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 ) 140 m , b ) 786 m , c ) 566 m , d ) 546 m , e ) 445 m | a | multiply(divide(multiply(56, const_1000), const_3600), 9) | a train running at the speed of 56 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 56 * 5 / 18 = 140 / 9 m / sec length of the train = speed * time = 140 / 9 * 9 = 140 m answer : a" | a = 56 * 1000
b = a / 3600
c = b * 9
|
a ) 69 , b ) 92 , c ) 94 , d ) 49 , e ) 96 | e | divide(divide(divide(960, divide(add(11, 9), const_2)), 9), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 11 m wide at the top and 9 m wide at the bottom and the area of cross - section is 960 sq m , the depth of cannel is ? | "1 / 2 * d ( 11 + 9 ) = 960 d = 96 answer : e" | a = 11 + 9
b = a / 2
c = 960 / b
d = c / 9
e = d / 2
|
a ) 9 , b ) 10 , c ) 11 , d ) 6 , e ) 13 | d | divide(18, divide(18, 6)) | suppose 6 monkeys take 6 minutes to eat 6 bananas . how many monkeys would it take to eat 18 bananas in 18 minutes ? | "one monkey takes 6 min to eat 1 banana , so in 18 mins 1 monkey will eat 3 bananas , so for 18 bananas in 18 min we need 18 / 3 = 6 monkeys answer : d" | a = 18 / 6
b = 18 / a
|
a ) a ) 15 , b ) b ) 20 , c ) c ) 25 , d ) d ) 40 , e ) e ) 45 | c | divide(subtract(subtract(240, 80), 60), const_4) | 6 ) a marketing firm determined that , of 240 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 ? | "160 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 160 = > 4 x = 100 = > x = 25 answer - c" | a = 240 - 80
b = a - 60
c = b / 4
|
a ) 3 : 1 , b ) 2 : 1 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3 | a | divide(divide(const_1, const_4), divide(20, const_100)) | if 20 % of a number is equal to three - fifth of another number , what is the ratio of first number to the second number ? | "let 20 % of a = 3 / 5 b . then , 20 a / 100 = 3 b / 5 = > 1 a / 5 = 3 b / 5 a / b = ( 3 / 5 * 5 / 1 ) = 3 / 1 a : b = 3 : 1 . answer : a" | a = 1 / 4
b = 20 / 100
c = a / b
|
a ) 2 years , b ) 4 years , c ) 6 years , d ) 8 years , e ) 10 years | e | divide(subtract(divide(80, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | the sum of the ages of 5 children born at the intervals of 3 years each is 80 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 80 5 x = 50 = > x = 10 . age of youngest child = x = 10 years . answer : e" | a = 5 / 3
b = 80 / a
c = 5 - 1
d = c * 3
e = b - d
f = e / 3
|
a ) 200 m , b ) 250 m , c ) 450 m , d ) 350 m , e ) 270 m | a | divide(multiply(80, divide(add(multiply(1, 3), 2), 3)), subtract(divide(add(multiply(1, 3), 2), 3), const_1)) | a runs 1 2 β 3 times as fast as b . if a gives b a start of 80 m , how far must the winning post be so that a and b might reach it at the same time ? | speed of a : speed of b = 5 β 3 : 1 = 5 : 3 i . e . , in a race of 5 m , a runs 5 m and b runs 3 m i . e . , a gains 2 m over b in a race of 5 m = > a gains 80 m over b in a race of 52 Γ 80 = 200 m hence , winning point should be 200 m away from the starting point . answer : a | a = 1 * 3
b = a + 2
c = b / 3
d = 80 * c
e = 1 * 3
f = e + 2
g = f / 3
h = g - 1
i = d / h
|
a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38 | e | divide(multiply(42, 24), 36) | in a division , a student took 42 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 42 = 24 . x = 24 * 42 . so correct answer would be , ( 24 * 42 ) / 36 = 38 . answer : e" | a = 42 * 24
b = a / 36
|
a ) 20 , b ) 18 , c ) 14 , d ) 15 , e ) 16 | d | multiply(5, const_3) | hyderabad chennai mumbai delhi bangalore trivandram ; hyderabad chennai mumbai delhi bangalore trivandram in the table above , what is the least number of table entries that are needed to show the mileage between each city and each of the other 5 cities ? | easy way to go about this problem is we have 6 * 6 = 36 enteries in table the least number of enteries would be ( 36 - 6 ) / 2 since 6 enteries represent the distances between same points . alternatively this can be solved as combination problem . correct answer d | a = 5 * 3
|
a ) 10.6 , b ) 10.2 , c ) 10.8 , d ) 10.4 , e ) 10.0 | e | multiply(add(20, 5), divide(24, const_60)) | the speed of a boat in still water in 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 5 ) = 25 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 25 = 10.0 km answer is e ." | a = 20 + 5
b = 24 / const_60
c = a * b
|
a ) 93 kg , b ) 50 kg , c ) 85 kg , d ) 80 kg , e ) 60 kg | a | add(multiply(8, 3.5), 65) | the average weight of 8 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 Γ£ β 3.5 = 28 if x is the weight of the new person , total increase in weight = x Γ’ Λ β 65 = > 28 = x - 65 = > x = 28 + 65 = 93 answer : option a" | a = 8 * 3
b = a + 65
|
a ) 4500 , b ) 6333 , c ) 6576 , d ) 6345 , e ) 2882 | b | divide(multiply(multiply(5, 19), multiply(4, const_1000)), multiply(const_1, const_60)) | a river 5 m deep and 19 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is ? | "( 4000 * 5 * 19 ) / 60 = 6333 m 3 answer : b" | a = 5 * 19
b = 4 * 1000
c = a * b
d = 1 * const_60
e = c / d
|
a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36 | e | divide(multiply(24, 3), subtract(3, const_1)) | if the number is divided by 3 , it reduced by 24 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 24 = > 2 x / 3 = 24 = > x = 36 answer : option e" | a = 24 * 3
b = 3 - 1
c = a / b
|
a ) $ 21 , b ) $ 27 , c ) $ 31 , d ) $ 43 , e ) it can not be determined | d | subtract(add(120, subtract(158.50, 120)), multiply(subtract(158.50, 120), 3)) | at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 158.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | "let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 158.5 subtracting the first equation from the second gives us b + 3 s = 38.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 43 . in any case , there is no necessity to know each item ' s price , just the sum . answer : d" | a = 158 - 50
b = 120 + a
c = 158 - 50
d = c * 3
e = b - d
|
a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | a | divide(40, multiply(subtract(const_1, divide(20, const_100)), add(divide(20, const_100), const_1))) | the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 20 % . if in july , a mixture containing equal quantities of tea and coffee costs 40 / kg . how much did a kg of coffee cost in june ? | "let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.8 x . in july the price of 1 / 2 kg ( 400 gm ) of tea and 1 / 2 kg ( 400 gm ) of coffee ( equal quantities ) = 40 1.2 x ( 1 / 2 ) + 0.8 x ( 1 / 2 ) = 40 = > x = 40 thus proved . . . option a ." | a = 20 / 100
b = 1 - a
c = 20 / 100
d = c + 1
e = b * d
f = 40 / e
|
a ) - 8 , b ) - 9 , c ) - 5 , d ) 9 , e ) 1 | d | divide(negate(add(37, 1)), 4) | solve below question 4 x + 1 = 37 | 1 . subtract 1 from both sides : 4 x + 1 - 1 = 37 - 1 2 . simplify both sides : 4 x = 36 3 . divide both sides by 4 : 4 . simplify both sides : x = 9 d | a = 37 + 1
b = negate / (
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | d | subtract(divide(add(220, 20), const_4), 20) | thabo owns exactly 220 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books does thabo own ? | "i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fiction books is 0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f 2 x + 40 0 nf x + 20 x total 3 x + 60 x 220 3 x + 60 + x = 220 x = 40 answer ( d . )" | a = 220 + 20
b = a / 4
c = b - 20
|
a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 15 | b | divide(multiply(10, 10), 5) | 10 men do a work in 10 days . how many men are needed to finish the work in 5 days ? | "men required to finish the work in 5 days = 10 * 10 / 5 = 20 answer is b" | a = 10 * 10
b = a / 5
|
a ) 15 , b ) 18 , c ) 21 , d ) 12 , e ) 19 | c | subtract(divide(subtract(multiply(6, 28), 7), const_4), 7) | the average age of a family of 6 members is 28 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 28 = 168 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 168 - 42 = 126 therefore , avg age at the birth of youngest member = 126 / 6 = 21 . answer : c" | a = 6 * 28
b = a - 7
c = b / 4
d = c - 7
|
a ) 20 , b ) 31 , c ) 42 , d ) 49 , e ) 64 | d | divide(add(84, 14), const_2) | there are 84 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 14 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ? | x + ( x - 14 ) = 84 - - > x = 49 . answer : d . | a = 84 + 14
b = a / 2
|
a ) 16 , b ) 25 , c ) 19 , d ) 64 , e ) 17 | d | power(divide(15, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | find out the square of a number which when doubled exceeds its one eighth by 15 ? | "let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 8 ) + 15 = > 16 p = p + 120 = > p = 8 p ^ 2 = 8 ^ 2 = 64 . answer : d" | a = 4 + 1
b = 1 / a
c = 2 - b
d = 15 / c
e = d ** 2
|
a ) 30 , b ) 28 , c ) 24 , d ) 20 , e ) 35 | a | multiply(5, 25) | the h . c . f . of two numbers is 5 and their l . c . m . is 150 . if one of the numbers is 25 , then the other is : | "product of two nos = product of hcf and lcm let one number = x = > 25 * x = 5 * 150 x = 5 * 150 / 25 = 30 answer a" | a = 5 * 25
|
a ) 47 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23 | a | divide(multiply(add(add(4, const_3), const_2), divide(4, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 4 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 5 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 4 ) / ( b + 4 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 5 ) / ( a + 5 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 47 / 5 ans : ( a )" | a = 4 + 3
b = a + 2
c = 4 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(270, add(subtract(190, multiply(6, divide(subtract(190, divide(220, 2)), 4))), divide(subtract(190, divide(220, 2)), 4))) | a sporting good store sells one type of baseball bat and one type of baseball . the cost for 2 bats and 4 balls is $ 220 . the cost for 1 bat and 6 balls is $ 190 , as well . if someone were to buy an equal number of bats and balls , at most how many bats can he purchase if he has a budget of $ 270 for the purchase ? | "imo it should be c that is 4 reason : formed an equation . . . bat = b ball = c 2 b + 4 c = 220 1 b + 6 c = 190 solving both we get b that is bat = 70 and c that is ball = 20 new equation 270 to be divided in equal 3 b + 3 c = 270 3 * 70 + 3 * 20 = 270 210 + 60 = 270" | a = 220 / 2
b = 190 - a
c = b / 4
d = 6 * c
e = 190 - d
f = 220 / 2
g = 190 - f
h = g / 4
i = e + h
j = 270 / i
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10 | e | divide(subtract(multiply(180, 2), multiply(30, 2)), subtract(add(add(30, 120), 180), multiply(100, const_3))) | in a coconut grove , ( x + 2 ) trees yield 30 nuts per year , x trees yield 120 nuts per year and ( x β 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) Γ 30 + x Γ 120 + ( x β 2 ) Γ 180 / ( x + 2 ) + x + ( x β 2 ) = 100 β 330 x β 300 / 3 x = 100 β 30 x = 300 β x = 10 answer e" | a = 180 * 2
b = 30 * 2
c = a - b
d = 30 + 120
e = d + 180
f = 100 * 3
g = e - f
h = c / g
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | add(subtract(multiply(const_4, 2), multiply(multiply(const_4, 5), 0.3)), 2) | a football player scores 2 goals in his fifth match thus increasing his average goals score by 0.3 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the answer is not correct , then you would have a gauge for whether you should gohigherorlowerwith your next test . here , i ' ll start with answer a = 4 goals if . . . . total goals = 4 goals 5 th game = 2 goals 1 st 4 games = 2 goals avg . for 1 st 4 games = 2 / 4 = 0.5 goal / game avg . for all 5 games = 4 / 5 = 0.8 goals / game this is an exact match for what we ' re told in the prompt , so answer a must be the answer ." | a = 4 * 2
b = 4 * 5
c = b * 0
d = a - c
e = d + 2
|
['a ) 0.550', 'b ) 0.350', 'c ) 0.450', 'd ) 0.700', 'e ) 0.375'] | e | divide(multiply(15, 25), 1000) | a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . if 25 cubic centimeters of the solution were poured into an empty container , how many grams of salt would be in the container ? | we are given that a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of salt is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of salt would be in the container when we have 25 cubic centimeters of solution , we can set up a proportion . we can say : β 15 grams of salt is to 1000 cubic centimeters of solution as x grams of salt is to 25 cubic centimeters of solution . β let β s now set up the proportion and solve for x . 15 / 1000 = x / 25 when we cross multiply we obtain : ( 15 ) ( 25 ) = 1000 x 375 = 1000 x 0.375 = x there are 0.375 grams of salt in the solution in the container . the answer is e . | a = 15 * 25
b = a / 1000
|
a ) $ 16 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 0 | a | subtract(120, add(add(multiply(120, divide(1, 5)), multiply(120, divide(1, 6))), multiply(120, divide(1, 2)))) | jennifer had $ 120 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | "1 / 5 x $ 120 = $ 24 for sandwich 1 / 6 x $ 120 = $ 20 for museum 1 / 2 x $ 120 = $ 60 for book $ 24 + $ 20 + $ 60 = $ 104 spent $ 120 - $ 104 = $ 16 left over correct answer a" | a = 1 / 5
b = 120 * a
c = 1 / 6
d = 120 * c
e = b + d
f = 1 / 2
g = 120 * f
h = e + g
i = 120 - h
|
a ) 55 / 601 , b ) 601 / 55 , c ) 11 / 120 , d ) 120 / 11 , e ) 120 / 13 | c | divide(divide(55, 5), 120) | if the sum of two numbers is 55 and the h . c . f . and l . c . m of these numbers are 5 and 120 respectively , then the sum of the reciprocals of the numbers is equal to : | "given a + b = 55 , hcm and lcm is 5 and 120 axb = lcmx hcf 600 = 120 x 5 1 / a + 1 / b = a + b / ab = > 55 / 600 = > 11 / 120 answer : c" | a = 55 / 5
b = a / 120
|
a ) rs . 72 , b ) rs . 65 , c ) rs . 54 , d ) rs . 50 , e ) none | b | divide(7.8, divide(12, const_100)) | the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.8 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.8 x 100 / 12 x 1 ) = rs . 65 . answer b" | a = 12 / 100
b = 7 / 8
|
a ) rs . 90 , b ) rs . 100 , c ) rs . 120 , d ) rs . 150 , e ) rs . 160 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served only for 9 months and receiveduniform and rs . 650 , find the priceof the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a = 1 * 900
b = 3 * 4
c = b * 9
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | multiply(5, divide(subtract(multiply(5, 4), multiply(3, 4)), subtract(multiply(5, 4), multiply(3, 4)))) | 2 men and 4 boys can complete a work in 4 days . 5 men and 6 boys can complete the same work in 3 days . the work done by 2 boys is equal to the work of how many men ? | exp : ( 2 m + 4 b ) β s one day β s work = 1 / 4 ( 5 m + 6 b ) β s one day β s work = 1 / 3 = > ( 8 m + 16 b ) β s one day β s work = ( 15 m + 18 b ) β s one day β s work = > 7 men β s work = 2 boy β s work so , we should be employ 5 more men to be completed the work . answer : b | a = 5 * 4
b = 3 * 4
c = a - b
d = 5 * 4
e = 3 * 4
f = d - e
g = c / f
h = 5 * g
|
a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 101.50 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(10092, multiply(3, 4)))), multiply(4, sqrt(divide(10092, multiply(3, 4))))), 25), const_100) | the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10092 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x Γ 4 x = 10092 β 12 x 2 = 10092 β x 2 = 841 β x = 29 . so , length = 87 m and breadth = 116 m . perimeter = [ 2 ( 87 + 116 ) ] m = 406 m . β΄ cost of fencing = rs . ( 0.25 Γ 406 ) = rs . 101.50 . answer d" | a = 3 * 4
b = 10092 / a
c = math.sqrt(b)
d = 3 * c
e = 3 * 4
f = 10092 / e
g = math.sqrt(f)
h = 4 * g
i = rectangle_perimeter * (
j = i / 25
|
a ) 120 , b ) 125 , c ) 130 , d ) 145 , e ) 155 | b | subtract(multiply(multiply(divide(1000, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(1000, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), subtract(const_1, divide(25, const_100)))) | of 1000 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 1000 = 1.25 b + b - 0.25 b - - > b = 500 , a = 1.25 b = 625 and { both } = 0.25 b = 125 . the number of students who read only book a is { a } - { both } = 625 - 125 = 500 ; the number of students who read only book b is { b } - { both } = 500 - 125 = 375 the difference is 500 - 375 = 125 . answer : b ." | a = 25 / 100
b = 20 / 100
c = a / b
d = c + 1
e = 25 / 100
f = d - e
g = 1000 / f
h = 25 / 100
i = 20 / 100
j = h / i
k = g * j
l = 20 / 100
m = 1 - l
n = k * m
o = 25 / 100
p = 20 / 100
q = o / p
r = q + 1
s = 25 / 100
t = r - s
u = 1000 / t
v = 25 / 100
w = 1 - v
x = u * w
y = n - x
|
a ) 4.75 days , b ) 3.75 days , c ) 6.50 days , d ) 7.50 days , e ) 5.50 days | b | divide(multiply(3, 15), subtract(15, 3)) | a and b finish the job in 15 days . while a , b and c can finish it in 3 days . c alone will finish the job in | "explanation : 3 = ( 15 * x ) / ( 15 + x ) 45 + 3 x = 15 x 12 x = 45 x = 3.75 answer : option b" | a = 3 * 15
b = 15 - 3
c = a / b
|
a ) 40 , b ) 45 , c ) 35 , d ) 50 , e ) 55 | c | add(add(multiply(inverse(const_3), 60), multiply(inverse(add(const_1, const_4)), 60)), 3) | the sum of 3 integers a , b and c is 60 . a is one third of the sum of b and c and b is one fifth of the sum of a and c . what is c ? | a is one third of the sum of b and c . . or a = 1 / 3 ( b + c ) . . . but a + b + c = 60 so 1 / 3 ( b + c ) + b + c = 60 4 ( b + c ) / 3 = 60 b + c = 60 * 3 / 4 = 45 a = 60 - 45 = 15 2 ) b is one fifth of the sum of a and c or b = 1 / 5 ( a + c ) . . . but a + b + c = 60 so 1 / 5 ( a + c ) + a + c = 60 a + c = 60 * 5 / 6 = 50 but a = 15 , so c = 50 - 15 = 35 answer : c | a = 1/(3)
b = a * 60
c = 1 + 4
d = 1/(c)
e = d * 60
f = b + e
g = f + 3
|
a ) 1 / 12 , b ) 1 / 9 , c ) 1 2 / 3 , d ) 2 / 3 , e ) 2 1 / 9 | d | subtract(add(divide(5, 11), divide(2, 3)), divide(25, 55)) | the instructions state that cheryl needs 5 / 11 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 25 / 55 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 5 / 11 + 2 / 3 left part 25 / 55 - - - > 5 / 11 so used part 5 / 11 + 2 / 3 - 5 / 11 = 2 / 3 ans d" | a = 5 / 11
b = 2 / 3
c = a + b
d = 25 / 55
e = c - d
|
a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54 | d | subtract(87, subtract(add(41, 22), 9)) | in a class of 87 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 87 54 + not = 87 not = 87 - 54 = 33 so answer is d | a = 41 + 22
b = a - 9
c = 87 - b
|
a ) 27556 , b ) 800 , c ) 796 , d ) 87 , e ) 166 | e | sqrt(subtract(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(2, 3), 4), 5), 6), 7), 8), 9), 11), 164)) | what is the least positive integer x for which ( x ^ 2 ) + 164 is divisible by 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , and 11 ? | 2 = 2 3 = 3 4 = 2 * 2 5 = 5 6 = 2 * 3 7 = 7 8 = 2 * 2 * 2 9 = 3 * 3 10 = 2 * 5 11 = 11 so ( x ^ 2 ) + 164 = ( 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 ) = 27720 . then ( x ^ 2 ) = 27556 . then x = 166 , which is option e . | a = math.lcm(2, 3)
b = math.lcm(a, 4)
c = math.lcm(b, 5)
d = math.lcm(c, 6)
e = math.lcm(d, 7)
f = math.lcm(e, 8)
g = math.lcm(f, 9)
h = math.lcm(g, 11)
i = h - 164
j = math.sqrt(i)
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a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 70 | e | divide(divide(divide(700, divide(add(16, 4), const_2)), 4), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 16 m wide at the top and 4 m wide at the bottom and the area of cross - section is 700 sq m , the depth of cannel is ? | "1 / 2 * d ( 16 + 4 ) = 700 d = 70 answer : e" | a = 16 + 4
b = a / 2
c = 700 / b
d = c / 4
e = d / 2
|
['a ) 300', 'b ) 150', 'c ) 75', 'd ) 40', 'e ) 20'] | c | divide(multiply(15, 20), const_4) | two equal triangles constitute rhombus a . if the diagonals of rhombus are 15 cm and 20 cm long . what is the area of each triangle ? | area of rhombus = 0.5 x 15 x 20 = 150 cm Γ’ Β² area of triangle = 150 / 2 = 75 cm Γ’ Β² answer : c | a = 15 * 20
b = a / 4
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a ) 4 / 7 , b ) 9 / 14 , c ) 11 / 14 , d ) 17 / 28 , e ) 23 / 28 | d | subtract(1, add(divide(1, 4), divide(1, 7))) | there is a total of 84 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 7 . what is the probability that the marble will be either red or blue ? | "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 28 / 28 - 7 / 28 - 4 / 28 = 17 / 28 the answer is d ." | a = 1 / 4
b = 1 / 7
c = a + b
d = 1 - c
|
['a ) 20', 'b ) 16', 'c ) 12', 'd ) 8', 'e ) 4'] | b | multiply(const_4, const_4) | if the side length of square b is four times that of square a , the area of square b is how many times the area of square a ? | let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( 4 x ) ^ 2 = 16 x ^ 2 . the answer is b . | a = 4 * 4
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a ) 70 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(multiply(36, divide(multiply(subtract(50, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 50 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 50 - 36 = 14 km / hr . = 14 * 5 / 18 = 35 / 9 m / sec . 2 x / 36 = 35 / 9 = > x = 70 . answer : a" | a = 50 - 36
b = a * 1000
c = b / 3600
d = 36 * c
e = d / 2
|
a ) 9 : 7 , b ) 9 : 9 , c ) 9 : 6 , d ) 9 : 1 , e ) 9 : 2 | a | divide(multiply(3, 3), 7) | a dog takes 3 leaps for every 7 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 7 leaps of hare = 9 : 7 . answer : a ) 9 : 7" | a = 3 * 3
b = a / 7
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a ) 150 , b ) 152 , c ) 154 , d ) 158 , e ) 144 | d | add(multiply(17, 9), 5) | what is the dividend . divisor 17 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 17 * 9 + 5 d = 153 + 5 d = 158" | a = 17 * 9
b = a + 5
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a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(2, subtract(divide(40, const_60), divide(10, const_60))) | circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 2 more revolutions than gear p ? | "gear q makes 30 more revolutions each 60 seconds . the time to make 2 = 30 / 15 more revolutions is 60 / 15 = 4 seconds . the answer is a ." | a = 40 / const_60
b = 10 / const_60
c = a - b
d = 2 / c
|
a ) 0.28 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | a | inverse(subtract(inverse(3), inverse(20))) | a and b together can do a piece of work in 3 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "b = 1 / 2 β 1 / 20 = 0.28 days answer : a" | a = 1/(3)
b = 1/(20)
c = a - b
d = 1/(c)
|
a ) 462 , b ) 744 , c ) 480 , d ) 504 , e ) 555 | b | multiply(62, 12) | the hcf of two numbers is 62 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 62 hcf will be always a factor of lcm 62 is factor of lcm other two factors are 11 & 12 then the numbers are ( 62 * 11 ) and ( 62 x 12 ) = 682 and 744 answer : option b" | a = 62 * 12
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a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | d | divide(subtract(const_1, divide(23, 40)), add(divide(const_1, 45), divide(const_1, 40))) | a and b can do a piece of work in 45 days and 40 days respectively . they began to do the work together but a leaves after some days and then b completed the remaining work in 23 days . the number of days after which a left the work was ? | let the total units of work to be done be 360 . the units of work done by a in a single day = 8 similarly , the units of work done by b in a single day = 9 . a and b β s one day work = 17 units a and b worked together for some days = 17 x ( assume ) b β s work alone for 23 days = 23 x 9 = 207 so , the work done by a and b together = ( 360 - 207 ) = 153 units therefore , 17 x = 153 = > 9 units therefore , the number of days after which a left the work was 9 days . answer : d | a = 23 / 40
b = 1 - a
c = 1 / 45
d = 1 / 40
e = c + d
f = b / e
|
a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 174.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(152, add(add(1, 1), 2)), add(126, 135)), 2) | teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 152 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 152 ) / \ x Γ’ Λ β 152 22.50 = > x Γ’ Λ β ( 152 / 22.50 ) = 1 . = > x Γ’ Λ β 152 = 22.50 . = > x = 174.50 rs . answer : c" | a = 1 + 1
b = a + 2
c = 152 * b
d = 126 + 135
e = c - d
f = e / 2
|
a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | b | subtract(15, reminder(5, 8)) | when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 8 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 15 ? | "when positive integer n is divided by 3 , the remainder is 1 i . e . , n = 3 x + 1 values of n can be one of { 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 . . . . . . . . . . . . . . 49 , 52 , 59 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 8 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can be one of { 5 , 13 , 21 , 29 , 37 , 45 , 53 , 61 . . . . . . . . } combining both the sets we get n = { 13 , 37 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 15 or 15 x in case of n = 13 p = 2 so for min value of p , we take min value of n . b is the answer ." | a = 15 - reminder
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a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 16 | c | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 402 * 502 * 604 * 646 * 547 * 448 * 349 ) ^ 2 | "( ( 301 * 402 * 502 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 2,4 , 6,7 , 8,9 ; 3 & 5 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,02 , 04,46 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 04 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 36 * 44 ) ^ 2 = 84 ^ 2 = 56 hence answer is c" | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | add(divide(multiply(60, const_100), 25), 60) | p runs 25 % faster than q so p gives q a 60 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ? | "let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.25 v = ( d - 60 ) / v d = 1.25 d - 75 0.25 d = 75 d = 300 meters . the answer is c ." | a = 60 * 100
b = a / 25
c = b + 60
|
a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | b | multiply(divide(divide(subtract(660, 560), 5), subtract(560, multiply(divide(subtract(660, 560), 5), 3))), const_100) | joe invested a certain sum of money in a simple interest bond whose value grew to $ 560 at the end of 3 years and to $ 660 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 560 - $ 60 = 500 . the interest rate is $ 20 / $ 500 = 4 % the answer is b ." | a = 660 - 560
b = a / 5
c = 660 - 560
d = c / 5
e = d * 3
f = 560 - e
g = b / f
h = g * 100
|
a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 50.9 , e ) 52.9 | d | divide(add(multiply(25, 40), multiply(30, 60)), add(25, 30)) | the average marks of a class of 25 students is 40 and that of another class of 30 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 25 students = 25 * 40 = 1000 sum of the marks for the class of 30 students = 30 * 60 = 1800 sum of the marks for the class of 55 students = 1000 + 1800 = 2800 average marks of all the students = 2800 / 55 = 50.9 . answer : d" | a = 25 * 40
b = 30 * 60
c = a + b
d = 25 + 30
e = c / d
|
['a ) 11236', 'b ) 11025', 'c ) 14400', 'd ) 12696', 'e ) can not be determined'] | a | power(add(divide(subtract(211, 1), const_2), 1), const_2) | a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 211 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cabbages did she produce this year ? | explanatory answer the shape of the area used for growing cabbages has remained a square in both the years . let the side of the square area used for growing cabbages this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cabbages last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cabbages grown has increased by 211 , the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space . hence , x 2 - y 2 = 211 ( x + y ) ( x - y ) = 211 . 211 is a prime number and hence it will have only two factors . i . e . , 211 and 1 . therefore , 211 can be expressed as product of 2 numbers in only way = 211 * 1 i . e . , ( x + y ) ( x - y ) = 211 * 1 so , ( x + y ) should be 211 and ( x - y ) should be 1 . solving the two equations we get x = 106 and y = 105 . therefore , number of cabbages produced this year = x 2 = 1062 = 11236 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice d . 12696 is not the square of any number . check choice a : if this year ' s produce is 11236 , last year ' s produce would have been 11236 - 211 = 11025 11025 is the square of 105 . so , 11236 is the answer . choice a is the correct answer . | a = 211 - 1
b = a / 2
c = b + 1
d = c ** 2
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a ) 0 , b ) 1 / 15 , c ) 2 / 15 , d ) 1 / 6 , e ) 2 / 10 | d | multiply(divide(1, 3), divide(subtract(1, divide(2, 3)), divide(2, 3))) | two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 3 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | "suppose the water in each jug is l liters cx x ( 1 / 3 ) = l cx = 3 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 3 l = 1 / 6 answer will be d" | a = 1 / 3
b = 2 / 3
c = 1 - b
d = 2 / 3
e = c / d
f = a * e
|
a ) 3 , b ) 6 , c ) 8 , d ) 16 , e ) 32 | c | add(const_2, const_2) | find the number of factors of 9321 ? | "9321 = 3 x 13 x 239 = 31 x 13 1 x 239 1 ; add one to all the powers i . e . , ( 1 + 1 ) x ( 1 + 1 ) x ( 1 + 1 ) = 8 ; we get eight factors 1 , 3 , 13 , 39 , 239 , 717 , 3107 , 9321 answer : c" | a = 2 + 2
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a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 55 | b | subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 60)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 60 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 70 = 35 b" | a = 20 + 40
b = a + 60
c = 5 * 3
d = c + 10
e = d + 60
f = b - e
|
a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | b | multiply(subtract(divide(power(39, const_2), 357), floor(divide(power(39, const_2), 357))), 357) | on dividing a number by 357 , we get 39 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 + 39 = ( 17 * 21 * y ) + ( 17 * 2 ) + 5 = 17 * ( 21 y + 2 ) + 5 . required number = 5 . answer is b" | a = 39 ** 2
b = a / 357
c = 39 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
|
a ) 60 , b ) 45 , c ) 38 , d ) 50 , e ) 39 | a | divide(subtract(12, multiply(12, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | a man Γ’ β¬ β’ s current age is ( 2 / 5 ) of the age of his father . after 12 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father Γ’ β¬ β’ s current age is a years . then , man Γ’ β¬ β’ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 12 ] = ( 1 / 2 ) ( a + 12 ) 2 ( 2 a + 60 ) = 5 ( a + 12 ) a = 60 a" | a = 1 / 2
b = 12 * a
c = 12 - b
d = 1 / 2
e = 2 / 5
f = d - e
g = c / f
|
a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 5 / 42 , e ) 7 / 42 | d | multiply(divide(subtract(subtract(5, const_1), const_1), subtract(subtract(9, const_1), const_1)), multiply(divide(5, 9), divide(subtract(5, const_1), subtract(9, const_1)))) | potato chips are on sale , buy 2 get 1 free . there are only 9 bags of chips left ( 4 plain and 5 bbq ) . a shopper randomly grabs 3 bags . what is the probability the shopper will grab exactly 3 bags of bbq chips ? | combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 3 bags from 9 bags = 9 c 3 = 9 ! / [ 3 ! ( 9 - 3 ) ! ] = 84 . to have 3 bbq there must be 0 plain , select 0 plain from 4 = 4 c 0 = 1 . and , select 3 bbq from 5 = 5 c 3 = 10 . 3 bbq and 0 plain = ( 1 ) ( 10 ) = 10 probability = ( number outcomes favorable ) / ( total number outcomes ) = 10 / 84 = 5 / 42 answer : d | a = 5 - 1
b = a - 1
c = 9 - 1
d = c - 1
e = b / d
f = 5 / 9
g = 5 - 1
h = 9 - 1
i = g / h
j = f * i
k = e * j
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a ) 14 , b ) 15 , c ) 16 , d ) 18 , e ) 20 | b | add(add(10, 12), 10) | two spherical balls lie on the ground touching . if one of the balls has a radius of 10 cm , and the point of contact is 12 cm above the ground , what is the radius of the other ball ( in centimeters ) ? | "a straight line will join the two centers and the point of contact , thus making similar triangles . 2 / 10 = ( r - 12 ) / r 2 r = 10 r - 120 r = 15 the answer is b ." | a = 10 + 12
b = a + 10
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a ) 5 / 8 , b ) 5 / 6 , c ) 5 / 1 , d ) 5 / 9 , e ) 5 / 2 | d | multiply(add(divide(const_3, add(6, 3)), divide(7, add(7, 2))), divide(const_1, const_2)) | there are two bags a and b . a contains 6 red flowers and 3 pink flowers . where as bag b contains 2 red flowers and 7 pink flowers . one flower is chosen from a bag randomly . what is the probability that the flower chosen is pink ? | bag a 6 red + 3 pink bag b 2 red + 7 pink probability of picking 1 pink from bag a is 3 / 9 probability of picking 1 pink from bag b is 7 / 9 total = 3 / 9 + 7 / 9 = 10 / 9 and probability of selecting a bag is 1 / 2 so finally probability of choosing a pink flower = 1 / 2 ( 10 / 9 ) = 5 / 9 answer : d | a = 6 + 3
b = 3 / a
c = 7 + 2
d = 7 / c
e = b + d
f = 1 / 2
g = e * f
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a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 10 | c | subtract(divide(multiply(50, 15), 10), 50) | a mixture of 50 litres of milk and water contains 10 % water . how many litres of water should be added to the mixture so that the mixture contains 15 % water ? | quantity of milk in the mixture = 90 / 100 ( 50 ) = 45 litres . after adding water , milk would form 60 % of the mixture . hence , if quantity of mixture after adding x liters of water , ( 60 ) / 100 x = 45 = > x = 75 hence 75 - 50 = 25 litres of water must be added . answer : c | a = 50 * 15
b = a / 10
c = b - 50
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a ) 36 , b ) 77 , c ) 27 , d ) 33 , e ) 99 | c | subtract(multiply(30, multiply(const_60.0, const_0_2778)), 100) | a train 100 m long crosses a platform 125 m long in 30 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 30 s = 225 / 30 * 18 / 5 = 27 kmph answer : c" | a = const_60 * 0
b = 30 * a
c = b - 100
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a ) $ 9.28 , b ) $ 11.30 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10 | a | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent computer b . computer b would , however , require 20 hours more than computer a to do the job . if either computer a , or computer b were rented the total cost to rent the computer would be $ 650.00 . what would be the approximate hourly charge to rent computer b ? | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 650 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 650 / 70 ~ 9.28 a" | a = 40 / 100
b = 4 + 3
c = 40 / 100
d = 20 / 4
e = d / 100
f = c + e
g = b + f
h = a + g
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | b | divide(subtract(add(multiply(10, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))), multiply(20, subtract(20, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))))), multiply(20, 16)), subtract(20, 16)) | each of the 20 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 16 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes must be removed ? | "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 20 boxes , 4 are 10 - pound boxes and 16 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 16 , the ratio of 10 - pound boxes : 20 - pound boxes should be 2 : 3 . the number of 10 pound boxes remains the same so we still have 4 of them . to get a ratio of 2 : 3 , the number of 20 - pound boxes must be 6 . we need to remove 10 of the 20 - pound boxes . the answer is b ." | a = 20 * 20
b = 20 * 18
c = a - b
d = 20 - 10
e = c / d
f = 10 * e
g = 20 * 20
h = 20 * 18
i = g - h
j = 20 - 10
k = i / j
l = 20 - k
m = 20 * l
n = f + m
o = 20 * 16
p = n - o
q = 20 - 16
r = p / q
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a ) 30 , b ) 25 , c ) 33.33 , d ) 44.44 , e ) 28 | c | multiply(divide(subtract(multiply(6, 5), multiply(5, 4)), multiply(6, 5)), const_100) | a television have side 6 * 5 . its screen size is 5 * 4 . calculate the percentage of unscreened part t . v . | unscreened part = 30 - 20 = 10 . % unscreened part = 10 / 30 * 100 = 33.33 % answer c | a = 6 * 5
b = 5 * 4
c = a - b
d = 6 * 5
e = c / d
f = e * 100
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a ) 31512 , b ) 32512 , c ) 33512 , d ) 34512 , e ) 37512 | e | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(5, 5)), multiply(const_4, const_2))), const_10) | how many 5 digit number contain number 3 ? | "total 5 digit no . = 9 * 10 * 10 * 10 * 10 = 90000 not containing 3 = 8 * 9 * 9 * 9 * 9 = 52488 total 5 digit number contain 3 = 90000 - 52488 = 37512 answer : e" | a = 1000 - 10
b = 5 * 5
c = 10 * b
d = 4 * 2
e = c * d
f = a - e
g = f + 10
|
a ) 150 m , b ) 300 m , c ) 400 m , d ) 180 m , e ) 160 m | a | divide(multiply(15, 100), subtract(25, 15)) | a train crosses a post in 15 seconds and a platform 100 m long in 25 seconds . its length is | explanation : assume x is the length of the train and v is the speed x / v = 15 = > v = x / 15 ( x + 100 ) / v = 25 = > v = ( x + 100 ) / 25 ie , x / 15 = ( x + 100 ) / 25 = > 5 x = 3 x + 300 = > x = 300 / 2 = 150 answer : option a | a = 15 * 100
b = 25 - 15
c = a / b
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a ) 2 , b ) 8 , c ) 24 , d ) 25 , e ) 36 | e | divide(49, multiply(const_10, const_2)) | how many factors does 49 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 6 * 6 = 36 answer e ." | a = 10 * 2
b = 49 / a
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a ) 56 , b ) 60 , c ) 62 , d ) 64 , e ) 65 | d | divide(power(power(25, const_2), const_3), power(25, const_3)) | a cube of side one meter length is cut into small cubes of side 25 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 100 / 25 = 4 along each edge 25 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 4 * 4 * 4 = 64 answer : d" | a = 25 ** 2
b = a ** 3
c = 25 ** 3
d = b / c
|
a ) 180 % , b ) 360 % , c ) 200 % , d ) 220 % , e ) 250 % | b | multiply(divide(multiply(divide(40, const_100), multiply(divide(subtract(const_100, 10), const_100), const_100)), multiply(divide(10, const_100), const_100)), const_100) | last week john spent 10 percent of his wages on recreation . this week , his wages are 10 percent less than last week ΚΌ s wages and he spent 40 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week | "say john ' s wages last week were $ 100 , so he spent 0.10 * 100 = $ 10 on recreation ; this week ' s wages is 0.90 * 100 = $ 90 , so he spends 0.4 * 90 = $ 36 on recreation ; 36 / 10 = 3.60 , hence the amount he spends on recreation this week is 360 % of the amount he spent on recreation last week : 10 * 3.6 = 36 answer : b" | a = 40 / 100
b = 100 - 10
c = b / 100
d = c * 100
e = a * d
f = 10 / 100
g = f * 100
h = e / g
i = h * 100
|
a ) 7 / 17 , b ) 14 / 31 , c ) 4 / 9 , d ) 17 / 35 , e ) 1 / 2 | c | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 8 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 8 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 8 / 5 = 8 - - > total time for both of them would be 10 + 8 = 18 - - > deluxe / total = 8 / 18 = 4 / 9 . answer : c ." | a = 3 + 2
b = a / 10
|
a ) 7 : 3 , b ) 6 : 3 , c ) 8 : 3 , d ) 2 : 3 , e ) 6 : 3 | a | divide(add(multiply(3, divide(subtract(multiply(const_2, add(10, 10)), add(10, 10)), subtract(3, const_2))), 10), add(divide(subtract(multiply(const_2, add(10, 10)), add(10, 10)), subtract(3, const_2)), 10)) | the age of mother one decade age was 3 times the age of her daughter . 10 years hence , mother β s age will be two times that of her daughter . the proposition of their current ages is : | a 7 : 3 let , ages of mother and daughter 1 decade ago was 3 a and a years correspondingly then , ( 3 a + 10 ) + 10 = 2 [ ( a + 10 ) + 10 ] 3 a + 20 = 2 a + 40 x = 20 needed ratio = ( 3 a + 10 ) : ( a + 10 ) = 70 : 30 = 7 : 3 | a = 10 + 10
b = 2 * a
c = 10 + 10
d = b - c
e = 3 - 2
f = d / e
g = 3 * f
h = g + 10
i = 10 + 10
j = 2 * i
k = 10 + 10
l = j - k
m = 3 - 2
n = l / m
o = n + 10
p = h / o
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | d | divide(divide(multiply(30, 40), multiply(6, 2)), const_10) | a line with slope 2 intersects a line with slope 6 at the point ( 40 , 30 ) . what is the distance between the x - intercepts of these lines ? | the line with slope 2 passing through ( 40 , 30 ) has x - intercept 40 β ( 30 / 2 ) . the line with slope 6 passing through ( 40 , 30 ) has x - intercept 40 β ( 30 / 6 ) . so the difference in the x - intercepts is ( 30 / 2 ) β ( 30 / 6 ) = 10 . correct answer d | a = 30 * 40
b = 6 * 2
c = a / b
d = c / 10
|
a ) 7 , b ) 8 , c ) 9 , d ) 18 , e ) 11 | d | multiply(divide(subtract(47, const_2), add(const_3, const_2)), const_2) | is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , then how old is b ? | let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 = > x = 9 hence , b ' s age = 2 x = 18 years . answer : d | a = 47 - 2
b = 3 + 2
c = a / b
d = c * 2
|
a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 200 , e ) $ 260 | d | divide(multiply(20, const_100), subtract(multiply(add(12, 1), 10), add(const_100, 20))) | right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al β s account were to increase by 10 % and eliot β s account were to increase by 20 % , then al would have exactly $ 20 more than eliot in his account . how much money does eliot have in his account right now ? | "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 20 + 1.2 e = = > 11 a = 200 + 12 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 13 e = 200 + 12 e = = > e = 200 d" | a = 20 * 100
b = 12 + 1
c = b * 10
d = 100 + 20
e = c - d
f = a / e
|
a ) 100000 , b ) 85236 , c ) 89562 , d ) 90083 , e ) 95842 | d | add(const_100, multiply(divide(9000, subtract(subtract(const_100.0, 45), 45)), 83)) | at an election 2 candidates are participated a candidate got 45 % of votes and defeated by 9000 votes . 83 votes are invalid . find the total polled votes ? | "winners votes = 100 - 45 = 55 % invalid votes = 83 polled votes = [ ( 100 * 9000 ) / ( 2 * 55 - 100 ) ] + 83 = 90083 answer is d" | a = 100 - 0
b = a - 45
c = 9000 / b
d = c * 83
e = 100 + d
|
a ) 5 / 6 , b ) 4 / 5 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 2 | b | subtract(const_1, divide(6, multiply(6, const_3))) | a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 6 seconds ? | "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 6 / 30 = 1 / 5 . the probability that he will stay in the dark is 1 - 1 / 5 = 4 / 5 the answer is b ." | a = 6 * 3
b = 6 / a
c = 1 - b
|
a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 40 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 12), subtract(const_100, 20)), const_100), subtract(const_100, 20)) | a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 20 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 112 - 80 ) / 80 * 100 = ~ 40 % answer : d ." | a = 100 + 12
b = 100 - 20
c = a - b
d = c * 100
e = 100 - 20
f = d / e
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | subtract(9, 3) | tough and tricky questions : number properties . if ( z + 3 ) / 9 is an integer , what is remainder when z is divided by 9 ? | assume the answer choices as the value of z 9 / 9 = 1 ans e | a = 9 - 3
|
a ) 300 , b ) 250 , c ) 1223.53 , d ) 1133.24 , e ) 776.47 | c | subtract(multiply(multiply(const_100, 9), const_2), divide(add(multiply(multiply(9, 8), const_2), 48), add(divide(9, const_100), divide(8, const_100)))) | if x dollars is invested at 9 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 9 percent investment will exceed the annual income from the 8 percent investment by $ 48 . if $ 2,000 is the total amount invested , how much is invested at 9 percent ? | "2 equations with 2 unknowns 9 x / 100 - 8 y / 100 = 48 and x + y = 2000 solving these 2 equations , x = 1223.53 and y = 776.47 answer - c" | a = 100 * 9
b = a * 2
c = 9 * 8
d = c * 2
e = d + 48
f = 9 / 100
g = 8 / 100
h = f + g
i = e / h
j = b - i
|
a ) 11 , b ) 14 , c ) 14 , d ) 12 , e ) 19 | e | divide(948, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 8)) | if the wheel is 8 cm then the number of revolutions to cover a distance of 948 cm is ? | "2 * 22 / 7 * 8 * x = 948 = > x = 18.9 answer : e" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 8
h = 948 / g
|
a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | divide(multiply(11, const_4), add(const_4, const_1)) | the cost price of 11 articles is equal to the selling price of 10 articles . what is the profit percent ? | "10 * sp = 11 * cp sp = 1.1 * cp the profit percent is 10 % . the answer is b ." | a = 11 * 4
b = 4 + 1
c = a / b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value d of z - x ? | "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] d = z - x = x + 9 - x = 9 ans : d" | a = 5 + 2
b = a + 2
|
a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | divide(multiply(multiply(add(const_2, const_3), const_1000), 8), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 70 / r years . if pat ' s parents invested $ 5,000 in a long - term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of the investment 18 years later , when pat is ready for college ? | since investment doubles in 70 / r years , then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,000 and in another 9 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a . | a = 2 + 3
b = a * 1000
c = b * 8
d = c / 2
|
a ) 9 , b ) 16 , c ) 17 , d ) none of these , e ) can not be determined | b | add(divide(subtract(90, 5), 5), const_1) | how many multiples of 5 are there between 5 to 90 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 . answer : b" | a = 90 - 5
b = a / 5
c = b + 1
|
a ) 97.765 % , b ) 99.765 % , c ) 96.765 % , d ) 95.765 % , e ) 98.765 % | e | subtract(const_100, multiply(divide(divide(9, 9), multiply(9, 9)), const_100)) | instead of multiplying a number by 9 , the number is divided by 9 . what is the percentage of error obtained ? | "let the number be x the right number is 9 x the wrong number is x / 9 error is ( 9 x - x / 9 ) = 80 x / 9 percentage of error is ( ( 80 x / 9 ) / 9 x ) * 100 = 98.765 % answer : e" | a = 9 / 9
b = 9 * 9
c = a / b
d = c * 100
e = 100 - d
|
a ) 12.9 , b ) 12.0 , c ) 14 , d ) 15 , e ) 16 | c | divide(multiply(17, 1500), add(1500, 320)) | 1500 men have provisions for 17 days . if 320 more men join them , for how many days will the provisions last now ? | "1500 * 17 = 1820 * x x = 14 answer : c" | a = 17 * 1500
b = 1500 + 320
c = a / b
|
a ) 6 meters , b ) 8 meters , c ) 9 meters , d ) 10 meters , e ) 12 meters | a | divide(multiply(divide(multiply(36, const_100), 30), divide(75, const_100)), 15) | the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paisa per meter is rs . 36 . the breadth of the room is ? | length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 β΄ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a | a = 36 * 100
b = a / 30
c = 75 / 100
d = b * c
e = d / 15
|
a ) 54.0 , b ) 54.9 , c ) 18.5 , d ) 57.0 , e ) 63.0 | c | add(add(multiply(divide(subtract(const_100, 20), const_100), 6), 6), multiply(divide(add(const_100, 25), const_100), 6)) | james took a 3 - hour bike ride . in the second hour he traveled 6 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 = 6 , x = 5 . now , the distance travelled in the 3 rd hour = 6 + 1 / 4 Γ’ Λ β 6 = 7.5 the only option ending with a 0.5 in the decimal place is c . answer : c" | a = 100 - 20
b = a / 100
c = b * 6
d = c + 6
e = 100 + 25
f = e / 100
g = f * 6
h = d + g
|
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 100), 9599) | what least number must be added to 9599 , so that the sum is completely divisible by 100 ? | "if we divide 9599 by 100 remainder is 99 100 - 99 = 1 answer : a" | a = 4 * 10
b = a + 2
c = b * 100
d = c - 9599
|
a ) 40 , b ) 52 , c ) 60 , d ) 80 , e ) 98 | c | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(28, const_100), 120)), subtract(const_1, divide(52, const_100))) | in august , a cricket team that played 120 matches won 28 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 33.6 + x by solving we get x = 60 answer : c" | a = 52 / 100
b = a * 120
c = 28 / 100
d = c * 120
e = b - d
f = 52 / 100
g = 1 - f
h = e / g
|
a ) 45 , b ) 55 , c ) 35 , d ) 65 , e ) 70 | a | divide(add(35, 55), const_2) | a fish can go upstream at 35 kmph and downstream at 55 kmph , and then find the speed of the fish in still water ? | us = 35 ds = 55 m = ( 35 + 55 ) / 2 = 45 answer : a | a = 35 + 55
b = a / 2
|
a ) 96 , b ) 20 , c ) 35 , d ) 78 , e ) 89 | a | subtract(const_100, multiply(divide(add(20, const_100), add(25, const_100)), const_100)) | two numbers are respectively 20 % and 25 % more than a third number . so find the percentage that is first of the second is ? | "first - 120 second - 125 third - 100 125 - - - - 120 100 - - - - - 96 answer a" | a = 20 + 100
b = 25 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 20029 , b ) 22500 , c ) 20289 , d ) 20027 , e ) 20026 | b | divide(450, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 450 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 450 = > x = 22500 answer : b" | a = 5 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 10 / 100
f = 20 / 100
g = e * f
h = d - g
i = 450 / h
|
a ) 150 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(126, const_100), subtract(const_100, 16)) | the number which exceeds 16 % of it by 126 is : | "solution solution let the number be x . x - 16 % of x = 126 x - 16 / 100 x = 126 x - 4 / 25 x = 126 21 / 25 x = 126 x = ( 126 x 25 / 21 ) = 150 answer a" | a = 126 * 100
b = 100 - 16
c = a / b
|
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