options
stringlengths 37
300
| correct
stringclasses 5
values | annotated_formula
stringlengths 7
727
| problem
stringlengths 5
967
| rationale
stringlengths 1
2.74k
| program
stringlengths 10
646
|
|---|---|---|---|---|---|
a ) 112 , b ) 64 , c ) 42 , d ) 24 , e ) 56 | e | multiply(add(7, 1), add(1, 6)) | in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 7 , 1 ) , ( 1 , 1 ) , ( 1 , - 6 ) and ( - 7 , - 6 ) ? | "length of side 1 = 7 + 1 = 8 length of side 2 = 6 + 1 = 7 area of rectangle = 7 * 8 = 56 e is the answer" | a = 7 + 1
b = 1 + 6
c = a * b
|
a ) 3387 , b ) 1000 , c ) 1200 , d ) 2787 , e ) 1121 | c | multiply(multiply(20, 12), 5) | in digging a pond 20 m * 12 m * 5 m the volumes of the soil extracted will be ? | "20 * 12 * 5 = 1200 answer : c" | a = 20 * 12
b = a * 5
|
a ) 28 , b ) 42 , c ) 41 , d ) 35 , e ) 49 | b | add(add(divide(subtract(300, 1), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(7, const_2)) | how many times digit 7 is used while writing numbers from 1 to 300 ? | "in 1 to 100 their are 14 seven ' s in 100 to 200 their are 14 seven ' s in 200 to 300 their are 14 seven ' s so total answer is 42 seven ' s correct option : b" | a = 300 - 1
b = a / 10
c = 10 + 1
d = 10 + 1
e = c * d
f = b + e
g = 7 * 2
h = f + g
|
a ) 7 , b ) 16 , c ) 10 , d ) 11 , e ) can not be determined | a | add(const_1, add(const_2, const_4)) | define a = the sum of digits of the number 7 ^ 100 and b = the sum of the digits of a . what is the sum of the digits of b ? | following approach will take you only couple of sec . \ ( 7 ^ { 100 } = ( 7 ^ { 50 } ) ^ 2 \ ) the sum of digits ( digital root ) of a perfect square can only be 1 , 4 , 7 or 9 . for example \ ( 7 ^ 2 = 49 \ ) = > dr = 4 + 9 = 13 = > 1 + 3 = 4 and etc . checking available options . answer a . | a = 2 + 4
b = 1 + a
|
a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | add(multiply(divide(subtract(add(add(add(multiply(const_4, const_100), multiply(5, const_10)), 5), const_1000), multiply(150, 5)), add(150, 85)), const_2), 5) | an optometrist charges $ 150 per pair for soft contact lenses and $ 85 per pair for hard contact lenses . last week she sold 5 more pairs of soft lenses than hard lenses . if her total sales for pairs of contact lenses last week were $ 1,925 , what was the total number of pairs of contact lenses that she sold ? | "( x + 5 ) * 150 + x * 85 = 1925 = > x = 5 total lens = 5 + ( 5 + 5 ) = 15 answer c" | a = 4 * 100
b = 5 * 10
c = a + b
d = c + 5
e = d + 1000
f = 150 * 5
g = e - f
h = 150 + 85
i = g / h
j = i * 2
k = j + 5
|
a ) 20 o , b ) 24 o , c ) 249 , d ) 30 o , e ) 325 | c | add(add(divide(1000, add(const_4, const_1)), divide(subtract(1000, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(1000, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros does 1000 ! end with ? | "according to above 1000 ! has 1000 / 5 + 1000 / 25 + 1000 / 125 + 1000 / 625 = 200 + 40 + 8 + 1 = 249 trailing zeros . answer : c" | a = 4 + 1
b = 1000 / a
c = 4 + 1
d = 1000 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 1000 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
|
a ) 9.03 pm , b ) 9.05 pm , c ) 9.07 pm , d ) 9.08 pm , e ) 9.10 pm | b | divide(add(divide(455, 60), divide(455, 45)), const_2) | train e and f , 455 miles apart , are traveling toward each other at constant rates and in the same time zone . if train e left at 4 pm traveling at a speed of 60 miles per hour , and train f left at 5 : 45 pm and traveling at 45 miles per hour , then at what time would they pass each other ? | first , since e has a headstart then in that 1 hr 45 min or 1.75 hrs he travels 105 miles then remaining distance to be traveled will be 455 - 105 = 350 miles now , using relative rates ( 105 ) ( t ) = 350 this gives 10 / 3 hours now 5.45 pm + 10 / 3 hours gives us 9.05 pm hence answer is b | a = 455 / 60
b = 455 / 45
c = a + b
d = c / 2
|
a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | c | add(add(const_2.0, 1), 3) | in the xy - plane , the point ( 1 , 3 ) is the center of a circle . the point ( - 2 , 3 ) lies inside the circle and the point ( 1 , - 2 ) lies outside the circle . if the radius r of the circle is an integer , then r = | "an easy way to solve this question will be just to mark the points on the coordinate plane . you ' ll see that the distance between the center ( 1 , 3 ) and the point inside the circle ( - 2 , 3 ) is 3 units ( both points are on y = 3 line so the distance will simply be 1 - ( - 2 ) = 3 ) so the radius must be more than 3 units . the distance between the center ( 1,3 ) and the point outside the circle ( 1 , - 2 ) is 5 units ( both points are on x = 1 line so the distance will simply be 3 - ( - 2 ) = 5 ) so the radius must be less than 5 units , which implies 3 < r < 5 , thus as r is an integer then r = 4 . answer : c ." | a = 2 + 0
b = a + 3
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | multiply(3, divide(negate(multiply(add(add(7, 5), 3), 5)), subtract(multiply(add(add(7, 5), 3), const_2), multiply(add(add(7, 5), 3), 5)))) | the ratio by weight , measured in pounds , of books to clothes to electronics in a suitcase initially stands at 7 : 5 : 3 . someone removes 5 pounds of clothing from the suitcase , thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? | "the weights of the items in the suitcase are 7 k , 5 k , and 3 k . if removing 5 pounds of clothes doubles the ratio of books to clothes , then 5 pounds represents half the weight of the clothes . 2.5 k = 5 pounds and then k = 2 pounds . the electronics weigh 3 ( 2 ) = 6 pounds . the answer is d ." | a = 7 + 5
b = a + 3
c = b * 5
d = negate / (
e = 7 + 5
f = e + 3
g = f * 2
h = 7 + 5
i = h + 3
j = i * 5
k = g - j
l = 3 * d
|
a ) 11 / 14 , b ) 13 / 18 , c ) 4 / 7 , d ) 3 / 7 , e ) 3 / 14 | a | divide(subtract(divide(2, add(1, 2)), subtract(divide(2, 9), multiply(divide(2, 9), divide(3, 4)))), divide(subtract(9, 2), 9)) | when 2 / 9 of the votes on a certain resolution have been counted , 3 / 4 of those counted are in favor of the resolution . what fraction w of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution ? | "if we use variable for total votes there will be too many fractions to manipulate with , so pick some smart # : let set total # of votes is 18 . 2 / 9 of the votes on a certain resolution have been counted - - > 4 counted and 18 - 4 = 14 votes left to be counted ; 3 / 4 of those counted are in favor of the resolution - - > 3 in favor and 1 against ; ratio of those who voted against to those who voted for to be 2 to 1 there should be total of 18 * 2 / 3 = 12 people who voted against , so in the remaining 14 votes there should be 12 - 1 = 11 people who voted against . thus w = 11 / 14 of the remaining votes must be against . answer : a ." | a = 1 + 2
b = 2 / a
c = 2 / 9
d = 2 / 9
e = 3 / 4
f = d * e
g = c - f
h = b - g
i = 9 - 2
j = i / 9
k = h / j
|
a ) 12 , b ) 27 , c ) 21 , d ) 17 , e ) 11 | d | multiply(17, const_1) | the total age of a and b is 17 years more than the total age of b and c . c is how many years younger than a ? | "( a + b ) - ( b + c ) = 17 a - c = 17 . c is younger than a by 17 years . answer : d" | a = 17 * 1
|
a ) 3 / 8 , b ) 21 / 25 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | b | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20.000), 6), 6), 3), 20.000) | a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 3 percent and 6 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | "x * 3 / 100 * 1 + ( 20000 - x ) * 8 / 100 * 1 = 1440 3 x - 8 x = 144000 - 160000 = > - 5 x = - 16000 = > x = 32000 so 16800 / 20000 = 21 / 25 answer - b" | a = 1 + 4
b = 2 * a
c = 100 * b
d = 4 * 100
e = c + d
f = 1 + 4
g = 2 * f
h = 4 * g
i = e + h
j = i / 20
k = j / 6
l = k + 6
m = l - 3
n = m / 20
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 80 | e | divide(add(multiply(10, 10), const_60), const_2) | if a = ( 10 ^ 10 ) - 47 , what is the sum of all the digit of a ? | start small and check for any pattern : 10 ^ 2 - 47 = 53 - > number of digits = power of 10 ; number of 9 s = ( power of 10 ) - 2 ; 53 is the last 2 digits 10 ^ 3 - 47 = 953 - > number of digits = power of 10 ; number of 9 s = ( power of 10 ) - 2 ; 53 is the last 2 digits . . . extrapolate 10 ^ 10 - 47 - > number of digits = 10 ; number of 9 s = 8 ; 53 will be the last 2 digits sum , a = 9 * 8 + 5 + 3 = 80 . answer ( e ) . | a = 10 * 10
b = a + const_60
c = b / 2
|
a ) 24 , b ) 17 , c ) 28 , d ) 42 , e ) it can not be determined from the information given . | d | subtract(multiply(7, const_10), multiply(const_4, 7)) | two sets of 7 consecutive positive integers have exactly one integer in common . the sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set ? | a = ( 1 , 2,3 , 4,5 , 6,7 ) , sum of this = 28 b = ( 7 , 8,9 , 10,11 , 12,13 ) , sum of this = 70 the differenct between 70 - 28 = 42 hence , 42 is the answer i . e . d | a = 7 * 10
b = 4 * 7
c = a - b
|
['a ) 12541', 'b ) 12542', 'c ) 12544', 'd ) 12545', 'e ) can not be determined'] | c | add(power(divide(subtract(223, 1), const_2), const_2), 223) | a lady grows cauliflower in her garden that is in the shape of a square . each cauliflower takes 1 square foot of area in her garden . this year , she has increased her output by 223 cauliflower when compared to last year . the shape of the area used for growing the cauliflower has remained a square in both these years . how many cauliflowers did she produce this year ? | explanatory answer the shape of the area used for growing cauliflower has remained a square in both the years . let the side of the square area used for growing cauliflower this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing cauliflower last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of cauliflower grown has increased by 223 , the area would have increased by 223 sq ft because each cauliflower takes 1 sq ft space . hence , x 2 - y 2 = 223 ( x + y ) ( x - y ) = 223 . 223 is a prime number and hence it will have only two factors . i . e . , 223 and 1 . therefore , 223 can be expressed as product of 2 numbers in only way = 223 * 1 i . e . , ( x + y ) ( x - y ) = 223 * 1 so , ( x + y ) should be 223 and ( x - y ) should be 1 . solving the two equations we get x = 112 and y = 111 . therefore , number of cauliflower produced this year = x 2 = 1122 = 12544 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice a , b and d . as 12541,12542 and 12545 are not the square of any number . check choice c : if this year ' s produce is 12544 , last year ' s produce would have been 12544 - 223 = 12321 12321 is the square of 111 . so , 12544 is the answer . choice c . | a = 223 - 1
b = a / 2
c = b ** 2
d = c + 223
|
a ) 150 , b ) 162 , c ) 154 , d ) 158 , e ) 144 | b | add(multiply(17, 9), 9) | 12 . what is the dividend . divisor 17 , the quotient is 9 and the remainder is 9 . | "d = d * q + r d = 17 * 9 + 9 d = 153 + 9 d = 162 answer b" | a = 17 * 9
b = a + 9
|
a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 5 hours , e ) none | c | divide(68, add(13, 4)) | a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr . find the time taken by the boat to go 68 km downstream ? | "solution speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 68 km downstream = ( 68 / 17 ) hrs = 4 hrs . answer c" | a = 13 + 4
b = 68 / a
|
a ) 138 , b ) 390 , c ) 393 , d ) 279 , e ) 222 | c | multiply(circumface(divide(42, const_2)), 3) | find the cost of fencing around a circular field of diameter 42 m at the rate of rs . 3 a meter ? | "2 * 22 / 7 * 21 = 131 131 * 3 = rs . 393 answer : c" | a = 42 / 2
b = circumface * (
|
a ) 3 : 5 , b ) 3 : 9 , c ) 3 : 2 , d ) 2 : 2 , e ) 3 : 8 | a | divide(divide(1, 15), power(divide(1, 2), 2)) | the volumes of two cones are in the ratio 1 : 15 and the radii of the cones are in the ratio of 1 : 2 . what is the length of the wire ? | "the volume of the cone = ( 1 / 3 ) Ο r 2 h only radius ( r ) and height ( h ) are varying . hence , ( 1 / 3 ) Ο may be ignored . v 1 / v 2 = r 1 ^ 2 . h 1 / r 2 ^ 2 . h 2 = > 1 / 15 = ( 1 ) ^ 2 h 1 / ( 2 ) ^ 2 h 2 = > h 1 / h 2 = 3 / 5 i . e . h 1 : h 2 = 3 : 5 answer : a" | a = 1 / 15
b = 1 / 2
c = b ** 2
d = a / c
|
a ) a ) 23 , b ) b ) 21 , c ) c ) 52 , d ) d ) 56 , e ) e ) 12 | d | add(7, divide(multiply(7, subtract(12000, 6750)), subtract(6750, 6000))) | the average salary of all the workers in a workshop is rs . 6750 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | "let the total number of workers be x . then , 6750 x = ( 12000 * 7 ) + 6000 ( x - 7 ) = > 750 x = 42000 = x = 56 . answer : d" | a = 12000 - 6750
b = 7 * a
c = 6750 - 6000
d = b / c
e = 7 + d
|
a ) 10 % , b ) 14 % , c ) 18 % , d ) 22 % , e ) 26 % | b | divide(multiply(subtract(const_100, 70), multiply(add(divide(const_2, const_0_25), const_2), const_4)), const_100) | each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , forty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 70 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on are incandescent ? | "let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.4 i + 0.8 f = 0.7 ( i + f ) 0.1 f = 0.3 i f = 3 i this means that for every 1 incandescent bulb , there are 3 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.4 i / ( 0.4 i + 0.8 f ) = 0.4 i / ( 0.4 i + 0.8 * 3 i ) = 0.4 i / 2.8 i = 1 / 7 which is about 14 % . the answer is b ." | a = 100 - 70
b = 2 / const_0_25
c = b + 2
d = c * 4
e = a * d
f = e / 100
|
a ) 12 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | floor(divide(109, 9)) | on dividinng 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | d = ( d - r ) / q = ( 109 - 1 ) / 9 = 108 / 9 = 18 a | a = 109 / 9
b = math.floor(a)
|
a ) 130 , b ) 620 , c ) 147 , d ) 120 , e ) none of these | c | subtract(multiply(700, power(add(const_1, divide(10, const_100)), 2)), 700) | what is the compound interest paid on a sum of rs . 700 for the period of 2 years at 10 % per annum . | "solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 / 100 = 11 total % of interest = 10 + 11 = 21 total interest = 21 % 700 = 700 * ( 21 / 100 ) = 147 answer c" | a = 10 / 100
b = 1 + a
c = b ** 2
d = 700 * c
e = d - 700
|
a ) 240 m , b ) 150 m , c ) 120 m , d ) 200 m , e ) 165 m | d | divide(15, subtract(divide(15, 10), 10)) | a train covers a distance of 15 km in 10 min . if it takes 10 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 15 / 10 * 60 ) km / hr = ( 90 * 5 / 18 ) m / sec = 25 m / sec . length of the train = 20 * 10 = 200 m . answer : d" | a = 15 / 10
b = a - 10
c = 15 / b
|
a ) 105.00 kg , b ) 115.60 kg , c ) 125.00 kg , d ) 130.50 kg , e ) 155.55 kg | e | multiply(divide(divide(multiply(subtract(const_100, 65), 400), const_100), subtract(const_100, 10)), const_100) | fresh grapes contain 65 % by weight while dried grapes contain 10 % water by weight . what is the weight of dry grapes available from 400 kg of fresh grapes ? | "from the question we know : 400 kg * 65 % = 260 kg of water in the fresh grapes 400 kg - 260 kg of water = 140 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 10 % of the weight of the dry graps is water and we know that 140 kg is non - water mass we can set up the following equation : x = 1 / 10 ( x ) + 140 kg 9 / 10 ( x ) = 140 kg x = 155.55 kg answer - e" | a = 100 - 65
b = a * 400
c = b / 100
d = 100 - 10
e = c / d
f = e * 100
|
a ) $ 382.10 , b ) $ 395.40 , c ) $ 404.30 , d ) $ 416.80 , e ) $ 426.50 | b | add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.70), multiply(multiply(4, 65), 0.70)), multiply(2.20, add(4, 2))) | in a fuel station the service costs $ 2.20 per vehicle and every liter of fuel costs $ 0.70 . assuming that you fill up 4 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ? | "the service cost of 4 vans and 2 trucks is 6 * 2.20 $ 13.20 the fuel in 4 vans is 4 * 65 = 260 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 546 liters the total fuel cost is 546 * 0.7 = $ 382.20 the total cost is $ 382.20 + $ 13.20 = $ 395.40 the answer is b ." | a = 120 * 65
b = a / 100
c = b + 65
d = c * 2
e = d * 0
f = 4 * 65
g = f * 0
h = e + g
i = 4 + 2
j = 2 * 20
k = h + j
|
a ) 66 , b ) 70 , c ) 74 , d ) 78 , e ) 82 | d | add(add(multiply(7, const_2), 4), add(multiply(7, multiply(const_2, 4)), 4)) | a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 7 with 4 students left over . what is the sum of the two smallest possible values of n ? | "n = 4 k + 1 = 7 j + 4 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 7 j + 4 . 1 , 5 , 9 , 13 , 17 , 21 , 25 = 7 ( 3 ) + 4 the next such number is 25 + 4 * 7 = 53 . 25 + 53 = 78 the answer is d ." | a = 7 * 2
b = a + 4
c = 2 * 4
d = 7 * c
e = d + 4
f = b + e
|
a ) 9 / 14 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | a | divide(add(divide(84, 2), divide(84, 8)), 84) | if an integer n is to be chosen at random from the integers 1 to 84 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? | "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 84 , there are 42 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 42 + 12 ) / 84 = 54 / 84 = 9 / 14 ans is a" | a = 84 / 2
b = 84 / 8
c = a + b
d = c / 84
|
a ) 120 , b ) 240 , c ) 360 , d ) 480 , e ) 720 | c | add(multiply(factorial(const_4), add(add(add(const_4, const_3), const_2), const_1)), factorial(subtract(6, const_1))) | an auto assembly plant performs 6 functions with each frame that arrives : add axles , add wheels to the axles , install the windshield to the frame , install the instrument panel , install the steering wheel , and install the interior seating . once those 6 tasks are performed , each car goes to a separate building for finishing touches . if these tasks can be arranged along a linear assembly line in any order , except that the axles must be installed before the wheels can be added , how many y ways can the assembly line be arranged ? | c ) 360 short way : there are 6 c ! ways to do the six tasks . half will have wheels before axles and half will have axles before wheels . so we want y = 6 c ! / 2 - > 720 / 2 = 360 | a = math.factorial(4)
b = 4 + 3
c = b + 2
d = c + 1
e = a * d
f = 6 - 1
g = math.factorial(f)
h = e + g
|
a ) 15 , b ) 17 , c ) 19 , d ) 20 , e ) 21 | c | divide(50, subtract(5, divide(const_1, 1))) | company x provides bottled water to its employees in 5 liter bottles , $ 3 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 50 fixed monthly rate and $ 1 for 20 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 20 l bottles will be x / 4 ( to equate the vol . ) since the total cost will be equal , 3 x = 50 + x / 4 so x = 18.18 or 19 . answer is ( c ) . | a = 1 / 1
b = 5 - a
c = 50 / b
|
a ) β 48 , b ) β 2 , c ) 2 , d ) 7 , e ) 48 | d | subtract(subtract(subtract(subtract(add(add(7, 8), subtract(7, 8)), const_1), const_1), const_1), const_1) | if a ( a - 7 ) = 8 and b ( b - 7 ) = 8 , where a β b , then a + b = | "i . e . if a = - 1 then b = 8 or if a = 8 then b = - 1 but in each case a + b = - 1 + 8 = 7 answer : option d" | a = 7 + 8
b = 7 - 8
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) s . 429 , b ) s . 480 , c ) s . 429 , d ) s . 488 , e ) s . 419 | d | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 30), const_100)) | by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 30 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 30 % = 30 % of 640 = rs . 192 sp = cp - loss = 640 - 192 = rs . 448 answer : d" | a = 800 * 100
b = 25 + 100
c = a / b
d = 800 * 100
e = 25 + 100
f = d / e
g = f * 30
h = g / 100
i = c - h
|
a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 233 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 15), 50)), divide(multiply(const_100, 15), 50))) | if the cost price of 50 articles is equal to the selling price of 15 articles , then the gain or loss percent is ? | "percentage of profit = 35 / 15 * 100 = 233 % answer : e" | a = 100 * 15
b = a / 50
c = 100 - b
d = 100 * 15
e = d / 50
f = c / e
g = 100 * f
|
a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(2, add(5, 4)) | if the function q is defined by the formula q = 5 e / ( 4 x ( z ^ 2 ) ) , by what factor will q be multiplied if e is quadrupled , x is doubled , and z is tripled ? | we just need to find the factor thats all , e - > quadrupled - > 4 e x - > doubled - > 2 x z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 e is in numerator , and x * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b | a = 5 + 4
b = 2 / a
|
a ) 76 kmph , b ) 6 kmph , c ) 9 kmph , d ) 8 kmph , e ) 4 kmph | c | divide(subtract(divide(95, 2), divide(60, 2)), const_2) | a man rows his boat 95 km downstream and 60 km upstream , taking 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 95 / ( 2 ) = 48 kmph speed upstream = d / t = 60 / ( 2 ) = 30 kmph the speed of the stream = ( 48 - 30 ) / 2 = 9 kmph answer : c" | a = 95 / 2
b = 60 / 2
c = a - b
d = c / 2
|
a ) 92.5 % , b ) 88.6 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(6, const_100)))), add(600, 400)), const_100) | a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 6 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ 600 = 9000 / 100 = 90 number of rotten bananas = 6 % of 400 = 24 therefore , total number of rotten fruits = 90 + 24 = 114 therefore number of fruits in good condition = 1000 - 114 = 886 therefore percentage of fruits in good condition = ( 886 / 1000 Γ 100 ) % = ( 88600 / 1000 ) % = 88.6 % answer : b" | a = 600 + 400
b = 15 / 100
c = 600 * b
d = 6 / 100
e = 400 * d
f = c + e
g = a - f
h = 600 + 400
i = g / h
j = i * 100
|
a ) 40 , b ) 60 , c ) 80 , d ) 90 , e ) 100 | b | multiply(add(const_2, const_3), multiply(4, 3)) | if a farmer sells 20 of his chickens , his stock of feed will last for 3 more days than planned , but if he buys 15 more chickens , he will run out of feed 4 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ? | let x = total feed required for the planned period n = number of chicken t = total time of the planned feed x = nt 1 ) x = ( n - 15 ) * ( t + 4 ) 2 ) x = ( n + 20 ) * ( t - 3 ) equating 1 & 2 ( n - 15 ) * ( t + 4 ) = ( n + 20 ) * ( t - 3 ) 7 n = 35 t n = 5 t x = n * n / 5 substituting this value in 1 n * n / 5 = ( n - 15 ) * ( n / 5 + 4 ) 5 n = 300 n = 60 b | a = 2 + 3
b = 4 * 3
c = a * b
|
a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 42 | e | subtract(456, reminder(987654, 456)) | which number can we add to 987654 to get a number exactly divisible by 456 ? | "987654 / 456 = 2165 with a remainder of 414 . we need to add 456 - 414 = 42 the answer is e ." | a = 456 - reminder
|
a ) 5 : 3 , b ) 4 : 3 , c ) 6 : 3 , d ) 3 : 4 , e ) 5 : 4 | b | divide(subtract(const_1, divide(4, multiply(const_3, const_2))), divide(25, const_100)) | if 25 % of a number is subtracted from a second number , the second number reduces to its 4 - sixth . what is the ratio of the first number to the second number ? | explanation : let the numbers be p & q q - p / 4 = 4 q / 6 = > q - 4 q / 6 = p / 4 = > 2 q / 6 = p / 4 = > p / q = 4 / 3 answer b | a = 3 * 2
b = 4 / a
c = 1 - b
d = 25 / 100
e = c / d
|
a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 5 / 9 | e | divide(subtract(60, 10), subtract(const_100, 10)) | the weight of a glass of jar is 10 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 10 g weight of coffee beans = 90 g weight of jar and remaining beans = 60 g weight of remaining beans = 50 g fraction remaining = 50 / 90 = 5 / 9 answer is e ." | a = 60 - 10
b = 100 - 10
c = a / b
|
a ) 16200 , b ) 16350 , c ) 16500 , d ) 16750 , e ) 16250 | c | multiply(800, multiply(5.5, 3.75)) | the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of rs . 800 per sq . metre . | "area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = rs . 800 hence , total cost = 5.5 Γ 3.75 Γ 800 = 5.5 Γ 3000 = rs . 16500 answer is c ." | a = 5 * 5
b = 800 * a
|
a ) 187 , b ) 193 , c ) 199 , d ) 205 , e ) 211 | e | add(multiply(105, const_2), const_1) | the sum of the even numbers between 1 and n is 105 * 106 , where n is an odd number , then n = ? | let n - 1 = 2 a . 2 + 4 + . . . + 2 a = 2 * ( 1 + 2 + . . . + a ) = 2 * ( a ) ( a + 1 ) / 2 = ( a ) ( a + 1 ) = 105 * 106 then a = 105 and n = 211 . the answer is e . | a = 105 * 2
b = a + 1
|
a ) $ 32 , b ) $ 50 , c ) $ 320 , d ) $ 400 , e ) $ 500 | c | divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 8), const_100) | in township k each property is taxed at 8 percent of its assessed value . if the assessed value of a property in township k is increased from $ 20,000 to $ 24,000 , by how much will the property tax increase ? | "property tax as a percentage of its assessed value = 8 % increase in assessed value of the property = 24000 - 20000 = 4,000 $ increase in property tax = ( 8 / 100 ) * 4000 = 320 $ answer c" | a = 3 + 4
b = a * 4
c = b * 1000
d = 4 + 1
e = d * 4
f = e * 1000
g = c - f
h = g * 8
i = h / 100
|
['a ) 68', 'b ) 69', 'c ) 70', 'd ) 71', 'e ) 72'] | b | sqrt(add(4321, power(multiply(const_2, const_10), const_2))) | if ( 20 ) Β² is subtracted from the square of a number , the answer so obtained is 4321 . what is the number ? | x ^ 2 = 4321 + 400 = 4721 4761 = 69 * 69 x = 69 answer : b | a = 2 * 10
b = a ** 2
c = 4321 + b
d = math.sqrt(c)
|
a ) 133 , b ) 143 , c ) 153 , d ) 163 , e ) 173 | c | subtract(10, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 1 and 10 ^ 17 are such that the sum of their digits is 2 ? | "the integers with a sum of 2 are : 2 , 20 , 200 , . . . , 2 * 10 ^ 16 and there are 17 integers in this list . also , these integers have a sum of 2 : 11 101 , 110 1001 , 1010 , 1100 etc . . . the number of integers in this list is 1 + 2 + . . . + 16 thus , the total number of integers is 1 + 2 + . . . + 16 + 17 = 17 * 18 / 2 = 153 the answer is c ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 10 - e
|
a ) 10 min , b ) 12 min , c ) 18 min , d ) 20 min , e ) 15 min | e | multiply(const_60, divide(subtract(60, 45), 60)) | excluding stoppages , the speed of a bus is 60 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 15 km less . time taken to cover 15 km = ( 15 / 60 x 60 ) min = 15 min answer : e" | a = 60 - 45
b = a / 60
c = const_60 * b
|
a ) 4050 , b ) 3500 , c ) 4000 , d ) 4500 , e ) 5000 | c | subtract(add(5050, 5200), 6250) | the average monthly income of p and q is rs . 5050 . the average monthly income of q and r is rs . 6250 and the average monthly income of p and r is rs . 5200 . the monthly income of p is : | let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 5050 x 2 ) = 10100 . . . . ( i ) q + r = ( 6250 x 2 ) = 12500 . . . . ( ii ) p + r = ( 5200 x 2 ) = 10400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 33000 or p + q + r = 16500 . . . . ( iv ) subtracting ( ii ) from ( iv ) , we get p = 4000 . p ' s monthly income = rs . 4000 . answer : c | a = 5050 + 5200
b = a - 6250
|
a ) 8 , b ) 12 , c ) 16 , d ) 24 , e ) 46 / 4 | e | divide(multiply(23, 2), const_4) | if ( 1 / 2 ) ^ 23 ( 1 / 81 ) ^ k = 1 / 18 ^ 23 , then k = | "i ' m going to focus on denominator only . . ( 2 ^ 23 ) . ( ( 3 ^ 4 ) ^ k = 18 ^ 23 ( 2 ^ 23 ) . ( ( 3 ^ 4 k ) = ( 2 . 3 ^ 2 ) ^ 23 ( 2 ^ 23 ) . ( ( 3 ^ 4 k ) = ( 2 ^ 23 ) . ( 3 ^ 2 ) ^ 23 hence 4 k = 46 k = 46 / 4 answer e i hope it ' s quite clear" | a = 23 * 2
b = a / 4
|
a ) $ 83 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 111 | e | subtract(multiply(90, 15), add(multiply(87, 7), multiply(90, 7))) | the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 90 per day . what was his wage on the 8 th day ? | "average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 90 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 90 * 7 ) = 1350 - ( 609 + 630 ) = 1350 - 1239 = 111 answer e" | a = 90 * 15
b = 87 * 7
c = 90 * 7
d = b + c
e = a - d
|
a ) 10.8 , b ) 20.8 , c ) 28.8 , d ) 18.8 , e ) 17.8 | c | multiply(6, add(const_1, divide(80, const_100))) | a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 80 % ? | "c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 480 % of re . 1 = rs . 24 / 5 for rs . 24 / 5 , toffees sold = 6 . for re . 1 , toffees sold = 6 x 24 / 5 = 28.8 . answer c" | a = 80 / 100
b = 1 + a
c = 6 * b
|
a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 15 | c | divide(add(add(11, const_4), subtract(34, const_4)), const_2) | find the average of all the numbers between 11 and 34 which are divisible by 10 ? | "average = ( 20 + 30 ) 2 = 50 / 2 = 25 answer is c" | a = 11 + 4
b = 34 - 4
c = a + b
d = c / 2
|
a ) $ 208 , b ) $ 96 , c ) $ 104 , d ) $ 120 , e ) $ 240 | a | multiply(add(divide(0.80, const_2), multiply(multiply(subtract(1.10, 0.80), const_10), 0.04)), 400) | a certain company expects quarterly earnings of $ 0.80 per share of stock , half of which will be distributed as dividends to shareholders while the rest will be used for research and development . if earnings are greater than expected , shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings . if quarterly earnings are $ 1.10 per share , what will be the dividend paid to a person who owns 400 shares of the company ' s stock ? | "eps actual > eps expected . each gets and additional . 12 per share . thus . 52 * 400 - - > $ 208 answer is a" | a = 0 / 80
b = 1 - 10
c = b * 10
d = c * 0
e = a + d
f = e * 400
|
['a ) 20', 'b ) 10', 'c ) 50', 'd ) 74', 'e ) 73'] | d | divide(volume_cylinder(divide(20, const_2), 10), volume_cylinder(divide(3, const_2), 6)) | the content of several smaller cylinders ( 3 meter diameter and 6 meter height ) were to be emptied into a larger cylinder ( 20 meter diameter and 10 meter height ) , how many smaller cylinders will fill up the larger cylinder ? | calculate the volume of the larger cylinder and divide it by the volume of the smaller cylinder . volume of cylinder = Ο r 2 h larger cylinder volume = 3141.59 smaller cylinder volume = 42.41 therefore the number of cylinders b that can be filled to capacity = 3141.59 / 42.41 = 74.1 answer is d 74 smaller cylinders can be filled to capacity . | a = 20 / 2
b = volume_cylinder / (
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | divide(divide(divide(divide(840, const_2), const_3), const_4), divide(const_10, const_2)) | if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 840 , what is the least possible value of n ? | 840 = 2 * 2 * 2 * 3 * 5 * 7 , so n must be at least 7 . the answer is c . | a = 840 / 2
b = a / 3
c = b / 4
d = 10 / 2
e = c / d
|
a ) 108 , b ) 42 , c ) 56 , d ) 32 , e ) 58 | a | divide(subtract(divide(30, divide(10, const_100)), 72), 2) | among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 10 % of listco ' s sales staff . listco employs 72 level - 1 employees , 30 of whom are college graduates . how many sales staff members without college degrees are level - 2 employees ? | "i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 72 and that c and l 1 = 0.10 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.10 x = 30 , therefore the grand total is 300 . now we know that l 2 is 300 - 72 = 228 . we also learn that c and no c are equally represented thus 150 each . therefore no c and no l 2 will be 150 - 42 = 108 . thus a is the correct answer choice" | a = 10 / 100
b = 30 / a
c = b - 72
d = c / 2
|
a ) 16 , b ) 25 , c ) 33 , d ) 46 , e ) 58 | d | add(divide(circumface(9), const_2), multiply(const_2, 9)) | a semicircle has a radius of 9 . what is the approximate perimeter of the semicircle ? | "the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r / 2 + 2 r = pi * r + 2 r the perimeter is pi * 9 + 2 * 9 which is about 46 . the answer is d ." | a = circumface / (
b = a + 2
|
a ) β 220 , b ) β 100 , c ) 100 , d ) 135 , e ) it can not be determined from the information given | c | subtract(multiply(const_60.0, const_2), multiply(210, const_2)) | if the average ( arithmetic mean ) of a and b is 210 , and the average of b and c is 160 , what is the value of a β c ? | "question : a - c = ? ( a + b ) / 2 = 210 = = = > a + b = 420 ( b + c ) / 2 = 160 = = = > b + c = 320 ( a + b ) - ( b + c ) = 420 - 320 = = = > a + b - b - c = 100 = = = > a - c = 100 answer : c" | a = const_60 * 0
b = 210 * 2
c = a - b
|
a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) 4691110842 | e | multiply(469158, 9999) | calculate 469158 x 9999 = ? | "answer 469158 x 9999 = 469157 x ( 10000 - 1 ) = 4691580000 - 469158 = 4691110842 . option : e" | a = 469158 * 9999
|
a ) 70 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | a | divide(multiply(add(33, divide(1, 3)), 210), const_100) | 33 1 / 3 % of 210 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 210 = 70 a )" | a = 1 / 3
b = 33 + a
c = b * 210
d = c / 100
|
a ) 33.3 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 66.6 % | a | divide(20, divide(subtract(const_100, 40), const_100)) | on a certain transatlantic crossing , 20 percent of a ship ' s passengers held round - trip tickets and also took their cars abroad the ship . if 40 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ' s passengers held round - trip tickets ? | "0.20 p = rt + c 0.4 ( rt ) = no c = > 0.60 ( rt ) had c 0.20 p = 0.60 ( rt ) rt / p = 1 / 3 = 33 % answer - a" | a = 100 - 40
b = a / 100
c = 20 / b
|
a ) 22 , b ) 23 , c ) 24 , d ) 26 , e ) 28 | e | divide(subtract(multiply(48, const_4), 136), const_2) | a man has some hens and cows . if the number of heads be 48 and the number of feet equals 136 , then the number of hens will be : | "let hens be x and cows be y now , feet : x * 2 + y * 4 = 136 heads : x * 1 + y * 1 = 48 implies , 2 x + 4 y = 136 and x + y = 48 solving these two equations , we get x = 28 and y = 20 therefore , hens are 28 . answer : e" | a = 48 * 4
b = a - 136
c = b / 2
|
a ) 14.0 , b ) 9.0 , c ) 10.0 , d ) 11.0 , e ) 12.0 | e | add(06, add(divide(multiply(divide(subtract(divide(500, const_100), const_1), divide(800, 1100)), subtract(divide(800, 1100), divide(500, 1100))), add(divide(500, 1100), const_1)), divide(subtract(divide(500, const_100), const_1), divide(800, 1100)))) | hillary and eddy are climbing to the summit of mt . everest from a base camp 5,100 ft from the summit . when they depart for the summit at 06 : 00 , hillary climbs at a rate of 800 ft / hr with eddy lagging behind at a slower rate of 500 ft / hr . if hillary stops 1100 ft short of the summit and then descends at a rate of 1,000 ft / hr , at what time do hillary and eddy pass each other on her return trip ? | "solution : h stopped 800 ft before reaching the final point , time taken to reach 4000 ft = 4000 / 800 = 5 hrs . this means she reached there at 11 : 00 . speed difference between them is 800 - 500 = 300 ft / hr so by the time h stops they have 1500 ft of distance so now here we use relative speed formula they both are travelling toward each other with speed of 1000 and 500 total 1500 ft / hr and distance between them is 1500 ft so time taken to meet = 1 hr from 11 : 00 means 12 : 00 is the answer . e" | a = 500 / 100
b = a - 1
c = 800 / 1100
d = b / c
e = 800 / 1100
f = 500 / 1100
g = e - f
h = d * g
i = 500 / 1100
j = i + 1
k = h / j
l = 500 / 100
m = l - 1
n = 800 / 1100
o = m / n
p = k + o
q = 6 + p
|
a ) 40 sec , b ) 42 sec , c ) 44 sec , d ) 48 sec , e ) 50 sec | c | divide(360, multiply(subtract(45, 190), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 190 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 190 = 550 m required time = 550 * 2 / 25 = 44 sec answer : c" | a = 45 - 190
b = a * const_0_2778
c = 360 / b
|
a ) 86 , b ) 88 , c ) 92 , d ) 94 , e ) 96 | a | add(multiply(add(multiply(6, const_3), 2), divide(add(multiply(6, const_3), 2), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder . the dividend is : | "divisor = ( 6 * 3 ) + 2 = 20 5 * quotient = 20 quotient = 4 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 4 ) + 6 = 86 . a" | a = 6 * 3
b = a + 2
c = 6 * 3
d = c + 2
e = d / 5
f = b * e
g = f + 6
|
a ) 21 years , b ) 22 years , c ) 20 years , d ) 24 years , e ) 26 years | c | divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 22 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 22 . = > 2 ( x + 2 ) = ( x + 22 + 2 ) = > 2 x + 4 = x + 24 = > x = 20 years option c" | a = 2 * 2
b = a - 2
c = 22 - b
d = 2 - 1
e = c / d
|
a ) 7 sec , b ) 5 sec , c ) 4 sec , d ) 6 sec , e ) 3 sec | e | multiply(const_3600, divide(divide(55, const_1000), add(60, 6))) | a train 55 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 55 * 3 / 55 ] sec = 3 sec answer : e" | a = 55 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 52 sec , b ) 45 sec , c ) 72 sec , d ) 25 sec , e ) 39 sec | c | divide(multiply(60, const_4), multiply(12, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 60 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 60 * 4 = 240 m time taken = 240 * 3 / 10 = 72 sec answer is c" | a = 60 * 4
b = 1000 / 3600
c = 12 * b
d = a / c
|
a ) 0.20 , b ) 0.24 , c ) 20 , d ) 24 , e ) 55 | d | multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.24, 100)) | in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 6 eggs . if the cost of each pound of rice is $ 0.24 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 24 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 / 12 * 24 = 24 cents . answer : d ." | a = 1 / 2
b = 6 / a
c = b / 12
d = 0 * 24
e = c * d
|
a ) 33 , b ) 360 / 7 , c ) 28 , d ) 27 / 8 , e ) 16 | b | divide(add(40, 60), 2) | murali travelled from city a to city b at a speed of 40 kmph and from city b to city c at 60 kmph . what is the average speed of murali from a to c given that the ratio of distances between a to b and b to c is 1 : 2 ? a . 48 kmph | let the distances between city a to b and b to c be x km and 2 x km respectively . total time taken to cover from a to c = ( x ) / 40 + ( 2 x ) / 60 = ( 3 x + 4 x ) / 120 = 7 x / 120 average speed = ( x + 2 x ) / ( 7 x / 120 ) = 360 / 7 kmph . answer : b | a = 40 + 60
b = a / 2
|
a ) 1 / 17 , b ) 13 / 30 , c ) 17 / 30 , d ) 11 / 30 , e ) 11 / 13 | b | divide(subtract(30, 17), 30) | a β s speed is 30 / 17 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a β s speed and b β s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 30 meters , b covers 17 meters in that time . so if the race is 30 meters long , when a reaches the finish line , b would be 13 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 13 meters so that he doesn β t need to cover that . in that case , the time required by a ( to cover 30 meters ) would be the same as the time required by b ( to cover 17 meters ) to reach the finish line . so b should get a head start of 13 / 30 th of the race . answer ( b )" | a = 30 - 17
b = a / 30
|
a ) 70 % , b ) 20 % , c ) 16 % , d ) 19 % , e ) 80 % | c | multiply(divide(subtract(780, 670), 670), const_100) | a cycle is bought for rs . 670 and sold for rs . 780 , find the gain percent ? | "explanation : 670 - - - - 110 100 - - - - ? = > 16 % answer : c" | a = 780 - 670
b = a / 670
c = b * 100
|
a ) 347.6 , b ) 987.6 , c ) 941.6 , d ) 637.6 , e ) 667.6 | c | add(add(multiply(4, multiply(divide(24, 10), divide(multiply(22, 6), 2))), multiply(3, divide(multiply(22, 6), 2))), multiply(5, 22)) | 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 $ 22 . 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 * 22 = 2 r a = 12 / 5 r and r = 66 a = 158.4 required total cost = 4 * 158.4 + 3 * 66 + 5 * 22 = 633.6 + 198 + 110 = $ 941.60 c" | a = 24 / 10
b = 22 * 6
c = b / 2
d = a * c
e = 4 * d
f = 22 * 6
g = f / 2
h = 3 * g
i = e + h
j = 5 * 22
k = i + j
|
a ) 60 % , b ) 27 % , c ) 30 % , d ) 56 % , e ) 73 % | b | subtract(divide(multiply(add(const_100, 20.65), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 20.65 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 120.65 let marked price be rs . x . then , 95 / 100 x = 120.65 x = 12065 / 95 = rs . 127 now , s . p . = rs . 127 , c . p . = rs . 100 profit % = 27 % . answer : b" | a = 100 + 20
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
|
a ) 12 sec , b ) 24 sec , c ) 30 sec , d ) 60 sec , e ) 62 sec | c | add(90, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 90 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 90 + 30 = 120 km / hr . 120 * 5 / 18 = 100 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 100 = 30 sec . answer : c" | a = 90 + 30
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | b | multiply(6, divide(7, 5)) | john was thrice as old as tom 6 years ago . john will be 7 / 5 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 7 / 5 * ( t + 6 ) 5 j + 30 = 7 t + 42 5 ( 3 t - 12 ) + 30 = 7 t + 42 8 t = 72 t = 9 the answer is b ." | a = 7 / 5
b = 6 * a
|
a ) 300 , b ) 1350 , c ) 1100 , d ) 1200 , e ) 1900 | b | subtract(subtract(2000, divide(2000, 10)), divide(subtract(2000, divide(2000, 10)), 4)) | workers at a campaign office have 2000 fliers to send out . if they send out 1 / 10 of them in the morning and 1 / 4 of the remaining ones out during the afternoon , how many are left for the next day ? | "( 1 / 10 ) * 2000 = 200 remaining = 2000 - 200 = 1800 ( 1 / 4 ) of remaining = ( 1 / 4 ) * 1800 = 450 remaining now = 1800 - 450 = 1350 answer : option b" | a = 2000 / 10
b = 2000 - a
c = 2000 / 10
d = 2000 - c
e = d / 4
f = b - e
|
a ) 5 / 8 , b ) 6 / 8 , c ) 4 / 8 , d ) 2 / 8 , e ) 3 / 8 | e | multiply(divide(3, 30), divide(const_1, multiply(3, add(divide(const_1, 30), add(divide(const_1, 20), divide(const_1, 20)))))) | three pipes a , b and c can fill a tank from empty to full in 20 minutes , 20 minutes and 30 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution q in the liquid in the tank after 3 minutes ? | "part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 20 + 1 / 20 + 1 / 30 ) = 6 / 15 part filled by b in 3 minutes = 3 / 20 required ratio = 3 / 20 * 15 / 6 = 3 / 8 answer : e" | a = 3 / 30
b = 1 / 30
c = 1 / 20
d = 1 / 20
e = c + d
f = b + e
g = 3 * f
h = 1 / g
i = a * h
|
a ) [ 33 ] , b ) [ 44 ] , c ) [ 45 ] , d ) [ 88 ] , e ) [ 90 ] | a | multiply(add(divide(5, 2), 1), add(multiply(2, 4), 1)) | when x is even , [ x ] = x / 2 + 1 , when x is odd [ x ] = 2 x + 1 then [ 5 ] * [ 4 ] = ? | "[ 5 ] * [ 4 ] = ( 2 * 5 + 1 ) ( 4 / 2 + 1 ) = [ 33 ] . ans - a" | a = 5 / 2
b = a + 1
c = 2 * 4
d = c + 1
e = b * d
|
a ) 6 , b ) 36 , c ) 19 , d ) 27 , e ) none of these | b | subtract(765, subtract(subtract(subtract(765, const_12), const_12), const_12)) | 765 chairs are to be arranged in a column in such a way that the number of chairs in each column should be equal to the columns . how many chairs will be excluded to make this arrangement possible ? | 27 ( 2 ) < 765 < 28 ( 2 ) β΄ required no . of chairs to be excluded = 765 β 729 = 36 answer b | a = 765 - 12
b = a - 12
c = b - 12
d = 765 - c
|
a ) 120 , b ) 150 , c ) 175 , d ) 160 , e ) 130 | a | multiply(divide(40, 20), const_60) | a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 40 percent . | 1 hr = 20 percent . thus 15 min = 5 percent . now to charge 40 percent 120 min . answer : a | a = 40 / 20
b = a * const_60
|
a ) 110 , b ) 106 , c ) 105 , d ) 107 , e ) 109 | c | subtract(divide(multiply(multiply(3500, 11), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 11 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 1 * 3 ) / 100 = > 105 answer : c" | a = 3500 * 11
b = a * 3
c = b / 100
d = 3500 * 10
e = d * 3
f = e / 100
g = c - f
|
a ) 10001000 , b ) 9001100 , c ) 1200800 , d ) 8001200 , e ) 1100 , 900 | e | multiply(divide(divide(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))), add(divide(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))), divide(add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))))), 2000) | x and y began business with rs . 10000 and rs . 8000 after 6 months , x advances rs . 2000 and y advances rs . 2000 more . at the end of the year , their profits amounted to rs . 2000 find the share of x and y respectively . | ( 10 * 6 + 12 * 6 ) : ( 8 * 6 + 10 * 6 ) 11 : 9 x share = 11 / 20 * 2000 = 1100 . y share = 9 / 20 * 2000 = 900 . x and y share respectively is 1100 , 900 . answer : e | a = 10000 * 6
b = 10000 + 2000
c = b * 6
d = a + c
e = 10000 * 6
f = 10000 + 2000
g = f * 6
h = e + g
i = 8000 * 6
j = 2000 + 2000
k = 8000 + j
l = k * 6
m = i + l
n = math.gcd(h, m)
o = d / n
p = 10000 * 6
q = 10000 + 2000
r = q * 6
s = p + r
t = 10000 * 6
u = 10000 + 2000
v = u * 6
w = t + v
x = 8000 * 6
y = 2000 + 2000
z = 8000 + y
A = z * 6
B = x + A
C = math.gcd(w, B)
D = s / C
E = 8000 * 6
F = 2000 + 2000
G = 8000 + F
H = G * 6
I = E + H
J = 10000 * 6
K = 10000 + 2000
L = K * 6
M = J + L
N = 8000 * 6
O = 2000 + 2000
P = 8000 + O
Q = P * 6
R = N + Q
S = math.gcd(M, R)
T = I / S
U = D + T
V = o / U
W = V * 2000
|
a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 18 | d | divide(30, const_2) | in a group of ducks and cows , the total number of legs are 30 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 30 = > 2 x = 30 = > x = 15 d" | a = 30 / 2
|
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 72 | e | divide(multiply(1, 48), 2) | the l . c . m of two numbers is 48 . the numbers are in the ratio 1 : 2 . the sum of numbers is : | "let the numbers be 1 x and 2 x . then , their l . c . m = 2 x . so , 2 x = 48 or x = 24 . the numbers are 24 and 48 . hence , required sum = ( 24 + 48 ) = 72 . answer : e" | a = 1 * 48
b = a / 2
|
a ) 1 / 35 , b ) 1 / 42 , c ) 1 / 49 , d ) 1 / 56 , e ) 1 / 63 | a | divide(1, 36) | a certain fraction has the same ratio to 1 / 36 , as 4 / 7 does to 5 / 9 . what is this certain fraction ? | x / ( 1 / 36 ) = ( 4 / 7 ) / ( 5 / 9 ) x = 4 * 9 * 1 / 36 * 7 * 5 = 1 / 35 the answer is a . | a = 1 / 36
|
a ) 75 , b ) 100 , c ) 125 , d ) 130 , e ) 225 | d | divide(subtract(multiply(divide(780, const_3), const_4), 780), const_2) | there are 780 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ? | "female = x male = 780 - x x / 2 + 780 - x / 4 = 1 / 3 * ( 780 ) = 260 x = 260 x / 2 = 130 is supposed to be the answer m is missing something correct option d" | a = 780 / 3
b = a * 4
c = b - 780
d = c / 2
|
a ) 228 , b ) 278 , c ) 192 , d ) 27 , e ) 21 | c | add(45, 30) | two goods trains each 2000 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 2000 + 2000 = 4000 m . required time = 4000 * 6 / 125 = 192 sec . answer : c" | a = 45 + 30
|
a ) 289 , b ) 500 , c ) 188 , d ) 666 , e ) 155 | b | divide(70, divide(multiply(4, add(3, divide(1, 2))), const_100)) | what sum of money will produce rs . 70 as simple interest in 4 years at 3 1 / 2 percent ? | "70 = ( p * 4 * 7 / 2 ) / 100 p = 500 answer : b" | a = 1 / 2
b = 3 + a
c = 4 * b
d = c / 100
e = 70 / d
|
a ) 16 , b ) 19 , c ) 15 , d ) 32 , e ) 40 | d | divide(subtract(34, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 34 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 34 ) + 2 = 2 ( x + 2 ) x + 36 = 2 x + 4 = > x = 32 . answer : d" | a = 2 * 2
b = a - 2
c = 34 - b
d = 2 - 1
e = c / d
|
a ) 28 sec , b ) 16 sec , c ) 37 sec , d ) 18 sec , e ) 17 sec | c | divide(add(240, 130), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 130 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 120 = 370 m . time taken = 370 / 10 = 37 sec . answer : c" | a = 240 + 130
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
|
a ) 44 , b ) 46 , c ) 48 , d ) 50 , e ) 52 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 360), multiply(36, const_0_2778)) | a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 360 - meter long platform ? | "let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 360 ) / 10 = 50 seconds the answer is d ." | a = 36 * const_0_2778
b = a * 14
c = b + 360
d = 36 * const_0_2778
e = c / d
|
a ) 6 , b ) 12 , c ) 24 , d ) 36 , e ) 48 | b | multiply(sqrt(divide(72, 2)), 2) | if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer m that must divide n is ? | "q : if n is a positive integer and n ^ 2 is divisible by 72 , then the largest positive integer m that must divide n is : a 6 , b 12 , c 24 , d 36 , e 48 n ^ 2 is divisible by 72 , but it must also be greater than 72 . if n is an integer , then n ^ 2 must be a perfect square . the factorization of 72 is ( 8 ) ( 9 ) , so if it is multiplied by 2 , it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144 , a perfect square . so n ^ 2 must be at least 144 or a multiple of 144 , which means that n must be 12 or a multiple of 12 . b" | a = 72 / 2
b = math.sqrt(a)
c = b * 2
|
a ) 29798 , b ) 27908 , c ) 78902 , d ) 25000 , e ) 27000 | e | divide(divide(divide(multiply(multiply(multiply(27, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5) | a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 27 m * 2 m * 0.75 m ? | 27 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 27 = 1 / 100 * x = > x = 27000 answer : e | a = 27 * 100
b = 2 * 100
c = a * b
d = 0 * 75
e = c * d
f = e / 20
g = f / 10
h = g / 7
|
a ) 13 , b ) 12 , c ) 9 , d ) 8 , e ) 4 | e | divide(subtract(subtract(add(add(divide(72, const_4), divide(72, const_4)), multiply(divide(72, 8), 5)), 72), const_1), const_2) | one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , and 5 / 8 of the workers have read the latest book by h . kureishi . the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book . if there are 72 workers in the palabras bookstore , how many of them have read both books ? | "there are total 72 workers . one - fourth of the workers in palabras bookstore have read the latest book by j . saramago , so 18 have read saramago . 5 / 8 of the workers have read the latest book by h . kureishi . so ( 5 / 8 ) * 72 = 45 have read kureishi the number of workers that have read neither book is one less than the number of the workers that have read the latest saramago book and have not read the latest kureishi book if b workers have read both books , 18 - b have read saramago but not kureishi . so , ( 18 - b - 1 ) have read neither . total = n ( a ) + n ( b ) - both + neither 72 = 18 + 45 - b + ( 18 - b - 1 ) b = 4 answer ( e )" | a = 72 / 4
b = 72 / 4
c = a + b
d = 72 / 8
e = d * 5
f = c + e
g = f - 72
h = g - 1
i = h / 2
|
a ) 10 , b ) 45 , c ) 30 , d ) 20 , e ) 40 | e | multiply(10, inverse(subtract(1, add(divide(1, 4), divide(1, 2))))) | in traveling from a dormitory to a certain city , a student went 1 / 4 of the way by foot , 1 / 2 of the way by bus , and the remaining 10 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 4 x + 1 / 2 x + 10 x - 1 / 4 x - 1 / 2 x = 10 x = 40 km option : e" | a = 1 / 4
b = 1 / 2
c = a + b
d = 1 - c
e = 1/(d)
f = 10 * e
|
a ) 3 , b ) 2 , c ) 1 , d ) 4 , e ) 8 | d | subtract(add(multiply(divide(1, 4), 16), multiply(divide(1, 5), 16)), divide(5, 16)) | if ( 1 / 5 ) ^ m * ( 1 / 4 ) ^ 4 = 1 / ( 16 * ( 10 ) ^ 4 ) , then m = | "( 1 / 5 ) ^ m ( 1 / 4 ) ^ 4 = 1 / { 16 ( 10 ) ^ 4 } 16 ( 10 ) ^ 4 / 4 ^ 4 = 5 ^ m [ 16 ( 2 ^ 4 ) ( 5 ^ 4 ) ] / [ 2 ^ 8 ] = 5 ^ m 5 ^ 4 = 5 ^ m m = 4 d" | a = 1 / 4
b = a * 16
c = 1 / 5
d = c * 16
e = b + d
f = 5 / 16
g = e - f
|
a ) 515 . , b ) 545 . , c ) 618 . , d ) 644 . , e ) 636 . | e | add(220, 423) | in the faculty of reverse - engineering , 220 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | answer is e : 636 solution : total number of students studying both are 423 + 220 - 134 = 509 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 509 , so 100 % is approx . 636 . | a = 220 + 423
|
a ) 6 , b ) 8 , c ) 10 , d ) 11 , e ) 12 | b | floor(add(divide(divide(multiply(1000, 85), const_100), divide(multiply(150, 75), const_100)), const_1)) | what would be the minimum number of pumps required to fill a tank of 1000 m 3 to 85 % in one hour ? the capacity of the pump is 150 m 3 / hour and operates at an efficiency of 75 % . | volume handled by pump in one hour = 150 x . 75 = 112.5 volume of the tank that needs to be filled = 1000 x . 85 = 850 number of pumps needed = 850 / 112.5 = 7.55 Γ’ β° Λ 8 answer : b | a = 1000 * 85
b = a / 100
c = 150 * 75
d = c / 100
e = b / d
f = e + 1
g = math.floor(f)
|
a ) 3250 , b ) 3450 , c ) 4000 , d ) 3750 , e ) 3800 | c | divide(360, multiply(divide(subtract(subtract(const_100, 60), 30), const_100), divide(90, const_100))) | of the land owned by a farmer , 90 percent was cleared for planting . of the cleared land , 60 percent was planted with grapes and 30 percent of the cleared land was planted with potato . if the remaining 360 acres of cleared land was planted with tomato , how many acres did the farmer own ? | 60 % of 90 % = 54 % , 30 % 0 f 90 % = 27 % so the remaining 90 - 54 - 27 = 9 % = 360 acres or 10 % of 90 % = 9 % - - > 360 / 9 * 100 = 4000 acres answer ( c ) | a = 100 - 60
b = a - 30
c = b / 100
d = 90 / 100
e = c * d
f = 360 / e
|
a ) 11 , b ) 10 , c ) 18 , d ) 16 , e ) 12 | d | multiply(multiply(divide(7, 5), divide(10, 7)), 8) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 8 months , find for how much time did q invest the money ? | "7 * 5 : 8 * x = 7 : 10 x = 16 answer : d" | a = 7 / 5
b = 10 / 7
c = a * b
d = c * 8
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.