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 ) 22 , b ) 28 , c ) 19 , d ) 12 , e ) 88 | c | divide(1672, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | if the wheel is 14 cm then the number of revolutions to cover a distance of 1672 cm is ? | "2 * 22 / 7 * 14 * x = 1672 = > x = 19 answer : c" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 14
h = 1672 / g
|
a ) 2017 , b ) 2088 , c ) 270 , d ) 1881 , e ) 1781 | a | add(1, 2016) | if f ( f ( n ) ) + f ( n ) = 2 n + 3 , f ( 0 ) = 1 then f ( 2016 ) = ? | f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 β β f ( 1 ) = 3 - 1 = 2 , f ( 1 ) = 2 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 β β f ( 2 ) = 5 - 2 = 3 , f ( 2 ) = 3 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 β β f ( 3 ) = 7 - 3 = 4 , f ( 3 ) = 4 . . . . . . . . . . . . . . f ( 2016 ) = 2017 ans : a | a = 1 + 2016
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a ) 8 , b ) 10 , c ) 2 , d ) 14 , e ) 16 | c | divide(subtract(100, multiply(12, const_3)), const_4) | a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 12 more blue balls than white balls , and six times as many red balls as blue balls , how many white balls does the box has ? | "x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 12 more blue balls than white balls . . . equation # 2 : y = 12 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 6 y solve equation # 2 for z : z = y - 12 now , we can replace both x and z with y in equation # 1 6 y + y + ( y - 12 ) = 100 8 y - 12 = 100 5 y = 112 y = 14 there are 14 blue balls . this is 12 more than the number of white balls , so z = 2 . that ' s the answer . just as a check , x = 84 , and 84 + 14 + 2 = 100 . answer = 2 , ( c )" | a = 12 * 3
b = 100 - a
c = b / 4
|
a ) 48 , b ) 180 / 7 , c ) 50.4 , d ) 60 , e ) 90 | a | divide(multiply(4, 9), divide(75, const_100)) | 75 percent of andrea ' s living room floor is covered by a carpet that is 4 feet by 9 feet . what is the area of her living room floor ? | 75 % of area of the floor = 4 * 9 square feet = 36 square feet i . e . 100 % area of floor = ( 36 / 75 ) * 100 = 48 square feet answer : option a | a = 4 * 9
b = 75 / 100
c = a / b
|
a ) 28 , b ) 32 , c ) 35 , d ) 30 , e ) 45 | a | divide(add(408, 264), multiply(multiply(multiply(const_2, const_2), const_2), const_3)) | there are 408 boys and 264 girls in a school which are to be divided into equal sections of either boys or girls alone . find the total number of sections thus formed . | "explanation : hcf ( 408 , 264 ) = 24 the number of boys or girls that can be placed in a section = 24 . thus the total number of sections is given by 408 / 24 + 264 / 24 = 17 + 11 = 28 answer : a" | a = 408 + 264
b = 2 * 2
c = b * 2
d = c * 3
e = a / d
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(500, multiply(add(120, 60), const_0_2778)) | the speed at which a woman can row a boat in still water is 120 kmph . if she rows downstream , where the speed of current is 60 kmph , what time will he take to cover 500 metres ? | speed of the boat downstream = 120 + 60 = 180 kmph = 180 * 5 / 18 = 50 m / s hence time taken to cover 500 m = 500 / 50 = 10 seconds . answer : c | a = 120 + 60
b = a * const_0_2778
c = 500 / b
|
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 9 | d | subtract(divide(factorial(subtract(divide(15, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(15, const_2), const_1)) | a company that ships boxes to a total of 15 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter ) | "back - solving is the best way to solve this problem . you basically need 12 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - d" | a = 15 / 2
b = a - 1
c = math.factorial(b)
d = math.factorial(3)
e = math.factorial(2)
f = d * e
g = c / f
h = 15 / 2
i = h - 1
j = g - i
|
a ) 5 / 16 , b ) 1 / 3 , c ) 9 / 20 , d ) 47 / 61 , e ) 5 / 7 | d | divide(subtract(divide(60, const_100), multiply(subtract(const_1, divide(61, const_100)), subtract(const_1, divide(2, 3)))), divide(61, const_100)) | at a small company , 61 percent of the employees are women , and 60 percent of the employees are married . if 2 / 3 of the men are single , what fraction of the women are married ? | "lets take total employees are 100 . given that , total women = 61 and total married = 60 . total men = 100 - 61 = 39 and single men = 2 / 3 * 39 = 26 . married men = total men - single men = 39 - 26 = 13 . married women = total married - married men = 60 - 13 = 47 . fraction of women are married = married women / total women = 47 / 61 = 5 / 7 . ans d" | a = 60 / 100
b = 61 / 100
c = 1 - b
d = 2 / 3
e = 1 - d
f = c * e
g = a - f
h = 61 / 100
i = g / h
|
a ) 229 , b ) 156 , c ) 144 , d ) 128 , e ) 121 | b | multiply(add(const_1, const_4), 39) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 39 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 39 4 / x = 1 / 39 = > x = 156 min . answer : b" | a = 1 + 4
b = a * 39
|
a ) 6.25 , b ) 6.22 , c ) 5.5 , d ) 6.39 , e ) 6.13 | c | divide(subtract(252, 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 252 runs ? | "required run rate = [ 252 - ( 3.2 * 10 ) ] / 40 = 220 / 40 = 5.5 answer : c" | a = 10 * 3
b = 252 - a
c = b / 40
|
a ) 122.5 , b ) 142.5 , c ) 162.5 , d ) 182.5 , e ) 112.5 | e | divide(multiply(add(85.5, divide(multiply(85.5, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 85.5 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 85.50 sp = 85.5 * ( 125 / 100 ) = 106.875 mp * ( 95 / 100 ) = 106.875 mp = 112.5 answer : e" | a = 85 * 5
b = a / 100
c = 85 + 5
d = c * 100
e = 100 - 5
f = d / e
|
a ) 26 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | d | divide(subtract(76, 12), const_2) | we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 12 , and the difference between the total number of strawberries and the total number of blueberries would increase by 76 . each blue box contains how many blueberries ? | "let x be the number of blueberries in each blue box . then there are x + 12 strawberries in each red box . x + ( x + 12 ) = 76 x = 32 the answer is d ." | a = 76 - 12
b = a / 2
|
a ) 27.57 , b ) 28.57 , c ) 29.57 , d ) 30.57 , e ) 32 | b | add(floor(divide(multiply(multiply(20, 8), multiply(15, 3)), multiply(multiply(21, 2), 6))), const_1) | 15 men take 20 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | 3 w = 2 m 15 m - - - - - - 20 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 15 * 20 * 8 = 14 * x * 6 x = 28.57 answer : b | a = 20 * 8
b = 15 * 3
c = a * b
d = 21 * 2
e = d * 6
f = c / e
g = math.floor(f)
h = g + 1
|
a ) 7.5 % , b ) 5.8 % , c ) 4.2 % , d ) 34.5 % , e ) 12.6 % | e | multiply(divide(multiply(divide(36, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | an association of mathematics teachers has 1,500 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 36 percent of the votes cast ? | "total umber of members = 1500 number of members that cast votes = 525 since , winning candidate received 36 percent of the votes cast number of votes for winning candidate = ( 36 / 100 ) * 525 = 189 percent of total membership that voted for winning candidate = ( 189 / 1260 ) * 100 = 12.6 % answer e" | a = 36 / 100
b = a * 525
c = 4 ** 2
d = 100 * c
e = b / d
f = e * 100
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | d | add(divide(subtract(358, 81), 10), const_1) | how many multiples of 10 are there between 81 and 358 ? | "10 * 9 = 90 10 * 35 = 350 total no of multiples = ( 35 - 9 ) + 1 = 27 answer d" | a = 358 - 81
b = a / 10
c = b + 1
|
a ) 7600 , b ) 8000 , c ) 8400 , d ) data inadequate , e ) none of these | b | divide(80, divide(subtract(7, 6), const_100)) | in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . states b had an equal number of candidates appeared and 7 % candidates got selected with 80 more candidates got selected than a . what was the number of candidates appeared from each state ? | "solution let the number of candidates appeared from each state be x . then , 7 % of x - 6 % of x = 80 β 1 % of x = 80 β x = 80 Γ 100 = 8000 . answer b" | a = 7 - 6
b = a / 100
c = 80 / b
|
a ) 287 , b ) 279 , c ) 90 , d ) 278 , e ) 379 | c | multiply(divide(multiply(45, const_1), subtract(90, 45)), 90) | a train leaves mumabai at 9 am at a speed of 45 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 90 kmph . when and at what distance from mumbai do the two trains meet ? | "when the second train leaves mumbai the first train covers 45 * 1 = 45 km so , the distance between first train and second train is 45 km at 10.00 am time taken by the trains to meet = distance / relative speed = 45 / ( 90 - 45 ) = 1 hours so , the two trains meet at 2 p . m . the two trains meet 1 * 90 = 90 km away from mumbai . answer : c" | a = 45 * 1
b = 90 - 45
c = a / b
d = c * 90
|
a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | d | add(1990, multiply(10, multiply(const_2, const_3))) | in 1990 the budgets for projects q and v were $ 660,000 and $ 860,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | "let the no of years it takes is x . 660 + 30 x = 860 - 10 x - - > 40 x = 200 and x = 5 . thus , it happens in 1995 . d ." | a = 2 * 3
b = 10 * a
c = 1990 + b
|
a ) 94 kmph , b ) 58 kmph , c ) 72 kmph , d ) 94 kmph , e ) 59 kmph | c | subtract(multiply(15, multiply(200, const_0_2778)), 100) | a train 100 m long crosses a platform 200 m long in 15 sec ; find the speed of the train ? | "d = 100 + 200 = 300 t = 15 s = 300 / 15 * 18 / 5 = 72 kmph answer : c" | a = 200 * const_0_2778
b = 15 * a
c = b - 100
|
a ) $ 19,250 , b ) $ 15,800 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 12,300 | b | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | the average salary of 15 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 20,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 20 k = 300 k 5 emp @ 25 k = 125 k 4 emp @ 20 k = 80 k remaing 6 emp sal = 300 k - 125 k - 80 k = 95 k average = 95 k / 6 = 15800 ans : b" | a = 5 * 4
b = 4 * 4
c = a * b
d = 5 * 5
e = d * 5
f = c - e
g = 4 * 15
h = f - g
i = 2 + 5
j = h / i
|
a ) 2 : 3 , b ) 3 : 4 , c ) 7 : 10 , d ) 20 : 3 , e ) 30 : 7 | c | divide(multiply(0.070, const_100), multiply(0.10, const_100)) | if 0.10 of a number is equal to 0.070 of another number , the ratio of the numbers is : | "0.10 a = 0.070 b - > a / b = 0.07 / 0.10 = 7 / 10 : . a : b = 7 : 10 answer : c" | a = 0 * 70
b = 0 * 10
c = a / b
|
a ) s 200 , b ) s 1500 , c ) s 300 , d ) s 450 , e ) s 550 | b | divide(600, subtract(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), divide(subtract(subtract(1, divide(1, 3)), divide(subtract(1, divide(1, 3)), 5)), 4))) | a person spends 1 / 3 rd of the money with him on clothes , 1 / 5 th of the remaining on food and 1 / 4 th of the remaining on travel . now , he is left with rs 600 . how much did he have with him in the beginning ? | "suppose the amount in the beginning was rs β x β money spent on clothes = rs 1 x / 3 balance = rs 2 x / 3 money spent on food = 1 / 5 of 2 x / 3 = rs 2 x / 15 balance = 2 x / 3 - 2 x / 15 = rs 8 x / 15 money spent on travel = 1 / 4 of 8 x / 15 = rs 2 x / 15 = 8 x / 15 - 2 x / 15 = 6 x / 15 = rs 2 x / 5 therefore 2 x / 5 = 100 = 1500 answer : b" | a = 1 / 3
b = 1 - a
c = 1 / 3
d = 1 - c
e = d / 5
f = b - e
g = 1 / 3
h = 1 - g
i = 1 / 3
j = 1 - i
k = j / 5
l = h - k
m = l / 4
n = f - m
o = 600 / n
|
a ) 120 , b ) 160 , c ) 190 , d ) 200 , e ) 400 | a | add(add(30, divide(multiply(30, 15), subtract(const_100, add(60, 15)))), divide(multiply(30, 60), subtract(const_100, add(60, 15)))) | a fruit drink is made of orange , watermelon , and grape juice , where 15 percent of the drink is orange juice and 60 percent is watermelon juice . if the drink is made with 30 ounces of grape juice , how many ounces is the drink total ? | let the total number of ounces in the drink be x . % of orange = 15 % % of watermelon = 60 % % of grape = 100 % - 75 % = 25 % 0.25 x = 30 x = 120 therefore there are a total of 120 ounces in the drink . the answer is a . | a = 30 * 15
b = 60 + 15
c = 100 - b
d = a / c
e = 30 + d
f = 30 * 60
g = 60 + 15
h = 100 - g
i = f / h
j = e + i
|
a ) 2554.0 , b ) 3387.0 , c ) 2503.0 , d ) 3307.5 , e ) 16537.11 | d | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 3000) | sam invested rs . 3000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 3000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 3000 * ( 1 + 5 / 100 ) 2 ] = ( 3000 * 21 / 20 * 21 / 20 ) = rs . 3307.50 answer : d" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 3000
|
a ) 10.2 , b ) 10.27 , c ) 10.23 , d ) 10.21 , e ) 10.25 | d | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(7.5, 6.5)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 7.5 cm and breadth 6.5 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 7.5 + 5.5 ) = 26 cm parameter of the square = 26 cm i . e . 4 a = 26 a = 6.5 diameter of the semicircle = 6.5 cm circimference of the semicircle = 1 / 2 ( Γ’ Λ Β ) ( 6.5 ) = 1 / 2 ( 22 / 7 ) ( 6.5 ) = 143 / 14 = 10.21 cm to two decimal places answer : d" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 480 , b ) 2,880 , c ) 4,800 , d ) 28,800 , e ) 43,200 | e | multiply(12, const_3600) | a space shuttle orbits the earth at about 12 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 12 * 3600 = 43,200 answer : e" | a = 12 * 3600
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a ) $ 180 , b ) $ 135 , c ) $ 108 , d ) $ 60 , e ) $ 54 | c | multiply(2, divide(multiply(36, divide(2, 3)), divide(const_1, 3))) | a collection of books went on sale , and 2 / 3 of them were sold for $ 1.50 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 1.50 ) or $ 108 . the best answer is therefore c . | a = 2 / 3
b = 36 * a
c = 1 / 3
d = b / c
e = 2 * d
|
a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these | a | multiply(0.16, 2.13) | if 213 Γ 16 = 3408 , then 0.16 Γ 2.13 is equal to : | "solution 0.16 Γ 2.13 = ( 16 / 100 x 213 / 100 ) = ( 16 x 213 / 10000 ) = 3408 / 10000 = 0.3408 . answer a" | a = 0 * 16
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a ) 16 , b ) 12 , c ) 47 , d ) 9 , e ) 12 | a | add(const_3, const_4) | what is the smallest positive integer x such that 80 - x is the cube of a positive integer | "given 80 - x is a perfect cube so we will take 64 = 4 * 4 * 4 80 - x = 64 x = 80 - 64 = 16 correct option is a" | a = 3 + 4
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a ) 10 , b ) 46 , c ) 70 , d ) 90 , e ) 100 | a | divide(multiply(5, 495), 10) | the l . c . m of two numbers is 495 and their h . c . f is 5 . if the sum of the numbers is 10 , then their difference is : | "let the numbers be x and ( 100 - x ) . then , x ( 100 - x ) = 5 * 495 x 2 - 100 x + 2475 = 0 ( x - 55 ) ( x - 45 ) = 0 x = 55 or 45 the numbers are 45 and 55 . required difference = 55 - 45 = 10 . answer : a" | a = 5 * 495
b = a / 10
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a ) 5 / 12 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 12 , e ) 2 / 3 | d | divide(const_6, multiply(const_6, const_6)) | if two dice are thrown together , the probability of getting a square number on both the dice is | the number of exhaustive outcomes is 36 . let e be the event of getting a sqaure number on both the dice . p ( e ) = 3 / 36 = 1 / 12 . d ) | a = 6 * 6
b = 6 / a
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a ) 288 , b ) 289 , c ) 200 , d ) 112 , e ) 392 | e | divide(square_area(28), const_2) | what is the area of a square field whose diagonal of length 28 m ? | "d 2 / 2 = ( 28 * 28 ) / 2 = 392 answer : e" | a = square_area / (
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a ) 3 % , b ) 61 1 / 9 % , c ) 25 % , d ) 33 1 / 3 % , e ) 60 % | b | subtract(const_100, divide(multiply(subtract(13.0, 12.0), const_100), subtract(18.8, 15.6))) | in 1982 and 1983 , company b β s operating expenses were $ 12.0 million and $ 13.0 million , respectively , and its revenues were $ 15.6 million and $ 18.8 million , respectively . what was the percent increase in company b β s profit ( revenues minus operating expenses ) from 1982 to 1983 ? | "profit in 1982 = 15.6 - 12 = 3.6 million $ profit in 1983 = 18.8 - 13 = 5.8 million $ percentage increase in profit = ( 5.8 - 3.6 ) / 3.6 * 100 % = 61 1 / 9 % answer b" | a = 13 - 0
b = a * 100
c = 18 - 8
d = b / c
e = 100 - d
|
a ) 10 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | c | divide(multiply(16, 10), 40) | 16 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 16 * 10 / 40 = 4 answer is c" | a = 16 * 10
b = a / 40
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a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8800 , d ) rs . 9000 , e ) none | c | divide(add(add(add(add(8000, 5000), 15000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 15000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 15000 + 7000 + 9000 / 5 = rs . 8800 correct option : c" | a = 8000 + 5000
b = a + 15000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(power(add(19, 2), 2), multiply(12, const_4)) | if n is a prime number greater than 19 , what is the remainder when n ^ 2 is divided by 12 ? | "there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 19 , square it and see what would be the remainder upon division of it by 12 . n = 23 - - > n ^ 2 = 529 - - > remainder upon division 529 by 12 is 1 . answer : b ." | a = 19 + 2
b = a ** 2
c = 12 * 4
d = b - c
|
a ) 22 , b ) 28 , c ) 27 , d ) 19 , e ) 11 | d | add(add(add(add(6, 3), 4), 3), 3) | total number of 4 digit number do not having the digit 3 or 6 . | answer : d | a = 6 + 3
b = a + 4
c = b + 3
d = c + 3
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a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 37 kg | e | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(43, const_3)) | the average weight of a , b and c is 43 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 129 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 37 . b ' s weight = 37 kg . e" | a = 40 * 2
b = 43 * 2
c = a + b
d = 43 * 3
e = c - d
|
a ) 360 , b ) 376 , c ) 299 , d ) 276 , e ) 111 | a | multiply(multiply(divide(80, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(80, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 80 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 80 l = 36 b = 10 lb = 36 * 10 = 360 a" | a = 3 * 2
b = 1 * 2
c = a + b
d = 80 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 80 / h
j = e * i
|
a ) 1 / 6 , b ) 2 / 9 , c ) 5 / 6 , d ) 7 / 9 , e ) 8 / 9 | c | divide(const_5, 6) | a dog breeder currently has 9 breeding dogs . 6 of the dogs have exactly 1 littermate , and 3 of the dogs have exactly 2 littermates . if 2 dogs are selected at random , what is the probability r that both selected dogs are not littermates ? | "we have three pairs of dogs for the 6 with exactly one littermate , and one triplet , with each having exactly two littermates . so , in fact there are two types of dogs : those with one littermate - say a , and the others with two littermates - b . work with probabilities : choosing two dogs , we can have either one dog of type b or none ( we can not have two dogs both of type b ) . the probability of choosing one dog of type b and one of type a is 3 / 9 * 6 / 8 * 2 = 1 / 2 ( the factor of 2 for the two possibilities ba and ab ) . the probability r of choosing two dogs of type a which are not littermates is 6 / 9 * 4 / 8 = 1 / 3 ( choose one a , then another a which is n ' t the previous one ' s littermate ) . the required probability is 1 / 2 + 1 / 3 = 5 / 6 . find the probability for the complementary event : choose aa or bb . probability of choosing two dogs of type a who are littermates is 6 / 9 * 1 / 8 = 1 / 12 . probability of choosing two dogs of type b ( who necessarily are littermates ) is 3 / 9 * 2 / 8 = 1 / 12 . again , we obtain 1 - ( 1 / 12 + 1 / 12 ) = 5 / 6 . answer : c" | a = 5 / 6
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a ) 350 m , b ) 278 m , c ) 876 m , d ) 150 m , e ) 267 m | a | multiply(divide(multiply(60, const_1000), const_3600), 21) | a train running at the speed of 60 km / hr crosses a pole in 21 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 21 ) m = 350 m . answer : a" | a = 60 * 1000
b = a / 3600
c = b * 21
|
a ) 7 , b ) 10 , c ) 11 , d ) 2 , e ) 3 | c | subtract(35, multiply(14, const_2)) | a number when divided by 214 gives a remainder 35 , what remainder will be obtained by dividing the same number 14 ? | "explanation : 214 + 35 = 249 / 14 = 11 ( remainder ) answer : c" | a = 14 * 2
b = 35 - a
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a ) 120 , b ) 180 , c ) 200 , d ) 220 , e ) none | a | divide(add(3, 3), subtract(divide(const_1, 3), divide(const_1, add(const_1, 3)))) | a number whose fifth part increased by 3 is equal to its fourth part diminished by 3 is ? | "answer let the number be n . then , ( n / 5 ) + 3 = ( n / 4 ) - 3 Γ’ β‘ β ( n / 4 ) - ( n / 5 ) = 6 Γ’ β‘ β ( 5 n - 4 n ) / 20 = 6 Γ’ Λ Β΄ n = 120 option : a" | a = 3 + 3
b = 1 / 3
c = 1 + 3
d = 1 / c
e = b - d
f = a / e
|
a ) 25 , b ) 67 , c ) 26 , d ) 29 , e ) 18 | a | add(divide(multiply(250, subtract(40, 15)), add(250, 50)), const_3) | a military camp has a food reserve for 250 personnel for 40 days . if after 15 days 50 more personnel are added to the camp , find the number of days the reserve will last for ? | explanation : as the camp has a reserve for 250 personnel that can last for 40 days , after 10 days the reserve left for 250 personnel is for 30 days . now 50 more personnel are added in the camp . hence , the food reserve for 300 personnel will last for : 250 : 300 : : x : 30 β¦ β¦ . . ( it is an indirect proportion as less men means more days ) x = ( 250 * 30 ) / 300 x = 25 days answer : a | a = 40 - 15
b = 250 * a
c = 250 + 50
d = b / c
e = d + 3
|
a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | c | subtract(12, multiply(2, 3)) | what is x if x + 2 y = 12 and y = 3 ? | "x = 12 - 2 y x = 12 - 6 . x = 6 answer : c" | a = 2 * 3
b = 12 - a
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a ) 13 , b ) 9 , c ) 11 , d ) 12 , e ) 10 | d | divide(multiply(divide(multiply(1000, 12), const_100), 2), multiply(divide(10, const_100), 200)) | in how many years rs 200 will produce the same interest at 10 % as rs . 1000 produce in 2 years at 12 % | explanation : clue : firstly we need to calculate the si with prinical 1000 , time 2 years and rate 12 % , it will be rs . 240 then we can get the time as time = ( 100 * 240 ) / ( 200 * 10 ) = 12 option d | a = 1000 * 12
b = a / 100
c = b * 2
d = 10 / 100
e = d * 200
f = c / e
|
a ) $ 2400 , b ) $ 2464 , c ) $ 2560 , d ) $ 2732 , e ) $ 2800 | c | multiply(2240, divide(add(const_100, 60), add(const_100, 40))) | a store β s selling price of $ 2240 for a certain computer would yield a profit of 40 percent of the store β s cost for the computer . what selling price would yield a profit of 60 percent of the computer β s cost ? | "1.4 x = 2240 x = 2240 / 1.4 so , 1.6 x = 2240 * 1.6 / 1.4 = 2560 answer : - c" | a = 100 + 60
b = 100 + 40
c = a / b
d = 2240 * c
|
a ) 112 , b ) 118 , c ) 121 , d ) 124 , e ) none of these | d | add(multiply(divide(subtract(110, multiply(18, 3)), subtract(10, 3)), 3), multiply(10, 10)) | a person buys 18 local tickets for rs 110 . each first class ticket costs rs 10 and each second class ticket costs rs 3 . what will another lot of 18 tickets in which the numbers of first class and second class tickets are interchanged cost ? | explanation : let , there are x first class ticket and ( 18 - x ) second class tickets . then , 110 = 10 x + 3 ( 18 β x ) . = > 110 = 10 x + 54 β 3 x . = > 7 x = 56 . = > x = 8 . if the number of the first class and second class tickets are interchanged , then the total cost would be 10 Γ 10 + 3 Γ 8 = 124 . answer : d | a = 18 * 3
b = 110 - a
c = 10 - 3
d = b / c
e = d * 3
f = 10 * 10
g = e + f
|
a ) 450 m , b ) 200 m , c ) 250 m , d ) 270 m , e ) 300 m | a | subtract(multiply(40, multiply(72, const_0_2778)), 250) | a train 250 m long running at 72 kmph crosses a platform in 40 sec . what is the length of the platform ? | "d = 72 * 5 / 18 = 40 = 800 Γ’ β¬ β 250 = 450 m answer : a" | a = 72 * const_0_2778
b = 40 * a
c = b - 250
|
a ) 92 , b ) 39 , c ) 87 , d ) 96 , e ) none | d | add(88, add(const_4, const_4)) | out of 3 given numbers , the first one is twice the second and 3 times the third . if the average of these numbers is 88 , then the difference between first and third is . | sum of three number is = 88 * 3 = 264 let three numbers are a , b , c and a is the highest and c is the lowest then , 2 b = a so b = a / 2 and 3 c = a so c = a / 3 we can write , a + b + c = 264 a + a / 2 + a / 3 = 264 11 a / 6 = 264 a = 144 so , c = 144 / 3 = 48 so there difference is = 144 - 48 = 96 answer d | a = 4 + 4
b = 88 + a
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a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 70.00 | e | multiply(subtract(10, divide(multiply(32, 8), const_100)), 10) | an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 32 % . also , if a customer will β buy at least 10 β items they will deduct an additional 8 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 40 % , thus he must pay slightly more than $ 60 . only answer choice e fits . answer : e ." | a = 32 * 8
b = a / 100
c = 10 - b
d = c * 10
|
a ) none , b ) two , c ) four , d ) five , e ) seven | e | subtract(subtract(200, const_4), const_4) | r is the set of positive odd integers less than 200 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive odd integers less than 200 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 1,3 , 5,7 , 9,11 , 13,15 . . . s = 1 , 9,25 , 49,81 . . . numbers : 1 , 9 , 25 , 49 , 81 , 121 , and 169 are odd integers ( less than 200 ) that are in both sets . solution : seven answer : e" | a = 200 - 4
b = a - 4
|
a ) $ 1.09 , b ) $ 1.67 , c ) $ 2.25 , d ) $ 2.36 , e ) $ 2.50 | d | divide(add(9.00, multiply(1.75, subtract(9, 2))), 9) | the cost to park a car in a certain parking garage is $ 9.00 for up to 2 hours of parking and $ 1.75 for each hour in excess of 2 hours . what is the average ( arithmetic mean ) cost per hour to park a car in the parking garage for 9 hours ? | "total cost of parking for 9 hours = 9 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 9 + 7 * 1.75 = 21.25 thus the average parking price = 21.25 / 9 = 2.36 $ d is the correct answer ." | a = 9 - 2
b = 1 * 75
c = 9 + 0
d = c / 9
|
a ) $ 645.56 , b ) $ 824.32 , c ) $ 954.26 , d ) $ 2472.96 , e ) $ 1020.45 | d | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 30000), 30000) | find the compound interest on $ 30000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 10000 rate = 2 % half yearly = 4 half years amount = 30000 * ( 1 + 2 / 100 ) ^ 4 = 30000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 32472.96 c . i . = 32472.96 - 10000 = $ 2472.96 answer is d" | a = 4 / 100
b = a / 2
c = b + 1
d = c ** 4
e = d * 30000
f = e - 30000
|
a ) 34 kg , b ) 40 kg , c ) 42 kg , d ) 41 kg , e ) 31 kg | e | subtract(multiply(40, const_2), subtract(multiply(45, const_3), multiply(43, const_2))) | the average waight of a , b , c is 45 kg . the avg wgt of a & b be 40 kg & that of b , c be 43 kg . find the wgt of b . | ". let a , b , c represent their individual wgts . then , a + b + c = ( 45 * 3 ) kg = 135 kg a + b = ( 40 * 2 ) kg = 80 kg & b + c = ( 43 * 2 ) kg = 86 kg b = ( a + b ) + ( b + c ) - ( a + b + c ) = ( 80 + 86 - 135 ) kg = 31 kg . answer is e ." | a = 40 * 2
b = 45 * 3
c = 43 * 2
d = b - c
e = a - d
|
a ) 150 , b ) 108 , c ) 42 , d ) 21 , e ) - 21 | d | divide(subtract(subtract(add(add(multiply(600, divide(35, const_100)), multiply(600, divide(40, const_100))), multiply(600, divide(50, const_100))), 600), multiply(divide(18, const_100), 600)), 2) | of the 600 residents of clermontville , 35 % watch the television show island survival , 40 % watch lovelost lawyers and 50 % watch medical emergency . if all residents watch at least one of these 3 shows and 18 % watch exactly 2 of these shows , then how many clermontville residents z watch all of the shows ? | oa is d . 100 = a + b + c - ab - ac - bc + abc , which is the same as the following formula 100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - 2 abc . the term between parantheses value 18 % so the equation to resolve is 100 = 35 + 40 + 50 - 18 - 2 abc therefore the value of abc is z = 3.5 % of 600 , is 21 . d is the correct answer | a = 35 / 100
b = 600 * a
c = 40 / 100
d = 600 * c
e = b + d
f = 50 / 100
g = 600 * f
h = e + g
i = h - 600
j = 18 / 100
k = j * 600
l = i - k
m = l / 2
|
a ) 70 , b ) 78 , c ) 80 , d ) 84 , e ) 90 | c | multiply(divide(subtract(82, 62), divide(1, 2)), 2) | john ' s marks wrongly entered as 82 instead of 62 . due to that the average marks for the class got increased by half ( 1 / 2 ) . the number of john in the class is ? | otal increase in marks = x x 1 = x 2 2 x / 2 = ( 82 - 62 ) x / 2 = 40 x = 80 . c | a = 82 - 62
b = 1 / 2
c = a / b
d = c * 2
|
a ) 150 , b ) 180 , c ) 250 , d ) 200 , e ) 225 | c | multiply(divide(multiply(90, const_1000), const_3600), 10) | a train running at the speed of 90 km / hr crosses a pole in 10 seconds . find the length of the train . | "speed = 90 * ( 5 / 18 ) m / sec = 25 m / sec length of train ( distance ) = speed * time 25 * 10 = 250 meter answer : c" | a = 90 * 1000
b = a / 3600
c = b * 10
|
['a ) 60 / Ο', 'b ) 24', 'c ) 40 / Ο', 'd ) 10', 'e ) 3 Ο'] | c | divide(divide(volume_rectangular_prism(4, 9, 10), 9), const_pi) | milk is poured from a full rectangular container with dimensions 4 inches by 9 inches by 10 inches into a cylindrical container with a diameter of 6 inches . if the milk does not overflow , how many inches high will the milk reach ? | let the height of level of milk in the cylinder = h since , volume of milk is constant . therefore 4 * 9 * 10 = Ο Ο * 3 ^ 2 * h = > Ο * h = 40 = > h = 40 / ( Ο ) answer c | a = volume_rectangular_prism / (
b = a / 9
|
a ) 65 , b ) 40 , c ) 55 , d ) 30 , e ) 60 | b | multiply(const_100, divide(add(23, multiply(divide(60, const_100), 5)), 65)) | if 65 % of a number is greater than 5 % of 60 by 23 , what is the number ? | "explanation : 65 / 100 * x - 5 / 100 * 60 = 23 65 / 100 * x - 3 = 23 65 / 100 * x = 26 x = 26 * 100 / 65 x = 40 answer : option b" | a = 60 / 100
b = a * 5
c = 23 + b
d = c / 65
e = 100 * d
|
a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 1 , d ) 2 : 3 , e ) none of these | a | divide(subtract(4, 2), subtract(8, 4)) | 8 men and 2 boys working together can do 4 times as much work as a man and a boy . working capacity of man and boy is in the ratio | explanation : let 1 man 1 day work = x 1 boy 1 day work = y then 8 x + 2 y = 4 ( x + y ) = > 4 x = 2 y = > x / y = 2 / 4 = > x : y = 1 : 2 option a | a = 4 - 2
b = 8 - 4
c = a / b
|
a ) 63 , b ) 6973 , c ) 5994 , d ) 6084 , e ) none of these | a | subtract(multiply(const_10, 7), 7) | the difference between the place value and the face value of 7 in the numeral 856973 is | ( place value of 7 ) - ( face value of 7 ) = ( 70 - 7 ) = 63 answer : option a | a = 10 * 7
b = a - 7
|
a ) 50 km , b ) 76 km , c ) 18 km , d ) 10 km , e ) 97 km | d | multiply(5, divide(20, subtract(15, 5))) | if a person walks at 15 km / hr instead of 5 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 5 = ( x + 20 ) / 15 10 x - 100 = > x = 10 km . answer : d" | a = 15 - 5
b = 20 / a
c = 5 * b
|
a ) 33 , b ) 45 , c ) 20 , d ) 77 , e ) 21 | c | divide(volume_cylinder(divide(4, const_2), 5), const_pi) | the diameter of a cylindrical tin is 4 cm and height is 5 cm . find the volume of the cylinder ? | "r = 2 h = 5 Ο * 2 * 2 * 5 = 20 Ο cc answer : c" | a = 4 / 2
b = volume_cylinder / (
|
a ) positive . , b ) divisible by 2 . , c ) divisible by 4 . , d ) divisible by 3 . , e ) divisible by 5 . | d | add(divide(subtract(3, const_1), multiply(const_2, const_1)), const_1) | the sum of 3 consecutive numbers is definitely | "if 1 st term is x : x + ( x + 1 ) + ( x + 2 ) = 3 x + 3 - - - > always divisible by 3 if 2 nd term is x : ( x - 1 ) + x + ( x + 1 ) = 3 x - - - > always divisible by 3 if 3 rd term is x : ( x - 2 ) + ( x - 1 ) + x = 3 x - 3 - - - > always divisible by 3 answer : d" | a = 3 - 1
b = 2 * 1
c = a / b
d = c + 1
|
a ) 1.8 , b ) 4 , c ) 6 , d ) 18 , e ) 60 | b | divide(divide(multiply(add(18, 6), add(divide(subtract(18, 6), const_2), const_1)), const_2), divide(multiply(add(divide(subtract(10, 6), const_2), const_1), add(6, 10)), const_2)) | for all even integers n , h ( n ) is defined to be the sum of the even integers between 6 and n , inclusive . what is the value of h ( 18 ) / h ( 10 ) ? | "concept : when terms are in arithmetic progression ( a . p . ) i . e . terms are equally spaced then mean = median = ( first + last ) / 2 and sum = mean * number of terms h ( 18 ) = [ ( 6 + 18 ) / 2 ] * 7 = 84 h ( 10 ) = ( 6 + 10 ) / 2 ] * 3 = 24 h ( 18 ) / h ( 10 ) = ( 84 ) / ( 24 ) ~ 4 answer : b" | a = 18 + 6
b = 18 - 6
c = b / 2
d = c + 1
e = a * d
f = e / 2
g = 10 - 6
h = g / 2
i = h + 1
j = 6 + 10
k = i * j
l = k / 2
m = f / l
|
a ) 36 % , b ) 35 % , c ) 34 % , d ) 33 % , e ) 32 % | a | multiply(divide(add(divide(multiply(60, 4), const_100), divide(multiply(20, 6), const_100)), add(6, 4)), const_100) | 20 % of a 6 litre solution and 60 % of 4 litre solution are mixed . what percentage of the mixture of solution | 20 % of 6 litre is ( 6 * 20 / 100 ) = 1.2 litre 60 % of 4 litre is ( 60 * 4 / 100 ) = 2.4 litre the mixture is 3.6 litre so the percentage is ( 3.6 * 100 / 10 ) = 36 % answer : a | a = 60 * 4
b = a / 100
c = 20 * 6
d = c / 100
e = b + d
f = 6 + 4
g = e / f
h = g * 100
|
a ) 5000 , b ) 4500 , c ) 4950 , d ) 1000 , e ) 2000 | a | divide(4455, multiply(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100)))) | population of a city decreases by 10 % at the end of first year and increases by 10 % at the end of second year and again decreases by 10 % at the end of third year . if the population of the city at the end of third year is 4455 , then what was the population of the city at the beginning of the first year ? | m . f = 90 / 100 * 110 * 100 * 90 / 100 = 81 * 11 / 1000 population before 3 yrs = i . q / m . f = 4455 * 1000 / 81 * 11 = 5000 answer : a | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 + c
e = b * d
f = 10 / 100
g = 1 - f
h = e * g
i = 4455 / h
|
a ) 8 , b ) 10 , c ) 15 , d ) 17 , e ) 19 | b | subtract(30, divide(add(multiply(7.5, 30), 425), add(7.5, 25))) | a contractor is engaged for 30 days on the condition thathe receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 425 in all . for how many days was he absent ? | 30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = 10 b | a = 7 * 5
b = a + 425
c = 7 + 5
d = b / c
e = 30 - d
|
a ) 4 miles , b ) 6 miles , c ) 9 miles , d ) 10 miles , e ) 12 mile | a | divide(multiply(divide(subtract(20, multiply(divide(add(1, 5), const_60), 20)), add(5, add(1, 5))), 5), const_2) | stalin and heather are 20 miles apart and walk towards each other along the same route . stalin walks at constant rate that is 1 mile per hour faster than heather ' s constant rate of 5 miles / hour . if heather starts her journey 20 minutes after stalin , how far from the original destination has heather walked when the two meet ? | original distance between s and h = 20 miles . speed of s = 5 + 1 = 6 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 20 / 60 = t + 2 / 6 hours . now , the total distances traveled by s and h = 20 miles - - - > 6 * ( t + 2 / 6 ) + 5 * t = 20 - - - > t = 8 / 11 hours . thus h has traveled for 8 / 11 hours giving you a total distance for h = 5 * 8 / 11 ~ 4 miles . a is thus the correct answer . p . s . : based on the wording of the question , you should calculatehow far from theoriginal destination has heather walkedwhen the two meet . ' original destination ' for h does not make any sense . original destination for h was situated at a distance of 20 miles . | a = 1 + 5
b = a / const_60
c = b * 20
d = 20 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
i = h / 2
|
a ) 10 % , b ) 25 % , c ) 20 % , d ) 50 % , e ) 45 % | d | multiply(divide(subtract(240, 120), 240), const_100) | a bag marked at $ 240 is sold for $ 120 . the rate of discount is ? | "rate of discount = 120 / 240 * 100 = 50 % answer is d" | a = 240 - 120
b = a / 240
c = b * 100
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a ) 70 kg , b ) 80 kg , c ) 81 kg , d ) 90 kg , e ) 91 kg | c | add(multiply(8, 2), 65) | the average weight of 8 person ' s increases by 2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 2 ) kg = 16 kg . weight of new person = ( 65 + 16 ) kg = 81 kg . c )" | a = 8 * 2
b = a + 65
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a ) 33 , b ) 77 , c ) 35 , d ) 88 , e ) 29 | b | sqrt(multiply(4, 9)) | the mean proportional between 4 and 9 is ? | "7 / 20 * 100 = 35 answer : b" | a = 4 * 9
b = math.sqrt(a)
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a ) 10 % , b ) 100 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(100, 50), 50), const_100) | a man buy a book in rs 50 & sale it rs 100 . what is the rate of profit ? ? ? | "cp = 50 sp = 100 profit = 100 - 50 = 50 % = 50 / 50 * 100 = 100 % answer : b" | a = 100 - 50
b = a / 50
c = b * 100
|
a ) 480 , b ) 287 , c ) 720 , d ) 270 , e ) 927 | c | multiply(24, 20) | a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 6 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 6 = 720 answer : c" | a = 24 * 20
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a ) 250 , b ) 225 , c ) 175 , d ) 195 , e ) 200 | e | divide(add(20, 60), divide(subtract(const_100, 60), const_100)) | a group of students was interviewed for that if it was asked whether or not they speak french and / or english . among those who speak french , 20 speak english well , while 60 of them do not speak english . if 60 % of students do not speak french , how many students were surveyed ? | "number of students who speak french are 60 + 20 = 80 of total students , the percentage of students who do not speak french was 60 % - - > percentage of who do is 40 % 80 - - - - - - - 40 % x - - - - - - - 100 % x = 80 * 100 / 40 = 200 = number of all students answer is e" | a = 20 + 60
b = 100 - 60
c = b / 100
d = a / c
|
a ) 4 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 21 kg , e ) none of these | a | subtract(multiply(add(29, const_1), 27.2), multiply(29, 28)) | the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.2 kg . the weight of the new student is | "exp . the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.2 weight of the new student = ( 30 * 27.2 β 29 * 28 ) = 816 - 812 = 4 answer : a" | a = 29 + 1
b = a * 27
c = 29 * 28
d = b - c
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a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | c | divide(add(add(multiply(80, 6), multiply(60, 4)), 30), add(6, 4)) | the average expenditure of a labourer for 6 months was 80 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 Γ 80 ) β debt = 480 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 750 average monthly income = 750 Γ· 10 = 75 answer c" | a = 80 * 6
b = 60 * 4
c = a + b
d = c + 30
e = 6 + 4
f = d / e
|
a ) 40 % , b ) 45 % , c ) 50 % , d ) 65 % , e ) 70 % | c | multiply(divide(subtract(multiply(const_3, divide(const_1, const_2)), const_1), const_1), const_100) | the length of a rectangle is halved , whileits breadth is tripled . wat is the % change in area ? | "let original length = x and original breadth = y . original area = xy . new length = x . 2 new breadth = 3 y . new area = x x 3 y = 3 xy . 2 2 increase % = 1 xy x 1 x 100 % = 50 % . 2 xy c" | a = 1 / 2
b = 3 * a
c = b - 1
d = c / 1
e = d * 100
|
a ) rs . 1000 , b ) rs . 2000 , c ) rs . 1500 , d ) rs . 3000 , e ) rs . 3100 | a | divide(multiply(175, const_100), subtract(const_100, add(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(10, const_100))))) | a man saves 25 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 175 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 75 savings = rs . 25 present expenditure 75 + 75 * ( 10 / 100 ) = rs . 82.5 present savings = 100 β 82.50 = rs . 17.50 if savings is rs . 17.50 , salary = rs . 100 if savings is rs . 175 , salary = 100 / 17.5 * 175 = 1000 answer : a" | a = 175 * 100
b = 100 - 25
c = 100 - 25
d = 10 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
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a ) s . 83.33 , b ) s . 110 , c ) s . 112 , d ) s . 125 , e ) s . 140 | d | multiply(divide(const_100, 12), 15) | a 15 % stock yielding 12 % is quoted at : | "income of rs 12 on investment of rs 100 income of rs 15 on investment of ? = ( 15 * 100 ) / 12 = 125 answer : d" | a = 100 / 12
b = a * 15
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a ) 18 , b ) 12 , c ) 20 , d ) 32 , e ) 36 | a | multiply(subtract(const_10, 1), 2) | on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 2 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on september 1 than were in stock on january 1 ? | "jan 1 = c feb 1 = c - 2 march 1 = c - 4 april 1 = c - 8 may 1 = c - 10 june 1 = c - 12 july 1 = c - 14 aug 1 = c - 16 sept 1 = c - 18 difference between stock on september 1 than were in stock on january 1 will be - c - ( c - 18 ) = 18 hence answer will be ( a )" | a = 10 - 1
b = a * 2
|
a ) $ 2.55 , b ) $ 2.85 , c ) $ 3.15 , d ) $ 3.45 , e ) $ 3.75 | d | divide(0.92, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5)))))) | having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.92 at the candy store . what is john β s weekly allowance ? | "x = 3 x / 5 + 1 / 3 * 2 x / 5 + 92 4 x / 15 = 92 x = 345 = $ 3.45 the answer is d ." | a = 3 / 5
b = 1 / 3
c = 3 / 5
d = 1 - c
e = b * d
f = a + e
g = 1 - f
h = 0 / 92
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a ) 12028 , b ) 12000 , c ) 12019 , d ) 12197 , e ) 18000 | c | divide(subtract(subtract(multiply(const_100, const_10), multiply(4000, divide(2.5, const_100))), 600), divide(2.5, const_100)) | a salesman β s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1300 plus 2.5 % commission on all sales exceeding rs . 4000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | [ 1300 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 18000 answer : c | a = 100 * 10
b = 2 / 5
c = 4000 * b
d = a - c
e = d - 600
f = 2 / 5
g = e / f
|
a ) 5768 , b ) 8925 , c ) 2345 , d ) 6474 , e ) 8935 | e | divide(divide(multiply(4020.75, const_100), 9), 5) | a sum fetched a total simple interest of 4020.75 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4020.75 ) / ( 9 x 5 ) = 402075 / 45 = 8935 . answer e" | a = 4020 * 75
b = a / 9
c = b / 5
|
a ) 457 km , b ) 444 km , c ) 547 km , d ) 588 km , e ) 653 km | d | add(multiply(divide(60, subtract(21, 28)), 28), multiply(divide(60, subtract(21, 28)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 28 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 28 + 21 = 49 t = 12 d = 49 * 12 = 588 answer : d" | a = 21 - 28
b = 60 / a
c = b * 28
d = 21 - 28
e = 60 / d
f = e * 21
g = c + f
|
a ) 21 , b ) 35 , c ) 33 , d ) 60 , e ) 65 | c | divide(multiply(multiply(7, 11), 3), 7) | exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 11 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ? | the total number of the people in the room must be a multiple of both 7 and 11 ( in order 3 / 7 and 5 / 11 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 11 , which is 77 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 77 people in the room . therefore there are 3 / 7 * 77 = 33 people in the room under the age of 21 . answer : c . | a = 7 * 11
b = a * 3
c = b / 7
|
a ) 90 , b ) 180 , c ) 270 , d ) 500 , e ) 1,000 | e | divide(divide(300, const_2), divide(15, const_100)) | one night 15 percent of the female officers on a police force were on duty . if 300 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 300 number of female officers on duty on that night = 300 / 2 = 150 ( 15 / 100 ) * f = 150 = > f = 1000 answer e" | a = 300 / 2
b = 15 / 100
c = a / b
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | subtract(subtract(812, 801), const_2) | with # andeach representing different digits in the problem below , the difference between # and # # is 801 . what is the value of ? # - # # ____ 812 | 100 x - ( 10 x + x ) - - - - - - - - - - - 89 x = 801 x = 9 c | a = 812 - 801
b = a - 2
|
a ) 11 : 15 , b ) 15 : 11 , c ) 16 : 15 , d ) 15 : 16 , e ) 11 : 16 | d | divide(multiply(divide(3, 4), 5), 4) | a cat leaps 5 leaps for every 4 leaps of a dog , but 3 leaps of the dog are equal to 4 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ? | "solution : given ; 3 dog = 4 cat ; or , dog / cat = 4 / 3 ; let cat ' s 1 leap = 3 meter and dogs 1 leap = 4 meter . then , ratio of speed of cat and dog = 3 * 5 / 4 * 4 = 15 : 16 . ' ' answer : option d" | a = 3 / 4
b = a * 5
c = b / 4
|
a ) 2 β 2 , b ) 2 β 3 , c ) 3 β 2 , d ) 3 β 3 , e ) 13 / 3 | e | divide(add(sqrt(97), sqrt(486)), sqrt(54)) | ( β 97 + β 486 ) / β 54 = ? | "( β 96 + β 486 ) / β 54 = ( 4 β 6 + 9 β 6 ) / 3 β 6 = 13 β 6 / 3 β 6 = 13 / 3 hence , the correct answer is e ." | a = math.sqrt(97)
b = math.sqrt(486)
c = a + b
d = math.sqrt(54)
e = c / d
|
a ) 25 min , b ) 32 min , c ) 33 min , d ) 34 min , e ) 35 min | e | multiply(add(2, 5), 5) | a secret can be told only 2 persons in 5 minutes . the same person tells to 2 more persons and so on . how long will take to tell it to 768 persons ? | at start one person will tell to 2 persons , it will take 5 min , now that 1 + 2 = 3 persons will tell this to next 6 persons , then 1 + 2 + 6 = 9 persons will tell to next 18 persons , then 1 + 2 + 6 + 18 = 27 persons to 54 similarly 1 + 2 + 6 + 18 + 54 = 81 persons will tell this to 162 persons similarly 1 + 2 + 6 + 18 + 54 + 162 = 243 persons will tell this to 486 persons , upto this step total persons who have listened this secret = 2 + 6 + 18 + 54 + 162 + 486 = 728 , and total time taken upto this step = 5 * 6 = 30 min , now next 5 min will be sufficient to tell this message to next 40 persons so total time taken = 35 min answer : e | a = 2 + 5
b = a * 5
|
a ) 2.35 sec , b ) 2.85 sec , c ) 7.5 sec , d ) 2.75 sec , e ) 1.5 sec | d | divide(110, multiply(144, const_0_2778)) | in what time will a train 110 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 110 / 40 = 2.75 sec . answer : d" | a = 144 * const_0_2778
b = 110 / a
|
a ) 6 , b ) 7 , c ) 5 , d ) 12 , e ) 14 | c | sqrt(add(power(multiply(6, divide(1, 2)), 2), power(multiply(subtract(6, 2), divide(1, 2)), 2))) | 9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 6 miles per hour while the second runner jogs due east at a constant rate that is 2 miles per hour faster than the first runner ' s rate . how far apart , to the nearest mile , will they be after 1 / 2 hour ? | "if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 6 m / h - - > 3 m in 1 / 2 hr side 2 = 8 m / h - - > 4 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 3 ^ 2 d ^ 2 = 25 = 5 answer option c" | a = 1 / 2
b = 6 * a
c = b ** 2
d = 6 - 2
e = 1 / 2
f = d * e
g = f ** 2
h = c + g
i = math.sqrt(h)
|
a ) 187 , b ) 197 , c ) 184 , d ) 219 , e ) 227 | c | subtract(multiply(13, 15), add(const_10, const_1)) | find the smallest number which when divided by 13 and 15 leaves respective remainders of 2 and 4 | let ' n ' is the smallest number which divided by 13 and 15 leaves respective remainders of 2 and 4 . required number = ( lcm of 13 and 15 ) - ( common difference of divisors and remainders ) = ( 195 ) - ( 11 ) = 184 . answer : c | a = 13 * 15
b = 10 + 1
c = a - b
|
a ) $ 420 , b ) $ 382 , c ) $ 385 , d ) $ 392 , e ) $ 399 | a | add(divide(406, add(const_1, divide(16, const_100))), multiply(divide(20, const_100), divide(406, add(const_1, divide(16, const_100))))) | if sharon ' s weekly salary increased by 16 percent , she would earn $ 406 per week . if instead , her weekly salary were to increase by 20 percent , how much would she earn per week ? | "( 406 / 116 ) 120 = 420 in this case long division does not take much time . ( 406 / 116 ) = 3.5 35 * 12 = 420 ( 350 + 70 ) answer a" | a = 16 / 100
b = 1 + a
c = 406 / b
d = 20 / 100
e = 16 / 100
f = 1 + e
g = 406 / f
h = d * g
i = c + h
|
a ) 12 , b ) 24 , c ) 36 , d ) 48 , e ) 38 | c | multiply(divide(divide(add(120, 120), 12), const_2), const_3_6) | two tains are running in opposite directions with the same speed . if the length of each train is 120 metres and they cross each other in 12 seconds , then the speed of each train ( in km / hr ) is : | sol . let the speed of each train be x m / sec . then , relative speed of the two trains = 2 x m / sec . so , 2 x = ( 120 + 120 ) / 12 β 2 x = 20 β x = 10 . β΄ speed of each train = 10 m / sec = [ 10 * 18 / 5 ] km / hr = 36 km / hr . answer c | a = 120 + 120
b = a / 12
c = b / 2
d = c * const_3_6
|
a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 2 miles , e ) 6 1 / 4 miles | b | divide(divide(multiply(subtract(4.90, 2.5), const_100), const_3), const_4) | a taxi company charges $ 2.5 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ? | "if we start out with $ 4.90 and have to spend $ 2.5 for the first quarter - mile , we will have $ 2.40 left to spend on quarter - mile intervals . since $ 2.40 / $ 0.15 = 16 , we can buy 16 more quarter - miles , and will travel 17 quarter miles in all : 17 Γ 1 / 4 = 4 1 / 4 miles . the correct answer is choice ( b ) ." | a = 4 - 90
b = a * 100
c = b / 3
d = c / 4
|
a ) 210 , b ) 280 , c ) 300 , d ) 320 , e ) 340 | c | divide(add(160, 20), divide(60, const_100)) | a student has to obtain 60 % of the total marks to pass . he got 160 marks and failed by 20 marks . the maximum marks are ? | "let the maximum marks be x then , 60 % of x = 160 + 20 60 x / 100 = 180 60 x = 180 * 100 = 18000 x = 300 answer is c" | a = 160 + 20
b = 60 / 100
c = a / b
|
a ) 215,000 , b ) 216,000 , c ) 217,000 , d ) 218,000 , e ) 219,000 | b | multiply(multiply(subtract(7, const_2), const_3600), const_12) | in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 7 - 2 = 5 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 5 = 216,000 . hence b . | a = 7 - 2
b = a * 3600
c = b * 12
|
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