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 ) 33 , b ) 878 , c ) 18 , d ) 88 , e ) 11 | c | inverse(subtract(6, divide(6, 9))) | a and b can do a piece of work in 9 days . with the help of c they finish the work in 6 days . c alone can do that piece of work in ? | "c = 1 / 6 â € “ 1 / 9 = 1 / 18 = > 18 days answer : c" | a = 6 / 9
b = 6 - a
c = 1/(b)
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['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) 14'] | d | sqrt(divide(507, const_3)) | the length of a rectangular garden is three times its width . if the area of the rectangular garden is 507 square meters , then what is the width of the rectangular garden ? | let x be the width of the garden . 3 x ^ 2 = 507 x ^ 2 = 169 x = 13 the answer is d . | a = 507 / 3
b = math.sqrt(a)
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a ) a ) 5.61 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 8.19 | e | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 30), divide(9, 55))), const_4) | due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 30 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 9 miles stretch = 9 * 60 / 55 = 9 * 12 / 11 = 9.81 new time in minutes to cross 9 miles stretch = 9 * 60 / 30 = 9 * 2 / 1 = 18 time difference = 8.19 ans : e" | a = 55 + 9
b = a - 1
c = 9 / 30
d = 9 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 8.8 , b ) 8.9 , c ) 6.6 , d ) 7.7 , e ) none | a | subtract(add(add(divide(220, 44), 44), const_4), 44) | 220 % of a number x is 44 . what is 44 % of x . | solution : given , 220 % of x = 44 or , x = 20 . thus , 44 % of 20 = ( 44 * 20 ) / 100 = 8.8 answer : option a | a = 220 / 44
b = a + 44
c = b + 4
d = c - 44
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a ) b - 3 , b ) b - 6 , c ) b - 1 , d ) b , e ) b + 1 | a | subtract(const_2, subtract(6, const_1)) | what is the smallest of 6 consecutive odd integers whose average ( arithmetic mean ) is b + 2 ? | since the numbers are consecutive odd integers , mean = median = 3 rd integer + 4 th integer / 2 and 1 st integer = 3 rd integer - 4 let ' s say 3 rd integer = n and 4 th integer = n + 2 2 n + 2 / 2 = b + 2 n = b + 1 1 st integer = b + 1 - 4 = b - 3 a is the answer | a = 6 - 1
b = 2 - a
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a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 9 , d ) 1 : 5 , e ) 1 : 1 | b | divide(divide(180, divide(180, 20)), divide(180, subtract(divide(180, 20), const_3))) | a motorcyclist goes from bombay to pune , a distance of 180 kms at an average of 20 kmph speed . another man starts from bombay by car 2 â ½ hours after the first , and reaches pune â ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | "t = 180 / 20 = 9 h t = 9 - 3 = 6 time ratio = 9 : 6 = 3 : 2 speed ratio = 2 : 3 answer : b" | a = 180 / 20
b = 180 / a
c = 180 / 20
d = c - 3
e = 180 / d
f = b / e
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a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 85 | e | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 19) | on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 19 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 19 more transactions than cal . jade = 66 + 19 = 85 jade handled = 85 transactions . answer : e" | a = 90 / 100
b = 10 + 100
c = a * b
d = c * 2
e = d / 3
f = e + 19
|
a ) 32 , b ) 34 , c ) 26 , d ) 28 , e ) 30 | b | add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2)) | in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of six metres is left from all sides of the boundary of the garden . what is the length of the garden ? | "between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 6 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 ã — 2 ) + 6 + 6 = 34 meter answer is b ." | a = 12 - 1
b = a * 2
c = 10 / 2
d = b + c
e = 10 / 2
f = d + e
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a ) 150 , b ) 450 , c ) 400 , d ) 600 , e ) 300 | e | subtract(divide(subtract(multiply(150, 31), multiply(150, 16)), 5), 150) | a garrison of 150 men has provisions for 31 days . at the end of 16 days , a reinforcement arrives , and it is now found that the provisions will last only for 5 days more . what is the reinforcement ? | "150 - - - - 31 150 - - - - 15 x - - - - - 5 x * 5 = 150 * 15 x = 450 150 - - - - - - - 300 answer : e" | a = 150 * 31
b = 150 * 16
c = a - b
d = c / 5
e = d - 150
|
a ) rs . 15,550 , b ) rs . 15,600 , c ) rs . 16,500 , d ) rs . 17,600 , e ) rs . 17,900 | d | multiply(800, multiply(5.5, 4)) | the lenght of a room is 5.5 m and width is 4 m . find the cost of paving the floor by slabs at the rate of rs . 800 per sq . metre . | area of the floor = ( 5.5 × 4 ) m 2 = 22 m 2 . cost of paving = rs . ( 800 × 22 ) = rs . 17600 . answer : option d | a = 5 * 5
b = 800 * a
|
a ) 1220 , b ) 1840 , c ) 2160 , d ) 2480 , e ) 2760 | c | multiply(multiply(2, 3), 2) | employees of a certain company are each to receive a unique 8 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , 5 , 6 , and 7 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ? | "there are 6 ! ways to make codes starting with 12 . there are 6 ! ways to make codes starting with 24 . there are 6 ! ways to make codes starting with 36 . the number of codes is 3 * 6 ! = 2160 . the answer is c ." | a = 2 * 3
b = a * 2
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a ) 88.8 , b ) 90.4 , c ) 88.0 , d ) 70.9 , e ) 71.2 | b | add(80, multiply(divide(13, const_100), 80)) | if x is 13 percent greater than 80 , then x = | "13 % of 80 = ( 80 * 0.13 ) = 10.4 11 % greater than 80 = 80 + 10.4 = 90.4 answer is clearly b ." | a = 13 / 100
b = a * 80
c = 80 + b
|
a ) 1 / 9 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | d | multiply(divide(divide(99, 3), 99), const_2.0) | an integer n between 1 and 99 , inclusive , is to be chosen at random . what is the probability that n ( n + 1 ) will be divisible by 3 ? | "n ( n + 1 ) to be divisible by 3 either n or n + 1 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 : 2 , 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 ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 1 / 2 | e | divide(const_1, const_2) | in a throw of a coin probability of getting a head . | "s = { h , t } e = { h } probability = n ( e ) / n ( s ) = 1 / 2 correct option is e" | a = 1 / 2
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a ) 13 , b ) 15 , c ) 16 , d ) 10 , e ) 18 | d | add(divide(subtract(multiply(floor(divide(40, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), const_1) | how many numbers from 10 to 40 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 . 10 numbers . 10 / 3 = 3 and 40 / 3 = 13 = = > 13 - 3 = 10 . therefore 10 digits d )" | a = 40 / 3
b = math.floor(a)
c = b * 3
d = 10 / 3
e = math.floor(d)
f = e + 1
g = f * 3
h = c - g
i = h / 3
j = i + 1
|
a ) 34 days , b ) 45 days , c ) 46 days , d ) 50 days , e ) none of these | b | multiply(divide(45, 45), 45) | in a dairy farm , 45 cows eat 45 bags of husk in 45 days . in how many days one cow will eat one bag of husk ? | "explanation : less cows , more days ( indirect proportion ) less bags , less days ( direct proportion ) [ cows 1 45 bags 45 1 ] : : 45 : x = > x ∗ 45 ∗ 1 = 45 ∗ 1 ∗ 45 = > x = 45 option b" | a = 45 / 45
b = a * 45
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a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | d | divide(multiply(17, const_100), 85) | how many pieces of 85 cm length can be cut from a rod of 17 meters long ? | number of pieces = 1700 / 85 = 20 the answer is d . | a = 17 * 100
b = a / 85
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a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 128 | e | add(add(const_1, 64), 64) | in the land of oz only one or two - letter words are used . the local language has 64 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | "the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 64 letters is 64 ; the number of 2 letter words ( xx ) that can be made from 64 letters is 64 * 64 , since each x can take 64 values . total : 64 + 64 * 64 . similarly : the number of 1 letter words ( x ) that can be made from 63 letters is 63 ; the number of 2 letter words ( xx ) that can be made from 63 letters is 63 * 63 , since each x can take 63 values . total : 63 + 63 * 63 . the difference is ( 64 + 64 * 64 ) - ( 63 + 63 * 63 ) = 128 . answer : e ." | a = 1 + 64
b = a + 64
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a ) 3 , b ) 3.4 , c ) 3.8 , d ) 4 , e ) 4.5 | b | divide(240, add(divide(240, 8), divide(240, 6))) | while working alone at their constant rates , computer x can process 240 files in 8 hours , and computer y can process 240 files in 6 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ? | "both computers together process files at a rate of 240 / 8 + 240 / 6 = 30 + 40 = 70 files per hour . the time required to process 240 files is 240 / 70 which is about 3.4 hours the answer is b ." | a = 240 / 8
b = 240 / 6
c = a + b
d = 240 / c
|
a ) 4.2 , b ) 4.4 , c ) 3.6 , d ) 5.6 , e ) 5.7 | c | divide(subtract(multiply(6, 3.95), add(multiply(2, 4.4), multiply(2, 3.85))), 2) | the average of 6 no . ' s is 3.95 . the average of 2 of them is 4.4 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 4.4 * 2 ) + ( 3.85 * 2 ) ] = 7.20 . required average = ( 7.2 / 2 ) = 3.6 . c" | a = 6 * 3
b = 2 * 4
c = 2 * 3
d = b + c
e = a - d
f = e / 2
|
a ) 12 , b ) 14 , c ) 16 , d ) 20 , e ) 13.3 | e | divide(add(120, 120), add(divide(120, 20), divide(120, 10))) | two tains of equal lengths take 10 seconds and 20 seconds respectively to cross a telegraph post . if the length of each train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | "sol . speed of the first train = [ 120 / 10 ] m / sec = 12 m / sec . speed of the second train = [ 120 / 20 ] m / sec = 6 m / sec . relative speed = ( 12 + 6 ) = m / sec = 18 m / sec . ∴ required time = ( 120 + 120 ) / 18 secc = 13.3 sec . answer e" | a = 120 + 120
b = 120 / 20
c = 120 / 10
d = b + c
e = a / d
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a ) 9 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | e | multiply(const_60, divide(subtract(120, 90), 120)) | the speed of a train including stoppages is 90 kmph and excluding stoppages is 120 kmph . of how many minutes does the train stop per hour ? | explanation : t = 30 / 120 * 60 = 15 answer : option e | a = 120 - 90
b = a / 120
c = const_60 * b
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a ) 45 % , b ) 47.5 % , c ) 35 % , d ) 32.5 % , e ) 30 % | b | subtract(const_100, multiply(divide(subtract(const_100, 30), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 30 per cent . the two reductions together are equal to a single reduction of | price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 30 percent is reduced which means the new price is merely 70 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 7 / 10 ) p = 21 / 40 p is the new price after both deduction which is 52.5 percent of the original value p . this implies this entire series of deduction is worth having discounted 47.5 % of p . so answer is b = 47.5 % | a = 100 - 30
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
|
a ) 114 , b ) 115 , c ) 116 , d ) 117 , e ) 118 | d | divide(15, 0.128) | a certain industrial loom weaves 0.128 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.128 : 15 : : 1 : x = > 0.128 * x = 15 = > x = 15 / 0.128 = > x = 117 answer : d | a = 15 / 0
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a ) – 16 , b ) – 44 , c ) 10 , d ) 16 , e ) 18 | a | divide(add(add(add(multiply(multiply(20, 2), 2), multiply(20, 2)), 20), 12), multiply(12, 2)) | # 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 = - 128 - - - > p = - 16 , a is the correct answer . | a = 20 * 2
b = a * 2
c = 20 * 2
d = b + c
e = d + 20
f = e + 12
g = 12 * 2
h = f / g
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a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 175 | e | divide(14, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 14 . if n / j = 134.08 , what is value of j ? | "1 ) we know that decimal part of decimal quotient = { remainder / divisor } so 0.08 , the decimal part of the decimal quotient , must equal the remainder , 14 , divided by the divisor j . 0.08 = 14 / j 0.08 * j = 14 j = 14 / 0.08 = 1400 / 8 = 700 / 4 = 350 / 2 = 175 so j = 175 , answer = e ." | a = 4 * 10
b = a + 2
c = 100 + b
d = 134 - 8
e = 14 / d
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a ) 8 , b ) 10 , c ) 15 , d ) 14 , e ) 16 | c | multiply(divide(10, 2), const_3) | a seller of used cars has 10 cars to sell and each of his clients selected 2 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 = 10 / 2 = 5 but since every car is being selected three times so no . of clients must be thrice as well = 5 * 3 = 15 answer : option c" | a = 10 / 2
b = a * 3
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a ) 11 : 14 , b ) 7 : 9 , c ) 17 : 25 , d ) 15 : 26 , e ) 2 : 3 | a | divide(add(const_100, 10), add(const_100, 40)) | two numbers are in respectively 10 % and 40 % more than a third number . the ratio of the two numbers is ? | let the 3 rd number be x then , first number = 110 % of x = 110 x / 100 = 11 x / 10 second number = 140 % of x = 140 x / 100 = 7 x / 5 ratio of first two numbers = 11 x / 10 : 7 x / 5 = 11 : 14 answer is a | a = 100 + 10
b = 100 + 40
c = a / b
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a ) 237 , b ) 270 , c ) 398.8 , d ) 166 , e ) 111 | c | floor(divide(8200, add(20, divide(2.8, const_100)))) | find the number of shares that can be bought for rs . 8200 if the market value is rs . 20 each with brokerage being 2.8 % . | "explanation : cost of each share = ( 20 + 2.8 % of 20 ) = rs . 20.56 therefore , number of shares = 8200 / 20.56 = 398.8 answer : c" | a = 2 / 8
b = 20 + a
c = 8200 / b
d = math.floor(c)
|
a ) 150 m , b ) 278 m , c ) 350 m , d ) 228 m , e ) 282 m | a | subtract(multiply(divide(300, 18), 27), 300) | a 300 meter long train crosses a platform in 27 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 27 = 50 / 3 3 ( x + 300 ) = 1350 è x = 150 m . answer : a" | a = 300 / 18
b = a * 27
c = b - 300
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a ) 15 , b ) 18 , c ) 21 , d ) 12 , e ) 27 | e | subtract(divide(subtract(multiply(6, 34), 7), const_4), 7) | the average age of a family of 6 members is 34 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 34 = 204 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 204 - 42 = 162 therefore , avg age at the birth of youngest member = 162 / 6 = 27 . answer : e" | a = 6 * 34
b = a - 7
c = b / 4
d = c - 7
|
a ) $ 19,250 , b ) $ 18,500 , c ) $ 18,000 , d ) $ 15,850 , e ) $ 22,600 | e | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | the average salary of 15 people in the shipping department at a certain firm is $ 20,000 . the salary of 5 of the employees is $ 20,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 20 k = 300 k 5 emp @ 20 k = 100 k 4 emp @ 16 k = 64 k remaing 6 emp sal = 300 k - 100 k - 64 k = 136 k average = 136 k / 6 = 22600 ans : e" | a = 5 * 4
b = 4 * 4
c = a * b
d = 5 * 5
e = d * 5
f = c - e
g = 4 * 15
h = f - g
i = 2 + 5
j = h / i
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | multiply(5, subtract(4, 3)) | find the value of x in ( x / 5 ) + 3 = 4 . | ( x / 5 ) + 3 = 4 . subtracting both sides with 3 gives x / 5 = 1 multiplying both sides by 5 gives x = 5 answer : e | a = 4 - 3
b = 5 * a
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a ) 10 , b ) 20 , c ) 5 , d ) 15 , e ) 9 | a | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 10))) | if a can do a work in 10 days and a and b can do a same piece of job in 5 days . in how days can do a work b alone ? | a ' s one day work = 1 / 10 a + b = 1 / 5 ( one day work ) b = ( 1 / 5 ) - a = ( 1 / 5 ) - ( 1 / 10 ) = 1 / 10 ( one day work ) b can done a work in 10 days . answer option a | a = 1 / 5
b = 1 / 10
c = a - b
d = 1 / c
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a ) 15 , b ) 17 , c ) 18 , d ) 56 , e ) 20 | e | multiply(add(add(1, 2), 2), divide(1000, subtract(add(200, multiply(50, 2)), multiply(25, 2)))) | pipe a and pipe b fill water into a tank of capacity 1000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated until the tank is filled . how long will it take to fill the tank ? | "tank capacity : 1000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 1000 l tank : 1000 / 250 = 4 time taken to fill : 4 * total time = 4 * 5 = 20 ( option e )" | a = 1 + 2
b = a + 2
c = 50 * 2
d = 200 + c
e = 25 * 2
f = d - e
g = 1000 / f
h = b * g
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a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 27 | a | divide(336, divide(subtract(480, 336), 6)) | a car traveled 480 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 6 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "let the speed in highway be h mpg and in city be c mpg . h = c + 6 h miles are covered in 1 gallon 462 miles will be covered in 462 / h . similarly c miles are covered in 1 gallon 336 miles will be covered in 336 / c . both should be same ( as car ' s fuel capacity does not change with speed ) = > 336 / c = 480 / h = > 336 / c = 480 / ( c + 6 ) = > 336 c + 336 * 6 = 480 c = > c = 336 * 6 / 144 = 14 answer a ." | a = 480 - 336
b = a / 6
c = 336 / b
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | subtract(multiply(add(floor(divide(1056, 26)), const_1), 26), 1056) | what is the least number should be added to 1056 , so the sum of the number is completely divisible by 26 ? | "( 1056 / 26 ) gives remainder 16 10 + 16 = 26 , so we need to add 10 a" | a = 1056 / 26
b = math.floor(a)
c = b + 1
d = c * 26
e = d - 1056
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a ) 148 , b ) 153 , c ) 158 , d ) 163 , e ) 168 | d | subtract(lcm(24, 8), 5) | what is the smallest number which , when increased by 5 , is divisible by 7 , 8 , and 24 ? | "lcm ( 7 , 8,24 ) = 24 x 7 = 168 so the least divisible number is 168 , and the number we are looking for is 168 - 5 = 163 . the answer is d ." | a = math.lcm(24, 8)
b = a - 5
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['a ) 10 %', 'b ) 20 %', 'c ) 17.6 %', 'd ) 30 %', 'e ) 35 %'] | c | multiply(divide(subtract(const_1, divide(subtract(const_100, 15), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | the length of a rectangle is reduced by 15 % . by what % would the width have to be increased to maintain the original area ? | sol . required change = ( 15 * 100 ) / ( 100 - 15 ) = 17.6 % c | a = 100 - 15
b = a / 100
c = 1 - b
d = 100 - 15
e = d / 100
f = c / e
g = f * 100
|
a ) 6 / 9 , b ) 5 / 3 , c ) 2 / 3 , d ) 4 / 3 , e ) 5 / 3 | c | subtract(const_1, divide(6, add(6, const_12))) | abi decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month . if abi ' s savings by the end of the year from these monthly savings were 6 times the amount she spent per month , what should be the fraction of her salary that she spent each month ? | let abi ' s monthly savings = s abi ' s monthly pay = p abi ' s monthly expenditure = p - s abi ' s savings by the end of the year from these monthly savings were six times the amount she spent per month 12 s = 6 * ( p - s ) = > 2 s = p - s = > p = 3 s abi ' s monthly expenditure = p - s = 3 s - s = 2 s fraction of her salary that abi spent each month = 2 s / 3 s = p - s / p = 2 / 3 . answer is c | a = 6 + 12
b = 6 / a
c = 1 - b
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a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 6 , d ) 1 / 8 , e ) 1 / 9 | a | multiply(const_2, subtract(const_1, add(divide(1, 3), divide(1, 2)))) | 3 people who work full - time are to work together on a project , but their total time on the project is to be equivalent to that of only one person working full - time . if one of the people is budgeted for 1 / 2 of his time to the project and a second person for 1 / 3 of her time , what part of the third worker ' s time should be budgeted to this project ? | explanation : one - half ( 1 / 2 ) plus one - third ( 1 / 3 ) equals five - sixths ( 5 / 6 ) , so one - sixth ( 1 / 6 ) of the third worker ' s time should be budgeted to the project in order for the total time to be equivalent to that of one person working full - time . if x 2 - x - 6 = 0 , then ( x + 2 ) ( x - 3 ) = 0 , so x = - 2 or x = 3 . correct answer : a | a = 1 / 3
b = 1 / 2
c = a + b
d = 1 - c
e = 2 * d
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a ) 1 : 9 , b ) 1 : 7 , c ) 1 : 2 , d ) 3 : 5 , e ) 1 : 4 | d | divide(divide(multiply(1, 3), multiply(2, 2)), divide(multiply(3, 4), multiply(2, 5))) | the compound ratio of 1 : 2 , 3 : 2 and 4 : 5 ? | "1 / 2 * 3 / 2 * 4 / 5 = 3 / 5 = 3 : 5 answer : d" | a = 1 * 3
b = 2 * 2
c = a / b
d = 3 * 4
e = 2 * 5
f = d / e
g = c / f
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a ) 45 , b ) 90 , c ) 80 , d ) 75 , e ) 70 | d | divide(add(180, 120), 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 120 questions and secures 180 marks , the number of questions he attempts correctly , is : | "let the number of correct answers be x . number of incorrect answers = ( 120 â € “ x ) . 3 x â € “ ( 120 â € “ x ) = 180 or 4 x = 300 or x = 75 answer : d" | a = 180 + 120
b = 3 + 1
c = a / b
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a ) 8 sec , b ) 16 sec , c ) 18 sec , d ) 14 sec , e ) 12 sec | a | multiply(divide(240, multiply(108, const_1000)), const_3600) | a train 240 m long , running with a speed of 108 km / hr will pass a tree in ? | "speed = 108 * 5 / 18 = 30 m / sec time taken = 240 * 1 / 30 = 8 sec answer : a" | a = 108 * 1000
b = 240 / a
c = b * 3600
|
a ) 35 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | d | divide(subtract(30, 10), subtract(22, 10)) | salad dressing p is made up of 30 % vinegar and 70 % oil , and salad dressing q contains 10 % vinegar and 90 % oil . if the two dressings are combined to produce a salad dressing that is 22 % vinegar , dressing p comprises what percentage of the new dressing ? | "let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.22 0.2 x = 0.12 x = 0.6 = 60 % the answer is d ." | a = 30 - 10
b = 22 - 10
c = a / b
|
a ) $ 440 , b ) $ 460 , c ) $ 480 , d ) $ 500 , e ) $ 520 | d | divide(350, subtract(const_1, divide(30, const_100))) | we had $ 350 left after spending 30 % of the money that we took for shopping . how much money did we start with ? | "let x be the amount of money we started with . 0.7 x = 350 x = 500 the answer is d ." | a = 30 / 100
b = 1 - a
c = 350 / b
|
a ) 28 % , b ) 37 % , c ) 32 % , d ) 36 % , e ) 72 % | b | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(30, const_100))), const_100)) | a baseball card decreased in value 30 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "consider the initial value of the baseball card as $ 100 after first year price = 100 * 0.7 = 70 after second year price = 70 * 0.9 = 63 final decrease = [ ( 100 - 63 ) / 100 ] * 100 = 37 % correct answer - b" | a = 10 / 100
b = 1 - a
c = 30 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
['a ) 9 v â € ™', 'b ) 4 v â € ™', 'c ) v â € ™ / 2', 'd ) v â € ™ / 4', 'e ) v â € ™'] | a | divide(power(12, const_2), power(4, const_2)) | when the oil from a circular cylinder leaked , the formula for the speed of the leak is v = kh ^ 2 , where h was the height of the remaining oil and k was constant . if the height of the cylinder is 4 , the speed of the leak is v â € ™ , when the height of the oil remaining is 12 , what was the speed of the leak , in terms of v â € ™ ? | when h = 4 v = v ' so v ' = k . 16 hence k = v ' / 16 now when h = 12 v = ( v ' / 16 ) . 12 ^ 2 v = v ' . 144 / 16 v = 9 v ' ans : a | a = 12 ** 2
b = 4 ** 2
c = a / b
|
a ) 10 , b ) 12 , c ) 14 , d ) 18 , e ) 20 | d | multiply(log(divide(500000, 1000)), 2) | the population of bacteria culture doubles every 2 minutes . approximately how many minutes will it take for the population to grow from 1000 to 500000 bacteria ? | 2000 ^ 12000 ^ 22000 ^ 32000 ^ 4 . . . . . . . 2000 ^ 9 population increases in this sequence taking 9 * 2 = 18 answer : d | a = 500000 / 1000
b = math.log(a)
c = b * 2
|
a ) 0.08 , b ) 0.8 , c ) 9 , d ) 90 , e ) none of the above | a | divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | what is 15 % of 2 / 3 of 0.8 ? | "the best way to solve these questions is to convert every term into fraction ( 15 / 100 ) * ( 2 / 3 ) * ( 8 / 10 ) = 240 / 3000 = 0.08 option a" | a = 3 + 2
b = a * 2
c = 3 * 4
d = c * 100
e = b * d
f = 3 + 4
g = 3 + 2
h = f * g
i = 3 + 2
j = i * 2
k = h * j
l = e + k
m = 3 + 3
n = l + m
o = 15 * n
p = o / 100
|
a ) 46 m , b ) 60 m , c ) 58 m , d ) 70 m , e ) 80 m | d | divide(add(divide(5300, 26.50), multiply(const_2, 40)), const_4) | length of a rectangular plot is 40 mtr more than its breadth . if the cost of fencin g the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 40 ) metres . perimeter = 5300 / 26.5 m = 200 m . 2 [ ( x + 40 ) + x ] = 200 2 x + 40 = 100 2 x = 60 x = 30 . hence , length = x + 40 = 70 m d" | a = 5300 / 26
b = 2 * 40
c = a + b
d = c / 4
|
a ) 6.6 kg , b ) 6.8 kg , c ) 7.48 kg , d ) 6.9 kg , e ) 7.8 kg | c | divide(multiply(8.5, 880), const_1000) | a envelop weight 8.5 gm , if 880 of these envelop are sent with an advertisement mail . how much wieght ? | "880 * 8.5 7480.0 gm 7.48 kg answer : c" | a = 8 * 5
b = a / 1000
|
a ) 320 , b ) 345 , c ) 375 , d ) 390 , e ) 400 | d | multiply(divide(624, 5), 3) | there are 624 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 624 . hence 3 x + 5 x = 624 solve for x 8 x = 624 x = 78 number of boys 3 x = 3 × 78 = 234 number of girls 5 x = 5 × 78 = 390 d" | a = 624 / 5
b = a * 3
|
a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 7 | e | divide(multiply(5, 2), add(multiply(5, 2), multiply(1, 4))) | at a loading dock , each worker on the night crew loaded 1 / 2 as many boxes as each worker on the day crew . if the night crew has 4 / 5 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 1 / 2 * 4 / 5 x = 2 / 5 x % loaded by day crew = x / ( x + 2 / 5 x ) = 5 / 7 answer e" | a = 5 * 2
b = 5 * 2
c = 1 * 4
d = b + c
e = a / d
|
a ) 6 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | d | add(const_4, add(floor(divide(53, add(const_4, const_3))), const_1)) | company z has 53 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | "say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 53 . now , plug options for x . d x > y as needed . answer : d ." | a = 4 + 3
b = 53 / a
c = math.floor(b)
d = c + 1
e = 4 + d
|
a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | a | subtract(multiply(add(20, const_1), 5), 50) | the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 50 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "20 x + 50 = 21 ( x – 5 ) x = 55 answer : a" | a = 20 + 1
b = a * 5
c = b - 50
|
a ) - 7 , b ) - 4 , c ) - 5 , d ) 1 , e ) 6 | c | divide(subtract(10, add(power(2, 2), multiply(3, 2))), 2) | if 2 is one solution of the equation x ^ 2 + 3 x + k = 10 , where k is a constant , what is the other solution ? | "the phrase “ 2 is one solution of the equation ” means that one value of x is 2 . thus , we first must plug 2 for x into the given equation to determine the value of k . so we have 2 ^ 2 + ( 3 ) ( 2 ) + k = 10 4 + 6 + k = 10 10 + k = 10 k = 0 next we plug 0 into the given equation for k and then solve for x . x ^ 2 + 3 x = 10 x ^ 2 + 3 x – 10 = 0 ( x + 5 ) ( x - 2 ) = 0 x = - 5 or x = 2 thus , - 5 is the other solution . answer c ." | a = 2 ** 2
b = 3 * 2
c = a + b
d = 10 - c
e = d / 2
|
a ) 7 , b ) 8 , c ) 9 , d ) 3 , e ) 0 | e | divide(divide(divide(lcm(2, 5452), 5452), const_4), const_4) | what is the least value of x , so that 2 x 5452 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 5 + 2 = 18 + x least value of x may be ' 0 ' , so that the total 18 + 0 = 18 is divisible by 9 . answer : option e" | a = math.lcm(2, 5452)
b = a / 5452
c = b / 4
d = c / 4
|
a ) 344 , b ) 218 , c ) 200 , d ) 388 , e ) 211 | c | multiply(multiply(multiply(5, const_4.0), 10), 1) | a man bought an article and sold it at a gain of 5 % . if he had bought it at 5 % less and sold it for re 1 less , he would have made a profit of 10 % . the c . p . of the article was | "explanation : let original cost price is x its selling price = ( 105 / 100 ) * x = 21 x / 20 new cost price = ( 95 / 100 ) * x = 19 x / 20 new selling price = ( 110 / 100 ) * ( 19 x / 20 ) = 209 x / 200 [ ( 21 x / 20 ) - ( 209 x / 200 ) ] = 1 = > x = 200 answer : c ) rs 200" | a = 5 * 4
b = a * 10
c = b * 1
|
['a ) b = 2 , l = 9', 'b ) b = 8 , l = 16', 'c ) b = 9 , l = 9', 'd ) b = 4 , l = 14', 'e ) b = 3 , l = 9'] | b | multiply(sqrt(divide(128, const_2)), const_2) | the length of a rectangle is twice the breadth . if the area is 128 cm 2 , determine the length and the breadth . | we are given length l = 2 b and l × b = 128 ∴ 2 b × b = 128 2 b ^ 2 = 128 b ^ 2 = 64 b = ± 8 but breadth must be positive , therefore b = 8 cm , and l = 2 b = 16 cm . answer is b . | a = 128 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | d | add(divide(subtract(120, add(add(add(2, add(2, 2)), add(add(2, 2), 2)), add(add(add(2, 2), 2), 2))), 5), add(add(add(2, 2), 2), 2)) | in a school with 5 classes , each class has 2 students less than the previous class . how many students are there in the largest class if the total number of students at school is 120 ? | "let x be the number of students in the largest class . then x + ( x - 2 ) + ( x - 4 ) + ( x - 6 ) + ( x - 8 ) = 120 5 x - 20 = 120 5 x = 140 x = 28 the answer is d ." | a = 2 + 2
b = 2 + a
c = 2 + 2
d = c + 2
e = b + d
f = 2 + 2
g = f + 2
h = g + 2
i = e + h
j = 120 - i
k = j / 5
l = 2 + 2
m = l + 2
n = m + 2
o = k + n
|
a ) 288 , b ) 277 , c ) 209 , d ) 992 , e ) 612 | d | multiply(power(add(const_1, divide(3, const_100)), 3), 800) | rs . 800 amounts to rs . 920 in 3 years at simple interest . if the interest is increased by 3 % , it would amount to how much ? | "( 800 * 3 * 3 ) / 100 = 72 920 + 72 = 992 answer : d" | a = 3 / 100
b = 1 + a
c = b ** 3
d = c * 800
|
a ) 151 , b ) 331 , c ) 511 , d ) 691 , e ) 871 | a | multiply(18, divide(subtract(add(multiply(15, 2), 1), 7), subtract(18, 15))) | when positive integer n is divided by positive integer p , the quotient is 18 , with a remainder of 7 . when n is divided by ( p + 2 ) , the quotient is 15 and the remainder is 1 . what is the value of n ? | "use the rule dividend = ( integer quotient ) * ( divisor ) + remainder to translate each sentence . the first sentence becomes n = 18 p + 7 . the second equation becomes n = ( p + 2 ) * 15 + 1 , which simplifies to n = 15 p + 31 . these are ordinary simultaneous equations since they both equal n already , let ’ s set them equal and solve for p . 18 p + 7 = 15 p + 31 3 p + 7 = 31 3 p = 24 p = 8 now that we know p = 8 , we can just plug in . the product 15 * 8 is particularly easy to do , without a calculator , by using the “ doubling and halving ” trick . double 15 to get 30 , and take half of 8 to get 4 — - 15 * 8 = 30 * 4 = 120 . so n = 15 ( 8 ) + 31 = 120 + 31 = 151 n = 151 , answer = a ." | a = 15 * 2
b = a + 1
c = b - 7
d = 18 - 15
e = c / d
f = 18 * e
|
a ) 7 : 3 , b ) 5 : 7 , c ) 9 : 7 , d ) 3 : 5 , e ) 11 : 6 | a | divide(multiply(35, 10), multiply(25, 6)) | car a runs at the speed of 35 km / hr & reaches its destination in 10 hr . car b runs at the speed of 25 km / h & reaches its destination in 6 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 36 × 10 = 350 km distance travelled by car b = 25 × 6 = 150 km ratio = 350 / 150 = 7 : 3 a" | a = 35 * 10
b = 25 * 6
c = a / b
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | add(7, divide(multiply(7, subtract(12000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . how many workers are there in the workshop ? | "explanation : let the number of workers = x given that average salary of all the workers = rs . 8000 then , total salary of all workers = 8000 x given that average salary of 7 technicians is rs . 12000 = > total salary of 7 technicians = 7 × 12000 = 84000 count of the rest of the employees = ( x - 7 ) average salary of the rest of the employees = rs . 6000 total salary of the rest of the employees = ( x - 7 ) ( 6000 ) 8000 x = 84000 + ( x - 7 ) ( 6000 ) = > 8 x = 84 + ( x - 7 ) ( 6 ) = > 8 x = 84 + 6 x - 42 = > 2 x = 42 = > x = 42 / 2 = 21 answer : option a" | a = 12000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
|
a ) 320 meter , b ) 225 meter , c ) 230 meter , d ) 235 meter , e ) none of these | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 180) | a 180 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 180 / 9 = 500 / 9 = > x + 180 = 500 = > x = 320 so the length of the train is 320 meters option a" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 180
|
a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | e | subtract(divide(multiply(divide(multiply(add(5, 6), divide(multiply(40, 5), add(5, 3))), 5), 6), add(5, 6)), divide(multiply(40, 3), add(5, 3))) | 40 liters of a mixture is created by mixing liquid p and liquid q in the ratio 5 : 3 . how many liters of liquid q must be added to make the ratio 5 : 6 ? | "let x be the amount of liquid q to be added . ( 3 / 8 ) * 40 + x = ( 6 / 11 ) * ( 40 + x ) 1320 + 88 x = 1920 + 48 x 40 x = 600 x = 15 the answer is e ." | a = 5 + 6
b = 40 * 5
c = 5 + 3
d = b / c
e = a * d
f = e / 5
g = f * 6
h = 5 + 6
i = g / h
j = 40 * 3
k = 5 + 3
l = j / k
m = i - l
|
a ) 7 , b ) 11 , c ) 14 , d ) 16 , e ) 20 | b | subtract(subtract(24, 7), const_2) | the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 100 inclusive is the digital sum a multiple of 7 ? | "is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 95 11 ways . . b" | a = 24 - 7
b = a - 2
|
a ) $ 1000 , b ) $ 1200 , c ) $ 1500 , d ) $ 1800 , e ) $ 2000 | c | multiply(multiply(subtract(4, 3), 500), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is d ' s share ? | let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 d ' s share = 3 x = 3 * $ 500 = $ 1500 the answer is c . | a = 4 - 3
b = a * 500
c = b * 3
|
a ) 17 / 36 , b ) 36 / 17 , c ) 17 / 6 , d ) 17 / 4 , e ) 85 / 18 | e | subtract(divide(add(multiply(const_10, const_2), 2), 3), divide(add(const_10, const_1), 4)) | what is 2 2 / 3 - 1 1 / 4 divided by 1 / 2 - 1 / 5 ? | 2 2 / 3 - 1 1 / 4 = 8 / 3 - 5 / 4 = ( 32 - 15 ) / 12 = 17 / 12 1 / 2 - 1 / 5 = ( 5 - 2 ) / 10 = 3 / 10 so 17 / 12 / 3 / 10 = 17 / 12 * 10 / 3 = 85 / 18 answer - e | a = 10 * 2
b = a + 2
c = b / 3
d = 10 + 1
e = d / 4
f = c - e
|
a ) 16000 , b ) 20000 , c ) 15000 , d ) 18000 , e ) 17000 | b | multiply(divide(subtract(24000, divide(multiply(24000, 10), const_100)), add(const_100, 8)), const_100) | mohit sold an article for $ 24000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 20000 - 10 / 100 ( 20000 ) = 20000 - 2000 = $ 21600 = > 1.08 x = 21600 = > x = 20000 b | a = 24000 * 10
b = a / 100
c = 24000 - b
d = 100 + 8
e = c / d
f = e * 100
|
a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | e | multiply(subtract(64, 10), 10) | what is the greatest positive integer x such that 4 ^ x is a factor of 64 ^ 10 ? | "64 ^ 10 = ( 4 ^ 3 ) ^ 10 = 4 ^ 30 answer : e" | a = 64 - 10
b = a * 10
|
a ) 4.05 , b ) 4.50045 , c ) 4.501 , d ) 4.5045 , e ) 4.5 | e | multiply(divide(9.009, 2.002), const_100) | 9.009 / 2.002 = | "9.009 / 2.002 = 9009 / 2002 = 9 ( 1001 ) / 2 ( 1001 ) = 9 / 2 = 4.5 the answer is e ." | a = 9 / 9
b = a * 100
|
a ) - 29 , b ) - 19 , c ) 20 , d ) 29 , e ) 39 | c | subtract(59, subtract(45, add(subtract(28, 37), 15))) | if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 59 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 59 = > 45 - [ 28 - { 37 - 15 + * } ] = 59 45 - [ 28 - 37 + 15 - * ] = 59 = > 45 [ 43 - 37 - * ] = 59 45 - [ 6 - * ] = 59 = > 45 - 6 + * = 59 39 + * = 59 = > * = 59 - 39 = 20 answer : c" | a = 28 - 37
b = a + 15
c = 45 - b
d = 59 - c
|
a ) 15 , b ) 20 , c ) 30 , d ) 40 , e ) 45 | d | subtract(subtract(subtract(200, 80), 60), divide(subtract(subtract(200, 80), 60), 3)) | a marketing firm determined that , of 200 households surveyed , 80 used neither brand w nor brand b soap , 60 used only brand w soap , and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 households surveyed used both brands of soap ? | solution for soap w and soap b ( d ) 40 | a = 200 - 80
b = a - 60
c = 200 - 80
d = c - 60
e = d / 3
f = b - e
|
a ) 18 , b ) 4 , c ) 6 , d ) 8 , e ) 19 | b | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 42)) | if the wheel is 42 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 42 * x = 1056 = > x = 4 answer : b" | a = 3 * 100
b = 1 * 10
c = a + b
d = c + 4
e = d / 100
f = 2 * e
g = f * 42
h = 1056 / g
|
a ) 22 , b ) 29 , c ) 19 , d ) 5 , e ) 2 | b | add(add(multiply(2, 11), 5), 2) | find the total number of prime factors in the expression ( 4 ) 11 x ( 7 ) 5 x ( 11 ) 2 . | "( 4 ) 11 x ( 7 ) 5 x ( 11 ) 2 = ( 2 x 2 ) 11 x ( 7 ) 5 x ( 11 ) 2 = 211 x 211 x 75 x 112 = 222 x 75 x 112 total number of prime factors = ( 22 + 5 + 2 ) = 29 . answer b 29" | a = 2 * 11
b = a + 5
c = b + 2
|
a ) 230 , b ) 234 , c ) 250 , d ) 547 , e ) 484 | b | floor(add(multiply(subtract(300, 125), divide(400, 300)), const_1)) | the average salary of workers in an industry is rs . 300 the average salary of technicians being rs . 400 and that of non - technicians being rs . 125 . what is the total number of workers ? | 400 125 \ / 300 / \ 175 100 7 : 4 7 - > 150 11 - > ? = > 234 answer : b | a = 300 - 125
b = 400 / 300
c = a * b
d = c + 1
e = math.floor(d)
|
a ) 6 only , b ) 6 and 12 , c ) 12 only , d ) 18 only , e ) 20 only | b | add(multiply(6, const_100), multiply(2, 6)) | if n is a natural number , t hen 6 n ^ 2 + 6 n is always divisible by ? | "6 n ^ 2 + 6 n = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . answer is b" | a = 6 * 100
b = 2 * 6
c = a + b
|
a ) 74 , b ) 84 , c ) 94 , d ) 49 , e ) 47 | a | divide(add(multiply(10, subtract(10, 20)), multiply(15, subtract(10, 30))), add(10, 15)) | a man buys 10 lts of liquid which contains 20 % of the liquid and the rest is water . he then mixes it with 15 lts of another mixture with 30 % of liquid . what is the % of water in the new mixture ? | "20 % in 10 lts is 2 . so water = 10 - 2 = 8 lts . 30 % of 15 lts = 4.5 . so water in 2 nd mixture = 15 - 4.5 = 10.5 lts . now total quantity = 10 + 15 = 25 lts . total water in it will be 8 + 10.5 = 18.5 lts . % of water = ( 100 * 18.5 ) / 25 = 74 answer : a" | a = 10 - 20
b = 10 * a
c = 10 - 30
d = 15 * c
e = b + d
f = 10 + 15
g = e / f
|
a ) 12 , b ) 20 , c ) 30 , d ) 15 , e ) 24 | e | multiply(multiply(divide(12, 4), const_2), 4) | a man can buy a new house in 4 years with all of his earnings . but he spend the same amount in household expenses in 8 years and for savings in 12 years . how many time needed for buying the house ? | his savings in 1 year = 1 / 4 - ( 1 / 8 + 1 / 12 ) = 1 / 24 he can buy a new house in 24 years answer : e | a = 12 / 4
b = a * 2
c = b * 4
|
a ) 5 , b ) 5 1 / 2 , c ) 7 1 / 2 , d ) 6 , e ) 9 1 / 2 | d | divide(multiply(6, 3), subtract(6, 3)) | a man can do a piece of work in 6 days , but with the help of his son , he can finish it in 3 days . in what time can the son do it alone ? | "son ' s 1 day work = 1 / 3 - 1 / 6 = 1 / 6 son alone can do the work in 6 days = 6 days answer is d" | a = 6 * 3
b = 6 - 3
c = a / b
|
a ) 3 , b ) 2 , c ) 1 , d ) 5 , e ) 6 | c | subtract(add(subtract(784839, const_1), 10), 784839) | what is the least whole number that should be added to 784839 if it is to be divisible by 10 ? | "a number is divisible by 10 if the last digit is 0 . here , 784839 = 9 , ( last digit is not 0 . 1 must be added to 784839 to make it divisible by 10 c" | a = 784839 - 1
b = a + 10
c = b - 784839
|
a ) 3 / 32 , b ) 7 / 64 , c ) 11 / 105 , d ) 15 / 127 , e ) 19 / 134 | c | multiply(divide(12, add(add(12, 16), 8)), divide(subtract(12, const_1), subtract(add(add(12, 16), 8), const_1))) | there are 12 slate rocks , 16 pumice rocks , and 8 granite rocks randomly distributed in a certain field . if 2 rocks are chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | 12 / 36 * 11 / 35 = 11 / 105 the answer is c . | a = 12 + 16
b = a + 8
c = 12 / b
d = 12 - 1
e = 12 + 16
f = e + 8
g = f - 1
h = d / g
i = c * h
|
a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6 | a | divide(multiply(4, 12), add(4, 12)) | a can do a work in 4 days . b can do the same work in 12 days . if both a & b are working together in how many days they will finish the work ? | a rate = 1 / 4 b rate = 1 / 16 ( a + b ) rate = ( 1 / 4 ) + ( 1 / 12 ) = 1 / 3 a & b finish the work in 3 days correct option is a | a = 4 * 12
b = 4 + 12
c = a / b
|
a ) 20 , b ) 31 , c ) 42 , d ) 43 , e ) 64 | d | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | how many positive integers less than 254 are multiple of 4 but not multiples of 6 ? | "252 / 4 = 63 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 63 - 20 = 43 answer : d" | a = 4 + 4
b = a - 1
c = math.factorial(b)
d = math.factorial(4)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 1600 , b ) 80 , c ) 160 , d ) 180 , e ) 240 | a | add(multiply(const_100, 4), const_100) | how many 4 - digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number ? | "prime digits 2 , 3,5 and 7 . three digit numbers _ _ _ 1 st place can be filled in 4 ways 2 nd place can be filled in 10 ways 3 rd place can be filled in 10 ways 3 rd place can be filled in 4 ways total = 4 * 10 * 10 * 4 = 1600 ans : a" | a = 100 * 4
b = a + 100
|
a ) 16000 , b ) 14567 , c ) 13787 , d ) 13456 , e ) none of these | c | divide(multiply(multiply(28, const_100), multiply(13, const_100)), multiply(22, 12)) | a courtyard is 28 meter long and 13 meter board is to be paved with bricks of dimensions 22 cm by 12 cm . the total number of bricks required is : | explanation : number of bricks = courtyard area / 1 brick area = ( 2800 ã — 1300 / 22 ã — 12 ) = 13787 option c | a = 28 * 100
b = 13 * 100
c = a * b
d = 22 * 12
e = c / d
|
a ) 53 , b ) 52 , c ) 51 , d ) 46 , e ) 49 | d | add(subtract(125, 80), const_1) | andy solves problems 80 to 125 inclusive in a math exercise . how many problems does he solve ? | "125 - 80 + 1 = 52 ' d ' is the answer" | a = 125 - 80
b = a + 1
|
a ) 20 , b ) 23 , c ) 21 , d ) 25 , e ) 26 | c | subtract(39, add(add(8, const_2), 8)) | set a of 8 positive integers may have the same element and have 39 . and set b of 8 positive integers must have different elements and have 39 . when m and n are the greatest possible differences between 39 and other elements ’ sums in set a and set b , respectively , m - n = ? | this is maximum - minimum . hence , 39 - ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) = 32 and 39 - ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) = 11 . so , 32 - 11 = 21 . the correct answer is c . | a = 8 + 2
b = a + 8
c = 39 - b
|
a ) 800 , b ) 500 , c ) 900 , d ) 1200 , e ) none | d | multiply(divide(multiply(multiply(3.6, 0.48), 2.50), multiply(multiply(0.12, 0.09), 0.5)), 1.5) | find the value of 1.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | "answer 1.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 1.5 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 1.5 x 4 x 4 x 50 = 1200 correct option : d" | a = 3 * 6
b = a * 2
c = 0 * 12
d = c * 0
e = b / d
f = e * 1
|
a ) $ 120 , b ) $ 100 , c ) $ 91 , d ) $ 83 , e ) $ 69 | d | multiply(100, divide(100, add(100, 20))) | a shopkeeper sold an article at $ 100 with 20 % profit . then find its cost price ? | "cost price = selling price * 100 / ( 100 + profit ) c . p . = 100 * 100 / 120 = $ 83 ( approximately ) answer is d" | a = 100 + 20
b = 100 / a
c = 100 * b
|
a ) 1 : 9 , b ) 1 : 7 , c ) 1 : 2 , d ) 2 : 5 , e ) 1 : 4 | d | divide(divide(multiply(1, 3), multiply(3, 2)), divide(multiply(3, 4), multiply(2, 5))) | the compound ratio of 1 : 3 , 3 : 2 and 4 : 5 ? | "1 / 3 * 3 / 2 * 4 / 5 = 2 / 5 = 2 : 5 answer : d" | a = 1 * 3
b = 3 * 2
c = a / b
d = 3 * 4
e = 2 * 5
f = d / e
g = c / f
|
a ) 250 , b ) 500 , c ) 450 , d ) 500 , e ) 520 | b | divide(550, add(const_1, divide(10, const_100))) | a number increased by 10 % gives 550 . the number is | "formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 10 % = 110 % 110 % - - - - - - - > 550 ( 110 × 5 = 550 ) 100 % - - - - - - - > 500 ( 100 × 5 = 500 ) b )" | a = 10 / 100
b = 1 + a
c = 550 / b
|
a ) 5 / 2 , b ) 10 / 3 , c ) 7 / 2 , d ) 14 / 3 , e ) 11 / 2 | e | subtract(divide(add(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2)), divide(subtract(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2))) | by how much does the larger root of the equation 2 z ^ 2 + 5 z = 12 exceed the smaller root ? | "for 2 z ^ 2 + 5 z = 12 roots are [ - 5 + sqrt ( 25 + 96 ) ] / 4 or [ - 5 - sqrt ( 25 + 96 ) ] / 4 = 1.5 or - 4 hence larger root 1.5 is 1.5 - ( - 4 ) = 5.5 = 11 / 2 greater than smaller root ( - 4 ) . hence option ( e ) ." | a = 5 ** 2
b = 2 * 12
c = b * 4
d = a + c
e = math.sqrt(d)
f = 5 + e
g = 2 * 2
h = f / g
i = 5 ** 2
j = 2 * 12
k = j * 4
l = i + k
m = math.sqrt(l)
n = 5 - m
o = 2 * 2
p = n / o
q = h - p
|
a ) 125 , b ) 977 , c ) 289 , d ) 1077 , e ) 111 | a | multiply(5, multiply(5, 5)) | . star question : if f ( 1 ) = 4 and f ( x + y ) = f ( x ) + f ( y ) + 7 xy + 4 , then f ( 2 ) + f ( 5 ) = ? | let x = 1 and y = 1 f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) + 7 x 1 x 1 + 4 ⇒ ⇒ f ( 2 ) = 19 let x = 2 and y = 2 f ( 2 + 2 ) = 19 + 19 + 7 x 2 x 2 + 4 ⇒ ⇒ f ( 4 ) = 70 let x = 1 and y = 4 f ( 1 + 4 ) = 4 + 70 + 28 + 4 = 106 f ( 2 ) + f ( 5 ) = 125 answer : a | a = 5 * 5
b = 5 * a
|
a ) rs . 4000 , b ) rs . 3750 , c ) rs . 5000 , d ) rs . 5500 , e ) rs . 6500 | b | divide(multiply(300, const_100), subtract(const_100, add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(30, const_100))))) | a man saves 30 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 30 % , he is only able to save rs . 300 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 70 savings = rs . 30 present expenditure 70 + 70 * ( 30 / 100 ) = rs . 91 present savings = 100 – 91 = rs . 8 if savings is rs . 8 , salary = rs . 100 if savings is rs . 300 , salary = 100 / 8 * 300 = 3750 answer : b" | a = 300 * 100
b = 100 - 30
c = 100 - 30
d = 30 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) 1 / 8 , b ) 1 / 4 , c ) 2 / 3 , d ) 3 / 4 , e ) 2 | b | multiply(subtract(1, divide(1, 3)), subtract(1, divide(1, 8))) | the probability that a man will be alive for 10 more yrs is 1 / 3 & the probability that his wife will alive for 10 more yrs is 5 / 8 . the probability that none of them will be alive for 10 more yrs , is | "sol . required probability = pg . ) x p ( b ) = ( 1 — d x ( 1 — i ) = : x 1 = 1 / 4 ans . ( b )" | a = 1 / 3
b = 1 - a
c = 1 / 8
d = 1 - c
e = b * d
|
a ) 16 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | subtract(add(add(20, 40), 60), add(add(multiply(8, const_3), 10), 70)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 8 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 8 = 32 sum of second list = 32 * 3 = 96 therefore the number = 96 - 80 = 16 answer : a" | a = 20 + 40
b = a + 60
c = 8 * 3
d = c + 10
e = d + 70
f = b - e
|
a ) 5 , b ) 5 / 4 , c ) 4 / 5 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 1 * / 4 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . based on this definition . . . . if . . . . k = 2 k * = ( 1 / 1 ) ( 1 / 2 ) = 1 / 2 if . . . . k = 3 k * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) = 1 / 6 we ' re asked to find the value of 5 * / 4 * now that we know how the symbol ' works ' , solving this problem should n ' t be too difficult . you can actually choose to do the math in a couple of different ways . . . . 5 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) do n ' t calculate this just yet though . . . . since we ' re dividing by 4 * , many of those fractions will ' cancel out . ' 4 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) we ' re looking for the value of : ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) / ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) since the first four fraction in the numerator and denominator cancel out , we ' re left with just one fraction : 4 / 5 a" | a = 1 / 3
b = a / 3
c = 1 + 4
d = b / c
|
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