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 Γ£ β ( 1 / 12 ) = 2 / 3 remaining work Γ’ β¬ β 1 Γ’ β¬ β 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 answer : e | 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 Γ 9 / 20 = 36000 p = ( 36000 Γ 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 transformation . ) | "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 β 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 β 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 Ζ orm in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the pla Ζ orm . | explanation : speed = distance / Ι΅ me = 300 / 18 = 50 / 3 m / sec let the length of the pla Ζ orm be x meters then distance = speed β timex + 300 = 503 β 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(divide(subtract(251, subtract(11, const_2)), 11), divide(subtract(add(divide(subtract(251, subtract(11, const_2)), 11), subtract(11, const_2)), subtract(11, const_2)), 11))) | 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 9 scraps . total 9 + 2 = 11 sscraps . by using 11 scraps , we make 1 soap . so , total = 22 + 2 + 1 = 25 soaps we produce . answer : b | 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
D = 11 - 2
E = C - D
F = E / 11
G = x + F
H = u + G
|
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 β 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 alternatively , percentage of women = 40 % percentage of women who did not hold full time jobs = ( 1 - 3 / 4 ) * 100 % = 25 % of all mothers did not have full time jobs 25 % of 40 % - - > 10 % of the total parents percentage of men = 60 % percentage of men who held full time jobs = ( 1 - 9 / 10 ) * 100 % = 10 % of all fathers did not have full time jobs 10 % of 60 % - - > 6 % of the total parents therefore , percentage of parents who did not hold full time jobs = 10 + 6 = 16 % answer e - 16 %" | 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 = > 15 answer c" | 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 slips ? | "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 had signed slips , so . 45 * 100 = 45 in total had signed slips thus 45 - 15 = 30 girl scouts had signed slips thus the percentage of scouts who were girl scouts arriving with signed permission slips is 30 / 100 = . 30 = 30 % e" | 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 , 4 ) , ( 5 , 6 ) , ( 6 , 1 ) , ( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 6 , 5 ) , ( 6 , 6 ) } n ( e ) = 27 . p ( e ) = n ( e ) / n ( s ) = 27 / 36 = 3 / 4 . answer : c ." | 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 Β± β ( 52 - 4 Γ 3 Γ 2 ) ) / ( 2 Γ 3 ) = ( - 5 Β± β ( 25 - 24 ) ) / 6 = ( - 5 Β± 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 β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ 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 ) = p ( a ) + p ( b ) - p ( ab ) . = > 4 / 5 = ( 3 / 4 ) + ( 2 / 5 ) - p ( ab ) . = > p ( ab ) = 7 / 20 . hence , the required probability is 7 / 20 . the answer is d ." | 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 of terms the median is just the average of the two most median terms ( here the 5 th and 6 th numbers in the set ) . this means that numbers 5 and 6 in this set are 153 and 155 . because we know that number 5 is 153 , we know that the smallest number is 4 odd numbers below this , which means that it is 4 * 2 = 8 below this ( every odd number is every other number ) . therefore 153 - 8 = 145 , answer choice c" | 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 β s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students β weights changed . what is the value of x ? | "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 = 6 students ( including the new one ) . so , initially there were 5 student . total weight = 5 x + 80 = 6 ( x - 1 ) - - > x = 92 pounds . answer : e ." | 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 that day ? | 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 bought 7 / 2 + 1 / 2 = 4 ( leaving 3 ) the third kid bought 3 / 2 + 1 / 2 = 2 ( leaving 1 ) the fourth kid bought 1 / 2 + 1 / 2 = 1 adding all of them together , we get 8 + 4 + 2 + 1 = 15 . | a = 4 * 2
b = 4 * 2
c = a + b
d = c - 1
|
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 Γ£ β 100 / 90 Γ£ β 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 , the ratio of the work done by a : b : c when they work together will be equal to 1 / 6 : 1 / 8 : 1 / 12 multiplying the numerator of all 3 fractions by 24 , the lcm of 6 , 8 and 12 will not change the relative values of the three values . we get 24 / 6 : 24 / 8 : 24 / 12 = 4 : 3 : 2 . i . e . , the ratio in which a : b : c will share $ 2340 will be 4 : 3 : 2 . hence , c ' s share will be 2 * 2340 / 9 = 520 . correct choice is ( b )" | 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 Γ 30 wages of 1 man for 1 day = 21600 Γ 2 / 40 Γ 30 wages of 1 man for 25 days = 21600 Γ 2 Γ 25 / 40 Γ 30 number of men = 14400 / ( 21600 Γ 2 Γ 25 / 40 Γ 30 ) = 144 / ( 216 Γ 50 / 40 Γ 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 s , the sum will have as many trailing 0 s as that in the lowest term . example : 20 + 2300 + 34210000 = 34212320 - > one 0 because lowest term 20 has one 0 . this case has an exception because of an additional 5 in 55 . answer is b | 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 = 82 . answer : c ." | 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 β ( 502 Γ· 100.4 ) = ? | "explanation : = 5020 β ( 502 / 1004 Γ 10 ) = 5020 β 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 increase in hours worked : wh = . 8 wx ( divide both sides by . 8 w ) 1.25 h = x we know that the new number of hours worked w will be 25 % greater than the original number . the answer is c ." | 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 = . . . 56 is 5 . the tens digit of 6 ^ 7 = . . . 36 is 3 again . etc . . . 18 has the form 5 n + 3 , so the tens digit of 6 ^ 18 is 1 . the answer is a ." | 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 each of them worked alone to complete the task ? | 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 dev works alone he will take 20 more hours to complete a task than if he worked with tina ' x - xy / ( x + y ) = 20 = > ( x ^ 2 ) / ( x + y ) = 20 . . . . . . . . . . . . . ( i ) 2 ) if tina works alone , she will take 5 more hours to complete the complete the task , then if she worked with dev y - xy / ( x + y ) = 5 = > ( y ^ 2 ) / ( x + y ) = 5 . . . . . . . . . . . . . . ( ii ) dividing ( i ) by ( ii ) x ^ 2 / y ^ 2 = 4 = > x / y = 2 ans : b hope it helps | 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 β 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 Γ’ Β β 15 = 3 Γ’ Β β 5 remaining job = 1 - 3 Γ’ Β β 5 = 2 Γ’ Β β 5 remaining job can be done by ashutosh in 2 Γ’ Β β 5 Γ£ β 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 speed of a passenger train s is half that of e . on a particular day , n leaves for b from a , 20 min behind the normal schedule . in order to maintain the schedule , both n and s increased their speeds . if the super fast train doubles its speed , what should be the ratio ( approximately ) of the speeds of passenger train to that of the super fast train so that the passenger train s reaches exactly at the scheduled time at a on that day ? | 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 answer is b . | 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 she drive in all ? | "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 / 50 - x / 50 = 3 / 2 - - > x / 50 = 3 / 2 - - > x = 75 . total distance = x + 25 = 100 miles . answer : a" | 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 β΄ x / 40 = ( x + 100 ) / 50 β 50 x = 40 x 4000 β 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 the remaining 20 will be females who are married . total females = 30 married females = 20 then single females = 30 - 20 = 10 we need to find the fraction of female students who are single i . e single female students / total female student = 10 / 30 = 1 / 3 [ a ] | 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 - $ 4 = $ 2 and his profit percent is ( 2.75 / 4 ) * 100 = 68.75 % answer : - c" | 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 done by krish and ram in a day = 3 w total amount of time needed by krish and ram to complete task = 30 w / 3 w = 10 days answer b" | 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 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 . since this represents all the grades in the class , it represents all the students in the class . thus we know : # a β s + # b β s + # c β s + # d β s = total number of students in the class 1 / 5 ( t ) + ΒΌ ( t ) + Β½ ( t ) + 5 = t we can multiply the entire equation by 20 to cancel out the denominators of the fractions and we have : 4 t + 5 t + 10 t + 100 = 20 t 19 t + 100 = 20 t 100 = t there are a total of 100 students in the math class . answer is d ." | 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 Γ’ β¬ β ( 26 + 40 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 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
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.