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mathqa_question_shift_test

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two mechanics were working on your car . one can complete the given job in six hours , but the new guy takes 10 hours . they worked together for the first two hours , but then the first guy left to help another mechanic on a different job . how long will it take the new guy to finish your car ?
"very easy question . 2 variables and 2 easy equations . xy = 4 - - - > x = 4 / y - ( i ) x / y = 16 - - - > replacing ( i ) here - - - > 4 / ( y ^ 2 ) = 16 - - - > y ^ 2 = 4 / 16 = 1 / 4 - - - > y = 1 / 2 or - 1 / 2 the question states that x and y are positive integers . therefore , y = 1 / 2 is the answer . answer a ."
a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6
a
sqrt(divide(4, 16))
divide(n0,n1)|sqrt(#0)|
general
if xy = 4 , x / y = 16 , for positive numbers x and y , y = ?
"explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 6 ; us = 2 x = ( 6 + 2 ) / 2 = 8 / 2 = 4 km / hr y = ( 6 - 2 ) / 2 = 4 / 2 = 2 km / hr answer : e"
a ) 4 , 3 , b ) 4 , 4 , c ) 3 , 3 , d ) 4 , 5 , e ) 4 , 2
e
divide(add(divide(48, 24), divide(24, 4)), const_2)
divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2)|
physics
if a boat is rowed downstream for 24 km in 4 hours and upstream for 48 km in 24 hours , what is the speed of the boat and the river ?
"when the storm deposited 115 billion gallons , volume of water in the reservoir = 280 + 120 = 400 billion gallons if this is only 80 % of the capacity of the reservoir , the total capacity of the reservoir = 400 / 0.5 = 800 billion gallons therefore percentage of reservoir that was full before the storm = ( 280 / 800 ) * 100 = 35 % option c"
a ) 45 % , b ) 48 % , c ) 54 % , d ) 58 % , e ) 65 %
c
multiply(divide(280, divide(add(120, 280), divide(50, const_100))), const_100)
add(n0,n2)|divide(n1,const_100)|divide(#0,#1)|divide(n2,#2)|multiply(#3,const_100)|
general
after a storm deposits 120 billion gallons of water into the city reservoir , the reservoir is 50 % full . if the original contents of the reservoir totaled 280 billion gallons , the reservoir was approximately what percentage full before the storm ?
"each row contains 14 plants . leaving 2 corner plants , 12 plants in between have ( 12 x 2 ) metres & 1 metre on each side is left . length = ( 24 + 2 ) m = 26 m . answer : d"
a ) 20 m , b ) 22 m , c ) 24 m , d ) 26 m , e ) 28 m
d
add(add(multiply(subtract(14, const_1), 2), divide(12, 2)), divide(12, 2))
divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)|
physics
in a garden , there are 12 rows and 14 columns of mango trees . the distance between two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is
"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 = 86 thus m = 688 - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 688 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 16 ans : e"
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 16
e
divide(subtract(multiply(90, 8), multiply(86, 8)), subtract(92, 90))
multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)|
general
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 86 , and the average of the female students ’ grades was 92 , how many female students took the test ?
"800 - - - - 180 100 - - - - ? = > 35 % answer : c"
a ) 22 , b ) 20 , c ) 35 , d ) 88 , e ) 11
c
multiply(divide(subtract(1080, 800), 800), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
a cycle is bought for rs . 800 and sold for rs . 1080 , find the gain percent ?
"solution for an income of rs . 1 in 12 % stock at 108 , investment = rs . ( 108 / 12 ) = rs . 9 . for an income of rs . 1 in 10 % stock at 60 , investment = rs . ( 60 / 10 ) = rs . 6 . ∴ ratio of investments = 9 : 6 = 3 : 2 answer b"
a ) 3 : 4 , b ) 3 : 2 , c ) 4 : 5 , d ) 16 : 15 , e ) none
b
divide(multiply(108, const_2), multiply(60, const_3))
multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)|
other
a man invests some money partly in 12 % stock at 108 and partly in 10 % stock at 60 . to obtain equal dividends from both , he must invest the money in the ratio :
": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 6 = 100 meter answer : c"
a ) 150 , b ) 278 , c ) 100 , d ) 776 , e ) 191
c
multiply(divide(multiply(60, const_1000), const_3600), 6)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
a train running at the speed of 60 km / hr crosses a pole in 6 seconds . find the length of the train .
"principal that is amount taken by laura at year beginning = 35 $ rate of interest = 8 % interest = ( 8 / 100 ) * 35 = 2.8 $ total amount that laura owes a year later = 35 + 2.8 = 37.8 $ answer d"
a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 37.8 , e ) $ 38.80
d
add(multiply(divide(8, const_100), 35), 35)
divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)|
general
laura took out a charge account at the general store and agreed to pay 8 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ?
explanation : 10 * 150 = 1500 3 * 100 + 4 * 150 = 900 1500 – 900 = 600 a
a ) a ) 600 , b ) b ) 350 , c ) c ) 450 , d ) d ) 470 , e ) e ) 500
a
subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(4, 150)))
multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)
general
a man purchased 3 blankets @ rs . 100 each , 4 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ?
"50 pages typed 1 x 40 pages typed 2 x ( original + one revision ) 10 pages typed 3 x ( original + two revisions ) 50 ( 6 ) + 40 ( 6 + 4 ) + 10 ( 6 + 4 + 4 ) = 300 + 400 + 140 = 840 answer - b"
a ) $ 850 , b ) $ 840 , c ) $ 860 , d ) $ 870 , e ) $ 880
b
add(add(multiply(100, 6), multiply(40, 4)), multiply(multiply(10, 4), const_2))
multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)|
general
rates for having a manuscript typed at a certain typing service are $ 6 per page for the first time a page is typed and $ 4 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ?
let x be the number of additional games the gambler needs to play . 0.4 ( 30 ) + 0.7 x = 0.6 ( x + 30 ) 0.1 x = 6 x = 60 the answer is e .
a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60
e
divide(subtract(multiply(30, divide(60, const_100)), multiply(30, divide(40, const_100))), subtract(divide(70, const_100), divide(60, const_100)))
divide(n3,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n1,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)
gain
a gambler has won 40 % of his 30 poker games for the week so far . if , all of a sudden , his luck changes and he begins winning 70 % of the time , how many more games must he play to end up winning 60 % of all his games for the week ?
"the rooms which were not rented is 1 / 5 the ac rooms which were not rented is ( 1 / 3 ) * ( 3 / 5 ) = 1 / 5 the percentage of unrented rooms which were ac rooms is ( 1 / 5 ) / ( 1 / 5 ) = 1 = 100 % the answer is e ."
a ) 60 % , b ) 70 % , c ) 80 % , d ) 90 % , e ) 100 %
e
multiply(divide(multiply(subtract(const_1, divide(2, 4)), multiply(divide(4, 5), const_100)), subtract(const_100, multiply(divide(4, 5), const_100))), const_100)
divide(n0,n1)|divide(n2,n0)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#2,#3)|subtract(const_100,#2)|divide(#4,#5)|multiply(#6,const_100)|
gain
one night a certain hotel rented 4 / 5 of its rooms , including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ?
"the ratio of times taken by a and b = 100 : 130 = 10 : 13 suppose b can do work in x days then 10 : 13 : : 23 : x x = ( 23 * 13 ) / 10 x = 299 / 10 a ' s 1 day ' s work = 1 / 23 b ' s 1 day ' s work = 10 / 299 ( a + b ) ' s 1 day ' s work = 1 / 23 + 10 / 299 = 23 / 299 = 1 / 13 a and b together can do work in 13 days answer ( b )"
a ) 25 days , b ) 13 days , c ) 14 days , d ) 20 days , e ) 15 days
b
inverse(add(divide(const_1, 23), divide(const_1, multiply(add(divide(30, const_100), const_1), 23))))
divide(const_1,n1)|divide(n0,const_100)|add(#1,const_1)|multiply(n1,#2)|divide(const_1,#3)|add(#0,#4)|inverse(#5)|
physics
a is 30 % more efficient than b . how much time they will working together take to complete a job which a alone could have done in 23 days ?
"100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 6 % reduction in employees lead to 94 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 94 employees is 94 * 1100 = 103400 now the new salary is more than previous salary by x % . x = ( 103400 / 100000 ) * 100 = 103.4 % so the answer is d"
a ) 98.5 % , b ) 100.0 % , c ) 102.8 % , d ) 103.4 % , e ) 105.0 %
d
divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 6), const_4)), multiply(const_100, 10))
multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)|
general
this year , mbb consulting fired 6 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool allocated to employees after headcount reduction is what percent of that before the headcount reduction ?
"total age of 10 students = 10 * 20 = 200 and the total age of 16 students = 21 * 15 = 315 . let the average age of 2 new students be x . therefore total age of the new students = 2 x . so total age of 12 student = 200 + 5 x = 315 , x = 23 hence , the correct answer is a ."
a ) 23 , b ) 21 , c ) 19.5 , d ) 20 , e ) 22.5
a
divide(subtract(multiply(add(20, 1), add(20, 1)), multiply(20, 10)), 2)
add(n1,n3)|multiply(n0,n1)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,n2)|
general
the average age of a group of 10 students is 20 years . if 2 more students join the group , the average age increases by 1 year . the average age of the new students is
"reduce 1 - 100 5 - 15 - 25 - 35 - 45 - - 55 - - 65 - - 75 - - 85 - - 95 are valid multiples . add them - - > 500 e"
a ) 180 , b ) 245 , c ) 320 , d ) 405 , e ) 500
e
divide(multiply(100, multiply(const_3, 5)), 5)
multiply(n2,const_3)|multiply(n1,#0)|divide(#1,n2)|
general
in the set of positive integers from 1 to 100 , what is the sum of all the odd multiples of 5 ?
"2 * 22 / 7 * 6 = 37.7 37.7 * 3 1 / 2 = rs . 131.95 answer : a"
a ) 131.95 , b ) 132.9 , c ) 140.33 , d ) 123.4 , e ) 190.4
a
multiply(circumface(divide(12, const_2)), 3.50)
divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|
physics
find the cost of fencing around a circular field of diameter 12 m at the rate of rs . 3.50 a meter ?
"explanation : as there is meal for 70 adults and 28 have their meal , the meal left can be catered to 42 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 42 adults = 54 children hence , the meal can be catered to 54 children . answer : b"
a ) 33 , b ) 54 , c ) 18 , d ) 17 , e ) 01
b
multiply(subtract(70, 28), divide(90, 70))
divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1)|
general
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 28 adults have their meal , find the total number of children that can be catered with the remaining food .
"let weight of side of beef before processing = x ( 75 / 100 ) * x = 540 = > x = ( 540 * 100 ) / 75 = 720 answer c"
a ) 191 , b ) 355 , c ) 720 , d ) 840 , e ) 1,560
c
divide(multiply(540, const_100), subtract(const_100, 25))
multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|
gain
a side of beef lost 25 percent of its weight in processing . if the side of beef weighed 540 pounds after processing , how many pounds did it weigh before processing ?
one case is : 1 / 3 * 1 / 3 * 2 / 3 * 2 / 3 * 2 / 3 = 2 ^ 3 / 3 ^ 5 we have 5 ! / 2 ! * 3 ! = 10 such cases so , 2 ^ 3 * 10 / 3 ^ 5 = 80 / 243 answer : d .
a ) 5 / 243 , b ) 25 / 243 , c ) 64 / 243 , d ) 80 / 243 , e ) 16 / 17
d
subtract(1, divide(const_2, const_3))
divide(const_2,const_3)|subtract(n0,#0)
general
when a random experiment is conducted , the probability that event a occurs is 1 / 3 . if the random experiment is conducted 5 independent times , what is the probability that event a occurs exactly twice ?
"940 * ( 30 / 100 ) = 282 - - - - 6 ? - - - - 1 = > rs . 47 answer : b"
a ) 72 , b ) 47 , c ) 40 , d ) 28 , e ) 20
b
divide(divide(multiply(940, 30), const_100), 6)
multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|
gain
a reduction of 30 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 940 , what is the reduced price for kg ?
"5 / 6 * 3 / 2 * 6 / 5 = 3 / 2 3 : 2 answer : d"
a ) 1 : 1 , b ) 1 : 87 , c ) 1 : 6 , d ) 3 : 2 , e ) 1 : 2
d
divide(divide(multiply(5, 3), multiply(6, 2)), divide(multiply(3, 6), multiply(2, 5)))
multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)|
other
the compound ratio of 5 : 6 , 3 : 2 and 6 : 5 ?
assume the revenue in 2000 to be 100 . then in 2003 it would be 130 and and in 2005 180 , so from 2003 to 2005 it increased by ( 180 - 130 ) / 130 = 50 / 130 = 39 % answer : e .
a ) 50 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 39 %
e
multiply(divide(subtract(add(const_1, divide(80, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(30, const_100))), const_100)
divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)
gain
a certain company reported that the revenue on sales increased 30 % from 2000 to 2003 , and increased 80 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ?
"95 % - - - - 8 152 % - - - - ? 95 / 152 * 8 = 5 answer : a"
a ) 5 , b ) 8 , c ) 7 , d ) 4 , e ) 2
a
divide(multiply(subtract(const_100, 5), 8), add(const_100, 52))
add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|
gain
if a man lost 5 % by selling oranges at the rate of 8 a rupee at how many a rupee must he sell them to gain 52 % ?
the total number of stations = 34 from 34 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 34 p 2 ways . 34 p 2 = 34 * 33 = 1122 answer : d
a ) 1800 , b ) 1820 , c ) 1150 , d ) 1122 , e ) 1900
d
multiply(add(32, const_2), subtract(add(32, const_2), const_1))
add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)
physics
there are 32 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ?
"explanation : a and b 1 day ' s work = 1 / 4 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 4 ) - ( 1 / 12 ) = 6 days answer : option a"
a ) 6 days , b ) 8 days , c ) 12 days , d ) 10 days , e ) 5 days
a
add(inverse(subtract(divide(const_1, 4), divide(const_1, 12))), divide(const_2, add(const_2, const_3)))
add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)|
physics
a and b together can do a work in 4 days . a alone can do it in 12 days . what time b will take to do the work alone ?
"relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 1.9 = 3 km = 3000 m . required time = 3000 * 3 / 125 = 72 sec . answer : b"
a ) 99 , b ) 72 , c ) 48 , d ) 96 , e ) 22
b
subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.9, const_1000), divide(multiply(90, const_1000), const_3600)))
multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)|
physics
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 1.9 km respectively . the time taken by the slower train to cross the faster train in seconds is ?
"ac = 11 and ab = 5 , so bc = 6 . bc = 2 cd so cd = 3 . the length of ae is ab + bc + cd + de = 5 + 6 + 3 + 5 = 19 the answer is a ."
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27
a
add(add(11, divide(subtract(11, 5), 2)), 5)
subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)|
physics
a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 5 , ab = 5 and ac = 11 , what is the length of ae ?
"clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 288 / 32 = 9 m / s distance covered by b in 4 seconds = speed ã — time = 9 ã — 4 = 36 metre i . e . , a beat b by 36 metre answer is e"
a ) 38 metre , b ) 28 metre , c ) 23 metre , d ) 15 metre , e ) 36 metre
e
subtract(288, multiply(divide(288, 32), 28))
divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|
physics
a can run 288 metre in 28 seconds and b in 32 seconds . by what distance a beat b ?
"= ( 417 ) ^ 2 + ( 383 ) ^ 2 = ( 400 + 17 ) ^ 2 + ( 400 - 17 ) ^ 2 = 2 [ ( 400 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 160000 + 289 ] = 2 x 160289 = 320578 answer is a"
a ) 320578 , b ) 80578 , c ) 80698 , d ) 81268 , e ) none of them
a
multiply(417, power(417, 383))
power(n1,n2)|multiply(n0,#0)|
general
417 x 417 + 383 x 383 = ?
"any x and y satisfying x / y = 7 / 4 should give the same value for ( x + y ) / ( x - y ) . say x = 7 and y = 4 , then ( x + y ) / ( x - y ) = ( 7 + 4 ) / ( 7 - 4 ) = 11 / 3 . answer : b ."
a ) 5 , b ) 11 / 3 , c ) - 1 / 6 , d ) - 1 / 5 , e ) - 5
b
divide(add(7, 4), subtract(7, 4))
add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|
general
if x / y = 7 / 4 , then ( x + y ) / ( x - y ) = ?
"in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.5 ) = $ 7 ( as the charge is 0.5 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 7 = 8 answer is e ."
a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 4.20 , e ) $ 8.00
e
add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.50))
divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)|
general
if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.50 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was
"let us assume x = 1 and y = 37.5 then 1.5 ( x ) = 0.04 ( y ) that is 1.5 ( 1 ) = 0.04 ( 37.5 ) = 1.5 = 1.5 therfore from given equation ( 37.5 - 1 ) ( 37.5 + 1 ) = ( 36.5 ) ( 38.5 ) = 1405.25 answer : a"
a ) 1405.25 , b ) 1305.25 , c ) 1205.25 , d ) 1505.25 , e ) 1605.25
a
divide(subtract(divide(multiply(1.5, const_100), const_2), const_2), add(divide(multiply(1.5, const_100), const_2), const_2))
multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)|
general
if 1.5 x = 0.04 y , then the value of ( y - x ) ( y + x ) is :
"say a 100 families existed in 1956 then the number of families owning a computer in 1956 - 25 number of families owning computer in 1960 = 25 * 115 / 100 = 28.75 number of families in 1960 = 105 the percentage = 28.75 / 105 * 100 = 27.38 % . option : e"
a ) 50 % , b ) 51.22 % , c ) 5.26 % , d ) 7.20 % , e ) 27.38 %
e
multiply(const_100, divide(divide(multiply(add(15, const_100), 25), const_100), add(const_100, 5)))
add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)|
general
of the families in city x in 1956 , 25 percent owned a personal computer . the number of families in city x owning a computer in 1960 was 15 percent greater than it was in 1956 , and the total number of families in city x was 5 percent greater in 1956 than it was in 1960 . what percent of the families in city x owned a personal computer in 1960 ?
"sol . sp = rs 100 : then cp = rs 91 : profit = rs 9 . profit = { ( 9 / 91 ) * 100 } % = 9.89 % answer is a ."
a ) 9.89 % , b ) 8.90 % , c ) 9.00 % , d ) 8.00 % , e ) 9.27 %
a
multiply(divide(subtract(const_100, 91), 91), const_100)
subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
if the cost price is 91 % of sp then what is the profit %
"a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 24 41 - - - - ? = > rs . 123 answer : d"
a ) 288 , b ) 262 , c ) 72 , d ) 123 , e ) 267
d
multiply(divide(24, 40), add(add(const_100, 65), 40))
add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)|
general
a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 24 , find the sum of money ?
"explanation : relative speed = ( 72 - 36 ) x 5 / 18 = 2 x 5 = 10 mps . distance covered in 32 sec = 32 x 10 = 320 m . the length of the faster train = 320 m . answer is d"
a ) 170 m , b ) 100 m , c ) 270 m , d ) 320 m , e ) 350 m
d
multiply(divide(subtract(72, 36), const_3_6), 32)
subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)|
physics
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a girl sitting at window seat in the slower train in 32 seconds . find the length of the faster train ?
number of ways first - place medal can be awarded to four contestants = 8 number of ways second - place medal can be awarded to contestants after awarding first - place medal = 3 therefore number of possibilities = 8 * 3 = 24 answer : e
a ) 6 , b ) 7 , c ) 12 , d ) 16 , e ) 24
e
multiply(8, subtract(4, const_1))
subtract(n1,const_1)|multiply(n0,#0)
general
tough and tricky questions : combinations . 8 contestants representing 4 different countries advance to the finals of a fencing championship . assuming all competitors have an equal chance of winning , how many possibilities are there with respect to how a first - place and second - place medal can be awarded ?
"explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 44000 ] = rs . 12000 answer : option a"
a ) rs . 12000 , b ) rs . 13375 , c ) rs . 11750 , d ) rs . 11625 , e ) none of these
a
multiply(44000, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1)))
divide(n1,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,const_1)|inverse(#3)|multiply(n3,#4)|
gain
a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 44000 . what is the share of c ?
"solution simple out of 400 20 % are male i . e 80 and 25 % are female i . e 100 , so total homeowner is 180 . now min number homeowner is 80 and max is 180 so question ask us to find least and 81 has least value among all option . so ans is 81 . answer : a"
a ) 81 , b ) 147 , c ) 145 , d ) 143 , e ) 141
a
add(multiply(multiply(divide(25, const_100), 20), multiply(divide(25, const_100), 20)), divide(subtract(400, 20), 20))
divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)|
general
in a 400 member association consisting of men and women , exactly 20 % of men and exactly 25 % women are homeowners . what is the least number of members who are homeowners ?
total amount of interest is i = p / 100 * 1 [ 3 / 12 + 9 / 12 + 27 / 12 … . 312 / 12 where p = 100 ; i = 1 / 12 ( 3 + 9 + … . . 312 ) i = 1 / 12 ( 3 ( 312 - 1 ) ) / 3 - 1 = 531440 * 3 / 12 * 2 = rs . 66430 answer : d
a ) rs . 797160 , b ) rs . 791160 , c ) rs . 65930 , d ) rs . 66430 , e ) rs . 67430
d
divide(multiply(subtract(power(3, const_12), const_1), 3), multiply(const_12, const_2))
multiply(const_12,const_2)|power(n1,const_12)|subtract(#1,const_1)|multiply(n1,#2)|divide(#3,#0)
general
a sum of rs . 100 is lent at simple interest of 3 % p . a . for the first month , 9 % p . a . for the second month , 27 % p . a . for the third month and so on . what is the total amount of interest earned at the end of the year approximately
the top and the bottom are each single faces formed by three equilateral triangles joining , as in the diagram on the left , to make an isosceles trapezoid . top = 1 face , and bottom = 1 face . this is a four - sided figure , so there are four rectangles extending from the bottom of this prism to the congruent figure at the top . notice , in particular , the larger vertical face in the “ back ” of the diagram to the right is formed by two faces of the original triangular prisms lying next to each other and smoothly joining . total = 1 top + 1 bottom + 4 sides = 6 faces . answer = b .
a ) 4 , b ) 6 , c ) 9 , d ) 10 , e ) 12
b
add(add(const_4, const_1), const_1)
add(const_1,const_4)|add(#0,const_1)|
geometry
suppose you have three identical prisms with congruent equilateral triangles as the end - polygons . suppose you attach them by the rectangular faces so they are perfectly aligned . there will be some large faces created by two or more co - planar faces of the individual prisms : count each such large face as one . given that , how many faces does the resultant solid have ?
the area of a square whose side is 6 – ( the area of a square whose side is 4 + the area of the semi - circle whose side is 4 ) = the area of the region shaded the correct answer is b .
['a ) 48 - 8 π', 'b ) 48 - 6 π', 'c ) 24 + 6 π', 'd ) 16 + 8 π', 'e ) 64 - 8 π']
b
subtract(multiply(const_3, multiply(const_4, const_4)), multiply(6, const_pi))
multiply(const_4,const_4)|multiply(n0,const_pi)|multiply(#0,const_3)|subtract(#2,#1)
geometry
the side of a square has the length of 6 . what is the area of the region shaded ?
"formula : ( constant ) / ( lcm of two nos ) = 100 / ( 4 * 5 ) = 5 answer : d"
a ) 50 , b ) 33 , c ) 16 , d ) 5 , e ) 14
d
divide(5, 4)
divide(n1,n0)|
general
how many positive integer solutions does the equation 4 x + 5 y = 100 have ?
as given , after lost , the remaining 8 crates total cost = $ 40 so , 1 crate cost = 40 / 8 = 5 to get 20 % profit , 1 crate cost should be = 5 + 5 * 20 / 100 = $ 6 answer : a
a ) $ 6 , b ) $ 8 , c ) $ 10 , d ) $ 12 , e ) $ 14
a
divide(add(40, multiply(40, divide(20, const_100))), subtract(10, 2))
divide(n3,const_100)|subtract(n0,n2)|multiply(n1,#0)|add(n1,#2)|divide(#3,#1)
gain
a man bought 10 crates of mangoes for $ 40 total . if he lost 2 of the crates , at what price would he have to sell each of the remaining crates in order to earn a total profit of 20 percent of the total cost ?
"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 32 = 15 = > x = 180 m . answer : e"
a ) 228 , b ) 240 , c ) 887 , d ) 166 , e ) 180
e
multiply(20, multiply(54, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics
a train passes a station platform in 32 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?
"9.18 x + 5.17 x + 2.05 x = 16.4 x = 170 cookies x = 170 / 16.4 = 10 ( approx ) so , irin baked 10 * 9.18 cookies or 91 cookies ( approx ) % share = 91 / 170 = 53.5 approx hence , answer is d ."
a ) 0.125 % , b ) 1.25 % , c ) 12.5 % , d ) 53.5 % , e ) 0.152 %
d
multiply(divide(divide(multiply(170, 5.17), add(add(9.18, 5.17), 2.05)), 170), const_100)
add(n0,n1)|multiply(n1,n3)|add(n2,#0)|divide(#1,#2)|divide(#3,n3)|multiply(#4,const_100)|
other
irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 170 cookies , what percent of the cookies did irin bake ?
"1 st digit can be filled up by the numbers - { 1 , 2 , 4 , 5 , 7 , 8 , 9 } = 7 ways 2 nd digit can be filled up by the numbers - { 0 , 1 , 2 , 4 , 5 , 7 , 8 , 9 } = 8 ways 3 rd digit can be filled up by the numbers - { 0 , 1 , 2 , 4 , 5 , 7 , 8 , 9 } = 8 ways so , total no of ways is 7 * 8 * 8 = > 3584 hence answer will be ( b )"
a ) 2401 , b ) 3584 , c ) 4096 , d ) 5040 , e ) 7200
b
multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 6)
multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)|
general
how many four - digit numbers that do not contain the digits 3 or 6 are there ?
"formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 30 % = 130 % 130 % - - - - - - - > 650 ( 130 ã — 5 = 650 ) 100 % - - - - - - - > 400 ( 100 ã — 5 = 500 ) option ' d '"
a ) 200 , b ) 300 , c ) 400 , d ) 500 , e ) 600
d
divide(650, add(const_1, divide(30, const_100)))
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
gain
a number increased by 30 % gives 650 . the number is ?
"my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 16.2 - answer b"
a ) 15.2 % , b ) 16.2 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 %
b
multiply(const_100, subtract(multiply(multiply(divide(add(25, const_100), const_100), divide(add(25, const_100), const_100)), divide(subtract(const_100, 25), const_100)), const_1))
add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#3,#3)|multiply(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)|
general
the positive numbers w , x , y , and z are such that x is 25 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ?
"they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 20 * 6 / 10 = 12 days . answer : a"
a ) 12 days . , b ) 17 days . , c ) 18 days . , d ) 19 days . , e ) 20 days .
a
subtract(20, multiply(divide(20, 10), 4))
divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)|
physics
arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 20 days .
120 and 60 are both divisible by 6 and 15 but also by 12 . so they are not the right answer . 36 and 54 are both clearly not divisible by 15 ( not correct ) 90 is both divisible by 6 and 15 but not by 12 . answer : ( d ) 90
a ) 120 , b ) 60 , c ) 36 , d ) 90 , e ) 54
d
multiply(6, 15)
multiply(n0,n1)
general
if ' x ' is a positive integer exactly divisible by 6 or 15 but not divisible by 12 . what could possibly be the value of ' x ' ?
"90 = ( 450 * 4 * r ) / 100 r = 5 % i = ( 450 * 6 * 5 ) / 100 = 135 450 + 135 = 585 answer : c"
a ) 227 , b ) 299 , c ) 270 , d ) 585 , e ) 271
c
subtract(multiply(subtract(540, 450), 6), subtract(540, 450))
subtract(n1,n0)|multiply(n3,#0)|subtract(#1,#0)|
gain
if rs . 450 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ?
"the difference between compound interest and simple interest on rs . p for 2 years at r % per annum = ( r ã — si ) / ( 2 ã — 100 ) difference between the compound interest and simple interest = 340 - 324 = 16 ( r ã — si ) / ( 2 ã — 100 ) = 16 ( r ã — 324 ) / ( 2 ã — 100 ) = 16 r = 9.87 % answer : option c"
a ) 15 % , b ) 14.25 % , c ) 9.87 % , d ) 10.5 % , e ) 11.5 %
c
divide(multiply(const_100, subtract(subtract(340, divide(324, 2)), divide(324, 2))), divide(324, 2))
divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)|
gain
on a sum of money , the simple interest for 2 years is rs . 324 , while the compound interest is rs . 340 , the rate of interest being the same in both the cases . the rate of interest is
"let t be the tier price , p be total price = 14000 per the given conditions : 0.10 t + 0.08 ( p - t ) = 1440 - - - - > t = 1600 . a is the correct answer ."
a ) $ 1600 , b ) $ 6000 , c ) $ 6050 , d ) $ 7050 , e ) $ 8000
a
divide(subtract(1440, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)|
general
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 10 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 14,000 imported car and ended up paying $ 1440 in taxes , what is the first tier ' s price level ?
"let the speed of bus be v - 20 , speed of taxi be v the bus travelled a total of 4 hrs and taxi a total of 2 hrs . hence 4 * ( v - 20 ) = 2 v 4 v - 80 = 2 v 2 v = 80 v = 40 mph e"
a ) 42 , b ) 48 , c ) 44 , d ) 50 , e ) 40
e
divide(add(multiply(2, 20), multiply(2, 20)), subtract(add(2, 2), 2))
add(n0,n0)|multiply(n0,n1)|add(#1,#1)|subtract(#0,n0)|divide(#2,#3)|
physics
a taxi leaves point a 2 hours after a bus left the same spot . the bus is traveling 20 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in two hours .
"when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 10 * 24 / 60 = 4 miles and the man walks the distance mm , which is 10 * 12 / 60 = 2 mile . the difference of 4 - 2 = 2 miles ( mw ) will be covered by the man in ( 2 ) / 9 = 2 / 9 of an hour , which is ~ 13 minutes . answer a ."
a ) 13 , b ) 10 , c ) 15 , d ) 11 , e ) 9
a
multiply(const_60, divide(multiply(divide(10, const_60), subtract(24, 12)), 12))
divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)|
physics
a man walking at a constant rate of 12 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 24 miles per hour . the woman stops to wait for the man 10 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ?
"| 5 x - 5 | = 100 5 x - 5 = 100 or 5 x - 5 = - 100 5 x = 105 or 5 x = - 95 x = 21 or x = - 19 sum = 21 - 19 = 2 answer is c"
a ) 1 , b ) - 2 , c ) 2 , d ) - 3 , e ) 4
c
subtract(subtract(subtract(100, 5), add(100, 5)), 5)
add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)|
general
if | 5 x - 5 | = 100 , then find the sum of the values of x ?
"sol . 39 ÷ 3 = 13 . answer : d"
a ) 20 lb , b ) 30 lb , c ) 10 lb , d ) 13 lb , e ) 5 lb
d
divide(39, const_1)
divide(n0,const_1)|
general
a bag of potatoes weighs 39 lbs divided by half of its weight . how much does the bag of potatoes weight ?
"actual price = 500 + 50 = $ 550 saving = 10 / 550 * 100 = 100 / 55 = 1.8 % approximately answer is e"
a ) 8 % , b ) 10 % , c ) 2 % , d ) 5 % , e ) 1.8 %
e
add(floor(multiply(divide(50, add(50, 500)), 500)), const_1)
add(n0,n1)|divide(n0,#0)|multiply(#1,n1)|floor(#2)|add(#3,const_1)|
general
a person saved $ 50 in buying an item on sale . if he spent $ 500 for the item , approximately how much percent he saved in the transaction ?
"let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 20 % of 850 = 170 . answer : b"
a ) 58 , b ) 170 , c ) 95 , d ) 100 , e ) 120
b
divide(multiply(510, add(const_4, const_1)), add(const_1, const_2))
add(const_1,const_4)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)|
general
the difference between a number and its two - fifth is 510 . what is 20 % of that number ?
let x be the number of guests . number of muffins prepared = 5 x + 2 number of muffins eaten + number of muffins remaining = number of muffins prepared 6 ( x - 1 ) + 6 = 5 x + 2 6 x = 5 x + 2 x = 2 answer : a
a ) 2 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 7 .
a
divide(subtract(add(6, 6), const_2), 5)
add(n1,n1)|subtract(#0,const_2)|divide(#1,n0)
general
monica planned her birthday party . she prepared 5 muffins for each of her guests and kept aside two additional muffins in case someone will want extra . after the party , it turned out that one of the guests did n ' t come but every one of the guests that did come ate 6 muffins and 6 muffins remained . how many guests did monica plan on ?
total distance traveled = 10 + 12 = 22 km / hr . total time taken = 10 / 12 + 12 / 10 = 61 / 30 hrs . average speed = 22 * 30 / 61 = 10.8 km / hr . answer : b
a ) 10.7 km / hr , b ) 10.8 km / hr , c ) 17.8 km / hr , d ) 10.5 km / hr , e ) 30.8 km / hr
b
divide(add(12, 10), const_2)
add(n0,n1)|divide(#0,const_2)
general
a boy rides his bicycle 10 km at an average speed of 12 km / hr and again travels 12 km at an average speed of 10 km / hr . his average speed for the entire trip is approximately ?
3 ^ x = 2 3 ^ 4 x = 2 ^ 4 3 ^ 4 x = 16 3 ^ ( 4 x + 3 ) = 3 ^ 4 x * 3 ^ 3 = 16 * 27 = 432 answer : c
a ) 429 , b ) 454 , c ) 432 , d ) 438 , e ) 108
c
power(3, add(multiply(4, divide(log(const_2), log(const_3))), 3))
log(const_2)|log(const_3)|divide(#0,#1)|multiply(n3,#2)|add(n0,#3)|power(n0,#4)
general
if 3 ^ x = 2 , then 3 ^ ( 4 x + 3 ) =
"they meet every 360 / 7 + 8 = 24 sec answer is d"
a ) 20 sec , b ) 15 sec , c ) 30 sec , d ) 24 sec , e ) 1 min
d
divide(360, add(8, 7))
add(n0,n1)|divide(n2,#0)|
physics
two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 360 meters , after what time will they meet at the starting point ?
"total savings = s amount spent on stereo = ( 1 / 3 ) s amount spent on television = ( 1 - 1 / 3 ) ( 1 / 3 ) s = ( 2 / 3 ) * ( 1 / 3 ) * s = ( 2 / 9 ) s ( stereo + tv ) / total savings = s ( 1 / 3 + 2 / 9 ) / s = 5 / 9 answer : c"
a ) 1 / 4 , b ) 2 / 7 , c ) 5 / 9 , d ) 1 / 2 , e ) 7 / 12
c
divide(1, 3)
divide(n0,n1)|
general
carol spends 1 / 3 of her savings on a stereo and 1 / 3 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ?
"y = 0.5 * 0.5 * x = x / 4 y % * x = 36 ( y / 100 ) * x = 36 ( x / 400 ) * x = 36 x ^ 2 = 36 * 400 x = 120 the answer is e ."
a ) 80 , b ) 96 , c ) 100 , d ) 108 , e ) 120
e
multiply(multiply(divide(50, 36), divide(50, 36)), const_1000)
divide(n0,n2)|divide(n1,n2)|multiply(#0,#1)|multiply(#2,const_1000)|
general
positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 36 . what is the value of x ?
"time = 6 distence = 540 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 540 / 9 = 60 kmph c )"
a ) 40 kmph , b ) 50 kmph , c ) 60 kmph , d ) 75 kmph , e ) 860 kmph
c
divide(540, multiply(divide(3, 2), 6))
divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)|
physics
a van takes 6 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ?
"p ( missing all 3 ) = ( 2 / 3 ) ^ 3 = 8 / 27 p ( success on at least one attempt ) = 1 - 8 / 27 = 19 / 27 the answer is d ."
a ) 1 / 3 , b ) 5 / 9 , c ) 7 / 9 , d ) 19 / 27 , e ) 23 / 27
d
subtract(const_1, power(divide(const_1, 3), 3))
divide(const_1,n0)|power(#0,n1)|subtract(const_1,#1)|
probability
on a ranch , a rancher can place a loop of rope , called a lasso , once in every 3 throws around a cow ’ s neck . what is the probability that the rancher will be able to place a lasso around a cow ’ s neck at least once in 3 attempts ?
"90 feet / 2 seconds = 45 feet / second ( 45 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 30.68 ( approximately ) the answer is a ."
a ) 30.68 , b ) 32.34 , c ) 34.56 , d ) 36.72 , e ) 38.95
a
divide(divide(90, 5280), multiply(2, divide(1, const_3600)))
divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)|
physics
if an object travels 90 feet in 2 seconds , what is the object ’ s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet )
"25 = e / ( 1000 - e ) * 100 1000 - e = 4 e 1000 = 5 e = > e = 200 1000 - 200 = 800 answer : c"
a ) 338 , b ) 278 , c ) 800 , d ) 269 , e ) 112
c
subtract(multiply(divide(const_100, 25), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100)
add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)|
gain
a dishonest dealer professes to sell goods at the cost price but uses a false weight and gains 25 % . find his false weight age ?
"16.02 ã — 0.1 = 1.602 the answer is c ."
a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these
c
multiply(divide(16.02, 0.1), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
16.02 ã — 0.1 = ?
"formula : ( after = 100 denominator ago = 100 numerator ) 13400 × 121 / 100 × 121 / 100 = 19619 c"
a ) 10000 , b ) 12000 , c ) 19619 , d ) 18989 , e ) 14400
c
multiply(13400, power(add(const_1, divide(21, const_100)), 2))
divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|
gain
the population of a village is 13400 . it increases annually at the rate of 21 % p . a . what will be its population after 2 years ?
"sol . sp = rs 100 : then cp = rs 90 : profit = rs 10 . profit = { ( 10 / 90 ) * 100 } % = 1.11 % answer is b ."
a ) 1.10 % , b ) 1.11 % , c ) 1.15 % , d ) 1.17 % , e ) 1.27 %
b
multiply(divide(subtract(const_100, 90), 90), const_100)
subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
if the cost price is 90 % of sp then what is the profit %
"least number of cubes will be required when the cubes that could fit in are biggest . 7 is the biggest number that could divide all three , 49 , 21 and 14 . thus side of cube must be 7 , and total number of cubes = 49 / 7 * 21 / 7 * 14 / 7 = 42 ans b it is ."
a ) 17 , b ) 42 , c ) 54 , d ) 108 , e ) 864
b
divide(multiply(multiply(49, 21), 14), volume_cube(divide(14, const_2)))
divide(n2,const_2)|multiply(n0,n1)|multiply(n2,#1)|volume_cube(#0)|divide(#2,#3)|
geometry
a box measuring 49 inches long by 21 inches wide by 14 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ?
"for first 2000 meters he does not get any discount . the price is 2 * 2000 = $ 4000 for next 1500 meters , he gets a 5 % discount . the price is 1.9 * 1500 = $ 2850 for the next 1500 meters , he gets a 7 % discount . the price is 1.86 * 2000 = $ 3720 the total price is $ 4000 + $ 2850 + $ 3720 = $ 10,570 the answer is d ."
a ) $ 7,360 , b ) $ 8,540 , c ) $ 9,720 , d ) $ 10,570 , e ) $ 11,680
d
multiply(multiply(2, const_3), const_100)
multiply(n4,const_3)|multiply(#0,const_100)|
gain
a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 5,500 meters of fabric ?
"( 7000 * 3 * 2 ) / 100 = 420 9200 - - - - - - - - 9620 answer : c"
a ) 9680 , b ) 2277 , c ) 9620 , d ) 2774 , e ) 1212
c
add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(9200, 7000), 3), 7000)), 7000), 3), 7000)
divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)|
gain
sonika deposited rs . 7000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ?
"we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 13 meters in that time . so if the race is 20 meters long , when a reaches the finish line , b would be 7 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 7 meters so that he doesn ’ t need to cover that . in that case , the time required by a ( to cover 20 meters ) would be the same as the time required by b ( to cover 13 meters ) to reach the finish line . so b should get a head start of 7 / 20 th of the race . answer ( d )"
a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 7 / 20 , e ) 3 / 10
d
divide(subtract(20, 13), 20)
subtract(n0,n1)|divide(#0,n0)|
general
a ’ s speed is 20 / 13 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ?
"explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 4 : 1 ( webs ) 1 : 9 } : : x : 3 â ‡ ’ 4 ã — 1 ã — 9 = 1 ã — 3 ã — x â ‡ ’ x = 12 answer : option b"
a ) 9 , b ) 12 , c ) 27 , d ) 24 , e ) 16
b
multiply(1, 4)
multiply(n0,n3)|
physics
if 4 spiders make 3 webs in 9 days , then how many days are needed for 1 spider to make 1 web ?
explanation : largest number of four digits = 9999 lcm of 15 , 25 , 40 and 75 = 600 9999 ÷ 600 = 16 , remainder = 399 hence , largest number of four digits which is divisible by 15 , 25 , 40 and 75 = 9999 - 399 = 9600 answer : a
a ) 9600 , b ) 5200 , c ) 362 , d ) 958 , e ) 258
a
subtract(subtract(multiply(const_100, const_100), const_1), reminder(subtract(multiply(const_100, const_100), const_1), lcm(lcm(lcm(15, 25), 40), 75)))
lcm(n1,n2)|multiply(const_100,const_100)|lcm(n3,#0)|subtract(#1,const_1)|lcm(n4,#2)|reminder(#3,#4)|subtract(#3,#5)
general
what is the largest number of 4 digits which is divisible by 15 , 25 , 40 and 75 ?
"let the side of the square plot be a ft . a 2 = 64 = > a = 8 length of the fence = perimeter of the plot = 4 a = 32 ft . cost of building the fence = 32 * 58 = rs . 1856 . answer : d"
a ) 3944 , b ) 2882 , c ) 2999 , d ) 1856 , e ) 2121
d
multiply(square_perimeter(sqrt(64)), 58)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
what will be the cost of building a fence around a square plot with area equal to 64 sq ft , if the price per foot of building the fence is rs . 58 ?
"20 % of 6 = 1.2 50 % of 6 = 3 shortage is 1.8 so we need to have 1.8 / 50 % to get 50 % alcohol content . = 3.6 d"
a ) a . 0.6 , b ) b . 1 , c ) c . 2.1 , d ) d . 3.6 , e ) e . 5.4
d
subtract(6, multiply(const_2, multiply(divide(20, const_100), 6)))
divide(n1,const_100)|multiply(n0,#0)|multiply(#1,const_2)|subtract(n0,#2)|
gain
a 6 litre sol is 20 % alcohol . how many litres of pure alcohol must be added to produce a sol that is 50 % alcohol ?
"( 45 / 100 ) z = ( 60 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 60 / 100 ) * ( 3 / 4 ) x i . e . z = ( 60 * 3 ) x / ( 45 * 4 ) i . e . z = ( 1 ) x = ( 100 / 100 ) x i . e . z is 100 % of x answer : option c"
a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50
c
multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(60, const_100))), const_100)
divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|divide(#1,#2)|divide(#0,#3)|multiply(#4,const_100)|
gain
if 45 % of z is 60 % of y and y is 75 % of x , what percent of x is z ?
in the range 800 - 900 : 1 choice for the first digit : 8 ; 5 choices for the third digit : 1 , 3 , 5 , 7 , 9 ( as the number is odd ) ; 8 choices for the second digit : 10 digits - first digit - third digit = 8 digits . 1 * 5 * 8 = 40 . in the range 900 - 999 : 1 choice for the first digit : 9 ; 4 choices for the third digit : 1 , 3 , 5 , 7 ( 9 is out as it ' s the first digit ) ; 8 choices for the second digit : 10 digits - first digit - third digit = 8 digits . 1 * 4 * 8 = 32 . total : 40 + 32 = 72 . answer : c
a ) 40 , b ) 56 , c ) 72 , d ) 81 , e ) 104
c
add(multiply(add(3, add(const_3, const_2)), const_4), multiply(multiply(const_1, add(const_3, const_2)), add(3, add(const_3, const_2))))
add(const_2,const_3)|add(n0,#0)|multiply(#0,const_1)|multiply(#1,const_4)|multiply(#1,#2)|add(#3,#4)
general
how many odd 3 - digit integers greater than 800 are there such that all their digits are different ?
"let 2 x be the number of spinsters . then 7 x is the number of cats . 7 x - 2 x = 40 x = 8 and the number of spinsters is 2 ( 8 ) = 16 . the answer is b ."
a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22
b
multiply(divide(40, subtract(7, 2)), 2)
subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|
other
the ratio of spinsters to cats is 2 to 7 . if there are 40 more cats than spinsters , how many spinsters are there ?
"33 1 / 3 % = 1 / 3 1 / 3 × 240 = 80 a )"
a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130
a
divide(multiply(add(33, divide(1, 3)), 240), const_100)
divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)|
gain
33 1 / 3 % of 240 ?
"price of 1 ticket = 20 $ revenue generated from sales of first 10 tickets = 10 * ( 60 / 100 * 20 ) = 10 * 12 = 120 revenue generated from sales of next 20 tickets = 20 * ( 85 / 100 * 20 ) = 20 * 17 = 340 revenue generated from sales of last 30 tickets = 30 * 26 = 780 revenue generated from sales of 60 tickets = 120 + 340 + 780 = 1240 $ answer e"
a ) $ 600 , b ) $ 740 , c ) $ 850 , d ) $ 980 , e ) $ 1,240
e
multiply(add(add(subtract(subtract(60, 20), 10), multiply(subtract(const_1, divide(40, const_100)), 10)), multiply(subtract(const_1, divide(15, const_100)), 20)), 20)
divide(n2,const_100)|divide(n4,const_100)|subtract(n5,n0)|subtract(const_1,#0)|subtract(#2,n1)|subtract(const_1,#1)|multiply(n1,#3)|multiply(n0,#5)|add(#6,#4)|add(#8,#7)|multiply(n0,#9)|
gain
tickets to a certain concert sell for $ 20 each . the first 10 people to show up at the ticket booth received a 40 % discount , and the next 20 received a 15 % discount . if 60 people bought tickets to the concert , what was the total revenue from ticket sales ?
"x ^ 2 + 5 x - 6 = ( x + 6 ) ( x - 1 ) = 0 then x = - 6 or x = 1 . x ^ 2 - 5 x + 4 = ( x - 4 ) ( x - 1 ) = 0 then x = 4 or x = 1 . thus x = 1 . the answer is b ."
a ) - 1 , b ) 1 , c ) 4 , d ) - 6 , e ) 6
b
divide(add(4, 6), add(2, 5))
add(n2,n6)|add(n0,n5)|divide(#0,#1)|
general
if x is a number such that x ^ 2 + 5 x - 6 = 0 and x ^ 2 - 5 x + 4 = 0 , then x =
"first pump drains 1 / 2 of the tank in 7 hours so 14 hours it will take to drain the full tank . let , 2 nd pump drains the full tank in a hours so both together can drain ( 1 / 14 + 1 / a ) part in 1 hour son in 1 / 2 hour they drain 1 / 2 * ( 1 / 14 + 1 / a ) part of the tank given that in 1 / 2 hour they drain 1 / 2 of the tank hence we can say 1 / 2 * ( 1 / 14 + 1 / a ) = 1 / 2 solving u get a = 14 / 13 = 1.1 hence answer is b"
a ) 1 hour , b ) 1.1 hour , c ) 3 hours , d ) 5 hours , e ) 6 hours
b
divide(const_1, subtract(const_1, divide(const_1, multiply(7, const_2))))
multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)|
physics
one pump drains one - half of a pond in 7 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ?
"18 * 18 = 324 sq m answer : d"
a ) 225 sq m , b ) 186 sq m , c ) 586 sq m , d ) 324 sq m , e ) 296 sq m
d
square_area(18)
square_area(n0)|
geometry
what is the area of square field whose side of length 18 m ?
"( 80 / 100 ) * 40 â € “ ( 4 / 5 ) * 30 32 - 24 = 8 answer : c"
a ) 12 , b ) 27 , c ) 8 , d ) 12 , e ) 81
c
subtract(multiply(40, divide(80, const_100)), multiply(divide(4, 5), 30))
divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)|
general
how much is 80 % of 40 is greater than 4 / 5 of 30 ?
"the odd numbers less than 60 are 1 , 3,5 , . . . 59 some of them are equal to product of 5 and odd number hence odd multiples of 5 which are less than 60 are 5,15 , 25,35 , 45,55 ( total = 6 ) ; answer : b"
a ) 4 , b ) 6 , c ) 11 , d ) 12 , e ) 15
b
divide(divide(60, 5), const_2)
divide(n0,n1)|divide(#0,const_2)|
general
how many unique positive odd integers less than 60 are equal to the product of a positive multiple of 5 and an odd number ?
"let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 18 - x ) + x + 2 = 30 37 - 2 x + x = 30 37 - x = 30 x = 7 so 7 members play both badminton and tennis . a"
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
a
subtract(add(add(17, 18), 2), 30)
add(n1,n2)|add(n3,#0)|subtract(#1,n0)|
other
in a sports club with 30 members , 17 play badminton and 18 play tennis and 2 do not play either . how many members play both badminton and tennis ?
"slope = ( y 2 - y 1 ) / ( x 2 - x 1 ) = > k = ( k + 4 ) / ( 3 + 3 ) = > 6 k = k + 4 = > k = 4 / 5 ans c it is !"
a ) 3 / 4 , b ) 1 , c ) 4 / 5 , d ) 2 , e ) 7 / 2
c
divide(4, 3)
divide(n1,n2)|
general
a line that passes through ( – 3 , – 4 ) and ( 3 , k ) has a slope = k . what is the value of k ?
"2 x will be the distance travelled by the train if the length of the train = the length of the platform = x as distance = speed * time distance = 2 x speed in kmph = 90 speed in mps = 90 * 5 / 18 as dist = speed * time 2 x = ( 90 * 5 / 18 ) * ( 60 sec ) on solving x = 900 mts answer : d"
a ) 500 , b ) 600 , c ) 750 , d ) 900 , e ) 950
d
divide(divide(multiply(90, const_1000), divide(const_60, const_1)), const_2)
divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)|
physics
the length of a train and that of a platform are equal . if with a speed of 90 kmph the train crosses the platform in one minute , then the length of the train in metres is
"4 x + 2 = 5 y + 2 . . . . . . . . . . . ie : 4 x - 5 y = 0 x , y must be > 1 and y is even ie ( 2 , 4,6 , . . etc ) if y = 2 thus x is fraction if y = 4 thus x = 5 n = 22 if y = 6 thus x = not possible fraction if y = 8 thus x = 10 n = 42 summation 22 + 42 = 64 c"
a ) 33 , b ) 46 , c ) 49 , d ) 53 , e ) 86
c
add(add(multiply(5, const_2), 2), add(multiply(5, multiply(const_2, 2)), 2))
multiply(n2,const_2)|multiply(const_2,n3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)|
general
a group of n students can be divided into equal groups of 4 with 2 student left over or equal groups of 5 with 2 students left over . what is the sum of the two smallest possible values of n ?
the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 140 tagged deers comprise 10 % of total # of deers - - > total # of deers = 140 * 10 = 1,400 . answer : d .
a ) 150 , b ) 750 , c ) 1,250 , d ) 1,400 , e ) 2,500
d
multiply(50, 5)
multiply(n1,n3)
general
in a forest 140 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ?
"machine a can produce 100 * 30 / 15 = 200 copies and , machine b can produce 150 * 30 / 10 = 450 copies total producing 650 copies . c is the answer"
a ) 250 , b ) 425 , c ) 650 , d ) 700 , e ) 750
c
multiply(add(divide(100, 15), divide(150, 10)), 30)
divide(n0,n1)|divide(n2,n3)|add(#0,#1)|multiply(n4,#2)|
physics
working at their respective constant rates , machine a makes 100 copies in 15 minutes and machine b makes 150 copies in 10 minutes . if these machines work simultaneously at their respective rates for 30 minutes , what is the total number of copies that they will produce ?
"age of the 15 th student = [ 15 * 16 - ( 14 * 5 + 16 * 9 ) ] = ( 240 - 214 ) = 26 years . answer : c"
a ) 11 years , b ) 17 years , c ) 26 years , d ) 14 years , e ) 12 years
c
subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16)))
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
general