Problem
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Rationale
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correct
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5 values
annotated_formula
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a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 8 % of bananas were rotten . find the percentage of fruits in good condition .
"otal number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ— 600 = 9000 / 100 = 90 number of rotten bananas = 8 % of 400 = 8 / 100 Γ— 400 = 3200 / 100 = 32 therefore , total number of rotten fruits = 90 + 32 = 122 therefore number of fruits in good condition = 1000 - 12...
a ) 87.8 % , b ) 87.9 % , c ) 37.8 % , d ) 87.7 % , e ) 22.8 %
a
multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(8, const_100)))), add(600, 400)), const_100)
add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)|
gain
reema took a loan of rs 1800 with simple interest for as many years as the rate of interest . if she paid rs . 632 as interest at the end of the loan period , what was the rate of interest .
"explanation : let rate = r % then time = r years . = > 1800 βˆ— r βˆ— r / 100 = 632 = > r 2 = 35.11 = > r = 5.925 % option a"
a ) 5.925 % , b ) 5.253 % , c ) 7 % , d ) 6.52 % , e ) 5.200 %
a
sqrt(divide(632, divide(1800, const_100)))
divide(n0,const_100)|divide(n1,#0)|sqrt(#1)|
gain
12 points lie on a circle . how many cyclic quadrilaterals can be drawn by using these points ?
explanation : for any set of 4 points we get a cyclic quadrilateral . number of ways of choosing 4 points out of 12 points is \ inline { \ color { black } 12 c _ 4 { } } = 495 . therefore , we can draw 495 quadrilaterals answer : c ) 495
['a ) 334', 'b ) 377', 'c ) 495', 'd ) 766', 'e ) 261']
c
divide(factorial(12), multiply(factorial(const_4), factorial(subtract(12, const_4))))
factorial(n0)|factorial(const_4)|subtract(n0,const_4)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)
geometry
there are 15 members in a family . it takes 180 days for a women to complete the work , however man completes in 120 days . they take 17 days to complete the work if men work on alternate days and women work every third day . if all of them started together on the 1 st day , then how many men are there in the family
"let the number of men be m and so the number of women = 15 - m out of 17 days , men and women work together on 1,4 , 7,10 , 13,16 th days ( total of 6 days ) from the start . men work on alternate days , that is 1 , 3,5 , 7,9 , 11,13 , 15,17 th days ( total of 9 days ) . let the total work be 360 units ( lcm of 180 an...
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
d
subtract(15, divide(180, 15))
divide(n1,n0)|subtract(n0,#0)|
physics
there is a train and car . the ratio between the speed of a train & a car is 16 : 15 respectively . also , a bus covered a distance of 480 km in 8 hours . the speed of the bus is 3 / 4 th of the speed of the train . how many kilometers will the car cover in 4 hours ?
"the speed of the bus is 480 / 8 = 60 km / hr the speed of the train is ( 60 * 4 ) / 3 = 80 km / hr the speed of the car is 80 / 16 * 15 = 75 km / hr the distance covered by the car in 4 hours is 75 Γ— 4 = 300 km the answer is c ."
a ) 200 , b ) 250 , c ) 300 , d ) 350 , e ) 400
c
multiply(divide(multiply(divide(multiply(divide(480, 4), 8), 3), 15), 16), 4)
divide(n2,n5)|multiply(n3,#0)|divide(#1,n4)|multiply(n1,#2)|divide(#3,n0)|multiply(n6,#4)|
physics
farm tax is levied on the 75 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :
"this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 75 ) = 16.667 % the...
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 0.2083 % , e ) 16.667 %
e
divide(multiply(multiply(divide(480, 3840), const_100), const_100), 75)
divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|
general
what is x if 2 x - y = 14 and y = 2 ?
substitute y by 2 in 2 x - y = 14 2 x - 2 = 14 2 x = 16 if we substitute x by 8 , we have 2 ( 8 ) = 16 . hence x = 8 correct answer b
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14
b
subtract(subtract(14, 2), multiply(2, 2))
multiply(n0,n0)|subtract(n1,n0)|subtract(#1,#0)
general
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
in what time will a train 200 m long cross an electric pole , it its speed be 216 km / hr ?
"speed = 216 * 5 / 18 = 60 m / sec time taken = 200 / 60 = 3.3 sec . answer : d"
a ) 2.5 sec , b ) 2.8 sec , c ) 8.5 sec , d ) 3.3 sec , e ) 4.5 sec
d
divide(200, multiply(216, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
if t = 1 / ( 1 * 2 ) + 1 / ( 2 * 3 ) + 1 / ( 3 * 4 ) + 1 / ( 4 * 5 ) + . . . . . . . . . 1 / [ n ( n + 1 ) ] , for some positive integer n , what is the smallest value of n such that t > 097 ?
we can break the t as : ( n + 1 - n ) / n ( n + 1 ) - - - - for n = 1 to n 1 / n - 1 / n + 1 plug value , 1,2 . . n we will get a pattern 1 - 1 / 2 - - - for 1 1 / 2 - 1 / 3 - - for 2 1 / 3 - 1 / 4 - - for 4 now work with options . . for 31 , t = 1 - 1 / 32 = 30 / 31 = 0.968 for 32 , t = 1 - 1 / 33 = 0,9696 for 33 , t ...
a ) 29 , b ) 30 , c ) 31 , d ) 32 , e ) 33
e
divide(97, subtract(const_100, 97))
subtract(const_100,n14)|divide(n14,#0)
general
jack and jill work at a hospital with 6 other workers . for an internal review , 2 of the 8 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ?
"1 / 8 c 2 = 1 / 28 . answer : d ."
a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 15 , d ) 1 / 28 , e ) 2 / 3
d
inverse(divide(factorial(8), multiply(factorial(2), factorial(6))))
factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)|
physics
a pupil ' s marks were wrongly entered as 83 instead of 70 . due to the average marks for the class got increased by half . the number of pupils in the class is ?
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 83 - 70 ) = > x / 2 = 13 = > x = 26 . answer : c"
a ) 18 , b ) 82 , c ) 26 , d ) 27 , e ) 29
c
multiply(subtract(83, 70), const_2)
subtract(n0,n1)|multiply(#0,const_2)|
general
calculate the average of all the numbers between 18 and 57 which are divisible by 7 .
"explanation : numbers divisible by 7 are 21,28 , 35,42 , 49,56 , average = ( 21 + 28 + 35 + 42 + 49 + 56 , ) / 6 = 231 / 6 = 38.5 answer : b"
a ) 36.9 , b ) 38.5 , c ) 18.5 , d ) 32.5 , e ) 28.5
b
multiply(divide(add(add(floor(divide(18, 7)), const_1), floor(divide(57, 7))), const_2), 7)
divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6)|
general
what is the square root of 6,561 ?
"a square root of a number is a number times itself equals the original number . ex ) the square root of 81 is 9 because 9 times itself ( 9 ) equals the original number ( 81 ) . 81 * 81 = 6,561 6561 divided by x = 81 x = 81 ( e ) 81"
a ) 18 , b ) 9 , c ) 45 , d ) 62 , e ) 81
e
circle_area(divide(6,561, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
other
how many positive integers less than 100 are neither multiples of 2 or 5 .
"to answer this q we require to know 1 ) multiples of 2 till 100 = 100 / 2 = 50 2 ) multiples of 5 till 100 = 100 / 5 = 20 add the two 50 + 20 = 70 ; subtract common terms that are multiple of both 2 and 5 . . lcm of 2 and 5 = 10 multiples of 10 till 100 = 100 / 10 = 10 so total multiples of 2 and 5 = 70 - 10 = 6 ans =...
a ) a ) 30 , b ) b ) 31 , c ) c ) 40 , d ) d ) 33 , e ) e ) 34
c
divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), factorial(subtract(const_4, const_1))))
add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|
general
a trained covered x km at 50 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km .
"total time taken = x / 50 + 2 x / 20 hours = 6 x / 50 = 3 x / 25 hours average speed = 3 x / ( 3 x / 25 ) = 25 kmph answer : c"
a ) 22 , b ) 99 , c ) 25 , d ) 66 , e ) 887
c
divide(multiply(50, 3), add(divide(50, 50), divide(multiply(2, 50), 20)))
divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)|
general
two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores is 1 / 3 , what is the value of y / x ?
"answer is d : 1 / 4 x - y = ( x + y ) / 3 solving for y / x = 1 / 4"
a ) 3 . , b ) 2 . , c ) 1 / 2 . , d ) 1 / 4 , e ) there is n ' t enough data to answer the question .
d
add(3, divide(multiply(3, 1), 3))
multiply(n0,n1)|divide(#0,n1)|add(n1,#1)|
general
a drink vendor has 60 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ?
"the number of liters in each can = hcf of 60 , 144 and 368 = 4 liters . number of cans of maaza = 60 / 4 = 15 number of cans of pepsi = 144 / 4 = 36 number of cans of sprite = 368 / 4 = 92 the total number of cans required = 15 + 36 + 92 = 143 cans . answer : c"
a ) 135 , b ) 137 , c ) 143 , d ) 310 , e ) 380
c
add(divide(368, gcd(gcd(60, 144), 368)), add(divide(60, gcd(gcd(60, 144), 368)), divide(144, gcd(gcd(60, 144), 368))))
gcd(n0,n1)|gcd(n2,#0)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)|
general
when 1 / 10 % of 4,000 is subtracted from 1 / 10 of 4,000 , the difference is
"1 / 10 % of 4000 = 4 1 / 10 of 4000 = 400 400 - 4 = 396 ans : c"
a ) 50 , b ) 200 , c ) 396 , d ) 398 , e ) 400
c
multiply(inverse(10), multiply(multiply(const_100, 10), add(const_4, const_4)))
add(const_4,const_4)|inverse(n4)|multiply(n4,const_100)|multiply(#0,#2)|multiply(#1,#3)|
general
if 15 % of 30 % of 50 % of a number is 99 , then what is the number ?
let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 99 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 99 = > a = 10 * 20 * 10 * 2 = 4400 . answer : c
a ) 4000 , b ) 3050 , c ) 4400 , d ) 4500 , e ) none of these
c
divide(99, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100)))
divide(n0,const_100)|divide(n2,const_100)|divide(n1,const_100)|multiply(#1,#2)|multiply(#0,#3)|divide(n3,#4)
gain
working together , printer a and printer b would finish the task in 24 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 4 pages a minute more than printer a ?
answer : b .
a ) 600 , b ) 480 , c ) 1000 , d ) 1200 , e ) 1500
b
multiply(divide(4, subtract(divide(60, 24), const_1)), 60)
divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)|
physics
if a farmer wants to plough a farm field on time , he must plough 100 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initial...
"let x be the number of days in the initial plan . therefore , the whole field is 100 Γ’ β€Ή … x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 100 x = 85 ( x + 2 ) + 40 15 x = 210 x = 14 so the farmer planned to have ...
a ) 600 , b ) 490 , c ) 720 , d ) 1400 , e ) 1679
d
multiply(divide(add(40, multiply(2, 85)), subtract(100, 85)), 100)
multiply(n1,n2)|subtract(n0,n1)|add(n3,#0)|divide(#2,#1)|multiply(n0,#3)|
physics
in a certain town , the ratio of ny yankees fans to ny mets fans is 3 : 2 , and the ratio of ny mets fans to boston red sox fans is 4 : 5 . if there are 330 baseball fans in the town , each of whom is a fan of exactly one of those 3 teams , how many ny mets fans are there in this town ?
the ratio of yankees : mets : red sox = 6 : 4 : 5 the mets fans are 4 / 15 of the population . ( 4 / 15 ) * 330 = 88 the answer is b .
a ) 72 , b ) 88 , c ) 95 , d ) 120 , e ) 135
b
multiply(divide(330, multiply(5, 3)), 4)
multiply(n0,n3)|divide(n4,#0)|multiply(n2,#1)
other
a retailer sells 5 shirts . he sells first 2 shirts for $ 30 and $ 20 . if the retailer wishes to sell the 5 shirts for an overall average price of over $ 30 , what must be the minimum average price of the remaining 3 shirts ?
first 2 shirts are sold for $ 30 and $ 20 = $ 50 . to get average price of $ 30 , total sale should be 5 * $ 30 = $ 150 so remaining 3 shirts to be sold for $ 150 - $ 50 = $ 100 answer should be 100 / 3 = $ 33.33 that is c
a ) $ 28.50 , b ) $ 30.50 , c ) $ 33.33 , d ) $ 40.50 , e ) $ 50.00
c
divide(subtract(multiply(30, 5), add(30, 20)), 3)
add(n2,n3)|multiply(n0,n2)|subtract(#1,#0)|divide(#2,n6)
general
ratio between rahul and deepak is 4 : 3 , after 10 years rahul age will be 26 years . what is deepak present age .
explanation : present age is 4 x and 3 x , = > 4 x + 10 = 26 = > x = 16 so deepak age is = 3 ( 4 ) = 12 option a
a ) 12 , b ) 15 , c ) 20 , d ) 22 , e ) 24
a
divide(multiply(subtract(26, 10), 3), 4)
subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)
other
difference of two numbers is 1760 . 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 = 1760 5 * y / 3 – y = 1760 2 * y / 3 = 1760 y = [ ( 1760 * 3 ) / 2 ] = 2640 . one number = 2640 , second number = 5 * y / 3 = 4400 . answer b ."
a ) 2640 , 1000 , b ) 3660 , 4400 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 4660 , 3000
b
multiply(divide(7.5, 12.5), divide(1760, subtract(const_1, divide(7.5, 12.5))))
divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2)|
gain
if a clock strikes 12 in 33 seconds , it will strike 6 in how many seconds ?
in order to strike 12 , there are 11 intervals of equal time = 33 ⁄ 11 = 3 seconds each therefore , to strike 6 it has 5 equal intervals , it requires 5 Γ— 3 = 15 sec . answer b
a ) 33 ⁄ 2 , b ) 15 , c ) 12 , d ) 22 , e ) none of these
b
multiply(subtract(6, const_1), divide(33, subtract(12, const_1)))
subtract(n0,const_1)|subtract(n2,const_1)|divide(n1,#0)|multiply(#2,#1)
physics
car a runs at the speed of 50 km / hr & reaches its destination in 8 hr . car b runs at the speed of 25 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ?
"sol . distance travelled by car a = 50 Γ— 8 = 400 km distance travelled by car b = 25 Γ— 4 = 100 km ratio = 400 / 100 = 4 : 1 answer : a"
a ) 4 : 1 , b ) 11 : 8 , c ) 13 : 7 , d ) 15 : 7 , e ) 16 : 9
a
divide(multiply(50, 8), multiply(25, 4))
multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)|
physics
from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 25 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ?
"1.25 x = 45 - - > 5 / 4 * x = 45 - - > x = 45 * 4 / 5 = 180 / 5 = 36 . answer : a ."
a ) 36 , b ) 39 , c ) 40 , d ) 41 , e ) 42
a
multiply(divide(const_100, add(const_100, 25)), 45)
add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)|
gain
if 3 people a , b , c can do a work simultaneously in 6 days . a and b can do work in 10 days . b and c can do work in 15 days . c and a can do the same work in 20 days . in how many days that a can complete the same work alone ?
a + b + c = 1 / 6 ( one day work ) a + b = 1 / 10 ; b + c = 1 / 15 ; c + a + 1 / 20 b = 1 / 15 - c ; c = 1 / 20 - a ; - - - - - - - - - ( 2 ) substitute equation 2 in a + b = 1 / 10 a = 1 / 10 - b ; a = 1 / 10 - 1 / 15 + c ; a = 1 / 10 - 1 / 15 + 1 / 20 - a ; 2 a = 1 / 10 - 1 / 15 + 1 / 20 ; a = 1 / 24 ( one day work )...
a ) 20 , b ) 25 , c ) 30 , d ) 24 , e ) 19
d
divide(const_1, subtract(divide(add(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 20)), const_2), divide(const_1, 15)))
divide(const_1,n2)|divide(const_1,n3)|divide(const_1,n4)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|subtract(#5,#1)|divide(const_1,#6)
physics
a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 170 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ?
"let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 170 = > . 1 x + 3 / 4 x = 170 = > x = 200 200 - 170 = pure gala = 30 . answer a"
a ) 30 , b ) 33 , c ) 55 , d ) 77 , e ) 88
a
subtract(divide(170, add(divide(10, const_100), divide(3, 4))), 170)
divide(n0,const_100)|divide(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,n1)|
general
a man is 46 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is :
"let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 46 ) + 2 = 2 ( x + 2 ) x + 48 = 2 x + 4 x = 44 . answer : a"
a ) 44 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) 16 years
a
divide(subtract(46, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1))
multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)|
general
pipe a that can fill a tank in an hour and pipe b that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe b is shut 10 minutes before the tank overflows . when will the tank overflow ?
the last 10 minutes only pipe a was open . since it needs 1 hour to fill the tank , then in 10 minutes it fills 1 / 5 th of the tank , thus 5 / 6 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 5 / 6 th of the tank they need ( time ) = ( work ) /...
a ) 37 mins , b ) 35 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins
a
subtract(add(divide(multiply(add(divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), divide(multiply(const_1, const_60), const_2)), divide(multiply(multiply(const_1, const_60), divide(multiply(const_1, const_60), const_2)), multiply(const_1, const_60))), 10), divide(multiply(mult...
multiply(const_1,const_60)|divide(#0,const_2)|multiply(#1,#0)|divide(#2,#1)|divide(#2,#0)|add(#3,#4)|multiply(n0,#5)|divide(#6,#4)|add(n0,#7)|subtract(#8,const_3)
physics
what is the maximum value of vx - yz . if the value of v , x , y , z have to be chosen from the set a where a ( - 3 , - 2 , - 1 , 01 , 23 )
explanation : to maximize the value of vx - yz , we make yz negative and vx as maximum as possible using given value . vx βˆ’ yz = ( βˆ’ 3 ) 2 βˆ’ ( βˆ’ 3 Γ— 2 ) vx βˆ’ yz = ( βˆ’ 3 ) 2 βˆ’ ( βˆ’ 3 Γ— 2 ) = 15 answer : a
a ) 15 , b ) 66 , c ) 25 , d ) 88 , e ) 17
a
subtract(power(negate(3), 2), multiply(negate(3), 2))
negate(n0)|multiply(n1,#0)|power(#0,n1)|subtract(#2,#1)
general
x and y are two towns . ganesh covers the distance from x to y at an average speed of 54 km / hr . however , he covers the distance from y to x at an average speed of 36 km / hr . his average speed during the whole journey in km / hr . is :
"solution : average speed = 2 xy / x + y = 2 * 54 * 36 / 54 + 36 = 43.2 answer : b"
a ) 34 , b ) 43.2 , c ) 40 , d ) 38 , e ) 29
b
divide(add(add(subtract(54, const_10), const_100), add(subtract(54, const_10), const_100)), add(divide(add(subtract(54, const_10), const_100), 54), divide(add(subtract(54, const_10), const_100), 36)))
subtract(n0,const_10)|add(#0,const_100)|add(#1,#1)|divide(#1,n0)|divide(#1,n1)|add(#3,#4)|divide(#2,#5)|
general
p runs 20 % faster than q so p gives q a 50 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ?
let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.2 v = ( d - 50 ) / v d = 1.2 d - 60 0.2 d = 60 d = 300 meters . the answer is b .
a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 450
b
add(divide(multiply(50, const_100), 20), 50)
multiply(n1,const_100)|divide(#0,n0)|add(n1,#1)
gain
suppose 10 monkeys take 10 minutes to eat 10 bananas . how many monkeys would it take to eat 70 bananas in 70 minutes ?
"one monkey takes 10 min to eat 1 banana , so in 70 mins 1 monkey will eat 7 bananas , so for 70 bananas in 70 min we need 70 / 7 = 10 monkeys answer : d"
a ) 9 , b ) 17 , c ) 11 , d ) 10 , e ) 13
d
divide(70, divide(70, 10))
divide(n3,n0)|divide(n3,#0)|
physics
a man walks at a rate of 10 mph . after every ten miles , he rests for 5 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 * 5 = 20 minutes . total time = 300 + 20 = 320 minutes . answer : c .
a ) 300 , b ) 318 , c ) 320 , d ) 324 , e ) 330
c
add(multiply(5, const_4), multiply(divide(50, 10), const_60))
divide(n2,n0)|multiply(n1,const_4)|multiply(#0,const_60)|add(#1,#2)
physics
on a map , 4.5 inches represent 45 miles . how many miles approximately is the distance if you measured 254 centimeters assuming that 1 - inch is 2.54 centimeters ?
"1 inch = 2.54 cm 4.5 inch = 2.54 * 4.5 cm 11.43 cm = 45 miles 254 cms = 45 / 11.43 * 254 = 1000 miles answer : a"
a ) 1000 , b ) 1100 , c ) 1200 , d ) 1250 , e ) 1300
a
multiply(divide(254, 2.54), divide(45, 4.5))
divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)|
physics
what is the total number of positive integers that are less than 400 and that have no positive factor in common with 400 other than 1 ?
"since 400 = 2 ^ 4 * 5 ^ 2 then a number can not have 2 and / or 5 as a factor . the odd numbers do not have 2 as a factor and there are 200 odd numbers from 1 to 400 . we then need to eliminate the 40 numbers that end with 5 , that is 5 , 15 , 25 , . . . , 395 . there are a total of 200 - 40 = 160 such numbers between...
a ) 100 , b ) 120 , c ) 150 , d ) 160 , e ) 180
d
subtract(subtract(400, 1), subtract(add(add(divide(subtract(subtract(400, const_2), const_2), const_2), 1), add(divide(subtract(subtract(400, divide(400, const_100)), divide(400, const_100)), divide(400, const_100)), 1)), add(divide(subtract(subtract(400, const_10), const_10), const_10), 1)))
divide(n0,const_100)|subtract(n0,n2)|subtract(n0,const_2)|subtract(n0,const_10)|subtract(#2,const_2)|subtract(n0,#0)|subtract(#3,const_10)|divide(#4,const_2)|divide(#6,const_10)|subtract(#5,#0)|add(#7,n2)|add(n2,#8)|divide(#9,#0)|add(#12,n2)|add(#10,#13)|subtract(#14,#11)|subtract(#1,#15)|
general
john bought 9.25 m of cloth for $ 416.25 . find the cost price per metre .
"cloth bought by john = 9.25 m cost of 9.25 m = $ 416.25 cost per metre = 416.25 Γ· 9.25 the cost of the cloth per metre = $ 45 answers : c"
a ) 46 , b ) 47 , c ) 45 , d ) 56 , e ) 54
c
divide(416.25, 9.25)
divide(n1,n0)|
physics
if p is a prime number greater than 3 , find the remainder when p ^ 2 + 12 is divided by 12 .
"every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 13 = 36 n ^ 2 + 12 n + 1 + 12 = 36 n ^ 2 + 12 n + 12 + 1 if p = 6 n - 1 , then p ^ 2 + 13 = 36 n ^ 2 - 12 n + 1 + 12 = 36 n ^ 2 - 12 n + 12 + 1 when divided by 12 , it must leave a remainder of 1 . the answer is b ."
a ) 6 , b ) 1 , c ) 0 , d ) 8 , e ) 7
b
subtract(add(12, power(add(const_1, const_4), 2)), multiply(12, 3))
add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)|
general
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 15 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ?
"1 h - - - - - 5 ? - - - - - - 60 12 h rs = 15 + 21 = 36 t = 12 d = 36 * 12 = 432 answer : a"
a ) 432 , b ) 444 , c ) 2881 , d ) 287 , e ) 221
a
add(multiply(divide(60, subtract(21, 15)), 15), multiply(divide(60, subtract(21, 15)), 21))
subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)|
physics
what is the rate percent when the simple interest on rs . 910 amount to rs . 260 in 4 years ?
"260 = ( 910 * 4 * r ) / 100 r = 7.14 % answer : a"
a ) 7.14 , b ) 6.14 , c ) 5.14 , d ) 3.14 , e ) 2.14
a
divide(multiply(const_100, 260), multiply(910, 4))
multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|
gain
in 1979 approximately 1 / 3 of the 37.3 million airline passengers traveling to or from the united states used kennedy airport . if the number of such passengers that used miami airport was 1 / 4 the number that used kennedy airport and 3 times the number that used logan airport , approximately how many millions of the...
"number of passengers using kennedy airport = 37 / 3 = ~ 12.43 passengers using miami airport = 12.43 / 4 = ~ 3.1 passengers using logan airport = 3.1 / 3 = ~ 1.03 so e"
a ) 18.6 , b ) 9.3 , c ) 6.2 , d ) 3.1 , e ) 1.03
e
divide(divide(37.3, 3), multiply(3, 4))
divide(n3,n2)|multiply(n5,n6)|divide(#0,#1)|
general
if $ 792 are divided between worker a and worker b in the ratio 5 : 6 , what is the share that worker b will get ?
"worker b will get 6 / 11 = 54.54 % the answer is a ."
a ) 54.54 % , b ) 55.55 % , c ) 56.56 % , d ) 57.57 % , e ) 58.58 %
a
divide(792, add(5, 6))
add(n1,n2)|divide(n0,#0)|
other
evaluate : 1182 - 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 1182 - 12 * 3 * 2 = 1182 - 8 = 174 correct answer c"
a ) 674 , b ) 574 , c ) 174 , d ) 274 , e ) 374
c
subtract(1182, multiply(multiply(12, 3), 2))
multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|
general
if a ( a - 4 ) = 5 and b ( b - 4 ) = 5 , where a β‰  b , then a + b =
"i . e . if a = - 1 then b = 5 or if a = 5 then b = - 1 but in each case a + b = - 1 + 5 = 4 answer : option c"
a ) βˆ’ 48 , b ) βˆ’ 4 , c ) 4 , d ) 9 , e ) 48
c
subtract(subtract(subtract(subtract(add(add(4, 5), subtract(4, 5)), const_1), const_1), const_1), const_1)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)|
general
in a software company 4 / 10 of people know c + + , 2 / 4 of them know java , 2 / 8 of them know neither , find the total possibility to know c + + ?
assume 100 persons in a company person knows c + + = 4 / 10 = 100 * 4 / 100 = 20 persons person knows java = 2 / 4 = 100 * 2 / 4 = 50 persons neither = 2 / 8 = 100 * 2 / 8 = 25 persons according to our assumption 100 person = 20 + 50 + 25 = 95 person covered ( 5 remaining ) so , total possibility = 20 person ( c + + ) ...
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
a
subtract(subtract(const_1, divide(2, 8)), divide(2, 4))
divide(n2,n5)|divide(n2,n0)|subtract(const_1,#0)|subtract(#2,#1)
general
the average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 10 years . find the average age of the women ?
"20 + 10 + 8 * 2 = 46 / 2 = 23 answer : e"
a ) 87 , b ) 98 , c ) 30 , d ) 28 , e ) 23
e
divide(add(add(20, 10), multiply(8, 2)), const_2)
add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)|
general
the number 0.8 is how much greater than 1 / 2 ?
"let x be the difference then . 8 - 1 / 2 = x 8 / 10 - 1 / 2 = x x = 3 / 10 ans b"
a ) Β½ , b ) 3 / 10 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500
b
subtract(0.8, divide(1, 2))
divide(n1,n2)|subtract(n0,#0)|
general
two trains running in opposite directions cross a man standing on the platform in 20 seconds and 18 seconds respectively and they cross each other in 19 seconds . the ratio of their speeds is :
"let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 20 x meters , and length of the second train = 18 y meters . ( 20 x + 15 y ) / ( x + y ) = 19 = = > 20 x + 18 y = 19 x + 19 y = = > x = y = = > x / y = 1 / 1 answer : option a"
a ) 1 : 1 , b ) 3 : 2 , c ) 3 : 8 , d ) 3 : 25 , e ) 3 : 4
a
divide(subtract(20, 19), subtract(19, 18))
subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)|
physics
cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 68 after it went on sale for 15 % off , what is the difference between the original price and t...
0.85 * { original price } = $ 68 - - > { original price } = $ 80 . { final price } = $ 68 * 1.25 = $ 85 . the difference = $ 85 - $ 85 = $ 5 . answer : d .
a ) $ 0.00 , b ) $ 1.00 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.80
d
subtract(multiply(68, divide(add(const_100, 25), const_100)), divide(68, divide(subtract(const_100, 15), const_100)))
add(n1,const_100)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#3)|multiply(n2,#2)|subtract(#5,#4)
general
the cost per pound of milk powder and coffee were the same in june . in july , the price of coffee shot up by 200 % and that of milk powder dropped by 60 % . if in july , a mixture containing equal quantities of milk powder and coffee costs $ 5.10 for 3 lbs , how much did a pound of milk powder cost in july ?
"lets assume price of coffee in june = 100 x price of tea in june = 100 x price of coffee in july = 300 x ( because of 200 % increase in price ) price of tea in july = 40 x ( because of 60 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of tea in july will be = 450 x + 60 x = 510 x as per question 510 x = ...
a ) $ 4 , b ) $ 0.4 , c ) $ 1 , d ) $ 3 , e ) $ 1.65
b
subtract(const_1, divide(60, const_100))
divide(n1,const_100)|subtract(const_1,#0)|
general
each of the 55 points is placed either inside or on the surface of a perfect sphere . if 60 % 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 60 % * 55 = 33 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 33 points , so at best we can form 16 pairs ( 32 ) . so , answer is ( d )
['a ) 7', 'b ) 11', 'c ) 13', 'd ) 16', 'e ) 38']
d
divide(multiply(55, divide(60, const_100)), const_2)
divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_2)
geometry
on the richter scale , which measures the total amount of energy released during an earthquake , a reading of x - 1 indicates one - tenth the released energy as is indicated by a reading of x . on that scale , the frequency corresponding to a reading of 2 is how many times as great as the frequency corresponding to a r...
"if richter scale reading goes from x - 1 to x it will be 10 if richter scale reading goes from 2 to 3 it will be 10 if richter scale reading goes from 3 to 4 it will be 10 if richter scale reading goes from 4 to 5 it will be 10 if richter scale reading goes from 5 to 6 it will be 10 similarly if richter scale reading ...
a ) 40 , b ) 50 , c ) 10 ^ 4 , d ) 10 ^ 5 , e ) 10 ^ 6
e
power(const_10, subtract(2, 8))
subtract(n1,n2)|power(const_10,#0)|
general
a train leaves mumabai at 9 am at a speed of 40 kmph . after one hour , another train leaves mumbai in the same direction as that of the first train at a speed of 60 kmph . when and at what distance from mumbai do the two trains meet ?
"when the second train leaves mumbai the first train covers 40 * 1 = 40 km so , the distance between first train and second train is 40 km at 10.00 am time taken by the trains to meet = distance / relative speed = 40 / ( 60 - 40 ) = 2 hours so , the two trains meet at 12 a . m . the two trains meet 2 * 60 = 120 km away...
a ) 187 , b ) 279 , c ) 120 , d ) 278 , e ) 379
c
multiply(divide(multiply(40, const_1), subtract(60, 40)), 60)
multiply(n1,const_1)|subtract(n2,n1)|divide(#0,#1)|multiply(n2,#2)|
physics
for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.30 – 6.30 * ?
"since z * is defined as the greatest positive even integer less than or equal to z , then 6.30 * = 6 ( the greatest positive even integer less than or equal to 6.30 is 6 ) . hence , 6.30 – 6.30 * = 6.30 - 6 = 0.30 answer : e ."
a ) 1.5 , b ) 0.5 , c ) 6.25 , d ) 0.25 , e ) 0.3
e
subtract(6.30, floor(6.30))
floor(n0)|subtract(n0,#0)|
general
the guests at a football banquet consumed a total of 406 pounds of food . if no individual guest consumed more than 2.5 pounds of food , what is the minimum number of guests that could have attended the banquet ?
"to minimize one quantity maximize other . 162 * 2.5 ( max possible amount of food a guest could consume ) = 405 pounds , so there must be more than 162 guests , next integer is 163 . answer : d ."
a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164
d
add(floor(divide(406, 2.5)), const_1)
divide(n0,n1)|floor(#0)|add(#1,const_1)|
general
in a group of 600 readers who read science fiction or literacy works or both , 200 read science fiction and 230 read literacy works . how many read both science fiction and literacy works ?
"consider total number of reader n ( s u l ) = 600 people who read science fiction n ( s ) = 400 people who read literacy works n ( l ) = 230 both science fiction and literacy n ( s Γ’ Λ† Β© l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s Γ’ Λ† Β© l ) 600 = 400 + 230 - n ( s Γ’ Λ† Β© l ) n ( s Γ’ Λ† Β© l ) = 630 - 600 n ( s Γ’ Λ† Β© ...
a ) 70 , b ) 30 , c ) 25 , d ) 100 , e ) 110
b
subtract(add(200, 230), 600)
add(n1,n2)|subtract(#0,n0)|
other
if x : y = 3 : 5 , find the value of ( 2 x + 4 y ) : ( 3 x – y )
"explanation : given : x / y = 3 / 5 ( 2 x + 4 y ) / ( 3 x – y ) = ( 2 * 3 + 4 * 5 ) : ( 3 * 3 – 5 ) = 26 : 4 = 13 : 2 answer : a"
a ) 13 : 2 , b ) 11 : 2 , c ) 13 : 3 , d ) 13 : 4 , e ) 13 : 5
a
divide(add(power(2, 4), 4), add(add(3, 5), 4))
add(n0,n1)|power(n2,n3)|add(n3,#1)|add(n3,#0)|divide(#2,#3)|
general
if daily wages of a man is double to that of a woman , how many men should work for 20 days to earn rs . 14400 ? given that wages for 40 women for 30 days are rs . 21600 .
"explanation : wages of 1 woman for 1 day = 21600 / 40 Γ£ β€” 30 wages of 1 man for 1 day = 21600 Γ£ β€” 2 / 40 Γ£ β€” 30 wages of 1 man for 20 days = 21600 Γ£ β€” 2 Γ£ β€” 20 / 40 Γ£ β€” 30 number of men = 14400 / ( 21600 Γ£ β€” 2 Γ£ β€” 20 / 40 Γ£ β€” 30 ) = 144 / ( 216 Γ£ β€” 40 / 40 Γ£ β€” 30 ) = 20 answer : option e"
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20
e
divide(14400, multiply(20, multiply(const_2, divide(21600, multiply(40, 30)))))
multiply(n2,n3)|divide(n4,#0)|multiply(#1,const_2)|multiply(n0,#2)|divide(n1,#3)|
physics
persons a and b . person a picks a random no . from 1 to 1000 . then person b picks a random no . from 1 to 1000 . what is the probability of b getting no . greater then what a has picked .
probability of a choosing 1 and b greater = ( 1 / 1000 ) * ( 999 / 1000 ) ; probability of a choosing 2 and b greater = ( 1 / 1000 ) * ( 998 / 1000 ) ; probability of a choosing 3 and b greater = ( 1 / 1000 ) * ( 997 / 1000 ) ; . . . ; probability of a choosing 998 and b greater = ( 1 / 1000 ) * ( 2 / 1000 ) ; probabil...
a ) 0.3995 , b ) 0.4995 , c ) 0.5995 , d ) 0.6995 , e ) 0.7995
b
divide(subtract(const_1, divide(1, 1000)), const_2)
divide(n0,n1)|subtract(const_1,#0)|divide(#1,const_2)
probability
cole drove from home to work at an average speed of 50 kmh . he then returned home at an average speed of 110 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ?
"let the distance one way be x time from home to work = x / 50 time from work to home = x / 110 total time = 2 hrs ( x / 50 ) + ( x / 110 ) = 2 solving for x , we get x = 275 / 4 time from home to work in minutes = ( 275 / 4 ) * 60 / 50 = 82.5 minutes ans = e"
a ) 66.5 , b ) 70.5 , c ) 72.5 , d ) 75.5 , e ) 82.5
e
multiply(divide(multiply(110, 2), add(50, 110)), const_60)
add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)|
physics
john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ?
"rakesh : rahul : ranjan : rohit = 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 5 x 60 ) = 12 rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : ...
a ) 59 , b ) 65 , c ) 70 , d ) 77 , e ) 98
a
add(add(add(divide(lcm(lcm(lcm(5, 1), 1), 3), 5), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 3))
lcm(n2,n3)|lcm(n3,#0)|lcm(n4,#1)|divide(#2,n2)|divide(#2,n3)|divide(#2,n4)|add(#3,#4)|add(#6,#4)|add(#7,#5)|
other
an equal number of desks and bookcases are to be placed along a library wall that is 15 meters long . each desk is 2 meters long , and each bookshelf is 1.5 meters long . if the maximum possible number of desks and bookcases are to be placed along the wall , then the space along the wall that is left over will be how m...
"let x be the number of desks and bookcases that are placed along the library wall . 2 x + 1.5 x < 15 3.5 x < 15 since x is a non negative integer , the largest number x can be is 4 . when x is 4 , the desks and bookcases take up 3.5 * 4 = 14 m , leaving 1 m of empty space . thus , i believe the answer is b ) 1"
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 3
b
subtract(15, add(multiply(const_4, 2), multiply(const_4, 1.5)))
multiply(n1,const_4)|multiply(n2,const_4)|add(#0,#1)|subtract(n0,#2)|
general
- 88 Γ— 39 + 312 = ?
- 88 Γ— 39 + 312 = - 88 Γ— ( 40 - 1 ) + 312 = - 88 Γ— 40 + 88 + 312 = - 3520 + 88 + 312 = - 3120 answer is c
a ) - 3200 , b ) 3120 , c ) - 3120 , d ) 3200 , e ) 3208
c
add(multiply(negate(88), 39), 312)
negate(n0)|multiply(n1,#0)|add(n2,#1)
general
what will be the cost of building a fence around a square plot with area equal to 25 sq ft , if the price per foot of building the fence is rs . 58 ?
"let the side of the square plot be a ft . a 2 = 25 = > a = 5 length of the fence = perimeter of the plot = 4 a = 20 ft . cost of building the fence = 20 * 58 = rs . 1160 . answer : a"
a ) 1160 , b ) 1251 , c ) 1102 , d ) 1352 , e ) 1450
a
multiply(square_perimeter(sqrt(25)), 58)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
a train 280 m long , running with a speed of 63 km / hr will pass a tree in
"solution speed = ( 63 x 5 / 18 ) m / sec = 35 / 2 msec time taken = ( 280 x 2 / 35 ) m / sec = 16 sec . answer b"
a ) 15 sec . , b ) 16 sec . , c ) 18 sec . , d ) 20 sec . , e ) none
b
multiply(divide(280, multiply(63, const_1000)), const_3600)
multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|
physics
a shop owner sells 15 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 15 m cloth ) – ( c . p . of 15 m cloth ) selling price of 5 m cloth = selling price of 15 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 5 m cloth = rs . 500 and s . p . of 5 m cloth = rs . rs . 15...
a ) 30 % , b ) 40 % , c ) 50 % , d ) 100 % , e ) 200 %
e
multiply(divide(10, subtract(15, 10)), const_100)
subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)|
gain
if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 7 k , what is one possible value for k ?
"first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 1...
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7
e
add(divide(power(2, 2), 2), const_1)
power(n1,n1)|divide(#0,n1)|add(#1,const_1)|
general
there were two candidates in an election . winner candidate received 62 % of votes and won the election by 348 votes . find the number of votes casted to the winning candidate ?
"w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 348 62 % - - - - - - - - ? = > 899 answer : c"
a ) 456 , b ) 744 , c ) 899 , d ) 1200 , e ) 1400
c
divide(multiply(divide(348, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100)
subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)|
gain
what is the greatest positive integer t such that 3 ^ t is a factor of 9 ^ 10 ?
"what is the greatest positive integer t such that 3 ^ t is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20"
a ) 5 , b ) t = 9 , c ) t = 10 , d ) t = 20 , e ) 30
d
multiply(subtract(9, 10), 10)
subtract(n1,n2)|multiply(n2,#0)|
other
in a ratio which is equal to 3 : 4 , if the antecedent is 12 , then the consequent is ?
"we have 3 / 4 = 12 / x 3 x = 48 x = 16 consequent = 16 answer is b"
a ) 12 , b ) 16 , c ) 20 , d ) 22 , e ) 25
b
add(12, divide(12, const_2))
divide(n2,const_2)|add(n2,#0)|
other
what is the rate percent when the simple interest on rs . 4000 amount to rs . 640 in 2 years ?
"interest for 1 year = 640 / 2 = 320 interest on rs 4000 p / a = 320 interest rate = 320 / 4000 * 100 = 8 % answer : b"
a ) 8.2 % , b ) 8 % , c ) 8.5 % , d ) 9 % , e ) 9.5 %
b
divide(multiply(const_100, 640), multiply(4000, 2))
multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)|
gain
if 2 x + y = 7 and x + 2 y = 8 , then ( x + y ) / 3 =
"we have two equations : 2 x + y = 7 x + 2 y = 8 notice that something nice happens when we add them . we get : 3 x + 3 y = 15 divide both sides by 3 to get : x + y = 5 so , ( x + y ) / 3 = 5 / 3 answer : e"
a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 5 / 3
e
divide(add(divide(subtract(multiply(7, 2), 8), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 8), subtract(multiply(2, 2), const_1))))), 3)
multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)|
general
farm tax is levied on the 90 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . william . mr . william paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is :
"this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to ( 480 / 3840 ) x 100 = 12.5 % in percentage terms . but the question asks ratio between his total land to total cultivated land . hence the answer is 12.5 % x ( 100 / 90 ) = 13.88 % the ...
a ) 15 % , b ) 25 % , c ) 0.125 % , d ) 13.88 % , e ) none
d
divide(multiply(multiply(divide(480, 3840), const_100), const_100), 90)
divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)|
general
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8500 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ?
"sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8500 = 34000 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 2000 salary of may is rs . 6500 salary of jan...
a ) 2177 , b ) 2876 , c ) 4500 , d ) 2981 , e ) 2711
c
subtract(multiply(8000, const_4), subtract(multiply(8500, const_4), 6500))
multiply(n0,const_4)|multiply(n1,const_4)|subtract(#1,n2)|subtract(#0,#2)|
general
don and his wife each receive an 8 percent annual raise . if don receives a raise rs . 800 and his wife receives a raise of rs . 840 , what is the difference between their annual income after their raises ?
don salary 8 % = 800 ; therefore , 100 % = 10000 his wife salary 8 % = 840 ; therefore , 100 % = 10500 after raise 8 % , don salary = 10000 + 800 = 10800 and his wife salary = 10500 + 840 = 11340 difference = 11340 - 10800 = 540 answer : d
a ) 40 , b ) 460 , c ) 500 , d ) 540 , e ) 150
d
subtract(add(divide(840, divide(8, const_100)), 840), add(divide(800, divide(8, const_100)), 800))
divide(n0,const_100)|divide(n2,#0)|divide(n1,#0)|add(n2,#1)|add(n1,#2)|subtract(#3,#4)
gain
if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 50 , what will be the compound interest on same values
"explanation : s . i . = p βˆ— r βˆ— t / 100 p = 50 βˆ— 100 / 5 βˆ— 2 = 500 amount = 500 ( 1 + 5100 ) 2 500 ( 21 / 20 βˆ— 21 / 20 ) = 551.25 c . i . = 551.25 βˆ’ 500 = 51.25 option c"
a ) rs . 51.75 , b ) rs 51.50 , c ) rs 51.25 , d ) rs 51 , e ) none of these
c
subtract(add(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(50, multiply(divide(5, const_100), 2)), 5), const_100), divide(50, multiply(divide(5, const_100), 2))), 5), const_100)), divide(50, m...
divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|subtract(#8,#2)|
gain
if x + y = 300 , x - y = 200 , for integers of x and y , y = ?
"x + y = 300 x - y = 200 2 x = 100 x = 50 y = 250 answer is b"
a ) 200 , b ) 250 , c ) 50 , d ) 115 , e ) 150
b
divide(add(300, 200), const_2)
add(n0,n1)|divide(#0,const_2)|
general
ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 66900 . what is his profit percent ?
"total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 66900 profit ( % ) = ( 66900 - 55000 ) / 55000 * 100 = 21.6 % answer : d"
a ) 22 , b ) 77 , c ) 18 , d ) 21.6 , e ) 88
d
multiply(divide(subtract(66900, add(42000, 13000)), add(42000, 13000)), const_100)
add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|
gain
a can do a work in 8 days . b can do the same work in 16 days . if both a & b are working together in how many days they will finish the work ?
"a rate = 1 / 8 b rate = 1 / 16 ( a + b ) rate = ( 1 / 8 ) + ( 1 / 16 ) = 3 / 16 a & b finish the work in 16 / 3 days correct option is b"
a ) 3 days , b ) 16 / 3 days , c ) 3 / 4 days , d ) 9 / 12 days , e ) 5 / 7 days
b
divide(multiply(8, 16), add(8, 16))
add(n0,n1)|multiply(n0,n1)|divide(#1,#0)|
physics
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 6 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the ...
"tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 4 / 6 ) = 1 / 2 hours peter worked for 3 / 2 hours so he painted 3 / 2 * 1 / 3 = 1 / 2 of the room . ...
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 2 / 3 , e ) 2 / 5
a
divide(const_4, add(multiply(const_4, 6), const_1))
multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)|
physics
if ( 1 – 1.25 ) n = 5 , then n =
"( 1 – 1.25 ) n = 5 simplify to get : - 0.25 n = 5 rewrite as ( - 1 / 4 ) n = 5 multiply both sides by - 4 to get : n = - 20 answer : a"
a ) βˆ’ 20 , b ) βˆ’ 140 , c ) βˆ’ 4 , d ) 4 , e ) 400
a
negate(multiply(5, const_4))
multiply(n2,const_4)|negate(#0)|
general
if in a race of 70 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by :
"explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 70 * 5 ) / 25 = 14 m hence , a beats b by 14 m . answer c"
a ) 20 m , b ) 16 m , c ) 14 m , d ) 10 m , e ) 15 m
c
multiply(divide(subtract(25, 20), 25), 70)
subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)|
physics
a certain social security recipient will receive an annual benefit of $ 12000 provided he has annual earnings of $ 9360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9360 . what amount of total annual earnings would result in a 65 percent reduction in the recipient ' s annual ...
for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12...
a ) $ 15,360 , b ) $ 17,360 , c ) $ 18,000 , d ) $ 21,960 , e ) $ 27,360
d
divide(multiply(divide(add(multiply(3, multiply(divide(subtract(65, multiply(const_3, const_10)), const_100), 12000)), 9360), add(multiply(3, multiply(divide(subtract(65, multiply(const_3, const_10)), const_100), 12000)), 9360)), const_3600), const_10)
multiply(const_10,const_3)|subtract(n5,#0)|divide(#1,const_100)|multiply(n0,#2)|multiply(n3,#3)|add(n1,#4)|divide(#5,#5)|multiply(#6,const_3600)|divide(#7,const_10)
general
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 5.2 hours , then the distance covered at 40 kmph is
"let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 5.2 or , x / 40 + ( 250 - x ) / 60 = 5.2 or , ( 3 x + 500 - 2 x ) / / 120 = 5.2 or 500 + x = 624 x = 124 answer : d"
a ) 130 km , b ) 120 km , c ) 100 km , d ) 124 km , e ) none of these
d
multiply(divide(subtract(multiply(60, 5.2), 250), negate(subtract(40, 60))), 40)
multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)|
physics
jane and ashley take 13 1 / 3 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an 8 day leave from work . this led to jane ' s w...
let us assume that the work is laying 40 bricks . jane = 3 bricks per day ashley = 1 brick per day together = 4 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 12 remaining bricks = 40 - 20 = 20 so together , they would take 20 / 4 = 5 total ...
a ) 10 days , b ) 15 days , c ) 16 days , d ) 17 days , e ) 20 days
d
subtract(multiply(inverse(multiply(4, divide(1, 3))), 40), 13)
divide(n1,n2)|multiply(n5,#0)|inverse(#1)|multiply(n3,#2)|subtract(#3,n0)
physics
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 82 , and the average of the female students ’ grades was 92 , how many female students took the test ?
"total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 82 thus m = 656 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 656 + 92 f ) / ( 8 + f ) = ...
a ) 8 , b ) 9 , c ) 10 , d ) 21 , e ) 32
e
divide(subtract(multiply(90, 8), multiply(82, 8)), subtract(92, 90))
multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|
general
if the length of the longest chord of a certain circle is 24 , what is the radius of that certain circle ?
"longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 24 = 2 * 12 so radius of the circle = 12 correct answer - a"
a ) 12 , b ) 5 , c ) 10 , d ) 15 , e ) 20
a
divide(24, const_2)
divide(n0,const_2)|
geometry
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 4 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ?
"1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 answer : b"
a ) 5768 , b ) 5760 , c ) 5762 , d ) 5766 , e ) 5712
b
divide(multiply(4, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1))
multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)|
physics
the class mean score on a test was 60 , and the standard deviation was 10 . if jack ' s score was within 2 standard deviations of the mean , what is the lowest score he could have received ?
1 sd from the mean is adding and subtrating the amount if standard deviation from the mean one time . 2 sd from the mean is adding and subtracting twice . 1 sd from the mean ranges from 70 to 50 , where 70 is within sd above the mean and 50 within 1 sd below the mean 2 sd = 10 twice = 20 from the the mean , which is 80...
a ) 30 , b ) 31 , c ) 45 , d ) 90 , e ) 20
e
multiply(2, 10)
multiply(n1,n2)
general
if an object travels at eight feet per second , how many feet does it travel in one hour ?
"explanation : if an object travels at 5 feet per second it covers 5 x 60 feet in one minute , and 5 x 60 x 60 feet in one hour . answer = 28800 answer : e ) 28800"
a ) 34880 , b ) 3778 , c ) 12788 , d ) 18000 , e ) 28800
e
multiply(multiply(const_3, const_60), const_60)
multiply(const_3,const_60)|multiply(#0,const_60)|
physics
the ratio of buses to cars on river road is 1 to 10 . if there are 90 fewer buses than cars on river road , how many cars are on river road ?
"b / c = 1 / 10 c - b = 90 . . . . . . . . . > b = c - 90 ( c - 90 ) / c = 1 / 10 testing answers . clearly eliminate abde put c = 100 . . . . . . . . . > ( 100 - 90 ) / 100 = 10 / 100 = 1 / 10 answer : c"
a ) 50 , b ) 40 , c ) 100 , d ) 30 , e ) 20
c
multiply(divide(90, subtract(10, 1)), 10)
subtract(n1,n0)|divide(n2,#0)|multiply(n1,#1)|
other
a merchant marks goods up by 75 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 57.5 % . what % discount did the merchant offer ?
"let p be the original price of the goods and let x be the rate after the markup . ( 1.75 p ) * x = 1.575 p x = 1.575 / 1.75 = 0.9 which is a discount of 10 % . the answer is a ."
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 %
a
multiply(divide(subtract(subtract(add(const_1, divide(75, const_100)), const_1), divide(57.5, const_100)), add(const_1, divide(75, const_100))), const_100)
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(#2,const_1)|subtract(#3,#1)|divide(#4,#2)|multiply(#5,const_100)|
gain
a certain roller coaster has 5 cars , and a passenger is equally likely to ride in any 1 of the 5 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 5 times , what is the probability that the passenger will ride in each of the 5 cars ?
"if he is to ride 5 times and since he can choose any of the 5 cars each time , total number of ways is = 5 * 5 * 5 * 5 * 5 = 3125 now the number of ways if he is to choose a different car each time is = 5 * 4 * 3 * 2 * 1 = 120 so the probability is = 120 / 3125 = 24 / 625 answer : b"
a ) 0 , b ) 24 / 625 , c ) 2 / 9 , d ) 1 / 3 , e ) 1
b
multiply(factorial(5), power(divide(1, 5), 5))
divide(n1,n0)|factorial(n0)|power(#0,n0)|multiply(#1,#2)|
general
jane and ashley take 10 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an eight day leave from work . this led to jane ' s wor...
let us assume that the work is laying 40 bricks . jane = 4 bricks per day ashley = 1 brick per day together = 5 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 16 remaining bricks = 40 - 24 = 16 so together , they would take 16 / 5 = 3.2 tota...
a ) 10.2 days , b ) 15.2 days , c ) 16.2 days , d ) 18 days , e ) 20 days
b
add(add(divide(subtract(subtract(const_1, multiply(const_4, divide(const_1, 10))), multiply(add(const_4, const_4), divide(const_1, 40))), add(divide(const_1, 10), divide(const_1, 40))), add(const_4, const_4)), const_4)
add(const_4,const_4)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|multiply(#1,const_4)|multiply(#0,#2)|subtract(const_1,#4)|subtract(#6,#5)|divide(#7,#3)|add(#0,#8)|add(#9,const_4)
physics
if john covers a certain distance in 1 hr . 24 min . by covering two third of the distance at 4 kmph and the rest at 5 kmph , then find the total distance .
"explanation : let the total distance be y km . then , ( 2 / 3 ) y / 4 + ( 1 / 3 ) y / 5 = 7 / 5 y / 6 + y / 15 = 7 / 5 7 y = 42 y = 6 km answer : d"
a ) 9 km , b ) 2 km , c ) 1 km , d ) 6 km , e ) 7 km
d
add(multiply(4, multiply(divide(const_2, const_3), add(1, divide(24, const_60)))), multiply(multiply(subtract(1, divide(const_2, const_3)), add(1, divide(24, const_60))), 5))
divide(n1,const_60)|divide(const_2,const_3)|add(n0,#0)|subtract(n0,#1)|multiply(#2,#1)|multiply(#2,#3)|multiply(n2,#4)|multiply(n3,#5)|add(#6,#7)|
physics
if m and n are positive integers and m ^ 2 + n ^ 2 = 10 , what is the value of m ^ 3 + n ^ 3 ?
"you need to integers which squared are equal 10 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 stop . the integers ca n ' t be greater than 3 or we will score above 10 . the second integer need to be picked up the same way . 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 the only pair that m...
a ) 28 , b ) 224 , c ) 320 , d ) 512 , e ) 1,600
a
gcd(10, const_4)
gcd(n2,const_4)|
general