Datasets:
shangdatalab-ucsd
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mathqa_choice_confusion

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Problem
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Rationale
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annotated_formula
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at what rate percent on simple interest will rs . 1750 amount to rs . 2000 in 5 years ?
"explanation : 250 = ( 1750 x 5 xr ) / 100 250 = ( 175 xr ) / 2 500 = ( 175 xr ) r = 500 / 175 r = 2 6 / 7 % answer : option c"
a ) 75 % , b ) $ 2000 , c ) 36 days , d ) 1.9632 , e ) 2 6 / 7 %
e
multiply(divide(divide(subtract(2000, 1750), 1750), 5), const_100)
subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)|
gain
tim and é lan are 60 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ?
tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice c meet the 2 : 1 ( tim : elan = 40 : 20 ) ratio of maintaining total distance traveled 60 miles so correct answer c
a ) 40 miles . , b ) 1 , c ) 150 m , d ) 27 / 128 , e ) 125177481
a
multiply(divide(10, add(5, 10)), 60)
add(n1,n2)|divide(n1,#0)|multiply(n0,#1)
physics
if y is 90 % greater than x , than x is what % less than y ?
"y = 1.9 x x = y / 1.9 = 10 y / 19 x is 9 / 19 less which is 47.4 % less than y . the answer is d ."
a ) 600 , b ) 683 , c ) 48 sec , d ) 47.4 % , e ) 53
d
multiply(divide(90, add(90, const_100)), const_100)
add(n0,const_100)|divide(n0,#0)|multiply(#1,const_100)|
general
what is the length of a bridge ( in meters ) , which a train 166 meters long and travelling at 45 km / h can cross in 40 seconds ?
"speed = 45 km / h = 45000 m / 3600 s = 25 / 2 m / s in 40 seconds , the train can travel 25 / 2 * 40 = 500 meters 500 = length of train + length of bridge length of bridge = 500 - 166 = 334 meters the answer is d ."
a ) 0 , b ) 165 , c ) 334 , d ) 22 , e ) 427.5
c
subtract(multiply(multiply(45, const_0_2778), 40), 166)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics
the average weight of 10 persons increases by 4.2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ?
"solution total weight increased = ( 10 x 4.2 ) kg = 42 kg . weight of new person = ( 65 + 42 ) kg = 107 kg . answer a"
a ) 176 , b ) 107 kg , c ) 65.80 % , d ) 240 , e ) 12
b
add(65, multiply(10, 4.2))
multiply(n0,n1)|add(n2,#0)|
general
if 5 < x < 9 and y = x + 3 , what is the greatest possible integer value of x + y ?
"x + y = x + x + 3 = 2 x + 3 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 8.5 then x + y = 8.5 + 11.5 = 20 . the answer is b ."
a ) 80 , b ) $ 1200 , c ) 50 % , d ) 3000000 , e ) 20
e
add(add(3, const_10), const_10)
add(n2,const_10)|add(#0,const_10)|
general
a certain class of students is being divided into teams . the class can either be divided into 8 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?
"let total no of students in the class be n so , we are told that n is divisible by both 8 24 so , lets find the least common multiple of 8 24 , ie 24 so our answer is ( d ) 24"
a ) 30 m , b ) 21.48 hours , c ) 16 , d ) 24 , e ) 40
d
lcm(8, 24)
lcm(n0,n1)|
general
a rectangular tiled patio is composed of 90 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 90 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ?
"r * c = 90 and ( 11 + 4 ) ( c - 2 ) = 90 - - > r = 11 and c = 8 . answer : d ."
a ) 16 , b ) 5 , c ) 11 , d ) 8 : 1 , e ) 63
c
divide(90, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(90, 2))))), 2))
multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)|
geometry
what is the units digit of ( 3 ^ 5 ) ( 4 ^ 13 ) ?
"- > the ones place of ( ~ 3 ) ^ n repeats after 4 times like 3  9  7  1  3 . the ones place of ( ~ 4 ) ^ n repeats after 2 times like 4  6  4 . then , 3 ^ 5 = 3 ^ 4 * 1 + 1  3 ^ 1 . , 4 ^ 13 = 4 ^ 2 * 6 + 1 = 4 ^ 1 = ~ 4 which is ( 3 ^ 5 ) ( 4 ^ 13 )  ( 3 ^ 1 ) ( ~ 4 ) = ( ~ 3 ) ( ~ 4 ) = ~ 2 . therefore , the answer is a ."
a ) 2 , b ) 53.5 % , c ) sqrt ( 2.88 / p , d ) 223 , e ) 7
a
add(add(const_4, const_3), const_2)
add(const_3,const_4)|add(#0,const_2)|
general
the sum of three consecutive even numbers is 63 . find the middle number of the three ?
"middle no = 63 / 3 = 21 ans e"
a ) 21 , b ) 6000 , c ) 6 , d ) 0 , e ) 12 days
a
add(add(power(add(add(divide(subtract(subtract(63, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(63, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(63, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(63, const_10), const_2), const_4), const_2), const_2)))
subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)|
physics
a cube is divided into 512 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly one colour ?
"each face of the cube has 8 x 8 = 64 cubelets . on each face , only the interior cubelets are painted one colour . on each side , 6 x 6 = 36 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 36 = 216 the answer is a ."
a ) 390 , b ) 144 min , c ) 3 / 10 , d ) 216 , e ) 240000
d
divide(subtract(512, multiply(multiply(const_4, const_2), const_3)), const_2)
multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)|
geometry
of the 150 employees at company x , 70 are full - time , and 100 have worked at company x for at least a year . there are 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ?
"150 employees 70 are full - time 100 have worked at company x for at least a year 20 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? 150 - 70 = 80 employees not full time 80 - 20 = 60 employees not full time who worked over a year 100 employees have worked at company x for at least a year - 60 employees not full time who worked over a year = 40 full - time employees of company x have worked at the company for at least a year ans b"
a ) 40 , b ) 270 , c ) 100 days , d ) 178 , e ) 40.46
a
subtract(subtract(150, 70), 20)
subtract(n0,n1)|subtract(#0,n3)|
general
paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 60 % of his income to paul and 40 % of his income to quentin , paul ' s new income would be what fraction of quentin ' s new income ?
"make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get p = 12 and q 8.8 so 12 / 8.8 = 3 / 2.2 ans : d"
a ) 36000 , b ) 15 / 11 , c ) 6.5 , d ) 72 , e ) 75 %
b
divide(add(multiply(40, const_100), multiply(40, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 40)))
add(n0,n1)|multiply(n4,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|multiply(n4,#3)|multiply(n4,#0)|add(#1,#4)|add(#2,#5)|divide(#6,#7)|
general
if 6 - 12 / x = 7 - 7 / x , then x =
"we ' re given the equation 6 - 12 / x = 7 - 7 / x . we ' re asked for the value of x . the common - denominator of these 4 numbers is x , so we need to multiply both sides of the equation by x , giving us . . . 6 x - 12 x / x = 7 x - 7 x / x we can then eliminate that denominator , which gives us . . . . 6 x - 12 = 7 x - 7 - 5 = x a"
a ) 8 , b ) 62 , c ) 30 m , d ) - 5 , e ) 75
d
divide(add(7, 12), subtract(6, 7))
add(n3,n1)|subtract(n0,n2)|divide(#0,#1)|
general
the sector of a circle has radius of 14 cm and central angle 135 o . find its perimeter ?
"perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 14 ) + 2 ( 14 ) = 33 + 28 = 61 cm answer : b"
a ) 5 % , b ) 59 , c ) 61 cm , d ) 5 , e ) 44
c
multiply(multiply(const_2, divide(multiply(subtract(14, const_3), const_2), add(const_4, const_3))), 14)
add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)|
physics
| x + 3 | – | 4 - x | = | 8 + x | how many s solutions will this equation have ?
"| x | = x when x > = 0 ( x is either positive or 0 ) | x | = - x when x < 0 ( note here that you can put the equal to sign here as well x < = 0 because if x = 0 , | 0 | = 0 = - 0 ( all are the same ) so the ' = ' sign can be put with x > 0 or with x < 0 . we usually put it with ' x > 0 ' for consistency . a"
a ) 0.125 days , b ) 9.09 % , c ) 19 , d ) 9 years , e ) 0
e
divide(multiply(add(4, 3), const_2), 8)
add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)|
general
in a games hour 4 different types of players came to the ground ? cricket 11 , hokey 15 , football 21 , softball 15 . how many players are present in the ground ?
"total number of players = 11 + 15 + 21 + 15 = 62 answer is c"
a ) 7 , b ) 51 , c ) 76.88 , d ) 1 / 2 , e ) 62
e
add(add(11, 15), add(21, 15))
add(n1,n2)|add(n3,n4)|add(#0,#1)|
physics
how long does a train 120 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ?
"d = 120 + 170 = 290 m s = 60 * 5 / 18 = 50 / 3 t = 290 * 3 / 50 = 17.4 sec answer : d"
a ) 17 th , b ) 270 m , c ) sec , d ) $ 6.80 , e ) 40 min
c
divide(add(120, 170), multiply(60, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
what is the ratio of 6 / 7 to the product 3 * ( 7 / 3 ) ?
"6 / 7 / 21 / 3 = 6 / 49 . . . imo option d ."
a ) 1000 , b ) $ 5 , c ) 216 , d ) 6 : 49 , e ) 32
d
divide(6, const_60)
divide(n0,const_60)|
general
a caterer ordered 125 ice - cream bars and 125 sundaes . if the total price was $ 275.00 and the price of each ice - cream bar was $ 0.60 , what was the price of each sundae ?
"let price of a sundae = s price of ice cream bar = . 6 $ 125 * . 6 + 125 * s = 275 = > 125 * s = 200 = > s = 1.6 answer e"
a ) $ 1.75 , b ) 1 / 9 , c ) $ 1.60 , d ) $ 875 , e ) 4.2
c
divide(subtract(275.00, multiply(125, 0.60)), 125)
multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)|
general
a batsman makes a score of 50 runs in the 6 th inning and thus increases his average by 2 . find his average after 6 th inning .
"let the average after 6 th inning = x then , average after 5 th inning = x - 2 5 ( x - 2 ) + 50 = 6 x x = 10 - 50 = 40 answer is a"
a ) 1400 , b ) 20 , c ) 40 , d ) 112 , e ) 2522
c
add(subtract(50, multiply(6, 2)), 2)
multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|
general
49 ã — 49 ã — 49 = 7 ^ ?
"49 ã — 49 ã — 49 = 7 ? or , 7 ( 2 ) ã — 7 ( 2 ) ã — 7 ( 2 ) = 7 ? or 7 ( 6 ) = 7 ? or , ? = 6 answer d"
a ) 28 , b ) 480 , c ) $ 23400 , d ) 6 , e ) 157
d
add(subtract(power(49, const_2), 49), subtract(power(49, const_2), 49))
power(n0,const_2)|power(n2,const_2)|subtract(#0,n0)|subtract(#1,n2)|add(#2,#3)|
general
a circle in the coordinate plane passes through points ( - 3 , - 2 ) and ( 1 , - 4 ) . what is the smallest possible area of that circle ?
"the distance between the two points is sqrt ( 20 ) . radius = sqrt ( 20 ) / 2 area = pi * ( sqrt ( 20 ) / 2 ) ^ 2 d . 5 π"
a ) 5 π , b ) 2 , c ) 9000 , d ) 683 , e ) 1380
a
square_area(divide(sqrt(add(multiply(add(3, 1), add(3, 1)), multiply(add(2, 4), add(2, 4)))), 2))
add(n0,n2)|add(n1,n3)|multiply(#0,#0)|multiply(#1,#1)|add(#2,#3)|sqrt(#4)|divide(#5,n1)|square_area(#6)|
geometry
the sum of the present ages of two persons a and b is 90 . if the age of a is twice that of b , find the sum of their ages 5 years hence ?
"a + b = 90 , a = 2 b 2 b + b = 90 = > b = 30 then a = 60 . 5 years , their ages will be 65 and 35 . sum of their ages = 65 + 35 = 100 . answer : e"
a ) 100 , b ) 8 , c ) 475 , d ) 17 hr , e ) $ 12,500
a
add(add(multiply(divide(90, 5), const_2), 5), add(divide(90, 5), 5))
divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)|
general
the distance between two cities a and b is 330 km . a train starts from a at 8 a . m . and travels towards b at 60 km / hr . another train starts from b at 9 a . m . and travels towards a at 75 km / hr . at what time do they meet ?
"suppose they meet x hrs after 8 a . m . then , ( distance moved by first in x hrs ) + [ distance moved by second in ( x - 1 ) hrs ] = 330 60 x + 75 ( x - 1 ) = 330 = > x = 3 so , they meet at ( 8 + 3 ) i . e . , 11 a . m . answer : c"
a ) 275 , b ) 3 : 5 , c ) 11 , d ) 26 % , e ) $ 0.40
c
add(divide(add(330, 75), add(60, 75)), 8)
add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)|
physics
a high school has 360 students 1 / 2 attend the arithmetic club , 5 / 8 attend the biology club and 3 / 4 attend the chemistry club . 3 / 8 attend all 3 clubs . if every student attends at least one club how many students attend exactly 2 clubs .
"basically , this question is asking you to figure out how many students are being double - counted . a - club has 180 members ( 1 / 2 of 360 ) b - club has 225 members ( 5 / 8 of 360 ) c - club has 270 members ( 3 / 4 of 360 ) we can create an equation to solve this : 180 + 225 + 270 = n + x + 2 y where n is the number of students , x is the number of students in two clubs , and y is the number of students in three clubs . the question provides y for us ( 135 ) . 180 + 225 + 270 = 360 + x + 270 x = 405 - 360 = 45 a"
a ) 1943236 , b ) 2140 , c ) 30 , d ) 45 , e ) 100
d
subtract(subtract(add(add(divide(multiply(360, 1), 2), divide(multiply(360, 5), 8)), divide(multiply(360, 3), 4)), multiply(divide(multiply(360, 3), 8), 2)), 360)
multiply(n0,n1)|multiply(n0,n3)|multiply(n0,n5)|divide(#0,n2)|divide(#1,n4)|divide(#2,n6)|divide(#2,n4)|add(#3,#4)|multiply(n2,#6)|add(#7,#5)|subtract(#9,#8)|subtract(#10,n0)|
general
the cash difference between the selling prices of an article at a profit of 8 % and 6 % is rs . 3 . the ratio of the two selling prices is ?
"let c . p . of the article be rs . x . then , required ratio = 108 % of x / 106 % of x = 108 / 106 = 54 / 53 = 54 : 53 answer : d"
a ) 98 m , b ) 2 : 3 , c ) 94 , d ) 54 : 53 , e ) 10
d
divide(add(const_100, 8), add(const_100, 6))
add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|
gain
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 588 as interest at the end of the loan period , what was the rate of interest ?
"let rate = r % and time = r years . then , 1200 x r x r / 100 = 588 12 r 2 = 588 r 2 = 49 r = 7 . answer : b"
a ) 32 square meters , b ) 1400 , c ) 0.39 , d ) 38 sec , e ) 7
e
sqrt(divide(multiply(588, const_100), 1200))
multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)|
gain
what is x if x + 5 y = 24 and y = 2 ?
"substitute y by 2 in x + 5 y = 24 x + 5 ( 2 ) = 24 x + 10 = 24 if we substitute x by 14 in x + 10 = 24 , we have 14 + 10 = 24 . hence x = 14 correct answer e"
a ) 250 m , b ) ( 40 , c ) 16 , d ) 225 , e ) 14
e
subtract(24, multiply(5, 2))
multiply(n0,n2)|subtract(n1,#0)|
general
the wages earned by robin is 20 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much percent is the wages earned by charles more than that earned by robin ?
"let wage of erica = 10 wage of robin = 1.2 * 10 = 12 wage of charles = 1.6 * 10 = 16 percentage by which wage earned by charles is more than that earned by robin = ( 16 - 12 ) / 12 * 100 % = 4 / 12 * 100 % = 33 % answer b"
a ) 33 % , b ) 6 , c ) 28 , d ) 51 days , e ) 1
a
multiply(divide(subtract(add(const_100, 60), add(const_100, 20)), add(const_100, 20)), const_100)
add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)|
general
there are two numbers . if 10 % of the first number is added to the second number , then the second number increases to its 6 - fifth . what is the ratio of the first number to the second number ?
let the two numbers be x and y . ( 1 / 10 ) * x + y = ( 6 / 5 ) * y ( 1 / 10 ) * x = ( 1 / 5 ) * y x / y = 2 / 1 = 2 / 1 the answer is e .
a ) 70 % , b ) 4 , c ) 5 , d ) 1984 , e ) 2 : 1
e
divide(divide(const_1, divide(10, const_2)), divide(const_1, 10))
divide(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|divide(#2,#1)
general
each child has 2 pencils and 13 skittles . if there are 6 children , how many pencils are there in total ?
2 * 6 = 12 . answer is b .
a ) 390 , b ) 3 , c ) 4 cm , d ) 12 , e ) 1.6
d
multiply(2, 6)
multiply(n0,n2)|
general
a reduction of 40 % in the price of bananas would enable a man to obtain 60 more for rs . 40 , what is reduced price per dozen ?
"40 * ( 40 / 100 ) = 16 - - - 60 ? - - - 12 = > rs . 3.2 answer : c"
a ) 98 m , b ) 21 , c ) 7 , d ) 28 , e ) 3.2
e
multiply(const_12, divide(multiply(40, divide(40, const_100)), 60))
divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)|
gain
in a fuel station the service costs $ 2.05 per car , every liter of fuel costs 0.6 $ . assuming that you fill up 3 mini - vans and 2 trucks , how much money will the fuel cost to all the cars owners total , if a mini - van ' s tank is 70 liters and a truck ' s tank is 120 % bigger and they are all empty - ?
"service cost of 3 van and 2 truck = 2.05 * ( 3 + 2 ) = 10.5 fuel in 3 van = 3 * 70 = 210 litre fuel in 2 trucks = 2 * 70 ( 1 + 120 / 100 ) = 308 total fuel ( van + truck ) = 518 litre total fuel cost = 518 * 0.6 = 310.8 total cost = fuel + service = 310.8 + 10.25 = 321.05 answer is b"
a ) 660 , b ) 0.7 , c ) 1 / 25 , d ) 10 v 2', ' , e ) 321.05 $
e
add(multiply(multiply(add(70, divide(multiply(70, 120), const_100)), 2), 0.6), multiply(multiply(70, 3), 0.6))
multiply(n4,n5)|multiply(n2,n4)|divide(#0,const_100)|multiply(n1,#1)|add(n4,#2)|multiply(n3,#4)|multiply(n1,#5)|add(#6,#3)|
general
a man two flats for $ 675958 each . on one he gains 16 % while on the other he loses 16 % . how much does he gain or lose in the whole transaction ?
"in such a case there is always a loss loss % = ( 16 / 10 ) ^ 2 = 64 / 25 = 2.56 % answer is b"
a ) 150 , b ) 38 , c ) 7600 , d ) 2.56 % , e ) 12000
d
multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 16)), 675958), multiply(divide(const_100, subtract(const_100, 16)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 16)), 675958), multiply(divide(const_100, subtract(const_100, 16)), 675958))), const_100)
add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100)|
gain
in an office , 60 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ?
"( 10 / 100 ) workers = 16 = > number of workers = 160 ( 60 / 100 ) * workers = x + 16 = > x = 96 answer a"
a ) 29 , b ) 60 , c ) 3.9 , d ) 8 , e ) 96
e
divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4))
divide(n3,const_100)|divide(const_1,const_2)|multiply(const_2,const_4)|divide(n2,#0)|multiply(n4,#3)|multiply(#3,#1)|divide(#4,const_100)|subtract(#6,#5)|divide(#7,#2)|
gain
a rectangular lawn of dimensions 120 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ?
"area = ( l + b – d ) d ( 120 + 60 – 10 ) 10 = > 1700 m 2 1700 * 3 = rs . 5100 answer : a"
a ) 14 , b ) 10 , c ) 3 , d ) 10.5 , e ) s . 5100
e
multiply(multiply(subtract(add(120, 60), 10), 10), 3)
add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)|
geometry
the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 502 to 600 inclusive ?
"2 + 4 + 6 + 8 + . . . + 100 = 2550 502 + 504 + . . . + 600 = 50 ( 500 ) + ( 2 + 4 + . . . + 100 ) = 25,000 + 2550 = 27,550 the answer is b ."
a ) 200 cm 2' , b ) 27,550 , c ) rs . 2500 , d ) 8.4 sec , e ) 0.0015
b
multiply(divide(add(600, 502), const_2), add(divide(subtract(600, 502), const_2), const_1))
add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)|
general
if n is the greatest positive integer for which 2 ^ n is a factor of 8 ! , then n = ?
"8 ! = 40320 e . 2 ^ 10 = 512 ( 40320 / 512 ) - this is not a factor of 8 ! d . 2 ^ 8 = 256 ( 40320 / 256 ) - this is not a factor of 8 ! c . 2 ^ 6 = 64 ( 720 / 64 ) - this is a factor of 8 ! c is answer"
a ) 6 , b ) 5 , c ) 14 , d ) 22 1 / 2 days , e ) 1 / 13
a
subtract(divide(8, const_2), const_1)
divide(n1,const_2)|subtract(#0,const_1)|
other
when x is multiplied by 3 , the result is 18 more than the result of subtracting x from 70 . what is the value of x ?
"the equation that can be formed is : 3 x - 18 = 70 - x or , 4 x = 88 or , x = 22 . e answer ."
a ) 22 , b ) 28 years , c ) 2 , d ) 180 m , e ) 40
a
divide(add(70, 18), add(3, const_1))
add(n1,n2)|add(const_1,n0)|divide(#0,#1)|
general
what is the average of xx , 2 x 2 x , and 66 ?
"by the definition of an average , we get : x + 2 x + 63 = 3 x + 63 x + 2 x + 63 = 3 x + 63 = 3 ( x + 2 ) 3 = x + 2 . = 3 ( x + 2 ) 3 = x + 2 . hence , the answer is x + 2 x + 2 or option c"
a ) x + 2 x + 2 , b ) 11 / 16 , c ) 135 ° , d ) 24 , e ) 27 / 64
a
multiply(divide(divide(multiply(2, add(2, const_1)), const_2), 2), 2)
add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|divide(#2,n0)|multiply(n1,#3)|
general
10 men can complete a work in 7 days . but 10 women need 14 days to complete the same work . how many days will 5 men and 10 women need to complete the work ?
work done by 10 men in 1 day = 1 / 7 work done by 1 man in 1 day = ( 1 / 7 ) / 10 = 1 / 70 work done by 10 women in 1 day = 1 / 14 work done by 1 woman in 1 day = 1 / 140 work done by 5 men and 10 women in 1 day = 5 × ( 1 / 70 ) + 10 × ( 1 / 140 ) = 5 / 70 + 10 / 140 = 1 / 7 = 5 men and 10 women can complete the work in 7 days answer : option c
a ) 3.6 km , b ) 29 , c ) 720 , d ) 625 , e ) 7
e
inverse(add(multiply(10, inverse(multiply(14, 10))), multiply(5, divide(inverse(7), 10))))
inverse(n1)|multiply(n0,n3)|divide(#0,n0)|inverse(#1)|multiply(n0,#3)|multiply(n4,#2)|add(#4,#5)|inverse(#6)
physics
vijay sells a cupboard at 10 % below cost price . had he got rs . 1500 more , he would have made a profit of 10 % . what is the cost price of the cupboard ?
"explanation : cost price = 1500 / ( 0.10 + 0.10 ) = 1500 / 0.20 = rs . 7500 answer b"
a ) 21.5 sec , b ) 46 , c ) 71.4 % , d ) 7500 , e ) 450 sq . m', '
d
divide(1500, divide(subtract(add(const_100, 10), subtract(const_100, 10)), const_100))
add(n0,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#2,const_100)|divide(n1,#3)|
gain
there has been successive increases of 20 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ?
"let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.2 * p = x * p y = x / ( 1.1 * 1.2 ) which is about 0.76 x which is a decrease of about 24 % . the answer is c ."
a ) 3 / 19 , b ) 7 , c ) 30 , d ) 24 % , e ) 504
d
multiply(subtract(const_1, divide(const_100, add(add(const_100, 20), divide(multiply(add(const_100, 20), 10), const_100)))), const_100)
add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100)|
general
the length of a rectangle is increased by 35 % and its breadth is decreased by 20 % . what is the effect on its area ?
"100 * 100 = 10000 135 * 80 = 10800 answer : a"
a ) 5 , b ) 25 , c ) 10800 , d ) 1 / 2 , e ) 11 / 48,000
c
multiply(add(35, const_100), subtract(const_100, 20))
add(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|
geometry
express 25 mps in kmph ?
"25 * 18 / 5 = 90 kmph answer : c"
a ) 26 % , b ) 833 , c ) 90 , d ) 8.4 , e ) 9 : 25', '
c
multiply(divide(25, const_1000), const_3600)
divide(n0,const_1000)|multiply(#0,const_3600)|
physics
a can finish a work in 24 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ?
"b ' s 10 day ' s work = ( 1 x 10 ) = 2 . 15 3 remaining work = ( 1 - 2 ) = 1 . 3 3 now , 1 work is done by a in 1 day . 24 therefore 1 work is done by a in ( 24 x 1 ) = 8 days . e"
a ) 60 , b ) 4096 , c ) 8 , d ) 4 , e ) 50
c
divide(multiply(multiply(divide(const_1, 15), 10), 24), const_2)
divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)|
physics
a satellite is composed of 30 modular units , each of which is equipped with a set of sensors , some of which have been upgraded . each unit contains the same number of non - upgraded sensors . if the number of non - upgraded sensors on one unit is 1 / 6 the total number of upgraded sensors on the entire satellite , what fraction of the sensors on the satellite have been upgraded ?
"let x be the number of upgraded sensors on the satellite . the number of non - upgraded sensors per unit is x / 6 . the number of non - upgraded sensors on the whole satellite is 30 ( x / 6 ) = 5 x . the fraction of sensors which have been upgraded is x / ( x + 5 x ) = x / 6 x = 1 / 6 the answer is c ."
a ) rs . 5845 , b ) 5 , c ) 1 / 6 , d ) 1 % , e ) 2
c
divide(30, add(30, multiply(30, multiply(divide(1, 6), 30))))
divide(n1,n2)|multiply(n0,#0)|multiply(n0,#1)|add(n0,#2)|divide(n0,#3)|
general
a certain class of students is being divided into teams . the class can either be divided into 16 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ?
"let total no of students in the class be n so , we are told that n is divisible by both 16 24 so , lets find the least common multiple of 16 24 , ie 48 so our answer is ( c ) 48"
a ) 48 , b ) 2 : 3 , c ) 30 , d ) 240 , e ) 2,000
a
lcm(16, 24)
lcm(n0,n1)|
general
if 5 % more is gained by selling an article for rs . 350 than by selling it for rs . 320 , the cost of the article is
"explanation : let c . p . be rs . x . then , 5 % of x = 350 - 320 = 30 x / 20 = 30 = > x = 600 answer : e"
a ) 80 , b ) 600 , c ) 4749 , d ) 54 , e ) 20
b
divide(subtract(350, 320), divide(5, const_100))
divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)|
gain
a hall 36 m long and 15 m broad is to be paved with stones , each measuring 5 dm by 5 dm . the number of stones required is :
"area of the hall = 3600 * 1500 area of each stone = ( 50 * 50 ) therefore , number of stones = ( 3600 * 1500 / 50 * 50 ) = 2160 answer : d"
a ) $ 330 , b ) 2160 , c ) 45 , d ) 90 , e ) 10
b
divide(multiply(36, 15), divide(multiply(5, 5), const_100))
multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#0,#2)|
physics
on the number line , the number p is twice as many units from - 2 as - 2 is from 3 . if p is less than – 2 , what is the value of p ?
"distance between - 2 and 3 = 5 since number p is twice as many units from - 2 as - 2 is from 3 , therefore p can be - 12 since , p is less than - 2 , the value of p = - 12 answer a"
a ) 1200 , b ) 350.5 , c ) 0.0012 , d ) – 12 , e ) 8.5
d
subtract(negate(2), multiply(2, subtract(2, negate(2))))
negate(n2)|subtract(n0,#0)|multiply(#1,n0)|subtract(#0,#2)|
general
before leaving home for the town of madison , pete checks a map which shows that madison is 6 inches from his current location , gardensquare . pete arrives in madison 2.5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ?
"pete covered 2.5 * 60 = 150 miles which correspond to 6 inches on the map - - > scale in inches per mile is 6 / 150 = 1 / 25 . answer : a ."
a ) 0.28 % , b ) 100 , c ) 6 , d ) 7 % , e ) 1 / 25
e
divide(const_1, multiply(divide(2.5, 6), 60))
divide(n1,n0)|multiply(#0,n2)|divide(const_1,#1)|
physics
a person lent a certain sum of money at 5 % per annum at simple interest and in 8 years the interest amounted to $ 480 less than the sum lent . what was the sum lent ?
"p - 480 = ( p * 5 * 8 ) / 100 p = 800 the answer is b ."
a ) 427.5 , b ) 800 , c ) 7 , d ) 24 , e ) q = 1600
b
divide(480, subtract(const_1, divide(multiply(5, 8), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
gain
all numbers from 1 to 200 ( in decimal system ) are written in base 6 and base 7 systems . how many of the numbers will have a non - zero units digit in both base 6 and base 7 notations ?
detailed solution if a number written in base 6 ends with a zero , it should be a multiple of 6 . in other words , the question wants us to find all numbers from 1 to 200 that are not multiples of 6 or 7 . there are 33 multiples of 6 less than 201 . there are 28 multiples of 7 less than 201 . there are 4 multiples of 6 & 7 ( or multiple of 42 ) from 1 to 200 . so , total multiples of 6 or 7 less than 201 = 33 + 28 - 4 = 57 . number of numbers with non - zero units digit = 200 - 57 = 143 . correct answer : a
a ) 4 , b ) 75 , c ) 143', ' , d ) rs . 1425 , e ) $ 154.1
c
subtract(200, subtract(add(divide(200, 6), divide(200, 7)), divide(200, multiply(6, 7))))
divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)
other
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 25 seconds . find the length of the faster train ?
"relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 25 sec = 25 * 10 = 250 m . the length of the faster train = 250 m . answer : b"
a ) 4 % , b ) 250 , c ) 16 , d ) 35 min , e ) 2000
b
multiply(divide(subtract(72, 36), const_3_6), 25)
subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)|
physics
an urn contains 6 red , 5 blue and 2 green marbles . if 2 marbles are picked at random , what is the probability that both are red ?
option ( b ) is correct p ( both are red ) , 6 c 2 / 13 c 2 = 5 / 26 answer b
a ) 87.86 , b ) 6400 , c ) 50 % , d ) 200 cm 2' , e ) 5 / 26
e
divide(divide(multiply(6, 5), const_2), divide(multiply(add(add(6, 5), 2), subtract(add(add(6, 5), 2), const_1)), const_2))
add(n0,n1)|multiply(n0,n1)|add(n2,#0)|divide(#1,const_2)|subtract(#2,const_1)|multiply(#2,#4)|divide(#5,const_2)|divide(#3,#6)
other
if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 30 % of m is equal to what percent of 200 % of j ?
"imo answer should be 350 . . . consider j = 10 , then k = 50 , l = 150 and m = 350 . . . . 30 % of 350 , comes out to be 105 . . . . 200 % of 10 is 20 . . . . ( 105 * 100 ) / 20 = 525 . . . . ans : c"
a ) 10 , b ) 19 , c ) 525 , d ) 44 . , e ) 55
c
multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 30), 200), const_100)
multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100)|
gain
solve below question 2 x + 1 = - 15
"2 x + 1 = - 15 x = - 8 a"
a ) s . 230 , b ) 11 , c ) 123 , d ) 17 , e ) - 8
e
divide(negate(add(15, 1)), 2)
add(n1,n2)|negate(#0)|divide(#1,n0)|
general
a salesman ' s income consists of a commission and a base salary of $ 350 per week . over the past 5 weeks , his weekly income totals have been $ 556 , $ 413 , $ 420 , $ 436 and $ 395 . what must his average ( arithmetic mean ) commission be per week over the next two weeks so that his average weekly income is $ 500 over the 7 - week period ?
"total weekly income over 5 weeks = $ 556 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2220 for avg weekly income to be $ 500 over 7 weeks , we need total weekly income over 7 weeks = $ 3500 now , $ 3500 - $ 2220 = $ 1280 from this , we subtract base salary for 2 weeks i . e $ 350 * 2 = $ 700 therefore , commission = $ 1280 - $ 700 = $ 580 for 2 weeks avg weekly commission = $ 290 answer b"
a ) 700 , b ) 18', ' , c ) $ 290 , d ) 89 % , e ) 60
c
subtract(divide(subtract(multiply(500, 7), add(add(add(556, 413), add(436, 420)), 395)), const_2), 350)
add(n2,n3)|add(n4,n5)|multiply(n7,n8)|add(#0,#1)|add(n6,#3)|subtract(#2,#4)|divide(#5,const_2)|subtract(#6,n0)|
general
a man can row 4.8 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ?
"speed of boat in still water ( b ) = 4.8 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.6 km / hr . answer : e"
a ) 1.6 , b ) 65 , c ) 2 / 9 , d ) 49 : 64 , e ) 10750
a
divide(subtract(multiply(4.8, const_2), 4.8), const_3)
multiply(n0,const_2)|subtract(#0,n0)|divide(#1,const_3)|
general
125 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ?
"milk = 3 / 5 * 125 = 75 liters water = 50 liters 75 : ( 50 + p ) = 3 : 4 150 + 3 p = 400 = > p = 50 50 liters of water are to be added for the ratio become 3 : 4 . answer : d"
a ) 50 liters , b ) 1 ⁄ 10 , c ) 480 , d ) 4 , e ) 39
a
multiply(divide(125, add(3, 2)), 2)
add(n1,n2)|divide(n0,#0)|multiply(n2,#1)|
general
a pupil ' s marks were wrongly entered as 35 instead of 23 . due to that the average marks for the class got increased by half . the number of pupils in the class is :
let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 . x / 2 = ( 35 - 23 ) = > x / 2 = 12 = > x = 24 . answer : e
a ) 40 , b ) 567 , c ) 24 , d ) 20 minutes , e ) 158
c
multiply(subtract(35, 23), const_2)
subtract(n0,n1)|multiply(#0,const_2)
general
there are 6 baskets numbered from 1 to 6 and filled with apples . 10 children are asked to pick apples one by one from each basket such that the number of apples picked by each child from each basket is equal to the number marked on the basket . if there were 1000 apples in total and the baskets were filled in such a way that none of the basket became empty in the process , how many apples were left in the end ?
"total number of apples in the basket initially = 1000 each child picks up = 1 + 2 + 3 + 4 + 5 + 6 = 6 * 7 / 2 = 21 total number of apples picked up by 10 children = 21 * 10 = 210 number of apples left in the end = 1000 - 210 = 790 answer is d"
a ) s . 129.76 , b ) 5.5 , c ) 790 , d ) 35 min , e ) 16
c
subtract(1000, multiply(add(add(multiply(add(1, 6), divide(10, const_2)), divide(10, const_2)), 1), 10))
add(n0,n1)|divide(n3,const_2)|multiply(#0,#1)|add(#1,#2)|add(#3,n1)|multiply(#4,n3)|subtract(n4,#5)|
general
in a lottery there are 100 prizes and 300 blanks . a lottery is drawn at random . what is the probability of getting a prize ?
"total draws = prizes + blanks = 100 + 300 = 400 probability of getting a prize = 100 / 400 = 1 / 4 correct option is e"
a ) 81 , b ) 12 hours , c ) 6.56 kg , d ) 47.4 % , e ) 1 / 4
e
divide(300, add(100, 300))
add(n0,n1)|divide(n1,#0)|
probability
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 400 resolutions ?
"in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 400 * 2 * 22 / 7 * 22.4 = 56320 cm = 563.2 m answer : e"
a ) – 12 , b ) 40 , c ) 1804 , d ) 9 , e ) 563.2 m
e
divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 400), const_100)
add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)|
physics
q is as much younger than r as he is older than t . if the sum of the ages of r and t is 50 years , what is definitely the difference between r and q ' s age ?
explanation : given r – q = q – t and r + t = 50 which gives q = 25 as the difference between r & q and q & t is same so answer is 25 years answer : c
a ) 25 , b ) 6 2 / 3 days , c ) 80 % , d ) 90 , e ) 3
a
divide(50, const_2)
divide(n0,const_2)
general
if x and y are integers , what is the least positive number of 24 x + 21 y ?
"24 x + 21 y = 3 ( 8 x + 7 y ) which will be a minimum positive number when 8 x + 7 y = 1 . 8 ( 1 ) + 7 ( - 1 ) = 1 then 3 ( 8 x + 7 y ) can have a minimum positive value of 3 . the answer is a ."
a ) 4 min , b ) 1.33 , c ) 72 , d ) 18 , e ) 3
e
subtract(24, 21)
subtract(n0,n1)|
general
the average of runs of a cricket player of 20 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 3 ?
"average = total runs / no . of innings = 32 so , total = average x no . of innings = 32 * 20 = 640 now increase in avg = 4 runs . so , new avg = 32 + 3 = 35 runs total runs = new avg x new no . of innings = 35 * 21 = 735 runs made in the 11 th inning = 735 - 640 = 95 answer : a"
a ) 95 , b ) 78 , c ) 500 m , d ) 0.3571 , e ) $ 1200
a
subtract(multiply(add(20, const_1), add(3, 32)), multiply(20, 32))
add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
general
a sum of money deposited at c . i . amounts to rs . 500 in 3 years and to rs . 650 in 4 years . find the rate percent ?
"500 - - - 150 100 - - - ? = > 30 % answer : b"
a ) 151 / 31 , b ) 2 : 1 , c ) 30 % , d ) 4.5 sec , e ) 5 / 9
c
multiply(divide(subtract(650, 500), 500), const_100)
subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 36400 , how many months will it take joan to pay back her mortgage ?
"joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 36400 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 364 so we have 1 , 3 , 9 , 27 . . . . . 36400 in gp we know that a = 1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula tn = a 3 ^ n - 1 . . . ) therefore to find the sum of n terms of a gp we use this formula : sn = a ( 1 - r ^ n ) / 1 - r using this and plugging in the information we get . . . 364 = 1 - 3 ^ n / 1 - 3 ; 1 - 3 ^ n / - 2 cross multiplying we get 364 x - 2 = 1 - 3 ^ n - 728 = 1 - 3 ^ n - 729 = - 3 ^ n 729 = 3 ^ n ( negatives cancel out ) 729 can also be re written as 3 ^ 6 therefore ; 3 ^ 6 = 3 ^ n thus n = 6 ( a )"
a ) 4 , b ) 18 , c ) 4264 , d ) rs . 960 , e ) 6
e
divide(log(add(divide(multiply(36400, const_2), 100), const_1)), log(const_3))
log(const_3)|multiply(n1,const_2)|divide(#1,n0)|add(#2,const_1)|log(#3)|divide(#4,#0)|
general
the average of first five multiples of 6 is
"solution average = 6 ( 1 + 2 + 3 + 4 + 5 ) / 5 = 90 / 5 . = 18 answer b"
a ) 1000 , b ) 18 , c ) 40 , d ) 11 , e ) $ 175
b
add(6, const_1)
add(n0,const_1)|
general
a big container is 35 % full with water . if 16 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container in liters ?
"16 liters is 40 % of the capacity c . 16 = 0.4 c c = 16 / 0.4 = 40 liters . the answer is b ."
a ) 40 , b ) $ 333.33 , c ) 1173.98 , d ) 1000 , e ) 88888883
a
divide(16, subtract(divide(3, 4), divide(35, const_100)))
divide(n2,n3)|divide(n0,const_100)|subtract(#0,#1)|divide(n1,#2)|
general
in objective test a correct ans score 4 marks and on a wrong ans 2 marks are - - - . a student score 480 marks from 150 question . how many ans were correct ?
let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = 150 = > 4 x - 2 y = 480 = > 6 x = 780 hence x = 130 answer : b
a ) 5 / 4 , b ) 4 , c ) 10 hr , d ) 130 , e ) 70400 yards
d
divide(add(480, multiply(150, 2)), add(4, 2))
add(n0,n1)|multiply(n1,n3)|add(n2,#1)|divide(#2,#0)
general
a number when divided by a certain divisor left remainder 251 , when twice the number was divided by the same divisor , the remainder was 112 . find the divisor ?
"easy solution : n = dq 1 + 251 2 n = 2 dq 1 + 502 - ( 1 ) 2 n = dq 2 + 112 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 390 d * some integer = 390 checking all options only ( c ) syncs with it . answer c"
a ) 22 , b ) 20.5 , c ) 390 , d ) 108 , e ) 2.25
c
subtract(multiply(251, const_2), 112)
multiply(n0,const_2)|subtract(#0,n1)|
general
excluding stoppages , the average speed of a bus is 120 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ?
"in 1 hr , the bus covers 120 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 120 - 40 ) km i . e 80 km at 120 km / hr . stoppage time = 80 / 120 hrs = 40 min answer : e"
a ) 130 , b ) 11 , c ) 10.6 , d ) 2 : 3 , e ) 40 min
e
subtract(multiply(const_1, const_60), multiply(divide(40, 120), const_60))
divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)|
general
the units digit of ( 10 ) ^ ( 87 ) + ( 93 ) ^ ( 46 ) is :
"first part will be zero always 2 nd part with power of three therefore , the second term has a units digit of 9 . of course 0 + 9 = 9 , a"
a ) 9 , b ) 26250 , c ) 1 , d ) 36.92 mph , e ) 350.5
a
add(reminder(multiply(reminder(46, const_4), 93), const_10), reminder(10, const_10))
reminder(n3,const_4)|reminder(n0,const_10)|multiply(n2,#0)|reminder(#2,const_10)|add(#3,#1)|
general
consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 9900 interest only for the period . the principal amount borrowed by her was
"explanation : principal = rs . ( 100 × 9900 / 12 × 3 ) = > rs . 27500 . answer : b"
a ) rs . 27500 , b ) 240 , c ) 380 , d ) 6 days , e ) 12
a
divide(9900, divide(multiply(3, 12), const_100))
multiply(n0,n1)|divide(#0,const_100)|divide(n2,#1)|
gain
if the cost price is 96 % of the selling price , then what is the profit percent ?
"let s . p . = $ 100 c . p . = $ 96 profit = $ 4 profit % = 4 / 96 * 100 = 25 / 6 = 4.17 % answer is a"
a ) 4.17 % , b ) 5 % , c ) 4 , d ) 2 , e ) 2 / 3
a
multiply(divide(subtract(const_100, 96), 96), const_100)
subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)|
gain
how many multiples of 5 are there between 70 and 358 ?
"5 * 14 = 70 5 * 71 = 355 total no of multiples = ( 71 - 14 ) + 1 = 57 + 1 = 58 answer is e ."
a ) 90 , b ) 6 , c ) 90 % , d ) 29 , e ) 58
e
add(divide(subtract(358, 70), 5), const_1)
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
general
a , b and c invests rs . 6000 , rs . 5000 and rs . 3000 in a business . after one year c removed his money ; a and b continued the business for two more year . if the net profit after 3 years be rs . 4206 , then c ' s share in the profit is ?
"6 * 36 : 5 * 36 : 3 * 12 6 : 5 : 1 1 / 12 * 4206 = 350.50 answer : d"
a ) 0.4 , b ) 351 , c ) 350.5 , d ) 270 , e ) 3
c
multiply(divide(6000, add(6000, add(multiply(5000, 3), multiply(3000, 3)))), 6000)
multiply(n1,n3)|multiply(n2,n3)|add(#0,#1)|add(n0,#2)|divide(n0,#3)|multiply(n0,#4)|
gain
a sum of money deposited at c . i . amounts to rs . 5000 in 5 years and to rs . 5750 in 6 years . find the rate percent ?
5000 - - - 750 100 - - - ? = > 15 % answer : d
a ) 135 % , b ) 2 , c ) 10 kmph , d ) 88.6 , e ) 15 %
e
multiply(divide(subtract(5750, 5000), 5000), const_100)
subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)
gain
renu can do a piece of work in 6 days , but with the help of her friend suma , she can do it in 5 days . in what time suma can do it alone ?
renu â € ™ s one day â € ™ s work = 1 / 6 suma â € ™ s one day â € ™ s work = 1 / 5 - 1 / 6 = 1 / 30 suma can do it alone in 30 days . answer : e
a ) − 120 , b ) 7 days , c ) 12 , d ) 5 : 6 , e ) 30
e
inverse(subtract(divide(const_1, 5), divide(const_1, 6)))
divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)
physics
if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 160 units ?
"here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 160 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 160 units is ' 8 times ' 20 , we need ' 8 times ' more time . ( 2.5 hours ) ( 8 times ) = 20 hours final answer : [ reveal ] spoiler : c"
a ) 10 , b ) 3 % , c ) 20 hours , d ) 11 , e ) 100 coins
c
divide(160, multiply(divide(divide(20, 10), 5), 20))
divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)|
physics
a policeman noticed a criminal from a distance of 200 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 3 minutes ?
"explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 3 / 60 ) km = 1 / 20 km = 50 m . . ' . distance between the criminal and policeman = ( 200 - 50 ) m = 150 m . answer : d"
a ) 150 m , b ) 5 , c ) 2.14 , d ) 2599980 , e ) 254 m 2
a
subtract(200, multiply(divide(3, const_60), const_1000))
divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)|
physics
the average of 6 no . ' s is 3.95 . the average of 2 of them is 4 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ?
"sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 4 * 2 ) + ( 3.85 * 2 ) ] = 8 required average = ( 8 / 2 ) = 4 e"
a ) 6 days , b ) 1.6021 , c ) 19.0 % , d ) 4 , e ) 30 %
d
divide(subtract(multiply(6, 3.95), add(multiply(2, 4), multiply(2, 3.85))), 2)
multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2)|
general
the population of a town increased from 50000 to 80000 in a decade . the average percent increase of population per year is :
solution increase in 10 year = ( 80000 - 50000 ) = 30000 . increase % = ( 30000 / 50000 x 100 ) % = 60 % â ˆ ´ required average = ( 60 / 10 ) % = 6 % answer c
a ) 2 , b ) 6 % , c ) 15 , d ) $ 214.16 , e ) 17 %
b
divide(multiply(divide(subtract(80000, 50000), 50000), const_100), const_10)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|divide(#2,const_10)
general
calculate the speed of a boat in still water ( in km / hr ) if in one hour , the boat goes 13 km / hr downstream and 10 km / hr upstream .
"speed in still water = ( 13 + 5 ) 1 / 2 kmph = 9 kmph . answer : b"
a ) 9 kmph , b ) 492 , c ) 240 , d ) $ 55.55 , e ) 5
a
divide(add(13, 10), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
in a single throw of a die , what is the probability of getting a number greater than 2 ?
"s = { 1,2 , 3,4 , 5,6 } e = { 3,4 , 5,6 } probability = 4 / 6 = 2 / 3 answer is e"
a ) 115 , b ) 2 / 3 , c ) 25 , d ) 75 , e ) 9216
b
divide(const_2, add(2, const_2))
add(n0,const_2)|divide(const_2,#0)|
probability
what is the probability that when a hand o f 6 cards is drawn from a well shuffled deck of 52 cards , it contains 2 queen ?
ncr = n ! / ( n - r ) ! r ! total number of possible hands = 52 c 6 52 c 6 = ( 52 ! ) / ( ( 52 - 6 ) ! × 6 ! ) 52 c 6 = 61075560 . number of hands with 2 queen and 4 non - queen cards = 4 c 2 × 48 c 4 4 c 2 = ( 4 ! ) / ( 2 ! × 2 ! ) = 6 . 48 c 4 = ( 48 ! ) / ( 44 ! × 4 ! ) = 3 × 47 × 46 × 45 = 291870 ( other 2 cards must be chosen from the rest 48 cards ) p ( 2 queen ) = ( 4 c 2 × 48 c 4 ) / 52 c 6 = 29187 / 1017926 c
a ) 12', ' , b ) 12 , c ) 29187 by 1017926 , d ) 60 , e ) 4 / 1
c
divide(multiply(choose(const_4, 2), choose(subtract(52, const_4), subtract(6, 2))), choose(52, 6))
choose(const_4,n2)|choose(n1,n0)|subtract(n1,const_4)|subtract(n0,n2)|choose(#2,#3)|multiply(#0,#4)|divide(#5,#1)
probability
carina has 130 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ?
"lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 130 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 130 - > y = 120 / 15 . = 8 d"
a ) 9 / 49 , b ) 20 sec , c ) 1,350 , d ) 8 , e ) 1200
d
divide(subtract(130, multiply(5, 2)), add(10, 5))
add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0)|
general
if x = 1 / q and y = ( 2 / q ) - 6 , then for what value of q , x is equal to y ?
explanation : x = y < = > 1 / q = ( 2 / q ) - 6 < = > 1 / q = 6 < = > q = 1 / 6 . answer : b
a ) 4 cm', ' , b ) 3 : 2 , c ) 60000 , d ) 1 / 6 , e ) 440
d
divide(subtract(2, 1), 6)
subtract(n1,n0)|divide(#0,n2)
general
a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ?
"explanation : 90 % 104 % - - - - - - - - 14 % - - - - 140 100 % - - - - ? = > rs . 1000 a )"
a ) 32.8 % , b ) 86400 , c ) rs . 1000 , d ) 2011 . , e ) 14
c
divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 10)))
add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)|
gain
arun and tarun can do a work in 10 days . after 4 days tarun went to his village . how many days are required to complete the remaining work by arun alone . arun can do the work alone in 50 days .
they together completed 4 / 10 work in 4 days . balance 6 / 10 work will be completed by arun alone in 50 * 6 / 10 = 30 days . answer : d
a ) 30 days . , b ) 500 , c ) 0 , d ) 778 , e ) 870
a
subtract(50, multiply(divide(50, 10), 4))
divide(n2,n0)|multiply(n1,#0)|subtract(n2,#1)
physics
a certain telescope increases the visual range at a particular location from 70 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ?
"original visual range = 70 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 70 ) / 70 * 100 % = 8 / 7 * 100 % = 114.28 % answer e"
a ) 114.28 % , b ) 7 days , c ) 1652 , d ) 2 : 3 , e ) 43
a
multiply(divide(subtract(150, 70), 70), const_100)
subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)|
general
circular gears l and r start to rotate at the same time at the same rate . gear l makes 10 complete revolutions per minute and gear r makes 40 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 9 more revolutions than gear l ?
"gear l - - 10 rotations per 60 seconds - - 1 rotation per 6 seconds . gear r - - 40 rotations per 60 seconds - - 4 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 3 rotations hence every 6 seconds the difference between the number of rotations of r and l gear is 3 units . required net difference should be 9 rotations = > 3 ( 6 seconds later ) = = > 18 seconds . answer : d ) ."
a ) 150 , b ) 5 % , c ) 6 , d ) 18 , e ) 3 : 2
d
divide(divide(9, subtract(divide(40, const_60), divide(10, const_60))), const_3)
divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|divide(#3,const_3)|
physics
find the simple interest on rs . 68,000 at 16 2 / 3 % per annum for 9 months .
"p = rs . 68000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 68,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8500 answer is a ."
a ) 310 , b ) 4 , c ) 4000 , d ) rs . 8500 , e ) 26 %
d
multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100))
add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)|
gain
three cubes of iron whose edges are 6 cm , 8 cm and 10 cm respectively are melted and formed into a single cube . the edge of the new cube formed is
sol . volume of the new cube = ( 63 + 83 + 103 ) cm 3 = 1728 cm 3 . let the edge of the new cube be a cm . ∴ a 3 = 1728 ⇒ a = 12 . answer b
a ) 1,350 , b ) 38 kg , c ) 28 , d ) 12 cm', ' , e ) 15 π
d
cube_edge_by_volume(add(volume_cube(10), add(volume_cube(6), volume_cube(8))))
volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|cube_edge_by_volume(#4)
physics
two trains are moving in the same direction at 126 kmph and 54 kmph . the faster train crosses a man in the slower train in 14 seconds . find the length of the faster train ?
"relative speed = ( 126 - 54 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 27 sec = 14 * 20 = 280 m . the length of the faster train = 280 m . answer : a"
a ) 45 , b ) 280 , c ) 9 , d ) 3 : 2 , e ) 240
b
multiply(divide(subtract(126, 54), const_3_6), 14)
subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)|
physics
if n is a prime number greater than 17 , what is the remainder when n ^ 2 is divided by 12 ?
"there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 17 , square it and see what would be the remainder upon division of it by 12 . n = 19 - - > n ^ 2 = 361 - - > remainder upon division 361 by 12 is 1 . answer : b ."
a ) 45 , b ) 1 , c ) 70 , d ) 10 / 3 , e ) 50
b
subtract(power(add(17, 2), 2), multiply(12, const_4))
add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)|
general