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 ) 31 % , b ) 45 % , c ) 52 % , d ) 61 % , e ) 25 % | a | multiply(divide(subtract(divide(const_100, divide(subtract(const_100, 20), const_100)), divide(const_100, divide(add(const_100, 30), const_100))), divide(const_100, divide(subtract(const_100, 20), const_100))), const_100) | certain stocks in january were 20 % less than they were in february and 30 % greater than they were in march . what was the percentage decrease in the stocks from february to march ? | "let , stock in february = 100 then , stock in january = 100 - ( 20 / 100 ) * 100 = 80 january = 30 % greater than march = 1.3 * stock in march i . e . stock in march = 90 / 1.3 = 69 approximately % decrease from february to march = ( 69 - 100 ) * 100 / 100 = 31 % approximately answer : option a" | a = 100 - 20
b = a / 100
c = 100 / b
d = 100 + 30
e = d / 100
f = 100 / e
g = c - f
h = 100 - 20
i = h / 100
j = 100 / i
k = g / j
l = k * 100
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a ) - 5 , b ) - 6 , c ) - 7 , d ) - 2 , e ) - 10 | d | add(divide(35, const_10), divide(35, divide(35, const_10))) | if a ( a + 2 ) = 35 and b ( b + 2 ) = 35 , where a β b , then a + b = | "i . e . if a = 5 then b = - 7 or if a = - 7 then b = 5 but in each case a + b = 5 - 7 = - 2 answer : d" | a = 35 / 10
b = 35 / 10
c = 35 / b
d = a + c
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a ) 63 , b ) 20 , c ) 42 , d ) 65 , e ) 84 | b | divide(80, multiply(divide(divide(20, 10), 5), 20)) | if 5 machines can produce 20 units in 10 hours , how long would it take 10 to produce 80 units ? | "5 machines would produce 80 units in 40 hours . increasing the amount of machines by 2 would mean dividing 40 hours by 2 . 40 / 2 = 20 answer : b" | a = 20 / 10
b = a / 5
c = b * 20
d = 80 / c
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a ) 7.6 % , b ) 8.3 % , c ) 9.6 % , d ) 10.3 % , e ) 11.5 % | d | multiply(divide(divide(26, const_2), add(26, const_100)), const_100) | jar a has 26 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ? | "an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 126 marbles . now , for both jars to have equal marbles we should move 13 marbles from a to b , which is 13 / 126 = ~ 10.3 % of a . answer : d ." | a = 26 / 2
b = 26 + 100
c = a / b
d = c * 100
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a ) 62 kg , b ) 60 kg , c ) 70 kg , d ) 72 kg , e ) none of these | d | subtract(92, multiply(5, 4)) | the average weight of 5 students decreases by 4 kg when one of them weighing 92 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 4 kg = > ( 92 - x ) / 5 = 4 = > x = 72 answer : d" | a = 5 * 4
b = 92 - a
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a ) 80 , b ) 110 , c ) 160 , d ) 500 , e ) 400 | d | divide(25, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 25 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 25 grades are d ' s . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a β s + # b β s + # c β s + # d β s = total number of students in the class 1 / 5 ( t ) + ΒΌ ( t ) + Β½ ( t ) + 25 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 500 = 20 t 19 t + 500 = 20 t 500 = t there are a total of 500 students in the math class . answer is d ." | a = 1 / 5
b = 1 / 4
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 25 / f
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a ) 2 / 3 , b ) 3 / 4 , c ) 1 / 5 , d ) 2 / 5 , e ) 3 / 5 | d | divide(square_perimeter(4), 10) | a 10 - meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | "the longer wire will form a square with an area more than 4 if the wire is cut at a point within two meters of either end . the probability of this is 4 / 10 = 2 / 5 . the answer is d ." | a = square_perimeter / (
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a ) 12.50 % , b ) 13.50 % , c ) 14 % , d ) 14.50 % , e ) none | e | multiply(subtract(subtract(add(const_1, divide(55, const_100)), multiply(add(const_1, divide(55, const_100)), divide(25, const_100))), const_1), const_100) | an uneducated retailer marks all his goods at 55 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 155 . s . p . = 75 % of rs . 155 = rs . 116.25 . β΄ gain % = 16.25 % . answer e" | a = 55 / 100
b = 1 + a
c = 55 / 100
d = 1 + c
e = 25 / 100
f = d * e
g = b - f
h = g - 1
i = h * 100
|
a ) 8 , b ) 9 , c ) 22 , d ) 17 , e ) 18 | c | divide(add(add(add(add(4, const_4), add(4, const_4)), add(const_4, const_4)), 70), 5) | the sum of ages of 5 children born 4 years different each is 70 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 4 ) , ( x + 8 ) , ( x + 12 ) and ( x + 16 ) years . then , x + ( x + 4 ) + ( x + 8 ) + ( x + 12 ) + ( x + 16 ) = 70 5 x = 30 x = 6 x + 16 = 6 + 16 = 22 answer : c" | a = 4 + 4
b = 4 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 70
g = f / 5
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a ) 8.25 , b ) 3.5 , c ) 9.0 , d ) 3.15 , e ) 2.0 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1400) | the difference between simple and compound interest on rs . 1400 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1400 * 10 * 1 ) / 100 = rs . 140 c . i . = [ 1400 * ( 1 + 5 / 100 ) 2 - 1400 ] = rs . 143.5 difference = ( 143.5 - 140 ) = rs . 3.50 answer : b" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = 10 / 100
f = e + 1
g = d - f
h = g * 1400
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a ) 8 , b ) 5 , c ) 11 , d ) 3 , e ) 4 | e | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(2, 1)) | if a * b * c = ( β ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 2 . | "6 * 15 * 2 = ( β ( 6 + 2 ) ( 15 + 3 ) ) / ( 2 + 1 ) = ( β 8 * 18 ) / 3 = ( β 144 ) / 3 = 12 / 3 = 4 . answer is e" | a = 6 + 2
b = 15 + 3
c = a * b
d = math.sqrt(c)
e = 2 + 1
f = d / e
|
a ) 150 , b ) 882 , c ) 277 , d ) 210 , e ) 281 | d | multiply(420, divide(420, add(add(5, 12), 13))) | the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 420 m , its area is ? | "5 x + 12 x + 13 x = 420 = > x = 14 a = 70 , b = 168 , c = 182 s = ( 70 + 168 + 182 ) / 2 = 210 answer : d" | a = 5 + 12
b = a + 13
c = 420 / b
d = 420 * c
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a ) 17 sec , b ) 16 sec , c ) 18 sec , d ) 10 sec , e ) 12 sec | d | multiply(divide(500, multiply(180, const_1000)), const_3600) | a train 500 m long , running with a speed of 180 km / hr will pass a tree in ? | speed = 180 * 5 / 18 = 50 m / sec time taken = 500 * 1 / 50 = 10 sec answer : d | a = 180 * 1000
b = 500 / a
c = b * 3600
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a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | c | divide(multiply(add(33, divide(1, 3)), 330), const_100) | 33 1 / 3 % of 330 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 330 = 110 c )" | a = 1 / 3
b = 33 + a
c = b * 330
d = c / 100
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a ) 1000 , b ) 800 , c ) 600 , d ) 700 , e ) 500 | b | divide(multiply(divide(80, const_100), multiply(multiply(40, 150), 10)), 60) | the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 40 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | "volume of pool = 40 * 150 * 10 cu . ft , 80 % full = 40 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 40 * 150 * 10 * 0.8 / 60 min = 800 min b" | a = 80 / 100
b = 40 * 150
c = b * 10
d = a * c
e = d / 60
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a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | a | subtract(multiply(add(22, 7), const_2), 19) | you collect pens . suppose you start out with 7 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 7 pens . mike gives you 22 pens : 7 + 22 = 29 pens . cindy doubles the number of pens you have : 29 Γ 2 = 58 pens . sharon takes 19 pens from you : 58 - 19 = 39 pens . so you have 39 at the end . correct answer : a" | a = 22 + 7
b = a * 2
c = b - 19
|
a ) 8.0 , b ) 3.25 , c ) 9.15 , d ) 3.13 , e ) 2.0 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1300) | the difference between simple and compound interest on rs . 1300 for one year at 10 % per annum reckoned half - yearly is ? | s . i . = ( 1300 * 10 * 1 ) / 100 = rs . 130 c . i . = [ 1300 * ( 1 + 5 / 100 ) 2 - 1300 ] = rs . 133.25 difference = ( 133.25 - 130 ) = rs . 3.25 answer : b | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = 10 / 100
f = e + 1
g = d - f
h = g * 1300
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a ) 10 , b ) 8 , c ) 14 , d ) 16 , e ) 20 | a | subtract(subtract(24, 7), const_2) | the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 90 inclusive is the digital sum a multiple of 7 ? | "is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 10 ways . . a" | a = 24 - 7
b = a - 2
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a ) 2 : 24 , b ) 5 : 24 , c ) 7 : 24 , d ) 1 : 24 , e ) 3 : 24 | d | divide(3000, 72000) | ravi and kavi start a business by investing Γ’ β ΒΉ 3000 and Γ’ β ΒΉ 72000 , respectively . find the ratio of their profits at the end of year . | ratio of profit = ratio of investments = 3000 : 72000 = 1 : 24 answer : d | a = 3000 / 72000
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a ) 40 , b ) 55 , c ) 68 , d ) 80 , e ) 98 | b | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(30, const_100), 120)), subtract(const_1, divide(52, const_100))) | in august , a cricket team that played 120 matches won 30 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 36 + x by solving we get x = 55 answer : b" | a = 52 / 100
b = a * 120
c = 30 / 100
d = c * 120
e = b - d
f = 52 / 100
g = 1 - f
h = e / g
|
a ) $ 5.20 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.40 | a | add(2.05, multiply(0.35, divide(3.6, divide(2, 5)))) | jim β s taxi service charges an initial fee of $ 2.05 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | let the fixed charge of jim β s taxi service = 2.05 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.05 + ( 3.6 / . 4 ) * . 35 = 2.05 + 9 * . 35 = 5.2 $ answer a | a = 2 / 5
b = 3 / 6
c = 0 * 35
d = 2 + 5
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a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | if a = 105 and a ^ 3 = 21 Γ 25 Γ 45 Γ w , what is the value of w ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 Γ 25 Γ 45 Γ w = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x w = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x w = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x w w = 7 ^ 2 = 49 answer d" | a = 105 ** 3
b = 21 * 25
c = b * 45
d = a / c
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a ) 42.85 , b ) 40 , c ) 35 , d ) 41 , e ) 39 | a | divide(add(multiply(50, const_10), divide(multiply(50, const_10), const_2)), add(divide(multiply(50, const_10), 40), divide(divide(multiply(50, const_10), const_2), 50))) | a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 40 miles per gallon , and from town b to town c , the car averaged 50 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ? | "step 1 ) took lcm of 40 and 50 . . came as 200 . step 2 ) 200 distance between b to c . . . do 200 / 50 hence 4 gallons used step 3 ) twice distance . . hence 200 * 2 = 400 . . . do as above . . 400 / 40 = 10 gallons used step 4 ) total gallons . . 4 + 10 = 14 gallons step ) total miles = 200 + 400 = 600 miles hence . . average of whole journey = 600 / 14 which comes to 42.85 answer : a" | a = 50 * 10
b = 50 * 10
c = b / 2
d = a + c
e = 50 * 10
f = e / 40
g = 50 * 10
h = g / 2
i = h / 50
j = f + i
k = d / j
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a ) 48 min , b ) 63 min , c ) 25 min , d ) 30 min , e ) 60 min | e | divide(30, 2) | a fill pipe can fill 2 / 5 of cistern in 30 minutes in how many minutes , it can fill 4 / 5 of the cistern ? | "2 / 5 of the cistern can fill in 30 min 4 / 5 of the cistern can fill in = 30 * 5 / 2 * 4 / 5 = 60 min answer is e" | a = 30 / 2
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a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 20 litres , e ) 8 litres | d | multiply(10, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 10 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 10 ) = 4 / 5 20 x = 4 ( 3 x + 10 ) 8 x = 40 x = 5 quantity of alcohol = ( 4 x 5 ) litres = 20 litres . answer is d ." | a = 10 * 1
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a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | a | add(divide(subtract(multiply(65, 4), multiply(45, 4)), subtract(75, 65)), 4) | a car averages 45 mph for the first 4 hours of a trip and averages 75 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ? | "let the time for which car averages 75 mph = t 65 * ( t + 4 ) = 45 * 4 + 75 t = > 10 t = 80 = > t = 8 total duration of the trip = 8 + 4 = 12 answer a" | a = 65 * 4
b = 45 * 4
c = a - b
d = 75 - 65
e = c / d
f = e + 4
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a ) 13 and 27 , b ) 16 and 24 , c ) 18 and 22 , d ) 11 and 29 , e ) none of these | b | subtract(40, divide(add(40, 8), const_3)) | the sum of the present age of henry and jill is 40 . what is their present ages if 8 years ago henry was twice the age of jill ? | "let the age of jill 8 years ago be x , age of henry be 2 x x + 8 + 2 x + 8 = 40 x = 8 present ages will be 16 and 24 answer : b" | a = 40 + 8
b = a / 3
c = 40 - b
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a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | b | divide(subtract(add(add(25, const_3), 25), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 25 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | let the average age of the whole team be x years . 11 x - ( 25 + 28 ) = 9 ( x - 1 ) = > 11 x - 9 x = 44 = > 2 x = 44 = > x = 22 . so , average age of the team is 22 years . b | a = 25 + 3
b = a + 25
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
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a ) 1.6 km , b ) 2 km , c ) 3.6 km , d ) 6.75 km , e ) none of these | d | multiply(divide(15, const_60), add(24, 3)) | the speed of a boat in still water is 24 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 15 minutes is | "explanation : speed downstreams = ( 24 + 3 ) kmph = 27 kmph . distance travelled = ( 27 x 15 / 60 ) km = 6.75 km option d" | a = 15 / const_60
b = 24 + 3
c = a * b
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a ) a ) 333 , b ) b ) 383 , c ) c ) 402 , d ) d ) 433 , e ) e ) 804 | e | divide(multiply(multiply(2000, add(const_1, divide(const_1, const_10))), add(const_1, divide(const_1, const_10))), 3) | louie takes out a 3 - month loan of $ 2000 . the lender charges him 10 % interest per month compounded monthly . the terms of the loan state that louie must repay the loan in 3 equal monthly payments . to the nearest dollar , how much does louie have to pay each month ? | here ' s the calculation for that case , assume monthly payment is x . after 1 st month : ( 2000 ) ( 1.1 ) - x = 2200 - x after 2 nd month : ( 2200 - x ) ( 1.1 ) - x = 2420 - 2.21 x after 3 rd month : ( 2420 - 2.21 x ) ( 1.1 ) - x = 2662 - 3.31 x now , the amount after the last payment in 3 rd month must bring the total to 0 . hence : 2662 - 3.31 x = 0 x = 2662 / 3.31 = 804 answer e | a = 1 / 10
b = 1 + a
c = 2000 * b
d = 1 / 10
e = 1 + d
f = c * e
g = f / 3
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a ) 6 , b ) 4 , c ) 7 , d ) 3 , e ) 5 | a | divide(power(50, 51), power(50, 11)) | 50 ^ 51 ^ 52 / 11 | "we know that 6 ^ 1 = 6 or 6 ^ 2 = 36 so for all power of 6 unit digit is 6 now que is 50 ^ 51 ^ 52 now if we divide 50 / 11 then rem is 6 means 6 ^ 51 = 6 at unit place same for 6 ^ 52 salso give 6 at unit place now finally 50 ^ 51 ^ 52 gives 6 at unit place now 6 / 11 = 6 answer : a" | a = 50 ** 51
b = 50 ** 11
c = a / b
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a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | b | divide(subtract(divide(45, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 45 % . what fraction of the original paint was replaced ? | "45 % is 20 % - points above 25 % and 5 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 1 : 4 . 1 / 5 of the original paint was replaced . the answer is b ." | a = 45 / 100
b = 50 / 100
c = a - b
d = 25 / 100
e = 50 / 100
f = d - e
g = c / f
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a ) 5 / 2 , b ) 7 / 2 , c ) 10 / 3 , d ) 15 / 4 , e ) 25 / 3 | a | divide(1, divide(add(divide(3, 5), multiply(divide(3, 5), divide(1, 3))), const_2)) | if a certain toy store ' s revenue in november was 3 / 5 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 3 d / 5 j = n / 3 = d / 5 the average of november and january is ( n + j ) / 2 = 4 d / 5 / 2 = 2 d / 5 d is 5 / 2 times the average of november and january . the answer is a ." | a = 3 / 5
b = 3 / 5
c = 1 / 3
d = b * c
e = a + d
f = e / 2
g = 1 / f
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a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 25 | d | divide(divide(divide(640, divide(add(10, 6), const_2)), 6), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 10 m wide at the top and 6 m wide at the bottom and the area of cross - section is 640 sq m , the depth of cannel is ? | "1 / 2 * d ( 10 + 6 ) = 640 d = 80 answer : d" | a = 10 + 6
b = a / 2
c = 640 / b
d = c / 6
e = d / 2
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a ) 5.29 min , b ) 3.84 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(528, const_1000), add(4.5, 3.75)), const_60) | the jogging track in a sports complex is 528 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 528 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 528 m apart , they take ( 100 / 825 * 528 / 1000 ) hrs = ( 528 / 8250 * 60 ) min = 3.84 min . answer : b" | a = 528 / 1000
b = 4 + 5
c = a / b
d = c * const_60
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | multiply(divide(multiply(const_10, 200), add(200, 230)), const_2) | a certain elevator has a safe weight limit of 2,500 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 200 pounds and the average weight of the others being 230 pounds ? | "lets assume there are 2 x people . half of them have average weight of 200 and other half has 230 . maximum weight is = 2500 so 200 * x + 230 * x = 2500 = > 430 x = 2500 = > x is approximately equal to 6 . so total people is 2 * 6 = 12 we are not taking 12 as answer because say 11 th person has minimum of 180 weight then 200 * 6 + 230 * 6 = 2580 ( which is more than 2500 ) answer e ." | a = 10 * 200
b = 200 + 230
c = a / b
d = c * 2
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a ) 3.5 : 6 , b ) 14 : 24 , c ) 7 : 12 , d ) 2.3 : 4 , e ) it can not be determined from the information given | c | divide(add(7, 3), add(12, 3)) | the ratio of two quantities is 7 to 12 . if each of the quantities is divided by 3 , what is the ratio of these 2 new quantities ? | "if both sides of a ratio are divided by the same number there is no change in the ratio . 5 : 6 means 7 x : 12 x . . . the ratio starts as 7 / ( 1 ) : 12 / ( 1 ) . so when you diveide by three it becomes 7 / ( 3 ) : 12 / ( 3 ) . . so if the xs are equal the ratio does not change . . answer : c" | a = 7 + 3
b = 12 + 3
c = a / b
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a ) 200 , b ) 120 , c ) 100 , d ) 65 , e ) 50 | b | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(72, const_100))), const_100) | if 45 % of z is 72 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 72 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 72 / 100 ) * ( 3 / 4 ) x i . e . z = ( 72 * 3 ) x / ( 45 * 4 ) i . e . z = ( 1.2 ) x = ( 120 / 100 ) x i . e . z is 120 % of x answer : option b" | a = 75 / 100
b = 45 / 100
c = 72 / 100
d = b / c
e = a / d
f = e * 100
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a ) 15 % , b ) 30 % , c ) 25 % , d ) 20 % , e ) 40 % | d | divide(divide(multiply(40, 70), multiply(80, 60)), add(divide(multiply(40, 70), multiply(80, 60)), 8)) | a certain car can travel 40 minutes on a gallon of gasoline at 70 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | let , tank capacity = t gallon used fuel = ( t - 8 ) gallons distance travelled ( @ 70 miles / hr ) = 80 miles distance travelled in 1 gallon = distance travelled in 40 mins ( @ 60 miles / hr ) = ( 70 / 70 ) * 40 = 40 miles fuel used to travel 80 miles = ( 80 / 40 ) = 2 gallon i . e . used fuel = ( t - 8 ) = 2 gallon i . e . t = 10 gallons i . e . used fuel = ( 2 / 10 ) * 100 = 20 % d | a = 40 * 70
b = 80 * 60
c = a / b
d = 40 * 70
e = 80 * 60
f = d / e
g = f + 8
h = c / g
|
a ) 600 , b ) 620 , c ) 500 , d ) 520 , e ) 840 | e | multiply(420, const_2) | on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 420 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 420 ) = 2 + 2 ( extra ) = > 2 x - 840 = x = > x = 840 . answer : e" | a = 420 * 2
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a ) 354354 , b ) 545454 , c ) 465785 , d ) 456573 , e ) 1300650 | e | multiply(1200000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1200000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city at the end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1300650 e" | a = 15 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1200000 * h
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a ) 4.85 , b ) 7.85 , c ) 6.85 , d ) 5.85 , e ) 6.15 | c | divide(add(111, 165), multiply(add(80, 65), const_0_2778)) | two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | t = ( 111 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 6.85 answer : c | a = 111 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
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a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 5), add(divide(33.33, const_100), const_1)), multiply(divide(const_1, 2), divide(50, const_100)))) | working alone , mary can pave a driveway in 5 hours and hillary can pave the same driveway in 2 hours . when they work together , mary thrives on teamwork so her rate increases by 33.33 % , but hillary becomes distracted and her rate decreases by 50 % . if they both work together , how many hours will it take to pave the driveway ? | "initial working rates : mary = 1 / 5 per hour hillary = 1 / 2 per hour rate when working together : mary = 1 / 5 + ( 1 / 3 * 1 / 5 ) = 1 / 4 per hour hillary = 1 / 2 - ( 1 / 2 * 1 / 2 ) = 1 / 4 per hour together they work 1 / 4 + 1 / 4 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the correct answer is a ." | a = 1 / 5
b = 33 / 33
c = b + 1
d = a * c
e = 1 / 2
f = 50 / 100
g = e * f
h = d + g
i = 1/(h)
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a ) 16 , b ) 8 β 2 , c ) 8 , d ) β 2 , e ) ( β 2 ) / 3 | d | sqrt(divide(multiply(2, 2), const_2)) | the two lines y = x and x = - 2 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 2 , - 2 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 2 Γ 2 ) = 2 ; so z = 2 . the next step requires us to find the length of a side of a cube that has a face area equal to 2 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 2 . since the area of a square is equal to s Β² , where s is the length of one of its side , we can write and solve the equation s Β² = 2 . clearly s = β 2 , oranswer choice ( d ) ." | a = 2 * 2
b = a / 2
c = math.sqrt(b)
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a ) 5 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | a | subtract(15, divide(subtract(15, divide(5, const_2)), const_2)) | sum of two numbers is 15 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 15 2 x β 3 y = 5 x = 10 y = 5 a )" | a = 5 / 2
b = 15 - a
c = b / 2
d = 15 - c
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a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | c | divide(subtract(divide(30, const_100), divide(50, const_100)), subtract(divide(20, const_100), divide(50, const_100))) | some of 50 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | 30 % is 10 % - points above 20 % and 20 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 2 : 1 . 2 / 3 of the original paint was replaced . the answer is c . | a = 30 / 100
b = 50 / 100
c = a - b
d = 20 / 100
e = 50 / 100
f = d - e
g = c / f
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a ) 72 , b ) 74 , c ) 76 , d ) 78 , e ) 80 | a | divide(divide(1800, add(multiply(10, const_0_2778), multiply(20, const_0_2778))), const_3) | two women started running simultaneously around a circular track of length 1800 m from the same point at speeds of 10 km / hr and 20 km / hr . when will they meet for the first time any where on the track if they are moving in opposite directions ? | time taken to meet for the first time anywhere on the track = length of the track / relative speed = 1800 / ( 30 + 60 ) 5 / 18 = 1800 * 18 / 90 * 5 = 72 seconds . answer : a | a = 10 * const_0_2778
b = 20 * const_0_2778
c = a + b
d = 1800 / c
e = d / 3
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a ) 31 , b ) 46 , c ) 88 , d ) 33.33 , e ) 12 | d | divide(add(multiply(40, 20), multiply(20, 10)), add(20, 10)) | the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 20 runs . find his average in all the 30 matches ? | "total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 200 . total score of the batsman in the 30 matches = 1000 . average score of the batsman = 1000 / 30 = 33.33 answer : d" | a = 40 * 20
b = 20 * 10
c = a + b
d = 20 + 10
e = c / d
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a ) 230 m , b ) 270 m , c ) 643 m , d ) 280 m , e ) 270 m | d | multiply(subtract(26, divide(240, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | a goods train runs at the speed of 72 kmph and crosses a 240 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 240 / 26 = 20 x + 240 = 520 x = 280 . answer : d" | a = const_0_2778 * 72
b = 240 / a
c = 26 - b
d = const_0_2778 * 72
e = c * d
|
a ) - 4 , b ) - 70 , c ) 0 , d ) 2 , e ) 4 | b | multiply(negate(multiply(divide(70, 2), 2)), 2) | if 9 a - b = 10 b + 70 = - 12 b - 2 a , what is the value of 2 a + 22 b ? | "this implies 9 a - b = 10 b + 70 , 9 a - b = - 12 b - 2 a , 10 b + 70 = - 12 b - 2 a manipulating the second equation gives us 10 b + 70 = - 12 b - 2 a = = > 2 a + 22 b = - 70 answer is b" | a = 70 / 2
b = a * 2
c = negate * (
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a ) 50 , b ) 80 , c ) 88 , d ) 76 , e ) 12 | b | divide(add(300, 500), multiply(subtract(72, 36), const_0_2778)) | how much time will a train of length 300 m moving at a speed of 72 kmph take to cross another train of length 500 m , moving at 36 kmph in the same direction ? | "the distance to be covered = sum of their lengths = 300 + 500 = 800 m . relative speed = 72 - 36 = 36 kmph = 36 * 5 / 18 = 10 mps . time required = d / s = 800 / 10 = 80 sec . answer : b" | a = 300 + 500
b = 72 - 36
c = b * const_0_2778
d = a / c
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a ) 3.40 % , b ) 4.40 % , c ) 5.40 % , d ) 6.40 % , e ) 7.40 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(4, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 8,000 gallons of a solution that is 4 percent sodium chloride by volume . if 3,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "we start with 8,000 gallons of a solution that is 4 % sodium chloride by volume . this means that there are 0.04 x 8,000 = 320 gallons of sodium chloride . when 3,000 gallons of water evaporate we are left with 5,000 gallons of solution . from here we can determine what percent of the 5,000 gallon solution is sodium chloride . ( sodium chloride / total solution ) x 100 = ? ( 320 / 5,000 ) x 100 = ? 0.064 x 100 = ? = 6.4 % answer is d ." | a = 100 * 100
b = 4 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
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a ) $ 1080.90 , b ) $ 1000.90 , c ) $ 1070.90 , d ) $ 1050.90 , e ) $ 1090.90 | e | multiply(divide(2000, add(add(inverse(2), inverse(4)), inverse(6))), inverse(2)) | x , y and z , each working alone can complete a job in 2 , 4 and 6 days respectively . if all three of them work together to complete a job and earn $ 2000 , what will be z ' s share of the earnings ? | the dollars earned will be in the same ratio as amount of work done 1 day work of z is 1 / 6 ( or 2 / 12 ) 1 day work of the combined workforce is ( 1 / 2 + 1 / 4 + 1 / 6 ) = 11 / 12 z ' s contribution is 2 / 9 of the combined effort translating effort to $ = 6 / 11 * 2000 = $ 1090.90 hence : e | a = 1/(2)
b = 1/(4)
c = a + b
d = 1/(6)
e = c + d
f = 2000 / e
g = 1/(2)
h = f * g
|
a ) 504 , b ) 536 , c ) 544 , d ) 548 , e ) 568 | d | multiply(54, const_10) | the least number , which when divided by 12 , 15 , 20 and 54 leaves in each case a remainder of 8 is : | "required number = ( l . c . m . of 12 , 15 , 20 , 54 ) + 8 = 540 + 8 = 548 . answer : option d" | a = 54 * 10
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a ) 15 seconds , b ) 18 seconds , c ) 25 seconds , d ) 30 seconds , e ) 45 seconds | a | divide(30, subtract(5, 3)) | nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 30 meter head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 30 for cristina = c = 5 * t @ t = 15 , n = 75 c = 75 right answer ans : a" | a = 5 - 3
b = 30 / a
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a ) 6400 , b ) 6410 , c ) 6440 , d ) 6500 , e ) 7040 | e | divide(multiply(multiply(multiply(8, const_100), multiply(6.6, const_100)), 22.5), multiply(multiply(25, 11.25), 6)) | how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6.6 m x 22.5 cm ? | number of bricks = volume of wall / volume of bricks = 800 x 660 x 22.5 / 25 x 11.25 x 6 = = 7040 answer : e | a = 8 * 100
b = 6 * 6
c = a * b
d = c * 22
e = 25 * 11
f = e * 6
g = d / f
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a ) 20 , b ) 30 , c ) 10 , d ) 40 , e ) 15 | b | divide(add(add(add(add(add(add(add(30, 5), 2), 2), 2), const_3), const_4), const_12), 2) | the total number of students in grades 1 and 2 is 30 more than the total number of students in grades 2 and 5 . how much lesser is the number of students in grade 5 as compared to grade 1 ? | ( grade 1 + grade 2 ) - ( grade 2 + grade 5 ) = 30 grade 1 - grade 5 = 30 answer : b | a = 30 + 5
b = a + 2
c = b + 2
d = c + 2
e = d + 3
f = e + 4
g = f + 12
h = g / 2
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a ) 14 , b ) 16 , c ) 18 , d ) 10 , e ) 21 | e | add(subtract(multiply(const_4, 5), multiply(multiply(const_4, 5), 0.2)), 5) | a football player scores 5 goals in his fifth match thus increasing his average goals score by 0.2 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the answer is not correct , then you would have a gauge for whether you should gohigherorlowerwith your next test . here , i ' ll start with answer e = 21 goals if . . . . total goals = 21 goals 5 th game = 5 goals 1 st 4 games = 16 goals avg . for 1 st 4 games = 16 / 4 = 4 goal / game avg . for all 5 games = 21 / 5 = 4.2 goals / game this is an exact match for what we ' re told in the prompt , so answer e must be the answer ." | a = 4 * 5
b = 4 * 5
c = b * 0
d = a - c
e = d + 5
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a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | d | divide(subtract(40, 22), subtract(40, 19)) | a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 22 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 - 0.21 x ) / 10 = 0.26 - - - > x = 20 / 3 ml . per the question , you need to find the x ml removed as a ratio of the initial volume - - - > ( 20 / 3 ) / 10 = 3 / 5 . hence , d is the correct answer ." | a = 40 - 22
b = 40 - 19
c = a / b
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a ) 148 kmph , b ) 152 kmph , c ) 106 kmph , d ) 103 kmph , e ) 165 kmph | d | divide(620, multiply(divide(3, 2), 4)) | a jeep takes 4 hours to cover a distance of 620 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 4 distance = 620 3 / 2 of 4 hours = 4 * 3 / 2 = 6 hours required speed = 620 / 6 = 103 kmph d )" | a = 3 / 2
b = a * 4
c = 620 / b
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a ) 36.7 , b ) 36.1 , c ) 36.5 , d ) 36.9 , e ) 36.3 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 60 was wrongly taken as 23 . the corrected new mean is ? | "correct sum = ( 36 * 50 + 60 - 23 ) = 1837 . correct mean = 1837 / 50 = 36.7 answer : a" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(add(sqrt(add(multiply(multiply(105, const_2), const_4), const_1)), const_1), const_2) | in a party every person shakes hands with every other person . if there were a total of 105 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n Γ’ Λ Β΄ total handshakes = nc 2 = 105 n ( n - 1 ) / 2 = 105 Γ’ Λ Β΄ n = 15 answer : a" | a = 105 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
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a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 3 , d ) 3 : 2 , e ) 2 : 1 | a | divide(subtract(multiply(add(const_1, divide(1.2, const_100)), const_1000), multiply(subtract(const_1, divide(4, const_100)), const_1000)), subtract(multiply(add(const_1, divide(12, const_100)), const_1000), multiply(add(const_1, divide(1.2, const_100)), const_1000))) | at a certain organisation , the number of male members went up by 12 % in the year 2001 from year 2000 , and the number of females members went down by 4 % in the same time period . if the total membership at the organisation went up by 1.2 % from the year 2000 to 2001 , what was the ratio of male members to female members in the year 2000 ? | "men increase by 12 % = = > 1.12 m = males in 2001 women decrease by 4 % = = > 0.96 f = women in 2001 total employees increase by 1.2 % = = > 1.012 * ( m + f ) = total number of employees in 2001 obviously ( males in 2001 ) + ( females in 2001 ) = total number of employees in 2001 1.12 m + 0.96 f = 1.012 * ( m + f ) 1.12 m + 0.96 f = 1.012 m + 1.012 f 1.12 m - 1.012 m = 1.012 f - 0.96 f 0.108 m = 0.052 f m / f = ( 0.052 ) / ( 0.108 ) = 52 / 108 = 1 / 2 answer = ( a )" | a = 1 / 2
b = 1 + a
c = b * 1000
d = 4 / 100
e = 1 - d
f = e * 1000
g = c - f
h = 12 / 100
i = 1 + h
j = i * 1000
k = 1 / 2
l = 1 + k
m = l * 1000
n = j - m
o = g / n
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(124, subtract(124, 124)) | what least value should be replaced by * in 842 * 124 so the number become divisible by 9 | "explanation : trick : number is divisible by 9 , if sum of all digits is divisible by 9 , so ( 8 + 4 + 2 + * + 1 + 2 + 4 ) = 21 + * should be divisible by 9 , 21 + 6 will be divisible by 9 , so that least number is 6 . answer : option d" | a = 124 - 124
b = 124 - a
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a ) 78 kg . , b ) 70 kg . , c ) 75 kg . , d ) 70 kg . , e ) 72 kg . | a | divide(add(multiply(7, 76), add(110, 60)), add(7, const_2)) | there are 7 players in a bowling team with an average weight of 76 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 76 * 7 + 110 + 60 ) / 9 = 78 kgs a is the answer" | a = 7 * 76
b = 110 + 60
c = a + b
d = 7 + 2
e = c / d
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a ) 14 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | b | divide(multiply(subtract(26, 2), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 2 years rahul age will be 26 years . what is deepak present age . | "explanation : present age is 4 x and 3 x , = > 4 x + 2 = 26 = > x = 6 so deepak age is = 3 ( 6 ) = 18 option b" | a = 26 - 2
b = a * 3
c = b / 4
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a ) 130 , b ) 135 , c ) 155 , d ) 195 , e ) 235 | c | add(add(add(add(25, const_2), add(add(25, const_2), const_2)), add(add(add(25, const_2), const_2), const_2)), 35) | the sum of all consecutive odd integers from β 25 to 35 , inclusive , is | "the sum of the odd numbers from - 25 to + 25 is 0 . let ' s add the remaining numbers . 27 + 29 + 31 + 33 + 35 = 5 ( 31 ) = 155 the answer is c ." | a = 25 + 2
b = 25 + 2
c = b + 2
d = a + c
e = 25 + 2
f = e + 2
g = f + 2
h = d + g
i = h + 35
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a ) 150 , b ) 887 , c ) 167 , d ) 197 , e ) 230 | e | subtract(multiply(250, divide(15, divide(15, const_3))), multiply(130, divide(20, divide(15, const_3)))) | a train crosses a platform of 130 m in 15 sec , same train crosses another platform of length 250 m in 20 sec . then find the length of the train ? | "length of the train be β x β x + 130 / 15 = x + 250 / 20 20 x + 2600 = 15 x + 3750 5 x = 1150 x = 230 m answer : e" | a = 15 / 3
b = 15 / a
c = 250 * b
d = 15 / 3
e = 20 / d
f = 130 * e
g = c - f
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a ) 6 , b ) 2.5 , c ) 9 , d ) 11 , e ) 13 | b | subtract(divide(1300, 20), divide(100, 4)) | one computer can upload 100 megabytes worth of data in 4 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 20 seconds . how long would it take for the second computer , working on its own , to upload 100 megabytes of data ? | "since the first computer can upload 100 megabytes worth of data in 4 seconds then in 4 * 5 = 20 seconds it can upload 5 * 100 = 500 megabytes worth of data , hence the second computer in 20 seconds uploads 1300 - 500 = 800 megabytes worth of data . the second computer can upload 100 megabytes of data in 2.5 seconds . answer : b ." | a = 1300 / 20
b = 100 / 4
c = a - b
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | c | log(power(2, const_10)) | the β length of integer x β refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 Γ 2 Γ 3 Γ 5 . ) what is the greatest possible length of integer z if z < 500 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 8 = 256 < 500 , then the greatest possible length of integer z is 8 . the answer is c ." | a = 2 ** 10
b = math.log(a)
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a ) 1400 , b ) 1240 , c ) 1200 , d ) 1100 , e ) 1500 | b | divide(620, subtract(const_1, divide(50, const_100))) | after decreasing 50 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 50 / 100 ) = 620 cp = 12.4 * 100 = > cp = 1240 answer : b" | a = 50 / 100
b = 1 - a
c = 620 / b
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a ) rs . 880 , b ) rs . 890 , c ) rs . 895 , d ) rs . 2100 , e ) none | d | divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 7), const_100) | if the compound interest on a certain sum of money for 7 years at 10 % per annum be rs . 993 , what would be the simple interest ? | "let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 7 yrs = ( 3000 x 10 x 7 ) / 100 = 2100 / - answer : d ." | a = 3 * 0
b = a * 10
c = b * 10
d = c * 7
e = d / 100
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a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(4, 3), const_2), const_4), 4), add(multiply(multiply(multiply(4, 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 * 4 + 3 and * 4 + 4 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 4 would be divisible by all numbers from 1 to 4 . = > * 4 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 4 is divisible by 3 ) in fact adding anyprimenumber between 1 to 4 to * 4 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 4 + 3 and * 4 + 11 then the answer would be 1 . for * 4 + 3 and * 4 + 13 , it is 2 and so on . . . a" | a = 4 * 3
b = a * 2
c = b * 4
d = c + 4
e = 4 * 3
f = e * 2
g = f * 4
h = g + 3
i = d - h
j = i - 1
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a ) 20 . , b ) 45 . , c ) 50 . , d ) 30 . , e ) 25 . | c | add(multiply(divide(subtract(const_100, 5), const_100), 50), multiply(divide(5, const_100), 50)) | in the hillside summer camp there are 50 children . 85 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | given there are 50 students , 84 % of 50 = 42 boys and remaining 8 girls . now here 84 % are boys and 16 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 8 % . . if we add 50 to existing 45 then the count will be 92 and the girls number will be 8 as it . now boys are 92 % and girls are 8 % . ( out of 100 students = 92 boys + 8 girls ) . imo option c is correct . | a = 100 - 5
b = a / 100
c = b * 50
d = 5 / 100
e = d * 50
f = c + e
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a ) 32 , b ) 22 , c ) 35 , d ) 27 , e ) 28 | c | subtract(negate(11), multiply(subtract(5, 7), divide(subtract(5, 7), subtract(4, 5)))) | 4 , 5 , 7 , 11 , 19 , ( . . . ) | "4 4 Γ 2 - 3 = 5 5 Γ 2 - 3 = 7 7 Γ 2 - 3 = 11 11 Γ 2 - 3 = 19 19 Γ 2 - 3 = 35 answer is c" | a = negate - (
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a ) 1 , b ) 0 , c ) 2 , d ) - 1 , e ) can not be determined | c | subtract(subtract(subtract(add(add(9, 8), 2), 2), 5), 8) | a set s = { x , - 8 , - 5 , - 2 , 2 , 6 , 9 , y } with elements arranged in increasing order . if the median and the mean of the set are the same , what is the value of | x | - | y | ? | "median of the set = ( - 2 + 2 ) / 2 = 0 as per statement , mean of the set = 0 mean of the set | y | - | x | + 17 - 15 = 0 ( where x is negative n y is positive ) | y | - | x | = - 2 so the absolute difference between two numbers is 2 answer c" | a = 9 + 8
b = a + 2
c = b - 2
d = c - 5
e = d - 8
|
a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many acres q were used for corn ? | consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn q = 4 * 94 = 376 correct option - c | a = 1000 * 1
b = 10 * 3
c = b + 4
d = a + c
e = 5 + 2
f = e + 4
g = d / f
h = g * 4
|
a ) $ 8829 , b ) $ 2840 , c ) $ 6578 , d ) $ 7782 , e ) $ 8925 | e | divide(divide(multiply(4016.25, const_100), 9), 5) | a sum fetched a total simple interest of $ 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ? | "e 8925 principal = $ 100 x 4016.25 / 9 x 5 = $ 401625 / 45 = $ 8925 ." | a = 4016 * 25
b = a / 9
c = b / 5
|
a ) 8 hrs , b ) 10 hrs , c ) 4 hrs , d ) 15 hrs , e ) 6 hrs | c | multiply(divide(1, 2), 8) | a train running at 1 / 2 of its own speed reached a place in 8 hours . how much time could be saved if the train would have run at its own speed ? | time taken if run its own speed = 1 / 2 * 8 = 4 hrs time saved = 8 - 4 = 4 hrs answer : c | a = 1 / 2
b = a * 8
|
a ) 12.5 % , b ) 40 % , c ) 120 % , d ) 125 % , e ) none | c | multiply(divide(6, 5), const_100) | the ratio 6 : 5 expressed as a percent equals | "solution 6 : 5 = 6 / 5 = ( 6 / 5 x 100 ) % . = 120 % . answer c" | a = 6 / 5
b = a * 100
|
a ) 198 cm 2 , b ) 320 cm 2 , c ) 279 cm 2 , d ) 128 cm 2 , e ) 297 cm 2 | b | multiply(20, 16) | find the area of a parallelogram with base 20 cm and height 16 cm ? | "area of a parallelogram = base * height = 20 * 16 = 320 cm 2 answer : b" | a = 20 * 16
|
a ) 8 : 5 , b ) 4 : 9 , c ) 4 : 3 , d ) 4 : 1 , e ) 4 : 2 | a | multiply(divide(16, const_100), 10) | a part of certain sum of money is invested at 10 % per annum and the rest at 16 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | 16 : 10 = 8 : 5 answer : a | a = 16 / 100
b = a * 10
|
a ) 255 , b ) 260 , c ) 265 , d ) 270 , e ) 275 | e | multiply(12, 288) | the h . c . f of two numbers is 12 and their l . c . m is 6600 . if one of the numbers is 288 , then the other is ? | "other number = ( 12 * 6600 ) / 288 = 275 . answer : e" | a = 12 * 288
|
a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | d | subtract(multiply(multiply(multiply(624, 708), 913), 463), subtract(multiply(multiply(multiply(624, 708), 913), 463), add(const_4, const_4))) | the unit digit in the product ( 624 * 708 * 913 * 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 * 8 * 3 * 3 ) = 8 answer : d" | a = 624 * 708
b = a * 913
c = b * 463
d = 624 * 708
e = d * 913
f = e * 463
g = 4 + 4
h = f - g
i = c - h
|
a ) 515 . , b ) 678 . , c ) 618 . , d ) 644 . , e ) 666 . | b | add(226, 450) | in the faculty of reverse - engineering , 226 second year students study numeric methods , 450 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | "answer is b : 678 solution : total number of students studying both are 450 + 226 - 134 = 542 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 542 , so 100 % is approx . 678 ." | a = 226 + 450
|
['a ) 77.77 %', 'b ) 75 %', 'c ) 67.5 %', 'd ) 87.5 %', 'e ) none of these'] | b | multiply(divide(subtract(divide(64, power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), divide(64, power(64, divide(const_1, const_3)))), const_100) | the volume of the sphere q is ( 37 / 64 ) % less than the volume of sphere p and the volume of sphere r is ( 19 / 27 ) % less than that of sphere q . by what is the surface area of sphere r less than the surface area of sphere p ? | explanation : let the volume of sphere p be 64 parts . therefore volume of sphere q = > 64 β ( 37 / 64 ) % of 64 . = > 64 β 37 = 27 parts . the volume of r is : - = > 27 β ( 19 / 27 ) Γ 27 . = > 27 β 19 = 8 parts . volume ratio : = > p : q : r = 64 : 27 : 8 . radius ratio : = > p : q : r = 4 : 3 : 2 . the surface area will be 16 : 9 : 5 . surface area of r is less than the surface area of sphere p . = > 16 k β 4 k = 12 k . now , = > ( 12 k / 16 k ) Γ 100 . = > 75 % . hence , the surface area of sphere r is less than the surface area of sphere p by 75 % . answer : b | a = 1 / 3
b = 64 ** a
c = 64 / b
d = 1 / 3
e = 64 ** d
f = c - e
g = 1 / 3
h = 64 ** g
i = 64 / h
j = f / i
k = j * 100
|
['a ) 2.5', 'b ) 2', 'c ) 1.5', 'd ) 1', 'e ) 0.5'] | e | divide(3.14, multiply(const_2, const_pi)) | what is the radius of a circle that has a circumference of 3.14 meters ? | circumference of a circle = 2 Ο r . given , circumference = 3.14 meters . therefore , 2 Ο r = circumference of a circle or , 2 Ο r = 3.14 . or , 2 ? 3.14 r = 3.14 , [ putting the value of pi ( Ο ) = 3.14 ] . or , 6.28 r = 3.14 . or , r = 3.14 / 6.28 . or , r = 0.5 . answer : 0.5 meter . correct answer e | a = 2 * math.pi
b = 3 / 14
|
a ) 8 , b ) 15 , c ) 14 , d ) 12 , e ) 10 | a | subtract(add(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2)), subtract(multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2), divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)))) | the difference between a two - digit number and the number obtained by interchanging the digit is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | since the number is greater than the number obtained on reversing the digits , so the ten ' s digit is greater than the unit ' s digit . let the ten ' s and unit ' s digits be 2 x and x respectively . then , ( 10 * 2 x + x ) - ( 10 x + 2 x ) = 36 9 x = 36 x = 4 required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 . answer a | a = 2 * 10
b = a + 1
c = b - 10
d = c - 2
e = 36 / d
f = 2 * 10
g = f + 1
h = g - 10
i = h - 2
j = 36 / i
k = j * 2
l = e + k
m = 2 * 10
n = m + 1
o = n - 10
p = o - 2
q = 36 / p
r = q * 2
s = 2 * 10
t = s + 1
u = t - 10
v = u - 2
w = 36 / v
x = r - w
y = l - x
|
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | b | multiply(divide(3, 20), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "first ratio = 3 / 20 = 15 % second ratio = ( 3 + 12 ) / ( 9 + 108 ) = 15 / 117 = 12,8 % decrease = 2,2 % 2,2 % / 15 % = 15 % approximately . answer : b" | a = 3 / 20
b = a * 100
|
a ) 31 , 20 , b ) 30 , 19 , c ) 29 , 18 , d ) 28 , 17 , e ) none | d | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | the difference of two numbers is 11 and one fifth of their sum is 9 . the numbers are : | "x β y = 11 , x + y = 5 Γ 9 x β y = 11 , x + y = 45 , y = 17 , x = 28 answer : d" | a = 10 / 2
b = a * 9
c = 10 / 2
d = c * 9
e = 11 + d
f = e / 2
g = b - f
h = 10 / 2
i = g + h
|
a ) 13 / 20 , b ) 15 / 20 , c ) 14 / 20 , d ) 12 / 20 , e ) 11 / 20 | a | divide(subtract(multiply(add(8, 20), divide(75, const_100)), 8), 20) | an internet recently hired 8 new network , in addvertisement 20 network already employed . all new network cam from university a . in addition 75 % of computer addvertisement came from same university a . what fraction of original 20 network addvertisement came from same univerity a ? pls help to solve | new networks = 8 already employed networks = 20 total networks = 28 computer advt . from uni . a = 28 * ( 75 / 100 ) = 21 out 28 networks 21 came from uni . a among these 21 we know 8 are newly hired so 21 - 8 = 13 networks among previously employed networks came from uni . a answer is : 13 / 20 answer : a | a = 8 + 20
b = 75 / 100
c = a * b
d = c - 8
e = d / 20
|
a ) 31 β 5 , b ) 16 β 5 , c ) 20 β 9 , d ) 5 / 2 , e ) 5 β 16 | d | multiply(divide(9, 3), divide(5, 6)) | dividing by 3 β 9 and then multiplying by 5 β 6 is the same as dividing by what number ? | "say x / 3 / 9 * 5 / 6 = x * 9 / 3 * 5 / 6 = x * 5 / 2 d" | a = 9 / 3
b = 5 / 6
c = a * b
|
a ) 1 β 3 , b ) 4 β 3 , c ) 80 , d ) 120 , e ) 360 | e | divide(multiply(2, const_60), divide(1, 3)) | if it takes a machine 1 β 3 minute to produce one item , how many items will it produce in 2 hours ? | "1 item takes 1 / 3 min so it takes 120 min to produce x x / 3 = 120 the x = 360 answer : e" | a = 2 * const_60
b = 1 / 3
c = a / b
|
a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many s apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons s = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a = 10 + 10
b = a + 2
c = 1430 / b
d = c + 45
e = 1430 / d
f = e + 9
|
a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | c | multiply(7700, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 7700 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 7700 x ( 2 / 11 ) ) = rs . 1400 . answer is c" | a = 2 * 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 3
f = 2 / e
g = 7700 * f
|
a ) 39 , b ) 9 , c ) 15 , d ) 19 , e ) 25 | a | add(add(power(5, const_2.0), multiply(2, 2)), 4) | if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 5 ] ] ? | "these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 5 ] ] = 5 ^ 2 + 2 * 5 + 4 = 39 . option a" | a = 5 ** 2
b = 2 * 2
c = a + b
d = c + 4
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 2 | b | subtract(add(38, 4), 36) | sobha ' s father was 38 years of age when she was born while her mother was 36 years old when her brother 4 years younger to her was born . what is the difference between the ages of her parents ? | age of sobha ' s father when sobha was born = 38 age of sobha ' s mother when sobha was born = 36 β 4 = 32 required difference of age = 38 β 32 = 6 answer is b . | a = 38 + 4
b = a - 36
|
a ) 25 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | c | divide(subtract(multiply(74, 2), subtract(16, 8)), add(2, 3)) | if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 74 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 74 a = 28 the answer is c ." | a = 74 * 2
b = 16 - 8
c = a - b
d = 2 + 3
e = c / d
|
a ) 2 : 3 , b ) 2 : 5 , c ) 3 : 7 , d ) 4 : 9 , e ) 5 : 6 | c | multiply(subtract(divide(5, 7), divide(const_1, const_2)), const_2) | the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 7 . what is the ratio of the smaller number to the larger number ? | "for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 7 , thus the smaller number must have a ratio of 3 . the ratio of the smaller number to the larger number is 3 : 7 . the answer is c ." | a = 5 / 7
b = 1 / 2
c = a - b
d = c * 2
|
a ) 18 , b ) 22 , c ) 28 , d ) 48 , e ) 98 | d | divide(multiply(120, 60), const_100) | find number which is 60 % less than 120 . | "explanation : 60 % less is 40 % of the given number therefore , 40 % of 120 is 48 . answer : d" | a = 120 * 60
b = a / 100
|
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