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pgilliar
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MNLP_M2_rag_dataset

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967
Rationale
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2.74k
options
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correct
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annotated_formula
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linear_formula
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answer
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the probability that a man will be alive for 10 more yrs is 1 / 4 & the probability that his wife will alive for 10 more yrs is 1 / 3 . the probability that none of them will be alive for 10 more yrs , is
sol . required probability = pg . ) x p ( b ) = ( 1 β€” d x ( 1 β€” i ) = : x 1 = 1 / 2 ans . ( a )
a ) 1 / 2 , b ) 1 , c ) 2 / 3 , d ) 3 / 5 , e ) 3 / 8
a
multiply(subtract(1, divide(1, 4)), subtract(1, divide(1, 3)))
divide(n1,n2)|divide(n1,n5)|subtract(n1,#0)|subtract(n1,#1)|multiply(#2,#3)
general
A
jose is uploading a file to the internet for a college project . the file weighs 160 megabytes . if jose ' s internet speed for uploading tops 8 megabytes per minute , how long will it take until the upload is completed ?
answer is ( c ) . jose calculates that if the files weighs 160 megabytes , and his internet uploads at 8 megabytes per minute , it would take 20 minutes to upload as 160 divided by 8 megabytes per minute equals 20 .
a ) 1 hour , b ) 45 minutes , c ) 20 minutes , d ) 5 minutes , e ) 1 hour and 10 minutes
c
divide(160, 8)
divide(n0,n1)
physics
C
a contractor undertook to do a piece of work in 10 days . he employed certain number of laboures but 5 of them were absent from the very first day and the rest could finish the work in only 13 days . find the number of men originally employed ?
"let the number of men originally employed be x . 10 x = 13 ( x Γ’ € β€œ 5 ) or x = 21.6 answer a"
a ) 21.6 , b ) 23.6 , c ) 22.6 , d ) 21.8 , e ) 21.3
a
divide(multiply(13, 5), subtract(13, 10))
multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|
physics
A
the average weight of a , b and c is 45 kg . if the average weight of a and b be 41 kg and that of b and c be 43 kg , then the weight of b is :
"a 33 kg let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 41 x 2 ) = 82 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 168 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 33 . b ' s weight = 33 kg . a"
a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 37 kg
a
subtract(add(multiply(41, const_2), multiply(43, const_2)), multiply(45, const_3))
multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|
general
A
the average weight of 6 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 47 kg . what might be the weight of the new person ?
"total weight increased = ( 6 x 3.5 ) kg = 21 kg . weight of new person = ( 47 + 21 ) kg = 68 kg option c"
a ) 60 kg , b ) 75 kg , c ) 68 kg , d ) 85 kg , e ) 90 kg
c
add(multiply(6, 3.5), 47)
multiply(n0,n1)|add(n2,#0)|
general
C
the sum of the numbers is 110 . if the first number be twice the second and third number be one - third of the first , then the second number is :
"let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 110 11 x / 3 = 110 x = 30 answer : a"
a ) 30 , b ) 54 , c ) 72 , d ) 84 , e ) 27
a
divide(multiply(110, const_3), add(const_10, const_1))
add(const_1,const_10)|multiply(n0,const_3)|divide(#1,#0)|
general
A
a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 5 x km .
"total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 5 x / ( x / 8 ) = 40 kmph answer : d"
a ) 16 , b ) 18 , c ) 24 , d ) 40 , e ) 12
d
divide(multiply(40, 5), add(divide(40, 40), divide(multiply(2, 40), 20)))
divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)|
general
D
patrick purchased 70 pencils and sold them at a loss equal to the selling price of 20 pencils . the cost of 70 pencils is how many times the selling price of 70 pencils ?
"say the cost price of 70 pencils was $ 70 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 70 - 70 p = 20 p - - > p = 7 / 9 . ( cost price ) / ( selling price ) = 1 / ( 7 / 9 ) = 9 / 7 = 1.28 . answer : d ."
a ) 0.75 , b ) 0.8 , c ) 1 , d ) 1.28 , e ) 1.35
d
inverse(divide(70, add(70, 20)))
add(n0,n1)|divide(n0,#0)|inverse(#1)|
general
D
what is the area of a square field whose diagonal of length 30 m ?
"d 2 / 2 = ( 30 * 30 ) / 2 = 450 answer : c"
a ) 287 , b ) 269 , c ) 450 , d ) 200 , e ) 230
c
divide(square_area(30), const_2)
square_area(n0)|divide(#0,const_2)|
geometry
C
at joel ’ s bookstore , the current inventory is 30 % historical fiction . of the historical fiction books , 30 % are new releases , while 40 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ?
"let there be 100 books in all historic fiction books = 30 % of total = 30 other books = 70 new historic fiction = 30 % of 30 = 9 other new books = 40 % of 70 = 28 total new books = 37 fraction = 9 / 37 ans : b"
a ) 4 / 25 , b ) 9 / 37 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 3
b
divide(divide(multiply(30, 30), const_100), add(divide(multiply(30, 30), const_100), divide(multiply(40, subtract(const_100, 30)), const_100)))
multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)|
gain
B
two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 20 hr . if all pipes are opened simultaneously , then the cistern will be filled in
"solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 βˆ’ 1 / 20 = 2 / 15 hence , tank will be filled in 15 / 2 = 7.5 hour option ( a )"
a ) 7.5 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these
a
inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 20)))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)|
physics
A
45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 40 meters deep by working for 6 hours ?
45 workers * 8 hours / 30 meters = x * 6 / 40 x = 80 total workers 80 - 45 = 35 the answer is d .
a ) 75 , b ) 80 , c ) 50 , d ) 35 , e ) 40
d
subtract(multiply(multiply(45, divide(8, 6)), divide(40, 30)), 45)
divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0)
physics
D
in one hour , a boat goes 13 km / hr along the stream and 9 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is :
"explanation : let the speed downstream be a km / hr and the speed upstream be b km / hr , then speed in still water = 1 / 2 ( a + b ) km / hr rate of stream = 1 / 2 ( a βˆ’ b ) km / hr speed in still water = 1 / 2 ( 13 + 9 ) kmph = 11 kmph . answer : option a"
a ) 11 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph
a
stream_speed(13, 9)
stream_speed(n0,n1)|
physics
A
the average ( arithmetic mean ) monthly income of 4 workers is $ 1000 . after one worker ’ s income increases by 50 percent the new average income is $ 1200 . what was the original income of the worker whose monthly income increased ?
increase in total income was 200 * 4 = $ 800 , we know that this increase was 50 % ( 1 / 2 ) of the workers original income , thus his / her original income was 800 * 2 = $ 1,600 . answer : d
a ) $ 1,800 , b ) $ 1,500 , c ) $ 1,300 , d ) $ 1,600 , e ) $ 1,100
d
subtract(subtract(subtract(multiply(1200, 4), multiply(subtract(4, const_1), 1000)), divide(subtract(multiply(1200, 4), multiply(subtract(4, const_1), 1000)), add(divide(50, const_100), const_1))), const_100)
divide(n2,const_100)|multiply(n0,n3)|subtract(n0,const_1)|add(#0,const_1)|multiply(n1,#2)|subtract(#1,#4)|divide(#5,#3)|subtract(#5,#6)|subtract(#7,const_100)
general
D
a shop owner sells 25 mtr of cloth and gains sp of 10 mtrs . find the gain % ?
"here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 25 m cloth ) – ( c . p . of 25 m cloth ) selling price of 15 m cloth = selling price of 25 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 15 m cloth = rs . 1500 and s . p . of 15 m cloth = rs . rs . 2500 profit % = 10 / 15 Γ— 100 = 66.67 % profit of 66.67 % was made by the merchant . d"
a ) 30 % , b ) 40 % , c ) 50 % , d ) 66.67 % , e ) 70 %
d
multiply(divide(10, subtract(25, 10)), const_100)
subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)|
gain
D
product of two natural numbers is 11 . then , the sum of reciprocals of their squares is
"explanation : if the numbers are a , b , then ab = 17 , as 17 is a prime number , so a = 1 , b = 17 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 11 ( 2 ) = 122 / 121 option b"
a ) 290 / 289 , b ) 122 / 121 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these
b
add(power(divide(const_1, const_1), const_2), power(divide(const_1, 11), const_2))
divide(const_1,const_1)|divide(const_1,n0)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)|
general
B
three solid cubes of sides 1 cm , 6 cm and 8 cm are melted to form a new cube . find the surface area of the cube so formed
explanation : volume of new cube = = edge of new cube = = 9 cm surface area of the new cube = ( 6 x 9 x 9 ) = 486 answer : a ) 486
['a ) 486', 'b ) 366', 'c ) 299', 'd ) 278', 'e ) 1888']
a
multiply(multiply(const_3, power(power(add(add(power(1, const_3), power(6, const_3)), power(8, const_3)), const_0_33), const_2)), const_2)
power(n0,const_3)|power(n1,const_3)|power(n2,const_3)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)|power(#5,const_2)|multiply(#6,const_3)|multiply(#7,const_2)
geometry
A
find the surface area of a cuboid 12 m long , 14 m broad and 7 m high
explanation : surface area = [ 2 ( 12 x 14 + 14 x 7 + 12 x 7 ) ] cm 2 = ( 2 x 350 ) cm 2 = 700 cm 2 . answer : c
['a ) 868 sq . cm', 'b ) 600 sq . cm', 'c ) 700 sq . cm', 'd ) 900 sq . cm', 'e ) none of these']
c
multiply(const_2, add(add(multiply(12, 14), multiply(12, 7)), multiply(14, 7)))
multiply(n0,n1)|multiply(n0,n2)|multiply(n1,n2)|add(#0,#1)|add(#3,#2)|multiply(#4,const_2)
geometry
C
what is the value of 4 ^ 5 + 4 ^ 8 ?
"4 ^ 5 + 4 ^ 8 = 4 ^ 5 ( 1 + 4 ^ 3 ) = 4 ^ 5 * 65 answer e"
a ) 4 ^ 12 , b ) 4 ^ 35 , c ) 17 ( 4 ^ 5 ) , d ) 8 ^ 12 , e ) 65 ( 4 ^ 5 )
e
divide(multiply(add(add(const_100, const_60), const_1), 4), const_100)
add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)|
general
E
richard traveled the entire 60 miles trip . if he did the first 2 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 48 miles per hour , what is the his average speed , in miles per hour ?
average speed = sum of distance / sum of time . if he travelled the first 2 miles at 24 miles / hr , it would take 0.083 hr . for the remaining trip , if he went at 48 miles / 1 hr , it would take 1 hour . then , the average speed is 60 miles / ( 0.083 + 1 ) hrs = 55 miles / 1 hr . therefore , the answer is e .
a ) 20 mph , b ) 24 mph , c ) 30 mph , d ) 32 mph , e ) 55 mph
e
divide(60, subtract(divide(subtract(60, 2), 48), divide(2, 24)))
divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|subtract(#2,#0)|divide(n0,#3)
physics
E
the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 130 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ?
if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 130 group # 2 = ( like understood ) = 3 / 4 ( 1 / 4 ) n = 130 n = 520 answer = ( e )
a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 520
e
divide(130, subtract(const_1, divide(3, 4)))
divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)
general
E
pipe x that can fill a tank in an hour and pipe y that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe y is shut 15 minutes before the tank overflows . when will the tank overflow ?
the last 15 minutes only pipe x was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 3 = 1 / 4 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : b
a ) 35 mins , b ) 30 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins
b
multiply(add(divide(subtract(const_1, divide(15, multiply(const_60, const_1))), add(const_1, const_2)), divide(15, multiply(const_60, const_1))), const_60)
add(const_1,const_2)|multiply(const_1,const_60)|divide(n0,#1)|subtract(const_1,#2)|divide(#3,#0)|add(#4,#2)|multiply(#5,const_60)
physics
B
in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 10 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?
"total fish = x percentage of second catch = ( 10 / 50 ) * 100 = 20 % so , x * 20 % = 50 x = 250 ans . c"
a ) 400 , b ) 625 , c ) 250 , d ) 2,500 , e ) 10,000
c
divide(50, divide(10, 50))
divide(n2,n1)|divide(n0,#0)|
gain
C
if k and w are the dimensions of a rectangle that has area 40 , and if k and w are integers such that k > w , what is the total number of possible values of k ?
"kw = 40 = 40 * 1 = 20 * 2 = 10 * 4 = 8 * 5 = - - > k can take 4 values , namely : 8 , 10,20 and 40 answer : c ."
a ) two , b ) three , c ) four , d ) five , e ) six
c
multiply(const_3, const_1)
multiply(const_1,const_3)|
geometry
C
if 5 / w + 5 / x = 5 / y and wx = y , then the average ( arithmetic mean ) of w and x is
"given : 5 / w + 5 / x = 5 / ywx = y find : ( w + x ) / 2 = ? 5 ( 1 / w + 1 / x ) = 5 ( 1 / y ) - divide both sides by 5 ( 1 / w + 1 / x ) = 1 / y ( x + w ) / wx = 1 / wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) / 2 = 1 / 2 ans : a"
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 4 , e ) 8
a
divide(5, add(5, 5))
add(n0,n0)|divide(n0,#0)|
general
A
the sale price sarees listed for rs . 550 after successive discount is 18 % and 12 % is ?
"550 * ( 88 / 100 ) * ( 82 / 100 ) = 396 answer : d"
a ) 298 , b ) 237 , c ) 342 , d ) 396 , e ) 291
d
subtract(subtract(550, divide(multiply(550, 18), const_100)), divide(multiply(subtract(550, divide(multiply(550, 18), const_100)), 12), const_100))
multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
D
the volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ?
the volume of the cone = ( 1 / 3 ) Ο€ r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) Ο€ may be ignored . v 1 / v 2 = r 12 h 1 / r 22 h 2 = > 1 / 10 = ( 1 ) 2 h 1 / ( 2 ) 2 h 2 = > h 1 / h 2 = 2 / 5 i . e . h 1 : h 2 = 2 : 5 answer : a
['a ) 2 : 5', 'b ) 2 : 9', 'c ) 2 : 2', 'd ) 2 : 9', 'e ) 2 : 1']
a
divide(divide(1, 10), power(divide(1, 2), const_2))
divide(n0,n1)|divide(n0,n3)|power(#1,const_2)|divide(#0,#2)
geometry
A
the price of a bushel of corn is currently $ 3.20 , and the price of a peck of wheat is $ 5.80 . the price of corn is increasing at a constant rate of 5 x 5 x cents per day while the price of wheat is decreasing at a constant rate of 2 √ βˆ— x βˆ’ x 2 βˆ— x βˆ’ x cents per day . what is the approximate price when a bushel of corn costs the same amount as a peck of wheat ?
"let yy be the # of days when these two bushels will have the same price . first let ' s simplify the formula given for the rate of decrease of the price of wheat : 2 √ βˆ— x βˆ’ x = 1.41 x βˆ’ x = 0.41 x 2 βˆ— x βˆ’ x = 1.41 x βˆ’ x = 0.41 x , this means that the price of wheat decreases by 0.41 x 0.41 x cents per day , in yy days it ' ll decrease by 0.41 xy 0.41 xy cents ; as price of corn increases 5 x 5 x cents per day , in yy days it ' ll will increase by 5 xy 5 xy cents ; set the equation : 320 + 5 xy = 580 βˆ’ 0.41 xy 320 + 5 xy = 580 βˆ’ 0.41 xy , solve for xyxy - - > xy = 48 xy = 48 ; the cost of a bushel of corn in yy days ( the # of days when these two bushels will have the same price ) will be 320 + 5 xy = 320 + 5 βˆ— 48 = 560320 + 5 xy = 320 + 5 βˆ— 48 = 560 or $ 5.6 . answer : e ."
a ) $ 4.50 , b ) $ 5.10 , c ) $ 5.30 , d ) $ 5.50 , e ) $ 5.60
e
add(3.20, multiply(divide(subtract(5.80, 3.20), add(5, subtract(sqrt(5), 2))), 5))
sqrt(n3)|subtract(n1,n0)|subtract(#0,n4)|add(n2,#2)|divide(#1,#3)|multiply(n2,#4)|add(n0,#5)|
gain
E
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.60 per glass on the first day , what was the price e per glass on the second day ?
"on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 3 * x - - > x = $ 0.4 . answer : d ."
a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45
d
divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2))
add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|
general
D
on a certain road 14 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ?
"answer is c . this question is in the og and thus well explained by ets . those who exceed : x so x = 14 % + 0,2 x id est x = 17,5 %"
a ) 10.5 % , b ) 12.5 % , c ) 17.5 % , d ) 22 % , e ) 30 %
c
divide(const_100, multiply(multiply(divide(14, const_100), divide(20, const_100)), const_100))
divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)|
gain
C
find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 35 as remainder
"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 35 5 x = 1330 x = 266 large number = 266 + 1365 = 1631 e"
a ) 1235 , b ) 1456 , c ) 1567 , d ) 1678 , e ) 1631
e
add(multiply(divide(subtract(1365, 35), subtract(6, const_1)), 6), 35)
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)|
general
E
what is the greatest value of x such that 4 ^ x is a factor of 21 ! ?
pretty simple , really . if m = 6 , then 4 m = 24 , which is 12 x 2 , both of which are included in 21 ! since 6 is the largest number here , its the answer . answer is b
a ) 5 , b ) 6 , c ) 3 , d ) 2 , e ) 4
b
add(divide(subtract(21, const_1), 4), const_1)
subtract(n1,const_1)|divide(#0,n0)|add(#1,const_1)
other
B
a train 820 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ?
"speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 820 + x ) / 60 = 65 / 3 x = 480 m . answer : option c"
a ) 510 , b ) 540 , c ) 480 , d ) 520 , e ) 589
c
divide(820, multiply(subtract(78, 1), const_0_2778))
subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
C
compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 342 grams of compound x ?
"total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 342 gms of x have 342 * 10 / 12 = 280 gms of b and 342 - 280 = 62 gms of a . cross check : - a / b = 62 / 280 = 2 / 10 ( as given ) ans e"
a ) 54 , b ) 162 , c ) 250 , d ) 270 , e ) 280
e
divide(multiply(342, 10), add(2, 10))
add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|
other
E
a certain car increased its average speed by 5 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 26 miles per hour , how many miles did the car travel in the third 5 - minute interval ?
in the third time interval the average speed of the car was 22 + 5 + 5 = 36 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 36 * 1 / 12 = 3 miles . answer : e .
a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0
e
multiply(add(add(26, 5), 5), divide(5, const_60))
add(n0,n3)|divide(n0,const_60)|add(n0,#0)|multiply(#2,#1)
physics
E
a man swims downstream 36 km and upstream 26 km taking 2 hours each time , what is the speed of the man in still water ?
"36 - - - 2 ds = 18 ? - - - - 1 26 - - - - 2 us = 13 ? - - - - 1 m = ? m = ( 18 + 13 ) / 2 = 15.5 answer : c"
a ) 9 , b ) 18.4 , c ) 15.5 , d ) 16.7 , e ) 13.4
c
divide(add(divide(26, 2), divide(36, 2)), const_2)
divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|
physics
C
at a certain bowling alley , it costs $ 1 to rent bowling shoes for the day and $ 1.25 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ?
"after renting bowling shoes the person is left with $ 12.80 - $ 1 = $ 11.80 , which is enough for 11.8 / 1.25 < 10 = ~ 9 . answer : e ."
a ) 7 , b ) 8 , c ) 5 , d ) 10 , e ) 9
e
divide(subtract(12.80, 1), 1.25)
subtract(n3,n0)|divide(#0,n1)|
physics
E
we invested a total of $ 1,000 . we invested one part of the money at 3 % and the rest of the money at 5 % . the total investment with interest at the end of the year was $ 1,046 . how much money did we invest at 3 % ?
"let x be the money invested at 3 % . 1.03 x + 1.05 ( 1000 - x ) = 1046 . 0.02 x = 1050 - 1046 . 0.02 x = 4 . 2 x = 400 . x = 200 . the answer is a ."
a ) $ 200 , b ) $ 240 , c ) $ 280 , d ) $ 320 , e ) $ 360
a
divide(subtract(multiply(multiply(const_100, multiply(add(const_2, const_3), const_2)), add(divide(5, const_100), const_1)), add(add(multiply(const_100, multiply(add(const_2, const_3), const_2)), multiply(multiply(add(const_2, const_3), const_2), 3)), multiply(const_2, const_3))), subtract(add(divide(5, const_100), const_1), add(divide(3, const_100), const_1)))
add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(const_2,const_3)|add(#1,const_1)|add(#2,const_1)|multiply(#0,const_2)|multiply(#6,const_100)|multiply(#6,const_4)|subtract(#4,#5)|add(#7,#8)|multiply(#4,#7)|add(#10,#3)|subtract(#11,#12)|divide(#13,#9)|
gain
A
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 22050 / - in 3 years at the same rate of interest . find the rate percentage ?
"explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 22050 / - - 17640 / - = rs . 4410 / - rate of interest = ( 4410 / 17640 ) Γ— ( 100 / 1 ) = > 25 % answer : option e"
a ) 5 % , b ) 7 % , c ) 9 % , d ) 11 % , e ) 25 %
e
multiply(divide(subtract(22050, 17640), 17640), const_100)
subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|
general
E
the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 4 . the sum of numbers is :
"let the numbers be 1 x and 4 x . then , their l . c . m = 4 x . so , 4 x = 48 or x = 12 . the numbers are 12 and 48 . hence , required sum = ( 12 + 48 ) = 60 . answer : e"
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 60
e
divide(multiply(1, 48), 4)
multiply(n0,n1)|divide(#0,n2)|
other
E
if 8 workers can build 8 cars in 8 days , then how many days would it take 5 workers to build 5 cars ?
"8 workers can build 1 car per day on average . 1 worker can build 1 / 8 of a car per day . 5 workers can build 5 / 8 car per day . the time required to build 5 cars is 5 / ( 5 / 8 ) = 8 days the answer is c ."
a ) 4 , b ) 5 , c ) 8 , d ) 10 , e ) 12
c
multiply(divide(multiply(8, 8), 8), divide(5, 5))
divide(n3,n3)|multiply(n0,n0)|divide(#1,n0)|multiply(#2,#0)|
physics
C
a and b can do a piece of work in 4 days , b and c in 6 days , c and a in 3 days . how long will c take to do it ?
"2 c = 1 / 6 + 1 / 3 – 1 / 4 = 1 / 4 c = 1 / 8 = > 8 days answer : a"
a ) 8 days , b ) 10 days , c ) 12 days , d ) 4 days , e ) 12 days
a
divide(multiply(4, const_3), subtract(divide(add(divide(multiply(4, const_3), 3), add(divide(multiply(4, const_3), 4), divide(multiply(4, const_3), 6))), const_2), divide(multiply(4, const_3), 4)))
multiply(n0,const_3)|divide(#0,n0)|divide(#0,n1)|divide(#0,n2)|add(#1,#2)|add(#4,#3)|divide(#5,const_2)|subtract(#6,#1)|divide(#0,#7)|
physics
A
in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 2,800 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 8,400 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?
"ratio of 2 bedroom apartment : 1 bedroom apartment = 2800 : 8400 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 4 ) * 100 - - - > 25 % answer : c"
a ) 30 % , b ) 35 % , c ) 25 % , d ) 40 % , e ) 50 %
c
divide(multiply(2,800, const_100), add(add(multiply(const_2, const_1000), const_100), 2,800))
multiply(n0,const_100)|multiply(const_1000,const_2)|add(#1,const_100)|add(n0,#2)|divide(#0,#3)|
general
C
the present population of a town is 10000 . population increase rate is 10 % p . a . find the population of town after 2 years ?
"p = 10000 r = 10 % required population of town = p ( 1 + r / 100 ) ^ t = 10000 ( 1 + 10 / 100 ) ^ 2 = 10000 ( 11 / 10 ) ^ 2 = 12100 answer is d"
a ) 10000 , b ) 11100 , c ) 15000 , d ) 12100 , e ) 14520
d
add(10000, divide(multiply(10000, 10), const_100))
multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|
gain
D
the ratio , by volume , of soap to alcohol to water in a certain solution is 2 : 40 : 70 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water will it contain ?
"soap : alcohol initial ratio soap : alcohol : water - - > 2 : 40 : 70 initial soap : alcohol = 2 / 40 = 2 : 40 after doubled soap : alcohol = 2 * 2 / 40 = 4 : 40 initial soap : water = 2 / 70 = 2 : 70 after halved soap : water : 1 / 2 * 2 / 70 = 1 / 70 = 1 : 70 after soap : alcohol : water - - > 4 : 40 : 280 - - > 1 : 10 : 70 given alcohol 100 cubic centimeter . ratio is 10 : 100 : 700 ( 1 : 10 : 70 ) for 100 cubic centimeter of alcohol - - - 700 cubic cm water is required . answer - d"
a ) 500 , b ) 600 , c ) 650 , d ) 700 , e ) 720
d
divide(divide(divide(divide(divide(volume_rectangular_prism(100, 70, 40), const_3), const_2), const_4), const_4.0), const_4)
volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,const_4)|divide(#3,const_4.0)|divide(#4,const_4)|
geometry
D
the average of marks obtained by 120 candidates was 35 . if the avg of marks of passed candidates was 39 & that of failed candidates was 39 and that of failed candidates was 15 , the no . of candidates who passed the examination is ?
"let the number of candidate who passed = y then , 39 y + 15 ( 120 - y ) = 120 x 35 β‡’ 24 y = 4200 - 1800 ∴ y = 2400 / 24 = 100 c"
a ) 80 , b ) 90 , c ) 100 , d ) 120 , e ) 140
c
divide(subtract(multiply(120, 35), multiply(120, 15)), subtract(39, 15))
multiply(n0,n1)|multiply(n0,n4)|subtract(n2,n4)|subtract(#0,#1)|divide(#3,#2)|
general
C
what is the value of 4 ^ 5 + 4 ^ 4 ?
"4 ^ 5 + 4 ^ 4 = 4 ^ 4 ( 4 + 1 ) = 4 ^ 4 * 5 answer b"
a ) 4 ^ 12 , b ) 5 ( 4 ^ 4 ) , c ) 17 ( 4 ^ 5 ) , d ) 8 ^ 12 , e ) 7 ( 4 ^ 5 )
b
divide(multiply(add(add(const_100, const_60), const_1), 4), const_100)
add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)|
general
B
in the biology lab of ` ` jefferson ' ' high school there are 0.037 * 10 ^ 5 germs , equally divided among 74000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ?
0.037 * 10 ^ 5 can be written as 3700 74000 * 10 ^ ( - 3 ) can be written as 74 required = 3700 / 74 = 50 answer : e
a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50
e
divide(multiply(multiply(const_1000, const_100), 0.037), divide(74000, const_1000))
divide(n3,const_1000)|multiply(const_100,const_1000)|multiply(n0,#1)|divide(#2,#0)
general
E
find the smallest number of 6 digits which is exactly divisible by 111 ?
"smallest number of 6 digits is 100000 . on dividing 100000 by 111 , we get 100 as remainder . number to be added = ( 111 - 100 ) - 11 . hence , required number = 100011 answer b"
a ) 12000 , b ) 15550 , c ) 100011 , d ) 158993 , e ) 100010
b
add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 111)
multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)|
general
B
a reduction of 40 % in the price of bananas would enable a man to obtain 66 more for rs . 40 , what is reduced price per dozen ?
"40 * ( 40 / 100 ) = 16 - - - 66 ? - - - 12 = > rs . 2.91 answer : b"
a ) 1.91 , b ) 2.91 , c ) 4.91 , d ) 3.91 , e ) 5.91
b
multiply(const_12, divide(multiply(40, divide(40, const_100)), 66))
divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)|
gain
B
in an office , 20 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ?
"( 10 / 100 ) workers = 16 = > number of workers = 160 ( 20 / 100 ) * workers = x + 16 = > x = 32 answer a"
a ) 32 , b ) 64 , c ) 50 , d ) 144 , e ) 160
a
divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4))
divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)|
gain
A
after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 4725 . what was the number of original inhabitants ?
"let the total number of original inhabitants be x . ( 75 / 100 ) * ( 90 / 100 ) * x = 4725 ( 27 / 40 ) * x = 4725 x = 4725 * 40 / 27 = 7000 the answer is b ."
a ) 5000 , b ) 7000 , c ) 4000 , d ) 8000 , e ) 9000
b
divide(4725, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100))))
divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#0)|multiply(#1,#2)|subtract(#2,#3)|divide(n2,#4)|
gain
B
in a graduating class of 232 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number t and the smallest possible number of students that could have taken both geometry and biology ?
"official solution : first of all , notice that since 144 took geometry and 119 took biology , then the number of students who took both geometry and biology can not be greater than 119 . { total } = { geometry } + { biology } - { both } + { neither } ; 232 = 144 + 119 - { both } + { neither } ; { both } = 31 + { neither } . { both } is minimized when { neither } is 0 . in this case { both } = 31 . the greatest possible number t of students that could have taken both geometry and biology , is 119 . thus , the answer is 119 - 31 = 88 . answer : d ."
a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31
d
subtract(119, subtract(add(144, 119), 232))
add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)|
other
D
at what rate percent on simple interest will rs . 750 amount to rs . 1050 in 5 years ?
"300 = ( 750 * 5 * r ) / 100 r = 8 % . answer : a"
a ) 8 , b ) 3 , c ) 4 , d ) 5 , e ) 6
a
multiply(divide(divide(subtract(1050, 750), 750), 5), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
A
a cycle is bought for rs . 900 and sold for rs . 1350 , find the gain percent ?
"900 - - - - 450 100 - - - - ? = > 50 % answer : d"
a ) 39 % , b ) 20 % , c ) 23 % , d ) 50 % , e ) 83 %
d
multiply(divide(subtract(1350, 900), 900), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
D
zachary is helping his younger brother , sterling , learn his multiplication tables . for every question that sterling answers correctly , zachary gives him 3 pieces of candy . for every question that sterling answers incorrectly , zachary takes away two pieces of candy . after 8 questions , if sterling had answered 2 more questions correctly , he would have earned 31 pieces of candy . how many of the 8 questions did zachary answer correctly ?
"i got two equations : 3 x - 2 y = 25 x + y = 8 3 x - 2 ( 8 - x ) = 25 3 x - 16 + 2 x = 25 5 x = 41 x = 8.2 or between 8 and 9 . 9 ans c )"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
c
divide(add(subtract(multiply(2, subtract(8, 2)), multiply(3, 2)), 31), add(3, 2))
add(n0,n2)|multiply(n0,n2)|subtract(n1,n2)|multiply(n2,#2)|subtract(#3,#1)|add(n3,#4)|divide(#5,#0)|
general
C
12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required
"that is , 1 work done = 12 Γ— 8 Γ— 10 then , 12 8 Γ— 10 = ? Γ— 15 Γ— 8 ? ( i . e . no . of men required ) = 12 Γ— 8 Γ— 10 / 15 Γ— 10 = 8 days . a"
a ) 8 days , b ) 3 days , c ) 7 days , d ) 5 days , e ) 6 days
a
divide(multiply(multiply(12, 10), 8), multiply(8, 15))
multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)|
physics
A
if there are 20 apples and the apples have to be shared equally among 3 babies . what number of apples are to be added ?
given there are 20 apples . if 1 extra apple is added , then it becomes 21 which can be divided equally that is 7 apples to each baby . option a is correct .
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
a
subtract(multiply(add(const_4, const_3), 3), 20)
add(const_3,const_4)|multiply(n1,#0)|subtract(#1,n0)
general
A
triangle xyz is an isosceles right triangle . if side xy is longer than side yz , and the area of the triangle is 36 , what is the measure of side xy ?
"ans c . . 12 . . xy being larger means it is the hyp . . area = ( 1 / 2 ) * ( yz ) ^ 2 = 36 or yz = 3 * \ sqrt { 2 } . . therefore hyp = xy = 12"
a ) 4 , b ) 4 √ 2 , c ) 12 , d ) 8 √ 2 , e ) can not be determined from the information provided
c
sqrt(add(power(multiply(sqrt(36), sqrt(const_2)), const_2), power(multiply(sqrt(36), sqrt(const_2)), const_2)))
sqrt(n0)|sqrt(const_2)|multiply(#0,#1)|power(#2,const_2)|add(#3,#3)|sqrt(#4)|
geometry
C
a box is in the shape of a semicircle with a radius of 12 . what is the approximate perimeter of the semicircle ?
perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 12 + 2 * 12 = 61.68 approximately 62 answer b
['a ) 54', 'b ) 62', 'c ) 25', 'd ) 34', 'e ) 60']
b
add(multiply(const_pi, 12), multiply(const_2, 12))
multiply(n0,const_pi)|multiply(n0,const_2)|add(#0,#1)
geometry
B
a man walks at a rate of 10 mph . after every ten miles , he rests for 8 minutes . how much time does he take to walk 50 miles ?
"to cover 50 miles the man needs ( time ) = ( distance ) / ( rate ) = 50 / 10 = 5 hours = 300 minutes . he will also rest 4 times ( after 10 , 20 , 30 and 40 miles ) , so total resting time = 4 * 8 = 32 minutes . total time = 300 + 32 = 332 minutes . answer : e ."
a ) 300 , b ) 318 , c ) 322 , d ) 324 , e ) 332
e
add(multiply(8, const_4), multiply(divide(50, 10), const_60))
divide(n2,n0)|multiply(n1,const_4)|multiply(#0,const_60)|add(#1,#2)|
physics
E
the mean of 50 observations was 36 . it was found later that an observation 44 was wrongly taken as 23 . the corrected new mean is
"solution correct sum = ( 36 x 50 + 44 - 23 ) = 1821 . Γ’ Λ† Β΄ correct mean = 1821 / 50 = 36.42 . answer d"
a ) 35.22 , b ) 36.12 , c ) 36.22 , d ) 36.42 , e ) none
d
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
D
if 30 oxen can plough 1 / 7 th of a field in 2 days , how many days 18 oxen will take to do the remaining work ?
solution : we will use work equivalence method , 30 / 18 = ( 1 / 7 ) / ( 6 / 7 ) * x / 2 ; 5 / 3 = ( 1 / 6 ) * x / 2 ; or , x = 60 / 3 = 20 days . answer : option b
a ) 30 days , b ) 20 days , c ) 15 days , d ) 18 days , e ) 21 days
b
divide(divide(30, 18), divide(divide(divide(1, 7), divide(subtract(7, const_1), 7)), 2))
divide(n0,n4)|divide(n1,n2)|subtract(n2,const_1)|divide(#2,n2)|divide(#1,#3)|divide(#4,n3)|divide(#0,#5)
physics
B
a student chose a number , multiplied it by 2 , then subtracted 152 from the result and got 102 . what was the number he chose ?
"solution : let x be the number he chose , then 2 * x * 152 = 102 2 x = 254 x = 127 correct answer a"
a ) 127 , b ) 100 , c ) 129 , d ) 160 , e ) 200
a
divide(add(102, 152), 2)
add(n1,n2)|divide(#0,n0)|
general
A
33 1 / 3 % of 390 ?
"33 1 / 3 % = 1 / 3 1 / 3 Γ— 390 = 130 e )"
a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130
e
divide(multiply(add(33, divide(1, 3)), 390), const_100)
divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)|
gain
E
machine p and machine q are each used to manufacture 220 sprockets . it takes machine p 10 hours longer to produce 220 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?
"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 220 / x = 220 / 1.1 x + 10 1.1 ( 220 ) = 220 + 11 x 11 x = 22 x = 2 the answer is b ."
a ) 5 , b ) 2 , c ) 55 , d ) 95 , e ) 125
b
divide(subtract(220, divide(220, add(divide(10, const_100), const_1))), 10)
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)|
gain
B
the average of numbers 0.64206 , 0.64207 , 0.64208 and 0.64209 is ?
"answer average = ( 0.64206 + 0.64207 + 0.64208 + 0.64209 ) / 4 = 2.5683 / 4 = 0.642075 correct option : d"
a ) 0.64202 , b ) 0.64204 , c ) 0.642022 , d ) 0.642075 , e ) none
d
divide(add(multiply(0.64206, 0.64207), multiply(0.64206, 0.64208)), 0.64206)
multiply(n0,n1)|multiply(n0,n2)|add(#0,#1)|divide(#2,n0)|
general
D
when 242 is divided by a certain divisor the remainder obtained is 4 . when 698 is divided by the same divisor the remainder obtained is 8 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 7 . what is the value of the divisor ?
"let that divisor be x since remainder is 4 or 8 it means divisor is greater than 8 . now 242 - 4 = 238 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 8 = 690 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 238 + 690 ) + 4 + 8 = x ( k + l ) + 12 when we divide this number by x then remainder will be equal to remainder of ( 12 divided by x ) = 7 hence x = 12 - 7 = 5 hence a"
a ) 5 , b ) 17 , c ) 13 , d ) 23 , e ) none of these
a
subtract(add(4, 8), 7)
add(n1,n3)|subtract(#0,n6)|
general
A
a sum of money deposited at c . i . amounts to rs . 3650 in 2 years and to rs . 4015 in 3 years . find the rate percent ?
"explanation : 3650 - - - - - - - - 365 100 - - - - - - - - ? ( 10 % ) answer : option a"
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 %
a
multiply(divide(subtract(4015, 3650), 3650), const_100)
subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
A
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1 / 2 m board ?
"30 * 33 / 2 = 5 / 2 * 2 * x = > x = 99 answer b"
a ) 90 , b ) 99 , c ) 97 , d ) 95 , e ) 65
b
divide(multiply(30, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2))
divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)|
general
B
a football team lost 5 yards and then gained 10 . what is the team ' s progress ?
"for lost , use negative . for gain , use positive . progress = - 5 + 10 = 5 yards c"
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8
c
subtract(10, 5)
subtract(n1,n0)|
gain
C
the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 60 ounces of this substance ?
"112 ounces of a substance has a volume of 48 cubic inches 60 ounces of a substance has a volume of ( 48 / 112 ) * 60 = 25 cubic inches answer a alternatively , we can use estimation 112 ounces of a substance has a volume of 48 cubic inches 56 ounces of a substance has a volume of 24 cubic inches therefore , 60 will have a volume a little more than 24 , that is 25 c"
a ) 27 , b ) 36 , c ) 25 , d ) 64 , e ) 147
c
multiply(divide(48, 112), 60)
divide(n0,n1)|multiply(n2,#0)|
geometry
C
two circular signs are to be painted . if the diameter of the larger sign is 7 times that of the smaller sign , how many times more paint is needed to paint the larger sign ? ( we can assume that a given amount of paint covers the same area on both signs . )
let r be the radius of the smaller sign . then the diameter of the smaller sign is 2 r , the diameter of the larger sign is 14 r , and the radius of the larger sign is 7 r . the area a of the smaller sign is a = pir ^ 2 . the area of the larger sign is pi ( 7 r ) ^ 2 = 49 pir ^ 2 = 49 a . since the area is 49 times larger , we need 49 times more paint for the larger sign . the answer is d .
a ) 7 , b ) 14 , c ) 42 , d ) 49 , e ) 98
d
power(7, const_2)
power(n0,const_2)
geometry
D
two trains , each 100 m long , moving in opposite directions , cross other in 10 sec . if one is moving twice as fast the other , then the speed of the faster train is ?
let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 10 = 3 x = > x = 20 / 3 . so , speed of the faster train = 40 / 3 = 40 / 3 * 18 / 5 = 48 km / hr . answer : c
a ) 76 km / hr , b ) 66 km / hr , c ) 48 km / hr , d ) 67 km / hr , e ) 22 km / hr
c
divide(multiply(multiply(divide(add(100, 100), multiply(10, add(const_1, const_2))), const_2), const_3600), const_1000)
add(n0,n0)|add(const_1,const_2)|multiply(n1,#1)|divide(#0,#2)|multiply(#3,const_2)|multiply(#4,const_3600)|divide(#5,const_1000)
physics
C
two cubes of their volumes in the ratio 8 : 125 . the ratio of their surface area is :
the ratio of their surface area is 8 : 125 2 : 5 answer is b .
['a ) 1.5 : 5', 'b ) 2 : 5', 'c ) 3 : 5', 'd ) 1 : 5', 'e ) 4 : 5']
b
divide(power(8, const_0_33), power(125, const_0_33))
power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1)
geometry
B
a person crosses a 1440 m long street in 12 minutes . what is his speed in km per hour ?
"speed = 1440 / ( 12 x 60 ) m / sec = 2 m / sec . converting m / sec to km / hr = 2 x ( 18 / 5 ) km / hr = 7.2 km / hr . answer : e"
a ) 4.1 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 7.2
e
divide(divide(1440, const_1000), divide(multiply(12, const_60), const_3600))
divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)|
physics
E
the perimeter of a semi circle is 216 cm then the radius is ?
"36 / 7 r = 216 = > r = 42 answer : e"
a ) 26 , b ) 48 , c ) 98 , d ) 37 , e ) 42
e
divide(216, add(const_2, const_pi))
add(const_2,const_pi)|divide(n0,#0)|
physics
E
a train 180 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ?
"speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 180 * 3 / 55 = 10 sec . answer b"
a ) 7 , b ) 10 , c ) 8 , d ) 2 , e ) 4
b
divide(180, multiply(add(60, 6), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
B
each of the 35 points is placed either inside or on the surface of a perfect sphere . if 88 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ?
"maximum number of points on the surface is 88 % * 35 = 30.8 . . . or 30 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 30 points , so at best we can form 15 pairs ( 30 ) . so , answer is ( e )"
a ) 7 , b ) 11 , c ) 13 , d ) 14 , e ) 15
e
divide(multiply(35, divide(88, const_100)), const_2)
divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_2)|
geometry
E
the spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls . the diameters of two of these are 1 Β½ cm and 2 cm respectively . the diameter of third ball is ?
4 / 3 Ο€ * 3 * 3 * 3 = 4 / 3 Ο€ [ ( 3 / 2 ) 3 + 23 + r 3 ] r = 1.25 d = 2.5 answer : b
a ) 2.66 cm , b ) 2.5 cm , c ) 3 cm , d ) 3.5 cm , e ) 4 cm
b
multiply(power(subtract(subtract(power(divide(3, 2), const_3), power(divide(add(1, divide(1, 2)), 2), const_3)), 1), inverse(3)), const_2)
divide(n0,n3)|divide(n2,n3)|inverse(n0)|add(n2,#1)|power(#0,const_3)|divide(#3,n3)|power(#5,const_3)|subtract(#4,#6)|subtract(#7,n2)|power(#8,#2)|multiply(#9,const_2)
physics
B
find the average marks of all the students in 2 separate classes , if the average marks of students in the first class of 55 students is 60 and that of another class of 48 students is 58
sum of the marks for the class of 55 students = 55 * 60 = 3300 sum of the marks for the class of 48 students = 48 * 58 = 2784 sum of the marks for the class of 103 students = 3300 + 2784 = 6084 average marks of all the students = 6084 / 103 = 59.1 answer : e
a ) 55.1 , b ) 51.1 , c ) 53.1 , d ) 52.1 , e ) 59.1
e
divide(add(multiply(55, 60), multiply(48, 58)), add(55, 48))
add(n1,n3)|multiply(n1,n2)|multiply(n3,n4)|add(#1,#2)|divide(#3,#0)
general
E
how many prime numbers are between 15 / 7 and 131 / 6 ?
"15 / 7 = 3 - 131 / 6 = 22 - prime numbers between 3 and 22 are 5 , 7 , 11 , 13 , 17 , and 19 - sign signifies that the number is marginally less . answer d"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
d
floor(const_2)
floor(const_2)|
general
D
c and d started a business by investing rs . 1000 / - and rs . 1500 / - respectively . find the d ’ s share out of a total profit of rs . 500 :
c = rs . 1000 / - d = rs . 1500 / - c share 2 parts & d share 3 parts total 5 parts - - - - - > rs . 500 / - - - - - > 1 part - - - - - - - > rs . 100 / - d share = 3 parts - - - - - > rs . 300 / - d
a ) 400 , b ) 300 , c ) 200 , d ) 100 , e ) 150
d
divide(500, add(divide(1000, 500), divide(1500, 500)))
divide(n0,n2)|divide(n1,n2)|add(#0,#1)|divide(n2,#2)
general
D
if greg buys 3 shirts , 4 trousers and 2 ties , the total cost is $ 90 . if greg buys 7 shirts , 2 trousers and 2 ties , the total cost is $ 50 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ?
"solution : 3 x + 4 y + 2 z = 90 7 x + 2 y + 2 z = 50 adding both the equations = 10 x + 6 y + 4 z = 140 5 x + 3 y + 2 z = 70 ans b"
a ) $ 60 , b ) $ 70 , c ) $ 75 , d ) $ 96 , e ) can not be determined
b
divide(add(50, 90), 2)
add(n3,n7)|divide(#0,n2)|
general
B
what is the sum of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 6 ?
"find the number , upon sum of 3 digits of a number gives a reminder 2 when it is divided by 6 seeing the options after dividing an finding the reminder of 2 my answer was a"
a ) 82,650 , b ) 64,850 , c ) 64,749 , d ) 49,700 , e ) 56,720
a
multiply(divide(add(divide(subtract(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)), 6), const_1), 2), add(subtract(const_1000, 3), add(add(multiply(multiply(3, 3), const_10), multiply(3, 3)), 3)))
multiply(n0,n0)|subtract(const_1000,n0)|multiply(#0,const_10)|add(#2,#0)|add(n0,#3)|add(#4,#1)|subtract(#1,#4)|divide(#6,n2)|add(#7,const_1)|divide(#8,n1)|multiply(#5,#9)|
general
A
a can do a piece of work in 24 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ?
"2 / 24 + 4 / 15 + x / 20 = 1 x = 13 answer : c"
a ) 10 , b ) 12 , c ) 13 , d ) 11 , e ) 15
c
add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(24), inverse(15))), 2))), inverse(15)), 4)
inverse(n1)|inverse(n2)|inverse(n0)|subtract(n4,n3)|add(#0,#1)|add(#2,#0)|add(#5,#1)|multiply(#4,#3)|multiply(n3,#6)|add(#7,#8)|subtract(const_1,#9)|divide(#10,#0)|add(n4,#11)|
physics
C
rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 170 . what is the share of rahul ?
"rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 170 * 2 / 5 = $ 68 answer is a"
a ) $ 68 , b ) $ 40 , c ) $ 60 , d ) $ 100 , e ) $ 90
a
multiply(divide(2, add(3, 2)), 170)
add(n0,n1)|divide(n1,#0)|multiply(n2,#1)|
physics
A
a total of 30 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 10 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ]
"let ' take the number of geese to be 100 . male = 30 . female = 70 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 10 % are male i . e 10 / 100 * 20 = 2 geese ( males ) and now we know out of the total 20 geese , 2 are male , then 18 have to be female . now the ratio part , male geese ratios = 2 / 30 = 1 / 15 . - a female geese ratios = 18 / 70 = 9 / 35 - b cross multiply equations a and b and you get = 7 / 27 . ans d"
a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 27 , e ) 8 / 7
d
divide(divide(divide(10, const_100), divide(30, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(30, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_4)|divide(#0,#1)|multiply(#2,const_10)|divide(#4,const_100)|divide(#5,#1)|divide(#3,#6)|
general
D
evaluate : 1222343 - 12 * 3 * 2 = ?
"according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 1222343 - 12 * 3 * 2 = 122343 - 8 = 122336 correct answer a"
a ) 122336 , b ) 145456 , c ) 122347 , d ) 126666 , e ) 383838
a
subtract(1222343, multiply(multiply(12, 3), 2))
multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|
general
A
if 12 men and 16 boys can do a piece of work in 5 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy ?
12 m + 16 b - - - - - 5 days 13 m + 24 b - - - - - - - 4 days 60 m + 80 b = 52 m + 96 b 8 m = 16 b = > 1 m = 2 b m : b = 2 : 1 answer : d
a ) 2 : 5 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 2
d
divide(subtract(multiply(4, 24), multiply(5, 16)), subtract(multiply(5, 12), multiply(4, 13)))
multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)
physics
D
if x / y = 9 / 8 , then ( 8 x + 7 y ) / ( 8 x Γ’ € β€œ 7 y ) = ?
"answer dividing numerator as well as denominator by y , we get given exp . = ( 8 x + 7 y ) / ( 8 x Γ’ € β€œ 7 y ) = ( 8 x / y + 7 ) / ( 8 x / y Γ’ € β€œ 7 ) since x / y = 9 / 8 this implies that = [ ( 8 * 9 ) / 8 + 7 ] / [ ( 8 * 9 ) / 8 - 7 ) ] = ( 9 + 7 ) / ( 9 - 7 ) = 8 option : a"
a ) 8 , b ) 7 , c ) 10 , d ) 9 , e ) 6
a
divide(add(9, 8), subtract(9, 8))
add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
general
A
the radius of the circle is reduced from 5 cm to 4 cm then the % change of area .
for 5 cm - > pi * r ^ 2 - - > 3.14 * 5 ^ 2 - > 78.539 for 4 cm - > pi * r ^ 2 - - > 3.14 * 4 ^ 2 - > 50.265 % change - > ( 1 - 50.265 / 78.539 ) * 100 = 36 ie 36 % answer : a
['a ) 36 %', 'b ) 37 %', 'c ) 35 %', 'd ) 38 %', 'e ) 39 %']
a
subtract(const_100, multiply(const_100, divide(circle_area(4), circle_area(5))))
circle_area(n1)|circle_area(n0)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)
geometry
A
there is 60 % increase in an amount in 6 yrs at si . what will be the ci of rs . 12,000 after 3 years at the same rate ?
"let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 12000 . t = 3 years and r = 10 % p . a . c . i . = rs . 12000 x 1 + 10 3 - 1 100 = rs . 12000 x 331 1000 = 3972 . e"
a ) 2354 , b ) 2450 , c ) 2540 , d ) 2650 , e ) 3972
e
subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)))
divide(n0,n1)|multiply(const_100,const_100)|divide(#0,const_100)|multiply(#0,const_100)|add(#2,const_1)|multiply(#3,n3)|add(#1,#5)|multiply(#4,#4)|multiply(#4,#7)|multiply(#6,#8)|subtract(#9,#6)|
gain
E
a 5 digit number divisible by 3 is to be formed using the digits 0 , 1 , 2 , 3 , 4 and 5 without repetitions . that total no of ways it can be done is ?
first step : we should determine which 5 digits from given 6 , would form the 5 digit number divisible by 3 . we have six digits : 0,1 , 2,3 , 4,5 . their sum = 15 . for a number to be divisible by 3 the sum of the digits must be divisible by 3 . as the sum of the six given numbers is 15 ( divisible by 3 ) only 5 digits good to form our 5 digit number would be 15 - 0 = ( 1 , 2,3 , 4,5 ) and 15 - 3 = ( 0 , 1,2 , 4,5 ) . meaning that no other 5 from given six will total the number divisible by 3 . second step : we have two set of numbers : 1 , 2,3 , 4,5 and 0 , 1,2 , 4,5 . how many 5 digit numbers can be formed using this two sets : 1 , 2,3 , 4,5 - - > 5 ! as any combination of these digits would give us 5 digit number divisible by 3 . 5 ! = 120 . 0 , 1,2 , 4,5 - - > here we can not use 0 as the first digit , otherwise number wo n ' t be any more 5 digit and become 4 digit . so , total combinations 5 ! , minus combinations with 0 as the first digit ( combination of 4 ) 4 ! - - > 5 ! - 4 ! = 96 120 + 96 = 216 answer : c .
a ) 122 , b ) 210 , c ) 216 , d ) 217 , e ) 220
c
add(factorial(5), multiply(factorial(subtract(5, const_1)), subtract(5, const_1)))
factorial(n0)|subtract(n0,const_1)|factorial(#1)|multiply(#2,#1)|add(#0,#3)
general
C
in what time will a train 120 m long cross an electric pole , it its speed be 144 km / hr ?
"speed = 144 * 5 / 18 = 40 m / sec time taken = 120 / 40 = 3 sec . answer : b"
a ) 2.5 sec , b ) 3 sec , c ) 8.9 sec , d ) 6.9 sec , e ) 2.9 sec
b
divide(120, multiply(144, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
B
a train covers a distance in 50 min , if it runs at a speed of 48 kmph on an average . the speed at which the train must run to reduce the time of journey to 40 min will be
"time = 50 / 60 hr = 5 / 6 hr speed = 48 mph distance = s * t = 48 * 5 / 6 = 40 km time = 40 / 60 hr = 2 / 3 hr new speed = 40 * 3 / 2 kmph = 60 kmph answer : b ."
a ) 45 min , b ) 60 min , c ) 55 min , d ) 70 min , e ) 80 min
b
divide(multiply(48, divide(50, const_60)), divide(40, const_60))
divide(n0,const_60)|divide(n2,const_60)|multiply(n1,#0)|divide(#2,#1)|
physics
B
machine a can finish a job in 6 hours , machine в can finish the job in 12 hours , and machine с can finish the job in 8 hours . how many hours will it take for a , b , and с together to finish the job ?
the combined rate is 1 / 6 + 1 / 12 + 1 / 8 = 9 / 24 of the job per hour . the time to complete the job is 24 / 9 = 8 / 3 hours . the answer is c .
a ) 6 / 5 , b ) 7 / 4 , c ) 8 / 3 , d ) 9 / 2 , e ) 12 / 5
c
inverse(add(add(inverse(6), inverse(12)), inverse(8)))
inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)
physics
C
a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 15 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ?
explanation : l * 30 = 1200 Γ¨ l = 40 40 + 30 + 50 = 120 120 * 15 = 1800 answer : option d
['a ) 1276', 'b ) 1200', 'c ) 2832', 'd ) 1800', 'e ) 1236']
d
multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, const_2), power(divide(1200, 30), const_2)))), 15)
divide(n1,n3)|power(n3,const_2)|add(n3,#0)|power(#0,const_2)|add(#1,#3)|sqrt(#4)|add(#2,#5)|multiply(n0,#6)
geometry
D
for how many values of q , is | | | q - 5 | - 10 | - 5 | = 2 ? ( those ls are mods )
i think its 8 | | | q - 5 | - 10 | - 5 | = 2 let | q - 5 | = a which makes above | | a - 10 | - 5 | = 2 let | a - 10 | = b which makes | b - 5 | = 2 now for the above b can take 3 , 7 for every b = 3 a can have 13 , 7 and for b = 7 a can have 17 and 3 so ' a ' has four solutions 13 , 7 , 17 and 3 for a = 13 ; q has 18 or - 8 thus has 2 for every combination hence 4 x 2 = 8 answer d
a ) 0 , b ) 2 , c ) 4 , d ) 8 , e ) more than 8
d
subtract(add(subtract(10, 5), 5), 2)
subtract(n1,n0)|add(n0,#0)|subtract(#1,n3)
general
D
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 2120 ?
"2120 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1892.85 answer : c"
a ) rs . 1000.15 , b ) rs . 1100.95 , c ) rs . 1892.85 , d ) rs . 1050.85 , e ) rs . 1200.25
c
divide(2120, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1))
multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|
general
C