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mathqa_question_shift_test

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in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.24 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]
"speed = 162 * ( 5 / 18 ) m / sec = 45 m / sec length of train ( distance ) = speed * time ( 45 ) * 9 = 405 meter answer : d"
a ) 150 meter , b ) 286 meter , c ) 186 meter , d ) 405 meter , e ) 265 meter
d
multiply(divide(multiply(162, const_1000), const_3600), 9)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|
physics
a train running at the speed of 162 km / hr crosses a pole in 9 seconds . find the length of the train .
"let the maximum marks be x then , 30 % of x = 150 + 30 30 x / 100 = 180 30 x = 180 * 100 = 18000 x = 600 answer is b"
a ) 750 , b ) 600 , c ) 650 , d ) 550 , e ) 500
b
divide(add(150, 30), divide(30, const_100))
add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|
general
a student has to obtain 30 % of the total marks to pass . he got 150 marks and failed by 30 marks . the maximum marks are ?
"1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 5 = 7200 . answer : c"
a ) 5729 , b ) 5760 , c ) 7200 , d ) 2870 , e ) 2799
c
divide(multiply(5, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1))
multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)|
physics
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 5 liters per minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours . the capacity of the tank is ?
"let b be the number of blue hats and let g be the number of green hats . b + g = 85 . b = 85 - g . 6 b + 7 g = 560 . 6 ( 85 - g ) + 7 g = 560 . 510 - 6 g + 7 g = 560 . g = 560 - 510 = 50 . the answer is b ."
a ) a ) 36 , b ) b ) 50 , c ) c ) 40 , d ) d ) 42 , e ) e ) 44
b
subtract(560, multiply(85, 6))
multiply(n0,n1)|subtract(n3,#0)|
general
we bought 85 hats at the store . blue hats cost $ 6 and green hats cost $ 7 . the total price was $ 560 . how many green hats did we buy ?
"i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 72 and that c and l 1 = 0.10 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.1 x = 20 , therefore the grand total is 200 . now we know that l 2 is 200 - 72 = 128 . we also learn that c and no c are equally represented thus 100 each . therefore no c and no l 2 will be 100 - 52 = 48 . thus b is the correct answer choice"
a ) 46 , b ) 48 , c ) 56 , d ) 32 , e ) 58
b
divide(subtract(divide(20, divide(10, const_100)), 72), 2)
divide(n3,const_100)|divide(n6,#0)|subtract(#1,n4)|divide(#2,n1)|
general
among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 10 % of listco ' s sales staff . listco employs 72 level - 1 employees , 20 of whom are college graduates . how many sales staff members without college degrees are level - 2 employees ?
"1498 - 1400 = 98 is the rate of interest on $ 1400 for one year . the rate of interest = ( 100 * 98 ) / ( 1400 ) = 7 % the answer is c ."
a ) 3 % , b ) 5 % , c ) 7 % , d ) 9 % , e ) 11 %
c
divide(multiply(subtract(1498, 1400), const_100), 1400)
subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)|
gain
the compound interest earned on a sum for the second and the third years are $ 1400 and $ 1498 respectively . what is the rate of interest ?
"m = 9 s = 3.1 ds = 12.1 us = 5.9 x / 12.1 + x / 5.9 = 1 x = 3.97 d = 3.97 * 2 = 7.94 answer : e"
a ) 2.21 , b ) 2.48 , c ) 9.24 , d ) 7.29 , e ) 7.94
e
multiply(divide(multiply(add(9, 3.1), subtract(9, 3.1)), add(add(9, 3.1), subtract(9, 3.1))), const_2)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|
physics
a man can row 9 kmph in still water . when the river is running at 3.1 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ?
"n case of stock 1 , if he invest rs . 105 , he will get a dividend of rs . 12 ( assume face value = 100 ) in case of stock 2 , if he invest rs . 88 , he will get a dividend of rs . 8 ( assume face value = 100 ) ie , if he invest rs . ( 88 * 12 ) / 8 , he will get a dividend of rs . 12 required ratio = 105 : ( 88 × 12 ) / 8 = 105 : ( 11 × 12 ) = 35 : ( 11 × 4 ) = 35 : 44 answer : option d"
a ) 31 : 44 , b ) 31 : 27 , c ) 16 : 15 , d ) 35 : 44 , e ) 35 : 27
d
divide(multiply(105, const_2), multiply(88, const_3))
multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)|
other
a man invests some money partly in 12 % stock at 105 and partly in 8 % stock at 88 . to obtain equal dividends from both , he must invest the money in the ratio :
"let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 15 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 15 , d = 15 b - 15 c ; every 5 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = 4 , d = 4 b + 4 c ; d = 15 b - 15 c = 5 b + 5 c , - - > b = 2 c , - - > d = 15 b - 15 b / 2 = 15 b / 2 . interval = \ frac { d } { b } = \ frac { 15 / 2 b } { b } = 15 / 2 answer : e ( 15 / 2 minutes ) ."
a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 15 / 2 minutes
e
divide(subtract(15, divide(15, divide(add(5, 15), subtract(15, 5)))), const_1)
add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|divide(n0,#2)|subtract(n0,#3)|divide(#4,const_1)|
physics
a man cycling along the road noticed that every 15 minutes a bus overtakes him and every 5 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ?
explanation : number of questions attempted correctly = ( 70 % of 10 + 40 % of 30 + 60 % of 35 ) = 7 + 12 + 21 = 40 . questions to be answered correctly for 60 % = 60 % of total quotations = 60 % of 75 = 45 . he would have to answer 45 - 40 = 5 answer : a
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
a
subtract(divide(multiply(75, 60), const_100), add(add(divide(multiply(10, 70), const_100), divide(multiply(30, 40), const_100)), divide(multiply(35, 60), const_100)))
multiply(n0,n6)|multiply(n1,n4)|multiply(n2,n5)|multiply(n3,n6)|divide(#0,const_100)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|add(#5,#6)|add(#8,#7)|subtract(#4,#9)
general
rakesh ' s mathematics test had 75 problems , 10 arithmetic , 30 algebra , 35 geometry problems . although he answered 70 % of arithmetic , 40 % of arithmetic and 60 % of geometry problems correctly , still he got less than 60 % problems right . how many more questions he would have to answer more to get passed ?
"sqrt ( 7 x / 3 ) to be perfect square x has to 7 / 3 ans : b"
a ) 25 / 9 , b ) 7 / 3 , c ) 5 / 3 , d ) 3 / 5 , e ) 9 / 25
b
divide(7, 3)
divide(n0,n1)|
general
a number x is multiplied by 7 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x if x ≠ 0 ?
"240 = 2 ^ 4 * 3 * 5 = ( 4 ) * 2 ^ 2 * 3 * 5 besides ( 4 ) , the exponents of 2 , 3 , and 5 are 2 , 1 , and 1 . there are ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 12 ways to make multiples of 4 . we must subtract 1 because one of these multiples is 240 . the answer is d ."
a ) 6 , b ) 8 , c ) 9 , d ) 11 , e ) 12
d
divide(divide(divide(240, 4), const_2), const_3)
divide(n0,n1)|divide(#0,const_2)|divide(#1,const_3)|
general
how many of the positive divisors of 240 are also multiples of 4 not including 240 ?
connie has a 1 6 chance of winning 6 dollars her first turn . she has a 5 / 6 1 / 6 chance of winning 1 dollar her second turn . next , she has a 25 36 1 / 6 chance of winning 1 / 6 dollars her third turn . generalizing , connie ' s expected earnings form a geometric series with initial term 1 / 6 * 6 = 1 and common ratio 5 / 6 * 1 / 6 = 5 / 36 . hence , connie ' s expected earnings are 1 / 1 - 5 / 36 = 36 / 31 correct answer d
a ) 32 / 31 , b ) 33 / 31 , c ) 34 / 31 , d ) 36 / 31 , e ) 0 / 31
d
divide(const_1, subtract(const_1, divide(divide(subtract(6, 1), 6), 6)))
subtract(n0,n4)|divide(#0,n0)|divide(#1,n0)|subtract(const_1,#2)|divide(const_1,#3)
general
one day , connie plays a game with a fair 6 - sided die . connie rolls the die until she rolls a 6 , at which point the game ends . if she rolls a 6 on her first turn , connie wins 6 dollars . for each subsequent turn , connie wins 1 6 of the amount she would have won the previous turn . what is connie ' s expected earnings from the game ?
"0.06 * 0.05 = 0.003 = 0.3 % the answer is b ."
a ) 0.125 % , b ) 0.3 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 %
b
multiply(6, divide(5, const_100))
divide(n1,const_100)|multiply(n0,#0)|
gain
in the manufacture of a certain product , 6 percent of the units produced are defective and 5 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ?
"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 26 = > 2 x = 26 = > x = 13 c"
a ) 11 , b ) 12 , c ) 13 , d ) 16 , e ) 18
c
divide(26, const_2)
divide(n0,const_2)|
general
in a group of ducks and cows , the total number of legs are 26 more than twice the no . of heads . find the total no . of buffaloes .
at 3 / 4 th of speed he is late by ' 2 hrs ' x - 3 / 4 ( x ) = 2 x = 8 so 8 - 2 = 6 hrs ( since 2 hrs late ) answer : c
a ) 5 hours , b ) 3 hours , c ) 6 hours , d ) 12 hours , e ) 15 hours
c
divide(multiply(multiply(multiply(divide(3, 4), 2), divide(3, 4)), 2), subtract(multiply(divide(3, 4), 2), multiply(multiply(divide(3, 4), 2), divide(3, 4))))
divide(n0,n1)|multiply(n2,#0)|multiply(#0,#1)|multiply(n2,#2)|subtract(#1,#2)|divide(#3,#4)
physics
a man walking at 3 / 4 th of the speed , reaches his office late by 2 hours . what is the usual time ?
because a + b = 1 , 2 a + 2 b = 2 ( a + b ) = 2 × 1 = 2 . correct answer d
a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 1
d
subtract(add(add(2, 1), 2), add(2, 1))
add(n0,n1)|add(n1,#0)|subtract(#1,#0)
general
given a + b = 1 , find the value of 2 a + 2 b . two solutions are presented below . only one is correct , even though both yield the correct answer .
"let the sums be p . now , 45 % of p = 4016.25 or , p = 8925 answer a"
a ) 8925 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these
a
divide(multiply(const_100, 4016.25), multiply(9, 5))
multiply(n0,const_100)|multiply(n1,n2)|divide(#0,#1)|
gain
a sum fetched total simple interest of 4016.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ?
"there are 6 possible arrangements of the three numbers . then each number will be in the hundreds , tens , and ones place two times each . the sum is 2 ( 222 ) + 2 ( 333 ) + 2 ( 555 ) = 2220 the answer is b ."
a ) 2210 , b ) 2220 , c ) 2230 , d ) 2240 , e ) 2250
b
multiply(add(add(multiply(add(add(5, 3), 2), const_100), multiply(add(add(const_4.0, 3), 2), const_10)), add(add(5, 3), 2)), 3)
add(n2,n3)|add(n1,#0)|multiply(#1,const_100)|multiply(#1,const_10)|add(#2,#3)|add(#4,#1)|multiply(#5,const_2)|
general
what is the sum of all possible 3 - digit numbers that can be constructed using the digits 2 , 3 , and 5 if each digit can be used only once in each number ?
"let ' take the number of geese to be 100 . male = 30 . female = 70 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 25 % are male i . e 25 / 100 * 20 = 5 geese ( males ) and now we know out of the total 20 geese , 5 are male , then 15 have to be female . now the ratio part , male geese ratios = 5 / 30 = 1 / 6 . - a female geese ratios = 15 / 70 = 3 / 14 - b cross multiply equations a and b and you get = 9 / 7 . ans e"
a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 8 , e ) 9 / 7
e
divide(divide(divide(25, const_100), divide(30, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(30, const_100)))
divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_4)|divide(#0,#1)|multiply(#2,const_10)|divide(#4,const_100)|divide(#5,#1)|divide(#3,#6)|
general
a total of 30 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 25 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ]
"other number = ( 11 * 7700 ) / 275 = 308 . answer : c"
a ) 288 , b ) 277 , c ) 308 , d ) 988 , e ) 112
c
multiply(11, 275)
multiply(n0,n2)|
physics
the h . c . f of two numbers is 11 and their l . c . m is 7700 . if one of the numbers is 275 , then the other is ?
"ac = 11 and ab = 5 , so bc = 6 . bc = 2 cd so cd = 3 . the length of ae is ab + bc + cd + de = 5 + 6 + 3 + 7 = 21 the answer is b ."
a ) 19 , b ) 21 , c ) 23 , d ) 25 , e ) 27
b
add(add(11, divide(subtract(11, 5), 2)), 7)
subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)|
physics
a , b , c , d and e are 5 consecutive points on a straight line . if bc = 2 cd , de = 7 , ab = 5 and ac = 11 , what is the length of ae ?
"between the 12 mango trees , there are 11 gaps and each gap has 2 meter length also , 4 meter is left from all sides of the boundary of the garden . hence , length of the garden = ( 11 ã — 2 ) + 4 + 4 = 30 meter answer is e ."
a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30
e
add(add(multiply(subtract(12, const_1), 2), divide(10, 2)), divide(10, 2))
divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)|
physics
in a garden , there are 10 rows and 12 columns of mango trees . the distance between the two trees is 2 metres and a distance of four metre is left from all sides of the boundary of the garden . what is the length of the garden ?
"out of the 17 integers : 9 are odd and 8 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 9 odd numbers first 2 . 10 th number will be the first even number so we need to withdraw at least 10 numbers to make sure that we get one even number and the product of all the integers picked is even . so , answer will be 10 . ( d )"
a ) 19 , b ) 12 , c ) 11 , d ) 10 , e ) 3
d
add(divide(17, const_2), 1)
divide(n1,const_2)|add(n0,#0)|
general
each of the integers from 1 to 17 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ?
sum of last 4 matches = ( ( 10 × 45 ) – ( 6 × 48 ) = 162 average = 162 / 4 = 40.5 answer : d
a ) 43.25 , b ) 43 , c ) 38 , d ) 40.5 , e ) 36
d
divide(subtract(multiply(45, 10), multiply(6, 48)), 4)
multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)
general
the averge score of a cricketer for 10 matches is 45 runs . if the average for the first 6 matches is 48 . then find the average for the last 4 matches ?
"explanation : let the cost price = rs 100 then , marked price = rs 140 required gain = 8 % , so selling price = rs 108 discount = 140 - 108 = 32 discount % = ( 32 / 140 ) * 100 = 22.85 % option b"
a ) 23.85 % , b ) 22.85 % , c ) 21.85 % , d ) 20.85 % , e ) none of these
b
subtract(const_100, multiply(divide(add(8, const_100), add(40, const_100)), const_100))
add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|
gain
a shopkeeper fixes the marked price of an item 40 % above its cost price . the percentage of discount allowed to gain 8 % is
"b 16.8 sec d = 60 + 80 = 140 m s = 60 * 5 / 18 = 50 / 3 t = 140 * 3 / 50 = 8.4 sec answer is b"
a ) 5.8 sec , b ) 8.4 sec , c ) 12.4 sec , d ) 6.8 sec , e ) 1.8 sec
b
divide(add(60, 80), multiply(60, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
how long does a train 60 m long travelling at 60 kmph takes to cross a bridge of 80 m in length ?
"the percent of the budget for transportation is 100 - ( 55 + 9 + 5 + 4 + 2 ) = 25 % 100 % of the circle is 360 degrees . then ( 25 % / 100 % ) * 360 = 90 degrees the answer is e ."
a ) 18 ° , b ) 36 ° , c ) 54 ° , d ) 72 ° , e ) 90 °
e
divide(multiply(const_360, subtract(const_100, add(add(add(add(55, 9), 5), 4), 2))), const_100)
add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|subtract(const_100,#3)|multiply(#4,const_360)|divide(#5,const_100)|
geometry
a circle graph shows how the budget of a certain company was spent : 55 percent for salaries , 9 percent for research and development , 5 percent for utilities , 4 percent for equipment , 2 percent for supplies , and the remainder for transportation . if the area of each sector of the graph is proportional to the percent of the budget it represents , how many degrees of the circle are used to represent transportation ?
"96 / 7 = 13 . xx we are not concerned about the exact value of 96 / 7 as we just need the integers . since the values are small , we can write down the integers . the different integers between 5 and 96 / 7 would be 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12,13 total number of integers = 9 option b"
a ) 7 , b ) 9 , c ) 10 , d ) 12 , e ) 15
b
add(subtract(divide(96, 7), 5), const_1)
divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)|
general
how many integers are between 5 and 96 / 7 , inclusive ?
"( 256 ) ^ 2 - ( 144 ) ^ 2 = ( 256 + 144 ) ( 256 - 144 ) = 400 x 112 = 44800 answer is b"
a ) 761200 , b ) 44800 , c ) 761800 , d ) 761500 , e ) none of them
b
add(multiply(256, 256), multiply(144, 144))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
general
simplify : 256 x 256 - 144 x 144
"perimeter of square = p side of square = p / 4 area of square = ( p ^ 2 ) / 16 = a given that a = 2 p + 20 ( p ^ 2 ) / 16 = 2 p + 20 p ^ 2 = 32 p + 320 p ^ 2 - 32 p - 320 = 0 p ^ 2 - 40 p + 8 p - 320 = 0 p ( p - 40 ) + 8 ( p + 40 ) = 0 ( p - 40 ) ( p + 8 ) = 0 p = 40 or - 8 discarding negative value , p = 40 answer is c"
a ) 28 , b ) 36 , c ) 40 , d ) 56 , e ) 64
c
subtract(subtract(add(const_10, multiply(20, 2)), const_0_25), const_0_25)
multiply(n0,n1)|add(#0,const_10)|subtract(#1,const_0_25)|subtract(#2,const_0_25)|
geometry
the area of a square garden is a square feet and the perimeter is p feet . if a = 2 p + 20 , what is the perimeter of the garden , in feet ?
first we need to figure out what numbers are exactly divisible by 7 , 12,10 . this will be the set { lcm , lcmx 2 , lcmx 3 , . . . } lcm ( 7 , 12,10 ) = 42 * 10 = 420 the numbers which will leave remainder 4 will be { 420 + 4 , 420 x 2 + 4 , , . . . } the largest such number less than or equal to 1856 is 420 * 4 + 4 or 1684 to obtain this you need to subtract 172 . b
a ) 168 , b ) 172 , c ) 182 , d ) 140 , e ) 160
b
subtract(1856, add(4, multiply(gcd(1856, lcm(lcm(7, 12), 10)), lcm(lcm(7, 12), 10))))
lcm(n1,n2)|lcm(n3,#0)|gcd(n0,#1)|multiply(#2,#1)|add(n4,#3)|subtract(n0,#4)
general
which is the least number that must be subtracted from 1856 so that the remainder when divided by 7 , 12 , 10 is 4 ?
"average = ( 5 + 10 + 15 ) / 3 = 10 answer is a"
a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 21
a
divide(add(add(add(3, const_1), add(add(3, const_1), const_2)), add(subtract(5, 3), subtract(5, const_2))), 3)
add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)|
general
find the average of first 3 multiples of 5 ?
"explanation : increase in 10 years = ( 242500 - 134800 ) = 107700 increase % = ( 107700 / 134800 x 100 ) % = 79 % . required average = ( 79 / 10 ) % = 7.9 % . answer : option b"
a ) 4.37 % , b ) 7.9 % , c ) 6.8 % , d ) 8.75 % , e ) none
b
add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4)
add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)|
general
the population of a town increased from 1 , 34,800 to 2 , 42,500 in a decade . the average percent increase of population per year is :
"to solve this type of question , simply divide the volume of wall with the volume of brick to get the numbers of required bricks so lets solve this number of bricks = volume of wall / volume of 1 brick = 800 ∗ 600 ∗ 22.5 / 25 ∗ 11.25 ∗ 6 = 6400 answer : a"
a ) 6400 , b ) 3777 , c ) 2679 , d ) 2667 , e ) 1997
a
divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6))
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|
physics
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
"65 % of 160 = x 0.65 * 160 = x 104 = x answer : c"
a ) 84 , b ) 94 , c ) 104 , d ) 114 , e ) 124
c
divide(multiply(65, 160), const_100)
multiply(n0,n1)|divide(#0,const_100)|
gain
a soccer team played 160 games and won 65 percent of them . how many games did it win ?
"explanation : 80 + 25 = 105 / 15 = 7 ( remainder ) d"
a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 9
d
subtract(subtract(subtract(80, 25), const_4), const_2)
subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)|
general
a certain no . when divided by 80 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
"sp of 1 m of cloth = 9000 / 80 = rs . 112.5 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 112.5 - rs . 23.5 = rs . 89 . answer : b"
a ) 22 , b ) 89 , c ) 90 , d ) 78 , e ) 80
b
subtract(divide(9000, 80), 23.5)
divide(n1,n0)|subtract(#0,n2)|
physics
a trader sells 80 meters of cloth for rs . 9000 at the profit of rs . 23.5 per metre of cloth . what is the cost price of one metre of cloth ?
chicken - ch cows - c sheep - s ch + c = 6 s c > ch and c > s each cow has 4 legs and 1 head each chicken has 2 legs and 1 head so 5 c + 3 ch = 100 ( sum of legs and head ) there are 2 possible solutions to this equation c = 11 and ch = 9 or c = 14 and ch = 10 since from first equation where ch + c = 6 s the sum of ch and c should be divisbile by 6 . 20 is not so the only possible solution is c = 14 and ch = 10 . so s = 4 answer : d
a ) 5 , b ) 8 , c ) 10 , d ) 4 , e ) 17
d
subtract(6, const_2)
subtract(n0,const_2)
general
a farm has chickens , cows and sheep . there are 6 times the number of chickens and cows than sheep . if there are more cows than chickens or sheep , and together , cows and chickens have a total of 100 feet and heads , how many sheep live at the farm ?
"say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.75 = 45 democrats are expected to vote for candidate a ; 40 * 0.25 = 10 republicans are expected to vote for candidate a . thus total of 45 + 10 = 55 registered voters are expected to vote for candidate a , which is 55 % of the total number of registered voters . answer : d ."
a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 57 %
d
add(multiply(60, divide(75, const_100)), multiply(subtract(const_100, 60), divide(25, const_100)))
divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4)|
gain
in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 75 percent of the registered voters who are democrats and 25 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are expected to vote for candidate a ?
"let x be the distance to big rock . time = x / 3 + x / 7 = 1 x = 21 / 10 = 2.1 km the answer is c ."
a ) 1.5 , b ) 1.8 , c ) 2.1 , d ) 2.4 , e ) 2.7
c
multiply(divide(subtract(5, 2), add(add(5, 2), subtract(5, 2))), add(5, 2))
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)|
physics
a rower can row 5 km / h in still water . when the river is running at 2 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ?
"24 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 24 = 1 / 100 * x = > x = 24000 answer : c"
a ) 29798 , b ) 27908 , c ) 24000 , d ) 25000 , e ) 27991
c
divide(divide(divide(multiply(multiply(multiply(24, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5)
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n5,const_100)|multiply(#0,#1)|multiply(#3,#2)|divide(#4,n0)|divide(#5,n1)|divide(#6,n2)|
physics
a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 24 m * 2 m * 0.75 m ?
"30 % of the members have passed the test , thus 70 % have not passed the test . we also know that 65 + 26 = 91 members have not passed the test , thus 0.7 * total = 91 - - > total = 130 . answer : d ."
a ) 60 , b ) 80 , c ) 100 , d ) 130 , e ) 140
d
divide(add(26, 65), divide(subtract(const_100, 65), const_100))
add(n0,n1)|subtract(const_100,n1)|divide(#1,const_100)|divide(#0,#2)|
gain
thirty percent of the members of a swim club have passed the lifesaving test . among the members who have not passed the test , 26 have taken the preparatory course and 65 have not taken the course . how many members are there in the swim club ?
"12 x = 16 y = 28 z 3 x = 4 y = 7 z 3 ( 4 * 7 ) = 4 ( 3 * 7 ) = 7 ( 3 * 4 ) addition = 28 + 21 + 12 = 61 answer would be multiple of 61 which is 122 answer : c"
a ) 52 , b ) 58 , c ) 122 , d ) 84 , e ) 168
c
divide(multiply(multiply(16, 28), 12), const_4)
multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_4)|
general
if 12 x = 16 y = 28 z , then what is a possible sum of positive integers x , y , and z ?
"first 100 machines = 3 % commission = 0.03 * 100 * 10000 = 30000 commission from sale of next machines = 34000 - 30000 = 4000 so 10 more machines . . total = 110 machines imo b . ."
a ) 90 , b ) 110 , c ) 105 , d ) 115 , e ) 120
b
add(100, divide(subtract(multiply(multiply(multiply(add(4, 3), multiply(3, const_2)), 100), multiply(add(4, const_1), const_2)), multiply(multiply(multiply(100, 100), divide(3, 100)), 100)), multiply(multiply(100, 100), divide(4, 100))))
add(n0,n2)|add(const_1,n2)|divide(n0,n1)|divide(n2,n1)|multiply(const_2,n0)|multiply(n1,n1)|multiply(#0,#4)|multiply(#1,const_2)|multiply(#2,#5)|multiply(#3,#5)|multiply(#6,n1)|multiply(n1,#8)|multiply(#10,#7)|subtract(#12,#11)|divide(#13,#9)|add(n1,#14)|
gain
a salesperson received a commission of 3 percent of the sale price for each of the first 100 machines that she sold and 4 percent of the sale price for each machine that she sold after the first 100 . if the sale price of each machine was $ 10,000 and the salesperson received a $ 32,000 commission , how many machines did she sell ?
"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 28 = > 2 x = 28 = > x = 14 c"
a ) 11 , b ) 12 , c ) 14 , d ) 16 , e ) 18
c
divide(28, const_2)
divide(n0,const_2)|
general
in a group of ducks and cows , the total number of legs are 28 more than twice the no . of heads . find the total no . of buffaloes .
"explanation : suppose he bought 5 kg and 3 kg of tea . cost price = rs . ( 5 x 18 + 3 x 20 ) = rs . 150 . selling price = rs . ( 8 x 26 ) = rs . 208 . profit = 208 - 150 = 58 so , profit % = ( 58 / 150 ) * 100 = 39 % option b"
a ) 12 % , b ) 39 % , c ) 14 % , d ) 15 % , e ) 16 %
b
divide(multiply(subtract(multiply(26, add(5, 3)), add(multiply(5, 18), multiply(3, 20))), const_100), add(multiply(5, 18), multiply(3, 20)))
add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|add(#1,#2)|multiply(n4,#0)|subtract(#4,#3)|multiply(#5,const_100)|divide(#6,#3)|
gain
a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 26 per kg , then his gain percent is
abel in the 2 days that he worked completed 1 / 5 of the job = 4 / 5 remains then if ben had to leave 4 days before the completion , this means that carla had to work alone for these 4 days in which she completed 1 / 4 of the job . now together , ben and carla completed the job in ( 1 / 12 + 1 / 15 ) ( t ) = 11 / 20 3 / 20 ( t ) = 11 / 20 - - - > t = 3 2 / 3 therefore , these 3 2 / 3 days worked plus the 4 days that carla had to work by herself add to 7 2 / 3 days answer : c
a ) 6 , b ) 7 , c ) 7 2 / 3 , d ) 9 , e ) 10
c
multiply(add(2, 4), 4)
add(n3,n4)|multiply(n4,#0)
physics
abel can complete a work in 10 days , ben in 12 days and carla in 16 days . all of them began the work together , but abel had to leave after 2 days and ben 4 days before the completion of the work . how long did the work last ?
"total paint initially = 360 gallons paint used in the first week = ( 1 / 4 ) * 360 = 90 gallons . remaning paint = 270 gallons paint used in the second week = ( 1 / 4 ) * 270 = 67 gallons total paint used = 157 gallons . option b"
a ) 18 , b ) 157 , c ) 175 , d ) 216 , e ) 250
b
add(multiply(divide(360, 4), 1), divide(subtract(360, multiply(divide(360, 4), 1)), 4))
divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)|
physics
joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 4 of all the paint . during the second week , he uses 1 / 4 of the remaining paint . how many gallons of paint has joe used ?
"0 = 1 * 1 - 1 2 = 2 * 2 - 2 6 = 3 * 3 - 3 12 = 4 * 4 - 4 20 = 5 * 5 - 5 30 = 6 * 6 - 6 42 = 7 * 7 - 7 similarly 8 * 8 - 8 = 56 answer : e"
a ) 55 , b ) 85 , c ) 59 , d ) 63 , e ) 56
e
subtract(negate(20), multiply(subtract(6, 12), divide(subtract(6, 12), subtract(0,2, 6))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
0,2 , 6 , 12 , 20 , 30 , 42 , ___
"distance covered in 1 min = ( 66 * 1000 ) / 60 = 1100 m circumference of the wheel = ( 2 * ( 22 / 7 ) * . 70 ) = 4.4 m no of revolution per min = 1100 / 4.4 = 250 answer : e"
a ) 210 , b ) 220 , c ) 230 , d ) 240 , e ) 250
e
divide(divide(multiply(66, const_1000), const_60), multiply(multiply(divide(add(66, const_2), add(const_4, const_3)), const_2), divide(divide(140, const_100), const_2)))
add(n1,const_2)|add(const_3,const_4)|divide(n0,const_100)|multiply(n1,const_1000)|divide(#3,const_60)|divide(#2,const_2)|divide(#0,#1)|multiply(#6,const_2)|multiply(#5,#7)|divide(#4,#8)|
physics
the diameter of the driving wheel of a bus in 140 cm . how many revolutions per minute must the wheel make in order to keep a speed of 66 kmph ?
"explanation : let the number of hens be x and the number of cows be y . then , x + y = 50 . . . . ( i ) and 2 x + 4 y = 160 x + 2 y = 80 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = 20 , y = 30 the required answer = 20 . answer : d"
a ) 22 , b ) 23 , c ) 24 , d ) 20 , e ) 28
d
divide(subtract(multiply(50, const_4), 160), const_2)
multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|
general
a man has some hens and cows . if the number of heads be 50 and the number of feet equals 160 , then the number of hens will be :
"we ' re told that the number of women in a town is equal to 50 % of the number of men in that town . men = 10 women = 5 we ' re asked for the number of men , as a percentage of the number of women . m / w % = 10 / 5 = 200 % answer is c"
a ) 100 % , b ) 120 % , c ) 200 % , d ) 150 % , e ) 180 %
c
multiply(divide(const_100, 50), const_100)
divide(const_100,n0)|multiply(#0,const_100)|
gain
if a population of women in a town is 50 % of men . what is the population of men as a percentage of population of women ?
"650 = 10 * 65 1300 = 20 * 65 the even multiples are 65 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ."
a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11
b
add(divide(subtract(1301, 649), multiply(65, const_2)), const_1)
multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)|
general
how many even multiples of 65 are there between 649 and 1301 ?
"explanation : relative speed = 70 + 90 = 160 km / hr ( since both trains are moving in opposite directions ) total distance = 1.1 + . 9 = 2 km time = 2 / 160 hr = 1 / 80 hr = 3600 / 80 seconds = = 45 seconds answer : option b"
a ) 56 , b ) 45 , c ) 47 , d ) 26 , e ) 25
b
multiply(divide(add(1.10, 0.9), add(70, 90)), const_3600)
add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(#2,const_3600)|
physics
two trains are moving in opposite directions with speed of 70 km / hr and 90 km / hr respectively . their lengths are 1.10 km and 0.9 km respectively . the slower train cross the faster train in - - - seconds
"0,08 r = x / 100 * 0.1 r answer a"
a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 %
a
multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(10, 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
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 10 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ?
"one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 26 * 2 = 52 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one class ( 52 - 32 = 20 ) . thus we get a number of the available extra - classes for teachers , and all that we need is just to count how many teachers could take 2 more classes , which is 20 / 2 = 10 . so the greatest possible value of the # of teachers who had 3 classes is 10 . only answer c has this option ."
a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 9 , e ) 2 and 8
c
divide(subtract(multiply(26, 2), 32), 2)
multiply(n0,n1)|subtract(#0,n2)|divide(#1,n0)|
general
a certain experimental mathematics program was tried out in 2 classes in each of 26 elementary schools and involved 32 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and greatest possible values of n , respectively , are
"( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 5 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 5 hence x = 7 . answer is a"
a ) 7 , b ) 11 , c ) 13 , d ) 15 , e ) 17
a
add(5, 2)
add(n0,n5)|
general
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 5 ) , what is the value of x ?
"1.8 hectares in ares 1 hectare = 100 ares therefore , 1.8 hectares = 1.8 × 100 ares = 180 ares . answer - c"
a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares .
c
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 1.8), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
convert 1.8 hectares in ares
explanation : number of students appeared from school ' p ' = 100 , say number of students qualified from school ' p ' = 70 and number of students appeared from school ' q ' = 130 number of students qualified from school ' q ' = 50 % more than those qualified from school ' p ' . = 70 + 35 = 105 % of students qualified to the number of students appeared from school b = 105 / 130 * 100 = 80.76 % answer : b
a ) 80.78 % , b ) 80.76 % , c ) 80.72 % , d ) 80.79 % , e ) 80.74 %
b
multiply(divide(multiply(divide(add(50, const_100), const_100), divide(70, const_100)), divide(add(30, const_100), const_100)), const_100)
add(n2,const_100)|add(n1,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#3,#2)|divide(#5,#4)|multiply(#6,const_100)
general
in an examination , the percentage of students qualified to the students appeared from school ' p ' is 70 % . in school ' q ' , the number of students appeared is 30 % more than the students appeared from school ' p ' and the number of students qualified from school ' q ' is 50 % more than the students qualified from school ' p ' . what is the % of students qualified to the number of students appeared from school ' q ' ?
let x be the height of the first person . then the heights are x , x + 2 , x + 4 , and x + 10 . 4 x + 16 = 4 ( 75 ) = 300 x = 71 and the fourth person has a height of 71 + 10 = 81 inches the answer is e .
a ) 73 , b ) 75 , c ) 77 , d ) 79 , e ) 81
e
add(divide(subtract(multiply(75, 4), add(6, add(4, 6))), 4), add(4, 6))
add(n0,n2)|multiply(n0,n3)|add(n2,#0)|subtract(#1,#2)|divide(#3,n0)|add(#0,#4)
general
there are 4 people of different heights standing in order of increasing height . the difference is 2 inches between the first person and the second person , and also between the second person and the third person . the difference between the third person and the fourth person is 6 inches and the average height is 75 . how tall is the fourth person ?
sum of interior angles of a polygon = ( n - 2 ) × 180 ° where n = number of sides there will be n angles which are in a . p . therefore , since we need to find maximum value of n , put minimum value for a and d . i . e . , take a = 20 and d = 20 then sum of the angles = n / 2 [ 2 × 20 + ( n − 1 ) 20 ] therefore , 10 n ( n + 1 ) = ( n − 2 ) 180 n = 14.52 since number of sides must be a positive integer , maximum value of n = 14 answer : b
a ) 12 , b ) 14 , c ) 21 , d ) 25 , e ) cant determine
b
divide(add(sqrt(subtract(power(subtract(20, const_3), const_2), multiply(const_4, multiply(divide(add(multiply(const_10, multiply(const_4, const_2)), const_100), const_10), const_2)))), subtract(20, const_3)), const_2)
multiply(const_2,const_4)|subtract(n0,const_3)|multiply(#0,const_10)|power(#1,const_2)|add(#2,const_100)|divide(#4,const_10)|multiply(#5,const_2)|multiply(#6,const_4)|subtract(#3,#7)|sqrt(#8)|add(#9,#1)|divide(#10,const_2)
general
if the angles of an n sided polygon are in a . p and a > = 20 and d > = 20 then wat is the maximum possible value of n ?
"sum of the 9 numbers = 207 if each number is increased by 4 , the total increase = 4 * 9 = 36 the new sum = 207 + 36 = 243 the new average = 243 / 9 = 27 . answer : c"
a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29
c
multiply(23, 4)
multiply(n1,n2)|
general
the average of 9 numbers is 23 . if each number is increased by 4 , what will the new average be ?
"shopkeeper sells 800 g instead of 1000 g . so , his gain = 1000 - 800 = 200 g . thus , % gain = 200 * 100 ) / 800 = 25 % . answer : option b"
a ) 10 % , b ) 25 % , c ) 11.11 % , d ) 12 % , e ) none of these
b
multiply(divide(add(multiply(const_2, const_100), divide(const_100, const_2)), 800), const_100)
divide(const_100,const_2)|multiply(const_100,const_2)|add(#0,#1)|divide(#2,n0)|multiply(#3,const_100)|
gain
a shopkeeper forced to sell at cost price , uses a 800 grams weight for a kilogram . what is his gain percent ?
"say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 54 . now , plug options for x . only b and e give an integer value for y . but only for b x > y as needed . answer : b ."
a ) 6 , b ) 10 , c ) 8 , d ) 9 , e ) 12
b
add(const_4, add(floor(divide(58, add(const_4, const_3))), const_1))
add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4)|
general
company z has 58 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday .
"four wheeler = 20 * 4 = 80 ( max ) 2 wheel = 1 so no of 4 wheeler = 20 answer : d"
a ) 11 , b ) 12 , c ) 13 , d ) 20 , e ) 25
d
divide(subtract(82, 2), 4)
subtract(n3,n0)|divide(#0,n1)|
general
if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 82 ?
"50 % of x = x / 2 ; 25 % of 2500 = 25 / 100 * 2500 = 625 given that , x / 2 = 625 - 25 = > x / 2 = 600 = > x = 1200 . answer : c"
a ) 1880 , b ) 2160 , c ) 1200 , d ) 8400 , e ) 1210
c
divide(subtract(multiply(2500, divide(25, const_100)), 25), divide(50, const_100))
divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)|
general
if 50 % of x is 25 less than 25 % of 2500 , then x is ?
"30 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 30 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 30 × 25 × 0.6 / 25 × 0.4 = 45 men answerc"
a ) 25 , b ) 30 , c ) 45 , d ) 20 , e ) none of these
c
divide(multiply(30, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10))
divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0)|
physics
a contractor undertakes to built a walls in 50 days . he employs 30 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ?
"pass percentage = 100 - ( 30 + 35 - 35 ) = 100 - 30 = 70 answer : d"
a ) 10 % , b ) 20 % , c ) 30 % , d ) 70 % , e ) 50 %
d
subtract(const_100, subtract(add(30, 35), 35))
add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|
general
in an examination , 30 % of total students failed in hindi , 35 % failed in english and 35 % in both . the percentage of these who passed in both the subjects is :
let the cp = 100 profit = ( 320 / 100 ) × 100 = 320 sp = cp + profit = 100 + 320 = 420 if the cost increases by 25 % , new cp = ( 125 / 100 ) × 100 = 125 selling price is constant , hence new sp = 420 profit = sp – cp = 420 – 125 = 295 required percentage = ( 295 / 420 ) × 100 = 2950 / 42 = 1475 / 21 ≈ 70 answer : e
a ) 180 % , b ) 120 % , c ) 90 % , d ) 80 % , e ) 70 %
e
multiply(divide(subtract(add(320, const_100), add(25, const_100)), add(320, const_100)), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)
gain
in a shop , the profit is 320 % of the cost . if the cost increases by 25 % but the selling price remains constant , find out approximately what percentage of the selling price is the profit ?
total expenditure = 40 + 25 + 15 + 10 = 90 % saving = ( 100 - 90 ) = 10 % 10 / 100 × salary = 1200 , salary = 12000 rs . answer : a
a ) 12000 , b ) 6000 , c ) 8000 , d ) 10000 , e ) none of these
a
divide(multiply(1200, const_100), 10)
multiply(n4,const_100)|divide(#0,n3)
gain
a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 10 % on conveyance . if his savings at the end of the month is rs . 1200 , then his salary per month in rupees is :
"explanation : lot the total number of workers be v then , 8 ooov = ( 12000 * 7 ) + 6000 ( v - 7 ) < = > 2000 v = 42000 < = > v = 21 . answer : b"
a ) 76 , b ) 21 , c ) 26 , d ) 28 , e ) 11
b
add(7, divide(multiply(7, subtract(12000, 8000)), subtract(8000, 6000)))
subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)|
general
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ?
"let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 108 x = 123 answer : a"
a ) 123 , b ) 267 , c ) 277 , d ) 267 , e ) 120
a
divide(add(108, 138), 2)
add(n1,n2)|divide(#0,n0)|
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 ?
"let the third number is x . then first number = ( 100 - 30 ) % of x = 70 % of x = 7 x / 10 second number is ( 63 x / 100 ) difference = 7 x / 10 - 63 x / 100 = 7 x / 10 so required percentage is , difference is what percent of first number ( 7 x / 100 * 10 / 7 x * 100 ) % = 10 % answer : b"
a ) 8 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 12 %
b
subtract(multiply(divide(subtract(37, 30), subtract(const_100, 30)), const_100), const_10)
subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|subtract(#3,const_10)|
gain
two numbers are less than third number by 30 % and 37 % respectively . how much percent is the second number less than by the first
"30 * 18 / 5 = 108 kmph answer : b"
a ) 122 , b ) 108 , c ) 110 , d ) 150 , e ) 100
b
multiply(divide(30, const_1000), const_3600)
divide(n0,const_1000)|multiply(#0,const_3600)|
physics
express 30 mps in kmph ?
"let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 8 = > 2 x = 8 = > x = 4 b"
a ) 5 , b ) 4 , c ) 6 , d ) 3 , e ) 2
b
divide(8, const_2)
divide(n0,const_2)|
general
in a group of ducks and cows , the total number of legs are 8 more than twice the no . of heads . find the total no . of buffaloes .
"for first 2000 meters he does not get any discount . the price is 2 * 2000 = $ 4000 for next 1500 meters , he gets a 5 % discount . the price is 1.9 * 1500 = $ 2850 for the next 1500 meters , he gets a 7 % discount . the price is 1.86 * 1500 = $ 2790 the total price is $ 4000 + $ 2850 + $ 2790 = $ 9640 the answer is e ."
a ) $ 8280 , b ) $ 8520 , c ) $ 8710 , d ) $ 8930 , e ) $ 9640
e
multiply(multiply(2, const_3), const_100)
multiply(n4,const_3)|multiply(#0,const_100)|
gain
a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 5,000 meters of fabric ?
"log 8 x + log 8 ( 1 / 6 ) = 1 / 3 = > ( log x / log 8 ) + ( log 1 / 6 / log 8 ) = log ( 81 / 3 ) = log 2 = > log x = log 2 – log 1 / 6 = log ( 2 * 6 / 1 ) = log 12 answer : a"
a ) 12 , b ) 16 , c ) 18 , d ) 24 , e ) 26
a
multiply(power(8, divide(1, 3)), 6)
divide(n2,n5)|power(n0,#0)|multiply(n3,#1)|
general
if log 8 x + log 8 1 / 6 = 1 / 3 , then the value of x is :
"sol . let the numbers be x and y . then , ( x + y ) = 12 and x 2 + y 2 = 124 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 12 ) 2 - 124 = 144 - 124 = 20 xy = 10 . answer a"
a ) 10 , b ) 44 , c ) 80 , d ) 88 , e ) 90
a
divide(subtract(power(12, const_2), 124), const_2)
power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|
general
if the sum of two numbers is 12 and the sum of their squares is 124 , then the product of the numbers is
"c . p . = rs . 5 x and s . p . = rs . 8 x . then , gain = rs . 3 x required ratio = 3 x : 5 x = 3 : 5 d"
a ) 23 , b ) 1 : 2 , c ) 2 : 5 , d ) 3 : 5 , e ) 25
d
divide(subtract(8, 5), 5)
subtract(n0,n1)|divide(#0,n1)|
other
the ratio between the sale price and the cost price of an article is 8 : 5 . what is the ratio between the profit and the cost price of that article ?
"b 90 sum = ( n x n ) + n hence , 9 x 9 = 81 + 9 = 90"
a ) 80 , b ) 90 , c ) 30 , d ) 70 , e ) 60
b
multiply(add(9, const_1), 9)
add(n0,const_1)|multiply(n0,#0)|
physics
tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 9 even numbers . his mother immediately worked out the answer . how much had he scored in math ?
"line k passes through the origin and has slope 1 / 7 means that its equation is y = 1 / 7 * x . thus : ( x , 1 ) = ( 7 , 1 ) and ( 7 , y ) = ( 7,1 ) - - > x - y = 7 - 1 = 6 . answer : a"
a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 3
a
multiply(multiply(7, 7), divide(1, 7))
divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)|
general
in the coordinate plane , points ( x , 1 ) and ( 7 , y ) are on line k . if line k passes through the origin and has slope 1 / 7 , then x - y =
"length = 15 cm , breadth = 13 cm perimeter of rectangle = 2 ( length + breadth ) = 2 ( 15 + 13 ) cm = 2 × 28 cm = 56 cm we know that the area of rectangle = length × breadth = ( 15 × 13 ) cm 22 = 195 cm 2 answer : d"
a ) 71 cm 2 , b ) 121 cm 2 , c ) 141 cm 2 , d ) 195 cm 2 , e ) 221 cm 2
d
square_area(15)
square_area(n0)|
geometry
find the perimeter and area of the rectangle of length 15 cm and breadth 13 cm .
"n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 400 - 50 = 350 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 350 = 150 . answer is b"
a ) 120 , b ) 150 , c ) 100 , d ) 180 , e ) 220
b
subtract(add(175, 325), subtract(400, 50))
add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|
other
out of 400 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ?
"sum of first 20 multiples of 12 are = ( 12 × 1 ) + ( 12 × 2 ) + ( 12 × 3 ) + . . . . . . + ( 12 × 19 ) + ( 12 × 20 ) . = 12 ( 1 + 2 + 3 + . . . . . + 20 ) use the formula : n ( n + 1 ) 2 ⇒ 12 × ( 20 × 21 ) 2 = 2520 . answer : a"
a ) 2520 , b ) 3878 , c ) 2778 , d ) 27 , e ) 911
a
add(divide(divide(20, divide(divide(divide(divide(divide(20, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(20, divide(divide(divide(divide(divide(20, const_2), const_2), const_2), const_2), const_2)), const_2))))
divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)|
general
find the sum of first 20 multiples of 12 .
total no . of handshakes = 49 + 48 + 47 + . . . + 3 + 2 + 1 = 19 * ( 19 + 1 ) / 2 = 1225 or , if there are n persons then no . of shakehands = nc 2 = 50 c 2 = 1225 answer : c
a ) 190 , b ) 200 , c ) 1225 , d ) 220 , e ) 230
c
multiply(subtract(50, const_1), divide(50, const_2))
divide(n0,const_2)|subtract(n0,const_1)|multiply(#0,#1)
general
50 men shake hands with each other . maximum no of handshakes without cyclic handshakes .
"since the population increases at the rate of 1 person every 15 seconds , it increases by 4 people every 60 seconds , that is , by 4 people every minute . thus , in 10 minutes the population increases by 10 x 4 = 40 people . answer . a ."
a ) 40 , b ) 100 , c ) 150 , d ) 240 , e ) 300
a
multiply(divide(const_60, 15), 10)
divide(const_60,n0)|multiply(n1,#0)|
physics
if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 10 minutes ?
"let the retail price be = x selling price of z = 0.75 x selling price of x = 0.95 * 0.75 x = 0.71 x selling price of y = ( ( 0.75 x + 0.71 x ) / 2 ) * 0.70 = 0.73 x * 0.75 = 0.55 x 0.55 x = k * 0.71 x k = 0.55 / 0.71 = 55 / 71 answer : b"
a ) 21 / 34 , b ) 55 / 71 , c ) 25 / 34 , d ) 26 / 34 , e ) 27 / 34
b
multiply(divide(divide(multiply(divide(add(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(subtract(const_100, 5), const_100))), const_2), subtract(const_100, 30)), const_100), multiply(subtract(const_100, 25), divide(subtract(const_100, 5), const_100))), const_10)
subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10)|
general
real - estate salesman z is selling a house at a 25 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 5 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 30 percent discount . salesman y ' s final price is what fraction of salesman x ' s final price ?
"p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 3 / 5 . 2 / 5 p ( a n b ) = 6 / 25 . a"
a ) 6 / 25 , b ) 3 / 25 , c ) 8 / 25 , d ) 2 / 13 , e ) 3 / 17
a
multiply(divide(3, 5), divide(2, 5))
divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|
general
if p ( a ) = 3 / 5 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events .
"on the first three flips , you must get heads . whats the probability of getting heads ? its 1 / 2 so for the first three flips , your probability is ( 1 / 2 ) ^ 3 = 1 / 8 now for the last two , you want to get tails only . whats the prob of getting tails ? well , its the same as prob of getting a heads , namely , 1 / 2 for the last two flips , your probability is ( 1 / 2 ) ^ 2 = 1 / 4 so your overall probability for the event in question is 1 / 8 * 1 / 4 = 1 / 32 answer : e"
a ) 3 / 5 , b ) 1 / 2 , c ) 1 / 5 , d ) 1 / 8 , e ) 1 / 32
e
power(divide(1, 2), 5)
divide(n0,n1)|power(#0,n2)|
probability
if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips ?
"x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] q = z - x = x + 9 - x = 9 ans : d"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
d
add(add(5, const_2), const_2)
add(n0,const_2)|add(#0,const_2)|
general
if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value q of z - x ?
originally let there be x men . less men , more days ( indirect ) : . ( x - 10 ) : x : : 100 : 110 or x - 10 / x = 100 / 110 or 11 x - 110 = 10 x or x = 110 so , originally there were 110 men . answer : d
a ) 75 , b ) 82 , c ) 100 , d ) 110 , e ) 120
d
divide(multiply(divide(add(100, 10), 10), 10), subtract(divide(add(100, 10), 10), 10))
add(n0,n1)|divide(#0,n1)|multiply(n1,#1)|subtract(#1,n1)|divide(#2,#3)
physics
a piece of work can finish by a certain number of men in 100 days . if however , there were 10 men less , it would take 10 days more for the work to be finished . how many men were there originally ?
d = 150 + 200 = 350 s = 54 * 5 / 18 = 15 mps t = 350 / 15 = 23.3 sec c ) 23.3 sec
a ) 17 sec , b ) 21 sec , c ) 23.3 sec , d ) 27.5 sec , e ) 29 sec
c
divide(add(200, 150), multiply(54, const_0_2778))
add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)
physics
how many seconds will a train 150 meters long take to cross a bridge 200 meters long if the speed of the train is 54 kmph ?
"percentage error in calculated area = ( 6 + 6 + ( 6 ã — 6 ) / 100 ) % = 12.36 % answer : e"
a ) 14.05 % , b ) 14.02 % , c ) 14 % , d ) 13 % , e ) 12.36 %
e
divide(multiply(subtract(square_area(add(const_100, 6)), square_area(const_100)), const_100), square_area(const_100))
add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)|
gain
an error 6 % in excess is made while measuring the side of a square . what is the percentage of error in the calculated area of the square ?
distance is same s 1 t 1 = s 2 t 2 40 t = 50 ( t - 1 ) t = 5 distance = speed * time 40 * 5 = 200 answer : b
a ) 150 , b ) 200 , c ) 450 , d ) 500 , e ) 600
b
multiply(divide(50, subtract(50, 40)), 40)
subtract(n1,n0)|divide(n1,#0)|multiply(n0,#1)
physics
liz drove from point a to point b at 40 km / h . on her way back she drove at 50 km / h and therefore her way back lasted one hour less . what is the distance ( in km ) between a and b ?
"first , you must find the total weight of the mixture given that 80 % of it will be dough . 80 % * total = 36 = > ( 8 / 10 ) total = 36 = > total = 360 / 8 = > total = 45 oz , from there , you must find 10 % of the total 40 oz of the mixture . 20 % * total = > ( 2 / 10 ) ( 45 ) = 9 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 10 - 9 = 1 oz chocolate left over . answer : e"
a ) 2 , b ) 4 , c ) 6 , d ) 3 , e ) 1
e
multiply(divide(20, const_100), 20)
divide(n2,const_100)|multiply(n2,#0)|
gain
uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 10 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 20 % chocolate ?
"a / b = 4 / 5 m / x = ( 2 / 5 ) * 5 / ( 7 / 4 ) * 4 = 2 / 7 the answer is e ."
a ) 2 / 5 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 4 , e ) 2 / 7
e
multiply(divide(subtract(const_100, 60), add(const_100, 75)), divide(5, 4))
add(n2,const_100)|divide(n1,n0)|subtract(const_100,n3)|divide(#2,#0)|multiply(#3,#1)|
general
the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 75 percent of a , and m equals b decreased by 60 percent of b , what is the value of m / x ?
ratio of 2 bedroom apartment : 1 bedroom apartment = 700 : 2100 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 4 ) * 100 - - - > 25 % answer : a
a ) 25 % , b ) 15 % , c ) 20 % , d ) 40 % , e ) 45 %
a
divide(multiply(700, const_100), add(add(multiply(const_2, const_1000), const_100), 700))
multiply(n0,const_100)|multiply(const_1000,const_2)|add(#1,const_100)|add(n0,#2)|divide(#0,#3)
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 $ 2,100 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?
"actual miles / gallon is = 480 / 4 = 12 miles / gallon . current engine miles / gallon is 8 miles / gallon . additional 4 miles / gallon is required to match the actual mileage . answer : b"
a ) 2 , b ) 4 , c ) 12 , d ) 40 , e ) 160
b
subtract(divide(480, 40), 8)
divide(n1,n0)|subtract(#0,n2)|
physics
the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 480 miles away . the pilot recognizes that the current engine , which can fly only 8 miles per gallon , will not get him there . by how many miles per gallon must the aircraft ’ s fuel efficiency be improved to make the flight to cleveland possible ?
"age of the teacher = ( 23 * 13 - 22 * 12 ) = 35 years . answer : c"
a ) 31 , b ) 36 , c ) 35 , d ) 53 , e ) 57
c
add(22, const_1)
add(n0,const_1)|
general