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 ) 2 , b ) 3 , c ) 4 , d ) 7 , e ) 6 | b | divide(124, 22) | what is the 10 th digit to the right of the decimal point in the decimal equivalent of 124 / 22 ? | "124 / 22 = 5.6363 . . . . 63 is non - terminating repeating decimal . the 10 th digit to the right of decimal point will be 3 . answer b" | a = 124 / 22
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a ) 100 , b ) 230 , c ) 300 , d ) 400 , e ) 500 | b | subtract(800, add(add(170, subtract(500, 170)), subtract(400, subtract(500, 170)))) | - - - - - - - - - - - - - - - - yes - - - - - - - - - no - - - - unsure subject m - - - - 500 - - - - - - - - 200 - - - - - 100 subject r - - - - 400 - - - - - - - - 100 - - - - - 300 a total of 800 students were asked whether they found two subjects , m and r , interesting . each answer was either yes or no or unsure , and the numbers of students who gave these answers are listed in the table above . if 170 students answered yes only for subject m , how many of the students did not answer yes for either subject ? | since 170 students answered yes only for subject m , then the remaining 330 students who answered yes for subject m , also answered yes for subject r . so , 330 students answered yes for both subjects . if 320 students answered yes for both subjects , then 400 - 330 = 70 students answered yes only for subject r . so , we have that : 200 students answered yes only for subject m ; 70 students answered yes only for subject r ; 300 students answered yes for both subjects ; therefore 800 - ( 200 + 70 + 300 ) = 230 students did not answer yes for either subject . answer : b . | a = 500 - 170
b = 170 + a
c = 500 - 170
d = 400 - c
e = b + d
f = 800 - e
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a ) 5 , b ) 8 , c ) 7 , d ) 10 , e ) 15 | d | divide(multiply(10, 5), subtract(10, 5)) | a man can do a piece of work in 10 days , but with the help of his son , he can finish it in 5 days . in what time can the son do it alone ? | "son ' s 1 day work = 1 / 5 - 1 / 10 = 1 / 10 son alone can do the work in 10 days answer is d" | a = 10 * 5
b = 10 - 5
c = a / b
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a ) 850 , b ) 1,100 , c ) 1,700 , d ) 2,200 , e ) 3,400 | d | divide(add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), add(add(divide(subtract(add(200, const_3600), multiply(add(add(const_1, const_2), const_3), 200)), 4), 200), 200)), subtract(multiply(const_2, const_10), const_1)) | alan buys 4 tvs , a 26 inch , a 28 inch , a 30 inch , and a 32 inch , for his new house . each tv costs $ 200 more than the size below it . alan spends a total of $ 4,400 . how much would he have spent if he had bought only the 28 inch and 30 inch tvs ? | "assume the cost of the least sized ( 26 inch ) tv = x cost of 28 inches tv = x + 200 cost of 30 inches tv = x + 400 cost of 32 inches tv = x + 600 total cost = 4 x + 1200 = 4400 therefore x = 3200 / 4 = 800 price of 28 inch + 30 inch = 1000 + 1200 = 2200 option d" | a = 200 + 3600
b = 1 + 2
c = b + 3
d = c * 200
e = a - d
f = e / 4
g = f + 200
h = 200 + 3600
i = 1 + 2
j = i + 3
k = j * 200
l = h - k
m = l / 4
n = m + 200
o = n + 200
p = g + o
q = 2 * 10
r = q - 1
s = p / r
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a ) 1345 , b ) 1250 , c ) 897 , d ) 900 , e ) 1635 | c | multiply(divide(subtract(750, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 750 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 750 = 6 x + 15 5 x = 735 x = 147 large number = 147 + 1365 = 897 c" | a = 750 - 15
b = 6 - 1
c = a / b
d = c * 6
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a ) 36 days , b ) 32 days , c ) 34 days , d ) 42 days , e ) 49 days | a | inverse(subtract(inverse(4), add(inverse(6), inverse(18)))) | a man , a woman and a boy can together complete a piece of work in 4 days . if a man alone can do it in 6 days and a boy alone in 18 days , how long will a woman take to complete the work ? | "explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 4 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 18 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 18 = 2 / 9 therefore , 1 woman β s 1 day β s work = 1 / 4 β 2 / 9 = 1 / 36 therefore , the woman alone can finish the work in 36 days . answer : option a" | a = 1/(4)
b = 1/(6)
c = 1/(18)
d = b + c
e = a - d
f = 1/(e)
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a ) $ 11.73 , b ) $ 12.72 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(210.00, add(const_100, 15)), const_100), divide(const_1, 15)) | the price of lunch for 15 people was $ 210.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 210 is for 115 then we have to calculate for 100 for 115 210 for 100 x so by cross multiplication we get 115 x = 100 * 210 = > x = 100 * 210 / 110 by simplifying we get x as 190.90 which is the price of lunch before gratuity so the gratuity is 19.10 so as the question ask the average price person excluding gratuity is 190.90 / 15 = 12.72 so our answer is b )" | a = 100 + 15
b = 210 / 0
c = b * 100
d = 1 / 15
e = c * d
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a ) 2 , b ) 6 , c ) 9 , d ) 11 , e ) 12 | b | add(multiply(subtract(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), multiply(floor(divide(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), const_10)), const_10)), 2), floor(divide(subtract(subtract(subtract(190, multiply(6, const_10)), 82), 7), const_10))) | 82 a 7 + 6 b ____ 190 if a and b represent positive single digits in the correctly worked computation above , what is the value of a + 2 b ? | adding the digits in unit ' s place , 2 + 7 + b = 10 ( since a and b are positive single digits ) = > b = 1 now adding the digits in ten ' s place , 8 + a + 6 + 1 = 19 ( 1 has been carried over from unit ' s place addition ) = > a = 4 a + 2 b = 4 + 2 * 1 = 6 answer b | a = 6 * 10
b = 190 - a
c = b - 82
d = c - 7
e = 6 * 10
f = 190 - e
g = f - 82
h = g - 7
i = h / 10
j = math.floor(i)
k = j * 10
l = d - k
m = l * 2
n = 6 * 10
o = 190 - n
p = o - 82
q = p - 7
r = q / 10
s = math.floor(r)
t = m + s
|
a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | d | divide(add(add(multiply(85, 6), multiply(60, 4)), 30), add(6, 4)) | the average expenditure of a labourer for 6 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 Γ 85 ) β debt = 510 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 780 average monthly income = 780 Γ· 10 = 78 answer d" | a = 85 * 6
b = 60 * 4
c = a + b
d = c + 30
e = 6 + 4
f = d / e
|
a ) 3 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | a | subtract(40, add(13, 24)) | in a certain alphabet , 13 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | "we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 13 + 24 = 37 ; next , we find the total # of letters without line : 40 - 37 = 3 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 3 - 0 = 3 . a" | a = 13 + 24
b = 40 - a
|
a ) 18 % , b ) 4 % , c ) 32 % , d ) 12 % , e ) 52 % | b | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 10)), const_100)) | the tax on a commodity is diminished by 10 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 80 * 120 = 9600 10000 - - - - - - - 400 100 - - - - - - - ? = 4 % answer : b" | a = 100 + 20
b = 100 - 10
c = a * b
d = c / 100
e = 100 - d
|
a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | c | add(add(const_4, const_3), const_2) | what is the units digit of ( 493 ) ( 915 ) ( 381 ) ( 756 ) | "just multiply the digits in the units place for each term and you will get the answer . it should be 0 . you got a 5 as a unit digit and an even number term . so the multiplication of this will definitely yield a 0 . answer has to be 0 . i also tried it using the calculator and the answer is 4 . imo c ." | a = 4 + 3
b = a + 2
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a ) 5 hours , b ) 6 hours , c ) 1 hours , d ) 10 hours , e ) 12 hours | c | divide(subtract(12, 10), 2) | two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 2 km apart , if they walk in the same direction ? | to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 2 = 1 hours answer is c | a = 12 - 10
b = a / 2
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a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 5 | c | add(subtract(add(10, 18), subtract(30, 3)), subtract(18, 10)) | of 30 applicants for a job , 10 had at least 4 years ' experience , 18 had degrees , and 3 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | c . 9 30 - 3 = 27 27 - 10 - 18 = - 9 then 9 are in the intersection between 4 years experience and degree . answer c | a = 10 + 18
b = 30 - 3
c = a - b
d = 18 - 10
e = c + d
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['a ) 8.76', 'b ) 8.79', 'c ) 8.75', 'd ) 8.74', 'e ) 8.72'] | c | divide(subtract(square_area(20), circle_area(divide(20, const_2))), const_10) | a paper is in a square form whose one side is 20 cm . two semi circles are drawn on its opposites as diameters . if these semi circles are cut down what is the area of the remaining paper ? | ( 5 * 3.5 ) / 2 = 8.75 answer : c | a = square_area - (
b = 20 / 2
c = a / circle_area
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a ) 54 , b ) 58 , c ) 63 , d ) 71 , e ) 92 | b | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 34)), 34) | of 70 players on a football team , 34 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 34 rest = 70 - 34 = 36 left handed = 36 / 3 = 12 right handed = 24 if all thrower are right handed then total right handed is 34 + 24 = 58 so b . 58 is the right answer" | a = 1 / 3
b = 1 - a
c = 70 - 34
d = b * c
e = d + 34
|
a ) 31 , b ) 0.31 , c ) 0.0031 , d ) 0.00031 , e ) 3.1 e - 05 | a | multiply(divide(multiply(multiply(multiply(10, 10), subtract(multiply(10, 10), const_1)), divide(31, subtract(multiply(10, 10), const_1))), const_1000), 10) | if the digits 31 in the decimal 0.00031 repeat indefinitely , what is the value of ( 10 ^ 5 - 10 ^ 3 ) ( 0.00031 ) ? | "99 * 0.31 = 30.69 approx . 31 answer : a" | a = 10 * 10
b = 10 * 10
c = b - 1
d = a * c
e = 10 * 10
f = e - 1
g = 31 / f
h = d * g
i = h / 1000
j = i * 10
|
a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 16 / 21 | e | divide(subtract(40, 24), subtract(40, 19)) | a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 24 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 - 0.21 x ) / 10 = 0.24 - - - > x = 160 / 21 ml . per the question , you need to find the x ml removed as a ratio of the initial volume - - - > ( 160 / 21 ) / 10 = 16 / 21 . hence , e is the correct answer ." | a = 40 - 24
b = 40 - 19
c = a / b
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a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 30 | e | divide(subtract(power(30, const_2), 840), const_2) | if the sum of two numbers is 30 and the sum of their squares is 840 , then the product of the numbers is | sol . let the numbers be x and y . then , ( x + y ) = 30 and x 2 + y 2 = 840 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 30 ) 2 - 840 = 900 - 840 = 60 xy = 30 . answer e | a = 30 ** 2
b = a - 840
c = b / 2
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a ) 298 km , b ) 278 km , c ) 278 km , d ) 300 km , e ) 267 km | d | multiply(inverse(add(inverse(multiply(const_2, 30)), inverse(multiply(25, const_2)))), 11) | pavan travelled for 11 hours . he covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph . find the distance travelled by pavan . | "let the distance travelled be x km . total time = ( x / 2 ) / 30 + ( x / 2 ) / 25 = 11 = > x / 60 + x / 50 = 11 = > ( 5 x + 6 x ) / 300 = 11 = > x = 300 km answer : d" | a = 2 * 30
b = 1/(a)
c = 25 * 2
d = 1/(c)
e = b + d
f = 1/(e)
g = f * 11
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(2, const_3), add(const_3, const_4)) | if 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) , then | a | + | b | + | c | + | d | | this is the hard one , definitely a 700 + level question . we need numbers a , b , c , and d such that 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) this means that ac = 6 , bd = β 12 , and ad + bc = 1 . the a & c pair could be ( 1 , 6 ) or ( 2 , 3 ) , in some order . the absolute values of the b & d pair could be ( 1 , 12 ) or ( 2 , 6 ) or ( 3 , 4 ) , and of course , in each case , one of the two would have to be negative . after some trial and error , we find : 6 x ^ 2 + x - 12 = ( 2 x + 3 ) ( 3 x - 4 ) thus , we see : | a | + | b | + | c | + | d | = 2 + 3 + 3 + 4 = 12 answer = b | a = 2 + 3
b = 3 + 4
c = a + b
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a ) 9.2 , b ) 10.2 , c ) 9.8 , d ) 9.6 , e ) 10.0 | d | multiply(add(20, 4), divide(24, const_60)) | the speed of a boat in still water in 20 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 24 minutes is : | "speed downstream = ( 20 + 4 ) = 24 kmph time = 24 minutes = 24 / 60 hour = 2 / 5 hour distance travelled = time Γ speed = 2 / 5 Γ 24 = 9.6 km answer is d ." | a = 20 + 4
b = 24 / const_60
c = a * b
|
a ) 2 / 7 , b ) 3 / 5 , c ) 3 / 13 , d ) 1 / 4 , e ) 7 / 16 | c | divide(multiply(choose(const_4.0, const_2), choose(add(const_3.0, 5), const_1)), choose(add(add(3, 5), 7), 3)) | a bag contains 3 red , 5 yellow and 7 green balls . 3 balls are drawn randomly . what is the probability that the balls drawn contain balls of different colours ? | "total number of balls = 3 + 5 + 7 = 15 n ( s ) = 15 c 3 = 455 n ( e ) = 3 c 1 * 5 c 1 * 7 c 1 = 105 probability = 105 / 455 = 3 / 13 answer is c" | a = math.comb(4, 0)
b = 3 + 0
c = math.comb(b, 1)
d = a * c
e = 3 + 5
f = e + 7
g = math.comb(f, 3)
h = d / g
|
a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | c | divide(add(multiply(100, const_4), multiply(divide(multiply(const_4, 100), add(const_1, const_4)), const_4)), add(100, divide(multiply(const_4, 100), add(const_1, const_4)))) | two cars are driving toward each other . the first car is traveling at a speed of 100 km / h , which is 25 % faster than the second car ' s speed . if the distance between the cars is 720 km , how many hours will it take until the two cars meet ? | the speed of the first car is 100 km / h . the speed of the second car is 100 / 1.25 = 80 km / h . the two cars complete a total of 180 km each hour . the time it takes the cars to meet is 720 / 180 = 4 hours . the answer is c . | a = 100 * 4
b = 4 * 100
c = 1 + 4
d = b / c
e = d * 4
f = a + e
g = 4 * 100
h = 1 + 4
i = g / h
j = 100 + i
k = f / j
|
a ) 6 m , b ) 5 m , c ) 4 m , d ) 3 m , e ) none of the above | b | divide(20, subtract(5, 1)) | a tiger is chasing its prey at a constant speed . it entire body passes a blade of grass in 1 second . it then runs above a fallen tree trunk that is 20 meters long in 5 seconds . what is the length of the tiger ? | sol . let the length of the tiger be x metres and its speed be y m / sec . then , x / y = 1 β y = x β΄ ( x + 20 ) / 5 = x β x = 5 m . answer b | a = 5 - 1
b = 20 / a
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a ) 5 % , b ) 7.5 % , c ) 10 % , d ) 12.5 % , e ) 15 % | d | subtract(subtract(25, 10), divide(25, 10)) | a couple who own an appliance store discover that if they advertise a sales discount of 10 % on every item in the store , at the end of one month the number of total items sold increases 25 % . their gross income from sales for one month increases by what percent ? | "let p be the original price and let x be the number of items sold originally . the original income is p * x . after the changes , the income is 0.9 p * 1.25 x = 1.125 * ( p * x ) , an increase of 12.5 % . the answer is d ." | a = 25 - 10
b = 25 / 10
c = a - b
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a ) 8 , b ) 9 , c ) 6 , d ) 2 , e ) 1 | c | multiply(multiply(subtract(const_1, divide(5, 8)), 10), divide(8, 5)) | 5 / 8 th of a job is completed in 10 days . if a person works at the same pace , how many days will he take to complete the job ? | explanation : solution : it is given that 5 / 8 th of the work is completed in 10 days . = > remaining work = 3 / 8 th of total applying unitary method : total work will be completed in 10 * 8 / 5 days = > it takes 16 days to complete total work = > hence , remaining work days = 16 - 10 = 6 days answer : c | a = 5 / 8
b = 1 - a
c = b * 10
d = 8 / 5
e = c * d
|
a ) 4 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | b | add(const_2, const_2) | find the number of different prime factors of 12800 | "explanation : l . c . m of 12800 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 2 , 5 number of different prime factors is 2 . answer : option b" | a = 2 + 2
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a ) 104 , b ) 114 , c ) 315 , d ) 200 , e ) 335 | d | multiply(divide(add(30, 10), add(const_1, const_1)), subtract(divide(add(30, 10), add(const_1, const_1)), 10)) | the sum of two numbers is 30 and their difference is 10 . find their product . | let the numbers be x and y . then x + y = 30 and x - y = 10 x = 20 ; y = 10 xy = 20 * 10 = 200 answer : d | a = 30 + 10
b = 1 + 1
c = a / b
d = 30 + 10
e = 1 + 1
f = d / e
g = f - 10
h = c * g
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a ) 2490 and 4150 , b ) 249 and 415 , c ) 2400 and 4100 , d ) 2290 and 4350 , e ) 229 and 435 | a | multiply(divide(7.5, 12.5), divide(1660, subtract(const_1, divide(7.5, 12.5)))) | difference of two numbers is 1660 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1660 5 * y / 3 β y = 1660 2 * y / 3 = 1660 y = [ ( 1660 * 3 ) / 2 ] = 2490 . one number = 2490 , second number = 5 * y / 3 = 4150 answer is a . | a = 7 / 5
b = 7 / 5
c = 1 - b
d = 1660 / c
e = a * d
|
a ) 396 m , b ) 267 m , c ) 180 m , d ) 200 m , e ) 250 m | a | multiply(divide(120, const_3_6), 12) | a car is running at a speed of 120 kmph . what distance will it cover in 12 sec ? | "speed = 120 kmph = 120 * 5 / 18 = 33 m / s distance covered in 12 sec = 33 * 12 = 396 m answer is a" | a = 120 / const_3_6
b = a * 12
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a ) 270 , b ) 300 , c ) 330 , d ) 360 , e ) 390 | d | multiply(divide(360, add(5, 7)), 12) | jack and christina are standing 360 feet apart on a level surface . their dog , lindy , is standing next to christina . at the same time , they all begin moving toward each other . jack walks in a straight line toward christina at a constant speed of 5 feet per second and christina walks in a straight line toward jack at a constant speed of 7 feet per second . lindy runs at a constant speed of 12 feet per second from christina to jack , back to christina , back to jack , and so forth . what is the total distance , in feet , that lindy has traveled when the three meet at one place ? | "the relative speed of jack and christina is 5 + 7 = 12 feet per second . the distance between them is 210 feet , hence they will meet in ( time ) = ( distance ) / ( relative speed ) = 360 / 12 = 30 seconds . for all this time lindy was running back and forth , so it covered ( distance ) = ( speed ) * ( time ) = 12 * 30 = 360 feet . answer : d ." | a = 5 + 7
b = 360 / a
c = b * 12
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a ) 5 : 1 , b ) 5 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 5 : 3 | e | divide(divide(subtract(multiply(450, const_100), multiply(6000, 6)), subtract(10, 6)), divide(subtract(multiply(450, const_100), multiply(6000, 6)), subtract(10, 6))) | rs . 6000 is lent out in two parts . one part is lent at 6 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 6 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 6 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 4 x / 100 = > 600 - 1 / 25 x = 450 = > x = 3750 amount lent at 10 % = 2250 required ratio = 3750 : 2250 = 75 : 45 = 15 : 9 = 5 : 3 answer : e" | a = 450 * 100
b = 6000 * 6
c = a - b
d = 10 - 6
e = c / d
f = 450 * 100
g = 6000 * 6
h = f - g
i = 10 - 6
j = h / i
k = e / j
|
a ) 1200 , b ) 150 , c ) 360 , d ) 240 , e ) none of these | b | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(52, add(32, 4))), 4) | average age of students of an adult school is 52 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | "explanation : let the original no . of students be x . according to situation , 52 x + 120 * 32 = ( x + 120 ) 36 β x = 30 so , required no . of students after joining the new students = x + 120 = 150 answer : b" | a = 32 + 4
b = a * 120
c = 120 * 32
d = b - c
e = 32 + 4
f = 52 - e
g = d / f
h = g * 4
|
a ) 77 , b ) 88 , c ) 99 , d ) 110 , e ) 120 | a | multiply(sqrt(divide(multiply(11, 385), 125)), 125) | the h . c . f and l . c . m of two numbers are 11 and 385 respectively . if one number lies between 75 and 125 , then that number is | "explanation : product of numbers = 11 x 385 = 4235 let the numbers be 11 a and 11 b . then , 11 a x 11 b = 4235 inline fn _ jvn rightarrow ab = 35 now , co - primes with product 35 are ( 1,35 ) and ( 5,7 ) so , the numbers are ( 11 x 1 , 11 x 35 ) and ( 11 x 5 , 11 x 7 ) since one number lies 75 and 125 , the suitable pair is ( 55,77 ) hence , required number = 77 . answer : a" | a = 11 * 385
b = a / 125
c = math.sqrt(b)
d = c * 125
|
a ) 150 meter , b ) 299 meter , c ) 135 meter , d ) 155 meter , e ) 144 meter | c | multiply(divide(multiply(54, const_1000), const_3600), 9) | a train running at the speed of 54 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 54 * ( 5 / 18 ) m / sec = 15 m / sec length of train ( distance ) = speed * time ( 15 ) * 9 = 135 meter answer : c" | a = 54 * 1000
b = a / 3600
c = b * 9
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a ) 48 , b ) 24 , c ) 56 , d ) 63 , e ) 14 | d | add(negate(subtract(multiply(divide(3, 4), 60), multiply(divide(8, 5), 60))), 12) | simplify 3 / 4 of 6 0 β 8 / 5 of 60 + ? = 12 | 60 Γ 3 / 4 = 45 , 60 Γ 8 / 5 = 96 45 β 96 + ? = 12 = > 96 + 12 = 108 β 45 = 63 [ 63 + 45 = 108 β 96 = 12 ] option d | a = 3 / 4
b = a * 60
c = 8 / 5
d = c * 60
e = b - d
f = negate + (
|
a ) 3 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | a certain junior class has 1000 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "there are 60 siblings in junior class and 60 their pair siblings in the senior class . we want to determine probability of choosing one sibling from junior class and its pair from senior . what is the probability of choosing any sibling from junior class ? 60 / 1000 ( as there are 60 of them ) . what is the probability of choosing pair of chosen sibling in senior class ? as in senior class there is only one pair of chosen sibling it would be 1 / 800 ( as there is only one sibling pair of chosen one ) . so the probability of that the 2 students selected will be a sibling pair is : 60 / 1000 β 1 / 800 = 3 / 40000 answer : a ." | a = 1 / 3
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a ) 90 , b ) 95 , c ) 70 , d ) 105 , e ) 110 | c | divide(subtract(160, multiply(10, const_2)), const_2) | what is the average ( arithmetic mean ) of all the multiples of 10 from 10 to 160 inclusive ? | 10 and 190 inclusive would mean there are 15 terms . no need to add up all the terms and further divide by 15 since the terms are evenly spaced in ascending order from 10 , 20 , 30 . . . . . . . . 160 the middle term is the average which is the tenth term = 70 c is the answer . | a = 10 * 2
b = 160 - a
c = b / 2
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a ) 43 , b ) 44 , c ) 45 , d ) 46 , e ) 47 | a | add(divide(40, multiply(7, const_2)), 40) | michael earns $ 7.00 per hour for the first 40 hours he works per week , and twice this rate for overtime . if michael earned $ 320 last week , how many hours did he work ? | $ 7 * 40 + $ 12 * x = $ 320 - - > x = 3 hours . total working hours = 40 + 3 = 43 . answer : a . | a = 7 * 2
b = 40 / a
c = b + 40
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a ) rs . 8082 , b ) rs . 7800 , c ) rs . 8100 , d ) rs . 8112 , e ) rs . 9000 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 2 years the amount to be paid is ? | "explanation : a = 7500 ( 26 / 25 ) 2 = 8112 answer is d" | a = 100 + 4
b = a / 100
c = 100 + 4
d = c / 100
e = b * d
f = 7500 * e
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a ) 68 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(80, 3), multiply(50, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 50 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ? | "distance traveled in 2 hours = 2 * 50 = 100 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 100 = 340 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 340 / 5 = 68 mph answer : a" | a = 80 * 3
b = 50 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
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a ) 85 , b ) 36 , c ) 72 , d ) 90 , e ) 108 | a | subtract(divide(153, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), divide(divide(153, add(add(1, divide(1, 3)), divide(1, multiply(3, const_2)))), multiply(3, const_2))) | pat , kate , and mark charged a total of 153 hours to a certain project . if pat charged twice as much time to the project as kate and 1 / 3 as much time as mark , how many more hours did mark charge to the project than kate ? | "85 all u do is do 2 : 1 : 6 = > 2 x + x + 6 x = 153 = > x = 17 34 : 17 : 102 102 - 17 = 85 answer a" | a = 1 / 3
b = 1 + a
c = 3 * 2
d = 1 / c
e = b + d
f = 153 / e
g = 1 / 3
h = 1 + g
i = 3 * 2
j = 1 / i
k = h + j
l = 153 / k
m = 3 * 2
n = l / m
o = f - n
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a ) one , b ) two , c ) three , d ) seven , e ) ten | b | add(1, 2) | if d = 1 / ( 2 ^ 3 * 5 ^ 8 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | "another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 8 ) = multiply and divide by 2 ^ 5 = 2 ^ 5 / ( 2 ^ 3 * 2 ^ 5 * 5 ^ 8 ) = 2 ^ 5 / 10 ^ 8 = > non zero digits are 32 = > ans b" | a = 1 + 2
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a ) 229 , b ) - 1 , c ) - 229 , d ) 90 , e ) - 90 | c | add(add(subtract(90, subtract(multiply(5, 90), 100)), divide(90, 3)), const_1) | the arithmetic mean of 5 negative integers is - 100 . if the average of 3 of these integers is - 90 , what is the least possible value that one of the other 2 integers can have ? | the arithmetic mean of 5 negative integers is ( a + b + c + d + e ) / 5 = - 100 so , the sum of 5 negative integers would be a + b + c + d + e = - 100 * 5 = - 500 similarly , the sum of 3 negative integers would be a + b + c = - 90 * 3 = - 270 thus , the sum of the other 2 integers is - 500 + 270 = - 230 . since we know that the integers are negative then the least one from these two can be - 229 , the other one being - 1 . hence , the correct ans is c | a = 5 * 90
b = a - 100
c = 90 - b
d = 90 / 3
e = c + d
f = e + 1
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a ) s 529 , b ) s 527 , c ) s 570 , d ) s 750 , e ) s 507 | d | multiply(5, divide(3150, add(add(5, 2), const_3))) | rs . 3150 is divided so that 5 times the first share , thrice the 2 nd share and six times the third share amount to the same . what is the value of the third share ? | "a + b + c = 3150 5 a = 2 b = 6 c = x a : b : c = 1 / 5 : 1 / 2 : 1 / 6 = 6 : 15 : 5 5 / 21 * 3150 = rs 750 answer : d" | a = 5 + 2
b = a + 3
c = 3150 / b
d = 5 * c
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10 | e | divide(add(94, 6), add(4, 6)) | maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 94 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he meets up with brad ? | "total distance = 94 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 90 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken to meet brad after brad started = 90 / 10 = 9 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 9 = 36 + 4 = 40 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell / maxwell ' s speed = 40 / 4 = 10 hrs . . . answer e" | a = 94 + 6
b = 4 + 6
c = a / b
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a ) 16 , b ) 17 , c ) 15 , d ) 10 , e ) 12 | d | subtract(const_60, multiply(const_60, divide(40, 48))) | excluding stoppages , the speed of a train is 48 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ? | "t = 8 / 48 * 60 = 10 answer : d" | a = 40 / 48
b = const_60 * a
c = const_60 - b
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a ) 450 min , b ) 540 min , c ) 630 min , d ) 360 min , e ) none of these | c | divide(add(multiply(const_60, divide(const_1, const_2)), 6), subtract(divide(const_1, 5), divide(const_1, 7))) | if a child walks at the rate of 5 m / min from his home , he is 6 minutes late for school , if he walks at the rate of 7 m / min he reaches half an hour earlier . how far is his school from his home ? | let the distance between home and school is x . let actual time to reach be t . thus , x / 5 = t + 6 - - - - ( 1 ) and x / 7 = t - 30 - - - - - ( 2 ) solving equation 1 and 2 x = 630 min answer : c | a = 1 / 2
b = const_60 * a
c = b + 6
d = 1 / 5
e = 1 / 7
f = d - e
g = c / f
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a ) 75 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | subtract(subtract(9, multiply(8, subtract(3, 12))), negate(subtract(5, 11))) | evaluate : | 9 - 8 ( 3 - 12 ) | - | 5 - 11 | = ? | according to order of operations , inner brackets first . hence | 9 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 9 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 9 + 72 | - | 5 - 11 | = | 81 | - | - 6 | = 81 - 6 = 75 correct answer a ) 75 | a = 3 - 12
b = 8 * a
c = 9 - b
d = 5 - 11
e = c - negate
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a ) 16 % , b ) 22 % , c ) 32 % , d ) 40 % , e ) 52 % | b | multiply(divide(subtract(64, 50), 64), const_100) | in town x , 64 percent of the population are employed , and 50 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 50 are employed males , hence 14 % are employed females . ( employed females ) / ( total employed people ) = 14 / 64 = 22 % answer : b ." | a = 64 - 50
b = a / 64
c = b * 100
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a ) 228 , b ) 240 , c ) 225 , d ) 166 , e ) 1811 | c | multiply(20, multiply(54, const_0_2778)) | a train passes a station platform in 35 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 35 = 15 = > x = 225 m . answer : c" | a = 54 * const_0_2778
b = 20 * a
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a ) 21 : 22 , b ) 52 : 33 , c ) 52 : 53 , d ) 45 : 53 , e ) 51 : 34 | a | divide(add(const_100, 5), add(const_100, 10)) | the cash difference between the selling prices of an book at a profit of 5 % and 10 % is $ 3 . the ratio of the two selling prices is : | "let c . p . of the book be $ x . then , required ratio = 105 % of x / 110 % of x = 105 / 110 = 21 / 22 = 21 : 22 a" | a = 100 + 5
b = 100 + 10
c = a / b
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a ) 2 , b ) 3 , c ) 7 , d ) 13 , e ) 17 | d | floor(divide(3, divide(6, const_2))) | what is the greatest prime factor of 3 ^ 6 - 1 ? | "3 ^ 6 - 1 = ( 3 ^ 3 ) ^ 2 - 1 = ( 27 ^ 2 ) - 1 dividing ( 27 ^ 2 ) - 1 by 13 will give us a reminder of 0 ( hint : ( 2 * 13 + 1 ) ^ 2 - 1 / 13 = ( reminder 1 ) - 1 = 0 hence the greatest prime factor must be 13 . answer : d" | a = 6 / 2
b = 3 / a
c = math.floor(b)
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a ) 1050 , b ) 1100 , c ) 1150 , d ) 1200 , e ) 1250 | d | divide(add(1200, multiply(800, 6)), const_3) | village p β s population is 1200 greater than village q ' s population . if village q β s population were reduced by 800 people , then village p β s population would be 6 times as large as village q ' s population . what is village q ' s current population ? | "p = q + 1200 . p = 6 ( q - 800 ) . 6 ( q - 800 ) = q + 1200 . 5 q = 6000 . q = 1200 . the answer is d ." | a = 800 * 6
b = 1200 + a
c = b / 3
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a ) 600 , b ) 277 , c ) 269 , d ) 261 , e ) 1080 | e | add(800, multiply(800, divide(35, const_100))) | a person buys an article at rs . 800 . at what price should he sell the article so as to make a profit of 35 % ? | "cost price = rs . 800 profit = 35 % of 800 = rs . 280 selling price = cost price + profit = 800 + 280 = 1080 answer : e" | a = 35 / 100
b = 800 * a
c = 800 + b
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | multiply(divide(18, 10), divide(40, 10)) | how many liters of water must be added to 18 liters of milk and water containing 10 % water to make it 40 % water ? | "by rule of alligation : 40 % - 10 % = 30 % 100 % - 40 % = 60 % quantity of pure water : quantity of the mixture = 1 : 2 there are 18 liters of mixture , so we need to add 9 liters of pure water . the answer is d ." | a = 18 / 10
b = 40 / 10
c = a * b
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a ) 79 , b ) 93 , c ) 88 , d ) 88 , e ) 75 | b | divide(add(add(add(add(96, 95), 82), 97), 95), divide(const_10, const_2)) | dacid obtained 96 , 95 , 82 , 97 and 95 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 96 + 95 + 82 + 97 + 95 ) / 5 = 465 / 5 = 93 . answer : b" | a = 96 + 95
b = a + 82
c = b + 97
d = c + 95
e = 10 / 2
f = d / e
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a ) 1 / 3 , b ) 2 / 10 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500 | a | subtract(0.5, divide(1, 6)) | the number 0.5 is how much greater than 1 / 6 ? | "let x be the difference then . 5 - 1 / 3 = x 5 / 10 - 1 / 3 = x x = 1 / 3 ans a" | a = 1 / 6
b = 0 - 5
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a ) 1180 , b ) 1320 , c ) 1540 , d ) 1760 , e ) 1920 | c | multiply(subtract(const_1, divide(3, 22)), 22) | there are 22 students in a class . in how many different ways can a committee of 3 students be formed ? | "22 c 3 = 22 * 21 * 20 / 6 = 1540 the answer is c ." | a = 3 / 22
b = 1 - a
c = b * 22
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a ) 12 , b ) 20 , c ) 88 , d ) 77 , e ) 14 | b | subtract(70, divide(70, add(divide(2, 5), const_1))) | a 70 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 5 = 5 : 2 2 / 7 * 70 = 20 answer : b" | a = 2 / 5
b = a + 1
c = 70 / b
d = 70 - c
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a ) 330 , b ) 159 , c ) 150 , d ) 200 , e ) 628 | e | multiply(multiply(const_pi, 10), 20) | the slant height of a cone is 20 cm and radius of the base is 10 cm , find the curved surface of the cone . | "Ο * 10 * 20 = 628 answer : e" | a = math.pi * 10
b = a * 20
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a ) 6819.59775 , b ) 7336.03775 , c ) 6918.59775 , d ) 6198.59775 , e ) 6891.59775 | b | subtract(6602.5, multiply(multiply(660.25, 66.025), 6.6025)) | evaluate : 6602.5 + 660.25 + 66.025 + 6.6025 + 0.66025 | "6602.5 660.25 66.025 6.6025 + 0.66025 - - - - - - - - - - - - - - - 7336.03775 answer is b ." | a = 660 * 25
b = a * 6
c = 6602 - 5
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a ) 117 , b ) 126 , c ) 252 , d ) 336 , e ) none of these | b | multiply(lcm(lcm(lcm(3, 4), 5), 6), const_2) | the least number which when divided by 3 , 4 , 5 and 6 leaves a remainder 6 , but when divided by 9 leaves no remainder , is | "explanation : l . c . m of 3 , 4 , 5 , 6 = 60 therefore required number is of the form 60 k + 6 . least value of k for which ( 60 k + 6 ) is divisible by 9 is k = 2 therefore required number = ( 60 x 2 + 6 ) = 126 . answer : b" | a = math.lcm(3, 4)
b = math.lcm(a, 5)
c = math.lcm(b, 6)
d = c * 2
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a ) 65 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | e | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 20 minutes after it was poured was how many degrees fahrenheit ? | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 20 minutes after it was poured is : f = 120 ( 2 ^ - 20 / 10 ) + 60 f = 120 * 1 / 4 + 60 f = 30 + 60 f = 90 e" | a = 60 / 10
b = 1 - 2
c = a * b
d = 2 ** c
e = d * 120
f = e + 60
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a ) 50 % , b ) 80 % , c ) 40 % , d ) 90 % , e ) 100 % | a | multiply(divide(10, 20), const_100) | the ratio 10 : 20 expressed as percent equals to | "explanation : actually it means 10 is what percent of 20 , which can be calculated as , ( 10 / 20 ) * 100 = 10 * 5 = 50 answer : option a" | a = 10 / 20
b = a * 100
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a ) 50 kmph , b ) 60 kmph , c ) 70 kmph , d ) 80 kmph , e ) 85 kmph | d | multiply(divide(40, const_60), 120) | the speed of a train is 120 kmph . what is the distance covered by it in 40 minutes ? | "120 * 40 / 60 = 80 kmph answer : d" | a = 40 / const_60
b = a * 120
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a ) 63 , b ) 336 , c ) 567 , d ) 3024 , e ) 5040 | d | subtract(factorial(7), multiply(multiply(multiply(subtract(7, 1), const_4), subtract(7, 1)), multiply(7, const_2))) | a right triangle aec has to be constructed in the xy - plane so that the right angle is at a and ae is parallel to x axis . the coordinates of a , e and c are integers and satisfy the inequalities - 1 β€ x β€ 7 and 1 β€ y β€ 7 . how many different triangles can be constructed with these properties ? | all the cordinates are integer hence , possible ae values are { - 10 } , { - 11 } . . . . { - 17 } : 8 ways . . . . { 7 , - 1 } , { 70 } . . . . . . { 76 } : 8 ways 9 * 8 ways = 72 for ac values can be { 12 } , { 13 } . . . . { 17 } : 6 ways . . . . . . { 71 } , { 72 } . . . . . { 76 } : 6 ways 7 * 6 = 42 total = 72 * 42 = 3024 hence d | a = math.factorial(7)
b = 7 - 1
c = b * 4
d = 7 - 1
e = c * d
f = 7 * 2
g = e * f
h = a - g
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a ) 10 , b ) 20 , c ) 30 , d ) 50 , e ) 70 | d | add(multiply(sqrt(25), const_10), const_3) | if a ^ 2 + b ^ 2 = 25 and ab = 10 , what is the value of the expression ( a - b ) ^ 2 + ( a + b ) ^ 2 ? | "( a - b ) ^ 2 = a ^ 2 + b ^ 2 - 2 ab = 25 - 20 = 5 ( a + b ) ^ 2 = a ^ 2 + b ^ 2 + 2 ab = 25 + 20 = 45 so ( a + b ) ^ 2 + ( a - b ) ^ 2 = 45 + 5 = 50 d" | a = math.sqrt(25)
b = a * 10
c = b + 3
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a ) 270 , b ) 540 , c ) 610 , d ) 510 , e ) 500 | b | multiply(divide(multiply(60, const_3), 25), 75) | if 25 typists can type 60 letters in 20 minutes , then how many letters will 75 typists working at the same rate complete in 1 hour ? | no . of letters typing by 25 typists in 20 minutes = 60 no . of letters typing by 25 typists in 60 minutes = 60 * 3 = 180 no . of letters typing by 75 typists in 60 minutes = 180 / 25 * 75 = 540 answer : b | a = 60 * 3
b = a / 25
c = b * 75
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a ) 18 % , b ) 20 % , c ) 17 % , d ) 19 % , e ) none of these | d | subtract(subtract(add(const_100, 20), multiply(add(const_100, 20), divide(10, const_100))), const_100) | a shopkeeper labeled the price of his articles so as to earn a profit of 20 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ? | "explanation : let the cp of the article = rs . 100 . then labeled price = rs . 120 . sp = rs . 120 - 10 % of 120 = rs . 120 - 13 = rs . 118 . gain = rs . 118 Γ’ β¬ β rs . 100 = rs . 18 therefore , gain / profit percent = 18 % . answer : option d" | a = 100 + 20
b = 100 + 20
c = 10 / 100
d = b * c
e = a - d
f = e - 100
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a ) 32 , b ) 28 , c ) 29 , d ) 54 , e ) 20 | c | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 6), add(add(9, 1), subtract(9, 1))) | a person can row at 9 kmph and still water . he takes 6 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 6.5 = > x / 10 + x / 8 = 13 / 2 = > ( 4 x + 5 x ) / 40 = 13 / 2 = > x = 29 km . answer : c" | a = 9 - 1
b = 9 + 1
c = a * b
d = c * 6
e = 9 + 1
f = 9 - 1
g = e + f
h = d / g
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a ) 4 . , b ) 6 . , c ) 7 . , d ) 10 . , e ) 8 . | d | add(multiply(18, divide(const_1, 2)), const_1) | in the junior basketball league there are 18 teams , 2 / 3 of them are bad and Β½ are rich . what ca n ' t be the number of teams that are rich and bad | "total teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . ans d ." | a = 1 / 2
b = 18 * a
c = b + 1
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a ) and 27 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | a | subtract(43, divide(add(43, 5), const_3)) | the sum of the present age of henry and jill is 43 . what is their present ages if 5 years ago henry was twice the age of jill ? | "let the age of jill 5 years ago be x , age of henry be 2 x x + 5 + 2 x + 5 = 43 x = 11 present ages will be 16 and 27 answer : a" | a = 43 + 5
b = a / 3
c = 43 - b
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a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | e | subtract(const_100, divide(subtract(const_100, 84), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 84 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to Β΅ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 84 % is one st . dev , then on either side of m we have 84 / 2 = 42 % . so , 84 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 42 = 92 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 92 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it e" | a = 100 - 84
b = a / 2
c = 100 - b
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a ) 1 : 1 , b ) 1 : 87 , c ) 3 : 4 , d ) 1 : 9 , e ) 1 : 2 | c | divide(divide(multiply(5, 3), multiply(8, 2)), divide(multiply(3, 4), multiply(2, 5))) | the compound ratio of 5 : 8 , 3 : 2 and 4 : 5 ? | "5 / 8 * 3 / 2 * 4 / 5 = 3 / 4 1 : 1 answer : c" | a = 5 * 3
b = 8 * 2
c = a / b
d = 3 * 4
e = 2 * 5
f = d / e
g = c / f
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a ) 10 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | a | multiply(divide(const_1, multiply(add(const_100, 22), divide(const_1, subtract(const_100, 22)))), 14) | by selling 14 pencils for a rupee a man loses 22 % . how many for a rupee should he sell in order to gain 22 % ? | "88 % - - - 14 122 % - - - ? 88 / 122 * 14 = 10 answer : a" | a = 100 + 22
b = 100 - 22
c = 1 / b
d = a * c
e = 1 / d
f = e * 14
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(multiply(60, 5), 240), subtract(60, 40)) | i travel the first part of my journey at 40 kmph and the second part at 60 kmph and cover the total distance of 240 km to my destination in 5 hours . how long did the first part of my journey last | the total time of journey = 5 hours . let ' x ' hours be the time that i travelled at 40 kmph therefore , 5 β x hours would be time that i travelled at 60 kmph . hence , i would have covered x Γ 40 + ( 5 β x ) 60 kms in the 5 hours = 240 kms . solving , for x in the equation 40 x + ( 5 β x ) Γ 60 = 240 , we get 40 x + 300 β 60 x = 240 β 20 x = 60 or x = 3 hours answer : c | a = 60 * 5
b = a - 240
c = 60 - 40
d = b / c
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a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 5 minutes | d | multiply(4, const_1) | if 4 cats can kill 4 rats in 4 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 4 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 4 minutes , so 100 cats can kill 100 rats in 4 minutes answer d" | a = 4 * 1
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a ) 25 , b ) 30 , c ) 99 , d ) 88 , e ) 61 | a | divide(500, multiply(subtract(75, 3), const_0_2778)) | how many seconds will a 500 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 75 km / hr ? | "speed of train relative to man = 75 - 3 = 72 km / hr . = 72 * 5 / 18 = 20 m / sec . time taken to pass the man = 500 * 1 / 20 = 25 sec . answer : a" | a = 75 - 3
b = a * const_0_2778
c = 500 / b
|
['a ) 486', 'b ) 255', 'c ) 2866', 'd ) 265', 'e ) 872'] | a | multiply(multiply(const_3, const_2), power(cube_edge_by_volume(729), const_2)) | if the volume of the cube is 729 , then the surface area of the cube will be | explanation : \ inline \ fn _ jvn a ^ 3 = 729 ; a = 9 surface area = ( 6 x 9 x 9 ) = 486 \ inline \ fn _ jvn cm ^ 2 answer : a ) 486 sq . cm | a = 3 * 2
b = cube_edge_by_volume ** (
c = a * b
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['a ) area', 'b ) circumference', 'c ) 4', 'd ) 2 Ο', 'e ) none of these'] | c | multiply(divide(multiply(const_2, const_pi), const_pi), const_2) | when the circumference and area of a circle are numerically equal , then the diameter is numerically equal to | according to question , circumference of circle = area of circle or Ο d = Ο ( d β 2 ) 2 [ where d = diameter ] β΄ d = 4 answer c | a = 2 * math.pi
b = a / math.pi
c = b * 2
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a ) - 16 , b ) - 14 , c ) 0 , d ) 14 , e ) 16 | a | add(sqrt(49), sqrt(64)) | if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 49 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ? | ( x + 1 ) ^ 2 < = 49 x < = 6 x > = - 8 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 8 Γ - 6 = 48 minimum possible value of xy is - 8 Γ 8 = - 64 - 64 + 48 = - 16 answer : a | a = math.sqrt(49)
b = math.sqrt(64)
c = a + b
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a ) 2500 , b ) 2700 , c ) 3000 , d ) 3100 , e ) nobe | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500) | in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , the number of valid votes that the other candidate got , was | solution number of valid votes = 80 % of 7500 = 6000 . valid votes polled by other candidates = 45 % of 6000 ( 45 / 100 Γ 6000 ) = 2700 . answer b | a = 20 / 100
b = 1 - a
c = 55 / 100
d = 1 - c
e = b * d
f = e * 7500
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a ) r = 200 , b ) r = 600 , c ) r = 800 , d ) r = 1600 , e ) r = 50 | d | multiply(power(const_2, 4), const_100) | cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost r is what percent of the original cost ? | original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600 | a = 2 ** 4
b = a * 100
|
a ) 23 , b ) 27 , c ) 30 , d ) 32 , e ) 34 | b | sqrt(divide(108, add(power(4, 3), add(power(5, 3), power(3, 3))))) | the ratio of three numbers is 5 : 3 : 4 and their sum is 108 . the second number of the three numbers is ? | "5 : 3 : 4 total parts = 12 12 parts - - > 108 1 part - - - - > 9 the second number of the three numbers is = 3 3 parts - - - - > 27 b )" | a = 4 ** 3
b = 5 ** 3
c = 3 ** 3
d = b + c
e = a + d
f = 108 / e
g = math.sqrt(f)
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a ) 56 , b ) 156 , c ) 256 , d ) 356 , e ) 456 | c | multiply(divide(subtract(subtract(power(multiply(5, divide(add(const_100, 20), const_100)), const_2), power(multiply(4, divide(subtract(const_100, 50), const_100)), const_2)), subtract(power(5, const_2), power(4, const_2))), subtract(power(5, const_2), power(4, const_2))), const_100) | there are two concentric circles with radii 5 and 4 . if the radius of the outer circle is increased by 20 % and the radius of the inner circle decreased by 50 % , by what percent does the area between the circles increase ? | the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 25 pi . the area of the small circle is 16 pi . the area a 1 between the circles is 9 pi . when the big circle ' s radius increases , the new area is 36 pi . when the small circle ' s radius decreases , the new area is 4 pi . the area a 2 between the circles is 32 pi . the ratio of a 2 / a 1 is 32 / 9 = 3.56 which is an increase of 256 % . the answer is c . | a = 100 + 20
b = a / 100
c = 5 * b
d = c ** 2
e = 100 - 50
f = e / 100
g = 4 * f
h = g ** 2
i = d - h
j = 5 ** 2
k = 4 ** 2
l = j - k
m = i - l
n = 5 ** 2
o = 4 ** 2
p = n - o
q = m / p
r = q * 100
|
a ) 500 , b ) 698 , c ) 780 , d ) 737 , e ) none | d | subtract(815, divide(multiply(subtract(854, 815), 2), 3)) | a sum of money at simple interest amounts to rs . 815 in 2 years and to rs . 854 in 3 years . the sum is : | "sol . s . i . for 1 year = rs . ( 854 - 815 ) = rs . 39 . s . i . for 2 years = rs . ( 39 * 2 ) = rs . 78 . Γ’ Λ Β΄ principal = rs . ( 815 - 78 ) = rs . 737 . answer d" | a = 854 - 815
b = a * 2
c = b / 3
d = 815 - c
|
['a ) 2 : 5', 'b ) 3 : 4', 'c ) 3 : 5', 'd ) 1 : 2', 'e ) 2 : 3'] | d | power(divide(const_1, sqrt(const_2)), const_2) | the ratio of the area of a square to that of the square drawn on its diagonal is ? | explanation : a 2 : ( a Γ’ Λ Ε‘ 2 ) 2 a 2 : 2 a 2 - > 1 : 2 answer is d | a = math.sqrt(2)
b = 1 / a
c = b ** 2
|
a ) $ 200 , b ) $ 240 , c ) $ 480 , d ) $ 960 , e ) $ 1,920 | a | multiply(8, 8) | a rectangular floor that measures 8 meters by 10 meters is to be covered with carpet squares that each measure 2 meters by 2 meters . if the carpet squares cost $ 10 apiece , what is the total cost for the number of carpet squares needed to cover the floor ? | "the width of the rectangular floor ( 8 m ) is a multiple of one side of the square ( 2 m ) , and the length of the floor ( 10 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 8 / 2 ) * ( 10 / 2 ) = 20 . the total cost is 20 * 10 = $ 200 . the answer is , therefore , a ." | a = 8 * 8
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a ) 500 / 1001 , b ) 503 / 1001 , c ) 303 / 1001 , d ) 301 / 1001 , e ) 505 / 1001 | e | divide(add(add(divide(factorial(6), multiply(factorial(4), factorial(const_2))), divide(factorial(8), multiply(factorial(4), factorial(4)))), multiply(divide(factorial(6), multiply(factorial(4), factorial(const_2))), divide(factorial(8), multiply(factorial(6), factorial(const_2))))), divide(factorial(add(6, 8)), multiply(factorial(subtract(add(6, 8), 4)), factorial(4)))) | there are 6 positive and 8 negative numbers . 4 numbers are choosen at random and multiplied . the probability that the product is positive is : | 6 c 4 / 14 c 4 + ( 6 c 2 * 8 c 2 ) / 14 c 4 + 8 c 4 / 14 c 4 = 505 / 1001 case 1 : only 4 positive no . case 2 : 2 positive and 2 negative no . case 3 : 4 negative no . answer : e | a = math.factorial(6)
b = math.factorial(4)
c = math.factorial(2)
d = b * c
e = a / d
f = math.factorial(8)
g = math.factorial(4)
h = math.factorial(4)
i = g * h
j = f / i
k = e + j
l = math.factorial(6)
m = math.factorial(4)
n = math.factorial(2)
o = m * n
p = l / o
q = math.factorial(8)
r = math.factorial(6)
s = math.factorial(2)
t = r * s
u = q / t
v = p * u
w = k + v
x = 6 + 8
y = math.factorial(x)
z = 6 + 8
A = z - 4
B = math.factorial(A)
C = math.factorial(4)
D = B * C
E = y / D
F = w / E
|
a ) 976374 , b ) 979923 , c ) 1009125 , d ) 2356677 , e ) 1083875 | c | multiply(1000000, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 35 % , then find the population of city atthe end of the year 2007 | required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 35 / 100 ) = 1009125 c | a = 15 / 100
b = 1 + a
c = 35 / 100
d = 1 - c
e = b * d
f = 35 / 100
g = 1 + f
h = e * g
i = 1000000 * h
|
a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | divide(subtract(2, multiply(multiply(3, 3), const_2)), subtract(const_1, multiply(3, 3))) | f ( x ) is a function such that f ( x ) + 3 f ( 8 - x ) = x for all real numbers x . find the value of f ( 2 ) . | f ( x ) + 3 f ( 8 - x ) = x : given f ( 2 ) + 3 f ( 6 ) = 2 : x = 2 above f ( 6 ) + 3 f ( 2 ) = 6 : x = 6 above f ( 6 ) = 6 - 3 f ( 2 ) : solve equation c for f ( 6 ) f ( 2 ) + 3 ( 6 - 3 f ( 2 ) ) = 2 : substitute f ( 2 ) = 2 : solve above equation . correct answer is d ) 2 | a = 3 * 3
b = a * 2
c = 2 - b
d = 3 * 3
e = 1 - d
f = c / e
|
a ) 20 coins , b ) 30 coins , c ) 40 coins , d ) 60 coins , e ) none of these | d | divide(105, add(add(inverse(const_4), inverse(const_2)), const_1)) | a bag contains an equal number of one rupee , 50 paise and 25 paise coins respectively . if the total value is 105 , how many coins of each type are there ? | "let number of each type of coin = x . then , 1 Γ x + . 50 Γ x + . 25 x = 105 β 1.75 x = 105 β x = 60 coins answer d" | a = 1/(4)
b = 1/(2)
c = a + b
d = c + 1
e = 105 / d
|
a ) 8 , b ) 10 , c ) 12 , d ) 16 , e ) 20 | b | add(5, divide(subtract(multiply(20, 10), multiply(10, 5)), add(10, 20))) | john can complete a given task in 20 days . jane will take only 10 days to complete the same task . john and jane set out to complete the task by beginning to work together . however , jane was indisposed 5 days before the work got over . in how many days did the work get over from the time john and jane started to work on it together ? | in such questions , you need to start from the end . last 5 days john works alone and completes 5 * ( 1 / 20 ) = 1 / 4 of the work . so 3 / 4 of the work should have been completed by the two of them together before jane left . their combined rate of work is 1 / 20 + 1 / 10 = 3 / 20 time taken to complete 3 / 4 of the work = ( 3 / 4 ) / ( 3 / 20 ) = 5 days . so total number of days taken to complete the work = 5 + 5 = 10 days . b | a = 20 * 10
b = 10 * 5
c = a - b
d = 10 + 20
e = c / d
f = 5 + e
|
a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | c | divide(original_price_before_loss(10, 90), divide(original_price_before_gain(20, 60), 20)) | a man sold 20 articles for $ 60 and gained 10 % . how many articles should he sell for $ 90 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 10 % ) / 20 = $ 2.70 required production costs for a loss of 20 % : $ 90 * ( 100 % + 20 % ) = $ 108 number of articles to be sold for $ 108 to incur a 20 % loss : $ 108 / $ 2.70 = 40 thus , solution c is correct ." | a = original_price_before_loss / (
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a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 4 | e | subtract(subtract(subtract(10, 4), const_1), const_1) | list k consists of 10 consecutive integers . if - 4 is the least integer in list k , what is the range of the positive integers in list k ? | "answer = e = 4 if least = - 4 , then largest = 5 range = 5 - 1 = 4" | a = 10 - 4
b = a - 1
c = b - 1
|
a ) 41 , b ) 63 , c ) 72 , d ) 63 , e ) 50 | c | divide(multiply(400, subtract(const_100, add(add(44, 28), 10))), const_100) | in a school of 400 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "44 + 28 + 10 = 82 % 100 β 82 = 18 % 400 * 18 / 100 = 72 answer : c" | a = 44 + 28
b = a + 10
c = 100 - b
d = 400 * c
e = d / 100
|
a ) 72 kmph , b ) 88 kmph , c ) 54 kmph , d ) 18 kmph , e ) 19 kmph | a | multiply(const_3_6, divide(120, 6)) | a train 120 m in length crosses a telegraph post in 6 seconds . the speed of the train is ? | "s = 120 / 6 * 18 / 5 = 72 kmph answer : a" | a = 120 / 6
b = const_3_6 * a
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a ) 1 / 12 , b ) 5 / 12 , c ) 1 / 6 , d ) 1 / 3 , e ) 3 / 15 | e | divide(add(21, 3), const_60) | two boats are heading towards each other at constant speeds of 3 miles / hr and 21 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ? | "the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 3 + 21 = 24 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 24 * 1 / 60 = 3 / 15 miles apart . answer : e ." | a = 21 + 3
b = a / const_60
|
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