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 ) 28.5 , b ) 27.675 , c ) 30 , d ) 50 , e ) none of these | d | subtract(divide(multiply(add(const_100, 42.5), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 42.5 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 ã — ( 100 + 42.5 ) / ( 100 â ˆ ’ 5 ) = 150 hence , required % profit = 150 â € “ 100 = 50 % answer d" | a = 100 + 42
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
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a ) 14 , b ) 7 , c ) 20 , d ) 70 , e ) 35 | c | divide(multiply(14, 10), 7) | pants , shorts and shirts in a dresser are in the ratio of 7 : 7 : 10 . if there are 14 pants , the number of shirts in the dresser is : | explanation : let pants = 7 x , shorts = 7 x & shirts = 10 x . now , 7 x = 14 hence x = 2 . number of apples = 10 x = 20 . answer : c | a = 14 * 10
b = a / 7
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a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 99 | c | multiply(multiply(11, const_2), const_2) | if x / ( 11 p ) is an even prime number , where x is a positive integer and p is a prime number , what is the least value of x ? | "x / ( 11 p ) = even prime number x = even prime number * 11 p least value of x = lowest even prime number * 11 * lowest value of p = 2 * 11 * 2 = 44 answer c" | a = 11 * 2
b = a * 2
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a ) 5 , b ) 7 , c ) 8 , d ) r = 9 , e ) 10 | d | subtract(multiply(add(divide(add(6, 4), 2), 3), 2), 7) | list i : { y , 2 , 4 , 7 , 10 , 11 } list ii : { 3 , 3 , 4 , 6 , 7 , 10 } if the median r of list i is equal to the sum of the median of list ii and the mode of list ii , then y equals | mode of list ii = 3 median r of list ii = 4 + 6 / 2 = 5 sum of mode + mean = 3 + 5 = 8 now to make 8 as the median we need to find a value of y such that if the no . of terms in list 1 are odd then y = 8 else if even then 7 + y / 2 = 8 here its even so 7 + y / 2 = 8 from this y = 9 ( d ) | a = 6 + 4
b = a / 2
c = b + 3
d = c * 2
e = d - 7
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a ) $ 28 , b ) $ 82 , c ) $ 110 , d ) $ 138 , e ) $ 194 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(4, const_100)), multiply(add(80, 4), 150)), 150) | a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 4 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 150 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "by the question , the equation would be 150 p - 84 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 194 . answer e for me ." | a = 4 * 4
b = a * 1000
c = 4 * 100
d = b + c
e = 80 + 4
f = e * 150
g = d + f
h = g / 150
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a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 60 % , e ) 66.6 % | b | divide(multiply(subtract(multiply(9, 9), multiply(6, 9)), const_100), multiply(9, 9)) | at a special sale , 9 tickets can be purchased for the price of 6 tickets . if 9 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 9 tickets ? | "let the price of a ticket be rs . 100 , so 6 tickets cost 600 & 9 tickets cost 900 9 tickets purchased at price of 6 tickets ie . , for 600 , so amount saved s rs . 300 , % of 9 tickets = ( 300 / 900 ) * 100 = 33.3 % answer : b" | a = 9 * 9
b = 6 * 9
c = a - b
d = c * 100
e = 9 * 9
f = d / e
|
a ) 6 , b ) 12 , c ) 14 , d ) 16 , e ) 28 | b | divide(multiply(subtract(35, 20), 8), 10) | how many gallons of milk that is 10 percent butter - fat must be added to 8 gallons of milk that is 35 percent butterfat to obtain milk that is 20 percent butterfat ? | "equate the fat : 0.1 x + 0.35 * 8 = 0.2 ( x + 8 ) - - > x = 12 . answer : b ." | a = 35 - 20
b = a * 8
c = b / 10
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a ) 3 , b ) 7 , c ) 17 , d ) 31 , e ) 38 | b | subtract(multiply(12, 12), add(add(add(add(add(add(add(add(add(add(3, 11), 7), 9), 15), 13), 8), 19), 17), 21), 14)) | a students was asked to find the arithmetic mean of the numbers 3 , 11 , 7 , 9 , 15 , 13 , 8 , 19 , 17 , 21 , 14 and x . he found the mean to be 12 . what should be the number in place of x ? | adding all the numbers we get 137 + x given - ( 137 + x ) / 12 = 12 137 + x = 12 * 12 x = 144 - 137 x = 7 answer : b | a = 12 * 12
b = 3 + 11
c = b + 7
d = c + 9
e = d + 15
f = e + 13
g = f + 8
h = g + 19
i = h + 17
j = i + 21
k = j + 14
l = a - k
|
a ) a ) 1190 , b ) b ) 1065 , c ) c ) 1070 , d ) d ) 1075 , e ) e ) 1080 | a | add(multiply(10, 70), multiply(9, 55)) | harkamal purchased 10 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 10 kg grapes = 70 × 10 = 700 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 700 + 490 = 1190 a" | a = 10 * 70
b = 9 * 55
c = a + b
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a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | a | multiply(subtract(divide(60, const_2), divide(multiply(60, 40), const_100)), const_2) | a chemical supply company has 60 liters of a 40 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 40 % hno 3 solution means hno 3 = 24 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 24 + x = 50 % of ( 60 + x ) or x = 12 . hence , a" | a = 60 / 2
b = 60 * 40
c = b / 100
d = a - c
e = d * 2
|
a ) 17 litres , b ) 24 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(36, 1399.45), multiply(36, 262.85)), subtract(3104.35, 1399.45)) | the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 36 litres of the mixed fruit juice . how many litres of the a ç ai berry juice does he need to add ? | "262.85 ( 36 ) + 3 , 104.35 x = 1 , 399.45 ( 36 + x ) solve the equation . 262.85 ( 36 ) + 3 , 104.35 x = 1 , 399.45 ( 36 + x ) 9 , 462.6 + 3 , 104.35 x = 50 , 380.2 + 1 , 399.45 x 9 , 462.6 + 1 , 704.9 x = 50 , 380.2 1 , 704.9 x = 40 , 917.6 x = 24 answer is b ." | a = 36 * 1399
b = 36 * 262
c = a - b
d = 3104 - 35
e = c / d
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a ) 1 % , b ) 8 % , c ) 12 % , d ) 5 % , e ) 4 % | b | divide(multiply(subtract(subtract(const_100, 35), 25), subtract(const_100, 80)), const_100) | a man spends 35 % of his income on food , 25 % on children education and 80 % of the remaining on house rent . what percent of his income he is left with ? | let the total income be x then , income left = ( 100 - 80 ) % of x - [ 100 - ( 35 + 25 ) ] % of x = 20 % of 40 % of x = 8 % of x answer is b | a = 100 - 35
b = a - 25
c = 100 - 80
d = b * c
e = d / 100
|
a ) 408 , b ) 475 , c ) 550 , d ) 625 , e ) 700 | a | multiply(divide(600, 11), 8) | a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 261 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 261 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 561 a + b + c = 51 per hour so for eight hrs = 51 * 8 = 408 = answer = a" | a = 600 / 11
b = a * 8
|
a ) 14.58 liters , b ) 20.16 liters , c ) 22.45 liters , d ) 26.95 liters , e ) 31.45 liters | a | subtract(subtract(subtract(20, 2), 2), 2) | a container contains 20 liters of milk , from this container 2 liters of milk was taken out and replaced by water . this process was repeated further 2 times . how much milk is now contained by the container ? | "amount of milk left after 3 operations = 20 ( 1 - 2 / 20 ) ^ 3 = 20 * 9 / 10 * 9 / 10 * 9 / 10 = 14.58 liters answer is a" | a = 20 - 2
b = a - 2
c = b - 2
|
a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | e | multiply(10, multiply(const_2, divide(sqrt(2025), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 2025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 2025 sq . units = > side of square = √ 2025 = 45 units the radius of the circle = side of the square = 45 units length of the rectangle = 2 / 5 * 45 = 18 units given that breadth = 10 units area of the rectangle = lb = 18 * 10 = 180 sq . units answer : option e" | a = math.sqrt(2025)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 57.6 , b ) 27 , c ) 25 , d ) 22 , e ) 72 | a | multiply(const_3_6, divide(add(250, 150), 25)) | a train of length 250 m crosses a bridge of length 150 m in 25 seconds . what is the speed of train ? | sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 150 ) = 25 x speed speed = 400 / 25 = 16 m / s = 57.6 km / h answer = a | a = 250 + 150
b = a / 25
c = const_3_6 * b
|
a ) 0.2 , b ) 1.2 , c ) 3 , d ) 4 , e ) 0.8 | e | subtract(subtract(multiply(500, power(add(const_1, divide(4, const_100)), 2)), 500), multiply(multiply(500, divide(4, const_100)), 2)) | indu gave bindu rs . 500 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "500 = d ( 100 / 4 ) 2 d = 0.8 answer : e" | a = 4 / 100
b = 1 + a
c = b ** 2
d = 500 * c
e = d - 500
f = 4 / 100
g = 500 * f
h = g * 2
i = e - h
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(add(const_2, const_3), const_2) | the product of the squares of two positive integers is 9 . how many pairs of positive integers satisfy this condition ? | "ans : b - 1 pairs ( x ˆ 2 ) ( y ˆ 2 ) = 9 [ square root both sides ] xy = 3 3 = 1 x 3 , 3 x 1 cancel the repeats this leaves us with exactly 1 options . hence , b" | a = 2 + 3
b = a - 2
|
a ) 9,000 , b ) 10,500 , c ) 11,750 , d ) 17,500 , e ) 19,500 | e | divide(divide(50, subtract(divide(subtract(divide(const_60, 52), inverse(2)), const_60), inverse(multiply(const_60, 2)))), const_1000) | with both inlets open , a water tank will be filled with water in 52 minutes . the first inlet alone would fill the tank in 2 hours . if in every minutes the second inlet admits 50 cubic meters of water than the first , what is the capacity of the tank ? | "the work done by inlet a and b together in 1 min = 1 / 52 the work done by inlet a ( first inlet ) in 1 min = 1 / 120 the work done by inlet b ( second inlet ) in 1 min = ( 1 / 52 ) - ( 1 / 120 ) = 1 / 92 difference of work done by b and a = b - a = 50 cubic meter i . e . ( 1 / 92 ) - ( 1 / 120 ) = 50 cubic meter = 19,500 cubic meter answer : option e" | a = const_60 / 52
b = 1/(2)
c = a - b
d = c / const_60
e = const_60 * 2
f = 1/(e)
g = d - f
h = 50 / g
i = h / 1000
|
['a ) 125 %', 'b ) 129 %', 'c ) 185 %', 'd ) 725 %', 'e ) 145 %'] | a | multiply(divide(subtract(multiply(power(divide(const_3, const_2), const_2), multiply(const_3, const_2)), multiply(const_3, const_2)), multiply(const_3, const_2)), const_100) | if each edge of a cube is increased by 50 % , find the percentage increase in its surface area | let the edge = a cm so increase by 50 % = = total surface area of original cube = tsa of new cube = = = increase in area = = increase % = = 125 % answer : a | a = 3 / 2
b = a ** 2
c = 3 * 2
d = b * c
e = 3 * 2
f = d - e
g = 3 * 2
h = f / g
i = h * 100
|
a ) 10 , b ) 15 , c ) 6 , d ) 8 , e ) 7 | c | multiply(4, divide(12, 8)) | 12 : 8 seconds : : ? : 4 minutes | "12 * 4 = 8 * x x = 6 answer : c" | a = 12 / 8
b = 4 * a
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a ) 180 , b ) 2700 , c ) 18 , d ) 18000 , e ) 1.8 | b | divide(multiply(36, 15), divide(multiply(4, 5), const_100)) | a hall 36 m long and 15 m broad is to be paved with stones , each measuring 4 dm by 5 dm . the number of stones required is : | "area of the hall = 3600 * 1500 area of each stone = ( 40 * 50 ) therefore , number of stones = ( 3600 * 1500 / 40 * 50 ) = 2700 answer : b" | a = 36 * 15
b = 4 * 5
c = b / 100
d = a / c
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a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | c | multiply(divide(10, subtract(subtract(const_100, 80), 10)), const_100) | jane makes toy bears . when she works with an assistant , she makes 80 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent r ? | "c . let ' s assume just jane 40 bears per 40 / hrs a week , so that is 1 bear / hr . with an assistant she makes 72 bears per 36 hours a week or 2 bears / hr ( [ 40 bears * 1.8 ] / [ 40 hrs * . 90 ] ) . r = [ ( 2 - 1 ) / 1 ] * 100 % = 100 % . c" | a = 100 - 80
b = a - 10
c = 10 / b
d = c * 100
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a ) $ 1.09 , b ) $ 1.67 , c ) $ 2.25 , d ) $ 2.37 , e ) $ 3.03 | e | divide(add(15, multiply(1.75, subtract(9, 2))), 9) | the cost to park a car in a certain parking garage is $ 15.00 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 = 15 $ for the first 2 hours and then 1.75 for ( 9 - 2 ) hours = 15 + 7 * 1.75 = 27.25 thus the average parking price = 27.25 / 9 = 3.03 $ e is the correct answer . | a = 9 - 2
b = 1 * 75
c = 15 + b
d = c / 9
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a ) 2 : 9 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | e | divide(subtract(4, 2), subtract(10, 4)) | cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 4 - 2 ) / ( 10 - 4 ) = 2 / 6 = 1 / 3 so they should be mixed in the ratio 1 : 3 answer - e" | a = 4 - 2
b = 10 - 4
c = a / b
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a ) 51 , b ) 88 , c ) 90 , d ) 42 , e ) 22 | a | subtract(divide(4950, 75), 15) | a trader sells 75 meters of cloth for rs . 4950 at the profit of rs . 15 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 4950 / 75 = rs . 66 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 66 - rs . 15 = rs . 51 answer : a" | a = 4950 / 75
b = a - 15
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a ) 586413 , b ) 587523 , c ) 614823 , d ) 615173 , e ) none of them | a | multiply(divide(587, 999), const_100) | 587 x 999 = ? | "= 587 x 999 = 587 x ( 1000 - 1 ) = 587 x 1000 - 587 - 1 = 587000 - 587 = 586413 answer is a" | a = 587 / 999
b = a * 100
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a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | b | subtract(add(14, 12), divide(14, divide(70, const_100))) | the contents of a certain box consist of 14 apples and 12 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ? | "the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 12 oranges , so to get to 6 oranges , we should remove 12 - 6 = 6 oranges . answer b" | a = 14 + 12
b = 70 / 100
c = 14 / b
d = a - c
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a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 95 | a | divide(110, const_2) | he total marks obtained by a student in physics , chemistry and mathematics is 110 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | "let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 110 + p c + m = 110 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 110 / 2 = 55 . answer : a" | a = 110 / 2
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['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) 12 cm'] | a | sqrt(square_area(multiply(4, 2))) | the diagonal of a square is 4 ? 2 cm . what will be the diameter of another square if their area is double that of the first square ? | d 1 = 4 ? 2 = > area = 1 / 2 d 12 = 1 / 2 * ( 4 ? 2 ) 2 = 16 cm 2 area of new square = ( 2 * 16 ) cm 2 = 32 cm 2 therefore 1 / 2 * d 22 = 32 = > d 22 = 64 = > d 2 = 8 cm answer a | a = 4 * 2
b = math.sqrt(square_area)
|
a ) 166 , b ) 178 , c ) 342 , d ) 199 , e ) 112 | c | subtract(subtract(400, divide(multiply(400, 10), const_100)), divide(multiply(subtract(400, divide(multiply(400, 10), const_100)), 5), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 10 % and 5 % is ? | "400 * ( 90 / 100 ) * ( 95 / 100 ) = 342 answer : c" | a = 400 * 10
b = a / 100
c = 400 - b
d = 400 * 10
e = d / 100
f = 400 - e
g = f * 5
h = g / 100
i = c - h
|
['a ) 2812', 'b ) 8231', 'c ) 2734', 'd ) 4222', 'e ) 4254'] | c | divide(10936, 4) | a volume of 10936 l water is in a container of sphere . how many hemisphere of volume 4 l each will be required to transfer all the water into the small hemispheres ? | a volume of 4 l can be kept in 1 hemisphere therefore , a volume of 10936 l can be kept in ( 10936 / 4 ) hemispheres ans . 2734 answer : c | a = 10936 / 4
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a ) 31 , b ) 36 , c ) 48 , d ) 45 , e ) 57 | c | add(31, const_1) | the average age of 31 students in a group is 16 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ? | "age of the teacher = ( 32 * 17 - 31 * 16 ) = 48 years . answer : c" | a = 31 + 1
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a ) 1.01 , b ) 4.44 , c ) 11.5 , d ) 1.75 , e ) 2.14 | b | divide(divide(multiply(5, 80), const_2), 45) | venki takes a trip and drives 5 hours from town x to town z at a rate of 80 miles per hour . on her way back , venki drives 45 miles per hour and stops in town y which is midway between town x and town z . how many hours does it take venki to drive from town z to town y ? | distance from x to z = 5 hr * 80 mph = 400 miles hence distance from y to z = 1 / 2 * 400 = 200 time = 200 / 45 = 4.44 hrs imo : b | a = 5 * 80
b = a / 2
c = b / 45
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['a ) 185', 'b ) 132', 'c ) 165', 'd ) 192', 'e ) 212'] | d | multiply(sqrt(subtract(power(20, const_2), power(16, const_2))), 16) | if a rectangle have length 16 cm and diagonal 20 cm . calculate its area . | breadth = √ 20 ^ 2 - 16 ^ 2 = √ 144 = 12 cm area = 16 * 12 = 192 sq . cm answer d | a = 20 ** 2
b = 16 ** 2
c = a - b
d = math.sqrt(c)
e = d * 16
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a ) 1 : 42 , b ) 1 : 34 , c ) 1 : 44 , d ) 1 : 54 , e ) 2 : 14 | a | divide(add(1, add(divide(const_12, const_100), add(add(multiply(multiply(divide(1, 5), divide(1, 5)), 2), multiply(divide(1, 5), divide(1, 5))), divide(1, 5)))), const_60) | a qualified worker digs a well in 5 hours . he invites 2 apprentices , both capable of working 3 / 5 as fast and 2 trainees both working 1 / 5 as fast as he . if the 5 - person team digs the same well , how much time does the team need to finish the job ? | and answer is correct as you have calculated : 100 / 59 hrs that is 1.71 hrs - 1 hr and 42 minutes , so 1 : 42 , hence , a . | a = 12 / 100
b = 1 / 5
c = 1 / 5
d = b * c
e = d * 2
f = 1 / 5
g = 1 / 5
h = f * g
i = e + h
j = 1 / 5
k = i + j
l = a + k
m = 1 + l
n = m / const_60
|
['a ) rs . 150', 'b ) rs . 132', 'c ) rs . 100', 'd ) rs . 125', 'e ) rs . 225'] | b | multiply(circumface(divide(28, const_2)), 1.5) | find the cost of fencing around a circular field of diameter 28 m at the rate of rs . 1.50 a meter ? | 2 * 22 / 7 * 14 = 88 88 * 1 1 / 2 = rs . 132 answer : b | a = 28 / 2
b = circumface * (
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a ) 5 hours , b ) 3 hours 20 minutes , c ) 3 hours , d ) 2 hours 45 minutes , e ) 8 hours | c | divide(add(14, 4), subtract(63, 57)) | truck x is 14 miles ahead of truck y , which is traveling the same direction along the same route as truck x . if truck x is traveling at an average speed of 57 miles per hour and truck y is traveling at an average speed of 63 miles per hour , how long will it take truck y to overtake and drive 4 miles ahead of truck x ? | "relative speed = 63 - 57 = 6 miles per hour dist required = 14 + 4 = 18 miles time taken to overtake = 18 / 6 = 3 hours . c is the answer ." | a = 14 + 4
b = 63 - 57
c = a / b
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a ) rs . 500 , b ) rs . 750 , c ) rs . 1250 , d ) rs . 1500 , e ) rs . 1520 | d | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | a and b ’ s salaries together amount to rs . 2,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a ’ s salary ? | "explanation : ( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 2000 4 b = 2000 = > b = 500 a = 1500 answer is d" | a = 2 + 3
b = 4 * a
c = 2 * b
d = c * 100
e = 85 / 100
f = 1 - e
g = d * f
h = 95 / 100
i = 1 - h
j = 85 / 100
k = 1 - j
l = i + k
m = g / l
|
a ) 1600 , b ) 3200 , c ) 800 , d ) 400 , e ) 200 | b | multiply(80, multiply(4, const_10)) | last year , for every 100 million vehicles that traveled on a certain highway , 80 vehicles were involved in accidents . if 4 billion vehicles traveled on the highway last year , how many of those vehicles were involved in accidents ? ( 1 billion = 1,000 , 000,000 ) | "to solve we will set up a proportion . we know that “ 100 million vehicles is to 80 accidents as 4 billion vehicles is to x accidents ” . to express everything in terms of “ millions ” , we can use 4,000 million rather than 4 billion . creating a proportion we have : 100 / 80 = 4,000 / x cross multiplying gives us : 100 x = 4,000 * 80 x = 40 * 80 = 3200 answer : b" | a = 4 * 10
b = 80 * a
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | add(divide(subtract(20, add(4, multiply(const_2, 4))), 4), 4) | on sunday , bill ran 4 more miles than he ran on saturday . julia did not run on saturday , but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of 20 miles on saturday and sunday , how many miles did bill run on sunday ? | let bill run x on saturday , so he will run x + 4 on sunday . . julia will run 2 * ( x + 4 ) on sunday . . totai = x + x + 4 + 2 x + 8 = 20 . . 4 x + 12 = 20 . . x = 2 . . ans = x + 4 = 2 + 4 = 6 answer b | a = 2 * 4
b = 4 + a
c = 20 - b
d = c / 4
e = d + 4
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(const_1, divide(8, const_4)) | log root ( 8 ) / log 8 is equal to | = > log ( 8 ) ^ 1 / 2 / log 8 = 1 / 2 log 8 / log 8 = 1 / 2 answer a | a = 8 / 4
b = 1 / a
|
a ) 15 , b ) 30 , c ) 35 , d ) 40 , e ) 48 | e | subtract(multiply(divide(40, const_100), 180), multiply(divide(const_1, const_3), multiply(divide(40, const_100), 180))) | one - third less than 40 % of 180 is equal to : | "lots of ways to tackle this . 40 % of 180 = 72 1 / 3 of 72 = 24 so , 1 / 3 less than 72 is equal to 72 - 24 = 48 answer : e" | a = 40 / 100
b = a * 180
c = 1 / 3
d = 40 / 100
e = d * 180
f = c * e
g = b - f
|
a ) 850 , b ) 1,100 , c ) 1,700 , d ) 2,550 , e ) 2,300 | e | divide(add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), 200)), subtract(multiply(const_2, const_10), const_1)) | alan buys 4 tvs , a 26 inch , a 28 inch , a 30 inch , and a 32 inch , for his new house . each tv costs $ 200 more than the size below it . alan spends a total of $ 4,600 . how much would he have spent if he had bought only the 28 inch and 30 inch tvs ? | "assume the cost of the least sized ( 26 inch ) tv = x cost of 28 inches tv = x + 200 cost of 30 inches tv = x + 400 cost of 32 inches tv = x + 600 total cost = 4 x + 1200 = 4600 therefore x = 3400 / 4 = 850 price of 28 inch + 30 inch = 1050 + 1250 = 2300 option e" | a = 200 + 3600
b = 1 + 2
c = b + 3
d = c * 200
e = a - d
f = e / 4
g = f + 200
h = 200 + 3600
i = 1 + 2
j = i + 3
k = j * 200
l = h - k
m = l / 4
n = m + 200
o = n + 200
p = g + o
q = 2 * 10
r = q - 1
s = p / r
|
a ) 4 days , b ) 5 days , c ) 6 days , d ) 7 days , e ) 8 days | a | divide(subtract(const_1, add(multiply(divide(const_1, 4), const_2), multiply(divide(const_1, 12), const_2))), divide(const_1, 12)) | a can finish a piece of work in 4 days . b can do it in 12 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 12 = 1 = > x = 4 days answer : a" | a = 1 / 4
b = a * 2
c = 1 / 12
d = c * 2
e = b + d
f = 1 - e
g = 1 / 12
h = f / g
|
a ) - 3 , b ) 3 , c ) - 4 , d ) 4 , e ) - 5 | b | subtract(multiply(5, divide(subtract(multiply(2, 4), 1), subtract(multiply(7, 2), multiply(2, 9)))), add(multiply(9, divide(subtract(multiply(2, 4), 1), subtract(multiply(7, 2), multiply(2, 9)))), 4)) | when positive integer x is divided by 9 , the quotient is y and the remainder is 4 . when 2 x is divided by 7 , the quotient is 2 y and the remainder is 1 . what is the value of 5 y – x ? | "( 1 ) x = 9 y + 4 ( 2 ) 2 x = 14 y + 1 ( 2 ) - ( 1 ) : x = 5 y - 3 5 y - x = 3 the answer is b ." | a = 2 * 4
b = a - 1
c = 7 * 2
d = 2 * 9
e = c - d
f = b / e
g = 5 * f
h = 2 * 4
i = h - 1
j = 7 * 2
k = 2 * 9
l = j - k
m = i / l
n = 9 * m
o = n + 4
p = g - o
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | divide(add(add(4, 3), 3), 2) | a school has 4 maths 3 physics and 3 chemistry teachers each teacher can teach 2 subjects max what is he minimum number of teachers required | "total subjects = 4 + 3 + 3 = 10 max subjects by 1 teacher = 2 so , min of teachers required = 10 / 2 = 5 answer : b" | a = 4 + 3
b = a + 3
c = b / 2
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 1 | e | add(8, divide(subtract(245, 20), add(25, 20))) | two stations a and b are 245 km apart on a straight line . one train starts from a at 7 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 7 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 245 45 x = 270 x = 6 . so , they meet at 1 p . m . answer : option e" | a = 245 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) one , b ) none , c ) two , d ) three , e ) four | b | subtract(subtract(add(multiply(multiply(multiply(6, 3), const_2), const_4), 6), add(multiply(multiply(multiply(6, 3), const_2), const_4), 3)), 1) | for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 6 + 3 and * 6 + 6 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 6 would be divisible by all numbers from 1 to 6 . = > * 6 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 6 is divisible by 3 ) in fact adding anyprimenumber between 1 to 6 to * 6 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 6 + 3 and * 6 + 11 then the answer would be 1 . for * 6 + 3 and * 6 + 13 , it is 2 and so on . . . b" | a = 6 * 3
b = a * 2
c = b * 4
d = c + 6
e = 6 * 3
f = e * 2
g = f * 4
h = g + 3
i = d - h
j = i - 1
|
a ) 18 % , b ) 72 % , c ) 32 % , d ) 25.25 % , e ) 52.25 % | d | subtract(const_100, divide(multiply(add(const_100, 15), subtract(const_100, 35)), const_100)) | the tax on a commodity is diminished by 35 % but its consumption is increased by 15 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 65 * 115 = 7475 10000 - - - - - - - 2525 100 - - - - - - - ? = 25.25 % answer : d" | a = 100 + 15
b = 100 - 35
c = a * b
d = c / 100
e = 100 - d
|
a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 32 | d | multiply(const_4, 4) | a university cafeteria offers 4 flavors of pizza - pepperoni , chicken , hawaiian and vegetarian . if a customer has an option ( but not the obligation ) to add extra cheese , mushrooms or both to any kind of pizza , how many different pizza varieties are available ? | there are 4 flavors of pizza and each can be : 1 . without cheese and mushrooms ; 2 . with cheese ; 3 . with mushrooms ; 4 . with cheese and mushrooms . so , the total number of pizza varieties is 4 * 4 = 16 . answer : d . | a = 4 * 4
|
a ) 1080 , b ) 10080 , c ) 10025 , d ) 11080 , e ) 12080 | b | add(divide(divide(multiply(floor(add(divide(subtract(multiply(const_10, const_1000), const_1), 15), const_1)), 15), const_4), const_100), const_2) | find the least number of five digits which is exactly divisible by 12 , 15 and 18 ? | "lcm = 180 180 ) 10000 ( 55 9900 - - - - - - - - - 100 10000 + 180 - 100 = 10080 answer : b" | a = 10 * 1000
b = a - 1
c = b / 15
d = c + 1
e = math.floor(d)
f = e * 15
g = f / 4
h = g / 100
i = h + 2
|
a ) 2.6 hours , b ) 3.6 hours , c ) 4.6 hours , d ) 5.6 hours , e ) none of these | b | divide(const_1, subtract(add(divide(const_1, 8), divide(const_1, 5)), divide(const_1, 20))) | a tank can be filled by pipe a in 5 hours and by pipe b in 8 hours , each pump working on its own . when the tank is full and a drainage hole is open , the water is drained in 20 hours . if initially the tank was empty and someone started the two pumps together but left the drainage hole open , how long does it take for the tank to be filled ? | let ' s first find the rates of the pumps and the drainage hole pump a : 1 / 5 , pump b : 1 / 8 , drainage hole : 1 / 20 let t be the time for the pumps to fill the tank . the pumps , add water into the tank however the drainage hole drains water out of the tank , hence t ( 1 / 5 + 1 / 8 - 1 / 20 ) = 1 solve for t t = 3.6 hours answer b | a = 1 / 8
b = 1 / 5
c = a + b
d = 1 / 20
e = c - d
f = 1 / e
|
a ) 8 : 8 , b ) 9 : 8 , c ) 9 : 7 , d ) 9 : 5 , e ) 9 : 10 | e | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(5000, add(6, 6))) | a and b invests rs . 3000 and rs . 5000 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 ) : ( 5 * 12 ) 54 : 60 = > 9 : 10 answer : e | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 5000 * e
g = d / f
|
a ) 120 mps , b ) 200 mps , c ) 181 mps , d ) 190 mps , e ) 125 mps | b | multiply(const_0_2778, 720) | express a speed of 720 kmph in meters per second ? | "720 * 5 / 18 = 200 mps answer : b" | a = const_0_2778 * 720
|
a ) 8.17 , b ) 7.14 , c ) 6.23 , d ) 4.35 , e ) 8.75 | e | divide(84, 12) | 84 / 12 / 0.8 = ? | "e 8.75 ? = 84 / ( 12 * 0.8 ) = 8.75" | a = 84 / 12
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a ) 2 / 13 , b ) 1 / 26 , c ) 5 / 26 , d ) 1 / 52 , e ) 17 / 52 | d | multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52)) | from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a king ? | "p ( club ) = 1 / 4 p ( king ) = 1 / 13 p ( club then a king ) = 1 / 4 * 1 / 13 = 1 / 52 the answer is d ." | a = 3 * 4
b = a + 1
c = b / const_52
d = 2 / const_52
e = c * d
|
['a ) 50 m', 'b ) 60 m', 'c ) 65 m', 'd ) 75 m', 'e ) 80 m'] | b | add(divide(subtract(divide(5300, 26.5), multiply(20, const_2)), const_4), 20) | the length of a rectangular plot is 20 mtr more than its breadth . if the cost of fencingthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | let breadth = x metres . then , length = ( x + 20 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 20 ) + x ] = 200 2 x + 20 = 100 2 x = 80 x = 40 . hence , length = x + 20 = 60 m b | a = 5300 / 26
b = 20 * 2
c = a - b
d = c / 4
e = d + 20
|
a ) − 2 π , b ) - 1 , c ) − π , d ) 2 π , e ) 4 π | c | subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(const_12, const_4)), const_1000), const_10), const_3), subtract(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(const_12, const_4)), const_1000), const_10), const_3), negate(add(divide(divide(add(multiply(add(const_10, const_4), const_100), add(const_12, const_4)), const_1000), const_10), const_3)))) | define r * by the equation r * = π - r . then ( ( − π ) * ) * = | for r * f ( f ( − π ) ) = f ( π − ( − π ) ) = f ( π + π ) = f ( 2 π ) = π − 2 π = − π = c | a = 10 + 4
b = a * 100
c = 12 + 4
d = b + c
e = d / 1000
f = e / 10
g = f + 3
h = 10 + 4
i = h * 100
j = 12 + 4
k = i + j
l = k / 1000
m = l / 10
n = m + 3
o = 10 + 4
p = o * 100
q = 12 + 4
r = p + q
s = r / 1000
t = s / 10
u = t + 3
v = n - negate
w = g - v
|
a ) 9 , b ) 18 , c ) 17 , d ) none of these , e ) can not be determined | b | add(divide(subtract(100, 5), 5), const_1) | how many multiples of 5 are there between 5 to 100 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 , 95 . answer : b" | a = 100 - 5
b = a / 5
c = b + 1
|
a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase r in sales ? | last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase r = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c . | a = 385 - 320
b = a / 320
c = b * 100
|
a ) 1 km , b ) 500 mts , c ) 900 mts , d ) 2 km , e ) 250 mts | c | multiply(multiply(divide(divide(11, const_60), add(add(divide(const_1, 3), divide(const_1, 6)), divide(const_1, 9))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 6 km / hr , 9 km / hr . and takes a total time of 11 minutes . find the total distance ? | "let the each distance be x km total distance = 3 x then total time , ( x / 3 ) + ( x / 6 ) + ( x / 9 ) = 11 / 60 x = 0.3 total distance = 3 * 0.3 = 0.9 km = 900 meters correct option is c" | a = 11 / const_60
b = 1 / 3
c = 1 / 6
d = b + c
e = 1 / 9
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 20 lb , b ) 14 lb , c ) 10 lb , d ) 15 lb , e ) 5 lb | b | divide(28, const_1) | a bag of potatoes weighs 28 lbs divided by half of its weight . how much does the bag of potatoes weight ? | "sol . 28 ÷ 2 = 14 . answer : b" | a = 28 / 1
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a ) 2 , b ) 5 , c ) 6 , d ) 4 , e ) 1 | d | divide(multiply(4, 2), subtract(4, 2)) | pipe a can fill a tank in 2 hours . due to a leak at the bottom , it takes 4 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 2 - 1 / x = 1 / 4 = > 1 / x = 1 / 2 - 1 / 4 = 1 / 4 = > x = 4 . answer : d" | a = 4 * 2
b = 4 - 2
c = a / b
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | multiply(15, const_1) | the total age of a and b is 15 years more than the total age of b and c . c is how many year younger than a | "explanation : given that a + b = 15 + b + c = > a ? c = 15 + b ? b = 15 = > c is younger than a by 15 years answer : option e" | a = 15 * 1
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a ) 24 , b ) 25 , c ) 34 , d ) 39 , e ) 43 | c | add(add(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), divide(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), const_10)), const_1) | 306 ã · ? ã — 15 + 270 = 405 | explanation : 306 ã · ? ã — 15 = 405 - 270 = 135 ( 306 ã — 15 ) / ? = 135 ? = ( 306 ã — 15 ) / 135 = 34 answer : option c | a = 405 - 270
b = a + 15
c = b * 2
d = c / 10
e = 405 - 270
f = e + 15
g = f * 2
h = g / 10
i = h / 10
j = d + i
k = j + 1
|
a ) 10 sec , b ) 11 sec , c ) 77 sec , d ) 99 sec , e ) 67 sec | a | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 50), inverse(multiply(const_0_2778, add(60, 40)))) | two trains of equal length , running with the speeds of 60 and 40 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 50 * 100 / 18 = 2500 / 9 rs = 60 + 40 = 100 * 5 / 18 t = 2500 / 9 * 18 / 500 = 10 sec answer : a" | a = 60 - 40
b = const_0_2778 * a
c = b * 50
d = 60 + 40
e = const_0_2778 * d
f = 1/(e)
g = c * f
|
a ) 8 / 3 , b ) 3 / 8 , c ) 8 / 5 , d ) 12 / 5 , e ) 5 / 3 | d | divide(divide(540, 3), divide(300, 4)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 540 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 540 km time taken by eddy = 3 hours average speed of eddy = 540 / 3 = 180 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 180 / 75 = 12 / 5 answer d" | a = 540 / 3
b = 300 / 4
c = a / b
|
a ) 4.25 , b ) 5 , c ) 6.2 , d ) 7 , e ) 7.5 | c | divide(subtract(282, multiply(10, 3.4)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.4 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 3.4 x 10 ) = 248 248 / 40 = 6.2 c )" | a = 10 * 3
b = 282 - a
c = b / 40
|
a ) 12 , b ) 11 , c ) 18 , d ) 24 , e ) 10 | b | add(subtract(32, add(20, 1)), 1) | the average weight of a group of boys is 20 kg . after a boy of weight 32 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 32 kg joins the group , total weight of boys = 20 x + 32 so 20 x + 32 = 21 ( x + 1 ) = > x = 11 . answer : b" | a = 20 + 1
b = 32 - a
c = b + 1
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | d | subtract(11, 5) | a football team lost 5 yards and then gained 11 . what is the team ' s progress ? | "for lost , use negative . for gain , use positive . progress = - 5 + 11 = 6 yards d" | a = 11 - 5
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a ) 17 sec , b ) 12 sec , c ) 16 sec , d ) 15 sec , e ) 18 sec | c | divide(multiply(160, const_2), add(speed(160, 15), speed(160, 10))) | two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 160 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 160 + 160 ) / 20 = 16 sec . answer : c" | a = 160 * 2
b = speed + (
c = a / b
|
a ) 126 % , b ) 128 % , c ) 130 % , d ) 132 % , e ) 134 % | d | multiply(const_100, divide(10, multiply(add(10, const_100), divide(12, const_100)))) | last year a worker saved 10 % of her annual salary . this year , she made 10 % more money than last year and she saved 12 % of her salary . the amount she saved this year was what percent of the amount she saved last year ? | "let x be the worker ' s salary last year . last year , she saved 0.1 x . this year , she saved 0.12 ( 1.1 x ) = 0.132 x 0.132 x / 0.1 x = 132 % the answer is d ." | a = 10 + 100
b = 12 / 100
c = a * b
d = 10 / c
e = 100 * d
|
a ) 60 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | a | divide(120, const_2) | what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 120 ? | "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 120 / 2 = 60 answer is a" | a = 120 / 2
|
a ) 15 , b ) 12 , c ) 17 , d ) 18 , e ) 19 | b | add(multiply(3, 4), const_1) | how many distinct integer values of n satisfy the inequality | | n - 3 | + 4 | ≤ 14 ? | "so i can write this as in - 3 i + 4 < = 14 or in - 3 i < = 10 so n can have - 7 to 13 = 12 true values . . . . b" | a = 3 * 4
b = a + 1
|
a ) 698 , b ) 387 , c ) 373 , d ) 593 , e ) 372 | a | subtract(815, divide(multiply(subtract(854, 815), 3), 4)) | a sum of money at simple interest amounts to 815 in 3 years and to 854 in 4 years . the sum is : | "a 698 s . i . for 1 year = ( 854 - 815 ) = 39 . s . i . for 3 years = ( 39 x 3 ) = 117 . principal = ( 815 - 117 ) = 698 ." | a = 854 - 815
b = a * 3
c = b / 4
d = 815 - c
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a ) 10058 , b ) 10068 , c ) 10078 , d ) 10088 , e ) 10038 | e | divide(5548, multiply(multiply(multiply(subtract(const_1, divide(15, const_100)), subtract(const_1, divide(15, const_100))), subtract(const_1, divide(10, const_100))), subtract(const_1, divide(15, const_100)))) | a woman spends 15 % of her income in house rent , 15 % of the rest on his children ’ s education , 10 % of the rest miscellaneous causes and 15 % in medical expenses . if she now posses rs . 5548 then her income is ? | x * ( 85 / 100 ) * ( 85 / 100 ) * ( 90 / 100 ) * ( 85 / 100 ) = 5548 x * 0.85 * 0.85 * 0.9 * 0.85 = 5548 x = 5548 / 0.5527 x = 10038 answer : e | a = 15 / 100
b = 1 - a
c = 15 / 100
d = 1 - c
e = b * d
f = 10 / 100
g = 1 - f
h = e * g
i = 15 / 100
j = 1 - i
k = h * j
l = 5548 / k
|
a ) 12 , b ) 16 , c ) 18 , d ) 24 , e ) 26 | b | divide(48, add(const_1, const_2)) | a student got twice as many sums wrong as he got right . if he attempted 48 sums in all , how many did he solve correctly ? | "explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 48 3 x = 48 x = 16 . answer : b" | a = 1 + 2
b = 48 / a
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a ) 570 , b ) 670 , c ) 1070 , d ) 470 , e ) 2070 | c | add(multiply(20, 50), multiply(35, multiply(20, divide(10, 50)))) | a worker is paid a regular rate of rs . 20 for completing a survey . the worker completes 50 surveys per week . for any survey involving the use of her cellphone , she is paid at a rate of that is 10 % higher than her regular rate . if she completed 35 surveys involving the use of her cellphone , how much did she get that week ? | "amount earned by using her cell phone = 35 * 22 = 770 amount earned as usual = 15 * 20 = 300 total = 770 + 300 = 1070 answer : c" | a = 20 * 50
b = 10 / 50
c = 20 * b
d = 35 * c
e = a + d
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a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 68 | e | divide(add(add(add(add(50, 60), 70), 80), 80), add(const_4, const_1)) | reeya obtained 50 , 60 , 70 , 80 and 80 out of 100 in different subjects , what will be the average | "explanation : ( 50 + 60 + 70 + 80 + 80 / 5 ) = 68 option e" | a = 50 + 60
b = a + 70
c = b + 80
d = c + 80
e = 4 + 1
f = d / e
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a ) 425 , b ) 375 , c ) 350 , d ) 400 , e ) 500 | d | divide(subtract(2500, multiply(10, 50)), 5) | 5 goats ’ s and 10 hens are brought for rs . 2500 . if the average price of a hen be rs . 50 . what is the average price of a goat . | explanation : average price of a hen = rs . 50 total price of 10 hens = 10 * 50 = rs . 500 but total price of 5 goats and 10 hens = rs . 2500 total price of 5 goats is = 2500 - 500 = 2000 average price of a goat = 2000 / 5 = rs . 400 answer : d | a = 10 * 50
b = 2500 - a
c = b / 5
|
a ) 10 miles , b ) 14 miles , c ) 12 miles , d ) 20 miles , e ) 30 miles | a | multiply(10, 1) | kareena , saif , and a puppy start walking down a road . they start at the same time , from the same point , in the same direction . the saif walks at 5 miles / hour kareena at 6 miles / hour . the puppy runs from saif to kareena and back again with a constant speed of 10 miles / hour . the puppy does not slow down on the turn . how far does the puppy travel in 1 hour ? | a 10 miles speed us 10 miles / hour | a = 10 * 1
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a ) 3 , b ) 6 , c ) 8 , d ) 12 , e ) 13 | e | multiply(const_3, divide(divide(39, const_3), const_3)) | if m is a positive integer and m ^ 2 is divisible by 39 , then the largest positive integer that must divide m is ? | "m ^ 2 is divisible by 48 so m ^ 2 must be multiple of 48 . if the value of m is multiples of 13 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is e but if m = 39 then answer should be 16 . is the question right ? or am i missing some thing ? e" | a = 39 / 3
b = a / 3
c = 3 * b
|
a ) 7.76 kms , b ) 8.76 kms , c ) 5.76 kms , d ) 3.76 kms , e ) 4.76 kms | c | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "explanation : speed in still water = 6 kmph stream speed = 1.2 kmph down stream = 7.2 kmph up stream = 4.8 kmph x / 7.2 + x / 4.8 = 1 x = 2.88 total distance = 2.88 x 2 = 5.76 kms answer : c" | a = 6 + 1
b = 6 - 1
c = a * b
d = 6 + 1
e = 6 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | b | divide(divide(divide(315, const_3), const_3), add(const_1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 315 what is the least possible value of k ? | 315 = 3 * 3 * 5 * 7 thus k must include numbers at least up to the number 7 . the answer is b . | a = 315 / 3
b = a / 3
c = 1 + 4
d = b / c
|
a ) 250 , b ) 420 , c ) 440 , d ) 650 , e ) 450 | e | divide(multiply(power(15, 2), power(8, 3)), 256) | find the value of a from ( 15 ) ^ 2 x 8 ^ 3 ã · 256 = a . | given exp . = ( 15 ) ^ 2 x 8 ^ 3 ã · 256 = a = 225 x 512 ã · 256 = 450 e | a = 15 ** 2
b = 8 ** 3
c = a * b
d = c / 256
|
a ) 85 , b ) 91 , c ) 115 , d ) 133 , e ) 166 | e | multiply(subtract(85, const_2), const_2) | the sum of two prime numbers is 85 . what is the product of these two prime numbers ? | since there is only one even prime number ( 2 ) , and the sum of two odd numbers is even , the product of these two numbers must be even . correct answer e | a = 85 - 2
b = a * 2
|
a ) 10 days , b ) 16 days , c ) 20 days , d ) 25 days , e ) 22 days | b | divide(const_1, divide(add(add(inverse(15), inverse(30)), inverse(45)), const_2)) | a and b can do a work in 15 days , b and c in 30 days , c and a in 45 days . if a , b and c work together , they will complete the work in ? | "a + b 1 day work = 1 / 15 b + c 1 day work = 1 / 30 c + a 1 day work = 1 / 45 adding we get 2 ( a + b + c ) = 1 / 15 + 1 / 30 + 1 / 45 = 11 / 90 a + b + c 1 day work = 11 / 180 a , b , c can finish the work in 180 / 11 days = 16 days approximately answer is b" | a = 1/(15)
b = 1/(30)
c = a + b
d = 1/(45)
e = c + d
f = e / 2
g = 1 / f
|
a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 900 , e ) $ 950 | c | subtract(multiply(multiply(5, const_2), const_100), subtract(multiply(subtract(add(multiply(multiply(5, const_2), const_100), add(multiply(multiply(5, const_2), 5), const_4)), multiply(multiply(5, const_2), const_100)), const_100), multiply(5, multiply(multiply(5, const_2), const_100)))) | a woman invested $ 1,000 , part at 5 % and the rest at 6 % . her total investment with interest at the end of the year was $ 1,053 . how much did she invest at 5 % ? | "et x be the portion invested at 5 % and let ( 1 - x ) be the rest which is invested at 6 % the question states that the return after 1 year is ( 1053 / 1000 ) - 1 = 0.053 = 5.3 % we want to find the dollar amount invested in x using our defined variables , put together the equation and solve for x ( the percentage of 1000 invested at 5 % ) 0.05 x + 0.06 ( 1 - x ) = 0.053 ( 0.05 ) x + 0.06 - ( 0.06 ) x = 0.053 - 0.01 x = - 0.007 x = - 0.007 / - 0.01 = 7 / 10 = 70 % so x = 70 % of the 1000 which is 700 answer : c" | a = 5 * 2
b = a * 100
c = 5 * 2
d = c * 100
e = 5 * 2
f = e * 5
g = f + 4
h = d + g
i = 5 * 2
j = i * 100
k = h - j
l = k * 100
m = 5 * 2
n = m * 100
o = 5 * n
p = l - o
q = b - p
|
a ) 120 , b ) 346 , c ) 356 , d ) 366 , e ) 396 | a | divide(factorial(6), factorial(subtract(6, 3))) | 6 ! / ( 6 - 3 ) ! = ? | 6 ! / 6 - 3 ) ! = 6 ! / 3 ! = 6 * 5 * 4 = 120 . hence , the correct answer is a . | a = math.factorial(6)
b = 6 - 3
c = math.factorial(b)
d = a / c
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | subtract(add(multiply(reminder(7, 500), 3), reminder(3, 500)), reminder(1, 500)) | x is a positive integer less than 500 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "when x divided by 7 and leave reminder 1 . x could be 1 , 8,15 , . . . . when divided by 3 and leave reminder 2 - x could be 2 , 5,8 , . . . . so x must be of the form 21 m + 8 where m = 0 , 1,2 , . . . . . first no will be 8 and last no will be 491 . so total no . less than 500 will be 24 . answer : d" | a = reminder * (
b = a + 3
c = b - reminder
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(add(3, add(const_2, const_2)), add(const_2, const_2)), add(3, add(const_2, const_2))) | the expression x # y denotes the product of the consecutive multiples of 3 between x and y , inclusive . what is the sum of the exponents in the prime factorization of 21 # 27 ? | "first , let ' s translate the expression 21 # 27 , using the definition given : 21 # 27 = 21 × 24 × 27 we need the prime factorization of this product . let ' s factor out 3 from each multiple . 21 # 27 = 3 ^ 3 ( 7 × 8 × 9 ) now let ' s replace each consecutive integer with its prime factorization : 21 # 27 = 3 ^ 3 ( 7 × 2 ^ 3 × 3 ^ 2 ) let ' s group the prime bases : 21 # 27 = 2 ^ 3 × 3 ^ 5 × 7 the sum of the exponents is 3 + 5 + 1 = 9 the answer is d ." | a = 2 + 2
b = 3 + a
c = 2 + 2
d = b + c
e = 2 + 2
f = 3 + e
g = d + f
|
a ) 27 , b ) 29 , c ) 11 , d ) data inadequate , e ) 28 | d | divide(add(16, 15), const_2) | the average age of the boys in a class is 16 years and that of the girls is 15 years . the average age for the whole class is | clearly , to find the average , we ought to know the number of boys , girls or students in the class , neither of which has been given . so , the data provided is inadequate answer : d | a = 16 + 15
b = a / 2
|
a ) $ 216 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | 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 $ 3 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 ( $ 3 ) or $ 216 . the best answer is therefore a . answer : a ." | a = 2 / 3
b = 36 * a
c = 1 / 3
d = b / c
e = 2 * d
|
a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | d | divide(subtract(add(26, add(26, 9)), multiply(9, 9)), const_2) | the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 9 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x â € “ ( 26 + 38 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer d" | a = 26 + 9
b = 26 + a
c = 9 * 9
d = b - c
e = d / 2
|
a ) 12 , b ) 8 , c ) 15 , d ) 16 , e ) 23 | a | power(divide(10, 5), const_4) | if ( 2 ^ 12 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ? | "given 2 ^ 12 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 12 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans a on comparing the power of 2 = > m = 12" | a = 10 / 5
b = a ** 4
|
a ) 794 m , b ) 704 m , c ) 454 m , d ) 186 m , e ) 176 m | e | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 125), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 125 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 125 resolutions . = 125 * 2 * 22 / 7 * 22.4 = 17600 cm = 176 m answer : e" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 125
i = h / 100
|
a ) 80 , b ) 86 , c ) 92 , d ) 98 , e ) 104 | c | divide(subtract(multiply(160, divide(add(const_100, 50), const_100)), 10), add(divide(add(const_100, 50), const_100), const_1)) | if leo gains 10 pounds , he will weigh 50 % more than his sister kendra . currently their combined weight is 160 pounds . what is leo ' s current weight ? | "l + k = 160 and so k = 160 - l l + 10 = 1.5 k = 1.5 ( 160 - l ) 2.5 l = 230 l = 92 the answer is c ." | a = 100 + 50
b = a / 100
c = 160 * b
d = c - 10
e = 100 + 50
f = e / 100
g = f + 1
h = d / g
|
a ) 270 , b ) 380 , c ) 260 , d ) 360 , e ) 240 | e | multiply(divide(160, divide(40, const_100)), divide(60, const_100)) | if 40 % of a certain number is 160 , then what is 60 % of that number ? | "explanation : 40 % = 40 * 4 = 160 60 % = 60 * 4 = 240 answer : option e" | a = 40 / 100
b = 160 / a
c = 60 / 100
d = b * c
|
a ) 95 , b ) 90 , c ) 88 , d ) 80 , e ) 75 | c | divide(22, divide(subtract(100, 75), 100)) | if 22 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | "let x be the number of students in the school . 0.22 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.22 x / 0.25 = 0.88 x students went to the camping which is 88 % . the answer is c" | a = 100 - 75
b = a / 100
c = 22 / b
|
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