Datasets:
ericsun153
/
mathqa_confusing_options_contamination_test

Dataset card Data Studio Files
xet
Community
1
Problem
stringlengths
5
628
Rationale
stringlengths
1
2.74k
options
stringlengths
39
113
correct
stringclasses
5 values
annotated_formula
stringlengths
6
848
linear_formula
stringlengths
7
357
category
stringclasses
6 values
in a 160 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is :
b runs 56 m in 7 sec . = > b runs 160 m in 7 / 56 * 160 = 20 seconds since a beats b by 7 seconds , a runs 160 m in ( 20 - 7 ) = 13 seconds hence , a ' s time over the course = 13 seconds answer : c
a ) 95 , b ) 13 seconds , c ) 80 % , d ) 18 : 5 , e ) 4 %
b
subtract(multiply(divide(7, 56), 160), 7)
divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)
physics
. 002 / ? = . 01
let . 002 / x = . 01 ; then x = . 002 / . 01 = . 2 / 1 = . 2 answer is a
a ) 16 , b ) . 2 , c ) 130 cm , d ) 24 , e ) 71
b
divide(divide(2, const_1000), divide(1, const_100))
divide(n0,const_1000)|divide(n1,const_100)|divide(#0,#1)
general
a cistern is filled by a tap in 7 1 / 2 hours . due to leak in the bottom of the cistern , it takes half an hour longer to fill the cistern . if the cistern is full how many hours will it take the leak to empty it ?
"filling rate - leak rate = net rate 1 / 7.5 - leak rate = 1 / 8 leak rate = 2 / 15 - 1 / 8 = 1 / 120 the answer is c ."
a ) 0.45 % , b ) 216 , c ) 1 : 88 , d ) $ 250 , e ) 120
e
divide(1, subtract(divide(const_1, add(7, divide(1, 2))), divide(1, const_4)))
divide(n1,n2)|divide(n1,const_4)|add(const_3.0,#0)|divide(n1,#2)|subtract(#3,#1)|divide(n1,#4)|
physics
by how much is 70 % of 120 greater than 35 % of 200 .
( 70 / 100 ) * 120 Γ’ € β€œ ( 35 / 100 ) * 200 84 - 70 = 14 answer : b
a ) 2340 , b ) $ 20 , c ) 14 , d ) 5.48 days , e ) rs . 1350
c
subtract(multiply(120, divide(70, const_100)), multiply(divide(35, const_100), 200))
divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)
gain
if 2 ^ k + 2 ^ k = ( 2 ^ 9 ) ^ ( 2 ^ 9 ) - 2 ^ k , then k = ?
"2 ^ k + 2 ^ k = ( 2 ^ 9 ) ^ 2 ^ 9 - 2 ^ k 2 * ( 2 ^ k ) = 2 ^ ( 4 * 3 ^ 9 ) = 2 ^ ( 2 ^ 2 * 2 ^ 9 ) = 2 ^ ( 2 ^ 11 ) 2 ^ k + 1 = 2 ^ ( 2 ^ 11 ) so k + 1 = 2 ^ 11 so k = 2 ^ 11 - 1 answer is c"
a ) 144 , b ) 56 , c ) 2 ^ 11 - 1 , d ) e = 121 , e ) 17 min .
c
subtract(power(2, add(9, const_2)), const_1)
add(n3,const_2)|power(n0,#0)|subtract(#1,const_1)|
general
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 5 years older than b , the present age of b is
"explanation : let b ' s age = x years . then , as age = ( x + 5 ) years . ( x + 5 + 10 ) = 2 ( x β€” 10 ) hence x = 35 . present age of b = 35 years answer : option a"
a ) 35 , b ) 69000 , c ) 2 : 1 , d ) 10 , e ) 5
a
add(multiply(const_2, 10), add(5, 10))
add(n0,n3)|multiply(n0,const_2)|add(#0,#1)|
general
if 2 + 7 = 57 ; 3 + 6 = 63 ; 5 + 9 = 206 then 5 + 8 = ?
"2 ^ 3 + 7 ^ 2 = 57 3 ^ 3 + 6 ^ 2 = 63 5 ^ 3 + 9 ^ 2 = 206 and 5 ^ 3 + 8 ^ 2 = 189 answer : e"
a ) 8 , b ) 189 , c ) 72.33 , d ) 5.48 days , e ) 81
b
add(power(5, 3), power(8, 2))
power(n9,n3)|power(n10,n0)|add(#0,#1)|
general
if a man lost 7 % by selling oranges at the rate of 21 a rupee at how many a rupee must he sell them to gain 42 % ?
"93 % - - - - 21 142 % - - - - ? 93 / 142 * 21 = 13.75 answer : e"
a ) 25 % , b ) 2 , c ) 13.75 , d ) 73 , e ) 13122
c
divide(multiply(subtract(const_100, 7), 21), add(const_100, 42))
add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)|
gain
what is the greatest prime factor of 5 ^ 6 - 1 ?
"5 ^ 6 - 1 = ( 5 ^ 3 - 1 ) ( 5 ^ 3 + 1 ) = 124 * 126 = 4 * 31 * 3 * 42 the answer is b ."
a ) 6 , b ) 31 , c ) 10 : 3 , d ) 1 / 3 , e ) 1225
b
floor(divide(5, divide(6, const_2)))
divide(n1,const_2)|divide(n0,#0)|floor(#1)|
general
a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.4 m and its walls are 5 cm thick . the thickness of the bottom is :
"explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) Γ— ( 260 - 10 ) Γ— ( 140 - x ) ] = 8000 Γ— 1000 = > 320 Γ— 250 Γ— ( 140 - x ) = 8000 Γ— 1000 = > ( 140 - x ) = 8000 Γ— 1000 / 320 = 100 = > x = 40 cm = 4 dm . answer : b"
a ) 4 dm , b ) 18 , c ) 5 / 9 , d ) 20 , e ) 50', '
a
subtract(multiply(multiply(3.3, 2.6), 1.4), divide(8000, const_1000))
divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0)|
physics
a and b can do a piece of work in 10 days . with the help of c they finish the work in 3 days . c alone can do that piece of work in ?
"c = 1 / 3 – 1 / 10 = 7 / 30 = > 4.3 days answer : b"
a ) 4.3 days , b ) 4 : 49 , c ) 11 sec , d ) 80 % , e ) 42
a
inverse(subtract(3, divide(3, 10)))
divide(n1,n0)|subtract(n1,#0)|inverse(#1)|
physics
a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 15 minutes . how long will it take to go 5 km in stationary water ?
"speed ( upstream ) = 2 / 1 = 2 kmhr speed ( downstream ) = 1 / ( 15 / 60 ) = 4 kmhr speed in still water = 1 / 2 ( 2 + 4 ) = 3 kmhr time taken in stationary = 5 / 3 = 1 hrs 40 min answer : e"
a ) 4 , b ) 528 , c ) 1 hour 40 min , d ) 3 , e ) 2 : 3
c
divide(5, divide(add(multiply(divide(1, 15), const_60), divide(2, 2)), const_2))
divide(n0,n0)|divide(n2,n3)|multiply(#1,const_60)|add(#0,#2)|divide(#3,const_2)|divide(n4,#4)|
physics
a can do a piece of work in 15 days . a does the work for 5 days only and leaves the job . b does the remaining work in 16 days . in how many days b alone can do the work ?
"explanation : a ’ s 5 day work = 5 * 1 / 15 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 b completes 2 / 3 work in 6 days b alone can do in x days 2 / 3 * x = 16 x = 24 days answer : option d"
a ) 3024 , b ) 24 days , c ) 324 , d ) 1659 , e ) s . 1665
b
inverse(multiply(inverse(16), subtract(const_1, multiply(5, inverse(15)))))
inverse(n2)|inverse(n0)|multiply(n1,#1)|subtract(const_1,#2)|multiply(#0,#3)|inverse(#4)|
physics
a and b complete a work in 10 days . a alone can do it in 40 days . if both together can do the work in how many days ?
"1 / 10 + 1 / 40 = 0.125 days answer : b"
a ) 16 , b ) 280 , c ) 3 , d ) 1 : 2', ' , e ) 0.125 days
e
inverse(add(inverse(10), inverse(40)))
inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)|
physics
a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 140 meter length .
"speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 140 = 500 meter time = distance / speed = 500 βˆ— 2 / 25 = 40 seconds answer : d"
a ) 6 , b ) 5 , c ) 48 , d ) 123 , e ) 40 seconds
e
divide(add(360, 140), divide(multiply(45, const_1000), const_3600))
add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|
physics
if a 2 - b 2 = 9 and a * b = 4 , find a 4 + b 4 .
"a 2 - b 2 = 9 : given a 4 + b 4 - 2 a 2 b 2 = 92 : square both sides and expand . a * b = 4 : given a 2 b 2 = 42 : square both sides . a 4 + b 4 - 2 ( 16 ) = 81 : substitute a 4 + b 4 = 113 correct answer c"
a ) 4 , b ) 113 , c ) 1 / 10 , d ) 6400 , e ) 50 %
b
add(power(9, 2), multiply(power(4, 2), 2))
power(n3,n0)|power(n2,n0)|multiply(#0,n0)|add(#2,#1)|
general
a work as fast as b . if b can complete a work in 20 days independently , the number of days in which a and b can together finish the work in ?
"ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 10 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 10 + 1 / 20 = 3 / 20 = > 20 / 3 = 6 2 / 3 a and b can finish the work in 6 2 / 3 days answer is e"
a ) 6 2 / 3 days , b ) 7 / 25 , c ) 33.3 % , d ) 22.8 , e ) 150
a
inverse(add(inverse(20), multiply(const_2, inverse(20))))
inverse(n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|
physics
there are 9 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ?
"number of ways of selecting 2 fiction books = 9 c 2 number of ways of selecting 2 non fiction books = 6 c 2 9 c 2 * 6 c 2 = 36 * 15 = 540 answer : c"
a ) 540 , b ) 9,600 , c ) 41 , d ) 58 , e ) 6
a
divide(multiply(multiply(9, const_4), multiply(6, 9)), power(factorial(2), 2))
factorial(n2)|multiply(n0,const_4)|multiply(n0,n1)|multiply(#1,#2)|power(#0,n2)|divide(#3,#4)|
general
a train 120 m long is running with a speed of 62 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going
"explanation : speed of the train relative to man = ( 62 - 8 ) kmph = ( 54 Γ— 5 / 18 ) m / sec = 15 m / sec time taken by the train to cross the man = time taken by it to cover 120 m at 15 m / sec = 120 Γ— 1 / 15 sec = 8 sec answer : option d"
a ) 24 days , b ) 8 sec , c ) 19 , d ) 22 Β½ days , e ) 50 %
b
divide(120, multiply(add(62, 8), const_0_2778))
add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)|
physics
a chemist mixes one liter of pure water with x liters of a 50 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ?
"concentration of salt in pure solution = 0 concentration of salt in salt solution = 50 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 50 - 15 ) / ( 15 - 0 ) = 7 / 3 1 / x = 7 / 3 x = 3 / 7 answer : e"
a ) 3 / 7 , b ) 63 , c ) 80 minutes , d ) 16', ' , e ) 81
a
divide(15, subtract(50, 15))
subtract(n0,n1)|divide(n1,#0)|
gain
the cyclist going at a constant rate of 18 miles per hour is passed by a motor - cyclist traveling in the same direction along the same path at 48 miles per hour . the motor - cyclist stops to wait for the cyclist 15 minutes after passing cyclist , while the cyclist continues to travel at constant rate , how many minutes must the motor - cyclist wait until the cyclist catches up ?
for the 15 minutes the motor - cyclist continues to overtake the cyclist , she is going at 30 miles per hour faster than the cyclist . once the motor - cyclist stops , the cyclist is going at 18 miles per hour while the motor - cyclist is at rest so the amount of time the cyclist will take to cover the distance between them is going to be in the ratio of the relative speeds . 30 / 18 * 15 or 25 minutes answer is ( a )
a ) 25 , b ) 52 % , c ) 22.4 hours , d ) 8.5 , e ) 100
a
divide(multiply(subtract(divide(48, const_4), divide(18, const_4)), const_60), 18)
divide(n1,const_4)|divide(n0,const_4)|subtract(#0,#1)|multiply(#2,const_60)|divide(#3,n0)
physics
a number is doubled and 9 is added . if the resultant is trebled , it becomes 81 . what is that number ?
"let the number be x . then , 3 ( 2 x + 9 ) = 81 2 x = 18 = > x = 9 answer : e"
a ) 40 , b ) 120 , c ) 9 , d ) 10 v 2', ' , e ) 28
c
divide(subtract(81, multiply(const_3, 9)), multiply(const_3, const_2))
multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)|
general
what is difference between biggest and smallest fraction among 1 / 3 , 3 / 4 , 4 / 5 and 5 / 6
"explanation : 1 / 3 = . 33 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 6 = . 833 so biggest is 5 / 6 and smallest is 1 / 3 their difference is 5 / 6 - 1 / 3 = 3 / 6 = 1 / 2 option b"
a ) rs . 240 , b ) 7 , c ) 1 / 2 , d ) 41.4 , e ) 21 - 61
c
subtract(divide(4, 5), divide(1, 3))
divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)|
general
in a renowned city , the average birth rate is 9 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day .
"every 2 seconds , 6 persons are added ( 9 - 3 ) . every second 3 persons are added . in a day 24 hrs = 24 * 60 minutes = 24 * 60 * 60 = 86400 seconds . 86400 * 3 = 259200 option e"
a ) $ 40000 , b ) 16 , c ) 259,200 , d ) 23 % . , e ) 3108
c
multiply(multiply(subtract(9, 3), const_3600), const_12)
subtract(n0,n1)|multiply(#0,const_3600)|multiply(#1,const_12)|
general
894.7 – 573.07 – 95.007 = ?
"solution given expression = 894.7 - ( 573.07 + 95.007 ) = 894.7 - 668.077 = 226.623 . answer a"
a ) 5 , b ) 226.623 , c ) 125177481 , d ) 600 , e ) 2,000
b
subtract(894.7, divide(573.07, 95.007))
divide(n1,n2)|subtract(n0,#0)|
general
three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 1800 . their h . c . f is ?
"let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 1800 x = 30 the numbers are 3 * 30 , 4 * 30 , 5 * 30 hence required h . c . f . = 30 answer is b"
a ) rs 20 , b ) 1070 , c ) 30 , d ) 384 , e ) 8100
c
add(multiply(multiply(3, 5), const_100), multiply(4, 5))
multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)|
other
john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 65 miles per hour , what is v in miles per hour ?
"hibunuel the question seems incorrect . it should not be 80 % at the speed of 80 . however if it ' s 20 % at the speed of 80 , answer comes out 55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it , which comes as : 0.8 d / 80 + 0.2 d / v = d / 65 = > v = 55 ( option d ) ."
a ) 30 , b ) 88 , c ) 55 , d ) 50 , e ) rs 9000
c
multiply(subtract(const_100, 80), subtract(divide(const_100, 65), divide(80, 80)))
divide(const_100,n2)|divide(n0,n0)|subtract(const_100,n0)|subtract(#0,#1)|multiply(#2,#3)|
physics
a man walking at a rate of 15 km / hr crosses a bridge in 35 minutes . the length of the bridge is ?
"speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 35 minutes = 15 / 18 * 35 * 60 = 1750 m answer is b"
a ) 12000 , b ) 87.86 , c ) 1750 m , d ) 13122 , e ) 45.6
c
multiply(divide(multiply(15, const_1000), const_60), 35)
multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)|
gain
the squared value of the diagonal of a rectangle is ( 64 + b 2 ) sq cm , where b is less than 8 cm . what is the breadth of that rectangle ?
diagonal 2 = 64 + b 2 or , 10 ( 2 ) = 64 + 6 ( 2 ) answer a
a ) 11 , b ) 0 , c ) 9620 , d ) 6 cm', ' , e ) 38
d
subtract(sqrt(64), const_2)
sqrt(n0)|subtract(#0,const_2)
geometry
the cost price of 60 articles is the same as the selling price of x articles . if the profit is 20 % , what is x ?
"let the cost price = y the cost price of 60 articles = 60 y the selling price of x articles = 1.20 y * x 1.20 y * x = 60 y x = 60 / 1.2 = 50 the answer is d ."
a ) 50 , b ) rs . 91.66 , c ) 1800 , d ) 5 , e ) 279 m
a
divide(multiply(60, const_4), add(const_4, const_1))
add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|
gain
in township k each property is taxed at 10 percent of its assessed value . if the assessed value of a property in township k is increased from $ 20,000 to $ 24,000 , by how much will the property tax increase ?
"increase in house value = $ 24,000 - $ 20,000 = $ 4000 so , tax increase = 10 % of $ 4000 = $ 400 answer : d"
a ) 10 , b ) 70 , c ) 11 , d ) 5 , e ) $ 400
e
divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 10), const_100)
add(const_3,const_4)|add(const_1,const_4)|multiply(#0,const_4)|multiply(#1,const_4)|multiply(#2,const_1000)|multiply(#3,const_1000)|subtract(#4,#5)|multiply(n0,#6)|divide(#7,const_100)|
general
on a certain transatlantic crossing , 40 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 20 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ?
"let t be the total number of passengers . let x be the number of people with round trip tickets . 0.4 t had round trip tickets and took their cars . 0.8 x had round trip tickets and took their cars . 0.8 x = 0.4 t x = 0.5 t the answer is c ."
a ) 86.4 % , b ) 50 % , c ) 180 m , d ) 10 Β° , e ) 20
b
divide(40, subtract(const_1, divide(20, const_100)))
divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|
gain
at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 3 ?
"salary grade of 5 is p ( 5 ) = 9.50 + 0.25 ( 5 – 1 ) = 9.50 + 0.25 * 4 ; salary grade of 3 is p ( 3 ) = 9.50 + 0.25 ( 3 – 1 ) = 9.50 + 0.25 * 2 ; p ( 5 ) - p ( 3 ) = 9.50 + 0.25 * 4 - 9.50 - 0.25 * 2 = 0.5 . answer : a ."
a ) 5 / 6 , b ) $ 0.50 , c ) 36', ' , d ) 3 , e ) 140
b
add(multiply(0.25, subtract(5, 1)), 0.25)
subtract(n1,n6)|multiply(n3,#0)|add(n3,#1)|
general
in a certain game , a large container is filled with red , yellow , green , and blue beads worth , respectively , 7 , 5 , 3 , and 2 points each . a number of beads are then removed from the container . if the product of the point values of the removed beads is 30 , 870000 , how many red beads were removed ?
30 , 870,000 = 2 ^ 4 * 5 ^ 4 * 3087 = 2 ^ 4 * 3 * 5 ^ 4 * 1029 = 2 ^ 4 * 3 ^ 2 * 5 ^ 4 * 343 = 2 ^ 4 * 3 ^ 2 * 5 ^ 4 * 7 ^ 3 the answer is c .
a ) 750 m , b ) 323.2 , c ) 3 , d ) 555681 , e ) 30
c
divide(multiply(3, const_1), const_1)
multiply(n2,const_1)|divide(#0,const_1)
general
# 88 a necklace is made by stringing q no individual beads together in the repeating pattern red bead , green bead , white bead , blue bead , and yellow bead . if the necklace design begins with a red bead and ends with a white bead , then q could equal
you can just write out the pattern and count : rgwbyrgwbyrgwby . . . but to save time a good test taker will just look for a pattern . min # is 3 , because w is the third one . then every 5 beads another white comes , so it must be 3 + 5 + 5 + 5 . . and so on . . . 3 + 5 = 8 3 + 5 + 5 = 13 3 + 5 + 5 + 5 = 18 3 + 5 + 5 + 5 + 5 = 23 so you see it ends in either 8 or 3 . pick an answer that ends in either 8 or 3 . only one answer does , b .
a ) 28 , b ) 2.6 units , c ) 8 , d ) 7200 , e ) 375
a
add(add(add(add(add(add(divide(88, 88), const_2), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3)), add(const_2, const_3))
add(const_2,const_3)|divide(n0,n0)|add(#1,const_2)|add(#2,#0)|add(#3,#0)|add(#4,#0)|add(#5,#0)|add(#6,#0)
general
a distributor sells a product through an online store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 16 per item . what is the price that the buyer observers online if the distributor wants to maintain a 20 % profit on the cost of the item ?
let x be the price that buyers see online . the distributor wants to receive 1.2 ( original price ) which should be 80 % of x . 1.2 ( 16 ) = 0.8 x x = 1.2 ( 16 ) / 0.8 = 1.5 ( 16 ) = $ 24 the answer is e .
a ) $ 24 , b ) 30 , c ) 320 km , d ) 91.7 , e ) 40 sec
a
divide(add(multiply(divide(20, const_100), 16), 16), divide(subtract(const_100, 20), const_100))
divide(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(n1,#0)|add(n1,#3)|divide(#4,#2)
gain
rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 3000 . then his monthly salary is
"sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 3000 Γ’ ‑ ” 20 / 100 x = 3000 Γ’ ‑ ” x = 3000 Γ£ β€” 5 = 15000 . answer a"
a ) 1,400 , b ) 3 , c ) rs . 15000 , d ) 1534 , e ) 8
c
multiply(3000, add(const_4, const_1))
add(const_1,const_4)|multiply(n4,#0)|
gain
the price of an item is discounted 6 percent on day 1 of a sale . on day 2 , the item is discounted another 6 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?
"let initial price be 100 price in day 1 after 6 % discount = 94 price in day 2 after 6 % discount = 88.36 price in day 3 after 10 % discount = 79.52 so , price in day 3 as percentage of the sale price on day 1 will be = 79.52 / / 94 * 100 = > 84.6 % answer will definitely be ( b )"
a ) 256 , b ) 24 days . , c ) 3 / 5 , d ) 40 , e ) 84.6 %
e
add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 6))), const_100), 2)
divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)|
gain
last year a certain bond price with a face value of 5000 yielded 9 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ?
"interest = 0.09 * 5000 = 0.065 * selling price - - > selling price = 0.09 * 5000 / 0.065 - - > selling price = ~ 6,923 answer : e ."
a ) 6923 , b ) 10 , c ) 30 , d ) 2 , e ) 9944
a
divide(multiply(5000, divide(9, const_100)), divide(6.5, const_100))
divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)|
gain
a train of 40 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ?
d = 40 * 60 + 1500 = 3900 m t = 3900 / 60 * 18 / 5 = 234 sec = 3.9 mins answer : d
a ) 3.9 , b ) 62 , c ) 19.0 % , d ) 9 / 14 , e ) 750 m
a
add(divide(multiply(add(40, const_1), 60), const_1000), 1.5)
add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)
physics
when a mobile is sold for rs . 7200 , the owner loses 20 % . at what price must that mobile be sold in order to gain 20 % ?
80 : 7200 = 120 : x x = ( 7200 x 120 ) / 80 = 10800 . hence , s . p . = rs . 10,800 . answer : option c
a ) 0 , b ) 10,800 , c ) 7 / 12 , d ) $ 96 , e ) 4.0 %
b
floor(multiply(divide(divide(divide(multiply(divide(multiply(7200, const_100), subtract(const_100, 20)), add(const_100, 20)), const_100), const_100), 20), const_2))
add(n1,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#1,#2)|multiply(#0,#3)|divide(#4,const_100)|divide(#5,const_100)|divide(#6,n1)|multiply(#7,const_2)|floor(#8)
gain
( 0.15 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.15 ) ( power 2 ) + 0.015 + ( 0.1 ) ( power 2 ) is :
"given expression = ( 0.15 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.15 ) ( power 2 ) + ( 0.15 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.15 - 0.1 ) = 0.05 answer is c ."
a ) 158 , b ) 5 , c ) 45 , d ) 83 % , e ) 0.05
e
divide(subtract(power(0.15, 3), power(0.1, 3)), add(add(power(0.15, 2), 0.015), power(0.1, 2)))
power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)|
general
a retailer sells 7 shirts . the first 2 he sells for $ 38 and $ 42 . if the retailer wishes to sell the 7 shirts for an overall average price of over $ 50 , what must be the minimum average price of the remaining 5 shirts ?
first 2 shirts are sold for $ 38 and $ 42 = $ 80 . to get average price of $ 50 , total sale should be 7 * $ 50 = $ 350 so remaining 5 shirts to be sold for $ 350 - $ 80 = $ 270 answer should be 270 / 5 = $ 54.00 that is a
a ) 9 , b ) 762 , c ) 64 : 27', ' , d ) $ 54.00 , e ) 2500
d
divide(subtract(multiply(7, 50), add(38, 42)), subtract(7, 2))
add(n2,n3)|multiply(n0,n5)|subtract(n0,n1)|subtract(#1,#0)|divide(#3,#2)
general
a person crosses a 500 m long street in 4 minutes . what is his speed in km per hour ?
"distance = 500 meter time = 4 minutes = 4 x 60 seconds = 240 seconds speed = distance / time = 500 / 240 = 2.08 m / s = 2.08 Γ£ β€” 18 / 5 km / hr = 7.5 km / hr answer : a"
a ) 40.6 , b ) 270 , c ) 7.5 , d ) 9 , e ) 29
c
divide(divide(500, const_1000), divide(multiply(4, const_60), const_3600))
divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)|
physics
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 160 , how much more remains to be paid ?
"10 / 100 p = 160 > > p = 160 * 100 / 10 = 1600 1600 - 160 = 1440 answer : e"
a ) 49.9 kg , b ) 3 ab / 11 , c ) $ 1,440 , d ) 88 , e ) 200 sq feet
c
subtract(multiply(160, divide(const_100, 10)), 160)
divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)|
general
if the perimeter of a rectangular garden is 600 m , its length when its breadth is 100 m is ?
"2 ( l + 100 ) = 600 = > l = 200 m answer : c"
a ) 85 , b ) 200 , c ) 5 , d ) $ 5 , e ) 2800
b
subtract(divide(600, const_2), 100)
divide(n0,const_2)|subtract(#0,n1)|
physics
a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 50 % while buying and by 10 % while selling . what is his percentage profit ?
"the owner buys 100 kg but actually gets 150 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 150 - 90 ) / 90 * 100 = 50 % answer : d ."
a ) 52500 , b ) 32 square inches , c ) 24 , d ) 50 % , e ) 45
d
divide(multiply(subtract(add(const_100, 50), subtract(const_100, 10)), const_100), subtract(const_100, 10))
add(n0,const_100)|subtract(const_100,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#1)|
gain
you need to pick any number from ' 1 , 3 , 5 , 7 , 9 , 11 , 13 and 15 ' to make below equation true . ( ) + ( ) + ( ) = 30 can you solve it ?
solution : 3 ! + 15 + 9 = 30 explanation : 3 ! = 3 * 2 * 1 = 6 6 + 15 + 9 = 30 answer b
a ) 30 , b ) 22 , c ) 84 , d ) 10 , e ) 990
a
add(add(11, factorial(3)), 13)
factorial(n1)|add(n5,#0)|add(n6,#1)
general
ram and shyam start a two - length swimming race at the same moment but from opposite ends of the pool . they swim in lanes at uniform speeds , but hardy is faster than andy . they 1 st pass at a point 18.5 m from the deep end and having completed one length each 1 is allowed to rest on the edge for exactly 45 sec . after setting off on the return length , the swimmers pass for the 2 nd time just 10.5 m from the shallow end . how long is the pool ?
let x = length of pool at first meeting , combined distance = x at second meeting , combined distance = 3 x if andy swims 18.5 m of x , then he will swim 3 * 18.5 = 55.5 m of 3 x andy ' s total distance to second meeting = x + 10.5 m x + 10.5 = 55.5 m x = 45 m e
a ) 11 / 30 , b ) 50 , c ) 80 , d ) 45 , e ) 720
d
subtract(add(multiply(18.5, const_2), 18.5), 10.5)
multiply(n1,const_2)|add(n1,#0)|subtract(#1,n5)
physics
jolene entered an 18 - month investment contract that guarantees to pay 2 percent interest at the end of 4 months , another 3 percent interest at the end of 12 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ?
"if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 920.24 answer : d ."
a ) s . 15000 , b ) 115 , c ) 40 % , d ) 288,889 , e ) $ 920.24
e
add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))))
divide(n1,const_100)|divide(n3,const_100)|divide(n5,const_100)|power(const_100,n1)|multiply(#0,#3)|add(#4,#3)|multiply(#5,#1)|add(#5,#6)|multiply(#7,#2)|add(#8,#6)|
gain
what least number must besubtracted from 427398 so that remaining no . is divisible by 15 ?
on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted answer : option a
a ) 18 , b ) 27 days , c ) 38 years , d ) 3.9 , e ) 725117481
e
subtract(subtract(subtract(multiply(multiply(multiply(427398, const_100), const_10), const_2), 427398), multiply(427398, const_100)), multiply(multiply(multiply(const_100, const_100), const_100), const_100))
multiply(n0,const_100)|multiply(const_100,const_100)|multiply(#0,const_10)|multiply(#1,const_100)|multiply(#2,const_2)|multiply(#3,const_100)|subtract(#4,n0)|subtract(#6,#0)|subtract(#7,#5)
general
one copy machine can make 30 copies a minute , and a second copy machine makes 20 copies a minute . if the two copiers work together , how long would it take them to make 2,000 copies ?
"total work done by both machines in a minute = 30 + 20 = 50 copies total number of copies required = 2000 time = 2000 / 50 = 40 mins answer b"
a ) 70 , b ) 1764713 , c ) 3 , d ) 40 minutes , e ) 36 / 31
d
divide(power(20, const_3), add(30, 20))
add(n0,n1)|power(n1,const_3)|divide(#1,#0)|
physics
little john had $ 8.50 . he spent $ 1.25 on sweets and gave to his two friends $ 1.20 and $ 2.20 . how much money was left ?
"john spent and gave to his two friends a total of 1.25 + 1.20 + 2.20 = $ 4.65 money left 8.50 - 4.65 = $ 3.85 correct answer is c ) $ 3.85"
a ) 22.37 , b ) 21 hours , c ) 50', ' , d ) $ 3.85 , e ) 14
d
subtract(8.50, add(1.25, add(1.20, 1.20)))
add(n2,n2)|add(n1,#0)|subtract(n0,#1)|
general
in covering a distance of 48 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is :
"let abhay ' s speed be x km / hr . then , 48 / x - 48 / 2 x = 3 6 x = 48 x = 8 km / hr . answer : option e"
a ) 120 , b ) 7 , c ) 40 , d ) 8 kmph , e ) 625
d
divide(subtract(48, divide(48, 2)), add(1, 2))
add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0)|
physics
find the number of different prime factors of 1250
"explanation : l . c . m of 1250 = 2 x 5 x 5 x 5 x 5 2 , 5 number of different prime factors is 2 . answer : option b"
a ) 20 , b ) 1 / 15 , c ) 2400 , d ) 2 , e ) 50
d
add(const_2, const_2)
add(const_2,const_2)|
other
the speed of a bus increases by 2 kmph after every one hour . if the distance travelled in the first one hour was 35 km , what was the total distance travelled in 12 hours ?
dist 1 st hr = 35 km speed of bus by 2 kmph 2 nd hr = 37 km 3 rd hr = 39 km tot = 35 + 37 + 39 + . . . . ( 12 terms ) 12 / 2 ( 2 * 35 + ( 12 - 1 ) 2 ] = 6 * 92 = 552 answer c
a ) 40 , b ) 75 , c ) 9 / 7 , d ) 552 , e ) 23 % .
d
multiply(divide(12, 2), add(multiply(subtract(12, const_1), 2), multiply(2, 35)))
divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#3,#1)|multiply(#4,#0)
physics
harkamal purchased 8 kg of grapes at the rate of 90 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ?
"cost of 8 kg grapes = 90 Γ— 8 = 720 . cost of 9 kg of mangoes = 55 Γ— 9 = 495 . total cost he has to pay = 720 + 495 = 1215 . b )"
a ) 1215 , b ) 28 , c ) 23 , d ) 104 miles , e ) 150 km
a
add(multiply(8, 90), multiply(9, 55))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
gain
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 16 terms of the arithmetic progression ?
"n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 16 . . . we get 15 / 2 [ 2 a + 14 d ] = 16 [ a + 7 d ] = 16 * 10 = 160 . . . answer is d . . ."
a ) 318 $ , b ) 30 , c ) 1 / 6 , d ) 4 , e ) 160
e
divide(multiply(20, 16), const_2)
multiply(n0,n1)|divide(#0,const_2)|
general
there are 3 red shoes & 7 green shoes . if two of red shoes are drawn what is the probability of getting red shoes
"taking 2 red shoe the probablity is 3 c 2 from 10 shoes probablity of taking 2 red shoe is 3 c 2 / 10 c 2 = 1 / 15 answer : d"
a ) 220 , b ) 1 / 15 , c ) rs . 1236 , d ) 12 cm , e ) 5.33 kmph
b
divide(choose(3, const_2), choose(add(3, 7), const_2))
add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)|
probability
it will take 16 days for mary to complete a certain task alone . she worked for 8 days before she was joined by her sister . both of them completed the remaining task in 2 and half days . if her sister had joined her when she started the task , how many days would it have taken ?
explanation : mary and her sister complete half work in 2.5 days = > they can complete whole work in 5 days answer : option d
a ) 11 / 16 , b ) 5 , c ) 360 , d ) 8 : 5 , e ) 6
b
add(divide(divide(const_1, 8), divide(const_1, 16)), const_3)
divide(const_1,n1)|divide(const_1,n0)|divide(#0,#1)|add(#2,const_3)
physics
on dividing 73 by a number , the quotient is 9 and the remainder is 1 . find the divisor ?
"d = ( d - r ) / q = ( 73 - 1 ) / 9 = 72 / 9 = 8 a )"
a ) 1200 , b ) $ 1600 , c ) 492 , d ) 140 , e ) 8
e
floor(divide(73, 9))
divide(n0,n1)|floor(#0)|
general
shruti purchased several number of 3 articles p , q and r in the proportion 3 : 2 : 3 . if the unit costs of the articles p , q and r are 200 , rs . 90 and rs . 60 respectively , how many articles of q must have been purchased in the total purchases of rs . 4800 ?
explanation : let the number of articles of types p , q and r be 3 a , 2 a and 3 a respectively . thus , we get , ( 200 x 3 a ) + ( 90 x 2 a ) + ( 60 x 3 a ) = 4800 960 a = 4800 a = 5 hence , the number of articles of type β€œ q ” = 2 x 5 = 10 answer b
a ) 400 , b ) 90 o , c ) 10 , d ) $ 22000 , e ) 30
c
multiply(divide(4800, add(add(multiply(3, 200), multiply(2, 90)), multiply(3, 60))), 2)
multiply(n0,n4)|multiply(n2,n5)|multiply(n0,n6)|add(#0,#1)|add(#3,#2)|divide(n7,#4)|multiply(n2,#5)
general
a distributor sells a product through an on - line store , which take a commission of 20 % of the price set by the distributor . the distributor obtains the product from a producer at the price of $ 15 per item . what is the price that the buyer observers on - line if the distributor wants to maintain a 40 % profit on the cost of the item ?
"producer price = $ 15 ; the distributor wants to maintain a 20 % profit on the cost of the item , thus he must get $ 15 * 1.2 = $ 18 after the store takes a commission of 40 % of the final price - - > ( final price ) * 0.6 = $ 18 - - > ( final price ) = $ 30 . answer : b ."
a ) 158.256 m , b ) 4 cm', ' , c ) 41 , d ) 24000 , e ) 30
e
multiply(multiply(15, divide(add(const_100, 40), const_100)), divide(add(const_100, 20), const_100))
add(n0,const_100)|add(n2,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(n1,#3)|multiply(#2,#4)|
gain
a boy multiplied 987 by a certain number and obtained 559981 as his answer . if in the answer both 9 are wrong and the other digits are correct , then the correct answer would be :
987 = 3 x 7 x 47 so , the required number must be divisible by each one of 3 , 7 , 47 553681 - > ( sum of digits = 28 , not divisible by 3 ) 555181 - > ( sum of digits = 25 , not divisible by 3 ) 555681 is divisible by 3 , 7 , 47 answer c
a ) 555681 , b ) 25 % , c ) 108 kmph , d ) 5120 , e ) 12
a
multiply(subtract(subtract(divide(559981, 987), const_4), const_0_33), 987)
divide(n1,n0)|subtract(#0,const_4)|subtract(#1,const_0_33)|multiply(n0,#2)
other
the least number which when increased by 5 each divisible by each one of 24 , 32 , 36 and 54 is :
solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 5 = 864 - 5 = 859 . answer b
a ) 4 % , b ) 95.07 square meter', ' , c ) 859 , d ) 69000 , e ) 64
c
subtract(lcm(lcm(lcm(24, 32), 36), 54), 5)
lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)
general
what is the compound interest on rs . 8500 at 7.5 % p . a . compounded half - yearly for 2 1 / 2 years .
"compound interest : a = p ( 1 + r / n ) nt a = 10 , 217.85 c . i . > > 10 , 217.85 - 8500 > > rs . 1717.85 answer : b"
a ) 20.5 , b ) 400,200 , c ) 158 , d ) 1717.85 , e ) 2520', '
d
multiply(8500, subtract(power(divide(add(divide(7.5, const_2), const_100), const_100), multiply(2, const_2)), const_1))
divide(n1,const_2)|multiply(n2,const_2)|add(#0,const_100)|divide(#2,const_100)|power(#3,#1)|subtract(#4,const_1)|multiply(n0,#5)|
gain
a certain company reported that the revenue on sales increased 40 % from 2000 to 2003 , and increased 80 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ?
"assume the revenue in 2000 to be 100 . then in 2003 it would be 140 and and in 2005 180 , so from 2003 to 2005 it increased by ( 180 - 140 ) / 140 = 40 / 140 = 2 / 7 = ~ 29 % . answer : e ."
a ) 280 , b ) 20 years , c ) 29 % , d ) 128 , e ) 1 / 32
c
multiply(divide(subtract(add(const_1, divide(80, const_100)), add(const_1, divide(40, const_100))), add(const_1, divide(40, const_100))), const_100)
divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)|
gain
the l . c . m of two numbers is 48 . the numbers are in the ratio 2 : 3 . the sum of numbers is ?
"let the numbers be 2 x and 3 x . then , their l . c . m = 6 x . so , 6 x = 48 or x = 8 . the numbers are 16 and 24 . hence , required sum = ( 16 + 24 ) = 40 . answer : c"
a ) 40 , b ) 8 , c ) 52 % , d ) 1500 , e ) 10
a
divide(multiply(2, 48), 3)
multiply(n0,n1)|divide(#0,n2)|
other
if the ratio of the sum of the first 6 terms of a g . p . to the sum of the first 3 terms of the g . p . is 65 , what is the common ratio of the g . p ?
65 = ( a 1 + a 2 + a 3 + a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) factorize the same terms 65 = 1 + ( a 4 + a 5 + a 6 ) / ( a 1 + a 2 + a 3 ) write every term with respect to r a 1 = a 1 a 2 = a 1 * r ^ 1 a 3 = a 1 * r ^ 2 . . . . . . . . . 65 = 1 + ( a 1 ( r ^ 3 + r ^ 4 + r ^ 5 ) ) / ( a 1 ( 1 + r ^ 1 + r ^ 2 ) ) 64 = ( r ^ 3 ( 1 + r ^ 1 + r ^ 2 ) ) / ( ( 1 + r ^ 1 + r ^ 2 ) ) 64 = r ^ 3 r = 4 a
a ) $ 488.9 , b ) rs . 145 , c ) 60 , d ) 144 hours , e ) 4
e
power(subtract(65, const_1), divide(const_1, const_3))
divide(const_1,const_3)|subtract(n2,const_1)|power(#1,#0)
other
when jessica withdrew $ 200 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 3 of the remaining balance , what will be the final balance in her bank account ?
"as per the question 200 = 2 a / 5 thus - a which is the total amount = 500 the amount thus left = 300 she then deposited 1 / 3 of 300 = 100 total amount in her account = 400 answer c"
a ) 21 years , b ) 3 , c ) 190 , d ) $ 205,000 , e ) 400
e
multiply(subtract(divide(200, subtract(1, divide(const_3, 5))), 200), add(1, divide(1, 3)))
divide(n3,n4)|divide(const_3,n2)|add(n3,#0)|subtract(n3,#1)|divide(n0,#3)|subtract(#4,n0)|multiply(#2,#5)|
general
a man traveled a total distance of 1200 km . he traveled one - third of the whole trip by plane and the distance traveled by train is two - thirds of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ?
total distance traveled = 1200 km . distance traveled by plane = 400 km . distance traveled by bus = x distance traveled by train = 2 x / 3 x + 2 x / 3 + 400 = 1200 5 x / 3 = 800 x = 480 km the answer is c .
a ) 63 , b ) 180 km , c ) 0.1388 , d ) 480 , e ) 100 %
d
divide(multiply(divide(multiply(1200, const_2), const_3), const_3), add(const_2, const_3))
add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)
physics
in town x , 64 percent of the population are employed , and 40 percent of the population are employed males . what percent of the employed people in town x are females ?
"we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 40 are employed males , hence 24 % are employed females . ( employed females ) / ( total employed people ) = 24 / 64 = 38 % answer : a ."
a ) 38 % , b ) 43426 , c ) 33 , d ) 36 kmph , e ) rs . 5084
a
multiply(divide(subtract(64, 40), 64), const_100)
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)|
gain
find the average of all prime numbers between 1 and 5 .
"sol . there are five prime numbers between 1 and 5 . they are 2 , 3 , 5 , 7 , 11 Γ’ Λ† Β΄ required average = [ 2 + 3 + 5 + 7 + 11 / 5 ] = 28 / 5 = 5.6 answer c"
a ) 1122 , b ) 466 , c ) 5.6 , d ) 3 : 5 , e ) 305
c
divide(add(add(add(1, const_1), add(add(1, const_1), const_2)), add(subtract(5, 1), subtract(5, const_2))), 1)
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
a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 237 for it , the cost price of the cricket bat for a is :
"125 % of 120 % of a = 237 125 / 100 * 120 / 100 * a = 237 a = 237 * 2 / 3 = 158 . answer c"
a ) 870 , b ) 800 , c ) 158 , d ) 85 , e ) rs . 315
c
divide(237, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|
gain
how many numbers between 100 and 756 are divisible by 2 , 3 , and 7 together ?
"explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 756 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 756 : 756 / 42 = 18 hence , 42 * 16 = 756 is the last number divisible by 42 within 756 hence , total numbers divisible by 2 , 3 , 7 together are ( 18 Γ’ € β€œ 2 ) = 16 answer : d"
a ) 16 , b ) 790 , c ) 6.56 kg , d ) 22.5 % , e ) 7 pm
a
subtract(divide(756, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7)))
multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)|
general
if the sides of a cube are in the ratio 9 : 5 . what is the ratio of their diagonals ?
"explanation : diagonal of a cube = a √ 3 where a is side a 1 : a 2 = 9 : 5 d 1 : d 2 = 9 : 5 where √ 3 cancelled both side answer : a"
a ) 1 / 10 , b ) 2800 , c ) 9 : 5 , d ) 131.6 , e ) 2 : 1
c
divide(9, 5)
divide(n0,n1)|
geometry
a and b together do a work in 20 days . b and c together in 15 days and c and a in 12 days . so a , b and c together finish same work in how many days ?
( a + b ) work in 1 day = 1 / 20 , ( b + c ) work in 1 days = 1 / 15 . , ( c + a ) work in 1 days = 1 / 12 ( 1 ) adding = 2 [ a + b + c ] in 1 day work = [ 1 / 20 + 1 / 15 + 1 / 12 ] = 1 / 5 ( a + b + c ) work in 1 day = 1 / 10 so , all three together finish work in 10 days answer d
a ) 12 , b ) 59.8 % , c ) 144 , d ) 10.5 , e ) 10
e
inverse(divide(add(inverse(12), add(inverse(20), inverse(15))), const_2))
inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|inverse(#5)
physics
if a train , travelling at a speed of 180 kmph , crosses a pole in 6 sec , then the length of train is ?
d = 180 * 5 / 18 * 6 = 300 m answer : a
a ) 1612.5 , b ) 7.5 , c ) 3 , d ) 300 , e ) 1000 m
d
multiply(multiply(180, const_0_2778), 6)
multiply(n0,const_0_2778)|multiply(n1,#0)
physics
a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 36 . find the profit percentage ?
"l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 180 ( 36 * 5 ) profit percentage = ( 180 - 100 ) / 100 * 100 = 80 % answer : a"
a ) 665 , b ) 80 % , c ) 40 , d ) 90 sq . cm' , e ) 1 / 6
b
subtract(multiply(36, add(const_4, const_1)), multiply(25, const_4))
add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)|
gain
a can do a piece of work 60 days . b can do work in 90 days . in how many days they will complete the work together ?
lcm = 180 , ratio = 60 : 90 = 2 : 3 no of days = 180 / ( 2 + 3 ) = 180 / 5 = 36 days answer : a
a ) 9.1 litres , b ) 36 days , c ) 4 / 5 , d ) 56.25 % , e ) 5
b
divide(const_1, add(divide(const_1, 60), divide(const_1, 90)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)
physics
what least number should be added to 1536 , so that the sum is completely divisible by 21 ?
"1536 Γ· 21 = 73 reminder - 3 3 + 18 = 21 hence 18 should be added to 1536 so that the sum will be divisible by 21 answer : c"
a ) 7 and 5 , b ) 24 seconds , c ) 1 / 9 , d ) 18 , e ) 41
d
subtract(21, reminder(1536, 21))
reminder(n0,n1)|subtract(n1,#0)|
general
find the simple interest on rs . 300 for 9 months at 6 paisa per month ?
"i = ( 300 * 9 * 6 ) / 100 = 162 answer : c"
a ) $ 336 , b ) 1260 , c ) 61 , d ) 162 , e ) βˆ’ 2 %
d
multiply(300, divide(9, const_100))
divide(n1,const_100)|multiply(n0,#0)|
gain
gary drove from point a to point b at 60 km / h . on his way back he took a train travelling at 110 km / h and therefore his trip back lasted 5 hours less . what is the distance ( in km ) between a and b ?
distance = speed * time d 1 = s 1 t 1 d 2 = s 2 t 2 the distance from point a to point b is the same for each trip so , d 1 = d 2 and t 2 = t 1 - 5 thus , s 1 t 1 = s 2 t 2 60 t 1 = s 2 ( t 1 - 5 ) t 1 = 11 60 * 11 = 660 answer : c
a ) 660 . , b ) 34 % , c ) 21 - 61 , d ) 16 , e ) $ 500
a
multiply(60, divide(multiply(110, 5), subtract(110, 60)))
multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)
physics
when n is divided by 19 , the remainder is 7 . find thee difference between previous remainder and the remainder when 18 n is divided by 9 ?
let n = 7 ( leaves a remainder of 7 when divided by 19 ) 18 n = 18 ( 7 ) = 126 , which leaves a remainder of 0 when divided by 9 . difference = 7 - 0 = 7 . answer a
a ) 7 , b ) 21 days , c ) 13 , d ) 17 , e ) 2 : 3
a
subtract(7, reminder(18, 9))
reminder(n2,n3)|subtract(n1,#0)
general
solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 75 percent carbonated water , what percent of the volume of the mixture is p ?
"75 % is 5 % - points below 80 % and 20 % - points above 55 % . so the ratio of solution p to solution q is 4 : 1 . mixture p is 4 / 5 = 80 % of the volume of mixture pq . the answer is d ."
a ) 80 % , b ) 2 , c ) $ 4000 , d ) 11 , e ) 8 days
a
multiply(divide(subtract(divide(75, const_100), divide(55, const_100)), add(subtract(divide(75, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(75, const_100)))), const_100)
divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#0)|add(#3,#4)|divide(#3,#5)|multiply(#6,const_100)|
gain
a can complete a work in 15 days and b can do the same work in 9 days . if a after doing 5 days , leaves the work , find in how many days b will do the remaining work ?
"the required answer = ( 15 - 5 ) * 9 / 15 = 40 / 10 = 6 days answer is c"
a ) 6 days , b ) 60 kg , c ) 51 , d ) 3.5 , e ) 25', '
a
add(multiply(15, 5), divide(15, 5))
divide(n0,n2)|multiply(n0,n2)|add(#0,#1)|
physics
the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km / hr and 15 km / hr respectively . they will meet for the first time in ?
"clearly , the two will meet when they are 1000 m apart to be 20 + 15 = 35 km apart , they take 1 hour to be 1000 m apart , they take 35 * 1000 / 1000 = 35 min . answer is c"
a ) 1 : 729 , b ) e = 121 , c ) 35 min , d ) $ 900 , e ) 1320
c
add(20, 15)
add(n1,n2)|
general
the average weight of a class of 24 students is 35 kg . if the weight of the teacher be included , the average rises by 400 g . the weight of the teacher is :
weight of the teacher = ( 35.4 x 25 - 35 x 24 ) kg = 45 kg . answer : a
a ) 2400 , b ) 9,600 , c ) $ 250 , d ) 45 , e ) 90
d
subtract(multiply(add(35, divide(400, const_1000)), add(24, const_1)), multiply(24, 35))
add(n0,const_1)|divide(n2,const_1000)|multiply(n0,n1)|add(n1,#1)|multiply(#3,#0)|subtract(#4,#2)
general
the contents of a certain box consist of 14 apples and 24 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ?
"the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 24 oranges , so to get to 6 oranges , we should remove 24 - 6 = 18 oranges . answer a"
a ) 4 , b ) 18 , c ) 49.9 kg , d ) 11 / 16 , e ) 350
b
subtract(add(14, 24), divide(14, divide(70, const_100)))
add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)|
general
mary ' s income is 60 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ?
"juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 96 ( 60 percent more than tim ' s income ) . thus , mary ' s income ( 96 ) is 96 % of juan ' s income ( 100 ) . answer : c"
a ) 85 , b ) 2000 , c ) 17 th , d ) 96 % , e ) 90 sq . cm'
d
multiply(multiply(subtract(const_1, divide(40, const_100)), add(const_1, divide(60, const_100))), const_100)
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|
general
a and b are two circles . the radius of a is four times as large as the diameter of b . what is the ratio between the areas of the circles ?
given : the radius of a is 4 times as large as the diameter of b . = > r ( a ) = 4 * d ( b ) = 4 * 2 * r ( b ) = 8 r ( b ) . the radius are in ratio of 1 : 8 thus the area will be in the ratio of square of radius . 1 : 64 . hence d .
a ) 60 , b ) 135 % , c ) 26.66 % , d ) 64 , e ) 1 : 64 .', '
e
divide(power(const_1, const_2), power(multiply(const_2, const_4), const_2))
multiply(const_2,const_4)|power(const_1,const_2)|power(#0,const_2)|divide(#1,#2)
geometry
a can run 192 metre in 28 seconds and b in 32 seconds . by what distance a beat b ?
"clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 192 / 32 = 6 m / s distance covered by b in 4 seconds = speed Γ£ β€” time = 6 Γ£ β€” 4 = 24 metre i . e . , a beat b by 24 metre answer is c"
a ) 305 , b ) 28 , c ) 17 min . , d ) 120 , e ) 24 metre
e
subtract(192, multiply(divide(192, 32), 28))
divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|
physics
a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 585.75 . how many boxes of plain cookies did she sell ?
"let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1585.75 = > x = 791 e"
a ) 600 , b ) 36 kmph , c ) - 39 , d ) 158.256 m , e ) 791
e
divide(add(const_1000, 585.75), const_2)
add(n4,const_1000)|divide(#0,const_2)|
other
three pipes a , b and c can fill a tank from empty to full in 30 minutes , 20 minutes and 10 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution q in the liquid in the tank after 3 minutes ?
"part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 30 + 1 / 20 + 1 / 10 ) = 11 / 20 part filled by b in 3 minutes = 3 / 20 required ratio = 3 / 20 * 20 / 11 = 3 / 11 answer : c"
a ) 131.6 , b ) 7 , c ) 3 / 11 , d ) 60 , e ) 28
c
multiply(divide(3, 10), divide(const_1, multiply(3, add(divide(const_1, 10), add(divide(const_1, 30), divide(const_1, 30))))))
divide(n3,n2)|divide(const_1,n0)|divide(const_1,n2)|add(#1,#1)|add(#3,#2)|multiply(n3,#4)|divide(const_1,#5)|multiply(#0,#6)|
physics
if a 5 cm cube is cut into 1 cm cubes , then what is the percentage increase in the surface area of the resulting cubes ?
"the area a of the large cube is 5 * 5 * 6 = 150 square cm . the area of the 125 small cubes is 125 * 6 = 750 = 5 a , an increase of 400 % . the answer is d ."
a ) 540 % , b ) rs . 90 , c ) 52 , d ) 1800 , e ) 400 %
e
multiply(const_100, divide(multiply(surface_cube(1), surface_cube(5)), surface_cube(5)))
surface_cube(n1)|surface_cube(n0)|multiply(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|
geometry
a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 550 ?
"slope of 2 and a y - intercept of 2 y - coordinate is 550 y = 2 x + 2 548 = 2 x x = 274 answer : e . 274"
a ) 45 , b ) 274 , c ) 7 , d ) 2,180 , e ) 868 cm ^ 2', '
b
divide(subtract(550, 2), 2)
subtract(n2,n0)|divide(#0,n0)|
general
one copy machine can make 30 copies a minute , and a second copy machine makes 15 copies a minute . if the two copiers work together , how long would it take them to make 900 copies ?
"total work done by both machines in a minute = 30 + 15 = 45 copies total number of copies required = 900 time = 900 / 45 = 20 mins answer b"
a ) rs . 960 , b ) 20 minutes , c ) 51 , d ) 9 / 19 , e ) 12
b
divide(power(15, const_3), add(30, 15))
add(n0,n1)|power(n1,const_3)|divide(#1,#0)|
physics
two mechanics were working on your car . one can complete the given job in six hours , but the new guy takes 10 hours . they worked together for the first two hours , but then the first guy left to help another mechanic on a different job . how long will it take the new guy to finish your car ?
"rate ( 1 ) = 1 / 6 rate ( 2 ) = 1 / 10 combined = 8 / 30 work done in 2 days = 8 / 15 work left = 7 / 15 rate * time = work left 1 / 8 * time = 7 / 15 time = 56 / 15 d"
a ) 20 , b ) 56 / 15 , c ) 169 , d ) 2700 , e ) 81
b
max(divide(subtract(const_1, multiply(add(divide(const_1, add(const_4, const_2)), divide(const_1, 10)), const_2)), divide(const_1, 10)), const_3)
add(const_2,const_4)|divide(const_1,n0)|divide(const_1,#0)|add(#2,#1)|multiply(#3,const_2)|subtract(const_1,#4)|divide(#5,#1)|max(#6,const_3)|
physics
if xy = 4 , x / y = 16 , for positive numbers x and y , y = ?
"very easy question . 2 variables and 2 easy equations . xy = 4 - - - > x = 4 / y - ( i ) x / y = 16 - - - > replacing ( i ) here - - - > 4 / ( y ^ 2 ) = 16 - - - > y ^ 2 = 4 / 16 = 1 / 4 - - - > y = 1 / 2 or - 1 / 2 the question states that x and y are positive integers . therefore , y = 1 / 2 is the answer . answer a ."
a ) 30 , b ) 12000 , c ) 923 , d ) 833 , e ) 1 / 2
e
sqrt(divide(4, 16))
divide(n0,n1)|sqrt(#0)|
general
if a boat is rowed downstream for 24 km in 4 hours and upstream for 48 km in 24 hours , what is the speed of the boat and the river ?
"explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x Γ’ € β€œ y x = ( ds + us ) / 2 y = ( ds Γ’ € β€œ us ) / 2 in the above problem ds = 6 ; us = 2 x = ( 6 + 2 ) / 2 = 8 / 2 = 4 km / hr y = ( 6 - 2 ) / 2 = 4 / 2 = 2 km / hr answer : e"
a ) 3 , b ) 150 , c ) 2 , d ) 4 , 2 , e ) 10
d
divide(add(divide(48, 24), divide(24, 4)), const_2)
divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2)|
physics