Datasets:
Cheukting
/
mathqa

Dataset card Data Studio Files
xet
 Community
1
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
how many positive integers less than 26 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ?
9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 3 odd multiples of 5 : { 5 , 15 , 25 } 10 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 } notice , that 5 is in two sets , thus total # of integers satisfying ...
a ) 27 , b ) 21 , c ) 24 , d ) 22 , e ) 20
b
subtract(subtract(subtract(26, 2), const_1), const_1)
subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)
general
how many liters of water must be evaporated from 50 liters of a 4 percent sugar solution to get a 5 percent sugar solution ?
"let x be the amount that needs to be evaporated . 0.04 ( 50 ) = 0.05 ( 50 - x ) 0.05 x = 2.5 - 2 x = 0.5 / 0.05 = 10 liters the answer is d ."
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
d
subtract(50, multiply(divide(50, const_100), 5))
divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)|
gain
a person lent a certain sum of money at 6 % per annum at simple interest and in 8 years the interest amounted to $ 520 less than the sum lent . what was the sum lent ?
"p - 520 = ( p * 6 * 8 ) / 100 p = 1000 the answer is b ."
a ) 800 , b ) 1000 , c ) 1200 , d ) 1400 , e ) 1600
b
divide(520, subtract(const_1, divide(multiply(6, 8), const_100)))
multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)|
gain
there are 30 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ?
"30 players are there . two players play one game with one another . so 30 c 2 = 30 * 29 / 2 = 435 so option c is correct"
a ) 10 , b ) 30 , c ) 435 , d ) 60 , e ) 90
c
divide(multiply(30, subtract(30, const_1)), const_2)
subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|
general
a wholesaler wishes to sell 100 pounds of mixed nuts at $ 1.50 a pound . she mixes peanuts worth $ 1.50 a pound with cashews worth $ 4.00 a pound . how many pounds of cashews must she use ?
"from the question stem we know that we need a mixture of 100 pounds of peanuts and cashews . if we represent peanuts as x and cashews as y , we get x + y = 100 . since the wholesaler wants to sell the mixture of 100 pounds @ $ 2.50 , we can write this as : $ 1.5 * ( x + y ) = $ 1.5 x + $ 4 y from the equation x + y = ...
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60
a
divide(multiply(subtract(4.00, 1.50), 100), 1.50)
subtract(n3,n1)|multiply(n0,#0)|divide(#1,n1)|
general
a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 90 , how many horses are there ?
"legs 18 * 4 = 72 now half on their horses so remaining on the walk so 9 men 9 men has 18 legs so , 18 + 72 = 90 legs walking answer : e"
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18
e
divide(90, add(const_4, divide(const_2, const_2)))
divide(const_2,const_2)|add(#0,const_4)|divide(n0,#1)|
general
240 is increased by 20 % . find the final number .
explanation final number = initial number + 20 % ( original number ) = 240 + 20 % ( 240 ) = 240 + 48 = 288 . answer d
a ) 200 , b ) 210 , c ) 180 , d ) 288 , e ) 220
d
add(240, multiply(240, divide(20, const_100)))
divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)|
gain
convert 20 miles into yards ?
"1 mile = 1760 yards 20 miles = 20 * 1760 = 35200 yards answer is b"
a ) 25630 yards , b ) 35200 yards , c ) 39520 yards , d ) 42560 yards , e ) 41520 yards
b
divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 20), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))
add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)|
physics
two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 25 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of
answer : c .
a ) 18 % , b ) 20 % , c ) 25 % , d ) 23 % , e ) can not be determined
c
add(divide(multiply(10, 4), 2), add(10, 5))
add(n4,n8)|multiply(n4,n6)|divide(#1,n1)|add(#0,#2)|
general
the average of 10 consecutive integers is 11 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ?
"the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ."
a ) 5 , b ) 5.5 , c ) 6 , d ) 6.5 , e ) 7
d
divide(subtract(multiply(10, 11), multiply(add(const_4, const_1), 9)), 10)
add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)|
general
let the number which when multiplied by 15 is increased by 196 .
"solution let the number be x . then , 15 x - x = 196 ‹ = › 14 x = 196 x ‹ = › 14 . answer a"
a ) 14 , b ) 20 , c ) 26 , d ) 28 , e ) 30
a
divide(196, subtract(15, const_1))
subtract(n0,const_1)|divide(n1,#0)|
general
running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 6 w widgets ?
"let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days w...
a ) 36 , b ) 6 , c ) 8 , d ) 10 , e ) 12
a
divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2)
multiply(n2,n3)|multiply(n0,#0)|subtract(#1,#0)|divide(#2,n0)|
general
a rectangular parking space is marked out by painting three of its sides . if the length of the unpainted side is 9 feet , and the sum of the lengths of the painted sides is 37 feet , then what is the area of the parking space in square feet ?
solution clearly , we have : l = 9 and l + 2 b = 37 or b = 14 . ∴ area = ( l × b ) = ( 9 × 14 ) sq . ft . = 126 sq . ft . answer c
['a ) 46', 'b ) 81', 'c ) 126', 'd ) 252', 'e ) none of these']
c
add(add(add(add(37, 37), 37), const_10), const_4)
add(n1,n1)|add(n1,#0)|add(#1,const_10)|add(#2,const_4)
geometry
lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 64,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is :
"total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 364000 x \ therefore required percentage = \ inline \ frac { 800 x } { 364000 } \ times 100 = 0.21978 c"
a ) 0.20833 , b ) 0.14544 , c ) 0.21978 , d ) 0.35466 , e ) 0.63435
c
multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100)
divide(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_4)|divide(n3,#0)|multiply(#1,const_1000)|multiply(#3,const_1000)|add(#5,#6)|divide(#4,#7)|multiply(#8,const_100)|
general
if p / q = 4 / 5 , then the value of 6 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ?
"answer given exp . = 6 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 6 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 6 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 6 / 7 + 6 / 14 = 6 / 7 + 3 / 7 = 10 / 7 correct option : c"
a ) 3 / 7 , b ) 34 , c ) 10 / 7 , d ) 2 , e ) 3
c
add(divide(6, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5))))
divide(n2,n3)|divide(n0,n1)|add(n4,#1)|subtract(n4,#1)|divide(#3,#2)|add(#0,#4)|
general
two persons start at the same point , walk in opposite directions with 6 km / hr and 7 km / hr respectively . what is the distance separated after 3 and half hrs ?
as the two persons are moving in the opposite direction , so they will be separateed in 1 hour = 6 + 7 = 13 km . they will be separated in 2.5 hours = 13 * 3.5 = 45.5 km answer : a
a ) 45.5 km , b ) 27.250 km , c ) 28.250 km , d ) 29.250 km , e ) 25.250 km
a
add(multiply(6, add(3, divide(const_2, const_4))), multiply(7, add(3, divide(const_2, const_4))))
divide(const_2,const_4)|add(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)
gain
on day one , a store sold 86 cups . on each of the next d days , the company sold 50 cups . if the average daily sales of cups over this time period ( including day one ) is 53 cups , what is the value of d ?
86 + 50 d = 53 ( d + 1 ) . 3 d = 33 . d = 11 . the answer is c .
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
c
subtract(add(divide(add(add(86, 50), 53), const_3), const_1), 53)
add(n0,n1)|add(n2,#0)|divide(#1,const_3)|add(#2,const_1)|subtract(#3,n2)
general
what is the value of â ˆ š 36 % ?
"explanation : br > â ˆ š 16 % = â ˆ š 36 / â ˆ š 100 = 6 / 10 = 60 / 100 = 60 % correct answer is d ) 60 %"
a ) 10 % , b ) 20 % , c ) 30 % , d ) 60 % , e ) 40 %
d
circle_area(divide(36, multiply(const_2, const_pi)))
multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)|
gain
there are two concentric circles with radii 8 and 4 . if the radius of the outer circle is increased by 25 % and the radius of the inner circle decreased by 25 % , by what percent does the area between the circles increase ?
"the area of a circle is pir ^ 2 , where r is the radius . the area of the big circle is 64 pi . the area of the small circle is 16 pi . the area a 1 between the circles is 48 pi . when the big circle ' s radius increases , the new area is 100 pi . when the small circle ' s radius decreases , the new area is 9 pi . the...
a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95
d
multiply(divide(subtract(subtract(power(multiply(8, divide(add(const_100, 25), const_100)), const_2), power(multiply(4, divide(subtract(const_100, 25), const_100)), const_2)), subtract(power(8, const_2), power(4, const_2))), subtract(power(8, const_2), power(4, const_2))), const_100)
add(n2,const_100)|power(n0,const_2)|power(n1,const_2)|subtract(const_100,n3)|divide(#0,const_100)|divide(#3,const_100)|subtract(#1,#2)|multiply(n0,#4)|multiply(n1,#5)|power(#7,const_2)|power(#8,const_2)|subtract(#9,#10)|subtract(#11,#6)|divide(#12,#6)|multiply(#13,const_100)|
gain
alice can bake a pie in 5 minutes . bob can bake a pie in 6 minutes . compute how many more pies alice can bake than bob in 60 minutes .
alice can bake 60 / 5 = 12 pies in 60 minutes . bob can bake 60 / 6 = 10 pies in 60 minutes . alice can therefore bake 12 - 10 = 2 more pies than bob . correct answer b
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
b
subtract(divide(60, 5), divide(60, 6))
divide(n2,n0)|divide(n2,n1)|subtract(#0,#1)
physics
a grocer has a sale of rs . 5420 , rs . 5660 , rs . 6200 , rs . 6350 and rs . 6500 for 5 consecutive months . find the sale he should have in the sixth month , so that he gets an average sale of rs . 6100 ?
"explanation : total sale for 5 months = rs . ( 5420 + 5660 + 6200 + 6350 + 6500 ) = rs . 30,130 therefore , required sale = rs . [ ( 6100 * 6 ) – 30,130 ] = rs . ( 36600 – 30,130 ) = rs . 6470 answer c"
a ) rs . 5870 , b ) rs . 5991 , c ) rs . 6470 , d ) rs . 6850 , e ) none of these
c
subtract(multiply(add(5, const_1), 6100), add(add(add(add(5420, 5660), 6200), 6350), 6500))
add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|
general
when a random experiment is conducted , the probability that event a occurs is 1 / 6 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ?
"one case is : 1 / 6 * 1 / 6 * 5 / 6 * 5 / 6 = 25 / 1296 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 25 / 1296 ) = 25 / 216 the answer is d ."
a ) 17 / 216 , b ) 19 / 216 , c ) 23 / 216 , d ) 25 / 216 , e ) 35 / 216
d
subtract(1, divide(const_2, 6))
divide(const_2,n1)|subtract(n0,#0)|
general
a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , c receives :
"let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d"
a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800
d
subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1)
add(n1,n2)|add(const_3,const_4)|multiply(n2,const_10)|multiply(n2,#1)|subtract(#2,n2)|subtract(#4,#0)|divide(#5,const_3)|add(#0,#6)|divide(#7,#2)|multiply(#8,#3)|divide(#9,const_1000)|floor(#10)|subtract(#11,const_1)|
general
a building contractor employs 20 male , 15 female and 5 child workers . to a male worker he pays rs . 35 per day , to a female worker rs . 20 per day and a child worker rs . 8 per day . the average wage per day paid by the contractor is ?
"20 15 5 35 20 8 500 + 300 + 40 = 1040 / 40 = 26 answer : e"
a ) 22 , b ) 23 , c ) 28 , d ) 29 , e ) 26
e
divide(add(add(multiply(20, 35), multiply(15, 20)), multiply(5, 8)), add(add(20, 15), 5))
add(n0,n1)|multiply(n0,n3)|multiply(n0,n1)|multiply(n2,n5)|add(#1,#2)|add(n2,#0)|add(#4,#3)|divide(#6,#5)|
physics
( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) what is m ?
( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / ( 2 * ( 2 * 5 ) ^ 35 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ 5 m = ( 1 / 5 ) ^ 35 ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / [ ( 2 ^ 36 ) * ( 5 ^ 35...
a ) 7 , b ) 18 , c ) 34 , d ) 35 , e ) 36
a
divide(log(divide(multiply(power(1, 18), multiply(power(10, 35), 2)), power(4, 18))), log(power(5, 5)))
power(n10,n11)|power(n0,n5)|power(n6,n5)|power(n1,n1)|log(#3)|multiply(n9,#0)|multiply(#5,#1)|divide(#6,#2)|log(#7)|divide(#8,#4)
general
eggs are sold in packages of 6 or 11 only . if doris bought 79 eggs exactly , what could be the number of large packs doris bought ?
no strategy involved . simple question demanding fast calculation . 11 x 5 = 55 = > 79 - 55 = 24 = > 24 / 6 is an integer ans d . 5 . good luck
a ) 6 . , b ) 2 . , c ) 3 . , d ) 5 . , e ) 4 .
d
divide(subtract(79, multiply(6, const_4)), 11)
multiply(n0,const_4)|subtract(n2,#0)|divide(#1,n1)
general
how long does a train 200 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ?
"speed = 72 * 5 / 18 = 20 m / sec total distance covered = 200 + 132 = 332 m . required time = 332 / 20 = 16.6 sec . answer : e"
a ) 82.1 sec , b ) 12.1 sec , c ) 16.1 sec , d ) 13.1 sec , e ) 16.6 sec
e
divide(add(200, 132), multiply(72, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
two tains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post . if the length of each train be 125 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ?
"sol . speed of the first train = [ 150 / 10 ] m / sec = 15 m / sec . speed of the second train = [ 150 / 15 ] m / sec = 10 m / sec . relative speed = ( 15 + 10 ) = m / sec = 25 m / sec . ∴ required time = ( 125 + 125 ) / 25 secc = 6 sec . answer a"
a ) 6 , b ) 14 , c ) 16 , d ) 20 , e ) 18
a
divide(add(125, 125), add(divide(125, 15), divide(125, 10)))
add(n2,n2)|divide(n2,n1)|divide(n2,n0)|add(#1,#2)|divide(#0,#3)|
physics
in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 6 , 0 ) , ( 0 , - 3.5 ) , ( - 6 , 0 ) ?
"ares of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 6 + 6 = 12 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 12 * 7 = 42 a is the answer"
a ) 42 , b ) 56 , c ) 88 , d ) 112 , e ) 116
a
rhombus_area(multiply(6, const_2), multiply(3.5, const_2))
multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)|
geometry
the perimeter of a triangle is 28 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle ?
area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 28 / 2 = 35 cm 2 answer : option e
['a ) 30', 'b ) 31', 'c ) 32', 'd ) 34', 'e ) 35']
e
triangle_area(2.5, 28)
triangle_area(n0,n1)
geometry
find out the c . i on rs . 5000 at 4 % p . a . compound half - yearly for 1 1 / 2 year ?
"a = 5000 ( 51 / 50 ) 3 = 5306.04 5000 - - - - - - - - - - - 306.04 answer : c"
a ) 306.06 , b ) 306.02 , c ) 306.04 , d ) 306.09 , e ) 306.12
c
subtract(multiply(5000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 5000)
divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)|
general
micheal and adam can do together a piece of work in 20 days . after they have worked together for 14 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately .
"rate of both = 1 / 20 together they do = 1 / 20 * 14 = 7 / 10 left work = 1 - 7 / 10 = 3 / 10 adam completes 3 / 10 work in 10 day so he took 10 * 10 / 3 = 100 / 3 days to complete the left work alone . thus the rate of adam is 3 / 100 rate of micheal = 1 / 20 - 3 / 100 = 1 / 50 thus micheal takes 50 days to complete ...
a ) 50 days , b ) 100 days , c ) 120 days , d ) 110 days , e ) 90 days
a
inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 14))), 10))))
inverse(n0)|multiply(n1,#0)|subtract(const_1,#1)|inverse(#2)|multiply(n2,#3)|inverse(#4)|subtract(#0,#5)|inverse(#6)|
physics
a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 3 or 17 ?
"number of multiples of 3 from 1 through 30 = 30 / 3 = 10 number of multiples of 17 from 1 through 30 = 30 / 17 = 1 number of multiples of 3 and 17 both from 1 through 30 = number of multiples of 17 * 3 ( = 51 ) = 0 total favourable cases = 10 + 1 - 0 = 11 probability = 11 / 30 answer : option e"
a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30
e
divide(add(floor(divide(30, 3)), divide(30, 17)), 30)
divide(n0,n2)|divide(n0,n1)|floor(#1)|add(#0,#2)|divide(#3,n0)|
probability
by selling an article at rs . 800 , a shopkeeper makes a profit of 35 % . at what price should he sell the article so as to make a loss of 25 % ?
"sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 35 % = 35 % of 640 = rs . 224 sp = cp - loss = 640 - 224 = rs . 416 answer : a"
a ) s . 416 , b ) s . 480 , c ) s . 429 , d ) s . 128 , e ) s . 419
a
subtract(divide(multiply(800, const_100), add(35, const_100)), divide(multiply(divide(multiply(800, const_100), add(35, const_100)), 25), const_100))
add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
a present value of a machine is $ 900 . its value depletion rate is 10 % per annum then find the machine value after 2 years ?
"p = $ 900 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 900 * 9 / 10 * 9 / 10 = $ 729 answer is e"
a ) $ 900 , b ) $ 810 , c ) $ 915 , d ) $ 715 , e ) $ 729
e
multiply(900, power(subtract(const_1, divide(10, const_100)), 2))
divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)|
gain
a motorist knows 6 different routes from bristol to birmingham . from birmingham to sheffield he knows 3 different routes and from sheffield to carlisle he knows two different routes . how many routes does he know from bristol to carlisle ?
explanation : total number of routes from bristol to carlisle = ( 6 x 3 x 2 ) = 36 . answer : e
a ) 4 , b ) 8 , c ) 12 , d ) 24 , e ) 36
e
multiply(multiply(6, 3), const_2)
multiply(n0,n1)|multiply(#0,const_2)
general
the grade point average of one third of the classroom is 54 ; the grade point average of the rest is 45 . what is the grade point average of the whole class ?
"let n = total students in class total points for 1 / 3 class = 54 n / 3 = 18 n total points for 2 / 3 class = 45 * 2 n / 3 = 30 n total points for whole class = 18 n + 30 n = 48 n 48 n total class points / n total students = 48 grade point average for total class answer : c"
a ) 52 , b ) 54 , c ) 48 , d ) 58 , e ) 60
c
add(multiply(divide(45, const_3), const_2), divide(54, const_3))
divide(n0,const_3)|divide(n1,const_3)|multiply(#1,const_2)|add(#0,#2)|
general
in what time will a train 100 m long cross an electric pole , it its speed be 72 km / hr ?
"speed = 72 * 5 / 18 = 20 m / sec time taken = 100 / 20 = 5 sec . answer : b"
a ) 2.5 , b ) 5 , c ) 2.4 , d ) 2.8 , e ) 2.1
b
divide(100, multiply(72, const_0_2778))
multiply(n1,const_0_2778)|divide(n0,#0)|
physics
if x is an integer that has exactly 3 positive divisors ( these include 1 and x ) , how many positive divisors does x ^ 3 have ?
since factors of ' 4 ' are { 1 , 2,4 } 4 ^ 3 = 64 number of factors / divisors of 64 = 2 ^ 6 we know that when a number is expressed as a product of the prime factors as below : n = a ^ x * b ^ y * c ^ z then no . of divisors = ( x + 1 ) * ( y + 1 ) * ( z + 1 ) then here ( 6 + 1 ) = 7 answer : d
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8
d
add(add(3, 3), 1)
add(n0,n0)|add(n1,#0)
general
three 6 faced dice are thrown together . the probability that all the three show the same number on them is -
"explanation : it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 and 666 . those are six in number . further the three dice can fall in 6 * 6 * 6 = 216 ways . hence the probability is 6 / 216 = 1 / 36 answer : b"
a ) 1 / 39 , b ) 1 / 36 , c ) 1 / 32 , d ) 1 / 31 , e ) 1 / 30
b
multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6))
divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)|
probability
how many bricks , each measuring 5 cm x 11 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 2 cm ?
"number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 2 ) / ( 5 x 11 x 6 ) = 2909 . answer : option c"
a ) 5600 , b ) 6000 , c ) 2909 , d ) 7200 , e ) 8600
c
divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 2), multiply(multiply(5, 11), 6))
multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)|
physics
a can do a piece of work in 7 days and b can do it in 6 days how long will they both work together to complete the work ?
"explanation : a ’ s one day work = 1 / 5 b ’ s one day work = 1 / 6 ( a + b ) ’ s one day work = 1 / 7 + 1 / 6 = 13 / 42 = > time = 42 / 13 = 3 3 / 13 days answer : option c"
a ) 6 / 11 , b ) 8 / 13 , c ) 3 / 13 , d ) 9 / 11 , e ) 10 / 13
c
divide(const_1, add(divide(const_1, 7), divide(const_1, 6)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|
physics
60 % of the population of a village is 23040 . the total population of the village is ?
answer ∵ 60 % of p = 23040 ∴ p = ( 23040 x 100 ) / 60 = 38400 correct option : a
a ) 38400 , b ) 24000 , c ) 24936 , d ) 25640 , e ) none
a
multiply(divide(const_100, 60), 23040)
divide(const_100,n0)|multiply(n1,#0)
general
when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 96.25 , what is the value of y ?
"guys , one more simple funda . 5 / 2 = 2.5 now . 5 x 2 = 1 is the remainder 25 / 4 = 6.25 now . 25 x 4 = 1 is the remainder 32 / 5 = 6.4 now . 4 x 5 = 2 is the remainder given x / y = 96.25 and remainder is 9 so . 25 x y = 9 hence y = 36 ans c"
a ) 96 , b ) 75 , c ) 36 , d ) 25 , e ) 12
c
divide(9, subtract(96.25, floor(96.25)))
floor(n1)|subtract(n1,#0)|divide(n0,#1)|
general
what least number should be subtracted from 13605 such that the remainder is divisible by 87 ?
13605 ÷ 87 = 156 , remainder = 33 hence 33 is the least number which can be subtracted from 13605 such that the remainder is divisible by 87 answer is e
a ) 27 , b ) 29 , c ) 28 , d ) 30 , e ) 33
e
reminder(13605, 87)
reminder(n0,n1)
general
378 × ? = 252
"explanation : 378 × ? = 252 ? = 252 / 378 = 126 / 189 = 14 / 21 = 2 / 3 answer is a"
a ) 2 / 3 , b ) 3 / 4 , c ) 1 / 2 , d ) 2 / 8 , e ) none of these
a
multiply(divide(378, 252), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 20 kg . what is the weight of the new person ?
total increase in weight = 8 × 2.5 = 20 if x is the weight of the new person , total increase in weight = x − 20 = > 20 = x - 20 = > x = 20 + 20 = 40 answer is d .
a ) 75 , b ) 65 , c ) 85 , d ) 40 , e ) 80
d
add(multiply(8, 2.5), 20)
multiply(n0,n1)|add(n2,#0)
general
liam is pulled over for speeding just as he is arriving at work . he explains to the police officer that he could not afford to be late today , and has arrived at work only four minutes before he is to start . the officer explains that if liam had driven 5 mph slower for his whole commute , he would have arrived at wor...
"let t be the number of hours he would need to reach office on time . when he is driving with over speed , he reached office 4 min earlier ! so the equation for this is s ( t - 4 / 60 ) = 20 where s is the speed and 20 is the distance . if he decreases his speed by 5 mph then he would have reached his office on time : ...
a ) 50 mph , b ) 45 mph , c ) 48 mph , d ) 52 mph , e ) 40 mph
e
add(20, multiply(5, const_4))
multiply(n0,const_4)|add(n1,#0)|
physics
rs . 1690 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ?
"a + b + c = 1690 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1690 = rs . 780 answer : c"
a ) s . 528 , b ) s . 542 , c ) s . 780 , d ) s . 540 , e ) s . 549
c
multiply(4, divide(1690, add(add(4, 2), const_3)))
add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)|
general
if 41 / 88 = 0.46590 , what is the 77 th digit to the right of the decimal point of the fraction ?
"we are not concerned what 41 / 88 means . . we have to look at the decimal . . 0.6590 means 0.465909090 . . . . so leaving girst , second and third digit to the right of decimal , all odd numbered are 0 and all even numbered are 9 . . here 77 is odd , so ans is 0 answer is d"
a ) 6 , b ) 2 , c ) 5 , d ) 0 , e ) 9
d
add(const_2, const_3)
add(const_2,const_3)|
general
a train running at the speed of 100 km / hr crosses a pole in 18 seconds . what is the length of the train ?
speed = ( 100 * 5 / 18 ) m / sec = ( 250 / 9 ) m / sec length of the train = ( speed x time ) = ( 250 / 9 * 18 ) m = 500 m . answer : c
a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700
c
multiply(divide(multiply(100, const_1000), const_3600), 18)
multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)
physics
there has been successive increases of 15 % 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 * 1.15 * p = x * p y = x / ( 1.1 * 1.15 ) which is about 0.79 x which is a decrease of about 21 % . the answer is d ."
a ) 12 % , b ) 15 % , c ) 18 % , d ) 21 % , e ) 24 %
d
multiply(subtract(const_1, divide(const_100, add(add(const_100, 15), divide(multiply(add(const_100, 15), 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
two numbers are less than a third number by 30 % and 37 % . how much percent is the second number is less than the first ?
let the third number be x first number = 70 % of x = 7 x / 10 ; second number = 63 % of x = 63 x / 100 ; difference = 7 x / 10 - 63 x / 100 = 7 x / 100 ; required percentage = 7 x / 100 * 10 / 7 x * 100 = 10 % answer is a
a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 %
a
multiply(subtract(add(30, const_1), 30), const_10)
add(n0,const_1)|subtract(#0,n0)|multiply(#1,const_10)
gain
there are 300 seniors at morse high school , and 40 % of them have cars . of the remaining grades ( freshmen , sophomores , and juniors ) , only 10 % of them have cars . if 15 % of all the students at morse have cars , how many students are in those other 3 lower grades ?
let the total no of students be n . the no of seniors having cars is 40 % of 300 i . e . 120 . the rest of the students with other three grades who have cars is 10 % of ( n - 300 ) . the total no of students with cars is 15 % of n . thus , 0.15 n = 120 + 0.1 ( n - 300 ) on solving this , we get n = 1800 . hence , the n...
a ) 600 , b ) 900 , c ) 1200 , d ) 1350 , e ) 1500
e
multiply(15, const_100)
multiply(n3,const_100)
gain
in an examination , a student scores 6 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 280 marks , the no of questions he attempts correctly is :
"let the number of correct answers be x . number of incorrect answers = ( 60 – x ) . 6 x – ( 60 – x ) = 280 = > 7 x = 340 = > x = 20 answer : a"
a ) a ) 20 , b ) b ) 38 , c ) c ) 90 , d ) d ) 88 , e ) e ) 37
a
divide(add(280, 60), add(6, 1))
add(n2,n3)|add(n0,n1)|divide(#0,#1)|
physics
a box measuring 90 inches long by 45 inches wide by 9 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ?
"least number of cubes will be required when the cubes that could fit in are biggest . 9 is the biggest number that could divide all three , 90 , 45 and 9 . thus side of cube must be 9 , and total number of cubes = 90 / 9 * 45 / 9 * 9 / 9 = 50 ans c it is ."
a ) 17 , b ) 18 , c ) 50 , d ) 108 , e ) 864
c
divide(multiply(multiply(90, 45), 9), volume_cube(divide(9, const_2)))
divide(n2,const_2)|multiply(n0,n1)|multiply(n2,#1)|volume_cube(#0)|divide(#2,#3)|
geometry
you collect baseball cards . suppose you start out with 11 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the en...
"solution start with 11 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 11 + 1 which is 6 , so you ' re left with 11 - 6 = 5 . peter takes 1 baseball card from you : 5 - 1 = 4 baseball cards . paul triples the number of baseball cards you have : 4 ã — 3 ...
a ) 12 , b ) 18 , c ) 19 , d ) 20 , e ) 21
a
multiply(subtract(subtract(subtract(11, const_4), const_4), 1), const_3)
subtract(n0,const_4)|subtract(#0,const_4)|subtract(#1,n1)|multiply(#2,const_3)|
general
a bullet train 130 m long is running with a speed of 50 kmph . in what time will it pass a man who is running at 10 kmph in the direction opposite to that in which the bullet train is going ?
"b 13 sec speed of the bullet train relative to man = ( 50 + 10 ) kmph = 60 * 5 / 18 m / sec = 30 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 130 m at ( 30 / 3 ) m / sec = ( 130 * 3 / 30 ) sec = 13 sec"
a ) 23 sec , b ) 13 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec
b
divide(130, divide(multiply(add(50, 10), const_1000), const_3600))
add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|
physics
find the smallest number of five digits exactly divisible by 22 , 33,66 and 44 .
"smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 22,33 , 66,44 i . e 132 , on dividing 10000 by 132 , we get 32 as remainder . therefore , required number = 10000 + ( 132 â € “ 32 ) = 10100 . answer is e ."
a ) 10101 , b ) 11000 , c ) 10110 , d ) 10111 , e ) 10100
e
add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 33,66)
multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)|
general
a man walking at a constant rate of 5 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 25 miles per hour . the woman stops to wait for the man 5 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman...
"when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 5 * 25 / 60...
a ) 16 , b ) 20 , c ) 24 , d ) 25 , e ) 28
b
multiply(const_60, divide(multiply(divide(5, const_60), subtract(25, 5)), 5))
divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)|
physics
the ages of two persons differ by 20 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively
explanation : let their ages be x and ( x + 20 ) years . 5 ( x - 5 ) = ( x + 20 - 5 ) or 4 x = 40 or x = 10 . their present ages are 30 years and 10 years option d
a ) 20,20 , b ) 20,10 , c ) 25,15 , d ) 30,10 , e ) none of these
d
subtract(add(divide(multiply(20, 5), subtract(5, const_1)), 5), 20)
multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)
general
p and q are the only two applicants qualified for a short - term research project that pays 300 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job ...
"let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 300 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 300 / 30 = $ 10 p ' s hourly wage is 300 / ( t - ...
a ) $ 5 , b ) $ 10 , c ) $ 15 , d ) $ 20 , e ) $ 25
a
subtract(divide(300, divide(10, subtract(divide(const_3, const_2), const_1))), divide(300, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10)))
divide(const_3,const_2)|subtract(#0,const_1)|divide(n2,#1)|add(n2,#2)|divide(n0,#2)|divide(n0,#3)|subtract(#4,#5)|
general
the average length of the sides of triangle abc is 12 . what is the perimeter of triangle abc ?
( average ) = ( perimeter ) / 3 ; 12 = ( perimeter ) / 3 ; ( perimeter ) = 36 . answer : e .
['a ) 4', 'b ) 6', 'c ) 12', 'd ) 24', 'e ) 36']
e
multiply(12, const_3)
multiply(n0,const_3)
geometry
a retailer buys 140 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ?
"let the market price of each pen be $ 1 then , cost price of 140 pens = $ 36 selling price of 140 pens = 99 % of $ 140 = $ 138.60 profit % = ( ( 102.60 * 100 ) / 36 ) % = 285 % answer e"
a ) 105 , b ) 155 , c ) 125 , d ) 185 , e ) 285
e
multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 140)), divide(36, 140)))
divide(n2,const_100)|divide(n1,n0)|subtract(n2,#0)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)|
gain
if 3 log ( 4 * 5 ^ 2 ) = x , find x
"3 ( log 2 ^ 2 * 5 ^ 2 ) = x 3 log ( 5 * 2 ) ^ 2 = x 3 * 2 log ( 5 * 2 ) = x 6 log 10 = x log 10 base 10 = 1 so 6 * 1 = x x = 6 answer : b"
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
b
power(multiply(4, power(5, 2)), 3)
power(n2,n3)|multiply(n1,#0)|power(#1,n0)|
general
two trains of equal lengths take 10 sec and 18 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ?
"speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 18 = 6.7 m / sec . relative speed = 12 + 6.7 = 18.7 m / sec . required time = ( 120 + 120 ) / 18.7 = 12.8 sec . answer : option e"
a ) 11 , b ) 9 , c ) 13 , d ) 14 , e ) 12.8
e
divide(multiply(120, const_2), add(speed(120, 18), speed(120, 10)))
multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)|
physics
if one - third of one fourth of a number is 15 , then 3 - tenth of that number is :
let the number be x . then , 1 / 3 of 1 / 4 of x = 15 x = 15 * 12 = 180 so , required number = ( 3 / 10 * 180 ) = 54 . answer : d
a ) 35 , b ) 25 , c ) 45 , d ) 54 , e ) 52
d
subtract(add(multiply(15, 3), const_12), const_3)
multiply(n0,n1)|add(#0,const_12)|subtract(#1,const_3)
general
r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ?
"r is the set of positive even integers less than 201 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , 36 , 64 , 100 , 144 , and 196 are even integers ( less than 201 ) that are in ...
a ) none , b ) two , c ) four , d ) five , e ) seven
e
subtract(const_4, const_1)
subtract(const_4,const_1)|
physics
on dividing a number by 7 , we get 3 as quotient and 0 as remainder . on dividing the same number by 3 , what will be the remainder ?
"number = 7 * 3 + 0 = 21 3 ) 21 ( 7 21 - - - - - - - - 0 required number = 0 . answer : d"
a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 4
d
reminder(multiply(7, 3), 3)
multiply(n0,n1)|reminder(#0,n3)|
general
the unit digit in the product ( 722 * 774 * 889 * 223 ) is :
"explanation : unit digit in the given product = unit digit in ( 2 * 4 * 9 * 3 ) = 6 answer : c"
a ) 2 , b ) 7 , c ) 6 , d ) 8 , e ) 1
c
subtract(multiply(multiply(multiply(722, 774), 889), 223), subtract(multiply(multiply(multiply(722, 774), 889), 223), add(const_4, const_4)))
add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)|
general
a and b invests rs . 3000 and rs . 7000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?
"( 3 * 6 + 6 * 6 ) : ( 7 * 12 ) 54 : 84 = > 9 : 14 . answer : c"
a ) 9 : 6 , b ) 9 : 8 , c ) 9 : 14 , d ) 9 : 9 , e ) 9 : 5
c
divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(7000, add(6, 6)))
add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)|
gain
cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 23 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ?
use the options . the options which after getting added to 23 shows a cube of a number could be right . here 64 + 23 = 87 72 + 23 = 95 86 + 23 = 109 98 + 23 = 121 102 + 23 = 125 - - - ( 5 * 5 * 5 ) so we have 102 as the answer ! e
['a ) 64', 'b ) 72', 'c ) 86', 'd ) 98', 'e ) 102']
e
subtract(power(add(const_1, add(const_1, const_3)), const_3), 23)
add(const_1,const_3)|add(#0,const_1)|power(#1,const_3)|subtract(#2,n0)
geometry
what is the % change in the area of a rectangle when its length increases by 25 % and its width decreases by 25 % ?
"( 125 / 10 ) * ( 75 / 10 ) = 9375 / 100 ~ 94 of original area 0.94 is a 6 % decrease from 100 / 100 - > d"
a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 6 % decrease , e ) insufficient data
d
subtract(const_100, divide(multiply(add(const_100, 25), subtract(const_100, 25)), const_100))
add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)|
geometry
bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 70 white marbles . how many red marbles could be in bag a ?
"6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 70 = 6 k + 4 k = > k = 7 total red ' s in bag a will be 2 k = 14 e"
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 14
e
divide(70, add(multiply(3, 2), 4))
multiply(n1,n2)|add(n5,#0)|divide(n6,#1)|
other
sandy gets 3 marks for each correct sum and loses 2 marks for each incorrect sum . sandy attempts 30 sums and obtains 60 marks . how many sums did sandy get correct ?
"let x be the correct sums and ( 30 - x ) be the incorrect sums . 3 x - 2 ( 30 - x ) = 60 5 x = 120 x = 24 the answer is d ."
a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26
d
divide(add(multiply(2, 30), 60), add(2, 3))
add(n0,n1)|multiply(n1,n2)|add(n3,#1)|divide(#2,#0)|
general
a car dealership has 40 cars on the lot , 20 % of which are silver . if the dealership receives a new shipment of 80 cars , 35 % of which are not silver , what percentage of total number of cars are silver ?
"the number of silver cars is 0.2 * 40 + 0.35 * 80 = 36 the percentage of cars which are silver is 36 / 120 = 30 % the answer is a ."
a ) 30 % , b ) 35 % , c ) 40 % , d ) 45 % , e ) 50 %
a
multiply(divide(add(multiply(40, divide(20, const_100)), multiply(80, divide(35, const_100))), add(40, 80)), const_100)
add(n0,n2)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)|
gain
at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 2...
"salary grade of 5 is p ( 5 ) = 9.50 + 0.25 ( 5 – 1 ) = 9.50 + 0.25 * 4 ; salary grade of 2 is p ( 2 ) = 9.50 + 0.25 ( 2 – 1 ) = 9.50 + 0.25 ; p ( 5 ) - p ( 2 ) = 9.50 + 0.25 * 4 - 9.50 - 0.25 = 0.75 . answer : b ."
a ) $ 0.50 , b ) $ 0.75 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75
b
add(multiply(0.25, subtract(5, 1)), 0.25)
subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)|
general
what is the product of all the possible values of x if x ^ 2 - 2 x - 8 ?
"explanation : = > y = x ^ 2 - 2 x - 8 = > y = ( x + 2 ) ( x - 4 ) = > x = - 2 , y = 4 product x = ( - 2 ) ( 4 ) = - 8 answer option - 8 answer : c"
a ) – 29 , b ) – 12 , c ) - 8 , d ) 29 , e ) 168
c
divide(8, const_1)
divide(n2,const_1)|
general
in a class of 30 students , 5 students did not borrow any books from the library , 10 students each borrowed 1 book , 8 students each borrowed 2 books , and the rest of the students each borrowed at least 4 books . if the average ( arithmetic mean ) number of books borrowed per student was 2 , what is the maximum numbe...
"the average number of books per student was 2 means that total of 2 * 30 = 60 books were borrowed ; 5 + 12 + 8 = 25 students borrowed total of 5 * 0 + 12 * 1 + 8 * 2 = 28 books ; so 60 - 28 = 32 books are left to distribute among 30 - 25 = 5 students , these 5 arethe rest who borrowed at least 4 books ; tomaximizethe ...
a ) 5 , b ) 8 , c ) 16 , d ) 18 , e ) 20
c
add(subtract(multiply(30, 2), add(add(10, multiply(8, 2)), multiply(5, 4))), 4)
multiply(n0,n5)|multiply(n4,n5)|multiply(n1,n6)|add(n2,#1)|add(#3,#2)|subtract(#0,#4)|add(n6,#5)|
general
in a office work is distribute between p persons . if 1 / 4 members are absent then work increased for each person is ?
"let total % of work is 100 % total person = p 1 / 4 person are absent of total person . so absent person is 1 / 4 p ie p / 4 . left person is , p - p / 4 = 3 p / 4 . p person do the work 100 % 1 person do the work 100 * p % 3 p / 4 person do the work ( 100 * p * 4 ) / 3 p % = 133.33 % work increased for each person is...
a ) 33.33 % , b ) 24.33 % , c ) 15.33 % , d ) 16.33 % , e ) 17.33 %
a
subtract(divide(multiply(const_100, 4), subtract(4, 1)), const_100)
multiply(n1,const_100)|subtract(n1,n0)|divide(#0,#1)|subtract(#2,const_100)|
general
john has $ 1,200 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ?
"suppose total money spent = x not spend ( money he still has ) = x - 800 x + x - 800 = 1200 x = 1000 money not spend = 1000 - 800 = 200 answer : a"
a ) $ 200 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1,200
a
subtract(add(multiply(const_100, const_10), 800), divide(add(800, add(multiply(const_100, const_10), 800)), const_2))
multiply(const_10,const_100)|add(n1,#0)|add(n1,#1)|divide(#2,const_2)|subtract(#1,#3)|
general
if a , b and c together can finish a piece of work in 2 days . a alone in 12 days and b in 18 days , then c alone can do it in ?
"c = 1 / 2 - 1 / 12 – 1 / 18 = 1 / 3 = > 3 days answer : c"
a ) 5 days , b ) 4 days , c ) 3 days , d ) 2 days , e ) 7 days
c
divide(const_1, subtract(subtract(const_0_25, divide(const_1, 12)), divide(const_1, 18)))
divide(const_1,n1)|divide(const_1,n2)|subtract(const_0_25,#0)|subtract(#2,#1)|divide(const_1,#3)|
physics
a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 3 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 6 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , ho...
"the tank is emptied at this rate : 9 + 6 - 3 = 12 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 12 = 4320 minutes . the answer is c ."
a ) 2110 , b ) 3210 , c ) 4320 , d ) 5430 , e ) 6540
c
divide(multiply(30, power(9, 3)), subtract(add(9, 6), 3))
add(n4,n5)|power(n8,n2)|multiply(n0,#1)|subtract(#0,n2)|divide(#2,#3)|
physics
a cistern is normally filled in 12 hours but takes two hours longer to fill because of a leak in its bottom . if the cistern is full , the leak will empty it in ?
"1 / 12 - 1 / x = 1 / 14 x = 84 answer : b"
a ) 33 , b ) 84 , c ) 40 , d ) 99 , e ) 11
b
inverse(subtract(divide(const_1, 12), divide(const_1, add(12, const_2))))
add(n0,const_2)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)|
physics
when positive integer n is divided by positive integer j , the remainder is 28 . if n / j = 154.04 , what is value of j ?
"when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 154.04 here 154 is the quotient . given that remainder = 28 so , 154.04 = 154 + 28 / j so , j = 700 answer : c"
a ) 400 , b ) 350 , c ) 700 , d ) 750 , e ) 800
c
divide(28, subtract(154.04, add(const_100, add(multiply(const_4, const_10), const_2))))
multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)|
general
what least number should be subtracted from 13602 such that the remainder is divisible by 87 ?
"13602 ÷ 87 = 156 , remainder = 30 hence 30 is the least number which can be subtracted from 13602 such that the remainder is divisible by 87 answer is d"
a ) 27 , b ) 29 , c ) 28 , d ) 30 , e ) 31
d
reminder(13602, 87)
reminder(n0,n1)|
general
45 litres of diesel is required to travel 600 km using a 800 cc engine . if the volume of diesel required to cover a distance varies directly as the capacity of the engine , then how many litres of diesel is required to travel 800 km using 1200 cc engine ?
"explanatory answer to cover a distance of 800 kms using a 800 cc engine , the amount of diesel required = 800 / 600 * 45 = 60 litres . however , the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity . i . e . , for instance , if the capacity...
a ) 80 litres , b ) 90 litres , c ) 120 litres , d ) 170 litres , e ) none of these
b
multiply(1200, multiply(800, divide(45, multiply(800, 600))))
multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n4,#2)|
physics
christine and siri have 21 rs between them . christine has 20 rs more than siri . how much does christine have .
let siri has x rs , then christine has 20 + x rs given x + ( 20 + x ) = 21 = > 2 x = 1 or x = 0.5 so christine and siri has 20.5 and 0.5 rs respectively . answer : e
a ) 16.5 , b ) 17.5 , c ) 18.5 , d ) 19.5 , e ) 20.5
e
add(divide(subtract(21, 20), const_2), 20)
subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)
general
the ratio between the length and the breadth of a rectangular park is 1 : 3 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 4 min , then the area of the park ( in sq . m ) is ?
"perimeter = distance covered in 4 min . = ( 12000 x 4 ) / 60 m = 800 m . let length = 1 x metres and breadth = 3 x metres . then , 2 ( 1 x + 3 x ) = 800 or x = 100 . length = 100 m and breadth = 300 m . area = ( 100 x 300 ) m 2 = 30000 m e"
a ) 800 m , b ) 12000 m , c ) 36000 m , d ) 80000 m , e ) 30000 m
e
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 4), const_1000), add(1, 3)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 4), const_1000), add(1, 3)), const_2), 3))
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
physics
in 10 years , a will be twice as old as b was 10 years ago . if a is now 4 years older than b the present age of b is
let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 4 = > 2 b - b - 4 = 30 b = 34 so the present age of b is 34 years answer : d
a ) 37 , b ) 38 , c ) 39 , d ) 34 , e ) 41
d
add(add(multiply(const_2, 10), 10), 4)
multiply(n0,const_2)|add(n0,#0)|add(n2,#1)
general
the cost of 10 kg of mangos is equal to the cost of 24 kg of rice . the cost of 6 kg of flour equals the cost of 2 kg of rice . the cost of each kg of flour is $ 23 . find the total cost of 4 kg of mangos , 3 kg of rice and 5 kg of flour ?
let the costs of each kg of mangos and each kg of rice be $ a and $ r respectively . 10 a = 24 r and 6 * 23 = 2 r a = 12 / 5 r and r = 69 a = 165.6 required total cost = 4 * 165.6 + 3 * 69 + 5 * 23 = 662.4 + 207 + 115 = $ 984.40 b
a ) 347.4 , b ) 984.4 , c ) 877.4 , d ) 637.4 , e ) 667.4
b
add(add(multiply(4, multiply(divide(24, 10), divide(multiply(23, 6), 2))), multiply(3, divide(multiply(23, 6), 2))), multiply(5, 23))
divide(n1,n0)|multiply(n2,n4)|multiply(n4,n7)|divide(#1,n3)|multiply(#0,#3)|multiply(n6,#3)|multiply(n5,#4)|add(#6,#5)|add(#7,#2)
general
if x is equal to the sum of the integers from 50 to 60 , inclusive , and y is the number of even integers from 50 to 60 , inclusive , what is the value of x + y ?
"sum s = n / 2 { 2 a + ( n - 1 ) d } = 11 / 2 { 2 * 50 + ( 11 - 1 ) * 1 } = 11 * 55 = 605 = x number of even number = ( 60 - 50 ) / 2 + 1 = 6 = y x + y = 605 + 6 = 611 e"
a ) 171 , b ) 281 , c ) 391 , d ) 491 , e ) 611
e
add(multiply(divide(add(50, 60), const_2), add(subtract(60, 50), const_1)), add(divide(subtract(60, 50), const_2), const_1))
add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)|
general
a number x is 7 times another number y . the percentage that y is less than x is
"say y = 1 and x = 7 . then y = 1 is less than x = 7 by ( 7 - 1 ) / 7 * 100 = 6 / 7 * 100 = 85.7 % . answer : b ."
a ) 12.5 % , b ) 85.7 % , c ) 80 % , d ) 11 % , e ) 1 %
b
multiply(divide(subtract(7, const_1), 7), const_100)
subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)|
general
an alloy weighing 48 ounces is 25 % gold . how many ounces of pure gold must be added to create an alloy that is 40 % gold ?
an alloy of 48 oz which is 25 % gold means there is 12 oz of gold . to get to an alloy that is 40 % gold , let ' s use this expression : ( 12 + x ) / ( 48 + x ) = 0.40 with x representing the amount of pure gold that must be added to get to 40 % . the expression we are using represents the new total weight of pure gold...
a ) 15 , b ) 20 , c ) 12 , d ) 14 , e ) 7
c
subtract(divide(multiply(48, subtract(40, 25)), const_12), 48)
subtract(n2,n1)|multiply(n0,#0)|divide(#1,const_12)|subtract(#2,n0)
gain
a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 420 . find the sum ?
"let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 420 = > 30 x / 100 - 24 x / 100 = 420 = > 6 x / 100 = 420 = > x = 7000 . answer : b"
a ) s . 5000 , b ) s . 7000 , c ) s . 14000 , d ) s . 17000 , e ) s . 27000
b
divide(420, divide(multiply(subtract(15, 12), const_2), const_100))
subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)|
gain
if the average ( arithmetic mean ) of 7 consecutive integers is 20 , then the product of the greatest and least integer is
n = smallest number n + 6 = largest number ( n + n + 6 ) / 2 = 20 = > ( 2 n + 6 ) / 2 = 20 = > n + 3 = 20 = > n = 17 so product of n and n + 6 = ( 17 ) ( 23 ) = 391 answer - d
a ) 343 , b ) 393 , c ) 363 , d ) 391 , e ) 409
d
multiply(divide(subtract(multiply(7, 20), multiply(const_3, 7)), 7), add(subtract(7, const_1), divide(subtract(multiply(7, 20), multiply(const_3, 7)), 7)))
multiply(n0,n1)|multiply(n0,const_3)|subtract(n0,const_1)|subtract(#0,#1)|divide(#3,n0)|add(#4,#2)|multiply(#5,#4)
general
a train , 120 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is
"explanation : assume the length of the bridge = x meter total distance covered = 120 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 120 + x ) / 30 m / s = > 45 ã — ( 10 / 36 ) = ( 120 + x ) / 30 = > 45 ã — 10 ã — 30 / 36 = 120 + x = > 45 ã — 10 ã — 10 / 12 = 120 + x = > 15 ã — ...
a ) 270 m , b ) 255 m , c ) 235 m , d ) 220 m , e ) 240 m
b
subtract(multiply(multiply(45, const_0_2778), 30), 120)
multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|
physics
find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 6 ) is
"required ratio = 2 / 3 * 6 / 11 * 11 / 6 = 2 / 1 = 2 : 3 answer is e"
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 2 : 3
e
multiply(divide(2, 3), multiply(divide(2, 3), divide(6, 3)))
divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)|
other
the number 90 can be written as the sum of the squares of 4 different positive integers . what is the sum of these 4 integers ?
"4 ^ 2 + 3 ^ 2 + 8 ^ 2 + 1 ^ 2 = 90 - - > 1 + 3 + 8 + 4 = 16 . b"
a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13
b
add(add(add(const_4, const_3), add(const_3, const_2)), 4)
add(const_3,const_4)|add(const_2,const_3)|add(#0,#1)|add(n1,#2)|
geometry
a train with 120 wagons crosses john who is going in the same direction , in 36 seconds . it travels for half an hour from the time it starts overtaking the john ( he is riding on the horse ) before it starts overtaking the mike ( who is also riding on his horse ) coming from the opposite direction in 24 seconds . in h...
"let the length of the train be l metres and speeds of the train arun and sriram be r , a and s respectively , then - - - - - - - - - - ( i ) and - - - - - - - - - ( ii ) from eq . ( i ) and ( ii ) 3 ( r - a ) = 2 ( r + k ) r = 3 a + 2 k in 30 minutes ( i . e 1800 seconds ) , the train covers 1800 r ( distance ) but th...
a ) 3576 s , b ) 3589 s , c ) 3596 s , d ) 3598 s , e ) 3576 s
e
subtract(divide(multiply(subtract(divide(add(36, 24), subtract(36, 24)), const_1), multiply(multiply(const_10, const_3), const_60)), const_2), 24)
add(n1,n2)|multiply(const_10,const_3)|subtract(n1,n2)|divide(#0,#2)|multiply(#1,const_60)|subtract(#3,const_1)|multiply(#4,#5)|divide(#6,const_2)|subtract(#7,n2)|
physics