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 ) 455 , b ) 570 , c ) 480 , d ) 520 , e ) 550 | a | divide(182, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 182 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 182 = > 4 x / 10 = 182 = > 4 x = 1820 = > x = 455 . answer : a" | a = 100 - 70
b = 70 - a
c = b / 100
d = 182 / c
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a ) 450 , b ) 451 , c ) 460 , d ) 462 , e ) none | d | add(457, subtract(11, reminder(457, 11))) | find the number which is nearest to 457 and is exactly divisible by 11 . | "solution on dividing 457 by 11 , remainder is 6 . required number is either 451 or 462 nearest to 456 is = 462 . answer d" | a = 11 - reminder
b = 457 + a
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a ) 18 , b ) 14 , c ) 20 , d ) 15 , e ) 15.55 | e | multiply(divide(subtract(520, 450), 450), const_100) | a cycle is bought for rs . 450 and sold for rs . 520 , find the gain percent ? | "450 - - - - 70 100 - - - - ? = > 15.55 % answer : e" | a = 520 - 450
b = a / 450
c = b * 100
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | 150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration r of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration r would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a = 150 * 30
b = a / 100
c = 400 * 12
d = c / 100
e = b + d
f = 150 + 400
g = e / f
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a ) 20.0015 , b ) 20.0105 , c ) 20.0115 , d ) 20.0114 , e ) none | d | subtract(add(12.1212, 17.0005), 9.1103) | 12.1212 + 17.0005 - 9.1103 = ? | solution given expression = ( 12.1212 + 17.0005 ) - 9.1103 = ( 29.1217 - 9.1103 ) = 20.0114 . answer d | a = 12 + 1212
b = a - 9
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a ) 123 , b ) 121 , c ) 277 , d ) 267 , e ) 120 | b | divide(add(104, 138), 2) | a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 104 . what was the number he chose ? | "let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 104 2 â ‹ … x â ˆ ’ 138 = 104 x = 121 answer : b" | a = 104 + 138
b = a / 2
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a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 10 and 10 | c | subtract(add(divide(18, 2), 1), 3) | one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 18 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x − 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x − 3 + 1 ) = ( x − 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) ⋅ ( x − 2 ) cm ... | a = 18 / 2
b = a + 1
c = b - 3
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 25 % , e ) 60 % | d | subtract(100, 75) | john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 75 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 75 = 25 . the â € œ original â € is our starting point ; in this case , it â € ™ s 100 . ( 25 / 100 ) * 100 = ( 0.25 ) * 100 = 25 % . d" | a = 100 - 75
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(floor(divide(log(const_3), log(add(const_1, divide(30, const_100))))), const_1) | an investment compounds annually at an interest rate of 30 % what is the smallest investment period by which time the investment will more than double in value ? | "1 year : 100 / 3 = 33.33 approx $ 34 : total : 134 2 nd year : 134 / 3 = 45 : total : 134 + 45 = 179 3 rd year : 179 / 3 = 60 : total : 179 + 60 = 239 > 2 ( 100 ) ; 3 years ; answer : a" | a = math.log(3)
b = 30 / 100
c = 1 + b
d = math.log(c)
e = a / d
f = math.floor(e)
g = f + 1
|
a ) 1960 , b ) 1965 , c ) 1970 , d ) 2050 , e ) 2060 | e | add(1950, add(1.25, add(const_4, const_3))) | in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.25 as old as robert ? | in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 ro = 10 , ri = 40 now for each year we can calculate : 1960 : ri = 50 , ro = 20 1965 : ri = 55 , ro = 25 2060 : ri = 120 , ro = 150 thus ans : e | a = 4 + 3
b = 1 + 25
c = 1950 + b
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a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord q , in feet ? | because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . q = pi * r = 30 . round pi = 3 , then r = 10... | a = 30 / 3
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a ) 1.6 , b ) 4.5 , c ) 8.2 , d ) 6.5 , e ) 2.9 | b | divide(divide(300, const_1000), divide(multiply(4, const_60), const_3600)) | a person crosses a 300 m long street in 4 minutes . what is his speed in km per hour ? | "distance = 300 meter time = 4 minutes = 4 x 60 seconds = 240 seconds speed = distance / time = 300 / 240 = 1.25 m / s = 1.25 ã — 18 / 5 km / hr = 4.5 km / hr answer : b" | a = 300 / 1000
b = 4 * const_60
c = b / 3600
d = a / c
|
a ) 51 , b ) 52 , c ) 53 , d ) 54 , e ) 56 | d | subtract(multiply(28, const_3), 28) | jerry and michelle play a card game . in the beginning of the game they have an equal number of cards . each player , at her turn , gives the other a third of her cards . michelle plays first , giving jerry a third of her cards . jerry plays next , and michelle follows . then the game ends . jerry ended up with 28 more... | "gamemichelle jerry initially 54 54 assume after game 1 36 72 after game 2 60 48 after game 3 40 68 now merry has 28 cards more than michelle . this option gives us exactly what number of cards they had initially . so the answer is d" | a = 28 * 3
b = a - 28
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a ) 643 , b ) 652 , c ) 660 , d ) 690 , e ) 693 | c | multiply(divide(693, add(const_1, divide(26, const_100))), add(const_1, divide(20, const_100))) | if albert ’ s monthly earnings rise by 26 % , he would earn $ 693 . if , instead , his earnings rise by only 20 % , how much ( in $ ) would he earn this month ? | = 693 / 1.26 ∗ 1.2 = 660 = 660 answer is c | a = 26 / 100
b = 1 + a
c = 693 / b
d = 20 / 100
e = 1 + d
f = c * e
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a ) $ 45 , b ) $ 48 , c ) $ 50 , d ) $ 54 , e ) $ 56 | c | divide(36, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100))) | what is the normal price of an article sold at $ 36 after two successive discounts of 10 % and 20 % ? | "0.8 * 0.9 * cost price = $ 36 cost price = $ 50 the answer is c ." | a = 100 - 10
b = a / 100
c = 100 - 20
d = c / 100
e = b * d
f = 36 / e
|
a ) - 16 , b ) - 12 , c ) - 8 , d ) - 6 , e ) 12 | d | add(3, 5) | if | x + 3 | = 5 , what is the sum of all the possible values of x ? | "there will be two cases x + 3 = 5 or x + 3 = - 5 = > x = 2 or x = - 8 sum of both the values will be - 8 + 2 = - 6 answer : d" | a = 3 + 5
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a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12 | c | subtract(14, const_4) | in a group of cows and hens , the number of legs are 14 more than twice the number of heads . the number of cows is : | "let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 14 2 x = 14 x = 7 . answer : c" | a = 14 - 4
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a ) 9 : 6 , b ) 3 : 8 , c ) 3 : 1 , d ) 9 : 9 , e ) 3 : 4 | e | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(6000, add(6, 6))) | a and b invests rs . 3000 and rs . 6000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 6 * 12 ) 18 : 24 = > 3 : 4 . answer : e" | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 6000 * e
g = d / f
|
a ) 1299 , b ) 2788 , c ) 2000 , d ) 2981 , e ) 2881 | c | multiply(20, const_100) | albert buys 4 horses and 9 cows for rs . 13,400 . if he sells the horses at 10 % profit and the cows at 20 % profit , then he earns a total profit of rs . 1880 . the cost of a horse is ? | explanation : let c . p . of each horse be rs . x and c . p . of each cow be rs . y . then , 4 x + 9 y = 13400 - - ( i ) and , 10 % of 4 x + 20 % of 9 y = 1880 2 / 5 x + 9 / 5 y = 1880 = > 2 x + 9 y = 9400 - - ( ii ) solving ( i ) and ( ii ) , we get : x = 2000 and y = 600 . cost price of each horse = rs . 2000 . answe... | a = 20 * 100
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a ) 20 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | divide(8, subtract(76.4, floor(76.4))) | when positive integer x is divided by positive integer y , the remainder is 8 . if x / y = 76.4 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 8 - - > x = qy + 8 ; x / y = 76.4 - - > x = 76 y + 0.4 y ( so q above equals to 76 ) ; 0.4 y = 8 - - > y = 20 . answer : a ." | a = math.floor(76, 4)
b = 76 - 4
c = 8 / b
|
a ) $ 1,000 , b ) $ 1,250 , c ) $ 1,400 , d ) $ 1,800 , e ) $ 2,200 | b | subtract(1,000, 850) | a family pays $ 850 per year for an insurance plan that pays 80 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total . | "upfront payment for insurance plan = 850 $ family needs to pay 20 % of first 1000 $ in expense = 200 $ total amount paid by family when medical expenses are equal to or greater than 1000 $ = 850 + 200 = 1050 $ total amount paid by insurance plan for first 1000 $ = 850 $ total amount paid by family will equal amount pa... | a = 1 - 0
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a ) 60 / 120 , b ) 15 / 120 , c ) 30 / 120 , d ) 80 / 120 , e ) 40 / 120 | c | divide(subtract(120, 90), 120) | in a group of 120 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ? | "number of people whose age is less than 20 is given by 120 - 90 = 30 probability p that a person selected at random from the group is less than 20 is gieven by 30 / 120 = 0.25 correct answer c" | a = 120 - 90
b = a / 120
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a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 70 % , e ) 81 % | e | add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 10))), const_100), 2) | the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 10 % , price = 81 - 8.1 = 72.9 which is 72.9 / 90 * 100 of the sale price on day 1 = ~ 81 % answer e" | a = 10 / 100
b = 1 / 10
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 1.9432 , 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 = 19432 , find the value of 2.68 x . 74 . | "solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 × . 74 = 1.9432 answer a" | a = 268 / 100
b = 74 / 100
c = a * b
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a ) 300 , b ) 350 , c ) 400 , d ) 450 , e ) 500 | d | multiply(divide(subtract(22, 10), subtract(30, 22)), 300) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 300 milliliters of solution x to create a solution that is 22 percent alcohol by volume ? | "22 % is 12 % - points higher than 10 % but 8 % - points lower than 30 % . thus there should be 2 parts of solution x for 3 parts of solution y . we should add 450 ml of solution y . the answer is d ." | a = 22 - 10
b = 30 - 22
c = a / b
d = c * 300
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a ) 50 min , b ) 20 min , c ) 1 hour , d ) 30 min , e ) 45 min | a | divide(10, subtract(inverse(divide(5, 6)), const_1)) | walking at 5 / 6 of its usual speed , a train is 10 minutes too late . find its usual time to cover the journey . | "new speed = 5 / 6 of the usual speed new time taken = 6 / 5 of the usual time taken so , ( 6 / 5 of the usual time ) - ( usual time ) = 10 min 1 / 5 of the usual time = 10 min usual time = 10 * 5 = 50 min correct option is a" | a = 5 / 6
b = 1/(a)
c = b - 1
d = 10 / c
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a ) 1 meter , b ) 2 meter , c ) 3 meter , d ) 4 meter , e ) 5 meter | a | subtract(subtract(divide(40, const_12), const_0_33), const_2) | in the rectangle below , the line mn cuts the rectangle into two regions . find x the length of segment nb so that the area of the quadrilateral mnbc is 40 % of the total area of the rectangle . | solution we first note that mc = 20 - 5 = 15 the quadrilateral mnbc is a trapezoid and its area a is given by a = ( 1 / 2 ) × 10 × ( x + mc ) = 5 ( x + 15 ) 40 % of the area of the rectangle is equal to 40 % × ( 20 × 10 ) = ( 40 / 100 ) × 200 = 80 since the area of mnbc is equal to 40 % the area of the rectangle , we c... | a = 40 / 12
b = a - const_0_33
c = b - 2
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a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec | c | divide(multiply(120, const_2), add(speed(120, 20), speed(120, 15))) | two trains of equal lengths take 15 sec and 20 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 15 = 8 m / sec . speed of the second train = 120 / 20 = 6 m / sec . relative speed = 8 + 6 = 14 m / sec . required time = ( 120 + 120 ) / 14 = 17 sec . answer : c" | a = 120 * 2
b = speed + (
c = a / b
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a ) 43 , b ) 42 , c ) 63 , d ) 65 , e ) 78 | b | subtract(negate(7), multiply(subtract(12,18, 25,33), divide(subtract(12,18, 25,33), subtract(3,7, 12,18)))) | 3,7 , 12,18 , 25,33 , . . . . . . . . . . . . . . 7 th terms | "3 + 4 = 7 7 + 5 = 12 12 + 6 = 18 18 + 7 = 25 25 + 8 = 33 33 + 9 = 42 answer : b" | a = negate - (
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a ) 144 m 2 , b ) 120 m 2 , c ) 108 m 2 , d ) 158 m 2 , e ) none of these | a | subtract(rectangle_area(add(multiply(2, 2), 20), add(12, multiply(2, 2))), rectangle_area(20, 12)) | the floor of a rectangular room is 20 m long and 12 m wide . the room is surrounded by a veranda of width 2 m on all its sides . the area of the veranda is : | "area of the outer rectangle = 24 ã — 16 = 384 m 2 area of the inner rectangle = 20 ã — 12 = 240 m 2 required area = ( 304 â € “ 180 ) = 144 m 2 answer a" | a = 2 * 2
b = a + 20
c = 2 * 2
d = 12 + c
e = rectangle_area - (
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a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 99 | e | add(add(90, 4), add(4, 1)) | the average weight of a class is x pounds . when a new student weighing 90 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the valu... | when the student weighs 90 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 90 = 20 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 20 / 5 =... | a = 90 + 4
b = 4 + 1
c = a + b
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a ) s . 550 , b ) s . 1000 , c ) s . 600 , d ) s . 200 , e ) s . 400 | e | divide(200, multiply(divide(5, const_100), 10)) | a sum was put at simple interest at a certain rate for 10 years . had it been put at 5 % higher rate , it would have fetched rs . 200 more . what was the sum ? | "at 5 % more rate , the increase in s . i for 10 years = rs . 200 ( given ) so , at 5 % more rate , the increase in si for 1 year = 200 / 10 = rs . 20 / - i . e . rs . 20 is 5 % of the invested sum so , 1 % of the invested sum = 20 / 5 therefore , the invested sum = 20 × 100 / 5 = rs . 400 answer : e" | a = 5 / 100
b = a * 10
c = 200 / b
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a ) 22 , b ) 12 , c ) 03 , d ) 04 , e ) 21 | e | subtract(divide(29, const_10), divide(power(const_2, const_3), const_10)) | the sum of a certain number and its reciprocal is equal to 29 . what is the absolute value of the difference of this number and its reciprocal ? | x + 1 / x = 2.9 x - 1 / x = y add two equations 2 x = 2.9 + y so when 2.9 and correct answer choice are added , the final digit should be even so eliminate a , b and d substitute y = 0.3 x = 1.6 1.6 + 1 / 1.6 is not equal to 2.9 substitute y = 2.1 so x = 2.5 2.5 + 1 / 2.5 = 2.9 answer : e | a = 29 / 10
b = 2 ** 3
c = b / 10
d = a - c
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a ) 50 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | c | subtract(subtract(500, const_1), subtract(add(add(divide(subtract(subtract(500, const_2), const_2), const_2), const_1), add(divide(subtract(subtract(500, divide(500, const_100)), divide(500, const_100)), divide(500, const_100)), const_1)), add(divide(subtract(subtract(500, const_10), const_10), const_10), 1))) | what is the total number of positive integers that are less than 500 and that have no positive factor in common with 500 other than 1 ? | since 500 = 2 ^ 2 * 5 ^ 3 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 250 odd numbers from 1 to 500 . we then need to eliminate the 50 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 495 . there are a total of 250 - 50 = 200 such numbers between ... | a = 500 - 1
b = 500 - 2
c = b - 2
d = c / 2
e = d + 1
f = 500 / 100
g = 500 - f
h = 500 / 100
i = g - h
j = 500 / 100
k = i / j
l = k + 1
m = e + l
n = 500 - 10
o = n - 10
p = o / 10
q = p + 1
r = m - q
s = a - r
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a ) 7.5 , b ) 7.4 , c ) 7.9 , d ) 7.2 , e ) 8.0 | e | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 8))) | a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 8 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 9 ) = 1 / 8 the cistern will be filled in 8 / 1 hrs i . e . , 8 hrs . answer : e" | a = 1 / 4
b = 1 / 8
c = a - b
d = 1 / c
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a ) 5 / 19 , b ) 5 / 12 , c ) 5 / 28 , d ) 5 / 10 , e ) 5 / 11 | b | divide(add(add(7, const_4), const_4), multiply(add(const_4, const_2), add(const_4, const_2))) | in a simultaneous throw of pair of dice . find the probability of getting the total more than 7 . | "explanation : here n ( s ) = ( 6 x 6 ) = 36 let e = event of getting a total more than 7 = { ( 2,6 ) , ( 3,5 ) , ( 3,6 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 ) , ( 5,3 ) , ( 5,4 ) , ( 5,5 ) , ( 5,6 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) } therefore , p ( e ) = n ( e ) / n ( s ) = 15 / 36 = 5 / 12 . answer : b ) 5 ... | a = 7 + 4
b = a + 4
c = 4 + 2
d = 4 + 2
e = c * d
f = b / e
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a ) rs . 78000 , b ) rs . 48000 , c ) rs . 6000 , d ) rs . 82000 , e ) rs . 32000 | a | multiply(6000, add(multiply(6, const_2), const_1)) | krishan and nandan jointly started a business . krishan invested 6 times as nandan did and invested his money for double time as compared to nandan . nandan earned rs . 6000 . if the gain is proportional to the money invested and the time for which the money is invested then the total gain was ? | 6 : 1 2 : 1 - - - - - - 12 : 1 1 - - - - - 6000 13 - - - - - ? = > rs . 78,000 answer : a | a = 6 * 2
b = a + 1
c = 6000 * b
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a ) 1700 , b ) 7570 , c ) 1576 , d ) 17576 , e ) 500 | d | power(add(add(add(const_10, const_10), const_3), const_3), subtract(5, const_2)) | consider the word rotor . whichever way you read it , from left to right or from right to left , you get the same word . such a word is known as palindrome . find the maximum possible number of 5 - letter palindromes | the first letter from the right can be chosen in 26 ways because there are 26 alphabets . having chosen this , the second letter can be chosen in 26 ways . = > the first two letters can be chosen in 26 ã — 26 = 67626 ã — 26 = 676 ways having chosen the first two letters , the third letter can be chosen in 26 ways . = >... | a = 10 + 10
b = a + 3
c = b + 3
d = 5 - 2
e = c ** d
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a ) 0.004 % , b ) 0.04 % , c ) 0.4 % , d ) 1.6 % , e ) 20 % | d | multiply(divide(multiply(0.04, 10), 25), const_100) | a glass was filled with 25 ounces of water , and 0.04 ounce of the water evaporated each day during a 10 - day period . what percent of the original amount of water evaporated during this period ? | "in 10 days 10 * 0.04 = 0.4 ounces of water evaporated , which is 0.4 / 25 â ˆ — 100 = 1.6 of the original amount of water . answer : d ." | a = 0 * 4
b = a / 25
c = b * 100
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a ) 15 , b ) 60 , c ) 180 , d ) 90 , e ) 105 | c | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 37.5 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 37.5 miles . teena ... | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) rs . 15000 , b ) rs . 6000 , c ) rs . 25000 , d ) rs . 10000 , e ) rs . 18000 | a | multiply(subtract(add(multiply(divide(10, const_100), 10000), divide(8, const_100)), 850), const_100) | vijay lent out an amount rs . 10000 into two parts , one at 8 % p . a . and the remaining at 10 % p . a . both on simple interest . at the end of the year he received rs . 850 as total interest . what was the amount he lent out at 8 % pa . a ? | let the amount lent out at 8 % p . a . be rs . a = > ( a * 8 ) / 100 + [ ( 10000 - a ) * 10 ] / 100 = 850 = > a = rs . 15000 . answer : a | a = 10 / 100
b = a * 10000
c = 8 / 100
d = b + c
e = d - 850
f = e * 100
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a ) 8 , b ) 16 , c ) 24 , d ) 36 , e ) 40 | b | divide(64, 4) | find k if 64 ÷ k = 4 . | "since 64 ÷ k = 4 and 64 ÷ 16 = 4 , then k = 16 correct answer b" | a = 64 / 4
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a ) 165 , b ) 185 , c ) 190 , d ) 191 , e ) 199 | d | subtract(200, 9) | the mean of 50 observations is 200 . but later he found that there is decrements of 9 from each observations . what is the the updated mean is ? | "191 answer is d" | a = 200 - 9
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a ) $ 124 , b ) $ 132 , c ) $ 140 , d ) $ 148 , e ) $ 156 | c | divide(multiply(divide(multiply(184.80, const_100), add(const_100, 20)), const_100), add(const_100, 10)) | a couple spent $ 184.80 in total while dining out and paid this amount using a credit card . the $ 184.80 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? | "let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 184.80 x = 140 the correct answer is c ." | a = 184 * 80
b = 100 + 20
c = a / b
d = c * 100
e = 100 + 10
f = d / e
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a ) 5 , b ) 5 / 4 , c ) 4 / 5 , d ) 1 / 4 , e ) 1 / 5 | d | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 5 * / 6 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all th... | a = 1 / 3
b = a / 3
c = 1 + 4
d = b / c
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a ) 1344 , b ) 3360 , c ) 8400 , d ) 12600 , e ) 67200 | d | multiply(multiply(560, divide(30, 8)), 6) | running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 30 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 30 machine produce 560 * 6 * ( 30 / 8 ) in 6 minutes 560 * 6 * 30 / 8 = 12600 answer is d ." | a = 30 / 8
b = 560 * a
c = b * 6
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a ) 6 , b ) 7 , c ) 8 , d ) 8 1 / 9 , e ) 10 | d | multiply(add(2, 5), 5) | abel can complete a work in 10 days , ben in 12 days and carla in 15 days . all of them began the work together , but abel had to leave after 2 days and ben 5 days before the completion of the work . how long did the work last ? | "abel in the 2 days that he worked completed 1 / 5 of the job = 4 / 5 remains then if ben had to leave 5 days before the completion , this means that carla had to work alone for these 5 days in which she completed 1 / 3 of the job . now together , ben and carla completed the job in ( 1 / 12 + 1 / 15 ) ( t ) = 7 / 15 3 ... | a = 2 + 5
b = a * 5
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a ) 13 / 2 , b ) 7 , c ) 15 / 2 , d ) 8 , e ) 9 | c | divide(multiply(5, 3), subtract(5, 3)) | a man can do a piece of work in 5 days , but with the help of his son , he can do it in 3 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 3 ) - ( 1 / 5 ) = 2 / 5 the son alone can do the work in 15 / 2 days answer is c" | a = 5 * 3
b = 5 - 3
c = a / b
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a ) 3.625 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | a | add(divide(subtract(divide(45, 10), 2.75), const_2), 2.75) | annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 45 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exac... | "set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.75 from which t = 27.5 mins . we know total journey time now is 45 + 27.5 = 72.5 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 72.5 = 7.25 km now the total journey would be halved as distance would be same in each dir... | a = 45 / 10
b = a - 2
c = b / 2
d = c + 2
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a ) 12 , b ) 26 , c ) 16 , d ) 97 , e ) 20 | e | add(20, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 20 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "10 * 20 + 36 – 26 = 200 / 10 = 20 answer : e" | a = 36 - 26
b = a / 10
c = 20 + b
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a ) 6.18 , b ) 7.12 , c ) 7.1 , d ) 8.09 , e ) 8.11 | d | divide(add(161, 165), multiply(add(80, 65), const_0_2778)) | two trains 161 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 161 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 8.09 answer : d" | a = 161 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
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a ) rs . 3932 , b ) rs . 3972 , c ) rs . 3372 , d ) rs . 3942 , e ) rs . 3772 | c | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multip... | there is 60 % increase in an amount in 6 years at s . i . what will be the c . i . of rs . 12,000 after 3 years at the same rate ? | "explanation : let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = ( 100 * 60 ) / ( 100 * 6 ) = 10 % p . a . now , p = rs . 12000 , t = 3 years and r = 10 % p . a . c . i . = [ 12000 * { ( 1 + 10 / 100 ) 3 - 1 } ] = 12000 * 331 / 1000 = rs . 3972 answer : option c" | a = 100 * 100
b = 60 / 6
c = 100 * b
d = c * 3
e = a + d
f = 60 / 6
g = f / 100
h = 1 + g
i = 60 / 6
j = i / 100
k = 1 + j
l = h * k
m = 60 / 6
n = m / 100
o = 1 + n
p = l * o
q = e * p
r = 100 * 100
s = 60 / 6
t = 100 * s
u = t * 3
v = r + u
w = q - v
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a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | c | multiply(divide(160, 2), const_3) | in a mixed college 160 students are there in one class . out of this 160 students 1 / 2 students are girls . how many boys are there ? | "total number of students : 160 total girls : 160 * 1 / 2 = 80 total boys : 160 - 80 = 80 answer is c" | a = 160 / 2
b = a * 3
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a ) 75 kg , b ) 50 kg , c ) 85 kg , d ) 80 kg , e ) 98.6 kg | e | add(multiply(8, 4.2), 65) | the average weight of 8 person ' s increases by 4.2 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 ã — 4.2 = 33.6 if x is the weight of the new person , total increase in weight = x â ˆ ’ 65 = > 33.6 = x - 65 = > x = 33.6 + 65 = 98.6 answer : option e" | a = 8 * 4
b = a + 65
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 20 | a | subtract(divide(add(add(add(add(6, 6), 6), multiply(add(6, 6), 9)), const_2), add(6, const_2)), add(6, 6)) | on sunday morning , pugsley and wednesday are trading pet spiders . if pugsley were to give wednesday two of his spiders , wednesday would then have 9 times as many spiders as pugsley does . but , if wednesday were to give pugsley 6 of her spiders , pugsley would now have 6 fewer spiders than wednesday had before they ... | if pugsley were to give wednesday two of his spiders , wednesday would then have nine times as many spiders as pugsley does : ( w + 2 ) = 9 ( p - 2 ) if wednesday were to give pugsley six of her spiders , pugsley would now have six fewer spiders than wednesday had before they traded : p + 6 = w - 6 solving gives p = 4 ... | a = 6 + 6
b = a + 6
c = 6 + 6
d = c * 9
e = b + d
f = e + 2
g = 6 + 2
h = f / g
i = 6 + 6
j = h - i
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a ) $ 1250 , b ) $ 1350 , c ) $ 1150 , d ) $ 1360 , e ) $ 1370 | b | add(add(multiply(100, 10), multiply(30, 5)), multiply(multiply(20, 5), const_2)) | rates for having a manuscript typed at a certain typing service are $ 10 per page for the first time a page is typed and $ 5 per page each time a page is revised . if a certain manuscript has 100 pages , of which 30 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total... | "50 pages typed 1 x 30 pages typed 2 x ( original + one revision ) 20 pages typed 3 x ( original + two revisions ) 50 ( 10 ) + 30 ( 10 + 5 ) + 20 ( 10 + 5 + 5 ) = 500 + 450 + 400 = 1350 answer - b" | a = 100 * 10
b = 30 * 5
c = a + b
d = 20 * 5
e = d * 2
f = c + e
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a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | b | subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 74), add(divide(10, const_100), divide(8, const_100)))) | if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 74 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | "0.1 x = 0.08 ( 2000 - x ) + 74 0.18 x = 234 x = 1300 then the amount invested at 8 % is $ 2000 - $ 1300 = $ 700 the answer is b ." | a = 100 * 10
b = a * 2
c = 10 * 8
d = c * 2
e = d + 74
f = 10 / 100
g = 8 / 100
h = f + g
i = e / h
j = b - i
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a ) 3.57 % , b ) 5.9 % , c ) 11.1 % , d ) 12.5 % , e ) 23.6 % | a | multiply(subtract(divide(29, 28), const_1), const_100) | at the opening of a trading day at a certain stock exchange , the price per share of stock k was $ 28 . if the price per share of stock k was $ 29 at the closing of the day , what was the percent increase in the price per share of stock k for that day ? | "opening = 28 closing = 29 rise in price = 1 so , percent increase = 1 / 28 * 100 = 3.57 answer : a" | a = 29 / 28
b = a - 1
c = b * 100
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a ) 24440 , b ) 24800 , c ) 28400 , d ) 28800 , e ) 28600 | d | add(20000, multiply(divide(multiply(20000, 20), const_100), 2)) | the population of a town is 20000 . it increases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "explanation : formula : ( after = 100 denominator ago = 100 numerator ) 20000 * 120 / 100 * 120 / 100 = 120 * 120 * 2 = 28800 answer : option d" | a = 20000 * 20
b = a / 100
c = b * 2
d = 20000 + c
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a ) 8 , b ) 10 , c ) 12 , d ) 19 , e ) 16 | d | divide(subtract(subtract(add(multiply(subtract(55, 10), 10), multiply(10, 20)), 360), multiply(gcd(55, 10), 20)), 10) | bookman purchased 55 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 360 , how many paperback copies were sold ? | the bookman had 10 hardback ad 55 - 10 = 45 paperback copies ; 14 copies were sold , hence 55 - 14 = 41 copies were left . let # of paperback copies left be p then 10 p + 20 ( 41 - p ) = 560 - - > 10 p = 260 - - > p = 26 # of paperback copies sold is 45 - 26 = 19 answer : d | a = 55 - 10
b = a * 10
c = 10 * 20
d = b + c
e = d - 360
f = math.gcd(55, 10)
g = f * 20
h = e - g
i = h / 10
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number r of employees who take the subway to work ? | "p / 3 + f / 4 = p / 3 + ( 48 - p ) / 4 = 12 + p / 2 p / 3 + f / 3 = ( p + f ) / 3 = 48 / 3 = 16 p / 4 + f / 4 = 12 p / 3 + f / 3 > p / 3 + f / 4 > p / 4 + f / 4 - - > 16 > 12 + p / 12 > 12 greatest possible r : 12 + p / 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer" | a = 48 * 3
b = 3 * 4
c = a / b
d = 3 * 4
e = 48 - d
f = 48 * 3
g = 3 * 4
h = f / g
i = e / h
j = c + i
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['a ) 13 / 15 ohms', 'b ) 15 / 13 ohms', 'c ) 5 / 8 ohms', 'd ) 3 / 7 ohms', 'e ) 45 / 18 ohms'] | a | add(divide(const_1, 6), add(divide(const_1, 2), divide(const_1, 5))) | in an electric circuit , three resistors with resistances 2 ohms , 5 ohms and 6 ohms are connected in parallel . in this case , if r is the combined resistance of these three resistors , then the reciprocal of r is equal to the sum of the reciprocals resistors . what is r value ? | the wording is a bit confusing , though basically we are told that 1 / r = 1 / 2 + 1 / 5 + 1 / 6 , from which it follows that r = 13 / 15 ohms option : a | a = 1 / 6
b = 1 / 2
c = 1 / 5
d = b + c
e = a + d
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a ) 7.2 kg . , b ) 10.8 kg . , c ) 12.4 kg . , d ) 18.0 kg , e ) none | b | divide(multiply(6, 23.4), 13) | if the weight of 13 meters long rod is 23.4 kg . what is the weight of 6 meters long rod ? | "answer ∵ weight of 13 m long rod = 23.4 kg ∴ weight of 1 m long rod = 23.4 / 13 kg ∴ weight of 6 m long rod = 23.4 x 6 / 13 = 10.8 kg option : b" | a = 6 * 23
b = a / 13
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a ) 4 % , b ) 2.50 % , c ) 6.20 % , d ) 5 % , e ) 2.85 % | e | divide(divide(const_100, add(const_1, const_4)), 7) | if the simple interest on a certain sum of money for 7 years is one – fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 7 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 7 ) = 20 / 7 = 2.85 % answer : option e" | a = 1 + 4
b = 100 / a
c = b / 7
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a ) 4 , b ) 5 , c ) 9 , d ) 10 , e ) 11 | b | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | how many positive integers less than 60 are equal to the product of a positive multiple of 5 and an even number ? | "number of positive multiples of 5 less than 60 = 11 5 * 1 = 5 5 * 2 = 10 5 * 3 = 15 5 * 4 = 20 5 * 5 = 25 5 * 6 = 30 5 * 7 = 35 5 * 8 = 40 5 * 9 = 45 5 * 10 = 50 5 * 11 = 55 only 5 of the above are a product of a positive multiple of 5 and an even number - 10 , 20,30 , 40,50 answer b" | a = 4 + 5
b = a - 1
c = math.factorial(b)
d = math.factorial(5)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 12 / 1024 , b ) 11 / 1024 , c ) 11 / 256 , d ) 12 / 256 , e ) 14 / 256 | b | add(divide(const_1, power(const_2, 10)), divide(10, power(const_2, 10))) | alex has to take a quiz , which has 10 true false - type question each carry 1 mark . what is the probability that alex ' s can score more than 8 mark in the quiz . given that he decides to guess randomly on the quiz . | each question has 2 options ( true or false ) . total number of ways of answering all 10 questions = ( 2 ) 10 ( 2 ) 10 more than 8 correct means either 9 correct or all correct . number of ways such that 9 correct = 10 ! / 9 ! = 10 number of ways of all correct = 1 therefore , probability of more than 8 correct = 11 / ... | a = 2 ** 10
b = 1 / a
c = 2 ** 10
d = 10 / c
e = b + d
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a ) 6 , b ) 4 , c ) 8 , d ) 8 , e ) none of these | a | divide(reminder(multiply(653, const_100), 80), const_10) | if x and y are the two digits of the number 653 xy such that this number is divisible by 80 , then x + y = ? | "80 = 2 x 5 x 8 since 653 xy is divisible by 2 and 5 both , so y = 0 . now , 653 x is divisible by 8 , so 13 x should be divisible by 8 . this happens when x = 6 . x + y = ( 6 + 0 ) = 6 . answer : a" | a = 653 * 100
b = reminder / (
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a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9 | b | subtract(24, 9) | if x and y are integers , what is the least positive number of 24 x + 9 y ? | "24 x + 9 y = 3 ( 8 x + 3 y ) which will be a minimum positive number when 8 x + 3 y = 1 . 8 ( - 1 ) + 3 ( 3 ) = 1 then 3 ( 8 x + 3 y ) can have a minimum positive value of 3 . the answer is b ." | a = 24 - 9
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a ) rs . 18.95 , b ) rs 16 , c ) rs . 21 , d ) rs . 25.71 , e ) none of these | c | divide(subtract(add(multiply(4, 18), multiply(4, 22)), 13), add(const_3, const_4)) | the average earning of a mechanic for the first - 4 days of a week is rs . 18 and for the last 4 days is rs . 22 . if he earns rs . 13 on the fourth day , his average earning for the whole week is ? | answer total earning for the week = sum of earning of first four days + sum of earning of last four days - earning of 4 th day = 4 x 18 + 4 x 22 - 13 = rs . 147 â ˆ ´ average earning = 147 / 7 = rs . 21 correct option : c | a = 4 * 18
b = 4 * 22
c = a + b
d = c - 13
e = 3 + 4
f = d / e
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(12, divide(multiply(multiply(4, multiply(multiply(add(4, const_3), 4), const_3)), 4), 12)) | six bells commence tolling together and toll at intervals of 4 , 6 , 8 , 10 , 12 and 14 seconds respectively . in 70 minutes , how many times do they toll together ? | "lcm of 4 , 6 , 8 10 , 12 and 14 is 840 . so , after each 840 seconds , they would toll together . hence , in 70 minutes , they would toll 70 * 60 seconds / 840 seconds = 5 times but then the question says they commence tolling together . so , they basically also toll at thebeginning ( 0 second ) . so , total tolls tog... | a = 4 + 3
b = a * 4
c = b * 3
d = 4 * c
e = d * 4
f = e / 12
g = 12 / f
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a ) 5 , b ) - 12 , c ) - 2 , d ) - 3 , e ) - 5 | b | subtract(negate(3), const_2) | can you please walk me through how to best approach this problem ? thanks if # p # = ap ^ 3 + bp – 1 where a and b are constants , and # - 5 # = 10 , what is the value of # 5 # ? | # p # = ap ^ 3 + bp - 1 # - 5 # = 10 putting p = - 5 in above equation - 125 a - ( 5 b + 1 ) = 10 or # - 5 # = ( 125 a + 5 b + 1 ) = - 10 therefore 125 a + 5 b = - 11 . . . . . ( 1 now putting p = 5 # 5 # = 125 a + 5 b - 1 using equation 1 ( 125 a + 5 b = - 11 ) # 5 # = - 11 - 1 = - 12 hence b | a = negate - (
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a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) none of them | c | subtract(multiply(multiply(3, 6), 3), subtract(multiply(multiply(3, 6), 3), const_4)) | what is the unit digit in the product ( 3 ^ 68 x 6 ^ 59 x 7 ^ 71 ) ? | "unit digit in 3 ^ 4 = 1 = > unit digit in ( 3 ^ 4 ) ^ 16 = 1 therefore , unit digit in 3 ^ 65 = unit digit in [ ( 3 ^ 4 ) ^ 16 x 3 ] = ( 1 x 3 ) = 3 unit digit in 6 ^ 59 = 6 unit digit in 7 ^ 4 = 1 = > unit digit in ( 7 ^ 4 ) ^ 17 is 1 unit digit in 7 ^ 71 = unit digit in [ ( 7 ^ 4 ) ^ 17 x 7 ^ 3 ] = ( ( 1 x 3 ) = 3 t... | a = 3 * 6
b = a * 3
c = 3 * 6
d = c * 3
e = d - 4
f = b - e
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a ) 78 , b ) 60 , c ) 26 , d ) 23 , e ) 12 | b | add(45, 15) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 15 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 15 = 60 km / hr . 60 * 5 / 18 = 50 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 50 = 60 sec . answer : b" | a = 45 + 15
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a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 231 | e | add(add(add(add(add(add(add(11, 10), add(11, const_2)), add(11, const_1)), 11), 10), const_2), const_1) | if two integers x , y ( x > y ) are selected from - 10 to 11 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 22 c 2 = 22 * 21 / 2 = 231 . therefore , the answer is e ." | a = 11 + 10
b = 11 + 2
c = a + b
d = 11 + 1
e = c + d
f = e + 11
g = f + 10
h = g + 2
i = h + 1
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a ) 160 , b ) 150 , c ) 100 , d ) 80 , e ) 200 | e | divide(subtract(multiply(200, divide(16, const_100)), 24), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 24 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 24 gallons then 0.12 a + 0.16 ( 200 - a ) = 24 - - > a =... | a = 16 / 100
b = 200 * a
c = b - 24
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
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a ) 3 hours , b ) 6 hours , c ) 7 hours , d ) 4 hours , e ) 8 hours | b | multiply(add(add(const_1, const_2), const_3), 2) | a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 2 days . b can do the work alone in ? | "b 6 hours suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 2 6 / x = 1 / 2 = > x = 12 so , b takes 6 hours to finish the work ." | a = 1 + 2
b = a + 3
c = b * 2
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a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 9 | b | subtract(multiply(divide(const_1, subtract(const_1, divide(10, const_100))), const_100), const_100) | a certain candy manufacturer reduced the weight of candy bar m by 10 percent buy left the price unchanged . what was the resulting percent increase in the price per ounce of candy bar m ? | "assume 1 oz candy cost $ 1 before . now price remain same $ 1 but weight of candy reduces to 0.9 oz new price of candy = 1 / 0.9 = 1.11 price increase 11 % b" | a = 10 / 100
b = 1 - a
c = 1 / b
d = c * 100
e = d - 100
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a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | a | divide(subtract(280, multiply(divide(subtract(12, 6), const_2), 10)), add(divide(subtract(12, 6), const_2), add(divide(subtract(12, 6), const_2), 6))) | a cyclist traveled for two days . on the second day the cyclist traveled 6 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | "solution : d = 280 mi t = 12 hrs đ â y 1 time = t 1 d â y 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs d à y 1 rate = r 1 d â y 2 rate = r 2 r 1 - r 2 = 10 mph í . ẹ . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 í . ẹ . 280 = 8 r 2 + 4 ( 10 + r... | a = 12 - 6
b = a / 2
c = b * 10
d = 280 - c
e = 12 - 6
f = e / 2
g = 12 - 6
h = g / 2
i = h + 6
j = f + i
k = d / j
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a ) 800 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(60, const_100))) | a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 60 % ? | "cost price = rs . 500 profit = 60 % of 500 = rs . 300 selling price = cost price + profit = 500 + 300 = 800 answer : a" | a = 60 / 100
b = 500 * a
c = 500 + b
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a ) 278 , b ) 277 , c ) 278 , d ) 450 , e ) 550 | e | subtract(multiply(const_10, 150), add(multiply(2, 100), multiply(5, 150))) | a man purchased 2 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 2 * 100 + 5 * 150 = 950 1500 – 1050 = 550 answer : e" | a = 10 * 150
b = 2 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) 118 , b ) 133 , c ) 112 , d ) 113 , e ) 115 | b | multiply(subtract(divide(divide(multiply(subtract(const_100, 5), add(const_100, 40)), const_100), const_100), const_1), const_100) | a trader bought a car at 5 % discount on its original price . he sold it at a 40 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 95 s = 95 * ( 140 / 100 ) = 133 100 - 133 = 5 % answer : b" | a = 100 - 5
b = 100 + 40
c = a * b
d = c / 100
e = d / 100
f = e - 1
g = f * 100
|
a ) 24.19 , b ) 24.12 , c ) 22.1 , d ) 24.1 , e ) 22.5 | e | subtract(subtract(multiply(1000, power(add(divide(6, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(6, const_100)), 4)) | what will be the difference between simple and compound interest at 6 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 6 * 4 ) / 100 = rs . 240 c . i . = [ 1000 * ( 1 + 6 / 100 ) 4 - 1000 ] = rs . 262.5 difference = ( 262.5 - 240 ) = rs . 22.5 answer : e" | a = 6 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 6 / 100
g = 1000 * f
h = g * 4
i = e - h
|
a ) 58 kg , b ) 60 kg , c ) 64 kg , d ) 70 kg , e ) none | c | add(multiply(divide(11, 9), 28.8), 28.8) | zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 28.8 kg of zinc has been consumed in it ? | sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 28.8 kg zinc , mixture , melted = [ 20 / 9 x 28.8 ] kg = 64 kg . answer c | a = 11 / 9
b = a * 28
c = b + 28
|
a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.5 kmph | e | divide(subtract(12, 9), const_2) | what is the speed of the stream if a canoe rows upstream at 9 km / hr and downstream at 12 km / hr | "sol . speed of stream = 1 / 2 ( 12 - 9 ) kmph = 1.5 kmph . answer e" | a = 12 - 9
b = a / 2
|
a ) 99 , b ) 100 , c ) 102 , d ) 288 , e ) 2761 | a | divide(495, 5) | . a car covers a distance of 495 km in 5 hours . find its speed ? | "495 / 5 = 99 kmph answer : a" | a = 495 / 5
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(divide(subtract(add(166, 1), 1), 2), subtract(divide(subtract(add(166, 1), 1), 2), 1)) | if a number p is prime , and 2 p + 1 = q , where q is also prime , then the decimal expansion of 1 / q will produce a decimal with q - 1 digits . if this method produces a decimal with 166 digits , what is the units digit of the product of p and q | "3 / 7 = 0.428571 . . . ( a repeating pattern one digit long ) b" | a = 166 + 1
b = a - 1
c = b / 2
d = 166 + 1
e = d - 1
f = e / 2
g = f - 1
h = c - g
|
a ) 60 kg , b ) 95.5 kg , c ) 80 kg , d ) 85 kg , e ) 90 kg | b | add(multiply(5, 5.5), 68) | the average weight of 5 person ' s increases by 5.5 kg when a new person comes in place of one of them weighing 68 kg . what might be the weight of the new person ? | "total weight increased = ( 5 x 5.5 ) kg = 27.5 kg . weight of new person = ( 68 + 27.5 ) kg = 95.5 kg option b" | a = 5 * 5
b = a + 68
|
a ) 6.5 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.13 | a | divide(subtract(292, multiply(10, 3.2)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 292 runs ? | "required run rate = [ 292 - ( 3.2 * 10 ) ] / 40 = 260 / 40 = 6.5 answer : a" | a = 10 * 3
b = 292 - a
c = b / 40
|
a ) 32.52 , b ) 32.48 , c ) 26.74 , d ) 32.9 , e ) 32.31 | c | add(divide(circumface(5.2), const_2), multiply(5.2, const_2)) | the radius of a semi circle is 5.2 cm then its perimeter is ? | "36 / 7 r = 5.2 = 26.74 answer : c" | a = circumface / (
b = a + 2
|
a ) rs . 65,000 , b ) rs . 70,000 , c ) rs . 75,000 , d ) rs . 90,000 , e ) rs . 60,000 | c | multiply(multiply(add(add(const_3, const_2), 2), add(const_3, const_2)), 2) | p and q invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if p invested rs . 50 , 000 , the amount invested by q is : | q invested = 50000 / 2 * 3 = 75000 answer : c | a = 3 + 2
b = a + 2
c = 3 + 2
d = b * c
e = d * 2
|
a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55 | b | divide(add(20, 20), add(divide(20, 15), divide(20, 30))) | eden drove an average speed of 15 miles per hour for the first 20 miles of a tripthen at a average speed of 30 miles / hr for the remaining 20 miles of the trip if she made no stops during the trip what was eden ' s avg speed in miles / hr for the entire trip | avg . speed = total distance / total time total distance = 40 miles total time = 20 / 15 + 15 / 30 = 2 avg . speed = 20 . answer - b | a = 20 + 20
b = 20 / 15
c = 20 / 30
d = b + c
e = a / d
|
a ) 65 degrees , b ) 73 degrees , c ) 36 degrees , d ) 34 degrees , e ) 74 degrees | c | subtract(44, subtract(multiply(48, const_4), multiply(46, const_4))) | the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 44 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 44 tu + w + th = 192 - 44 = 148 f = 184 – 148 = 36 answer : c" | a = 48 * 4
b = 46 * 4
c = a - b
d = 44 - c
|
a ) 11 , b ) 44 , c ) 48 , d ) 28 , e ) 26 | e | subtract(add(floor(divide(subtract(88, 49), 3)), divide(subtract(88, 49), 2)), floor(divide(subtract(88, 49), multiply(2, 3)))) | if w is the set of all the integers between 49 and 88 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 87 , the smallest is 51 ) : 87 - 51 = 36 step 2 . divide by 3 : 36 / 3 = 12 step 3 . add 1 : 12 + 1 = 13 . so there are 13 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc ... | a = 88 - 49
b = a / 3
c = math.floor(b)
d = 88 - 49
e = d / 2
f = c + e
g = 88 - 49
h = 2 * 3
i = g / h
j = math.floor(i)
k = f - j
|
a ) 6 , b ) - 1 , c ) - 2 , d ) - 6 , e ) - 10 | b | divide(subtract(7, 5), subtract(sqrt(add(9, 7)), sqrt(add(5, 9)))) | in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , - 5 ) and ( t , - 7 ) . what is the least possible value of the slope of line l ? | we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , - 5 ) . hence , x = p and y = - 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , - 7 ) should satisfy the equation . hence x = t and y = - 7 . - 7 = - t 2 + 9 t = + ... | a = 7 - 5
b = 9 + 7
c = math.sqrt(b)
d = 5 + 9
e = math.sqrt(d)
f = c - e
g = a / f
|
a ) 35 , b ) 26 , c ) 76 , d ) 87 , e ) 24 | b | divide(multiply(52, 8), 16) | two numbers n and 16 have lcm = 52 and gcf = 8 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 * n = 52 * 8 n = 52 * 8 / 16 = 26 correct answer b" | a = 52 * 8
b = a / 16
|
a ) 1.8 , b ) 1.6 , c ) 2.775 . , d ) 1.9 , e ) 3.5 . | a | multiply(divide(divide(1000, const_1000), divide(30, const_60)), subtract(const_1, divide(10, const_100))) | an ant walks an average of 1000 meters in 30 minutes . a beetle walks 10 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | the ant walks an average of 1000 meters in 30 minutes 1000 meters in 1 / 2 hours the beetle walks 10 % less distance = 1000 - 10 = 900 meters in 30 minutes 0.900 km in 30 / 60 = 1 / 2 hours speed = 0.900 * 2 = 1.8 km / h i guess option a should be 1.8 | a = 1000 / 1000
b = 30 / const_60
c = a / b
d = 10 / 100
e = 1 - d
f = c * e
|
a ) 7.19 , b ) 7.18 , c ) 7.16 , d ) 7.84 , e ) 7.12 | d | divide(add(151, 165), multiply(add(80, 65), const_0_2778)) | two trains 151 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 151 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.84 answer : d" | a = 151 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
|
a ) 1 , b ) 2 , c ) none , d ) 3 / 7 , e ) 3 / 4 | d | divide(3, add(4, 3)) | joe has candys on his pocket , 4 of those candies have lemon flavor , and 3 have caramel flavor . if joe take one , what is the probability that he picks one caramel flavor ? | the total number of candies is 7 and the number of candies with lemon flavor is 3 then , in one chance the probability to pick one candie with lemon flavor is 3 / 7 answer d | a = 4 + 3
b = 3 / a
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(divide(65, const_100), 24) | john bought a total of 24 mangoes and oranges . each mango costs 80 cents and each orange costs 60 cents . if the average price of the 24 mangoes and oranges that john originally purchased was 65 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | "let number of mangoes be x , number of oranges be 24 - x 0.80 x + ( 24 - x ) 0.60 / 24 = 0.65 solving for x , we get x = 6 - - > mangoes 6 , oranges 18 now , number of oranges to be returned be y 0.80 * 6 + ( 18 - y ) * 0.60 / 24 - y = 0.72 solving for y , y = 14 ans : e" | a = 65 / 100
b = a * 24
|
a ) 25 years , b ) 30 years , c ) 35 years , d ) 40 years , e ) none | c | add(subtract(10, subtract(10, add(const_3, const_2))), multiply(subtract(10, add(const_3, const_2)), const_2)) | the difference between the ages of two persons is 10 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the elder person is | "sol . let their ages of x years and ( x + 10 ) years respectively . then , ( x + 10 ) - 15 = 2 ( x - 15 ) ⇔ x - 5 = 2 x - 30 ⇔ x = 25 . ∴ present age of the elder person = ( x + 10 ) = 35 years . answer c" | a = 3 + 2
b = 10 - a
c = 10 - b
d = 3 + 2
e = 10 - d
f = e * 2
g = c + f
|
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