Problem
stringlengths 5
967
| Rationale
stringlengths 1
2.74k
| options
stringlengths 37
300
| correct
stringclasses 5
values | annotated_formula
stringlengths 7
6.48k
| linear_formula
stringlengths 8
925
| category
stringclasses 6
values |
|---|---|---|---|---|---|---|
a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 15 per metre of cloth . what is the cost price of one metre of cloth ?
|
"sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 15 = rs . 90 . answer : c"
|
a ) rs . 90 , b ) rs . 76 , c ) rs . 98 , d ) rs . 18 , e ) rs . 02
|
c
|
subtract(divide(8925, 85), 15)
|
divide(n1,n0)|subtract(#0,n2)|
|
physics
|
20 % of a number is added to 120 , the result is the same number . find the number ?
|
"( 20 / 100 ) * x + 120 = x 2 x = 300 x = 150 answer : a"
|
a ) 150 , b ) 288 , c ) 1687 , d ) 997 , e ) 121
|
a
|
divide(120, divide(120, const_100))
|
divide(n1,const_100)|divide(n1,#0)|
|
gain
|
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5830 , what is the length of the plot in metres ?
|
"let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 5830 [ 4 l - 40 ] = 5830 / 26.50 = 220 4 l = 260 l = 260 / 4 = 65 meters . answer : e"
|
a ) 333 , b ) 200 , c ) 288 , d ) 276 , e ) 220
|
e
|
subtract(divide(divide(5830, 26.50), const_2), multiply(const_2, 20))
|
divide(n2,n1)|multiply(n0,const_2)|divide(#0,const_2)|subtract(#2,#1)|
|
physics
|
there is a train and car . the ratio between the speed of a train & a car is 16 : 15 respectively . also , a bus covered a distance of 480 km in 8 hours . the speed of the bus is 3 / 4 th of the speed of the train . how much distance will the car cover in 5 hours ?
|
"the speed of the bus is 480 / 8 = 60 km / hr the speed of the train is ( 60 * 4 ) / 3 = 80 km / hr the speed of the car is 80 / 16 * 15 = 75 km / hr the distance covered by the car in 5 hours is 75 Γ 5 = 375 km the answer is c ."
|
a ) 300 km , b ) 325 km , c ) 375 km , d ) 425 km , e ) 450 km
|
c
|
multiply(divide(multiply(divide(multiply(divide(480, 4), 8), 3), 15), 16), 5)
|
divide(n2,n5)|multiply(n3,#0)|divide(#1,n4)|multiply(n1,#2)|divide(#3,n0)|multiply(n6,#4)|
|
physics
|
if a town of 65,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ?
|
1 % is quite small and hence the answer is a )
|
a ) a . 66000 , b ) b . 67000 , c ) c . 68000 , d ) d . 69000 , e ) e . 70000
|
a
|
multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3))
|
add(const_3,const_4)|add(#0,const_3)|multiply(n2,#0)|multiply(#2,const_100)|multiply(#1,#3)|
|
gain
|
two trains , a and b , started simultaneously from opposite ends of a 75 - mile route and traveled toward each other on parallel tracks . train a , traveling at a constant rate , completed the 75 - mile trip in 3 hours ; train b , traveling at a constant rate , completed the 75 - mile trip in 2 hours . how many miles had train a traveled when it met train b ?
|
"as the ratio of the rates of a and b is 2 to 3 then the distance covered at the time of the meeting ( so after traveling the same time interval ) would also be in that ratio , which means that x would cover 2 / ( 2 + 3 ) = 2 / 5 of 75 miles : 75 * 2 / 5 = 30 miles . answer : c ."
|
a ) 20 , b ) 22.5 , c ) 30 , d ) 37.5 , e ) 45
|
c
|
multiply(divide(2, add(3, 2)), 75)
|
add(n2,n4)|divide(n4,#0)|multiply(n0,#1)|
|
physics
|
in one alloy there is 12 % chromium while in another alloy it is 8 % . 15 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy .
|
"the amount of chromium in the new 15 + 30 = 45 kg alloy is 0.12 * 15 + 0.08 * 30 = 4.2 kg , so the percentage is 4.2 / 45 * 100 = 9.33 % . answer : b ."
|
a ) 9.4 % , b ) 9.33 % , c ) 9.6 % , d ) 9.8 % , e ) 10 %
|
b
|
multiply(divide(add(divide(multiply(12, 15), const_100), divide(multiply(8, 30), const_100)), add(15, 30)), const_100)
|
add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|
|
gain
|
what is the remainder when 135 ^ 77 is divided by 7 ?
|
"( 15 ^ 77 * 9 ^ 77 ) / 7 ; ( 15 ^ 77 ) / 7 * ( 9 ^ 77 ) / 7 ; 1 * 2 ^ 77 / 7 ; ( ( 2 ^ 3 ) ^ 25 * 2 ^ 2 ) / 7 ; 1 * 4 = 4 answer : e"
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
|
e
|
subtract(divide(const_100.0, const_2), multiply(135, 135))
|
divide(const_100.0,const_2)|multiply(n0,n0)|subtract(#0,#1)|
|
general
|
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 45 . what is the difference between the two digits of that number ?
|
"sol . let the ten β s digit be x and unit β s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 45 β 9 ( x - y ) = 45 β x - y = 5 answer b"
|
a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9
|
b
|
divide(45, subtract(const_10, const_1))
|
subtract(const_10,const_1)|divide(n0,#0)|
|
general
|
if 42.30 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ?
|
"42.30 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.30 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.30 = km / 50 . . . . . . 40 / 100 = 3 m / 50 . . . . . . m = 5 k + m = 3 + 5 = 8 answer : c"
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
|
c
|
add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.30, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.30, subtract(const_0_25, divide(const_1, const_100))), 14))
|
divide(const_1,const_100)|subtract(const_0_25,#0)|multiply(n2,#1)|subtract(n0,#1)|divide(#3,n1)|divide(#2,#4)|add(#5,#4)|
|
general
|
a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many j cubes are painted on only 2 faces ?
|
"1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all on the edge of the cube , in themiddleof the edge . there are 4 in front , 4 in back and 4 more on thestripthat runs around the left / top / right / bottom of the cube . j = 4 + 4 + 4 = 12 . answer a"
|
a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16
|
a
|
multiply(const_4, power(27, divide(const_1, const_3)))
|
divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)|
|
geometry
|
at present the ratio between the ages of arun and deepak is 5 : 7 . after 6 years , arun ' s age will be 36 years . what is the age of deepak at present ?
|
"let the present ages of arun and deepak be 5 x years and 7 x years respectively 5 x + 6 = 36 5 x = 30 x = 6 deepak ' s age = 7 x = 42 years answer is c"
|
a ) 35 years , b ) 25 years , c ) 42 years , d ) 39 years , e ) 40 years
|
c
|
divide(multiply(subtract(36, 6), 7), 5)
|
subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|
|
other
|
the average age of 15 students of a class is 16 years . out of these , the average age of 5 students is 14 years and that of the other 10 students is 16 years . tee age of the 15 th student is ?
|
"age of the 15 th student = [ 15 * 16 - ( 14 * 5 + 16 * 10 ) ] = ( 240 - 230 ) = 10 years . answer : e"
|
a ) 18 years , b ) 16 years , c ) 14 years , d ) 12 years , e ) 10 years
|
e
|
subtract(multiply(15, 15), add(multiply(5, 14), multiply(10, 16)))
|
multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)|
|
general
|
excluding stoppages , the speed of a bus is 75 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ?
|
"due to stoppages , it covers 30 km less . time taken to cover 30 km = ( ( 30 / 75 ) Γ£ β 60 ) = 24 min . option ( b ) is correct"
|
a ) 15 , b ) 24 , c ) 17 , d ) 19 , e ) none of these
|
b
|
multiply(const_60, divide(subtract(75, 45), 75))
|
subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)|
|
physics
|
the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 130 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ?
|
if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 130 group # 2 = ( like understood ) = 3 / 4 ( 1 / 4 ) n = 130 n = 520 answer = ( e )
|
a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 520
|
e
|
divide(130, subtract(const_1, divide(3, 4)))
|
divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)
|
general
|
when a certain perfect square is increased by 148 , the result is another perfect square . what is the value of the original perfect square ?
|
let β s call the two perfect squares x ^ 2 and y ^ 2 , respectively . then the given information translates as x ^ 2 + 148 = y ^ 2 . subtracting x ^ 2 gives 148 = y ^ 2 β x ^ 2 , a difference of squares . this , in turn , factors as ( y + x ) ( y β x ) = 148 . the next step is tricky . it begins with factoring 148 , which breaks down as 2 β 2 β 37 . since we β re dealing with perfect squares , x and y are positive integers , and ( y + x ) and ( y β x ) must be paired integer factors of 148 . the options are 148 β 1,74 β 2 , and 37 β 4 . but our number properties establish that ( y + x ) and ( y β x ) must be either both odd or both even , so only 74 β 2 is an actual possibility . and because for any positive integers ( y + x ) > ( y β x ) , we can conclude that y + x = 74 and y β x = 2 . solving by elimination , 2 y = 76 , y = 38 , and x = 36 . finally , we just need to square 36 . but rather than multiplying it out , note that 36 ^ 2 ends in 6 β as does only one answer , choice a . this answer must be the one we want .
|
['a ) 1296', 'b ) 1369', 'c ) 1681', 'd ) 1764', 'e ) 2500']
|
a
|
multiply(multiply(multiply(const_3, const_3), const_4), multiply(multiply(const_3, const_3), const_4))
|
multiply(const_3,const_3)|multiply(#0,const_4)|multiply(#1,#1)
|
geometry
|
train a is 13 miles ahead of train b , which is traveling the same direction along the same route as train a . if train a is traveling at an average speed of 37 miles per hour and train b is traveling at an average speed of 43 miles per hour , how long will it take train b to overtake and drive 17 miles ahead of train a ?
|
relative speed = 43 - 37 = 6 miles per hour dist required = 13 + 17 = 30 miles time taken to overtake = 30 / 6 = 5 hours . d is the answer .
|
a ) 4 hours , b ) 2 hours , c ) 2 hours 30 minutes , d ) 5 hours , e ) 3 hours 20 minutes
|
d
|
divide(add(13, 17), subtract(43, 37))
|
add(n0,n3)|subtract(n2,n1)|divide(#0,#1)
|
physics
|
x , y , and z are all unique numbers . if x is chosen randomly from the set { 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ?
|
"p ( x is prime ) = 1 / 3 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 3 * 1 / 4 = 1 / 12 the answer is b ."
|
a ) 1 / 20 , b ) 1 / 12 , c ) 2 / 15 , d ) 3 / 10 , e ) 1 / 7
|
b
|
multiply(divide(const_1, const_2), divide(const_1, const_4))
|
divide(const_1,const_2)|divide(const_1,const_4)|multiply(#0,#1)|
|
probability
|
on increasing the price of t . v . sets by 30 % , their sale decreases by 20 % . what is the effect on the revenue receipts of the shop ?
|
"explanation : let the price be = rs . 100 , and number of units sold = 100 then , sale value = rs . ( 100 Γ 100 ) = rs . 10000 new sale value = rs . ( 130 Γ 80 ) = rs . 10400 increase % = 4001000040010000 Γ 100 = 4 % answer : a"
|
a ) 4 , b ) 5 , c ) 56 , d ) 2 , e ) 7
|
a
|
subtract(add(const_100, 30), add(divide(multiply(add(const_100, 30), 20), const_100), const_100))
|
add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#2,const_100)|subtract(#0,#3)|
|
gain
|
if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 10 percent , how much would she earn per week ?
|
"( 348 / 116 ) 110 = 330 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 110 = 330 ( 300 + 30 ) answer b"
|
a ) $ 374 , b ) $ 330 , c ) $ 385 , d ) $ 392 , e ) $ 399
|
b
|
add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(10, const_100), divide(348, add(const_1, divide(16, const_100)))))
|
divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4)|
|
general
|
if the charge of staying in a student youth hostel $ 18.00 / day for the first week , and $ 14.00 / day for each additional week , how much does it cost to stay for 23 days ?
|
"total number of days of stay = 23 charge of staying in first week = 18 * 7 = 126 $ charge of staying for additional days = ( 23 - 7 ) * 14 = 16 * 14 = 224 $ total charge = 126 + 224 = 350 $ answer b"
|
a ) $ 160 , b ) $ 350 , c ) $ 282 , d ) $ 274 , e ) $ 286
|
b
|
add(multiply(18.00, add(const_3, const_4)), multiply(14.00, subtract(23, add(const_3, const_4))))
|
add(const_3,const_4)|multiply(n0,#0)|subtract(n2,#0)|multiply(n1,#2)|add(#1,#3)|
|
general
|
mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 125 mangoes .
|
"mr . john used to purchase certain number of mangoes for $ 360 since the price of mangoes is reduced by 10 % he got 12 more mangoes today . find the original price of 120 mangoes . method 1 : let price per mango = x . let number of mangoes be n . then , nx = 360 . now price = 0.9 x ; number of mangoes = n + 12 . total amount = 0.9 x * ( n + 12 ) = 360 . nx = 0.9 nx + 10.8 x = > 0.1 nx = 10.8 x = > n = 108 = > x = 360 / 108 = 3.33 original price of 125 mangoes = 125 * 3.33 = 416 . answer d"
|
a ) 360 , b ) 380 , c ) 400 , d ) 416 , e ) 412
|
d
|
multiply(divide(subtract(divide(360, subtract(const_1, divide(const_1, 10))), 360), 12), 125)
|
divide(const_1,n1)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|divide(#3,n2)|multiply(n3,#4)|
|
gain
|
calculate the time it will take for a full tank of water to become completely empty due to a leak given that the tank could be filled in 10 hours , but due to the leak in its bottom it takes 11 hours to be filled ?
|
part filled without leak in 1 hour = 1 / 10 part filled with leak in 1 hour = 1 / 11 work done by leak in 1 hour = 1 / 10 Γ’ Λ β 1 / 11 = 110 hours answer : e
|
a ) 90 hours , b ) 80 hours , c ) 50 hours , d ) 120 hours , e ) 110 hours
|
e
|
divide(multiply(10, 11), subtract(divide(multiply(10, 11), 10), divide(multiply(10, 11), 11)))
|
multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|subtract(#1,#2)|divide(#0,#3)
|
physics
|
in the first 25 overs of a cricket game , the run rate was only 5.4 . what should be the run rate in the remaining 25 overs to reach the target of 400 runs ?
|
required run rate = 400 - ( 5.4 x 25 ) / 25 = 265 / 25 = 10.6 option c
|
a ) 16.5 , b ) 13.2 , c ) 10.6 , d ) 11.2 , e ) 7.25
|
c
|
divide(subtract(400, multiply(25, 5.4)), 25)
|
multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)|
|
gain
|
if one positive integer is greater than another positive integer by 3 , and the difference of their cubes is 279 , what is their sum ?
|
"1 ^ 3 = 1 2 ^ 3 = 8 3 ^ 3 = 27 4 ^ 3 = 64 5 ^ 3 = 125 6 ^ 3 = 216 7 ^ 3 = 343 the two numbers are 4 and 7 . the answer is d ."
|
a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 13
|
d
|
add(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(279, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), subtract(divide(add(power(3, 3), sqrt(add(power(power(3, 3), const_2), power(subtract(279, power(3, 3)), const_2)))), multiply(power(3, const_2), const_2)), 3))
|
power(n0,n0)|power(n0,const_2)|multiply(#1,const_2)|power(#0,const_2)|subtract(n1,#0)|power(#4,const_2)|add(#3,#5)|sqrt(#6)|add(#0,#7)|divide(#8,#2)|subtract(#9,n0)|add(#9,#10)|
|
general
|
a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 12100 after a year ?
|
"explanation : 6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 12100 = 3630 answer : a"
|
a ) 3630 , b ) 3877 , c ) 2667 , d ) 2977 , e ) 2378
|
a
|
multiply(divide(6300, add(add(6300, 4200), 10500)), 12100)
|
add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)|
|
gain
|
what is the unit digit in 4 ^ 100 ?
|
unit digit in 4 ^ 100 = unit digit in [ ( 4 ^ 4 ) ^ 25 ] but unit digit in 4 ^ 4 = 6 unit digit 4 ^ 100 = 6 answer is c
|
a ) 0 , b ) 1 , c ) 6 , d ) 3 , e ) 5
|
c
|
reminder(power(4, const_2), const_10)
|
power(n0,const_2)|reminder(#0,const_10)
|
general
|
the average age of 37 students in a group is 14 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ?
|
"explanation : age of the teacher = ( 38 * 15 - 37 * 14 ) years = 52 years . answer : d"
|
a ) 35 years , b ) 45 years , c ) 51 years , d ) 52 years , e ) none of these
|
d
|
add(37, const_1)
|
add(n0,const_1)|
|
general
|
pipe a can fill a tank in 12 hours . due to a leak at the bottom , it takes 18 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ?
|
"let the leak can empty the full tank in x hours 1 / 12 - 1 / x = 1 / 18 = > 1 / x = 1 / 12 - 1 / 18 = ( 3 - 2 ) / 36 = 1 / 36 = > x = 36 . answer : a"
|
a ) 36 , b ) 87 , c ) 40 , d ) 37 , e ) 86
|
a
|
divide(multiply(18, 12), subtract(18, 12))
|
multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)|
|
physics
|
if one and a half women , drink one and a half tea in one and a half minutes . how many tea can 9 women drink in 3 minutes ?
|
e 18 explanation : more minutes implies more tea . more women implies more tea . the time become twice ( 3 minutes / 1.5 minutes ) . women become six times ( 9 / 1.5 ) . number of tea = 2 * 6 * original ( 1.5 )
|
a ) 11 , b ) 8 , c ) 37 , d ) 15 , e ) 18
|
e
|
divide(multiply(9, 3), divide(const_3, const_2))
|
divide(const_3,const_2)|multiply(n0,n1)|divide(#1,#0)
|
physics
|
a can run 2 km distance in 2 min 20 seconds , while b can run this distance in 3 min 5 sec . . by how much distance can a beat b ?
|
a takes time 2.20 minutes = 140 sec b takes time 3 minutes = 185 sec difference = 185 - 140 = 45 sec now we are to find distance covered in 40 sec by b 180 sec = 2000 m 1 sec = 100 / 9 m 45 sec = 45 x 100 / 9 = 500 m answer : b
|
a ) 900 m , b ) 500 m , c ) 120 m , d ) 180 m , e ) 190 m
|
b
|
subtract(multiply(2, const_1000), multiply(divide(multiply(2, const_1000), add(multiply(3, const_60), 5)), add(multiply(2, const_60), 20)))
|
multiply(n0,const_1000)|multiply(n0,const_60)|multiply(n3,const_60)|add(n2,#1)|add(n4,#2)|divide(#0,#4)|multiply(#3,#5)|subtract(#0,#6)
|
physics
|
the price of 10 chairs is equal to that of 4 tables . the price of 15 chairs and 2 tables together is rs . 4000 . the total price of 12 chairs and 3 tables is :
|
"let the cost of a chair and that of a table be rs . x and rs . y respectively . then , 10 x = 4 y or y = 5 x . 2 15 x + 2 y = 4000 15 x + 2 x 5 x = 4000 2 20 x = 4000 x = 200 . so , y = 5 x 200 = 500 . 2 hence , the cost of 12 chairs and 3 tables = 12 x + 3 y = rs . ( 2400 + 1500 ) = rs . 3900 . a"
|
a ) rs . 3900 , b ) rs . 3990 , c ) rs . 4000 , d ) rs . 4100 , e ) rs . 4200
|
a
|
add(multiply(divide(4000, add(multiply(15, divide(4, 10)), 2)), 3), multiply(multiply(divide(4, 10), divide(4000, add(multiply(15, divide(4, 10)), 2))), 12))
|
divide(n1,n0)|multiply(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n6,#3)|multiply(#0,#3)|multiply(n5,#5)|add(#4,#6)|
|
general
|
if the remainder is 15 when the integer n is divided by 28 , what is the remainder when 2 n is divided by 14 ?
|
"n = 28 k + 15 2 n = 2 ( 28 k + 15 ) = 4 k * 14 + 30 = 4 k * 14 + 2 * 14 + 2 = 14 j + 2 the answer is c ."
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
|
c
|
subtract(multiply(15, 2), 28)
|
multiply(n0,n2)|subtract(#0,n1)|
|
general
|
rice weighing 25 / 2 pounds was divided equally and placed in 4 containers . how many ounces of rice were in each container ? ( note that 1 pound = 16 ounces )
|
"25 / 2 Γ· 4 = 25 / 8 pounds in each container 25 / 8 pounds * 16 ounces / pound = 50 ounces in each container the answer is b ."
|
a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80
|
b
|
divide(multiply(divide(25, 2), 16), 4)
|
divide(n0,n1)|multiply(n4,#0)|divide(#1,n2)|
|
general
|
a man sold 18 toys for rs . 21000 , gaining thereby the cost price of 3 toy find the cost price of a toy
|
"let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 21000 . gain = sp Γ’ β¬ β cp 3 x = 21000 Γ’ β¬ β 18 x 21 x = 21000 x = rs . 1000 . answer : option e"
|
a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1000
|
e
|
divide(21000, add(18, 3))
|
add(n0,n2)|divide(n1,#0)|
|
gain
|
in a school of 700 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ?
|
45 + 23 + 15 = 83 % 100 β 83 = 17 % 700 * 17 / 100 = 119 the answer is b .
|
a ) 109 , b ) 119 , c ) 129 , d ) 139 , e ) 149
|
b
|
subtract(700, add(add(multiply(divide(45, const_100), 700), multiply(divide(23, const_100), 700)), multiply(divide(15, const_100), 700)))
|
divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)
|
gain
|
two brother x and y appeared for an exam . the probability of selection of x is 1 / 7 and that of b is 2 / 9 . find the probability that both of them are selected .
|
"explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 7 , p ( b ) = 2 / 9 . let c be the event that both are selected . p ( c ) = p ( a ) Γ p ( b ) as a and b are independent events : = ( 1 / 7 ) Γ ( 2 / 9 ) = 2 / 63 answer : c ) 2 / 63"
|
a ) 2 / 69 , b ) 2 / 60 , c ) 2 / 63 , d ) 2 / 29 , e ) 2 / 10
|
c
|
multiply(divide(1, 7), divide(2, 9))
|
divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|
|
general
|
150 metres long yard , 11 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees
|
"11 trees have 10 gaps between them , required distance ( 150 / 10 ) = 15 b"
|
a ) 10 , b ) 15 , c ) 14 , d ) 16 , e ) 17
|
b
|
divide(150, add(subtract(11, 2), const_1))
|
subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)|
|
physics
|
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 $ 231 for it , the cost price of the cricket bat for a is :
|
125 % of 120 % of a = 231 125 / 100 * 120 / 100 * a = 231 a = 231 * 2 / 3 = 154 . answer c
|
a ) 150 , b ) 120 , c ) 154 , d ) 160 , e ) 210
|
c
|
divide(231, 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
|
the length of the bridge , which a train 150 m long and traveling at 45 km / hr can cross in 30 sec is ?
|
"speed = 45 * 5 / 18 = 25 / 2 m / sec . time = 30 sec let the length of bridge be x meters . then , ( 150 + x ) / 30 = 25 / 2 x = 225 m . answer : b"
|
a ) 377 , b ) 225 , c ) 237 , d ) 245 , e ) 267
|
b
|
subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 150)
|
speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|
|
physics
|
a boat running up stram takes 4 hours to cover a certain distance , while it takes 8 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ?
|
"explanation : let speed of boat is x km / h and speed stream is y km / hr 4 ( x + y ) = 8 ( x - y ) 4 x + 4 y = 8 x - 8 y 12 y = 4 x 3 y = x x / y = 3 / 1 3 : 1 answer : option d"
|
a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 3 : 1 , e ) 8 : 1
|
d
|
subtract(8, 4)
|
subtract(n1,n0)|
|
physics
|
a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many w acres were used for corn ?
|
"consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn w = 4 * 94 = 376 correct option - c"
|
a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517
|
c
|
multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4)
|
add(n1,n2)|multiply(const_10,const_3)|multiply(const_1,const_1000)|add(n3,#1)|add(n3,#0)|add(#3,#2)|divide(#5,#4)|multiply(n3,#6)|
|
other
|
find the l . c . m of 22 , 25 , 44 and 20 .
|
"explanation : 2 x 2 x 11 x 5 x 5 = 1100 answer : option e"
|
a ) 1240 , b ) 1600 , c ) 1320 , d ) 1150 , e ) 1100
|
e
|
multiply(multiply(power(const_3, const_3), multiply(power(const_2, const_3), power(add(const_4, const_1), const_2))), divide(divide(divide(divide(divide(44, const_2), const_2), const_3), add(const_4, const_1)), add(const_4, const_1)))
|
add(const_1,const_4)|divide(n2,const_2)|power(const_2,const_3)|power(const_3,const_3)|divide(#1,const_2)|power(#0,const_2)|divide(#4,const_3)|multiply(#2,#5)|divide(#6,#0)|multiply(#7,#3)|divide(#8,#0)|multiply(#10,#9)|
|
physics
|
3 models ( f , g , and h ) of cars are distributed among 3 showrooms . the number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom . there are 19 cars of model f , 17 cars of model g , and 15 cars of model h . what is the maximum number of cars of model f in any showroom ?
|
the total number of cars is 51 . so each showroom has 17 cars ( since the number of cars in each showrooms should be equal 51 / 3 = 17 ) . moreover that the number of model f is maximum means that the numbers of model g and h should be minimum . since each model must be represented by at least one car in every showroom that minimum number should be 1 . so maximum number of model f is 17 - 2 = 15 . the answer is ( c )
|
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13
|
c
|
subtract(divide(add(add(19, 17), 15), 3), subtract(3, const_1))
|
add(n2,n3)|subtract(n0,const_1)|add(n4,#0)|divide(#2,n0)|subtract(#3,#1)
|
general
|
the timing of a college is from 12 p . m to 5.00 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture .
|
"explanation : total time a student spends in college = 5 hours 00 minutes = 300 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 300 Γ’ β¬ β 20 ) / 5 = 56 minutes answer a"
|
a ) 56 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes
|
a
|
divide(multiply(5.00, const_60), 5)
|
multiply(n1,const_60)|divide(#0,n2)|
|
physics
|
if q = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x is between
|
my attempt : if q = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x would be when y is minimum . let y ^ 4 be 0 . now x ^ 4 = 100 . x should be definitely greater than 3 but less than 4 . the only option that fits this range is b hence answer is - - b ) 3 and 6 .
|
a ) 0 and 3 , b ) 3 and 6 , c ) 6 and 9 , d ) 9 and 12 , e ) 12 and 15
|
b
|
add(multiply(const_10, const_3), add(4, const_2))
|
add(n0,const_2)|multiply(const_10,const_3)|add(#0,#1)
|
general
|
what is the units digit of 29 ! + 50 ! + 3 ! + 3 ! ?
|
"for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 6 + 6 = 12 the units digit is 2 . the answer is b ."
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
|
b
|
add(add(const_4, const_3), const_2)
|
add(const_3,const_4)|add(#0,const_2)|
|
other
|
a train 285 m long , running with a speed of 54 km / hr will pass a tree in
|
"sol . speed = ( 54 x 5 / 18 ) m / sec . = 15 m / sec . time taken = ( 285 x 1 / 15 ) sec = 19 sec answer a"
|
a ) 19 sec , b ) 15 sec , c ) 16 sec , d ) 20 sec , e ) none
|
a
|
multiply(divide(285, multiply(54, const_1000)), const_3600)
|
multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)|
|
physics
|
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 7 adults have their meal , find the total number of children that can be catered with the remaining food .
|
"explanation : as there is meal for 70 adults and 7 have their meal , the meal left can be catered to 63 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 63 adults = 81 children hence , the meal can be catered to 81 children . answer : c"
|
a ) 33 , b ) 54 , c ) 81 , d ) 17 , e ) 01
|
c
|
multiply(subtract(70, 7), divide(90, 70))
|
divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1)|
|
general
|
if 42 / 330 = 0.1272 , what is the 32 nd digit to the right of the decimal point of the fraction ?
|
"we are not concerned what 42 / 330 means . . we have to look at the decimal . . 0 . . 1272 means 0.1272727 . . . so leaving first digit to the right of decimal , all odd numbered are 7 and all even numbered are 2 . here 32 is even , so ans is 2 answer is e"
|
a ) 1 , b ) 4 , c ) 5 , d ) 7 , e ) 2
|
e
|
add(const_2, const_3)
|
add(const_2,const_3)|
|
general
|
a began business with rs . 36000 and was joined afterwards by b with rs . 54000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ?
|
"36 * 12 : 54 * x = 2 : 1 x = 4 12 - 4 = 8 answer : d"
|
a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 2
|
d
|
subtract(multiply(const_4, const_3), divide(divide(multiply(36000, multiply(const_4, const_3)), 54000), 2))
|
multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3)|
|
other
|
find the value of β ( 25 / 16 ) .
|
"β ( 25 / 16 ) = β 25 / β 16 = 5 / 4 answer is e ."
|
a ) 5 / 2 , b ) 3 / 4 , c ) 2 / 3 , d ) 1 / 3 , e ) 5 / 4
|
e
|
multiply(subtract(16, const_4), 25)
|
subtract(n1,const_4)|multiply(#0,n0)|
|
general
|
how many of the positive factors of 24 are not factors of 25
|
"factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 25 - 1,5 , 25 comparing both , we have 1 factor of 24 which are not factors of 25 - 1 answer : d"
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5
|
d
|
divide(25, 24)
|
divide(n1,n0)|
|
other
|
a man can do a piece of work in 4 days , but with the help of his daughter , he can do it in 3 days . in what time can his daughter do it alone ?
|
explanation : daughter β s 1 day β s work = ( 1 / 3 β 1 / 4 ) = 1 / 12 daughter alone can do the work in 12 / 1 = 12 days answer : a
|
a ) 12 , b ) 28 , c ) 26 , d ) 19 , e ) 112
|
a
|
divide(multiply(4, 3), subtract(4, 3))
|
multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)
|
physics
|
john had a stock of 900 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ?
|
"let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 900 - n = 900 - 402 = 498 percentage books not sold / total number of books = 498 / 900 = 0.553 = 55.3 % correct answer e"
|
a ) 50.4 % , b ) 23.8 % , c ) 66.5 % , d ) 34.9 % , e ) 55.3 %
|
e
|
multiply(divide(subtract(900, add(add(add(75, 50), add(64, 78)), 135)), 900), const_100)
|
add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)|
|
gain
|
a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 854 in 4 years . the sum is :
|
"s . i . for 1 year = rs . ( 854 - 825 ) = rs . 29 . s . i . for 3 years = rs . ( 29 x 3 ) = rs . 87 . principal = rs . ( 825 - 87 ) = rs . 738 answer : b"
|
a ) 647 , b ) 738 , c ) 654 , d ) 847 , e ) 976
|
b
|
subtract(825, divide(multiply(subtract(854, 825), 3), 4))
|
subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|
|
gain
|
the age of father 7 years ago was 5 times the age of his son . 5 years hence , father ' s age will be thrice that of his son . the ratio of their present ages is :
|
let the ages of father and son 7 years ago be 5 x and x years respectively . then , ( 5 x + 7 ) + 5 = 3 [ ( x + 7 ) + 5 ] 5 x + 12 = 3 x + 36 x = 12 . required ratio = ( 5 x + 7 ) : ( x + 7 ) = 67 : 19 . answer : option a
|
a ) 67 : 19 , b ) 65 : 17 , c ) 61 : 19 , d ) 63 : 15 , e ) 67 : 17
|
a
|
divide(subtract(multiply(5, divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3))), subtract(multiply(7, 5), 7)), divide(subtract(add(multiply(5, const_3), subtract(multiply(7, 5), 7)), 5), subtract(5, const_3)))
|
multiply(n1,const_3)|multiply(n0,n1)|subtract(n1,const_3)|subtract(#1,n0)|add(#0,#3)|subtract(#4,n1)|divide(#5,#2)|multiply(n1,#6)|subtract(#7,#3)|divide(#8,#6)
|
general
|
huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 14 inches on a side and costs $ 10 , and a large pizza that measures 21 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ?
|
"in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 14 * 14 = 196 ) for the small pizza and ( 21 * 21 = 441 ) for the large pizza . in total sq inches that would be ( 196 + 441 ) * 2 = 1274 sq inches . in the second case if they pool their money together they can buy 3 large pizzas . in terms of square inches that would be 3 * 441 = 1323 sq inches . hence , the difference is 49 square inches more ( 1323 - 1274 ) . the correct answer is c"
|
a ) 5 square inches , b ) 10 square inches , c ) 49 square inches , d ) 25 square inches , e ) 350 square inches
|
c
|
subtract(multiply(power(21, const_2), const_3), add(add(power(14, const_2), power(21, const_2)), add(power(14, const_2), power(21, const_2))))
|
power(n2,const_2)|power(n0,const_2)|add(#1,#0)|multiply(#0,const_3)|add(#2,#2)|subtract(#3,#4)|
|
geometry
|
if the lcm of a and b is 72 and their hcf is 6 , what is the product of a & b ?
|
the most important property of lcm ( the least common multiple ) and gcd ( the greatest common divisors ) is : for any positive integers x and y , x β y = gcd ( x , y ) β lcm ( x , y ) so ab = 72 * 6 = 432 . answer : a
|
a ) 432 , b ) 433 , c ) 434 , d ) 435 , e ) 436
|
a
|
multiply(6, 72)
|
multiply(n0,n1)
|
physics
|
in a 120 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 120 m in 7 / 56 * 120 = 15 seconds since a beats b by 7 seconds , a runs 120 m in ( 15 - 7 ) = 8 seconds hence , a ' s time over the course = 8 seconds answer : d"
|
a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 8 seconds , e ) 28 seconds
|
d
|
subtract(multiply(divide(7, 56), 120), 7)
|
divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)|
|
physics
|
if x is a randomly chosen integer between 1 and 20 , inclusive , and y is a randomly chosen integer between 21 and 40 , inclusive , what is the probability that xy is a multiple of 4 ?
|
there are two ways that i see of getting a multiple of 4 1 ) x and / or y is a multiple of 4 2 ) x is a multiple of 2 and y is a multiple of 2 for 1 we have p ( option 1 ) = ( 5 / 20 ) ( 20 / 20 ) + ( 5 / 20 ) ( 20 / 20 ) - ( 5 / 20 ) ( 5 / 20 ) = 1 / 2 - 1 / 16 since 5 of the numbers in each range are divisible by 4 , and if either x or y is selected to be divisible by 4 the selection of the other variable does not matter . notice the last subtraction term which corrects the fact that we have counted the possibility of both being divisible by 4 twice ( e . g . p ( a union b ) = p ( a ) + p ( b ) - p ( a and b ) ) . notice at this point that the probability of option one is already 7 / 16 , the second largest available option . if we can come up with any other possibilities for getting xy divisble by 4 we must go with option e . of course this is the case because if we select a multiple of 2 that is not a multiple of 4 from both ranges of numbers , the product is divisible by 4 . we can formally compute this as : p ( option 2 ) = ( 5 / 20 ) * ( 5 / 20 ) = 1 / 16 since we do not wish to double count possibilities that fall into category 1 , we only count the probability of selecting even numbers from each set that are not divisible by 4 ( e . g . the other 5 even integers in each set ) . total probability is then p ( option 1 ) + p ( option 2 ) = 1 / 2
|
a ) 1 / 4 , b ) 1 / 3 , c ) 3 / 8 , d ) 7 / 16 , e ) Β½
|
e
|
subtract(1, add(multiply(multiply(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)), add(multiply(divide(const_1, const_2), divide(const_1, const_2)), multiply(multiply(divide(const_1, const_2), divide(const_1, const_2)), divide(const_1, const_2)))))
|
divide(const_1,const_2)|multiply(#0,#0)|multiply(#0,#1)|add(#1,#2)|add(#3,#2)|subtract(n0,#4)
|
general
|
the average of 20 number is zero of them at the most , how many may be greater than zero ?
|
solution average of 20 numbers = 0 . β΄ sum of 20 numbers = ( 0 Γ 20 ) = 0 . it is quite possible that 19 of these number may be positive and if their sum is a , then 20 th number is ( - a ) . answer d
|
a ) 0 , b ) 1 , c ) 10 , d ) 19 , e ) 20
|
d
|
subtract(20, const_1)
|
subtract(n0,const_1)
|
general
|
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1792 ?
|
"1792 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1600 . answer : a"
|
a ) 1600 , b ) 2777 , c ) 2889 , d ) 27670 , e ) 2771
|
a
|
divide(1792, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1))
|
multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)|
|
general
|
of the 13 employees in a certain department , 1 has an annual salary of 38,000 , 2 have an annual salary of 45,800 each , 2 have an annual salary of 42,500 each , 3 have an annual salary of 40,000 each and 5 have an annual salary of 48,500 each . what is the median annual salary for the 13 employees ?
|
"median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.8 so , answer is d ."
|
a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,800 , e ) 48,500
|
d
|
multiply(multiply(const_100, const_100), const_4)
|
multiply(const_100,const_100)|multiply(#0,const_4)|
|
general
|
there are 200 female managers in a certain company . find the total number r of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers .
|
"{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c ."
|
a ) r = 300 , b ) r = 400 , c ) r = 500 , d ) 600 , e ) none of these
|
c
|
divide(200, divide(2, 5))
|
divide(n1,n2)|divide(n0,#0)|
|
general
|
in a certain boys camp , 20 % of the total boys are from school a and 30 % of those study science . if there are 49 boys in the camp that are from school a but do not study science then what is the total number of boys in the camp ?
|
"since 30 % of the boys from school a study science , then 70 % of the boys from school a do not study science and since 20 % of the total number of boys are from school a , then 0.2 * 0.7 = 0.14 , or 14 % of the boys in the camp are from school a and do not study science . we are told that this number equals to 49 , so 0.14 * { total } = 49 - - > { total } = 350 . answer : e ."
|
a ) 70 , b ) 245 , c ) 150 , d ) 35 , e ) 350
|
e
|
divide(49, multiply(divide(subtract(const_100, 30), const_100), divide(20, const_100)))
|
divide(n0,const_100)|subtract(const_100,n1)|divide(#1,const_100)|multiply(#2,#0)|divide(n2,#3)|
|
gain
|
how many litres of pure acid are there in 8 litres of a 20 % solution ?
|
"explanation : quantity of pure acid = 8 Γ ( 20 / 100 ) = 1.6 answer : option d"
|
a ) 2 litres , b ) 1.4 litres , c ) 1 litres , d ) 1.6 litres , e ) none of these
|
d
|
multiply(divide(20, const_100), 8)
|
divide(n1,const_100)|multiply(n0,#0)|
|
gain
|
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 500 m long ?
|
"l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1700 m t = d / s t = 1700 / 10 t = 170 sec answer : b"
|
a ) 266 sec , b ) 170 sec , c ) 776 sec , d ) 166 sec , e ) 997 sec
|
b
|
divide(add(1200, 500), divide(1200, 120))
|
add(n0,n2)|divide(n0,n1)|divide(#0,#1)|
|
physics
|
what is the least number which when divided by 6 , 7 , 9 and 18 leaves remainder 4 in each care ?
|
"explanation : lcm of 6 , 7 , 9 and 18 is 126 required number = 126 + 4 = 130 answer : option a"
|
a ) 130 , b ) 140 , c ) 136 , d ) 156 , e ) 166
|
a
|
add(lcm(lcm(6, 7), lcm(9, 18)), 4)
|
lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)|
|
general
|
the profit earned by selling an article for rs . 832 is equal to the loss incurred when the same article is sold for rs . 448 . what should be the sale price for making 50 % profit ?
|
"let c . p . = rs . x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 150 % of rs . 640 = 150 / 100 * 640 = rs . 960 . answer : b"
|
a ) 277 , b ) 960 , c ) 277 , d ) 266 , e ) 121
|
b
|
multiply(subtract(832, divide(subtract(832, 448), const_2)), add(const_1, divide(50, const_100)))
|
divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)|
|
gain
|
a room 14 m 21 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor .
|
"explanation : area of the room = ( 1421 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1421 cm and 777 cm = 7 cm . area of 1 tile = ( 7 x 7 ) cm 2 . number of tiles required = ( 1147 Γ 777 ) / ( 7 Γ 7 ) = 22533 answer : option d"
|
a ) 22636 , b ) 22640 , c ) 22647 , d ) 22533 , e ) 22675
|
d
|
divide(multiply(add(multiply(14, const_100), 21), add(multiply(7, const_100), 77)), multiply(subtract(21, add(multiply(const_2, const_4), const_2)), subtract(21, add(multiply(const_2, const_4), const_2))))
|
multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)|
|
physics
|
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 5 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ?
|
"distance = relative speed * time = ( 5 + 3 ) * 4 = 32 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : a"
|
a ) 32 km , b ) 20 km , c ) 65 km , d ) 18 km , e ) 16 km
|
a
|
multiply(add(5, 3), 4)
|
add(n0,n1)|multiply(n2,#0)|
|
physics
|
the number which exceeds 16 % of it by 105 is :
|
solution solution let the number be x . x - 16 % of x = 105 x - 16 / 100 x = 105 x - 4 / 25 x = 105 21 / 25 x = 105 x = ( 105 x 25 / 21 ) = 125 answer a
|
a ) 125 , b ) 52 , c ) 58 , d ) 60 , e ) 62
|
a
|
divide(multiply(105, const_100), subtract(const_100, 16))
|
multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)
|
gain
|
sachin is younger than rahul by 7 years . if the ratio of their ages is 6 : 9 , find the age of sachin
|
"explanation : if rahul age is x , then sachin age is x - 7 , so , 9 x - 63 = 6 x 3 x = 63 x = 21 so sachin age is 21 - 7 = 14 answer : b ) 14"
|
a ) 24.5 , b ) 14 , c ) 24.21 , d ) 27 , e ) 25
|
b
|
multiply(divide(7, subtract(9, 6)), 6)
|
subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)|
|
other
|
a man has some hens and cows . if the number of heads be 44 and the number of feet equals 140 , then the number of hens will be :
|
"let hens be x and cows be y now , feet : x * 2 + y * 4 = 140 heads : x * 1 + y * 1 = 44 implies , 2 x + 4 y = 140 and x + y = 44 solving these two equations , we get x = 18 and y = 26 therefore , hens are 26 . answer : c"
|
a ) 22 , b ) 23 , c ) 18 , d ) 26 , e ) 28
|
c
|
divide(subtract(multiply(44, const_4), 140), const_2)
|
multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|
|
general
|
in the x - y plane , there are 4 points ( 00 ) , ( 04 ) , ( 54 ) , and ( 50 ) . if these 4 points makes a rectangle , what is the probability that x + y < 4 ?
|
the line y = - x + 4 intersects the rectangle and these three points of intersection ( 0,0 ) , ( 0,4 ) and ( 4,0 ) form a triangle . the points below the line y = - x + 4 satisfy x + y < 4 . the area of this triangle is ( 1 / 2 ) ( 4 ) ( 4 ) = 8 the area of the rectangle is 20 . p ( x + y < 4 ) = 8 / 20 = 2 / 5 the answer is b .
|
a ) 2 / 3 , b ) 2 / 5 , c ) 3 / 4 , d ) 3 / 7 , e ) 5 / 8
|
b
|
divide(divide(4, const_2), divide(50, const_10))
|
divide(n0,const_2)|divide(n4,const_10)|divide(#0,#1)
|
general
|
the probability of two events a and b are 0.25 and 0.35 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________
|
"we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 25 + . 35 - . 15 = . 45 but the probability of neither a nor b = 1 - . 45 = 0.55 answer : a"
|
a ) 0.55 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6
|
a
|
subtract(const_1, subtract(add(0.25, 0.35), 0.15))
|
add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)|
|
other
|
in a certificate by mistake a candidate gave his height as 25 % more than actual height . in the interview panel , he clarified that his height was 5 feet 8 nches . find the % correction made by the candidate from his stated height to his actual height ?
|
"his height was = 5 feet 8 inch = 8 + 60 = 68 inch . required % correction = 68 * ( 1.25 - 1 ) = 17 d"
|
a ) 10 , b ) 20 , c ) 30 , d ) 17 , e ) 50
|
d
|
multiply(divide(subtract(add(25, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(25, add(const_0_25, const_0_25)), 5))), const_100)
|
add(const_0_25,const_0_25)|add(n0,#0)|subtract(#1,n1)|add(#2,const_100)|divide(#2,#3)|multiply(#4,const_100)|
|
gain
|
the tens digit of a two digit number is 4 times of its unit digit and the sum of the digits be 10 . then the first number is
|
let the number be 10 x + y according to the question x = 4 y x + y = 10 4 y + y = 10 or y = 2 and x = 8 so required number is = 10 * 8 + 2 = 82 answer : d
|
a ) 67 , b ) 73 , c ) 53 , d ) 82 , e ) 89
|
d
|
multiply(multiply(4, divide(10, add(const_1, 4))), const_10)
|
add(n0,const_1)|divide(n1,#0)|multiply(n0,#1)|multiply(#2,const_10)
|
general
|
a policeman noticed a criminal from a distance of 180 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 = ( 180 - 50 ) m = 130 m . answer : c
|
a ) 100 m , b ) 120 m , c ) 130 m , d ) 150 m , e ) none of these
|
c
|
subtract(180, multiply(divide(3, const_60), const_1000))
|
divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)
|
physics
|
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 40 liters and they are all empty , how much money total will it cost to fuel all cars ?
|
"total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 40 ) = 21 + 312 = > 333 hence answer will be ( a ) 333"
|
a ) 333 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $
|
a
|
multiply(multiply(0.65, 40), 12)
|
multiply(n1,n3)|multiply(n2,#0)|
|
general
|
in a mixture 60 litres , the ratio of milk and water 2 : 1 . if this ratio is to be 1 : 2 , then the quanity of water to be further added is
|
"quantity of milk = 60 x 2 litres = 40 litres . 3 quantity of water in it = ( 60 - 40 ) litres = 20 litres . new ratio = 1 : 2 let quantity of water to be added further be x litres . then , milk : water = 40 . 20 + x now , 40 = 1 20 + x 2 20 + x = 80 x = 60 . quantity of water to be added = 60 liters . c"
|
a ) 50 liters , b ) 55 liters , c ) 60 liters , d ) 70 liters , e ) 75 liters
|
c
|
subtract(multiply(subtract(60, divide(60, add(2, 1))), 2), divide(60, add(2, 1)))
|
add(n1,n2)|divide(n0,#0)|subtract(n0,#1)|multiply(n1,#2)|subtract(#3,#1)|
|
general
|
a 280 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train .
|
"relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 280 / 9 = 500 / 9 x = 220 ans is 220 m answer : b"
|
a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 250 m
|
b
|
subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 280)
|
add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)|
|
physics
|
find the average of all the numbers between 11 and 34 which are divisible by 5 ?
|
"average = ( 15 + 20 + 25 + 30 ) 4 = 90 / 4 = 22 answer is c"
|
a ) 10 , b ) 20 , c ) 22 , d ) 30 , e ) 15
|
c
|
divide(add(add(11, const_4), subtract(34, const_4)), const_2)
|
add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)|
|
general
|
x and y can do a piece of work in 20 days and 12 days respectively . x started the work alone and then after 4 days y joined him till the completion of work . how long did the work last ?
|
"let the total units of work to be completed be 60 . units of work completed by x in a single day = 60 / 20 = 3 units . units of work completed by y in a single day = 60 / 12 = 5 units . thus , 8 units of work can be done by them in a single day working together . x works for 4 days alone = 12 units of work . remaining work = ( 60 - 12 ) = 48 units . now , x and y works together to complete the remaining work = 48 / 8 = 6 days hence , total days taken to complete the work = ( 6 + 4 ) = 10 days . answer : b"
|
a ) 8 days , b ) 10 days , c ) 12 days , d ) 13 days , e ) 14 days
|
b
|
add(4, divide(multiply(12, multiply(20, subtract(const_1, divide(4, 20)))), add(20, 12)))
|
add(n0,n1)|divide(n2,n0)|subtract(const_1,#1)|multiply(n0,#2)|multiply(n1,#3)|divide(#4,#0)|add(n2,#5)|
|
physics
|
if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 8 ?
|
"because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 1 * 1 / 4 = 1 / 4 answer : a"
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
|
a
|
divide(const_2, 8)
|
divide(const_2,n3)|
|
general
|
find the curved surface area , if the radius of a cone is 7 m and slant height is 14 m ?
|
"cone curved surface area = Γ― β¬ rl 22 / 7 Γ£ β 7 Γ£ β 14 = 22 Γ£ β 14 = 308 m ( power 2 ) answer is a ."
|
a ) 308 , b ) 369 , c ) 434 , d ) 476 , e ) 513
|
a
|
multiply(const_pi, multiply(7, 14))
|
multiply(n0,n1)|multiply(#0,const_pi)|
|
geometry
|
if | x | = 9 x - 16 , then x = ?
|
substituted option c i . e x = 2 . inequality satisfied . this eliminates all options except answer is c
|
a ) 1 , b ) 1 / 2 , c ) 2 , d ) 4 , e ) - 1 / 2
|
c
|
divide(16, subtract(9, const_1))
|
subtract(n0,const_1)|divide(n1,#0)
|
general
|
sonika deposited rs . 8000 which amounted to rs . 10200 after 3 years at simple interest . had the interest been 2 % more . she would get how much ?
|
"( 8000 * 3 * 2 ) / 100 = 480 10200 - - - - - - - - 10680 answer : b"
|
a ) 9680 , b ) 10680 , c ) 2999 , d ) 2774 , e ) 1212
|
b
|
add(multiply(multiply(add(divide(2, const_100), divide(divide(subtract(10200, 8000), 3), 8000)), 8000), 3), 8000)
|
divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)|
|
gain
|
the average monthly salary of 24 employees in an organisation is rs . 2400 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?
|
"manager ' s monthly salary = rs . ( 2500 * 25 - 2400 * 24 ) = rs . 4900 answer : c"
|
a ) rs . 3601 , b ) rs . 3618 , c ) rs . 4900 , d ) rs . 3619 , e ) rs . 3610
|
c
|
subtract(multiply(add(2400, 100), add(24, const_1)), multiply(2400, 24))
|
add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
|
general
|
the s . i . on a certain sum of money for 6 years at 16 % per annum is half the c . i . on rs . 8000 for 2 years at 20 % per annum . the sum placed on s . i . is ?
|
c . i . = [ 8000 * ( 1 + 20 / 100 ) 2 - 8000 ] = ( 8000 * 6 / 5 * 6 / 5 - 8000 ) = rs . 3520 . sum = ( 3520 * 100 ) / ( 6 * 16 ) = rs . 1833.33 answer : c
|
a ) 3000.33 , b ) 2802.33 , c ) 1833.33 , d ) 2990.33 , e ) 2982.33
|
c
|
divide(divide(subtract(multiply(8000, power(add(const_1, divide(20, const_100)), 2)), 8000), 2), multiply(6, divide(16, const_100)))
|
divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)|
|
gain
|
a farmer wishes to start a 100 square metres rectangular vegetable garden . since he has only 30 m barbed wire , he fences three sides of the garden letting his house compound wall act as the fourth side fencing . the dimension of the garden is :
|
we have : 2 b + l = 30 β l = 30 β 2 b . area = 100 m 2 β l Γ b = 100 β b ( 30 β 2 b ) = 100 β b 2 β 15 b + 50 = 0 β ( b β 10 ) ( b β 5 ) = 0 β b = 10 or b = 5 . when b = 10 , l = 10 and when b = 5 , l = 20 . since the garden is rectangular , so its dimension is 20 m Γ 5 m . answer b
|
['a ) 15 m Γ 6.67 m', 'b ) 20 m Γ 5 m', 'c ) 30 m Γ 3.33 m', 'd ) 40 m Γ 2.5 m', 'e ) none of these']
|
b
|
divide(100, divide(30, const_3))
|
divide(n1,const_3)|divide(n0,#0)
|
physics
|
if z is a multiple of 9402 , what is the remainder when z ^ 2 is divided by 9 ?
|
"the sum of the digits is 9 + 4 + 2 = 15 . thus 3 is a factor of 9402 , so 3 is a factor of z . then 3 ^ 3 = 9 is a factor of z ^ 2 . then the remainder when z ^ 2 is divided by 9 is 0 . the answer is a ."
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8
|
a
|
multiply(subtract(divide(power(divide(9402, const_3), 2), 9), add(add(add(multiply(multiply(const_1000, subtract(9, 2)), const_100), multiply(multiply(const_1000, subtract(9, 2)), const_10)), multiply(2, const_100)), subtract(const_100, 2))), 9)
|
divide(n0,const_3)|multiply(n1,const_100)|subtract(n2,n1)|subtract(const_100,n1)|multiply(#2,const_1000)|power(#0,n1)|divide(#5,n2)|multiply(#4,const_100)|multiply(#4,const_10)|add(#7,#8)|add(#9,#1)|add(#10,#3)|subtract(#6,#11)|multiply(n2,#12)|
|
general
|
a techno company has 14 machines of equal efficiency in its factory . the annual manufacturing costs are rs 42000 and establishment charges are rs 12000 . the annual output of the company is rs 70000 . the annual output and manufacturing costs are directly proportional to the number of machines . the shareholders get 12.5 profit , which is directly proportional to the annual output of the company . if 7.14 machines remain closed throughout the year , then the percentage decrease in the amount of profit of the shareholders would be :
|
explanation : the original profit is : - = > 70000 β 42000 β 12000 = 16000 . if 7.14 % of 14 i . e . one of the machines closed throughout the year , then the change in profit will be : - = > ( 13 / 14 ) Γ [ 70000 β 42000 ] . = > 14000 . thus , decrease in the profit % is : - = > ( 2000 / 16000 ) Γ 100 . = > 12.5 % . answer : b
|
a ) 12 % , b ) 12.5 % , c ) 13.5 % , d ) 13 % , e ) 14 %
|
b
|
multiply(divide(subtract(multiply(divide(12.5, const_100), subtract(subtract(70000, 42000), 12000)), multiply(subtract(multiply(divide(subtract(14, const_1), 14), subtract(70000, 42000)), 12000), divide(12.5, const_100))), multiply(divide(12.5, const_100), subtract(subtract(70000, 42000), 12000))), const_100)
|
divide(n4,const_100)|subtract(n3,n1)|subtract(n0,const_1)|divide(#2,n0)|subtract(#1,n2)|multiply(#0,#4)|multiply(#3,#1)|subtract(#6,n2)|multiply(#0,#7)|subtract(#5,#8)|divide(#9,#5)|multiply(#10,const_100)
|
general
|
p and q can complete a work in 80 days and 48 days respectively . p alone started the work and q joined him after 8 days till the completion of the work . how long did the work last ?
|
"explanation : work done by p in 1 day = 1 / 80 work done by q in 1 day = 1 / 48 work done by p in 8 days = 8 Γ£ β ( 1 / 80 ) = 1 / 10 remaining work = 1 Γ’ β¬ β 1 / 10 = 9 / 10 work done by p and q in 1 day = 1 / 80 + 1 / 48 = 1 / 30 number of days p and q take to complete the remaining work = ( 9 / 10 ) / ( 1 / 30 ) = 27 total days = 8 + 27 = 35 answer : option a"
|
a ) 35 days , b ) 10 days , c ) 14 days , d ) 22 days , e ) 26 days
|
a
|
add(8, divide(subtract(const_1, divide(8, 80)), add(inverse(80), inverse(48))))
|
divide(n2,n0)|inverse(n0)|inverse(n1)|add(#1,#2)|subtract(const_1,#0)|divide(#4,#3)|add(n2,#5)|
|
physics
|
a and b together can complete work in 6 days . a alone starts working and leaves it after working for 3 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ?
|
explanation : ( a + b ) one day work = 1 / 6 now a does half of the work in 3 days so a can complete the whole work in 6 days a β s one day work = 1 / 6 b β s one day work = 1 / 6 - 1 / 6 = 1 / 6 b alone can complete the work in 6 days so half of the work in 3 days answer : option a
|
a ) 3 , b ) 8 , c ) 5 , d ) 6 , e ) 7
|
a
|
multiply(6, divide(3, 6))
|
divide(n1,n0)|multiply(n0,#0)
|
physics
|
let f ( x ) = x ^ 2 + bx + c . if f ( 4 ) = 0 and f ( - 3 ) = 0 , then b + c =
|
"f ( x ) = x ^ 2 + bx + c . if f ( 4 ) = 0 and f ( - 3 ) = 0 , then b + c = f ( 4 ) = 0 = 16 + 4 b + c - - - taking 16 to the other side - > 4 b + c = - 16 f ( - 3 ) = 0 = 9 - 3 b + c - - - taking - 3 b + c to the other side - > 3 b - c = 9 when we add these 2 equations , we get 7 b = - 7 - - - > b = - 1 and while substituting b = - 1 we get c = - 12 . b + c = - 13 - - - answer c"
|
a ) 18 , b ) 15 , c ) - 13 , d ) - 21 , e ) - 24
|
c
|
negate(divide(subtract(power(3, 2), 4), add(3, 4)))
|
add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)|
|
general
|
how many even number in the range between 25 to 100 inclusive are not divisible by 3
|
"we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 26 , 28,30 . . . 100 using ap formula , we can say 100 = 10 + ( n - 1 ) * 2 or n = 38 . for 6 , 36 , 42 , . . . 96 using ap formula , we can say 96 = 12 + ( n - 1 ) * 6 or n = 11 . hence , only divisible by 2 but not 3 = 38 - 11 = 27 . hence , answer a"
|
a ) 27 , b ) 30 , c ) 31 , d ) 33 , e ) 46
|
a
|
subtract(divide(subtract(subtract(100, 25), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(100, const_2), multiply(3, const_4)), 3), 3), 3), const_2))
|
multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)|
|
general
|
sum of two numbers is 30 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ?
|
"explanation : x + y = 30 2 x β 3 y = 5 x = 19 y = 11 c )"
|
a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15
|
c
|
subtract(30, divide(subtract(30, divide(5, const_2)), const_2))
|
divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)|
|
general
|
the sum of three consecutive multiples of 3 is 72 . what is the larget number ?
|
"sol . let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 72 β 9 x = 63 β x = 7 . β΄ largest number = 3 x + 6 = 27 . answer b"
|
a ) 26 , b ) 27 , c ) 46 , d ) 57 , e ) none
|
b
|
add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, 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)|
|
general
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.