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if the selling price of 50 articles is equal to the cost price of 45 articles , then the loss or gain percent is :
"c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 45 . loss % = 5 / 50 * 100 = 10 % answer b"
a ) 45 % , b ) 10 % , c ) 20 % , d ) 60 % , e ) 56 %
b
subtract(50, 45)
subtract(n0,n1)|
gain
the average marks of a class of 20 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ?
"sum of the marks for the class of 20 students = 20 * 40 = 800 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 70 students = 800 + 3000 = 3800 average marks of all the students = 3800 / 70 = 54.28 answer : b"
a ) 44.28 , b ) 54.28 , c ) 34.28 , d ) 64.28 , e ) 74.28
b
divide(add(multiply(20, 40), multiply(50, 60)), add(20, 50))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|
general
in a certain circle there are 9 points . what is the number of the triangles connecting 3 points of the 9 points ?
"imo : b here we have to select 3 points out of 9 points . order is not important so the answer will be 9 c 3 = 84 answer b"
a ) 12 , b ) 84 , c ) 108 , d ) 120 , e ) 132
b
divide(factorial(9), multiply(factorial(divide(const_10, const_2)), factorial(3)))
divide(const_10,const_2)|factorial(n0)|factorial(n1)|factorial(#0)|multiply(#3,#2)|divide(#1,#4)|
geometry
in shop there are 10 bulbs , a total of 4 are defective . if a customer buys 4 bulbs selected at random from the box , what is the probability that neither bulbs will be defective ?
first , there are 6 c 4 ways you can select 6 good bulbs from 4 good ones . second , there are 10 c 4 ways you select 4 bulbs from 10 ones in the box . then , the probability that neither bulb will be defective is : 6 c 4 / 10 c 4 = 15 / 210 = 1 / 15 answer is d
a ) 2 / 19 , b ) 3 / 29 , c ) 4 / 15 , d ) 1 / 15 , e ) 1 / 2
d
subtract(divide(4, subtract(10, 4)), divide(subtract(10, 4), 10))
subtract(n0,n1)|divide(n1,#0)|divide(#0,n0)|subtract(#1,#2)
general
if x + | x | + y = 2 and x + | y | - y = 6 what is x + y = ?
"if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions on...
a ) 1 , b ) - 1 , c ) 3 , d ) 5 , e ) 13
a
multiply(6, const_2)
multiply(n1,const_2)|
general
employees of a certain company are each to receive a unique 5 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , and 4 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ?
there are 3 ! ways to make codes starting with 12 . there are 3 ! ways to make codes starting with 24 . the number of codes is 2 * 3 ! = 12 . the answer is b .
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
b
multiply(multiply(const_2, const_3), 2)
multiply(const_2,const_3)|multiply(n3,#0)
general
36 men can complete a piece of work in 18 days . in how many days will 72 men complete the same work ?
"explanation : less men , means more days { indirect proportion } let the number of days be x then , 72 : 36 : : 18 : x x = 9 answer : d ) 9 days"
a ) 24 , b ) 17 , c ) 18 , d ) 9 , e ) 11
d
divide(multiply(18, 36), 72)
multiply(n0,n1)|divide(#0,n2)|
physics
in a room filled with 7 people , 4 people have exactly 1 sibling in the room and 3 people have exactly 3 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ?
"there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssds ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssdssibl ∈ gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and...
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 32 / 35
e
add(multiply(divide(4, 7), divide(subtract(7, 3), subtract(7, 1))), multiply(divide(4, 7), divide(subtract(7, 4), subtract(7, 1))))
divide(n1,n0)|subtract(n0,n4)|subtract(n0,n2)|subtract(n0,n1)|divide(#1,#2)|divide(#3,#2)|multiply(#0,#4)|multiply(#0,#5)|add(#6,#7)|
general
if p is the product of the integers from 1 to 34 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of p ?
34 ! has 3 , 6 , 9 , . . . . 30 , 33 as factors , which are 11 multiples of 3 . we need to add 4 more to these 11 because of 9 , 18 , and 27 . the greatest integer of k is 15 . the answer is c .
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19
c
add(multiply(3, const_4), 3)
multiply(n2,const_4)|add(n2,#0)
general
a present value of a machine is $ 2500 . its value depletiation rate is 5 % per annum then find the machine value after 2 years ?
"p = $ 2500 r = 5 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 2500 * 19 / 20 * 19 / 20 = $ 2256 approximately answer is c"
a ) $ 2100 , b ) $ 2546 , c ) $ 2256 , d ) $ 2451 , e ) $ 2345
c
multiply(2500, power(subtract(const_1, divide(5, const_100)), 2))
divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)|
gain
the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1280 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ?
"you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 1...
a ) 2480 , b ) 3490 , c ) 6785 , d ) 8255 , e ) 9255
d
subtract(divide(multiply(multiply(30, add(30, 1)), add(multiply(2, 30), 1)), add(3, 3)), 1280)
add(n17,n1)|add(n5,n5)|multiply(n17,n2)|add(#2,n1)|multiply(n17,#0)|multiply(#3,#4)|divide(#5,#1)|subtract(#6,n9)|
general
the diameter of the wheel of a car is 120 m . how many revolution / min must the wheel makein order to keep a speed of 60 km / hour approximately ?
distance to be covered in 1 min . = ( 60 x 1000 ) / ( 60 ) m = 1000 m . circumference of the wheel = ( 2 x ( 22 / 7 ) x 0.60 ) m = 3.77 m . number of revolutions per min . = ( 1000 / 3.77 ) = 265 e
a ) 210 , b ) 245 , c ) 230 , d ) 254 , e ) 265
e
divide(divide(multiply(60, const_1000), const_60), multiply(divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3)), divide(120, const_100)))
add(const_3,const_4)|divide(n0,const_100)|multiply(n1,const_1000)|multiply(const_10,const_2)|add(#3,const_2)|divide(#2,const_60)|divide(#4,#0)|multiply(#6,#1)|divide(#5,#7)
physics
each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , thirty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 70 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on...
let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.3 i + 0.8 f = 0.7 ( i + f ) 0.1 f = 0.4 i f = 4 i this means that for every 1 incandescent bulb , there are 4 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.3 i / ( 0.3 i + 0.8 f ) = 0.3...
a ) 4.5 % , b ) 8.6 % , c ) 12.3 % , d ) 16.7 % , e ) 20 %
b
multiply(divide(multiply(subtract(multiply(divide(divide(subtract(multiply(multiply(const_4, const_2), multiply(add(const_4, const_1), const_2)), 70), subtract(70, multiply(const_3, multiply(add(const_4, const_1), const_2)))), add(const_4, const_1)), multiply(add(const_4, const_1), const_2)), divide(multiply(const_3, m...
add(const_1,const_4)|multiply(const_2,const_4)|multiply(#0,const_2)|multiply(#2,const_3)|multiply(#1,#2)|divide(#3,const_100)|subtract(#4,n0)|subtract(n0,#3)|divide(#6,#7)|divide(#8,#0)|multiply(#9,#2)|subtract(#10,#5)|multiply(#3,#11)|divide(#12,n0)|multiply(#13,const_100)
gain
a chemist mixes one liter of pure water with x liters of a 40 % salt solution , and the resulting mixture is a 20 % salt solution . what is the value of x ?
"concentration of salt in pure solution = 0 concentration of salt in salt solution = 40 % concentration of salt in the mixed solution = 20 % the pure solution and the salt solution is mixed in the ratio of - - > ( 40 - 20 ) / ( 20 - 0 ) = 1 / 1 1 / x = 1 / 1 x = 1 answer : d"
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3
d
divide(20, subtract(40, 20))
subtract(n0,n1)|divide(n1,#0)|
gain
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.5 years . the ration of the number of boys to the number of girls in the class is ?
"let the ratio be k : 1 . then , k * 16.4 + 1 * 15.5 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.5 ) = k = 0.3 / 0.6 = 1 / 2 required ratio = 1 / 2 : 1 = 1 : 2 . answer : e"
a ) 2 : 6 , b ) 2 : 3 , c ) 2 : 5 , d ) 2 : 1 , e ) 1 : 2
e
divide(subtract(15.8, 15.5), subtract(16.4, 15.8))
subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)|
general
consider a right circular cone of base radius 4 and height 10 cm . a cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone . find the largest possible total surface area ( in sq cm ) of the cylinder .
explanation : let the radius of cylinder be r and its height is h . so , the height remained above cylinder is 10 - h also , triangles abc and ade are similar . so ratio of corresponding sides must be equal i . e ab / ad = bc / de . i . e . ( 10 - h ) / 10 = r / 4 . so 40 - 4 h = 10 r ( by cross multiply ) h = ( 40 - 1...
['a ) 100 Ο€ / 3', 'b ) 80 Ο€ / 3', 'c ) 120 Ο€ / 7', 'd ) 130 Ο€ / 9', 'e ) 110 Ο€ / 7']
a
multiply(const_pi, multiply(power(multiply(divide(divide(10, 4), subtract(divide(10, 4), const_1)), const_2), const_2), const_3))
divide(n1,n0)|subtract(#0,const_1)|divide(#0,#1)|multiply(#2,const_2)|power(#3,const_2)|multiply(#4,const_3)|multiply(#5,const_pi)
geometry
a football field is 10800 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 3600 square yards ?
"answer a ) 10800 yards need 1200 lbs 1 yard will need 1200 / 10800 = 1 / 9 lbs 3600 yards will need 1 / 9 * 3600 yards = 400 lbs"
a ) 400 , b ) 600 , c ) 750 , d ) 2400 , e ) 3200
a
multiply(3600, divide(1200, 10800))
divide(n1,n0)|multiply(n2,#0)|
geometry
two same glasses are respectively 1 / 4 th 1 / 5 th full of milk . they are then filled with water and the contents mixed in a tumbler . the ratio of milk and water in the tumbler is ?
1 / 4 : 3 / 4 = ( 1 : 3 ) 5 = 5 : 15 1 / 5 : 4 / 5 = ( 1 : 4 ) 4 = 4 : 16 - - - - - - 9 : 31 answer : b
a ) 9 : 39 , b ) 9 : 31 , c ) 9 : 36 , d ) 9 : 32 , e ) 9 : 34
b
divide(add(5, 4), add(multiply(const_3, const_10), 1))
add(n1,n3)|multiply(const_10,const_3)|add(n0,#1)|divide(#0,#2)
general
the area of a rectangle is 15 square centimeters and the perimeter is 16 square centimeters . what are the dimensions of the rectangle ?
let x and y be the length and width of the rectangle . using the formulas for the area and the perimeter , we can write two equations . 15 = x y and 16 = 2 x + 2 y solve the second equation for x x = 8 - y substitute x in the equation 15 = x y by 8 - y to rewrite the equation as 15 = ( 8 - y ) y solve for y to find y =...
['a ) 3 and 5', 'b ) 3 and 6', 'c ) 3 and 7', 'd ) 3 and 8', 'e ) 3 and 9']
a
divide(subtract(15, sqrt(subtract(power(15, const_2), multiply(const_4, 16)))), const_2)
multiply(n1,const_4)|power(n0,const_2)|subtract(#1,#0)|sqrt(#2)|subtract(n0,#3)|divide(#4,const_2)
geometry
sales price is $ 91 , gross profit is 160 % of cost , what is the value of gross profit ?
"cost + profit = sales cost + ( 160 / 100 ) cost = 91 cost = 35 profit = 91 - 35 = 56 answer ( e )"
a ) 32 , b ) 33 , c ) 39 , d ) 40 , e ) 56
e
subtract(91, divide(91, add(const_1, divide(160, const_100))))
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|
gain
a number exceeds by 25 from its 3 / 8 part . then the number is ?
"explanation : x Γ’ € β€œ 3 / 8 x = 25 x = 40 answer : d"
a ) 38 , b ) 29 , c ) 27 , d ) 40 , e ) 91
d
divide(multiply(25, 8), subtract(8, 3))
multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)|
general
in an it company , there are a total of 70 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ?
you could pick 70 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 25 . b
a ) 10 , b ) 25 , c ) 55 , d ) 35 , e ) 65
b
add(subtract(80, 70), subtract(50, 35))
subtract(n2,n0)|subtract(n1,n3)|add(#0,#1)
general
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 272 runs ?
"required run rate = [ 272 - ( 3.2 * 10 ) ] / 40 = 240 / 40 = 6.00 answer : e"
a ) 6.25 , b ) 6.22 , c ) 6.29 , d ) 6.39 , e ) 6.0
e
divide(subtract(272, multiply(10, 3.2)), 40)
multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|
gain
a certain basketball team that has played 2 / 3 of its games has a record of 16 wins and 4 losses . what is the greatest number of the remaining games that the team can lose and still win at least 3 / 4 of all of its games ?
"16 wins , 4 losses - total 20 games played . the team has played 2 / 3 rd of all games so total number of games = 30 3 / 4 th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games . it has already lost 4 games so it can lose another 3 at most . answer ( e )"
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3
e
subtract(multiply(divide(3, 2), add(16, 4)), add(floor(multiply(divide(3, 4), multiply(divide(3, 2), add(16, 4)))), const_1))
add(n2,n3)|divide(n1,n0)|divide(n1,n5)|multiply(#0,#1)|multiply(#2,#3)|floor(#4)|add(#5,const_1)|subtract(#3,#6)|
general
sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 06 : 00 and starts climbing at a constant pace . ken starts climbing only at 08 : 00 , when sari is already 800 meters ahead of him . nevertheless , ken climbs at a constant pace of 500...
both sari and ken climb in the same direction . speed of sari = 800 / 2 = 400 meters / hr ( since she covers 800 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 800 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / r...
a ) 16.3 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30
a
add(divide(add(800, 50), subtract(500, divide(800, const_2))), 8)
add(n4,n6)|divide(n4,const_2)|subtract(n5,#1)|divide(#0,#2)|add(n2,#3)
physics
boy sells a book for rs . 800 he gets a loss of 20 % , to gain 10 % , what should be the sp ?
cost price = 800 / 80 x 100 = 1000 to gain 10 % = 1000 x 10 / 100 = 100 sp = cp + gain = 1000 + 100 = 1100 answer : a
a ) 1100 , b ) 450 , c ) 550 , d ) 590 , e ) 600
a
add(divide(800, subtract(const_1, divide(20, const_100))), multiply(divide(800, subtract(const_1, divide(20, const_100))), divide(10, const_100)))
divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n0,#2)|multiply(#3,#1)|add(#3,#4)
gain
the age of somu is one - third his father ' s . 9 years back he was one - fifth of his father ' s age . what is his persent age ?
explanation : let somu ' s age be x and that of his father be 3 x . so , x - 9 = 3 x - 9 / 5 = x = 18 answer : option b
a ) 11 , b ) 18 , c ) 14 , d ) 12 , e ) 10
b
divide(subtract(multiply(add(const_4, const_1), 9), 9), subtract(add(const_4, const_1), const_3))
add(const_1,const_4)|multiply(n0,#0)|subtract(#0,const_3)|subtract(#1,n0)|divide(#3,#2)
general
at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 11 pencils and 16 charts have been brought to the meeting , how man...
"say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 11 total charts are 11 so a + 2 b = 16 add both : 3 a + 3 b = 27 so a + b = 9 total number of people = 9 d"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
d
divide(add(16, 11), add(2, 1))
add(n4,n5)|add(n0,n1)|divide(#0,#1)|
general
a garrison of 400 men had a provision for 31 days . after 20 days 180 persons re - enforcement leave the garrison . find the number of days for which the remaining ration will be sufficient ?
"400 - - - 31 400 - - - 11 220 - - - ? 400 * 11 = 220 * x = > x = 20 days . answer : c"
a ) 65 days , b ) 45 days , c ) 20 days , d ) 16 days , e ) 18 days
c
divide(multiply(subtract(31, 20), 400), 180)
subtract(n1,n2)|multiply(n0,#0)|divide(#1,n3)|
other
a goods train leaves a station at a certain time and at a fixed speed . after ^ hours , an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph . this train catches up the goods train in 4 hours . find the speed of the goods train .
let the speed of the goods train be x kmph . distance covered by goods train in 10 hours = distance covered by express train in 4 hours 10 x = 4 x 90 or x = 36 . so , speed of goods train = 36 kmph . ans : b
a ) 15 kmph , b ) 36 kmph , c ) 21 kmph , d ) 17 kmph , e ) 19 kmph
b
divide(multiply(90, 4), add(4, add(const_3, const_3)))
add(const_3,const_3)|multiply(n0,n1)|add(n1,#0)|divide(#1,#2)
physics
ram and shakil run a race of 2000 meters . first , ram gives shakil a start of 200 meters and beats him by one minute . if , ram gives shakil a start of 6 minutes ram is beaten by 1000 meters . find the time in minutes in which ram and shakil can run the races separately .
let x and y are the speeds of ram and shakil . . then by problem we got following equation 2000 / x = ( 1800 / y ) - 1 1000 / x = ( 2000 / y ) - 6 solve equation 1 and 2 , we get x = 250 and y = 200 therefore , time taken by ram and shakilk to complete a race of 2000 m is 8 min and 10 min answer : b
a ) 7,10 , b ) 8,10 , c ) 9,10 , d ) 10,10 , e ) 10,12
b
subtract(divide(divide(subtract(2000, 200), divide(1000, 200)), 6), multiply(divide(1000, 200), const_10))
divide(n3,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#0,const_10)|divide(#2,n2)|subtract(#4,#3)
physics
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 $ 355 . what is the share of rahul ?
"rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 355 * 2 / 5 = $ 142 answer is b"
a ) $ 50 , b ) $ 142 , c ) $ 60 , d ) $ 100 , e ) $ 90
b
multiply(divide(2, add(3, 2)), 355)
add(n0,n1)|divide(n1,#0)|multiply(n2,#1)|
physics
a certain fruit stand sold apples for $ 0.70 each and cherry for $ 0.50 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ?
some multiple of 7 + some multiple of 5 should yield 63 . to get to a some multiple of 5 , we should ensure that a 3 or 8 ( 5 + 3 ) should be a multiple of 7 . 63 is a direct multiple of 7 , however in this case there wo n ' t be any cherry . hence the next option is to look for a multiple of 7 that has 8 as the unit d...
a ) 12 , b ) 11 , c ) 13 , d ) 14 , e ) 15
b
add(multiply(0.7, const_10), const_4)
multiply(n0,const_10)|add(#0,const_4)
other
the element being searched for is not found in an array of 100 elements . what is the average number of comparisons needed in a sequential search to determine that the element is not there , if the elements are completely unordered ?
let us assume k be the element we need to search then let array be n 1 , n 2 , n 3 . . . . . . . . . . . . . . n 100 now to prove that the element searched is not there in the array we have to compare the element k with every element in the array . so , totally we need 100 comparisons . answer : d
a ) 25 , b ) 50 , c ) 75 , d ) 100 , e ) 125
d
multiply(const_2, floor(divide(multiply(add(100, const_1), divide(100, const_2)), 100)))
add(n0,const_1)|divide(n0,const_2)|multiply(#0,#1)|divide(#2,n0)|floor(#3)|multiply(#4,const_2)
general
what is 15 % of 2 / 3 of 0.7 ?
"the best way to solve these questions is to convert every term into fraction ( 15 / 100 ) * ( 2 / 3 ) * ( 7 / 10 ) = 210 / 3000 = 0.07 option a"
a ) 0.07 , b ) 0.9 , c ) 9 , d ) 90 , e ) none of the above
a
divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|
gain
what is the greatest integer m for which the number 50 ! / 14 ^ m is an integer ?
"14 ^ m = 2 ^ m * 7 ^ m . let ' s figure out how many 7 ' s are in the prime factorization of 50 ! the multiples of 7 are : 7 , 14 , 21 , 28 , 35 , 42 , 7 * 7 . thus 7 ^ 8 will divide 50 ! but 7 ^ 9 will not . clearly 2 ^ 8 will divide 50 ! so m = 8 is the largest possible integer . the answer is d ."
a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 7
d
floor(divide(50, divide(14, const_2)))
divide(n1,const_2)|divide(n0,#0)|floor(#1)|
general
two printing presses begin printing currency at the same time and at constant speeds . press f produces 5 - dollar bills at the rate of 1,000 bills per minute . press t produces 20 - dollar bills at the rate of 200 bills per minute . once the machines have begun printing , how many seconds does it take for press f to p...
press f produces $ 5 bills @ 1000 per minute ( 60 sec ) so $ 500 in 6 secs press t produces $ 20 bills 200 per minute ( 60 sec ) so $ 400 in 6 secs . . so we can see f produces $ 100 in 6 secs or will print $ 50 in 3 secs . . ans 3 secs . . b
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
b
multiply(const_60, divide(50, multiply(const_100, const_10)))
multiply(const_10,const_100)|divide(n4,#0)|multiply(#1,const_60)
physics
last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 9 % of her salary . the amount saved this year was what percent of the amount she saved last year ?
let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.09 * 1.1 x = 0.099 x 0.099 x / 0.06 x = 99 / 60 = 1.65 = 165 % the answer is c .
a ) 140 % , b ) 152 % , c ) 165 % , d ) 176 % , e ) 190 %
c
multiply(divide(multiply(divide(9, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100)
add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)
gain
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
the ratio between sumit ' s and prakash ' s age at present is 2 : 3 . sumit is 6 years younger than prakash . the ratio of sumit ' s age to prakash ' s age after 6 years will be :
explanation : let their ages be 2 x and 3 x years . 3 x - 2 x = 6 or x = 6 sumit ' s age = 12 years , prakash ' s age = 24 years ratio of their ages = 18 : 24 = 3 : 4 . correct option : c
a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 3 : 8 , e ) none
c
divide(add(multiply(6, const_2), 6), add(add(multiply(6, const_2), 6), 6))
multiply(n2,const_2)|add(n2,#0)|add(n2,#1)|divide(#1,#2)
general
if a person walks at 12 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is :
"let the actual distance travelled be x km . x / 10 = ( x + 20 ) / 12 12 x = 10 x + 200 2 x = 200 x = 100 km . answer : b"
a ) 50 km , b ) 100 km , c ) 60 km , d ) 70 km , e ) 80 km
b
multiply(10, divide(20, subtract(12, 10)))
subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)|
general
cara leaves her home and walks toward don ' s house . two hours later , don leaves his home and walks toward cara ' s house . the distance between their homes is 45 kilometers , cara ' s walking speed is 6 km / h , and don ' s walking speed is 5 km / h . how many kilometers will cara walk before she meets don ?
cara walks 12 km in the first two hours so there are 33 km remaining . when don starts walking , they complete a total of 11 km per hour . they will meet three hours after don starts walking . since cara walks for 5 hours , she walks 30 km . the answer is e .
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30
e
add(multiply(6, const_2), multiply(6, divide(subtract(45, multiply(6, const_2)), add(6, 5))))
add(n1,n2)|multiply(n1,const_2)|subtract(n0,#1)|divide(#2,#0)|multiply(n1,#3)|add(#1,#4)
physics
in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 3 / 4 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and curved
good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 3 / 4 of the garden is green , so , since all the flowers must be either green or yellow , we know that 1 / 4 are yellow . we ' re also told there is an equal pr...
a ) 1 / 8 , b ) 1 / 6 , c ) 1 / 4 , d ) 3 / 4 , e ) 7 / 8
a
multiply(subtract(const_1, divide(3, 4)), divide(1, 2))
divide(n2,n3)|divide(n0,n1)|subtract(const_1,#1)|multiply(#0,#2)
probability
if n is a positive integer , what is the remainder when ( 7 ^ ( 3 n + 3 ) ) ( 6 ^ n ) is divided by 10 ?
"this one took me bout 3 1 / 2 min . just testin numbers and what not . first notice that n is positive . save time by noticing thati worked out one solution where n = 0 only to find that thats not an option : p . 1 - 7 stands for ^ 1 thru 7 1 : 7 * 1 = 7 2 : 7 * 7 = 9 3 : 7 * 9 = 3 4 : 7 * 3 = 1 5 : 7 * 1 = 7 6 : 7 * ...
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
d
reminder(multiply(multiply(const_3, const_3), 6), 10)
multiply(const_3,const_3)|multiply(n3,#0)|reminder(#1,n4)|
general
10 / 21 of 3 / 5 of a number is greater than 4 / 7 of 1 / 5 of the same number by 36 . what is half of that number ?
"let no . be x 10 / 21 * 3 / 5 * x - 4 / 7 * 1 / 5 * x = 8 by further solving 30 x / 105 - 4 x / 35 = 36 18 x / 105 = 36 x = 210 we have to find x / 2 = 210 / 2 = 105 answer : a"
a ) 105 , b ) 180 , c ) 210 , d ) 240 , e ) 315
a
divide(divide(36, subtract(multiply(divide(const_4.0, 21), divide(5, 5)), multiply(divide(const_2.0, 3), divide(10, 7)))), 1)
divide(n0,n1)|divide(n7,n3)|divide(n6,n7)|divide(n0,n5)|multiply(#0,#1)|multiply(#2,#3)|subtract(#4,#5)|divide(n8,#6)|divide(#7,n6)|
general
the speed of a bus increases by 2 km after every one hour . if the distance travelling in the first one hour was 35 km . what was the total distance travelled in 12 hours ?
given that distance travelled in 1 st hour = 35 km and speed of the bus increases by 2 km after every one hour hence distance travelled in 2 nd hour = 37 km hence distance travelled in 3 rd hour = 39 km . . . total distance travelled = [ 35 + 37 + 39 + . . . ( 12 terms ) ] this is an arithmetic progression ( ap ) with ...
a ) 233 , b ) 552 , c ) 376 , d ) 287 , e ) 166
b
multiply(add(35, add(35, multiply(subtract(12, const_1), 2))), divide(12, 2))
divide(n2,n0)|subtract(n2,const_1)|multiply(n0,#1)|add(n1,#2)|add(n1,#3)|multiply(#4,#0)
physics
a tailor trims 11 feet from opposite edges of a square piece of cloth , and 7 feet from the other two edges . if 84 square feet of cloth remain , what was the length of a side of the original piece of cloth ?
"let the original side of the square be x . ( x - 22 ) * ( x - 14 ) = 84 = 6 * 14 x = 28 the answer is c ."
a ) 26 , b ) 27 , c ) 28 , d ) 29 , e ) 30
c
divide(add(add(multiply(const_2, 11), multiply(const_2, const_3.0)), sqrt(add(multiply(11, subtract(84, multiply(multiply(const_2, 7), multiply(const_2, 11)))), power(add(multiply(const_2, 11), multiply(const_2, 7)), const_2)))), const_2)
multiply(n0,const_2)|multiply(const_3.0,const_2)|add(#0,#1)|multiply(#1,#0)|power(#2,const_2)|subtract(n2,#3)|multiply(#5,n0)|add(#6,#4)|sqrt(#7)|add(#2,#8)|divide(#9,const_2)|
geometry
an order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth . what be the cost of a carpet whose length and breadth are 40 % more and 25 % more respectively than the first carpet . given that the ratio of carpet is rs . 45 per sq . m ?
explanation : length of the first carpet = ( 1.44 ) ( 6 ) = 8.64 cm area of the second carpet = 8.64 ( 1 + 40 / 100 ) 6 ( 1 + 25 / 100 ) = 51.84 ( 1.4 ) ( 5 / 4 ) sq m = ( 12.96 ) ( 7 ) sq m cost of the second carpet = ( 45 ) ( 12.96 x 7 ) = 315 ( 13 - 0.04 ) = 4095 - 12.6 = rs . 4082.40 answer is a
a ) rs . 4082.40 , b ) rs . 1024.21 , c ) rs . 2810.6 , d ) rs . 3214 , e ) none of these
a
multiply(multiply(multiply(multiply(6, 1.44), divide(add(const_100, 40), const_100)), multiply(6, divide(add(const_100, 25), const_100))), 45)
add(n2,const_100)|add(n3,const_100)|multiply(n0,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#3,#2)|multiply(n0,#4)|multiply(#5,#6)|multiply(n4,#7)
general
in a office work is distribute between p persons . if 1 / 7 members are absent then work increased for each person is ?
"let total % of work is 100 % total person = p 1 / 7 person are absent of total person . so absent person is 1 / 7 p ie p / 7 . left person is , p - p / 7 = 6 p / 7 . p person do the work 100 % 1 person do the work 100 * p % 6 p / 7 person do the work ( 100 * p * 7 ) / 6 p % = 116.67 % work increased for each person is...
a ) 13.67 % , b ) 14.67 % , c ) 15.67 % , d ) 16.67 % , e ) 17.67 %
d
subtract(divide(multiply(const_100, 7), subtract(7, 1)), const_100)
multiply(n1,const_100)|subtract(n1,n0)|divide(#0,#1)|subtract(#2,const_100)|
general
a courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ?
"number of bricks = courtyard area / 1 brick area = ( 2500 Γ— 1600 / 20 Γ— 10 ) = 20000 answer : c"
a ) 22877 , b ) 27778 , c ) 20000 , d ) 27999 , e ) 17799
c
divide(multiply(multiply(25, const_100), multiply(16, const_100)), multiply(20, 10))
multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)|
physics
in a school of 700 boys , 44 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ?
"44 + 28 + 10 = 82 % 100 – 82 = 18 % 700 * 18 / 100 = 126 answer : b"
a ) 173 , b ) 126 , c ) 153 , d ) 143 , e ) 133
b
divide(multiply(700, subtract(const_100, add(add(44, 28), 10))), const_100)
add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)|
gain
a shop keeper marked 30 % above the cost price and offered 20 % discount then find it ' s net profit ?
"net profit = 30 - 20 + ( 30 * ( - 25 ) / 100 ) = 2.5 % answer is b"
a ) 2 % , b ) 2.5 % , c ) 3.7 % , d ) 4.6 % , e ) 5 %
b
subtract(subtract(add(const_100, 30), divide(multiply(add(const_100, 30), 20), const_100)), const_100)
add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)|
gain
the speed of a car is 100 km in the first hour and 80 km in the second hour . what is the average speed of the car ?
"s = ( 100 + 80 ) / 2 = 90 kmph answer : e"
a ) 12 , b ) 75 , c ) 88 , d ) 54 , e ) 90
e
divide(add(100, 80), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
the average marks of 10 students in a class is 100 . but a student mark is wrongly noted as 60 instead of 10 then find the correct average marks ?
"correct avg marks = 100 + ( 10 - 60 ) / 10 avg = 100 - 5 = 95 answer is c"
a ) a ) 78 , b ) b ) 82 , c ) c ) 95 , d ) d ) 91 , e ) e ) 85
c
divide(add(subtract(multiply(100, 10), 60), 10), 10)
multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|
general
there are 300 seniors at morse high school , and 50 % of them have cars . of the remaining grades ( freshmen , sophomores , and juniors ) , only 10 % of them have cars . if 20 % of all the students at morse have cars , how many students are in the 3 lower grades ?
let x be the number of students in the lower three grades . 0.1 x + 150 = 0.2 ( x + 300 ) 0.1 x = 900 x = 900 the answer is b .
a ) 600 , b ) 900 , c ) 1200 , d ) 1350 , e ) 1500
b
divide(subtract(multiply(300, 50), multiply(300, 20)), subtract(20, 10))
multiply(n0,n1)|multiply(n0,n3)|subtract(n3,n2)|subtract(#0,#1)|divide(#3,#2)
gain
carrie likes to buy t - shirts at the local clothing store . they cost $ 9.95 each . one day , she bought 20 t - shirts . how much money did she spend ?
$ 9.95 * 20 = $ 199 . answer is c .
a ) $ 150 , b ) $ 248.75 , c ) $ 199 , d ) $ 171.6 , e ) $ 190
c
floor(multiply(20, 9.95))
multiply(n0,n1)|floor(#0)|
general
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 90 km / hr . what is the length of the platform ?
"speed = 90 * 5 / 18 = 25 m / sec . length of the train = 25 * 20 = 500 m . let the length of the platform be x m . then , ( x + 500 ) / 36 = 25 = > x = 400 m . answer : c"
a ) 229 , b ) 240 , c ) 400 , d ) 277 , e ) 221
c
multiply(20, multiply(90, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics
in a mixed college 160 students are there in one class . out of this 160 students 5 / 8 students are girls . how many boys are there ?
"total number of students : 160 total girls : 160 * 5 / 8 = 100 total boys : 160 - 100 = 60 answer is b"
a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140
b
multiply(divide(160, 8), const_3)
divide(n0,n3)|multiply(#0,const_3)|
general
a ferry can transport 85 tons of vehicles . automobiles range in weight from 1,800 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ?
"to get maximum vehicles we must take into consideration the minimum weight i . e 1800 pounds here since , 1 ton = 2000 pounds 85 tons will be 170,000 pounds from the answer choices : let max number of vehicles be 86 total weight will be = 90 * 1800 = 162000 pounds , which is lesser than the maximum weight allowed . an...
a ) 23 , b ) 41 , c ) 48 , d ) 90 , e ) 86
d
divide(multiply(multiply(85, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100))
add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)|
general
if 10 gallons of grape juice are added to 50 gallons of a mixture , which contains 10 percent grape juice then what percent of the resulting mixture is grape juice ?
"official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 50 Γ— 0.10 = 5 gallons of grape juice . 50 Γ— 0.90 = 45 gallons of other components . if we add 10 gallons of grape juice , we will end up with 15 gallons of grape juice and 45 gallons of other components , and we will...
a ) 14 % , b ) 25 % , c ) 28 % , d ) 34 % , e ) 50 %
b
multiply(divide(add(multiply(divide(10, const_100), 50), 10), add(50, 10)), const_100)
add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|add(n0,#2)|divide(#3,#0)|multiply(#4,const_100)|
general
an artist who needs to buy only paint and canvas finds that the cost of canvas has decreased by 40 percent and the cost of paint has decreased by 60 percent . if paint previously accounted for 4 times as much of the expense of painting as did canvas , by what percent did the total cost for paint and canvas decrease ?
paint : canvas cost is 4 : 1 . so , paint accounts for 80 % of cost and canvas accounts for 20 % . canvas , after decrease by 40 % will be ( 100 - 40 ) % of 20 % = > 60 % of 20 % = > 12 % of original paint , after decrease by 60 % will be ( 100 - 60 ) % of 80 % = > 40 % of 80 % = > 32 % of original new cost as a % of o...
a ) 40 % , b ) 44 % , c ) 50 % , d ) 56 % , e ) 60 %
d
multiply(subtract(const_1, divide(add(multiply(divide(40, const_100), 4), divide(60, const_100)), add(4, const_1))), const_100)
add(n2,const_1)|divide(n1,const_100)|divide(n0,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|subtract(const_1,#5)|multiply(#6,const_100)
general
a man can row upstream at 55 kmph and downstream at 65 kmph , and then find the speed of the man in still water ?
"us = 55 ds = 65 m = ( 55 + 65 ) / 2 = 60 answer : c"
a ) 27 , b ) 70 , c ) 60 , d ) 80 , e ) 24
c
divide(add(55, 65), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 60 paise per sq . metre
"area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 60 = 51000 p in rs = 51000 / 100 = rs 510 answer : a"
a ) s 510 , b ) s 780 , c ) s 880 , d ) s 480 , e ) s 980
a
divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 60), const_100)
multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)|
physics
a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length
"explanation : speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 βˆ— 2 / 25 = 40 seconds option c"
a ) 20 seconds , b ) 30 seconds , c ) 40 seconds , d ) 50 seconds , e ) none of these
c
divide(add(360, 140), divide(multiply(45, const_1000), const_3600))
add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|
physics
in some quantity of ghee , 60 % is pure ghee and 40 % is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was :
"explanation : let the original quantity be x kg . vanaspati ghee in x kg = ( 40 x / 100 ) kg = ( 2 x / 5 ) kg . now , ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 < = > 2 x / ( 5 x + 50 ) = 1 / 5 < = > 5 x = 50 < = > x = 10 . answer : a ) 10 kg"
a ) 10 , b ) 27 , c ) 87 , d ) 23 , e ) 93
a
divide(multiply(20, 10), subtract(40, 20))
multiply(n2,n3)|subtract(n1,n3)|divide(#0,#1)|
gain
evaluate 926 + 9.026 + 0.926 + 9.0026
"926 + 9.026 + 0.926 + 9.0026 = 944.9546 option e"
a ) 944.9755 , b ) 944.9655 , c ) 944.9565 , d ) 946.9565 , e ) 944.9546
e
divide(926, divide(9.026, 926))
divide(n1,n0)|divide(n0,#0)|
general
what is the units digit of the expression 17 ^ 7 - 2 ?
"7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) 7 ^ 5 = 7 ( last digit ) and the cycle repeats after every 4 powers therefore , last digit of 17 ^ 7 = 3 3 - 2 = 1 answer a"
a ) 1 , b ) 2 , c ) 3 , d ) 7 , e ) 9
a
add(add(const_4, const_3), const_2)
add(const_3,const_4)|add(#0,const_2)|
general
if v is the product of the integers from 1 to 30 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of v ?
v = 30 ! 8 v = 30 x 29 x 28 x 27 x 26 x 25 x 24 x 24 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 out of these 30 , 27 , 24 , 21 , 18 , 15 , 12 , 09 , 06 , 3 are factors of 3 3 x 10 , 3 x 3 x 3 , 3 x 8 , 3 x 3 x 2 , 3 x 5 , 3 x 4 , 3 x 3 x 3 , 3 x 2 , 3 so we hav...
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
c
add(add(divide(30, 3), floor(divide(30, power(3, const_2)))), floor(divide(30, power(3, 3))))
divide(n1,n2)|power(n2,const_2)|power(n2,n2)|divide(n1,#1)|divide(n1,#2)|floor(#3)|floor(#4)|add(#0,#5)|add(#7,#6)
general
find the side of the largest possible square slabs which can be paved on the floor of a room 2 m 50 cm long and 1 m 50 cm broad . also find the number of such slabs to pave the floor .
explanation : hcf ( 250 , 150 ) = 50 cm the number slabs = 250 * 150 / 50 * 50 = 15 answer : c
a ) 25 , 20 , b ) 30 , 15 , c ) 50 , 15 , d ) none of these , e ) 75,35
c
multiply(add(const_2, const_3), multiply(2, const_3))
add(const_2,const_3)|multiply(n0,const_3)|multiply(#0,#1)
physics
a train speeds past a pole in 25 seconds and a platform 200 m long in 30 seconds . its length is :
"let the length of the train be x meters and its speed be y m / sec . they , x / y = 25 = > y = x / 25 x + 200 / 30 = x / 25 x = 1000 m . answer : option d"
a ) 1200 m , b ) 2000 m , c ) 1500 m , d ) 1000 m , e ) 4000 m
d
multiply(200, subtract(const_2, const_1))
subtract(const_2,const_1)|multiply(n1,#0)|
physics
two cubes have their volumes in the ratio 1 : 27 . find the ratio of their surface areas .
let their edges be a and b . then , = a ^ 3 / b ^ 3 = 1 / 27 ( or ) ( a / b ) ^ 3 = ( 1 / 3 ) 3 ( or ) ( a / b ) = ( 1 / 3 ) . therefore , ratio of their surface area = 6 a ^ 2 / 6 b ^ 2 = a ^ 2 / b ^ 2 = ( a / b ) ^ 2 = 1 / 9 , i . e . 1 : 9 . answer is a .
['a ) 1 : 9', 'b ) 2 : 9', 'c ) 3 : 9', 'd ) 5 : 9', 'e ) none of them']
a
power(power(divide(1, 27), inverse(const_3)), const_2)
divide(n0,n1)|inverse(const_3)|power(#0,#1)|power(#2,const_2)
geometry
one fourth of a solution that was 14 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ?
"consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.14 + 25 * x = 100 * 0.16 - - > x = 0.22 . answer : c ."
a ) 24 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 %
c
multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(14, const_100), const_4), divide(14, const_100))), const_100)
divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)|
gain
a certain fraction has the same ratio to 1 / 18 , as 2 / 5 does to 2 / 9 . what is this certain fraction ?
"x / ( 1 / 18 ) = ( 2 / 5 ) / ( 2 / 9 ) x = 2 * 9 * 1 / 18 * 5 * 2 = 1 / 10 the answer is c ."
a ) 1 / 5 , b ) 1 / 7 , c ) 1 / 10 , d ) 1 / 12 , e ) 1 / 15
c
divide(1, 18)
divide(n0,n1)|
other
how many odd numbers between 10 and 2000 are the squares of integers ?
"the numbers are the squares of 5 , 7 , 9 , . . . , 43 which includes 20 numbers . the answer is e ."
a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20
e
add(10, const_4)
add(n0,const_4)|
geometry
for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score e for the exam ?
"a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean e = 74 . answer : a ."
a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82
a
divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3))
add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)|
general
walking at the rate of 3 kmph a man cover certain distance in 3 hr 20 min . running at a speed of 9 kmph the man will cover the same distance in .
"distance = speed * time 3 * 10 / 3 = 10 km new speed = 9 kmph therefore time = d / s = 10 / 9 = 54 min answer : e ."
a ) 45 min , b ) 50 min , c ) 40 min , d ) 48 min , e ) 54 min
e
multiply(divide(multiply(3, add(3, divide(20, const_60))), 9), const_60)
divide(n2,const_60)|add(n1,#0)|multiply(n0,#1)|divide(#2,n3)|multiply(#3,const_60)|
physics
what percent is 25 gm of 1 kg ?
"1 kg = 1000 gm 25 / 1000 Γ— 100 = 2500 / 1000 = 5 / 2 = 2.5 % d )"
a ) 0.5 % , b ) 1 % , c ) 1.5 % , d ) 2.5 % , e ) 3 %
d
multiply(divide(25, 1), const_100)
divide(n0,n1)|multiply(#0,const_100)|
gain
an airplane covers 50 miles in 1 / 5 hours . how many miles can the airplane cover in 5 hours ?
divide 5 by 1 / 5 = 5 Γ· 1 / 5 = 5 / 1 Γ· 1 / 5 = 5 / 1 * 5 / 1 = 25 . then , multiply 50 by 25 to get 1250 in 5 hours , the airplane will cover 1250 miles answer is c .
a ) 800 , b ) 600 , c ) 1250 , d ) 950 , e ) 1000
c
multiply(multiply(50, 5), 5)
multiply(n0,n2)|multiply(n2,#0)
physics
evaluate 35 / . 07
"explanation : 35 / . 07 = 3500 / 7 = 500 option b"
a ) 400 , b ) 500 , c ) 505 , d ) none of these , e ) 506
b
divide(const_100.0, divide(07, 35))
divide(n1,const_100)|divide(n0,#0)|
general
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 19.7 % . what would have been the percentage of profit earned if no discount was offered ?
"let c . p . be rs . 100 . then , s . p . = rs . 19.70 let marked price be rs . x . then , 95 / 100 x = 119.70 x = 11970 / 95 = rs . 126 now , s . p . = rs . 126 , c . p . = rs . 100 profit % = 26 % . answer : b"
a ) 60 % , b ) 26 % , c ) 30 % , d ) 56 % , e ) 73 %
b
subtract(divide(multiply(add(const_100, 19.7), const_100), subtract(const_100, 5)), const_100)
add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)|
gain
0.003 x 0.2 = ?
"3 x 2 = 6 . sum of decimal places = 4 0.003 x 0.2 = 0.0006 answer : option a"
a ) 0.0006 , b ) 0.006 , c ) 0.06 , d ) 0.6 , e ) 6.0
a
multiply(divide(0.003, 0.2), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
a candidate who gets 30 % of the marks fails by 60 marks . but another candidate who gets 45 % marks gets 30 marks more than necessary for passing . find the number of marks for passing ?
"30 % - - - - - - - - - - - - 60 45 % - - - - - - - - - - - - 30 - - - - - - - - - - - - - - - - - - - - - - 15 % - - - - - - - - - - - - - 90 30 % - - - - - - - - - - - - - - 180 180 + 60 = 240 marks answer : b"
a ) 100 marks , b ) 240 marks , c ) 280 marks , d ) 371 marks , e ) 827 marks
b
add(multiply(divide(add(60, 30), subtract(divide(45, const_100), divide(30, const_100))), divide(30, const_100)), 60)
add(n1,n3)|divide(n2,const_100)|divide(n0,const_100)|subtract(#1,#2)|divide(#0,#3)|multiply(#4,#2)|add(n1,#5)|
gain
in may , the groundskeeper at spring lake golf club built a circular green with an area of 80 Ο€ square feet . in august , the groundskeeper doubled the distance from the center of the green to the edge of the green . what is the total area of the renovated green ?
"area = Ο€ r ^ 2 , so doubling the radius results in an area that is 4 times the original area . 4 ( 80 Ο€ ) = 320 Ο€ the answer is b ."
a ) 160 Ο€ , b ) 320 Ο€ , c ) 400 Ο€ , d ) 540 Ο€ , e ) 640 Ο€
b
multiply(power(multiply(sqrt(80), const_2), const_2), const_pi)
sqrt(n0)|multiply(#0,const_2)|power(#1,const_2)|multiply(#2,const_pi)|
geometry
80 pie square inches of material of negligible thickness are required to construct a 1 / 16 scale model of a cylindrical barrel . if the diameter of the base of the barrel is 160 inches , then what is the volume of the barrel , to the nearest cubic foot ?
radius of barrel , r = 80 radius of model , r = 80 / 16 = 5 surface area of the model = 2 βˆ— pi βˆ— r ( r + h ) = 80 βˆ— pi which gives h = 3 and h = 3 βˆ— 16 = 48 so , volume of the barrel = pi βˆ— r 2 βˆ— h = pi βˆ— ( 80 ) 2 βˆ— ( 48 ) in cubic inches converting in cubic ft results : volume = pi βˆ— ( 80 ) 2 βˆ— ( 48 ) / ( 12 ) 3 = 178...
['a ) 2 pie', 'b ) 11 pie', 'c ) 178 pie', 'd ) 711 pie', 'e ) 1280 pie']
c
floor(divide(volume_cylinder(divide(160, const_2), multiply(subtract(divide(divide(80, const_2), divide(80, 16)), divide(80, 16)), 16)), power(add(const_10, const_2), const_3)))
add(const_10,const_2)|divide(n3,const_2)|divide(n0,const_2)|divide(n0,n2)|divide(#2,#3)|power(#0,const_3)|subtract(#4,#3)|multiply(n2,#6)|volume_cylinder(#1,#7)|divide(#8,#5)|floor(#9)
geometry
the average weight of 20 girls increases by 2 kg when a new girl comes in place of one of them weighing 40 kg . what might be the weight of the new girl ?
"total weight increased = 20 x 2 kg = 40 kg . weight of new person = 40 + 40 kg = 80 kg answer : c"
a ) 85 kg , b ) 90 kg , c ) 80 kg , d ) 75 kg , e ) 70 kg
c
add(multiply(20, 2), 40)
multiply(n0,n1)|add(n2,#0)|
general
a ( 5 , w ^ 3 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 + 2 . what is the value of w ?
"y = x ^ 2 + 2 x = 5 y = 5 ^ 2 + 2 = 27 y is in the form w ^ 3 w ^ 3 = 27 w = 3 answer : a"
a ) 3 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 9
a
power(subtract(power(5, 2), 2), divide(2, 5))
divide(n3,n0)|power(n0,n2)|subtract(#1,n3)|power(#2,#0)|
general
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 + { neith...
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
a certain fruit stand sold apples for $ 0.80 each and bananas for $ 0.70 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.50 , what total number of apples and bananas did the customer purchase ?
let ' s start with 1 apple for $ 0.80 . let ' s subtract $ 0.80 from $ 6.50 until we get a multiple of $ 0.70 . $ 6.50 , $ 5.70 , $ 4.90 = 7 * $ 0.70 the customer purchased 7 bananas and 2 apples . the answer is b .
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12
b
add(divide(subtract(6.5, multiply(0.8, const_2)), 0.7), const_2)
multiply(n0,const_2)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_2)
other
what is the thousandths digit in the decimal equivalent of 62 / 5000 ?
"62 / 5000 = 62 / ( 5 * 10 ^ 3 ) = ( 62 / 5 ) * 10 ^ - 3 = 12.4 * 10 ^ - 3 = . 0124 thousandths digit = 2 answer c"
a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 6
c
floor(multiply(const_100, divide(62, 5000)))
divide(n0,n1)|multiply(#0,const_100)|floor(#1)|
general
in how many seconds will a train 180 meters long pass an oak tree , if the speed of the train is 36 km / hr ?
"speed = 36 * 5 / 18 = 10 m / s time = 180 / 10 = 18 seconds the answer is c ."
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22
c
divide(180, multiply(const_0_2778, 36))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
what number should replace the question mark ? 10 , 30 , 14 , 25 , 18 , 20 , - - - ?
"answer : d 10 , 30 , 14 , 25 , 18 , 20 , 22 ? there are two alternate sequences : + 4 and - 5 ."
a ) 19 , b ) 21 , c ) 25 , d ) 22 , e ) 20
d
add(add(add(const_2.0, 10), 10), 10)
add(const_2.0,n0)|add(n0,#0)|add(#1,n0)|
general
a train of length 250 m crosses a bridge of length 180 m in 20 seconds . what is the speed of train ?
"sol : ( length of train + length of bridge ) = speed of train x time ( 250 + 180 ) = 20 x speed speed = 430 / 20 = 21.5 m / s = 77.4 km / h answer = b"
a ) 33 , b ) 77.4 , c ) 25 , d ) 22 , e ) 72
b
multiply(const_3_6, divide(add(250, 180), 20))
add(n0,n1)|divide(#0,n2)|multiply(#1,const_3_6)|
physics
what is the perimeter , in meters , of a rectangular garden 24 meters wide that has the same area as a rectangular playground 16 meters long and 12 meters wide ?
"say length of garden is l meters . then ( l * 24 ) = 16 * 12 [ given area is same . area = l * b ] therefore l works out to 8 meters . perimeter of garden = 2 * ( 8 + 24 ) = 64 meter . answer is c"
a ) 48 , b ) 56 , c ) 64 , d ) 76 , e ) 192
c
rectangle_perimeter(divide(rectangle_area(16, 12), 24), 24)
rectangle_area(n1,n2)|divide(#0,n0)|rectangle_perimeter(n0,#1)|
geometry
a can do a work in 12 days . when he had worked for 3 days , b joinedhim . if they complete the work in 3 more days , in how manydays can balone finish the work ?
sax work done by afar 3 days : i j . . remzming war β€” 1 . work done by ( a + b ) for 1 day : . work done by a for 1 day 6 days d
a ) 2 days , b ) 4 days , c ) 5 days , d ) 6 days , e ) 9 days
d
divide(12, subtract(const_3, const_1))
subtract(const_3,const_1)|divide(n0,#0)
physics
if k is an integer and 0.0010101 x 10 ^ k is greater than 100 , what is the least possible value of k ?
"0.0010101 * 10 ^ k > 100 we need to move the decimal point to the right 5 places to get 101.01 this is equivalent to multiplying by 10 ^ 5 . the answer is d ."
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
d
divide(log(divide(100, 0.0010101)), log(10))
divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|
general
gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 18 times as heavy as water ?
"g = 19 w c = 9 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 19 w + 9 wx = x + 1 * 18 w 19 + 9 x = 18 ( x + 1 ) x = 1 / 9 ratio of gold with copper = 1 : 1 / 9 = 9 : 1 answer is c"
a ) 1 : 2 , b ) 3 : 2 , c ) 9 : 1 , d ) 5 : 2 , e ) 6 : 5
c
divide(subtract(18, 9), subtract(19, 18))
subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|
general
find the ones digit of 73 ^ 350
"3,9 , 7,1 repeats after every four times therefore 350 / 4 = 87 + 2 remainder 9 is the ones digit so e is correct"
a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9
e
subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(350, multiply(floor(divide(350, const_4)), const_4))), multiply(const_2, const_10))
add(const_3,const_4)|divide(n1,const_4)|multiply(const_10,const_2)|floor(#1)|multiply(#0,const_10)|multiply(#3,const_4)|subtract(n0,#4)|subtract(n1,#5)|power(#6,#7)|subtract(#8,#2)|
general
a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served onlyfor 9 months and received uniform & rs . 650 , find the price of the uniform ?
"9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b"
a ) rs . 90 , b ) rs . 100 , c ) rs . 150 , d ) rs . 190 , e ) rs . 200
b
multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4)
multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)|
general
a 600 meter long train crosses a signal post in 20 seconds . how long will it take to cross a 3 kilometer long bridge , at the same speed ?
"s = 600 / 20 = 1530 mps s = 3600 / 30 = 120 sec = 2 min answer : b"
a ) 4 min , b ) 2 min , c ) 8 min , d ) 9 min , e ) 5 min
b
divide(divide(multiply(3, const_1000), speed(600, 20)), const_60)
multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)|
physics
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 5 : 3 . what is the ratio of luxury cars to sport utility vehicles ?
"the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 15 : 10 . the ratio of economy cars to sport utility vehicles is 5 : 3 - - > e : s = 5 : 3 = 15 : 9 . thus , l : s = 10 : 9 . answer : e ."
a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 10 : 9
e
divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4)))
multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)|
other