Datasets:
pgilliar
/
MNLP_M2_rag_dataset

Dataset card Data Studio Files
xet
Community
Problem
stringlengths
5
967
Rationale
stringlengths
1
2.74k
options
stringlengths
37
164
correct
stringclasses
5 values
annotated_formula
stringlengths
7
1.65k
linear_formula
stringlengths
8
925
category
stringclasses
6 values
answer
stringclasses
5 values
there are 14 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 ?
"10 players are there . two players play one game with one another . so 14 c 2 = 14 * 13 / 2 = 91 so option a is correct"
a ) 91 , b ) 30 , c ) 45 , d ) 60 , e ) 90
a
divide(multiply(14, subtract(14, const_1)), const_2)
subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|
general
A
in how many years will a sum of money doubles itself at 30 % per annum on simple interest ?
"p = ( p * 30 * r ) / 100 r = 3 % answer : b"
a ) 7 % , b ) 3 % , c ) 5 % , d ) 8 % , e ) 2 %
b
divide(const_100, 30)
divide(const_100,n0)|
gain
B
a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 14 km ?
m = 4 s = 2 us = 4 - 2 = 2 d = 14 t = 14 / 2 = 7 answer : e
a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 7
e
divide(14, subtract(4, 2))
subtract(n0,n1)|divide(n2,#0)
physics
E
a man sitting in a train which is traveling at 30 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ?
"relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 30 ) kmph = 82 kmph . answer : b ."
a ) 50 kmph , b ) 82 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph
b
subtract(multiply(divide(280, 9), const_3_6), 30)
divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)|
physics
B
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 60 only . men Γ’ € β„’ s wages are ?
"5 m = xw = 8 b 5 m + xw + 8 b - - - - - 60 rs . 5 m + 5 m + 5 m - - - - - 60 rs . 15 m - - - - - - 60 rs . = > 1 m = 4 rs . answer : c"
a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 3 rs
c
divide(60, multiply(const_3, 5))
multiply(n0,const_3)|divide(n2,#0)|
general
C
how many 2 x 2 x 2 cubes could fit in a box of 7 x 9 x 4 ?
the answer is a ) 31 . a 2 x 2 x 2 cube has an area of 8 . a 7 x 9 x 4 box has an area of 252 . if you divide 252 by 8 , you get 31.5 . since that means you can only fit 31 entire cubes in the box , the answer is 31 .
a ) 31 , b ) 32 , c ) 30 , d ) 33 , e ) 29
a
volume_rectangular_prism(7, 9, 4)
volume_rectangular_prism(n3,n4,n5)|
geometry
A
simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on rs . 4000 for 2 years at 10 % per annum . the sum placed on simple interest is
"explanation : c . i . = ( 4000 Γ— ( 1 + 10 / 100 ) 2 βˆ’ 4000 ) = 4000 βˆ— 11 / 10 βˆ— 11 / 10 βˆ’ 4000 = 840 so s . i . = 840 / 2 = 420 so sum = s . i . βˆ— 100 / r βˆ— t = 420 βˆ— 100 / 3 βˆ— 8 = rs 1750 option b"
a ) rs 1650 , b ) rs 1750 , c ) rs 1850 , d ) rs 1950 , e ) none of these
b
divide(multiply(divide(divide(add(divide(multiply(4000, 10), const_100), divide(multiply(add(4000, divide(multiply(4000, 10), const_100)), 10), const_100)), 2), 3), const_100), 8)
multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1)|
gain
B
village x has a population of 76000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ?
"let the population of two villages be equal after p years then , 76000 - 1200 p = 42000 + 800 p 2000 p = 34000 p = 17 answer is b ."
a ) 15 , b ) 17 , c ) 11 , d ) 18 , e ) 13
b
divide(subtract(76000, 42000), add(800, 1200))
add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|
general
B
on a sum of money , the s . i . for 2 years is $ 660 , while the c . i . is $ 693 , the rate of interest being the same in both the cases . the rate of interest is ?
"difference in c . i . and s . i for 2 years = $ 693 - $ 660 = $ 33 s . i for one year = $ 330 s . i . on $ 330 for 1 year = $ 33 rate = ( 100 * 33 ) / ( 330 ) = 10 % the answer is a ."
a ) 10 % , b ) 32 % , c ) 72 % , d ) 14 % , e ) 82 %
a
divide(multiply(const_100, subtract(693, 660)), divide(660, 2))
divide(n1,n0)|subtract(n2,n1)|multiply(#1,const_100)|divide(#2,#0)|
gain
A
a batch of cookies was divided amomg 4 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin
"this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 4 b = 1 / 4 we can solve for g which is 5 / 12 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 5 / 12 + 1 / 4 ) = 5 / 12 x = 5 / 8 answer : e"
a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 7 / 5 , e ) 5 / 8
e
add(subtract(1, divide(2, 3)), subtract(divide(2, 3), divide(1, 4)))
divide(n1,n2)|divide(n3,n4)|subtract(n3,#0)|subtract(#0,#1)|add(#2,#3)|
general
E
the perimeter of a triangle is 20 cm and the inradius of the triangle is 3 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 = 3 * 20 / 2 = 30 cm 2 answer : b"
a ) 22 , b ) 30 , c ) 77 , d ) 54 , e ) 23
b
triangle_area(3, 20)
triangle_area(n0,n1)|
geometry
B
a horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long . over how much area of the field can it graze ?
area of the shaded portion = 1 ⁄ 4 Γ— Ο€ Γ— ( 14 ) 2 = 154 m 2 answer a
['a ) 154 cm 2', 'b ) 308 m 2', 'c ) 150 m 2', 'd ) 407 m 2', 'e ) none of these']
a
divide(multiply(power(14, const_2), const_pi), const_4)
power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)
geometry
A
the average weight of a group of persons increased from 48 kg to 51 kg , when two persons weighing 88 kg and 93 kg join the group . find the initial number of members in the group ?
"let the initial number of members in the group be n . initial total weight of all the members in the group = n ( 48 ) from the data , 48 n + 88 + 93 = 51 ( n + 2 ) = > 51 n - 48 n = 79 = > n = 26 therefore there were 26 members in the group initially . answer : d"
a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27
d
divide(subtract(add(88, 93), multiply(51, const_2)), subtract(51, 48))
add(n2,n3)|multiply(n1,const_2)|subtract(n1,n0)|subtract(#0,#1)|divide(#3,#2)|
general
D
x can finish a work in 30 days . y can finish the same work in 15 days . yworked for 10 days and left the job . how many days does x alone need to finish the remaining work ?
"work done by x in 1 day = 1 / 30 work done by y in 1 day = 1 / 15 work done by y in 10 days = 10 / 15 = 2 / 3 remaining work = 1 – 2 / 3 = 1 / 3 number of days in which x can finish the remaining work = ( 1 / 3 ) / ( 1 / 30 ) = 10 b"
a ) 3 , b ) 10 , c ) 6 , d ) 8 , e ) 9
b
divide(subtract(const_1, multiply(10, divide(const_1, 15))), divide(const_1, 30))
divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|subtract(const_1,#2)|divide(#3,#1)|
physics
B
a worker can load one truck in 6 hours . a second worker can load the same truck in 8 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ?
"the workers fill the truck at a rate of 1 / 6 + 1 / 8 = 14 / 48 = 7 / 24 of the truck per hour . then the time to fill one truck is 24 / 7 which is about 3.4 hours . the answer is e ."
a ) 2.6 , b ) 2.8 , c ) 3.0 , d ) 3.2 , e ) 3.4
e
inverse(add(divide(const_1, 6), divide(const_1, 8)))
divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|
physics
E
what is the difference between the place values of two sevens in the numeral 54179759 ?
explanation : required difference = 70000 - 700 = 69300 answer is d
a ) 699990 , b ) 99990 , c ) 99980 , d ) 69300 , e ) none of these
d
subtract(multiply(multiply(add(const_3, const_4), const_10), const_1000), multiply(add(const_3, const_4), const_10))
add(const_3,const_4)|multiply(#0,const_10)|multiply(#1,const_1000)|subtract(#2,#1)
general
D
if k is the greatest positive integer such that 4 ^ k is a divisor of 32 ! then k =
"32 / 4 = 8 32 / 16 = 2 8 + 2 = 10 = k answer : e"
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10
e
add(const_1, divide(32, 4))
divide(n1,n0)|add(#0,const_1)|
general
E
mixture a is 40 percent alcohol , and mixture b is 80 percent alcohol . if the two are poured together to create a 4 - gallon mixture that contains 50 percent alcohol , approximately how many gallons of mixture a are in the mixture ?
"( 80 - 50 ) / ( 50 - 40 ) = qa / qb 30 / 10 = qa / qb 3 / 1 = qa / qb qa = ( 3 / 5 ) * 4 = 12 / 5 = 2.4 approx answer : e"
a ) 1.5 , b ) 1.7 , c ) 2.3 , d ) 2.5 , e ) 2.4
e
divide(divide(multiply(40, 80), 80), const_2)
multiply(n0,n1)|divide(#0,n1)|divide(#1,const_2)|
general
E
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 1 / 3 hours to fill the tank . the leak can drain all the water of the tank in ?
"work done by the tank in 1 hour = ( 1 / 2 - 1 / 3 ) = 1 / 14 leak will empty the tank in 14 hrs . answer : d"
a ) 17 hr , b ) 19 hr , c ) 10 hr , d ) 14 hr , e ) 16 hr
d
inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 3), 1), 3))))
divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)|
physics
D
the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs . 8748 , its purchase price was :
"explanation : = rs . 12000 answer : b ) 12000"
a ) 12003 , b ) 12000 , c ) 12002 , d ) 12289 , e ) 12019
b
divide(8748, subtract(const_1, multiply(divide(10, const_100), 3)))
divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2)|
gain
B
a can run 4 times as fast as b and gives b a start of 69 m . how long should the race course be so that a and b might reach in the same time ?
"speed of a : speed of b = 4 : 1 means in a race of 4 m a gains 3 m . then in a race of 69 m he gains 69 * ( 4 / 3 ) i . e 92 m answer : e"
a ) 70 m , b ) 60 m , c ) 80 m , d ) 65 m , e ) 92 m
e
add(multiply(4, divide(divide(69, 4), subtract(4, const_1))), 69)
divide(n1,n0)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2)|add(n1,#3)|
physics
E
the manufacturing cost of a shoe is rs . 200 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains
"explanation : total cost of a watch = 200 + ( 500 / 100 ) = 205 . gain = 20 % = > sp = 1.2 cp = 1.2 x 205 = 246 answer : b"
a ) s 222 , b ) s 246 , c ) s 220 , d ) s 210 , e ) s 217
b
add(add(200, divide(500, 100)), multiply(divide(20, 100), add(200, divide(500, 100))))
divide(n1,n2)|divide(n3,n2)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)|
gain
B
tickets numbered from 1 to 20 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bearsa number which is a multiple of 3 ?
"here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 ] p ( e ) = n ( e ) / n ( s ) = 6 / 20 = 3 / 10 c"
a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7
c
divide(divide(20, 3), 20)
divide(n1,n2)|divide(#0,n1)|
general
C
in some quantity of ghee , 60 % is pure ghee and 40 % is vanaspati . if 10 kg of pure ghee is added , then the strength of vanaspati ghee becomes 20 % . the original quantity was ?
let the original quantity be x then , vanaspati ghee in xkg = 40 x / 100 kg = 2 x / 5 kg ( 2 x / 5 ) / ( x + 10 ) = 20 / 100 2 x / ( 5 x + 50 ) = 1 / 5 x = 10 answer is a
a ) 10 , b ) 15 , c ) 20 , d ) 18 , e ) 22
a
divide(multiply(20, 10), subtract(40, 20))
multiply(n2,n3)|subtract(n1,n3)|divide(#0,#1)
gain
A
the ratio of a to b is 4 to 5 , where a and b are positive . if x equals a increased by 25 percent of a , and m equals b decreased by 80 percent of b , what is the value of m / x ?
"a / b = 4 / 5 m / x = ( 1 / 5 ) * 5 / ( 5 / 4 ) * 4 = 1 / 5 the answer is a ."
a ) 1 / 5 , b ) 3 / 4 , c ) 4 / 5 , d ) 5 / 4 , e ) 3 / 2
a
multiply(divide(subtract(const_100, 80), add(const_100, 25)), divide(5, 4))
add(n2,const_100)|divide(n1,n0)|subtract(const_100,n3)|divide(#2,#0)|multiply(#3,#1)|
general
A
14 men can complete a piece of work in 32 days . in how many days can 16 men complete that piece of work ?
"14 * 32 = 16 * x = > x = 27 1 / 2 days answer : d"
a ) 23 , b ) 27 3 / 4 , c ) 20 1 / 2 , d ) 27 1 / 2 , e ) 11
d
divide(multiply(32, 14), 16)
multiply(n0,n1)|divide(#0,n2)|
physics
D
8.008 / 2.002
"answer is 4 , move the decimal forward three places for both numerator and denominator or just multiply both by a thousand . the result is 8008 / 2002 = 4 answer c"
a ) 0.004 , b ) 0.04 , c ) 4 , d ) 40 , e ) 400
c
multiply(divide(8.008, 2.002), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
C
rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 9 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ?
"let the amount lent at 7 % be rs . x amount lent at 9 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 9 / 100 ( 6000 - x ) = 540 - 2 x / 100 = > 540 - 1 / 50 x = 450 = > x = 4500 amount lent at 10 % = 1500 required ratio = 4500 : 1500 = 9 : 3 answer : a"
a ) 9 : 3 , b ) 9 : 5 , c ) 5 : 8 , d ) 5 : 4 , e ) 9 : 2
a
divide(divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7)), divide(subtract(multiply(450, const_100), multiply(6000, 7)), subtract(9, 7)))
multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)|
gain
A
anand and deepak started a business investing rs . 22,500 and rs . 35,000 respectively . out of a total profit of rs . 13,800 , deepak ’ s share is _____
explanation : ratio of their investments = 22500 : 35000 = 9 : 14 so deepak ' s share = 923923 Γ— 13800 = rs . 5,400 answer : a
a ) 5400 , b ) 3797 , c ) 27877 , d ) 2772 , e ) 9911
a
subtract(multiply(add(add(multiply(multiply(add(const_2, const_3), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), multiply(const_3, multiply(const_100, multiply(add(const_2, const_3), const_2)))), multiply(multiply(const_2, const_4), const_100)), divide(add(multiply(multiply(multiply(add(const_2, const_3), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), const_3), multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2))), add(add(multiply(multiply(multiply(add(const_2, const_3), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), const_3), multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2))), add(add(multiply(multiply(const_100, const_2), const_100), multiply(const_2, multiply(const_100, multiply(add(const_2, const_3), const_2)))), multiply(add(const_2, const_3), const_100))))), multiply(const_3, multiply(const_100, multiply(add(const_2, const_3), const_2))))
add(const_2,const_3)|multiply(const_2,const_4)|multiply(const_100,const_2)|multiply(#0,const_2)|multiply(#1,const_100)|multiply(#0,const_100)|multiply(#2,const_100)|multiply(#3,const_100)|multiply(#5,#3)|multiply(#3,#7)|multiply(#7,const_3)|multiply(#7,const_2)|add(#9,#10)|add(#6,#11)|multiply(#9,const_3)|add(#12,#4)|add(#14,#8)|add(#13,#5)|add(#16,#17)|divide(#16,#18)|multiply(#15,#19)|subtract(#20,#10)
gain
A
ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 60900 . what is his profit percent ?
"total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 60900 profit ( % ) = ( 60900 - 55000 ) / 55000 * 100 = 10.7 % answer : a"
a ) 10.7 % , b ) 19 % , c ) 18 % , d ) 14 % , e ) 16 %
a
multiply(divide(subtract(60900, add(42000, 13000)), add(42000, 13000)), const_100)
add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)|
gain
A
a work crew of 4 men takes 5 days to complete one - half of a job . if 11 men are then added to the crew and the men continue to work at the same rate , how many days will it take the enlarged crew to do the rest of the job ?
"suppose 1 man can do work in x days . . so 4 men will do in . . 4 / x = 1 / 5 * 1 / 2 as half job is done x = 40 now 11 more are added then 15 / 40 = 1 / 2 * 1 / d for remaining half job d = 1 1 / 3 number of days c"
a ) 2 , b ) 3 , c ) 1 1 / 3 , d ) 4 , e ) 4 4 / 5
c
add(5, divide(multiply(4, 5), add(4, 11)))
add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|add(n1,#2)|
physics
C
if the ratio of s . i earned on certain amount on same rate is 4 : 5 . what is the ratio of time ?
s . i 1 / s . i 2 = [ ( p * r * t 1 ) / 100 ] / [ ( p * r * t 2 ) / 100 ] 4 / 5 = t 1 / t 2 ratio = 4 : 5 answer c
a ) 1 : 2 , b ) 6 : 9 , c ) 4 : 5 , d ) 2 : 3 , e ) not possible to calculate
c
divide(4, 5)
divide(n0,n1)
other
C
sum of 24 odd numbers is ?
"sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 24 odd numbers = 24 ^ 2 = 576 answer : e"
a ) 572 , b ) 573 , c ) 574 , d ) 575 , e ) 576
e
multiply(multiply(24, const_2), divide(24, const_2))
divide(n0,const_2)|multiply(n0,const_2)|multiply(#0,#1)|
general
E
the ages of two persons differ by 12 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 + 12 ) years . 5 ( x - 5 ) = ( x + 12 - 5 ) or 4 x = 32 or x = 8 . their present ages are 20 years and 8 years option b"
a ) 20,20 , b ) 20,8 , c ) 25,15 , d ) 30,10 , e ) none of these
b
subtract(add(divide(multiply(12, 5), subtract(5, const_1)), 5), 12)
multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)|
general
B
solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 60 percent carbonated water , what percent of the volume of the mixture is p ?
"60 % is 20 % - points below 80 % and 5 % - points above 55 % . so the ratio of solution p to solution q is 1 : 4 . mixture p is 1 / 5 = 20 % of the volume of mixture pq . the answer is a ."
a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 %
a
multiply(divide(subtract(divide(60, const_100), divide(55, const_100)), add(subtract(divide(60, const_100), divide(55, const_100)), subtract(divide(80, const_100), divide(60, const_100)))), const_100)
divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#0)|add(#3,#4)|divide(#3,#5)|multiply(#6,const_100)|
gain
A
the measurements obtained for the interior dimensions of a rectangular box are 150 cm by 150 cm by 225 cm . if each of the three measurements has an error of at most 1 centimeter , which of the following is the closes maximum possible difference , in cubic centimeters , between the actual capacity of the box and the capacity computed using these measurements ?
the options are well spread so we can approximate . changing the length by 1 cm results in change of the volume by 1 * 150 * 225 = 33,750 cubic centimeters ; changing the width by 1 cm results in change of the volume by 150 * 1 * 225 = 33,750 cubic centimeters ; changing the height by 1 cm results in change of the volume by 150 * 150 * 1 = 22,500 cubic centimeters . so , approximate maximum possible difference is 33,750 + 33,750 + 22,500 = 90,000 cubic centimeters . answer : a
['a ) 90,000', 'b ) 95,000', 'c ) 93,000', 'd ) 92,000', 'e ) 91,000']
a
add(add(multiply(150, 150), multiply(150, 225)), multiply(225, 150))
multiply(n0,n0)|multiply(n0,n2)|add(#0,#1)|add(#2,#1)
physics
A
the mean of 50 observations was 36 . it was found later that an observation 90 was wrongly taken as 23 . the corrected new mean is ?
"correct sum = ( 36 * 50 + 90 - 23 ) = 1867 . correct mean = 1825 / 50 = 37.3 answer : a"
a ) 37.3 , b ) 36.1 , c ) 36.5 , d ) 36.9 , e ) 36.3
a
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
A
5.005 / 2.002 =
"5.005 / 2.002 = 5005 / 2002 = 5 ( 1001 ) / 2 ( 1001 ) = 5 / 2 = 2.5 the answer is e ."
a ) 2.05 , b ) 2.50025 , c ) 2.501 , d ) 2.5025 , e ) 2.5
e
multiply(divide(5.005, 2.002), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
E
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 36 , what is the total amount ?
"x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 36 35 - - - ? = > 140 answer : b"
a ) 166 , b ) 140 , c ) 178 , d ) 177 , e ) 169
b
add(add(multiply(divide(const_100, 45), 36), multiply(divide(30, 45), 36)), 36)
divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)|
general
B
find the cost of fencing around a circular field of diameter 40 m at the rate of rs . 3 a meter ?
"2 * 22 / 7 * 20 = 125.66 125.66 * 3 = rs . 376.98 answer : c"
a ) 400 , b ) 370.4 , c ) 376.98 , d ) 340.9 , e ) 350.03
c
multiply(circumface(divide(40, const_2)), 3)
divide(n0,const_2)|circumface(#0)|multiply(n1,#1)|
physics
C
in the junior basketball league there are 18 teams , 2 / 3 of them are bad and Β½ are rich . what ca n ' t be the number of teams that are rich and bad ?
otal teams = 18 bad teams = ( 2 / 3 ) * 18 = 12 rich teams = 9 so maximum value that the both rich and bad can take will be 9 . so e = 10 can not be that value . answer : e
a ) 4 . , b ) 6 . , c ) 7 . , d ) 8 . , e ) 10
e
add(multiply(18, divide(const_1, const_2)), const_1)
divide(const_1,const_2)|multiply(n0,#0)|add(#1,const_1)
general
E
the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years k in which the population would be doubled assuming there is no emigration or immigration is
"suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 / 100 ) ) ^ n n = 35 answer is d ."
a ) 20 , b ) k = 25 , c ) 30 , d ) k = 35 , e ) 40
d
divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100))
multiply(const_10,const_3)|subtract(n1,n2)|divide(#1,n0)|subtract(const_100,#0)|multiply(#2,const_100)|divide(#3,#4)|
general
D
a boat goes 100 km downstream in 10 hours , and 200 km upstream in 25 hours . the speed of the stream is ?
"100 - - - 10 ds = 10 ? - - - - 1 200 - - - - 30 us = 8 ? - - - - - 1 s = ( 10 - 8 ) / 2 = 1 kmph . answer : a"
a ) 1 , b ) 22 1 / 7 , c ) 2 , d ) 22 1 / 2 , e ) 3
a
divide(subtract(divide(100, 10), divide(200, 25)), const_2)
divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2)|
physics
A
at deluxe paint store , fuchsia paint is made by mixing 5 parts of red paint with 3 parts of blue paint . mauve paint is made by mixing 3 parts of red paint with 6 parts blue paint . how many liters of blue paint must be added to 16 liters of fuchsia to change it to mauve paint ?
"in 16 liters , red = 5 / 8 * 16 = 10 and blue = 6 so , 10 / ( 6 + x ) = 3 / 6 or , x = 14 ( answer c )"
a ) 9 , b ) 12 , c ) 14 , d ) 16 , e ) 18
c
subtract(multiply(divide(multiply(16, divide(5, add(5, 3))), divide(3, add(5, 3))), divide(6, add(5, 3))), subtract(16, multiply(16, divide(5, add(5, 3)))))
add(n0,n1)|divide(n0,#0)|divide(n2,#0)|divide(n3,#0)|multiply(n4,#1)|divide(#4,#2)|subtract(n4,#4)|multiply(#5,#3)|subtract(#7,#6)|
general
C
last year department store x had a sales total for december that was 7 times the average ( arithmetic mean ) of the monthly sales totals for january through november . the sales total for december was what fraction of the sales total for the year ?
"let avg for 11 mos . = 10 therefore , dec = 70 year total = 11 * 10 + 70 = 180 answer = 70 / 180 = 7 / 18 = d"
a ) 1 / 4 , b ) 4 / 15 , c ) 1 / 3 , d ) 7 / 18 , e ) 4 / 5
d
divide(7, add(subtract(const_12, const_1), 7))
subtract(const_12,const_1)|add(n0,#0)|divide(n0,#1)|
general
D
a can run 160 metre in 28 seconds and b in 32 seconds . by what distance a beat b ?
"clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 160 / 32 = 5 m / s distance covered by b in 4 seconds = speed Γ£ β€” time = 5 Γ£ β€” 4 = 20 metre i . e . , a beat b by 20 metre answer is c"
a ) 38 metre , b ) 28 metre , c ) 20 metre , d ) 15 metre , e ) 28 metre
c
subtract(160, multiply(divide(160, 32), 28))
divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|
physics
C
the simple interest on a sum of money will be rs . 900 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ?
"p - - - 10 - - - - 900 p - - - 5 - - - - - 450 3 p - - - 5 - - - - - 1350 - - - - - - = > 1800 answer : a"
a ) 1800 , b ) 2888 , c ) 1200 , d ) 2699 , e ) 2771
a
add(multiply(multiply(divide(900, 10), 5), const_3), multiply(divide(900, 10), 5))
divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)|
general
A
the product of three consecutive numbers is 504 . then the sum of the smallest two numbers is ?
"product of three numbers = 504 504 = 7 * 8 * 9 . so , the three numbers are 7 , 8 and 9 . and sum of smallest of these two = 7 + 8 = 15 . answer : option b"
a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56
b
multiply(power(const_2, 504), factorial(const_4))
factorial(n0)|power(const_2,const_4.0)|multiply(#0,#1)|
general
B
a number is said to be prime saturated if the product of all the different positive prime factors of e is less than the square root of e . what is the greatest two digit prime saturated integer ?
"e = 96 = 3 * 32 = 3 * 2 ^ 5 answer is d ."
a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95
d
add(multiply(const_2, const_3), subtract(const_100, const_10))
multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)|
other
D
two trains 161 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ?
"t = ( 161 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 8.09 answer : d"
a ) 6.18 , b ) 7.12 , c ) 7.1 , d ) 8.09 , e ) 8.11
d
divide(add(161, 165), multiply(add(80, 65), const_0_2778))
add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|
physics
D
a block of wood has dimensions 10 cm x 10 cm x 50 cm . the block is painted red and then cut evenly at the 25 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ?
"the area of each half is 100 + 4 ( 250 ) + 100 = 1200 the area that is not painted is 100 . the fraction that is not painted is 100 / 1200 = 1 / 12 = 8.3 % the answer is b ."
a ) 5.5 % , b ) 8.3 % , c ) 11.6 % , d ) 14.2 % , e ) 17.5 %
b
multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100)
multiply(const_100,const_4)|multiply(#0,const_4)|add(#1,const_100)|add(#2,const_100)|divide(const_100,#3)|multiply(#4,const_100)|
geometry
B
if p is a prime number greater than 3 , find the remainder when p ^ 2 + 17 is divided by 12 .
"take any prime number greater than 3 and check - say p = 5 so , p ^ 2 + 17 = 25 + 17 = > 42 42 / 12 will have remainder as 6 say p = 7 so , p ^ 2 + 17 = 49 + 17 = > 66 66 / 12 will have remainder as 6 so answer will be ( a ) 6"
a ) 6 , b ) 1 , c ) 0 , d ) 8 , e ) 7
a
subtract(add(17, power(add(const_1, const_4), 2)), multiply(12, 3))
add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)|
general
A
a man purchased 15 pens , 12 books , 10 pencils and 5 erasers . the cost of each pen is rs . 36 , each book is rs . 45 , each pencil is rs . 8 , and the cost of each eraser is rs . 40 less than the combined costs of pen and pencil . find the total amount spent ?
explanation : cost of each eraser = ( 36 + 8 - 40 ) = rs . 4 required amount = 15 * 36 + 12 * 45 + 10 * 8 + 5 * 4 540 + 540 + 80 + 20 = rs . 1180 answer : e
a ) 2388 , b ) 2337 , c ) 1192 , d ) 2827 , e ) 1180
e
add(add(multiply(15, 36), multiply(12, 45)), multiply(10, 8))
multiply(n0,n4)|multiply(n1,n5)|multiply(n2,n6)|add(#0,#1)|add(#3,#2)
general
E
a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points w that an individual player might have scored ?
"general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points w for 12 th player is 100 - 77 = 23 . answer : e ."
a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23
e
add(subtract(100, multiply(12, 7)), 7)
multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)|
general
E
if x : y = 4 : 7 , find the value of ( 6 x + 2 y ) : ( 5 x – y )
explanation : given : x / y = 4 / 7 ( 6 x + 2 y ) : ( 5 x – y ) = ( 6 * 4 + 2 * 7 ) : ( 5 * 4 – 7 ) 38 : 13 answer : e
a ) 7 : 13 , b ) 38 : 7 , c ) 3 : 13 , d ) 2 : 3 , e ) 38 : 13
e
divide(add(power(6, 2), 2), add(add(4, 7), 2))
add(n0,n1)|power(n2,n3)|add(n3,#1)|add(n3,#0)|divide(#2,#3)
general
E
let f ( x ) = x ^ 2 + bx + c . if f ( 1 ) = 0 and f ( - 8 ) = 0 , then f ( x ) crosses the y - axis at what y - coordinate ?
"when x = 1 and when x = - 8 , the expression f ( x ) = x Β² + bx + c equals 0 . then f ( x ) = ( x - 1 ) ( x + 8 ) f ( 0 ) = - 8 the answer is a ."
a ) - 8 , b ) - 1 , c ) 0 , d ) 1 , e ) 8
a
negate(divide(subtract(power(8, 2), 1), add(8, 1)))
add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)|
general
A
a particular library has 150 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 65 percent of books that were loaned out are returned and there are 122 books in the special collection at that time , how many books of the special collection were loaned out during that month ?
"the total number of books is 150 . let x be the number of books which were loaned out . 65 % of books that were loaned out are returned . 35 % of books that were loaned out are not returned . now , there are 122 books , thus the number of un - returned books is 150 - 122 = 28 books . 0.35 x = 28 x = 80 the answer is c ."
a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120
c
divide(subtract(150, 122), subtract(const_1, divide(65, const_100)))
divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)|
gain
C
the salary of a worker is first increased by 40 % and afterwards reduced by 40 % . what is the net change in the worker ' s salary ?
"let x be the original salary . the final salary is 0.6 ( 1.4 x ) = 0.84 x the answer is c ."
a ) 8 % decrease , b ) 8 % increase , c ) 16 % decrease , d ) 16 % increase , e ) no change
c
subtract(const_100, subtract(add(40, const_100), divide(multiply(add(40, const_100), 40), const_100)))
add(n0,const_100)|multiply(n0,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)|
gain
C
machine p and machine q are each used to manufacture 990 sprockets . it takes machine p 10 hours longer to produce 990 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?
"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 990 / x = 990 / 1.1 x + 10 1.1 ( 990 ) = 990 + 11 x 11 x = 99 x = 9 the answer is e ."
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9
e
divide(subtract(990, divide(990, add(divide(10, const_100), const_1))), 10)
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)|
gain
E
the number 523 fbc is divisible by 7 , 89 . then what is the value of f * b * c
lcm of 7 , 8 and 9 is 504 , thus 523 fbc must be divisible by 504 . 523 fbc = 523000 + fbc 523000 divided by 504 gives a remainder of 352 . hence , 352 + fbc = k * 504 . k = 1 fbc = 152 - - > f * b * c = 10 k = 2 fbc = 656 - - > f * b * c = 180 as fbc is three digit number k can not be more than 2 . two answers ? well only one is listed in answer choices , so d . answer : d .
a ) 504 , b ) 532 , c ) 210 , d ) 180 , e ) 280
d
add(divide(add(divide(523, const_4), divide(523, const_4)), const_2), multiply(add(const_4, const_1), const_10))
add(const_1,const_4)|divide(n0,const_4)|add(#1,#1)|multiply(#0,const_10)|divide(#2,const_2)|add(#4,#3)
general
D
the youngest of 4 children has siblings who are 2 , 7 , and 11 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 25 , what is the age of the youngest sibling ?
"x + ( x + 2 ) + ( x + 7 ) + ( x + 11 ) = 100 4 x + 20 = 100 4 x = 80 x = 20 the answer is d ."
a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21
d
divide(subtract(multiply(const_4.0, 25), add(add(4, 7), 11)), 4)
add(n1,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)|
general
D
set a contains all the even numbers between 22 and 70 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ?
"each term in set b is 40 more than the corresponding term in set a . the difference of the sums = 25 * 40 = 1000 . the answer is c ."
a ) 600 , b ) 800 , c ) 1000 , d ) 1200 , e ) 1400
c
multiply(subtract(62, 22), add(divide(subtract(70, 22), const_2), const_1))
subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)|
general
C
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 72 kmph . after how much time will the trains meet ?
"they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 54 + 72 ) * 5 / 18 = 7 * 5 = 35 mps . the time required = d / s = 100 / 35 = 20 / 7 sec . answer : c"
a ) 20 / 8 sec , b ) 20 / 4 sec , c ) 20 / 7 sec , d ) 22 / 7 sec , e ) 60 / 7 sec
c
divide(100, multiply(add(54, 72), const_0_2778))
add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)|
physics
C
a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , with 17 balls per box . if 94 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ?
"there is no way to store 94 balls without leftovers : 94 βˆ’ 0 βˆ— 25 = 94 - 94 βˆ’ 2 βˆ— 25 = 44 , 94 βˆ’ 3 βˆ— 25 = 19 are not divisible by 17 . 93 balls can be stored successfully : 93 βˆ’ 1 βˆ— 25 = 68 is divisible by 17 . thus , 93 = 1 βˆ— 25 + 4 βˆ— 17 and we need 1 big box and 4 small boxes . answer : b"
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
b
subtract(25, 17)
subtract(n0,n1)|
general
B
jim drove 923 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ?
the number of miles to drive to finish his journey is given by 1200 - 923 = 277 miles correct answer b
a ) 113 miles , b ) 277 miles , c ) 456 miles , d ) 887 miles , e ) 767 miles
b
subtract(1200, 923)
subtract(n1,n0)
physics
B
i flew my tiny seaplane to visit my mother . on the flight up , i flew at 110 mph . on the way home , i flew 72 mph . what was my average speed for the trip ?
"( 110 mph + 72 mph ) / 2 = 91 mph correct answer is : b"
a ) 198 mph , b ) 91 mph , c ) 88 mph , d ) 100 mph , e ) 99 mph
b
divide(add(110, 72), const_2)
add(n0,n1)|divide(#0,const_2)|
physics
B
a trained covered x km at 65 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 3 x km .
"total time taken = x / 65 + 2 x / 20 hours = 3 x / 26 hours average speed = 3 x / ( 3 x / 26 ) = 26 kmph answer : c"
a ) 22 , b ) 99 , c ) 26 , d ) 66 , e ) 887
c
divide(multiply(65, 3), add(divide(65, 65), divide(multiply(2, 65), 20)))
divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)|
general
C
two boats are heading towards each other at constant speeds of 4 miles / hr and 20 miles / hr respectively . they begin at a distance 20 miles from each other . how far are they ( in miles ) one minute before they collide ?
"the question asks : how far apart will they be 1 minute = 1 / 60 hours before they collide ? since the combined rate of the boats is 4 + 20 = 25 mph then 1 / 60 hours before they collide they ' ll be rate * time = distance - - > 24 * 1 / 60 = 6 / 15 miles apart . answer : b ."
a ) 1 / 12 , b ) 6 / 15 , c ) 1 / 6 , d ) 1 / 3 , e ) 1 / 5
b
divide(add(20, 4), const_60)
add(n0,n1)|divide(#0,const_60)|
physics
B
9.8 , 9.8 , 9.9 , 9.9 , 10.0 , 10.0 , 10.1 , 10.5 the mean and the standard deviation of the 8 numbers shown above is 10 and 0.212 respectively . what percent of the 8 numbers are within 1 standard deviation of the mean ?
within 1 standard deviation of the mean - means in the range { mean - 1 * sd ; mean + 1 * sd } = { 10 - 1 * 0.212 ; 10 + 0.3 } = { 9.788 ; 10.212 } . from the 8 listed numbers , 7 are within this range so 6 / 8 = 87.5 % . answer : b .
a ) 90.5 % , b ) 87.5 % , c ) 80.5 % , d ) 77.5 % , e ) 70.5 %
b
multiply(divide(subtract(8, const_1), 8), const_100)
subtract(n8,const_1)|divide(#0,n8)|multiply(#1,const_100)
general
B
a man can row 9 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ?
"m = 9 s = 1.2 ds = 10.2 us = 7.8 x / 10.2 + x / 7.8 = 1 x = 4.42 d = 4.42 * 2 = 8.84 answer : c"
a ) 6.24 km , b ) 6 km , c ) 8.84 km , d ) 5.66 km , e ) 10 km
c
multiply(divide(multiply(add(9, 1.2), subtract(9, 1.2)), add(add(9, 1.2), subtract(9, 1.2))), const_2)
add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)|
physics
C
the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 13 . if p invested the money for 5 months , find for how much time did q invest the money ?
"7 * 5 : 5 * x = 7 : 13 x = 13 answer : c"
a ) 19 , b ) 17 , c ) 13 , d ) 10 , e ) 12
c
multiply(multiply(divide(7, 5), divide(13, 7)), 5)
divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)|
gain
C
joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 35 . he was allowed to drop his lowest score . after doing so , his average test score improved to 40 . what is the lowest test score that was dropped ?
the arithmetic mean of 4 equally weighted tests was 35 . so what we can assume is that we have 4 test scores , each 35 . he dropped his lowest score and the avg went to 40 . this means that the lowest score was not 35 and other three scores had given the lowest score 5 each to make it up to 35 too . when the lowest score was removed , the other 3 scores got their 5 back . so the lowest score was 3 * 5 = 15 less than 35 . so the lowest score = 35 - 15 = 20 answer ( a )
a ) 20 , b ) 25 , c ) 55 , d ) 65 , e ) 80
a
subtract(multiply(35, 4), multiply(40, const_3))
multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)
general
A
a card game called β€œ high - low ” divides a deck of 52 playing cards into 2 types , β€œ high ” cards and β€œ low ” cards . there are an equal number of β€œ high ” cards and β€œ low ” cards in the deck and β€œ high ” cards are worth 2 points , while β€œ low ” cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β€œ high ” and β€œ low ” cards to earn 5 points if you must draw exactly 3 β€œ low ” cards ?
to get a 5 , you need one high and three lows ( you could have had 2 highs and one low , but the constraint is that you must have three low cards ) hlll = 4 ! 3 ! = 4 4 ! is the number of ways you can arrange these four spaces . divide by 3 ! because you you repeat three low cards ans : d
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
d
divide(multiply(multiply(multiply(const_4, 3), 2), 1), multiply(multiply(3, 2), 1))
multiply(n5,const_4)|multiply(n1,n5)|multiply(n1,#0)|multiply(n3,#1)|multiply(n3,#2)|divide(#4,#3)
general
D
what least number must be subtracted from 427398 so that remaining number is divisible by 15
"explanation : on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted answer : option a"
a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) none of these
a
subtract(427398, multiply(floor(divide(427398, 15)), 15))
divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)|
general
A
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 50 .
"first , let us say i have a number n which is divisible by 5 and by 7 . we all agree that it will be divisible by 35 , the lcm of 5 and 7 . now , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1 , we can say the number is of the form n = 5 a + 1 e . g . 5 + 1 , 10 + 1 , 15 + 1 , 20 + 1 , 25 + 1 , 30 + 1 , 35 + 1 etc and n = 7 b + 1 e . g . 7 + 1 , 14 + 1 , 21 + 1 , 28 + 1 , 35 + 1 etc so when it is divided by the lcm , 35 , it will give 1 as remainder ( as is apparent above ) next , if i have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3 , we can say the number is of the form n = 5 a + 1 and n = 7 b + 3 now , the only thing you should try to understand here is that when n is divided by 5 and if i say the remainder is 1 , it is the same as saying the remainder is - 4 . e . g . when 6 is divided by 5 , remainder is 1 because it is 1 more than a multiple of 5 . i can also say it is 4 less than the next multiple of 5 , ca n ' t i ? 6 is one more than 5 , but 4 less than 10 . therefore , we can say n = 5 x - 4 and n = 7 y - 4 ( a remainder of 3 when divided by 7 is the same as getting a remainder of - 4 ) now this question is exactly like the question above . so when you divide n by 50 , remainder will be - 4 i . e . n will be 4 less than a multiple of 50 . so you must add 12 to n to make it a multiple of 50 c"
a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35
c
subtract(50, reminder(3, 7))
reminder(n3,n2)|subtract(n4,#0)|
general
C
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 7 minutes , then the area of the park ( in sq . m ) is :
"perimeter = distance covered in 7 min . = ( 12000 / 60 ) x 7 m = 1400 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1400 or x = 140 . length = 420 m and breadth = 280 m . area = ( 420 x 280 ) m 2 = 117600 m 2 . answer : d"
a ) 153601 , b ) 153600 , c ) 153602 , d ) 117600 , e ) 153604
d
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 7), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 7), const_1000), add(3, 2)), const_2), 2))
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
D
water is leaking out from a cylinder container at the rate of 0.31 m ^ 3 per minute . after 10 minutes , the water level decreases 4 meters . what is value of the radius ?
"10 * 0.31 = 3.1 m ^ 3 = pi * r ^ 2 * h r ^ 2 = 3.1 / ( pi * 4 ) which is about 1 / 4 r = 1 / 2 the answer is a ."
a ) 0.5 , b ) 1.0 , c ) 1.5 , d ) 2.0 , e ) 2.5
a
divide(multiply(10, 0.31), 4)
multiply(n0,n2)|divide(#0,n3)|
physics
A
a sum fetched a total simple interest of rs . 100 at the rate of 5 p . c . p . a . in 4 years . what is the sum ?
"sol . principal = rs . [ 100 * 100 / 5 * 4 ] = rs . [ 10000 / 20 ] = rs . 500 . answer c"
a ) 800 , b ) 600 , c ) 500 , d ) 1000 , e ) 300
c
divide(divide(multiply(100, const_100), 5), 4)
multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)|
gain
C
if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 16 ?
"@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 16 . then 4 n < 16 and n < 4 . the largest possible integer is n = 3 . the answer is c ."
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
c
floor(divide(16, subtract(5, const_1)))
subtract(n0,const_1)|divide(n1,#0)|floor(#1)|
general
C
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
E
of 3 numbers , the third is 4 times the second and the second is two times the first . if their average is 165 , the smallest of the 3 numbers is :
explanation : let first number be x . so , 2 nd no . = 2 x & 3 rd no . = 8 x . so , x + 2 x + 8 x = 165 Γ— 3 = 495 11 x = 495 / 11 x = 495 / 11 hence , smallest number x = 45 answer : c
a ) 18 , b ) 19 , c ) 45 , d ) 21 , e ) 22
c
divide(multiply(165, 3), add(multiply(4, const_2), 3))
multiply(n0,n2)|multiply(n1,const_2)|add(n0,#1)|divide(#0,#2)
general
C
what is the range of all the roots of | x ^ 2 - 6 | = x ?
"we get 2 quadratic equations here . . 1 ) x ^ 2 - x - 2 = 0 . . . . . . . roots 2 , - 1 2 ) x ^ 2 + x - 2 = 0 . . . . . . . . roots - 2 , 1 inserting each root in given equation , it can be seen that - 1 and - 2 do not satisfy the equations . so value of x for given equation . . . . x = 6 or x = 1 i guess range is 6 - 1 = 5 e"
a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 5
e
sqrt(6)
sqrt(n1)|
general
E
the total number of digits used in numbering the pages of a book having 266 pages is
"total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 167 ) = ( 9 + 180 + 501 ) = 690 . answer : e"
a ) 732 , b ) 990 , c ) 109 , d ) 130 , e ) 690
e
subtract(subtract(multiply(266, const_3), subtract(const_100, const_1)), subtract(const_10, const_1))
multiply(n0,const_3)|subtract(const_100,const_1)|subtract(const_10,const_1)|subtract(#0,#1)|subtract(#3,#2)|
general
E
find the area of a parallelogram with base 12 cm and height 10 cm ?
"area of a parallelogram = base * height = 12 * 10 = 120 cm 2 answer : c"
a ) 297 cm 2 , b ) 384 cm 2 , c ) 120 cm 2 , d ) 267 cm 2 , e ) 186 cm 2
c
multiply(12, 10)
multiply(n0,n1)|
geometry
C
in a classroom , 12 students brought apples and 8 students brought bananas . if exactly 10 students brought only one of the two types of fruits , how many students brought both types of fruits ?
say x students brought both fruits . ( 12 - x ) + ( 8 - x ) = 10 - - > x = 5 . answer : a .
a ) 5 , b ) 6 , c ) 7 , d ) 12 , e ) 14
a
divide(add(12, 8), const_4)
add(n0,n1)|divide(#0,const_4)
other
A
calculate the share of y , if rs . 2690 is divided among x , y and z in the ratio 5 : 7 : 9 ?
"5 + 7 + 9 = 21 2690 / 21 = 128.1 y ' s share = 7 * 128.1 = 896.7 answer : d"
a ) 890.7 , b ) 826.7 , c ) 895.7 , d ) 896.7 , e ) 816.7
d
multiply(divide(2690, add(add(5, 7), 9)), 5)
add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n1,#2)|
general
D
a glass was filled with 15 ounces of water , and 0.05 ounce of the water evaporated each day during a 15 - day period . what percent of the original amount of water evaporated during this period ?
"in 15 days 15 * 0.05 = 0.75 ounces of water evaporated , which is 0.75 / 15 Γ’ Λ† β€” 100 = 5 of the original amount of water . answer : d ."
a ) 0.005 % , b ) 0.05 % , c ) 0.5 % , d ) 5 % , e ) 25 %
d
multiply(divide(multiply(0.05, 15), 15), const_100)
multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)|
gain
D
2056 x 987 = ?
"= 2056 x 987 = 2056 x ( 1000 - 13 ) = 2056 x 1000 - 2056 x 13 = 2056000 - 26728 = 2029272 answer is b"
a ) 1936372 , b ) 2029272 , c ) 1896172 , d ) 1923472 , e ) none of them
b
multiply(divide(2056, 987), const_100)
divide(n0,n1)|multiply(#0,const_100)|
general
B
in a kilometer race , a beats b by 40 meters or 15 seconds . what time does a take to complete the race ?
"time taken by b run 1000 meters = ( 1000 * 15 ) / 50 = 300 sec . time taken by a = 300 - 15 = 285 sec . answer : b"
a ) 277 sec , b ) 285 sec , c ) 267 sec , d ) 167 sec , e ) 276 sec
b
subtract(divide(multiply(const_1, const_1000), divide(40, 15)), 15)
divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)|
physics
B
rain is falling at a rate of 3 centimeters per hour all over springfield . somewhere downtown in springfield a group of pigeons is waiting for the rain to stop . if the rain filled a round puddle the with a base area of 350 square centimeters and a depth of 13.5 centimeters , how long did the pigeons wait for the rain to stop ?
the volume of the puddle is irrelevant and only height matters since rain fell all over the city . thus , it takes only . 13.5 / 3 = 4.5 hours of rain to fill the puddle answer is : b , 4 hours and 30 mins
['a ) 3 hours and 12 minutes .', 'b ) four hours and 30 minutes', 'c ) four hours and 45 minutes', 'd ) five hours and 10 minutes', 'e ) five hours and 30 minutes']
b
divide(13.5, 3)
divide(n2,n0)
geometry
B
if the average ( arithmetic mean ) of a and b is 100 , and c – a = 120 , what is the average of b and c ?
"a + b / 2 = 100 = > a + b = 200 a = c - 120 . . . sub this value c - 120 + b = 200 = > c + b = 320 = > c + b / 2 = 160 answer : c"
a ) 150 , b ) 140 , c ) 160 , d ) 170 , e ) 180
c
subtract(multiply(120, const_2), multiply(100, const_2))
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
general
C
company z has 53 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday .
"say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 53 . now , plug options for x . d x > y as needed . answer : d ."
a ) 6 , b ) 7 , c ) 8 , d ) 11 , e ) 12
d
add(const_4, add(floor(divide(53, add(const_4, const_3))), const_1))
add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4)|
general
D
of the 200 employees at company x , 50 are full - time , and 150 have worked at company x for at least a year . there are 10 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ?
"200 employees 50 are full - time 150 have worked at company x for at least a year 10 employees at company x who aren ’ t full - time and haven ’ t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ? 200 - 50 = 150 employees not full time 150 - 10 = 140 employees not full time who worked over a year 150 employees have worked at company x for at least a year - 140 employees not full time who worked over a year = 10 full - time employees of company x have worked at the company for at least a year ans c"
a ) 20 , b ) 30 , c ) 10 , d ) 80 , e ) 100
c
subtract(subtract(200, 50), 10)
subtract(n0,n1)|subtract(#0,n3)|
general
C
p and q invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if p invested rs . 60 , 000 , the amount invested by q is :
"suppose q invested rs . y . then , y = 60,000 / y = 2 / 3 or y = 60000 / 2 * 3 = 90,000 . answer : d"
a ) rs . 65,000 , b ) rs . 70,000 , c ) rs . 80,000 , d ) rs . 90,000 , e ) rs . 60,000
d
multiply(multiply(add(add(3, 2), 2), add(3, 2)), 2)
add(n0,n1)|add(n0,#0)|multiply(#1,#0)|multiply(n0,#2)|
gain
D
sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 30 ps per meter is ?
"3 x * 2 x = 3750 = > x = 25 2 ( 75 + 30 ) = 210 m 210 * 1 / 2 = rs . 105 answer : b"
a ) s . 122 , b ) s . 105 , c ) s . 125 , d ) s . 120 , e ) s . 121
b
divide(multiply(30, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100)
multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)|
physics
B
the average of 25 results is 50 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ?
"solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 50 or , x = 878 . answer : option c"
a ) 741 , b ) 752 , c ) 878 , d ) 785 , e ) 458
c
subtract(subtract(multiply(25, 50), multiply(12, 17)), multiply(12, 14))
multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)|
general
C
the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 12 , and k < m < r < s < t . if t is 20 , what is the greatest possible value of the median of the 5 integers ?
"we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 12 which means that in effect , all numbers are equal to 12 . if the largest number is 20 , it is 8 more than 12 . we need r to be maximum so k and m should be as small as possible to get the average of 12 . since all the numbers are positive integers , k and m can not be less than 1 and 2 respectively . 1 is 11 less than 12 and 2 is 10 less than 12 which means k and m combined are 21 less than the average . 20 is already 8 more than 12 and hence we only have 21 - 8 = 13 extra to distribute between r and s . since s must be greater than r , r can be 12 + 6 = 18 and s can be 12 + 7 = 19 . so r is 18 . answer ( d )"
a ) 16 , b ) 18 , c ) 19 , d ) 20 , e ) 22
d
subtract(divide(subtract(multiply(12, 5), 20), const_2), const_2)
multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|subtract(#2,const_2)|
general
D
what is the least number of square tiles required to pave the floor of a room 11 m 50 cm long and 1 m 50 cm broad ?
"solution length of largest tile = h . c . f . of 1150 cm & 150 cm = 50 cm . area of each tile = ( 50 x 50 ) cm 2 ∴ required number of tiles = [ 1150 x 150 / 50 x 50 ] = 69 . answer c"
a ) 724 , b ) 804 , c ) 69 , d ) 844 , e ) none
c
divide(multiply(add(multiply(1, const_100), 50), multiply(11, const_100)), power(50, const_2))
multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2)|
physics
C
a reduction of 40 % in the price of bananas would enable a man to obtain 65 more for rs . 40 , what is reduced price per dozen ?
"40 * ( 40 / 100 ) = 16 - - - 65 ? - - - 12 = > rs . 2.95 answer : b"
a ) 1.95 , b ) 2.95 , c ) 4.95 , d ) 3.95 , e ) 5.95
b
multiply(const_12, divide(multiply(40, divide(40, const_100)), 65))
divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)|
gain
B
two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team e . if team a ' s average speed was 5 mph greater than team e ' s , what was team e ' s average mph ?
this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team e ' s speed was s . then team e ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team e . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team e is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct .
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25
d
divide(divide(300, 5), 3)
divide(n0,n2)|divide(#0,n1)
physics
D