two opposite surfaces and then cut in to 125 smaller cubes then find how many number of cubes have pink color ? | answer : b | a = math.floor(cube_edge_by_volume)
|
a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 10 , e ) e ) 45 | d | divide(subtract(subtract(180, 80), 60), const_4) | 6 ) a marketing firm determined that , of 180 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | 100 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 100 = > 4 x = 40 = > x = 10 answer - d | a = 180 - 80
b = a - 60
c = b / 4
|
a ) 2.00 seconds , b ) 2.20 seconds , c ) 2.50 seconds , d ) 2.95 seconds , e ) 3.00 seconds | d | inverse(add(inverse(multiply(3, const_60)), inverse(3))) | one machine takes 3 minutes to check fault in 100 km fiber optic cable . another machine does the same job in 3 seconds . what time will it take to do the same job when both machines are put to work ? | by guess it is clear that the time taken will be less than 3 seconds and very near to it . therefore , answer 2.95 seconds will be correct . answer - d | a = 3 * const_60
b = 1/(a)
c = 1/(3)
d = b + c
e = 1/(d)
|
a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 150 % | e | multiply(divide(subtract(150, 60), 60), const_100) | a certain telescope increases the visual range at a particular location from 60 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | "original visual range = 60 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 60 ) / 60 * 100 % = 3 / 2 * 100 % = 150 % answer e" | a = 150 - 60
b = a / 60
c = b * 100
|
a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | b | divide(181, divide(subtract(const_100, 13), const_100)) | a small company reduced its faculty by approximately 13 percent to 181 employees . what was the original number of employees ? | "if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 181 x = 208 so the original number of employees is 208 correct answer - b" | a = 100 - 13
b = a / 100
c = 181 / b
|
a ) 64 , b ) 150 , c ) 152 , d ) 216 , e ) 256 | b | subtract(volume_cube(add(cube_edge_by_volume(66), const_2)), 66) | 66 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "66 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 66 = 150 hence , b is the answer ." | a = cube_edge_by_volume + (
b = volume_cube - (
|
a ) 0.063 , b ) 0.0063 , c ) 0.63 , d ) 0.00063 , e ) 63 | b | multiply(0.9, 0.007) | 0.9 * 0.007 = Β Β Β Β Β Β Β Β Β Β _________ | 9 * 7 = 63 sum of decimal places = 4 so , 0.9 * 0.007 = 0.0063 answer : b | a = 0 * 9
|
a ) 228 , b ) 240 , c ) 165 , d ) 166 , e ) 1811 | c | multiply(20, multiply(54, const_0_2778)) | a train passes a station platform in 31 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 31 = 15 = > x = 165 m . answer : c" | a = 54 * const_0_2778
b = 20 * a
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | sqrt(divide(768, const_3)) | the length of a rectangular garden is three times its width . if the area of the rectangular garden is 768 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 768 x ^ 2 = 256 x = 16 the answer is a ." | a = 768 / 3
b = math.sqrt(a)
|
a ) $ 900 , b ) $ 1275 , c ) $ 1200 , d ) $ 1300 , e ) $ 1060 | b | divide(multiply(multiply(divide(add(multiply(multiply(4, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), const_4), 9), multiply(const_3, const_4)) | mr . hernandez , who was a resident of state x for only 9 months last year , had a taxable income of $ 42,500 for the year . if the state tax rate were 4 percent of the year β s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez β s state x tax for last year ? | total tax for the year = 42,500 x 4 % = 1700 as stated annual tax is prorated as per the duration of stay . prorated tax = 1700 ( 9 / 12 ) = 1275 answer b | a = 4 * 100
b = a * 100
c = 100 * const_0_25
d = c * 100
e = b + d
f = e / 100
g = f * 4
h = g * 9
i = 3 * 4
j = h / i
|
a ) 6 , b ) 8 , c ) 7 , d ) 4 , e ) 2 | a | divide(multiply(subtract(const_100, 10), 10), add(const_100, 50)) | if a man lost 10 % by selling oranges at the rate of 10 a rupee at how many a rupee must he sell them to gain 50 % ? | "90 % - - - - 10 150 % - - - - ? 90 / 150 * 10 = 6 answer : a" | a = 100 - 10
b = a * 10
c = 100 + 50
d = b / c
|
a ) 110 , b ) 120 , c ) 130 , d ) 152 , e ) 150 | d | divide(add(42, 22), divide(42, const_100)) | a candidate appearing for an examination has to secure 42 % marks to pass paper i . but he secured only 42 marks and failed by 22 marks . what is the maximum mark for paper i ? | "he secured 42 marks nd fail by 22 marks so total marks for pass the examinatn = 64 let toal marks x x * 42 / 100 = 64 x = 152 answer : d" | a = 42 + 22
b = 42 / 100
c = a / b
|
a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | b | divide(12, 3) | if ( - 3 ) ^ ( 2 x ) = 3 ^ ( 12 - x ) and x is an integer , what is the value of x ? | since x is an integer , ( - 3 ) ^ ( 2 x ) is always positive . so , 3 ^ 2 x = 3 ^ ( 12 - x ) 2 x = 12 - x 3 x = 12 x = 4 answer : b | a = 12 / 3
|
a ) 5336 , b ) 5242 , c ) 6524 , d ) 3526 , e ) 4526 | a | multiply(multiply(5, 46), divide(multiply(5, 46), 10)) | in a division sum , the divisor is 10 times the quotient and 5 times the remainder . if the remainder is 46 , what is the dividend ? | "divisor = ( 5 * 46 ) = 230 10 * quotient = 230 quotient = 230 / 10 = 23 . dividend = ( divisor * quotient ) + remainder = ( 230 * 23 ) + 46 = ( 5290 + 46 ) = 5336 . answer is a" | a = 5 * 46
b = 5 * 46
c = b / 10
d = a * c
|
a ) 6 , b ) 3 , c ) 2 , d ) 4 , e ) 5 | c | floor(add(divide(20, add(const_10, const_1)), const_1)) | a company plans to gradually replace its fleet of 20 cars with newer models . at the end of every year starting with 2000 it retires 5 of its old cars and buys 6 new models . how many years did it take before the number of older cars left in the company ' s fleet was less than 50 percent of the fleet ? | total fleet - - 20 end of year 2000 - - ( retires 5 and buys 6 ) total fleet - - 21 ( 15 old 6 new ) . end of year 2001 - - ( retires 5 and buys 6 ) total fleet - - 22 ( 10 old 12 new ) . answer is c - - after 2 years the company has a total of 22 cars ( 10 old 12 new ) , the old cars are already less than 50 % of its fleet . | a = 10 + 1
b = 20 / a
c = b + 1
d = math.floor(c)
|
a ) 2 : 1 , b ) 4 : 9 , c ) 8 : 77 , d ) 4 : 121 , e ) none | b | power(divide(2, 3), 2) | the ratio of radius of a circle and the side of a square is 2 : 3 . find the ratio of their areas : | "radius / side = 2 / 3 Γ’ β‘ β area of circle / area of square = 4 / 9 answer : b" | a = 2 / 3
b = a ** 2
|
a ) $ 138.14 , b ) $ 150.12 , c ) $ 142.65 , d ) $ 152 , e ) $ 164.12 | a | subtract(multiply(500, power(add(const_1, divide(5, const_100)), 5)), 500) | find the compound interest on $ 500 in 5 years at 5 % per annum , the interest being compounded yearly ? | "principle = $ 500 rate = 5 % yearly amount = 500 * ( 1 + 5 / 100 ) ^ 5 = 500 * 21 / 20 * 21 / 20 * 21 / 20 * 21 / 20 * 21 / 20 = $ 638.14 c . i . = 638.14 - 500 = $ 138.14 answer is a" | a = 5 / 100
b = 1 + a
c = b ** 5
d = 500 * c
e = d - 500
|
a ) 132,000 , b ) 235,000 , c ) 175,000 , d ) 200,000 , e ) 250,000 | a | multiply(200, divide(subtract(2007, 2003), const_2)) | during 2003 , a company produced an average of 1,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 200 % over its 2003 average ? | "company produced 12 * 1000 = 12,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 12,000 . the gives the average of ( x + 12,000 ) / 4 . this average needs to be 200 % higher than that in 2003 . in math terms , 12,000 + 200 % ( 12,000 ) = 36,000 . so : ( x + 12,000 ) / 4 = 36,000 x + 12,000 = 144,000 x = 132,000 the answer is a ." | a = 2007 - 2003
b = a / 2
c = 200 * b
|
a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 45 | b | divide(add(45, 5), const_2) | the sum of two numbers is 45 . their difference is 5 . the bigger number is | x + y = 45 x - y = 5 adding 2 x = 50 = = > x = 25 and y = 45 - 25 = 20 bigger number is x = 25 answer b . | a = 45 + 5
b = a / 2
|
a ) rs . 312.5 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 158 , e ) none | a | divide(338, power(add(divide(4, const_100), const_1), 2)) | the present worth of rs . 338 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 338 / ( 1 + 4 / 100 ) Β² ] = rs . ( 338 x 25 / 26 x 25 / 26 ) = rs . 312.5 answer a" | a = 4 / 100
b = a + 1
c = b ** 2
d = 338 / c
|
a ) 4 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | a | subtract(multiply(subtract(11, 1), add(6, 1)), multiply(11, 6)) | the average of 6 observations is 11 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 66 + x ) / 7 = 10 = > x = 4 . hence , the seventh observation is 4 . answer : a" | a = 11 - 1
b = 6 + 1
c = a * b
d = 11 * 6
e = c - d
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | c | add(divide(lcm(350, 200), 350), divide(lcm(350, 200), 200)) | a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 350 and $ 200 per unit , respectively . what is the fewest number of computer printers that the company can purchase ? | "the smallest amount that the company can spend is the lcm of 350 and 200 , which is 1400 for each , which is total 2800 . the number of 1 st type of computers which costing $ 350 = 1400 / 350 = 4 . the number of 2 nd type of computers which costing $ 200 = 1400 / 200 = 7 . total = 4 + 7 = 11 answer is c ." | a = math.lcm(350, 200)
b = a / 350
c = math.lcm(350, 200)
d = c / 200
e = b + d
|
a ) 1200 , b ) 1750 , c ) 1050 , d ) 1570 , e ) 1005 | b | divide(1050, subtract(const_1, divide(40, const_100))) | after decreasing 40 % in the price of an article costs rs . 1050 . find the actual cost of an article ? | "cp * ( 60 / 100 ) = 1050 cp = 17.5 * 100 = > cp = 1750 answer : b" | a = 40 / 100
b = 1 - a
c = 1050 / b
|
a ) 1 / 6 , b ) 7 / 216 , c ) 4 / 36 , d ) 9 / 216 , e ) 11 / 216 | c | divide(subtract(5, const_1), power(6, 2)) | a fair sided dice labeled 1 to 6 is tossed 2 times . what is the probability the sum of the 2 throws is 5 ? | the total possible number of combination if the dice is thrown 2 times is 36 . there are 4 possible outcomes that the sum of the number is 5 which is ( 2 + 3 ) ( 3 + 2 ) ( 1 + 4 ) and ( 4 + 1 ) answer : c | a = 5 - 1
b = 6 ** 2
c = a / b
|
a ) 21 % , b ) 25 % , c ) 69 % , d ) 35 % , e ) 19 % | d | subtract(subtract(add(50, const_100), divide(multiply(add(50, const_100), 10), const_100)), const_100) | a merchant marks his goods up by 50 % and then offers a discount of 10 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 150 after a 50 % markup . now a discount of 10 % on 150 . profit = 135 - 100 35 % answer d" | a = 50 + 100
b = 50 + 100
c = b * 10
d = c / 100
e = a - d
f = e - 100
|
a ) 337.33 , b ) 280.33 , c ) 299.33 , d ) 268.33 , e ) 289.33 | d | divide(multiply(460, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | i bought two books ; for rs . 460 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 460 - x ) 119 / 100 x = 280 answer : d" | a = 100 + 19
b = 460 * a
c = 100 - 15
d = 100 + 19
e = c + d
f = b / e
|
a ) 15 , b ) 20 , c ) 40 , d ) 44 , e ) 54 | a | divide(rectangle_area(multiply(6, const_100), multiply(75, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | what is the least number of squares tiles required to pave the floor of a room 6 m 75 cm long and 4 m 5 cm broad ? | "length of largest tile = h . c . f . of 675 cm and 405 cm = 135 cm . area of each tile = ( 135 x 135 ) cm 2 . required number of tiles = 675 x 405 / ( 135 ^ 2 ) = 15 answer : a" | a = 6 * 100
b = 75 * 100
c = rectangle_area / (
|
a ) 25 years , b ) 28 years , c ) 10 years , d ) 20 years , e ) 21 years | d | divide(const_100, 5) | in how many years will a sum of money doubles itself at 5 % per annum simple interest | "let the initial sum be x then , amount = 2 x ( because sum will be double ) amount = principal + s . i 2 x = x + s . i s . i = x s . i = ( p * r * t ) / 100 x = ( x * 5 * t ) / 100 therefore , t = 20 years answer : d" | a = 100 / 5
|
a ) 500 , b ) 625 , c ) 750 , d ) 550 , e ) 600 | b | add(divide(subtract(subtract(50, 1), add(1, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of odd all the integers between 1 and 50 ? | "numbers are 1 , 3,5 - - - - - - - , 47,49 average of the set : ( largest + smallest ) / 2 = ( 49 + 1 ) / 2 = 25 ; # of terms : ( largest - smallest ) / 2 + 1 = ( 49 - 1 ) / 2 + 1 = 25 the sum = 25 * 25 = 625 answer : b" | a = 50 - 1
b = 1 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | d | multiply(subtract(divide(power(41, const_2), 357), floor(divide(power(41, const_2), 357))), 357) | on dividing a number by 357 , we get 41 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 41 = ( 17 * 21 * y ) + ( 17 * 2 ) + 7 = 17 * ( 21 y + 2 ) + 7 . required number = 7 . answer is d" | a = 41 ** 2
b = a / 357
c = 41 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
|
a ) 8 , b ) 12 , c ) 15 , d ) 48 , e ) 24 | d | divide(multiply(12, divide(add(6, 12), const_2.0)), 12) | machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 12 hours . how many hours does it take machines x and y , working simultaneously , to produce 12 k bottles ? | "x rate = k / 6 y rate = k / 12 k / 6 + k / 12 = 12 k / t solving t = 48 answer d" | a = 6 + 12
b = a / 2
c = 12 * b
d = c / 12
|
a ) 3 / 16,000 , b ) 1 / 8,100 , c ) 1 / 600 , d ) 1 / 90 , e ) 2 / 45 | c | multiply(divide(80, 800), divide(const_1, 600)) | a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 80 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ? | "solution provided by stanford 2012 is correct : 80 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 80 / 800 * 1 / 600 = 1 / 6,000 . answer : c ." | a = 80 / 800
b = 1 / 600
c = a * b
|
a ) 1 / 140 , b ) 11 / 12 , c ) 14 / 8 , d ) 13 / 12 , e ) 57 / 120 | d | add(add(divide(1, 2), divide(1, 4)), divide(1, 3)) | in a race where 8 cars are running , the chance that car x will win is 1 / 2 , that y will win is 1 / 4 and that z will win is 1 / 3 . assuming that a dead heat is impossible , find the chance that one of them will win . | required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 2 + 1 / 4 + 1 / 3 = 13 / 12 answer : d | a = 1 / 2
b = 1 / 4
c = a + b
d = 1 / 3
e = c + d
|
a ) $ 1220 , b ) $ 1160 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 4), 30), multiply(subtract(100, 30), 10)) | at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 4 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . if those visits were the only source of revenue for the salon , what was the revenue for the last calendar month at the salon ? | "i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 4 ) = $ 160 total revenue : 1000 + 160 = $ 1160 the answer is b ." | a = 10 + 4
b = a * 30
c = 100 - 30
d = c * 10
e = b + d
|
a ) 10 , b ) 12 , c ) 8 , d ) 15 , e ) 18 | b | divide(multiply(multiply(3, 8), 2), 4) | 3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day should 4 pumps work in order to empty the tank in 1 day ? | "explanation : let the required hours needed be x more pumps , less hours ( indirect proportion ) more days , less hours ( indirect proportion ) hence we can write as ( pumps ) 3 : 4 ( days ) 2 : 1 } : : x : 8 β 3 Γ 2 Γ 8 = 4 Γ 1 Γ x β 3 Γ 2 Γ 2 = x β x = 12 answer : option b" | a = 3 * 8
b = a * 2
c = b / 4
|
a ) 9 : 7 , b ) 7 : 12 , c ) 12 : 7 , d ) 5 : 6 , e ) 7 : 9 | b | divide(divide(subtract(42, 35), subtract(54, 35)), subtract(const_1, divide(subtract(42, 35), subtract(54, 35)))) | in what ratio must rice of rs . 54 per kg be mixed with rice of rs . 35 per kg so that cost of mixture is rs . 42 per kg ? | ( 42 - 35 ) / ( 54 - 42 ) = 7 / 12 = 7 : 12 answer : b | a = 42 - 35
b = 54 - 35
c = a / b
d = 42 - 35
e = 54 - 35
f = d / e
g = 1 - f
h = c / g
|
a ) 48 , b ) 54 , c ) 72 , d ) 84 , e ) 27 | c | divide(multiply(264, const_3), add(const_10, const_1)) | the sum of the numbers is 264 . if the first number be twice the second and third number be one - third of the first , then the second number is : | let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 264 11 x / 3 = 264 x = 72 answer : c | a = 264 * 3
b = 10 + 1
c = a / b
|
a ) rs . 114.80 , b ) rs . 86.70 , c ) rs . 55.40 , d ) rs . 51.60 , e ) none | d | subtract(multiply(subtract(83.20, 64), 86), subtract(83.20, 64)) | if rs . 64 amount to rs . 83.20 in 2 years , what will rs . 86 amount to in 4 years at the same rate percent per annum ? | "solution p = rs . 64 , s . i . = rs . ( 83.20 - 64 ) = rs . 19.20 , t = 2 years . so , rate = ( 100 x 19.20 / 64 x 2 ) = 15 % . now , p = rs . 86 , r = 15 % , t = 4 years . β΄ s . i . = rs . ( 86 x 15 x 4 / 100 ) rs . 51.60 . answer d" | a = 83 - 20
b = a * 86
c = 83 - 20
d = b - c
|
a ) 243 cm 2 , b ) 239 cm 2 , c ) 267 cm 2 , d ) 230 cm 2 , e ) 228 cm 2 | e | divide(multiply(12, add(20, 18)), const_2) | nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 12 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 12 ) = 228 cm 2 answer : e" | a = 20 + 18
b = 12 * a
c = b / 2
|
a ) 6 ^ 36 , b ) 6 ^ 38 , c ) 6 ^ 72 , d ) 6 ^ 216 , e ) 6 ^ 432 | b | multiply(power(6, 36), 36) | if x = 6 ^ 36 and x ^ x = 6 ^ k , what is k ? | "solution : we know that x = 6 ^ 36 which implies x ^ x = ( 6 ^ 36 ) ^ ( 6 ^ 36 ) = 6 ^ ( 36 * 6 ^ 36 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 6 ^ ( 6 ^ 2 * 6 ^ 36 ) = 6 ^ ( 6 ^ ( 2 + 36 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 6 ^ ( 6 ^ 38 ) given that x ^ x = 6 ^ k so 6 ^ ( 6 ^ 38 ) = 6 ^ k since the base is same the exponent will also be same therefore k = 6 ^ 38 answer : b" | a = 6 ** 36
b = a * 36
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.