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a ) 1000 , b ) 3000 , c ) 1225 , d ) 6000 , e ) 9000 | c | divide(multiply(50, subtract(50, const_1)), const_2) | there are 50 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "50 players are there . two players play one game with one another . so 50 c 2 = 50 * 49 / 2 = 1225 so option c is correct" | a = 50 - 1
b = 50 * a
c = b / 2
|
a ) 12500 , b ) 37500 , c ) 29977 , d ) 26777 , e ) 19871 | a | divide(subtract(multiply(100000, divide(add(9, divide(1, 4)), const_100)), multiply(100000, divide(9, const_100))), subtract(divide(11, const_100), divide(9, const_100))) | an amount of rs . 100000 is invested in two types of shares . the first yields an interest of 9 % p . a and the second , 11 % p . a . if the total interest at the end of one year is 9 1 / 4 % , then the amount invested at 11 % was ? | "let the sum invested at 9 % be rs . x and that invested at 11 % be rs . ( 100000 - x ) . then , ( x * 9 * 1 ) / 100 + [ ( 100000 - x ) * 11 * 1 ] / 100 = ( 100000 * 37 / 4 * 1 / 100 ) ( 9 x + 1100000 - 11 x ) = 37000 / 4 = 925000 x = 87500 sum invested at 9 % = rs . 87500 sum invested at 11 % = rs . ( 100000 - 87500 ) = rs . 12500 . answer : a" | a = 1 / 4
b = 9 + a
c = b / 100
d = 100000 * c
e = 9 / 100
f = 100000 * e
g = d - f
h = 11 / 100
i = 9 / 100
j = h - i
k = g / j
|
a ) a ) 45 , b ) b ) 33 , c ) c ) 48 , d ) d ) 55 , e ) e ) 61 | c | multiply(subtract(divide(multiply(24, 4), 3), 24), divide(add(subtract(35, 25), 50), subtract(35, 25))) | a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late . had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late . what is the speed y of the train . | "let y be the balance distance to be covered and x be the former speed . a train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late so , y / ( 3 x / 4 ) - y / x = 35 / 60 4 y / 3 x - y / x = 7 / 12 y / x ( 4 / 3 - 1 ) = 7 / 12 y / x * 1 / 3 = 7 / 12 y / x = 7 / 4 4 y - 7 x = 0 . . . . . . . . 1 had the accident occurred 24 km farther , it would have reached the destination only 25 minutes late so , ( y - 24 ) / ( 3 x / 4 ) - ( y - 24 ) / x = 25 / 60 4 ( y - 24 ) / 3 x - ( y - 24 ) / x = 5 / 12 ( y - 24 ) / x ( 4 / 3 - 1 ) = 5 / 12 ( y - 24 ) / x * 1 / 3 = 5 / 12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96 . . . . . . . 2 eq 2 - eq 1 2 x = 96 x = 48 = y ans = c" | a = 24 * 4
b = a / 3
c = b - 24
d = 35 - 25
e = d + 50
f = 35 - 25
g = e / f
h = c * g
|
a ) 77 days , b ) 55 days , c ) 12 days , d ) 33 days , e ) 11 days | c | divide(multiply(10, const_3), subtract(divide(add(divide(multiply(10, const_3), 15), add(divide(multiply(10, const_3), 10), divide(multiply(10, const_3), 5))), const_2), divide(multiply(10, const_3), 10))) | a and b can do a piece of work in 10 days , b and c in 5 days , c and a in 15 days . how long will c take to do it ? | 2 c = 1 / 5 + 1 / 15 – 1 / 10 = 1 / 6 c = 1 / 12 = > 12 days answer : c | a = 10 * 3
b = 10 * 3
c = b / 15
d = 10 * 3
e = d / 10
f = 10 * 3
g = f / 5
h = e + g
i = c + h
j = i / 2
k = 10 * 3
l = k / 10
m = j - l
n = a / m
|
a ) 10 , b ) 100 , c ) 999 , d ) 99 , e ) 9 | d | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | when 1 / 10 percent of 1,000 is subtracted from 1 / 10 of 1,000 , the difference is | "( 1 / 10 ) * 1000 - ( 1 / 10 ) % * 1000 = 100 - ( 1 / 1000 ) * 1000 = 100 - 1 = 99 the answer is d ." | a = 3 + 4
b = a * 1000
c = 1 / 10
d = b * c
e = 1 / 10
f = e / 100
g = 3 + 4
h = g * 1000
i = f * h
j = d - i
|
['a ) 14 cms', 'b ) 21 cms', 'c ) 42 cms', 'd ) none of these', 'e ) 56 cms'] | b | multiply(divide(divide(divide(divide(multiply(3.78, const_100), const_2), const_3), const_3), divide(divide(divide(divide(multiply(3.78, const_100), const_2), const_3), const_3), const_3)), divide(divide(divide(divide(multiply(3.78, const_100), const_2), const_3), const_3), const_3)) | a rectangular courtyard 3.78 meters long 5.25 meters wide is to be paved exactly with square tiles , all of the same size . what is the largest size of the tile which could be used for the purpose ? | explanation : 3.78 meters = 378 cm = 2 × 3 × 3 × 3 × 7 5.25 meters = 525 cm = 5 × 5 × 3 × 7 hence common factors are 3 and 7 hence lcm = 3 × 7 = 21 hence largest size of square tiles that can be paved exactly with square tiles is 21 cm . answer : b | a = 3 * 78
b = a / 2
c = b / 3
d = c / 3
e = 3 * 78
f = e / 2
g = f / 3
h = g / 3
i = h / 3
j = d / i
k = 3 * 78
l = k / 2
m = l / 3
n = m / 3
o = n / 3
p = j * o
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number e of integers that can be negative ? | "minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning 5 out of 6 also possible as you will see e = 5 or 1 will give you same equation . . ans d" | a = 10 / 2
b = a + 0
|
a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares . | a | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 1.3), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 1.3 hectares in ares | "1.3 hectares in ares 1 hectare = 100 ares therefore , 1.3 hectares = 1.3 × 100 ares = 130 ares . answer - a" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 1
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 10.5 % , b ) 12.5 % , c ) 16.7 % , d ) 25 % , e ) 30 % | c | multiply(divide(10, subtract(const_100, 40)), const_100) | on a certain road , 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 40 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 10 % of them exceeded the speed limit and received the ticket , i . e . x / 10 . again , suppose total no . of motorists who exceeded the speed limit are y . 40 % of y exceeded the speed limit but did n ' t received the ticket , i . e . 2 y / 5 . it means 3 y / 5 received the ticket . hence , 3 y / 5 = x / 10 or y / x = 1 / 6 or y / x * 100 = 1 / 6 * 100 = 16.7 % c" | a = 100 - 40
b = 10 / a
c = b * 100
|
a ) 18.75 , b ) 19.75 , c ) 20.75 , d ) 33 , e ) 22.75 | d | subtract(multiply(divide(const_100, 54), divide(multiply(60, 84), const_100)), 60) | there is 60 lit of milk and water in which milk forms 84 % . howmuch water must be added to this solution to make it solution in which milk forms 54 % | 60 * 84 / 100 = 50.40 lit milk that is 9.60 lit water let x lit water will be added then ( 60 + x ) * 54 / 100 = 50.40 so x = 33.33 answer : d | a = 100 / 54
b = 60 * 84
c = b / 100
d = a * c
e = d - 60
|
a ) 54 , b ) 51 , c ) 58 , d ) 55 , e ) 52 | c | add(40, divide(subtract(998, multiply(14, 40)), divide(multiply(14, add(const_100, 75)), const_100))) | a certain bus driver is paid a regular rate of $ 14 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 $ 998 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 14 = 560 excess = 998 - 560 = 438 for extra hours = . 75 ( 14 ) = 10.5 + 14 = 24.5 number of extra hrs = 438 / 24.5 = 17.8 = 18 approx . total hrs = 40 + 18 = 58 answer c" | a = 14 * 40
b = 998 - a
c = 100 + 75
d = 14 * c
e = d / 100
f = b / e
g = 40 + f
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 2 | e | subtract(6, reminder(3, 8)) | when n is divided by 24 , the remainder is 6 . what is the remainder when 3 n is divided by 8 ? | "let n = 6 ( leaves a remainder of 6 when divided by 24 ) 3 n = 3 ( 6 ) = 18 , which leaves a remainder of 2 when divided by 8 . answer e" | a = 6 - reminder
|
a ) 442.5 , b ) 450 , c ) 465 , d ) 468 , e ) 475 | a | divide(add(multiply(20, 410), multiply(add(410, 65), subtract(40, 20))), 40) | a certain debt will be paid in 40 installments from january 1 to december 31 of a certain year . each of the first 20 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 20 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? | "total number of installments = 40 payment per installment for the first 20 installments = 410 payment per installment for the remaining 20 installments = 410 + 65 = 475 average = ( 20 * 410 + 20 * 475 ) / 40 = 442.50 answer a" | a = 20 * 410
b = 410 + 65
c = 40 - 20
d = b * c
e = a + d
f = e / 40
|
a ) 200 , b ) 177 , c ) 188 , d ) 166 , e ) 129 | a | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 20)), 1000), multiply(divide(const_100, add(const_100, 20)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 20)), 1000), multiply(divide(const_100, add(const_100, 20)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 20 % profit on the first article and 20 % loss on second article . find the net profit or loss percent ? | "profit on first article = 20 % of 1000 = 200 . answer : a" | a = 100 - 20
b = 100 / a
c = b * 1000
d = 100 + 20
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 20
l = 100 / k
m = l * 1000
n = 100 + 20
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 50 , b ) 20 , c ) 30 , d ) 100 , e ) 15 | d | divide(multiply(20, 20), 4) | 20 men do a work in 20 days . how many men are needed to finish the work in 4 days ? | "men required to finish the work in 4 days = 20 * 20 / 4 = 100 answer is d" | a = 20 * 20
b = a / 4
|
a ) 50 % , b ) 100 % , c ) 25 % , d ) 15 % , e ) 60 % | b | multiply(divide(10, 10), const_100) | the cost of 20 articles is equal to selling price of 10 . find the gain or loss percent ? | c . p . of each article be $ 1 c . p . of 10 articles = 10 gain % = 10 / 10 * 100 = 100 % answer is b | a = 10 / 10
b = a * 100
|
a ) 150000 , b ) 16000 , c ) 15000 , d ) 190000 , e ) 13000 | e | divide(6500, multiply(divide(2, 3), divide(3, 4))) | a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 6500 rs , what is the value of buzness ? | "if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 6500 - > x = 13000 answer : e" | a = 2 / 3
b = 3 / 4
c = a * b
d = 6500 / c
|
a ) $ 30 , b ) $ 54 , c ) $ 48.32 , d ) $ 44.10 , e ) $ 9.60 | c | multiply(7.35, 6) | johnny makes $ 7.35 per hour at his work . if he works 6 hours , how much money will he earn ? | 4.75 * 6 = 28.50 . answer is c . | a = 7 * 35
|
a ) rs . 50 , b ) rs . 70 , c ) rs . 100 , d ) rs . 80 , e ) rs . 60 | a | multiply(200, divide(25, const_100)) | find the 25 % of rs . 200 . | explanation : 25 % of 200 = > 25 / 100 * 200 = rs . 50 answer : a | a = 25 / 100
b = 200 * a
|
a ) 60 , b ) 120 , c ) 240 , d ) 360 , e ) 720 | c | lcm(lcm(add(const_10, const_2), subtract(multiply(const_3, const_10), const_3)), 16) | what is the least common multiple of 15 , 16 , and 24 ? | "let us first write the numbers in the form of prime factors : 15 = 3 * 5 16 = 2 ^ 4 24 = 2 * 17 ^ 1 the lcm would be the largest powers of the prime numbers from all these three numbers . hence lcm = 240 option c" | a = 10 + 2
b = 3 * 10
c = b - 3
d = math.lcm(a, c)
e = math.lcm(d, 16)
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(800, const_10) | how many integers from 101 to 800 , inclusive , remains the value unchanged when the digits were reversed ? | "first digit possibilities - 1 through 7 = 7 8 is not possible here because it would result in a number greater than 8 ( i . e 808 , 818 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 7 * 10 = 70 answer is c ." | a = 800 / 10
|
a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 19 | e | divide(subtract(multiply(add(14, 25), divide(70, const_100)), 14), divide(70, const_100)) | a bowl of fruit contains 14 apples and 25 oranges . how many oranges must be removed so that 70 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 25 let number of oranges that must be removed so that 70 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 14 + ( 25 - x ) = 39 - x 14 / ( 39 - x ) = 7 / 10 = > 20 = 39 - x = > x = 19 answer e" | a = 14 + 25
b = 70 / 100
c = a * b
d = c - 14
e = 70 / 100
f = d / e
|
a ) 90 % , b ) 99 % , c ) 100 % , d ) 101 % , e ) 110 % | b | multiply(10, 10) | on july 1 of last year , total employees at company e was decreased by 10 percent . without any change in the salaries of the remaining employees , the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease . the total of the combined salaries of all the employees at company e after july 1 last year was what percent r of thatbeforejuly 1 last year ? | the total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before r = [ 0.9 n ( 1.1 x ) ] / xn = 0.99 or 99 % . b | a = 10 * 10
|
a ) 98.5 kgs , b ) 80 kgs , c ) 76.5 kgs , d ) 67.5 kgs , e ) 58.2 kgs | b | divide(multiply(add(const_1, 3), 140), 7) | 3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 140 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 140 / 7 = 80 kgs answer : b" | a = 1 + 3
b = a * 140
c = b / 7
|
a ) 2 : 4 , b ) 3 : 6 , c ) 4 : 12 , d ) 4 : 9 , e ) 1 : 3 | d | power(divide(8, 27), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 8 : 27 what is the ratio of their total surface areas ? | "ratio of the sides = â ³ â ˆ š 8 : â ³ â ˆ š 27 = 2 : 3 ratio of surface areas = 4 : 9 answer : d" | a = 8 / 27
b = 1 / 3
c = a ** b
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | c | add(divide(subtract(21, 3), 3), const_2) | what is the greatest of 3 consecutive integers whose sum is 21 ? | "the sum of three consecutive integers can be written as n + ( n + 1 ) + ( n + 2 ) = 3 n + 3 if the sum is 24 , we need to solve the equation 3 n + 3 = 21 ; = > 3 n = 18 ; = > n = 6 the greatest of the three numbers is therefore 6 + 2 = 8 answer : c" | a = 21 - 3
b = a / 3
c = b + 2
|
a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 5 : 2 | e | divide(multiply(45000, const_12), multiply(24000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 24000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 × 12 : 24000 × 9 = 45 × 12 : 24 × 9 = 15 × 12 : 8 × 9 = 5 : 2 answer is e ." | a = 45000 * 12
b = 4 + 3
c = 24000 * b
d = a / c
|
a ) 4 : 1 , b ) 1 : 4 , c ) 3 : 2 , d ) 2 : 3 , e ) 2 : 1 | e | divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), divide(multiply(multiply(multiply(const_3, const_2), const_100), const_100), multiply(add(const_2, const_3), const_2))) | if a and b get profits of rs . 4,000 and rs . 2,000 respectively at the end of year then ratio of their investments are | "ratio = 4000 / 2000 = 2 : 1 answer : e" | a = 3 * 2
b = a * 100
c = b * 100
d = 3 * 2
e = d * 100
f = e * 100
g = 2 + 3
h = g * 2
i = f / h
j = c / i
|
a ) $ 15,360 , b ) $ 17,360 , c ) $ 18,000 , d ) $ 21,360 , e ) $ 29,160 | e | add(multiply(const_100, 3), const_60) | a certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360 . what amount of total annual earnings would result in a 45 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) | "for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12000 - 0.45 * 12000 = 12000 - 5400 x = 6600 earning becomes 9360 + 3 x = 9360 + 3 * 6600 = 29160 ans : e" | a = 100 * 3
b = a + const_60
|
a ) 48 hours , b ) 51 hours , c ) 36 hours , d ) 56 hours , e ) none | d | add(divide(210, add(8, 2)), divide(210, subtract(8, 2))) | speed of a boat in standing water is 8 kmph and the speed of the stream is 2 kmph . a man rows to place at a distance of 210 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 6 kmph ; speed downstream = 10 kmph . ∴ total time taken = [ 210 / 6 + 210 / 10 ] hours = 56 hours . answer d" | a = 8 + 2
b = 210 / a
c = 8 - 2
d = 210 / c
e = b + d
|
a ) 550 , b ) 1000 , c ) 2500 , d ) 300 , e ) 4000 | b | divide(multiply(6800, 20), 136) | find the annual income derived by investing $ 6800 in 20 % stock at 136 . | "by investing $ 136 , income obtained = $ 20 . by investing $ 6800 , income obtained = $ [ ( 20 / 136 ) * 6800 ] = $ 1000 . answer b ." | a = 6800 * 20
b = a / 136
|
a ) 741 / 357 , b ) 456 / 789 , c ) 636 / 735 , d ) 564 / 263 , e ) - 636 / 735 | c | subtract(multiply(7, divide(12, 5)), power(divide(6, 7), 7)) | if x = - 12 / 5 and y = - 6 / 7 , what is the value of the expression - 2 x – y ^ 2 ? | "x = - 12 / 15 and y = - 6 / 7 = = > - 2 ( - 12 / 15 ) - ( - 6 / 7 ) ^ 2 = 24 / 15 - 36 / 49 = 636 / 735 ans : c" | a = 12 / 5
b = 7 * a
c = 6 / 7
d = c ** 7
e = b - d
|
a ) 5 kmph , b ) 6 kmph , c ) 7 kmph , d ) 4 kmph , e ) 9 kmph | a | divide(subtract(16, 6), const_2) | a man can row his boat with the stream at 16 km / h and against the stream in 6 km / h . the man ' s rate is ? | "explanation : ds = 16 us = 6 s = ? s = ( 16 - 6 ) / 2 = 5 kmph answer : a" | a = 16 - 6
b = a / 2
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a ) rs . 880 , b ) rs . 890 , c ) rs . 1200 , d ) rs . 900 , e ) none | c | divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 4), const_100) | if the compound interest on a certain sum of money for 4 years at 10 % per annum be rs . 993 , what would be the simple interest ? | "let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 4 yrs = ( 3000 x 10 x 4 ) / 100 = 1200 / - answer : c ." | a = 3 * 0
b = a * 10
c = b * 10
d = c * 4
e = d / 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(4, reminder(16, 7)) | when n is divided by 27 , the remainder is 4 . what is the remainder when n + 16 is divided by 7 ? | "assume n = 23 remainder ( n / 27 ) = 4 n + 16 = 39 remainder ( 39 / 7 ) = 4 option c" | a = 4 - reminder
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a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 13 % | b | multiply(divide(divide(16, const_2), add(16, const_100)), const_100) | jar a has 16 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ? | an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 116 marbles . now , for both jars to have equal marbles we should move 8 marbles from a to b , which is 8 / 116 = ~ 7 % of a . answer : b . | a = 16 / 2
b = 16 + 100
c = a / b
d = c * 100
|
a ) 7600 , b ) 7700 , c ) 7800 , d ) 7900 , e ) 8000 | e | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 14,000), multiply(14,000, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 21,000 and rs . 28,000 respectively . out of a total profit of rs . 14,000 , b ’ s share is : | "ratio of their shares = 21000 : 28000 = 3 : 4 . b ’ s share = rs . 14000 * 4 / 7 = rs . 8000 answer : e" | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 14
f = 3 * 3
g = f + 1
h = 14 * 0
i = e / h
|
a ) 20 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | d | multiply(divide(subtract(subtract(add(const_1, divide(75, const_100)), const_1), divide(5, const_100)), add(const_1, divide(75, const_100))), const_100) | a merchant marks goods up by 75 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 5 % . what % discount did the merchant offer ? | "let p be the original price of the goods and let x be the rate after the markup . ( 1.75 p ) * x = 1.05 p x = 1.05 / 1.75 = 0.6 which is a discount of 40 % . the answer is d ." | a = 75 / 100
b = 1 + a
c = b - 1
d = 5 / 100
e = c - d
f = 75 / 100
g = 1 + f
h = e / g
i = h * 100
|
a ) 2998 , b ) 2799 , c ) 2890 , d ) 1683 , e ) 2780 | d | multiply(volume_cylinder(divide(3, const_2), 14), 17) | find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 17 per cubic meter ? | "22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 17 = 1683 answer : d" | a = 3 / 2
b = volume_cylinder * (
|
a ) - 180 , b ) 6530 , c ) 6630 , d ) 6730 , e ) 6830 | a | subtract(negate(6430), multiply(subtract(1320, 1070), divide(subtract(1320, 1070), subtract(1370, 1320)))) | 1370 , 1320 , 1070 , x , - 6430 | "1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 answer : a ." | a = negate - (
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a ) 12 , b ) - 12 , c ) 1 / 12 , d ) - 1 / 12 , e ) 3 | a | divide(add(27, 18), subtract(divide(27, 4), 3)) | if ( 27 / 4 ) x - 18 = 3 x + 27 what is the value of x ? | ( 27 / 4 ) x - 18 = 3 x + 27 = > 27 x - 72 = 12 x + 108 = > 15 x = 180 = > x = 12 answer : a | a = 27 + 18
b = 27 / 4
c = b - 3
d = a / c
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a ) 53 , b ) 53.5 , c ) 54.68 , d ) 55.6 , e ) 67.2 | c | divide(add(add(multiply(55, 50), multiply(60, 55)), multiply(45, 60)), add(60, add(55, 50))) | if the average marks of 3 classes of 55 , 60 and 45 students respectively is 50 , 55 , 60 , then find the average marksof all the students is | required average = 55 x 50 + 60 x 55 + 45 x 60 55 + 60 + 45 = 2750 + 3300 + 2700 160 = 8750 160 = 54.68 c | a = 55 * 50
b = 60 * 55
c = a + b
d = 45 * 60
e = c + d
f = 55 + 50
g = 60 + f
h = e / g
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a ) 179 , b ) 208 , c ) 210 , d ) 223 , e ) 248 | e | divide(add(150, subtract(multiply(63, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 63 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | explanation : total runs scored by the batsman = 63 * 46 = 2898 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2898 – 2552 = 346 runs . let the highest score be x , hence the lowest score = x – 150 x + ( x - 150 ) = 346 2 x = 496 x = 248 runs answer e | a = 63 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
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a ) 107 , b ) 108 , c ) 130 , d ) 150 , e ) 145 | c | divide(multiply(multiply(10, 13), 5), 5) | a lady builds 10 cm length , 13 cm width , and 5 cm height box using 5 cubic cm cubes . what is the minimum number of cubes required to build the box ? | "number of cubes required = volume of box / volume of cube = 10 * 13 * 5 / 5 = 130 cubes answer : c" | a = 10 * 13
b = a * 5
c = b / 5
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a ) 8 , b ) 6 , c ) 5 , d ) 15 , e ) 1 | d | divide(275, multiply(add(60, 6), const_0_2778)) | a train 275 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 275 * 3 / 55 = 15 sec . answer : d" | a = 60 + 6
b = a * const_0_2778
c = 275 / b
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a ) 285 , b ) 305 , c ) 315 , d ) 350 , e ) 245 | b | add(multiply(multiply(5, 6), const_10), 2) | when 2 + 3 = 65 , 3 + 4 = 125 , 4 + 5 = 205 , then 5 + 6 = ? | "2 + 3 = > 2 ã — 3 = 6 = > 6 ã — 10 + 5 = 65 3 + 4 = > 3 ã — 4 = 12 = > 12 ã — 10 + 5 = 125 4 + 5 = > 4 ã — 5 = 20 = > 20 ã — 10 + 5 = 205 then 5 + 6 = > 5 ã — 6 = 30 = > 30 ã — 10 + 5 = 305 answer : b" | a = 5 * 6
b = a * 10
c = b + 2
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a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 6 , e ) 2 and 8 | d | divide(subtract(multiply(22, 2), 31), 2) | a certain experimental mathematics program was tried out in 2 classes in each of 22 elementary schools and involved 31 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and greatest possible values of n , respectively , are | "one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 22 * 2 = 44 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one class ( 44 - 31 = 13 ) . thus we get a number of the available extra - classes for teachers , and all that we need is just to count how many teachers could take 2 more classes , which is 13 / 2 = 6.5 . so the greatest possible value of the # of teachers who had 3 classes is 6 . only answer d has this option ." | a = 22 * 2
b = a - 31
c = b / 2
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a ) 28 days . , b ) 16 days . , c ) 12 days . , d ) 24 days . , e ) 36 days . | e | inverse(divide(inverse(24), add(const_2, const_1))) | a is twice as good a work man as b and together they finish a piece of work in 24 days . the number of days taken by a alone to finish the work is : | "solution ( a ’ s 1 day ’ s work ) : ( b ’ s 1 day ’ s work ) = 2 : 1 . ( a + b ) ' s 1 day ’ s work = 1 / 24 divide 1 / 14 in the ratio 2 : 1 . ∴ a ’ s 1 day ’ s work = ( 1 / 24 x 2 / 3 ) = 1 / 36 hence , a alone can finish the work in 36 days . answer e" | a = 1/(24)
b = 2 + 1
c = a / b
d = 1/(c)
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a ) 94.19 , b ) 94.12 , c ) 93.5 , d ) 64.1 , e ) 64.11 | c | subtract(subtract(multiply(1000, power(add(divide(12, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(12, const_100)), 4)) | what will be the difference between simple and compound interest at 12 % per annum on a sum of rs . 1000 after 4 years ? | "s . i . = ( 1000 * 12 * 4 ) / 100 = rs . 480 c . i . = [ 1000 * ( 1 + 12 / 100 ) 4 - 1000 ] = rs . 573.5 difference = ( 573.5 - 480 ) = rs . 93.5 answer : c" | a = 12 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 12 / 100
g = 1000 * f
h = g * 4
i = e - h
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a ) 40 sec , b ) 52 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec | b | divide(360, multiply(subtract(45, 290), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 290 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 290 = 650 m required time = 650 * 2 / 25 = 52 sec answer : b" | a = 45 - 290
b = a * const_0_2778
c = 360 / b
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a ) 6 , b ) 106 , c ) 206 , d ) 306 , e ) 406 | d | multiply(subtract(multiply(subtract(multiply(3, 7), 3), 3), 3), subtract(7, const_1)) | q ' = 3 q - 3 , what is the value of ( 7 ' ) ' ? | "( 7 ' ) ' = ( 3 * 7 - 3 ) ' = 18 ' = 18 * 18 - 18 = 306 answer d" | a = 3 * 7
b = a - 3
c = b * 3
d = c - 3
e = 7 - 1
f = d * e
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a ) 2 : 0 , b ) 2 : 3 , c ) 2 : 1 , d ) 8 : 9 , e ) 2 : 8 | d | divide(subtract(6.30, 5.50), subtract(7.20, 6.30)) | find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.50 a kg to produce a mixture worth rs . 6.30 a kg | "by the rule of alligation : cost of 1 kg rice of 1 st kind cost of 1 kg rice of 2 nd kind required ratio = 80 : 90 = 8 : 9 answer : d" | a = 6 - 30
b = 7 - 20
c = a / b
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a ) s . 1200 , b ) s . 5400 , c ) s . 5400 , d ) s . 4999 , e ) s . 5000 | a | subtract(multiply(add(5, const_1), 5600), add(add(add(add(5400, 9000), 6300), 7200), 4500)) | a grocer has a sale of rs . 5400 , rs . 9000 , rs . 6300 , rs . 7200 and rs . 4500 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5600 ? | "total sale for 5 months = rs . ( 5400 + 9000 + 6300 + 7200 + 4500 ) = rs . 32400 . required sale = rs . [ ( 5600 x 6 ) - 32400 ] = rs . ( 33600 - 32400 ) = rs . 1200 . option a" | a = 5 + 1
b = a * 5600
c = 5400 + 9000
d = c + 6300
e = d + 7200
f = e + 4500
g = b - f
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a ) 20 , b ) 50 , c ) 60 , d ) 80 , e ) 100 | a | divide(subtract(16000, 12000), 200) | marginal cost is the cost of increasing the quantity produced ( or purchased ) by one unit . if the fixed cost for n products is $ 12000 and the marginal cost is $ 200 , and the total cost is $ 16000 , what is the value of n ? | total cost for n products = fixed cost for n products + n * marginal cost - - > $ 16,000 = $ 12,000 + n * $ 200 - - > n = 20 . answer : a . | a = 16000 - 12000
b = a / 200
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a ) $ 212 , b ) $ 216 , c ) $ 220 , d ) $ 224 , e ) $ 228 | d | multiply(divide(392, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | $ 392 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | "let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 392 x = 56 4 x = 224 the answer is d ." | a = 1 / 2
b = a + 1
c = b + 2
d = 392 / c
e = d * 2
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a ) 9 , b ) 10 , c ) 12 , d ) 15 , e ) 18 | e | multiply(divide(multiply(subtract(60, 15), divide(6, const_60)), 15), const_60) | the cyclist walking at a constant rate of 15 miles per hour is passed by a car traveling in the same direction along the same path at 60 miles per hour . the car stops to wait for the cyclist for 6 minutes after passing her , while the cyclist continues to go at her constant rate , how many minutes must the car wait until the cyclist catches up ? | for the 6 minutes the car continues to overtake the cyclist , she is going at 45 miles per hour faster than the cyclist . once the car stops , the cyclist is going at 15 miles per hour while the car is at rest so the amount of time the cyclist will take to cover the distance between them is going to be in the ratio of the relative speeds . 45 / 15 * 6 or 18 minutes answer is ( e ) | a = 60 - 15
b = 6 / const_60
c = a * b
d = c / 15
e = d * const_60
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a ) 0.5 , b ) 4.63 , c ) 4.9 , d ) 7.7 , e ) 49.1 | c | divide(floor(multiply(divide(multiply(0.889, 55), 9.97), const_10)), const_10) | the value for d = ( 0.889 × 55 ) / 9.97 to the nearest tenth is | if we read the q , we can easily home on to the answer , , the numerator is clearly between 40 and 50 . . denomiator is close to 10 . . so d = ( 0.889 × 55 ) / 9.97 is between 4 and 5 . . it may tempt us to solve it since ther are two values between 4 and 5 . . but the catch is innearest tenth 4.63 can be called nearest to hundreth and no tenth , so can be eliminated . . 4.9 is our answer . . . c | a = 0 * 889
b = a / 9
c = b * 10
d = math.floor(c)
e = d / 10
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a ) 259 , b ) 261 , c ) 263 , d ) 265 , e ) 267 | b | add(2, lcm(37, 7)) | find the least number which when divided by 37 and 7 leaves a remainder of 2 in each case . | the least number which when divided by different divisors leaving the same remainder in each case = lcm ( different divisors ) + remainder left in each case . hence the required least number = lcm ( 37 , 7 ) + 2 = 261 . answer : b | a = math.lcm(37, 7)
b = 2 + a
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['a ) 1 / 2', 'b ) 1 / 3', 'c ) 1 / 4', 'd ) 1 / 5', 'e ) 1 / 6'] | c | divide(16, volume_cube(sqrt(16))) | a cubical tank is filled with water to a level of 1 foot . if the water in the tank occupies 16 cubic feet , to what fraction of its capacity is the tank filled with water ? | the volume of water in the tank is h * l * b = 16 cubic feet . since h = 1 , then l * b = 16 and l = b = 4 . since the tank is cubical , the capacity of the tank is 4 * 4 * 4 = 64 . the ratio of the water in the tank to the capacity is 16 / 64 = 1 / 4 the answer is c . | a = math.sqrt(16)
b = 16 / volume_cube
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a ) 723 % , b ) 156 % , c ) 316 % , d ) 37 % , e ) 29 % | c | multiply(divide(subtract(5, divide(2, 5)), divide(2, 5)), const_100) | by approximately what percent is x greater than 2 / 5 if ( 3 / 5 ) ( x ) = 1 ? | "what percent is x greater than 2 / 5 if ( 3 / 5 ) ( x ) = 1 ? = > x = 5 / 3 % change = [ ( 5 / 3 - 2 / 5 ) / ( 2 / 5 ) ] * 100 = 316.5 % = 316 % approx ans , c" | a = 2 / 5
b = 5 - a
c = 2 / 5
d = b / c
e = d * 100
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a ) 10 % , b ) 15 % , c ) 20 % , d ) 28 % , e ) 50 % | d | subtract(multiply(multiply(subtract(const_1, divide(20, const_100)), add(const_1, divide(60, const_100))), const_100), const_100) | a furniture store owner decided to drop the price of her recliners by 20 % to spur business . by the end of the week she had sold 60 % more recliners . what is the percentage increase of the gross ? | "say a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 % , she will earn $ 80 per recliner and she sells 60 % more ie . , 160 recliners hence her sales tields 160 * 80 = $ 12800 increase in sales = 12800 - 10000 = $ 2800 so % increase = 2800 * 100 / 10000 = 28 % d is the answer" | a = 20 / 100
b = 1 - a
c = 60 / 100
d = 1 + c
e = b * d
f = e * 100
g = f - 100
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a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | b | subtract(add(10, add(25, 5)), 5) | 10 play kabadi , 25 play kho kho only , 5 play both gmaes . then how many in total ? | "10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 25 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 25 + 5 = 30 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 30 - 5 = 35 answer : b" | a = 25 + 5
b = 10 + a
c = b - 5
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a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 18 | d | divide(power(8, const_2), 4) | if the area of a square with sides of length 8 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 8 ^ 2 = l * 4 = > l = 64 / 4 = 16 answer d" | a = 8 ** 2
b = a / 4
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a ) $ 19,250 , b ) $ 18,500 , c ) $ 21,000 , d ) $ 15,850 , e ) $ 12,300 | c | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | the average salary of 15 people in the shipping department at a certain firm is $ 21,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 21 k = 315 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 6 emp sal = 315 k - 125 k - 64 k = 126 k average = 126 k / 6 = 21000 ans : c" | a = 5 * 4
b = 4 * 4
c = a * b
d = 5 * 5
e = d * 5
f = c - e
g = 4 * 15
h = f - g
i = 2 + 5
j = h / i
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a ) 40 min , b ) 25 min , c ) 54 min , d ) 65 min , e ) 58 min | c | multiply(divide(multiply(3, add(6, divide(40, const_60))), 18), const_60) | walking at the rate of 3 kmph a man cover certain distance in 6 hr 40 min . running at a speed of 18 kmph the man will cover the same distance in . | "distance = speed * time 3 * 20 / 3 = 20 km new speed = 18 kmph therefore time = 20 / 18 = 10 / 9 = 54 min answer : c ." | a = 40 / const_60
b = 6 + a
c = 3 * b
d = c / 18
e = d * const_60
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['a ) 15', 'b ) 30', 'c ) 45', 'd ) 75', 'e ) 90'] | d | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 5) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of thesportformulation contains 5 ounces of corn syrup , how many ounces of water does it contain ? | standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 5 . so , x = 325 / 4 ounces . water = ( 60 / 65 ) * ( 325 / 4 ) = 60 * 3 / 4 = 75 ounces . ans d | a = 1 / 12
b = a * 3
c = 1 / 30
d = c / 2
e = b / d
f = e * 5
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a ) 3 , b ) 2 , c ) 1 / 2 , d ) 1 / 3 , e ) there is n ' t enough data to answer the question . | d | divide(const_1, const_3) | two brothers took the gmat exam , the higher score is u and the lower one is v . if the difference between the two scores is equal to their average , what is the value of v / u ? | answer is d : 1 / 3 u - v = ( u + v ) / 2 solving for v / u = 1 / 3 d | a = 1 / 3
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a ) 7.16 , b ) 7.16 , c ) 7.12 , d ) 7.35 , e ) 7.11 | d | divide(add(131, 165), multiply(add(80, 65), const_0_2778)) | two trains 131 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 131 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.35 answer : d" | a = 131 + 165
b = 80 + 65
c = b * const_0_2778
d = a / c
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | b | divide(2, 6) | in the rectangular coordinate system , if the line x = 6 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ? | "x = 6 y + 5 , and thus y = x / 6 - 5 / 6 the slope is 1 / 6 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) / ( m + 2 - m ) = p / 2 p / 2 = 1 / 6 and thus p = 1 / 3 the answer is b ." | a = 2 / 6
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a ) 65 , b ) 63.75 , c ) 80 , d ) 85 , e ) 90 | b | add(multiply(power(2, multiply(divide(60, 6), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 6 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 30 minutes after it was poured was how many degrees fahrenheit ? | "first , we have to find a . we know that after t = 6 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 6 a ) + 60 60 = 120 * ( 2 ^ - 6 a ) 60 / 120 = 2 ^ - 6 a 1 / 2 = 2 ^ - 6 a 2 ^ - 1 = 2 ^ - 6 a - 1 = - 6 a 1 / 6 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 6 * 30 ) + 60 f = 120 * ( 2 ^ - 5 ) + 60 f = 120 * ( 1 / 2 ^ 5 ) + 60 f = 120 * 1 / 32 + 60 f = 3.75 + 60 = 63.75 answer b !" | a = 60 / 6
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
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a ) 4 kmph , b ) 2 kmph , c ) 7 kmph , d ) 8 kmph , e ) 3 kmph | a | divide(subtract(16, 8), const_2) | a man can row his boat with the stream at 16 km / h and against the stream in 8 km / h . the man ' s rate is ? | "ds = 16 us = 8 s = ? s = ( 16 - 8 ) / 2 = 4 kmph answer : a" | a = 16 - 8
b = a / 2
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a ) 2.09 % , b ) 20.9 % , c ) 209 % , d ) 0.209 % , e ) none of these | c | multiply(2.09, const_100) | 2.09 can be expressed in terms of percentage as | "explanation : while calculation in terms of percentage we need to multiply by 100 , so 2.09 * 100 = 209 answer : option c" | a = 2 * 9
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a ) 5.5 , b ) 11 , c ) 47.5 , d ) 66 , e ) 49.5 | e | divide(subtract(multiply(6, 36.5), multiply(subtract(6, const_2), 30)), const_2) | johnny bought 6 peanut butter cans at an average price ( arithmetic mean ) of 36.5 ¢ . if johnny returned two cans to the retailer , and the average price of the remaining cans was 30 ¢ , then what is the average price , in cents , of the two returned peanut butter cans ? | total price of six cans = 6 * 36.5 = 219 total price of 4 cans = 4 * 30 = 120 total rice of two cans = 219 - 120 = 99 average price of two cans = 99 / 2 = 49.5 c another way to do it is this : assume that the four leftover cans were of 30 c each . the avg was 36.5 c initially because the two cans were 36.5 c each and were providing another 6.5 c of cost to other 4 cans . so cost of the two cans = 2 * 36.5 + 4 * 6.5 = 99 avg cost of the two cans = 99 / 2 = 49.5 c answer ( e ) | a = 6 * 36
b = 6 - 2
c = b * 30
d = a - c
e = d / 2
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a ) 180 o , b ) 270 o , c ) 360 o , d ) 45 o , e ) 135 o | b | subtract(divide(const_3600, const_10), divide(divide(const_3600, const_10), const_4)) | a clockwise rotation around point z ( that is , a rotation in the direction of the arrow ) transforms the shaded quadrilateral to the unshaded quadrilateral . the angle of rotation is approximately | the measure of \ pzq formed by the bottom edge of the shaded quadrilateral , zq , and this same edge after the rotation , zp , is approximately 90 . the transformation of the shaded quadrilateral to the unshaded quadrilateral is a clockwise rotation around point z through an an - gle equal to the measure of re ex angle pzq . the measure of \ pzq added to the measure of re ex \ pzq is equal to the measure of one complete rotation , or 360 . therefore , the measure of re ex angle pzq is approximately 360 o - 90 o or 270 o . thus the clockwise rotation around point z is through an angle of approximately 270 o correct answer b | a = 3600 / 10
b = 3600 / 10
c = b / 4
d = a - c
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a ) 2532.12 , b ) 2552.2 , c ) 9524.32 , d ) 10403.25 , e ) 20512.54 | d | subtract(multiply(66,000, power(add(const_1, divide(9, const_100)), 20)), 66,000) | find the compound interest on $ 66,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 66000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 66000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 76403.25 ci . = $ ( 76403.25 - 66000 ) = $ 10403.25 answer d ." | a = 9 / 100
b = 1 + a
c = b ** 20
d = 66 * 0
e = d - 66
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a ) 3 , b ) 1 / 2 , c ) 1 and 1 / 2 , d ) - 1 / 2 , e ) - 1 | a | divide(1, subtract(4, const_1)) | if | x | = 4 x - 1 , then x = ? | "answer : a approach : substituted option a i . e x = 1 . inequality satisfied ." | a = 4 - 1
b = 1 / a
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a ) 12.50 % , b ) 13.50 % , c ) 5 % , d ) 14.50 % , e ) none | c | multiply(subtract(subtract(add(const_1, divide(40, const_100)), multiply(add(const_1, divide(40, const_100)), divide(25, const_100))), const_1), const_100) | an uneducated retailer marks all his goods at 40 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ? | sol . let c . p . = rs . 100 . then , marked price = rs . 140 . s . p . = 75 % of rs . 140 = rs . 105 . ∴ gain % = 5 % . answer c | a = 40 / 100
b = 1 + a
c = 40 / 100
d = 1 + c
e = 25 / 100
f = d * e
g = b - f
h = g - 1
i = h * 100
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a ) 20 % , b ) 70 % , c ) 50 % , d ) 30 % , e ) 40 % | b | multiply(subtract(const_1, subtract(const_2, add(divide(10, const_100), const_1))), const_100) | dick and jane each saved $ 4,000 in 1989 . in 1990 dick saved 10 percent more than in 1989 , and together he and jane saved a total of $ 6,500 . approximately what percent less did jane save in 1990 than in 1989 ? | "1990 dick saved = $ 4400 jane saved = $ 2100 ( jane saved $ 2800 less than she did the prior year ) jane saved approximately $ 2800 / 4000 $ ( 70 % ) less in 1990 answer : b" | a = 10 / 100
b = a + 1
c = 2 - b
d = 1 - c
e = d * 100
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['a ) 0.35 cm 2', 'b ) 17.5 cm 2', 'c ) 8.75 cm 2', 'd ) 55 cm 2', 'e ) 50 cm 2'] | c | multiply(divide(const_1, const_2), multiply(5, 3.5)) | the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is ? | explanation : ( 5 * 3.5 ) / 2 = 8.75 answer is c | a = 1 / 2
b = 5 * 3
c = a * b
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(189, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), subtract(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(189, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), 3)) | if one positive integer is greater than another positive integer by 3 , and the difference of their cubes is 189 , what is their sum ? | "1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 the two numbers are 3 and 6 . the answer is d ." | a = 3 ** 3
b = 3 ** 3
c = b ** 2
d = 3 ** 3
e = 189 - d
f = e ** 2
g = c + f
h = math.sqrt(g)
i = a + h
j = 3 ** 2
k = j * 2
l = i / k
m = 3 ** 3
n = 3 ** 3
o = n ** 2
p = 3 ** 3
q = 189 - p
r = q ** 2
s = o + r
t = math.sqrt(s)
u = m + t
v = 3 ** 2
w = v * 2
x = u / w
y = x - 3
z = l + y
|
a ) 20 , b ) 33 , c ) 12 , d ) 17 , e ) 25 | a | add(16, 5) | there are 16 bees in the hive , then 5 more fly . how many bees are there in all ? | 16 + 5 = 20 . answer is a . | a = 16 + 5
|
a ) $ 450,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 400,000 , e ) $ 2 , 200,000 | a | multiply(multiply(200, const_100), const_10) | a special municipal payroll tax charges not tax on a payroll less than $ 250,000 and only 0.1 % on a company ’ s payroll above $ 250,000 . if belfried industries paid $ 200 in this special municipal payroll tax , then they must have had a payroll of | "answer : a , ( with different approach ) : the 200 paid is 0.1 % of the additional amount above 250,000 . let it be x now 0.1 % of x = 200 therefore x = 200,000 total = 250,000 + x = 450,000" | a = 200 * 100
b = a * 10
|
a ) $ 115,000 , b ) $ 160,000 , c ) $ 210,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(16, 10), 10) | an auction house charges a commission of 16 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.16 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 210,000 ( 16 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c ." | a = 16 * 10
b = a + 10
|
a ) 11 , b ) 408 , c ) 2881 , d ) 287 , e ) 221 | b | add(multiply(divide(60, subtract(21, 13)), 13), multiply(divide(60, subtract(21, 13)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 13 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 13 + 21 = 34 t = 12 d = 34 * 12 = 408 answer : b" | a = 21 - 13
b = 60 / a
c = b * 13
d = 21 - 13
e = 60 / d
f = e * 21
g = c + f
|
a ) 12 , b ) 17 , c ) 18 , d ) 77 , e ) 2 | e | subtract(4, reminder(1202, 4)) | what should be the least number to be added to the 1202 number to make it divisible by 4 ? | answer : 2 option : e | a = 4 - reminder
|
a ) 30 % , b ) 10 % , c ) 20 % , d ) 17 % , e ) 50 % | d | multiply(divide(subtract(3830, 3250), 3250), const_100) | a sum of money deposited at c . i . amounts to rs . 3250 in 2 years and to rs . 3830 in 3 years . find the rate percent ? | "3250 - - - 580 100 - - - ? = > 17 % answer : d" | a = 3830 - 3250
b = a / 3250
c = b * 100
|
a ) - 45 , b ) 56 , c ) - 62 , d ) 35 , e ) - 30 | b | subtract(subtract(subtract(150, 10), add(150, 10)), 10) | if | 20 x - 10 | = 150 , then find the product of the values of x ? | "| 20 x - 10 | = 150 20 x - 10 = 150 or 20 x - 10 = - 150 20 x = 160 or 20 x = - 140 x = 8 or x = - 7 product = - 7 * 8 = - 56 answer is b" | a = 150 - 10
b = 150 + 10
c = a - b
d = c - 10
|
a ) 2 , b ) 3 , c ) 6 , d ) 8 , e ) 10 | b | subtract(divide(180, 10), divide(180, 12)) | a certain car uses 12 gallons of gasoline in traveling 180 miles . in order for the car to travel the same distance using 10 gallons of gasoline , by how many miles per gallon must the car ’ s gas mileage be increased ? | "180 / 10 = 18 . the difference is 18 - 15 = 3 . answer b" | a = 180 / 10
b = 180 / 12
c = a - b
|
a ) 23 years , b ) 22 years , c ) 21 years , d ) 20 years , e ) 30 years | b | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 24 years older than his son . in two years , his age will be twice the age of his son . what is the present age of his son ? | "let present age of the son = x years then present age the man = ( x + 24 ) years given that in 2 years man ' s age will be twice the age of his son x = 22 answer b" | a = 2 * 2
b = a - 2
c = 24 - b
d = 2 - 1
e = c / d
|
a ) 50 . , b ) 45 . , c ) 40 . , d ) 60 . , e ) 25 . | d | add(multiply(divide(subtract(const_100, 5), const_100), 60), multiply(divide(5, const_100), 60)) | in the hillside summer camp there are 60 children . 90 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 5 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 60 students , 90 % of 60 = 54 boys and remaining 6 girls . now here 90 % are boys and 10 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 5 or 5 % . . if we add 60 to existing 54 then the count will be 114 and the girls number will be 6 as it . now boys are 95 % and girls are 5 % . ( out of 120 students = 114 boys + 6 girls ) . imo option d is correct ." | a = 100 - 5
b = a / 100
c = b * 60
d = 5 / 100
e = d * 60
f = c + e
|
a ) - 30 , b ) - 20 , c ) - 5 , d ) 5 , e ) 0 | e | subtract(power(multiply(36, 3), const_0_33), divide(36, power(multiply(36, 3), const_0_33))) | if x and y are integers such that x ^ 2 = 3 y and xy = 36 , then x – y = ? | here x and y are integers . x ^ 2 = 3 y , xy = 36 . substitute ( x ^ 2 ) / 3 = y in xy = > x ^ 3 = 36 * 3 = > x ^ 3 = 108 . here x ^ 3 is positive , x is also positive . x = 6 then y = 6 . x - y = 0 so option e is correct | a = 36 * 3
b = a ** const_0_33
c = 36 * 3
d = c ** const_0_33
e = 36 / d
f = b - e
|
a ) 2 days , b ) 5 / 2 days , c ) 13 / 5 days , d ) 3 days , e ) none of these | c | multiply(multiply(divide(49, 196), divide(8, 5)), divide(13, 2)) | 49 pumps can empty a reservoir in 13 / 2 days , working 8 hours a day . if 196 pumps are used for 5 hours each day , then the same work will be completed in : | explanation : let the required number of days be x . then , more pumps , less days ( indirect proportion ) less working hrs / day , more days ( indirect proportion ) pumps 196 : 49 working hrs / day 5 : 8 : : 13 / 2 : x 96 x 5 x x = 49 x 8 x 13 / 2 x = 49 x 8 x 13 / 2 x 1 / ( 196 x 5 ) x = 13 / 5 answer c | a = 49 / 196
b = 8 / 5
c = a * b
d = 13 / 2
e = c * d
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 / 8 , e ) 4 | d | subtract(2, multiply(divide(factorial(3), factorial(1)), power(divide(2, 1), 3))) | a couple decides to have 3 children . if they succeed in having 4 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 2 girls and 1 boy ? | "sample space = 2 ^ 3 = 8 . favourable events = { bgg } , { bgb } , { bbb } , { ggg } , { gbg } probability = 5 / 8 = 5 / 8 . ans ( d ) ." | a = math.factorial(3)
b = math.factorial(1)
c = a / b
d = 2 / 1
e = d ** 3
f = c * e
g = 2 - f
|
a ) 25 , b ) 37 , c ) 50 , d ) 30 , e ) 75 | d | multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100) | working together at their respective constant rates , machine a and machine b can produce 600 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 600 unit total would it produce ? | "1 / a + 1 / b = 1 / t 1 / a + 1 / 12 = 1 / 8 ( 50 % more of 8 is 12 ) 1 / a = 1 / 24 machine a can produce 600 units in 24 hrs , so it can produce 600 * 8 / 24 = 200 units is 8 hrs . 200 is 30 % of 600 . d is the answer" | a = 4 * 4
b = a * 100
c = b / 8
d = 4 * 4
e = d * 100
f = 8 * 50
g = f / 100
h = 8 + g
i = e / h
j = c - i
k = j * 8
l = 4 * 4
m = l * 100
n = k / m
o = n * 100
|
a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 2 / 5 | a | divide(40, add(50, 30)) | some of the 50 % solution of acid was removed and this was replaced with an equal amount of 30 % solution of acid . as a result , a 40 % solution of acid was obtained . what fraction of the original solution was replaced ? | "let x be the fraction of the original solution that was replaced . 0.5 * ( 1 - x ) + 0.3 ( x ) = 0.4 0.2 x = 0.1 x = 1 / 2 the answer is a ." | a = 50 + 30
b = 40 / a
|
a ) 42 , b ) 35 , c ) 40 , d ) 33 , e ) 38 | e | divide(subtract(subtract(subtract(subtract(200, const_1), const_2), const_3), const_4), 5) | in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 5 rows in all and a total of 200 bricks in the wall , how many bricks does the bottom row contain ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 = 200 5 x - 10 = 200 5 x = 190 x = 38 answer : e" | a = 200 - 1
b = a - 2
c = b - 3
d = c - 4
e = d / 5
|
a ) 3 / 40000 , b ) 3 / 20000 , c ) 3 / 4000 , d ) 9 / 400 , e ) 6 / 130 | a | multiply(multiply(divide(1, 500), divide(1, 800)), 30) | a certain business school has 500 students , and the law school at the same university has 800 students . among these students , there are 30 sibling pairs consisting of 1 business student and 1 law student . if 1 student is selected at random from both schools , what is the probability that a sibling pair is selected ? | probability of selecting 1 student from harvard ' s business school - - - 1 / 500 probability of selecting 1 student from harvard ' s law school - - - - - - - - - 1 / 800 probability that these two students are siblings - - - - ( 1 / 500 * 1 / 800 ) since there are 30 siblings , hence ( 1 / 500 * 1 / 800 ) * 30 . 3 / 40000 answer : a | a = 1 / 500
b = 1 / 800
c = a * b
d = c * 30
|
a ) 250 , b ) 500 , c ) 450 , d ) 500 , e ) 520 | b | divide(700, add(const_1, divide(40, const_100))) | a number increased by 40 % gives 700 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 40 % = 140 % 140 % - - - - - - - > 700 ( 140 × 5 = 700 ) 100 % - - - - - - - > 700 ( 100 × 5 = 500 ) b )" | a = 40 / 100
b = 1 + a
c = 700 / b
|
a ) 15 % , b ) 32 % , c ) 40 % , d ) 68 % , e ) 80 % | d | multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100)))), const_1)) | the profits of qrs company rose 40 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ? | "assume 100 in march , then 140 in april as 40 % increase , then 112 in may as 20 % decrease from april , and then 168 in june which is 150 % of 112 . so overall increase is from 100 to 168 is 68 % answer d" | a = 50 / 100
b = 1 + a
c = 40 / 100
d = 1 + c
e = 20 / 100
f = 1 - e
g = d * f
h = b * g
i = h - 1
j = 100 * i
|
a ) 800 , b ) 3200 , c ) 900 , d ) 1600 , e ) none | b | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 4) | find the value of 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 4 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 4 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 4 x 4 x 4 x 50 = 3200 correct option : b" | a = 3 * 6
b = a * 2
c = 0 * 12
d = c * 0
e = b / d
f = e * 4
|
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