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a ) 4 , b ) 9 , c ) 6 , d ) 8 , e ) 10 | b | add(add(const_2, add(const_1, const_4)), const_2) | an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 100000 people who applied for the army and 175 were finally selected , what was the value of n ? | "fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 16000 = 8000 step ( 4 ) 50 % of 8000 = 4000 step ( 5 ) 50 % of 4000 = 2000 step ( 6 ) 50 % of 2000 = 1000 step ( 7 ) 50 % of 1000 = 500 step ( 8 ) 50 % of 500 = 250 step ( 9 ) 70 % of 350 = 175 ( our target ) total 9 steps required . ans b" | a = 1 + 4
b = 2 + a
c = b + 2
|
a ) 3.0 , b ) 3.36 , c ) 24.64 , d ) 25.0 , e ) 31.36 | d | divide(multiply(28, const_100), add(const_100, 12)) | from the sale of sleeping bags , a retailer made a gross profit of 12 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | cost price * 1.12 = selling price - - > cost price * 1.12 = $ 28 - - > cost price = $ 25 . answer : d . actually even without any math only c and d make any sense , but since 24.64 * 1.12 wo n ' t be an integer ( $ 28 ) then only answer choice d remains . | a = 28 * 100
b = 100 + 12
c = a / b
|
a ) 30 days , b ) 40 days , c ) 50 days , d ) 59 days , e ) 65 days | d | divide(multiply(700, 25), 300) | there is enough provisions for 700 men in an army camp for 25 days . if there were 300 men less , how long will the provision last ? | "exp : we have , m 1 d 1 = m 2 d 2 700 * 25 = 300 * d 2 d 2 = 700 * 25 / 300 = 59 days . answer : d" | a = 700 * 25
b = a / 300
|
a ) 11 , b ) 23 3 / 4 , c ) 16 1 / 2 , d ) 9 1 / 2 , e ) 98 | d | add(8, divide(140, add(24, 46))) | the distance between delhi and mathura is 140 kms . a starts from delhi with a speed of 24 kmph at 7 a . m . for mathura and b starts from mathura with a speed of 46 kmph at 8 a . m . from delhi . when will they meet ? | "d = 140 β 24 = 116 rs = 46 + 24 = 70 t = 116 / 70 = 1.6 hours 8 a . m . + 1.6 = 9 1 / 2 a . m . . answer : d" | a = 24 + 46
b = 140 / a
c = 8 + b
|
a ) 5 , b ) 6 , c ) 3 , d ) 4 , e ) 7 | d | add(3, const_1) | 3 distinct single - digit numbers . a , b , c are in gp . if abs ( x ) for real x is the absolute value of x ( x if x is + ve or 0 and - x if x is - ve ) , then the number of different possible values of abs ( a + b + c ) is | a , b , c may be ( 1 , 2,4 ) & ( 4 , 2,1 ) ( 1 , 3,9 ) & ( 9 , 3,1 ) ( 2 , 4,8 ) & ( 8 , 4,2 ) ( 4 , 6,9 ) & ( 9 , 6,4 ) find abs ( a + b + c ) for these 8 gps 7,13 , 15,19 so , we get 4 different values answer : d | a = 3 + 1
|
a ) 15 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 6 % | e | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 25), subtract(const_100, 20)), const_100)), multiply(divide(subtract(const_100, 25), subtract(const_100, 20)), const_100)), const_100) | the charge for a single room at hotel p is 25 percent less than the charge for a single room at hotel r and 20 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "let rate in r = 100 x then p = 75 x g = 100 y p = 80 y thus 75 x = 80 y or x = 1.06 y ans r = 106 y so increase = 6 % answer : e" | a = 100 - 25
b = 100 - 20
c = a / b
d = c * 100
e = 100 - d
f = 100 - 25
g = 100 - 20
h = f / g
i = h * 100
j = e / i
k = j * 100
|
a ) 52 , b ) 53 , c ) 54 , d ) 55 , e ) 51 | e | subtract(multiply(4, const_2), multiply(2, const_2)) | if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 53 , what is the value of x ? | "am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 53 x = 51 answer : e" | a = 4 * 2
b = 2 * 2
c = a - b
|
a ) 16000 , b ) 18000 , c ) 20000 , d ) 30000 , e ) none of these | d | divide(multiply(multiply(18, const_100), multiply(12, const_100)), multiply(12, 6)) | a courtyard is 18 meter long and 12 meter board is to be paved with bricks of dimensions 12 cm by 6 cm . the total number of bricks required is : | "explanation : number of bricks = courtyard area / 1 brick area = ( 1800 Γ£ β 1200 / 12 Γ£ β 6 ) = 30000 option d" | a = 18 * 100
b = 12 * 100
c = a * b
d = 12 * 6
e = c / d
|
a ) 1 km , b ) 2 km , c ) 3.3 km , d ) 4 km , e ) 5 km | c | multiply(multiply(divide(divide(52, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000) | a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 52 minutes . the total distance is ? | "c 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 52 / 60 47 x / 60 = 52 / 60 = > x = 1.1 total distance = 3 * 1.1 = 3.3 km ." | a = 52 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 9 / 7 | c | divide(subtract(divide(2, 3), divide(1, 4)), add(subtract(divide(2, 3), divide(1, 4)), divide(1, 3))) | a batch of cookies was divided among 3 tins : 2 / 3 of all the cookies were placed in either the blue tin or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction w of the cookies that were placed in the other tins were placed in the green tin ? | blue tin or red tin : 2 / 3 ( n ) red tin : ( 1 / 3 ) n blue tin : ( 1 / 4 ) n what the last statment meant , is it wants this fraction : ( # of cookies in green tin ) / ( # of cookies in red and green tin ) # of cookies in green tin = 2 n / 3 - n / 4 = 8 n - 3 n / 12 = 5 n / 12 # of cookies in red and green tin = n / 3 + 5 n / 12 = 9 n / 12 fraction w = 5 n / 12 * 12 / 9 n = 5 / 9 ( c ) | a = 2 / 3
b = 1 / 4
c = a - b
d = 2 / 3
e = 1 / 4
f = d - e
g = 1 / 3
h = f + g
i = c / h
|
a ) 22 , b ) 13 , c ) 77 , d ) 99 , e ) 21 | b | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 12))) | two trains of equal lengths take 12 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 12 = 10 m / sec . speed of the second train = 120 / 15 = 8 m / sec . relative speed = 10 + 8 = 18 m / sec . required time = ( 120 + 120 ) / 18 = 13 sec . answer : b" | a = 120 * 2
b = speed + (
c = a / b
|
a ) 28 and 17 , b ) 27 and 18 , c ) 25 and 15 , d ) 27 and 15 , e ) 25 and 18 | a | add(subtract(multiply(divide(const_10, const_2), 9), divide(add(11, multiply(divide(const_10, const_2), 9)), const_2)), divide(const_10, const_2)) | the difference of two numbers is 11 and one - fifth of their sum is 9 . find the numbers . | "let the number be x and y . then , x β y = 11 - - - - ( i ) and 1 / 5 ( x + y ) = 9 = > x + y = 45 - - - - ( ii ) adding ( i ) and ( ii ) , we get : 2 x = 56 or x = 28 . putting x = 28 in ( i ) , we get : y = 17 . hence , the numbers are 28 and 17 . answer is a ." | a = 10 / 2
b = a * 9
c = 10 / 2
d = c * 9
e = 11 + d
f = e / 2
g = b - f
h = 10 / 2
i = g + h
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | b | multiply(40, const_1) | 40 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 40 % of companies which offer wireless internet also offer snacks . the answer is b ." | a = 40 * 1
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a ) 2 kmph , b ) 3 kmph , c ) 4 kmph , d ) 5 kmph , e ) 6 kmph | a | divide(subtract(divide(1, divide(6, const_60)), divide(1, divide(10, const_60))), const_2) | a boat moves down stream at the rate of 1 km in 6 minutes and upstream at the rate of 1 km in 10 minutes . the speed of current is | if speed ( in kmph ) of the boat = b and current = c , then in downstream time taken ( in hrs . ) = 1 / ( b + c ) = 6 / 60 = 1 / 10 or b + c = 10 - - - ( i ) & in upstream time taken ( in hrs . ) = 1 / ( b - c ) = 10 / 60 = 1 / 6 or b - c = 6 - - - ( ii ) from ( i ) & ( ii ) , b = 8 , c = 2 answer : a | a = 6 / const_60
b = 1 / a
c = 10 / const_60
d = 1 / c
e = b - d
f = e / 2
|
a ) 9 : 8 , b ) 8 : 15 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | b | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4))) | a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 5 : 2 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 5 : 2 - - > e : l = 5 : 2 = 20 : 8 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 20 : 15 . thus , l : s = 8 : 15 . answer : b ." | a = 4 * 3
b = 3 * 3
c = a / b
d = 3 * 4
e = 2 * 4
f = d / e
g = c / f
|
a ) 89.7 , b ) 91.4 , c ) 91.7 , d ) 91.3 , e ) 91.1 | a | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 130 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 130 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 47.7 + 42 = 89.7 cm answer : a" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 9 % , b ) 9.27 % , c ) 22.50 % , d ) 12 % , e ) none of these | c | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(8, 1)), subtract(8, 1)), const_100), subtract(8, 1))), subtract(8, 1)), const_100), add(divide(multiply(add(const_100, subtract(8, 1)), subtract(8, 1)), const_100), subtract(8, 1))) | the population of a city increases by 8 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 7 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 7 / 100 ) ^ 3 = 122.5043 β΄ required percentage = 22.50 % answer : c" | a = 8 - 1
b = 100 + a
c = 8 - 1
d = b * c
e = d / 100
f = 8 - 1
g = e + f
h = 100 + g
i = 8 - 1
j = h * i
k = j / 100
l = 8 - 1
m = 100 + l
n = 8 - 1
o = m * n
p = o / 100
q = 8 - 1
r = p + q
s = k + r
|
a ) 4.5 , b ) 17 , c ) 18 , d ) 19 , e ) 12 | a | divide(multiply(9, 3), subtract(9, 3)) | pipe a can fill a tank in 3 hours . due to a leak at the bottom , it takes 9 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 3 - 1 / x = 1 / 9 = > 1 / x = 1 / 3 - 1 / 9 = ( 3 - 1 ) / 9 = 2 / 9 = > x = 9 / 2 = 4.5 . answer : a" | a = 9 * 3
b = 9 - 3
c = a / b
|
a ) 32.51 , b ) 32.4 , c ) 32.1 , d ) 32.92 , e ) 24.69 | e | add(divide(circumface(4.8), const_2), multiply(4.8, const_2)) | the radius of a semi circle is 4.8 cm then its perimeter is ? | "36 / 7 r = 4.8 = 24.69 answer : e" | a = circumface / (
b = a + 2
|
a ) 11 , b ) 33 , c ) 45 , d ) 67 , e ) 80 | a | divide(add(multiply(33, 6), multiply(33, 5)), add(6, 5)) | the number of stamps that p and q had were in the ratio of 9 : 2 respectively . after p gave q 33 stamps , the ratio of the number of p ' s stamps to the number of q ' s stamps was 6 : 5 . as a result of the gift , p had how many more stamps than q ? | "p started with 9 k stamps and q started with 2 k stamps . ( 9 k - 33 ) / ( 2 k + 33 ) = 6 / 5 45 k - 12 k = 178 + 165 33 k = 343 k = 11 p has 9 ( 11 ) - 33 = 66 stamps and q has 2 ( 11 ) + 33 = 55 stamps . the answer is a ." | a = 33 * 6
b = 33 * 5
c = a + b
d = 6 + 5
e = c / d
|
a ) 2 , b ) 8 , c ) 1 , d ) 6 , e ) 5 | a | subtract(10, divide(40, 5)) | a man can row a distance of 5 km in 60 min with the help of the tide . the direction of the tide reverses with the same speed . now he travels a further 40 km in 10 hours . how much time he would have saved if the direction of tide has not changed ? | explanation : he covered 5 km in 1 hour , so he might cover 40 km in 8 hours . but he took 10 hours . he would have saved 10 Γ’ β¬ β 8 = 2 hours . answer : a | a = 40 / 5
b = 10 - a
|
a ) 16 , b ) 80 , c ) 160 , d ) 180 , e ) 400 | e | add(multiply(const_100, 3), const_100) | how many 3 - digit numerals end with a digit that represents a prime number ? | "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 4 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 10 ways total = 4 * 10 * 10 = 400 ans : e" | a = 100 * 3
b = a + 100
|
a ) 15 , b ) 17 , c ) 19 , d ) 21 , e ) 24 | c | divide(add(multiply(6, const_60), 20), add(add(add(const_1, 3), add(3, const_1)), multiply(add(3, const_1), 3))) | machine a and machine b process the same work at different rates . machine c processes work as fast as machines a and b combined . machine d processes work 3 times as fast as machine c ; machine d β s work rate is also exactly 4 times machine b β s rate . assume all 4 machines work at fixed unchanging rates . if machine a works alone on a job , it takes 6 hours and 20 minutes . if all 4 machines work together on the same job simultaneously , how many minutes will it take all of them to complete it ? | c = a + b d = 3 c = 3 ( a + b ) = 4 b then b = 3 a and c = 4 a the combined rate of the four machines is a + 3 a + 4 a + 12 a = 20 a machine a can complete the work in 380 minutes , so its rate is 1 / 380 of the work per minute . the combined rate is 20 / 380 = 1 / 19 so the work will be completed in 19 minutes . the answer is c . | a = 6 * const_60
b = a + 20
c = 1 + 3
d = 3 + 1
e = c + d
f = 3 + 1
g = f * 3
h = e + g
i = b / h
|
a ) 320 $ , b ) 380 $ , c ) 410 $ , d ) 430 $ , e ) 447 $ | e | multiply(multiply(0.65, 55), 12) | in a fuel station the service costs $ 1.50 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.50 + 0.65 * 12 * 55 = 447 hence - e | a = 0 * 65
b = a * 12
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a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | c | multiply(divide(multiply(90, 2), add(60, 90)), const_60) | cole drove from home to work at an average speed of 60 kmh . he then returned home at an average speed of 90 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "let the distance one way be x time from home to work = x / 60 time from work to home = x / 90 total time = 2 hrs ( x / 60 ) + ( x / 90 ) = 2 solving for x , we get x = 72 time from home to work in minutes = ( 72 ) * 60 / 60 = 72 minutes ans = c" | a = 90 * 2
b = 60 + 90
c = a / b
d = c * const_60
|
a ) rs . 185 , b ) rs . 410 , c ) rs . 285 , d ) rs . 385 , e ) rs . 475 | a | multiply(divide(40, 65), add(add(const_100, 80), 65)) | a certain sum of money is divided among a , b and c so that for each rs . a has 80 paisa , b has 65 paisa and c 40 paisa . if c ' s share is rs . 40 , find the sum of money ? | "a : b : c = 80 : 65 : 40 = 16 : 13 : 8 8 - - - - 40 37 - - - - ? = > rs . 185 answer : a" | a = 40 / 65
b = 100 + 80
c = b + 65
d = a * c
|
a ) 4 / 15 , b ) 6 / 17 , c ) 8 / 19 , d ) 10 / 21 , e ) 12 / 23 | d | divide(multiply(choose(5, const_2), choose(5, const_2)), choose(add(5, 5), 6)) | from a group of 5 boys and 5 girls , 6 children are to be randomly selected . what is the probability that 3 boys and 3 girls will be selected ? | "the total number of ways to choose 6 children from 10 is 10 c 6 = 210 the number of ways to choose 3 boys and 3 girls is 5 c 3 * 5 c 3 = 10 * 10 = 100 p ( 3 boys and 3 girls ) = 100 / 210 = 10 / 21 the answer is d ." | a = math.comb(5, 2)
b = math.comb(5, 2)
c = a * b
d = 5 + 5
e = math.comb(d, 6)
f = c / e
|
a ) 70 , 1515 , b ) 50 , 2525 , c ) 60 , 2020 , d ) 40 , 3030 , e ) 20 , 3030 | b | multiply(multiply(subtract(100, 70), subtract(100, 95)), const_10) | 3 types of tea the a , b , c costs rs . 95 / kg , 100 / kg and 70 / kg respectively . how many kgs of each should be blended to produce 100 kg of mixture worth rs . 90 / kg , given that the quantities of band c are equal | given quantities of b and c are equal . therefore instead of considering them as a different quantities take average of both and consider it as a single entity . so the cost of the mixture ` ` d ' ' ( equal quantities of ` ` b ' ' and ` ` c ' ' ) is 170 / 2 = 85 rs / kg now the tea contains only ` ` a ' ' of 95 rs / per and ` ` d ' ' of 85 rs / kg . it is clearly seen that ` ` a ' ' is 5 more and ` ` d ' ' is 5 less than the required cost per kg . therefore 50 / 50 will go well . a = 50 kg and d = 50 kg . ( d = b & c in equal ratio ) . therefore b & c = 25 kg each . ( a : b : c ) = ( 50 , 2525 ) answer : b | a = 100 - 70
b = 100 - 95
c = a * b
d = c * 10
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a ) β 108 , b ) β 44 , c ) 10 , d ) 16 , e ) 18.25 | e | divide(add(add(20, multiply(20, 6)), 6), multiply(2, const_4)) | # p is defined as 2 p - 20 for any number p . what is p , if # ( # ( # p ) ) = 6 ? | # p = 2 p - 20 - - - > # ( # p ) = 2 ( 2 p - 20 ) - 20 = 4 p - 60 and thus # ( 4 p - 60 ) = 2 ( 4 p - 60 ) - 20 = 8 p - 140 = 6 - - - > 8 p = 146 - - - > p = 18.25 , e is the correct answer . | a = 20 * 6
b = 20 + a
c = b + 6
d = 2 * 4
e = c / d
|
a ) 1000 , b ) 2876 , c ) 1977 , d ) 2778 , e ) 1692 | e | divide(multiply(220, const_100), subtract(add(const_100, 4), subtract(const_100, 9))) | a watch was sold at a loss of 9 % . if it was sold for rs . 220 more , there would have been a gain of 4 % . what is the cost price ? | "91 % 104 % - - - - - - - - 13 % - - - - 220 100 % - - - - ? = > rs . 1692 answer : e" | a = 220 * 100
b = 100 + 4
c = 100 - 9
d = b - c
e = a / d
|
a ) 1 / 8 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 5 / 3 | d | multiply(10, add(divide(const_1, 30), divide(const_1, 60))) | two persons a and b can complete a piece of work in 30 days and 60 days respectively . if they work together , what part of the work will be completed in 10 days ? | a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 60 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 60 = 1 / 20 the part of the work completed in 10 days = 10 ( 1 / 20 ) = 1 / 2 . answer : d | a = 1 / 30
b = 1 / 60
c = a + b
d = 10 * c
|
a ) 19 kmph , b ) 15 kmph , c ) 12 kmph , d ) 20 kmph , e ) 23 kmph | b | divide(subtract(multiply(30, const_3), divide(multiply(multiply(30, const_3), const_2), const_3)), subtract(const_3, const_1)) | by travelling at 30 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "let the time taken to reach the destination be 3 x hours . total distance = 30 * 3 x = 90 x km he covered 2 / 3 * 90 x = 60 x km in 1 / 3 * 3 x = x hours so , the remaining 30 x km , he has to cover in 2 x hours . required speed = 30 x / 2 x = 15 kmph . answer : b" | a = 30 * 3
b = 30 * 3
c = b * 2
d = c / 3
e = a - d
f = 3 - 1
g = e / f
|
a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | b | divide(multiply(add(33, divide(1, 3)), 270), const_100) | 33 1 / 3 % of 270 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 270 = 90 b )" | a = 1 / 3
b = 33 + a
c = b * 270
d = c / 100
|
a ) $ 900 , b ) $ 720 , c ) $ 600 , d ) $ 300 , e ) $ 60 | a | divide(multiply(multiply(divide(add(multiply(multiply(6, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), 6), 8), multiply(const_3, 6)) | mr . hernandez , who was a resident of state x for only 8 months last year , had a taxable income of $ 22,500 for the year . if the state tax rate were 6 percent of the year β s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez β s state x tax for last year ? | "total tax for the year = 22,500 x 6 % = 1350 as stated annual tax is prorated as per the duration of stay . prorated tax = 1350 ( 8 / 12 ) = 900 answer a" | a = 6 * 100
b = a * 100
c = 100 * const_0_25
d = c * 100
e = b + d
f = e / 100
g = f * 6
h = g * 8
i = 3 * 6
j = h / i
|
a ) 546 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1300) | in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1300 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1300 female employees , 60 % of 1300 . = ( 60 * 1300 ) / 100 = 780 . then male employees , = 520 50 % of male are computer literate , = 260 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1300 ) / 100 = 806 computer literate . thus , female computer literate = 806 - 260 = 546 . answer : option a" | a = 62 / 100
b = 60 / 100
c = 1 - b
d = 50 / 100
e = c * d
f = a - e
g = f * 1300
|
a ) 48 , b ) 66 , c ) 76 , d ) 78 , e ) 72 | e | subtract(multiply(multiply(multiply(5, 4), 5), const_2), multiply(multiply(multiply(4, 5), const_2), const_2)) | each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 5 and the 4 can not be adjacent digits in the integer , how many 5 - digit integers are possible ? | "number of arrangements using 5 distinct digits = 5 ! number of arrangements in which 4 and 5 are adjacent - consider 4 and 5 together as one group . now you have 4 numbers / groups to arrange which can be done in 4 ! ways . in each of these arrangements , 4 and 5 can be arranged as 45 or 54 . number of arrangements in which 3 and 4 are not adjacent = 5 ! - 2 * 4 ! = 72 answer : e" | a = 5 * 4
b = a * 5
c = b * 2
d = 4 * 5
e = d * 2
f = e * 2
g = c - f
|
a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55 | b | divide(add(40, 20), add(divide(40, 15), divide(20, 60))) | susan drove an average speed of 15 miles per hour for the first 40 miles of a tripthen at a average speed of 60 miles / hr for the remaining 20 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles / hr for the entire trip | avg . speed = total distance / total time total distance = 60 miles total time = 40 / 15 + 20 / 60 = 3 avg . speed = 20 . answer - b | a = 40 + 20
b = 40 / 15
c = 20 / 60
d = b + c
e = a / d
|
a ) 150 , b ) 240 , c ) 277 , d ) 261 , e ) 281 | b | multiply(480, divide(480, add(add(5, 12), 13))) | the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 480 m , its area is ? | "5 x + 12 x + 13 x = 480 = > x = 16 a = 80 , b = 192 , c = 208 s = ( 80 + 192 + 208 ) / 2 = 240 answer : b" | a = 5 + 12
b = a + 13
c = 480 / b
d = 480 * c
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a ) 17 , b ) 28 , c ) 19 , d ) 11 , e ) 22 | e | divide(113, add(const_2, const_pi)) | the perimeter of a semi circle is 113 cm then the radius is ? | "36 / 7 r = 113 = > r = 22 answer : e" | a = 2 + math.pi
b = 113 / a
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a ) 12 kg , b ) 120 kg , c ) 72 kg , d ) 96 kg , e ) none of these | b | multiply(multiply(multiply(4, 3), divide(1, const_100)), const_1000) | a boat having a length 4 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "explanation : volume of water displaced = ( 4 x 3 x 0.01 ) m 3 = 0.12 m 3 . β΄ mass of man = volume of water displaced x density of water = ( 0.12 x 1000 ) kg = 120 kg . answer : b" | a = 4 * 3
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 271 , b ) 266 , c ) 350 , d ) 277 , e ) 300 | e | subtract(multiply(25, multiply(108, const_0_2778)), 450) | a train 450 m long running at 108 kmph crosses a platform in 25 sec . what is the length of the platform ? | "length of the platform = 108 * 5 / 18 * 25 = 750 β 450 = 300 answer : e" | a = 108 * const_0_2778
b = 25 * a
c = b - 450
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a ) 20 kg , b ) 25 kg , c ) 30 kg , d ) 35 kg , e ) none of these | b | multiply(divide(add(const_100, 10), add(const_100, 32)), 30) | when the price of sugar was increased by 32 % , a family reduced its consumption in such a way that the expenditure on sugar was only 10 % more than before . if 30 kg were consumed per month before , find the new monthly consumption . | since , expenditure = price Γ consumption β΄ 110 % of 30 = 132 β 100 Γ new consumption β 110 β 100 Γ 30 = 132 β 100 Γ new consumption β new consumption = 25 kg answer b | a = 100 + 10
b = 100 + 32
c = a / b
d = c * 30
|
a ) 145 , b ) 38 , c ) 173 , d ) 156 , e ) 189 | c | add(multiply(subtract(25, const_1), 7), 5) | find the 25 th term of an arithmetic progression whose first term is 5 and the common difference is 7 . | n th term of a . p = a + ( n - 1 ) * d = 5 + ( 25 - 1 ) * 7 , = 5 + 168 = 173 . answer : c | a = 25 - 1
b = a * 7
c = b + 5
|
a ) 0.56 , b ) 0.62 , c ) 0.68 , d ) 0.74 , e ) 0.8 | b | subtract(const_1, add(multiply(divide(40, const_100), divide(50, const_100)), multiply(divide(30, const_100), subtract(const_1, divide(40, const_100))))) | in a graduating class , 40 percent of the students are male . in this class , 50 percent of the male students and 30 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | let x be the total number of students . the number students who are younger than 25 is 0.5 * 0.4 x + 0.7 * 0.6 x = 0.62 x the answer is b . | a = 40 / 100
b = 50 / 100
c = a * b
d = 30 / 100
e = 40 / 100
f = 1 - e
g = d * f
h = c + g
i = 1 - h
|
a ) 1 , b ) 5 , c ) 7 , d ) 8 , e ) 11 | d | add(add(5, 2), const_1) | the function g ( a ) is defined for integers a such that if a is even , g ( a ) = a / 2 and if a is odd , g ( a ) = a + 5 . given that g ( g ( g ( g ( g ( a ) ) ) ) ) = 19 , how many possible values for a would satisfy this equation ? | let me define terms : in g ( a ) = r a is argument , r is result , g ( ) is function , in g ( g ( g ( g ( g ( a ) ) ) ) ) , g 1 is inner most , g 5 is outermost for identification . from definition of function g , we can deduce that : if result is even then two possibilities for argument = 1 even 1 odd if result is odd then one possibility for argument = 1 even since final result = 19 = odd possibilities : g 1 : 1 even g 2 : 1 * ( even , odd ) = 1 even 1 odd g 3 : 1 * ( even , odd ) + 1 even = 2 even 1 odd g 4 : 2 * ( even , odd ) + 1 even = 3 even 2 odd g 5 : 3 * ( even , odd ) + 2 even = 5 even 3 odd = total 8 ans d it is ! | a = 5 + 2
b = a + 1
|
a ) 7 , b ) 33 , c ) 12 , d ) 17 , e ) 25 | e | add(16, 9) | there are 16 bees in the hive , then 9 more fly . how many bees are there in all ? | 16 + 9 = 25 . answer is e . | a = 16 + 9
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a ) 10 , b ) 12 , c ) 21 , d ) 18 , e ) 20 | c | add(20, 1) | seller selling an apple for rs . 20 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 20 loss = cp 21 loss = cp β sp = cp β 20 β cp 21 = cp β 20 β 20 cp 21 = 20 β cp 21 = 1 β cp = 21 c" | a = 20 + 1
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a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 00 | e | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(const_2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 402 * 503 * 604 * 645 * 547 * 448 * 349 ) ^ 2 | ( ( 301 * 402 * 503 * 604 * 645 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 5,7 , 8,9 ; 6 is missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,45 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 45 ) ( 56 * 49 ) ) ^ 2 = ( 00 * 44 ) ^ 2 = 00 ^ 2 = 00 hence answer is e | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) $ 22 , b ) $ 27.6 , c ) $ 30 , d ) $ 36 , e ) $ 40 | b | multiply(divide(multiply(1.15, 8), subtract(8, 6)), 6) | nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 1.15 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "b its is . let price = x ( x - 1.15 ) 8 = 6 x x = 4.6 hence total money = 6 * 4.6 = 27.6" | a = 1 * 15
b = 8 - 6
c = a / b
d = c * 6
|
a ) β 5 % , b ) 5 % , c ) 12 % , d ) 20 % , e ) 80 % | c | multiply(subtract(multiply(add(const_1, divide(60, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 60 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | the actual answer is obtained by multiplying 160 % by 70 % and subtracting 100 % from this total . that is : 160 % Γ 70 % = 112 % ; 112 % β 100 % = 12 % . answer : c | a = 60 / 100
b = 1 + a
c = 30 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 3 / 4 , b ) 3 / 2 , c ) 1 / 2 , d ) 3 / 4 , e ) 1 / 4 | c | divide(add(add(add(choose(3, const_2), choose(3, const_3)), choose(3, const_4)), choose(3, 3)), power(const_2, 3)) | a fair coin is tossed 3 times . what is the probability of getting at least 2 tails ? | "let ' s find the probability of 2 tails , 3 tails p ( ttt ) = ( ( 1 / 2 ) ^ 3 = 1 / 8 . p ( htt ) = ( 3 ! / 2 ! ) * ( 1 / 2 ) ^ 3 = 3 / 8 total probablity = 1 / 8 + 3 / 8 = 1 / 2 answer ( c )" | a = math.comb(3, 2)
b = math.comb(3, 3)
c = a + b
d = math.comb(3, 4)
e = c + d
f = math.comb(3, 3)
g = e + f
h = 2 ** 3
i = g / h
|
a ) 12.5 % , b ) 14 % , c ) 15.5 % , d ) 17 % , e ) 18.5 % | a | add(3.5, multiply(divide(10, const_100), 75)) | a 75 - liter solution of cool - drink is made from 10 % jasmine water . if 3.5 liters of jasmine and 9.5 liters of water were added to the solution , what percent of the solution is jasmine ? | "the percent of jasmine in the resulting solution is : ( amount of jasmine ) / ( total volume ) ( 0.1 ( 75 ) + 3.5 ) / 88 = 11 / 88 = 1 / 8 = 12.5 % the answer is a ." | a = 10 / 100
b = a * 75
c = 3 + 5
|
a ) 200 sec , b ) 240 sec , c ) 167 sec , d ) 197 sec , e ) 179 sec | b | divide(add(1200, 1200), divide(1200, 120)) | a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 1200 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 2400 m t = d / s t = 2400 / 10 t = 240 sec answer : b" | a = 1200 + 1200
b = 1200 / 120
c = a / b
|
a ) 310 meter , b ) 335 meter , c ) 345 meter , d ) 350 meter , e ) none of these | d | subtract(multiply(divide(300, 18), 39), 300) | a 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform . | "explanation : speed = distance / time = 300 / 18 = 50 / 3 m / sec let the length of the platform be x meters then distance = speed β time x + 300 = 50 / 3 β 39 = > 3 ( x + 300 ) = 1950 = > x = 350 meters option d" | a = 300 / 18
b = a * 39
c = b - 300
|
a ) 120 sec , b ) 198 sec , c ) 178 sec , d ) 665 sec , e ) 20 sec | e | divide(100, subtract(multiply(36, const_0_2778), multiply(18, const_0_2778))) | a and b go around a circular track of length 100 m on a cycle at speeds of 18 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 100 / ( 18 * 5 / 18 ) , 100 / ( 36 * 5 / 18 ) } = lcm ( 20 , 10 ) = 20 sec . answer : e" | a = 36 * const_0_2778
b = 18 * const_0_2778
c = a - b
d = 100 / c
|
a ) $ 3 , b ) $ 2 , c ) $ 9 , d ) $ 1 , e ) $ 4 | d | subtract(reminder(add(multiply(7, multiply(const_1000, const_1000)), 348340), add(5, const_1)), const_1) | zoey won the lottery and got $ 7 , 348340 . she wants to split it among herself and 5 friends evenly . how much money must she add if she wants to split it evenly ? | zoey and her 5 friends is 6 people in all . $ 7 , 348340 must be divisible by 6 if she wants to split it evenly . the money is divisible by 6 if it ' s divisible by 2 and 3 . 7 + 3 + 4 + 8 + 3 + 4 + 0 = 29 . 2 and 3 cant go into 29 . adding 1 dollar makes it 30 . 2 and 3 can go into 30 so 6 can also . the answer is d . | a = 1000 * 1000
b = 7 * a
c = b + 348340
d = 5 + 1
e = reminder - (
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a ) 2 , b ) 6278 , c ) 277 , d ) 281 , e ) 22 | a | multiply(const_2, const_1) | in a certain code language , ' book ' is coded as ' pencil ' , ' pencil ' is coded as ' mirror ' , ' mirror ' is coded as ' board ' . then what is useful to write on a paper ? | we use pencil to write on a paper but here pencil is coded as mirror . so , the answer is mirror . answer : a | a = 2 * 1
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a ) 65 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 100 deg | e | subtract(multiply(40, multiply(const_3, const_2)), 8) | what is the angle between the hands of a clock when time is 8 : 40 ? | "angle between two hands = 40 h - 11 / 2 m = 40 * 8 - 40 * 11 / 2 = 320 - 220 = 100 deg answer : e" | a = 3 * 2
b = 40 * a
c = b - 8
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(50, const_10) | how many integers from 0 to 50 inclusive have a remainder of 3 when divided by 6 ? | "the numbers should be of the form 6 c + 3 . the minimum is 3 when c = 0 . the maximum is 45 when c = 7 . there are 8 such numbers . the answer is d ." | a = 50 / 10
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a ) 1 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 | e | divide(add(4, 4), 4) | a box contains 4 red balls and 4 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 4 black balls be bbbband 4 red balls be rrrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 8 ! ) / ( 4 ! x 4 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 7 slots and 3 black ( bbb ) and 4 red ( rrrr ) they can be arranged in ( 7 ! ) / ( 4 ! x 3 ! ) hence required probability = [ ( 8 ! ) / ( 4 ! x 4 ! ) ] / [ ( 7 ! ) / ( 4 ! x 3 ! ) ] = 2 e" | a = 4 + 4
b = a / 4
|
a ) 20 % , b ) 30 % , c ) 25 % , d ) 15 % , e ) 50 % | c | multiply(divide(subtract(50, 30), add(50, 30)), const_100) | if 50 % of ( x - y ) = 30 % of ( x + y ) , then what percent of x is y ? | "50 % of ( x - y ) = 30 % of ( x + y ) 50 / 100 ( x - y ) = 30 / 100 ( x + y ) x = 4 y required percentage = y / x * 100 = y / 4 y * 100 = 25 % answer is c" | a = 50 - 30
b = 50 + 30
c = a / b
d = c * 100
|
a ) 5 ^ 5 , b ) 5 ^ 6 , c ) 3 ^ 6 , d ) 6 ^ 3 , e ) 15 ^ 3 | a | divide(power(30, 10), power(30, 5)) | 30 ^ 10 / 180 ^ 5 = ? | "30 ^ 10 / 180 ^ 5 = ? a . 5 ^ 5 b . 5 ^ 6 c . 3 ^ 6 d . 6 ^ 3 e . 15 ^ 3 - > 30 ^ 10 / 180 ^ 5 = ( 30 ^ 10 ) / ( 6 ^ 5 ) ( 30 ^ 5 ) = ( 30 ^ 5 ) / ( 6 ^ 5 ) = ( 6 ^ 5 ) ( 5 ^ 5 ) / ( 6 ^ 5 ) = 5 ^ 5 . thus , a is the answer ." | a = 30 ** 10
b = 30 ** 5
c = a / b
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | e | add(add(const_2, add(const_1, const_4)), const_2) | an army β s recruitment process included n rounds of selection tasks . for the first a rounds , the rejection percentage was 60 percent per round . for the next b rounds , the rejection percentage was 50 percent per round and for the remaining rounds , the selection percentage was 70 percent per round . if there were 100000 people who applied for the army and 245 were finally selected , what was the value of n ? | fastly i reduce 60 % till it gets closer to our required 2000 candidates step ( 1 ) 40000 accepted . step ( 2 ) another 40 % of 40000 = 16000 accepted . here it is quiet observable that if we further deduct candidate by 60 % it would change our probablity of easy going 2000 candidate . so i would get to second stage of recruitment where 50 % is accepted step ( 3 ) 50 % of 16000 = 8000 step ( 4 ) 50 % of 8000 = 4000 step ( 5 ) 50 % of 4000 = 2000 step ( 6 ) 50 % of 2000 = 1000 step ( 7 ) 50 % of 1000 = 500 step ( 8 ) 70 % of 500 = 350 step ( 9 ) 70 % of 350 = 245 ( our target ) total 9 steps required . ans e | a = 1 + 4
b = 2 + a
c = b + 2
|
a ) 26 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 30)), const_100), const_100) | when the price of an article was reduced by 30 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 30 % , the new price is 0.7 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.7 p ) ( 1.8 n ) = 1.26 np , which is 1.26 times the old revenue , so is 26 % greater . answer : a" | a = 80 + 100
b = 100 - 30
c = a * b
d = c / 100
e = d - 100
|
a ) 4 % , b ) 7 % , c ) 6 % , d ) 5 % , e ) 3 % | d | divide(divide(const_100, add(const_1, const_4)), 4) | if the simple interest on a certain sum of money for 4 years is one β fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 4 years rate of interest per annum ( r ) = ( 100 Γ si ) / pt = ( 100 Γ ( x / 5 ) / ( x Γ 4 ) = 20 / 4 = 5 % answer : option d" | a = 1 + 4
b = 100 / a
c = b / 4
|
a ) 50 , b ) 52 , c ) 65 , d ) 42 , e ) 66 | e | subtract(add(multiply(6, 58), multiply(6, 63)), multiply(11, 60)) | the average of 11 results is 60 . if the average of first 6 results is 58 and that of the last 6 is 63 , find the sixth result ? | sixth result = 58 * 6 + 63 * 6 - 60 * 11 = 66 answer is e | a = 6 * 58
b = 6 * 63
c = a + b
d = 11 * 60
e = c - d
|
a ) 76 , b ) 77 , c ) 78 , d ) 79 , e ) 80 | d | subtract(116, divide(subtract(116, 5), const_3)) | if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 116 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 116 3 x = 111 x = 37 kg then the student ' s weight is 79 kg . the answer is d ." | a = 116 - 5
b = a / 3
c = 116 - b
|
a ) rs . 48 , b ) rs . 64 , c ) rs . 44 , d ) rs . 46 , e ) rs . 34 | c | multiply(25, subtract(circle_area(add(13, 2)), circle_area(13))) | a circular path of 13 m radius has marginal walk 2 m wide all round it . find the cost of leveling the walk at 25 p per m 2 ? | "Ο ( 152 - 132 ) = 176 176 * 1 / 4 = rs . 44 answer : c" | a = 13 + 2
b = circle_area - (
c = 25 * b
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a ) 12 , b ) 20 , c ) 24 , d ) 30 , e ) 36 | b | divide(multiply(60, 36), multiply(36, const_3)) | tom drives from town q to town b , driving at a constant speed of 60 miles per hour . from town b tom immediately continues to town c . the distance between q and b is twice the distance between b and c . if the average speed of the whole journey was 36 mph , then what is tom ' s speed driving from b to c in miles per hour ? | let ' s assume that it takes 4 hours to go from point q to b . then the distance between them becomes 240 which makes distance between b and c 120 . ( 240 + 120 ) / ( 4 + x ) gives us the average speed which is 36 . you find x = 6 . so the question simplifies itself to 120 / 6 = 20 hence the answer is b . | a = 60 * 36
b = 36 * 3
c = a / b
|
a ) 1000 , b ) 2998 , c ) 3000 , d ) 2788 , e ) 2991 | c | divide(multiply(540, const_100), subtract(add(const_100, 8), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 540 more , there would have been a gain of 8 % . what is the cost price ? | "90 % 108 % - - - - - - - - 18 % - - - - 540 100 % - - - - ? = > rs . 3000 answer : c" | a = 540 * 100
b = 100 + 8
c = 100 - 10
d = b - c
e = a / d
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a ) rs . 2000 , b ) rs . 10,500 , c ) rs . 15,500 , d ) rs . 22,500 , e ) none | d | divide(divide(multiply(5400, const_100), multiply(3, 8)), multiply(divide(const_60, const_4), const_100)) | a man took a loan from a bank at the rate of 8 % p . a . simple interest . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was : | solution principal = rs . ( 100 x 5400 / 8 x 3 ) = rs . 22,500 . answer d | a = 5400 * 100
b = 3 * 8
c = a / b
d = const_60 / 4
e = d * 100
f = c / e
|
a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | what is the remainder when the number q = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | 14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus q = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a ) | a = 5 / 5
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a ) 2 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 2 | d | multiply(divide(3, subtract(add(add(3, 3), 1), 1)), divide(add(3, 3), add(add(3, 3), 1))) | a jar contains 3 black , 3 white and 1 green balls . if you pick two balls at the same time , what ' s the probability that one ball is black and one is white ? | p ( 1 st black , 2 nd white ) = 3 / 7 * 3 / 6 = 9 / 42 ; p ( 1 st white , 2 nd black ) = 3 / 7 * 3 / 6 = 9 / 42 . p = 9 / 42 + 9 / 42 = 18 / 42 = 3 / 7 . answer : d . | a = 3 + 3
b = a + 1
c = b - 1
d = 3 / c
e = 3 + 3
f = 3 + 3
g = f + 1
h = e / g
i = d * h
|
['a ) 201 m ( power ) 2', 'b ) 220 m ( power ) 2', 'c ) 211 m ( power ) 2', 'd ) 219 m ( power ) 2', 'e ) 205 m ( power ) 2'] | a | subtract(square_area(add(1, sqrt(multiply(const_10, const_1000)))), multiply(const_10, const_1000)) | of the two square fields , the area of the one is 1 hectare , while anothe one is broader by 1 % . there differences in area is : | area of one square field = 10000 m ( power ) 2 10000 Γ 1 = 10000 side of this field = β 10000 m = 100 m side of another square = 101 m difference of areas = [ 101 ( power ) 2 - 100 ( power ) 2 ] m ( power ) 2 [ 101 + 100 ] [ 101 - 100 ] m ( power ) 2 ( 201 ) ( 1 ) m 2 = 201 m ( power ) 2 answer is a . | a = 10 * 1000
b = math.sqrt(a)
c = 1 + b
d = square_area - (
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | add(4, 8) | if there are 4 peanuts in a box and mary puts 8 more peanuts inside , how many peanuts are in the box ? | "8 + 4 = 12 correct answer is e ) 12" | a = 4 + 8
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a ) 5 : 1 , b ) 5 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 5 : 2 | b | divide(divide(subtract(multiply(450, const_100), multiply(6000, 5)), subtract(10, 5)), divide(subtract(multiply(450, const_100), multiply(6000, 5)), subtract(10, 5))) | rs . 6000 is lent out in two parts . one part is lent at 5 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | let the amount lent at 5 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 5 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 1 x / 20 = > 600 - 1 / 20 x = 450 = > x = 3000 amount lent at 10 % = 3000 required ratio = 3000 : 3000 = 5 : 5 answer : b | a = 450 * 100
b = 6000 * 5
c = a - b
d = 10 - 5
e = c / d
f = 450 * 100
g = 6000 * 5
h = f - g
i = 10 - 5
j = h / i
k = e / j
|
a ) 250 , b ) 270 , c ) 310 , d ) 330 , e ) 350 | b | divide(divide(1440, add(add(divide(1, multiply(3, 4)), divide(1, 4)), 1)), 4) | if rs . 1440 / - are divided among a , b and c so that a receives 1 / 3 rd as much as b and b receives 1 / 4 th as much as c . the amount b received is ? | a : b : c = 1 : 3 : 12 total parts = 16 b ' s share is = 3 parts 16 - - - - - > 1440 1 - - - - - > 90 3 - - - - - > 270 ( b ' s share is 270 ) b ) | a = 3 * 4
b = 1 / a
c = 1 / 4
d = b + c
e = d + 1
f = 1440 / e
g = f / 4
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 15 | e | divide(375, add(subtract(26, 2), const_1)) | 375 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 375 / 25 ) = 15 e" | a = 26 - 2
b = a + 1
c = 375 / b
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(multiply(7, add(const_2, const_4)), add(lcm(12, 18), 3)) | when toys are grouped into 12 and 18 , always 3 toys remain ungrouped . what is the minimum number of toys to be added so that when they are grouped into 7 , no toys will remain ungrouped ? | t = 12 a + 3 = 18 b + 3 2 a = 3 b - - > min ( a , b ) = ( 3,2 ) so minimum number of toys - - > 12 * 3 + 3 = 39 when toys are grouped into 7 , 4 will remain ungrouped so in order to make another group of 7 toys - - > 7 - 4 = 3 toys at least to be added . answer : d | a = 2 + 4
b = 7 * a
c = math.lcm(12, 18)
d = c + 3
e = b - d
|
a ) 25 % , b ) 44 % , c ) 56 % , d ) 62 % , e ) 70 % | b | subtract(add(62, 54), subtract(const_100, 28)) | in a particular state , 62 % of the counties received some rain on monday , and 54 % of the counties received some rain on tuesday . no rain fell either day in 28 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ? | "62 + 54 + 28 = 144 % the number is 44 % above 100 % because 44 % of the counties were counted twice . the answer is b ." | a = 62 + 54
b = 100 - 28
c = a - b
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a ) 65 degrees , b ) 73 degrees , c ) 37 degrees , d ) 34 degrees , e ) 33 degrees | e | subtract(41, subtract(multiply(48, const_4), multiply(46, const_4))) | the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 41 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 41 tu + w + th = 192 - 41 = 151 f = 184 β 151 = 33 answer : e" | a = 48 * 4
b = 46 * 4
c = a - b
d = 41 - c
|
a ) 22877 , b ) 27778 , c ) 20000 , d ) 27999 , e ) 14400 | e | divide(multiply(multiply(18, const_100), multiply(16, const_100)), multiply(20, 10)) | a courtyard is 18 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ? | "number of bricks = courtyard area / 1 brick area = ( 1800 Γ 1600 / 20 Γ 10 ) = 14400 answer : e" | a = 18 * 100
b = 16 * 100
c = a * b
d = 20 * 10
e = c / d
|
a ) 6 , b ) 5 , c ) 5.5 , d ) 7 , e ) 8 | d | divide(multiply(multiply(divide(const_1, 15), 10), 21), const_2) | a can finish a work in 21 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 x 10 ) = 2 . 15 3 remaining work = ( 1 - 2 ) = 1 . 3 3 now , 1 work is done by a in 1 day . 21 therefore 1 work is done by a in ( 21 x 1 ) = 7 days . d" | a = 1 / 15
b = a * 10
c = b * 21
d = c / 2
|
a ) 12 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none of these | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of man is : | "solution volume of water displaced = ( 3 Γ 2 Γ 0.01 ) m 3 = 0.06 m 3 . β΄ mass of man = volume of water displaced Γ density of water = ( 0.06 Γ 100 ) kg = 60 kg . answer b" | a = 3 * 2
b = 1 / 100
c = a * b
d = c * 1000
|
a ) 26 % , b ) 28 % , c ) 27 % , d ) 40 % , e ) 70 % | a | subtract(divide(subtract(const_100, 16), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 16 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 84 2 / 3 sp 1 = 84 sp 1 = 126 100 - - - 26 = > 26 % answer : a" | a = 100 - 16
b = 2 / 3
c = a / b
d = c - 100
|
a ) 66 % , b ) 64 % , c ) 68 % , d ) 52 % , e ) 72 % | d | subtract(const_100, add(multiply(4, 6), multiply(6, 4))) | uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought less of toyota than honda at the ratio of 4 : 6 . if 40 % of the toyota bought and 60 % of the honda bought were suv Γ£ Β’ Γ’ β Β¬ Γ’ β Β’ s . how many suv Γ£ Β’ Γ’ β Β¬ Γ’ β Β’ s did uba capital buy in the aforementioned purchase ? | "let total no of vehicles bought be 100 , toyota 40 and honda 60 , so total number of suv ' s bought for toyota and honda respectively 40 * 40 / 100 = 16 and 60 * 60 / 100 = 36 so total 52 suv ' s were bought out of 100 vehicles bought . . so required % is 52 % answer : d" | a = 4 * 6
b = 6 * 4
c = a + b
d = 100 - c
|
a ) 52 , b ) 75 , c ) 60 , d ) 136 , e ) 184 | c | divide(multiply(300, divide(40, subtract(56, 36))), divide(add(36, 56), subtract(56, 36))) | an investor can sell her microtron stock for 36 $ per share and her dynaco stock for 56 $ per share , if she sells 300 shares altogether , some of each stock , at an average price per share of 40 $ , how many shares of dynaco stock has she sold ? | "w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) = ( 56 - 40 ) / ( 40 - 36 ) = 16 / 4 = 4 / 1 = number of microtron stocks / number of dynaco stocks so for every 4 microtron stock , she sold 1 dynaco stock . so out of 300 total stocks , ( 1 / 5 ) th i . e . 300 / 5 = 60 must be dynaco stock . answer ( c )" | a = 56 - 36
b = 40 / a
c = 300 * b
d = 36 + 56
e = 56 - 36
f = d / e
g = c / f
|
a ) 42 , b ) 44 , c ) 48 , d ) 45 , e ) 47 | b | multiply(const_100, divide(subtract(power(add(const_100, 20), const_3), power(const_100, const_3)), power(const_100, const_3))) | if each edge of cube increased by 20 % , the percentage increase in | "100 Γ ( 120 ) / 100 Γ ( 120 ) / 100 = 144 = > 44 % answer is b ." | a = 100 + 20
b = a ** 3
c = 100 ** 3
d = b - c
e = 100 ** 3
f = d / e
g = 100 * f
|
a ) 12 , b ) 8 , c ) 15 , d ) 20 , e ) 16 | e | divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100)) | if 15 % of 40 is greater than 25 % of a number by 2 , then the number is ? | "let the number be x 50 % of 40 - 25 % of x = 2 25 / 100 x = ( 15 / 100 * 40 ) - 2 x / 4 = 4 x = 16 answer is e" | a = 15 / 100
b = a * 40
c = b - 2
d = 25 / 100
e = c / d
|
a ) 65 deg , b ) 75 deg , c ) 50 deg , d ) 15 deg , e ) 30 deg | c | subtract(multiply(20, multiply(const_3, const_2)), 8) | what is the angle between the hands of a clock when time is 8 : 20 ? | "angle between two hands = 20 h - 11 / 2 m = 20 * 8 - 20 * 11 / 2 = 160 - 110 = 50 deg answer : c" | a = 3 * 2
b = 20 * a
c = b - 8
|
a ) 1.8 , b ) 2.2 , c ) 2 , d ) 1.9 , e ) 2.4 | c | multiply(3, divide(2, 3)) | month average price per dozen april $ 1.2 may $ 1.20 june $ 3.00 the table above shows the average ( arithmetic mean ) price per dozen of the large grade a eggs sold in a certain store during 3 successive months . if 2 / 3 as many dozen were sold in april as in may , and twice as many were sold in june as in april , what was the average price per dozen of the eggs sold over the 3 - month period ? | say x dozen eggs were sold in may . then 2 x / 3 dozen eggs were sold in april and 2 * 2 x / 3 = 4 x / 3 dozen eggs were sold in may . total x + 2 x / 3 + 4 x / 3 = 3 x dozen eggs were sold in three months . total sales : 2 x / 3 * 1.2 + x * 1.2 + 4 x / 3 * 3.00 = $ 6.00 x . the average price per dozen 6.00 x / 3 x = $ 2.00 c | a = 2 / 3
b = 3 * a
|
a ) 22 , b ) 28 , c ) 99 , d ) 49 , e ) 66 | d | multiply(7, 7) | walking 7 / 6 of his usual rate , a boy reaches his school 7 min early . find his usual time to reach the school ? | "speed ratio = 1 : 7 / 6 = 6 : 7 time ratio = 7 : 6 1 - - - - - - - - 7 7 - - - - - - - - - ? 49 m answer : d" | a = 7 * 7
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | d | multiply(divide(subtract(divide(2, const_10), divide(8, 100)), divide(2, const_10)), 100) | a company received $ 2 million in royalties on the first $ 10 million in sales and then $ 8 million in royalties on the next $ 100 million in sales . by what percent did the ratio of royalties to sales decrease from the first $ 10 million in sales to the next $ 100 million in sales ? | first find out ( ratio of royalties to sales ) : 2 : 10 first 0.2 8 : 100 second 0.08 decrease = 0.2 - 0.08 = 0.12 percent = ( 0.12 / 0.20 ) = 60 % . answer : d | a = 2 / 10
b = 8 / 100
c = a - b
d = 2 / 10
e = c / d
f = e * 100
|
a ) 18 , b ) 16 , c ) 26 , d ) 17 , e ) 20 | e | multiply(divide(subtract(1500, 1200), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1200 , find the loss % ? | "explanation : 1500 - - - - 300 100 - - - - ? = > 20 % answer : e" | a = 1500 - 1200
b = a / 1500
c = b * 100
|
a ) 100 , b ) 625 , c ) 600 , d ) 7500 , e ) 375 | a | multiply(6, 100) | a trailer carries 3 , 4 and 6 crates on a trip . each crate weighs no less than 100 kg . what is the maximum weight of the crates on a single trip ? | "max no . of crates = 6 . max weight = 100 kg max . weight carried = 6 * 100 = 100 kg = a ." | a = 6 * 100
|
a ) 14 , b ) 15 , c ) 16 , d ) 28 , e ) 29 | c | add(multiply(divide(multiply(divide(13.00, multiply(subtract(0.70, 0.30), const_2)), const_2), const_10), const_2), multiply(divide(13.00, multiply(subtract(0.70, 0.30), const_2)), const_2)) | the toll for crossing a certain bridge is $ 0.70 each crossing . drivers who frequently use the bridge may instead purchase a sticker each month for $ 13.00 and then pay only $ 0.30 each crossing during that month . if a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day , what is the least value of x for which this driver can save money by using the sticker ? | "option # 1 : $ 0.75 / crossing . . . . cross twice a day = $ 1.5 / day option # 2 : $ 0.30 / crossing . . . . cross twice a day = $ 0.6 / day + $ 13 one time charge . if we go down the list of possible answers , you can quickly see that 14 days will not be worth purchasing the sticker . 1.5 x 14 ( 21 ) is cheaper than 0.6 x 14 + 13 ( 21.4 ) . . . it ' s pretty close so let ' s see if one more day will make it worth it . . . if we raise the number of days to 15 , the sticker option looks like a better deal . . . 1.5 x 15 ( 22.5 ) vs 0.6 x 15 + 13 ( 22 ) . answer : c" | a = 0 - 70
b = a * 2
c = 13 / 0
d = c * 2
e = d / 10
f = e * 2
g = 0 - 70
h = g * 2
i = 13 / 0
j = i * 2
k = f + j
|
a ) 78 kg , b ) 60 kg , c ) 64 kg , d ) 70 kg , e ) none | a | add(multiply(divide(11, 9), 35.1), 35.1) | zinc and copper are melted together in the ratio 9 : 11 . what is the weight of melted mixture , if 35.1 kg of zinc has been consumed in it ? | "sol . for 9 kg zinc , mixture melted = ( 9 + 11 ) kg . for 35.1 kg zinc , mixture , melted = [ 20 / 9 x 35.1 ] kg = 78 kg . answer a" | a = 11 / 9
b = a * 35
c = b + 35
|
a ) 5980 , b ) 5990 , c ) 6000 , d ) 6010 , e ) 6020 | c | divide(add(multiply(add(add(18,000, const_3), const_3), multiply(add(18,000, const_3), 18,000)), add(18,000, const_3)), add(multiply(const_3, multiply(add(18,000, const_3), 18,000)), add(18,000, const_3))) | p and q started a business investing rs . 54,000 and rs . 36,000 respectively . out of a total profit of rs . 18,000 , q ' s share is : | "ratio of their shares = 54000 : 27000 = 6 : 3 . q β s share = rs . 18000 * 3 / 9 = rs . 6000 answer : c" | a = 18 + 0
b = a + 3
c = 18 + 0
d = c * 18
e = b * d
f = 18 + 0
g = e + f
h = 18 + 0
i = h * 18
j = 3 * i
k = 18 + 0
l = j + k
m = g / l
|
a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | b | subtract(const_100, multiply(divide(add(71, const_100), add(80, const_100)), const_100)) | two numbers are respectively 71 % and 80 % more than a third number . the percentage that is first of the second is ? | "i ii iii 171 180 100 180 - - - - - - - - - - 171 100 - - - - - - - - - - - ? = > 5 % answer : b" | a = 71 + 100
b = 80 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 25 , b ) 30 , c ) 40 , d ) 45 , e ) 50 | c | multiply(divide(add(multiply(1, const_60), 20), 40), 20) | a machine , working at a constant rate , manufactures 20 candles in 40 minutes . how many candles does it make in 1 hr 20 min ? | "change 1 hr 20 min to 80 min . for this , we need to set up a simple proportion of staplers per time 20 / 40 = s / 80 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead , cancel before you multiply . for what we can and can β t cancel in a proportion , seethis post . we can cancel the factor of 10 in the 20 and 40 . 2 / 4 = s / 80 1 / 2 = s / 80 ; 1 / 2 * 2 = s / 80 * 2 1 / 1 = s / 40 now that the fraction is entirely simplified , we can cross - multiply . s = 1 * 40 = 40 the machine would be 40 candles in 1 hr 20 min . answer : c ." | a = 1 * const_60
b = a + 20
c = b / 40
d = c * 20
|
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