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 ) 276 , b ) 289 , c ) 480 , d ) 400 , e ) 278 | c | multiply(divide(4000, add(4000, add(multiply(3000, 2), multiply(2000, 2)))), 4000) | a , b and c invests rs . 4000 , rs . 3000 and rs . 2000 in a business . after one year b removed his money ; a and c continued the business for one more year . if the net profit after 2 years be rs . 2400 , then b ' s share in the profit is ? | "4 * 24 : 3 * 12 : 2 * 24 8 : 3 : 4 3 / 15 * 2400 = 480 answer : c" | a = 3000 * 2
b = 2000 * 2
c = a + b
d = 4000 + c
e = 4000 / d
f = e * 4000
|
a ) 10 hr , b ) 14 hr , c ) 12 hr , d ) 9 hr , e ) 15 hr | b | inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 3), 1), 3)))) | a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 3 hours to fill the tank . the leak can drain all the water in ? | "work done by the leak in 1 hour = 1 / 2 - 3 / 7 = 1 / 14 leak will empty the tank in 14 hrs answer is b" | a = 1 / 2
b = 2 * 3
c = b + 1
d = c / 3
e = 1/(d)
f = a - e
g = 1/(f)
|
a ) rs . 32 , b ) rs . 28 , c ) rs . 36 , d ) rs . 25 , e ) none of these | e | multiply(multiply(16, const_4), divide(352, add(16, 4))) | the cost of 16 pens and 8 pencils is rs . 352 and the cost of 4 pens and 4 pencils is rs . 96 . find the cost of each pen ? | "let the cost of each pen and pencil be ' p ' and ' q ' respectively . 16 p + 8 q = 352 - - - ( 1 ) 4 p + 4 q = 96 8 p + 8 q = 192 - - - ( 2 ) ( 1 ) - ( 2 ) = > 8 p = 160 = > p = 20 answer : e" | a = 16 * 4
b = 16 + 4
c = 352 / b
d = a * c
|
a ) 80 , b ) 100 , c ) 150 , d ) 180 , e ) 220 | e | multiply(divide(const_60, 15), 55) | if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 55 minutes ? | "since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 55 minutes the population increases by 55 x 4 = 220 people . answer . e ." | a = const_60 / 15
b = a * 55
|
a ) 1 / 32 , b ) 1 / 1296 , c ) 1 / 33 , d ) 1 / 38 , e ) 1 / 34 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | five 6 faced dice are thrown together . the probability that all the three show the same number on them is ? | the three dice can fall in 6 * 6 * 6 * 6 * 6 = 7776 ways . hence the probability is 6 / 7776 = 1 / 1296 answer : b | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 11 . | e | add(multiply(21, divide(const_1, 1)), const_1) | in the junior basketball league there are 21 teams , 1 / 3 of them are bad and Β½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 21 bad teams = ( 1 / 3 ) * 21 = 7 rich teams = 10 so maximum value that the both rich and bad can take will be 10 . so e = 11 can not be that value . ans e ." | a = 1 / 1
b = 21 * a
c = b + 1
|
a ) 500 , b ) 600 , c ) 750 , d ) 800 , e ) none of them | b | multiply(divide(const_1, add(add(const_4, 3), const_1)), 1200) | a and b undertake to do a piece of work for rs . 1200 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of a . | "c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a β s share = rs . ( 1200 * 4 / 8 ) = rs . 600 , b ' s share = rs . ( 1200 * 3 / 8 ) = rs . 450 c ' s share = rs . [ 1200 - ( 300 + 225 Β» ) = rs . 150 . answer is b" | a = 4 + 3
b = a + 1
c = 1 / b
d = c * 1200
|
a ) 56 days , b ) 420 days , c ) 46 days , d ) 560 days , e ) 96 days | a | add(inverse(multiply(60, const_2)), inverse(multiply(35, const_3))) | a can do a half of certain work in 60 days and b one third of the same in 35 days . they together will do the whole work in . | "a = 120 days b = 105 days 1 / 120 + 1 / 105 = 45 / 2520 = 1 / 56 = > 56 days answer : a" | a = 60 * 2
b = 1/(a)
c = 35 * 3
d = 1/(c)
e = b + d
|
a ) 6 , b ) 4 , c ) 0 , d ) 8 , e ) 7 | b | subtract(add(15, power(add(const_1, const_4), 2)), multiply(12, 3)) | if p is a prime number greater than 3 , find the remainder when p ^ 2 + 15 is divided by 12 . | "every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 15 = 36 n ^ 2 + 12 n + 1 + 15 = 36 n ^ 2 + 12 n + 12 + 4 if p = 6 n - 1 , then p ^ 2 + 15 = 36 n ^ 2 - 12 n + 1 + 15 = 36 n ^ 2 - 12 n + 12 + 4 when divided by 12 , it must leave a remainder of 4 . the answer is b ." | a = 1 + 4
b = a ** 2
c = 15 + b
d = 12 * 3
e = c - d
|
a ) 8 % , b ) 14 % , c ) 26 % , d ) 15 % , e ) 17 % | c | subtract(subtract(add(40, const_100), divide(multiply(add(40, const_100), 10), const_100)), const_100) | a merchant marks his goods up by 40 % 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 140 after a 40 % markup . now a discount of 10 % on 126 . profit = 126 - 100 26 % answer c" | a = 40 + 100
b = 40 + 100
c = b * 10
d = c / 100
e = a - d
f = e - 100
|
a ) 0 and 3 , b ) 3 and 4 , c ) 4 and 5 , d ) 5 and 7 , e ) 7 and 9 | c | add(multiply(floor(power(100, inverse(3))), const_10), add(floor(power(100, inverse(3))), const_1)) | if a and b are positive numbers , and a ^ 3 + b ^ 3 = 100 , then the greatest possible value of a is between : | "if a = 4.5 and b is a bit more than 2 , then a ^ 3 + b ^ 3 can equal 100 . if a > 5 , then a ^ 3 + b ^ 3 > 100 . the answer is c ." | a = 1/(3)
b = 100 ** a
c = math.floor(b)
d = c * 10
e = 1/(3)
f = 100 ** e
g = math.floor(f)
h = g + 1
i = d + h
|
a ) 9 , b ) 31 , c ) 56 , d ) 72 , e ) 90 | c | subtract(subtract(multiply(18, 15), multiply(5, 14)), multiply(9, 16)) | the average age of 18 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ? | age of the 15 th student = 18 * 15 - ( 14 * 5 + 16 * 9 ) = 270 - 214 = 56 years answer is c | a = 18 * 15
b = 5 * 14
c = a - b
d = 9 * 16
e = c - d
|
a ) $ 16,200 , b ) $ 5,600 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 22,500 | a | multiply(1800, power(const_3, divide(28, divide(112, 8)))) | money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 1800 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 1800 triples to $ 5400 , in 28 years , it will again triple to $ 5400 * 3 = $ 16,200 answer a" | a = 112 / 8
b = 28 / a
c = 3 ** b
d = 1800 * c
|
a ) s . 240 , b ) s . 140 , c ) s . 340 , d ) s . 50 , e ) s . 90 | b | subtract(divide(divide(700, add(divide(1, 2), divide(1, 3))), 2), divide(divide(700, add(divide(1, 2), divide(1, 3))), 3)) | a profit of rs . 700 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 700 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 300 and 200 . difference in profit share = 420 - 280 = 140 answer : b" | a = 1 / 2
b = 1 / 3
c = a + b
d = 700 / c
e = d / 2
f = 1 / 2
g = 1 / 3
h = f + g
i = 700 / h
j = i / 3
k = e - j
|
a ) 28.5 , b ) 27.675 , c ) 30 , d ) data inadequate , e ) none of these | c | subtract(divide(multiply(add(const_100, 23.5), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 23.5 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 Γ ( 100 + 23.5 ) / ( 100 β 5 ) = 100 Γ 247 / 190 = 130 hence , required % profit = 130 β 100 = 30 % answer c" | a = 100 + 23
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 2 : 7 , b ) 2 : 6 , c ) 17 : 6 , d ) 2 : 3 , e ) 7 : 8 | d | divide(10000, 15000) | p and q started a business investing rs 10000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 10000 : 15000 = 10 : 15 = 2 : 3 option d" | a = 10000 / 15000
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a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.11111 | e | multiply(divide(0.99999, 0.11112), const_100) | 0.99999 + 0.11112 = ? | "0.99999 + 0.11112 = 0.99999 + 0.11111 + 0.00001 = ( 0.99999 + 0.00001 ) + 0.11111 = 1 + 0.11111 = 1.11111 e" | a = 0 / 99999
b = a * 100
|
a ) 11.2 β 2 , b ) 13 β 2 , c ) 23 β 2 , d ) 12 β 4 , e ) 13 β 9 | a | sqrt(multiply(add(power(divide(40, const_4), const_2), power(divide(20, const_4), const_2)), const_2)) | the perimeter of one square is 40 cm and that of another is 20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | "4 a = 40 4 a = 20 a = 10 a = 5 a 2 = 100 a 2 = 25 combined area = a 2 = 125 = > a = 11.2 d = 11.2 β 2 answer : a" | a = 40 / 4
b = a ** 2
c = 20 / 4
d = c ** 2
e = b + d
f = e * 2
g = math.sqrt(f)
|
a ) s . 650 , b ) s . 690 , c ) s . 680 , d ) s . 700 , e ) s . 720 | c | subtract(815, divide(multiply(subtract(860, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 860 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 860 - 815 ) = rs . 45 . s . i . for 3 years = rs . ( 45 x 3 ) = rs . 135 . principal = rs . ( 815 - 135 ) = rs . 680 . answer : option c" | a = 860 - 815
b = a * 3
c = b / 4
d = 815 - c
|
a ) s . 650 , b ) s . 638 , c ) s . 698 , d ) s . 700 , e ) s . 760 | b | subtract(815, divide(multiply(subtract(874, 815), 3), 4)) | a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 874 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 874 - 815 ) = rs . 59 . s . i . for 3 years = rs . ( 59 x 3 ) = rs . 177 . principal = rs . ( 815 - 177 ) = rs . 638 . answer : option b" | a = 874 - 815
b = a * 3
c = b / 4
d = 815 - c
|
a ) 7 / 98 , b ) 1 / 48 , c ) 1 / 98 , d ) 1 / 96 , e ) 3 / 42 | e | inverse(multiply(2, 7)) | the compound ratio of 2 / 3 , 6 / 7 , 1 / 3 and 3 / 8 is given by ? | 2 / 3 * 6 / 7 * 1 / 3 * 3 / 8 = 36 / 504 = 3 / 42 answer : e | a = 2 * 7
b = 1/(a)
|
a ) 3.5 , 8.5 , b ) 7 , 5 , c ) 7.5 , 4.5 , d ) 8 , 4 , e ) 8 , 5 | c | floor(subtract(12, divide(subtract(multiply(12, 15), multiply(12, 12)), subtract(20, 12)))) | find the quantities of two types of rice to be mixed , first variety is of rs . 12 per kg and second variety is of rs . 20 per kg . to get 12 kg of rice mixture worth rs . 15 per kg ? ( in kg ) | from allegation & mixture ratio = c - m / m - d = 20 - 15 / 15 - 12 = 5 / 3 . from the option c r = 7.5 / 4.5 = 5 / 3 answer : c | a = 12 * 15
b = 12 * 12
c = a - b
d = 20 - 12
e = c / d
f = 12 - e
g = math.floor(f)
|
a ) 5 , b ) 1 , c ) 3 , d ) 6 , e ) 7 | d | add(choose(3, 2), choose(3, 2)) | there are 3 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair . what is the number of pair that can be formed . | "first of all you should remember that there is a difference in left and right sock . now no . of way to select any of the sock = 3 and for second = 2 so total methods = 3 * 2 = 6 answer : d" | a = math.comb(3, 2)
b = math.comb(3, 2)
c = a + b
|
a ) 50 km , b ) 45 km , c ) 33 km , d ) 60 km , e ) 54 km | d | multiply(add(divide(add(multiply(2, 10), 20), subtract(20, 10)), 2), 10) | a boy is traveling from his house to school at 10 km / hr and reached school 2 hours late . next day he traveled 20 km / hr and reached 1 hour early . then find the distance between house and school ? | let distance be x s 1 = 10 km / hr s 2 = 20 km / hr t 1 = x / 10 hr t 2 = x / 20 hr difference in time = 2 + 1 = 3 hr ( x / 10 ) - ( x / 20 ) = 3 x = 60 km answer is d | a = 2 * 10
b = a + 20
c = 20 - 10
d = b / c
e = d + 2
f = e * 10
|
a ) rs . 200 , b ) rs . 350 , c ) rs . 275 , d ) rs . 415 , e ) none of these | d | divide(subtract(multiply(30, 350), multiply(15, 285)), 15) | the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 285 , then the mean profit for the last 15 days would be | "average would be : 350 = ( 285 + x ) / 2 on solving , x = 415 . answer : d" | a = 30 * 350
b = 15 * 285
c = a - b
d = c / 15
|
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(divide(subtract(7.20, multiply(0.80, const_2)), 0.50), const_2) | a certain fruit stand sold apples for $ 0.80 each and bananas for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 7.20 , what total number of apples and bananas did the customer purchase ? | "let ' s start with 1 apple for $ 0.80 . let ' s subtract $ 0.80 from $ 7.20 until we get a multiple of $ 0.50 . $ 7.20 , $ 6.40 , $ 5.60 , $ 4.80 , $ 4.00 = 8 * $ 0.50 the customer purchased 8 bananas and 4 apples . the answer is c ." | a = 0 * 80
b = 7 - 20
c = b / 0
d = c + 2
|
a ) 6.25 , b ) 8.42 , c ) 8.3 , d ) 8.1 , e ) 6.21 | b | divide(subtract(282, multiply(7, 4.2)), 30) | in the first 7 overs of a cricket game , the run rate was only 4.2 . what should be the rate in the remaining 30 overs to reach the target of 282 runs ? | "required run rate = [ 282 - ( 4.2 * 7 ) ] / 30 = 252.60 / 40 = 8.42 answer : b" | a = 7 * 4
b = 282 - a
c = b / 30
|
a ) 51.6 , b ) 51.5 , c ) 51.7 , d ) 51.1 , e ) 51.0 | b | divide(subtract(add(add(30, 35), 42), const_2), const_2) | an empty tank be filled with an inlet pipe β a β in 42 minutes . after 12 minutes an outlet pipe β b β is opened which can empty the tank in 30 minutes . after 6 minutes another inlet pipe β c β opened into the same tank , which can fill the tank in 35 minutes and the tank is filled . find the time taken to fill the tank ? | assume total tank capacity = 210 liters now capacity of pipe a = 210 / 42 = 5 liters capacity of b = 210 / 30 = - 7 liters capacity of c = 210 / 35 = 6 min assume tank gets filled in x min after the third pipe got opened . so x Γ 5 + 6 Γ ( β 2 ) + 4 x = 210 x Γ 5 + 6 Γ ( β 2 ) + 4 x = 210 β 48 + 4 x = 210 β 4 x = 162 β x = 40.5 β 48 + 4 x = 210 β 4 x = 162 β x = 40.5 total time taken to fill the tank = 40.5 + 12 + 6 = 51.5 answer : b | a = 30 + 35
b = a + 42
c = b - 2
d = c / 2
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | d | subtract(50, subtract(add(25, 23), 21)) | in a group of 50 people , 25 have visited iceland and 23 have visited norway . if 21 people have visited both iceland and norway , how many people have visited neither country ? | "this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 50 = 25 + 23 - 21 + ( # in neither group ) = 23 the prompt gives you all of the numbers you need to get to the correct answer . just plug in and solve . d" | a = 25 + 23
b = a - 21
c = 50 - b
|
a ) 4 / 3 , b ) 5 / 3 , c ) 7 / 3 , d ) 8 / 3 , e ) 10 / 3 | d | subtract(divide(4, divide(3, 5)), 4) | a container holds 4 quarts of alcohol and 4 quarts of water . how many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume ? | let the number of quarts that should be added to get the required ratio = x so total quarts of water = ( x + 4 ) but the original number of quarts of alcohol remains the same , so we have : 4 / ( x + 4 ) = 3 / 5 cross - multiply : now we have 20 = 3 x + 12 = > 3 x = 20 - 12 = > x = 8 / 3 answer - d | a = 3 / 5
b = 4 / a
c = b - 4
|
a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 26 | b | subtract(37, add(add(8, const_2), 8)) | set a of 8 positive integers may have the same element and have 37 . and set b of 8 positive integers must have different elements and have 37 . when m and n are the greatest possible differences between 37 and other elements β sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 37 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 30 and 37 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 9 . so , 30 - 9 = 21 . the correct answer is b . | a = 8 + 2
b = a + 8
c = 37 - b
|
a ) 2 , b ) 4 , c ) 7 , d ) 14 , e ) 56 | b | lcm(6, 8) | if 6 and 8 are factors of 60 n , what is the minimum value of n ? | "60 n / 6 * 8 should be integer = > 2 * 2 * 3 * 5 * n / 2 * 3 * 2 * 2 * 2 = 5 * n / 4 must be an integer for this to be true n must multiple of 4 , thus min of n = 4 hence b" | a = math.lcm(6, 8)
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a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3 | b | subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5))) | an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how many meters q long ? | "let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m = q , leaving 1 m of empty space . thus , i believe the answer is b ) 1" | a = 4 * 2
b = 4 * 1
c = a + b
d = 15 - c
|
a ) 20 , b ) 25 , c ) 30 , d ) 32 , e ) 42 | a | multiply(8, const_2) | if the average of 8 x and 8 y is greater than 120 , and x is twice y , what is the least integer value of x ? | "substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 x / 2 = 4 x now average of 8 x and 4 x is greater than 120 . average of 8 x and 4 x is 6 x . so , 6 x > 120 x > 20 answer : a" | a = 8 * 2
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a ) 179 , b ) 202 , c ) 210 , d ) 223 , e ) 229 | b | divide(add(150, subtract(multiply(61, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 61 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 61 * 46 = 2806 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2806 β 2552 = 254 runs . let the highest score be x , hence the lowest score = x β 150 x + ( x - 150 ) = 254 2 x = 404 x = 202 runs answer b" | a = 61 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
|
a ) 377 , b ) 126 , c ) 111 , d ) 727 , e ) 121 | c | sqrt(add(power(sqrt(subtract(37, multiply(const_2, 4107))), const_2), multiply(const_4, 4107))) | the product of two numbers is 4107 . if the h . c . f of these numbers is 37 , then the greater number is ? | "let the numbers be 37 a and 37 b . then , 37 a * 37 b = 4107 = > ab = 3 now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 * 1 , 37 * 3 ) i . e . , ( 1 , 111 ) . greater number = 111 . answer : c" | a = 2 * 4107
b = 37 - a
c = math.sqrt(b)
d = c ** 2
e = 4 * 4107
f = d + e
g = math.sqrt(f)
|
a ) 6 / Ο , b ) 9 / Ο , c ) 6 , d ) 9 , e ) 12 | d | add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi))) | the surface area of a sphere is 4 Ο r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area t of that hemisphere ? | "given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 / pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 / pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 / 2 = 6 ( curved part , not including the flat rounded base ) . but the total surface area = 6 + area of the base of a hemisphere . = 6 + 3 = 9 . answer is d ! !" | a = 4 * math.pi
b = 3 / math.pi
c = a * b
d = c / 2
e = 3 / math.pi
f = math.pi * e
g = d + f
|
a ) 26 , b ) 32 , c ) 35 , d ) 30 , e ) 45 | a | divide(add(408, 216), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | there are 408 boys and 216 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | "explanation : hcf ( 408 , 216 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 216 / 24 = 17 + 9 = 26 answer : a" | a = 408 + 216
b = 2 * 2
c = b * 2
d = c * 3
e = a / d
|
['a ) 18', 'b ) 22', 'c ) 21', 'd ) 20', 'e ) 19'] | b | multiply(const_2, sqrt(divide(250, const_2))) | the number 250 can be written as sum of the squares of 3 different positive integers . what is the sum of these 3 different integers ? | sum of the squares of 3 different positive integers = 250 15 ^ 2 + 3 ^ 2 + 4 ^ 2 = 250 now , sum of these 3 different integers = 15 + 3 + 4 = 22 ans - b | a = 250 / 2
b = math.sqrt(a)
c = 2 * b
|
a ) 7.33 % , b ) 9 % , c ) 9.67 % , d ) 11 % , e ) 11.5 % | a | multiply(const_100, divide(multiply(divide(16, const_100), 11), add(13, 11))) | 13 ltr of water is added with 11 ltr of a solution containing 16 % of alcohol in the water . the % of alcohol in the new mixture is ? | "we have an 11 litre solution containing 16 % of alcohol in the water . = > quantity of alcohol in the solution = 11 Γ£ β 16 / 100 now 13 litre of water is added to the solution . = > total quantity of the new solution = 11 + 13 = 24 percentage of alcohol in the new solution = 11 Γ£ β 16 / 100 24 Γ£ β 100 = 11 Γ£ β 1610024 Γ£ β 100 = 11 Γ£ β 0.67 / 100 = 7.33 % a" | a = 16 / 100
b = a * 11
c = 13 + 11
d = b / c
e = 100 * d
|
a ) 0.5 , b ) 0.51 , c ) - 0.52 , d ) 0.31 , e ) - 0.49 | c | divide(subtract(12, 49), 71) | a straight line in the xy - plane has y - intercept of 49 . on this line the x - coordinate of the point is 71 and y - coordinate is 12 then what is the slope of the line ? | "eq of line = y = mx + c c = 49 x = 71 y = 12 substitute given : m = ( y - c ) / x = ( 12 - 49 ) / 71 = - 37 / 71 = - 0.52 correct option is c" | a = 12 - 49
b = a / 71
|
a ) 36 , b ) 57 , c ) 44 , d ) 48 , e ) 52 | b | add(40, divide(subtract(1116, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1116 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 1116 - 640 = 476 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 476 / 28 = 17 total hrs = 40 + 17 = 57 answer b 57" | a = 16 * 40
b = 1116 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 271 , b ) 250 , c ) 350 , d ) 277 , e ) 232 | b | subtract(multiply(20, multiply(126, const_0_2778)), 450) | a train 450 m long running at 126 kmph crosses a platform in 20 sec . what is the length of the platform ? | "length of the platform = 126 * 5 / 18 * 20 = 700 β 450 = 250 answer : b" | a = 126 * const_0_2778
b = 20 * a
c = b - 450
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a ) 200 , b ) 210 , c ) 250 , d ) 190 , e ) 220 | c | add(200, multiply(200, divide(25, const_100))) | 200 is increased by 25 % . find the final number . | explanation final number = initial number + 25 % ( original number ) = 200 + 25 % ( 200 ) = 200 + 50 = 250 . answer c | a = 25 / 100
b = 200 * a
c = 200 + b
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a ) 1.9332 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none | a | multiply(divide(268, const_100), divide(74, const_100)) | given that 268 x 74 = 19332 , find the value of 2.68 x . 74 . | "solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 Γ . 74 = 1.9332 answer a" | a = 268 / 100
b = 74 / 100
c = a * b
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a ) 21 % , b ) 22.5 % , c ) 25 % , d ) 30 % , e ) 40 % | d | multiply(divide(subtract(add(add(100, 30), 100), multiply(2, 100)), 100), 100) | a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 30 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ? | "1 st year income = i 1 st year savings = s 1 st year expense = e 1 2 nd year income = 1.3 i 2 nd year savings = 2 s ( 100 % increase ) 2 nd year expense = e 2 e 1 + e 2 = 2 e 1 e 2 = e 1 that means expenses are same during both years . with increase of 30 % income the savings increased by 100 % . or s = . 3 i or s = 30 % of income d is the answer" | a = 100 + 30
b = a + 100
c = 2 * 100
d = b - c
e = d / 100
f = e * 100
|
a ) 8 , b ) 14 , c ) 15 , d ) 17 , e ) 30 | d | add(multiply(8, const_2), const_1) | the average age of applicants for a new job is 31 , with a standard deviation of 8 . the hiring manager is only willing to accept applications whose age is within one standard deviation of the average age . what is the maximum number of different ages of the applicants ? | "within one standard deviation of the average age means 31 + / - 7 23 - - 31 - - 39 number of dif . ages - 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 total = 17 d" | a = 8 * 2
b = a + 1
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a ) 12 , b ) 18 , c ) 24 , d ) 30 , e ) 36 | e | multiply(12, multiply(12, divide(4, multiply(4, 4)))) | 4 weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 12 weavers in 12 days ? | "1 weaver can weave 1 mat in 4 days . 12 weavers can weave 12 mats in 4 days . 12 weavers can weave 36 mats in 12 days . the answer is e ." | a = 4 * 4
b = 4 / a
c = 12 * b
d = 12 * c
|
a ) 420 , b ) 520 , c ) 620 , d ) 620 , e ) none of these | b | add(200, multiply(divide(200, 5), 8)) | in a college , the ratio of the number of boys to girls is 8 : 5 . if there are 200 girls , the total number of students in the college is | explanation : let the boy are 8 x and girls are 5 x = > 5 x = 200 = > x = 40 total students = 8 x + 5 x = 13 x = 13 ( 40 ) = 520 answer : b | a = 200 / 5
b = a * 8
c = 200 + b
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a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio q of milk and water in the resultant mixture ? | he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } Β = ( 1 / 2 ) ^ 2 Β = 1 / 4 therefore q , w _ r Β = { w _ o } / 4 = 8 / 4 Β = 2 litres . b | a = 3 + 2
b = 3 / a
c = 20 - 10
d = b * c
e = d + 10
f = e / 20
g = 20 - 10
h = f * g
i = h + 10
j = 3 + 2
k = 2 / j
l = 20 - 10
m = k * l
n = m / 20
o = 20 - 10
p = n * o
q = i / p
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a ) 6 hr , b ) 2 hr , c ) 4 hr , d ) 1 hr , e ) none of these | b | divide(8, subtract(subtract(divide(96, 8), 4), 4)) | a boatman can row 96 km downstream in 8 hr . if the speed of the current is 4 km / hr , then find in what time will be able to cover 8 km upstream ? | "explanation : speed downstream = 96 β 8 = 12 kmph speed of current = 4 km / hr speed of the boatman in still water = 12 - 4 = 8 kmph speed upstream = 8 - 4 = 4 kmph time taken to cover 8 km upstream = 8 β 4 = 2 hours . answer : option b" | a = 96 / 8
b = a - 4
c = b - 4
d = 8 / c
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a ) 9 , b ) 11 , c ) 15 , d ) 12 , e ) 26 | e | subtract(subtract(multiply(16, 15), multiply(5, 14)), multiply(9, 16)) | the average age of 16 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ? | "age of the 15 th student = 16 * 15 - ( 14 * 5 + 16 * 9 ) = 240 - 214 = 26 years answer is e" | a = 16 * 15
b = 5 * 14
c = a - b
d = 9 * 16
e = c - d
|
a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 45 | c | divide(50, add(divide(25, 60), divide(subtract(50, 25), 30))) | a driver goes on a trip of 50 kilometers , the first 25 kilometers at 60 kilometers per hour and the remaining distance at 30 kilometers per hour . what is the average speed of the entire trip in kilometers per hour ? | "the time for the first part of the trip was 25 / 60 = 5 / 12 hours . the time for the second part of the trip was 25 / 30 = 5 / 6 hours . the total time for the trip was 5 / 12 + 5 / 6 = 15 / 12 = 5 / 4 hours . the average speed for the trip was 50 / ( 5 / 4 ) = 40 kph the answer is c ." | a = 25 / 60
b = 50 - 25
c = b / 30
d = a + c
e = 50 / d
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | b | divide(multiply(const_12, log(2)), log(2)) | if 2 ^ ( 2 w ) = 8 ^ ( w β 2 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w β 2 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w β 2 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 6 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 6 w = 6 the answer is b ." | a = math.log(2)
b = 12 * a
c = math.log(2)
d = b / c
|
a ) 227.623 , b ) 224.777 , c ) 233.523 , d ) 414.637 , e ) none of these | a | subtract(895.7, divide(573.07, 95.007)) | 895.7 β 573.07 β 95.007 = ? | "solution given expression = 895.7 - ( 573.07 + 95.007 ) = 895.7 - 668.077 = 227.623 . answer a" | a = 573 / 7
b = 895 - 7
|
a ) $ 1.40 , b ) $ 2.50 , c ) $ 4.10 , d ) $ 4.70 , e ) $ 8.20 | a | divide(subtract(9.65, 6.85), const_2) | a train ride from two p to town q costs $ 6.85 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.65 . what is the cost of a bus ride from town p to town q ? | "let x be the cost of a bus ride . x + ( x + 685 ) = 965 2 x = 280 x = $ 1.40 the answer is a ." | a = 9 - 65
b = a / 2
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a ) 18 , b ) 91 , c ) 11 , d ) 17.4 , e ) 12 | d | subtract(divide(multiply(60, 50), const_100), divide(multiply(42, 30), const_100)) | how much 60 % of 50 is greater than 42 % of 30 ? | "( 60 / 100 ) * 50 β ( 42 / 100 ) * 30 30 - 12.6 = 17.4 answer : d" | a = 60 * 50
b = a / 100
c = 42 * 30
d = c / 100
e = b - d
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(const_1000, const_100), subtract(250, subtract(250, 150))) | matt gets a $ 1,000 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 250 , how many sales has matt made ? | let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1000 new average = ( xy + 1000 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1000 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1000 ) = ( xy + x + 150 y + 150 ) i . e . ( 850 ) = ( x + 150 y ) new commission = 250 = 150 + x i . e . x = 100 i . e . y = 5 new sales = y + 1 = 6 answer : option d | a = 1000 - 100
b = 250 - 150
c = 250 - b
d = a / c
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | multiply(divide(divide(divide(3, power(divide(3, const_10), const_0_33)), power(divide(3, const_10), const_0_33)), divide(divide(3, power(divide(3, const_10), const_0_33)), power(divide(3, const_10), const_0_33))), const_100) | what is the smallest positive integer x for which x 3 + 5 xx 3 + 5 x has a value more than 80 ? | "4 ^ 3 + 5 * 4 = 64 + 20 = 84 that ' s pretty close to 80 already so i can cancel d and e since they ' ll be more than 80 . b just needs to be checked just in case so . . . 3 ^ 3 + 5 * 3 = 27 + 15 = not close to 80 . therefore , 4 ( c ) is the smallest positive integer x for which x ^ 3 + 5 x has a value more than 80 . answer : c" | a = 3 / 10
b = a ** const_0_33
c = 3 / b
d = 3 / 10
e = d ** const_0_33
f = c / e
g = 3 / 10
h = g ** const_0_33
i = 3 / h
j = 3 / 10
k = j ** const_0_33
l = i / k
m = f / l
n = m * 100
|
a ) 12 , b ) 72 , c ) 48 , d ) 99 , e ) 11 | b | subtract(divide(multiply(1.75, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.25, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.75 km and 1.25 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.75 + 1.25 = 3 km = 3000 m . required time = 3000 * 3 / 125 = 72 sec . answer : b" | a = 1 * 75
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 25
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 4 , b ) 6 , c ) 7.2 , d ) 7.8 , e ) 9 | a | multiply(divide(20, const_100), 20) | uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 13 ounces of chocolate . how many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ? | answer is a . x / x + 36 = 1 / 5 x = 9 13 - 9 = 4 | a = 20 / 100
b = a * 20
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['a ) 23.12', 'b ) 24.12', 'c ) 25.12', 'd ) 26.12', 'e ) 27.12'] | c | multiply(add(divide(2, const_2), 3), multiply(const_2, const_pi)) | a cylinder is inserted in a sphere d / h = 2 / 3 . find the surface area of the cylinder ? | take r = 1 , h = 3 = 2 * 3.14 * r * r + 2 * 3.14 * r * h = 2 * 3.14 * r ( r + h ) = 2 * 3.14 * 1 ( 1 + 3 ) = 2 * 3.14 ( 4 ) = 25.12 answer : c | a = 2 / 2
b = a + 3
c = 2 * math.pi
d = b * c
|
['a ) 90,000', 'b ) 95,000', 'c ) 93,000', 'd ) 92,000', 'e ) 91,000'] | a | add(add(multiply(150, 150), multiply(150, 225)), multiply(225, 150)) | the measurements obtained for the interior dimensions of a rectangular box are 150 cm by 150 cm by 225 cm . if each of the three measurements has an error of at most 1 centimeter , which of the following is the closes maximum possible difference , in cubic centimeters , between the actual capacity of the box and the capacity computed using these measurements ? | the options are well spread so we can approximate . changing the length by 1 cm results in change of the volume by 1 * 150 * 225 = 33,750 cubic centimeters ; changing the width by 1 cm results in change of the volume by 150 * 1 * 225 = 33,750 cubic centimeters ; changing the height by 1 cm results in change of the volume by 150 * 150 * 1 = 22,500 cubic centimeters . so , approximate maximum possible difference is 33,750 + 33,750 + 22,500 = 90,000 cubic centimeters . answer : a | a = 150 * 150
b = 150 * 225
c = a + b
d = 225 * 150
e = c + d
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['a ) 1 : 1', 'b ) 1 : 2', 'c ) 2 : 3', 'd ) 2 : 1', 'e ) none of these'] | d | divide(rectangle_area(const_1, const_1), triangle_area(rectangle_area(const_1, const_1), rectangle_area(const_1, const_1))) | what will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ? | explanation : as far as questions of area or volume and surface area are concerned , it is all about formulas and very little logic . so its a sincere advice to get all formulas remembered before solving these questions . lets solve this , area of rectangle = l β b area of triangle = 1 / 2 l β b ratio = l β b : 1 / 2 l β b = 1 : 1 / 2 = 2 : 1 option d | a = rectangle_area / (
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a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | c | add(divide(add(power(const_3.0, 4), sqrt(add(power(power(4, 4), const_2), power(subtract(316, power(4, 4)), const_2)))), multiply(power(4, const_2), const_2)), subtract(divide(add(power(4, 4), sqrt(add(power(power(4, 4), const_2), power(subtract(316, power(4, 4)), const_2)))), multiply(power(4, const_2), const_2)), 4)) | if one positive integer is greater than another positive integer by 4 , and the difference of their cubes is 316 , what is their sum ? | "1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 7 ^ 3 = 343 the two numbers are 3 and 7 . the answer is c ." | a = 3 ** 0
b = 4 ** 4
c = b ** 2
d = 4 ** 4
e = 316 - d
f = e ** 2
g = c + f
h = math.sqrt(g)
i = a + h
j = 4 ** 2
k = j * 2
l = i / k
m = 4 ** 4
n = 4 ** 4
o = n ** 2
p = 4 ** 4
q = 316 - p
r = q ** 2
s = o + r
t = math.sqrt(s)
u = m + t
v = 4 ** 2
w = v * 2
x = u / w
y = x - 4
z = l + y
|
a ) 10 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | divide(subtract(multiply(10, subtract(40, 4)), multiply(10, 32)), 4) | the average age of an adult class is 40 years . 10 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 10 x 32 = ( y + 10 ) x 36 Γ’ β‘ β 40 y + 320 = 36 y + 360 Γ’ β‘ β 4 y = 40 Γ’ Λ Β΄ y = 10 a" | a = 40 - 4
b = 10 * a
c = 10 * 32
d = b - c
e = d / 4
|
a ) 1 , b ) 499 , c ) 500 , d ) 999 , e ) 10 | e | divide(1000, const_100) | a king has 1000 bottles of wine , a queen wants to kill the king and sends a servant to poison the wine . fortunately the king ' s guard ' s catch d servant after he has only poisoned one bottle and the guard do n ' t know which bottle is poisoned . furthermore it takes one month to have an effect , and there is an anniversary party coming up in 5 weeks time ! the king decides he will get some of the prisoners in his vast dungeons to drink the wine . how many minimum prisoners does it required to sample the wine to find out the poisoned bottle so that the guests will still be able to drink the rest of the wine at his anniversary party in 5 weeks time ? | since you need 10 digits to represent 1000 numbers in binary , you will need 10 prisoners . now the logic is this : every bottle has a unique number from 1 to 1000 and a unique binary representation from 0000000001 to 1111101000 . now you have 10 prisoners . each bottle will be tasted by the prisoners depending on the binary representation of the bottle . wherever there is 1 , the bottle will be tasted by that prisoner ( assuming units digit represents the 1 st prisoner , tens digit the 2 nd prisoner and so on . . . ) 1 st bottle is 0000000001 so it will be tasted by only the 1 st prisoner . 2 nd bottle is 0000000010 so it will be tasted by only the 2 nd prisoner . 3 rd bottle is 0000000011 so it will be tasted by the 1 st and 2 nd prisoners . and so on . . . 1000 th bottle is 1111101000 so it will be tasted by 4 th , 6 th , 7 th , 8 th , 9 th and 10 th prisoners . at the end of a month , we see which prisoners die . say , only the 3 rd prisoner dies . this means the poisoned bottle number is 0000000100 = 3 . say 1 st , 7 th and 9 th prisoners die . this means the poisoned bottle number is 0101000001 = 321 ans : e | a = 1000 / 100
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['a ) 120 cm', 'b ) 200 cm', 'c ) 88 cm', 'd ) 666 cm', 'e ) 776 cm'] | b | multiply(sqrt(divide(120, 30)), const_100) | 30 square stone slabs of equal size were needed to cover a floor area of 120 sq . m . find the length of each stone slab ? | area of each slab = 120 / 30 m 2 = 4 m 2 length of each slab β 4 = 2 m = 200 cm b | a = 120 / 30
b = math.sqrt(a)
c = b * 100
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a ) 14 , b ) 28 , c ) 36 , d ) 84 , e ) 252 | c | add(12, const_1) | if x and y are positive integers and 18 x = 12 y what is the least possible value of xy ? | "18 x = 12 y = > x / y = 2 / 3 = > 3 x = 2 y 3 ( 3 ) = 2 ( 3 ) = > x * y = 9 but it is not given 3 ( 6 ) = 2 ( 6 ) = > x * y = 36 c" | a = 12 + 1
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a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 6468 | e | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.90) | the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.90 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.90 ) = rs . 6468 . answer : e ) 6468" | a = 13 / 86
b = math.sqrt(a)
c = b * 100
d = circumface * (
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a ) 6 : 5 , b ) 7 : 5 , c ) 9 : 8 , d ) 8 : 9 , e ) 8 : 7 | c | divide(subtract(divide(9, add(7, 9)), divide(6, add(6, 6))), subtract(divide(6, add(6, 6)), divide(4, add(5, 4)))) | two containers of milk contain mixtures of water and milk in ratio 5 : 4 and 7 : 9 . in what ratio they should be mixed so that mixture is of 6 : 6 ratio ? | 6 / 6 - 5 / 4 = = > - 1 / 4 7 / 9 - 6 / 6 = = > - 2 / 9 now , - 1 / 4 = - 2 / 9 = = > 1 / 4 = 2 / 9 = = > 9 : 8 answer : c | a = 7 + 9
b = 9 / a
c = 6 + 6
d = 6 / c
e = b - d
f = 6 + 6
g = 6 / f
h = 5 + 4
i = 4 / h
j = g - i
k = e / j
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a ) 25 , b ) 36 , c ) 23 , d ) 49 , e ) 27 | c | subtract(negate(11), multiply(subtract(5, 7), divide(subtract(5, 7), subtract(3, 5)))) | 3 , 5 , 7 , 11 , 13 , 17 , 19 , ____ | "the sequence is a series of prime numbers , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 . . . . answer : c ." | a = negate - (
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a ) 6 minutes , b ) 5 minutes , c ) 7 minutes , d ) 3 minutes , e ) 4 minutes | a | divide(360, add(36, divide(36, divide(const_3, const_2)))) | earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 360 envelopes | "earl takes 1 min . for 36 envelopes . ellen takes 3 / 2 mins for the same . so ellen can stuff ( ( 36 ) / ( 3 / 2 ) ) in 1 min . i . e . , 24 envlpes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 360 envelopes they will take 360 / 60 mins . i . e . , 6 mins . answer : a" | a = 3 / 2
b = 36 / a
c = 36 + b
d = 360 / c
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a ) 31 , b ) 35 , c ) 30 , d ) 40 , e ) 28 | a | subtract(multiply(multiply(4, 4), 4), divide(subtract(14, const_1), const_2)) | if x / 4 years ago roger was 14 years old and x / 4 years from now he will be 4 x years old , how old will he be 4 x years from now ? | "assume the current age = a a - x / 4 = 14 ( i ) a + x / 4 = 4 x or a = 15 x / 4 ( ii ) putting the value of a from ( ii ) in ( i ) 15 x / 4 - x / 4 = 14 or 14 x / 4 = 14 therefore x = 4 and a = 15 4 x years from now , age will be 15 + 4 * 4 = 31 option a" | a = 4 * 4
b = a * 4
c = 14 - 1
d = c / 2
e = b - d
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a ) 100 , b ) 120 , c ) 200 , d ) 242 , e ) 250 | d | add(220, divide(multiply(220, 10), const_100)) | the present population of a town is 220 . population increase rate is 10 % p . a . find the population of town after 1 years ? | "p = 220 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 220 * ( 1 + 10 / 100 ) = 220 * ( 11 / 10 ) = 242 answer is d" | a = 220 * 10
b = a / 100
c = 220 + b
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a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | if a = 105 and a ^ 3 = 21 Γ 25 Γ 45 Γ b , what is the value of b ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 Γ 25 Γ 45 Γ b = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x b = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x b = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x b b = 7 ^ 2 = 49 answer d" | a = 105 ** 3
b = 21 * 25
c = b * 45
d = a / c
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a ) 3 / 4 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2)) | if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 2 strokes ? | "left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 so removed = 1 - 1 / 4 = 3 / 4" | a = 1 / 2
b = 1 / 2
c = b / 2
d = a + c
e = 1 / 2
f = e / 2
g = f / 2
h = d + g
i = 1 / 2
j = i / 2
k = j / 2
l = k / 2
m = h + l
n = 1 / 2
o = n / 2
p = o / 2
q = p / 2
r = q / 2
s = m + r
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a ) a ) 300 , b ) b ) 350 , c ) c ) 450 , d ) d ) 470 , e ) e ) 500 | a | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(6, 150))) | a man purchased 3 blankets @ rs . 100 each , 6 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 . 150 . find the unknown rate of two blankets ? | "explanation : 10 * 150 = 1500 3 * 100 + 6 * 150 = 1200 1500 β 1200 = 300 a" | a = 10 * 150
b = 3 * 100
c = 6 * 150
d = b + c
e = a - d
|
a ) 2200 , b ) 2520 , c ) 2600 , d ) 3354 , e ) none of these | b | add(divide(270, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 270) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 270 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 270 . x Γ£ β 16 Γ£ β ( 9 / 12 ) Γ£ β ( 1 / 100 ) = 270 or x = 2250 . p . w . = rs . 2250 sum due = p . w . + t . d . = rs . ( 2250 270 ) = rs . 2520 answer : b" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 270 / d
f = e + 270
|
a ) 0.05 , b ) 0.36 , c ) 0.2 , d ) 0.25 , e ) 0.6 | b | subtract(divide(subtract(25, 8), 25), divide(8, 25)) | for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 25 people , 8 of whom are females , by how much does the index for the females exceed the index for the males in the group ? | "index for females = ( 25 - 8 ) / 25 = 17 / 25 = 0.68 index for males = ( 25 - 17 / 25 = 8 / 25 = 0.32 index for females exceeds males by 0.68 - 0.32 = 0.36 answer : b" | a = 25 - 8
b = a / 25
c = 8 / 25
d = b - c
|
a ) 62 , b ) 171 , c ) 475 , d ) 513 , e ) 684 | c | divide(multiply(57, 3), divide(36, const_100)) | at the end of year x , automobile installment credit accounted for 36 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 57 billion of credit , or 1 / 3 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at that time ? | "total automobile instalment credit = 57 * 3 = 171 total consumer instalment credit = 171 β 100 / 36 = 475 ; answer = c . 475" | a = 57 * 3
b = 36 / 100
c = a / b
|
a ) 17 , b ) 28 , c ) 24 , d ) 11 , e ) 12 | c | divide(122, add(const_2, const_pi)) | the perimeter of a semi circle is 122 cm then the radius is ? | "36 / 7 r = 122 = > r = 24 answer : c" | a = 2 + math.pi
b = 122 / a
|
a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | b | divide(add(multiply(factorial(493), factorial(915)), multiply(factorial(493), factorial(756))), 493) | what is the units digit of ( 493 ) ( 915 ) ( 381 ) ( 756 ) ( 28 ) | "just multiply the digits in the units place for each term and you will get the answer . it should be 0 . you got a 5 as a unit digit and an even number term . so the multiplication of this will definitely yield a 0 . answer has to be 0 . i also tried it using the calculator and the answer is 1 . imo b ." | a = math.factorial(493)
b = math.factorial(915)
c = a * b
d = math.factorial(493)
e = math.factorial(756)
f = d * e
g = c + f
h = g / 493
|
a ) β 220 , b ) β 100 , c ) 120 , d ) 135 , e ) it can not be determined from the information given | c | subtract(multiply(90, const_2), multiply(150, const_2)) | if the average ( arithmetic mean ) of a and b is 150 , and the average of b and c is 90 , what is the value of a β c ? | "a + b = 300 b + c = 180 a - c = 120 . imo option c ." | a = 90 * 2
b = 150 * 2
c = a - b
|
a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | a number is said to be prime saturated if the product of all the different positive prime factors of e is less than the square root of e . what is the greatest two digit prime saturated integer ? | "e = 96 = 3 * 32 = 3 * 2 ^ 5 answer is d ." | a = 2 * 3
b = 100 - 10
c = a + b
|
a ) 39 , b ) 68 , c ) 54 , d ) 57 , e ) 60 | b | add(subtract(92, 25), const_1) | claire has a total of 92 pets consisting of gerbils and hamsters only . one - quarter of the gerbils are male , and one - third of the hamsters are male . if there are 25 males altogether , how many gerbils does claire have ? | "g + h = 92 . . . 1 ; g / 4 + h / 3 = 25 . . . . 2 or 3 g + 4 h = 25 * 12 = 300 g = 92 - h or 3 ( 92 - h ) + 4 h = 300 h = 300 - 276 = 24 then g = 92 - 24 = 68 b" | a = 92 - 25
b = a + 1
|
a ) 10 , b ) 12 , c ) 16 , d ) 18 , e ) 20 | e | add(divide(60, add(0.75, 5)), divide(60, add(0.75, 5))) | a car ferry can hold up to 60 tons of cargo . what is the greatest number of vehicles that the ferry can carry if half the vehicles are cars with an average ( arithmetic mean ) weight of 0.75 tons and half of the vehicles are trucks with an average ( arithmetic mean ) weight of 5 tons ? | "the weight of one car and one truck is 5.75 tons . 60 / 5.75 = 10 plus a remainder the ferry could carry 10 cars and 10 trucks for a total of 20 vehicles . the answer is e ." | a = 0 + 75
b = 60 / a
c = 0 + 75
d = 60 / c
e = b + d
|
a ) 3 km , b ) 4 km , c ) 5 km , d ) 6 km , e ) 7 km | b | multiply(4, divide(add(add(multiply(4, 6), multiply(5, 6)), 6), const_60)) | if a man walks at the rate of 4 kmph , he misses a train by only 6 min . however , if he walks at the rate of 5 kmph he reaches the station 6 minutes before the arrival of the train . the distance covered by him to reach the station is | let the distance covered be d and time taken to reach the station in time t d / 4 = t + 6 / 60 - - - ( 1 ) d / 5 = t - 6 / 60 - - - ( 2 ) ( 1 ) - ( 2 ) gives d / 4 - d / 5 = ( 6 + 6 ) / 60 d / 20 = 12 / 60 d = 12 * 20 / 60 = 4 km answer : b | a = 4 * 6
b = 5 * 6
c = a + b
d = c + 6
e = d / const_60
f = 4 * e
|
a ) 20200 , b ) 20000 , c ) 40400 , d ) 40000 , e ) 44000 | a | multiply(add(divide(subtract(subtract(301, 99), const_2), const_2), 99), divide(add(subtract(301, 99), const_2), const_2)) | what is the sum of all even integers between 99 and 301 ? | "a = 100 , tn = 300 a + ( n - 1 ) d = 300 = > n = 101 sn = n / 2 ( a + l ) = 101 / 2 ( 100 + 300 ) sn = 20200 answer : a" | a = 301 - 99
b = a - 2
c = b / 2
d = c + 99
e = 301 - 99
f = e + 2
g = f / 2
h = d * g
|
a ) 18 , b ) 32 , c ) 27 , d ) 26 , e ) 19 | b | add(multiply(4, 4), multiply(4, 4)) | a person was asked to state his age in years . his reply was , ` ` take my age 4 years hence , multiply it by 4 and subtract 4 times my age 4 years ago and you will know how old i am . ' ' what was the age of the person ? | "explanation : let the present age of person be x years . then , 4 ( x + 4 ) - 4 ( x - 4 ) = x < = > ( 4 x + 16 ) - ( 4 x - 16 ) = x < = > x = 32 . . answer : b" | a = 4 * 4
b = 4 * 4
c = a + b
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | c | subtract(divide(add(24, 12), const_2), divide(add(48, 24), const_2)) | the average ( arithmetic mean ) of the even integers from 24 to 48 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 12 to 24 inclusive ? | "so , according to a mean of a set of even numbers from 24 to 48 = ( 24 + 48 ) / 2 = 36 and mean of a set of even numbers from 12 to 24 = ( 12 + 24 ) / 2 = 18 difference = 36 - 18 = 18 answer : c ." | a = 24 + 12
b = a / 2
c = 48 + 24
d = c / 2
e = b - d
|
a ) 100 minute , b ) 110 minute , c ) 120 minute , d ) 130 minute , e ) 140 minute | b | multiply(divide(5, subtract(divide(440, const_10), const_4)), multiply(440, 2)) | 2 friends a and b running up hill and then to get down length of road - 440 yads a on his return journey met b going up at 20 yards from top if a has finished race 5 minute earlier than b then how much time a had taken to complete the race ? | total journey = 440 * 2 = 880 a meet b 20 yards from top in getting down it means he has covered 440 + 20 = 460 yards while b is 420 yards . so he is 40 yards ahead of b which is equals to 5 minute . so 40 yards in 5 min 880 yards will be in 5 * 880 / ( 40 ) = 110 minute answer : b | a = 440 / 10
b = a - 4
c = 5 / b
d = 440 * 2
e = c * d
|
a ) 1150 m , b ) 2250 m , c ) 1450 m , d ) 1350 m , e ) 1250 m | e | multiply(500, subtract(const_2, const_1)) | a train speeds past a pole in 25 seconds and a platform 500 m long in 35 seconds . its length is : | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 25 = > y = x / 25 x + 500 / 35 = x / 25 x = 1250 m . answer : option e" | a = 2 - 1
b = 500 * a
|
a ) 22 , b ) 27 , c ) 28 , d ) 26 , e ) 56 | e | subtract(add(add(multiply(40, 15), 16), 40), multiply(40, 15)) | the average age of 40 students in a class is 15 years . if the age of teacher is also included , the average becomes 16 years , find the age of the teacher . | "explanation : if teacher ' s age is 15 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 16 . age of teacher = average age of all + total increase in age = 16 + ( 1 x 40 ) = 56 years answer : e" | a = 40 * 15
b = a + 16
c = b + 40
d = 40 * 15
e = c - d
|
a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 2548 | e | multiply(divide(divide(factorial(14), factorial(subtract(14, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | 14 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | "no . of ways of picking 2 biology books ( from 14 books ) = 14 c 2 = ( 14 * 13 ) / 2 = 91 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 91 * 28 = 2548 ( option e )" | a = math.factorial(14)
b = 14 - 2
c = math.factorial(b)
d = a / c
e = d / 2
f = math.factorial(8)
g = 8 - 2
h = math.factorial(g)
i = f / h
j = i / 2
k = e * j
|
a ) 9 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | a | subtract(multiply(5, const_2), const_1) | the average of non - zero number and its square is 5 times the number . the number is : | let the number be x . then , ( x + x 2 ) / 2 = 5 x = > x 2 - 9 x = 0 = > x ( x - 9 ) = 0 = > x = 0 or x = 9 so , the number is 9 . answer : a | a = 5 * 2
b = a - 1
|
a ) 44 , b ) 54 , c ) 16 , d ) 27 , e ) 30 | a | subtract(50, multiply(multiply(12, 4), 2)) | evaluate : 50 - 12 Γ· 4 Γ 2 = | "according to order of operations , 12 Γ· 4 Γ 2 ( division and multiplication ) is done first from left to right 12 Γ· 4 Γ 2 = 3 Γ 2 = 6 hence 50 - 12 Γ· 4 Γ 2 = 50 - 6 = 44 correct answer a ) 44" | a = 12 * 4
b = a * 2
c = 50 - b
|
a ) 80 , b ) 150 , c ) 75 , d ) 90 , e ) none of these | b | divide(36, multiply(divide(40, const_100), divide(3, 5))) | if 40 % of 3 / 5 of a number is 36 , then the number is ? | "let the number be x . then 40 % of 3 / 5 of x = 36 40 / 100 * 3 / 5 * x = 36 x = ( 36 * 50 / 12 ) = 150 required number = 150 . correct option : b" | a = 40 / 100
b = 3 / 5
c = a * b
d = 36 / c
|
a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 22 | e | sqrt(subtract(subtract(power(24, const_2), multiply(23, const_2)), multiply(23, const_2))) | the sum and the product of two numbers are 24 and 23 respectively , the difference of the number is ? | explanation : x + y = 24 xy = 23 ( x - y ) 2 = ( x + y ) 2 - 4 xy ( x - y ) 2 = 576 - 92 = > ( x - y ) = 22 answer : e | a = 24 ** 2
b = 23 * 2
c = a - b
d = 23 * 2
e = c - d
f = math.sqrt(e)
|
a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | d | add(multiply(0.25, subtract(7, 1)), 0.25) | at a certain company , each employee has a salary grade s that is at least 1 and at most 7 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s β 1 ) . an employee with a salary grade of 7 receives how many more dollars per hour than an employee with a salary grade of 1 ? | "salary grade of 7 is p ( 7 ) = 9.50 + 0.25 ( 7 β 1 ) = 9.50 + 0.25 * 6 ; salary grade of 1 is p ( 1 ) = 9.50 + 0.25 ( 1 β 1 ) = 9.50 ; p ( 7 ) - p ( 1 ) = 9.50 + 0.25 * 6 - 9.50 = 1.5 . answer : d ." | a = 7 - 1
b = 0 * 25
c = b + 0
|
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