options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 6693.75 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these | a | divide(multiply(const_100, 4016.25), multiply(12, 5)) | a sum fetched total simple interest of 4016.25 at the rate of 12 p . c . p . a . in 5 years . what is the sum ? | "let the sums be p . now , 60 % of p = 4016.25 or , p = 6693.75 answer a" | a = 100 * 4016
b = 12 * 5
c = a / b
|
a ) 5.3 % , b ) 7.6 % , c ) 10.6 % , d ) 12 % , e ) none of these | b | multiply(const_100, divide(subtract(multiply(50, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | a man buys 50 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 50 pens = rs . 46 s . p . of 50 pens = 99 % of rs . 50 = rs . 49.50 profit % = ( profit / c . p . ) x 100 profit % = ( 3.50 / 46 ) x 100 = 7.6 % answer b" | a = 100 - 1
b = 50 * a
c = 46 * 100
d = b - c
e = 46 * 100
f = d / e
g = 100 * f
|
a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord w , in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . w = pi * r = 30 . round pi = 3 , then r = 10 . chord is about 10 feet long . d" | a = 30 / 3
|
a ) 7 , b ) 8 , c ) 4 , d ) 3 , e ) 2 | c | divide(divide(divide(lcm(2, 5475), 5475), const_4), const_4) | what is the least value of x , so that 2 x 5475 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 5 = 23 + x least value of x may be ' 4 ' , so that the total 23 + 4 = 27 is divisible by 9 . answer : option c" | a = math.lcm(2, 5475)
b = a / 5475
c = b / 4
d = c / 4
|
a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 3 / 7 | d | divide(2, 5) | in the rectangular coordinate system , if the line x = 5 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ? | "x = 5 y + 5 , and thus y = x / 5 - 1 the slope is 1 / 5 . 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 / 5 and thus p = 2 / 5 the answer is d ." | a = 2 / 5
|
a ) 3387 , b ) 1000 , c ) 2866 , d ) 1500 , e ) 1121 | d | multiply(multiply(20, 15), 5) | in digging a pond 20 m * 15 m * 5 m the volumes of the soil extracted will be ? | "20 * 15 * 5 = 1500 answer : d" | a = 20 * 15
b = a * 5
|
a ) 10 , b ) 100 , c ) 50 , d ) 200 , e ) 300 | c | multiply(multiply(subtract(divide(subtract(multiply(2, divide(9, const_3600)), multiply(4, divide(10, const_3600))), subtract(divide(9, const_3600), divide(10, const_3600))), 2), divide(9, const_3600)), const_1000) | a train overtakes two persons who are walking in the same direction in which the train is going , at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . the length of the train is : | 2 kmph = ( 2 x 5 / 18 ) m / sec = 5 / 9 m / sec . 4 kmph = ( 4 x 5 / 18 ) m / sec = 10 / 9 m / sec . let the length of the train be x metres and its speed by y m / sec . then , ( x / y - 5 / 9 ) = 9 and ( x / y - 10 / 9 ) = 10 . 9 y - 5 = x and 10 ( 9 y - 10 ) = 9 x = > 9 y - x = 5 and 90 y - 9 x = 100 . on solving , we get : x = 50 . therefore length of the train is 50 m . answer : c | a = 9 / 3600
b = 2 * a
c = 10 / 3600
d = 4 * c
e = b - d
f = 9 / 3600
g = 10 / 3600
h = f - g
i = e / h
j = i - 2
k = 9 / 3600
l = j * k
m = l * 1000
|
a ) 55 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | a | add(add(multiply(divide(45, 5), const_2), 5), add(divide(45, 5), 5)) | the sum of the present ages of two persons a and b is 45 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 45 , a = 2 b 2 b + b = 45 = > b = 15 then a = 30 . 5 years , their ages will be 35 and 20 . sum of their ages = 35 + 20 = 55 . answer : a" | a = 45 / 5
b = a * 2
c = b + 5
d = 45 / 5
e = d + 5
f = c + e
|
a ) 71.5 , b ) 72 , c ) 72.5 , d ) 73 , e ) 85.7 | e | divide(1, divide(add(multiply(const_3600, divide(1, 90)), 2), const_3600)) | a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 90 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 90 kilometers per hour is 1 / 90 hours = 3,600 / 90 seconds = 40 seconds ; time to cover this distance at regular speed is 40 + 2 = 42 seconds = 42 / 3,600 hours = 7 / 600 hours ; so , we get that to cover 1 kilometer 1 / ( 7 / 600 ) hours is needed - - > regular speed 85.7 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : e ." | a = 1 / 90
b = 3600 * a
c = b + 2
d = c / 3600
e = 1 / d
|
a ) 5 / 6 , b ) 5 / 12 , c ) 7 / 12 , d ) 11 / 12 , e ) 11 / 15 | a | subtract(const_1, divide(5, multiply(5, const_3))) | a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 5 / 30 = 1 / 6 . the probability that he will stay in the dark is 1 - 1 / 6 = 5 / 6 the answer is a ." | a = 5 * 3
b = 5 / a
c = 1 - b
|
a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | on a map , 1 inch represents 28 miles . how many v inches would be necessary to represent a distance of 383.6 miles ? | "v inches necessary to represent a distance of 383.6 miles = 383.6 / 28 = 13.7 answer c" | a = 383 / 6
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a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 5)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 5))) | 36 welders work at a constant rate they complete an order in 5 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 5 = 180 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 5 of the job . so 24 workers have to finish only 4 / 5 of the job . 5 . total time taken by 24 workers can be got from formula used at ( 2 ) . i . e . , 24 * total time taken = 180 . total time taken by 6 workers to finish the complete job is 180 / 24 = 7.5 days . 6 . time taken by 24 workers to finish 4 / 5 of the job is 4 / 5 * 7.5 = 6 days . the answer is choice e" | a = 36 * 5
b = 1 / a
c = b * 36
d = 1 - c
e = 36 - 12
f = d / e
g = 36 * 5
h = 1 / g
i = f / h
|
a ) 1254 , b ) 1482 , c ) 1771 , d ) 1875 , e ) 1923 | c | multiply(subtract(const_1, divide(3, 23)), 23) | there are 23 students in a class . in how many different ways can a committee of 3 students be formed ? | "23 c 3 = 23 * 22 * 21 / 6 = 1771 the answer is c ." | a = 3 / 23
b = 1 - a
c = b * 23
|
['a ) 40 %', 'b ) 50 %', 'c ) 75 %', 'd ) 80 %', 'e ) 90 %'] | c | multiply(divide(subtract(add(subtract(add(multiply(power(64, divide(const_1, const_3)), power(27, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), add(subtract(add(multiply(power(64, divide(const_1, const_3)), power(27, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3)))), const_100) | the volume of the sphere qq is ( dfrac { 37 } { 64 } % ) less than thevolume of sphere pp and the volume of sphere rr is ( dfrac { 19 } { 27 } % ) less than that of sphere qq . by what is the surface areaof sphere rr less than the surfacearea of sphere pp ? | let the volume of sphere pp be 64 parts . therefore volume of sphere qq = 64 β 3764 % = 64 β 3764 % of 6464 = 64 β 37 = 27 = 64 β 37 = 27 parts . the volume of rr = 27 β 1927 Γ 27 = 27 β 1927 Γ 27 = 27 β 19 = 8 = 27 β 19 = 8 parts . volume ratio : = p : q : r = 64 : 27 : 8 = p : q : r = 64 : 27 : 8 radius ratio : = p : q : r = 4 : 3 : 2 = p : q : r = 4 : 3 : 2 the surface area will be 16 : 9 : 516 : 9 : 5 surface area of rr is less than the surface area of sphere pp 16 k β 4 k = 12 k 16 k β 4 k = 12 k now , = 12 k 16 k Γ 100 = 12 k 16 k Γ 100 = 75 % = 75 % thus surface area of sphere rr is less than the surface area of sphere p by 75 % c | a = 1 / 3
b = 64 ** a
c = 1 / 3
d = 27 ** c
e = b * d
f = 1 / 3
g = 64 ** f
h = e + g
i = 1 / 3
j = 64 ** i
k = h - j
l = 1 / 3
m = 64 ** l
n = k + m
o = 1 / 3
p = 64 ** o
q = n - p
r = 1 / 3
s = 64 ** r
t = 1 / 3
u = 27 ** t
v = s * u
w = 1 / 3
x = 64 ** w
y = v + x
z = 1 / 3
A = 64 ** z
B = y - A
C = 1 / 3
D = 64 ** C
E = B + D
F = q / E
G = F * 100
|
a ) 32.44 hours , b ) 31.44 hours , c ) 30.44 hours , d ) 29.44 hours , e ) 28.44 hours | b | add(multiply(add(add(7, 1.5), subtract(7, 1.5)), 105), multiply(subtract(add(divide(105, add(7, 1.5)), divide(105, subtract(7, 1.5))), add(add(7, 1.5), subtract(7, 1.5))), const_60)) | speed of a boat in standing water is 7 kmph and speed of the stream is 1.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 5.5 kmph speed downstream = 8.5 kmph total time taken = 105 / 5.5 + 105 / 8.5 = 31.44 hours answer is b" | a = 7 + 1
b = 7 - 1
c = a + b
d = c * 105
e = 7 + 1
f = 105 / e
g = 7 - 1
h = 105 / g
i = f + h
j = 7 + 1
k = 7 - 1
l = j + k
m = i - l
n = m * const_60
o = d + n
|
a ) 73 , b ) 83 , c ) 12 , d ) 83 , e ) 11 | e | subtract(const_60, multiply(const_60, divide(37, 45))) | excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 37 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 8 / 45 * 60 = 11 answer : option e" | a = 37 / 45
b = const_60 * a
c = const_60 - b
|
a ) 700 , b ) 800 , c ) 900 , d ) 1000 , e ) 1100 | b | divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3)) | a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is one - half of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ? | "total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = x / 2 x + x / 2 + 600 = 1800 3 x / 2 = 1200 x = 800 km the answer is b ." | a = 1800 * 2
b = a / 3
c = b * 3
d = 2 + 3
e = c / d
|
a ) 5.5 min . , b ) 6 min . , c ) 4.9 min . , d ) 5.28 min . , e ) 5 min . | d | divide(726, divide(add(multiply(4.5, const_1000), multiply(3.75, const_1000)), const_60)) | the jogging track in a sports complex is 726 m in circumference . suresh and his wife start from the same point and walk in opposite direction at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in : | "let both of them meet after t min 4500 m are covered by suresh in 60 m . in t min he will cover 4500 t / 60 likewise , in t min suresh ' s wife will cover 3750 t / 60 given , ( 4500 t / 60 ) + ( 3750 t / 60 ) = 726 t = 5.28 minutes answer : d" | a = 4 * 5
b = 3 * 75
c = a + b
d = c / const_60
e = 726 / d
|
a ) rs . 3400 , b ) rs . 2700 , c ) rs . 1720 , d ) rs . 7200 , e ) rs . 8200 | d | multiply(divide(subtract(divide(multiply(4, 1800), 3), 1600), subtract(divide(multiply(4, 5), 3), 6)), 6) | a and b have monthly incomes in the ratio 5 : 6 and monthly expenditures in the ratio 3 : 4 . if they save rs . 1800 and rs . 1600 respectively , find the monthly income of b | explanation : incomes of a and b = 5 x and 6 x and expenses of a and b = 3 y and 4 y then , savings of a = 5 x - 3 y = 1800 β ? ( 1 ) savings of b = 6 x - 4 y = 1600 β ? ( 2 ) by solving equations ( 1 ) and ( 2 ) y = 1400 monthly income of b = expenses of b + savings of b = 4 y + 1600 = 4 ( 1400 ) + 1600 = rs . 7200 answer : d | a = 4 * 1800
b = a / 3
c = b - 1600
d = 4 * 5
e = d / 3
f = e - 6
g = c / f
h = g * 6
|
a ) 10 % , b ) 12 % , c ) 15 % , d ) 17 % , e ) 20 % | d | multiply(divide(subtract(divide(50, const_100), divide(40, const_100)), subtract(const_1, divide(40, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 40 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "lets assume that candidate got 40 % votes and total votes is 100 . candidate won = 40 remaining = 60 to get 50 % , candidate requires 10 votes from 100 which is 10 % and 10 votes from 60 . 10 / 60 = 1 / 6 = . 166 = 16.67 % which is approx 17 % . hence the answer is d" | a = 50 / 100
b = 40 / 100
c = a - b
d = 40 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) 1 / 10 , b ) 13 / 50 , c ) 1 / 2 , d ) 7 / 10 , e ) 9 / 10 | b | divide(subtract(100, subtract(add(50, 40), 16)), 100) | a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 16 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ? | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 16 - 2 cmb = 100 b - cb - mb - 2 cmb = 26 hence the probability = 26 / 100 = 13 / 50 . b" | a = 50 + 40
b = a - 16
c = 100 - b
d = c / 100
|
a ) 5 : 32 , b ) 5 : 36 , c ) 5 : 48 , d ) 5 : 54 , e ) 5 : 58 | a | divide(5, add(const_60, subtract(14, reminder(add(multiply(multiply(2, const_1000), const_1000), 880702), const_60)))) | if it is 5 : 14 in the evening on a certain day , what time in the morning was it exactly 2 , 880702 minutes earlier ? ( assume standard time in one location . ) | 5 : 14 minus 2 , 880,702 must end with 2 , the only answer choice which ends with 2 is a . answer : a . | a = 2 * 1000
b = a * 1000
c = b + 880702
d = 14 - reminder
e = const_60 + d
f = 5 / e
|
a ) 2.29 , b ) 2.88 , c ) 2.2 , d ) 2.12 , e ) 2.14 | b | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . how far is the place ? | "m = 6 s = 1.2 ds = 6 + 1.2 = 7.2 us = 6 - 1.2 = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 answer : b" | a = 6 + 1
b = 6 - 1
c = a * b
d = 6 + 1
e = 6 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(100, reminder(const_4.0, 6)) | when positive integer n is divided by 3 , the remainder is 2 . when n is divided by 6 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and next multiple of 7 will be 84 + 21 > 100 . . ans 4 . . d" | a = 100 - reminder
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a ) 55 , b ) 100 , c ) 490 , d ) 500 , e ) 980 | a | divide(539, 9.8) | a sports equipment store sold ping pong rackets for a total of $ 539 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | "average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 539 x = 55 pairs were sold . answer : a" | a = 539 / 9
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a ) $ 80 , b ) $ 100 , c ) $ 120 , d ) 135 , e ) 160 | c | divide(multiply(add(80, divide(multiply(80, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 20)) | a retailer bought a machine at a wholesale price of $ 80 and later on sold it after a 20 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the wholesale price , what is the retail price of the machine ? | "the price after the discount was 1.2 * 80 = $ 96 let x be the retail price . 0.8 x = $ 96 x = 96 / 0.8 = $ 120 the answer is c ." | a = 80 * 20
b = a / 100
c = 80 + b
d = c * 100
e = 3 * 3
f = e * 20
g = d / f
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a ) 15 , b ) 16 , c ) 28 , d ) 42 , e ) 64 | d | divide(multiply(7, subtract(7, const_1)), const_2) | there are 7 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 6 teams . . . so total no of matches = 7 x 6 = 42 . now , each match is played twice = > 42 x 2 but 2 teams play a match = > 42 x 2 / 2 = 42 . answer : d" | a = 7 - 1
b = 7 * a
c = b / 2
|
a ) 0.35 % , b ) 0.4 % , c ) 0.045 % , d ) 0.6 % , e ) none of these | c | multiply(divide(divide(multiply(add(const_3, const_4), const_4), const_100), 61.472), const_10) | in expressing a length of 61.472 km as nearly as possible with the 3 significant digits , find the percentage error ? | explanation : error = ( 61.5 - 61.472 ) = 0.028 required percentage = 0.028 / 61.472100 = 0.045 answer : c | a = 3 + 4
b = a * 4
c = b / 100
d = c / 61
e = d * 10
|
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(1, const_4) | if x + y + z = 1 . then xy + yz + zx is | 1 / 2 + 1 / 2 + 0 = 1 xy = 1 / 2 * 1 / 2 = 1 / 4 answer : c | a = 1 / 4
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a ) 600 , b ) 1200 , c ) 1500 , d ) 3600 , e ) 6000 | e | multiply(divide(5, 2), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 2 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | "sugar : flour = 5 : 2 = 25 : 10 ; flour : soda = 10 : 1 = 10 : 1 ; thus we have that sugar : flour : soda = 25 x : 10 x : 1 x . also given that 10 x / ( 1 x + 60 ) = 8 / 1 - - > x = 240 - - > sugar = 25 x = 6000 answer : e ." | a = 5 / 2
b = 8 * 60
c = 10 - 8
d = b / c
e = 10 * d
f = a * e
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | e | divide(9800, subtract(divide(6000, 2.5), subtract(divide(3000, 1), divide(6000, 3)))) | machine a can process 6000 envelopes in 3 hours . machines b and c working together but independently can process the same number of envelopes in 2.5 hours . if machines a and c working together but independently process 3000 envelopes in 1 hour , then how many hours would it take machine b to process 9800 envelopes . | "you can either take the amount of work done as the same as karishma has done or take the work done by each in the same time . i will do the latter 1 . work done in 1 hr by a is 2000 envelopes 2 . work done in 1 hr by a and c is 3000 envelopes 3 . so work done in 1 hr by c is 1000 envelopes 4 . work done in 1 hr by b and c is 2400 envelopes 5 . so work done in 1 hr by b is 1400 envelopes 6 . so to process 9800 envelopes b will take 9800 / 1400 hrs = 7 hrs so the answer is choice e" | a = 6000 / 2
b = 3000 / 1
c = 6000 / 3
d = b - c
e = a - d
f = 9800 / e
|
a ) 5,1 , b ) 7,3 , c ) 9,8 , d ) 11,6 , e ) 13,7 | d | subtract(subtract(8, 4), const_1) | difference between two numbers is 5 , 6 times of the smaller lacks by 8 from the 4 times of the greater . find the numbers ? | explanation : x β y = 5 4 x β 6 y = 8 x = 11 y = 6 d ) | a = 8 - 4
b = a - 1
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a ) 2 , b ) 3 , c ) 4 , d ) 9 , e ) 12 | d | multiply(divide(subtract(600, divide(480, const_2)), 40), const_2) | angelina walked 600 meters from her home to the grocery at a constant speed . she then walked 480 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 = 600 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 600 / x - 240 / x = 40 . . 360 / x = 40 . . x = 9 m / secs . . so grocery to gym = 2 * 9 = 18 m / s . . . d" | a = 480 / 2
b = 600 - a
c = b / 40
d = c * 2
|
a ) 450 m , b ) 500 m , c ) 550 m , d ) 600 m , e ) 650 m | d | multiply(divide(multiply(180, const_1000), const_3600), 36) | a train running at the speed of 180 km / hr crosses a pole in 36 seconds . what is the length of the train ? | "speed = ( 180 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 36 ) m = 600 m . answer : d" | a = 180 * 1000
b = a / 3600
c = b * 36
|
a ) 3 . , b ) 2 . , c ) 1 / 5 , d ) 1 / 3 . , e ) there is n ' t enough data to answer the question . | c | add(4, divide(multiply(4, 1), 4)) | two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores 1 / 4 , what is the value of y / x ? | "answer is c : 1 / 5 x - y = ( x + y ) / 4 solving for y / x = 1 / 5" | a = 4 * 1
b = a / 4
c = 4 + b
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a ) 12 , b ) 99 , c ) 27 , d ) 28 , e ) 20 | a | multiply(4, 3) | walking 3 / 2 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | "speed ratio = 1 : 3 / 2 = 2 : 3 time ratio = 3 : 2 1 - - - - - - - - 3 4 - - - - - - - - - ? 12 m . answer : a" | a = 4 * 3
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a ) 32 , b ) 150 , c ) 60 , d ) 27 , e ) 11 | b | divide(5, divide(divide(const_1, multiply(const_3, add(const_2, 5))), const_2)) | when 5 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "explanation : let the number be 5 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 5 = a / 30 = > a = 150 answer : b" | a = 2 + 5
b = 3 * a
c = 1 / b
d = c / 2
e = 5 / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(5, const_2) | in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is 440 , how many purple chips were selected ? | "440 = 1 * 5 * 8 * 11 the factor of 8 must come from the purple point value , so there is 1 purple chip . the answer is a ." | a = 5 - 2
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a ) 144 , b ) 131 , c ) 99 , d ) 90 , e ) 45 | c | add(divide(multiply(9, subtract(9, const_1)), const_2), multiply(9, 7)) | 9 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | there are 9 business exec and in each handshake 2 business execs are involved . hence 9 c 2 = 36 also , each of 9 exec will shake hand with every 7 other chairmen for total of 63 handshake . total = 36 + 63 = 99 ans : c | a = 9 - 1
b = 9 * a
c = b / 2
d = 9 * 7
e = c + d
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a ) 2 / 13 , b ) 3 / 13 , c ) 1 / 26 , d ) 1 / 52 , e ) 3 / 52 | c | multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52)) | from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a club and the second card is a king or queen ? | "p ( club ) = 1 / 4 p ( king or queen ) = 2 / 13 p ( club then a king / queen ) = 1 / 4 * 2 / 13 = 1 / 26 the answer is c ." | a = 3 * 4
b = a + 1
c = b / const_52
d = 2 / const_52
e = c * d
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a ) 29.16 l , b ) 29.19 l , c ) 29.12 l , d ) 29.11 l , e ) 29.14 l | a | subtract(40, multiply(4, 3)) | a container has 40 l milk . 4 l milk is taken out and replaced with water , the process done 3 times . what is the amount of milk now left ? | explanation : in regard to the problem let after n times liquid in the container is x and liquid removed is y = > liquid in container = x = > liquid removed = y after one trial , quantity of pure liquid = x - ( 1 - y / x ) after n trials quantity of pure liquid = x - ( 1 - y / x ) n therefore , putting the values ( x = 40 l y = 4 l n = 3 ) and solving = > 40 - ( 1 - 4 / 40 ) 3 = 29.16 l answer : a | a = 4 * 3
b = 40 - a
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['a ) 145', 'b ) 135', 'c ) 75', 'd ) 140', 'e ) none of these'] | b | multiply(sqrt(divide(6075, const_3)), const_3) | if the length of rectangle is three times of its breadth . if the area of rectangle is 6075 sq . m , then calculate the length ? | let breadth = x , length = 3 x . area of rectangle = length * breadth = 3 x ^ 2 = 6075 x ^ 2 = 2025 , x = β 2025 = 45 m length = 135 m answer b | a = 6075 / 3
b = math.sqrt(a)
c = b * 3
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | subtract(add(multiply(10, 4), multiply(9, 4)), multiply(9, 9)) | the average of 9 observations was 9 , that of the 1 st of 4 being 10 and that of the last 4 being 8 . what was the 5 th observation ? | "1 to 9 = 9 * 9 = 81 1 to 4 = 4 * 10 = 40 4 to 9 = 6 * 8 = 48 5 th = 48 + 40 = 88 β 81 = 7 answer : d" | a = 10 * 4
b = 9 * 4
c = a + b
d = 9 * 9
e = c - d
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a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) none of these | b | divide(20, add(1, divide(1, 2))) | how much water must be added to 48 litres of milk at 1 1 β 2 litres for 20 so as to have a mixture worth 10 2 β 3 a litre ? | "c . p . of 1 litre of milk = ( 20 Γ 2 β 3 ) = 40 β 3 β΄ ratio of water and milk = 8 β 3 : 32 β 3 = 8 : 32 = 1 : 4 β΄ quantity of water to be added to 48 litres of milk = ( 1 β 4 Γ 48 ) litres = 12 litres . answer b" | a = 1 / 2
b = 1 + a
c = 20 / b
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a ) s . 14,000 , b ) s . 12,000 , c ) s . 32,000 , d ) s . 40,000 , e ) s . 50,000 | c | subtract(multiply(5, const_4), const_12) | a salesman Γ’ β¬ β’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 100 more than that by the previous schema , his sales were worth ? | "[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 100 x = 32000 answer : c" | a = 5 * 4
b = a - 12
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a ) 3 / 6 , b ) 3 / 2 , c ) 7 / 3 , d ) 3 / 5 , e ) 5 / 2 | c | divide(subtract(27, 24), subtract(24, 17)) | two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 24 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 24 = = > 27 x + 17 y = 24 x + 24 y = = > 3 x = 7 y = = > x / y = 7 / 3 . answer : c" | a = 27 - 24
b = 24 - 17
c = a / b
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a ) 25 , b ) 30 , c ) 50 , d ) 55 , e ) 80 | e | multiply(divide(32, 40), const_100) | 32 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ? | "i came up with ( e ) 80 think of 100 people total : from the first fact , 32 of these are women with fair hair . from the second fact , these 30 women make up 40 % of the total fair haired population . we can then make a ratio of 60 : 40 fair haired men to fair haired women . this means that ( 60 / 40 ) * 32 equals the number of fair haired men , which is 48 men with fair hair . add this 48 to the 32 women and get 80 fair haired men and women out of 100 total men and women . 80 % e" | a = 32 / 40
b = a * 100
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a ) 127 , b ) 196 , c ) 116 , d ) 140 , e ) 234 | b | add(subtract(100, const_4), const_100) | while visiting a small town in the united states , i lost my overcoat in a bus . when i reported the matter to the bus company i was asked the number of the bus . though i did not remember the exact number i did remember that the bus number bad a certain peculiarity about it . the number plate showed the bus number as a perfect square and also if the plate was turned upside down . ? the number would still be a perfect square β of course it was not ? i came to know from the bus company they had only 5 100 buses numbered from 1 to 500 . from this i was able to deduce the bus number . can you tell what was the other number . | b 196 the only numbers that can be turned upside down and still read as a number are 0 , 1 , 6 , 8 and 9 . the numbers 0 , 1 and 8 remain 0 , 1 and 8 when turned over , but 6 becomes 9 and 9 becomes 6 . therefore the possible numbers on the bus were 9 , 16 , 81 , 100 , 169 or 196 . however , the number 196 is the only number which becomes a perfect square when turned over because 961 is the perfect square of 31 . therefore 196 is the correct answer . | a = 100 - 4
b = a + 100
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a ) 24 . , b ) 34 . , c ) 36 . , d ) 42 . , e ) 44 . | b | inverse(add(inverse(51), divide(inverse(51), const_2))) | a car traveled from san diego to san francisco at an average speed of 51 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 51 x total distance traveled = 2 * 51 x = 102 x total time taken = x + 2 x = 3 x average speed = 102 x / 3 x = 34 answer : b" | a = 1/(51)
b = 1/(51)
c = b / 2
d = a + c
e = 1/(d)
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a ) 2.42 , b ) 6.42 , c ) 5.42 , d ) 4.42 , e ) 3.42 | d | divide(150, multiply(122, const_0_2778)) | in what time will a train 150 metres long cross an electric pole , if its speed be 122 km / hr ? | "solution speed = ( 122 x 5 / 18 ) m / sec = 33.88 m / sec time taken = ( 150 / 33.88 ) sec = 4.42 sec . answer d" | a = 122 * const_0_2778
b = 150 / a
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a ) 148 , b ) 222 , c ) 144 , d ) 157 , e ) 164 | a | divide(multiply(16, 36), const_4) | what is the sum of the greatest common factor and the lowest common multiple of 16 and 36 ? | "prime factorization of the given numbers 16 = 2 ^ 4 36 = 2 ^ 2 * 3 * 2 greatest common factor = 2 ^ 2 = 4 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 4 + 144 = 148 answer a" | a = 16 * 36
b = a / 4
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a ) 14 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | subtract(add(multiply(reminder(7, 100), 3), reminder(3, 100)), reminder(1, 100)) | x is a positive integer less than 100 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 100 n = 14.42 or n = 14 answer : - a" | a = reminder * (
b = a + 3
c = b - reminder
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a ) 600 , b ) 500 , c ) 400 , d ) 300 , e ) 800 | b | divide(add(125, 40), divide(33, const_100)) | a student has to obtain 33 % of the total marks to pass . he got 125 marks and failed by 40 marks . the maximum mark is | "the student got 125 marks and still failed by 40 marks = 125 + 40 = 165 therefore 33 % of the total marks = 165 100 % of the total = 165 * 100 / 33 = 55 max mark = 500 answer b" | a = 125 + 40
b = 33 / 100
c = a / b
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a ) 27 , b ) 33 , c ) 72 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 9) | the number of boxes in a warehouse can be divided evenly into 9 equal shipments by boat or 24 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 9 and 24 = 72 answer c" | a = 2 * 2
b = a * 2
c = b * 3
d = c * 9
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a ) 8600 litres , b ) 200 litres , c ) 12800 litres , d ) 6400 litres , e ) 13200 litres | d | multiply(divide(multiply(multiply(8, const_60), add(5, 3)), 3), 5) | an outlet pipe empties a tank which is full in 5 hours . if the inlet pipe is kept open , which lets water in at the rate of 8 litres / min then outlet pipe would take 3 hours longer . find the capacity of the tank . | "let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 8 * 60 = 480 liters / hour ; net outflow rate when both pipes operate would be x - 480 liters / hour . capacity of the tank = x * 5 hours = ( x - 480 ) * ( 5 + 3 ) hours 5 x = ( x - 480 ) * 8 - - > x = 1280 - - > capacity = 5 x = 6400 liters . answer : d" | a = 8 * const_60
b = 5 + 3
c = a * b
d = c / 3
e = d * 5
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a ) 2 hours , b ) 5 hours , c ) 6 hours , d ) 4 hours , e ) 3 hours | e | divide(const_1, subtract(divide(const_1, 2), subtract(divide(const_1, 2), divide(const_1, 3)))) | a can do a piece of work in 3 hours ; b and c together can do it in 2 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ? | "a ' s 1 hour work = 1 / 3 ; ( b + c ) ' s 1 hour work = 1 / 2 ; ( a + b ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 3 + 1 / 2 ) = 5 / 6 c ' s 1 hour work = ( 5 / 6 - 1 / 2 ) = 1 / 3 c alone will take 3 hours to do the work . answer : e" | a = 1 / 2
b = 1 / 2
c = 1 / 3
d = b - c
e = a - d
f = 1 / e
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a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 9 | b | divide(multiply(7, 9), add(7, 9)) | tony alone can paint a wall in 7 days and his friend roy alone can paint the same wall in 9 days . in how many days they can paint the wall working together ? round off the answer to the nearest integer . | sol . use formula ( xy / x + y ) so nearest value for 3.93 = 4 answer : option b | a = 7 * 9
b = 7 + 9
c = a / b
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a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8 | b | subtract(9, 9) | when the number 9 y 30012 is exactly divisible by 11 , what is the smallest whole number that can replace y ? | "the given number = 9 y 30012 sum of the odd places = 2 + 0 + 3 + 9 = 14 sum of the even places = 1 + 0 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 14 - ( 1 + y ) = divisible by 11 13 οΏ½ y = divisible by 11 . y must be 2 , to make given number divisible by 11 . b" | a = 9 - 9
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a ) $ 20 , b ) $ 21 , c ) $ 24 , d ) $ 25 , e ) $ 28 | d | divide(35, add(const_1, divide(40, const_100))) | a worker ' s daily wage is increased by 40 % and the new wage is $ 35 per day . what was the worker ' s daily wage before the increase ? | "let x be the daily wage before the increase . 1.4 x = $ 35 x = $ 25 the answer is d ." | a = 40 / 100
b = 1 + a
c = 35 / b
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a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 5 | e | subtract(divide(subtract(90, 60), subtract(65, 60)), const_1) | for the past n days , the average ( arithmetic mean ) daily production at a company was 60 units . if today ' s production of 90 units raises the average to 65 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 60 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 60 n + 90 = 65 * ( n + 1 ) - - > n = 5 . or as 30 extra units increased the average for n + 1 days by 5 units per day then 30 / ( n + 1 ) = 5 - - > n = 5 . answer : e ." | a = 90 - 60
b = 65 - 60
c = a / b
d = c - 1
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a ) 9 / 16 , b ) 1 / 2 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 6 | d | divide(1, subtract(7, 3)) | a 7 - digit combination lock on a safe has zero exactly 3 times , does not have the digit 1 at all . what is the probability that exactly 3 of its digits are odd ? | remaining number of digits after three ' 0 ' digits = 4 constraint - no more ' 0 ' and no more ' 1 ' well , these four digits can be odd or even 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 ways to choose 3 odd digits out of total 4 digits = 4 c 3 = 4 4 * 1 / 16 = 1 / 4 d is the answer hi ! since we are talking about remaining 4 digits . each digit can either be odd or even ( total ways to select = 2 . desired outcome = 1 ) so let ' s say we want to choose first 3 digits as odd digits and last digit as even . the probability will be : - 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 but out of remaining 4 digit , any 3 can be odd and one even as there are no constraints in the question . therefore , total ways of choosing 3 odd digits out of 4 total digits = 4 c 3 so , the probability = 4 * 1 / 16 = 1 / 4 answer is d | a = 7 - 3
b = 1 / a
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['a ) 250 meter', 'b ) 100 meter', 'c ) 50 β 2 meter', 'd ) 50 meter', 'e ) 35 meter'] | b | sqrt(multiply(const_2, multiply(divide(1, 2), multiply(const_100, const_100)))) | area of a square is 1 / 2 hectare . the diagonal of the square is ? | area = 1 / 2 hectare = 10000 / 2 m 2 = 5000 m 2 again area = 1 / 2 x ( diagonal ) 2 so 1 / 2 x ( diagonal ) 2 = 5000 m 2 diagonal 2 = 10000 diagonal = 100 answer : b | a = 1 / 2
b = 100 * 100
c = a * b
d = 2 * c
e = math.sqrt(d)
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a ) 67 % , b ) 45 % , c ) 25 % , d ) 70 % , e ) 50 % | c | multiply(subtract(divide(5, 4), const_1), const_100) | the ratio of the cost price and the selling price is 4 : 5 . the profit percent is : | "c c . p . = rs . 4 x . then , s . p . = rs . 5 x gain = ( 5 x - 4 x ) = rs . x gain % = ( x * 100 ) / 4 x = 25 % ." | a = 5 / 4
b = a - 1
c = b * 100
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a ) 120 , b ) 100 , c ) 140 , d ) 80 , e ) 90 | b | multiply(add(add(1, 2), 2), divide(5000, subtract(add(200, multiply(50, const_2)), multiply(25, 2)))) | pipe a and pipe b fill water into a tank of capacity 5000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated until the tank is filled . how long will it take to fill the tank ? | tank capacity : 5000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 5000 l tank : 1000 / 250 = 20 time taken to fill : 20 * total time = 20 * 5 = 100 ( option b ) | a = 1 + 2
b = a + 2
c = 50 * 2
d = 200 + c
e = 25 * 2
f = d - e
g = 5000 / f
h = b * g
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a ) 15 days , b ) 10 days , c ) 12 days , d ) 19 days , e ) 13 days | b | add(divide(subtract(const_1, add(multiply(5, divide(const_1, 10)), multiply(5, divide(const_1, 20)))), divide(const_1, 20)), 5) | a can do a piece of work in 10 days and b in 20 days . they began the work together but 5 days before the completion of the work , a leaves . the work was completed in ? | b ( x Γ’ β¬ β 5 ) / 10 + x / 20 = 1 x = 10 days | a = 1 / 10
b = 5 * a
c = 1 / 20
d = 5 * c
e = b + d
f = 1 - e
g = 1 / 20
h = f / g
i = h + 5
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a ) 8.5 gallons , b ) 7.5 gallons , c ) 6.5 gallons , d ) 5.5 gallons , e ) 4.5 gallons | c | divide(130, 20) | a car gets 20 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 130 kilometers ? | "each 20 kilometers , 1 gallon is needed . we need to know how many 20 kilometers are there in 130 kilometers ? 130 / 20 = 6.5 * 1 gallon = 6.5 gallons correct answer c" | a = 130 / 20
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a ) 40 m 2 , b ) 44 m 2 , c ) 52.8 m 2 , d ) 36 m 2 , e ) none of these | c | multiply(6, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 6 revolutions ? ( use Γ― β¬ = 22 Γ’ Β β 7 ) | "required area covered in 5 revolutions = 6 Γ£ β 2 Γ― β¬ rh = 6 Γ£ β 2 Γ£ β 22 Γ’ Β β 7 Γ£ β 0.7 Γ£ β 2 = 52.8 m 2 answer c" | a = 22 / 7
b = 2 * a
c = 1 / 4
d = b * c
e = d * 2
f = 6 * e
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a ) 62.5 , b ) 62.0 , c ) 266.67 , d ) 62.1 , e ) 62.2 | c | divide(multiply(add(200, divide(multiply(200, 20), const_100)), const_100), subtract(const_100, 10)) | at what price must an article costing rs . 200 be marked in order that after deducting 10 % from the list price . it may be sold at a profit of 20 % on the cost price ? | "cp = 200 sp = 200 * ( 120 / 100 ) = 240 mp * ( 90 / 100 ) = 240 mp = 266.67 answer : c" | a = 200 * 20
b = a / 100
c = 200 + b
d = c * 100
e = 100 - 10
f = d / e
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | add(11, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | evaluate : 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = ? | "according to order of operations , inner brackets first where 6 Γ 4 Γ· 3 is first calculated since it has a multiplication and a division . 6 Γ 4 Γ· 3 = 24 Γ· 3 = 8 hence 11 + sqrt ( - 4 + 6 Γ 4 Γ· 3 ) = 11 + sqrt ( - 4 + 8 ) = 11 + sqrt ( 4 ) = 11 + 2 = 13 correct answer d ) 13" | a = 6 * 4
b = a / 3
c = b - 4
d = math.sqrt(c)
e = 11 + d
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a ) 1000 , b ) 10 , c ) 567 , d ) 57 , e ) 678 | a | divide(multiply(1000, 100), 1100) | the lcm and hcf of two numbers are 1000 and 100 respectively . find the larger of the two numbers if their sum is 1100 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 1000 * 100 = 100000 so the answer which we are looking for has to be a factor of 100000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 100000 . and then take the highest number as the answer as the question asks abt the highest number . answer is a" | a = 1000 * 100
b = a / 1100
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a ) 17608 , b ) 17606 , c ) 17604 , d ) 18400 , e ) 117601 | d | divide(multiply(add(const_100, 15), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 15 % if no discount was offered ? | "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 15 % if no discount was offered = 16000 * 115 / 100 = rs . 18400 . answer : d" | a = 100 + 15
b = 12500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
|
a ) 5 : 3 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(10, const_100), divide(1, 6)) | in a school 10 % of the boys are same in number as 1 / 6 th of the girls . what is the ratio of boys to the girls in the school ? | "10 % of b = 1 / 6 g 10 b / 100 = g / 6 b = 5 g / 3 b / g = 5 / 3 b : g = 5 : 3 answer is a" | a = 10 / 100
b = 1 / 6
c = a / b
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a ) 654 , b ) 584 , c ) 384 , d ) 364 , e ) 394 | d | divide(factorial(subtract(15, const_1)), multiply(factorial(subtract(const_4, const_1)), factorial(subtract(subtract(15, const_1), subtract(const_4, const_1))))) | the number of positive integer solutions for the equation x + y + z + t = 15 is | the number of positive integer solutions for the equatio fx 1 + x 2 + β― + xn = k ( k - 1 ) c ( n - 1 ) - where k is the number and n is number of variable in the equation . 15 - 1 c 4 - 1 = 14 c 3 = 364 answer : d | a = 15 - 1
b = math.factorial(a)
c = 4 - 1
d = math.factorial(c)
e = 15 - 1
f = 4 - 1
g = e - f
h = math.factorial(g)
i = d * h
j = b / i
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a ) 400 , b ) 625 , c ) 1250 , d ) 2500 , e ) 10 000 | c | divide(50, divide(2, 50)) | in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "if x is total number of fish in the pond : 4 = 50 / x * 100 = > x = 1250 so answer is c" | a = 2 / 50
b = 50 / a
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a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(const_100, divide(8, subtract(12, 8))) | if the cost price of 12 pencils is equal to the selling price of 8 pencils , the gain percent is : | c 50 % let c . p . of each pencil be $ 1 . then , c . p . of 8 pencils = $ 8 ; s . p . of 8 pencils = $ 12 . gain % = 4 / 8 * 100 = 50 % | a = 12 - 8
b = 8 / a
c = 100 / b
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a ) 10 , b ) 99 , c ) 18 , d ) 55 , e ) 21 | c | add(inverse(subtract(divide(const_1, 9), divide(const_1, 18))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 9 days . if a alone can do it in 18 days . in how many days can b alone do it ? | "1 / 9 β 1 / 18 = 1 / 18 = > 18 answer : c" | a = 1 / 9
b = 1 / 18
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
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a ) - 8 , b ) - 2 , c ) - 5 , d ) - 4 , e ) 1 | b | divide(negate(add(13, 1)), 6) | solve below question 6 x - 1 = - 13 | 1 . subtract 1 from both sides : 6 x - 1 + 1 = - 13 + 1 2 . simplify both sides : 6 x = - 12 3 . divide both sides by 6 : 4 . simplify both sides : x = - 2 b | a = 13 + 1
b = negate / (
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a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | subtract(reminder(add(power(reminder(32, const_10), reminder(43, const_4)), power(reminder(43, const_10), const_4)), const_100), multiply(const_2, const_3)) | if n = ( 32 ) ^ 43 + ( 43 ) ^ 33 what is the units digit of n ? | "first of all , the units digit of ( 32 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 33 is the same as that of 3 ^ 33 . so , we need to find the units digit of 3 ^ 43 + 3 ^ 33 . next , the units digit of 3 in positive integer power repeats in blocks of four { 3 , 9 , 7 , 1 } : 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ) . . . thus : the units digit of 3 ^ 43 is the same as the units digit of 3 ^ 3 , so 7 ( 43 divided by the cyclicity of 4 gives the remainder of 3 ) . the units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1 , so 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ) . therefore the units digit of ( 23 ) ^ 43 + ( 43 ) ^ 33 is 5 + 3 = 0 . answer : e ." | a = reminder ** (
b = a + reminder
c = reminder - (
|
a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) - 3 | e | divide(log(divide(1, 125)), log(5)) | if 125 ( 5 ^ y ) = 1 then y = | 5 ^ x = 1 / 125 5 ^ x = 1 / 5 ^ 3 5 ^ x = 5 ^ - 3 x = - 3 e | a = 1 / 125
b = math.log(a)
c = math.log(5)
d = b / c
|
a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none | b | divide(add(add(add(add(8000, 5000), 11000), 7000), 9000), add(const_4, const_1)) | the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 11000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 11000 + 7000 + 9000 / 5 = rs . 8000 correct option : b" | a = 8000 + 5000
b = a + 11000
c = b + 7000
d = c + 9000
e = 4 + 1
f = d / e
|
a ) 6 , b ) 78 , c ) 8 , d ) 9 , e ) 10 | a | add(divide(subtract(multiply(70, 4), multiply(75, 4)), subtract(60, 70)), 4) | a car averages 75 mph for the first 4 hours of a trip and averages 60 mph for each additional hour . the average speed for the entire trip was 70 mph . how many hours long is the trip ? | "let the time for which car averages 60 mph = t 70 * ( t + 4 ) = 75 * 4 + 60 t = > 10 t = 20 = > t = 2 total duration of the trip = 4 + 2 = 6 answer a" | a = 70 * 4
b = 75 * 4
c = a - b
d = 60 - 70
e = c / d
f = e + 4
|
a ) $ 840 , b ) $ 800 , c ) $ 1000 , d ) $ 700 , e ) $ 1500 | a | divide(210, subtract(const_1, divide(3, 4))) | linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 210 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 210 . so 1 / 4 of her savings is $ 210 . so her original savings are 4 times $ 210 = $ 840 correct answer a" | a = 3 / 4
b = 1 - a
c = 210 / b
|
a ) 1 : 2 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13 | c | divide(power(1, const_2), power(4, const_2)) | the duplicate ratio of 1 : 4 is ? | "1 ^ 2 : 4 ^ 2 = 2 : 16 = 1 : 8 answer : c" | a = 1 ** 2
b = 4 ** 2
c = a / b
|
a ) 287 , b ) 94.2 , c ) 156 , d ) 158 , e ) 267 | b | multiply(circumface(divide(20, const_2)), 1.50) | find the cost of fencing around a circular field of diameter 20 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 10 = 62.8 62.8 * 1 1 / 2 = rs . 94.2 answer : b" | a = 20 / 2
b = circumface * (
|
a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(6, multiply(divide(0.375, 5), 60)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(60, divide(subtract(6, multiply(divide(0.375, 5), 60)), subtract(divide(1.375, 10), divide(0.375, 5))))) | a 60 kg metal bar made of alloy of tin and silver lost 6 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 6 kg when 60 kg alloy is immersed . this is a loss of ( 6 / 60 ) * 100 = 10 % . this is aavg the loss of tin is 1.375 kg for every 10 kg . this means it loses ( 1.375 / 10 ) * 100 = 13.75 % of its weight in water . this is a 1 . the loss of silver is . 375 kg for every 5 kg . this means it loses ( . 375 / 5 ) * 100 = 7.5 % of its weight in water . this is a 2 . weight of tin / weight of silver = ( silver ' s loss - avg loss ) / ( avg loss - tin ' s loss ) x / y = ( 7.5 - 10 ) / ( 10 - 13.75 ) = 2 / 3 e" | a = 0 / 375
b = a * 60
c = 6 - b
d = 1 / 375
e = 0 / 375
f = d - e
g = c / f
h = 0 / 375
i = h * 60
j = 6 - i
k = 1 / 375
l = 0 / 375
m = k - l
n = j / m
o = 60 - n
p = g / o
|
a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km | e | multiply(10, 20) | a walks at 10 kmph and 10 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 10 hours . x / 10 - x / 20 = 10 x = 200 km answer is e" | a = 10 * 20
|
a ) 387 , b ) 429 , c ) 457 , d ) 499 , e ) 536 | c | add(add(add(add(add(add(add(60, const_1), add(add(60, const_1), const_1)), add(add(add(60, const_1), const_1), const_2)), add(add(add(add(60, const_1), const_1), const_2), const_1)), add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1), const_1), const_1)) | the sum of the non - prime numbers between 60 and 70 , non - inclusive , is | "sum of consecutive integers from 61 to 69 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 61 + 69 ) / 2 * 9 = 65 * 9 = 585 sum of non - prime numbers b / w 60 and 70 , non inclusive = = = > 585 - 128 ( i . e . , 61 + 67 , being the prime # s in the range ) = 457 answer : c" | a = 60 + 1
b = 60 + 1
c = b + 1
d = a + c
e = 60 + 1
f = e + 1
g = f + 2
h = d + g
i = 60 + 1
j = i + 1
k = j + 2
l = k + 1
m = h + l
n = 60 + 1
o = n + 1
p = o + 2
q = p + 1
r = q + 1
s = m + r
t = 60 + 1
u = t + 1
v = u + 2
w = v + 1
x = w + 1
y = x + 1
z = s + y
A = 60 + 1
B = A + 1
C = B + 2
D = C + 1
E = D + 1
F = E + 1
G = F + 1
H = z + G
|
a ) 15 s , b ) 20 s , c ) 25 s , d ) 30 s , e ) 35 s | d | divide(500, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 500 meter long train moving with a speed of 63 km / hr , take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | distance = 500 m speed = 63 - 3 km / hr = 60 km / hr = 600 / 36 m / s = 50 / 3 m / s time taken = distance / speed = 500 / ( 50 / 3 ) = 30 s answer is d . | a = 63 - 3
b = a * const_0_2778
c = 500 / b
|
a ) 232 , b ) 450 , c ) 252 , d ) 262 , e ) 272 | b | divide(50, divide(450, 50)) | evaluate 50 % of 450 + 45 % of 500 | "explanation : = ( 50 / 100 ) * 450 + ( 45 / 100 ) * 500 = 450 option b" | a = 450 / 50
b = 50 / a
|
a ) - 10 , b ) 10 , c ) - 19 , d ) 20 , e ) 21 | a | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of b - a ? | "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 b - a = 100 - 110 = - 10 ans : a" | a = 20 / 2
b = 20 / 2
c = b + 1
d = a * c
e = 19 + 1
f = e / 2
g = 19 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
|
a ) 2.3 m , b ) 4.6 m , c ) 7.8 m , d ) 9.2 m , e ) 10.1 | d | multiply(4.6, const_2) | the angle of elevation of a ladder leaning against a wall is 60 ΒΊ and the foot of the ladder is 4.6 m away from the wall . the length of the ladder is : | "let ab be the wall and bc be the ladder . then , acb = 60 ΒΊ and ac = 4.6 m . ac / bc = cos 60 ΒΊ = 1 / 2 bc = 2 x ac = ( 2 x 4.6 ) m = 9.2 m . answer : option d" | a = 4 * 6
|
a ) s . 6420 , b ) s . 3563 , c ) s . 2400 , d ) s . 5480 , e ) s . 5076 | e | multiply(multiply(94, 54), divide(50, 10)) | what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 94 m & 54 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 94 * 54 = 5076 m . total cost of levelling = rs . 5076 e" | a = 94 * 54
b = 50 / 10
c = a * b
|
a ) - 13 , b ) 13 , c ) 5 , d ) 7 , e ) 8 | b | add(8, 5) | what is the minimum value of | x - 4 | + | x + 8 | + | x - 5 | ? | "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 8 | > = 0 - - > minimum occurs at x = - 8 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 8 - - > result = 12 + 0 + 13 = 25 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 12 + 1 = 13 x = 5 - - > result = 1 + 13 + 0 = 14 x = 8 - - > result = 4 + 16 + 3 = 23 so minimum value of the expression occurs at x = 4 and the resultant value = 13 answer : b" | a = 8 + 5
|
a ) 350 , b ) 375 , c ) 320 , d ) 530 , e ) 230 | c | divide(1600, const_3) | divide rs . 1600 among a , b and c so that a receives 1 / 4 as much as b and c together and b receives 3 / 4 as a and c together . a ' s share is ? | "a + b + c = 1600 a = 1 / 4 ( b + c ) ; b = 3 / 4 ( a + c ) a / ( b + c ) = 1 / 4 a = 1 / 5 * 1600 = > 320 answer : c" | a = 1600 / 3
|
a ) 13300 , b ) 13350 , c ) 13225 , d ) 13500 , e ) 13600 | c | add(10000, multiply(divide(multiply(10000, 15), const_100), 2)) | the population of a town is 10000 . it increases annually at the rate of 15 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ£ β 115 / 100 Γ£ β 115 / 100 = 13225 c" | a = 10000 * 15
b = a / 100
c = b * 2
d = 10000 + c
|
a ) 9 , b ) 9.6 , c ) 11 , d ) 12 , e ) 13 | b | subtract(subtract(16, 3.2), 3.2) | a man ' s speed with the current is 16 km / hr and the speed of the current is 3.2 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 16 - 3.2 ) km / hr = 12.8 km / hr . man ' s rate against the current = ( 12.8 - 3.2 ) km / hr = 9.6 km / hr . answer : b" | a = 16 - 3
b = a - 3
|
a ) 1 , b ) 20 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(multiply(multiply(const_4, 3), 2), 1), multiply(multiply(3, 2), 1)) | a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β high β and β low β cards to earn 6 points if you must draw exactly 3 β low β cards ? | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 3 times . as a result , the formula is : 5 ! / 3 ! which represents ( 5 * 4 * 3 * 2 * 1 ) / ( 3 * 2 * 1 ) which simplifies to just 20 , giving you answer b ." | a = 4 * 3
b = a * 2
c = b * 1
d = 3 * 2
e = d * 1
f = c / e
|
a ) 9 / 13 , b ) 5 / 12 , c ) 3 / 13 , d ) 6 / 17 , e ) 1 / 13 | a | divide(subtract(52, multiply(const_4, const_4)), 52) | a card is drawn from a pack of 52 cards . the probability of not getting a face card ? | clearly in the 52 cards other than face cards = 52 - 16 = 36 probability of not getting a face card = 36 / 52 = 9 / 13 correct option is a | a = 4 * 4
b = 52 - a
c = b / 52
|
a ) 128 , b ) 225 , c ) 288 , d ) 324 , e ) 336 | a | divide(64, subtract(const_1, divide(const_1, const_4))) | what number is 64 more than two - fourth of itself ? | "2 / 4 x + 64 = x that means 64 = 2 / 4 x x = 64 * 2 = 128 a is the answer" | a = 1 / 4
b = 1 - a
c = 64 / b
|
a ) 37 , b ) 36 , c ) 39 , d ) 40 , e ) 41 | b | add(add(multiply(const_2, 10), 10), 6) | in 10 years , a will be twice as old as b was 10 years ago . if a is now 6 years older than b the present age of b is | "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 6 = > 2 b - b - 6 = 30 b = 36 so the present age of b is 36 years answer : b" | a = 2 * 10
b = a + 10
c = b + 6
|
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