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a ) 21 st , b ) 22 nd , c ) 23 rd , d ) 24 th , e ) none of these | b | subtract(multiply(2, 12), 1) | a monkey ascends a greased pole 12 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | in 2 minutes , he ascends = 1 metre β΄ 10 metres , he ascends in 20 minutes . β΄ he reaches the top in 21 st minute . answer b | a = 2 * 12
b = a - 1
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | power(17, 17) | what will be the remainder when 17 ^ 200 is divided by 18 | "17 ^ 200 % 18 ( 17 - 18 ) ^ 200 % 18 ( - 1 ) ^ 200 % 18 1 % 18 = 1 answer : a" | a = 17 ** 17
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a ) rs . 4000 , b ) rs . 4500 , c ) rs . 5500 , d ) rs . 6000 , e ) rs . 6500 | a | divide(subtract(1360, multiply(divide(15, 125), 12000)), subtract(divide(12, 120), divide(15, 125))) | sakshi invests a part of rs . 12000 in 12 % stock at rs . 120 and the remainder in 15 % stock at rs . 125 . if his total dividend per annum is rs . 1360 , how much does he invest in 12 % stock at rs . 120 ? | let investment in 12 % stock be rs . x . then , investment in 15 % stock = rs . ( 12000 - x ) . 12 / 120 * x + 15 / 125 * ( 12000 - x ) = 1360 x / 10 + 3 / 25 ( 12000 - x ) = 1360 5 x + 72000 - 6 x = 1360 x 50 x = 4000 . answer = a | a = 15 / 125
b = a * 12000
c = 1360 - b
d = 12 / 120
e = 15 / 125
f = d - e
g = c / f
|
a ) 11 / 8 , b ) 3 / 80 , c ) 9 / 64 , d ) 5 / 64 , e ) 3 / 64 | b | multiply(multiply(divide(1, 5), divide(1, 2)), subtract(1, divide(5, 8))) | xavier , yvonne , and zelda each try independently to solve a problem . if their individual probabilities for success are 1 / 5 , 1 / 2 and 5 / 8 , respectively , what is the probability that xavier and yvonne , but not zelda , will solve the problem ? | "p ( xavier will solve ) = 1 / 5 p ( yvonne will solve ) = 1 / 2 p ( zelda will not solve ) = 1 - 5 / 8 = 3 / 8 . now , we need to multiply all this ps to find an answer : p = ( 1 / 5 ) * ( 1 / 2 ) * ( 3 / 8 ) = 3 / 80 . ans . b ." | a = 1 / 5
b = 1 / 2
c = a * b
d = 5 / 8
e = 1 - d
f = c * e
|
a ) 1040 , b ) 920 , c ) 980 , d ) 720 , e ) 820 | a | add(400, multiply(divide(400, 5), 8)) | in a college , the ratio of the number of boys to girls is 8 : 5 . if there are 400 girls , the total number of students in the college is | "explanation : let the boy are 8 x and girls are 5 x = > 5 x = 400 = > x = 80 total students = 8 x + 5 x = 13 x = 13 ( 80 ) = 1040 option a" | a = 400 / 5
b = a * 8
c = 400 + b
|
a ) 2 % , b ) 8.7 % , c ) 5.7 % , d ) 6.7 % , e ) 1 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(18, const_100))), const_100)) | the tax on a commodity is diminished by 18 % and its consumption increased by 15 % . the effect on revenue is ? | "100 * 100 = 10000 82 * 115 = 9430 - - - - - - - - - - - 10000 - - - - - - - - - - - 570 100 - - - - - - - - - - - ? = > 5.7 % decrease answer : c" | a = 15 / 100
b = 1 + a
c = 18 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 21 Β½ days , b ) 22 Β½ days , c ) 23 Β½ days , d ) 15 days , e ) none of these | d | add(divide(const_1, 10), divide(const_1, 30)) | a can do a job in 10 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "explanation : 1 / 10 + 1 / 30 = 4 / 30 = 2 / 15 15 / 2 = 15 / 2 * 2 = 15 days answer : d" | a = 1 / 10
b = 1 / 30
c = a + b
|
a ) rs 7500 , b ) rs 4000 , c ) rs 2000 , d ) rs 8000 , e ) rs 7000 | d | multiply(divide(5, add(7, 5)), 20000) | if jamie ' s income is rs 20000 , calculate his savings given that his income and expenditure are in the ratio 7 : 5 . ? | let the income and the expenditure of the person be rs . 7 x and rs . 7 x respectively . income , 7 x = 28000 = > x = 4000 savings = income - expenditure = 7 x - 5 x = 2 x so , savings = rs . 8000 answer : d | a = 7 + 5
b = 5 / a
c = b * 20000
|
a ) 82762 , b ) 12877 , c ) 82882 , d ) 80925 , e ) 80325 | e | divide(divide(multiply(4016.25, const_100), 1), 5) | a sum fetched a total simple interest of rs . 4016.25 at the rate of 1 % p . a . in 5 years . what is the sum ? | "principal = ( 100 * 4016.25 ) / ( 1 * 5 ) = rs . 80325 . answer : e" | a = 4016 * 25
b = a / 1
c = b / 5
|
a ) 92 , b ) 80 , c ) 72 , d ) 82 , e ) 28 | d | add(divide(multiply(add(90, divide(90, 10)), 2), 3), 16) | on thrusday mabel handled 90 transactions . anthony handled 10 % more transaction than mabel , cal handled 2 / 3 rds of the transactions that anthony handled and jade handled 16 more transaction than cal . how much transaction did jade handled ? | mabel handled 90 transaction , anthony 10 % more transaction the mabel . anthony = 90 + 90 * 10 % , = 90 + 90 * 0.10 , = 90 + 9 , = 99 , cal handled 2 / 3 rds than anthony handled , cal = 2 / 3 * 99 = 66 , jade handled 16 more than cal . jade = 66 + 16 = 82 , jade handled = 82 tranasactions . correct answer : ( d ) | a = 90 / 10
b = 90 + a
c = b * 2
d = c / 3
e = d + 16
|
a ) 5 / 6 , b ) 5 / 12 , c ) 7 / 12 , d ) 7 / 18 , e ) 11 / 18 | c | add(divide(1, 3), multiply(divide(1, 3), divide(1, 3))) | jar x is 1 / 3 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | "let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 3 + p / 4 = 7 p / 12 the answer is c ." | a = 1 / 3
b = 1 / 3
c = 1 / 3
d = b * c
e = a + d
|
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 36 | a | divide(10, subtract(1, add(divide(1, 3), divide(1, 4)))) | a cement mixture is composed of 3 elements . by weight , 1 / 3 of the mixture is sand , 1 / 4 of the mixture is water , and the remaining 10 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ? | "let the total weight be x . sand content = ( 1 / 3 ) x water content = ( 1 / 4 ) x gravel = x - ( 1 / 3 ) x - ( 1 / 4 ) x = ( 5 / 12 ) x = 10 x = 24 then answer will be a = 24" | a = 1 / 3
b = 1 / 4
c = a + b
d = 1 - c
e = 10 / d
|
a ) 1 % , b ) 10 % , c ) 100 % , d ) 50 % , e ) 25 % | b | multiply(divide(0.01, 0.1), const_100) | 0.01 is what percent of 0.1 ? | "required percentage = 0.01 * 100 / 0.1 = 100 / 10 = 10 % answer is b" | a = 0 / 1
b = a * 100
|
a ) 389 m , b ) 150 m , c ) 289 m , d ) 299 m , e ) 219 m | b | subtract(multiply(speed(300, 26), 39), 300) | a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 26 sec . what is the length of the platform ? | "speed = 300 / 26 = 150 / 13 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 150 / 13 = > x = 150 m . answer : b" | a = speed * (
b = a - 39
|
a ) 117 , b ) 180 , c ) 150 , d ) 200 , e ) 250 | a | multiply(9, divide(add(multiply(8, 12), 8), subtract(9, const_1))) | 9 persons went to a hotel for taking their meals . 8 of them spent rs 12 each on their meals and the ninth spent rs . 8 more than the average expenditure of all the 9 . what was the total money spent by them ? | exp . let the total expenditure be x , then the average = x / 9 , 8 * 12 + [ x / 9 + 8 ] = x or [ x - x / 9 ] = 104 . 8 x / 9 = 104 , x = 104 * 9 / 8 = 117 . answer : a | a = 8 * 12
b = a + 8
c = 9 - 1
d = b / c
e = 9 * d
|
a ) 342 km , b ) 352 km , c ) 252 km , d ) 742 km , e ) 382 km | c | divide(add(add(30, multiply(2, 7)), 30), 2) | a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 7 hours of the journey . | "the total distance covered by the car in the first 7 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 = sum of 7 terms in ap whose first term is 30 and last term is 42 = 7 / 2 [ 30 + 42 ] = 252 km . answer : c" | a = 2 * 7
b = 30 + a
c = b + 30
d = c / 2
|
a ) 55 , b ) 69 , c ) 100 , d ) 109 , e ) 115 | b | subtract(multiply(40, 4), multiply(45, 2)) | the average ( arithmetic mean ) of 4 positive integers is 40 . if the average of 2 of these integers is 45 , what is the greatest possible value that one of the other 2 integers can have ? | a + b + c + d = 160 a + b = 90 c + d = 70 greatest possible = 69 ( just less than 1 ) answer = b | a = 40 * 4
b = 45 * 2
c = a - b
|
a ) 1010 , b ) 804 , c ) 814 , d ) 844 , e ) none | a | divide(multiply(add(multiply(1, const_100), 20), multiply(12, const_100)), power(20, const_2)) | what is the least number of square tiles required to pave the floor of a room 12 m 12 cm long and 1 m 20 cm broad ? | "solution length of largest tile = h . c . f . of 1212 cm & 120 cm = 12 cm . area of each tile = ( 12 x 12 ) cm 2 β΄ required number of tiles = [ 1212 x 120 / 12 x 12 ] = 1010 . answer a" | a = 1 * 100
b = a + 20
c = 12 * 100
d = b * c
e = 20 ** 2
f = d / e
|
a ) 3 / 5 , b ) 5 / 3 , c ) 15 , d ) 25 , e ) none | b | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | if 2 x = 3 β 32 , then x is equal to : | "sol . 2 x = 3 β 32 β 2 x = ( 32 ) 1 / 3 = ( 25 ) 1 / 3 = 25 / 3 β x = 5 / 3 . answer b" | a = 2 * 100
b = a / 2
c = b - 2
d = 2 * 100
e = d / 2
f = e + 2
g = c / f
|
a ) 120 meter , b ) 299 meter , c ) 666 meter , d ) 155 meter , e ) 144 meter | a | multiply(divide(multiply(48, const_1000), const_3600), 9) | a train running at the speed of 48 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 48 * ( 5 / 18 ) m / sec = 40 / 3 m / sec length of train ( distance ) = speed * time ( 40 / 3 ) * 9 = 120 meter answer : a" | a = 48 * 1000
b = a / 3600
c = b * 9
|
a ) 64 , b ) 72 , c ) 101 , d ) 111 , e ) 128 | c | divide(subtract(multiply(subtract(20.5, 0.5), 5), 52), 0.5) | a cricket player whose bowling average was 20.5 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.5 . what will be the number of wickets taken by him till the last match ? | "average = total runs / total wickets total runs after last match = 20.5 w + 52 total wickets after last match = w + 5 ( 20.5 w + 52 ) / ( w + 5 ) = 20.5 - 0.5 = 20 w = 96 so total wickets aftr last match = w + 5 = 101 answer : c" | a = 20 - 5
b = a * 5
c = b - 52
d = c / 0
|
a ) 1.21 % , b ) 1.74 % , c ) 1.84 % , d ) 1.47 % , e ) 1.24 % | a | divide(multiply(11, 11), const_100) | if a trader sold two cars each at rs . 325475 and gains 11 % on the first and loses 11 % on the second , then his profit or loss percent on the whole is ? | "sp of each car is rs . 325475 , he gains 11 % on first car and losses 11 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 11 ) ( 11 ) / 100 % = 1.21 % answer : a" | a = 11 * 11
b = a / 100
|
a ) 15 days , b ) 18 days , c ) 21 days , d ) 24 days , e ) 27 days | b | divide(multiply(36, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 15), multiply(divide(add(24, 40), 4), 3)), multiply(divide(8, 2), 3))) | 3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 36 days . if 40 more cows and 15 more buffaloes are brought in , how many days will the fodder last ? | "2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 36 days . 40 cows + 15 buffaloes = 20 oxen + 10 oxen = 30 oxen the new total is equal to 60 oxen instead of 30 oxen . 30 * 36 / 60 oxen = 18 days the answer is b ." | a = 24 / 4
b = a * 3
c = 15 + b
d = 8 / 2
e = d * 3
f = c + e
g = 36 * f
h = 15 + 15
i = 24 + 40
j = i / 4
k = j * 3
l = h + k
m = 8 / 2
n = m * 3
o = l + n
p = g / o
|
a ) ab - 4 , b ) 2 b - ab - 4 , c ) ab + 4 , d ) ab - 2 b + 4 , e ) 4 - ab | a | subtract(subtract(const_1, const_1), 4) | for any non - zero a and b that satisfy | ab | = ab and | a | = a | b - 4 | + | ab - b | = ? | | ab | = ab and | a | = - a = > b is also negative . so | b - 4 | + | ab - b | = ( b - 4 ) + ab - b ( as ab is positive and - b is positive , so expression is totally positive ) = b - 4 + ab - b = ab - 4 answer - a | a = 1 - 1
b = a - 4
|
a ) 106 , b ) 107 , c ) 126 , d ) 109 , e ) 110 | c | add(90, divide(90, 3)) | p software has coding line 5 % more than n , n software has coding line 1 / 3 more than m . m software has 90 lines of coding . find p lines . | m s / w has 90 line of code n s / w has = 90 + 90 * 1 / 3 = 120 line of code p s / w 5 % more n ' code 120 + 6 = 126 answer : c | a = 90 / 3
b = 90 + a
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a ) 120 sec , b ) 165 sec , c ) 186 sec , d ) 167 sec , e ) 360 sec | e | divide(1800, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | a and b go around a circular track of length 1800 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 1800 / ( 36 * 5 / 18 ) , 1800 / ( 54 * 5 / 18 ) } = lcm ( 180 , 120 ) = 360 sec . answer : e | a = 54 * const_0_2778
b = 36 * const_0_2778
c = a - b
d = 1800 / c
|
a ) 8 , b ) 10 , c ) 12 , d ) 9 , e ) 6 | d | add(add(4, 4), 1) | the greatest common divisor of ( b - 2 ) ! , ( b + 1 ) ! , and ( b + 4 ) ! is 5040 . what is the value of b ? | gcd of ( b - 2 ) ! , ( b + 1 ) ! , and ( b + 4 ) ! is 5040 . find x . best way to find n is substitute each of the answer and find out the correct one . only option d gave the correct answer . option d : 9 and find gcd or 7 ! , 10 ! and 13 ! . 7 ! itself is 5040 and 7 ! which is contained in the other two factorials . this satisfies the condition , hence answer is 9 . answer is d . | a = 4 + 4
b = a + 1
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 60 | e | multiply(10, multiply(const_3, const_2)) | the length of a side of a hexagon is 10 inches . what is the perimeter ? | "hexagon . it means 6 equal sides . p = 6 ( 10 ) = 60 inches answer e" | a = 3 * 2
b = 10 * a
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a ) 15 % , b ) 17 % , c ) 24 % , d ) 30 % , e ) 33 % | e | multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100) | in february wilson β s earnings were 40 percent of his family β s total income . in march wilson earned 25 percent less than in february . if the rest of his family β s income was the same in both months , then , in march , wilson β s earnings were approximately what percent s of his family β s total income ? | "lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb = 60 x thus wilson ' s earning as % of total family income in march s = 30 x / ( 30 + 60 ) x = 30 x / 90 x = 33.33 % thus answer is e" | a = 40 / 100
b = 25 / 100
c = 40 / 100
d = b * c
e = a - d
f = 25 / 100
g = 40 / 100
h = f * g
i = 1 - h
j = e / i
k = j * 100
|
a ) 5 / 46 , b ) 7 / 36 , c ) 1 / 36 , d ) 5 / 36 , e ) 5 / 37 | d | divide(multiply(multiply(1, 13), 5), multiply(multiply(4, 9), 13)) | if a : b = 1 : 4 , b : c = 13 : 9 , c : d = 5 : 13 , find a : d ? | a / d = ( 1 / 4 ) * ( 13 / 9 ) * ( 5 / 13 ) = > 5 / 36 answer : d | a = 1 * 13
b = a * 5
c = 4 * 9
d = c * 13
e = b / d
|
['a ) 20', 'b ) 30', 'c ) 40', 'd ) 55', 'e ) 48'] | b | multiply(add(5, 1), add(1, 4)) | in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 5 , 1 ) , ( 1 , 1 ) , ( 1 , - 4 ) and ( - 5 , - 4 ) ? | length of side 1 = 5 + 1 = 6 length of side 2 = 4 + 1 = 5 area of rectangle = 6 * 5 = 30 b is the answer | a = 5 + 1
b = 1 + 4
c = a * b
|
a ) 976374 , b ) 971750 , c ) 980241 , d ) 2356677 , e ) 1083875 | b | multiply(1000000, 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 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 30 % , then find the population of city atthe 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 + 30 / 100 ) = 971750 b" | 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 = 1000000 * h
|
a ) 180 , b ) 160 , c ) 140 , d ) 150 , e ) 120 | a | subtract(100, subtract(power(2, divide(100, 90)), const_1)) | what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 90 ? | "put in simple words , we need to find the highest power of 2 in 100 ^ 90 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 90 = ( 2 ^ 2 * 5 ^ 2 ) ^ 90 = 2 ^ 180 * 5 ^ 180 answer : a" | a = 100 / 90
b = 2 ** a
c = b - 1
d = 100 - c
|
a ) 680 , b ) 810 , c ) 800 , d ) 730 , e ) 750 | b | divide(multiply(multiply(multiply(9, const_100), sqrt(const_100)), 9), const_100) | alex takes a loan of $ 9,000 to buy a used truck at the rate of 9 % simple interest . calculate the annual interest to be paid for the loan amount . | from the details given in the problem principle = p = $ 9,000 and r = 9 % or 0.09 expressed as a decimal . as the annual interest is to be calculated , the time period t = 1 . plugging these values in the simple interest formula , i = p x t x r = 9,000 x 1 x 0.09 = 810.00 annual interest to be paid = $ 810 answer : b | a = 9 * 100
b = math.sqrt(100)
c = a * b
d = c * 9
e = d / 100
|
a ) 1 / 6 , b ) 1 / 2 , c ) 5 / 8 , d ) 5 / 6 , e ) 15 / 16 | b | divide(subtract(0.75, 0.7), subtract(0.8, 0.7)) | last month , john rejected 0.7 % of the products that he inspected and jane rejected 0.8 percent of the products that she inspected . if total of 0.75 percent of the products produced last month were rejected , what fraction of the products did jane inspect ? | "x - fraction of products jane inspected ( 1 - x ) - fraction of products john inspected 0.8 ( x ) + 0.7 ( 1 - x ) = 0.75 0.1 x = 0.75 - 0.7 x = 0.05 / 0.1 x = 1 / 2 therefore the answer is b : 1 / 2 ." | a = 0 - 75
b = 0 - 8
c = a / b
|
a ) 100 , b ) 120 , c ) 125 , d ) 115 , e ) 90 | a | subtract(multiply(80, const_2), multiply(60, const_2)) | if the average ( arithmetic mean ) of x and y is 60 , and z β x = 80 , what is the average of y and z ? | "x + y / 2 = 60 = > x + y = 120 x = z - 80 . . . sub this value z - 80 + y = 120 = > z + y = 200 = > z + y / 2 = 100 answer : a" | a = 80 * 2
b = 60 * 2
c = a - b
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | what is the smallest integer z for which 27 ^ z > 3 ^ 24 ? | "27 ^ z > 3 ^ 24 converting into the same bases : 27 ^ z > 27 ^ 8 therefore for the equation to hold true , z > 8 or z = 9 option c" | a = 24 / 3
b = math.floor(a)
c = b + 1
|
a ) 23 , b ) 12 , c ) 77 , d ) 92 , e ) 61 | b | inverse(divide(inverse(8), add(const_2, const_1))) | a is twice as good a work man as b and together they finish the work in 8 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 8 = > x = 1 / 24 2 x = 1 / 12 a can do the work in 12 days . answer : b" | a = 1/(8)
b = 2 + 1
c = a / b
d = 1/(c)
|
a ) 48 , b ) 41 , c ) 46 , d ) 43 , e ) 96 | d | divide(subtract(527, power(21, const_2)), const_2) | the sum of two numbers is 21 and the sum of their squares is 527 . find the product of the numbers . | let a and b be the two numbers ( a + b ) ^ 2 = a ^ 2 + 2 ab + b ^ 2 given ( a + b ) = 21 a ^ 2 + b ^ 2 = 527 so , 21 ^ 2 = 527 + 2 ab 2 ab = 527 - 441 2 ab = 86 ab = 43 ans d | a = 21 ** 2
b = 527 - a
c = b / 2
|
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 55 | e | subtract(divide(add(280, 20), const_4), 20) | thabo owns exactly 280 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books does thabo own ? | "i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fiction books is 0 . double - matrix : p = paperback ; h = hardcover ; f = fiction ; nf = nonfiction p h total f 2 x + 40 0 nf x + 20 x total 3 x + 60 x 280 3 x + 60 + x = 280 x = 55 answer ( e . )" | a = 280 + 20
b = a / 4
c = b - 20
|
a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 25 | d | multiply(divide(55, divide(110, const_10)), subtract(divide(55, divide(110, const_10)), const_2)) | in the number 100 ab , a and b represent the tens and units digits , respectively . if 110 ab is divisible by 55 , what is the greatest possible value of b Γ a ? | you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 5 = 15 ( next number wo n ' t have 110 as the first 3 digits so we have only two options 0 and 25 ) . answer : e . ! please post ps questions in the ps subforum : gmat - problem - solving - ps - 140 / please post ds questions in the ds subforum : gmat - data - sufficiency - ds - 141 / no posting of ps / ds questions is allowed in the mainmath forum . d | a = 110 / 10
b = 55 / a
c = 110 / 10
d = 55 / c
e = d - 2
f = b * e
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a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 88 sec , e ) 20 sec | e | multiply(divide(350, multiply(63, const_1000)), const_3600) | a train 350 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 350 * 2 / 35 = 20 sec answer : e" | a = 63 * 1000
b = 350 / a
c = b * 3600
|
a ) 78 kmph , b ) 85 kmph , c ) 35 kmph , d ) 24 kmph , e ) 74 kmph | c | divide(multiply(divide(42, add(1, divide(add(multiply(40, const_60), 48), const_3600))), 7), 5) | a car travelling with 5 / 7 km of its actual speed covers 42 km in 1 hr 40 min 48 sec find the actual speed of the car ? | time taken = 1 hr 40 min 48 sec = 126 / 75 hrs let the actual speed be x kmph then 5 / 7 x * 126 / 75 = 42 x = ( 42 * 7 * 75 ) / 5 * 126 = 35 kmph answer ( c ) | a = 40 * const_60
b = a + 48
c = b / 3600
d = 1 + c
e = 42 / d
f = e * 7
g = f / 5
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a ) 2 : 1 , b ) 1 : 2 , c ) 1 : 3 , d ) 3 : 1 , e ) 2 : 3 | b | divide(90, multiply(divide(50, const_1000), const_3600)) | compare the rates of two trains , one travelling at 90 km / hr and other is at 50 m / s ? | speed of the 1 st train = 90 km / hr speed of the 2 nd train = 50 m / s = 50 * 18 / 5 = 180 km / hr ratio of the speeds of the train = 90 : 180 = 1 : 2 answer is b | a = 50 / 1000
b = a * 3600
c = 90 / b
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a ) 122 , b ) 112 , c ) 134 , d ) 108 , e ) 101 | b | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(2, const_100)), multiply(add(100, 2), 1000)), 1000) | a computer manufacturer produces a certain electronic component at a cost of $ 100 per component . shipping costs for delivering the components are $ 2 per unit . further , the manufacturer has costs of $ 10,000 a month related to the electronic component regardless of how many it produces . if the manufacturer produces and sells 1000 components a month , what is the lowest price it can sell them for such that the costs do n ' t exceed the revenues ? | "$ 10000 is a fixed cost each component is $ 102 ( $ 100 to produce , $ 2 to ship ) manufacturer will be producing and selling 1000 components so therefore the equation to find price would be 1000 * p = 10000 + ( 1000 * 100 ) + ( 1000 * 2 ) p = ( 10000 + 100000 + 2000 ) / 1000 p = 112 answer : b" | a = 4 * 4
b = a * 1000
c = 2 * 100
d = b + c
e = 100 + 2
f = e * 1000
g = d + f
h = g / 1000
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a ) 13 : 00 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | d | add(divide(add(700, 50), subtract(500, divide(700, const_2))), 10) | sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 08 : 00 and starts climbing at a constant pace . ken starts climbing only at 10 : 00 , when sari is already 700 meters ahead of him . nevertheless , ken climbs at a constant pace of 500 meters per hour , and reaches the summit before sari . if sari is 50 meters behind ken when he reaches the summit , at what time did ken reach the summit ? | "both sari and ken climb in the same direction . speed of sari = 700 / 2 = 350 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 700 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / relative speed = ( 700 + 50 ) / ( 500 - 350 ) = 5 hrs starting from 10 : 00 , in 5 hrs , the time will be 15 : 00 answer ( d )" | a = 700 + 50
b = 700 / 2
c = 500 - b
d = a / c
e = d + 10
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a ) 14 kg , b ) 15 kg , c ) 18 kg , d ) 19 kg , e ) 16 kg | e | subtract(add(multiply(25, const_2), multiply(28, const_2)), multiply(30, const_3)) | the average weight of a , b and c is 30 kg . if the average weight of a and b be 25 kg and that of b and c be 28 kg , then the weight of b is : | explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 30 x 3 ) = 90 Γ’ β¬ Β¦ . ( i ) a + b = ( 25 x 2 ) = 50 Γ’ β¬ Β¦ . ( ii ) b + c = ( 28 x 2 ) = 56 Γ’ β¬ Β¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 106 Γ’ β¬ Β¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 16 . b Γ’ β¬ β’ s weight = 16 kg . answer e | a = 25 * 2
b = 28 * 2
c = a + b
d = 30 * 3
e = c - d
|
a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | a | multiply(subtract(divide(192000, 120000), const_1), const_100) | the cost of a one - family home was $ 120000 in 1980 . in 1988 , the price had increased to $ 192000 . what was the percent increase in the cost of the home ? | increase = 192000 - 120000 = 72000 % increase = 72000 * 100 / 120000 = 60 % answer : option a | a = 192000 / 120000
b = a - 1
c = b * 100
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a ) a ) 140 , b ) b ) 150 , c ) c ) 180 , d ) d ) 240 , e ) e ) 220 | d | divide(multiply(divide(20, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | one fourth of one third of two fifth of a number is 20 . what will be 40 % of that number | explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 20 then x = 20 * 30 = 600 40 % of 600 = 240 answer : option d | a = 1 / 4
b = 1 / 3
c = a * b
d = 2 + 3
e = 2 / d
f = c * e
g = 20 / f
h = g * 40
i = h / 100
|
a ) 135 , b ) 136 , c ) 137 , d ) 138 , e ) 139 | b | multiply(35, const_4) | a card board of 34 * 14 has to be attached to a wooden box and a total of 35 pins are to be used on the each side of the cardbox . find the total number of pins used . | ( 35 * 4 ) - 4 = 136 answer : b | a = 35 * 4
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a ) 39.9 , b ) 37.88 , c ) 35 , d ) 38.88 , e ) 38.09 | e | subtract(multiply(divide(200, 42), 50), 200) | a 200 meter long train crosses a platform in 50 seconds while it crosses a signal pole in 42 seconds . what is the length of the platform ? | "speed = [ 200 / 42 ] m / sec = 100 / 21 m / sec . let the length of the platform be x meters . then , x + 200 / 50 = 100 / 21 21 ( x + 200 ) = 5000 Γ¨ x = 38.09 m . answer : e" | a = 200 / 42
b = a * 50
c = b - 200
|
a ) 25 liters , b ) 29 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters | b | subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 11) | how much water should be added to 11 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ? | "let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 11 + x ) ) ) 2 = 5 % * ( 11 + x ) - - - - - - - - > x = 29 ans b" | a = 2 * 100
b = 75 / 100
c = b * 20
d = 20 - c
e = a / d
f = e - 11
|
a ) 80 , b ) 84 , c ) 88 , d ) 92 , e ) 96 | a | divide(600, multiply(divide(3, 2), 5)) | a van takes 5 hours to cover a distance of 600 km . what speed in kph should the van maintain to cover the same distance in 3 / 2 of the previous time ? | "( 3 / 2 ) * 5 = 7.5 hours 600 / 7.5 = 80 kph the answer is a ." | a = 3 / 2
b = a * 5
c = 600 / b
|
a ) 17 , b ) 25 , c ) 27 , d ) 52 , e ) 60 | e | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 12) | the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of bren 12 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren ' s age 12 years hence = 40 + 12 = 52 answer = e" | a = 8 * 3
b = a + 76
c = 5 + 8
d = c + 7
e = b / d
f = e * 8
g = f + 12
|
a ) $ 200 , b ) $ 150 , c ) $ 300 , d ) $ 25 , e ) $ 100 | a | multiply(divide(multiply(multiply(2, 10), inverse(5)), add(add(multiply(multiply(2, 10), subtract(inverse(2), add(inverse(5), inverse(10)))), multiply(multiply(2, 10), inverse(5))), multiply(multiply(2, 10), inverse(10)))), 500) | a and b undertake to do a piece of work for $ 500 . a alone can do it in 5 days while b alone can do it in 10 days . with the help of c , they finish it in 2 days . find the share of c ? | "c ' s 1 day work = ( 1 / 2 ) - ( 1 / 5 + 1 / 10 ) = 1 / 5 a : b : c = 1 / 5 : 1 / 10 : 1 / 5 = 2 : 1 : 2 b ' s share = 500 * 2 / 5 = $ 200 answer is a" | a = 2 * 10
b = 1/(5)
c = a * b
d = 2 * 10
e = 1/(2)
f = 1/(5)
g = 1/(10)
h = f + g
i = e - h
j = d * i
k = 2 * 10
l = 1/(5)
m = k * l
n = j + m
o = 2 * 10
p = 1/(10)
q = o * p
r = n + q
s = c / r
t = s * 500
|
a ) $ 168.58 , b ) $ 187.54 , c ) $ 158.74 , d ) $ 120.56 , e ) $ 200.84 | c | divide(multiply(80, divide(multiply(add(80, 85), divide(add(80, 15), const_2)), 80)), subtract(divide(add(80, 15), const_2), 15)) | a clock store sold a certain clock to a collector for 15 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 45 percent of what the collector had paid . the shop then sold the clock again at a profit of 85 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 80 , for how much did the shop sell the clock the second time ? | "now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 15 % profit . this means the clocks ' selling price was c ( 1.15 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 45 % the price at which the collector bought it . thus , you get = 1.15 * 0.45 * c = 0.5175 c furthermore , the store sells the clock for the second time for 85 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.85 = 1.85 * 0.5175 c finally given that c - 0.5175 c = 80 - - - - > c = 165.80 $ thus , the cost of the clock the second time around = 1.85 * 0.5175 c = 1.85 * 0.5175 * 165.80 = 158.74 $ . hence c is the correct answer ." | a = 80 + 85
b = 80 + 15
c = b / 2
d = a * c
e = d / 80
f = 80 * e
g = 80 + 15
h = g / 2
i = h - 15
j = f / i
|
a ) 2 , b ) 3 , c ) 0.5 , d ) 6 , e ) 12 | c | multiply(divide(subtract(100, divide(180, const_2)), 40), const_2) | angelina walked 100 meters from her home to the grocery at a constant speed . she then walked 180 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | "let the speed be x . . . so time taken from home to grocery = 100 / x . . the speed to gym = 2 x . . so time taken = 180 / 2 x = 90 / x . . its given 100 / x - 90 / x = 40 . . 10 / x = 40 . . x = 0.25 m / secs . . so grocery to gym = 2 * 0.25 = 0.5 m / s . . . answer : c" | a = 180 / 2
b = 100 - a
c = b / 40
d = c * 2
|
a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50 | d | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(39, const_100))), const_100) | if 45 % of z is 39 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 39 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 39 / 100 ) * ( 3 / 4 ) x i . e . z = ( 39 * 3 ) x / ( 45 * 4 ) i . e . z = ( 0.65 ) x = ( 65 / 100 ) x i . e . z is 65 % of x answer : option d" | a = 75 / 100
b = 45 / 100
c = 39 / 100
d = b / c
e = a / d
f = e * 100
|
a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 5200 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 130000), multiply(multiply(multiply(10, 16), 6), 10)) | 10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 130000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 130000 p = rs . [ ( 24 * 4 * 4 * 130000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 49920000 / 9600 ) = > p = rs . 5200 answer : d" | a = 24 * 4
b = a * 4
c = b * 130000
d = 10 * 16
e = d * 6
f = e * 10
g = c / f
|
a ) 40 , b ) 54 , c ) 45 , d ) 50 , e ) none | d | divide(150, 3) | a worker makes a toy in every 3 h . if he works for 150 h , then how many toys will he make ? | no . of toys = 150 / 3 = 50 answer : d | a = 150 / 3
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a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 7 | c | divide(add(11, 5), const_2) | in one hour , a boat goes 11 km along the stream and 5 km against the stream . the speed of the boat in still water in ( km / hr ) is | "solution speed in still water = 1 / 2 ( 11 + 5 ) km / hr = 8 km / hr . answer c" | a = 11 + 5
b = a / 2
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a ) 75 , b ) 80 , c ) 50 , d ) 35 , e ) 40 | d | subtract(multiply(multiply(45, divide(8, 6)), divide(40, 30)), 45) | 45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 40 meters deep by working for 6 hours ? | 45 workers * 8 hours / 30 meters = x * 6 / 40 x = 80 total workers 80 - 45 = 35 the answer is d . | a = 8 / 6
b = 45 * a
c = 40 / 30
d = b * c
e = d - 45
|
a ) rs 1950 , b ) rs 2000 , c ) rs 745 , d ) rs 765 , e ) none | a | divide(subtract(multiply(add(multiply(const_4, const_10), const_3), const_1000), const_100), 22) | if an amount of rs 42,900 is distributed equally amongst 22 persons , how much amount would each person get ? | required amount = 42900 / 22 = rs 1950 answer a | a = 4 * 10
b = a + 3
c = b * 1000
d = c - 100
e = d / 22
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a ) 9 ^ 2 , b ) - 18 ^ 16 , c ) 63 ( 9 ^ 7 ) , d ) - 80 ( 9 ^ 7 ) , e ) 80 ( 9 ^ 7 ) | e | divide(multiply(add(add(const_100, const_60), const_1), 9), const_100) | what is the value of : - 9 ^ 7 + 9 ^ 9 ? | - 9 ^ 7 + 9 ^ 9 = 9 ^ 7 ( - 1 + 9 ^ 2 ) = 9 ^ 7 ( - 1 + 81 ) = 80 ( 9 ^ 7 ) , answer e . | a = 100 + const_60
b = a + 1
c = b * 9
d = c / 100
|
['a ) 360', 'b ) 450', 'c ) 560', 'd ) 670', 'e ) 780'] | b | multiply(multiply(multiply(5, const_3), const_2), multiply(5, const_3)) | let the sides of a rectangular prism are consecutive multiples of 5 . which among the following could be the base area | length = 5 * a breadth = 5 * ( a + 1 ) or 5 * ( a + 2 ) base area - > multiple of 25 450 = 25 * 18 answer : b | a = 5 * 3
b = a * 2
c = 5 * 3
d = b * c
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 210 | e | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 45 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 45 miles . teena can cover 7.5 miles in 30 mins teena will cover 45 miles in 180 mins so answer 30 + 180 = 210 mins . ( answer e )" | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | subtract(subtract(divide(80, 4), 10), 10) | the sum of ages of 4 children born at the intervals of 10 years each is 80 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 10 ) + ( x + 20 ) + ( x + 30 ) = 80 4 x = 20 x = 5 . age of the youngest child = x = 5 years . answer : b" | a = 80 / 4
b = a - 10
c = b - 10
|
a ) 15 / 56 , b ) 41 / 56 , c ) 13 / 28 , d ) 15 / 28 , e ) 5 / 14 | d | divide(multiply(3, 5), divide(multiply(add(3, 5), add(5, const_2)), const_2)) | a bag contains 3 blue and 5 white marbles . one by one , marbles are drawn out randomly until only two are left in the bag . what is the probability x that out of the two , one is white and one is blue ? | "the required probability x = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls . ways to select 6 out of total 8 = 8 c 6 ways to select 4 out of 5 white balls = 5 c 4 ways to select 2 out of 3 blue balls = 3 c 2 thus the required probability = ( 5 c 4 * 3 c 2 ) / 8 c 6 = 15 / 28 . d is thus the correct answer ." | a = 3 * 5
b = 3 + 5
c = 5 + 2
d = b * c
e = d / 2
f = a / e
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a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 8 | e | multiply(divide(subtract(17, 3), add(3, 4)), 4) | one hour after yolanda started walking from x to y , a distance of 17 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob Ρ ' s was 4 miles per hour , how many miles had bob walked when they met ? | when b started walking y already has covered 3 miles out of 17 , hence the distance at that time between them was 17 - 3 = 14 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 14 / 7 = 2 hours . in 6 hours b walked 2 * 4 = 8 miles . answer : e . | a = 17 - 3
b = 3 + 4
c = a / b
d = c * 4
|
a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | c | multiply(divide(multiply(subtract(add(add(10, multiply(divide(9, const_100), 10)), add(10, multiply(divide(6, const_100), 10))), add(10, 10)), const_100), add(10, 10)), const_2) | a dozen eggs and 10 pounds of oranges are currently at the same price . if the price of a dozen eggs rises by 9 percent and the price of oranges rises by 6 % . how much more will it cost to buy a dozen eggs and 10 pounds of oranges . | "say currently both a dozen eggs and 10 pounds of oranges cost $ 100 ( they are at the same price ) . so , to buy a dozen eggs and 10 pounds of oranges we need $ 100 . after the increase , the price of a dozen eggs will be $ 109 and the price of 10 pounds of oranges will be $ 106 . so after the increase , to buy a dozen eggs and 10 pounds of oranges we ' ll need $ 215 . increase = 15 % . answer : c ." | a = 9 / 100
b = a * 10
c = 10 + b
d = 6 / 100
e = d * 10
f = 10 + e
g = c + f
h = 10 + 10
i = g - h
j = i * 100
k = 10 + 10
l = j / k
m = l * 2
|
a ) 62.5 , b ) 69.5 , c ) 67.5 , d ) 62.5 , e ) 70 | e | divide(multiply(add(47.50, divide(multiply(47.50, 40), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 40 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 140 / 100 ) = 66.5 mp * ( 95 / 100 ) = 66.5 mp = 70 answer : e" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 5
f = d / e
|
a ) 54 , b ) 57 , c ) 60 , d ) 36 , e ) 96 | b | multiply(19, divide(divide(36, 2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 19 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | "b 57 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 19 mph = 57 miles" | a = 36 / 2
b = a / 6
c = 19 * b
|
a ) 10 hours , b ) 12 hours , c ) 14 hours , d ) 18 hours , e ) none of these | d | add(divide(48, subtract(6, 2)), divide(48, add(6, 2))) | in a river flowing at 2 km / hr , a boat travels 48 km upstream and then returns downstream to the starting point . if its speed in still water be 6 km / hr , find the total journey time . | "explanation : speed of the boat = 6 km / hr speed downstream = ( 6 + 2 ) = 8 km / hr speed upstream = ( 6 - 2 ) = 4 km / hr distance traveled downstream = distance traveled upstream = 48 km total time taken = time taken downstream + time taken upstream = ( 48 / 8 ) + ( 48 / 4 ) = 18 hr . answer : option d" | a = 6 - 2
b = 48 / a
c = 6 + 2
d = 48 / c
e = b + d
|
a ) 50 % , b ) 40 % , c ) 25 % , d ) 20 % , e ) 10 % | c | multiply(divide(subtract(multiply(2, divide(add(3, 1), add(3, 2))), subtract(3, multiply(3, divide(add(3, 1), add(3, 2))))), add(3, 1)), const_100) | in smithtown , the ratio of right - handed people to left - handed people is 3 to 1 and the ratio of men to women is 3 to 2 . if the number of right - handed men is maximized , then what x percent of all the people in smithtown are left - handed women ? | "looking at the ratio we can take total number of people = 20 . . ans 5 / 20 or 25 % c" | a = 3 + 1
b = 3 + 2
c = a / b
d = 2 * c
e = 3 + 1
f = 3 + 2
g = e / f
h = 3 * g
i = 3 - h
j = d - i
k = 3 + 1
l = j / k
m = l * 100
|
a ) 38 , b ) 27 , c ) 75 , d ) 28 , e ) 21 | c | multiply(multiply(add(floor(2.3), multiply(divide(subtract(2.3, floor(2.3)), const_60), const_100)), const_3), const_10) | the number of degrees that the hour hand of a clock moves through between noon and 2.30 in the afternoon of the same day is ? | explanation : the hour hand moves from pointing to 12 to pointing to half way between 2 and 3 . the angle covered between each hour marking on the clock is 360 / 12 = 30 . since the hand has covered 2.5 of these divisions the angle moved through is 75 . answer : c | a = math.floor(2, 3)
b = math.floor(2, 3)
c = 2 - 3
d = c / const_60
e = d * 100
f = a + e
g = f * 3
h = g * 10
|
a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | a | divide(30, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(30, 2))))), 2)) | a rectangular tiled patio is composed of 30 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 30 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 30 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 30 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 30 ( 2 r - 4 ) * r = 30 2 r ^ 2 - 4 r - 30 = 0 r can not be negative so r = 5 and c = 6 so , answer will be a" | a = negate + (
b = 4 ** 2
c = 30 * 2
d = 4 * c
e = b + d
f = math.sqrt(e)
g = a / f
h = 30 / g
|
a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | b | multiply(42, const_1) | a row of ants comprised of 102 ants . a man walked upon it and all except 42 were crushed beneath his foot . how many ants are left alive ? | solution : all except 42 were crushed means that all the ants except 42 were crushed beneath the man ' s foot . hence , 42 ants are alive now answer b | a = 42 * 1
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(subtract(add(multiply(divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40)), 40), multiply(subtract(10, divide(subtract(multiply(60, 10), multiply(54, 10)), subtract(60, 40))), 60)), multiply(45, 10)), subtract(60, 45)) | at a certain food stand , the price of each apple is 40 Β’ and the price of each orange is 60 Β’ . mary selects a total of 10 apples and oranges from the food stand , and the average ( arithmetic mean ) price of the 10 pieces of fruit is 54 Β’ . how many oranges must mary put back so that the average price of the pieces of fruit that she keeps is 45 Β’ ? | "if the average price is 54 , then mary has 7 oranges and 3 apples ( a ratio of 7 : 3 ) . for the average price to be 45 , the ratio should be 1 : 3 . mary must put back 6 oranges . the answer is e ." | a = 60 * 10
b = 54 * 10
c = a - b
d = 60 - 40
e = c / d
f = e * 40
g = 60 * 10
h = 54 * 10
i = g - h
j = 60 - 40
k = i / j
l = 10 - k
m = l * 60
n = f + m
o = 45 * 10
p = n - o
q = 60 - 45
r = p / q
|
a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 5 % | e | multiply(divide(50, 1), const_100) | what percent is 50 gm of 1 kg ? | "1 kg = 1000 gm 50 / 1000 Γ 100 = 5000 / 1000 = 5 % e )" | a = 50 / 1
b = a * 100
|
['a ) 10 : 17', 'b ) 2 : 5', 'c ) 5 : 16', 'd ) 10 : 19', 'e ) 32 : 25'] | d | divide(const_10, add(add(multiply(8, const_2), const_2), const_1)) | the dimensions of a rectangular solid are 6 inches , 5 inches , and 8 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ? | answer : d . | a = 8 * 2
b = a + 2
c = b + 1
d = 10 / c
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | e | divide(subtract(1300, multiply(5, add(20, 20))), add(add(20, 20), 20)) | john purchased some shirts and trousers for $ 1300 . he paid $ 500 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ? | "given that the total purchase of two items cost 1300 so the average purchase of one item will cost 1300 / 2 = 650 . its given as total shirt cost 500 $ less . hence total shirt cost = 650 - 250 and total trouser cost = 650 + 250 5 shirts = 400 $ = = > one shirt = 80 $ one trouser = 80 + 20 = 100 $ total trousers = 900 / 100 = 9 . e" | a = 20 + 20
b = 5 * a
c = 1300 - b
d = 20 + 20
e = d + 20
f = c / e
|
a ) 75 men , b ) 80 men , c ) 70 men , d ) 60 men , e ) 50 men | c | subtract(120, 50) | a contractor undertakes to complete the construction of a tunnel 720 meters long in 240 days and employs 50 men for the purpose . after 120 days , he finds that only 240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? | "in 120 days , only 240 m of the tunnel is constructed by 50 men . the remaining 120 days , 480 m of the tunnel can be constructed by 120 men . additional number of men required = 120 - 50 = 70 men . answer : c" | a = 120 - 50
|
a ) 60 , b ) 116 , c ) 210 , d ) 263 , e ) 360 | e | multiply(add(add(add(add(add(add(add(multiply(const_4, 2), subtract(multiply(const_4, 2), const_1)), subtract(multiply(const_4, 2), const_2)), subtract(multiply(const_4, 2), const_3)), subtract(multiply(const_4, 2), const_4)), subtract(multiply(const_4, 2), subtract(multiply(const_4, 2), const_3))), subtract(multiply(const_4, 2), subtract(multiply(const_4, 2), const_2))), subtract(multiply(const_4, 2), subtract(multiply(const_4, 2), const_1))), add(add(add(multiply(2, const_2), subtract(multiply(2, const_2), const_1)), subtract(multiply(2, const_2), const_2)), subtract(multiply(2, const_2), const_3))) | for all integers x > 1 . < x > = 2 x + ( 2 x - 1 ) + ( 2 x - 2 ) + . . . . . . 2 + 1 . what is the value of < 2 > * < 4 > ? | since < x > = 2 x + ( 2 x - 1 ) + ( 2 x - 2 ) + . . . . . . 2 + 1 = 1 + 2 + . . + ( 2 x - 2 ) + ( 2 x - 1 ) + 2 x then < x > is basically the sum of all integers from 1 to 2 x , inclusive . hence < 2 > is the sum of all integers from 1 to 2 * 2 = 4 and < 4 > is the sum of all integers from 1 to 2 * 4 = 8 - > < 4 > = 36 and < 2 > = 10 - - > < 3 > * < 2 > = 21 * 10 = 360 answer : e | a = 4 * 2
b = 4 * 2
c = b - 1
d = a + c
e = 4 * 2
f = e - 2
g = d + f
h = 4 * 2
i = h - 3
j = g + i
k = 4 * 2
l = k - 4
m = j + l
n = 4 * 2
o = 4 * 2
p = o - 3
q = n - p
r = m + q
s = 4 * 2
t = 4 * 2
u = t - 2
v = s - u
w = r + v
x = 4 * 2
y = 4 * 2
z = y - 1
A = x - z
B = w + A
C = 2 * 2
D = 2 * 2
E = D - 1
F = C + E
G = 2 * 2
H = G - 2
I = F + H
J = 2 * 2
K = J - 3
L = I + K
M = B * L
|
['a ) 16 - 4 pi', 'b ) 4 - pi', 'c ) 2 pi - 4', 'd ) 8 pi - 16', 'e ) pi ^ 2 - 4'] | d | add(divide(multiply(const_3, const_pi), const_2), 4) | two identical circles intersect such that their centers and the points at which they intersect form a square of side 4 cm . what is the area of the region where the two circles intersect ? | drawing a diagram , we see that the radius of the circle equals the side of the square . the area of intersection is : pi * r ^ 2 / 4 - ( r ^ 2 - pi * r ^ 2 / 4 ) = pi * r ^ 2 / 2 - r ^ 2 = 8 pi - 16 the answer is d . | a = 3 * math.pi
b = a / 2
c = b + 4
|
a ) 300 , b ) 600 , c ) 800 , d ) 500 , e ) 900 | d | divide(add(150, 25), divide(35, const_100)) | pradeep has to obtain 35 % of the total marks to pass . he got 150 marks and failed by 25 marks . the maximum marks are | "explanation : let their maximum marks be x . then , 35 % of x = 150 + 25 = > 35 / 100 x = 175 x = ( 17500 / 35 ) x = 500 . answer : d" | a = 150 + 25
b = 35 / 100
c = a / b
|
a ) 1 / 33 , b ) 2 / 33 , c ) 1 / 3 , d ) 16 / 33 , e ) 11 / 12 | d | multiply(multiply(multiply(divide(multiply(6, const_2), multiply(6, const_2)), divide(multiply(4, 4), subtract(multiply(6, const_2), const_1))), divide(subtract(multiply(4, 4), const_2), multiply(4, 4))), divide(subtract(subtract(multiply(4, 4), const_2), const_2), subtract(multiply(4, 4), const_1))) | if 4 people are selected from a group of 6 married couples , what is the probability that none of them would be married to each other ? | "each couple can send only one ` ` representative ' ' to the committee . we can choose 4 couples ( as there should be 4 members ) to send only one ` ` representatives ' ' to the committee in c 46 c 64 # of ways . but these 4 chosen couples can send two persons ( either husband or wife ) : 2 β 2 β 2 β 2 = 242 β 2 β 2 β 2 = 24 . so # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is : c 46 β 24 c 64 β 24 . total # of ways to choose 4 people out of 12 is c 412 c 124 . p = c 46 β 24 c 412 = 1633 p = c 64 β 24 c 124 = 1633 answer : d ." | a = 6 * 2
b = 6 * 2
c = a / b
d = 4 * 4
e = 6 * 2
f = e - 1
g = d / f
h = c * g
i = 4 * 4
j = i - 2
k = 4 * 4
l = j / k
m = h * l
n = 4 * 4
o = n - 2
p = o - 2
q = 4 * 4
r = q - 1
s = p / r
t = m * s
|
['a ) 4 % increase', 'b ) 6 % increase', 'c ) 5 % decrease', 'd ) 4 % decrease', 'e ) none of these'] | d | multiply(subtract(const_1, multiply(add(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100)))), const_100) | if the length of a rectangle is increased by 20 % and the breadth is reduced by 20 % , what will be the effect on its area ? | percentage change = x - y - xy β 100 = 20 β 20 β 20 Γ 20 / 100 = β 4 % = 4 % decrease answer d | a = 20 / 100
b = 1 + a
c = 20 / 100
d = 1 - c
e = b * d
f = 1 - e
g = f * 100
|
a ) 6.7 % , b ) 10.3 % , c ) 14.5 % , d ) 18.2 % , e ) 23.6 % | a | divide(multiply(subtract(const_100, 75), multiply(add(divide(const_2, const_0_25), const_2), const_4)), const_100) | each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , forty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 75 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on are incandescent ? | "let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.4 i + 0.8 f = 0.75 ( i + f ) 0.05 f = 0.35 i f = 7 i this means that for every 1 incandescent bulb , there are 7 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.4 i / ( 0.4 i + 0.8 f ) = 0.4 i / ( 0.4 i + 0.8 * 7 i ) = 0.4 i / 6 i = 1 / 15 which is about 6.7 % . the answer is a ." | a = 100 - 75
b = 2 / const_0_25
c = b + 2
d = c * 4
e = a * d
f = e / 100
|
a ) 1 days , b ) 2 days , c ) 3 days , d ) 4 days , e ) 5 days | b | divide(subtract(const_1, add(divide(1, 6), divide(1, 2))), divide(1, 6)) | a and b can do a piece of work in 2 days and 6 days respectively . both work for 1 days and then a goes away . find how long will b take to complete the remaining work ? | 1 / 2 + ( 1 + x ) / 6 = 1 x = 2 days answer : b | a = 1 / 6
b = 1 / 2
c = a + b
d = 1 - c
e = 1 / 6
f = d / e
|
a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many h miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance traveled by tom when karen is 4 miles ahead of him h : t * ( t + 4 / 60 ) = 45 ( 7 / 15 + 4 / 60 ) = 24 miles . d is the correct answer ." | a = 45 / 60
b = a * 4
c = b + 4
d = 60 / 60
e = 45 / 60
f = d - e
g = c / f
h = g - 4
|
a ) rs . 300 , b ) rs . 400 , c ) rs . 480 , d ) rs . 490 , e ) rs . 500 | b | multiply(divide(1210, add(add(divide(5, 4), divide(10, 9)), const_1)), divide(10, 9)) | rs . 1210 were divided among 3 person p , q , r so that p : q = 5 : 4 and q : r = 9 : 10 . then r gets the amount . | "p : q = 5 : 4 , q : r = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9 . so , p : q : r = 5 : 4 : 40 / 9 = 45 : 36 : 40 sum of ratio terms is = ( 45 + 36 + 40 ) = 121 . r share of amount is rs ( 1210 x 40 / 121 ) = rs . 400 . b" | a = 5 / 4
b = 10 / 9
c = a + b
d = c + 1
e = 1210 / d
f = 10 / 9
g = e * f
|
a ) 19 : 10 , b ) 19 : 11 , c ) 11 : 19 , d ) 12 : 18 , e ) 11 : 20 | b | divide(divide(subtract(14.50, 9.75), subtract(14.50, 7)), subtract(const_1, divide(subtract(14.50, 9.75), subtract(14.50, 7)))) | in what ratio should a variety of rice costing rs . 7 per kg be mixed with another variety of rice costing rs . 14.50 per kg to obtain a mixture costing rs . 9.75 per kg ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 14.50 - 9.75 ) / ( 9.75 - 7 ) = 31 / 15 answer : b" | a = 14 - 50
b = 14 - 50
c = a / b
d = 14 - 50
e = 14 - 50
f = d / e
g = 1 - f
h = c / g
|
a ) - 3.5 , b ) - 3.7 , c ) - 5 , d ) - 9 , e ) 11 | b | divide(subtract(multiply(10, 5), 13), subtract(5, multiply(const_3, 5))) | if 10 is added to thrice a number and this sum is multiplied by 5 , the result is the same as if the number is multiplied by 5 and 13 is added to the product . what is the number ? | "let the number be x ; 5 ( 10 + 3 x ) = 5 x + 13 ; x = - 3.7 answer : b" | a = 10 * 5
b = a - 13
c = 3 * 5
d = 5 - c
e = b / d
|
a ) 19 , b ) 20 , c ) 25 , d ) 30 , e ) 64 | b | sqrt(add(390, multiply(5, const_2))) | if a 2 + b 2 + c 2 = 390 and ab + bc + ca = 5 , then a + b + c is | by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 390 and ab + bc + ca = 5 , ( a + b + c ) ^ 2 = 390 + 2 ( 5 ) = 400 = 20 ^ 2 therefore : a + b + c = 20 answer : b | a = 5 * 2
b = 390 + a
c = math.sqrt(b)
|
a ) 630 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1500) | in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1500 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1500 female employees , 60 % of 1500 . = ( 60 * 1500 ) / 100 = 900 . then male employees , = 600 50 % of male are computer literate , = 300 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1500 ) / 100 = 930 computer literate . thus , female computer literate = 930 - 300 = 630 answer : option a" | a = 62 / 100
b = 60 / 100
c = 1 - b
d = 50 / 100
e = c * d
f = a - e
g = f * 1500
|
a ) 22 , b ) 20 , c ) 24 , d ) 26 , e ) 28 | d | power(27, 27) | what will be the remainder when ( 27 ^ 27 ) + 27 is divided by 28 ? | "x ^ n + 1 will be divisible by x + 1 only when n is odd 27 ^ 27 + 1 will be divisible by 27 + 1 ( 27 ^ 27 + 1 ) + 26 , when divided by 28 will give 26 as remainder answer is d" | a = 27 ** 27
|
a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 18 | c | divide(subtract(multiply(48, const_4), 140), const_2) | a man has some hens and cows . if the number of heads be 48 and the number of feet equals 140 , then the number of hens will be | let number of hens = h number of cows = c number of heads = 48 β h + c = 48 β― ( 1 ) number of feet = 140 β 2 h + 4 c = 140 β h + 2 c = 70 β― ( 2 ) ( 2 ) β ( 1 ) gives 2 c β c = 70 β 48 β c = 22 substituting the value of c in ( 1 ) , we get h + 22 = 48 β h = 48 β 22 = 26 h + 22 = 48 β h = 48 β 22 = 26 i . e . , number of hens = 26 answer : c | a = 48 * 4
b = a - 140
c = b / 2
|
a ) 120.25 , b ) 120.35 , c ) 106.5 , d ) 156 , e ) 240 | a | subtract(120.50, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 120.50 , brokerage being 1 / 4 % is | "explanation : cash realised = rs . ( 120.50 - 0.25 ) = rs . 120.25 . answer : a" | a = 1 / 4
b = 120 - 50
|
a ) 77 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23 | a | divide(multiply(add(add(7, const_3), const_2), divide(7, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 7 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 7 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let bucket a be a and bucket b be b scenario 1 a - 7 = 1 / 3 ( b + 7 ) - - - - > 3 a - 21 = b + 7 scenario 2 b - 7 = 1 / 2 ( a + 7 ) - - - - - > 2 b - 14 = a + 7 from scenario 1 , b = 3 a - 28 substitute b with this information in stmt 2 2 ( 3 a - 28 ) - 14 = a + 7 - - - - - - > 6 a - 56 - 14 = a + 7 - - - - - - > 6 a - a = 70 + 7 - - - > 5 a = 77 a = 77 / 5 , answer choice a" | a = 7 + 3
b = a + 2
c = 7 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
|
a ) 20 , b ) 25 , c ) 30 , d ) 32 , e ) 35 | c | subtract(50, multiply(divide(50, const_100), 5)) | how many liters of water must be evaporated from 50 liters of a 2 percent sugar solution to get a 5 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.02 ( 50 ) = 0.05 ( 50 - x ) 0.05 x = 2.5 - 1 x = 1.5 / 0.05 = 30 liters the answer is c ." | a = 50 / 100
b = a * 5
c = 50 - b
|
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