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a ) a ) 35 , b ) b ) 40 , c ) c ) 90 , d ) d ) 88 , e ) e ) 37 | b | divide(add(130, 30), add(3, 1)) | in an examination , a student scores 3 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 30 questions and secures 130 marks , the no of questions he attempts correctly is : | let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 3 x β ( 30 β x ) = 130 = > 4 x = 160 = > x = 40 answer : b | a = 130 + 30
b = 3 + 1
c = a / b
|
a ) 2.0 , b ) 2.5 , c ) 3.0 , d ) 3.5 , e ) 4.0 | a | multiply(divide(add(2.3, divide(100, const_1000)), 72), const_60) | a train with a length of 100 meters , is traveling at a speed of 72 km / hr . the train enters a tunnel 2.3 km long . how many minutes does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges ? | 72 km / hr = 1.2 km / min the total distance is 2.4 km . 2.4 / 1.2 = 2 minutes the answer is a . | a = 100 / 1000
b = 2 + 3
c = b / 72
d = c * const_60
|
a ) 1888 , b ) 2766 , c ) 2999 , d ) 2000 , e ) 1712 | d | subtract(divide(subtract(multiply(2000, 40), multiply(2000, 20)), 10), 2000) | a garrison of 2000 men has provisions for 40 days . at the end of 20 days , a reinforcement arrives , and it is now found that the provisions will last only for 10 days more . what is the reinforcement ? | 2000 - - - - 40 2000 - - - - 20 x - - - - - 10 x * 10 = 2000 * 20 x = 4000 2000 - - - - - - - 2000 answer : d | a = 2000 * 40
b = 2000 * 20
c = a - b
d = c / 10
e = d - 2000
|
a ) 11 , b ) 10 , c ) 15 , d ) 16 , e ) 17 | a | divide(10, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 10)) | five bells commence tolling together and toll at intervals of 2 , 4 , 6 , 8 10 seconds respectively . in 20 minutes , how many times do they toll together ? | "l . c . m of 2 , 4,6 , 8,10 is 240 . i . e after each 2 min they will toll together . so in 20 min they will toll 10 times . as they have initially tolled once , the answer will be 10 + 1 = 11 . answer : a" | a = 2 + 3
b = a * 2
c = b * 3
d = 2 * c
e = d * 2
f = e / 10
g = 10 / f
|
a ) a ) 36 , b ) b ) 90 , c ) c ) 40 , d ) d ) 42 , e ) e ) 44 | b | subtract(600, multiply(85, 6)) | we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 600 . how many green hats did we buy ? | "let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 600 . 6 ( 85 - g ) + 7 g = 600 . 510 - 6 g + 7 g = 600 . g = 600 - 510 = 90 . the answer is b ." | a = 85 * 6
b = 600 - a
|
a ) 1160 , b ) 1190 , c ) 1220 , d ) 1250 , e ) 1280 | d | divide(750, divide(60, const_100)) | if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 750 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "let x be the minimum number of questionnaires to be mailed . 0.6 x = 750 x = 1250 the answer is d ." | a = 60 / 100
b = 750 / a
|
a ) rs . 2500 , b ) rs . 2300 , c ) rs . 2200 , d ) rs . 1400 , e ) none of these | a | multiply(multiply(49, const_12), divide(1517.25, 357)) | if the price of 357 apples is rs . 1517.25 , what will be the approximate price of 49 dozens of such apples | "let the required price be x more apples , more price ( direct proportion ) hence we can write as apples 357 : ( 49 Γ 12 ) } : : 1517.25 : x β 357 x = ( 49 Γ 12 ) Γ 1517.25 β x = ( 49 Γ 12 Γ 1517.25 ) / 357 = ( 7 Γ 12 Γ 1517.25 ) / 51 = ( 7 Γ 4 Γ 1517.25 ) / 17 = 7 Γ 4 Γ 89.25 β 2500 . answer a" | a = 49 * 12
b = 1517 / 25
c = a * b
|
a ) 127 , b ) 237 , c ) 387 , d ) 400 , e ) 481 | d | divide(multiply(36, 30), divide(multiply(9, 24), 80)) | if 9 men can reap 80 hectares in 24 days , then how many hectares can 36 men reap in 30 days ? | "explanation : let the required no of hectares be x . then men - - - hectares - - - days 9 - - - - - - - - - 80 - - - - - - - - - 24 36 - - - - - - - - - x - - - - - - - - - 30 more men , more hectares ( direct proportion ) more days , more hectares ( direct proportion ) x = 36 / 9 * 30 / 24 * 80 x = 400 answer : d" | a = 36 * 30
b = 9 * 24
c = b / 80
d = a / c
|
a ) 82 m , b ) 150 m , c ) 172 m , d ) 200 m , e ) none of these | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 144), const_2) | two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 144 seconds . the length of each train is : | "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 Γ 518 = 259 m / s in 144 secs the total distance traveled is 144 x 25 / 9 = 400 m . therefore the length of each train is = 400 / 2 = 200 m . answer d" | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 144
e = d / 2
|
a ) $ 986.25 , b ) $ 988.25 , c ) $ 990.25 , d ) $ 992.25 , e ) $ 994.25 | d | multiply(900, power(add(const_1, divide(divide(10, const_2), const_100)), const_2)) | if an investor puts $ 900 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ? | 1.05 * 1.05 * 900 = $ 992.25 the answer is d . | a = 10 / 2
b = a / 100
c = 1 + b
d = c ** 2
e = 900 * d
|
a ) 8 , b ) 27 , c ) 10 , d ) 7 , e ) 19 | a | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 21)) | if the wheel is 21 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 21 * x = 1056 = > x = 8 answer : a" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 21
h = 1056 / g
|
a ) $ 80 , b ) $ 90 , c ) $ 100 , d ) $ 110 , e ) $ 120 | b | subtract(multiply(2.75, 60), multiply(1.25, 60)) | for each color copy , print shop x charges $ 1.25 and print shop y charges $ 2.75 . how much greater is the charge for 60 color copies at print shop y than at print shop x ? | "the difference in the two prices is $ 2.75 - $ 1.25 = $ 1.50 for each color copy . each color copy will cost an extra $ 1.50 at print shop y . 60 * $ 1.50 = $ 90 the answer is b ." | a = 2 * 75
b = 1 * 25
c = a - b
|
a ) 687 , b ) 638 , c ) 683 , d ) 726 , e ) 267 | c | subtract(multiply(40, multiply(75, const_0_2778)), 180) | a train 180 m long running at 75 kmph crosses a platform in 40 sec . what is the length of the platform ? | "d = 75 * 5 / 18 = 40 = 833 Γ’ β¬ β 150 = 683 answer : c" | a = 75 * const_0_2778
b = 40 * a
c = b - 180
|
a ) 32.7 , b ) 41 , c ) 50.29 , d ) 40.41 , e ) 20.28 | d | divide(add(327, 122), divide(multiply(40, const_1000), const_3600)) | a train is 327 meter long is running at a speed of 40 km / hour . in what time will it pass a bridge of 122 meter length ? | "speed = 40 km / hr = 40 * ( 5 / 18 ) m / sec = 100 / 9 m / sec total distance = 327 + 122 = 449 meter time = distance / speed = 449 * ( 9 / 100 ) = 40.41 seconds . answer : d" | a = 327 + 122
b = 40 * 1000
c = b / 3600
d = a / c
|
a ) - 4.5 . , b ) - 2 . , c ) - 0.5 . , d ) 3 . , e ) 2.5 . | c | multiply(21, 2) | if x + y = 2 x - 2 z , x - 2 y = 4 z and x + y + z = 21 , what is the value of z / y ? | "x + y = 2 x - 2 z x - y = 2 z - - - - - - - - - - 1 x - 2 y = 4 z - - - - - - - - - 2 subtracting equation 1 from equation 2 2 z = - y z / y = - 0.5 c is the answer" | a = 21 * 2
|
a ) 50 km , b ) 56 km , c ) 60 km , d ) 70 km , e ) 33.3 km | e | multiply(10, divide(20, subtract(16, 10))) | if a person walks at 16 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is : | "let the actual distance travelled be x km . x / 10 = ( x + 20 ) / 16 16 x = 10 x + 200 6 x = 200 x = 33.3 km . answer : e" | a = 16 - 10
b = 20 / a
c = 10 * b
|
a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 105 | d | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | if teena is driving at 55 miles per hour and is currently 7.5 miles behind coe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 15 miles ahead of coe ? | this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . coe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 15 miles . teena can cover 7.5 miles in 30 mins teena will cover 15 miles in 60 mins so answer 30 + 60 = 90 mins . d | a = 55 - 40
b = a + 7
c = 55 - 40
d = b / c
e = d * const_60
|
a ) 60 m , b ) 20 m , c ) 43 m , d ) 20 m , e ) 100 m | e | subtract(500, divide(multiply(subtract(500, 300), 4), 2)) | in a 500 m race , the ratio of the speeds of two contestants a and b is 2 : 4 . a has a start of 300 m . then , a wins by : | "to reach the winning post a will have to cover a distance of ( 500 - 300 ) m , i . e . , 200 m . while a covers 2 m , b covers 4 m . while a covers 200 m , b covers 4 x 200 / 2 m = 400 m . thus , when a reaches the winning post , b covers 400 m and therefore remains 100 m behind . a wins by 100 m . answer : e" | a = 500 - 300
b = a * 4
c = b / 2
d = 500 - c
|
a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | b | divide(multiply(8, 8), const_2) | the two lines y = x and x = - 8 intersect on the coordinate plane . what is the value of the area of the figure formed by the intersecting lines and the x - axis ? | "the point of intersection is ( - 8 , - 8 ) . the triangle has a base of length 8 and a height of 8 . area = ( 1 / 2 ) * base * height = ( 1 / 2 ) * 8 * 8 = 32 the answer is b ." | a = 8 * 8
b = a / 2
|
a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | b | divide(multiply(divide(multiply(divide(40, const_100), 400), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 400 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 40 % of them donated $ 16000 ( 400 * 40 ) $ 16000 is 60 % of total amount . so total amount = 16000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 16000 * ( 100 / 60 ) * ( 40 / 100 ) = 32000 / 3 this amount has to be divided by 60 ( remaining people are 60 ) so per head amount is 32000 / 3 / 60 = 32000 / 180 = 1600 / 9 = $ 177.78 ; answer : b" | a = 40 / 100
b = a * 400
c = 60 / 100
d = b / c
e = 40 / 100
f = d * e
g = 60 / 100
h = f / g
|
a ) 78 , b ) 65 , c ) 58 , d ) 62 , e ) 48 | a | subtract(add(multiply(6, 58), multiply(6, 65)), multiply(11, 60)) | the average of 11 numbers is 60 . out of 11 numbers the average of first 6 no . is 58 , and last 6 numbers is 65 then find 6 th number ? | "6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 58 + 6 * 65 - 11 * 60 = 78 answer is a" | a = 6 * 58
b = 6 * 65
c = a + b
d = 11 * 60
e = c - d
|
a ) 157 , b ) 213 , c ) 221 , d ) 223 , e ) 226 | a | divide(multiply(100, 30), subtract(30, divide(multiply(add(divide(multiply(100, 30), 16), 30), 5), 100))) | each month a retailer sells 100 identical items . on each item he makes a profit of $ 30 that constitutes 16 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ? | for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 16 % of the cost : sell price = cost + $ 20 sell price = $ 125 + $ 20 thus , the sell price is $ 145 for each item . selling all 100 items gives the retailer . . . 100 ( $ 20 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 145 ) = $ 7.25 discount $ 137.75 = $ 125 + $ 12.75 now , the retailer makes a profit of just $ 12.75 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 12.75 ( x ) = $ 2,000 x = 2,000 / 12.75 x = 156.8627 = 157 items a | a = 100 * 30
b = 100 * 30
c = b / 16
d = c + 30
e = d * 5
f = e / 100
g = 30 - f
h = a / g
|
a ) 1 / 213 , b ) 1 / 215 , c ) 1 / 216 , d ) 2 / 113 , e ) 3 / 114 | c | multiply(divide(const_3, add(const_3, const_3)), divide(const_3, add(const_3, const_3))) | 4 dice are thrown simultaneously on the board . find the probability show the same face . | "the total number of elementary events associated to the random experiments of throwing four dice simultaneously is : = 6 Γ 6 Γ 6 Γ 6 = 64 = 6 Γ 6 Γ 6 Γ 6 = 64 n ( s ) = 64 n ( s ) = 64 let xx be the event that all dice show the same face . x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 , 5,5 ) , ( 6,6 , 6,6 ) } x = { ( 1,1 , 1,1 , ) , ( 2,2 , 2,2 ) , ( 3,3 , 3,3 ) , ( 4,4 , 4,4 ) , ( 5,5 , 5,5 ) , ( 6,6 , 6,6 ) } n ( x ) = 6 n ( x ) = 6 hence required probability , = n ( x ) n ( s ) = 664 = n ( x ) n ( s ) = 664 = 1 / 216 c" | a = 3 + 3
b = 3 / a
c = 3 + 3
d = 3 / c
e = b * d
|
a ) 40 % , b ) 100 % , c ) 150 % , d ) 200 % , e ) 300 % | d | multiply(const_100, divide(multiply(surface_cube(1), surface_cube(3)), surface_cube(3))) | if a 3 cm cube is cut into 1 cm cubes , then what is the percentage increase in the surface area of the resulting cubes ? | "the area a of the large cube is 3 * 3 * 6 = 54 square cm . the area of the 27 small cubes is 27 * 6 = 54 * 3 = 3 a , an increase of 200 % . the answer is d ." | a = surface_cube * (
b = a / surface_cube
c = 100 * b
|
a ) 5100 , b ) 5120 , c ) 5200 , d ) 5400 , e ) 6480 | e | subtract(subtract(8000, multiply(8000, divide(10, const_100))), multiply(subtract(8000, multiply(8000, divide(10, const_100))), divide(10, const_100))) | the population of a town is 8000 . it decreases annually at the rate of 10 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 8000 Γ£ β 90 / 100 Γ£ β 90 / 100 = 6480 answer : e" | a = 10 / 100
b = 8000 * a
c = 8000 - b
d = 10 / 100
e = 8000 * d
f = 8000 - e
g = 10 / 100
h = f * g
i = c - h
|
a ) $ 3.15 , b ) $ 4.45 , c ) $ 3.60 , d ) $ 5.05 , e ) $ 5.40 | c | add(2.25, multiply(0.15, divide(3.6, divide(2, 5)))) | jim β s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.15 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim β s taxi service = 2.25 $ and charge per 2 / 5 mile ( . 4 mile ) = . 15 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 / . 4 ) * . 15 = 2.25 + 9 * . 15 = 3.6 $ answer c" | a = 2 / 5
b = 3 / 6
c = 0 * 15
d = 2 + 25
|
a ) 65 % , b ) 52 % , c ) 85 % , d ) 69 % , e ) 75 % | a | add(multiply(70, divide(80, const_100)), multiply(subtract(const_100, 70), divide(30, const_100))) | in a certain city , 70 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 80 percent of the registered voters who are democrats and 30 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ? | "say there are total of 100 registered voters in that city . thus 70 are democrats and 30 are republicans . 70 * 0.80 = 56 democrats are expected to vote for candidate a ; 30 * 0.30 = 9 republicans are expected to vote for candidate a . thus total of 56 + 9 = 65 registered voters are expected to vote for candidate a , which is 65 % of the total number of registered voters . answer : a" | a = 80 / 100
b = 70 * a
c = 100 - 70
d = 30 / 100
e = c * d
f = b + e
|
a ) 33 , b ) 9 1 / 3 , c ) 30 , d ) 88 , e ) 11 | b | inverse(subtract(4, divide(4, 7))) | a and b can do a piece of work in 7 days . with the help of c they finish the work in 4 days . c alone can do that piece of work in ? | "c = 1 / 4 Γ’ β¬ β 1 / 7 = 3 / 28 = > 28 / 3 = 9 1 / 3 days answer : b" | a = 4 / 7
b = 4 - a
c = 1/(b)
|
a ) β 5 % , b ) 5 % , c ) 15 % , d ) 26 % , e ) 80 % | d | multiply(subtract(multiply(add(const_1, divide(80, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 80 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | "the actual answer is obtained by multiplying 180 % by 70 % and subtracting 100 % from this total . that is : 180 % Γ 70 % = 126 % ; 126 % β 100 % = 26 % . answer : d" | a = 80 / 100
b = 1 + a
c = 30 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | multiply(divide(inverse(add(divide(const_1, 16), divide(const_1, 24))), 4), 5) | pipe a can fill a tank in 16 hrs and pipe b can fill it in 24 hrs . if both the pipes are opened in the empty tank . in how many hours will it be fill 5 / 4 th of that tank ? | part filled a in 1 hr = ( 1 / 16 ) part filled b in 1 hr = ( 1 / 24 ) part filled by ( a + b ) together in 1 hr = ( 1 / 16 ) + ( 1 / 24 ) = 5 / 48 so , the tank will be full in 48 / 5 hrs time taken to fill exact quarter tank = ( 48 / 5 ) * ( 5 / 4 ) = 12 hrs answer : b | a = 1 / 16
b = 1 / 24
c = a + b
d = 1/(c)
e = d / 4
f = e * 5
|
a ) 7000 , b ) 7029 , c ) 2778 , d ) 4200 , e ) 2791 | d | divide(504, 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 . 504 . find the sum ? | "let the sum be rs . x . ( x * 18 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 504 = > 36 x / 100 - 24 x / 100 = 504 = > 12 x / 100 = 840 = > x = 4200 . answer : d" | a = 18 - 12
b = a * 2
c = b / 100
d = 504 / c
|
a ) 22 , b ) 30 , c ) 99 , d ) 38 , e ) 27 | b | add(divide(add(21, 23), const_2), multiply(const_1, 8)) | the average age of 8 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 8 * 2 ) years = 16 years . sum of ages of two new men = ( 21 + 23 + 16 ) years = 60 years average age of two new men = ( 60 / 2 ) years = 30 years . answer : b" | a = 21 + 23
b = a / 2
c = 1 * 8
d = b + c
|
a ) 5 km , b ) 7 km , c ) 9 km , d ) 10 km , e ) 12 km | d | divide(multiply(add(5, divide(48, const_60)), divide(multiply(multiply(12.5, 2), const_2), add(12.5, 2))), const_2) | a boy traveled from the village to the post - office at the rate of 12.5 kmph and walked back at the rate of 2 kmph . if the whole journey took 5 hours 48 minutes , find the distance of the post - office from the village . | explanation : solution : average speed = 2 xy / ( x + y ) km / hr = ( 2 * 12.5 * 2 ) / ( 12.5 + 2 ) km / hr = 50 / 14.5 km / hr . total distance = ( 50 / 14.5 * 29 / 5 ) km . = 20 km . required distance = 20 / 2 = 10 km . answer : d | a = 48 / const_60
b = 5 + a
c = 12 * 5
d = c * 2
e = 12 + 5
f = d / e
g = b * f
h = g / 2
|
a ) 12 , b ) 20 , c ) 45 , d ) 60 , e ) 52 | c | divide(multiply(177, lcm(3, 4)), add(multiply(9, 3), add(multiply(4, 5), multiply(4, 3)))) | the sum of the numbers is 177 . if the ratio between the first and the second be 5 : 3 and that between the second and third be 4 : 9 , then find the second number ? | "given ratios 5 : 3 4 : 9 20 : 12 : 27 the second number = 177 / ( 20 + 12 + 27 ) * 15 = 45 answer is c" | a = math.lcm(3, 4)
b = 177 * a
c = 9 * 3
d = 4 * 5
e = 4 * 3
f = d + e
g = c + f
h = b / g
|
a ) 2 . , b ) 3.5 . , c ) 4 . , d ) 5.5 . , e ) 6 . | e | subtract(subtract(divide(108, const_2), 7.5), divide(multiply(108, divide(const_3, const_4)), const_2)) | a green lizard can travel from the green cave to the blue cave in 108 minutes ; the blue lizard can travel from the blue cave to the green cave in 25 % less time . if the green lizard started to travel 7.5 minutes before the blue lizard , how many minutes after the blue lizard , will the green lizard pass the middle line ? | time take by the green lizard to cover the distance between the caves = 108 mins time take by the green lizard to cover half the distance = 108 / 2 = 54 mins time take by the blue lizard to cover the distance between the caves = 108 x 3 / 4 = 81 mins time take by the blue lizard to cover half the distance = 81 / 2 = 40.5 mins now the green lizard had been travelling for 7.5 mins when the blue lizard started . therefore when the blue lizard starts to move the green lizard will have to cover 54 - 7.5 = 46.5 mins worth of distance at its current speed . difference in time when they both reach the mid point = 46.5 - 40.5 = 6 mins . e is the correct answer . | a = 108 / 2
b = a - 7
c = 3 / 4
d = 108 * c
e = d / 2
f = b - e
|
a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 144 / 17 , e ) 11 / 4 | d | divide(12, divide(const_1, add(divide(const_1, 16), divide(const_1, 18)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 16 , and 18 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 16 + 1 / 18 = 17 / 144 therefore they take 144 / 17 ratio = 144 / 17 = d" | a = 1 / 16
b = 1 / 18
c = a + b
d = 1 / c
e = 12 / d
|
a ) 3472 , b ) 8222 , c ) 4370 , d ) 26777 , e ) 8222 | c | subtract(8100, divide(subtract(9100, multiply(add(const_1, divide(10, const_100)), 8100)), subtract(add(const_1, divide(15, const_100)), add(const_1, divide(10, const_100))))) | the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked from each cow each year ? | let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 β x ) ( 8100 β x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % 15 % or 10 % 10 % . so 8100 + 15100 β
x + 10100 β
( 8100 β x ) = 91008100 + 15100 β
x + 10100 β
( 8100 β x ) = 9100 therefore 8100 + 320 x + 110 ( 8100 β x ) = 91008100 + 320 x + 110 ( 8100 β x ) = 9100 120 x = 190120 x = 190 x = 3800 x = 3800 therefore , the cows produced 3800 and 4300 litres of milk the first year , and 4370 and 4730 litres of milk the second year , respectively . answer : c | a = 10 / 100
b = 1 + a
c = b * 8100
d = 9100 - c
e = 15 / 100
f = 1 + e
g = 10 / 100
h = 1 + g
i = f - h
j = d / i
k = 8100 - j
|
a ) 9 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | b | divide(15, divide(add(2.5, 1.5), const_2)) | sara bought both german chocolate and swiss chocolate for some cakes she was baking . the swiss chocolate cost $ 2.5 per pound , and german chocolate cost $ 1.5 per pound . if the total the she spent on chocolate was $ 15 and both types of chocolate were purchased in whole number of pounds , how many total pounds of chocolate she purchased ? | "if there were all the expensive ones 2.5 . . . . there would be 15 / 2.5 or 6 of them but since 1.5 $ ones are also there , answer has to be > 6 . . . . if all were 1.5 $ ones , there will be 15 / 1.5 or 10 . . . so only 8 is left ans b . ." | a = 2 + 5
b = a / 2
c = 15 / b
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a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | multiply(divide(divide(99, const_3), 99), 2) | an integer n between 1 and 99 , inclusive , is to be chosen at random . what is the probability that n ( n + 2 ) will be divisible by 3 ? | n ( n + 2 ) to be divisible by 3 either n or n + 2 must be a multiples of 3 . in each following group of numbers : { 1 , 2 , 3 } , { 4 , 5 , 6 } , { 7 , 8 , 9 } , . . . , { 97 , 98 , 99 } there are exactly 2 numbers out of 3 satisfying the above condition . for example in { 1 , 2 , 3 } n can be : 1 , or 3 . thus , the overall probability is 2 / 3 . answer : d . | a = 99 / 3
b = a / 99
c = b * 2
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a ) 2998 , b ) 2799 , c ) 2890 , d ) 1485 , e ) 2780 | d | multiply(volume_cylinder(divide(3, const_2), 14), 15) | find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 15 per cubic meter ? | "22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 15 = 1485 answer : d" | a = 3 / 2
b = volume_cylinder * (
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a ) 37 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(28, multiply(2, 10)), 28), 2) | a car started running at a speed of 28 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | "a 37 km the total distance covered by the car in the first 10 hours = 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 = sum of 10 terms in ap whose first term is 28 and last term is 46 = 10 / 2 [ 28 + 46 ] = 370 km ." | a = 2 * 10
b = 28 + a
c = b + 28
d = c / 2
|
a ) 26 , b ) 22 , c ) 25 , d ) 27 , e ) 29 | d | add(add(multiply(2, 11), 3), 2) | find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 3 x ( 11 ) ^ 2 | "( 4 ) ^ 11 x ( 7 ) ^ 3 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 3 x ( 11 ) ^ 2 = 2 ^ 11 x 2 ^ 11 x 7 ^ 3 x 11 ^ 2 = 2 ^ 22 x 7 ^ 3 x 11 ^ 2 total number of prime factors = ( 22 + 3 + 2 ) = 27 . answer is d ." | a = 2 * 11
b = a + 3
c = b + 2
|
a ) 0.05 , b ) 0.0625 , c ) 0.2 , d ) 0.3 , e ) 0.6 | d | subtract(divide(subtract(20, 7), 20), divide(7, 20)) | for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 7 of whom are females , by how much does the index for the females exceed the index for the males in the group ? | "index for females = ( 20 - 7 ) / 20 = 13 / 20 = 0.65 index for males = ( 20 - 13 / 20 = 7 / 20 = 0.35 index for females exceeds males by 0.65 - 0.35 = 0.3 answer : d" | a = 20 - 7
b = a / 20
c = 7 / 20
d = b - c
|
a ) 20 , b ) 98 , c ) 37 , d ) 26 , e ) 17 | a | add(multiply(sqrt(divide(subtract(138, 131), const_2)), const_100), sqrt(subtract(138, divide(subtract(138, 131), const_2)))) | the sum of the squares of three numbers is 138 , while the sum of their products taken two at a time is 131 . their sum is : | "explanation : let the numbers be a , b and c . then , a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) 138 + 2 * 131 = 400 ( a + b + c ) = Γ’ Λ Ε‘ 400 = 20 answer : a" | a = 138 - 131
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 138 - 131
f = e / 2
g = 138 - f
h = math.sqrt(g)
i = d + h
|
a ) 34.375 % , b ) 48 % , c ) 50.5 % , d ) 52 % , e ) 56 % | a | multiply(subtract(divide(subtract(const_100, 14), 64), const_1), const_100) | the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 14 % ? | "explanation : let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 86 gain % = 22 / 64 * 100 = 34.375 % . answer : option a" | a = 100 - 14
b = a / 64
c = b - 1
d = c * 100
|
['a ) 77.14 cm', 'b ) 47.14 cm', 'c ) 84.92 cm', 'd ) 94.94 cm', 'e ) 23.57 cm'] | e | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(16, 14)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | let the side of the square be a cm . parameter of the rectangle = 2 ( 16 + 14 ) = 60 cm parameter of the square = 60 cm i . e . 4 a = 60 a = 15 diameter of the semicircle = 15 cm circimference of the semicircle = 1 / 2 ( β ) ( 15 ) = 1 / 2 ( 22 / 7 ) ( 15 ) = 330 / 14 = 23.57 cm to two decimal places answer : e | a = square_edge_by_perimeter / (
b = circumface / (
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a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 20 , e ) e ) 25 | c | add(divide(50000, 20000), add(multiply(divide(50000, 20000), const_3), divide(100000, 20000))) | in a graduating class , the difference between the highest and lowest salaries is $ 100000 . the median salary is $ 50000 higher than the lowest salary and the average salary is $ 20000 higher than the median . what is the minimum number of students e in the class ? | the difference between the highest and lowest salaries is $ 100000 . so there are at least 2 people - say one with salary 0 and the other with 100 k . no salary will be outside this range . median = 50 k more than lowest . so median is right in the center of lowest and highest since lowest and highest differ by 100 k . in our example , median = 50 k . since there are more than 2 people , there would probably be a person at 50 k . mean = 20 k more than median so in our example , mean salary = 70 k on the number line , 0 . . . . . . . . 50 k ( median ) . . . . . . . . 100 k mean = 70 k so there must be people more toward 100 k to bring the mean up to 70 k . since we want to add minimum people , we will add people at 100 k to quickly make up the right side deficit . 0 and 50 k are ( 70 k + 20 k ) = 90 k away from 70 k . 100 k is 30 k away from 70 k . to bring the mean to 70 k , we will add two people at 100 k each to get : 0 . . . . 50 k . . . . . 100 k , 100 k , 100 k but when we add more people to the right of 70 k , the median will shift to the right . we need to keep the median at 50 k . so every time we add people to the right of 70 k , we need to add people at 50 k too to balance the median . 50 k is 20 k less than 70 k while 100 k is 30 k more than 70 k . to keep the mean same , we need to add 2 people at 100 k for every 3 people we add at 50 k . so if we add 3 people at 50 k and 2 people at 100 k , we get : 0 , . . . 50 k , 50 k , 50 k , 50 k , . . . 100 k , 100 k , 100 k , 100 k , 100 k the median is not at 50 k yet . add another 3 people at 50 k and another 2 at 100 k to get 0 , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 50 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k , 100 k now the median is 50 k and mean is 70 k . total number of people is 15 . answer ( c ) | a = 50000 / 20000
b = 50000 / 20000
c = b * 3
d = 100000 / 20000
e = c + d
f = a + e
|
a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 71.17 , e ) 71.13 | a | multiply(320, divide(const_1, add(divide(160, 64), divide(160, 80)))) | a car travels first 160 km at 64 km / hr and the next 160 km at 80 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 64 km / hr time taken to travel first 160 km = distancespeed = 16064 car travels next 160 km at 80 km / hr time taken to travel next 160 km = distancespeed = 16080 total distance traveled = 160 + 160 = 2 Γ 160 total time taken = 16064 + 16080 average speed = total distance traveledtotal time taken = 2 Γ 16016064 + 16080 = 2164 + 180 = 2 Γ 64 Γ 8080 + 64 = 2 Γ 64 Γ 80144 = 2 Γ 8 Γ 8018 = 6409 = 71.11 km / hr answer : a" | a = 160 / 64
b = 160 / 80
c = a + b
d = 1 / c
e = 320 * d
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a ) $ 200 , b ) $ 220 , c ) $ 280 , d ) $ 300 , e ) $ 340 | e | subtract(subtract(800, divide(multiply(800, 2), 5)), multiply(subtract(subtract(20, divide(multiply(20, 3), 5)), const_1), 20)) | a prize of $ 800 is to be distributed among 20 winners , each of whom must be awarded at least $ 20 . if 2 / 5 of the prize will be distributed to 3 / 5 of the winners , what is the greatest possible individual award ? | "total value of the prize = $ 800 number of people = 20 2 / 5 of 800 ( = $ 320 ) should be distributed among 3 / 5 of 20 ( = 12 people ) with each getting $ 20 each . remaining money = 800 - 320 = $ 480 . now in order to ' maximize ' 1 prize , we need to minimise the others and we have been given that each should get $ 20 . thus , minimising the remaining 7 people ( = 20 - 12 - 1 . ' - 1 ' to exclude 1 that needs to be maximised ) = 7 * 20 = 140 . thus the maximum award can be = 480 - 140 = $ 340 , hence e is the correct answer ." | a = 800 * 2
b = a / 5
c = 800 - b
d = 20 * 3
e = d / 5
f = 20 - e
g = f - 1
h = g * 20
i = c - h
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a ) rs . 6500 , b ) rs . 1947.4 , c ) rs . 973.7 , d ) rs . 2000 , e ) none | e | divide(multiply(divide(divide(add(divide(multiply(6500, 14), const_100), divide(multiply(add(6500, divide(multiply(6500, 14), const_100)), 14), const_100)), 2), 3), const_100), 10) | simple interest on a certain sum of money for 3 years at 10 % per annum is half the compound interest on rs . 6500 for 2 years at 14 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 6500 x ( 1 + 14 / 100 ) Γ’ Β² - 6500 ] rs . ( 6500 x 114 / 100 x 114 / 100 - 6500 ) = rs . 1947.4 . sum = rs . [ 973.7 x 100 / 3 x 10 ] = rs . 3245.67 . answer e" | a = 6500 * 14
b = a / 100
c = 6500 * 14
d = c / 100
e = 6500 + d
f = e * 14
g = f / 100
h = b + g
i = h / 2
j = i / 3
k = j * 100
l = k / 10
|
a ) 300 , b ) 420 , c ) 460 , d ) 320 , e ) 400 | c | add(multiply(50, 8), multiply(subtract(50, 20), multiply(8, divide(25, const_100)))) | mary works in a restaurant a maximum of 50 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | "mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 30 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 30 ) = $ 300 . the total amount is $ 160 + $ 300 = $ 460 answer c 460 ." | a = 50 * 8
b = 50 - 20
c = 25 / 100
d = 8 * c
e = b * d
f = a + e
|
a ) 8600 litres , b ) 3200 litres , c ) 12800 litres , d ) 11200 litres , e ) 13200 litres | b | multiply(divide(multiply(multiply(4, const_60), add(5, 3)), 3), 5) | an outlet pipe empties a tank which is full in 5 hours . if the inlet pipe is kept open , which lets water in at the rate of 4 litres / min then outlet pipe would take 3 hours longer . find the capacity of the tank . | let the rate of outlet pipe be x liters / hour ; rate of inlet pipe is 8 litres / min , or 4 * 60 = 240 liters / hour ; net outflow rate when both pipes operate would be x - 240 liters / hour . capacity of the tank = x * 5 hours = ( x - 240 ) * ( 5 + 3 ) hours 5 x = ( x - 240 ) * 8 - - > x = 640 - - > capacity = 5 x = 3200 liters . answer : b . | a = 4 * const_60
b = 5 + 3
c = a * b
d = c / 3
e = d * 5
|
a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 25 | e | sqrt(625) | you buy a piece of land with an area of Γ’ Λ Ε‘ 625 , how long is one side of the land plot ? | "try filling the numbers into the answer y x y = find the closest to 625 . answer e" | a = math.sqrt(625)
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a ) 72 , b ) 90 , c ) 100 , d ) 110 , e ) 120 | c | add(multiply(add(multiply(4, const_3), 2), divide(add(multiply(4, const_3), 2), 2)), 4) | in a division sum , the remainder is 4 and the divisor is 2 times the quotient and is obtained by adding 2 to the thrice of the remainder . the dividend is : | "diver = ( 4 * 3 ) + 2 = 14 2 * quotient = 14 quotient = 7 dividend = ( divisor * quotient ) + remainder dividend = ( 14 * 7 ) + 2 = 100 c" | a = 4 * 3
b = a + 2
c = 4 * 3
d = c + 2
e = d / 2
f = b * e
g = f + 4
|
a ) 2 , b ) 4 , c ) 3 , d ) 1 , e ) 0 | b | sqrt(subtract(power(42, const_2), multiply(const_4, 437))) | if the sum of two numbers is 42 and their product is 437 , then find the absolute difference between the numbers . | let the numbers be x and y . then , x + y = 42 and xy = 437 x - y = sqrt [ ( x + y ) 2 - 4 xy ] = sqrt [ ( 42 ) 2 - 4 x 437 ] = sqrt [ 1764 β 1748 ] = sqrt [ 16 ] = 4 . required difference = 4 . answer is b . | a = 42 ** 2
b = 4 * 437
c = a - b
d = math.sqrt(c)
|
a ) 300 , b ) 310 , c ) 320 , d ) 330 , e ) 340 | b | add(240, add(divide(240, 8), divide(240, 6))) | for dinner , cara ate 240 grams of bread which was 8 times as much bread as she ate for lunch , and 6 times as much bread as she ate for breakfast . how much bread did cara eat in total ? | for breakfast , cara ate 240 / 6 = 40 grams . for lunch , cara ate 240 / 8 = 30 grams . for dinner , cara ate 240 grams . the total is 40 + 30 + 240 = 310 grams . the answer is b . | a = 240 / 8
b = 240 / 6
c = a + b
d = 240 + c
|
['a ) 3 m', 'b ) 4 m', 'c ) 5 m', 'd ) 6 m', 'e ) 2 m'] | a | divide(subtract(const_100, sqrt(subtract(power(const_100, const_2), multiply(subtract(multiply(60, 40), 2109), const_4)))), const_2) | a rectangular box 60 m long and 40 m wide has two concrete crossroads running in the middle of the box and rest of the box has been used as a lawn . the area of the lawn is 2109 sq . m . what is the width of the road ? | total area of cross roads = 60 x + 40 x β x 2 but total area of the cross roads = 291 m 2 hence , 60 x + 40 x β x 2 = 291 β 100 x β x 2 = 291 β x 2 β 100 x + 291 = 0 β ( x β 97 ) ( x β 3 ) = 0 β x = 3 answer : a | a = 100 ** 2
b = 60 * 40
c = b - 2109
d = c * 4
e = a - d
f = math.sqrt(e)
g = 100 - f
h = g / 2
|
a ) 2 : 5 , b ) 1 : 4 , c ) 3 : 7 , d ) 6 : 11 , e ) 2 : 3 | e | divide(divide(const_1, const_4), divide(50, const_100)) | if 50 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ? | "let 50 % of a = 1 / 3 b then 50 a / 100 = 1 b / 3 a / 2 = b / 3 a / b = 2 / 3 a : b = 2 : 3 answer is e" | a = 1 / 4
b = 50 / 100
c = a / b
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 50 % | e | multiply(divide(180, divide(const_3600, const_10)), const_100) | the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 180 Β° of the circle , what percentage of megatek employees are in manufacturing ? | "answer : e 180 Β° divided by 360 Β° equals 0.5 , therefore the sector is equal to 50 % of the total" | a = 3600 / 10
b = 180 / a
c = b * 100
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a ) 18 seconds , b ) 27 seconds , c ) 26 seconds , d ) 12 seconds , e ) 8 seconds | e | divide(80, multiply(add(25, 11), const_0_2778)) | the speed at which a man can row a boat in still water is 25 kmph . if he rows downstream , where the speed of current is 11 kmph , what time will he take to cover 80 metres ? | "speed of the boat downstream = 25 + 11 = 36 kmph = 36 * 5 / 18 = 10 m / s hence time taken to cover 80 m = 80 / 10 = 8 seconds . answer : e" | a = 25 + 11
b = a * const_0_2778
c = 80 / b
|
a ) 10 , b ) 5 , c ) 2 , d ) 3 , e ) 1 | c | subtract(multiply(10, divide(50, const_100)), 3) | for a particular american football game , the probability of a team ' s quarterback throwing a completed pass on each throw is 3 / 10 . what is the least number of times that the quarterback should throw the ball that will increase the probability of getting a completed pass at least once to more than 50 % . | rule of subtraction : p ( a ) = 1 - p ( a ' ) rule of multiplication : p ( a β© b ) = p ( a ) p ( b ) the probability that the quarterback throws a completed pass at least once in 2 throws is 1 - ( 7 / 10 ) ^ 2 = 1 - 49 / 100 = 51 / 100 > 50 % answer : c | a = 50 / 100
b = 10 * a
c = b - 3
|
a ) 9.6 days , b ) 7.6 days , c ) 4 days , d ) 8.6 days , e ) 6.6 days | a | multiply(inverse(add(inverse(8), inverse(12))), const_2) | a and b can do a work in 8 hours and 12 hours respectively . a starts the work at 6 am and they work alternately for one hour each . when will the work be completed ? | work done by a and b in the first two hours , working alternately = first hour a + second hour b = 1 / 8 + 1 / 12 = 5 / 24 . total time required to complete the work = 2 * 24 / 5 = 9.6 days . answer : a | a = 1/(8)
b = 1/(12)
c = a + b
d = 1/(c)
e = d * 2
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a ) 225 , b ) 275 , c ) 325 , d ) 350 , e ) 700 | e | divide(add(1800, 300), const_3) | in a weight - lifting competition , the total weight of joe ' s two lifts was 1800 pounds . if twice the weight of his first lift was 300 pounds more than the weight of his second lift , what was the weight , in pounds , of his first lift ? | this problem is a general word translation . we first define variables and then set up equations . we can define the following variables : f = the weight of the first lift s = the weight of the second lift we are given that the total weight of joe ' s two lifts was 1800 pounds . we sum the two variables to obtain : f + s = 1800 we are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift . we express this as : 2 f = 300 + s 2 f β 300 = s we can now plug in ( 2 f β 300 ) for s into the first equation , so we have : f + 2 f β 300 = 1800 3 f = 2100 f = 700 answer is e . | a = 1800 + 300
b = a / 3
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a ) 10 , b ) 20 , c ) 30 , d ) 15 , e ) 45 | b | divide(12, subtract(divide(12, 10), 1)) | a train covers a distance of 12 km in 10 min . if it takes 1 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 1 = 20 m . answer : option b" | a = 12 / 10
b = a - 1
c = 12 / b
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a ) 54 , b ) 162 , c ) 250 , d ) 275 , e ) 322 | d | divide(multiply(330, 10), add(2, 10)) | compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 330 grams of compound x ? | total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 330 gms of x have 330 * 10 / 12 = 275 gms of b and 330 - 275 = 55 gms of a . cross check : - a / b = 55 / 275 = 2 / 10 ( as given ) ans d | a = 330 * 10
b = 2 + 10
c = a / b
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a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 14 | e | add(7, divide(multiply(7, subtract(12000, 9000)), subtract(9000, 6000))) | the average salary of all the workers in a workshop is rs . 9000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | "let the total number of workers be x . then , 9000 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = > 3000 x = 42000 = x = 14 . answer : e" | a = 12000 - 9000
b = 7 * a
c = 9000 - 6000
d = b / c
e = 7 + d
|
a ) 470 , b ) 468 , c ) 465 , d ) 463 , e ) 485 | c | add(divide(divide(30, divide(divide(divide(divide(divide(30, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(30, divide(divide(divide(divide(divide(30, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 30 natural numbers | "explanation : sum of n natural numbers = n ( n + 1 ) 2 = 30 ( 30 + 1 ) 2 = 30 ( 31 ) 2 = 465 answer : option c" | a = 30 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 30 / e
g = f / 2
h = 30 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 30 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | d | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(5, add(add(5, 3), 3))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(3, add(add(5, 3), 3)))) | a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 5 : 3 : 1 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 4 : 3 , respectively . how many more acres of land were planted with tobacco under the new system ? | "we have ratio in 5 : 3 : 1 and this is shifted to 2 : 4 : 3 ( c : s : t ) if we observe the ratio of the tobacco increased by 2 times ( 3 - 1 ) t = 1 / 9 * 1350 = 150 . since tobacco increased by 2 times . . we get 150 * 2 = 300 . answer d is the answer . ." | a = 100 * 10
b = 3 * 100
c = a + b
d = 5 * 10
e = c + d
f = 5 + 3
g = f + 3
h = 5 / g
i = e * h
j = 100 * 10
k = 3 * 100
l = j + k
m = 5 * 10
n = l + m
o = 5 + 3
p = o + 3
q = 3 / p
r = n * q
s = i - r
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a ) 100 % , b ) 120 % , c ) 150 % , d ) 200 % , e ) 300 % | d | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 70), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 70), subtract(const_100, 10)), const_100)), const_100) | the charge for a single room at hotel p is 70 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "p = 0.3 r = 0.9 g r = 0.9 g / 0.3 = 3 g thus r is 200 % greater than g . the answer is d ." | a = 100 - 70
b = 100 - 10
c = a / b
d = c * 100
e = 100 - d
f = 100 - 70
g = 100 - 10
h = f / g
i = h * 100
j = e / i
k = j * 100
|
a ) a ) 70 , b ) b ) 76 , c ) c ) 78 , d ) d ) 80 , e ) e ) 88 | c | subtract(multiply(add(34, 4), add(10, const_1)), multiply(10, 34)) | average of 10 matches is 34 , how many runs one should should score to increase his average by 4 runs . | explanation : average after 11 innings should be 38 so , required score = ( 11 * 38 ) - ( 10 * 34 ) = 418 - 340 = 78 answer : option c | a = 34 + 4
b = 10 + 1
c = a * b
d = 10 * 34
e = c - d
|
a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 5 | a | subtract(25, reminder(1021, 25)) | what least number should be added to 1021 , so that the sum is completely divisible by 25 ? | "1021 Γ£ Β· 25 = 40 with remainder = 21 21 + 4 = 25 . hence 4 should be added to 1021 so that the sum will be divisible by 25 answer : option a" | a = 25 - reminder
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a ) 6.25 , b ) 6.5 , c ) 6.75 , d ) 7 , e ) none | a | divide(subtract(0, multiply(10, 3.2)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 40 0 vers to reach the target of 282 runs ? | "sol . required run rate = 282 β ( 3.2 Γ 10 / 40 ) = 240 / 40 = 6.25 . answer a" | a = 10 * 3
b = 0 - a
c = b / 40
|
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | b | divide(factorial(subtract(add(const_4, 6), const_1)), multiply(factorial(6), factorial(subtract(const_4, const_1)))) | how many positive integers less than 50 are multiples of 6 but not multiples of 8 ? | "the lcm of 6 and 8 is 24 . if x < 50 and x is divisible by 6 not by 8 - - > x is not divisible by 24 . from 1 - - > 50 , we have 2 numbers which is divisible by 24 : 24 , 48 . from 1 - - > 50 , we have ( 48 - 6 ) / 6 + 1 = 8 numbers divisible by 6 . therefore , our answer is 8 - 2 = 6 numbers . b" | a = 4 + 6
b = a - 1
c = math.factorial(b)
d = math.factorial(6)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) s . 42 , b ) s . 46 , c ) s . 40 , d ) s . 41 , e ) s . 16 | e | divide(divide(multiply(800, 10), const_100), 5) | a reduction of 10 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 10 / 100 ) = 80 - - - - 5 ? - - - - 1 = > rs . 16 answer : e" | a = 800 * 10
b = a / 100
c = b / 5
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a ) rs . 571200 , b ) rs . 180000 , c ) rs . 201600 , d ) rs . 504000 , e ) none of these | a | multiply(multiply(multiply(17000, add(const_1, divide(12, const_100))), divide(5, 2)), 12) | the monthly incomes of a and b are in the ratio 5 : 2 . b ' s monthly income is 12 % more than c ' s monthly income . if c ' s monthly income is rs . 17000 , then find the annual income of a ? | "b ' s monthly income = 17000 * 112 / 100 = rs . 19040 b ' s monthly income = 2 parts - - - - > rs . 19040 a ' s monthly income = 5 parts = 5 / 2 * 19040 = rs . 47600 a ' s annual income = rs . 47600 * 12 = rs . 571200 answer : a" | a = 12 / 100
b = 1 + a
c = 17000 * b
d = 5 / 2
e = c * d
f = e * 12
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a ) 1 / 15 , b ) 1 / 16 , c ) 1 / 11 , d ) 1 / 10 , e ) 1 / 12 | a | subtract(add(divide(const_1, 18), divide(const_1, 30)), divide(const_1, 45)) | two pipes can fill a tank in 18 minutes and 30 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | part of the filled by all the three pipes in one minute = 1 / 18 + 1 / 30 - 1 / 45 = 1 / 15 so , the tank becomes full in 15 minutes . answer : a | a = 1 / 18
b = 1 / 30
c = a + b
d = 1 / 45
e = c - d
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a ) 154 , b ) 184 , c ) 240 , d ) 272 , e ) 306 | d | multiply(17, subtract(17, const_1)) | 17 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "2 * 17 c 2 = 2 * 136 = 272 the answer is d ." | a = 17 - 1
b = 17 * a
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a ) 15 , b ) 20 , c ) 20 / 3 , d ) 25 , e ) 10 | c | multiply(add(const_2, 4), multiply(1, 4)) | if a farmer sells 5 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 20 more chickens , he will run out of feed 1 day earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | "let x = total feed required for the planned period n = number of chicken t = total time of the planned feed x = nt 1 ) x = ( n - 5 ) * ( t + 4 ) 2 ) x = ( n + 20 ) * ( t - 1 ) equating 1 & 2 ( n - 5 ) * ( t + 4 ) = ( n + 20 ) * ( t - 1 ) 5 n = 25 t n = 5 t x = n * n / 5 substituting this value in 1 n * n / 5 = ( n - 5 ) * ( n / 5 + 4 ) 15 n = 100 n = 20 / 3 c" | a = 2 + 4
b = 1 * 4
c = a * b
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a ) 16 , b ) 14 , c ) 12 , d ) 18 , e ) 21 | c | subtract(add(power(13, 13), 13), multiply(14, floor(divide(add(power(13, 13), 13), 14)))) | what will be the reminder when ( 13 ^ 13 + 13 ) is divided by 14 ? | ( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 13 ^ 13 + 1 ) will be divisible by ( 13 + 1 ) ; ( 13 ^ 13 + 1 ) + 12 when divided by 14 will give 12 as remainder . correct option : c | a = 13 ** 13
b = a + 13
c = 13 ** 13
d = c + 13
e = d / 14
f = math.floor(e)
g = 14 * f
h = b - g
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a ) 4 : 10 , b ) 2 : 3 , c ) 1 : 3 , d ) 2 : 4 , e ) 3 : 5 | d | divide(divide(subtract(divide(40, const_2), const_10), const_2), add(divide(40, const_2), const_10)) | two whole numbers whose sum is 40 can not be in the ratio | d ) 2 : 4 | a = 40 / 2
b = a - 10
c = b / 2
d = 40 / 2
e = d + 10
f = c / e
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a ) 76 , b ) 101 , c ) 126 , d ) 151 , e ) 176 | e | multiply(25, 7) | the remainder when positive integer nn ( n > 1 n > 1 ) is divided by 25 is 1 and the remainder when nn is divided by 7 is also 1 . what is the least value of nn ? | we ' re asked to find the least number that gives us a remainder of 1 when divided by 25 and gives us a remainder of 1 when divided by 7 . you might notice , rather quickly , that all of the answers gives us a remainder when divided by 25 . . . 76 = 75 + 1 101 = 100 + 1 126 = 125 + 1 151 = 150 + 1 176 = 175 + 1 so the real question is ` ` which of these 5 is the least number that has a remainder of 1 when divided by 7 ? ' ' from here , you can just do the ` ` brute force ' ' math : a : 76 / 7 = 10 r 6 not a match b : 101 / 7 = 14 r 3 not a match c : 126 / 7 = 18 r 0 not a match d : 151 / 7 = 21 r 4 not a match none of these numbers ` ` fits ' ' the description , so the answer must be e . e : 176 / 7 = 25 r 1 ; answer : e | a = 25 * 7
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a ) 56 hours , b ) 28 hours , c ) 55 hours , d ) 66 hours , e ) 47 hours | a | multiply(divide(add(add(const_4, const_2), const_1), const_2), add(multiply(3, const_4), const_4)) | a tank is filled in sixteen hours by 3 pipes a , b and c . pipe a is twice as fast as pipe b , and b is twice as fast as c . how much time will pipe b alone take to fill the tank ? | 1 / a + 1 / b + 1 / c = 1 / 16 ( given ) also given that a = 2 b and b = 2 c = > 1 / 2 b + 1 / b + 2 / b = 1 / 16 = > ( 1 + 2 + 4 ) / 2 b = 1 / 16 = > 2 b / 7 = 16 = > b = 56 hours . answer : a | a = 4 + 2
b = a + 1
c = b / 2
d = 3 * 4
e = d + 4
f = c * e
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a ) 24 , b ) 34 , c ) 44 , d ) 54 , e ) 64 | d | divide(multiply(48, 18), 16) | two numbers n and 16 have lcm = 48 and gcf = 18 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 16 Γ n = 48 Γ 18 n = 48 Γ 18 / 16 = 54 correct answer d" | a = 48 * 18
b = a / 16
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(5, reminder(16, 5)) | when n is divided by 20 , the remainder is 5 . what is the remainder when n + 16 is divided by 5 ? | "assume n = 15 remainder ( n / 20 ) = 5 n + 16 = 31 remainder ( 31 / 5 ) = 1 option a" | a = 5 - reminder
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a ) 105 , b ) 245 , c ) 360 , d ) 480 , e ) 542 | d | add(divide(add(add(add(add(10, 15), 18), 22), 36), add(const_4, const_1)), 18) | if a = { 10 , 15 , 18 , 22 , 36 , 44 } , what is the product of mean and median of the numbers in a ? | "mean = ( 10 + 15 + 18 + 22 + 36 + 43 ) / 6 = 24 median = ( 18 + 22 ) / 2 = 20 product = 24 * 20 = 480 option d" | a = 10 + 15
b = a + 18
c = b + 22
d = c + 36
e = 4 + 1
f = d / e
g = f + 18
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a ) 25 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.7 % | e | multiply(divide(multiply(divide(1, 6), subtract(1, divide(1, 3))), add(multiply(divide(1, 6), subtract(1, divide(1, 3))), subtract(1, divide(1, 6)))), const_100) | of the 3,600 employees of company x , 1 / 6 are clerical . if the clerical staff were to be reduced by 1 / 3 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 6 are clerical out of 3600 so 600 are clerical 600 reduced by 1 / 3 is 600 * 1 / 3 so it reduced 200 people , so there is 400 clerical people left but since 200 people left , it also reduced from the total of 3600 so there are 3400 people total since 400 clerical left / 3400 people total you get ( a ) 11.7 % answer : e" | a = 1 / 6
b = 1 / 3
c = 1 - b
d = a * c
e = 1 / 6
f = 1 / 3
g = 1 - f
h = e * g
i = 1 / 6
j = 1 - i
k = h + j
l = d / k
m = l * 100
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a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | multiply(divide(divide(multiply(50, 30), const_60), subtract(80, 50)), const_60) | two trains leave a station traveling in the same direction . train a leaves traveling at a constant speed of 50 mph , while train b leaves traveling at a constant speed of 80 mph . if train b left the station 30 minutes after train a left , in how many minutes will train b overtake train a ? | we can use a form of the equation d = rt [ distance = rate * time ] train a will have traveled for 30 minutes longer when train b overtakes it so time of train a : t + 30 minutes = t + 1 / 2 hours ( switch to hours since the rates are in hours ) time of train b : t rate of train a : 50 mph rate of train b : 80 mph the distance traveled by each will be the same when b overtakes a so set the right side of d = rt equal to each other for the two trains 50 * ( t + 1 / 2 ) = 80 * t 50 t + 25 = 80 t 25 = 30 t 25 / 30 = t 5 / 6 hours = t which is 5 / 6 * 60 = 50 minutes c | a = 50 * 30
b = a / const_60
c = 80 - 50
d = b / c
e = d * const_60
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a ) 24 , b ) 22 , c ) 28 , d ) 30 , e ) 18 | a | divide(add(multiply(6, 4), 6), 36) | if 4 : 6 : : x : 36 , then find the value of x | explanation : treat 4 : 6 as 4 / 6 and x : 36 as x / 36 , treat : : as = so we get 4 / 6 = x / 36 = > 6 x = 144 = > x = 24 option a | a = 6 * 4
b = a + 6
c = b / 36
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a ) 4 ^ 12 , b ) 4 ^ 35 , c ) 17 ( 4 ^ 6 ) , d ) 8 ^ 12 , e ) 7 ( 4 ^ 5 ) | c | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | what is the value of 4 ^ 6 + 4 ^ 8 ? | "4 ^ 6 + 4 ^ 8 = 4 ^ 6 ( 1 + 4 ^ 2 ) = 4 ^ 6 * 17 answer c" | a = 100 + const_60
b = a + 1
c = b * 4
d = c / 100
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(add(add(const_1, 2), const_1), const_1) | let the least number of 6 digits , which when divided by 4 , 610 and 15 leaves in each case the same remainder of 2 , be n . the sum of the digits in n is : | solution least number of 6 digital is 100000 . l . c . m . of 4 , 610 and 15 = 60 . on dividing 100000 by 60 , the remainder obtained is 40 . so , least number of 6 digits divisible by 4 , 610 and 15 = 100000 + ( 60 - 40 ) = 100020 . so , n = ( 100020 + 2 ) = 100022 . sum of digits in n = ( 1 + 2 + 2 ) = 5 . answer c | a = 1 + 2
b = a + 1
c = b + 1
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a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 63 | e | divide(power(105, 3), multiply(multiply(21, 25), 35)) | if a = 105 and a ^ 3 = 21 * 25 * 35 * b , what is the value of b ? | "first step will be to break down all the numbers into their prime factors . 105 = 3 * 5 * 7 21 = 7 * 3 25 = 5 * 5 35 = 7 * 5 so , ( 105 ) ^ 3 = 3 * 7 * 5 * 5 * 7 * 5 * b therefore ( 3 * 5 * 7 ) ^ 3 = 3 * 5 ^ 3 * 7 ^ 2 * b therefore , b = 3 ^ 3 * 5 ^ 3 * 7 ^ 3 / 3 * 5 ^ 3 * 7 ^ 2 b = 3 ^ 2 * 7 = 9 * 7 = 63 correct answer e ." | a = 105 ** 3
b = 21 * 25
c = b * 35
d = a / c
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a ) 5 , b ) 100 , c ) 16 , d ) 20 , e ) 25 | b | divide(multiply(20, 20), multiply(2, 2)) | what is the maximum number of pieces of birthday cake of size 2 β by 2 β that can be cut from a cake 20 β by 20 β ? | the prompt is essentially asking for the maximum number of 2 x 2 squares that can be cut from a larger 20 by 20 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 10 ' pieces ' each , we have ( 10 ) ( 10 ) = 100 total smaller squares ( at maximum ) . b | a = 20 * 20
b = 2 * 2
c = a / b
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a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 20 | e | divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), factorial(subtract(const_4, const_1)))) | how many positive integers less than 50 are multiples of 2 but not multiples of 5 ? | "imo the answer is c ( 8 numbers ) the lcm of 2 and 5 is 10 . if x < 50 and x is divisible by 2 not by 5 - - > x is not divisible by 10 . from 1 - - > 50 , we have 5 numbers which is divisible by 10 : 10 , 20 , 30 , 40 , 50 . from 1 - - > 50 , we have ( 50 - 2 ) / 2 + 1 = 25 numbers divisible by 2 . therefore , our answer is 25 - 5 = 20 numbers . e" | a = 4 + 2
b = a - 1
c = math.factorial(b)
d = math.factorial(2)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 2 , b ) - 7 , c ) 4 , d ) - 5 , e ) 6 | b | divide(subtract(46, 14), 4) | if | 4 x + 14 | = 46 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 14 = 46 or 4 x + 14 = - 46 = > x = 8 or x = - 15 sum of both the values will be - 15 + 8 = - 7 answer is b" | a = 46 - 14
b = a / 4
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a ) 0.0007 , b ) 0.0042 , c ) 0.0002 , d ) 0.0003 , e ) 0.0004 | a | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0282) | what is the least number . which should be added to 0.0282 to make it a perfect square ? | "0.0282 + 0.0007 = 0.0289 ( 0.17 ) ^ 2 answer : a" | a = 12 + 4
b = a + 2
c = b / 100
d = 12 + 4
e = d + 2
f = e / 100
g = c * f
h = g - 0
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a ) 444 , b ) 432 , c ) 346 , d ) 422 , e ) 448 | d | add(subtract(302, divide(112, const_2)), subtract(232, divide(112, const_2))) | at the faculty of aerospace engineering , 302 students study random - processing methods , 232 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "302 + 232 - 112 ( since 112 is counted twice ) = 422 d is the answer" | a = 112 / 2
b = 302 - a
c = 112 / 2
d = 232 - c
e = b + d
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a ) 60 , b ) 61 , c ) 62 , d ) 63 , e ) 64 | d | subtract(100, subtract(add(55, 43), 61)) | in a group of 100 people , 55 have visited iceland and 43 have visited norway . if 61 people have visited both iceland and norway , how many people have visited neither country ? | "this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 100 = 55 + 43 - 61 + ( # in neither group ) = 63 the prompt gives you all of the numbers you need to get to the correct answer . just plug in and solve . d" | a = 55 + 43
b = a - 61
c = 100 - b
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a ) 80 . , b ) 75 , c ) 70 , d ) 65 . , e ) 54 . | a | divide(add(multiply(divide(20, add(20, 40)), 60), multiply(divide(40, add(20, 40)), 90)), divide(add(20, 40), const_60)) | a car was driving at 60 km / h for 20 minutes , and then at 90 km / h for another 40 minutes . what was its average speed ? | "driving at 60 km / h for 20 minutes , distance covered = 60 * 1 / 3 = 20 km driving at 90 km / h for 40 minutes , distance covered = 90 * 2 / 3 = 60 km average speed = total distance / total time = 80 / 1 = 80 km / h answer : a" | a = 20 + 40
b = 20 / a
c = b * 60
d = 20 + 40
e = 40 / d
f = e * 90
g = c + f
h = 20 + 40
i = h / const_60
j = g / i
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a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 1 . | e | subtract(add(5, 3), 9) | if ( a + b ) = 5 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | given a + b = 5 b + c = 9 c + d = 3 ( a + b ) - ( b + c ) + ( c + d ) = ( a + d ) = > 5 - 9 + 3 = - 1 . option e . . . | a = 5 + 3
b = a - 9
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a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 104 % | e | multiply(divide(multiply(13, subtract(const_1, divide(20, const_100))), 10), const_100) | in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 13 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,13 = 13 / 100 x = 80 * 13 / 100 x = 10.4 10.4 / 10 = 1,04 = 104 % , answer e | a = 20 / 100
b = 1 - a
c = 13 * b
d = c / 10
e = d * 100
|
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