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a train 110 m long is running at 50 kmph . in how much time will it pass a platform 210 m long ?
distance travelled = 110 + 210 m = 320 m speed = 50 * 5 / 8 = 125 / 4 m time = 320 * 4 / 125 = 10.24 seconds answer : e .
a ) 14.15 seconds , b ) 15.24 seconds , c ) 16.28 seconds , d ) 13.30 seconds , e ) 10.24 seconds
e
divide(add(110, 210), multiply(50, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
a man buys an article for $ 100 . and sells it for $ 115 . find the gain percent ?
"c . p . = $ 100 s . p . = $ 115 gain = $ 15 gain % = 15 / 100 * 100 = 25 % answer is b"
a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 %
b
subtract(divide(115, divide(100, const_100)), const_100)
divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)|
gain
there was a cycle race going on . 1 / 5 th of the those in front of a person and 5 / 6 th of those behind him gives the total number of participants . how many people took part in the race ?
explanation : let total participants be ( x + 1 ) then = > ( 1 / 5 ) * x + ( 5 / 6 ) * x = x + 1 = > 31 x / 30 = x + 1 = > x = 30 so total participants ( x + 1 ) = 31 hence ( a ) is the correct answer . answer : a
a ) 31 , b ) 39 , c ) 30 , d ) 33 , e ) 34
a
add(multiply(5, 6), const_1)
multiply(n1,n3)|add(#0,const_1)
general
at what rate percent on simple interest will rs . 1750 amount to rs . 2000 in 2 years ?
"explanation : 250 = ( 1750 x 2 xr ) / 100 r = 7.14 % answer : option b"
a ) 4 % , b ) 7.14 % , c ) 2 6 / 7 % , d ) 5 % , e ) 6 %
b
multiply(divide(divide(subtract(2000, 1750), 1750), 2), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
what is the least number which when divided by 5 , 6 , 7 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder ?
"lcm of 5 , 6 , 7 and 8 = 840 hence the number can be written in the form ( 840 k + 3 ) which is divisible by 9 . if k = 1 , number = ( 840 × 1 ) + 3 = 843 which is not divisible by 9 . if k = 2 , number = ( 840 × 2 ) + 3 = 1683 which is divisible by 9 . hence 1683 is the least number which when divided by 5 , 6 , 7 an...
a ) 1108 , b ) 1683 , c ) 2007 , d ) 3363 , e ) 1436
b
add(lcm(lcm(5, 6), lcm(7, 8)), 3)
lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)|
general
what is the difference between the largest number and the least number written with the digits 7 , 3 , 1 , 4 ?
"1347 7431 - - - - - - - - - - - - 6084 answer : a"
a ) 6084 , b ) 6184 , c ) 5084 , d ) 6048 , e ) 6049
a
subtract(add(add(add(multiply(multiply(7, const_100), const_10), multiply(4, const_100)), multiply(3, const_10)), 1), add(add(add(const_1000, multiply(3, const_100)), multiply(4, const_10)), 7))
multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)|
general
if a light flashes every 30 seconds , how many times will it flash in 1 / 2 of an hour ?
"there are 60 minutes in an hour . in 1 / 2 of an hour there are ( 60 * 1 / 2 ) minutes = 30 minutes . in 1 / 2 of an hour there are ( 60 * 30 ) seconds = 1800 seconds . light flashed for every 30 seconds . in 1800 seconds 1800 / 30 = 60 times . the count start after the first flash , the light will flashes 61 times in...
a ) 30 , b ) 42 , c ) 61 , d ) 59 , e ) 45
c
divide(multiply(divide(const_3600, const_4), const_3), 30)
divide(const_3600,const_4)|multiply(#0,const_3)|divide(#1,n0)|
physics
what is the smallest positive integer k such that the product of 720 x k is a perfect square ?
"a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 720 = 3 ^ 2 * 4 ^ 2 * 5 , so if k = 5 then 720 k = ( 3 * 4 * 5 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of...
a ) 5 , b ) 9 , c ) 15 , d ) 25 , e ) 63
a
add(const_3, const_4)
add(const_3,const_4)|
general
the ratio of two numbers is 2 : 3 and their h . c . f . is 6 . their l . c . m . is
explanation : let the numbers be 2 x and 3 x . then , their h . c . f . = x . so , x = 6 . so , the numbers 12 and 18 l . c . m . of 12 and 18 = 36 option c
a ) 48 , b ) 22 , c ) 36 , d ) 27 , e ) 67
c
multiply(6, multiply(2, 3))
multiply(n0,n1)|multiply(n2,#0)
other
the speed at which a girl can row a boat in still water is 90 kmph . if she rows downstream , where the speed of current is 18 kmph , what time will he take to cover 330 metres ?
"speed of the boat downstream = 90 + 18 = 108 kmph = 108 * 5 / 18 = 30 m / s hence time taken to cover 330 m = 330 / 30 = 11 seconds . answer : a"
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
a
divide(multiply(330, const_3_6), add(90, 18))
add(n0,n1)|multiply(n2,const_3_6)|divide(#1,#0)|
physics
one hour after matthew started waking from q to y , a distance of 45 km , johnny started walking along the same road from y to q . matthew ' s walking rate was 3 km per hour and johnny ' s was 4 km per hour , how many km had johnny walked when they met ?
just an alternative method . . . after the first hour the distance q is 42 km ( 45 - 3 ) . now the problem can be treated as if bothof them started at the same time . since the speeds are in the ratio 3 : 4 , the distances will also be in the same ratio . splitting 42 in that ratio we get 18 : 24 . so answer is 24 . he...
a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 19.5
a
multiply(divide(subtract(45, 3), add(3, 4)), 4)
add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)
physics
if 2 / w + 2 / x = 2 / y and wx = y , then the average ( arithmetic mean ) of w and x is
"given : 2 / w + 2 / x = 2 / ywx = y find : ( w + x ) / 2 = ? 2 ( 1 / w + 1 / x ) = 2 ( 1 / y ) - divide both sides by 2 ( 1 / w + 1 / x ) = 1 / y ( x + w ) / wx = 1 / wx - sub ' d in y = wx x + w - 1 = 0 x + w = 1 therefore ( w + x ) / 2 = 1 / 2 ans : a"
a ) 1 / 2 , b ) 1 , c ) 2 , d ) 4 , e ) 8
a
divide(2, add(2, 2))
add(n0,n0)|divide(n0,#0)|
general
a rectangular field is to be fenced on three sides leaving a side of 30 feet uncovered . if the area of the field is 810 sq . feet , how many feet of fencing will be required ?
"explanation we have : l = 30 ft and lb = 810 sq . ft . so , b = 27 ft . length of fencing = ( l + 2 b ) = ( 30 + 54 ) ft = 84 ft . answer d"
a ) 34 , b ) 40 , c ) 68 , d ) 84 , e ) none
d
add(multiply(divide(810, 30), const_2), 30)
divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)|
geometry
the food in a camp lasts for 40 men for 45 days . if fifteen more men join , how many days will the food last ?
one man can consume the same food in 40 * 45 = 1800 days . 15 more men join , the total number of men = 55 the number of days the food will last = 1800 / 55 = 33 days . answer : b
a ) 80 days , b ) 33 days , c ) 35 days , d ) 16 days , e ) 15 days
b
divide(multiply(45, 40), add(add(const_12, const_3), 40))
add(const_12,const_3)|multiply(n0,n1)|add(n0,#0)|divide(#1,#2)
physics
lionel left his house and walked towards walt ' s house , 48 miles away . four hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 3 miles per hour and walt ' s 6 miles per hour , how many miles had lionel walked when he met walt ?
"in the first 4 hours lionel at the rate of 3 miles per hour covered distance = rate * time = 4 * 3 = 12 miles . so , the distance between him and walt was 48 - 12 = 36 miles when walt left his house . now , their combined rate to cover this distance was 3 + 6 = 9 miles per hour , hence they will meet ( they will cover...
a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28
d
multiply(const_2.0, add(divide(subtract(48, multiply(3, 3)), add(6, 3)), 3))
add(const_2.0,n2)|multiply(n1,n1)|subtract(n0,#1)|divide(#2,#0)|add(#3,n1)|multiply(n1,#4)|
physics
a certain car can travel 72 kilometers on a liter of fuel . if the fuel tank ’ s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed , how fast is the car moving , in miles per hour ? ( 1 gallon = 3.8 liters ; 1 mile = 1.6 kilometers )
"fuel used 3.9 gallons ; convert to liters - - > 3.9 x 3.8 liters time = 5.7 hours 1 mile = 1.6 kilometers ; convert to miles - - > 1 km = 1 / 1.6 mile speed ( km / hour ) = d / t = 72 ( km * ) x 3.9 x 3.8 / 5.7 replace ( km * ) to miles ; multiply by 1 / 1.6 mile speed ( miles / hour ) = 72 x 3.9 x 3.8 / 5.7 x 1.6 = 7...
a ) 52 , b ) 65 , c ) 117 , d ) 91 , e ) 104
c
divide(multiply(divide(multiply(72, 3.8), 1.6), 3.9), 5.7)
multiply(n0,n4)|divide(#0,n6)|multiply(n1,#1)|divide(#2,n2)|
physics
find the probability that a leap year selected at random will have 53 sundays
"there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 sundays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 sundays we should have either { sat , sun } or { sun , mon ...
a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7
e
divide(const_2, add(const_3, const_4))
add(const_3,const_4)|divide(const_2,#0)|
probability
a tank is 25 m long , 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 75 paise per sq . m , is :
explanation : area to be plastered = [ 2 ( l + b ) x h ] + ( l x b ) = { [ 2 ( 25 + 12 ) x 6 ] + ( 25 x 12 ) } m 2 = ( 444 + 300 ) m 2 = 744 m 2 . cost of plastering = rs . 744 x 75 / 100 = rs . 558 answer : c
a ) rs . 456 , b ) rs . 458 , c ) rs . 558 , d ) rs . 568 , e ) none of these
c
multiply(add(multiply(multiply(add(25, 12), const_2), 6), multiply(25, 12)), divide(75, const_100))
add(n0,n1)|divide(n3,const_100)|multiply(n0,n1)|multiply(#0,const_2)|multiply(n2,#3)|add(#4,#2)|multiply(#5,#1)
physics
if p is the product of the integers from 1 to 38 , inclusive , what is the greatest integer k for which 3 ^ k is a factor of p ?
"38 ! has 3 , 6 , 9 , . . . . 30 , 33 , 36 as factors , which are 12 multiples of 3 . we need to add 5 more to these 12 because of 9 , 18 , 27 , and 36 . the greatest integer of k is 17 . the answer is c ."
a ) 13 , b ) 15 , c ) 17 , d ) 19 , e ) 21
c
add(multiply(3, const_4), 3)
multiply(n2,const_4)|add(n2,#0)|
general
find the average increase rate , if increase in the population in the first year is 30 % and that in the second year is 40 % .
solution : let 100 be the original population . 1 st year ' s population increased = 30 % so , population after first year = ( 100 + 30 % of 100 ) = 130 . population in second year increases by 40 % , then population = ( 130 + 40 % of 130 ) = 182 . the final population become 182 which was originally at 100 . it means ...
a ) 41 , b ) 56 , c ) 40 , d ) 38 , e ) 39
a
add(subtract(40, 30), 30)
subtract(n1,n0)|add(n0,#0)
general
rs . 6000 is lent out in two parts . one part is lent at 4 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ?
"let the amount lent at 4 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 4 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 3 x / 100 = > 600 - 3 / 100 x = 450 = > x = 2500 amount lent at 10 % = 3500 required ratio = 5 : 7 answer : b"
a ) 5 : 1 , b ) 5 : 7 , c ) 5 : 2 , d ) 5 : 8 , e ) 5 : 2
b
divide(divide(subtract(multiply(450, const_100), multiply(6000, 4)), subtract(10, 4)), divide(subtract(multiply(450, const_100), multiply(6000, 4)), subtract(10, 4)))
multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)|
gain
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 800 per sq . metre .
"c $ 16500 area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 800 hence , total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = $ 16500"
a ) $ 12500 , b ) $ 13500 , c ) $ 16500 , d ) $ 14500 , e ) $ 18500
c
multiply(800, multiply(5.5, 3.75))
multiply(n0,n1)|multiply(n2,#0)|
physics
? % of 360 = 115.2
? % of 360 = 115.2 or , ? = 115.2 × 100 / 360 = 32 answer d
a ) 277 , b ) 36 , c ) 64 , d ) 32 , e ) none of these
d
divide(multiply(115.2, const_100), 360)
multiply(n1,const_100)|divide(#0,n0)
gain
walking 7 / 6 of his usual rate , a boy reaches his school 2 min early . find his usual time to reach the school ?
"speed ratio = 1 : 7 / 6 = 6 : 7 time ratio = 7 : 6 1 - - - - - - - - 7 2 - - - - - - - - - ? 14 m answer : a"
a ) 14 , b ) 28 , c ) 99 , d ) 77 , e ) 66
a
multiply(2, 7)
multiply(n0,n2)|
gain
what least number must be subtracted from 427398 so that remaining no . is divisible by 11
"explanation : on dividing 427398 by 11 we get the remainder 4 , so 4 should be subtracted answer : option b"
a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 8
b
subtract(427398, multiply(floor(divide(427398, 11)), 11))
divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|
general
4 milkmen rented a pasture . a grazed 24 cows for 3 months , b 10 cows for 5 months , c 35 cows for 4 months and d 21 cows for 3 months . if a ’ s share of rent is rs . 720 , find the total rent of the field .
ratio of shares of a , b , c , d = ( 24 * 3 ) : ( 10 * 5 ) : ( 35 * 4 ) : ( 21 * 3 ) = 72 : 50 : 140 : 63 let total rent be rs . x . then , a ’ s share = rs . 72 x / 325 . therefore , 72 x / 325 = 720 = > x = 720 * 325 / 72 = 3250 . hence , total rent of the field is rs . 3250 . answer : a
a ) rs . 3250 , b ) rs . 3350 , c ) rs . 3450 , d ) rs . 3550 , e ) rs . 3650
a
divide(multiply(720, add(add(multiply(24, 3), multiply(10, 5)), add(multiply(35, 4), multiply(21, 3)))), multiply(24, 3))
multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n5)|multiply(n2,n7)|add(#0,#1)|add(#2,#3)|add(#4,#5)|multiply(n9,#6)|divide(#7,#0)
general
500 , 250 , 750 , 375 . . .
"500 / 2 = 250 x 3 = 750 / 2 = 375 x 3 = 1125 answer is d"
a ) 875 , b ) 1050 , c ) 950 , d ) 1125 , e ) 925
d
subtract(negate(375), multiply(subtract(250, 750), divide(subtract(250, 750), subtract(500, 250))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
at present , the ratio between the ages of shekhar and shobha is 4 : 3 . after 6 years , shekhar ' s age will be 26 years . find out the age of shobha at present ?
after 6 years , shekhar ' s age will be 26 years therefore , present age of shekhar = 26 − 6 = 20 let present age of shobha = x then , 20 x = 43 x = 20 × 34 = 15 answer is d .
a ) 10 , b ) 12 , c ) 14 , d ) 15 , e ) 20
d
multiply(divide(subtract(26, 6), 4), 3)
subtract(n3,n2)|divide(#0,n0)|multiply(n1,#1)
other
if ( a - b - c + d = 12 ) and ( a + b - c - d = 6 ) , what is the value of ( b - d ) ^ 2 ?
"eq 1 : a - b - c + d = 12 eq 2 : a + b - c - d = 6 ( 1 ) subtract eq 1 from eq 2 a - b - c + d = 12 - a + b - c - d = 6 - - - - - - - - - - - - - - - - - - - - - - - - - 2 b + 2 d = 6 ( 2 ) simplify - b + d = 3 b - d = - 3 ( b - d ) ^ 2 = ( - 3 ) ^ 2 = 9 my answer : b"
a ) 4 . , b ) 9 . , c ) 12 . , d ) 16 . , e ) 64 .
b
power(subtract(6, divide(add(12, 6), 2)), 2)
add(n0,n1)|divide(#0,n2)|subtract(n1,#1)|power(#2,n2)|
general
ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 1000 then loss of pyarelal ?
"let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 1000 , loss of pyarelal = 1000 * 9 x / 10 x = 900 answer : d"
a ) 600 , b ) 700 , c ) 800 , d ) 900 , e ) 1000
d
divide(multiply(9, 1000), add(1, 9))
add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|
gain
what will be the cost of building a fence around a square plot with area equal to 81 sq ft , if the price per foot of building the fence is rs . 58 ?
"let the side of the square plot be a ft . a 2 = 81 = > a = 9 length of the fence = perimeter of the plot = 4 a = 36 ft . cost of building the fence = 36 * 58 = rs . 2088 . answer : d"
a ) 3944 , b ) 2287 , c ) 2977 , d ) 2088 , e ) 1298
d
multiply(square_perimeter(sqrt(81)), 58)
sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)|
geometry
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 16 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?
"0,112 r = x / 100 * 0.1 r answer d"
a ) 80 % , b ) 105 % , c ) 120 % , d ) 112 % , e ) 138 %
d
multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(16, const_100)), divide(10, const_100)), const_100)
divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)|
gain
while driving from a - ville to b - town , harriet drove at a constant speed of 105 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 145 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive fr...
"5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give sa...
a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 174
e
multiply(subtract(5, divide(multiply(105, 5), add(105, 145))), const_60)
add(n0,n1)|multiply(n0,n2)|divide(#1,#0)|subtract(n2,#2)|multiply(#3,const_60)|
physics
in a competitive examination in a state a , 10 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 12 % candidates got selected with 100 more candidates got selected than a . what was the number of candidates appeared from each state ?
"let the number of candidates appeared be x then , 12 % of x - 10 % of x = 100 2 % of x = 100 x = 100 * 100 / 2 = 5000 answer is b"
a ) 4000 , b ) 5000 , c ) 5500 , d ) 6500 , e ) 4200
b
multiply(multiply(subtract(12, 10), const_100), 100)
subtract(n1,n0)|multiply(#0,const_100)|multiply(n2,#1)|
general
if it is assumed that 80 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ?
"minimum no of mail to be sent for getting 300 responses at 80 % = 300 / 0.8 = 375 option a"
a ) 375 , b ) 420 , c ) 480 , d ) 500 , e ) 600
a
divide(300, divide(80, const_100))
divide(n0,const_100)|divide(n1,#0)|
gain
when a number is divided by 6 & then multiply by 12 the answer is 13 what is the no . ?
"if $ x $ is the number , x / 6 * 12 = 13 = > 2 x = 13 = > x = 6.5 a"
a ) 6.5 , b ) 5 , c ) 5.5 , d ) 5.8 , e ) 6
a
multiply(divide(13, 12), 6)
divide(n2,n1)|multiply(n0,#0)|
general
in arun ' s opinion , his weight is greater than 66 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 69 kg . if all of them are correct in their estimation , what is...
"let arun ' s weight be x kg . according to arun , 66 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 69 . the values satisfying all the above conditions are 67 and 68 . required average = ( 67 + 68 ) / 2 = 67.5 kg answer : b"
a ) 86.5 kg , b ) 67.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg
b
divide(add(69, add(66, const_1)), const_2)
add(n0,const_1)|add(n4,#0)|divide(#1,const_2)|
general
find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 20 as remainder
"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 20 5 x = 1345 x = 269 large number = 269 + 1365 = 1634 c"
a ) 1235 , b ) 1456 , c ) 1634 , d ) 1678 , e ) 1635
c
add(multiply(divide(subtract(1365, 20), subtract(6, const_1)), 6), 20)
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)|
general
by selling 9 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ?
"80 % - - - 9 120 % - - - ? 80 / 120 * 9 = 6 answer : e"
a ) 8 , b ) 9 , c ) 7 , d ) 4 , e ) 6
e
multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 9)
add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)|
gain
in a public show 62 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ?
"75 % of 600 = 62 / 100 × 600 = 372 therefore , the number of vacant seats = 600 - 372 = 228 . answer : e"
a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 228
e
divide(multiply(600, subtract(const_100, 62)), const_100)
subtract(const_100,n0)|multiply(n1,#0)|divide(#1,const_100)|
gain
there is a 30 % increase in the price of an article in the first year , a 20 % decrease in the second year and a 10 % increase in the next year . if the final price of the article is rs . 4576 , then what was the price of the article initially ?
"let the price of the article , four years age be rs . 100 in the 1 st year , price of the article = 100 + 30 = rs . 130 . in the 2 nd year , price = 130 - 20 % of 130 = 130 - 26 = rs . 104 . in the 3 rd year , price = 104 + 10 % of 104 = 104 + 10.4 = rs . 114.40 . but present price of the article is rs . 4576 for 114....
a ) rs . 5000 , b ) rs . 3000 , c ) rs . 4000 , d ) rs . 2000 , e ) rs . 6000
c
divide(4576, add(const_1, divide(add(subtract(30, 20), 10), const_100)))
subtract(n0,n1)|add(n2,#0)|divide(#1,const_100)|add(#2,const_1)|divide(n3,#3)|
general
50 boys and 100 girls are examined in a test ; 50 % of the boys and 40 % of the girls pass . the percentage of the total who failed is ?
"total number of students = 50 + 100 = 150 number of students passed = ( 50 % of 50 + 40 % of 100 ) = 25 + 40 = 65 number of failures = 85 * 100 / 150 = 56.7 % answer is c"
a ) 42.5 % , b ) 62.1 % , c ) 56.7 % , d ) 39.9 % , e ) 51.4 %
c
multiply(const_100, divide(subtract(add(50, 100), add(multiply(50, divide(50, const_100)), multiply(100, divide(40, const_100)))), add(50, 100)))
add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)|
general
if ( 5 ^ 5 ) ( 9 ^ 3 ) = 3 ( 15 ^ x ) , what is the value of x ?
"( 5 ^ 5 ) ( 9 ^ 3 ) = 3 ( 15 ^ x ) = > 5 ^ 5 * 3 ^ 6 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 5 * 3 ^ 6 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 5 answer a"
a ) 5 , b ) 9 , c ) 11 , d ) 13 , e ) 15
a
divide(log(divide(multiply(power(5, 5), power(9, 3)), 3)), log(15))
log(n5)|power(n0,n1)|power(n2,n3)|multiply(#1,#2)|divide(#3,n4)|log(#4)|divide(#5,#0)|
general
5356 x 51 = ?
"5356 x 51 = 5356 x ( 50 + 1 ) = 5356 x 50 + 5356 x 1 = 267800 + 5356 = 273156 . a"
a ) 273156 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273258
a
multiply(divide(5356, 51), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
marts income is 60 percent more than tims income and tims income is 60 percent less than juans income . what percentage of juans income is marts income
"m = ( 160 / 100 ) t t = ( 40 / 100 ) j = > m = ( 64 / 100 ) j answer b ."
a ) 124 % , b ) b ) 64 % , c ) c ) 96 % , d ) d ) 80 % , e ) e ) 64 %
b
multiply(divide(add(const_100, 60), multiply(divide(const_100, subtract(const_100, 60)), const_100)), const_100)
add(n0,const_100)|subtract(const_100,n1)|divide(const_100,#1)|multiply(#2,const_100)|divide(#0,#3)|multiply(#4,const_100)|
general
a man swims downstream 16 km and upstream 10 km taking 2 hours each time , what is the speed of the man in still water ?
"16 - - - 2 ds = 8 ? - - - - 1 10 - - - - 2 us = 5 ? - - - - 1 m = ? m = ( 8 + 5 ) / 2 = 6.5 answer : c"
a ) 7.0 , b ) 8.2 , c ) 6.5 , d ) 7 , e ) 6
c
divide(add(divide(10, 2), divide(16, 2)), const_2)
divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|
physics
the manufacturing cost of a shoe is rs . 190 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains
"explanation : total cost of a watch = 190 + ( 500 / 100 ) = 195 . gain = 20 % = > sp = 1.2 cp = 1.2 x 195 = 234 answer : a"
a ) s 234 , b ) s 216 , c ) s 220 , d ) s 210 , e ) s 217
a
add(add(190, divide(500, 100)), multiply(divide(20, 100), add(190, divide(500, 100))))
divide(n1,n2)|divide(n3,n2)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)|
gain
what is the area of a circle having the same perimeter as that of a square whose area is 121 .
side of the square = sqrt ( 12 ) = 11 perimeter of the square = 44 2 * pie * r = 44 r = 44 / 2 * pie area of the circle = pie * r ^ 2 = 308 answer : c
['a ) 306', 'b ) 307', 'c ) 308', 'd ) 309', 'e ) 310']
c
multiply(multiply(multiply(divide(divide(multiply(sqrt(121), const_4), const_pi), const_2), divide(divide(multiply(sqrt(121), const_4), const_pi), const_2)), const_pi), const_2)
sqrt(n0)|multiply(#0,const_4)|divide(#1,const_pi)|divide(#2,const_2)|multiply(#3,#3)|multiply(#4,const_pi)|multiply(#5,const_2)
geometry
the manager of a produce market purchased a quantity of tomatoes for $ 0.80 per pound . due to improper handling , 20 percent of the tomatoes , by weight , were ruined and discarded . at what price per pound should the manager sell the remaining tomatoes if she wishes to make a profit on the sale of the tomatoes equal ...
"assume the manager bought 100 tomatoes . cost price = 80 given : 20 % are damaged - - > available tomatoes to sell = 80 80 * x - 80 = 0.08 * 80 80 x - 80 = 6.4 80 x = 86.64 x = 86.64 / 80 = 87 / 80 ( approx ) = 1.0875 x is slightly under 1.0875 = 1.09 answer : e"
a ) $ 0.94 , b ) $ 0.96 , c ) $ 0.98 , d ) $ 1.00 , e ) $ 1.09
e
multiply(add(const_1, divide(8, const_100)), divide(0.80, divide(subtract(const_100, 20), const_100)))
divide(n2,const_100)|subtract(const_100,n1)|add(#0,const_1)|divide(#1,const_100)|divide(n0,#3)|multiply(#2,#4)|
gain
a sum fetched a total simple interest of rs . 4016.25 at the rate of 3 % p . a . in 5 years . what is the sum ?
"principal = ( 100 * 4016.25 ) / ( 3 * 5 ) = rs . 26775 . answer : a"
a ) 26775 , b ) 28772 , c ) 28821 , d ) 89255 , e ) 28871
a
divide(divide(multiply(4016.25, const_100), 3), 5)
multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|
gain
what is the units digit of ( 498 ) ( 915 ) ( 381 ) ( 756 ) ( 29 )
"just multiply the digits in the units place for each term and you will get the answer . it should be 0 . you got a 5 as a unit digit and an even number term . so the multiplication of this will definitely yield a 0 . answer has to be 0 . i also tried it using the calculator and the answer is 2 . imo c ."
a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 6
c
divide(add(multiply(factorial(498), factorial(915)), multiply(factorial(498), factorial(756))), 498)
factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)|
general
a certain store sells all maps at one price and all books at another price . on monday the store sold 12 maps and 10 books for a total of $ 42.00 , and on tuesday the store sold 20 maps and 15 books for a total of $ 70.00 . at this store , how much less does a map sell for than a book ?
"12 x + 10 y = 42 20 x + 15 y = 70 multiply eq 1 by 5 and eq 2 by 3 60 x + 50 y = 42 * 5 60 x + 45 y = 70 * 3 subtracting 1 from 2 5 y = 210 - 210 y = 0 therefore x = 3.5 difference in price = 3.5 c"
a ) $ 0.25 , b ) $ 0.50 , c ) $ 3.5 , d ) $ 2.5 , e ) $ l . 25
c
divide(subtract(70.00, negate(multiply(15, divide(subtract(multiply(42.00, const_4), multiply(70.00, const_2)), subtract(multiply(10, const_4), multiply(15, const_2)))))), 20)
multiply(n2,const_4)|multiply(n5,const_2)|multiply(n1,const_4)|multiply(n4,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n4,#6)|negate(#7)|subtract(n5,#8)|divide(#9,n3)|
general
angelo and isabella are both salespersons . in any given week , angelo makes $ 590 in base salary plus 8 percent of the portion of his sales above $ 1000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of mo...
let the weekly sales of both = x 590 + ( x − 1000 ) 8 / 100 = 10 / 100 x x = 25500 answer : c
a ) 23,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500
c
subtract(divide(divide(subtract(590, multiply(divide(8, const_100), 1000)), subtract(divide(10, const_100), divide(8, const_100))), const_1000), const_1)
divide(n1,const_100)|divide(n3,const_100)|multiply(n2,#0)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,const_1000)|subtract(#6,const_1)
general
the least number which when divided by 2 , 3 , 5 and 6 leaves a remainder 3 , but when divided by 9 leaves no remainder , is
"explanation : l . c . m of 2 , 3 , 5 , 6 = 30 therefore required number is of the form 30 k + 3 . least value of k for which ( 30 k + 3 ) is divisible by 9 is k = 2 therefore required number = ( 30 x 2 + 3 ) = 63 . answer : b"
a ) 16 , b ) 63 , c ) 73 , d ) 83 , e ) none of these
b
multiply(lcm(lcm(lcm(2, 3), 5), 6), const_2)
lcm(n0,n1)|lcm(n2,#0)|lcm(n3,#1)|multiply(#2,const_2)|
general
how many 4 - digit numbers ( abcd ) can be formed such that | a – d | = 2 ?
digit a can be filled in 9 ways ( excluding 0 ) digit b can be filled in 10 ways digit c can be filled in 10 ways digit d can be filled in 2 ways ( because of modulus a - d ) the answer will be 9 * 10 * 10 * 2 = 1800 ans c
a ) 2000 , b ) 1900 , c ) 1800 , d ) 1600 , e ) 1,500
c
multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 2)
multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)
general
find the smallest number of 6 digits which is exactly divisible by 111 ?
"smallest number of 6 digits is 100000 . on dividing 100000 by 111 , we get 100 as remainder . number to be added = ( 111 - 100 ) - 11 . hence , required number = 100011 answer b"
a ) 12000 , b ) 15550 , c ) 100011 , d ) 158993 , e ) 100010
b
add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 111)
multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)|
general
if the mean of numbers 28 , x , 42 , 78 and 104 is 90 , then what is the mean of 128 , 255 , 511 , 1023 and x ?
the mean of numbers 28 , x , 42 , 78 and 104 is 62 : 28 + x + 42 + 78 + 104 = 90 * 5 - - > x = 198 ; so , the mean of 128 , 255 , 511 , 1023 and x is ( 128 + 255 + 511 + 1023 + 198 ) / 5 = 423 . answer : d .
a ) 395 , b ) 275 , c ) 355 , d ) 423 , e ) 365
d
divide(add(add(add(add(subtract(multiply(90, add(const_4, const_1)), add(add(add(28, 42), 78), 104)), 128), 255), 511), 1023), add(const_4, const_1))
add(const_1,const_4)|add(n0,n1)|add(n2,#1)|multiply(n4,#0)|add(n3,#2)|subtract(#3,#4)|add(n5,#5)|add(n6,#6)|add(n7,#7)|add(n8,#8)|divide(#9,#0)
general
a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 12 minutes . how long will it take to go 5 km in stationary water ?
"speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 12 / 60 ) = 5 kmhr speed in still water = 1 / 2 ( 2 + 5 ) = 3.5 kmhr time taken in stationary = 5 / 3.5 = 1 hrs 25 min answer : c"
a ) 40 minutes , b ) 1 hour , c ) 1 hour 25 min , d ) 1 hour 30 min , e ) 1 hour 10 min
c
divide(5, divide(add(multiply(divide(1, 12), const_60), divide(2, 2)), const_2))
divide(n0,n0)|divide(n2,n3)|multiply(#1,const_60)|add(#0,#2)|divide(#3,const_2)|divide(n4,#4)|
physics
the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 23 hrs . how much distance will the car cover in 7 hrs if the speed of the bike is twice speed of the tractor ?
"sol . average speed of a tractor = 25 km / h the speed of a bus in an hour = 25 × 2 = 50 km the speed of a car in an hour = 9 / 5 * 50 = 90 km so , the distance covered by car in 7 h is 90 × 7 = 630 km ans . ( e )"
a ) 400 km , b ) 500 km , c ) 360 km , d ) 550 km , e ) 630 km
e
multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 23))), 7)
divide(n1,n2)|divide(n3,n4)|add(n0,#0)|multiply(#1,const_2)|multiply(#2,#3)|multiply(n5,#4)|
general
the average of 64 numbers is 48 . if two numbers namely 35 and 45 are discarded , the average of remaining numbers is ?
total of 50 numbers = 64 * 48 = 3072 average of 62 numbers = 3072 - ( 45 + 55 ) / 62 = 2972 / 62 = 47.9 answer is e .
a ) 35.5 , b ) 37.5 , c ) 39.5 , d ) 34.5 , e ) 47.9
e
divide(add(64, 35), const_2)
add(n0,n2)|divide(#0,const_2)
general
a caterer ordered 125 ice - cream bars and 125 sundaes . if the total price was $ 250.00 and the price of each ice - cream bar was $ 0.60 , what was the price of each sundae ?
"let price of a sundae = s price of ice cream bar = . 6 $ 125 * . 6 + 125 * s = 250 = > 125 * s = 175 = > s = 1.4 answer e"
a ) $ 0.60 , b ) $ 0.80 , c ) $ 1.00 , d ) $ 1.20 , e ) $ 1.40
e
divide(subtract(250.00, multiply(125, 0.60)), 125)
multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)|
general
the sum of ages of 5 children born at the intervals of 3 years each is 65 years . what is the age of the youngest child ?
"let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 65 5 x = 35 x = 7 . age of the youngest child = x = 7 years . d )"
a ) 3 years , b ) 4 years , c ) 6 years , d ) 7 years , e ) 8 years
d
subtract(subtract(divide(65, 5), 3), 3)
divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)|
general
a toy train produces at least 10 different tunes when it moves around a circular toy track of radius 5 meters , at 10 meters per minute . however , the toy train is defective and it now produces only two different tunes at random . what is the probability that the toy train produces 3 music tunes of the same type ( 1 i...
1 / 2 ^ 3 = 1 / 8 but here der r two types of tone and we need to find the odds of d train of two tones to form 3 tones consequently so 2 * 1 / 8 = 1 / 4 so ans is 4 answer : d
a ) 3 , b ) 9 , c ) 8 , d ) 4 , e ) 5
d
inverse(multiply(power(divide(1, const_2), 3), const_2))
divide(n4,const_2)|power(#0,n3)|multiply(#1,const_2)|inverse(#2)
physics
evaluate : 56 - 12 * 3 * 2 = ?
"according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 56 - 12 * 3 * 2 = 56 - 8 = 48 correct answer b"
a ) 42 , b ) 48 , c ) 62 , d ) 72 , e ) 82
b
subtract(56, multiply(multiply(12, const_2.0), 2))
multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)|
general
if a - b = 3 and a ^ 2 + b ^ 2 = 31 , find the value of ab .
"2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 31 - 9 = 22 ab = 11 . answer is d ."
a ) 12 , b ) 15 , c ) 10 , d ) 11 , e ) 13
d
multiply(multiply(add(3, divide(subtract(sqrt(31), 3), 2)), divide(subtract(sqrt(31), 3), 2)), 2)
sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)|
general
if x , y , and z are positive real numbers such that x ( y + z ) = 132 , y ( z + x ) = 152 , and z ( x + y ) = 180 , then xyz is
xy + xz = 132 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 ) yz + yx = 152 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ) xz + zy = 180 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3 ) re - writing equation 3 as follows...
a ) 40 , b ) 60 , c ) 160 , d ) 100 , e ) 120
c
subtract(180, subtract(152, 132))
subtract(n1,n0)|subtract(n2,#0)
general
a polling company surveyed a certain country , and it found that 35 % of that country ’ s registered voters had an unfavorable impression of both of that state ’ s major political parties and that 20 % had a favorable impression only of party q . if one registered voter has a favorable impression of both parties for ev...
s = 100 not ( q and b ) = 35 only q = 20 ( q and b ) / b = 1 / 2 let ( q and b ) = x only b = 2 x so now , 20 + 35 + x + 2 x = 100 x = 15 a ans
a ) 15 , b ) 20 , c ) 30 , d ) 35 , e ) 45
a
divide(subtract(const_100, add(35, 20)), const_3)
add(n0,n1)|subtract(const_100,#0)|divide(#1,const_3)
gain
there is enough provisions for 100 girls in a hostel for 50 days . if there were 20 men less , how long will the provision last ?
we have , m 1 d 1 = m 2 d 2 100 * 50 = 20 * d 2 d 2 = 100 * 50 / 20 = 250 days . answer : a
a ) 250 , b ) 255 , c ) 260 , d ) 265 , e ) 270
a
add(divide(multiply(100, 20), const_10), 50)
multiply(n0,n2)|divide(#0,const_10)|add(n1,#1)
physics
at present the ratio between the ages of arun and deepak is 4 : 3 . after 6 years arun ' s age will be 26 years . what is the age of the deepak ?
"let the present age of arun and deepak be 4 x and 3 x 4 x + 6 = 26 4 x = 20 x = 5 deepak ' s age = 3 x = 15 years answer is a"
a ) 15 , b ) 20 , c ) 18 , d ) 22 , e ) 35
a
divide(multiply(subtract(26, 6), 3), 4)
subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|
other
what is the greatest integer m for which the number 40 ! / 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 40 ! the multiples of 7 are : 7 , 14 , 21 , 28 , 35 . thus 7 ^ 5 will divide 40 ! but 7 ^ 6 will not . clearly 2 ^ 5 will divide 40 ! so m = 5 is the largest possible integer . the answer is d .
a ) 8 , b ) 7 , c ) 6 , d ) 5 , e ) 4
d
floor(divide(40, divide(14, const_2)))
divide(n1,const_2)|divide(n0,#0)|floor(#1)
general
the price of a book is increased from $ 300 to $ 480 . what is the % of increase in its price ?
"explanation : change in the price = rs 480 â € “ rs 300 = rs 180 percentage of increase = change in the price initial price * 100 . percentage increase in price = ( 180 300 ) * 100 = 60 % e"
a ) 10 % , b ) 20 % , c ) 40 % , d ) 50 % , e ) 60 %
e
multiply(divide(subtract(480, 300), 300), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 21 ) both lie on the line defined by the equation x = ( y / 7 ) - ( 2 / 5 ) , what is the value of p ?
x = ( y / 7 ) - ( 2 / 5 ) , and so y = 7 x + 14 / 5 . the slope is 7 . ( n + 21 - n ) / ( m + p - m ) = 7 p = 3 the answer is c .
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
c
divide(21, 7)
divide(n0,n1)
general
let a be the event that a randomly selected two digit number is divisible by 10 and let b be the event that a randomly selected two digit number is divisible by 5 . what is p ( a and b ) ?
p ( a and b ) = 1 / 10 * 1 / 5 = 1 / 50 the answer is a .
a ) 1 / 50 , b ) 1 / 30 , c ) 1 / 15 , d ) 1 / 10 , e ) 1 / 5
a
multiply(divide(subtract(10, const_1), multiply(subtract(10, const_1), 10)), divide(multiply(subtract(10, const_1), const_2), multiply(subtract(10, const_1), 10)))
subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)
physics
in a group of cows and hens , the number of legs are 12 more than twice the number of heads . the number of cows is :
"let no of cows be x , no of hens be y . so heads = x + y legs = 4 x + 2 y now , 4 x + 2 y = 2 ( x + y ) + 12 2 x = 12 x = 6 . answer : b"
a ) 5 , b ) 6 , c ) 7 , d ) 10 , e ) 12
b
subtract(12, const_4)
subtract(n0,const_4)|
general
find the least number which when divided by 12,24 , 36,42 leaves remainder 9,21 , 33,39 respectively .
"the difference of 12 - 9 = 3 , 24 - 21 = 3 , 36 - 33 = 3 , 42 - 39 = 3 lcm of 12,24 , 36,42 = 504 required number = 504 - 3 = 501 answer is b"
a ) 499 , b ) 501 , c ) 452 , d ) 365 , e ) 325
b
add(9,21, lcm(12,24, 36,42))
lcm(n0,n1)|add(n2,#0)|
general
the difference between a 6 digit number and the number formed by reversing its digit is not a multiple of
another approach is to test a number . let ' s say the original number is 123004 so , the reversed number is 400321 the difference = 400321 - 123004 no check the answer choices 277317 is a multiple of 3,92439 , 30813,10271 10272 is not a multiple of 277317 answer : e
a ) 3 , b ) 92439 , c ) 30813 , d ) 10271 , e ) 10272
e
power(add(6, const_4), const_4)
add(n0,const_4)|power(#0,const_4)
general
sandy is younger than molly by 12 years . if their ages are in the respective ratio of 7 : 9 , how old is molly ?
"s = m - 12 s / m = 7 / 9 9 s = 7 m 9 ( m - 12 ) = 7 m m = 54 the answer is c ."
a ) 36 , b ) 45 , c ) 54 , d ) 63 , e ) 72
c
multiply(divide(7, subtract(9, 7)), 12)
subtract(n2,n1)|divide(n1,#0)|multiply(n0,#1)|
other
two pipes can fill a tank in 20 minutes and 15 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ?
"part of the filled by all the three pipes in one minute = 1 / 20 + 1 / 15 - 1 / 45 = ( 9 + 12 - 4 ) / 180 = 17 / 180 so , the tank becomes full in 5 minutes . answer : a"
a ) 5 minutes , b ) 17 minutes , c ) 15 minutes , d ) 10 minutes , e ) 12 minutes
a
subtract(add(divide(const_1, 20), divide(const_1, 15)), divide(const_1, 45))
divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|
physics
the output of a factory was increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 10 % . by approximately what percent would the output now have to be decreased in order to restore the original output ?
"the original output increases by 10 % and then 10 % . total % change = a + b + ab / 100 total % change = 10 + 10 + 10 * 10 / 100 = 21 % now , you want to change it to 0 , so , 0 = 21 + x + 21 x / 100 x = - 21 ( 100 ) / 121 = 17 % approximately answer is e"
a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 17 %
e
divide(multiply(subtract(add(add(const_100, 10), multiply(add(const_100, 10), divide(10, const_100))), const_100), const_100), add(add(const_100, 10), multiply(add(const_100, 10), divide(10, const_100))))
add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|add(#0,#2)|subtract(#3,const_100)|multiply(#4,const_100)|divide(#5,#3)|
general
a company recently conducted a survey and found that 25,000 of its customers live in rural areas . if the number of customers who live in urban areas is 120 percent greater than the number of customers who live in rural areas , how many customers live in urban areas ?
"the number of customers in urban areas is 25,000 + 1.2 * 25,000 = 55,000 . the answer is b ."
a ) 50,000 , b ) 55,000 , c ) 60,000 , d ) 65,000 , e ) 70,000
b
subtract(const_100, multiply(const_4, const_2))
multiply(const_2,const_4)|subtract(const_100,#0)|
gain
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 142 m length ?
"speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 142 = 252 m . required time = 252 / 20 = 12.6 sec . answer : b"
a ) 82.1 sec . , b ) 12.6 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec .
b
divide(add(110, 142), multiply(72, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration q of the acid in the mixture ?
"do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration q would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d"
a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5
d
divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400))
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|
general
in measuring the sides of a rectangle , one side is taken 10 % in excess and other 5 % in deficit . find the error percentage in the area calculated from these measurements .
"say both sides of the rectangle are equal to 100 ( so consider that we have a square ) . in this case the area is 100 * 100 = 10,000 . now , the area obtained with wrong measurements would be 110 * 95 = 10,450 , which is 4.5 % greater than the actual area . answer : d ."
a ) 3.5 % , b ) 2.4 % , c ) 3 % , d ) 4.5 % , e ) 2.6 %
d
subtract(subtract(10, 5), divide(multiply(10, 5), const_100))
multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)|
geometry
if 2 x + y = 7 and x + 2 y = 5 , then 4 xy / 3 = ?
"2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 4 * 3 * 1 ) / 3 = 12 / 3 = 4 e"
a ) a ) 1 , b ) b ) 2 , c ) c ) 17 / 5 , d ) d ) 18 / 5 , e ) e ) 4
e
divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 4)
multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)|
general
rates for having a manuscript typed at a certain typing service are $ 5 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 30 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total ...
50 pages typed 1 x 30 pages typed 2 x ( original + one revision ) 20 pages typed 3 x ( original + two revisions ) 50 ( 5 ) + 30 ( 5 + 4 ) + 20 ( 5 + 4 + 4 ) = 250 + 270 + 260 = 780 answer - e
a ) $ 760 , b ) $ 750 , c ) $ 810 , d ) $ 800 , e ) $ 780
e
add(add(multiply(100, 5), multiply(30, 4)), multiply(multiply(20, 4), const_2))
multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)
general
0.2 x 0.7 = ?
"explanation : 2 x 7 = 14 . sum of decimal places = 2 0.2 x 0.7 = 0.14 answer - d"
a ) 0.0001 , b ) 0.001 , c ) 0.01 , d ) 0.14 , e ) none of these
d
multiply(divide(0.2, 0.7), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
the cost price of a radio is rs . 1860 and it was sold for rs . 1340 , find the loss % ?
"1860 - - - - 520 100 - - - - ? = > 27 % answer : d"
a ) 18 % , b ) 12 % , c ) 22 % , d ) 27 % , e ) 21
d
multiply(divide(subtract(1860, 1340), 1860), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
in a kilometer race , a beats b by 16 meters or 10 seconds . what time does a take to complete the race ?
"time taken by b run 1000 meters = ( 1000 * 10 ) / 16 = 625 sec . time taken by a = 625 - 10 = 615 sec . answer : e"
a ) 167 sec , b ) 190 sec , c ) 176 sec , d ) 716 sec , e ) 615 sec
e
subtract(divide(multiply(const_1, const_1000), divide(16, 10)), 10)
divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|
physics
if 15 students in a class average 75 % on an exam and 10 students average 95 % on the same exam , what is the average in percent for all 25 students ?
"( 15 * 75 + 10 * 95 ) / 25 = 83 % the answer is c ."
a ) 81 % , b ) 82 % , c ) 83 % , d ) 84 % , e ) 85 %
c
divide(add(multiply(15, 75), multiply(10, 95)), 25)
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)|
general
a boat crossed a lake from north to west at the speed of 5 km / h , entered a river and covered twice as much distance going upstream at 4 km / h . it then turned around and stopped at the south shore of the lake . if it averaged 3.6 km / h that day , what was its approximate downstream speed ?
one way of solving this is : speed of boat on still water ( lake ) = 5 kmph speed upstream = 4 kmph = speed in still water - speed of river = > speed of river = 1 kmph = > speed downstream = speed in still water + speed of river = 5 + 1 = 6 kmph ans is c
a ) 2 , b ) 3 , c ) 6 , d ) 7 , e ) 8
c
subtract(multiply(5, const_2), 4)
multiply(n0,const_2)|subtract(#0,n1)
general
fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 50...
"x discount on pony jeans , ( 0.5 - x ) discount on fox jeans . set the equation : 3 * 15 ( 0.5 - x ) + 2 * 18 x = 9 - - > x = 1.5 = 150 % answer : e ."
a ) 9 % , b ) 10 % , c ) 110 % , d ) 120 % , e ) 150 %
e
multiply(subtract(divide(50, const_100), divide(subtract(9, multiply(divide(50, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100)
divide(n6,const_100)|multiply(n1,n5)|multiply(n0,n4)|multiply(#0,#1)|subtract(#2,#1)|subtract(n2,#3)|divide(#5,#4)|subtract(#0,#6)|multiply(#7,const_100)|
gain
a shopkeeper forced to sell at cost price , uses a 750 grams weight for a kilogram . what is his gain percent ?
shopkeeper sells 750 g instead of 1000 g . so , his gain = 1000 - 750 = 250 g . thus , % gain = ( 250 * 100 ) / 750 = 33.33 % . answer : option c
a ) 10 % , b ) 9 % , c ) 33.33 % , d ) 12 % , e ) none of these
c
multiply(divide(add(multiply(const_2, const_100), divide(const_100, const_2)), 750), const_100)
divide(const_100,const_2)|multiply(const_100,const_2)|add(#0,#1)|divide(#2,n0)|multiply(#3,const_100)
gain
uba capital recently bought brand new vehicles for office use . uba capital only went for toyota and honda and bought more of toyota than honda at the ratio of 9 : 1 . if 90 % of the toyota bought and 10 % of the honda bought were suv ã ¢ â ‚ ¬ â „ ¢ s . how many suv ã ¢ â ‚ ¬ â „ ¢ s did uba capital buy in the aforeme...
"let total no of vehicles bought be 100 , toyota 90 and honda 10 , so total number of suv ' s bought for toyota and honda respectively 90 * 90 / 100 = 81 and 10 * 10 / 100 = 1 so total 82 suv ' s were bought out of 100 vehicles bought . . so required % is 82 % answer : a"
a ) 82 % , b ) 64 % , c ) 28 % , d ) 69 % , e ) 80 %
a
subtract(const_100, add(multiply(9, 1), multiply(1, 9)))
multiply(n0,n1)|add(#0,#0)|subtract(const_100,#1)|
other
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 171 games will be played during the tournament . find the number of participants .
"let p be the number of participants . pc 2 = 171 ( p ) ( p - 1 ) = 342 = 19 * 18 p = 19 the answer is e ."
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19
e
divide(add(sqrt(add(multiply(multiply(171, const_2), const_4), const_1)), const_1), const_2)
multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)|
general
a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 10 that a person who randomly guesses the answer to each question will be a winner ?
a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . 5...
a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000
d
multiply(const_1000, divide(1, 10))
divide(n0,n1)|multiply(#0,const_1000)
general
a pipe takes a hours to fill the tank . but because of a leakage it took 3 times of its original time . find the time taken by the leakage to empty the tank
"pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 180 x = 90 min answer : c"
a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 70 min
c
multiply(const_10, multiply(const_1, 3))
multiply(n0,const_1)|multiply(#0,const_10)|
physics
the average of 13 numbers is 60 . average of the first 7 of them is 56 and that of the last 7 is 61 . find the 8 th number ?
"sum of all the 13 numbers = 13 * 60 = 780 sum of the first 7 of them = 7 * 56 = 392 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 392 - 780 = 39 . answer : a"
a ) 39 , b ) 83 , c ) 45 , d ) 53 , e ) 64
a
subtract(multiply(8, 61), subtract(multiply(13, 60), multiply(56, 7)))
multiply(n5,n6)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3)|
general
a student got twice as many sums wrong as he got right . if he attempted 75 sums in all , how many did he solve correctly ?
"explanation : suppose the boy got x sums right and 2 x sums wrong . then , x + 2 x = 75 3 x = 75 x = 25 . answer : a"
a ) 25 , b ) 16 , c ) 18 , d ) 24 , e ) 26
a
divide(75, add(const_1, const_2))
add(const_1,const_2)|divide(n0,#0)|
general
how much water should be added to 12 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ?
let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 12 + x ) ) ) 2 = 5 % * ( 12 + x ) - - - - - - - - > x = 28 ans a
a ) 28 liters , b ) 27 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters
a
subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 12)
divide(n2,const_100)|multiply(const_100,const_2)|multiply(n1,#0)|subtract(n1,#2)|divide(#1,#3)|subtract(#4,n0)
general
a sum of money lent out at s . i . amounts to rs . 710 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ?
"s . i for 5 years = ( 1020 - 710 ) = rs . 310 . s . i . for 2 years = 310 / 5 * 2 = rs . 124 . principal = ( 710 - 124 ) = rs . 586 . answer : b"
a ) s . 440 , b ) s . 586 , c ) s . 540 , d ) s . 740 , e ) s . 840
b
subtract(710, multiply(divide(subtract(1020, 710), 5), 2))
subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|
general