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a ) a ) 3800 , b ) b ) 4200 , c ) c ) 4400 , d ) d ) 4500 , e ) e ) 4600 | c | floor(divide(3168, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 20), const_100)))) | 10 % people of a village in sri lanka died by bombardment , 20 % of the remainder left the village on account of fear . if now the population is reduced to 3168 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 80 / 100 ) = 3168 x = 4400 answer : c" | a = 100 - 10
b = a / 100
c = 100 - 20
d = c / 100
e = b * d
f = 3168 / e
g = math.floor(f)
|
a ) 190 , b ) 200 , c ) 210 , d ) 1770 , e ) 230 | d | multiply(subtract(60, const_1), divide(60, const_2)) | 60 men shake hands with each other . maximum no of handshakes without cyclic handshakes . | "1 st person will shake hand with 59 people 2 nd person will shake hand with 58 people 3 rd person will shake hand with 57 people . . . . . . total no . of handshakes = 59 + 58 + 57 + . . . + 3 + 2 + 1 = 19 * ( 19 + 1 ) / 2 = 1770 or , if there are n persons then no . of shakehands = nc 2 = 60 c 2 = 1770 answer : d" | a = 60 - 1
b = 60 / 2
c = a * b
|
a ) s . 2.04 , b ) s . 2.08 , c ) s . 1.63 , d ) s . 2.83 , e ) s . 2.42 | c | subtract(multiply(4000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(4000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | what is the difference between the c . i . on rs . 4000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 4000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 4000 * 26 / 25 * 51 / 50 = rs . 4243.2 c . i . when interest is compounded half - yearly = [ 4000 * ( 1 + 2 / 100 ) 2 ] = ( 4000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 4244.83 difference = ( 4244.83 - 4243.2 ) = rs . 1.63 . answer : c" | a = 2 / 100
b = 1 + a
c = 2 / 100
d = 1 + c
e = b * d
f = 2 / 100
g = 1 + f
h = e * g
i = 4000 * h
j = 2 / 100
k = 1 + j
l = 4 / 100
m = 1 + l
n = k * m
o = 4000 * n
p = i - o
|
a ) a ) 899.015 , b ) b ) 752.804 , c ) c ) 389.884 , d ) d ) 629.906 , e ) of these | c | subtract(multiply(divide(657.987, const_100), 56.84), multiply(divide(const_1, const_3), multiply(divide(657.987, const_100), 56.84))) | 657.987 - ? + 56.84 = 324.943 | "explanation : 389.884 answer : option c" | a = 657 / 987
b = a * 56
c = 1 / 3
d = 657 / 987
e = d * 56
f = c * e
g = b - f
|
a ) 0.0066 % , b ) 0.066 % , c ) 0.66 % , d ) 6.6 % , e ) 66 % | c | divide(divide(divide(53.42, const_3), const_3), subtract(const_10, const_1)) | approximately what percentage of the world β s forested area is represented by finland given that finland has 53.42 million hectares of forested land of the world β s 8.076 billion hectares of forested land . | since 1 billion is 1000 million soc changing all quantities to million finland forest area = 53.42 million hect . worlds ' s forest area = 8.076 billion hect . % = 53.42 * 100 / 8.076 * 1000 = 0.66 % answer c | a = 53 / 42
b = a / 3
c = 10 - 1
d = b / c
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | a | add(8, divide(subtract(65, 20), add(25, 20))) | two stations a and b are 65 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 ) = 65 45 x = 90 x = 2 . so , they meet at 9 a . m . answer : option a | a = 65 - 20
b = 25 + 20
c = a / b
d = 8 + c
|
a ) s . 800 , b ) s . 200 , c ) s . 600 , d ) s . 500 , e ) s . 900 | b | divide(800, const_3) | divide rs . 800 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 800 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 800 = > 200 answer : b" | a = 800 / 3
|
a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 90,000 , e ) none | d | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3) | x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 60,000 . the amount invested by y is | "solution suppose y invested rs . y then , 60000 / y = 2 / 3 Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = ( 60000 Γ£ β 3 / 2 ) . Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ y = 90000 . answer d" | a = 1 + 4
b = a * 1000
c = b * 2
d = c / 3
|
a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,00 , d ) $ 4,500 , e ) $ 15,000 | e | divide(add(divide(subtract(360, multiply(divide(6, const_100), const_1000)), subtract(divide(10, const_100), divide(6, const_100))), divide(subtract(360, multiply(divide(6, const_100), const_1000)), subtract(divide(10, const_100), divide(6, const_100)))), const_1000) | salesperson a ' s compensation for any week is $ 360 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 10 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | sometime , setting up an equation is an easy way to go with : 360 + 0.06 ( x - 1000 ) = 0.1 x x = 15,000 ans : e | a = 6 / 100
b = a * 1000
c = 360 - b
d = 10 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1000
j = 360 - i
k = 10 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1000
|
a ) 3 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | what is the smallest positive integer k such that the product of 61347 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 61347 = 13 ^ 2 * 11 ^ 2 * 3 , so if k = 3 then 61347 k = ( 13 * 11 * 3 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of 7 to even power as powers of other primes are already even ) . answer : a ." | a = 3 + 4
|
a ) 16 , b ) 12 , c ) 15 , d ) 20 , e ) 13 | e | multiply(13, const_1) | the total age of a and b is 13 years more than the total age of b and c . c is how many years younger than a . ? | "( a + b ) - ( b - c ) = 13 a - c = 13 answer is e" | a = 13 * 1
|
a ) 30 , b ) 80 , c ) 44 , d ) 25 , e ) 26 | c | multiply(subtract(67, 45), const_2) | a pupil ' s marks were wrongly entered as 67 instead of 45 . due to that the average marks for the class got increased by half . the number of pupils in the class is : | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 . x / 2 = ( 67 - 45 ) = > x / 2 = 22 = > x = 44 . answer : c" | a = 67 - 45
b = a * 2
|
a ) sec , b ) sec , c ) sec , d ) sec , e ) sec | d | divide(add(100, 170), multiply(60, const_0_2778)) | how long does a train 100 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ? | "d = 100 + 170 = 270 m s = 60 * 5 / 18 = 50 / 3 t = 270 * 3 / 50 = 16.2 sec answer : d" | a = 100 + 170
b = 60 * const_0_2778
c = a / b
|
a ) 1260 , b ) 2600 , c ) 3600 , d ) 4200 , e ) 5200 | b | multiply(divide(add(multiply(multiply(const_3, const_3), const_1000), const_100), 7), 2) | a marketing survey of anytown found that the ratio of trucks to sedans to motorcycles was 3 : 7 : 2 , respectively . given that there are 9,100 sedans in anytown , how many motorcycles are there ? | let the total number of trucks = 3 x total number of sedans = 7 x total number of motorcycles = 2 x total number of sedans = 9100 = > 7 x = 9100 = > x = 1300 total number of motorcycles = 2 x = 2 * 1300 = 2600 answer b | a = 3 * 3
b = a * 1000
c = b + 100
d = c / 7
e = d * 2
|
a ) 12.5 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 67.5 % | b | subtract(add(60, 55), subtract(const_100, 25)) | in a particular state , 60 % of the counties received some rain on monday , and 55 % of the counties received some rain on tuesday . no rain fell either day in 25 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ? | "60 + 55 + 25 = 140 % the number is 40 % above 100 % because 40 % of the counties were counted twice . the answer is b ." | a = 60 + 55
b = 100 - 25
c = a - b
|
a ) 1 / 9 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(2, 2), add(multiply(divide(2, 5), const_60), multiply(divide(2, 10), const_60))) | triathlete dan runs along a 2 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 5 miles per hour , what is his average rate for the entire trip in miles per minute ? | dan travels 4 miles round trip . running part : ( 2 / 10 = 1 / 5 * 60 = 12 minutes ) swimming part : ( 2 / 5 = 2 / 5 * 60 = 24 minutes ) 4 miles in ( 12 + 24 ) minutes 4 / 36 = 1 / 9 mile per minute answer : 1 / 9 mile per minute | a = 2 + 2
b = 2 / 5
c = b * const_60
d = 2 / 10
e = d * const_60
f = c + e
g = a / f
|
a ) 837 , b ) 947 , c ) 1027 , d ) 1155 , e ) 1231 | a | add(lcm(lcm(12, 15), lcm(35, 40)), 3) | what is the smallest number which when increased by 3 is divisible by 12 , 15 , 35 , and 40 ? | "factor each of the numbers 8 , 15 , 35 , and 40 into primes : 12 = 2 * 2 * 3 ; 15 = 3 * 5 ; 35 = 5 * 7 ; 40 = 2 * 2 * 2 * 5 the smallest number divisible by all of them is thus 2 * 2 * 2 * 3 * 5 * 7 = 840 837 + 3 = 840 a" | a = math.lcm(12, 15)
b = math.lcm(35, 40)
c = math.lcm(a, b)
d = c + 3
|
a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number r of races we need to conduct to get the top 3 fastest horses ? | "r = 7 is the correct answer . good solution buneul . b" | a = 25 / 5
b = a + 2
|
a ) 1 / 4 , b ) 1 / 2 , c ) 10 / 3 , d ) 2 , e ) 4 | c | divide(1, divide(add(divide(2, 5), multiply(divide(2, 5), divide(1, 2))), const_2)) | if a certain toy store ' s revenue in november was 2 / 5 of its revenue in december and its revenue in january was 1 / 2 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 = 2 d / 5 j = n / 2 = d / 5 the average of november and january is ( n + j ) / 2 = 3 d / 5 / 2 = 3 d / 10 d is 10 / 3 times the average of november and january . the answer is c ." | a = 2 / 5
b = 2 / 5
c = 1 / 2
d = b * c
e = a + d
f = e / 2
g = 1 / f
|
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | c | divide(subtract(multiply(30, 7), multiply(5, 7)), 7) | 7 people average age is 30 . youngest person age is 5 . find average of the people when youngest was born . | average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 35 = 161 so average age would be 175 / 7 = 25 answer : c | a = 30 * 7
b = 5 * 7
c = a - b
d = c / 7
|
a ) 8 , b ) 16 , c ) 31 , d ) 18 , e ) 34 | c | add(subtract(add(35, 30), subtract(35, 30)), const_1) | the average age of applicants for a new job is 35 , with a standard deviation of 30 . the hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age . assuming that all applicants ' ages are integers and that the endpoints of the range are included , what is the maximum number of different ages of the applicants ? | "minimum age = average - 1 standard deviation = 35 - 30 = 5 maximum age = average + 1 standard deviation = 35 + 30 = 35 maximum number of different ages of the applicants = 35 - 5 + 1 = 31 answer c" | a = 35 + 30
b = 35 - 30
c = a - b
d = c + 1
|
a ) $ 900 , b ) $ 300 , c ) $ 600 , d ) $ 1000 , e ) $ 800 | d | divide(400, subtract(const_1, divide(3, 5))) | linda spent 3 / 5 of her savings on furniture and the rest on a tv . if the tv cost her $ 400 , what were her original savings ? | "if linda spent 3 / 5 of her savings on furnitute , the rest 5 / 5 - 3 / 5 = 2 / 5 on a tv but the tv cost her $ 400 . so 2 / 5 of her savings is $ 400 . so her original savings are 5 / 2 times $ 400 = $ 2000 / 2 = $ 1000 correct answer d" | a = 3 / 5
b = 1 - a
c = 400 / b
|
a ) rs . 660 , b ) rs . 760 , c ) rs . 860 , d ) rs . 1040 , e ) none of these | d | multiply(1300, subtract(const_1, divide(20, const_100))) | a man buys an item at rs . 1300 and sells it at the loss of 20 percent . then what is the selling price of that item | explanation : here always remember , when ever x % loss , it means s . p . = ( 100 - x ) % of c . p when ever x % profit , it means s . p . = ( 100 + x ) % of c . p so here will be ( 100 - x ) % of c . p . = 80 % of 1300 = 80 / 100 * 1300 = 1040 option d | a = 20 / 100
b = 1 - a
c = 1300 * b
|
a ) 12 , b ) 15 , c ) 17 , d ) w = 18 , e ) 20 | d | multiply(multiply(3, const_2), 3) | two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer w between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | "since we are concerned with integers w between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 11,12 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11,13 ) , ( 7,17 ) , ( 5 , 19 ) 2 , 3 , 5 , 7 , 11 , 13 , 15,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 13 , 17 ) , ( 11 , 19 ) , ( 7 , 23 ) 2 , 3 , 5 , 7 , 11 , 13,17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11 , 23 ) , ( 5 , 29 ) , ( 3 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 18,19 , 23 , 29 , 31 , 37 - four pairs ( 17 , 19 ) , ( 13 , 23 ) , ( 7 , 29 ) , ( 5 , 31 ) 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 20,23 , 29 , 31 , 37 - definitely can not be more than 4 since there are only 4 primes more than 20 . so must be less than 4 pairs . ignore . answer ( d ) ." | a = 3 * 2
b = a * 3
|
a ) s . 440 , b ) s . 500 , c ) s . 540 , d ) s . 782 , e ) s . 840 | d | subtract(850, multiply(divide(subtract(1020, 850), 5), 2)) | a sum of money lent out at s . i . amounts to rs . 850 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 850 ) = rs . 170 . s . i . for 2 years = 170 / 5 * 2 = rs . 68 . principal = ( 850 - 68 ) = rs . 782 . answer : d" | a = 1020 - 850
b = a / 5
c = b * 2
d = 850 - c
|
a ) 743 , b ) 154 , c ) 852 , d ) 176 , e ) 785 | d | divide(subtract(15698, 14), 89) | on dividing 15698 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 15698 - 14 ) / 89 = 176 answer ( d )" | a = 15698 - 14
b = a / 89
|
a ) 23,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | b | floor(divide(divide(subtract(570, multiply(1,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 1,000)) | angelo and isabella are both salespersons . in any given week , angelo makes $ 570 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "let the weekly sales of both = x 570 + ( x β 1000 ) 8 / 100 = 10 / 100 x x = 24500 answer : b" | a = 8 / 100
b = 1 * 0
c = 570 - b
d = 10 / 100
e = 8 / 100
f = d - e
g = c / f
h = g / 1
i = math.floor(h)
|
a ) rs . 3600 , b ) rs . 3603 , c ) rs . 3639 , d ) rs . 3632 , e ) rs . 5000 | e | subtract(20000, multiply(divide(3, 4), 20000)) | income and expenditure of a person are in the ratio 4 : 3 . if the income of the person is rs . 20000 , then find his savings ? | "let the income and the expenditure of the person be rs . 4 x and rs . 3 x respectively . income , 4 x = 20000 = > x = 5000 savings = income - expenditure = 4 x - 3 x = x so , savings = rs . 5000 answer : e" | a = 3 / 4
b = a * 20000
c = 20000 - b
|
a ) 1.9832 , b ) 1.0025 , c ) 1.5693 , d ) 1.0266 , e ) none | a | multiply(divide(268, const_100), divide(74, const_100)) | given that 268 x 74 = 19832 , find the value of 2.68 x . 74 . | solution sum of decimals places = ( 2 + 2 ) = 4 . therefore , = 2.68 Γ . 74 = 1.9832 answer a | a = 268 / 100
b = 74 / 100
c = a * b
|
a ) 71 , b ) 65 , c ) 61 , d ) 45 , e ) 36 | a | subtract(add(add(multiply(60, 10), 1), 60), multiply(60, 10)) | the average age of 60 students in a class is 10 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ? | "total age of 50 students = 60 * 10 = 600 total age of 51 persons = 61 * 11 = 671 age of teacher = 671 - 600 = 71 years answer is a" | a = 60 * 10
b = a + 1
c = b + 60
d = 60 * 10
e = c - d
|
a ) 28 , b ) 10 , c ) 288 , d ) 277 , e ) 211 | b | multiply(multiply(divide(7, 5.00001), divide(10, 7.00001)), 5) | the ratio of investments of two partners p and q is 7 : 5.00001 and the ratio of their profits is 7.00001 : 10 . if p invested the money for 5 months , find for how much time did q invest the money ? | 7 * 5 : 5 * x = 7 : 10 x = 10 answer : b | a = 7 / 5
b = 10 / 7
c = a * b
d = c * 5
|
a ) 9.25 , b ) 5.25 , c ) 7.25 , d ) 6.25 , e ) 5.1 | d | multiply(divide(15, const_60), add(20, 5)) | the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance traveled downstream in 15 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 15 / 60 ) ) km = 6.25 km . answer : d" | a = 15 / const_60
b = 20 + 5
c = a * b
|
a ) 25 % , b ) 30 % , c ) 35 % , d ) 40 % , e ) 50 % | a | multiply(divide(20, 80), const_100) | by selling 80 pens , a trader gains the cost of 20 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 80 pens = rs . 80 profit = cost of 20 pens = rs . 20 profit % = 20 / 80 * 100 = 25 % answer : a" | a = 20 / 80
b = a * 100
|
a ) 1750 , b ) 2789 , c ) 2500 , d ) 1550 , e ) 2000 | d | divide(divide(multiply(add(1000, add(1000, multiply(100, add(const_10, const_1)))), const_12), const_2), const_12) | a salt manufacturing company produced a total of 1000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its average monthly production for that year ? | total production of salt by the company in that year = 1000 + 1100 + 1200 + . . . . + 2100 = 18600 . average monthly production of salt for that year = 18600 / 12 = 1550 . answer : d | a = 10 + 1
b = 100 * a
c = 1000 + b
d = 1000 + c
e = d * 12
f = e / 2
g = f / 12
|
a ) 58 , b ) 59 , c ) 68 , d ) 69 , e ) 78 | b | add(subtract(69, multiply(1, 11)), 1) | a batsman in his 11 th inning makes a score of 69 and their by increasing his average by 1 . what is his average after the 11 th inning ? | "10 x + 69 = 11 ( x + 1 ) x = 58 + 1 = 59 answer : b" | a = 1 * 11
b = 69 - a
c = b + 1
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a ) 15 % , b ) 16.66 % , c ) 17.8 % , d ) 19 % , e ) 37.5 % | e | multiply(divide(subtract(55, 40), 40), const_100) | john makes $ 40 a week from his job . he earns a raise andnow makes $ 55 a week . what is the % increase ? | "increase = ( 15 / 40 ) * 100 = 37.5 % . e" | a = 55 - 40
b = a / 40
c = b * 100
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a ) always even , b ) always odd , c ) odd only when x is odd , d ) even only when y is even , e ) odd only when xy is odd | a | add(1, const_1) | if x and y are positive integers and y β 1 , then xy ( y β 1 ) is | "expression is xy ( y - 1 ) . we can ignore x and only work with y . if y = even - > entire expression is even since anything multiplied by even is even if y = odd , y - 1 - > even - > entire expression is even since anything multiplied by even is even hence , entire expression will always be even . answer ( a ) ." | a = 1 + 1
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a ) 15 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | a | multiply(subtract(subtract(subtract(13, const_4), const_4), 1), const_3) | you collect baseball cards . suppose you start out with 13 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the end ? | "solution start with 13 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 13 + 1 which is 7 , so you ' re left with 13 - 7 = 6 . peter takes 1 baseball card from you : 6 - 1 = 5 baseball cards . paul triples the number of baseball cards you have : 5 Γ£ β 3 = 15 baseball cards . so you have 15 at the end . correct answer : a" | a = 13 - 4
b = a - 4
c = b - 1
d = c * 3
|
a ) 8 % , b ) 7 % , c ) 10 % , d ) 2 % , e ) 4 % | d | multiply(divide(subtract(subtract(70, multiply(70, divide(10, const_100))), 61.74), subtract(70, multiply(70, divide(10, const_100)))), const_100) | the list price of an article is rs . 70 . a customer pays rs . 61.74 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "explanation : 70 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 61.74 x = 2 % option d" | a = 10 / 100
b = 70 * a
c = 70 - b
d = c - 61
e = 10 / 100
f = 70 * e
g = 70 - f
h = d / g
i = h * 100
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | what is the smallest integer b for which 27 ^ b > 3 ^ 24 ? | "27 ^ b > 3 ^ 24 converting into the same bases : 27 ^ b > 27 ^ 8 therefore for the equation to hold true , b > 8 or b = 9 option c" | a = 24 / 3
b = math.floor(a)
c = b + 1
|
a ) $ 50000 , b ) $ 60000 , c ) $ 60500 , d ) $ 65000 , e ) can not be determined | e | multiply(add(const_1, divide(10, const_100)), 50000) | the median annual household income in a certain community of 21 households is $ 50000 . if the mean r income of a household increases by 10 % per year over the next 2 years , what will the median income in the community be in 2 years ? | answer is e , because there are different numbers in the set and we are not sure which side of the numbers in the set will be increased so the mean r is increase by 10 % . it could be the case that small number of higher end incomes increased a little or many low end incomes increased - it can not be identified . | a = 10 / 100
b = 1 + a
c = b * 50000
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 65 | b | subtract(add(80, 30), divide(80, divide(50, const_100))) | the contents of a certain box consist of 80 apples and 30 oranges . how many oranges must be added to the box so that exactly 50 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.5 80 = ( 30 + 80 + x ) * 0.5 x = 50 answer : b" | a = 80 + 30
b = 50 / 100
c = 80 / b
d = a - c
|
a ) 15 , b ) 30 , c ) 40 , d ) 50 , e ) none of these | a | add(multiply(sqrt(divide(subtract(125, 50), const_2)), const_100), sqrt(subtract(125, divide(subtract(125, 50), const_2)))) | the sum of the squares of three numbers is 125 , while the sum of their products taken two at a time is 50 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 125 xy + yz + zx = 50 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 125 + ( 2 * 50 ) ( x + y + z ) ^ 2 = 225 so x + y + z = 15 answer : a" | a = 125 - 50
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 125 - 50
f = e / 2
g = 125 - f
h = math.sqrt(g)
i = d + h
|
a ) 76 , b ) 88 , c ) 100 , d ) 112 , e ) 124 | b | divide(multiply(add(add(1, 8), add(2, 8)), subtract(14, 3)), 2) | in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 3 , - 1 ) , ( 3 , 8 ) , ( 14 , 2 ) , ( 14 , - 5 ) ? | "by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) Γ ( base 1 + base 2 ) Γ ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 Γ ( 9 + 7 ) Γ 11 = 88 . the answer is b ." | a = 1 + 8
b = 2 + 8
c = a + b
d = 14 - 3
e = c * d
f = e / 2
|
a ) 7 days , b ) 5 days , c ) 3 days , d ) 4 days , e ) 8 days | a | divide(subtract(multiply(30, 10), 216), add(10, 2)) | a man was engaged on a job for 30 days on the condition that he would get a wage of rs . 10 for the day he works , but he have to pay a fine of rs . 2 for each day of his absence . if he gets rs . 216 at the end , he was absent for work for . . . days . | the equation portraying the given problem is : 10 * x β 2 * ( 30 β x ) = 216 where x is the number of working days . solving this we get x = 23 number of days he was absent was 7 ( 30 - 23 ) days . answer : a | a = 30 * 10
b = a - 216
c = 10 + 2
d = b / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | e | divide(16, const_2) | if 9 ^ y = 3 ^ 16 , what is y ? | "9 ^ y = 3 ^ 2 y = 3 ^ 16 2 y = 16 y = 8 the answer is e ." | a = 16 / 2
|
a ) 6 , b ) 9 , c ) 16 , d ) 4 , e ) 21 | e | multiply(7, divide(multiply(15, 3), 15)) | the average age of 3 boys is 15 years and their ages are in proportion 3 : 5 : 7 . what is the age in years of the eldest boy ? | 3 x + 5 x + 7 x = 45 x = 3 7 x = 21 answer : e | a = 15 * 3
b = a / 15
c = 7 * b
|
a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 2 , d ) 7 / 10 , e ) 29 / 100 | e | divide(subtract(100, subtract(add(50, 40), 19)), 100) | a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 19 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ? | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 19 - 2 cmb = 100 b - cb - mb - 2 cmb = 29 hence the probability = 29 / 100 . e" | a = 50 + 40
b = a - 19
c = 100 - b
d = c / 100
|
a ) 24 % , b ) 40.5 % , c ) 44 % , d ) 54 % , e ) 64 % | b | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), subtract(const_100, 15))), const_100) | a towel , when bleached , lost 30 % of its length and 15 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( β 30 β 15 + ( 30 Γ 15 ) / 100 ) % = β 40.5 % i . e . , area is decreased by 40.5 % answer : b" | a = 100 * 100
b = 100 - 30
c = 100 - 15
d = b * c
e = a - d
f = e / 100
|
a ) 1 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | c | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 12)), divide(const_1, 24))) | if a , b and c together can finish a piece of work in 4 days . a alone in 12 days and b in 24 days , then c alone can do it in ? | c = 1 / 4 - 1 / 12 β 1 / 24 = 1 / 8 = > 8 days ' answer : c | a = 1 / 12
b = const_0_25 - a
c = 1 / 24
d = b - c
e = 1 / d
|
a ) 20 % , b ) 25 % , c ) 62.5 % , d ) 75 % , e ) 80 % | c | multiply(divide(divide(multiply(subtract(120, divide(120, const_3)), subtract(subtract(divide(120, const_3), 15), 15)), add(15, subtract(subtract(divide(120, const_3), 15), 15))), subtract(120, divide(120, const_3))), const_100) | at the end of the day , february 14 th , a florist had 120 roses left in his shop , all of which were red , white or pink in color and either long or short - stemmed . a third of the roses were short - stemmed , 15 of which were white and 15 of which were pink . the percentage of pink roses that were short - stemmed equaled the percentage of red roses that were short - stemmed . if none of the long - stemmed roses were white , what percentage of the long - stemmed roses were red ? | "r + w + p = 120 s + l = 120 1 / 3 * 120 = 40 short - stemmed white = 15 short - stemmed pink = 15 = > short - stemmed red = 10 15 / p = 10 / r = > r = 2 r / 3 so total long stemmed = 80 and long stemmed red + long stemmed pink = 80 so long stemmed red / long stemmed = ? total white = 20 ( as no long stemmed white ) = > r + 2 r / 3 + 20 = 120 = > 5 r = 300 and r = 60 long stemmed r = 60 - 10 = 50 so long stemmed red / r = 50 / 80 = 62.5 % answer - c" | a = 120 / 3
b = 120 - a
c = 120 / 3
d = c - 15
e = d - 15
f = b * e
g = 120 / 3
h = g - 15
i = h - 15
j = 15 + i
k = f / j
l = 120 / 3
m = 120 - l
n = k / m
o = n * 100
|
a ) 120 kg , b ) 130 kg , c ) 137 kg , d ) 190 kg , e ) none | c | add(65, multiply(10, 7.2)) | the average weight of 10 persons increases by 7.2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "solution total weight increased = ( 10 x 7.2 ) kg = 72 kg . weight of new person = ( 65 + 72 ) kg = 137 kg . answer c" | a = 10 * 7
b = 65 + a
|
a ) 3 : 1 , b ) 3 : 2 , c ) 3 : 5 , d ) 3 : 8 , e ) 3 : 7 | b | divide(subtract(15, 9), subtract(19, 15)) | a fold density is 19 times greater than the water and for copper it is 9 times . at what ratio you can mix gold and copper to get 15 times denser than water . ? | suppose x units of gold are mixed with y units of copper to make ( x + y ) units of alloy which is 15 times denser than water . then 19 * x + 9 * y = 15 * ( x + y ) = > 19 * x - 15 * x = 15 * y - 9 * y = > 4 * x = 6 * y = > x / y = 3 / 2 so gold and copper should be mixed respectively in the ratio of 3 : 2 answer : b | a = 15 - 9
b = 19 - 15
c = a / b
|
a ) none , b ) two , c ) three , d ) five , e ) seven | c | subtract(const_4, const_1) | r is the set of positive even integers less than 50 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "squares < 50 { 1 , 4,9 , 16,25 , 36,49 } s = { 1,4 , 16,36 } r = { 2 , . . . . . 48 } hence c ." | a = 4 - 1
|
a ) 115 , b ) 116 , c ) 195 , d ) 118 , e ) 119 | c | add(add(multiply(divide(const_100, 45), 45), multiply(divide(50, 45), 45)), 45) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 45 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 45 39 - - - ? = > 195 answer : c" | a = 100 / 45
b = a * 45
c = 50 / 45
d = c * 45
e = b + d
f = e + 45
|
a ) 14 lts , b ) 18 lts , c ) 20 lts , d ) 16 lts , e ) 26 lts | d | divide(multiply(add(10, 10), 4), add(1, 4)) | a jar contains a mixture of two liquids acid ( nitric acid ) and base ( ammonium chloride ) in the ratio 4 : 1 . when 10 litres of the mixture is taken out and 10 litres of liquid base is poured into the jar , the ratio becomes 2 : 3 . how many litres of liquid acid was contained in the jar ? | % age of liquid base in the original mixture = 1 / 5 x 100 = 20 % in the final mixture % of the liquid base = 3 / 5 x 100 = 60 % now using the rule of allegation hence reduced quantity of the first mixture and the quantity of mixture b which is to be added are the same . total mixture = 10 + 10 = 20 liters and quantity of liquid a = 20 / 5 x 4 = 16 lts answer : d | a = 10 + 10
b = a * 4
c = 1 + 4
d = b / c
|
a ) 20 , b ) 15 , c ) 25 , d ) 13 , e ) 42 | a | divide(add(4, 36), const_2) | if x + y = 4 , x - y = 36 , for integers of x and y , x = ? | "x + y = 4 x - y = 36 2 x = 40 x = 20 answer is a" | a = 4 + 36
b = a / 2
|
a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 125 % , e ) 250 % | d | multiply(divide(divide(25, const_100), divide(15, const_100)), const_100) | if c is 25 % of a and 15 % of b , what percent of a is b ? | "answer = d 25 a / 100 = 20 b / 100 b = 25 a / 20 = 125 a / 100 = 125 %" | a = 25 / 100
b = 15 / 100
c = a / b
d = c * 100
|
a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | d | add(40, divide(subtract(864, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 864 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 864 - 640 = 224 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 224 / 28 = 56 / 7 = 8 total hrs = 40 + 8 = 48 answer d 48" | a = 16 * 40
b = 864 - a
c = 100 + 75
d = 16 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) rs . 7500 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) rs . 6000 | d | divide(divide(3500, subtract(const_1, divide(5, const_12))), divide(2, 3)) | praveen starts business with rs . 3500 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari β s contribution in the capital ? | "let hari β s capital be rs . x . then , 3500 * 12 / 7 x = 2 / 3 = > 14 x = 126000 = > x = 9000 . answer : d" | a = 5 / 12
b = 1 - a
c = 3500 / b
d = 2 / 3
e = c / d
|
a ) a ) 1000 , b ) b ) 1010 , c ) c ) 1065 , d ) d ) 1075 , e ) e ) 1080 | b | add(multiply(8, 70), multiply(9, 50)) | harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 50 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 Γ 8 = 560 . cost of 9 kg of mangoes = 50 Γ 9 = 450 . total cost he has to pay = 560 + 450 = 1010 . b )" | a = 8 * 70
b = 9 * 50
c = a + b
|
a ) 268 , b ) 180 , c ) 150 , d ) 18 , e ) 88 | d | subtract(add(180, 88), 250) | in a group of 250 readers who read science fiction or literacy works or both , 180 read science fiction and 88 read literacy works . how many read both science fiction and literacy works ? | "consider total number of reader n ( s u l ) = 250 people who read science fiction n ( s ) = 180 people who read literacy works n ( l ) = 88 both science fiction and literacy n ( s Γ’ Λ Β© l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s Γ’ Λ Β© l ) 250 = 180 + 88 - n ( s Γ’ Λ Β© l ) n ( s Γ’ Λ Β© l ) = 268 - 250 n ( s Γ’ Λ Β© l ) = 18 so people read both science fiction and literacy works are 18 answer : d" | a = 180 + 88
b = a - 250
|
a ) 80 , b ) 96 , c ) 108 , d ) 120 , e ) 252 | e | divide(factorial(10), multiply(factorial(divide(const_10, const_2)), factorial(5))) | in a certain circle there are 10 points . what is the number of the triangles connecting 5 points of the 10 points ? | "imo : e here we have to select 5 points out of 10 points . order is not important so the answer will be 10 c 5 = 252 answer e" | a = math.factorial(10)
b = 10 / 2
c = math.factorial(b)
d = math.factorial(5)
e = c * d
f = a / e
|
a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | a | multiply(divide(subtract(14, 10), subtract(30, 14)), 200) | solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 200 milliliters of solution x to create a solution that is 14 percent alcohol by volume ? | "14 % is 4 % - points higher than 10 % but 16 % - points lower than 30 % . thus there should be 4 parts of solution x for 1 part of solution y . we should add 50 ml of solution y . the answer is a ." | a = 14 - 10
b = 30 - 14
c = a / b
d = c * 200
|
a ) 973 , b ) 6973 , c ) 5994 , d ) 9554 , e ) none of them | c | subtract(multiply(const_10, 6), 6) | the difference between the place value and the face value of 6 in the numerical 856973 is | "= ( place value of 6 ) - ( face value of 6 ) = ( 6000 - 6 ) = 5994 answer is c" | a = 10 * 6
b = a - 6
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | e | divide(log(5), log(power(8, 8))) | if n = 8 ^ 8 β 5 , what is the units digit of n ? | "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 5 , we get unit digit as 5 answer : e" | a = math.log(5)
b = 8 ** 8
c = math.log(b)
d = a / c
|
a ) 0.75 , b ) 1.2 , c ) 1.8 , d ) 2.4 , e ) 4.25 | b | inverse(divide(80, add(80, 16))) | patrick purchased 80 pencils and sold them at a loss equal to the selling price of 16 pencils . the cost of 80 pencils is how many times the selling price of 80 pencils ? | "say the cost price of 80 pencils was $ 80 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 80 - 80 p = 16 p - - > p = 5 / 6 . ( cost price ) / ( selling price ) = 1 / ( 5 / 6 ) = 6 / 5 = 1.2 . answer : b ." | a = 80 + 16
b = 80 / a
c = 1/(b)
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | e | add(10, const_4) | how many odd numbers between 10 and 1,200 are the squares of integers ? | "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 16 numbers . answer : e ." | a = 10 + 4
|
a ) 25 min , b ) 20 min , c ) 28 min , d ) 15 min , e ) 10 min | b | divide(const_60, divide(18, 6)) | if 6 women take an hour to dig a ditch , then how long should 18 women take a dig to ditch of the same type ? | if 6 women take an hour to dig a ditch , then 18 men will take 6 * 60 / 18 = 20 mins to dig a ditch of the same type . answer : b | a = 18 / 6
b = const_60 / a
|
a ) 20 % decrease , b ) 10 % increase , c ) 10 % decrease , d ) 15 % increase , e ) 25 % decrease | a | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 50), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 60), const_100)), const_100)), const_100)) | if the price of a book is first decreased by 50 % and then increased by 60 % , then the net change in the price will be ? | "let the original price be $ 100 new final price = 160 % of ( 50 % of $ 100 ) = 160 / 100 * 50 / 100 * 100 = $ 80 decrease is 20 % answer is a" | a = 100 * 50
b = a / 100
c = 100 - b
d = c / 100
e = 100 * 60
f = e / 100
g = 100 + f
h = g / 100
i = d * h
j = i * 100
k = 100 - j
|
a ) 20 , b ) 50 , c ) 40 , d ) 30 , e ) 60 | d | multiply(add(const_2, 4), multiply(6, 4)) | if a farmer sells 15 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 10 more chickens , he will run out of feed 6 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | "let x = total feed required for the planned period n = number of chicken t = total time of the planned feed x = nt 1 ) x = ( n - 15 ) * ( t + 4 ) 2 ) x = ( n + 10 ) * ( t - 6 ) equating 1 & 2 ( n - 15 ) * ( t + 4 ) = ( n + 10 ) * ( t - 6 ) or nt + 4 n - 15 t - 60 = nt - 6 n + 10 t - 60 10 n = 25 t n = 5 / 2 * t or t = 2 n / 5 x = n * 2 n / 5 substituting this value in 1 n * 2 n / 5 = ( n - 15 ) * ( 2 n / 5 + 4 ) 2 n = 60 n = 30 d" | a = 2 + 4
b = 6 * 4
c = a * b
|
a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability q that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability q = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b" | a = math.comb(10, 3)
b = 10 + 10
c = math.comb(b, 3)
d = a / c
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | add(divide(21, 7), 1) | find the value of x from this equation ? 7 ( x - 1 ) = 21 | 1 . divide both sides by 7 : 2 . simplify both sides : x - 1 = 3 3 . add 1 to both sides : x - 1 + 1 = 3 + 1 4 . simplify both sides : x = 4 b | a = 21 / 7
b = a + 1
|
a ) $ 100 , b ) $ 122 , c ) $ 150 , d ) $ 210 , e ) $ 170 | b | divide(100, subtract(const_1, divide(10, const_100))) | a person incurs 10 % loss by selling a watch for $ 100 . at what price should the watch be sold to earn 10 % profit ? | "let the new selling price be $ x ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) ( 100 - 10 ) / 100 = ( 100 + 10 ) / x x = 110 * 100 / 90 = 122 approximately answer is b" | a = 10 / 100
b = 1 - a
c = 100 / b
|
a ) 388 , b ) 105 , c ) 288 , d ) 266 , e ) 281 | b | add(add(multiply(divide(const_100, 45), 27), multiply(divide(30, 45), 27)), 27) | a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 27 , what is the total amount ? | x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 27 35 - - - ? = > 105 answer : b | a = 100 / 45
b = a * 27
c = 30 / 45
d = c * 27
e = b + d
f = e + 27
|
a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 14 % | e | multiply(divide(multiply(divide(40, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | an association of mathematics teachers has 1,500 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 40 percent of the votes cast ? | "total umber of members = 1500 number of members that cast votes = 525 since , winning candidate received 40 percent of the votes cast number of votes for winning candidate = ( 40 / 100 ) * 525 = 210 percent of total membership that voted for winning candidate = ( 210 / 1500 ) * 100 = 14 % answer e" | a = 40 / 100
b = a * 525
c = 4 ** 2
d = 100 * c
e = b / d
f = e * 100
|
a ) 41.66667 , b ) 60.5 , c ) 63.5 , d ) 62.5 , e ) 64.5 | a | divide(multiply(25, add(const_4, const_1)), const_2) | to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to 3 / 5 of its present ? | "let capacity of 1 bucket = x capacity of the tank = 25 x new capacity of the bucket = 3 x / 5 hence , number of buckets needed = 25 x / ( 3 x / 5 ) = ( 25 Γ 5 ) / 3 = 41.66667 answer is a ." | a = 4 + 1
b = 25 * a
c = b / 2
|
a ) $ 21,000 , b ) $ 12,000 , c ) $ 15,000 , d ) $ 4,500 , e ) $ 4,000 | b | divide(add(divide(subtract(300, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(300, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 300 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | "sometime , setting up an equation is an easy way to go with : 300 + 0.06 ( x - 1000 ) = 0.08 x x = 12,000 ans : b" | a = 6 / 100
b = a * 1
c = 300 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 300 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
|
a ) 17.36 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these | a | multiply(divide(add(multiply(const_2, multiply(multiply(const_2, add(const_1, const_4)), const_100)), multiply(add(const_1, const_4), const_100)), divide(add(multiply(multiply(const_2, add(const_1, const_4)), const_100), multiply(add(const_2, const_4), const_100)), divide(12.5, const_100))), const_100) | ritesh and co . generated revenue of rs . 1,800 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ? | "explanation : given , ritesh and co . generated revenue of rs . 1,800 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1800 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) Γ 1800 . = > 14,400 . hence , the total revenue by end of 2007 is rs . 14,400 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 14400 ) Γ 100 . = > 17.36 % . answer : a" | a = 1 + 4
b = 2 * a
c = b * 100
d = 2 * c
e = 1 + 4
f = e * 100
g = d + f
h = 1 + 4
i = 2 * h
j = i * 100
k = 2 + 4
l = k * 100
m = j + l
n = 12 / 5
o = m / n
p = g / o
q = p * 100
|
a ) 870 , b ) 970 , c ) 1070 , d ) 1000 , e ) 800 | a | multiply(multiply(const_3, const_10), 28) | there are 28 stations between kolkatta and chennai . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | the total number of stations = 30 from 30 stations we have to choose any two stations and the direction of travel ( i . e . , kolkatta to chennai is different from chennai to kolkatta ) in 30 p 2 ways . 30 p 2 = 30 * 29 = 870 answer : a | a = 3 * 10
b = a * 28
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a ) 8 , b ) 10 , c ) 15 , d ) 17 , e ) 19 | a | subtract(30, divide(add(multiply(30, 7.5), 490), add(25, 7.5))) | a contractor is engaged for 30 days on the condition that he receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 490 in all . for how many days was he absent ? | 30 * 25 = 750 490 - - - - - - - - - - - 260 25 + 7.50 = 32.5 260 / 32.5 = 8 a | a = 30 * 7
b = a + 490
c = 25 + 7
d = b / c
e = 30 - d
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a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(multiply(2, 10), 4) | albert is 2 times mary β s age and 4 times as old as betty . mary is 10 years younger than albert . how old is betty ? | "a = 2 m = m + 10 m = 10 a = 20 a = 4 b , and so b = 5 the answer is a ." | a = 2 * 10
b = a / 4
|
a ) 56.5 rupees , b ) 58.5 rupees , c ) 57.5 rupees , d ) 59.5 rupees , e ) 55.5 rupees | c | multiply(add(add(add(add(const_1, divide(const_1, const_3)), divide(const_2, const_3)), divide(const_1, multiply(const_2, const_4))), divide(const_3, 4)), 20) | a worker is paid rs . 20 / - for a full days work . he works 11 / 32 / 31 / 8.3 / 4 days in a week . what is the total amount paid for that worker ? | the total days worked = 1 + 0.333 + 0.667 + 0.125 + 0.75 = 2.875 days amount for 2.875 days is ( 2.875 * 20 ) 57.5 rupees answer : c | a = 1 / 3
b = 1 + a
c = 2 / 3
d = b + c
e = 2 * 4
f = 1 / e
g = d + f
h = 3 / 4
i = g + h
j = i * 20
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | b | add(add(add(const_4, const_2), const_1), const_1) | how many odd factors does 270 have ? | "start with the prime factorization : 270 = 2 * 3 * 5 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 270 . answer : b ." | a = 4 + 2
b = a + 1
c = b + 1
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | divide(add(15, const_10), subtract(7, const_2)) | find a number such that when 15 is subtracted from 7 times the number , the results is more than twice the number ? | solution let the number be x . = then , 7 x - 15 = 2 x + 10 = 5 x = 25 βΉ = βΊ x = 5 . hence the required number is 5 . answer d | a = 15 + 10
b = 7 - 2
c = a / b
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a ) $ 2.75 , b ) $ 3.25 , c ) $ 3.75 , d ) $ 4.25 , e ) $ 4.75 | c | divide(multiply(multiply(3, 5), 1.00), const_4) | having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.00 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 100 cents ( 4 / 15 ) x = 100 x = $ 3.75 the answer is c ." | a = 3 * 5
b = a * 1
c = b / 4
|
a ) 1 / 2 , b ) 2 , c ) 5 / 2 , d ) 4 , e ) 5 | d | add(subtract(2, divide(5, 2)), add(divide(5, 2), 2)) | if 2 | 2 β e | = 5 , what is the sum of all the possible values for e ? | if e < 2 , then 2 - e = 5 / 2 ; e = - 1 / 2 if e > 2 , then e - 2 = 5 / 2 ; e = 9 / 2 9 / 2 - 1 / 2 = 8 / 2 = 4 = d | a = 5 / 2
b = 2 - a
c = 5 / 2
d = c + 2
e = b + d
|
a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 6,400 , e ) $ 800 | d | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(divide(10, const_100), const_2))) | the market value of a certain machine decreased by 10 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | d market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 10 / 100 ) = 8000 - 800 = $ 7200 market value in 1984 = market value in 1983 - ( 10 % of $ 8000 ) = 7200 - 800 = $ 6400 | a = 100 * 10
b = 2 * 4
c = a * b
d = 100 * 10
e = 2 * 4
f = d * e
g = 10 / 100
h = g * 2
i = f * h
j = c - i
|
a ) 9 % , b ) 10 % , c ) 34.4 % , d ) 12 % , e ) 15 % | c | multiply(subtract(divide(18, const_100), divide(subtract(5, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 5 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ? | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 5 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 5 or 45 * 18 - 45 * x + 36 * x = 5 * 100 or 9 x = - 5 * 100 + 45 * 18 x = 310 / 9 = 34.4 c" | a = 18 / 100
b = 18 / 100
c = 18 * 2
d = b * c
e = 5 - d
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
|
a ) - 14 , b ) - 13 , c ) - 8 , d ) - 12 , e ) - 10 | e | add(subtract(8, 5), subtract(negate(8), 5)) | if | y + 5 | = 8 , what is the sum of all the possible values of y ? | "there will be two cases y + 5 = 8 or y + 5 = - 8 = > y = 3 or y = - 13 sum of both the values will be - 13 + 3 = - 10 answer : e" | a = 8 - 5
b = negate - (
c = a + b
|
a ) 5.8 days , b ) 4.78 days , c ) 6.25 days , d ) 3.15 days , e ) 2.75 days | a | inverse(add(inverse(10), inverse(14))) | a and b complete a work in 10 days . a alone can do it in 14 days . if both together can do the work in how many days ? | 1 / 10 + 1 / 14 = 6 / 35 35 / 6 = 5.8 days answer : a | a = 1/(10)
b = 1/(14)
c = a + b
d = 1/(c)
|
a ) 131 , b ) 135 , c ) 113 , d ) 147 , e ) 188 | c | add(multiply(divide(subtract(153, 33), const_3), const_2), 33) | if jake loses 33 pounds , he will weigh twice as much as his sister . together they now weigh 153 pounds . what is jake ' s present weight , in pounds ? | "j = jake β s current weight , in pounds s = sister β s current weight , in pounds we are told that β if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j β 33 = 2 s j = 2 s + 33 ( equation 1 ) next , we are told that β together they now weigh 153 pounds . β we can also put this into an equation . j + s = 153 ( equation 2 ) to solve this equation , we can substitute 2 s + 8 from equation 1 for the variable j in equation 2 : 2 s + 33 = 153 - s 3 s = 120 s = 40 j + 40 = 153 j = 113 answer : c" | a = 153 - 33
b = a / 3
c = b * 2
d = c + 33
|
a ) $ 250640 , b ) $ 430640 , c ) $ 120640 , d ) $ 110640 , e ) $ 150640 | d | add(multiply(multiply(subtract(1, divide(24, const_100)), subtract(1, divide(24, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(24, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 24 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.76 * 0.76 * 1 , 50,000 = $ 86,640 sp such that a profit of $ 24,000 is made = 86,640 + 24,000 = $ 1 , 10,640 d" | a = 24 / 100
b = 1 - a
c = 24 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 24 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00 | a | add(divide(add(10, 14), const_2), add(const_0_25, const_0_25)) | a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price v per t - shirt that the vendor paid for these purchases ? | "correct answer : a explanation : the relevant formula for this problem is average v = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ' s cost was $ 165 , since 11 x 15 = 165 . the grand sum is then $ 140 + $ 165 , which equals $ 305 . the total number of shirts purchased was 25 , so to get the average price per shirt , we divide 305 by 25 , which equals $ 12.20 . as a result , the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of 11 and 14 is 12.5 , but the vendor sells more shirts at the lower price than at the higher price , the weighted average must be less than $ 12.50 ; only answer choice a is a possibility ." | a = 10 + 14
b = a / 2
c = const_0_25 + const_0_25
d = b + c
|
a ) 50 hours , b ) 40 hours , c ) 15 hours , d ) 12 hours , e ) 8 hours | c | divide(120, multiply(divide(divide(20, 10), 5), 20)) | if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 120 units ? | "here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 120 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 120 units is ' 6 times ' 20 , we need ' 6 times ' more time . ( 2.5 hours ) ( 6 times ) = 15 hours final answer : [ reveal ] spoiler : c" | a = 20 / 10
b = a / 5
c = b * 20
d = 120 / c
|
['a ) 76 m', 'b ) 80 m', 'c ) 84 m', 'd ) 88 m', 'e ) none of these'] | c | divide(66, multiply(power(divide(1, const_2), const_2), const_pi)) | 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length if the wire in meters will be : | explanation : let the length of the wire be h radius = 1 / 2 mm = 120 cm Ο r 2 h = 66 22 / 7 β 1 / 20 β 1 / 20 β h = 66 = > h = 66 β 20 β 20 β 7 / 22 = 8400 cm = 84 m option c | a = 1 / 2
b = a ** 2
c = b * math.pi
d = 66 / c
|
a ) 288 , b ) 262 , c ) 246 , d ) 205 , e ) 267 | c | multiply(divide(48, 40), add(add(const_100, 65), 40)) | a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 48 , find the sum of money ? | "a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 48 41 - - - - ? = > rs . 246 answer : c" | a = 48 / 40
b = 100 + 65
c = b + 40
d = a * c
|
a ) 10 , b ) 12 , c ) 22 , d ) 28 , e ) 20 | b | inverse(add(inverse(divide(60, const_4)), inverse(60))) | a is four times as fast as b . if b alone can do a piece of work in 60 days , in what time can a and b together complete the work ? | a can do the work in 60 / 4 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 60 = ( 4 + 1 ) / 60 = 1 / 12 so a and b together can do the work in 12 days . answer : b | a = 60 / 4
b = 1/(a)
c = 1/(60)
d = b + c
e = 1/(d)
|
a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 10 children together complete the piece of work ? | explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 10 men β s 1 day β s work = 1 Γ 10 / 96 = 5 / 48 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 10 men + 4 women + 10 children ) β s 1 day β s work = 5 / 48 + 1 / 48 + 1 / 24 = 8 / 48 = 1 / 6 the required no . of days = 6 days answer : option d | a = 24 * 10
b = 1/(a)
c = 10 * b
d = 12 * 8
e = 1/(d)
f = e * 10
g = 48 * 4
h = 1/(g)
i = h * 4
j = f + i
k = c + j
l = 1/(k)
|
a ) 3 , b ) 2 , c ) can not be determined , d ) 4 , e ) 1 | e | divide(211, 211) | if a positive integer n has 211 factors , then how many prime factors does n have ? | let the factor be a ^ x * b ^ y . . . . . so ( x + 1 ) ( y + 1 ) . . . . . = 211 . . now 211 is a prime number . . . so only one variable x or y is 210 and the others are 0 . . ( 210 + 1 ) ( 0 + 1 ) . . . so the number becomes a ^ 210 . . thus only one prime factor is there . | a = 211 / 211
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