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 ) 9000 cubic meters , b ) 10500 cubic meters , c ) 11750 cubic meters , d ) 12000 cubic meters , e ) 12500 cubic meters | d | multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1)))) | with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity e of the pool ? | d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) / ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or , 120 x = 80 ( x + 50 ) . or x = 100 . hence , the capacity of pool = 120 x = 12000 cubic meters . | a = 48 / const_60
b = 4 + 1
c = a * b
d = const_60 * c
e = 50 * d
|
a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72 | c | add(divide(subtract(multiply(18, 24), multiply(12, 16)), const_2), 24) | if p # q denotes the least common multiple of p and q , then t = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ? | "there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , t = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c ." | a = 18 * 24
b = 12 * 16
c = a - b
d = c / 2
e = d + 24
|
a ) 0.8 , b ) 0.5 , c ) 1.0 , d ) 1.1 , e ) 1.2 | b | divide(22, multiply(divide(90, const_3600), multiply(subtract(add(90, 90), const_4), const_10))) | how many seconds will it take for a car that is traveling at a constant rate of 90 miles per hour to travel a distance of 22 yards ? ( 1 mile = 1,160 yards ) | speed = 90 miles / hr = 44 yard / s distance = 22 yards time = distance / speed = 22 / 44 = 0.5 sec ans - b | a = 90 / 3600
b = 90 + 90
c = b - 4
d = c * 10
e = a * d
f = 22 / e
|
a ) 3 : 2 , b ) 2 : 1 , c ) 3 : 3 , d ) 1 : 2 , e ) 6 : 1 | d | divide(subtract(const_2, 1), subtract(4, const_2)) | if 4 men and 1 boy working together , can do two times as much work per hour as a man and a boy together . find the ratio of the work done by a man and that of a boy for a given time ? | 4 m + 1 b = 2 ( 1 m + 1 b ) 4 m + 1 b = 2 m + 2 b 2 m = 1 b the required ratio of work done by a man and a boy = 1 : 2 d | a = 2 - 1
b = 4 - 2
c = a / b
|
a ) 65 sec , b ) 54 sec , c ) 48 sec , d ) 33 sec , e ) 12 sec | b | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.15, const_1000), divide(multiply(90, const_1000), const_3600))) | two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 1.15 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 1.15 = 2.25 km = 2250 m . required time = 2250 * 3 / 125 = 54 sec . answer : b" | a = 1 * 10
b = 60 * 1000
c = b / 3600
d = a / c
e = 1 * 15
f = 90 * 1000
g = f / 3600
h = e / g
i = d - h
|
a ) 2.4 . , b ) 3.6 . , c ) 4.2 . , d ) 8 . , e ) 6.4 | d | multiply(multiply(divide(divide(2, divide(5, 2)), 2), 4), 5) | two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 4 carpenters build in 5 hours ? | w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 × r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 4 carpenters in 5 hrs = 4 × rate of each carpenter x time = 4 × 2 / 5 × 5 = 8 desks d is the correct answer . | a = 5 / 2
b = 2 / a
c = b / 2
d = c * 4
e = d * 5
|
a ) 40 , b ) 65 , c ) 72 , d ) 78 , e ) 20 | b | multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 60)), divide(36, 60))) | a retailer buys 60 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ? | let the market price of each pen be $ 1 then , cost price of 60 pens = $ 36 selling price of 60 pens = 99 % of $ 60 = $ 59.40 profit % = ( ( 23.40 * 100 ) / 36 ) % = 65 % answer b | a = 1 / 100
b = 1 - a
c = 36 / 60
d = b - c
e = 36 / 60
f = d / e
g = 100 * f
|
a ) 45 % , b ) 40 % , c ) 35 % , d ) 30 % , e ) 25 % | b | subtract(add(add(65, 55), 20), const_100) | if 65 percent of a class answered the first question on a certain test correctly , 55 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "65 % answered the first question correctly and 20 % answered neither correctly . then 15 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 55 % - 15 % = 40 % . the answer is b ." | a = 65 + 55
b = a + 20
c = b - 100
|
a ) 20 km , b ) 40 km , c ) 60 km , d ) 80 km , e ) 26.6 km | e | divide(multiply(multiply(multiply(5, const_2), multiply(4, const_2)), 6), add(multiply(5, const_2), multiply(4, const_2))) | a person starts walking at a speed of 5 km / hr through half the distance , rest of the distance he covers with aspeed 4 km / hr . total time of travel is 6 hours . what is the maximum distance he can cover ? | t = d / s , so 6 = x / 2 * 1 / 5 + x / 2 * 1 / 4 ( because half distance with 5 km / ph , and remaining half with 4 km / hr ) 6 = x ( 18 / 80 ) x = 26.6 km answer : e | a = 5 * 2
b = 4 * 2
c = a * b
d = c * 6
e = 5 * 2
f = 4 * 2
g = e + f
h = d / g
|
a ) 12300 , b ) 12350 , c ) 12000 , d ) 12500 , e ) 12600 | c | add(10000, multiply(divide(multiply(10000, 20), const_100), 1)) | the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 1 year ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 ã — 120 / 100 = 12000 c" | a = 10000 * 20
b = a / 100
c = b * 1
d = 10000 + c
|
a ) 50 , b ) 43 , c ) 45 , d ) 55 , e ) 60 | a | subtract(95, subtract(add(floor(divide(95, add(const_4, const_1))), floor(divide(95, const_4))), floor(divide(95, add(const_10, add(const_4, const_1)))))) | two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fifth widget , starting with the fifth , and steven inspects every third , starting with the third , how many of the 95 widgets produced in the first hour of operation are not inspected by either inspector ? | "widgets inspected by lauren : ( ( 95 - 5 ) / 5 ) + 1 = 18 + 1 = 19 widgets inspected by steven : ( ( 93 - 3 ) / 3 ) + 1 = 31 + 1 = 32 widgets inspected by both : ( ( 90 / 15 ) + 1 = 6 total : 19 + 32 - 6 = 45 hence , widgets not inspected : 95 - 45 = 50 option a" | a = 4 + 1
b = 95 / a
c = math.floor(b)
d = 95 / 4
e = math.floor(d)
f = c + e
g = 4 + 1
h = 10 + g
i = 95 / h
j = math.floor(i)
k = f - j
l = 95 - k
|
a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 3 % | b | multiply(divide(10, 1), const_100) | what percent is 10 gm of 1 kg ? | "1 kg = 1000 gm 10 / 1000 × 100 = 1000 / 1000 = 1 % b )" | a = 10 / 1
b = a * 100
|
a ) $ 80 , b ) $ 81 , c ) $ 82 , d ) $ 83 , e ) $ 84 | e | add(add(divide(add(multiply(6, 11), 6), subtract(7, const_1)), 6), multiply(6, 11)) | 7 people went to a restaurant for dinner . 6 of them spent $ 11 each on their meals and the seventh person spent $ 6 more than the average expenditure of all 7 people . what was the total amount of money that the 7 people spent ? | let t be the total amount of money spent . 6 * 11 + ( t / 7 + 6 ) = t 72 = 6 t / 7 t = ( 7 * 72 ) / 6 = 84 the answer is e . | a = 6 * 11
b = a + 6
c = 7 - 1
d = b / c
e = d + 6
f = 6 * 11
g = e + f
|
a ) 8 , b ) 12 , c ) 20 , d ) 32 , e ) 36 | d | multiply(subtract(const_10, 1), 4) | on june 1 a bicycle dealer noted that the number of bicycles in stock had decreased by 4 for each of the past 5 months . if the stock continues to decrease at the same rate for the rest of the year , how many fewer bicycles will be in stock on september 1 than were in stock on january 1 ? | "jan 1 = c feb 1 = c - 4 march 1 = c - 8 april 1 = c - 12 may 1 = c - 16 june 1 = c - 20 july 1 = c - 24 aug 1 = c - 28 sept 1 = c - 32 difference between stock on september 1 than were in stock on january 1 will be - c - ( c - 32 ) = 32 hence answer will be ( d )" | a = 10 - 1
b = a * 4
|
a ) 272 , b ) 167 , c ) 157 , d ) 352 , e ) 380 | a | multiply(add(15, const_1), add(add(15, const_1), const_1)) | there are 15 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 17 from 17 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 17 p ₂ ways . 17 p ₂ = 17 * 16 = 272 . answer : a" | a = 15 + 1
b = 15 + 1
c = b + 1
d = a * c
|
a ) 6.99 , b ) 5.99 , c ) 8.99 , d ) 4.99 , e ) 9.99 | c | multiply(divide(divide(170, const_1000), 68), const_3600) | how long will it take a train travelling at 68 kmph to pass an electric pole if the train is 170 m long | "sol . speed = [ 68 x 5 / 18 ] m / sec = 18.9 m / sec . time taken = ( 170 / 18.9 ) sec = 8.99 sec . answer c" | a = 170 / 1000
b = a / 68
c = b * 3600
|
a ) 270 , b ) 289 , c ) 267 , d ) 255 , e ) 333 | e | multiply(divide(1073, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 1073 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 1073 = 333 answer : e" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 1073 / e
g = 16 * 9
h = f * g
|
a ) 80 % , b ) 100 % , c ) 116.7 % , d ) 120 % , e ) 160 % | e | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(10, subtract(10, const_1)), subtract(const_100, 10)), const_100), 15), 25)), const_100), multiply(subtract(const_1, divide(add(15, divide(multiply(add(10, const_4), subtract(const_100, 10)), const_100)), 20)), const_100)), const_100) | during a special promotion , a certain filling station is offering a 10 percent discount on gas purchased after the first 15 gallons . if kim purchased 20 gallons of gas , and isabella purchased 25 gallons of gas , then isabella ‟ s total per - gallon discount is what percent of kim ‟ s total per - gallon discount ? | "kim purchased 20 gallons of gas . she paid for 15 + 0.9 * 5 = 19.5 gallons , so the overall discount she got was 0.5 / 20 = 0.025 . isabella purchased 25 gallons of gas . she paid for 15 + 0.9 * 10 = 24 gallons , so the overall discount she got was 1 / 25 = 0.04 . 4 / 2.5 * 100 = 160 % answer : e ." | a = 10 - 1
b = 10 + a
c = 100 - 10
d = b * c
e = d / 100
f = e + 15
g = f / 25
h = 1 - g
i = h * 100
j = 10 + 4
k = 100 - 10
l = j * k
m = l / 100
n = 15 + m
o = n / 20
p = 1 - o
q = p * 100
r = i / q
s = r * 100
|
a ) 40 , b ) 50 , c ) 62 , d ) 45 , e ) 56 | b | add(subtract(100, multiply(11, 5)), 5) | a batsman makes a score of 100 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 100 = 11 x x = 100 - 50 = 50 answer is b" | a = 11 * 5
b = 100 - a
c = b + 5
|
a ) 10 , b ) 4 , c ) 21 , d ) 3 , e ) 5 | e | divide(add(7, 3), const_2) | sam wrote 7 letters on tuesday and 3 letters on wednesday . how many letters did he write on average per day ? | ( 7 + 3 ) / 2 = 5 the correct answer is e | a = 7 + 3
b = a / 2
|
a ) 30 , b ) 25.4 , c ) 40 , d ) 35.25 , e ) 26.55 | d | divide(add(multiply(36, 20), subtract(subtract(20, const_2), 40)), 20) | the mean of 20 observations was 36 . it was found later that an observation 25 was wrongly taken as 40 . the corrected new mean is | "explanation : correct sum = ( 36 * 20 + 25 - 40 ) = 705 correct mean = = 705 / 20 = 35.25 answer : d" | a = 36 * 20
b = 20 - 2
c = b - 40
d = a + c
e = d / 20
|
a ) 1 / 3 , b ) 2 / 5 , c ) 2 / 7 , d ) 2 / 9 , e ) 3 / 10 | d | divide(multiply(4, divide(2, 3)), 12) | an outlet pipe can empty 2 / 3 of a cistern in 12 minutes . in 4 minutes , what part of the cistern will be emptied ? | "4 / 12 * 2 / 3 = 2 / 9 the answer is d ." | a = 2 / 3
b = 4 * a
c = b / 12
|
a ) a . 43 , b ) b . 45 , c ) c . 55 , d ) d . 68 , e ) e . 60 | b | subtract(divide(multiply(multiply(30, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 30) | an engineer undertakes a project to build a road 15 km long in 300 days and employs 30 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 30 * 12.5 * 100 ) / ( 2.5 * 200 ) thus , x = 75 thus , more men needed to finish the task = 75 - 30 = 45 hence answer is b" | a = 15 - 2
b = 30 * a
c = b * 100
d = 300 - 100
e = 2 * 5
f = c / e
g = f - 30
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 6 | e | divide(26, 24) | how many of the positive factors of 24 are not factors of 26 | "factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 26 - 1 , 2 , 13 , 26 comparing both , we have 6 factors of 24 which are not factors of 26 - 3 , 4 , 6 , 8 , 12,24 answer : e" | a = 26 / 24
|
a ) 75 kg , b ) 55 kg , c ) 45 kg , d ) 80 kg , e ) 25 kg | d | add(multiply(2.5, 6), 65) | he average weight of 6 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 6 x 2.5 ) kg = 15 kg . weight of new person = ( 65 + 15 ) kg = 80 kg . answer : d" | a = 2 * 5
b = a + 65
|
a ) 150 , b ) 200 , c ) 250 , d ) 350 , e ) 400 | c | divide(subtract(multiply(500, divide(99, const_100)), multiply(500, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | there are 500 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ? | "there are 495 managers and 5 others . the 5 others would compose 2 % of the total number of people if there were 250 people in the room . thus 250 managers must leave . the answer is c ." | a = 99 / 100
b = 500 * a
c = 98 / 100
d = 500 * c
e = b - d
f = 98 / 100
g = 1 - f
h = e / g
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(100, multiply(add(30, 6), const_0_2778)) | the speed at which a woman can row a boat in still water is 30 kmph . if he rows downstream , where the speed of current is 6 kmph , what time will he take to cover 100 metres ? | "speed of the boat downstream = 30 + 6 = 36 kmph = 36 * 5 / 18 = 10 m / s hence time taken to cover 100 m = 100 / 10 = 10 seconds . answer : c" | a = 30 + 6
b = a * const_0_2778
c = 100 / b
|
a ) 1.26 , b ) 2.54 , c ) 5.23 , d ) 4.23 , e ) 3.33 | e | divide(const_100, 30) | a company decreased the price of its main product by 30 % . subsequently , the number of units sold increased such that the total revenue remained unchanged . what was the ratio of the percent increase in the units sold to the percent decrease of the original price for this product ? | for the total revenue to remain the same when the price is 3 / 10 , the number of products sold must 3 / 10 times . therefore increase in the number of products sold is 100 % = > the required ratio = 100 % / 30 % = 3.33 answer : e | a = 100 / 30
|
a ) 3.5 , b ) 4.5 , c ) 5.5 , d ) 6.5 , e ) 7.5 | b | divide(110, divide(add(80, 8), const_3_6)) | a train 110 m long is running with a speed of 80 km / h . in how many seconds will the train pass a man who is running at 8 km / h in the direction opposite to that in which the train is going ? | "the speed of the train relative to the man = 80 + 8 = 88 km / h . 88000 m / h * 1 h / 3600 s = ( 880 / 36 ) m / s ( 110 m ) / ( 880 / 36 m / s ) = ( 110 * 36 ) / 880 = 36 / 8 = 4.5 seconds the answer is b ." | a = 80 + 8
b = a / const_3_6
c = 110 / b
|
a ) 7500 , b ) 7800 , c ) 8100 , d ) 8400 , e ) 8700 | b | divide(5265, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100)))) | after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 5265 . what was the number of original inhabitants ? | "let the total number of original inhabitants be x . ( 75 / 100 ) * ( 90 / 100 ) * x = 5265 ( 27 / 40 ) * x = 5265 x = 5265 * 40 / 27 = 7800 the answer is b ." | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 - c
e = 25 / 100
f = d * e
g = b - f
h = 5265 / g
|
a ) 700 kg , b ) 500 kg , c ) 600 kg , d ) 800 kg , e ) 900 kg | c | multiply(multiply(add(const_2, const_3), const_2), multiply(subtract(add(divide(14, const_100), const_1), add(const_1, divide(8, const_100))), 1000)) | a merchant has 1000 kg of sugar part of which he sells at 8 % profit and the rest at 18 % profit . he gains 14 % on the whole . the quantity sold at 18 % profit is ? | "by the rule of alligation : profit of first part profit of second part so , ratio of 1 st and 2 nd parts = 4 : 6 = 2 : 3 . quantity of 2 nd kind = ( 3 / 5 x 1000 ) kg = 600 kg answer : c" | a = 2 + 3
b = a * 2
c = 14 / 100
d = c + 1
e = 8 / 100
f = 1 + e
g = d - f
h = g * 1000
i = b * h
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | divide(add(subtract(8, const_0.5), add(8, 4)), const_2) | a rower whose speed is 8 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 4 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ? | "time upstream = d / 4 time downstream = d / 12 total time = d / 4 + d / 12 = d / 3 average speed = 2 d / ( d / 3 ) = 6 km / hr the answer is b ." | a = 8 - 0
b = 8 + 4
c = a + b
d = c / 2
|
a ) 20 , b ) 15 , c ) 40 , d ) 60 , e ) 25 | d | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(25, 40))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | how many pounds of salt at 50 cents / lb must be mixed with 40 lbs of salt that costs 25 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 25 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt 1 ) = ( 50 - 40 ) / ( 40 - 25 ) = 2 / 3 we know that weight of salt 1 is 40 lbs . weight of salt 2 must be 60 lbs . answer ( d )" | a = 48 * 40
b = 100 + 20
c = b / 100
d = 25 * 40
e = c * d
f = a - e
g = 100 + 20
h = g / 100
i = 50 * h
j = i - 48
k = f / j
|
a ) 14 % , b ) 25 % , c ) 28 % , d ) 34 % , e ) 50 % | c | multiply(divide(add(multiply(divide(10, const_100), 40), 10), add(40, 10)), const_100) | if 10 gallons of grape juice are added to 40 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ? | "mixture contains 4 gallons of grape juice out of 40 gallons after addition : 14 gallons of grape juice out of 50 gallons = = > 28 % answer c" | a = 10 / 100
b = a * 40
c = b + 10
d = 40 + 10
e = c / d
f = e * 100
|
a ) 23 days , b ) 36 days , c ) 22 days , d ) 29 days , e ) 20 days | b | divide(multiply(30, 18), 15) | 18 men can complete a piece of work in 30 days . in how many days can 15 men complete that piece of work ? | "b 36 days 18 * 30 = 15 * x = > x = 36 days" | a = 30 * 18
b = a / 15
|
a ) rs . 156.66 , b ) rs . 150.23 , c ) rs . 123.75 , d ) rs . 135.06 , e ) none of these | c | subtract(multiply(add(divide(add(4, divide(1, 4)), const_100), 1), divide(4500, add(divide(add(divide(1, 2), 1), const_100), 1))), 4500) | if the sales tax be reduced from 4 ( 1 / 4 ) % to 1 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 4500 ? | explanation : required difference = ( 4 ( 1 / 4 ) of rs . 4500 ) - ( 1 ( 1 / 2 ) of rs . 4500 ) = ( 17 / 4 – 3 / 2 ) % of rs . 4500 = ( 11 / 4 ) x ( 1 / 100 ) x 4500 = rs . 123.75 answer c | a = 1 / 4
b = 4 + a
c = b / 100
d = c + 1
e = 1 / 2
f = e + 1
g = f / 100
h = g + 1
i = 4500 / h
j = d * i
k = j - 4500
|
a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) 9 : 14 | b | sqrt(divide(16, 9)) | two , trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively . the ratio of their speeds is : | let the slower train have speed of 1 unit and faster train a speed of r units of speed . let t = time to when the trains meet . rt 1 t howrah | - - - - - - - - - - - - - - - - - - - - - - - | - - - - - - - - - - | patna meet they meet rt units from howrah and 1 t units from patna for the second part of the journey the fast train at speed r has 1 t unit of distance and 9 hours to do it so 1 t / r = 9 slow train has rt units of distance at speed of 1 unit and 16 hours to do it . rt / 1 = 16 so rt / 1 16 - - - - - = - - - - giving r ^ 2 = 16 / 9 1 t / r 9 r = 4 / 3 answer : b | a = 16 / 9
b = math.sqrt(a)
|
a ) 1 : 9 , b ) 1 : 6 , c ) 1 : 3 , d ) 1 : 1 , e ) 1 ratio 3 | e | divide(subtract(multiply(divide(8, add(8, 27)), 70), multiply(divide(1, add(1, 4)), multiply(divide(3, add(4, 3)), 70))), subtract(multiply(divide(4, add(4, 3)), 70), subtract(multiply(divide(8, add(8, 27)), 70), multiply(divide(1, add(1, 4)), multiply(divide(3, add(4, 3)), 70))))) | in a colony of 70 resident s , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of education to uneducated persons is 8 : 27 , then find the ratio of the number of educated and uneducated men in the colony ? | number of men in the colony = 4 / 7 ( 70 ) = 40 number of women in the colony = 3 / 7 ( 70 ) = 30 number of educated women in the colony = 1 / 5 ( 30 ) = 6 number of uneducated women in the colony = 4 / 5 ( 30 ) = 24 number of educated persons in the colony = 8 / 35 ( 70 ) = 16 as 6 females are educated , remaining 10 educated persons must be men . number of uneducated men in the colony = 40 - 10 = 30 number of educated men and uneducated men are in the ratio 10 : 30 = > 1 : 3 answer : e | a = 8 + 27
b = 8 / a
c = b * 70
d = 1 + 4
e = 1 / d
f = 4 + 3
g = 3 / f
h = g * 70
i = e * h
j = c - i
k = 4 + 3
l = 4 / k
m = l * 70
n = 8 + 27
o = 8 / n
p = o * 70
q = 1 + 4
r = 1 / q
s = 4 + 3
t = 3 / s
u = t * 70
v = r * u
w = p - v
x = m - w
y = j / x
|
a ) 155 , b ) 195 , c ) 200 , d ) 205 , e ) 210 | a | divide(divide(multiply(add(10, 300), add(divide(subtract(300, 10), 10), const_1)), const_2), add(divide(subtract(300, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 300 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 300 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 30 ( 10 ) = 300 so we know that there are 30 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 300 = 310 20 + 290 = 310 30 + 280 = 310 etc . since there are 30 total terms , this pattern will create 15 ' pairs ' of 310 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 15 ) ( 310 ) / ( 30 ) = 155 answer : a" | a = 10 + 300
b = 300 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 300 - 10
h = g / 10
i = h + 1
j = f / i
|
a ) 1.12 , b ) 1.25 , c ) 1.20 , d ) 1.3 , e ) none | c | divide(multiply(0.75, 8), 5) | if 0.75 : x : : 5 : 8 , then x is equal to | "sol . ( x × 5 ) = ( 0.75 × 8 ) ⇒ x = 6 / 5 = 1.20 . answer c" | a = 0 * 75
b = a / 5
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a ) 126 , b ) 136 , c ) 146 , d ) 156 , e ) 169 | e | divide(multiply(13, 312), 24) | the reciprocal of the hcf and lcm of two are 1 / 13 and 1 / 312 . if one of the number is 24 then other no . is | reciprocal of the hcf and lcm of two are 1 / 13 and 1 / 312 so , hcf = 13 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 13 * 312 / 24 = 169 answer : e | a = 13 * 312
b = a / 24
|
a ) 3 : 2 , b ) 4 : 3 , c ) 5 : 1 , d ) 6 : 5 , e ) 7 : 1 | e | add(4, 3) | the ratio of two no . addition and subtraction be 4 : 3 . the what is the ratio of numbers ? | ( x + y ) / ( x - y ) = 4 / 3 dividing numerator and denominator by y ( x / y ) + 1 / ( x / y ) - 1 = 4 / 3 let x / y be z z + 1 / z - 1 = 4 / 3 3 z + 3 = 4 z - 4 z = 7 x / y = 7 answer e | a = 4 + 3
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a ) 30 , b ) 90 , c ) 120 , d ) 172 , e ) 188 | b | subtract(300, subtract(add(multiply(divide(300, const_100), 56), multiply(divide(300, const_100), 44)), multiply(divide(300, const_100), 30))) | among 300 students , 56 % study sociology , 44 % study mathematics and 40 % study biology . if 30 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | "i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 90 using your analysis : we see that the total number of students who study either maths or sociology = 132 + 168 - 90 = 210 so , in the image we know that the number of students in the zone with the black boundary = 210 let ' s assume the number of students who studyonlybiology to beb ( this is the number that we have to maximize ) and , let ' s assume the number of students who study none of the three subjects , that is the number of students in the white space = w since the total number of students = 300 , we can write : 210 + b + w = 300 or , b + w = 300 - 210 = 90 that is , b = 90 - w so , the maximum value ofbwill happen forw = 0 this is how we get , the maximum value ofb = 90 b" | a = 300 / 100
b = a * 56
c = 300 / 100
d = c * 44
e = b + d
f = 300 / 100
g = f * 30
h = e - g
i = 300 - h
|
a ) 0.004 , b ) 0.16666 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(6, 0.03), divide(6, 0.37))) | 6 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 6 / 36 = . 16666 answer ( b )" | a = 6 / 0
b = 6 / 0
c = a + b
d = 1/(c)
|
a ) 2800 , b ) 2100 , c ) 3600 , d ) 4200 , e ) 5200 | a | multiply(divide(add(multiply(multiply(3, 3), const_1000), const_100), 7), 2) | a marketing survey of anytown found that the ratio of trucks to sedans to motorcycles was 3 : 7 : 2 , respectively . given that there are 9,800 sedans in anytown , how many motorcycles are there ? | "let the total number of trucks = 3 x total number of sedans = 7 x total number of motorcycles = 2 x total number of sedans = 9800 = > 7 x = 9800 = > x = 1400 total number of motorcycles = 2 x = 2 * 1400 = 2800 answer a" | a = 3 * 3
b = a * 1000
c = b + 100
d = c / 7
e = d * 2
|
a ) 14 , b ) 44 , c ) 48 , d ) 50 , e ) 56 | a | add(add(multiply(5, divide(2, subtract(multiply(divide(2.00001, 5.00001), 5), 1))), divide(2, subtract(multiply(divide(2.00001, 5.00001), 5), 1))), 2) | in a can , there is a mixture of milk and water in the ratio 1 : 5 . if it is filled with an additional 2 litres of milk the can would be full and ratio of milk and water would become 2.00001 : 5.00001 . find the capacity of the can ? | let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 1 / 6 ( t - 2 ) after adding milk , quantity of milk in the mixture = 2 / 7 t . 2 t / 7 - 2 = 1 / 6 ( t - 2 ) 5 t = 84 - 14 = > t = 14 . answer : a | a = 2 / 1
b = a * 5
c = b - 1
d = 2 / c
e = 5 * d
f = 2 / 1
g = f * 5
h = g - 1
i = 2 / h
j = e + i
k = j + 2
|
a ) 252 kms , b ) 152 kms , c ) 552 kms , d ) 792 kms , e ) 152 kms | d | multiply(add(multiply(2, 55), multiply(subtract(12, const_1), 2)), divide(12, 2)) | the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 55 kms . what was the total distance travelled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 55 + 57 + 59 + . . . . . upto 12 terms ) this is an a . p with first term , a = 55 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 55 + { 12 - 1 ) x 2 ] = 6 ( 132 ) = 792 kms . answer : d" | a = 2 * 55
b = 12 - 1
c = b * 2
d = a + c
e = 12 / 2
f = d * e
|
a ) 16.2 , b ) 7 , c ) 5 , d ) 500 , e ) none | b | divide(multiply(divide(35, const_100), 900), 45) | 45 x ? = 35 % of 900 | "answer let 45 x a = ( 35 x 900 ) / 100 ∴ a = ( 35 x 9 ) / 45 = 7 correct option : b" | a = 35 / 100
b = a * 900
c = b / 45
|
['a ) 2 : 1', 'b ) 4 : 7', 'c ) 8 : 77', 'd ) 4 : 169', 'e ) none'] | d | power(divide(2, 13), const_2) | the ratio of radius of a circle and the side of a square is 2 : 13 . find the ratio of their areas : | radius / side = 2 / 13 â ‡ ’ area of circle / area of square = 4 / 169 answer : d | a = 2 / 13
b = a ** 2
|
a ) 60 , b ) 62 , c ) 64 , d ) 65 , e ) 66 | d | subtract(multiply(60, const_10), add(add(add(add(add(add(add(add(54, 55), 57), 58), 59), 62), 62), 63), 65)) | if the average of 54 , 55 , 57 , 58 , 59 , 62 , 62 , 63 , 65 and x is 60 , what is the value of x ? | "sum of the deviations of the numbers in the set from the mean is always zero 54 , 55 , 57 , 58 , 59 , 62 , 62 , 63 , 65 mean is 60 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 60 + 5 = 65 d" | a = 60 * 10
b = 54 + 55
c = b + 57
d = c + 58
e = d + 59
f = e + 62
g = f + 62
h = g + 63
i = h + 65
j = a - i
|
a ) 18 , b ) 21 , c ) 38 , d ) 40 , e ) 57 | d | add(add(lcm(6, multiply(divide(6, 2), 3)), 4), lcm(6, multiply(divide(6, 2), 3))) | a ranch has both horses and ponies . exactly 5 / 6 of the ponies have horseshoes , and exactly 2 / 3 of the ponies with horseshoes are from iceland . if there are 4 more horses than ponies , what is the minimum possible combined number of horses and ponies on the ranch ? | "5 / 6 * p have horseshoes , so p is a multiple of 6 . 2 / 3 * 5 / 6 * p = 5 / 9 * p are icelandic ponies with horseshoes , so p is a multiple of 9 . the minimum value of p is 18 . then h = p + 4 = 22 . the minimum number of horses and ponies is 40 . the answer is d ." | a = 6 / 2
b = a * 3
c = math.lcm(6, b)
d = c + 4
e = 6 / 2
f = e * 3
g = math.lcm(6, f)
h = d + g
|
a ) 72 , b ) 80 , c ) 88 , d ) 68 , e ) 78 | b | subtract(negate(34), multiply(subtract(22, 26), divide(subtract(22, 26), subtract(20, 22)))) | 20 , 22 , 26 , 34 , 41 , 46 , 56 , 67 , __ ? | "next no . = previous + ( sum of digits ) 20 + ( 2 + 0 ) = 22 22 + ( 2 + 2 ) = 26 26 + ( 2 + 6 ) = 34 - - - - - - - - 56 + ( 5 + 6 ) = 67 67 + ( 6 + 7 ) = 80 answer : b" | a = negate - (
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a ) 321 , b ) 276 , c ) 342 , d ) 265 , e ) 198 | e | subtract(subtract(278, divide(multiply(278, 18), const_100)), divide(multiply(subtract(278, divide(multiply(278, 18), const_100)), 19), const_100)) | the sale price sarees listed for rs . 278 after successive discount is 18 % and 19 % is ? | "explanation : 278 * ( 88 / 100 ) * ( 81 / 100 ) = 198 answer : e" | a = 278 * 18
b = a / 100
c = 278 - b
d = 278 * 18
e = d / 100
f = 278 - e
g = f * 19
h = g / 100
i = c - h
|
a ) 9 , b ) 19 , c ) 23 , d ) 27 , e ) 29 | e | subtract(36, subtract(add(36, 30), 59)) | in a group of 90 students , 36 are taking history , and 30 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ? | "number of students taking history = h = 36 number of students taking statistics = s = 30 total number of students = t = 90 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 90 - 59 = 31 letxbe the number of students taking both history and statistics . then t = h + s + n - x or 90 = 36 + 30 + 31 - x or x = 7 now , number of students taking only history will be h - x or 36 - 7 = 29 answer : - e" | a = 36 + 30
b = a - 59
c = 36 - b
|
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(427398, multiply(floor(divide(427398, 14)), 14)) | what least number must be subtracted from 427398 so that remaining no . is divisible by 14 | "explanation : on dividing 427398 by 14 we get the remainder 6 , so 6 should be subtracted answer : option c" | a = 427398 / 14
b = math.floor(a)
c = b * 14
d = 427398 - c
|
a ) 25 , b ) 30 , c ) 40 , d ) 35 , e ) 45 | d | subtract(multiply(30, const_2), multiply(20, const_2)) | if the average ( arithmetic mean ) of a and b is 20 , and c – a = 30 , what is the average of b and c ? | "a + b / 2 = 20 = > a + b = 40 a = c - 30 . . . sub this value c - 30 + b = 40 = > c + b = 70 = > c + b / 2 = 35 answer : d" | a = 30 * 2
b = 20 * 2
c = a - b
|
a ) 87 , b ) 42 , c ) 78 , d ) 76 , e ) 26 | b | subtract(subtract(multiply(divide(subtract(multiply(2450, 6), multiply(1250, 6)), subtract(1250, 950)), const_2), const_4), const_2) | the average monthly salary of laborers and supervisors in a factory is rs . 1250 per month ; where as the average monthly salary of 6 supervisors is rs . 2450 . if the average monthly salary of the laborers is rs . 950 find the number of laborers ? | 5 x 6 x 2 x 50 25 100 250 x + 150 x + 200 x = 4200 600 x = 4200 x = 7 = > 6 x = 42 answer : b | a = 2450 * 6
b = 1250 * 6
c = a - b
d = 1250 - 950
e = c / d
f = e * 2
g = f - 4
h = g - 2
|
a ) 20 % , b ) 24 % , c ) 30 % , d ) 35 % , e ) 79 % | d | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 40), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 40), const_100))), const_100) | the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 40 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 40 % then output equals 154 to restore new holidy demand output to original 100 . final - initial / final * 100 = 54 / 154 * 100 = 35 % approxiamately . option d is correct ." | a = 100 + 10
b = a / 100
c = 100 + 40
d = c / 100
e = b * d
f = e - 1
g = 100 + 10
h = g / 100
i = 100 + 40
j = i / 100
k = h * j
l = f / k
m = l * 100
|
a ) 1176 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(424, multiply(const_4, const_100)), add(multiply(subtract(424, multiply(const_4, const_100)), 2), const_1)) | balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 424 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did . how many balls made up the triangle ? | as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 424 ) ( x - 8 ) ^ 2 = t + 424 - - - - - ( b ) now the hardest part of the question will be to solve these 2 equations and this looks like time consuming but the easy way will be plug and play . also , we ' ve to find a value of t ( from 5 optiosn given below ) which can make a square of a a number . one we know this , it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - 1176 + 424 = 1600 = 40 ^ 2 = ( x - 8 ) ^ 2 hence , x = 48 . cross check by putting in eq ( a ) = x ( x + 1 ) / 2 = t = > 48 * 49 / 2 = 1176 hence , answer is a . | a = 4 * 100
b = 424 - a
c = 4 * 100
d = 424 - c
e = d * 2
f = e + 1
g = b * f
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | divide(multiply(subtract(4.00, 2.50), 100), 2.50) | a wholesaler wishes to sell 100 pounds of mixed nuts at $ 2.50 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ? | "cashews / peanuts = ( mean price - peanuts price ) / ( cashew price - mean price ) = ( 2.5 - 1.5 ) / ( 4 - 2.5 ) = 2 / 3 cashew = ( 2 / 5 ) * 100 = 40 kg answer is a ." | a = 4 - 0
b = a * 100
c = b / 2
|
a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | b | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 44 hours . if harry and annie were paid the same amount last week , how many hours did harry work last week ? | "annie earned 40 x + 4 ( 2 x ) = 48 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 48 x ( 1.5 x ) ( h ) = 63 x h = 42 hours the answer is b ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
|
a ) 6 min , b ) 7 min , c ) 8 min , d ) 10 min , e ) 17 min | c | add(multiply(9, subtract(const_1, multiply(add(inverse(9), inverse(18)), 24))), 24) | two pipes a and b can fill a cistern in 9 and 18 minutes respectively , and a third pipe c can empty it in 24 minutes . how long will it take to fill the cistern if all the three are opened at the same time ? | "1 / 9 + 1 / 18 - 1 / 24 = 1 / 8 8 / 1 = 8 answer : c" | a = 1/(9)
b = 1/(18)
c = a + b
d = c * 24
e = 1 - d
f = 9 * e
g = f + 24
|
a ) $ 2.10 , b ) $ 36.75 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80 | b | add(multiply(divide(5, const_100), 35), 35) | laura took out a charge account at the general store and agreed to pay 5 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 5 % interest = ( 5 / 100 ) * 35 = 1.75 $ total amount that laura owes a year later = 35 + 1.75 = 36.75 $ answer b" | a = 5 / 100
b = a * 35
c = b + 35
|
a ) 8 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | c | multiply(divide(subtract(multiply(divide(3, 20), const_100), multiply(divide(9, 108), const_100)), multiply(divide(3, 20), const_100)), const_100) | a pharmaceutical company received $ 3 million in royalties on the first $ 20 million in sales of the generic equivalent of one of its products and then $ 9 million in royalties on the next $ 108 million in sales . by approximately what percent did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ? | "first $ 20 million : royalties / sales ratio = 3 / 20 = 36 / 240 next $ 108 million : royalties / sales ratio = 9 / 108 = 1 / 12 = 20 / 240 answer : c" | a = 3 / 20
b = a * 100
c = 9 / 108
d = c * 100
e = b - d
f = 3 / 20
g = f * 100
h = e / g
i = h * 100
|
a ) 0 , b ) 5 , c ) 195 , d ) 875 , e ) 965 | c | divide(multiply(194, 195), const_4) | what is the sum of the integers from - 194 to 195 inclusive ? | "sum / n = average . sum = ( average ) ( n ) average = a + b / 2 = 194 + 195 / 2 = 0.5 number of items ( n ) = b - a + 1 = 195 - ( - 194 ) + 1 = 195 + 195 = 390 . sum = average * n = 0.5 * 390 = 195 . answer is c" | a = 194 * 195
b = a / 4
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a ) 35.4 , b ) 42 , c ) 44.4 , d ) 57 , e ) 67 | c | add(30, const_1) | the average of first seven prime numbers which are between 30 and 70 is | "explanation : first seven prime numbers which are between 30 and 70 = 31 , 37 , 41 , 43 , 47 , 53 , 59 average = ( 31 + 37 + 41 + 43 + 47 + 53 + 59 ) / 7 = 44.4 answer : c" | a = 30 + 1
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a ) 25 , b ) 31 , c ) 27 , d ) 29 , e ) 42 | e | add(subtract(90, multiply(17, 3)), 3) | a batsman makes a score of 90 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 90 = total runs scored in 17 innings = > 16 ( x - 3 ) + 90 = 17 x = > 16 x - 48 + 90 = 17 x = > x = 42 answer is e ." | a = 17 * 3
b = 90 - a
c = b + 3
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a ) y / z , b ) xy , c ) y ² / z ² , d ) 36 y ² / z ² , e ) 15 y ² / z ² | d | divide(add(3, 4), subtract(3, 4)) | if x / y = 3 / z , then 4 x ^ 2 = | "this question is most easily solved by isolating y in the equation and substituting into the expression 4 x ² : x / y = 3 / z x = 3 y / z if we substitute 3 y / z into the expression for x , we get : 4 ( 3 y / z ) ² = 4 ( 9 y ² / z ² ) = 36 y ² / z ² . the correct answer is choice ( d ) ." | a = 3 + 4
b = 3 - 4
c = a / b
|
a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 24 years | c | divide(subtract(add(25, add(25, 3)), multiply(3, 3)), const_2) | the captain of a cricket team of 11 members is 25 years old and the wicket keeper is 3 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 € “ ( 25 + 28 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 44 2 x = 44 x = 22 . so , average age of the team is 22 years . answer c" | a = 25 + 3
b = 25 + a
c = 3 * 3
d = b - c
e = d / 2
|
a ) $ 0.15 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.6 | e | divide(multiply(add(const_1, const_1), 0.90), add(const_1, const_2)) | on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.90 per glass on the first day , what was the price per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.9 = 3 * x - - > x = $ 0.6 . answer : e ." | a = 1 + 1
b = a * 0
c = 1 + 2
d = b / c
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a ) 30 , b ) 25 , c ) 20.67 , d ) 12.33 , e ) 19.48 | c | multiply(10, divide(31, subtract(25, 10))) | if a person walks at 25 km / hr instead of 10 km / hr , he would have walked 31 km more . the actual distance traveled by him is ? | "let the actual distance traveled is ' x ' km then by given conditions , we have x / 10 = ( x + 31 ) / 25 25 x = 10 x + 310 x = 20.67 km ans - c" | a = 25 - 10
b = 31 / a
c = 10 * b
|
a ) 14 minutes , b ) 4 minutes , c ) 24 minutes , d ) 28 minutes , e ) 23 minutes | b | subtract(8, 1) | the speed of a boat in still water is 8 kmph . if it can travel 1 km upstream in 1 hr , what time it would take to travel the same distance downstream ? | "speed of the boat in still water = 8 km / hr speed upstream = 1 ⁄ 1 = 1 km / hr speed of the stream = 8 - 1 = 7 km / hr speed downstream = ( 8 + 7 ) = 15 km / hr answer is b time taken to travel 1 km downstream = 1 / 15 hr = ( 1 × 60 ) / 15 = 4 minutes" | a = 8 - 1
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a ) – 108 , b ) 19 , c ) 10 , d ) 16 , e ) 18 | b | divide(add(add(20, multiply(20, 12)), 12), multiply(2, const_4)) | # p is defined as 2 p - 20 for any number p . what is p , if # ( # ( # p ) ) = 12 ? | # p = 2 p - 20 - - - > # ( # p ) = 2 ( 2 p - 20 ) - 20 = 4 p - 60 and thus # ( 4 p - 60 ) = 2 ( 4 p - 60 ) - 20 = 8 p - 140 = 12 - - - > 8 p = 152 - - - > p = 19 , b is the correct answer . | a = 20 * 12
b = 20 + a
c = b + 12
d = 2 * 4
e = c / d
|
a ) 100 , b ) 1000 , c ) 1080 , d ) 1020 , e ) 1200 | c | divide(multiply(90, 600), 50) | if 90 percent of 600 is 50 percent of x , then x = ? | "0.9 * 600 = 0.5 * x x = 9 / 5 * 600 = 1080" | a = 90 * 600
b = a / 50
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a ) 800 , b ) 700 , c ) 500 , d ) 900 , e ) 600 | d | divide(1080, add(const_1, divide(20, const_100))) | a number increased by 20 % gives 1080 . the number is ? | "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 20 % = 120 % 120 % - - - - - - - > 1080 ( 120 ã — 9 = 1080 ) 100 % - - - - - - - > 900 ( 100 ã — 9 = 900 ) option ' d '" | a = 20 / 100
b = 1 + a
c = 1080 / b
|
a ) 87 kmph , b ) 97 kmph , c ) 72 kmph , d ) 90 kmph , e ) 19 kmph | d | divide(divide(500, const_1000), divide(20, const_3600)) | a train 500 m long can cross an electric pole in 20 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 500 / 20 s = 25 m / sec speed = 25 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 90 kmph answer : d" | a = 500 / 1000
b = 20 / 3600
c = a / b
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a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | d | add(add(3, const_10), const_10) | if 5 < x < 12 and y = x + 3 , what is the greatest possible integer value of x + y ? | "x + y = x + x + 3 = 2 x + 3 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 11.5 then x + y = 11.5 + 14.5 = 26 . the answer is d ." | a = 3 + 10
b = a + 10
|
a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | c | multiply(subtract(divide(power(const_3.0, const_2), 357), floor(divide(power(42, const_2), 357))), 357) | on dividing a number by 357 , we get 42 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 42 = ( 17 * 21 * y ) + ( 17 * 2 ) + 8 = 17 * ( 21 y + 2 ) + 8 . required number = 8 . answer is c" | a = 3 ** 0
b = a / 357
c = 42 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
|
a ) 16 , b ) 10 , c ) 4 , d ) 8 , e ) 6 | d | subtract(20, multiply(multiply(16, 4), 3)) | evaluate : 20 - 16 ÷ 4 × 3 = | "according to order of operations , 16 ÷ 4 × 3 ( division and multiplication ) is done first from left to right 16 ÷ 4 × 3 = 4 × 3 = 12 hence 20 - 16 ÷ 4 × 3 = 20 - 12 = 8 correct answer d ) 8" | a = 16 * 4
b = a * 3
c = 20 - b
|
a ) 287 , b ) 298 , c ) 396 , d ) 726 , e ) 267 | c | subtract(multiply(50, multiply(102, const_0_2778)), 1020) | a train 1020 m long running at 102 kmph crosses a platform in 50 sec . what is the length of the platform ? | "d = 102 * 5 / 18 = 50 = 1416 â € “ 1020 = 396 answer : c" | a = 102 * const_0_2778
b = 50 * a
c = b - 1020
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | e | multiply(divide(15, 3), const_3) | a seller of used cars has 15 cars to sell and each of his clients selected 3 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ? | "ifno caris selected more than once then the number of clients = 15 / 3 = 5 but since every car is being selected three times so no . of clients must be thrice as well = 5 * 3 = 15 answer : option e" | a = 15 / 3
b = a * 3
|
a ) 24,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 24 , 000,000 , e ) 264 , 000,000 | e | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(6, 100)), multiply(11, 100)) | if a rectangular room measures 11 meters by 6 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "e . 264 , 000,000 11 * 100 * 6 * 100 * 4 * 100 = 264 , 000,000" | a = 4 * 100
b = 1 / 10
c = a * b
d = 6 * 100
e = c * d
f = 11 * 100
g = e * f
|
a ) 134.8 feet , b ) 136.2 feet , c ) 132.7 feet , d ) 132.2 feet , e ) 115.3 feet | c | multiply(16, multiply(divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)), divide(subtract(sqrt(add(multiply(1100, 1100), multiply(multiply(1100, 3), multiply(const_4, 16)))), 1100), multiply(16, const_2)))) | a rock is dropped into a well and the distance traveled is 16 t 2 feet , where t is the time . if the water splash is heard 3 seconds after the rock was dropped , and that the speed of sound is 1100 ft / sec , approximate the height of the well | let t 1 be the time it takes the rock to reach the bottom of the well . if h is the height of the well , we can write h = 16 t 1 2 let t 2 be the time it takes sound wave to reach the top of the well . we can write h = 1100 t 2 the relationship between t 1 and t 2 is t 1 + t 2 = 3 eliminate h and combine the equations h = 16 t 1 2 and h = 1100 t 2 to obtain 16 t 1 2 = 1100 t 2 we now substitute t 2 by 3 - t 1 in the above equation 16 t 1 2 = 1100 ( 3 - t 1 ) the above is a quadratic equation that may be written as 16 t 1 2 + 1100 t 1 - 3300 = 0 the above equations has two solutions and only one of them is positive and is given by t 1 = 2.88 seconds ( 2 decimal places ) we now calculate the height h of the well h = 16 t 1 2 = 16 ( 2.88 ) 2 = 132.7 feet answer c | a = 1100 * 1100
b = 1100 * 3
c = 4 * 16
d = b * c
e = a + d
f = math.sqrt(e)
g = f - 1100
h = 16 * 2
i = g / h
j = 1100 * 1100
k = 1100 * 3
l = 4 * 16
m = k * l
n = j + m
o = math.sqrt(n)
p = o - 1100
q = 16 * 2
r = p / q
s = i * r
t = 16 * s
|
a ) 1 kmph , b ) 3 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | d | divide(subtract(18, 4), const_2) | a man can row his boat with the stream at 18 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 18 us = 4 s = ? s = ( 18 - 4 ) / 2 = 7 kmph answer : d" | a = 18 - 4
b = a / 2
|
a ) 0.2 , b ) 0.02 , c ) 0.002 , d ) 0.0002 , e ) 2 | b | divide(2, const_100) | how is 2 % expressed as a decimal fraction ? | 2 / 100 = 0.02 answer : b | a = 2 / 100
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers v are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a = 1 + 1
|
a ) 0.008416667 , b ) 0.110773333 , c ) 0.12526234 , d ) 0.01072333 , e ) 0.12725002 | a | divide(25.25, 3000) | 25.25 / 3000 is equal to : | "25.25 / 3000 = 2525 / 300000 = 0.008416667 answer : a" | a = 25 / 25
|
a ) 2 , 9 , b ) 4 , 9 , c ) 8 , 9 , d ) 19 , 9 , e ) 7 , 9 | d | divide(divide(add(28, 10), const_2), const_2) | a man can row downstream at 28 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 28 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 38 = > x = 19 , y = 9 . answer : d" | a = 28 + 10
b = a / 2
c = b / 2
|
a ) 13 sec . , b ) 10.4 sec . , c ) 12 sec . , d ) 17 sec . , e ) 19 sec . | b | divide(add(120, 120), add(speed(120, 8), speed(120, 15))) | two bullet trains of equal lengths take 8 seconds and 15 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | speed of the first bullet train = 120 / 8 m / sec = 15 m / sec . speed of the second bullet train = 120 / 15 m / sec = 8 m / sec . relative speed = ( 15 + 8 ) = 23 m / sec . required time = ( 120 + 120 ) / 23 sec = 10.4 sec . b | a = 120 + 120
b = speed + (
c = a / b
|
a ) 15 , b ) 30 , c ) 31 , d ) 33 , e ) 54 | e | subtract(divide(subtract(subtract(170, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(170, const_2), multiply(3, const_4)), 3), 3), 3), const_2)) | how many even number in the range between 10 to 170 inclusive are not divisible by 3 | "we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 170 using ap formula , we can say 170 = 10 + ( n - 1 ) * 2 or n = 81 . for 6 , 12,18 , . . . 168 using ap formula , we can say 168 = 12 + ( n - 1 ) * 6 or n = 27 . hence , only divisible by 2 but not 3 = 81 - 27 = 54 . hence , answer e" | a = 170 - 10
b = a - 2
c = b / 2
d = 170 - 2
e = 3 * 4
f = d - e
g = f - 3
h = g - 3
i = h / 3
j = i / 2
k = c - j
|
a ) 80 feet , b ) 70 feet , c ) 60 feet , d ) 50 feet , e ) 20 feet | a | add(multiply(divide(600, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 600 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 600 = > breadth = 30 feet area to be fenced = 2 b + l = 2 * 30 + 20 = 80 feet answer : option a" | a = 600 / 20
b = a * 2
c = b + 20
|
a ) 96 , b ) 75 , c ) 50 , d ) 25 , e ) 12 | c | divide(8, subtract(96.16, floor(96.16))) | when positive integer x is divided by positive integer y , the remainder is 8 . if x / y = 96.16 , what is the value of y ? | "by the definition of a remainder , the remainder here is equal to 8 / y . the remainder in decimal form is given as . 16 therefore , 8 / y = . 16 solve for y and get 50 . c" | a = math.floor(96, 16)
b = 96 - 16
c = 8 / b
|
a ) 151 , b ) 140.33 , c ) 152 , d ) 148 , e ) none of the above | b | divide(add(multiply(30, 140), subtract(145, 135)), 30) | the mean of 30 values was 140 . it was detected on rechecking that one value 145 was wrongly copied as 135 for the computation of the mean . find the correct mean . | "corrected mean = 140 × 30 − 135 + 145 / 30 = 4200 − 135 + 145 / 30 = 4210 / 30 = 140.33 answer b" | a = 30 * 140
b = 145 - 135
c = a + b
d = c / 30
|
a ) 7.2 , b ) 7.85 , c ) 6.85 , d ) 5.85 , e ) 6.15 | a | divide(add(110, 180), multiply(add(80, 65), const_0_2778)) | two trains 110 meters and 180 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 110 + 180 ) / ( 80 + 65 ) * 18 / 5 t = 7.20 answer : a" | a = 110 + 180
b = 80 + 65
c = b * const_0_2778
d = a / c
|
['a ) 28', 'b ) 29', 'c ) 30', 'd ) 31', 'e ) 32'] | c | sqrt(900) | you buy a piece of land with an area of √ 900 , how long is one side of the land plot ? | try filling the numbers into the answer y x y = find the closest to 900 . answer c | a = math.sqrt(900)
|
a ) 250 , b ) 350 , c ) 450 , d ) 550 , e ) 525 | e | multiply(21, 25) | a large box contains 21 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 21 * 25 = 525 correct answer e" | a = 21 * 25
|
a ) 62 % , b ) 63 % , c ) 64 % , d ) 65 % , e ) 66 % | c | multiply(divide(multiply(subtract(20, multiply(2, const_2)), subtract(30, multiply(3, const_2))), multiply(20, 30)), const_100) | a typist uses a sheet measuring 20 cm by 30 cm lenghtwise . if a margin of 2 cm is left on each side and a 3 cm margin on the top and bottom , then what is the percentage of page used by the typist ? | area of the sheet = ( 20 * 30 ) cm 2 = 600 cm 2 area used for typing = ( [ 20 - 4 ] * [ 30 - 6 ] ) cm 2 = 384 cm 2 therefore required percentage = ( 384 / 600 ) * 100 = 64 % answer c | a = 2 * 2
b = 20 - a
c = 3 * 2
d = 30 - c
e = b * d
f = 20 * 30
g = e / f
h = g * 100
|
a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 25.83 % | e | multiply(divide(multiply(divide(62, const_100), 525), multiply(const_100, power(const_4, const_2))), const_100) | an association of mathematics teachers has 1,260 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 62 percent of the votes cast ? | "total number of members = 1260 number of members that cast votes = 525 since , winning candidate received 62 percent of the votes cast number of votes for winning candidate = ( 62 / 100 ) * 525 = 325.5 percent of total membership that voted for winning candidate = ( 325.5 / 1260 ) * 100 = 25.83 % answer e" | a = 62 / 100
b = a * 525
c = 4 ** 2
d = 100 * c
e = b / d
f = e * 100
|
a ) 2670 , b ) 2980 , c ) 3550 , d ) 4030 , e ) 4540 | d | add(divide(subtract(subtract(161, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 161 ? | "100 + 102 + . . . + 160 = 100 * 31 + ( 2 + 4 + . . . + 60 ) = 100 * 31 + 2 * ( 1 + 2 + . . . + 30 ) = 100 * 31 + 2 ( 30 ) ( 31 ) / 2 = 100 * 31 + 30 * 31 = 130 ( 31 ) = 4030 the answer is d ." | a = 161 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 45 minutes , b ) 55 minutes , c ) 35 minutes , d ) 25 minutes , e ) 40 minutes | c | multiply(add(const_3, const_4), 5) | a clock shows the time as 11 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 6 p . m . ? | "there are 7 hours in between 11 a . m . to 6 p . m . 7 * 5 = 35 minutes . answer : c" | a = 3 + 4
b = a * 5
|
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