Problem
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628
| Rationale
stringlengths 1
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| options
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| correct
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848
| linear_formula
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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 ?
|
"1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 5 = 7200 . answer : c"
|
a ) 7 , b ) 10 % , c ) 786858 , d ) 22.4 hours , e ) 7200
|
e
|
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
|
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 ?
|
"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 ) 44 seconds , b ) 20 minutes , c ) 50 , d ) 2 , e ) 40
|
c
|
subtract(560, multiply(85, 6))
|
multiply(n0,n1)|subtract(n3,#0)|
|
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 ?
|
"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 ) 8985 , b ) 6666 , c ) 48 , d ) 28.57 % , e ) 165
|
c
|
divide(subtract(divide(20, divide(10, const_100)), 72), 2)
|
divide(n3,const_100)|divide(n6,#0)|subtract(#1,n4)|divide(#2,n1)|
|
general
|
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 ?
|
"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 ) 36 + 36 * sqrt ( , b ) 7 % , c ) 4 % , d ) 405 meter , e ) 72
|
b
|
divide(multiply(subtract(1498, 1400), const_100), 1400)
|
subtract(n1,n0)|multiply(#0,const_100)|divide(#1,n0)|
|
gain
|
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 ?
|
"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 ) 10 : 3 , b ) 3 , c ) 7.94 , d ) 0.0375 days , e ) 3 / 4
|
c
|
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 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 :
|
"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 ) 130 , b ) 350.5 , c ) rs . 2000 , d ) 35 : 44 , e ) 0.05
|
d
|
divide(multiply(105, const_2), multiply(88, const_3))
|
multiply(n1,const_2)|multiply(n3,const_3)|divide(#0,#1)|
|
other
|
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 ?
|
"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 ) it can not be determined from the information given , b ) 12,526 , c ) 555681 , d ) 15 / 2 minutes , e ) 12
|
d
|
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
|
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 ?
|
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 ) 125 , c ) 1173.98 , d ) 37.5 % , e ) 6825
|
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
|
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 ?
|
"sqrt ( 7 x / 3 ) to be perfect square x has to 7 / 3 ans : b"
|
a ) 75 % , b ) 7 / 3 , c ) 48 days , d ) 50 % , e ) 180 m
|
b
|
divide(7, 3)
|
divide(n0,n1)|
|
general
|
how many of the positive divisors of 240 are also multiples of 4 not including 240 ?
|
"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 ) 11 , b ) rs . 126 , c ) 6 : 1 , d ) 3 , e ) 190
|
a
|
divide(divide(divide(240, 4), const_2), const_3)
|
divide(n0,n1)|divide(#0,const_2)|divide(#1,const_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 ?
|
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 ) $ 37.8 , b ) 36 / 31 , c ) 200 , d ) $ 20 , e ) 50
|
b
|
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
|
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 ?
|
"0.06 * 0.05 = 0.003 = 0.3 % the answer is b ."
|
a ) 1 , b ) 5.5 mph , c ) 40 , d ) 0.3 % , e ) 3 .
|
d
|
multiply(6, divide(5, const_100))
|
divide(n1,const_100)|multiply(n0,#0)|
|
gain
|
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 .
|
"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 ) 14 , b ) 2 / 9 , c ) 9.87 % , d ) 13 , e ) $ 24
|
d
|
divide(26, const_2)
|
divide(n0,const_2)|
|
general
|
a man walking at 3 / 4 th of the speed , reaches his office late by 2 hours . what is the usual time ?
|
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 ) 10.6 , b ) 200 , c ) 10 % , d ) 6 hours , e ) 10.8 km / hr
|
d
|
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
|
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 .
|
because a + b = 1 , 2 a + 2 b = 2 ( a + b ) = 2 × 1 = 2 . correct answer d
|
a ) 192 , b ) 128 , c ) 2 , d ) 16 days , e ) 4 / 9
|
c
|
subtract(add(add(2, 1), 2), add(2, 1))
|
add(n0,n1)|add(n1,#0)|subtract(#1,#0)
|
general
|
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 ?
|
"let the sums be p . now , 45 % of p = 4016.25 or , p = 8925 answer a"
|
a ) 3280 , b ) 8925 , c ) 20 , d ) $ 1.80 , e ) 9
|
b
|
divide(multiply(const_100, 4016.25), multiply(9, 5))
|
multiply(n0,const_100)|multiply(n1,n2)|divide(#0,#1)|
|
gain
|
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 ?
|
"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 ) 2220 , b ) $ 154.1 , c ) 432 , d ) 11.1 % , e ) 135 %
|
a
|
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
|
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 ) ]
|
"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 ) 2 , b ) 70400 yards , c ) 6400 , d ) 9 / 7 , e ) 55 cm 2
|
d
|
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
|
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 ?
|
"other number = ( 11 * 7700 ) / 275 = 308 . answer : c"
|
a ) 0.6 , b ) 445 , c ) 391 , d ) 60 , e ) 308
|
e
|
multiply(11, 275)
|
multiply(n0,n2)|
|
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 ?
|
"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 ) 80 , b ) 7 , c ) 21 , d ) 5 ⁄ 3 , e ) 50 %
|
c
|
add(add(11, divide(subtract(11, 5), 2)), 7)
|
subtract(n4,n0)|divide(#0,n1)|add(n4,#1)|add(n2,#2)|
|
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 ?
|
"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 ) 1 / 2 , b ) 30 , c ) 54 : 53 , d ) 3108 , e ) 200
|
b
|
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
|
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 ?
|
"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 ) 1 / 15 , b ) 6 / 5 , c ) 7 , d ) 10 , e ) 3
|
d
|
add(divide(17, const_2), 1)
|
divide(n1,const_2)|add(n0,#0)|
|
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 ?
|
sum of last 4 matches = ( ( 10 × 45 ) – ( 6 × 48 ) = 162 average = 162 / 4 = 40.5 answer : d
|
a ) 0.25 , b ) 151 / 31 , c ) 3 : 5 , d ) 40.5 , e ) 85
|
d
|
divide(subtract(multiply(45, 10), multiply(6, 48)), 4)
|
multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)
|
general
|
a shopkeeper fixes the marked price of an item 40 % above its cost price . the percentage of discount allowed to gain 8 % is
|
"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 ) rs . 145 , b ) 22.85 % , c ) 5 : 7 , d ) 15 , e ) 20
|
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
|
how long does a train 60 m long travelling at 60 kmph takes to cross a bridge of 80 m in length ?
|
"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 ) 16 , b ) 8.4 sec , c ) 49 , d ) 30 , e ) 642
|
b
|
divide(add(60, 80), multiply(60, const_0_2778))
|
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
|
physics
|
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 ?
|
"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 ) 6400 , b ) 6 days , c ) 90 ° , d ) 27 , e ) 5
|
c
|
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
|
how many integers are between 5 and 96 / 7 , inclusive ?
|
"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 ) 27 / 19 , b ) 576 cm 2', ' , c ) 9 , d ) rs . 145 , e ) 975
|
c
|
add(subtract(divide(96, 7), 5), const_1)
|
divide(n1,n2)|subtract(#0,n0)|add(#1,const_1)|
|
general
|
simplify : 256 x 256 - 144 x 144
|
"( 256 ) ^ 2 - ( 144 ) ^ 2 = ( 256 + 144 ) ( 256 - 144 ) = 400 x 112 = 44800 answer is b"
|
a ) 195 cm 2 , b ) s : 1300 , c ) 42 , d ) 44800 , e ) 120 miles
|
d
|
add(multiply(256, 256), multiply(144, 144))
|
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
|
general
|
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 ?
|
"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 ) 40 , b ) 80 , c ) 54 °', ' , d ) 5 kmph , e ) 0.358
|
a
|
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
|
which is the least number that must be subtracted from 1856 so that the remainder when divided by 7 , 12 , 10 is 4 ?
|
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 ) 172 , b ) 8 and 20 , c ) 6 km , d ) 66.67 % , e ) 7
|
a
|
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
|
find the average of first 3 multiples of 5 ?
|
"average = ( 5 + 10 + 15 ) / 3 = 10 answer is a"
|
a ) 67 % , b ) 6666 , c ) 10 , d ) 60 , e ) 2
|
c
|
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
|
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 :
|
"explanation : increase in 10 years = ( 242500 - 134800 ) = 107700 increase % = ( 107700 / 134800 x 100 ) % = 79 % . required average = ( 79 / 10 ) % = 7.9 % . answer : option b"
|
a ) 5 cm , b ) 20 , c ) 4 % , d ) 94 , e ) 7.9 %
|
e
|
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
|
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
|
"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 ) 16 , b ) 21 % , c ) $ 54,000 , d ) 6400 , e ) 315
|
d
|
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
|
a soccer team played 160 games and won 65 percent of them . how many games did it win ?
|
"65 % of 160 = x 0.65 * 160 = x 104 = x answer : c"
|
a ) 2 , b ) 104 , c ) 870 , d ) 315 , e ) 88888883
|
b
|
divide(multiply(65, 160), const_100)
|
multiply(n0,n1)|divide(#0,const_100)|
|
gain
|
a certain no . when divided by 80 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
|
"explanation : 80 + 25 = 105 / 15 = 7 ( remainder ) d"
|
a ) 5 , b ) 12 , c ) $ 54,000 , d ) 7 , e ) 30
|
d
|
subtract(subtract(subtract(80, 25), const_4), const_2)
|
subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)|
|
general
|
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 ?
|
"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 ) 1.8 , b ) 69000 , c ) 7 / 3 , d ) 3 : 2 , e ) 89
|
e
|
subtract(divide(9000, 80), 23.5)
|
divide(n1,n0)|subtract(#0,n2)|
|
physics
|
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 ?
|
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 ) 3 / 4 , b ) 0.7 , c ) 8 , d ) 7 , e ) 4
|
e
|
subtract(6, const_2)
|
subtract(n0,const_2)
|
general
|
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 ?
|
"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 ) 55 % , b ) 0 % , c ) 10 ° , d ) 21 , e ) 101 kg
|
a
|
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
|
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 ?
|
"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 ) 68.75 % , b ) 2 sqrt ( 6 , c ) 2.1 , d ) 996004 , e ) 5.75
|
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 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 ?
|
"24 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 24 = 1 / 100 * x = > x = 24000 answer : c"
|
a ) 24000 , b ) 7 / 15 , c ) 5 , d ) 8 , e ) 10.6
|
a
|
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
|
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 ?
|
"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 ) 30 % , b ) 130 , c ) rs . 6725 . , d ) $ 1.96 , e ) 39
|
b
|
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
|
if 12 x = 16 y = 28 z , then what is a possible sum of positive integers x , y , and z ?
|
"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 ) $ 120 , b ) 500 , c ) 122 , d ) $ 1,440 , e ) 65.80 %
|
c
|
divide(multiply(multiply(16, 28), 12), const_4)
|
multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_4)|
|
general
|
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 ?
|
"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 ) 110 , b ) $ 620 , c ) 252 , d ) 36', ' , e ) 88.2 %
|
a
|
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
|
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 .
|
"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 ) 14 , b ) 220010 , c ) 3 , d ) 10 v 2', ' , e ) 544
|
a
|
divide(28, const_2)
|
divide(n0,const_2)|
|
general
|
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
|
"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 ) 6 , b ) 39 % , c ) 5 , d ) 14 , e ) 25 days
|
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
|
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 ?
|
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 ) 1014 , b ) 70 , c ) 7 2 / 3 , d ) 70 m , e ) 20 hrs
|
c
|
multiply(add(2, 4), 4)
|
add(n3,n4)|multiply(n4,#0)
|
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 ?
|
"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 ) 1 / 13 , b ) 2.5 , c ) 157 , d ) 2 / 5 , e ) 120
|
c
|
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
|
0,2 , 6 , 12 , 20 , 30 , 42 , ___
|
"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 ) 6 , b ) 56 , c ) 2,120 , d ) 36 , e ) 481
|
b
|
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
|
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 ?
|
"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 ) 26250 , b ) 17 hr , c ) 12 % , d ) 6.7 kg . , 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
|
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 :
|
"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 ) 1.25 hours , b ) 42 , c ) 20 , d ) 18 feet , e ) 548
|
c
|
divide(subtract(multiply(50, const_4), 160), const_2)
|
multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|
|
general
|
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 ?
|
"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 ) 7 , b ) 552 km , c ) 1 / 81 , d ) 45 , e ) 200 %
|
e
|
multiply(divide(const_100, 50), const_100)
|
divide(const_100,n0)|multiply(#0,const_100)|
|
gain
|
how many even multiples of 65 are there between 649 and 1301 ?
|
"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 ) 3 , b ) 1804 , c ) 6 , d ) 14 , e ) 75 %
|
c
|
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
|
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
|
"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 ) 5 : 8 , b ) 4 , c ) 45 , d ) 650 , e ) rs . 145
|
c
|
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
|
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 ?
|
"0,08 r = x / 100 * 0.1 r answer a"
|
a ) 12 % , b ) 5 pm . , c ) 5 % , d ) 10,800 , e ) 80 %
|
e
|
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
|
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
|
"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 ) 1 and 10 , b ) 1 , c ) - 4 , d ) 60 , e ) 7 / 15
|
a
|
divide(subtract(multiply(26, 2), 32), 2)
|
multiply(n0,n1)|subtract(#0,n2)|divide(#1,n0)|
|
general
|
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 5 ) , what is the value of x ?
|
"( 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 ) 60 sec , c ) 144 , d ) 1235 , e ) $ 7717.50
|
a
|
add(5, 2)
|
add(n0,n5)|
|
general
|
convert 1.8 hectares in ares
|
"1.8 hectares in ares 1 hectare = 100 ares therefore , 1.8 hectares = 1.8 × 100 ares = 180 ares . answer - c"
|
a ) 180 ares . , b ) 45 , c ) 312 , d ) 462 cm ²', ' , e ) 20
|
a
|
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
|
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 ' ?
|
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 ) 5 hours , b ) 9 , c ) 1.11 % , d ) 80.76 % , e ) 31 st
|
d
|
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
|
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 ?
|
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 ) $ 600 , b ) 96 % , c ) 70 , d ) 4966 , 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
|
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 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 ) 7550 , b ) 60 , c ) 14 , d ) 6 , e ) 0.28 %
|
c
|
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
|
the average of 9 numbers is 23 . if each number is increased by 4 , what will the new average be ?
|
"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 ) 999936 , b ) 40 miles . , c ) 21 - 61 , d ) 24 days . , e ) 27
|
e
|
multiply(23, 4)
|
multiply(n1,n2)|
|
general
|
a shopkeeper forced to sell at cost price , uses a 800 grams weight for a kilogram . what is his gain percent ?
|
"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 ) 200 , b ) rs . 1350 , c ) 25 % , d ) 40 , e ) 180
|
c
|
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
|
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 .
|
"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 ) 33.33 % , b ) 240000 , c ) 10 , d ) 6 hours , e ) 4.8
|
c
|
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
|
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 ?
|
"four wheeler = 20 * 4 = 80 ( max ) 2 wheel = 1 so no of 4 wheeler = 20 answer : d"
|
a ) 9 , b ) 20 , c ) 2520', ' , d ) 8915 , e ) 12
|
b
|
divide(subtract(82, 2), 4)
|
subtract(n3,n0)|divide(#0,n1)|
|
general
|
if 50 % of x is 25 less than 25 % of 2500 , then x is ?
|
"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 ) 28 min , b ) 15552 , c ) 3 , d ) s . 120 , e ) 1200
|
e
|
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
|
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 ?
|
"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 ) 1122 , b ) 45 , c ) 60 , d ) 8 , e ) 150
|
b
|
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
|
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 :
|
"pass percentage = 100 - ( 30 + 35 - 35 ) = 100 - 30 = 70 answer : d"
|
a ) 70 % , b ) 46 , c ) 60 , d ) 8 , e ) 90
|
a
|
subtract(const_100, subtract(add(30, 35), 35))
|
add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)|
|
general
|
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 ?
|
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 ) 1 / 9 , b ) 17 , c ) 70 % , d ) 56 , e ) 27
|
c
|
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
|
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 :
|
total expenditure = 40 + 25 + 15 + 10 = 90 % saving = ( 100 - 90 ) = 10 % 10 / 100 × salary = 1200 , salary = 12000 rs . answer : a
|
a ) 198', ' , b ) 8 , c ) 12000 , d ) 42 , e ) 27 / 20
|
c
|
divide(multiply(1200, const_100), 10)
|
multiply(n4,const_100)|divide(#0,n3)
|
gain
|
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 ?
|
"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 ) 60 , b ) 178 , c ) 21 , d ) rs . 98.56 , e ) $ 20,800
|
c
|
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
|
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 xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 108 x = 123 answer : a"
|
a ) 4 , b ) 12 , c ) 11 years , d ) 123 , e ) 24 .
|
d
|
divide(add(108, 138), 2)
|
add(n1,n2)|divide(#0,n0)|
|
general
|
two numbers are less than third number by 30 % and 37 % respectively . how much percent is the second number less than by the first
|
"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 ) 10 % , b ) 20 % , c ) 900 , d ) 5.6 , e ) 9
|
a
|
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
|
express 30 mps in kmph ?
|
"30 * 18 / 5 = 108 kmph answer : b"
|
a ) 4 years , b ) 108 , c ) 400 , d ) 7.5 , e ) 800
|
b
|
multiply(divide(30, const_1000), const_3600)
|
divide(n0,const_1000)|multiply(#0,const_3600)|
|
physics
|
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 .
|
"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 ) 2 , b ) 75 minutes , c ) 83 % , d ) 2520 , e ) 4
|
e
|
divide(8, const_2)
|
divide(n0,const_2)|
|
general
|
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 ?
|
"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 ) $ 9640 , b ) 27 / 128 , c ) 24 , d ) 4.8 , e ) - 5
|
a
|
multiply(multiply(2, const_3), const_100)
|
multiply(n4,const_3)|multiply(#0,const_100)|
|
gain
|
if log 8 x + log 8 1 / 6 = 1 / 3 , then the value of x is :
|
"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 ) 4 , c ) 7 , d ) 9.9 % , e ) 23 years
|
a
|
multiply(power(8, divide(1, 3)), 6)
|
divide(n2,n5)|power(n0,#0)|multiply(n3,#1)|
|
general
|
if the sum of two numbers is 12 and the sum of their squares is 124 , then the product of the numbers 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 ) 20 , b ) 9 / 19 , c ) 10 , d ) 308 , e ) 40
|
c
|
divide(subtract(power(12, const_2), 124), const_2)
|
power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|
|
general
|
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 ?
|
"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 ) 1 / 24 , b ) 3 : 5 , c ) 46 , d ) 2991 , e ) 982.14', '
|
b
|
divide(subtract(8, 5), 5)
|
subtract(n0,n1)|divide(#0,n1)|
|
other
|
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 ?
|
"b 90 sum = ( n x n ) + n hence , 9 x 9 = 81 + 9 = 90"
|
a ) 100 days , b ) 12.5 , c ) 50 km , d ) 12 kmph , e ) 90
|
e
|
multiply(add(9, const_1), 9)
|
add(n0,const_1)|multiply(n0,#0)|
|
physics
|
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 =
|
"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 ) $ 1600 , b ) 6 , c ) 3 , d ) 9.09 % , e ) 4
|
b
|
multiply(multiply(7, 7), divide(1, 7))
|
divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)|
|
general
|
find the perimeter and area of the rectangle of length 15 cm and breadth 13 cm .
|
"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 ) 10 , b ) 250 , c ) 8.33 km , d ) 3 , e ) 195 cm 2
|
e
|
square_area(15)
|
square_area(n0)|
|
geometry
|
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 ?
|
"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 ) 150 , b ) $ . 90 , c ) s : 14172.48 , d ) 21 % , e ) 3.75
|
a
|
subtract(add(175, 325), subtract(400, 50))
|
add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)|
|
other
|
find the sum of first 20 multiples of 12 .
|
"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 ) 9 , b ) 101 kg , c ) 10750 , d ) 2520 , e ) 6
|
d
|
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
|
50 men shake hands with each other . maximum no of handshakes without cyclic handshakes .
|
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 ) 28 , b ) 1 / 2 , c ) 20 , d ) 200 % , e ) 1225
|
e
|
multiply(subtract(50, const_1), divide(50, const_2))
|
divide(n0,const_2)|subtract(n0,const_1)|multiply(#0,#1)
|
general
|
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 ?
|
"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 ) 2 , b ) 1 : 49 , c ) 21 , d ) 40 , e ) 8
|
d
|
multiply(divide(const_60, 15), 10)
|
divide(const_60,n0)|multiply(n1,#0)|
|
physics
|
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 ?
|
"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 ) 16 days , b ) 21.48 hours , c ) 323.2 , d ) 4.37 sec , e ) 55 / 71
|
e
|
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
|
if p ( a ) = 3 / 5 and p ( b ) = 2 / 5 , find p ( a n b ) if a and b are independent events .
|
"p ( a n b ) = p ( a ) . p ( b ) p ( a n b ) = 3 / 5 . 2 / 5 p ( a n b ) = 6 / 25 . a"
|
a ) 55 % , b ) 192 , c ) $ 1,240 , d ) 6 / 25 , e ) 60 %
|
d
|
multiply(divide(3, 5), divide(2, 5))
|
divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|
|
general
|
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 ?
|
"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 ) 0 , b ) 1 / 32 , c ) 3 b , d ) 82 , e ) 309400
|
b
|
power(divide(1, 2), 5)
|
divide(n0,n1)|power(#0,n2)|
|
probability
|
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 ?
|
"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 ) 192 , b ) 9 , c ) − 120 , d ) 30 , e ) 5 / 4
|
b
|
add(add(5, const_2), const_2)
|
add(n0,const_2)|add(#0,const_2)|
|
general
|
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 ?
|
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 ) 40 days , b ) 24 , c ) 110 , d ) 58 , e ) 30
|
c
|
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
|
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 ?
|
d = 150 + 200 = 350 s = 54 * 5 / 18 = 15 mps t = 350 / 15 = 23.3 sec c ) 23.3 sec
|
a ) $ 10,570 , b ) 23.3 sec , c ) 244907.04 m , d ) 100 , e ) 20
|
b
|
divide(add(200, 150), multiply(54, const_0_2778))
|
add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)
|
physics
|
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 ?
|
"percentage error in calculated area = ( 6 + 6 + ( 6 ã — 6 ) / 100 ) % = 12.36 % answer : e"
|
a ) 50 , b ) 331 times , c ) 150 , d ) 12.36 % , e ) 0.0012
|
d
|
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
|
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 ?
|
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 ) 146 1 / 7 min , b ) 200 , c ) 7 , d ) 40 , e ) 4
|
b
|
multiply(divide(50, subtract(50, 40)), 40)
|
subtract(n1,n0)|divide(n1,#0)|multiply(n0,#1)
|
physics
|
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 ?
|
"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 ) 1 , b ) 49 , c ) 180 ares . , d ) 273054 , e ) 10
|
a
|
multiply(divide(20, const_100), 20)
|
divide(n2,const_100)|multiply(n2,#0)|
|
gain
|
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 ?
|
"a / b = 4 / 5 m / x = ( 2 / 5 ) * 5 / ( 7 / 4 ) * 4 = 2 / 7 the answer is e ."
|
a ) 2 / 7 , b ) 266 cm 2 , c ) 14 , d ) 113 , e ) 40
|
a
|
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
|
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 ?
|
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 ) 12 hours , b ) rs . 6725 . , c ) 25 % , d ) 27 sec , e ) 7
|
c
|
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
|
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 ?
|
"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 ) 4 , b ) 30 , c ) 36 , d ) $ 44.33 , e ) 270 m
|
a
|
subtract(divide(480, 40), 8)
|
divide(n1,n0)|subtract(#0,n2)|
|
physics
|
the average age of 22 students in a group is 12 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ?
|
"age of the teacher = ( 23 * 13 - 22 * 12 ) = 35 years . answer : c"
|
a ) 110 , b ) 35 , c ) 160 , d ) 500 cm 2 , e ) 6 hours
|
b
|
add(22, const_1)
|
add(n0,const_1)|
|
general
|
a square is drawn inside a right - angled triangle with the two perpendicular sides as 12 cm and 8 cm . what is the side of the largest possible square that can be drawn ?
|
area of triangle is 1 / 2 * 12 * 8 = 48 side of square = x the entire triangle split into two right angled triangle and one square with dimensions as follows i ) square with side x ii ) right angled triangle with perpendicular sides x and 12 - x iii ) right angled triangle with perpendicular sides 8 - x and x sum of area of all three = 48 = x 2 + 1 / 2 * x * ( 12 - x ) + 1 / 2 * x * ( 8 - x ) = 48 = x = 4.8 cm answer : a
|
a ) 4.8 cm', ' , b ) 975 , c ) - 8 , d ) 3 , e ) 43
|
a
|
divide(divide(multiply(12, 8), const_2), const_10)
|
multiply(n0,n1)|divide(#0,const_2)|divide(#1,const_10)
|
geometry
|
the membership of a committee consists of 3 english teachers , 4 mathematics teachers , and 2 social studies teachers . if 2 committee members are to be selected at random to write the committee ’ s report , what is the probability that the two members selected will both be social teachers ?
|
"probability of first member an english teacher = 3 / 9 probability of second member an english teacher = 2 / 8 probability of both being english teacher = 3 / 9 x 2 / 8 = 1 / 12 ( b )"
|
a ) $ 620 , b ) 1 / 12 , c ) 3 / 4 , d ) 10 ° , e ) 31
|
b
|
multiply(divide(3, add(add(3, 4), 2)), divide(2, subtract(add(add(3, 4), 2), const_1)))
|
add(n0,n1)|add(n2,#0)|divide(n0,#1)|subtract(#1,const_1)|divide(n2,#3)|multiply(#2,#4)|
|
probability
|
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