Problem
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stringlengths 1
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| options
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| correct
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848
| linear_formula
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find the remainder of the division ( 2 ^ 14 ) / 7 .
|
b 425 ( 68 ã · 16 ) ã — 100
|
a ) 336 , b ) 425 , c ) 275 , d ) 235 , e ) 689
|
b
|
subtract(multiply(multiply(68, const_4), const_2), const_100)
|
multiply(n0,const_4)|multiply(#0,const_2)|subtract(#1,const_100)
|
general
|
7 m - 20 = 2 m , then m + 7 is equal to ?
|
drop = 12 inches / day 5 days ago = w , means now it ' s equal w - 60 and in 4 days = w - 60 - 48 = w - 108 answer a
|
a ) w − 108 , b ) w − 56 , c ) w − 14 , d ) w + 14 , e ) w + 126
|
a
|
multiply(12, divide(4, 5))
|
divide(n2,n1)|multiply(n0,#0)
|
gain
|
each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 3 and the 24 can not be adjacent digits in the integer , how many 5 - digit integers are possible ?
|
sol . apples 250 each carries 25 = 250 / 25 = 10 answer : d
|
a ) a ) 9 , b ) b ) 5 , c ) c ) 7 , d ) d ) 10 , e ) e ) none of the above
|
d
|
divide(250, 25)
|
divide(n0,n1)
|
general
|
what is the sum of all digits for the number 10 ^ 29 - 41 ?
|
"15 km / hr = 15000 m / 3600 s = ( 150 / 36 ) m / s = ( 25 / 6 ) m / s time = 600 / ( 25 / 6 ) = 144 seconds the answer is c ."
|
a ) 128 , b ) 136 , c ) 144 , d ) 152 , e ) 160
|
c
|
divide(600, multiply(15, const_0_2778))
|
multiply(n1,const_0_2778)|divide(n0,#0)|
|
physics
|
a truck covers a distance of 376 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?
|
"total profit = 1000 ratio = 600 / 300 = 2 : 1 answer : e"
|
a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 2 : 1
|
e
|
divide(600, 300)
|
divide(n0,n1)|
|
other
|
two employees x and y are paid a total of rs . 650 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?
|
"correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer a"
|
a ) 36.5 , b ) 35 , c ) 34 , d ) 33 , e ) 32.5
|
a
|
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
|
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
|
general
|
village a ’ s population is 300 greater than village b ' s population . if village b ’ s population were reduced by 600 people , then village a ’ s population would be 4 times as large as village b ' s population . what is village b ' s current population ?
|
"let us suppose there are 100 people . 50 % of them donated $ 25000 ( 500 * 50 ) $ 25000 is 60 % of total amount . so total amount = 25000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 25000 * ( 100 / 60 ) * ( 40 / 100 ) = 50000 / 3 this amount has to be divided by 50 ( remaining people are 50 ) so per head amount is 50000 / 3 / 50 = 32000 / 180 = 333.33 ; answer : b"
|
a ) $ 200 , b ) $ 333.33 , c ) $ 100.25 , d ) $ 277.78 , e ) $ 377.00
|
b
|
divide(multiply(divide(multiply(divide(50, const_100), 500), divide(60, const_100)), divide(50, const_100)), divide(60, const_100))
|
divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)|
|
general
|
in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 700 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 2100 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?
|
"if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 18 times . answer : d . similar questions to practice : hope it helps ."
|
a ) 6.33 , b ) 7.5 , c ) 10 , d ) 18 , e ) 19
|
d
|
subtract(19, divide(19, 19))
|
divide(n0,n0)|subtract(n0,#0)|
|
physics
|
two men a and b start from place x walking at 4 ½ kmph and 5 ¾ kmph respectively . how many km apart they are at the end of 4 ½ hours if they are walking in the same direction ?
|
circumference = 2 * pi * r = 2 * pi * 4 / pi = > 8 a
|
['a ) 8', 'b ) 4 π', 'c ) 4', 'd ) 6', 'e ) 5']
|
a
|
circumface(divide(4, const_pi))
|
divide(n0,const_pi)|circumface(#0)
|
geometry
|
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 36 minutes , then the slower pipe alone will be able to fill the tank in ?
|
"given exp . = 0.3 * 0.3 + ( 0.3 * 0.3 ) = 0.09 + 0.09 = 0.18 answer is c ."
|
a ) 0.52 , b ) 0.42 , c ) 0.18 , d ) 0.64 , e ) 0.46
|
c
|
add(multiply(0.3, 0.3), multiply(0.3, 0.3))
|
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
|
general
|
how many terminating zeroes r does 200 ! have ?
|
straight a way lets exclude all the even numbers between 260 and 280 . so now the number starts from 261 to 279 ( only odd ) 261 is a divisible of 3 and next odd divisible by 3 will be 261 + 6 = 267 + 6 = 273 + 6 = 279 . also we can eliminate numbers ending with ' 5 ' so in odd , the excluded numbers are 261 , 265,267 , 273,279 , which leave us with 263,269 , 271,277 . checked the above listed four numbers are divisible by any numbers till 20 . answer : e
|
a ) none , b ) one , c ) two , d ) three , e ) four
|
e
|
subtract(divide(subtract(280, 260), const_4), const_1)
|
subtract(n1,n0)|divide(#0,const_4)|subtract(#1,const_1)
|
general
|
what is the units digit of ( 5 ! * 5 ! + 6 ! * 5 ! ) / 3 ?
|
"first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 = 5 / 2 so they complete one robot in 2 / 5 th of an hour - ( 3 ) now the 5 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 3 so they complete one robot in 1 / 3 th of an hour - ( 4 ) now the 6 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 7 / 2 so they complete one robot in 2 / 7 th of an hour - ( 5 ) now the 7 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 9 / 2 so they complete one robot in 2 / 9 th of an hour - ( 6 ) and now we have 8 robots so total = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ( 5 ) + ( 6 ) = 146 1 / 7 minutes answer - e"
|
a ) 70 min , b ) 94 min , c ) 110 min , d ) 131 1 / 7 min , e ) 146 1 / 7 min
|
e
|
add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2))))
|
inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)|
|
physics
|
find large number from below question the difference of two numbers is 1380 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder
|
x = - 2.5 prob = 1 / 12 answer - a
|
a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2
|
a
|
divide(1, multiply(6, 2))
|
multiply(n1,n14)|divide(n5,#0)
|
general
|
a train speeds past a pole in 10 seconds and a platform 50 m long in 20 seconds . its length is :
|
"the distance between the nest and the ditch is 300 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 300 = 9000 . d = st 9000 / 1.5 = t , i think we can take 9000 meters as 9 km , then only we get t = 6 . ( 1000 meters = 1 km ) d )"
|
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
|
d
|
divide(divide(multiply(300, multiply(15, const_2)), const_1000), divide(15, const_10))
|
divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)|
|
physics
|
125 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ?
|
"now given that q is set the consecutive integers between a and b . and q contains 9 multiples of 9 let take a as 36 . then 36 45 54 63 72 81 90 99 108 . . . so b will 108 . now let ' s check the multiples of 4 among this set 108 - 36 / 4 + 1 = > 18 + 1 = > 19 ans option b ."
|
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22
|
b
|
subtract(multiply(9, const_2), const_1)
|
multiply(n1,const_2)|subtract(#0,const_1)|
|
physics
|
if the average of 5 positive integers is 65 and the difference between the largest and the smallest of these 5 numbers is 10 , what is the maximum value possible for the largest of these 5 integers ?
|
"20 c 20 = 1 ( 20 c 2 ) * ( 20 c 20 ) = 20 ! * 1 / 18 ! = 20 * 19 * 18 ! / 18 ! = 20 * 19 * 1 = 380 answer : b"
|
a ) 400 , b ) 380 , c ) 360 , d ) 350 , e ) 330
|
b
|
multiply(add(divide(18, 20), 20), 20)
|
divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|
|
general
|
if the average marks of 3 batches of 55 , 60 and 45 students respectively is 40 , 62 , 58 , then the average marks of all the students is
|
"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 30 % - 50 % = 20 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.30 x 95 + 0.50 x 79 + 0.20 x 60 = 80 the class average ( rounded ) is 80 % final answer e ) 80 %"
|
a ) 76 % , b ) 77 % , c ) 78 % , d ) 79 % , e ) 80 %
|
e
|
divide(add(add(multiply(30, 95), multiply(50, 79)), multiply(subtract(const_100, add(30, 50)), 60)), const_100)
|
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)|
|
gain
|
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 .
|
"the profit on the first kind of vodka = x % ; the profit on the second kind of vodka = y % . when they are mixed in the ratio 1 : 2 ( total of 3 parts ) the average profit is 10 % : ( x + 2 y ) / 3 = 10 . when they are mixed in the ratio 2 : 1 ( total of 3 parts ) the average profit is 20 % : ( 2 x + y ) / 3 = 20 . solving gives : x = 30 % and y = 0 % . after the individual profit percent on them areincreased by 4 / 3 and 5 / 3 times respectively the profit becomes 40 % and 0 % , on the first and te second kinds of vodka , respectively . if they are mixed in equal ratio ( 1 : 1 ) , then the mixture will fetch the profit of ( 40 + 0 ) / 2 = 20 % . answer : a"
|
a ) 20 % , b ) 40 % , c ) 18 % , d ) 23 % , e ) can not be determined
|
a
|
add(divide(multiply(10, 4), 6), add(10, 5))
|
add(n4,n8)|multiply(n4,n6)|divide(#1,const_2.0)|add(#0,#2)|
|
general
|
if a > x > y > z on the number line , y is halfway between x and z , and x is halfway between w and z , then ( y - x ) / ( y - a ) =
|
if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? a . 2 b . 3 c . 4 d . 5 e . 6 - > if m = y = 60 , 2 ( m + y ) = 240 is derived . if m = y = 69 , 2 ( m + y ) = 276 is derived , which makes 4,5 , 6,7 possible for the tens digit . therefore , the answer is c .
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
|
c
|
subtract(6, 2)
|
subtract(n0,n1)
|
physics
|
if 3 < x < 6 < y < 11 , then what is the greatest possible positive integer difference of x and y ?
|
"number of pens = 848 number of pencils = 630 required number of students = h . c . f . of 848 and 630 = 2 answer is c"
|
a ) 10 , b ) 4 , c ) 2 , d ) 14 , e ) 16
|
c
|
gcd(848, 630)
|
gcd(n0,n1)|
|
general
|
how many single - digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6 ?
|
1 : 2 answer : a
|
['a ) 1 : 2', 'b ) 2 : 3', 'c ) 2 : 9', 'd ) 2 : 1', 'e ) 2 : 2']
|
a
|
divide(1, 2)
|
divide(n0,n1)
|
geometry
|
sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 350 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 3 of the days were hot , then what was the price of 1 cup on a hot day ?
|
"you have 6 digits : 1 , 2 , 3 , 7 , 8 , 9 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 1 and 9 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 2 * * and 3 * * ) . make the 2 * * number as large as possible and make the 3 * * number as small as possible . 298 and 317 ( difference is 19 ) or try and make a pair with ( 7 * * and 8 * * ) . make the 7 * * number as large as possible and make the 8 * * number as small as possible . we get 793 and 812 ( difference is 19 ) a"
|
a ) 19 , b ) 49 , c ) 58 , d ) 113 , e ) 131
|
a
|
subtract(subtract(const_100, multiply(subtract(8, 1), const_10)), const_1)
|
subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)|
|
general
|
after decreasing 90 % in the price of an article costs rs . 320 . find the actual cost of an article ?
|
"explanation : let the average age of the whole team by x years . 11 x â € “ ( 29 + 32 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer e"
|
a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 26 years
|
e
|
divide(subtract(add(29, add(29, 3)), multiply(3, 3)), const_2)
|
add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)|
|
general
|
a bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is - .
|
"the triangle with sides 31 cm , 29 cm and 15 cm is right angled , where the hypotenuse is 31 cm . area of the triangle = 1 / 2 * 29 * 15 = 217.5 cm 2 answer : e"
|
a ) 220.75 cm 2 , b ) 258 cm 2 , c ) 225.50 cm 2 , d ) 222.25 cm 2 , e ) 217.5 cm 2
|
e
|
divide(multiply(29, 15), const_2)
|
multiply(n1,n2)|divide(#0,const_2)|
|
geometry
|
find the least number of complete years in which a sum of money put out at 45 % compound interest will be more than double of itself ?
|
"let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 6.5 ) = 5 / 4 answer : c"
|
a ) 5 / 6 , b ) 5 / 9 , c ) 5 / 4 , d ) 5 / 3 , e ) 7 / 6
|
c
|
divide(divide(subtract(8.75, 7.50), subtract(8.75, 6.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6.5))))
|
subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|
|
other
|
in the game of dubblefud , red chips , blue chips and green chips are each worth 2 , 4 and 5 points respectively . in a certain selection of chips , the product of the point values of the chips is 16000 . if the number of blue chips in this selection doubles the number of green chips , how many red chips are in the selection ?
|
"drawing two balls of same color from seven green balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = ⁷ c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 7 / 15 answer : e"
|
a ) 7 / 16 , b ) 7 / 12 , c ) 7 / 19 , d ) 7 / 12 , e ) 7 / 15
|
e
|
add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 8), const_1))))
|
add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)|
|
other
|
in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.815 per pound ?
|
"explanation : manager ' s monthly salary rs . ( 1500 * 21 - 1400 * 20 ) = rs . 3500 . answer : e"
|
a ) 3600 , b ) 3890 , c ) 88798 , d ) 2789 , e ) 3500
|
e
|
subtract(multiply(add(1400, 100), add(20, const_1)), multiply(1400, 20))
|
add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
|
general
|
the value of x + x ( xx ) when x = 7
|
"let 1 man ' s 1 day work = x and 1 woman ' s 1 day work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving these two equations , we get : x = 11 / 400 and y = 1 / 400 1 woman ' s 1 day work = ( 1 / 400 * 10 ) = 1 / 40 . hence , 10 women will complete the work in 40 days . answer : b"
|
a ) 21 days , b ) 40 days , c ) 27 days , d ) 18 days , e ) 17 days
|
b
|
inverse(multiply(divide(subtract(divide(const_1, 10), multiply(3, divide(subtract(divide(const_1, 8), multiply(divide(6, 7), divide(const_1, 10))), subtract(4, multiply(3, divide(6, 7)))))), 7), 8))
|
divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)|
|
physics
|
a cat leaps 6 leaps for every 5 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ?
|
"purchase price = 180 selling price = x 180 + 0.4 * x = x 0.6 * x = 180 x = 300 profit = 300 - 180 = 120 answer : d"
|
a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 120 , e ) $ 100
|
d
|
divide(multiply(subtract(divide(180, subtract(const_1, divide(40, const_100))), 180), const_100), 180)
|
divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)|
|
gain
|
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 15 % . what was its profit on the items sold in february ?
|
"length = speed * time speed = l / t s = 400 / 10 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : b"
|
a ) 165 kmph , b ) 144 kmph , c ) 172 kmph , d ) 175 kmph , e ) 178 kmph
|
b
|
divide(divide(400, const_1000), divide(10, const_3600))
|
divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|
|
physics
|
two pipes a and b can separately fill a tank in 10 and 15 minutes respectively . a third pipe c can drain off 20 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ?
|
"ds = 30 us = 14 s = ? s = ( 30 - 14 ) / 2 = 8 kmph answer : e"
|
a ) 1 kmph , b ) 4 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph
|
e
|
divide(subtract(30, 14), const_2)
|
subtract(n0,n1)|divide(#0,const_2)|
|
gain
|
45 pupil , out of them 12 in debate only and 22 in singing only . then how many in both ?
|
explanation : work done by 4 men and 6 women in 1 day = 1 / 8 work done by 3 men and 7 women in 1 day = 1 / 10 let 1 man does m work in 1 day and 1 woman does w work in 1 day . the above equations can be written as 4 m + 6 w = 1 / 8 - - - ( 1 ) 3 m + 7 w = 1 / 10 - - - ( 2 ) solving equation ( 1 ) and ( 2 ) , we get m = 11 / 400 and w = 1 / 400 amount of work 10 women can do in a day = 10 × ( 1 / 400 ) = 1 / 40 ie , 10 women can complete the work in 40 days answer : option b
|
a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10
|
b
|
inverse(multiply(divide(subtract(divide(const_1, 8), multiply(4, divide(subtract(divide(const_1, 10), multiply(divide(7, 6), divide(const_1, 8))), subtract(3, multiply(4, divide(7, 6)))))), 6), 10))
|
divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)
|
physics
|
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 920 gm . for a kg . his gain is … % .
|
log ( 0.0000134 ) . since there are four zeros between the decimal point and the first significant digit , the characteristic is – 5 . answer : b
|
a ) 5 , b ) - 5 , c ) 6 , d ) - 6 , e ) 7
|
b
|
floor(divide(log(0.0000134), log(const_10)))
|
log(n0)|log(const_10)|divide(#0,#1)|floor(#2)
|
other
|
a , b and c started a business with a total investment of rs . 72000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find a ' s share .
|
this is equivalent to : - 2 x * 4 y * 5 z = 16000 y / 2 = z ( given ) 2 x * 4 y * 5 y / 2 = 16000 2 x * y ^ 2 = 16000 / 10 2 x * y ^ 2 = 1600 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 2 as 2 * 2 * y ^ 2 = 1600 y ^ 2 = 400 y = 20 number of red chips = 2 hence b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
|
b
|
divide(multiply(multiply(power(2, 4), power(2, const_3)), power(5, const_3)), multiply(power(const_2, multiply(2, const_3)), power(5, const_3)))
|
multiply(n0,const_3)|power(n0,n1)|power(n0,const_3)|power(n2,const_3)|multiply(#1,#2)|power(const_2,#0)|multiply(#4,#3)|multiply(#5,#3)|divide(#6,#7)
|
general
|
find the simple interest on $ 10000 at 6 % per annum for 12 months ?
|
"10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 2000 [ 1 + ( 5 / 100 ) ] ^ 3 - 2000 = 2000 [ ( 21 / 20 ) ^ 3 - 1 ] = 2000 ( 9261 - 8000 ) / 8000 = 2 * 1261 / 8 = 315 answer : d"
|
a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 315 , e ) none of the above
|
d
|
subtract(multiply(2000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 2000)
|
divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)|
|
gain
|
the maximum number of students among them 1234 pens and 874 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is :
|
"cone curved surface area = ï € rl 22 / 7 ã — 49 ã — 35 = 154 ã — 35 = 5390 m ( power 2 ) answer is b ."
|
a ) 5160 , b ) 5390 , c ) 6430 , d ) 6720 , e ) 7280
|
b
|
volume_cone(49, 35)
|
volume_cone(n0,n1)|
|
geometry
|
a train covers a distance of 11 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ?
|
"perimeter = distance covered in 10 min . = ( 12000 / 60 ) x 10 m = 2000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2000 or x = 200 . length = 600 m and breadth = 400 m . area = ( 600 x 400 ) m 2 = 240000 m 2 . answer : e"
|
a ) 153601 , b ) 153600 , c ) 153602 , d ) 153603 , e ) 240000
|
e
|
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), 2))
|
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
|
physics
|
a train 450 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in :
|
"speed = ( 11 / 10 * 60 ) km / hr = ( 66 * 5 / 18 ) m / sec = 55 / 3 m / sec . length of the train = 55 / 3 * 6 = 110 m . answer : c"
|
a ) m , b ) m , c ) m , d ) m , e ) m
|
c
|
divide(11, subtract(divide(11, 10), 6))
|
divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)|
|
physics
|
speed of a boat in standing water is 10 kmph and speed of the stream is 2.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ?
|
( 38 g + 20 b ) / ( g + b ) = 30 38 g + 20 b = 30 ( g + b ) 8 g = 10 b b / g = 4 / 5 the answer is d .
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6
|
d
|
divide(30, 38)
|
divide(n2,n1)
|
general
|
paul sells encyclopedias door - to - door . he earns $ 150 on every paycheck , regardless of how many sets he sells . in addition , he earns commission as follows : commission sales 10 % $ 0.00 - $ 10 , 000.00 5 % $ 10 , 000.01 - - - > he does not earn double commission . that is , if his sales are $ 12,000 , he earns 10 % on the first $ 10,000 and 5 % on the remaining $ 2,000 . his largest paycheck of the year was $ 1,320 . what were his sales for that pay period ?
|
"i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 5 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 5 = 1000 ( m * n ) water available is 2000 ( p ) it will last 2 months ( p / m * n ) ans : d"
|
a ) 9 , b ) 5 , c ) 6 , d ) 2 , e ) 4
|
d
|
divide(2000, multiply(200, 5))
|
multiply(n0,n1)|divide(n2,#0)|
|
gain
|
a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ' s contribution in the capital ?
|
"as two numbers are prime , only two options satisfy ie option c and d . but option c will not make the product of numbers i . e 99 answer : d"
|
a ) 8 and 12 , b ) 14 and 6 , c ) 19 and 1 , d ) 11 and 9 , e ) 12 and 9
|
d
|
add(99, 20)
|
add(n0,n1)|
|
physics
|
a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 10 - sided polygon have ?
|
"s = 180 / 20 * 18 / 5 = 32 kmph answer : c"
|
a ) 37 kmph , b ) 35 kmph , c ) 32 kmph , d ) 38 kmph , e ) 36 kmph
|
c
|
multiply(const_3_6, divide(180, 20))
|
divide(n0,n1)|multiply(#0,const_3_6)|
|
physics
|
a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 300,000 ?
|
"solution s . i . = rs . ( 956 - 925 ) = rs . 31 rate = ( 100 x 31 / 925 x 3 ) = 124 / 111 % new rate = ( 124 / 111 + 4 ) % = 568 / 111 % new s . i . = rs . ( 925 x 568 / 111 x 3 / 100 ) rs . 142 ∴ new amount = rs . ( 925 + 142 ) = rs . 1067 . answer c"
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1067 , d ) data inadequate , e ) none of these
|
c
|
add(925, divide(multiply(multiply(925, add(divide(multiply(subtract(956, 925), const_100), multiply(925, 3)), 4)), 3), const_100))
|
multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|
|
gain
|
simplify : 0.3 * 0.3 + 0.3 * 0.3
|
"the total number of the people in the room must be a multiple of both 7 and 12 ( in order 3 / 7 and 5 / 12 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 12 , which is 84 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 84 people in the room . therefore there are 3 / 7 * 84 = 36 people in the room under the age of 21 . answer : c ."
|
a ) 21 , b ) 35 , c ) 36 , d ) 60 , e ) 65
|
c
|
divide(multiply(multiply(7, 12), 3), 7)
|
multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)|
|
general
|
3 pounds of 05 grass seed contain 1 percent herbicide . a different type of grass seed , 20 , which contains 20 percent herbicide , will be mixed with 3 pounds of 05 grass seed . how much grass seed of type 20 should be added to the 3 pounds of 05 grass seed so that the mixture contains 15 percent herbicide ?
|
"3 x * 2 x = 2460 = > x = 20.24 2 ( 79.76 + 50 ) = 259.52 m 259.52 * 1 / 2 = rs . 129.76 answer : c"
|
a ) s . 122 , b ) s . 129 , c ) s . 129.76 , d ) s . 120 , e ) s . 121
|
c
|
divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(2460, 2), 3)), divide(2460, sqrt(divide(multiply(2460, 2), 3))))), const_100)
|
multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)|
|
physics
|
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 12 % 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 9 % . if ron imported a $ 18,000 imported car and ended up paying $ 1950 in taxes , what is the first tier ' s price level ?
|
"to solve this problem , all you have to do is take every even number between 24 and 50 and add them together . so we have 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 , which is 481 . final answer : b"
|
a ) 592 , b ) 481 , c ) 330 , d ) 475 , e ) 483
|
b
|
add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 24), add(const_2, const_4))
|
add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)|
|
general
|
of the 150 employees at company x , 70 are full - time , and 100 have worked at company x for at least a year . there are 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ?
|
r + b + c + 14 + 15 = 12 * 5 = 60 = > r + b + c = 60 - 29 = 31 r + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
|
d
|
divide(add(subtract(multiply(add(const_4, const_1), 12), add(14, 15)), 29), const_4)
|
add(const_1,const_4)|add(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|add(n3,#3)|divide(#4,const_4)
|
general
|
what is the tens digit of 36 ^ 5 ?
|
expl : 15 % of a i = 30 % of b = 15 a / 100 = 30 b / 100 = 2 / 1 = 2 : 1 answer : e
|
a ) 1 : 4 , b ) 4 : 3 , c ) 6 : 7 , d ) 3 : 5 , e ) 2 : 1
|
e
|
divide(divide(30, const_100), divide(15, const_100))
|
divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)
|
gain
|
if n is a positive integer and n ^ 2 is divisible by 200 , then what is the largest positive integer that must divide n ?
|
"p ( a and b ) = 1 / 2 * 1 / 5 = 1 / 10 the answer is b ."
|
a ) 1 / 18 , b ) 1 / 10 , c ) 1 / 5 , d ) 1 / 3 , e ) 1 / 2
|
b
|
multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2)))
|
subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)|
|
physics
|
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 ?
|
"required angle = 240 – 24 × ( 11 / 2 ) = 240 – 132 = 108 ° answer d"
|
a ) 100 ° , b ) 107 ° , c ) 106 ° , d ) 108 ° , e ) none of these
|
d
|
subtract(multiply(8, multiply(const_3, const_2)), 2)
|
multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)|
|
geometry
|
a man can row upstream at 10 kmph and downstream at 20 kmph , and then find the speed of the man in still water ?
|
let x per minute be the speed of c and y per minute be the speed of d . after meeting at a point , c travels for 32 mins and d travels for 50 mins . so distance covered by each of them post point of crossing c = 32 x and d = 50 y the distance covered by c and d before they cross each would be distance covered by d and c post crossing respectively . therefore distance covered by d before he meets c = 32 x time taken by d cover 32 x distance = 32 x / y mins therefore total time taken by d = 32 x / y + 50 mins . . . . . . . . . . . . . . . . . i we need to find value of x in terms of y to arrive at final answer . total distance = 32 x + 50 y combined speed of c and d = x + y therefore time taken before c and d meet en - route = ( 32 x + 50 y ) / ( x + y ) time taken by d reach destination after meeting c = 50 mins total travel time for d = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 mins . . . . . . . . . . . . . . . . . . . ii equate i and ii 32 x / y + 50 = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 ( 32 x + 50 y ) / y = ( 82 x + 100 y ) / ( x + y ) 32 x ^ 2 + 50 xy + 32 xy + 50 y ^ 2 = 82 xy + 100 y ^ 2 32 x ^ 2 + 82 xy - 82 xy + 50 y ^ 2 - 100 y ^ 2 = 0 32 x ^ 2 - 50 y ^ 2 = 0 32 x ^ 2 = 50 y ^ 2 16 x ^ 2 = 25 y ^ 2 taking square root . . ( since x and y denote speed , square root ca n ' t be negative ) 4 x = 5 y y = 4 x / 5 . . . . . . . . . . . . iii substitute in i = 32 x / ( 4 x / 5 ) + 50 = 32 x * 5 / 4 x + 50 = 40 + 50 = 90 mins a
|
a ) 90 , b ) 80 , c ) 75 , d ) 60 , e ) 65
|
a
|
add(sqrt(multiply(50, 32)), 50)
|
multiply(n0,n1)|sqrt(#0)|add(n1,#1)
|
physics
|
the wages earned by robin is 40 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ?
|
"a 150 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 ."
|
a ) 150 , b ) 120 , c ) 130 , d ) 160 , e ) 210
|
a
|
divide(225, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))
|
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|
|
gain
|
if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 15 - m ) then m = ?
|
"let x be the total worker then 0.5 x = female worker and 0.5 x is male worker then 20 male worker added 05 x / ( 0.5 x + 20 ) = 50 / 100 or 50 x = 50 * ( 0.5 x + 100 ) = 25 x + 5000 or 25 x = 5000 , x = 5000 / 25 = 200 total worker = 200 + 20 = 220 b"
|
a ) 225 , b ) 220 , c ) 230 , d ) 235 , e ) 240
|
b
|
add(divide(multiply(divide(50, const_100), 20), subtract(divide(const_60.0, const_100), divide(50, const_100))), 20)
|
divide(n2,const_100)|divide(const_60.0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)|
|
gain
|
alex takes a loan of $ 8,000 to buy a used truck at the rate of 9 % simple interest . calculate the annual interest to be paid for the loan amount .
|
"explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 5 ^ 2 = 9 / 25 answer : e"
|
a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 25
|
e
|
divide(multiply(1, power(3, const_2)), power(5, const_2))
|
power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)|
|
general
|
what is the difference between the largest number and the least number written with the figures 3 , 4 , 7 , 0 , 3 ?
|
notice that 7 play both baseball and cricket does not mean that out of those 7 , some does not play football too . the same for cricket / football and baseball / football . [ color = # ffff 00 ] { total } = { baseball } + { cricket } + { football } - { hc + ch + hf } + { all three } + { neither } for more checkadvanced overlapping sets problems [ / color ] 50 = 20 + 15 + 11 - ( 7 + 4 + 5 ) + { all three } + 18 - - > { all three } = 2 ; those who play only baseball and cricket are 7 - 2 = 5 ; those who play only cricket and football are 4 - 2 = 2 ; those who play only baseball and football are 5 - 2 = 3 ; hence , 5 + 2 + 3 = 10 students play exactly two of these sports . answer : a .
|
a ) 10 , b ) 46 , c ) 67 , d ) 68 , e ) 446
|
a
|
add(subtract(5, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), add(subtract(7, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), subtract(4, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18)))))
|
add(n1,n2)|add(n4,n5)|add(n3,#0)|add(n6,#1)|subtract(#2,#3)|add(n7,#4)|subtract(n0,#5)|subtract(n4,#6)|subtract(n5,#6)|subtract(n6,#6)|add(#7,#8)|add(#10,#9)
|
other
|
john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 65 miles per hour , what is v in miles per hour ?
|
"90 * 10 / 60 = 15 kmph answer : a"
|
a ) 15 , b ) 87 , c ) 99 , d ) 77 , e ) 55
|
a
|
multiply(divide(10, const_60), 90)
|
divide(n1,const_60)|multiply(n0,#0)|
|
physics
|
find the number of different prime factors of 1250
|
"speed in still water = ( 11 + 5 ) / 2 = 8 kmph ans - c"
|
a ) 2 kmph , b ) 3 kmph , c ) 8 kmph , d ) 9 kmph , e ) 7 kmph
|
c
|
stream_speed(11, 5)
|
stream_speed(n0,n1)|
|
physics
|
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 15 % . what was its profit on the items sold in february ?
|
"sq rt ( 5 x / 3 ) = x = > 5 x / 3 = x ^ 2 = > x = 5 / 3 ans - d"
|
a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 5 / 3 , e ) 1 / 2
|
d
|
divide(5, 3)
|
divide(n0,n1)|
|
general
|
if 36 men can do a piece of work in 25 hours , in how mwny hours will 15 men do it ?
|
"suppose commodity x will cost 40 paise more than y after z years . then , ( 4.20 + 0.40 z ) - ( 6.30 + 0.15 z ) = 0.40 0.25 z = 0.40 + 2.10 z = 2.50 / 0.25 = 250 / 25 = 10 . therefore , x will cost 40 paise more than y 10 years after 2001 i . e . , 2011 . answer is d ."
|
a ) 2010 , b ) 2001 , c ) 2012 , d ) 2011 , e ) 2009
|
d
|
add(2001, divide(add(divide(40, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), divide(15, const_100))))
|
divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5)|
|
general
|
a 240 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ?
|
"a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 5 years , their ages will be 48 and 28 . sum of their ages = 48 + 28 = 76 . answer : d"
|
a ) 50 , b ) 60 , c ) 70 , d ) 76 , e ) 90
|
d
|
add(add(multiply(divide(60, 8), const_2), 8), add(divide(60, 8), 8))
|
divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)|
|
general
|
if it takes a tub 5 minutes to drain 5 / 7 of its content , how much more time will it take for the tub to be empty ?
|
"explanation : capital = rs . x , then 5 / 7 x = 61.5 x = 87.86 answer : b ) rs . 87.86"
|
a ) 22.378 , b ) 87.86 , c ) 246.0 , d ) 78.88 , e ) 127.71
|
b
|
divide(61.50, divide(const_4, 7))
|
divide(const_4,n3)|divide(n4,#0)|
|
gain
|
on a certain transatlantic crossing , 40 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 20 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ?
|
"let the numbers be x and y . then , xy = 468 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 468 ) = 1225 x + y = 35 . option e"
|
a ) a ) 23 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35
|
e
|
sqrt(add(power(sqrt(subtract(289, multiply(const_2, 468))), const_2), multiply(const_4, 468)))
|
multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)|
|
general
|
a 4 digit number divisible by 7 becomes divisible by 3 when 19 is added to it . the largest such number is :
|
"soap : alcohol initial ratio soap : alcohol : water - - > 4 : 20 : 60 initial soap : alcohol = 4 / 20 = 4 : 20 after doubled soap : alcohol = 2 * 4 / 20 = 8 : 20 initial soap : water = 4 / 60 = 4 : 60 after halved soap : water : 1 / 2 * 4 / 60 = 2 / 60 = 2 : 60 after soap : alcohol : water - - > 8 : 20 : 240 - - > 2 : 5 : 60 given alcohol 100 cubic centimeter . ratio is 40 : 100 : 1200 ( 2 : 5 : 60 ) for 100 cubic centimeter of alcohol - - - 1200 cubic cm water is required ."
|
a ) 1200 , b ) 1250 , c ) 1300 , d ) 1400 , e ) 1450
|
a
|
divide(divide(divide(divide(divide(volume_rectangular_prism(100, 60, 20), const_3), const_2), 4), 4), 4)
|
volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,n0)|divide(#3,n0)|divide(#4,n0)|
|
geometry
|
rs . 385 were divided among x , y , z in such a way that x had rs . 20 more than y and z had rs 15 more than x . how much was y ’ s share ?
|
"since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 13 then f ( 5 ) = 13 - 5 = 8 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 8 - 6 = 2 . answer : d . questions on funtions to practice :"
|
a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 4
|
d
|
subtract(subtract(13, add(1, 4)), 6)
|
add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)|
|
general
|
the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 π · r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams t ?
|
area of semicircle = ½ π r 2 = ½ × 22 ⁄ 7 × 7 × 7 = 77 m 2 answer b
|
['a ) 154 sq metres', 'b ) 77 sq metres', 'c ) 308 sq metres', 'd ) 22 sq metres', 'e ) none of these']
|
b
|
divide(circle_area(divide(14, const_2)), const_2)
|
divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)
|
geometry
|
a dealer offers a cash discount of 16 % and still makes a profit of 25 % when he further allows 60 articles to be sold at the cost price of 50 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 2200 ( 100 / 110 ) = rs . 2000 answer : b"
|
a ) 2299 , b ) 2000 , c ) 2670 , d ) 6725 , e ) 2601
|
b
|
divide(2200, add(const_1, divide(10, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 20 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ?
|
"say the second solution ( which was 1 / 4 th of total ) was x % salt , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : b ."
|
a ) 24 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 %
|
b
|
multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(10, const_100), const_4), divide(10, const_100))), const_100)
|
divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)|
|
gain
|
a man speaks truth 3 out of 4 times . he throws a die and reports it to be a 6 . what is the probability of it being a 6 ?
|
"the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 170 - - 2 subtracting 1 from 2 we have c - a = 80 ; answer : b"
|
a ) 25 , b ) 80 , c ) 90 , d ) 140 , e ) it can not be determined from the information given
|
b
|
subtract(multiply(85, const_2), multiply(45, const_2))
|
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
|
general
|
if a and b are positive integers , and a = 5 b + 20 , the greatest common divisor of a and b can not be
|
the required number of working hours per day x , more pumps , less working hours per day ( indirect ) less days , more working hours per day ( indirect ) pumps 4 : 3 , days 1 : 2 } : : 8 : x therefore 4 * 1 * x = 3 * 2 * 8 , x = ( 3 * 2 * 8 ) / 4 x = 12 correct answer ( d )
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
|
d
|
divide(multiply(multiply(3, 8), 2), 4)
|
multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)
|
physics
|
a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction q of the sum of the 21 numbers in the list ?
|
since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 810 = 44 interest earned for 3 years = 44 * 3 = 132 amount invested = 815 - 132 = 683 answer : b
|
a ) 670 , b ) 683 , c ) 698 , d ) 744 , e ) 700
|
b
|
subtract(810, multiply(divide(subtract(854, 810), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100)))
|
divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5)
|
gain
|
the mass of 1 cubic meter of a substance is 200 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters )
|
"800 * 9.2 7360.0 gm 7.36 kg answer : e"
|
a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.9 kg , e ) 7.36 kg
|
e
|
divide(multiply(9.2, 800), const_1000)
|
multiply(n0,n1)|divide(#0,const_1000)|
|
general
|
a camera lens filter kit containing 5 filters sells for $ 67.50 . if the filters are purchased individually , 2 of them are priced at $ 7.45 each , 2 at $ 10.05 each , 1 at $ 14.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ?
|
"1022 ã · 25 = 40 with remainder = 22 22 + 3 = 25 . hence 3 should be added to 1022 so that the sum will be divisible by 25 answer : option b"
|
a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 5
|
b
|
subtract(25, reminder(1022, 25))
|
reminder(n0,n1)|subtract(n1,#0)|
|
general
|
a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 108 . what was the number he chose ?
|
"a hostel had provisions for 250 men for 44 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 44 250 × 44 = 200 x 5 × 44 = 4 x x = 5 × 11 = 55 answer a"
|
a ) 55 , b ) 40 , c ) 50 , d ) 60 , e ) 65
|
a
|
divide(multiply(250, 44), subtract(250, 50))
|
multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)|
|
gain
|
a truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes . what is the ratio of their speed ?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 8339 ( 100 / 124 ) = rs . 6725 . answer : c"
|
a ) rs . 6825 , b ) rs . 6721 , c ) rs . 6725 . , d ) rs . 4298 , e ) rs . 6729
|
c
|
divide(8339, add(const_1, divide(24, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
x and y are both integers . if x / y = 59.60 , then what is the sum of all the possible two digit remainders of x / y ?
|
"63 + 5 * 12 / ( 180 / 3 ) = 63 + 5 * 12 / ( 60 ) = 63 + ( 5 * 12 ) / 60 = 63 + 1 = 64 . answer : d"
|
a ) 22 , b ) 77 , c ) 29 , d ) 64 , e ) 21
|
d
|
add(63, divide(multiply(5, 12), divide(180, 3)))
|
divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)|
|
general
|
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 37 , the how old is b ?
|
12 * 1.75 + 0.45 * 12 * 55 = 318 hence - a
|
a ) 318 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $
|
a
|
multiply(multiply(0.45, 55), 12)
|
multiply(n1,n3)|multiply(n2,#0)|
|
general
|
a candidate got 35 % of the votes polled and he lost to his rival by 2430 votes . how many votes were cast ?
|
"since x is an integer , ( - 6 ) ^ 2 x is always positive . so , 6 ^ 2 x = 6 ^ ( 7 + x ) 2 x = 7 + x x = 7 answer : e"
|
a ) 5 , b ) 4 , c ) 3 , d ) 8 , e ) 7
|
e
|
divide(7, 6)
|
divide(n3,n0)|
|
general
|
the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if x 20 = 190 , what is the value of x ?
|
"given ; 4 duck = 3 hen ; or , duck / hen = 3 / 4 ; let hen ' s 1 leap = 4 meter and ducks 1 leap = 3 meter . then , ratio of speed of hen and duck = 4 * 3 / 3 * 2 = 2 : 1 ' ' answer : 2 : 1 ;"
|
a ) 2 : 1 , b ) 3 : 4 , c ) 4 : 3 , d ) 1 : 4 , e ) 5 : 6
|
a
|
divide(divide(3, 2), divide(3, 4))
|
divide(n0,n1)|divide(n3,n2)|divide(#0,#1)|
|
other
|
by travelling at 60 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ?
|
"let hari ’ s capital be rs . x . then , 3640 * 12 / 7 x = 2 / 3 = > 14 x = 131040 = > x = 9360 . answer : e"
|
a ) s . 7500 , b ) s . 8000 , c ) s . 8500 , d ) s . 9000 , e ) s . 9360
|
e
|
divide(divide(3640, subtract(const_1, divide(5, const_12))), divide(2, 3))
|
divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)|
|
other
|
in a class of 42 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 10 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ?
|
"let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.20 0.2 x = 0.10 x = 0.5 = 50 % the answer is b ."
|
a ) 60 % , b ) 50 % , c ) 40 % , d ) 30 % , e ) 20 %
|
b
|
divide(subtract(30, 10), subtract(20, 10))
|
subtract(n0,n2)|subtract(n4,n2)|divide(#0,#1)|
|
gain
|
two trains are moving in the same direction at 126 kmph and 54 kmph . the faster train crosses a man in the slower train in 14 seconds . find the length of the faster train ?
|
1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e .
|
a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26
|
e
|
subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), floor(add(subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(add(const_10, 5), 2)), const_1))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5)))
|
add(n4,const_10)|divide(n1,n2)|divide(n0,n2)|divide(n1,n3)|divide(n0,n3)|divide(n1,n4)|divide(n0,n4)|subtract(n1,n0)|add(#7,const_1)|add(n2,#0)|subtract(#1,#2)|subtract(#3,#4)|subtract(#5,#6)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|subtract(#8,#13)|subtract(#14,#9)|subtract(#15,#0)|add(#17,const_1)|floor(#19)|subtract(#16,#20)|subtract(#21,#18)
|
other
|
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 8 years hence ?
|
"solution s . i . = rs . ( 956 - 850 ) = rs . 106 rate = ( 100 x 106 / 850 x 3 ) = 212 / 51 % new rate = ( 212 / 51 + 4 ) % = 416 / 51 % new s . i . = rs . ( 850 x 416 / 51 x 3 / 100 ) rs . 208 . ∴ new amount = rs . ( 850 + 208 ) = rs . 1058 . answer c"
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1058 , d ) data inadequate , e ) none of these
|
c
|
add(850, divide(multiply(multiply(850, add(divide(multiply(subtract(956, 850), const_100), multiply(850, 3)), 4)), 3), const_100))
|
multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|
|
gain
|
what is the sum of all the odd numbers between 24 and 50 , inclusive ?
|
"let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 4 gms in 600,000 cm ^ 3 , y = 4 * 600,000 = 2 , 400,000 gms total gms in 1 m ^ 3 = 1 , 000,000 + 2 , 400,000 = 3 , 400,000 answer : a"
|
a ) 3 , 400,000 , b ) 2 , 800,000 , c ) 55,000 , d ) 28,000 , e ) 280
|
a
|
subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 4)), 4)), volume_cube(100))
|
multiply(const_2,n3)|volume_cube(n5)|divide(#1,const_10)|multiply(#2,n2)|multiply(#2,#0)|multiply(#3,n2)|multiply(#4,n3)|add(#5,#6)|subtract(#7,#1)|
|
geometry
|
solving a linear equation with several occurrences of the variable , solve for w . simplify answer as much as possible . ( 7 w + 6 ) / 6 + ( 9 w + 8 ) / 2 = 22
|
"the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 10 ) = 1 / 600 of the video per second . the combined rate is 1 / 15 + 1 / 600 = 41 / 600 of the video per second . the time required to load the video is 600 / 41 = 14.63 seconds . the answer is d ."
|
a ) 13.42 , b ) 13.86 , c ) 14.25 , d ) 14.63 , e ) 14.88
|
d
|
subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))
|
add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)|
|
physics
|
what is the greatest prime factor of 5 ^ 6 - 1 ?
|
since the starting point is given as the $ 4000 scholarship , assume $ 4000 scholarships to be x by the given information , $ 2500 scholarships = 2 x and $ 1250 scholarships = 6 x gievn : total $ 1250 scholarships = $ 75000 6 x * 1250 = 75000 solve for x = 10 option d
|
a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 15
|
d
|
divide(divide(75000, 1250), multiply(const_2, 3))
|
divide(n8,n0)|multiply(n5,const_2)|divide(#0,#1)
|
general
|
the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is :
|
"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 / 12 * 33 = 33 cents . answer : c ."
|
a ) 0.33 , b ) 0.44 , c ) 33 , d ) 44 , e ) 55
|
c
|
multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.33, 100))
|
divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|
|
general
|
how many bricks , each measuring 25 cm * 11.25 cm * 6 cm , will be needed to build a wall 8 m * 6 m * 22.5 m
|
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 40 ) = > x / 2 = 33 = > x = 66 . answer : c"
|
a ) 18 , b ) 82 , c ) 66 , d ) 27 , e ) 77
|
c
|
multiply(subtract(73, 40), const_2)
|
subtract(n0,n1)|multiply(#0,const_2)|
|
general
|
a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length .
|
"let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 25 / 100 ) * 100 = 75 after second year , value of baseball card = ( 1 - 10 / 100 ) * 75 = 67.5 total percent decrease of the card ' s value over the two years = ( 100 - 67.5 ) / 100 * 100 % = 31.5 % answer c"
|
a ) 28 % , b ) 30 % , c ) 32.5 % , d ) 36 % , e ) 72 %
|
c
|
subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(25, const_100))), const_100))
|
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)|
|
gain
|
determine the value of ( 27 / 31 * 31 / 27 ) * 3
|
"total nos of ways in which we can choose n = 96 n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? case 1 : n = odd then n + 2 = odd & n + 1 will be even i . e this needs get divided by 8 , hence is a multiple of 8 so we have 8 . . 96 = 12 multiples to fill the n + 1 pos hence 12 ways case 2 : n is even then n + 2 will be even & the product will be divisible by 24 & thus 8 so nos of values that can be used for n = 2 . . . . 96 ( all even nos ) i . e 48 nos total = 48 + 12 = 60 ways so reqd p = 60 / 96 = 5 / 8 ; answer : d"
|
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4
|
d
|
divide(add(divide(96, 2), divide(96, 8)), 96)
|
divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|
|
general
|
jane makes toy bears . when she works with an assistant , she makes 100 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ?
|
"let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 40 m per hour = 1.5 hours distance travelled by car a = 40 * 1.5 = 60 miles since car b catches up car a at 10 : 30 , time = 60 mins = 1 hour speed of car b = 60 / ( 1 ) = 60 miles per hour answer b"
|
a ) 45 , b ) 60 , c ) 53 , d ) 55 , e ) 50
|
b
|
divide(60, divide(add(multiply(subtract(10, 9), const_60), subtract(40, 10)), const_60))
|
subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)|
|
physics
|
village a ’ s population is 300 greater than village b ' s population . if village b ’ s population were reduced by 600 people , then village a ’ s population would be 4 times as large as village b ' s population . what is village b ' s current population ?
|
"1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ? 150 = 4 check the fourth eqn . answer : c"
|
a ) 1 , b ) 255 , c ) 4 , d ) 445 , e ) 235
|
c
|
divide(subtract(subtract(225, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2)
|
add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)|
|
general
|
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 ?
|
"given cash discount - 16 % profit - 25 % items sold - 60 price sold at = list price of 50 assume list price = $ 10 total invoice = $ 500 - 16 % cash discount = $ 420 let cost price of 60 items be x so total cost = 60 * x given the shopkeeper had a profit of 25 % 60 * x * 125 / 100 = 420 or x = $ 7 * 4 / 5 = $ 28 / 5 which means his products were listed at $ 10 which is a 78 + ( 4 / 7 ) % markup over $ 28 / 5 answer e"
|
a ) 50 % , b ) 60 % , c ) 70 % , d ) 75 % , e ) 78 + ( 4 / 7 ) %
|
e
|
multiply(subtract(divide(divide(divide(add(const_100, 25), const_100), subtract(const_1, divide(subtract(60, 50), 60))), divide(subtract(const_100, 16), const_100)), const_1), const_100)
|
add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)|
|
gain
|
a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 390 m long platform ?
|
as per question = > n = 8 p + 1 for some integer p hence 3 n = > 24 q + 3 = > remainder = > 3 for some integer q hence b
|
a ) 1 , b ) 3 , c ) 7 , d ) 5 , e ) 6
|
b
|
multiply(3, 1)
|
multiply(n1,n2)
|
general
|
a shopkeeper sold an article at $ 1050 and gained a 20 % profit . what was the cost price ?
|
a 3 = 2197 = > a = 13 6 a 2 = 6 * 13 * 13 = 1014 answer : b
|
a ) 864 , b ) 1014 , c ) 1299 , d ) 1268 , e ) 1191
|
b
|
surface_cube(cube_edge_by_volume(2197))
|
cube_edge_by_volume(n0)|surface_cube(#0)|
|
geometry
|
find large number from below question the difference of two numbers is 1380 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder
|
"there are 15 business exec and in each handshake 2 business execs are involved . hence 15 c 2 = 105 also , each of 15 exec will shake hand with every 3 other chairmen for total of 45 handshake . total = 45 + 105 = 150 ans : a"
|
a ) 150 , b ) 131 , c ) 115 , d ) 90 , e ) 45
|
a
|
add(divide(multiply(15, subtract(15, const_1)), const_2), multiply(15, 3))
|
multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)|
|
geometry
|
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 3 feet . how many feet does the squirrel travels if the post is 18 feet tall and 3 feet in circumference ?
|
"1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . mumbo is 43426 . . . answer : a"
|
a ) 43426 , b ) 14236 , c ) 13436 , d ) 14263 , e ) 15263
|
a
|
divide(79523, add(const_3, const_3))
|
add(const_3,const_3)|divide(n0,#0)|
|
general
|
suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 540 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 720 . how much more do pairs of boots cost than pairs of shoes at suzie ’ s discount footwear ?
|
"a can do the work in 18 / 2 i . e . , 9 days . a and b ' s one day ' s work = 1 / 9 + 1 / 18 = ( 2 + 1 ) / 18 = 1 / 6 so a and b together can do the work in 6 days . answer : d"
|
a ) 10 , b ) 16 , c ) 18 , d ) 6 , e ) 12
|
d
|
inverse(add(divide(const_1, 18), multiply(divide(const_1, 18), const_2)))
|
divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|
|
physics
|
what is the probability that the sum of two dice will yield a 6 , and then when both are thrown again , their sum will again yield a 6 ? assume that each die has 5 sides with faces numbered 1 to 5 .
|
"1 / 2 * 5.5 * 6 = 16.5 m 2 answer : b"
|
a ) 11 m 2 , b ) 16.5 m 2 , c ) 18.5 m 2 , d ) 19.5 m 2 , e ) 12 m 2
|
b
|
triangle_area(5.5, 6)
|
triangle_area(n0,n1)|
|
geometry
|
evaluate : 60 - 12 * 3 * 2 = ?
|
"15 * a + 15 * b = x pages in 15 mins printer a will print = 15 / 45 * x pages = 1 / 3 * x pages thus in 15 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 3 more pages per min that printer a . in 15 mins printer b will print 45 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 45 = > x = 135 pages answer : b"
|
a ) 125 , b ) 135 , c ) 145 , d ) 155 , e ) 165
|
b
|
multiply(divide(3, subtract(divide(45, 15), const_1)), 45)
|
divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)|
|
physics
|
the radius of a cone is 49 m , slant height is 35 m . find the curved surface area ?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 125 ) = rs . 6720 . answer : d"
|
a ) rs . 5725 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6720 , e ) none of these
|
d
|
divide(8400, add(const_1, divide(25, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
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