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 ) 67 , b ) 26 , c ) 89 , d ) 26 , e ) 75 | c | divide(add(add(add(add(86, 85), 92), 87), 95), divide(const_10, const_2)) | dacid obtained 86 , 85 , 92 , 87 and 95 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 86 + 85 + 92 + 87 + 95 ) / 5 = 445 / 5 = 89 . answer : c" | a = 86 + 85
b = a + 92
c = b + 87
d = c + 95
e = 10 / 2
f = d / e
|
a ) 30 days , b ) 60 days , c ) 70 days , d ) 80 days , e ) 24 days | e | multiply(divide(8, subtract(12, 8)), 12) | david and andrew can finish the work 12 days if they work together . they worked together for 8 days and then andrew left . david finished the remaining work in another 8 days . in how many days david alone can finish the work ? | amount of work done by david and andrew in 1 day = 1 / 12 amount of work done by david and andrew in 8 days = 8 ã — ( 1 / 12 ) = 2 / 3 remaining work â € “ 1 â € “ 2 / 3 = 1 / 3 david completes 1 / 3 work in 8 days amount of work david can do in 1 day = ( 1 / 3 ) / 8 = 1 / 24 = > david can complete the work in 24 days ... | a = 12 - 8
b = 8 / a
c = b * 12
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | multiply(divide(45, subtract(multiply(34, 2), 63)), 2) | a firm is comprised of partners and associates in a ratio of 2 : 63 . if 45 more associates were hired , the ratio of partners to associates would be 1 : 34 . how many partners are currently in the firm ? | "the ratio 1 : 34 = 2 : 68 so the ratio changed from 2 : 63 to 2 : 68 . 68 - 63 = 5 which is 1 / 9 of the increase in 45 associates . the ratio changed from 18 : 567 to 18 : 612 . thus the number of partners is 18 . the answer is d ." | a = 34 * 2
b = a - 63
c = 45 / b
d = c * 2
|
a ) 70000 , b ) 60000 , c ) 80000 , d ) 90000 , e ) 50000 | c | add(add(multiply(divide(27000, 81000), 36000), multiply(divide(72000, 81000), 36000)), 36000) | a , b and c started a partnership business by investing rs . 27000 , rs . 72000 , rs . 81000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | a : b : c = 27000 : 72000 : 81000 = 3 : 8 : 9 let total profit = p then p × 9 / 20 = 36000 p = ( 36000 × 20 ) / 9 = 80000 answer is c . | a = 27000 / 81000
b = a * 36000
c = 72000 / 81000
d = c * 36000
e = b + d
f = e + 36000
|
a ) s . 7,000 , b ) s . 7,200 , c ) s . 7,400 , d ) s . 7,700 , e ) s . 7,800 | b | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30,000 , c receives : | "let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 30000 x 12 / 50 ) = rs . 7,200 . b" | a = 10 * 5000
b = a - 5000
c = 4000 + 5000
d = b - c
e = d / 3
f = 4000 + 5000
g = e + f
h = 10 * 5000
i = g / h
j = 3 + 4
k = j * 5000
l = i * k
m = l / 1000
n = math.floor(m)
o = n - 1
|
a ) $ 25 , b ) $ 31.25 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 20), divide(50, const_2)), const_100) | a pair of articles was bought for $ 50 at a discount of 20 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 80 % of x = 25 x = 25 * 100 / 80 = $ 31.25 answer is b" | a = 100 - 20
b = 50 / 2
c = a * b
d = c / 100
|
a ) 12 , b ) 14.11 , c ) 18.33 , d ) 20 , e ) 21 | c | multiply(divide(multiply(const_2, divide(60, const_10)), 60), const_100) | what percentage of numbers from 1 to 60 have 1 or 9 in the unit ' s digit ? | "clearly , the numbers which have 1 or 9 in the unit ' s digit , have squares that end in the digit 1 . such numbers from 1 to 60 are 1 , 9 , 11 , 19 , 21 , 29 , 31 , 39 , 41 , 49 , 51 , 59 number of such number = 11 answer : c" | a = 60 / 10
b = 2 * a
c = b / 60
d = c * 100
|
a ) 955 , b ) 550 , c ) 600 , d ) 700 , e ) 750 | a | divide(650, subtract(const_1, divide(32, const_100))) | shop offered 32 % offer for every shirt , smith bought a shirt at rs . 650 . and what was the shop ' s original selling price ? | "sp * ( 68 / 100 ) = 650 sp = 9.55 * 100 = > cp = 955 answer : a" | a = 32 / 100
b = 1 - a
c = 650 / b
|
a ) 2 , b ) 25 , c ) 92 , d ) 96 , e ) 98 | b | multiply(divide(subtract(100, 99), subtract(100, 96)), const_100) | each of the cucumbers in 100 pounds of cucumbers is composed of 99 % water , by weight . after some of the water evaporates , the cucumbers are now 96 % water by weight . what is the new weight of the cucumbers , in pounds ? | out of 100 pounds 99 % or 99 pounds is water and 1 pound is non - water . after somewaterevaporates the cucumbers become 96 % water and 4 % of non - water , so now 1 pound of non - water composes 4 % of cucucmbers , which means that the new weight of cucumbers is 1 / 0.04 = 25 pounds . answer : b . | a = 100 - 99
b = 100 - 96
c = a / b
d = c * 100
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 16 | e | multiply(divide(multiply(const_10, 150), add(150, 160)), const_2) | a certain elevator has a safe weight limit of 2,500 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 150 pounds and the average weight of the others being 160 pounds ? | "lets assume there are 2 x people . half of them have average weight of 150 and other half has 160 . maximum weight is = 2500 so 150 * x + 160 * x = 2500 = > 310 x = 2500 = > x is approximately equal to 8 . so total people is 2 * 8 = 16 answer e ." | a = 10 * 150
b = 150 + 160
c = a / b
d = c * 2
|
a ) 900 km , b ) 300 km , c ) 700 km , d ) 800 km , e ) 100 km | a | add(add(divide(multiply(100, 40), subtract(50, 40)), 100), divide(multiply(100, 40), subtract(50, 40))) | two trains start from p and q respectively and travel towards each other at a speed of 50 km / hr and 40 km / hr respectively . by the time they meet , the first train has traveled 100 km more than the second . the distance between p and q is ? | a 900 km at the time of meeting , let the distance traveled by the second train be x km . then , distance covered by the first train is ( x + 100 ) km . x / 40 = ( x + 100 ) / 50 50 x = 40 x + 4000 = > x = 400 so , distance between p and q = ( x + x + 100 ) km = 900 km . | a = 100 * 40
b = 50 - 40
c = a / b
d = c + 100
e = 100 * 40
f = 50 - 40
g = e / f
h = d + g
|
a ) 50 , b ) 54 , c ) 55 , d ) 56 , e ) 60 | a | divide(multiply(divide(subtract(const_100, 20), const_100), 5), divide(subtract(const_100, 92), const_100)) | if grapes are 92 % water and raisins are 20 % water , then how many kilograms did a quantity of raisins , which currently weighs 5 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transform... | "let x be the original weight . the weight of the grape pulp was 0.08 x . since the grape pulp is 80 % of the raisins , 0.08 x = 0.8 ( 5 ) . then x = 50 kg . the answer is a ." | a = 100 - 20
b = a / 100
c = b * 5
d = 100 - 92
e = d / 100
f = c / e
|
a ) 55 , b ) 90 , c ) 73 , d ) 82 , e ) 91 | b | subtract(100, divide(subtract(100, 70), const_3)) | a teacher grades students ’ tests by subtracting twice the number of incorrect responses from the number of correct responses . if student a answers each of the 100 questions on her test and receives a score of 70 , how many questions did student a answer correctly ? | "let the number of correct responses be x then the number of incorrect responses = 100 - x according to question x - 2 ( 100 - x ) = 70 ( subtracting twice of incorrect from correct ) 3 x = 270 x = 90 answer : b" | a = 100 - 70
b = a / 3
c = 100 - b
|
a ) 45 , b ) 55 , c ) 65 , d ) 75 , e ) 85 | b | subtract(factorial(subtract(8, 4)), add(multiply(4, const_3), const_3)) | there are 8 executives , including the ceo and cfo , that are asked to form a small team of 4 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | "the total number of ways to form a team of 4 is 8 c 4 = 70 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 6 c 2 = 15 . the number of ways to form an acceptable team is 70 - 15 = 55 . the answer is b ." | a = 8 - 4
b = math.factorial(a)
c = 4 * 3
d = c + 3
e = b - d
|
a ) $ 4500 , b ) $ 3500 , c ) $ 5500 , d ) $ 5000 , e ) $ 6300 | d | multiply(subtract(50, multiply(20, const_2)), 500) | redo ’ s manufacturing costs for sets of horseshoes include a $ 10,000 initial outlay , and $ 20 per set . they can sell the sets $ 50 . if profit is revenue from sales minus manufacturing costs , and the company producessells 500 sets of horseshoes , what was their profit ? | total manufacturing cost = 10000 + 500 * 20 = 20000 total selling cost = 500 * 50 = 25000 profit = 25000 - 20000 = 5000 answer : d | a = 20 * 2
b = 50 - a
c = b * 500
|
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 pla ƞ orm in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the pla ƞ orm . | explanation : speed = distance / ɵ me = 300 / 18 = 50 / 3 m / sec let the length of the pla ƞ orm be x meters then distance = speed ∗ timex + 300 = 503 ∗ 39 = > 3 ( x + 300 ) = 1950 = > x = 350 meters answer : d | a = 300 / 18
b = a * 39
c = b - 300
|
a ) 153 , b ) 308 , c ) 121 , d ) 96 , e ) 511 | a | divide(multiply(1, const_2), 1) | what is the smallest positive integer x such that ( x + 1 ) ^ 2 is divisible by 28 , 98 , 242 , and 308 ? | "28 = 2 * 2 * 7 98 = 2 * 7 * 7 242 = 2 * 11 * 11 308 = 2 * 2 * 7 * 11 so ( x + 1 ) ^ 2 = 2 * 2 * 7 * 7 * 11 * 11 , which means ( x + 1 ) = 2 * 7 * 11 = 154 , which means x = 153 , which is option a" | a = 1 * 2
b = a / 1
|
a ) 190 , b ) 153 , c ) 210 , d ) 220 , e ) 230 | b | multiply(subtract(18, const_1), divide(18, const_2)) | 18 men shake hands with each other . maximum no of handshakes without cyclic handshakes . | "1 st person will shake hand with 17 people 2 nd person will shake hand with 16 people 3 rd person will shake hand with 15 people . . . . . . total no . of handshakes = 17 + . . . + 3 + 2 + 1 = 17 * ( 17 + 1 ) / 2 = 153 or , if there are n persons then no . of shakehands = nc 2 = 18 c 2 = 153 answer : b" | a = 18 - 1
b = 18 / 2
c = a * b
|
a ) 42.8 % , b ) 25 % , c ) 55 % , d ) 28 % , e ) 55 % | a | subtract(multiply(divide(const_100, 700), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 700 grams per kg , what is his percent ? | "700 - - - 300 100 - - - ? = > 42.8 % answer : a" | a = 100 / 700
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
|
a ) 545 , b ) 685 , c ) 865 , d ) 495 , e ) 346 | e | divide(multiply(173, 240), 120) | ? x 120 = 173 x 240 | let y x 120 = 173 x 240 then y = ( 173 x 240 ) / 120 = 346 . answer : e | a = 173 * 240
b = a / 120
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | b | add(divide(add(subtract(11, const_2), divide(subtract(add(divide(subtract(251, subtract(11, const_2)), 11), subtract(11, const_2)), subtract(11, const_2)), 11)), add(subtract(11, const_2), divide(subtract(add(divide(subtract(251, subtract(11, const_2)), 11), subtract(11, const_2)), subtract(11, const_2)), 11))), add(di... | in a soap company a soap is manufactured with 11 parts . for making one soap you will get 1 part as scrap . at the end of the day you have 251 such scraps . from that how many soaps can be manufactured ? | using 251 scraps we make = 22 soaps . ( i . e . , 11 * 22 = 242 ) remaining scraps = 9 . again , 22 soaps produce = 22 scraps . so , now we have 22 + 9 = 31 scraps remaining . using 31 scraps we make 2 soaps and 2 scraps remaining . ( i . e . , 2 * 11 = 22 ) again , that 2 soaps produce = 2 scraps . and already we have... | a = 11 - 2
b = 11 - 2
c = 251 - b
d = c / 11
e = 11 - 2
f = d + e
g = 11 - 2
h = f - g
i = h / 11
j = a + i
k = 11 - 2
l = 11 - 2
m = 251 - l
n = m / 11
o = 11 - 2
p = n + o
q = 11 - 2
r = p - q
s = r / 11
t = k + s
u = j / t
v = 11 - 2
w = 251 - v
x = w / 11
y = 11 - 2
z = 251 - y
A = z / 11
B = 11 - 2
C = A + B
... |
a ) 1888 , b ) 1640 , c ) 2768 , d ) 2976 , e ) 2691 | b | subtract(2665, divide(multiply(multiply(3, 5), 2665), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2665 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2665 - x ) * 3 * 5 ) / 100 24 x / 100 = 39975 / 100 - 15 x / 100 39 x = 39975 = > x = 1025 second sum = 2665 – 1025 = 1640 answer : b" | a = 3 * 5
b = a * 2665
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2665 - f
|
a ) 27 % , b ) 21 % , c ) 19 % , d ) 18 % , e ) 16 % | e | add(subtract(subtract(const_100, 40), multiply(divide(9, 10), subtract(const_100, 40))), subtract(40, multiply(divide(3, 4), 40))) | in a survey of parents , exactly 3 / 4 of the mothers and 9 / 10 of the fathers held full - time jobs . if 40 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "let the total number of parents = 100 number of women = 40 number of men = 60 number of mothers who held full time jobs = 3 / 4 * 40 = 30 number of fathers who held full time jobs = 9 / 10 * 60 = 54 total number of parents who held full time jobs = 84 total number of parents who did not hold jobs = 100 - 84 = 16 alter... | a = 100 - 40
b = 9 / 10
c = 100 - 40
d = b * c
e = a - d
f = 3 / 4
g = f * 40
h = 40 - g
i = e + h
|
a ) 1850 , b ) 2960 , c ) 3000 , d ) 1110 , e ) 1712 | d | subtract(divide(subtract(multiply(1850, 28), multiply(1850, 12)), 10), 1850) | a garrison of 1850 men has provisions for 28 days . at the end of 12 days , a reinforcement arrives , and it is now found that the provisions will last only for 10 days more . what is the reinforcement ? | "1850 - - - - 28 1850 - - - - 16 x - - - - - 10 x * 10 = 1850 * 16 x = 2960 1850 - - - - - - - 1110 answer : d" | a = 1850 * 28
b = 1850 * 12
c = a - b
d = c / 10
e = d - 1850
|
a ) 7000 , b ) 7029 , c ) 2778 , d ) 2800 , e ) 2000 | e | divide(240, divide(multiply(subtract(18, 12), const_2), const_100)) | a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 240 . find the sum ? | "let the sum be rs . x . ( x * 18 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 240 = > 36 x / 100 - 24 x / 100 = 240 = > 12 x / 100 = 240 = > x = 2000 . answer : e" | a = 18 - 12
b = a * 2
c = b / 100
d = 240 / c
|
a ) 5 sec , b ) 10 sec , c ) 12 sec , d ) 18 sec , e ) 15 sec | b | divide(150, multiply(subtract(62, 8), const_0_2778)) | a train 150 m long is running at a speed of 62 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "answer : b . speed of the train relative to man = ( 62 - 8 ) kmph = ( 54 * 5 / 18 ) m / sec = 15 m / sec time taken by the train to cross the man = time taken by it to cover 150 m at 15 m / sec = 150 * 1 / 15 sec = 10 sec" | a = 62 - 8
b = a * const_0_2778
c = 150 / b
|
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | a | divide(divide(divide(588, const_3), const_3), const_4) | the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 588 sq meters , then what is the width ( in meters ) of the rectangular plot ? | "area = l * w = 3 w ^ 2 = 588 w ^ 2 = 196 w = 14 the answer is a ." | a = 588 / 3
b = a / 3
c = b / 4
|
a ) 6 hours , b ) 8 hours , c ) 15 hours , d ) 24 hours , e ) 32 hours | c | multiply(5, const_3) | jamshid can paint a fence in 50 percent less time than taimour can when each works alone . when they work together , they can paint the fence in 5 hours . how long would it take taimour to paint the fence alone ? | "i believe the answer is c . please see below for explanation . if jamshid can paint a dence in 50 percent less time then taimour we can infer the following rate j = 2 t if working together they can do the job in 8 hours we can infer 1 = 2 t + t * 5 = > 1 / 15 working alone taimour can do the job in 1 = 1 / 15 * hours ... | a = 5 * 3
|
a ) 20 , b ) 21 , c ) 25 , d ) 28 , e ) 30 | e | multiply(divide(subtract(45, divide(multiply(60, 25), const_100)), subtract(const_100, 60)), const_100) | a group of boy scouts and girls scouts is going on a rafting trip . 45 % of the scouts arrived with signed permission slips . if 60 % of the scouts were boy scouts and 25 % of the boy scouts arrived with signed permission slips , then what percentage of the scouts were girl scouts who arrived with signed permission sli... | "we do n ' t know how many scouts went on the trip , so let ' s assume 100 scouts went on the trip 60 % were boy scouts so 60 % of 100 = 60 were boy scouts 25 % of the boy scouts brought their permission slips signed , so . 25 * 60 = 15 boy scouts had signed slips 60 - 15 = 45 boy scouts did not 45 % of all the scouts ... | a = 60 * 25
b = a / 100
c = 45 - b
d = 100 - 60
e = c / d
f = e * 100
|
a ) 1 / 2 , b ) 3 / 8 , c ) 3 / 4 , d ) 5 / 16 , e ) 9 / 16 | c | multiply(divide(const_3, add(const_3, const_3)), divide(const_3, add(const_3, const_3))) | two dice are thrown simultaneously . what is the probability of getting two numbers whose product is even ? | "in a simultaneous throw of two dice , we have n ( s ) = ( 6 x 6 ) = 36 . then , e = { ( 1 , 2 ) , ( 1 , 4 ) , ( 1 , 6 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 2 ) , ( 3 , 4 ) , ( 3 , 6 ) , ( 4 , 1 ) , ( 4 , 2 ) , ( 4 , 3 ) , ( 4 , 4 ) , ( 4 , 5 ) , ( 4 , 6 ) , ( 5 , 2 ) , ( 5 ,... | a = 3 + 3
b = 3 / a
c = 3 + 3
d = 3 / c
e = b * d
|
a ) 20 % , b ) 40 % , c ) 50 % , d ) 80 % , e ) 100 % | e | subtract(const_100, divide(subtract(const_100, 40), add(const_1, divide(20, const_100)))) | when sold at a 40 % discount , a sweater nets the merchant a 20 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | "et the marked up price = 100 . . selling price = 100 - 40 % of 100 = 60 . . profit = 20 % . . therefore the wholesale purchase cost = x . . . . 1.2 x = 60 or x = 50 . . . marked price was 100 so 50 over 50 . . . so answer is 100 % . . answer : e" | a = 100 - 40
b = 20 / 100
c = 1 + b
d = a / c
e = 100 - d
|
a ) 27 , b ) 29 , c ) 30 , d ) 20 , e ) 24 | d | divide(add(15, 25), const_2) | a man can row upstream at 15 kmph and downstream at 25 kmph , and then find the speed of the man in still water ? | "us = 15 ds = 25 m = ( 15 + 25 ) / 2 = 20 answer : d" | a = 15 + 25
b = a / 2
|
a ) 220 , b ) 230 , c ) 1200 , d ) 560 , e ) 590 | c | divide(multiply(multiply(subtract(3.60, 3), const_1000), const_100), 50) | workers decided to raise rs . 3 lacs by equal contribution from each . had they contributed rs . 50 eachextra , the contribution would have been rs . 3.60 lacs . how many workers were they ? | "n * 50 = ( 360000 - 300000 ) = 60000 n = 60000 / 50 = 1200 c" | a = 3 - 60
b = a * 1000
c = b * 100
d = c / 50
|
a ) 93700 , b ) 97300 , c ) 93800 , d ) 98300 , e ) none of them | d | subtract(983, multiply(multiply(207, 983), 107)) | evaluate : 983 x 207 - 983 x 107 | "983 x 207 - 983 x 107 = 983 x ( 207 - 107 ) = 983 x 100 = 98300 . answer is d ." | a = 207 * 983
b = a * 107
c = 983 - b
|
a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(add(add(3, 3), divide(multiply(3, 3), const_2)), const_10) | find value of x : 3 x ^ 2 + 5 x + 2 = 0 | "a = 3 , b = 5 , c = 2 x 1,2 = ( - 5 ± √ ( 52 - 4 × 3 × 2 ) ) / ( 2 × 3 ) = ( - 5 ± √ ( 25 - 24 ) ) / 6 = ( - 5 ± 1 ) / 6 x 1 = ( - 5 + 1 ) / 6 = - 4 / 6 = - 2 / 3 x 2 = ( - 5 - 1 ) / 6 = - 6 / 6 = - 1 a" | a = 3 + 3
b = 3 * 3
c = b / 2
d = a + c
e = d * 10
|
a ) 4 / 11 , b ) 5 / 12 , c ) 6 / 17 , d ) 7 / 20 , e ) 11 / 30 | d | subtract(add(divide(3, 4), subtract(const_1, divide(3, 5))), divide(5, 5)) | the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 3 / 5 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ? | "let , a ≡ event of getting hardware contract b ≡ event of getting software contract ab ≡ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 3 / 5 ) = 2 / 5 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b )... | a = 3 / 4
b = 3 / 5
c = 1 - b
d = a + c
e = 5 / 5
f = d - e
|
a ) 0.375 , b ) 0.25 , c ) 0.325 , d ) 0.5 , e ) 0.666 | a | multiply(power(divide(const_1, const_2), 3), 3) | if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land tails up exectly twice in 3 consecutive flips ? | "total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 * 2 = 8 ways for 2 t and 1 th thus probability is = p ( htt ) + p ( tth ) + p ( tht ) = 1 / 8 + 1 / 8 + 1 / 8 = 3 / 8 = . 375 answer : a" | a = 1 / 2
b = a ** 3
c = b * 3
|
a ) a ) 51 , b ) b ) 58 , c ) c ) 145 , d ) d ) 190 , e ) e ) 210 | c | add(subtract(154, 10), const_1) | if the average ( arithmetic mean ) of 10 consecutive odd integers is 154 , then the least of these integers is | "a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 154 . there are 10 numbers in the set , and in a set with an even number... | a = 154 - 10
b = a + 1
|
a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | e | add(add(100, 4), add(4, 1)) | the average weight of a class is x pounds . when a new student weighing 100 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the val... | "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 ... | a = 100 + 4
b = 4 + 1
c = a + b
|
a ) 150000 , b ) 160000 , c ) 170000 , d ) 190000 , e ) 250000 | a | divide(75000, multiply(divide(2, 3), divide(3, 4))) | a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 75000 rs , what is the value of buzness ? | "if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 75000 - > x = 150000 answer : a" | a = 2 / 3
b = 3 / 4
c = a * b
d = 75000 / c
|
a ) 15 , b ) 20 , c ) 22 , d ) 12 , e ) 8 | a | subtract(add(multiply(4, const_2), multiply(4, const_2)), const_1) | on a certain day , joey , the ice - cream seller sold his ice creams to 4 different kids in a manner that each of the kids purchased half of the remaining ice creams and half ice - cream more . if we tell you that the fourth kid bought just a single ice cream , can you find out how many ice creams were sold by joey tha... | a joey sold 15 ice creams that day . the fourth kid bought a single ice cream . therefore , we have the following equation : total - ( total / 2 + 1 / 2 ) = 1 solving it , we get the total as 3 . let ' s work this method till we reach the first kid . the first kid bought 15 / 2 + 1 / 2 = 8 ( leaving 7 ) the second kid ... | a = 4 * 2
b = 4 * 2
c = a + b
d = c - 1
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a ) 9750 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000 | a | subtract(subtract(7900, multiply(7900, divide(10, const_100))), multiply(subtract(7900, multiply(7900, divide(10, const_100))), divide(10, const_100))) | the population of a town is 7900 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ? | "formula : ( after = 100 denominator ago = 100 numerator ) 7900 ã — 100 / 90 ã — 100 / 90 = 9753 a )" | a = 10 / 100
b = 7900 * a
c = 7900 - b
d = 10 / 100
e = 7900 * d
f = 7900 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) 48 kmph , b ) 54 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 12), const_3_6) | a 160 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 160 / 12 * 18 / 5 = 48 kmph answer : a" | a = 160 / 12
b = a * const_3_6
|
a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | d | divide(log(multiply(9, 9)), log(const_10)) | 9 log 9 ( 5 ) = ? | "exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 5 ) = 5 correct answer d" | a = 9 * 9
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 2 , b ) 7 , c ) 6 , d ) 8 , e ) 1 | c | subtract(multiply(multiply(multiply(124, 812), 816), 467), subtract(multiply(multiply(multiply(124, 812), 816), 467), add(const_4, const_4))) | the unit digit in the product ( 124 * 812 * 816 * 467 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 * 2 * 6 * 7 ) = 6 answer : c" | a = 124 * 812
b = a * 816
c = b * 467
d = 124 * 812
e = d * 816
f = e * 467
g = 4 + 4
h = f - g
i = c - h
|
a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 48 | c | multiply(divide(divide(600, 1000), 72), const_3600) | if a truck is traveling at a constant rate of 72 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 72 km / hr = > 72,000 m / hr in one minute = > 72000 / 60 = 1200 meters in one sec = > 1200 / 60 = 20 meters time = total distance need to be covered / avg . speed = > 600 / 20 = 30 and hence the answer : c" | a = 600 / 1000
b = a / 72
c = b * 3600
|
a ) $ 200 , b ) $ 300 , c ) $ 480 , d ) $ 500 , e ) $ 600 | c | multiply(multiply(4, 3), divide(40, subtract(4, 3))) | the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 4 people , the cost per person would have been $ 40 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 3 * p c = 4 * p - 160 substituting the value of p from the first equation onto the second we get p = 160 . plugging in the value of p in the first equation , we get c = 480 . which leads us to answer choice c" | a = 4 * 3
b = 4 - 3
c = 40 / b
d = a * c
|
a ) 107 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 150), add(5, 2)) | a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 150 pounds of the mixture ? | "almonds : walnuts = 5 : 2 total mixture has 7 parts in a 150 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 150 = 107 pounds answer ( a )" | a = 5 * 150
b = 5 + 2
c = a / b
|
a ) $ 1100 , b ) $ 520 , c ) $ 1080 , d ) $ 1170 , e ) $ 630 | b | multiply(2340, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2340 , what will be c ' s share of the earnings ? | "explanatory answer a , b and c will share the amount of $ 2340 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , t... | a = 1/(8)
b = 1/(12)
c = 1/(6)
d = 1/(8)
e = c + d
f = b + e
g = a / f
h = 2340 * g
|
a ) 4 and 1 , b ) 1 and 5 , c ) 5 and 1 , d ) 3 and 5 , e ) 5 and 3 | c | multiply(multiply(5, 5), divide(1, 5)) | in the coordinate plane , points ( x , 1 ) and ( 5 , y ) are on line k . if line k passes through the origin and has slope 1 / 5 , then what are the values of x and y respectively ? | "line k passes through the origin and has slope 1 / 5 means that its equation is y = 1 / 5 * x . thus : ( x , 1 ) = ( 5 , 1 ) and ( 5 , y ) = ( 5,1 ) - - > x = 5 and y = 1 answer : c" | a = 5 * 5
b = 1 / 5
c = a * b
|
a ) 25 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 % | a | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(13, const_100), const_4), divide(13, const_100))), const_100) | one fourth of a solution that was 13 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.13 + 25 * x = 100 * 0.16 - - > x = 0.25 . answer : a ." | a = 16 / 100
b = a * 4
c = 13 / 100
d = c * 4
e = 13 / 100
f = d - e
g = b - f
h = g * 100
|
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | divide(14400, multiply(25, multiply(const_2, divide(21600, multiply(40, 30))))) | if daily wages of a man is double to that of a woman , how many men should work for 25 days to earn rs . 14400 ? given that wages for 40 women for 30 days are rs . 21600 . | explanation : wages of 1 woman for 1 day = 21600 / 40 × 30 wages of 1 man for 1 day = 21600 × 2 / 40 × 30 wages of 1 man for 25 days = 21600 × 2 × 25 / 40 × 30 number of men = 14400 / ( 21600 × 2 × 25 / 40 × 30 ) = 144 / ( 216 × 50 / 40 × 30 ) = 144 / 9 = 16 answer : option c | a = 40 * 30
b = 21600 / a
c = 2 * b
d = 25 * c
e = 14400 / d
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 7 / 9 | d | divide(multiply(const_3, const_2), multiply(divide(const_26, const_2), multiply(multiply(const_3, const_2), const_2))) | what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular hexagon ? | "it has two vertices on sides , which do not make a diagonal but a side . . so remaining 3 vertices make diagonals . . . only the opposite vertex will make largest diagonal and other two smaller ones . . so prob = 2 / ( 2 + 1 ) = 2 / 3 answer : d" | a = 3 * 2
b = const_26 / 2
c = 3 * 2
d = c * 2
e = b * d
f = a / e
|
a ) 501.13 , b ) 502.13 , c ) 503.13 , d ) 504.13 , e ) 505.13 | c | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(const_4, const_3)))), const_100) | what is the compound interest on rs : 30,000 for 4 months at the rate of 5 % per annum | it is monthly compound rate = 5 / 12 % per month 30000 * ( 1 + 5 / 1200 ) ^ 4 - 30000 = 503.13 answer : c | a = 3 * 100
b = a * 100
c = 4 * 3
d = 4 / c
e = 5 * d
f = b * e
g = f / 100
|
a ) 12 , b ) 75 , c ) 88 , d ) 65 , e ) 15 | d | divide(add(90, 40), const_2) | the speed of a car is 90 km in the first hour and 40 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 40 ) / 2 = 65 kmph answer : d" | a = 90 + 40
b = a / 2
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(divide(add(54, const_1), add(const_1, const_4)), const_2) | how many trailing zeroes does 53 ! + 54 ! have ? | 53 ! + 54 ! = 53 ! + 54 * 53 ! = 53 ! ( 1 + 54 ) = 53 ! * 55 number of trailing 0 s in 53 ! = number of 5 s in the expansion of 53 ! = 10 + 2 = 12 there is 1 more 5 in 55 . hence , total number of trailing 0 s = 12 + 1 = 13 answer ( b ) . in most cases , when we are adding multiple terms , all of which have trailing 0 ... | a = 54 + 1
b = 1 + 4
c = a / b
d = c + 2
|
a ) 160 , b ) 150 , c ) 82 , d ) 80 , e ) 50 | c | divide(subtract(multiply(208, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | an empty fuel tank with a capacity of 208 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 208 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 208 - a gallons of fuel b is 0.16 ( 208 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 208 - a ) = 30 - - > a =... | a = 16 / 100
b = 208 * a
c = b - 30
d = 16 / 100
e = 12 / 100
f = d - e
g = c / f
|
a ) 15 , b ) 20 , c ) 5015 , d ) 25 , e ) 35 | c | subtract(5020, divide(502, 100.4)) | 5020 − ( 502 ÷ 100.4 ) = ? | "explanation : = 5020 − ( 502 / 1004 × 10 ) = 5020 − 5 = 5015 option c" | a = 502 / 100
b = 5020 - a
|
a ) 48 , b ) 50 , c ) 54 , d ) 66 , e ) 86 | b | divide(add(subtract(multiply(46, 10), 25), 65), 10) | the average of 10 numbers is calculated as 46 . it is discovered later on that while calculating the average , the number 65 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 40 . then the average will increase by 40 / 10 = 4 . the correct average is 50 . the answer is b ." | a = 46 * 10
b = a - 25
c = b + 65
d = c / 10
|
a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent w should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the inc... | a = 10 * 4
b = 100 - 20
c = b / 100
d = c * 10
e = a / d
f = e - 4
g = f / 4
h = g * 100
|
a ) $ 7.2 , b ) $ 7.5 , c ) $ 8 , d ) $ 9 , e ) $ 10 | c | add(divide(320, add(divide(subtract(320, multiply(4, 40)), 4), 40)), 4) | john spends $ 320 buying his favorite dolls . if he buys only small monkey dolls , which are $ 4 cheaper than the large monkey dolls , he could buy 40 more dolls than if he were to buy only large monkey dolls . how much does a large monkey doll cost ? | a and b is not an integer . so we start with c if large doll costs $ 8 , then he can buy 320 / 8 = 40 large dolls and 320 / 4 = 80 small dolls . difference is 40 , which is we wanted . answer c . | a = 4 * 40
b = 320 - a
c = b / 4
d = c + 40
e = 320 / d
f = e + 4
|
a ) 10 , b ) 12 , c ) 18 , d ) 14 , e ) 17 | c | multiply(const_2, sqrt(divide(110, const_2))) | the number 110 can be written as sum of the squares of 3 different positive integers . what is the sum of these 3 different integers ? | "sum of the squares of 3 different positive integers = 110 5 ^ 2 + 6 ^ 2 + 7 ^ 2 = 110 now , sum of these 3 different integers = 5 + 6 + 7 = 18 ans - c" | a = 110 / 2
b = math.sqrt(a)
c = 2 * b
|
a ) 26 km / hr , b ) 40 km / hr , c ) 60 km / hr , d ) 77 km / hr , e ) 46 km / hr | b | divide(multiply(multiply(divide(add(100, 100), multiply(12, add(const_1, const_2))), const_2), const_3600), const_1000) | two trains , each 100 m long , moving in opposite directions , cross other in 12 sec . if one is moving twice as fast the other , then the speed of the faster train is ? | "let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 12 = 3 x = > x = 50 / 9 . so , speed of the faster train = 100 / 9 = 100 / 9 * 18 / 5 = 40 km / hr . answer : b" | a = 100 + 100
b = 1 + 2
c = 12 * b
d = a / c
e = d * 2
f = e * 3600
g = f / 1000
|
a ) s . 200 , b ) s . 100 , c ) s . 300 , d ) s . 50 , e ) s . 90 | a | subtract(divide(divide(1000, add(divide(1, 2), divide(1, 3))), 2), divide(divide(1000, add(divide(1, 2), divide(1, 3))), 3)) | a profit of rs . 1000 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 1000 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 600 and 400 . difference in profit share = 600 - 400 = 200 answer : a" | a = 1 / 2
b = 1 / 3
c = a + b
d = 1000 / c
e = d / 2
f = 1 / 2
g = 1 / 3
h = f + g
i = 1000 / h
j = i / 3
k = e - j
|
a ) 1 / 6 , b ) 1 / 4 , c ) 2 / 7 , d ) 1 / 3 , e ) 5 / 12 | e | subtract(1, add(divide(1, 4), divide(1, 3))) | there is a total of 120 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 3 . what is the probability that the marble will be either red or blue ? | "total marbles in the box = 120 white marbles = 120 / 4 = 30 green marbles = 120 / 3 = 40 w + g = 70 red + blue = 50 p ( red or blue ) = 50 / 120 = 5 / 12 answer : e" | a = 1 / 4
b = 1 / 3
c = a + b
d = 1 - c
|
a ) 08 hours , b ) 12 hours , c ) 03 hours , d ) 04 hours , e ) 21 hours | b | inverse(multiply(divide(const_2, const_3), divide(const_1, 8))) | three pipes of same capacity can fill a tank in 8 hours . if there are only two pipes of same capacity , the tank can be filled in ? | "the part of the tank filled by three pipes in one hour = 1 / 8 = > the part of the tank filled by two pipes in 1 hour = 2 / 3 * 1 / 8 = 1 / 12 . the tank can be filled in 12 hours . answer : b" | a = 2 / 3
b = 1 / 8
c = a * b
d = 1/(c)
|
a ) 44 , b ) 56 , c ) 16 , d ) 32 , e ) 31 | e | sqrt(add(941, multiply(10, 2))) | if a 2 + b 2 + c 2 = 941 and ab + bc + ca = 10 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 941 and ab + bc + ca = 10 , ( a + b + c ) ^ 2 = 941 + 2 ( 10 ) = 961 = 31 ^ 2 therefore : a + b + c = 31 answer : e" | a = 10 * 2
b = 941 + a
c = math.sqrt(b)
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | floor(divide(reminder(power(6, reminder(18, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 6 ^ 18 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . .... | a = 4 + 1
b = 6 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
a ) 4 : 1 , b ) 2 : 1 , c ) 10 : 1 , d ) 3 : 1 , e ) 1 : 2 | b | divide(add(20, sqrt(multiply(20, 5))), add(5, sqrt(multiply(20, 5)))) | if dev works alone he will take 20 more hours to complete a task than if he worked with tina to complete the task . if tina works alone , she will take 5 more hours to complete the complete the task , then if she worked with dev to complete the task ? what is the ratio of the time taken by dev to that taken by tina if ... | let time taken by dev to complete the work alone be x days and that by tina be y days work done by dev in 1 day = 1 / x work done by tina in 1 day = 1 / y work done by devtina in 1 day = 1 / x + 1 / y = ( x + y ) / xy thus working together they will complete the work in xy / ( x + y ) days acc . to the ques : 1 ) if de... | a = 20 * 5
b = math.sqrt(a)
c = 20 + b
d = 20 * 5
e = math.sqrt(d)
f = 5 + e
g = c / f
|
a ) 8.1 , b ) 7.0 , c ) 9.33 , d ) 8 , e ) 9 | c | divide(140, multiply(add(50, 4), const_0_2778)) | a train 140 m long is running with a speed of 50 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 50 + 4 = 54 km / hr . = 54 * 5 / 18 = 15 m / sec . time taken to pass the men = 140 * 1 / 15 = 9.33 sec . answer : option c" | a = 50 + 4
b = a * const_0_2778
c = 140 / b
|
a ) 5 , b ) 6 , c ) 7.5 , d ) 8.8 , e ) 9 | d | divide(110, multiply(add(40, 5), const_0_2778)) | a train 110 m long is running with a speed of 40 km / hr . in what time will it pass a man who is running at 5 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 40 + 5 = 45 km / hr . = 45 * 5 / 18 = 25 / 2 m / sec . time taken to pass the men = 110 * 2 / 25 = 8.8 sec . answer : option d" | a = 40 + 5
b = a * const_0_2778
c = 110 / b
|
a ) 13 seconds , b ) 17 seconds , c ) 26 seconds , d ) 34 seconds , e ) 42.5 seconds | e | divide(add(divide(multiply(3.8, add(15, 2)), subtract(4.2, 3.8)), add(15, 2)), 4.2) | john and steve are speed walkers in a race . john is 15 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m / s , while steve maintains a blistering 3.8 m / s speed . if john finishes the race 2 meters ahead of steve , how long was john ’ s final push ? | "let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.8 t + 15 + 2 - - - > 0.4 t = 17 - - - > t = 42.5 seconds . e is the correct answer ." | a = 15 + 2
b = 3 * 8
c = 4 - 2
d = b / c
e = 15 + 2
f = d + e
g = f / 4
|
a ) 14 , b ) 14.5 , c ) 15 , d ) 15.5 , e ) 16 | d | divide(subtract(multiply(10, 20), multiply(add(const_4, const_1), 9)), 10) | the average of 10 consecutive integers is 20 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a = 10 * 20
b = 4 + 1
c = b * 9
d = a - c
e = d / 10
|
a ) 40 , b ) 80 , c ) 70 , d ) 60 , e ) 90 | e | subtract(100, 10) | a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 10 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ? | "200 men do the rest of the work in 100 - 10 = 90 days 100 men can do the rest of the work in 90 * 200 / 100 = 180 days required number of days = 180 - 90 = 90 days answer is e" | a = 100 - 10
|
a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) none of these | d | multiply(subtract(const_1, divide(9, 15)), 30) | suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 30 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | "the part of job that suresh completes in 9 hours = 9 â „ 15 = 3 â „ 5 remaining job = 1 - 3 â „ 5 = 2 â „ 5 remaining job can be done by ashutosh in 2 â „ 5 ã — 30 = 12 hours answer d" | a = 9 / 15
b = 1 - a
c = b * 30
|
a ) 24,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 200 , 000,000 , e ) 240 , 000,000 | d | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(5, 100)), multiply(10, 100)) | if a rectangular room measures 10 meters by 5 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "d . 200 , 000,000 10 * 100 * 5 * 100 * 4 * 100 = 200 , 000,000" | a = 4 * 100
b = 1 / 10
c = a * b
d = 5 * 100
e = c * d
f = 10 * 100
g = e * f
|
a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 1 : 6 , e ) 1 : 7 | d | divide(const_2, subtract(subtract(20, divide(const_12, const_2)), const_2)) | only a single rail track exists between stations a and b on a railway line . one hour after the north bound super fast train n leaves station a for station b , a south - bound passenger train s reaches station a from station b . the speed of the super fast train is twice that of a normal express train e , while the spe... | explanation : if speed of n = 4 , speed of s = 1 . = > average speed = ( 2 x 4 x 1 ) / ( 4 + 1 ) = 1.6 . because time available is 2 / 3 , speed = 3 / 2 . now average speed = 2.4 now speed of n = 8 . speed of s = y . = > ( 2 x 8 x y ) / ( 8 + y ) = 2.4 = > y = 1.3 the required ratio is 1.3 : 8 i . e 1 : 6 . answer : d | a = 12 / 2
b = 20 - a
c = b - 2
d = 2 / c
|
a ) 18 , b ) 21 , c ) 24 , d ) 27 , e ) 30 | b | add(divide(subtract(const_1, multiply(divide(const_1, 30), 15)), add(divide(const_1, 20), divide(const_1, 30))), 15) | a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 15 days before the project is completed , in how many days total will the project be completed ? | a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 15 days , b can do 1 / 2 of the project . thus a and b must complete 1 / 2 of the project , which takes 6 days . the total number of days is 6 + 15 = 21 . the answ... | a = 1 / 30
b = a * 15
c = 1 - b
d = 1 / 20
e = 1 / 30
f = d + e
g = c / f
h = g + 15
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | d | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | how many positive integers less than 200 are there such that they are multiples of 13 or multiples of 14 ? | "200 / 13 = 15 ( plus remainder ) so there are 15 multiples of 13 200 / 14 = 14 ( plus remainder ) so there are 14 multiples of 14 we need to subtract 1 because 13 * 14 is a multiple of both so it was counted twice . the total is 15 + 14 - 1 = 28 the answer is d ." | a = 4 + 13
b = a - 1
c = math.factorial(b)
d = math.factorial(13)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 30 kg , b ) 37 kg , c ) 42 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(36, 30), multiply(24, 30)), add(36, 24)) | there are 2 sections a and b in a class , consisting of 36 and 24 students respectively . if the average weight of section a is 30 kg and that of section b is 30 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 36 * 30 + 24 * 30 = 1800 average weight of the class is = 1800 / 60 = 30 kg answer is a" | a = 36 * 30
b = 24 * 30
c = a + b
d = 36 + 24
e = c / d
|
a ) 100 , b ) 175 , c ) 210 , d ) 245 , e ) 280 | a | subtract(30, subtract(25, 30)) | sarah is driving to the airport . after driving at 25 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 50 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did s... | "after driving at 25 miles per hourfor one hour , this distance left to cover is d - 25 . say this distance is x miles . now , we know that the difference in time between covering this distance at 25 miles per hour and 50 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 25 - x / 50 = 3 / 2 - - > 2 x / ... | a = 25 - 30
b = 30 - a
|
a ) 500 km , b ) 630 km , c ) 900 km , d ) 660 km , e ) none | c | add(add(divide(multiply(100, 40), subtract(50, 40)), 100), divide(multiply(100, 40), subtract(50, 40))) | two trains start from p and q respectively and travel towards each other at a speed of 50 km / hr and 40 km / hr respectively . by the time they meet , the first train has travelled 100 km more than the second . the distance between p and q is : | "sol . at the time of meeting , let the distane travelled byb the second train be x km . then , distance covered by the first train is ( x + 100 ) km ∴ x / 40 = ( x + 100 ) / 50 ⇔ 50 x = 40 x 4000 ⇔ x = 400 . so , distance between p and q = ( x + x + 100 ) km = 900 km . answer c" | a = 100 * 40
b = 50 - 40
c = a / b
d = c + 100
e = 100 * 40
f = 50 - 40
g = e / f
h = d + g
|
a ) 15 days , b ) 10 days , c ) 9 days , d ) 8 days , e ) 7 days | a | divide(multiply(const_3, 10), const_2) | a is thrice as good a workman as b and takes 10 days less to do a piece of work than b takes . b alone can do the whole work in | explanation : ratio of times taken by a and b = 1 : 3 means b will take 3 times which a will do in 1 time if difference of time is 2 days , b takes 3 days if difference of time is 10 days , b takes ( 3 / 2 ) * 10 = 15 days option a | a = 3 * 10
b = a / 2
|
a ) 8.0 , b ) 3.0 , c ) 9.5 , d ) 3.75 , e ) 2.15 | d | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1500) | the difference between simple and compound interest on rs . 1500 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1500 * 10 * 1 ) / 100 = rs . 150 c . i . = [ 1500 * ( 1 + 5 / 100 ) 2 - 1500 ] = rs . 153.75 difference = ( 153.75 - 150 ) = rs . 3.75 answer : d" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = 10 / 100
f = e + 1
g = d - f
h = g * 1500
|
a ) 12 , b ) 15 , c ) 17 , d ) 18 , e ) 22 | c | subtract(add(12, 8), 3) | in a class , 12 students like to play basketball and 8 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 12 + 8 - 3 = 17 c )" | a = 12 + 8
b = a - 3
|
a ) 1 / 3 , b ) 1 / 7 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | a | divide(const_10, 30) | in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if one - sevenths of the male students are married , what fraction of the female students is single ? | let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 1 / 7 of 70 = 10 males are married if 30 is the total number of students who are married and out of that 10 are males then th... | a = 10 / 30
|
a ) $ 1.75 , b ) $ 2.50 , c ) $ 4.10 , d ) $ 4.70 , e ) $ 8.20 | a | divide(subtract(9.85, 6.35), const_2) | a train ride from two p to town q costs $ 6.35 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.85 . what is the cost of a bus ride from town p to town q ? | "let x be the cost of a bus ride . x + ( x + 635 ) = 985 2 x = 350 x = $ 1.75 the answer is a ." | a = 9 - 85
b = a / 2
|
a ) 135 , b ) 261 , c ) 422 , d ) 430 , e ) 438 | b | add(divide(368, gcd(gcd(10, 144), 368)), add(divide(10, gcd(gcd(10, 144), 368)), divide(144, gcd(gcd(10, 144), 368)))) | a drink vendor has 10 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | "the number of liters in each can = hcf of 10 , 144 and 368 = 2 liters . number of cans of maaza = 10 / 2 = 5 number of cans of pepsi = 144 / 2 = 72 number of cans of sprite = 368 / 2 = 184 the total number of cans required = 5 + 72 + 184 = 261 cans . answer : b" | a = math.gcd(10, 144)
b = math.gcd(a, 368)
c = 368 / b
d = math.gcd(10, 144)
e = math.gcd(d, 368)
f = 10 / e
g = math.gcd(10, 144)
h = math.gcd(g, 368)
i = 144 / h
j = f + i
k = c + j
|
a ) 270 , b ) 380 , c ) 120 , d ) 360 , e ) 180 | c | multiply(divide(160, divide(40, const_100)), divide(30, const_100)) | if 40 % of a certain number is 160 , then what is 30 % of that number ? | "explanation : 40 % = 40 * 4 = 160 30 % = 30 * 4 = 120 answer : option c" | a = 40 / 100
b = 160 / a
c = 30 / 100
d = b * c
|
a ) 55 , b ) 5 , c ) 10 , d ) 45 , e ) 4 | b | add(floor(divide(45, 11)), const_1) | if the sum of two numbers is 45 and the l . c . m and sum of the reciprocal of the numbers are 120 and 11 / 120 then hcf of numbers is equal to : | let the numbers be a and b and hcf is x , then , a + b = 45 and ab = x * 120 = 120 x required sum = 1 / a + 1 / b = ( a + b ) / ab = 45 / 120 x = 11 / 120 . x = 5 . answer : b | a = 45 / 11
b = math.floor(a)
c = b + 1
|
a ) 12 , b ) 120 , c ) 24 , d ) 240 , e ) 36 | d | divide(0.24, divide(0.1, const_100)) | find the missing figures : 0.1 % of ? = 0.24 | "let 0.1 % of x = 0.24 . then , 0.1 * x / 100 = 0.24 x = [ ( 0.24 * 100 ) / 0.1 ] = 240 . answer is d ." | a = 0 / 1
b = 0 / 24
|
a ) 23 % , b ) 27 % , c ) 31 % , d ) 35 % , e ) 39 % | b | multiply(subtract(const_1, divide(const_100, add(add(const_100, 25), divide(multiply(add(const_100, 25), 10), const_100)))), const_100) | there has been successive increases of 25 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | "let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.25 * p = x * p y = x / ( 1.1 * 1.25 ) which is about 0.73 x which is a decrease of about 27 % . the answer is b ." | a = 100 + 25
b = 100 + 25
c = b * 10
d = c / 100
e = a + d
f = 100 / e
g = 1 - f
h = g * 100
|
a ) 60 min , b ) 45 min , c ) 90 min , d ) 70 min , e ) 30 min | b | inverse(add(divide(const_1, 180), divide(const_1, 60))) | pipe a can fill a tank in 60 min . there is a second pipe in the bottom of the cistern to empty it . if all the two pipes are simultaneously opened , then the cistern is full in 180 min . in how much time , the second pipe alone can empty the cistern ? | work done by the third pipe in 1 min = 1 / 180 - ( 1 / 60 ) = - 1 / 45 . [ - ve sign means emptying ] the third pipe alone can empty the cistern in 45 min . answer : b | a = 1 / 180
b = 1 / 60
c = a + b
d = 1/(c)
|
a ) 44 % , b ) 40 % , c ) 68.75 % , d ) 56.25 % , e ) 36 % | c | multiply(subtract(divide(add(const_100, 35), subtract(const_100, 20)), const_1), const_100) | a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 35 % impurities to the product . find the net profit percentage of the dealer ? | "the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 35 % impurities to the product . it means that he is selling $ 5 worth of product for $ 6.75 . so his profit is $ 6.75 - $... | a = 100 + 35
b = 100 - 20
c = a / b
d = c - 1
e = d * 100
|
a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 48 | e | multiply(divide(multiply(2, 8), subtract(8, 6)), 6) | nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 2 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | e its is . let price = x ( x - 2 ) 8 = 6 x x = 8 hence total money = 6 * 8 = 48 | a = 2 * 8
b = 8 - 6
c = a / b
d = c * 6
|
a ) 1 / 14 , b ) 4 / 49 , c ) 18 / 19 , d ) 45 / 49 , e ) 13 / 14 | c | divide(subtract(choose(20, const_2), choose(subtract(20, multiply(20, divide(3, 4))), const_2)), choose(20, const_2)) | according to a recent student poll , 3 / 4 out of 20 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "15 students are interested , 5 are not interested prob = 1 - 5 c 2 / 20 c 2 = 1 - ( 5 * 4 / ( 20 * 19 ) ) = 1 - 1 / 19 = 18 / 19 answer : c" | a = math.comb(20, 2)
b = 3 / 4
c = 20 * b
d = 20 - c
e = math.comb(d, 2)
f = a - e
g = math.comb(20, 2)
h = f / g
|
a ) 1 / 3 , b ) 1 / 14 , c ) 3 / 14 , d ) 1 / 15 , e ) 1 / 16 | c | divide(choose(4, const_2), choose(add(4, 4), const_2)) | there are 4 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probability is 4 c 2 from 8 shoes probability of taking 2 red shoes is 4 c 2 / 8 c 2 = 3 / 14 answer : c" | a = math.comb(4, 2)
b = 4 + 4
c = math.comb(b, 2)
d = a / c
|
a ) 16 days , b ) 10 days , c ) 8 days , d ) 6 days , e ) 18 days | b | inverse(add(divide(const_1, 30), divide(const_1, divide(30, const_2)))) | ram , who is half as efficient as krish , will take 30 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ? | "number of days taken by ram to complete task = 30 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 30 days is 30 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work ... | a = 1 / 30
b = 30 / 2
c = 1 / b
d = a + c
e = 1/(d)
|
a ) 80 , b ) 110 , c ) 160 , d ) 100 , e ) 400 | d | divide(5, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 5 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain mat... | a = 1 / 5
b = 1 / 4
c = a + b
d = 1 / 2
e = c + d
f = 1 - e
g = 5 / f
|
a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | e | divide(subtract(add(26, add(26, 11)), multiply(11, 11)), const_2) | the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 11 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x â € “ ( 26 + 40 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 54 2 x = 54 x = 27 . so , average age of the team is 27 years . answer e" | a = 26 + 11
b = 26 + a
c = 11 * 11
d = b - c
e = d / 2
|
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