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 |
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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 2 , which is 18 cm 2 more than the first area . therefore a 1 + 18 = a 2 x ( x − 3 ) + 18 = ( x + 1 ) ( x − 2 ) x 2 − 3 x + 18 = x 2 + x − 2 x − 2 2 x = 20 x = 10 . so , the sides of the rectangle are 10 cm and ( 10 − 3 ) = 7 cm long . so answer is c ." | a = 18 / 2
b = a + 1
c = b - 3
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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
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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 . chord is about 10 feet long . d | 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
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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 cards than michelle . how many cards did each player have originally ? | "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
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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
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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 . answer : c | 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
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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 paid by plan when medical expense = 1250 $ ( since insurance plan will pay 100 % of amount that exceeds 1050 $ ) answer b" | 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
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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 can write 5 ( x + 15 ) = 80 5 x + 75 = 80 5 x = 5 x = 1 meter answer is a | 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 value of x ? | 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 = 4 students ( including the new one ) . so , initially there were 5 student . total weight = 5 x + 90 = 6 ( x - 1 ) - - > x = 96 pounds . answer : e . | 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 1 and 500 . the answer is c . | 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 / 12" | 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 . = > all the three letters can be chosen in 676 ã — 26 = 17576676 ã — 26 = 17576 ways . it implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter . answer d | 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 can cover 7.5 miles in 30 mins teena will cover 37.5 miles in 150 mins so answer 30 + 150 = 180 mins . ( answer c )" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
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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
|
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 the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . based on this definition . . . . if . . . . k = 2 k * = ( 1 / 1 ) ( 1 / 2 ) = 1 / 2 if . . . . k = 3 k * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) = 1 / 6 we ' re asked to find the value of 5 * / 4 * now that we know how the symbol ' works ' , solving this problem should n ' t be too difficult . you can actually choose to do the math in a couple of different ways . . . . 5 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) do n ' t calculate this just yet though . . . . since we ' re dividing by 4 * , many of those fractions will ' cancel out . ' 4 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) we ' re looking for the value of : ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) / ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) since the first four fraction in the numerator and denominator cancel out , we ' re left with just one fraction : 1 / 4 d" | 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 / 20 ( t ) = 7 / 15 - - - > t = 3 1 / 9 therefore , these 3 1 / 9 days worked plus the 5 days that carla had to work by herself add to 8 1 / 9 days answer : d" | 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 exactly 45 minutes , for how many kilometers total did she hike east ? | "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 direction . 7.25 / 2 = 3.625 . a ." | 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(multiply(const_100, divide(60, 6)), 3))) | 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 traded . how many pet spiders does pugsley have before the trading game commences ? | 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 and w = 16 . answer : a . | 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
|
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 cost of having the manuscript typed ? | "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 / ( 2 ) 1011 / ( 2 ) 10 = 11 / 1024 answer : b | 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 together = 5 + 1 = 6 answer : e" | a = 4 + 3
b = a * 4
c = b * 3
d = 4 * c
e = d * 4
f = e / 12
g = 12 / f
|
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 therefore , required digit = unit digit in ( 3 x 6 x 3 ) = 4 answer is c" | a = 3 * 6
b = a * 3
c = 3 * 6
d = c * 3
e = d - 4
f = b - e
|
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 = 200 . answer : e ." | 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
|
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 2 ) í . ẹ . r 2 = 20 mph answer : a" | 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
|
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
|
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 . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 88 , the smallest is 50 ) : 88 - 50 = 38 step 2 . divide by 2 : 38 / 2 = 19 step 3 . add 1 : 19 + 1 = 20 . so there are 19 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 13 multiples of 3 and the 19 multiples of 2 : 13 + 19 = 32 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 13 + 19 = 32 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 32 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 84 - 54 = 30 step 2 . divide by 6 : 30 / 6 = 5 step 3 . add 1 : 5 + 1 = 6 . so there are 6 multiples of 6 within the range : we counted 6 numbers twice . subtract the 6 multiples of 6 from the sum of the multiples of 2 and 3 : = 13 + 19 - 6 = 32 - 6 = 26 therefore , the final number of multiples of 2 , 3 or 6 is 26 . hence , this is the correct answer . ( e )" | 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 = + or - 4 considering p = - 2 and t = 4 , the least slope is ( - 7 + 5 ) / ( 4 - 2 ) = - 1 imo option b is correct answer . | 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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