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 ) 88 % , b ) 90 % , c ) 80 % , d ) 87 % , e ) 20 % | c | multiply(divide(divide(360, multiply(multiply(divide(subtract(const_100, 10), const_100), divide(add(const_100, 25), const_100)), divide(subtract(const_100, 20), const_100))), 500), const_100) | at an examination in which full marks were 500 . a got 10 % less than b , b got 25 % more than c and c got 20 % less than d . if a got 360 marks , what percentage of full marks was obtained by d ? | a b c d 90 100 80 100 a d 90 - - - - - 100 360 - - - - - - ? = 400 500 - - - - - - 400 100 - - - - - - - ? = > 80 % answer : c | a = 100 - 10
b = a / 100
c = 100 + 25
d = c / 100
e = b * d
f = 100 - 20
g = f / 100
h = e * g
i = 360 / h
j = i / 500
k = j * 100
|
a ) 8 , b ) 12 , c ) 20 , d ) 24 , e ) 28 | b | subtract(subtract(40, divide(subtract(multiply(6, 40), 212), subtract(6, 4))), divide(subtract(multiply(6, 40), 212), subtract(6, 4))) | a local restaurant recently renovated its dining space , purchasing new tables and chairs to use in addition to the original tables and chairs . the new tables each seat 6 customers , while the original tables each seat 4 customers . altogether , the restaurant now has 40 tables and is capable of seating 212 customers . how many more new tables than original tables does the restaurant have ? | if all the tables seated 4 , the number of customers could be 4 * 40 = 160 . 212 - 160 = 52 , so 52 / 2 = 26 tables must be tables seating 6 people . the number of tables seating 4 people is 40 - 26 = 14 . the number of new tables is 26 - 14 = 12 more than the number of old tables . the answer is b . | a = 6 * 40
b = a - 212
c = 6 - 4
d = b / c
e = 40 - d
f = 6 * 40
g = f - 212
h = 6 - 4
i = g / h
j = e - i
|
['a ) area is quadrupled', 'b ) area is tripled', 'c ) area is doubles', 'd ) area become half', 'e ) none of these'] | a | divide(divide(power(multiply(const_pi, 8), const_2), multiply(const_pi, 4)), divide(power(multiply(const_pi, 4), const_2), multiply(const_pi, 4))) | if the circumference of a circle increases from 4 pi to 8 pi , what change occurs in the area ? | explanation : 2 π r 1 = 4 π = > r 1 = 22 π r 2 = 8 π > r 2 = 4 original area = 4 π ∗ 22 = 16 π new area = 4 π ∗ 42 = 64 π so the area quadruples . option a | a = math.pi * 8
b = a ** 2
c = math.pi * 4
d = b / c
e = math.pi * 4
f = e ** 2
g = math.pi * 4
h = f / g
i = d / h
|
a ) 21 st , b ) 22 nd , c ) 23 rd , d ) 24 th , e ) 37 th | e | subtract(multiply(2, 20), 1) | a monkey ascends a greased pole 20 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ? | "in 2 minutes , he ascends = 1 metre â ˆ ´ 18 metres , he ascends in 36 minutes . â ˆ ´ he reaches the top in 37 th minute . answer e" | a = 2 * 20
b = a - 1
|
a ) 91.5 , b ) 92 , c ) 93 , d ) 94 , e ) 124.5 | e | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 225 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 225 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 124.5 cm answer : option e" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
['a ) 0', 'b ) 1', 'c ) 2', 'd ) 3', 'e ) 4'] | d | subtract(add(const_2, const_3), const_2) | the product of the squares of two positive integers is 400 . how many pairs of positive integers satisfy this condition ? | first break down 200 into 20 * 20 and further into the prime factors 2 * 2 * 5 * 2 * 2 * 5 . now we are looking for all the possible pairs ( 2 numbers ) of squares whose product results in 400 . 1 st : 2 ^ 2 * 10 ^ 2 ( i . e . the first two 2 ' s and two times 2 * 5 = 10 ) 2 nd : 4 ^ 2 * 5 ^ 2 ( i . e . two times 2 * 2 = 4 = 4 ^ 2 and 5 ^ 2 ) . 3 rd : 1 ^ 2 * 20 ^ 2 ( i . e . two times 2 * 2 * 5 and 1 ^ 2 = 1 ) answer d . | a = 2 + 3
b = a - 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | d | divide(subtract(subtract(subtract(175, multiply(175, divide(3, 7))), multiply(subtract(175, multiply(175, divide(3, 7))), divide(3, 7))), 3), subtract(subtract(subtract(175, multiply(175, divide(3, 7))), multiply(subtract(175, multiply(175, divide(3, 7))), divide(3, 7))), 3)) | george baked a total of 175 pizzas for 7 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ? | "3 / 5 of the 175 pizzas cooked on saturday = 105 pizzas 3 / 5 of the remaining pizzas on sunday = 42 pizzas we ' re left with ( 175 - 105 - 42 ) = 28 pizzas for the remaining 5 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w th f 8 7 6 4 3 = 28 w = 6 d" | a = 3 / 7
b = 175 * a
c = 175 - b
d = 3 / 7
e = 175 * d
f = 175 - e
g = 3 / 7
h = f * g
i = c - h
j = i - 3
k = 3 / 7
l = 175 * k
m = 175 - l
n = 3 / 7
o = 175 * n
p = 175 - o
q = 3 / 7
r = p * q
s = m - r
t = s - 3
u = j / t
|
a ) 39 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | a | subtract(subtract(const_100, multiply(subtract(8, 2), const_10)), const_1) | n and m are each 3 - digit integers . each of the numbers 2 , 3,4 , 6 , 7 , and 8 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | you have 6 digits : 2 , 3 , 4 , 6 , 7 , 8 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 2 and 8 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 3 * * and 4 * * ) . make the 3 * * number as large as possible and make the 4 * * number as small as possible . 387 and 426 ( difference is 39 ) or try and make a pair with ( 6 * * and 7 * * ) . make the 6 * * number as large as possible and make the 7 * * number as small as possible . we get 684 and 723 ( difference is 39 ) a | a = 8 - 2
b = a * 10
c = 100 - b
d = c - 1
|
a ) 1,072 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(12, 10), multiply(9, add(const_4, const_1)))), const_1000)) | gary ’ s gas station serves an average of 12 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary ’ s station serve over the course of a typical week ? | "6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 12 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 12 + 10 ) = 16 * 67 = 1072 answer : a" | a = 6 + 10
b = 12 + 10
c = 4 + 1
d = 9 * c
e = b + d
f = a * e
g = f / 1000
h = math.floor(g)
|
a ) 80 , b ) 90 , c ) 95 , d ) 121.5 , e ) 108 | d | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 35), const_100)) | from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 35 percent . what percent of the stock â € ™ s 2007 starting price was the price of the stock at the end of 2009 ? | assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 35 % = 90 * 1.35 = 121.5 thus ratio = 121.5 / 100 = 1.215 ( in % terms = 121.5 % ) . thus d is the correct answer . | a = 100 + 20
b = a / 100
c = 100 * b
d = 100 - 25
e = d / 100
f = c * e
g = 100 + 35
h = g / 100
i = f * h
|
a ) 13.3 % , b ) 9.22 % , c ) 9 % , d ) 14 % , e ) 12 % | a | multiply(subtract(const_1, multiply(subtract(const_1, divide(15, const_100)), add(const_1, divide(2, const_100)))), const_100) | you enter a weight loss challenge game and manage to lose 15 % of your body weight . for the final weigh in you are forced to wear clothes that add 2 % to your weight . what percentage of weight loss is measured at the final weigh in ? | "( 100 % - 15 % ) * ( 100 % + 2 % ) = 0.85 * 1.02 = 13.3 % the weigh in records your weight loss at 13.3 % ! the answer is a" | a = 15 / 100
b = 1 - a
c = 2 / 100
d = 1 + c
e = b * d
f = 1 - e
g = f * 100
|
a ) 25 , b ) 30 , c ) 28 , d ) 34 , e ) 42 | d | divide(subtract(multiply(89, 2), subtract(16, 8)), add(2, 3)) | if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 89 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 89 a = 34 the answer is d ." | a = 89 * 2
b = 16 - 8
c = a - b
d = 2 + 3
e = c / d
|
a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | a | divide(subtract(multiply(divide(10, const_100), 2000), 92), add(divide(8, const_100), divide(10, const_100))) | if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 92 . if $ 2000 is the total amount invested , how much is invested at 8 percent ? | 0.1 x = 0.08 ( 2000 - x ) + 92 0.18 x = 252 x = 1400 then the amount invested at 8 % is $ 2000 - $ 1400 = $ 600 the answer is a . | a = 10 / 100
b = a * 2000
c = b - 92
d = 8 / 100
e = 10 / 100
f = d + e
g = c / f
|
a ) 1 / 248 , b ) 1 / 216 , c ) 1 / 144 , d ) 1 / 200 , e ) 1 / 242 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | four 6 faced dice are thrown together . the probability that all the four show the same number on them is ? | "it all 4 numbers have to be same basically we want quadruplets . 1111 , 2222 , 3333 , 4444 , 5555 and 6666 . those are six in number . further the four dice can fall in 6 * 6 * 6 * 6 = 1296 ways . hence the probability is 6 / 1296 = 1 / 216 answer : b" | a = 1 / 6
b = 1 / 6
c = a * b
d = 1 / 6
e = c * d
f = 1 / 6
g = e * f
|
a ) 12 % decrease , b ) 18 % decrease , c ) 19 % decrease , d ) 13 % decrease , e ) 12.5 % increase | e | subtract(const_100, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | the tax on a commodity is diminished by 10 % and its consumption increased by 25 % . the effect on revenue is ? | "100 * 100 = 10000 90 * 125 = 11250 - - - - - - - - - - - 10000 - - - - - - - - - - - 1250 100 - - - - - - - - - - - ? = > 12.5 % decrease answer : e" | a = 25 / 100
b = 1 + a
c = 10 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 35 % , b ) 25 % , c ) 20 % , d ) 30 % , e ) none of these | b | subtract(const_100, multiply(divide(add(const_100, 50), const_100), 50)) | a man ’ s wages were decreased by 50 % . again , the reduced wages were increased by 50 % . he has a loss of ? | here , x = - 50 and y = 50 therefore , the net % change in value = ( x + y + xy / 100 ) % = [ - 50 + 50 + ( - 50 x 50 ) / 100 ] % or - 25 % since the sign is negative , there is loss of 25 % answer : b | a = 100 + 50
b = a / 100
c = b * 50
d = 100 - c
|
a ) 48 , b ) 32 , c ) 24 , d ) 56 , e ) 12 | d | multiply(7, multiply(const_2, const_4)) | a cubical block of metal weighs 7 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 7 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 7 pounds = 56 pounds so answer is d and similar but more theoretical approach : if we have sides a and b than they have equal ration with their areas : a / b = a ^ 2 / b ^ 2 and they have equal ration with their volumes : a / b = a ^ 3 / b ^ 3 we have two sides 1 / 2 so their volume will be in ratio 1 / 8 weight of one cube * volume of another cube 7 * 8 = 56 so answer is d" | a = 2 * 4
b = 7 * a
|
['a ) 1.5 : 5', 'b ) 2 : 5', 'c ) 3 : 5', 'd ) 1 : 5', 'e ) 4 : 5'] | b | divide(power(8, const_0_33), power(125, const_0_33)) | two cubes of their volumes in the ratio 8 : 125 . the ratio of their surface area is : | the ratio of their surface area is 8 : 125 2 : 5 answer is b . | a = 8 ** const_0_33
b = 125 ** const_0_33
c = a / b
|
a ) 16 : 27 , b ) 12 : 13 , c ) 13 : 14 , d ) 14 : 15 , e ) 31 : 27 | a | multiply(divide(4, 3), multiply(divide(4, 3), divide(1, 3))) | find the compound ratio of ( 4 : 3 ) , ( 1 : 3 ) and ( 2 : 3 ) is | required ratio = 4 / 3 * 1 / 3 * 2 / 3 = 16 / 27 = 16 : 27 answer is a | a = 4 / 3
b = 4 / 3
c = 1 / 3
d = b * c
e = a * d
|
a ) 220 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | a | subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 19)) | a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 19 seconds . what is the length of the platform in meters ? | "speed of the train in metres / sec = 72000 / 3600 = 20 distance travelled by train to cross the platform = 30 * 20 = 600 = length of train + length of platform distance travelled by train to cross the man = 19 * 20 = 380 = length of train length of platform = 600 - 380 = 220 answer : a" | a = 72 * 1000
b = a / 3600
c = b * 30
d = 72 * 1000
e = d / 3600
f = e * 19
g = c - f
|
a ) 50 % , b ) 40 % , c ) 60 % , d ) 57 % , e ) 80 % | d | multiply(divide(subtract(140, add(multiply(3, 8), multiply(8, 3))), 140), const_100) | a batsman scored 140 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "number of runs made by running = 140 - ( 3 x 4 + 8 x 6 ) = 140 - ( 60 ) = 80 now , we need to calculate 80 is what percent of 140 . = > 80 / 140 x 100 = 57 % answer : d" | a = 3 * 8
b = 8 * 3
c = a + b
d = 140 - c
e = d / 140
f = e * 100
|
a ) 14 sec , b ) 10 sec , c ) 12 sec , d ) 8 sec , e ) 5 sec | b | divide(140, add(11, 3)) | an escalator moves towards the top level at the rate of 11 ft . sec and its length is 140 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length . | explanation : time taken to cover the entire length = tot . dist / resultant speed = 140 / ( 11 + 3 ) = 10 sec answer : b | a = 11 + 3
b = 140 / a
|
a ) 1 , b ) 3 , c ) 5 , d ) 2 , e ) 9 | d | divide(log(25), log(add(const_4, const_1))) | for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 25 true ? | answer d ( a + b ) ( a - b ) = 25 5 cases for ( a + b ) , ( a - b ) 25 , 1 5 , 5 answer d | a = math.log(25)
b = 4 + 1
c = math.log(b)
d = a / c
|
a ) 6.6 , b ) 6.7 , c ) 6.8 , d ) 7.0 , e ) 6.9 | d | divide(add(multiply(10, 6.2), 8), 10) | set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 8 , while all other numbers remain the same , what is the new average of set s ? | old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 8 then total increased to 62 + 8 = 70 new avg - 70 / 10 = 7.0 . hence answer is d . | a = 10 * 6
b = a + 8
c = b / 10
|
a ) 23 , b ) 27 3 / 4 , c ) 20 1 / 2 , d ) 27 1 / 2 , e ) 11 | d | divide(multiply(32, 14), 16) | 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 = 32 * 14
b = a / 16
|
a ) 2 , b ) 3 , c ) 3 1 / 3 , d ) 4 , e ) 4 2 / 3 | e | add(10, divide(multiply(7, 10), add(7, 8))) | a work crew of 7 men takes 10 days to complete one - half of a job . if 8 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 7 men will do in . . 7 / x = 1 / 10 * 1 / 2 as half job is done x = 140 now 8 more are added then 15 / 140 = 1 / 2 * 1 / d for remaining half job d = 4 2 / 3 number of days e | a = 7 * 10
b = 7 + 8
c = a / b
d = 10 + c
|
a ) 255 , b ) 205 , c ) 502 , d ) 225 , e ) 235 | d | multiply(divide(multiply(90, const_1000), const_3600), 6) | a train running at the speed of 90 km / hr crosses a pole in 6 seconds . what is the length of the train ? | "speed = ( 90 * 5 / 18 ) m / sec = ( 25 ) m / sec length of the train = ( speed x time ) = ( 25 * 9 ) m = 225 m . answer : d" | a = 90 * 1000
b = a / 3600
c = b * 6
|
a ) 2.3 m , b ) 4.6 m , c ) 7.8 m , d ) 9.2 m , e ) none | d | multiply(4.6, const_2) | the angle of elevation of a ladder leaning against a wall is 60 ° and the foot of the ladder is 4.6 m away from the wall . the length of the ladder is | solution let ab be the wall and bc be the ladder . then , < abc = 60 ° ac = 4.6 m . ; ac / bc = cos 60 ° = 1 / 2 ‹ = › bc = 2 × ac = ( 2 × 4.6 ) m = 9.2 m answer d | a = 4 * 6
|
a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio of those accepted to those rejected would be 99 to 1 . how many t eggs does the plant process per day ? | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the no . of eggs inspected remain and 12 more eggs get accepted that means there t = 384 + 12 = 396 eggs accepted and 4 rejected . . . and the ratio will be 99 : 1 bingo . . . this is what the questions says . . . . its always a good idea to start with c ." | a = 99 - 96
b = 12 / a
c = b * 100
|
a ) 298 , b ) 237 , c ) 342 , d ) 381 , e ) 291 | d | subtract(subtract(510, divide(multiply(510, 12), const_100)), divide(multiply(subtract(510, divide(multiply(510, 12), const_100)), 15), const_100)) | the sale price sarees listed for rs . 510 after successive discount is 12 % and 15 % is ? | 510 * ( 88 / 100 ) * ( 85 / 100 ) = 381 answer : d | a = 510 * 12
b = a / 100
c = 510 - b
d = 510 * 12
e = d / 100
f = 510 - e
g = f * 15
h = g / 100
i = c - h
|
a ) 2400 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2050 | a | add(3456, divide(multiply(3456, 20), const_100)) | the present population of a town is 3456 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 3456 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3456 / ( 1 + 20 / 100 ) ^ 2 = 3456 / ( 6 / 5 ) ^ 2 = 2400 ( approximately ) answer is a" | a = 3456 * 20
b = a / 100
c = 3456 + b
|
a ) 42.85 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | a | divide(150, subtract(divide(150, 30), divide(90, 90))) | the distance from city a to city b is 150 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 90 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ? | "the time it takes cara to drive to city b is 150 / 30 = 5 hours . dan needs to take less than 3.5 hours for the trip . dan needs to exceed a constant speed of 150 / 3.5 = 42.85 miles per hour . the answer is a ." | a = 150 / 30
b = 90 / 90
c = a - b
d = 150 / c
|
a ) rs . 20.25 , b ) rs . 22.50 , c ) rs . 25 , d ) rs . 42.75 , e ) none | b | subtract(divide(multiply(divide(202.50, divide(4.5, const_100)), 5), const_100), 202.50) | the interest on a certain deposit at 4.5 % p . a . is rs . 202.50 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a . ? | "solution s . i . = rs . 202.50 , r = 4.5 % , t = 1 year . principal = rs . ( 100 x 202.50 / 4.5 x 1 ) = rs . 4500 . now , p = rs . 4500 , r = 5 % , t = 1 year . s . i . = rs . ( 4500 x 5 x 1 / 1000 = rs . 225 . ∴ difference in interest = rs . ( 225 - 202.50 ) = rs . 22.50 . answer b" | a = 4 / 5
b = 202 / 50
c = b * 5
d = c / 100
e = d - 202
|
a ) − 48 , b ) − 5 , c ) 4 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 12), subtract(4, 12)), const_1), const_1), const_1), const_1) | if a ( a - 4 ) = 12 and b ( b - 4 ) = 12 , where a ≠ b , then a + b = | i . e . if a = 6 then b = - 2 or if a = - 2 then b = 6 but in each case a + b = - 2 + 6 = 4 answer : option c | a = 4 + 12
b = 4 - 12
c = a + b
d = c - 1
e = d - 1
f = e - 1
g = f - 1
|
a ) 200 , b ) 75 , c ) 60 , d ) 33 1 ⁄ 3 , e ) 25 | a | multiply(divide(60, 120), const_100) | what percent of 60 is 120 ? | "% of 60 is 120 ? = 120 / 60 = 21 = 200 % thus a is the correct answer ." | a = 60 / 120
b = a * 100
|
a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | c | add(2001, divide(add(divide(65, const_100), subtract(6.30, 4.20)), subtract(divide(45, const_100), subtract(6.30, 4.20)))) | the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 6.30 . in which year will the price of commodity x be 65 cents more than the price of commodity y ? | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 65 cents more than commodity y . $ 2.75 / 25 cents = 11 years the answer is 2001 + 11 years = 2012 . the answer is c ." | a = 65 / 100
b = 6 - 30
c = a + b
d = 45 / 100
e = 6 - 30
f = d - e
g = c / f
h = 2001 + g
|
a ) 60 , b ) 90 , c ) 120 , d ) 100 , e ) 240 | d | add(divide(multiply(400, 3), 5), multiply(3, 5)) | the number of timeshare condos available at sunset beach is 3 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 400 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 3 / 5 x = 400 hence , x = 250 . so , number of timeshare condos available at playa del mar = 250 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 3 / 5 x = 2 / 5 x = 2 / 5 ( 250 ) = 100 the correct answer is d" | a = 400 * 3
b = a / 5
c = 3 * 5
d = b + c
|
a ) 229 , b ) 288 , c ) 600 , d ) 720 , e ) 121 | d | multiply(divide(multiply(30, add(const_3, 3)), subtract(40, 30)), 40) | a train leaves delhi at 9 a . m . at a speed of 30 kmph . another train leaves at 3 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ? | "d = 30 * 6 = 180 rs = 40 – 30 = 10 t = 180 / 10 = 18 d = 40 * 18 = 720 km answer : d" | a = 3 + 3
b = 30 * a
c = 40 - 30
d = b / c
e = d * 40
|
a ) $ 2420 , b ) $ 2610 , c ) $ 2860 , d ) $ 3050 , e ) $ 3270 | b | add(1,000, divide(112.70, divide(7, const_100))) | when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 112.70 , what was the total value of the item ? | "let x be the value of the item . 0.07 * ( x - 1000 ) = 112.70 x = 2610 the answer is b ." | a = 7 / 100
b = 112 / 70
c = 1 + 0
|
a ) 54 , b ) 162 , c ) 250 , d ) 270 , e ) 322 | c | divide(multiply(300, 10), add(2, 10)) | compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 300 grams of compound x ? | "total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 300 gms of x have 300 * 10 / 12 = 250 gms of b and 300 - 250 = 50 gms of a . cross check : - a / b = 50 / 250 = 2 / 10 ( as given ) ans c" | a = 300 * 10
b = 2 + 10
c = a / b
|
a ) 122.55 , b ) 132 , c ) 156 , d ) 158 , e ) 267 | a | multiply(circumface(divide(26, const_2)), 1.50) | find the cost of fencing around a circular field of diameter 26 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 13 = 81.7 81.7 * 1 1 / 2 = rs . 122.55 answer : a" | a = 26 / 2
b = circumface * (
|
a ) 200 seconds , b ) 140 seconds , c ) 220 seconds , d ) 190 seconds , e ) none | d | subtract(multiply(divide(10, 50), const_1000), 10) | a can give b a start of 50 metres or 10 seconds in a kilometer race . how long does a take to complete the race ? | solution : a can give b a start of 50 metres or 10 seconds in a 1000 m race . that is , b takes 10 seconds to run 50 metres . therefore , b will take ( 10 / 50 ) * 1000 = 200 seconds to run 1000 metres . a who can give b a start of 10 seconds will take 10 seconds lesser to run the 1000 m . hence , the time taken by a = 190 seconds . answer d | a = 10 / 50
b = a * 1000
c = b - 10
|
a ) 133 % , b ) 145 % , c ) 158 % , d ) 170 % , e ) 183 % | e | multiply(divide(multiply(divide(10, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100) | last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 10 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | "let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.1 * 1.1 x = 0.11 x 0.11 x / 0.06 x = 11 / 6 = 1.83 = 183 % the answer is e ." | a = 10 / 100
b = 100 + 10
c = a * b
d = 6 * 100
e = d / 100
f = c / e
g = f * 100
|
a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 56 , d ) rs . 55.33 , e ) none of the above | c | divide(add(multiply(10, 50), multiply(5, 68)), add(10, 5)) | if 10 litres of an oil of rs . 50 per litres be mixed with 5 litres of another oil of rs . 68 per litre then what is the rate of mixed oil per litre ? | "50 * 10 = 500 68 * 5 = 340 840 / 15 = 56 answer : c" | a = 10 * 50
b = 5 * 68
c = a + b
d = 10 + 5
e = c / d
|
a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 80 % | a | multiply(subtract(multiply(add(const_1, divide(90, const_100)), subtract(const_1, divide(50, const_100))), const_1), const_100) | a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 90 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 50 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | "the actual answer is obtained by multiplying 140 % by 70 % and subtracting 100 % from this total . that is : 190 % × 50 % = 95 % ; 95 % − 100 % = - 5 % . answer : a" | a = 90 / 100
b = 1 + a
c = 50 / 100
d = 1 - c
e = b * d
f = e - 1
g = f * 100
|
a ) 4 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | c | subtract(23, reminder(1056, 23)) | what least number should be added to 1056 , so that the sum is completely divisible by 23 | "explanation : ( 1056 / 23 ) gives remainder 21 21 + 2 = 23 , so we need to add 2 answer : option c" | a = 23 - reminder
|
a ) 39 ° c , b ) 44 ° c , c ) 37 ° c , d ) 47 ° c , e ) none of these | d | subtract(multiply(52, const_3), subtract(multiply(54, const_3), 53)) | the average temperature for tuesday , wednesday and thursday was 52 ° c . the average temperature for wednesday , thursday and friday was 54 ° c . if the temperature on friday be 53 ° c , what was the temperature on tuesday ? | explanation : t + w + t = 52 × 3 = 156 ° c w + t + f = 54 × 3 = 162 ° c also , temperature on friday = 53 ° c temperature on tuesday = 156 + 53 - 162 = 47 ° c answer : option d | a = 52 * 3
b = 54 * 3
c = b - 53
d = a - c
|
a ) 1008 , b ) 1014 , c ) 1022 , d ) 1032 , e ) 1043 | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 6) | the smallest number which when diminished by 6 , is divisible by 12 , 16 , 18 , 21 and 28 is | "required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 6 = 1008 + 6 = 1014 answer : b" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 6
|
a ) 2200 , b ) 2240 , c ) 1600 , d ) 1354 , e ) none of these | b | add(divide(240, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 240) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 240 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 240 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 240 or x = 2000 . p . w . = rs . 2000 . sum due = p . w . + t . d . = rs . ( 2000 240 ) = rs . 2240 . answer : b" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 240 / d
f = e + 240
|
a ) 92.33 % , b ) 91.33 % , c ) 95.33 % , d ) 93.33 % , e ) 94.33 % | d | subtract(const_100, multiply(divide(divide(5, 5), multiply(3, 5)), const_100)) | instead of multiplying a number by 3 , the number is divided by 5 . what is the percentage of error obtained ? | "let the number be x the right number is 3 x the wrong number is x / 5 error is ( 3 x - x / 5 ) = 14 x / 5 percentage of error is ( ( 14 x / 5 ) / 3 x ) * 100 = 93.33 % answer : d" | a = 5 / 5
b = 3 * 5
c = a / b
d = c * 100
e = 100 - d
|
a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1100 | e | divide(23100, add(18, 3)) | a man sold 18 toys for rs . 23100 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 23100 . gain = sp â € “ cp 3 x = 23100 â € “ 18 x 21 x = 23100 x = rs . 1100 . answer : option e" | a = 18 + 3
b = 23100 / a
|
a ) 6 , b ) 2 , c ) 3 , d ) 7 , e ) 5 | d | divide(17, 3) | if ( a – b ) is 17 more than ( c + d ) and ( a + b ) is 3 less than ( c – d ) , then ( a – c ) is : | "( a – b ) – ( c + d ) = 17 and ( c – d ) – ( a + b ) = 3 = > ( a – c ) – ( b + d ) = 17 and ( c – a ) – ( b + d ) = 3 = > ( b + d ) = ( a – c ) – 17 and ( b + d ) = ( c – a ) – 3 = > ( a – c ) – 17 = ( c – a ) – 3 = > 2 ( a – c ) = 14 = > ( a – c ) = 7 answer : d" | a = 17 / 3
|
a ) 150 , b ) 155 , c ) 140 , d ) 120 , e ) 165 | d | divide(add(multiply(150, const_2), 420), add(4, 2)) | in an objective competitive exam , a correct answer score 4 marks and on a wrong answer 2 marks are negatively added . a student scores 420 marks from 150 question . how many answers were correct ? | let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = 150 . . . . . ( 1 ) = > 4 x - 2 y = 420 . . . . . ( 2 ) by solving ( 1 ) & ( 2 ) , we get = > 6 x = 720 hence x = 120 therefore , number of correct answers are 120 . answer d | a = 150 * 2
b = a + 420
c = 4 + 2
d = b / c
|
a ) 4 , b ) 6 , c ) 8 , d ) 24 , e ) 12 | d | divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2) | running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 4 w widgets ? | "let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 4 * 6 = 24 days to produce 4 w widgets . answer : d" | a = 3 * 4
b = a * 2
c = 3 * 4
d = b - c
e = d / 2
|
a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 50 | e | divide(subtract(add(50, 50), 75), subtract(divide(add(50, 50), 50), divide(75, 50))) | a car traveled 75 % of the way from town a to town b at an average speed of 50 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 50 miles per hour . what is s ? | "total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip → 75 / 50 = 1.5 2 nd part of the trip → 25 / s = t total trip → ( 75 + 25 ) / 50 = 1.5 + t » 100 / 50 = 1.5 + t » 2 = 1.5 + t » t = 0.5 back to 2 nd part of the trip formula : 25 / s = 0.5 » s = 50 ans e" | a = 50 + 50
b = a - 75
c = 50 + 50
d = c / 50
e = 75 / 50
f = d - e
g = b / f
|
a ) 22 , b ) 30 , c ) 77 , d ) 20 , e ) 99 | b | subtract(multiply(const_1, const_60), multiply(divide(30, 60), const_60)) | excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 30 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 30 km with stoppages . stoppage time = time take to travel ( 60 - 30 ) km i . e 30 km at 60 km / hr . stoppage time = 30 / 60 hrs = 30 min . answer : b" | a = 1 * const_60
b = 30 / 60
c = b * const_60
d = a - c
|
a ) 22 , b ) 99 , c ) 50 , d ) 29 , e ) 11 | c | divide(divide(factorial(add(13, const_2)), factorial(13)), const_2) | if a + b + c = 13 , then find the ab + bc + ca : | answer : c ) 50 | a = 13 + 2
b = math.factorial(a)
c = math.factorial(13)
d = b / c
e = d / 2
|
a ) 5 , b ) 2 , c ) 10 , d ) 5 , e ) 1 | c | subtract(subtract(multiply(6250, power(add(const_1, divide(4, const_100)), 2)), 6250), multiply(multiply(6250, divide(4, const_100)), 2)) | indu gave bindu rs . 6250 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | 6250 = d ( 100 / 4 ) 2 d = 10 answer : c | a = 4 / 100
b = 1 + a
c = b ** 2
d = 6250 * c
e = d - 6250
f = 4 / 100
g = 6250 * f
h = g * 2
i = e - h
|
a ) 3.208 , b ) 3.202 , c ) 3.209 , d ) 3.204 , e ) 3.2112 | d | add(multiply(const_4, 0.301), divide(log(const_100), log(const_10))) | if log 102 = 0.3010 , what is the value of log 101600 ? | explanation : log 101600 = log 10 ( 16 × 100 ) = log 10 ( 16 ) + log 10 ( 100 ) = log 10 ( 24 ) + log 10 ( 102 ) = 4 log 10 ( 2 ) + 2 = ( 4 × 0.3010 ) + 2 = 1.204 + 2 = 3.204 answer : option d | a = 4 * 0
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 8 kmph , b ) 9 kmph , c ) 2 kmph , d ) 4 kmph , e ) 1 kmph | c | divide(subtract(16, 12), const_2) | a man can row his boat with the stream at 16 km / h and against the stream in 12 km / h . the man ' s rate is ? | "ds = 16 us = 14 s = ? s = ( 16 - 2 ) / 2 = 2 kmph answer : c" | a = 16 - 12
b = a / 2
|
a ) 0.35 , b ) 3.5 , c ) 35 , d ) 350 , e ) 3500 | d | multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 20), 200), const_100) | if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 200 % of j ? | first of all , let us write the information in form of equations and numbers 125 j = 5 k or 5 j = k ( i ) 150 k = 50 l or 3 k = l ( ii ) 175 l = 75 m or 7 l = 3 m ( iii ) we need to find a relation between j and m from ( i ) and ( ii ) , 15 j = l multiplying this by 7 , 105 j = 7 l = 3 m hence , 35 j = m now , we are asked : 20 % of m is equal to what percent of 200 % of j assume j = 1 , hence m = 35 20 % of m = 7 and 200 % of j = 2 20 % of m = 3.5 ( 200 % of j ) = ( 350 / 100 ) * ( 200 % of j ) hence 20 % of m = 350 % of 200 % of j . option d | a = 125 * 150
b = a * 175
c = 25 * 50
d = c * 75
e = b / d
f = e * 20
g = f / 200
h = g * 100
|
a ) 250 , b ) 276 , c ) 280 , d ) 295 , e ) none | d | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 570), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | a library has an average of 570 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "since the month begins with sunday , to there will be five sundays in the month average required = ( 570 x 5 + 240 x 25 ) / 30 ) = 295 answer : option d" | a = 3 + 4
b = 30 / a
c = math.floor(b)
d = c + 1
e = d * 570
f = 3 + 4
g = 30 / f
h = math.floor(g)
i = h + 1
j = 30 - i
k = j * 240
l = e + k
m = l / 30
|
a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 21 , d ) 5 / 7 , e ) 16 / 21 | c | add(multiply(divide(1, 7), divide(subtract(7, 2), subtract(7, 1))), multiply(divide(1, 7), divide(subtract(7, 1), subtract(7, 1)))) | in a room filled with 7 people , 1 people have exactly 1 sibling in the room and 6 people have exactly 2 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | "there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssds ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssdssibl ∈ gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and viceversa ) . . . now remaning efg are 6 people who have exactly 2 siblings . . . . ( e has f and g as his / her sibling and so on . . ) there are now 3 different set of siblings ( a and b ) ( c and d ) ; ( efg ) now first selecting 2 people out of 7 is 7 c 2 = 21 first sibling pair - - - - ( a and b ) - - selecting 2 people - - 2 c 2 = 1 second sibling pair ( c and d ) - - selecting 2 people - - 2 c 2 = 1 third sibling pair ( e f g ) - - selecting 2 out of 6 - - 6 c 2 = 15 total = 1 + 1 + 15 = 17 but , a / c to formula p ( success ) - 1 - p ( fail ) here , p ( failure ) is selecting 2 people who are siblings = 17 / 21 ( 21 is 7 c 2 ) = 1 - 17 / 21 = 4 / 21 ans c" | a = 1 / 7
b = 7 - 2
c = 7 - 1
d = b / c
e = a * d
f = 1 / 7
g = 7 - 1
h = 7 - 1
i = g / h
j = f * i
k = e + j
|
a ) 39 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 . | a | subtract(multiply(sqrt(divide(676, 4)), 4), sqrt(divide(676, 4))) | the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 676 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 676 x ^ 2 = 169 x = 13 length = 4 * 13 = 52 difference = 52 - 13 = 39 a is the answer" | a = 676 / 4
b = math.sqrt(a)
c = b * 4
d = 676 / 4
e = math.sqrt(d)
f = c - e
|
['a ) 32 m', 'b ) 12 m', 'c ) 20 m', 'd ) 22 m', 'e ) 25 m'] | d | sqrt(divide(360, subtract(const_1, divide(25, const_100)))) | the area of a rectangular plot is 360 square metres . if the length is 25 % less than the breadth , what is the breadth of the plot ? | length = 75 % of breadth . length × breadth = 360 m 2 ⇒ 75 % of breadth × breadth = 360 m 2 ⇒ 75 / 100 × breadth × breadth = 360 m 2 ⇒ breadth × breadth = 480 m 2 ⇒ breadth = 22 m answer : d | a = 25 / 100
b = 1 - a
c = 360 / b
d = math.sqrt(c)
|
a ) a ) 73 , b ) b ) 20 , c ) c ) 83 , d ) d ) 17.1 , e ) e ) 52 | d | subtract(60, divide(60, add(divide(2, 5), const_1))) | a 60 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "explanation : 1 : 2 / 5 = 5 : 2 2 / 7 * 60 = 20 answer : option d" | a = 2 / 5
b = a + 1
c = 60 / b
d = 60 - c
|
a ) 24 , b ) 16 , c ) 12 , d ) 8 , e ) 4 | a | add(12, multiply(24, divide(50, const_100))) | one week , a certain truck rental lot had a total of 24 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 12 trucks on the lot that saturday morning , what is the greatest number of different trucks that could have been rented out during the week ? | n - not rented trucks ; r - rented trucks n + r = 24 n + r / 2 = 12 r = 24 a | a = 50 / 100
b = 24 * a
c = 12 + b
|
a ) 90 , b ) 85 , c ) 80 , d ) 75 , e ) 70 | a | multiply(divide(1455, add(470, 500)), const_60) | if two projectiles are launched at the same moment from 1455 km apart and travel directly towards each other at 470 km per hour and 500 km per hour respectively , how many minutes will it take for them to meet ? | "the projectiles travel a total of 970 km per hour . the time to meet is 1455 / 970 = 1.5 hours = 90 minutes the answer is a ." | a = 470 + 500
b = 1455 / a
c = b * const_60
|
a ) 32 , b ) 45 , c ) 59 , d ) 73 , e ) 87 | e | add(add(multiply(36, const_2), divide(36, 4)), add(const_3, const_3)) | if the least common addition of two prime numbers x and y is 36 , where x < y , then the value of 4 x + y is | ( x + y ) = 36 and both x an y are prime . the only values of x and y can be 17 and 19 ( x = 17 and y = 19 ) 4 x + y = 4 * 17 + 19 = 87 correct option : e | a = 36 * 2
b = 36 / 4
c = a + b
d = 3 + 3
e = c + d
|
a ) 6.3 , b ) 6.9 , c ) 5.3 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 10), divide(const_1, 20)))) | a man can do a job in 10 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | "1 day work of the three persons = ( 1 / 10 + 1 / 20 + 1 / 25 ) = 19 / 100 so , all three together will complete the work in 100 / 19 = 5.3 days . answer : c" | a = 1 / 25
b = 1 / 10
c = 1 / 20
d = b + c
e = a + d
f = 1 / e
|
a ) 178 , b ) 180 , c ) 182 , d ) 184 , e ) 186 | b | divide(divide(2500, const_1000), divide(50, const_3600)) | a train 2500 m long can cross an electric pole in 50 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 2500 / 50 s = 50 m / sec speed = 50 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 180 kmph answer : b" | a = 2500 / 1000
b = 50 / 3600
c = a / b
|
a ) 72 , b ) 76 , c ) 80 , d ) 84 , e ) 88 | a | add(divide(circumface(14), const_2), multiply(const_2, 14)) | a semicircle has a radius of 14 . 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 * 14 + 2 * 14 which is about 72 . the answer is a ." | a = circumface / (
b = a + 2
|
a ) 740 , b ) 228 , c ) 690 , d ) 780 , e ) 458 | b | subtract(subtract(multiply(25, 24), multiply(12, 17)), multiply(12, 14)) | the average of 25 results is 24 . 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 * 24 or , x = 228 . answer : option b" | a = 25 * 24
b = 12 * 17
c = a - b
d = 12 * 14
e = c - d
|
a ) 24887 , b ) 20778 , c ) 16200 , d ) 9000 , e ) 2811 | c | divide(multiply(multiply(4500, const_12), 3), multiply(subtract(const_12, 7), 2)) | a starts business with rs . 4500 and after 7 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b â € ™ s contribution in the capital ? | "explanation : a invested rs . 3500 for 12 months . let b joined with investment x . and he invested for 12 - 7 = 5 months . so there profit ratio = ( 4500 ã — 12 ) : ( 5 x ) = 2 : 3 â ‡ ’ x = 16200 answer : c" | a = 4500 * 12
b = a * 3
c = 12 - 7
d = c * 2
e = b / d
|
a ) 2.3 , b ) 2.5 , c ) 3.5 , d ) 3.6 , e ) 4.6 | e | divide(subtract(multiply(6, 3.95), add(multiply(2, 3.4), multiply(2, 3.85))), 2) | the average of 6 no . ' s is 3.95 . the average of 2 of them is 3.4 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 3.4 * 2 ) + ( 3.85 * 2 ) ] = 23.70 - ( 6.8 + 7.7 ) = 23.70 - 14.5 = 9.20 . required average = ( 9.2 / 2 ) = 4.6 . e" | a = 6 * 3
b = 2 * 3
c = 2 * 3
d = b + c
e = a - d
f = e / 2
|
a ) 9 , b ) 8 , c ) 4 , d ) 3 , e ) 1 | d | multiply(multiply(12, divide(const_1, const_2)), divide(const_1, const_2)) | some persons can do a piece of work in 12 days . two times the number of such persons will do half of that work in | "explanation : let x men can do the in 12 days and the required number of days be z more men , less days [ indirect proportion ] less work , less days [ direct proportion ] answer : d ) 3 days" | a = 1 / 2
b = 12 * a
c = 1 / 2
d = b * c
|
a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | d | add(add(floor(divide(173, add(const_1, const_4))), floor(divide(173, power(add(const_1, const_4), const_2)))), floor(divide(173, power(add(const_1, const_4), const_3)))) | find the number of zero ’ s in 173 ! ( 173 factorial ) ? | no of zeroes in 173 ! is 173 / 5 = 34 ( quotient ) 34 / 5 = 6 ( quotient ) 6 / 5 = 1 ( quotient ) 34 + 6 + 1 = 41 ans 41 zeroes answer : d | a = 1 + 4
b = 173 / a
c = math.floor(b)
d = 1 + 4
e = d ** 2
f = 173 / e
g = math.floor(f)
h = c + g
i = 1 + 4
j = i ** 3
k = 173 / j
l = math.floor(k)
m = h + l
|
a ) 24 , b ) 25 , c ) 30 , d ) 39 , e ) 43 | c | add(add(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), divide(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), const_10)), const_1) | 270 ã · ? ã — 15 + 270 = 405 | "explanation : 270 ã · ? ã — 15 = 405 - 270 = 135 ( 270 ã — 15 ) / ? = 135 ? = ( 270 ã — 15 ) / 135 = 30 answer : option c" | a = 405 - 270
b = a + 15
c = b * 2
d = c / 10
e = 405 - 270
f = e + 15
g = f * 2
h = g / 10
i = h / 10
j = d + i
k = j + 1
|
a ) 495 , b ) 360 , c ) 555 , d ) 600 , e ) 605 | b | divide(multiply(55, 65), const_4) | what is the sum of the odd integers from 55 to 65 , inclusive ? | "the mean is 60 . sum = mean ( # of elements ) there are 6 odd numbers between 55 - 65 inclusive . 6 * 60 = 360 b" | a = 55 * 65
b = a / 4
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), const_1) | when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of e ? | a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a person aged 39 is added to a group of n people , the average age increases by 2 . e * 23 + 39 = 25 * ( e + 1 ) e = 7 ans . ( a ) | a = 39 - 15
b = 2 + 1
c = a / b
d = c - 1
|
a ) 88 , b ) 77 , c ) 33 , d ) 44 , e ) 27 | e | add(multiply(3, 8), 3) | the present age of a father is 3 years more than 3 times the age of his son . 3 years hence , father as age will be 8 years more than twice the age of the son . find the present age of the father . | explanation : let the present age be ' x ' years . then father ' s present age is 3 x + 3 years . three years hence ( 3 x + 3 ) + 3 = 2 ( x + 3 ) + 8 x = 8 hence father ' s present age = 3 x + 3 = [ ( 3 x 8 ) + 3 ] = 27 years . answer : e | a = 3 * 8
b = a + 3
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | log(power(2, const_10)) | the “ length of integer x ” refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 × 2 × 3 × 5 . ) what is the greatest possible length of integer z if z < 2500 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 11 = 2048 < 2500 , then the greatest possible length of integer z is 11 . the answer is c ." | a = 2 ** 10
b = math.log(a)
|
a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.8 % | b | subtract(subtract(6, 5), divide(multiply(6, 5), const_100)) | in measuring the sides of a rectangle , one side is taken 6 % in excess , and the other 5 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 53 / 50 ) x ( 19 / 20 ) y = ( 144 / 143 ) ( xy ) error in measurement = ( 144 / 143 ) xy - xy = ( 1 / 143 ) xy error percentage = [ ( 1 / 143 ) xy ( 1 / xy ) 100 ] % = ( 7 / 10 ) % = 0.7 % . answer is b ." | a = 6 - 5
b = 6 * 5
c = b / 100
d = a - c
|
a ) 980 , b ) 1170 , c ) 1530 , d ) 1720 , e ) 1960 | c | divide(add(subtract(1.95, divide(65, 100)), multiply(divide(divide(10, 100), 100), 250)), divide(divide(10, 100), 100)) | a courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof . what could be the weight in grams of a package for which the charge is $ 1.95 ? | "the charge is 65 cents for the first 250 grams . this leaves a charge of $ 1.95 - $ 0.65 = $ 1.30 the charge for the next 1200 grams is $ 1.20 which leaves a charge of $ 0.10 the weight is somewhere between 1450 and 1550 . the answer is c ." | a = 65 / 100
b = 1 - 95
c = 10 / 100
d = c / 100
e = d * 250
f = b + e
g = 10 / 100
h = g / 100
i = f / h
|
a ) 35 % , b ) 25 % , c ) 20 % , d ) 15 % , e ) 10 % | b | divide(multiply(circle_area(50), const_100), circle_area(const_100)) | if the diameter of circle r is 50 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ? | "let diameter of circle r , dr = 50 and diameter of circle s , ds = 100 radius of circle r , rr = 25 radius of circle s , rs = 50 area of circle r / area of circle s = ( pi * rr ^ 2 ) / ( pi * rs ^ 2 ) = ( 25 / 50 ) ^ 2 = ( 5 / 10 ) ^ 2 = 25 % answer : b" | a = circle_area * (
b = a / 100
|
a ) 18 , b ) 77 , c ) 66 , d ) 54 , e ) 12 | d | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 9)))), divide(const_1, 9))), add(9, 8)) | a can do a piece of work in 9 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 9 + 6 / x = 1 x = 54 days answer : d" | a = 1 / 8
b = 1 / 9
c = 2 * b
d = 1 - c
e = a * d
f = 1 / 9
g = e - f
h = 1/(g)
i = 9 + 8
j = h - i
|
a ) 100000 , b ) 8500 , c ) 9000 , d ) 9500 , e ) 10000 | a | add(divide(multiply(50000, 50), const_100), 50000) | john ' s bank ' s saving amount is decreased 50 % due to loan payment and current balance is rs . 50000 . find the actual balance before deduction ? | "50 % decreased 50 % balance = 50000 100 % = 50000 / 50 * 100 = 100000 answer : a" | a = 50000 * 50
b = a / 100
c = b + 50000
|
a ) 8 , b ) 5 , c ) 9 , d ) 6 , e ) 12 | e | divide(subtract(192, multiply(3, 40)), multiply(3, const_2)) | a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 192 , how many hours overtime did he work ? | "at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 192 - $ 120 = $ 72 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 72 / $ 6 = 12 ans is e" | a = 3 * 40
b = 192 - a
c = 3 * 2
d = b / c
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | d | divide(divide(divide(divide(294, 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 294 , what is the least possible value of n ? | "294 = 2 * 3 * 7 * 7 , so n must be at least 14 . the answer is d ." | a = 294 / 2
b = a / 3
c = b / 4
d = 10 / 2
e = c / d
|
a ) 82 , b ) 16 , c ) 12 , d ) 82 , e ) 18 | b | subtract(const_60, multiply(const_60, divide(40, 55))) | excluding stoppages , the speed of a train is 55 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 15 / 55 * 60 = 16 answer : option b" | a = 40 / 55
b = const_60 * a
c = const_60 - b
|
a ) 80 kmph , b ) 69 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | b | divide(add(210, 270), add(3, 4)) | a train travels 210 km in 3 hours and 270 km in 4 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 210 + 270 = 480 km thus , total speed = 7 hrs or , average speed = 480 / 7 or , 69 kmph . answer : b" | a = 210 + 270
b = 3 + 4
c = a / b
|
a ) 272 , b ) 77 , c ) 168 , d ) 56 , e ) 59 | b | add(subtract(15, 2), multiply(divide(4, divide(const_1, const_2)), 8)) | 15 - 2 + 4 ÷ 1 / 2 × 8 = ? | explanation : 15 - 2 + 4 ÷ 1 / 2 × 8 = 15 - 2 + 4 × 2 × 8 = 15 - 2 + 64 = 77 answer : option b | a = 15 - 2
b = 1 / 2
c = 4 / b
d = c * 8
e = a + d
|
a ) $ 10000 , b ) $ 50000 , c ) $ 15200 , d ) $ 12500 , e ) $ 15000 | a | divide(multiply(multiply(const_100, const_100), 500), const_100) | the difference between the compound interest and simple interest on a certain sum at 20 % per annum for 2 years is $ 500 . find the sum ? | "let the sum be $ x c . i . = x ( 1 + 20 / 100 ) ^ 2 - x = 35 x / 100 s . i . = ( x * 20 * 2 ) / 100 = 2 x / 5 c . i . - s . i . = ( 35 x / 100 ) - ( 2 x / 5 ) = 5 x / 100 5 x / 100 = 500 x = 10000 answer is a" | a = 100 * 100
b = a * 500
c = b / 100
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | sqrt(divide(432, const_3)) | the length of a rectangular garden is three times its width . if the area of the rectangular garden is 432 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 432 x ^ 2 = 144 x = 12 the answer is e ." | a = 432 / 3
b = math.sqrt(a)
|
a ) 35.2 , b ) 36.1 , c ) 36.5 , d ) 39.1 , e ) none | e | divide(add(multiply(39, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 39 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 39 × 50 + 48 – 23 ) = 1975 . therefore correct mean = 1975 / 50 = 39.5 . answer e" | a = 39 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | c | subtract(10, reminder(5, 7)) | when positive integer n is divided by 2 , the remainder is 1 . when n is divided by 7 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 10 ? | "when positive integer n is divided by 2 , the remainder is 1 i . e . , n = 2 x + 1 values of n can be one of { 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17,19 , . . . . . . . . . . . . . . 31 , 33,35 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 7 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n can be one of { 5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 . . . . . . . . } combining both the sets we get n = { 19 , 52 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 21 or 21 x in case of n = 5 p = 5 so for min value of p , we take min value of n . c is the answer ." | a = 10 - reminder
|
a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | add(add(add(const_4, const_2), const_1), const_1) | how many odd factors does 160 have ? | "start with the prime factorization : 160 = 2 * 5 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 } plus 1 to each = { 2 } product = 2 therefore , there are 2 odd factors of 160 . answer : a ." | a = 4 + 2
b = a + 1
c = b + 1
|
['a ) 250 %', 'b ) 300 %', 'c ) 500 %', 'd ) 650 %', 'e ) 700 %'] | c | multiply(subtract(multiply(const_2, const_3), const_1), const_10) | the length of a rectangle is doubled while its width is tripled . what is the % change in area ? | the original area is l * w the new area is 2 l * 3 w = 6 * l * w = l * w + 5 * l * w the area increased by 500 % . the answer is c . | a = 2 * 3
b = a - 1
c = b * 10
|
a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | subtract(100, add(add(add(subtract(100, 81), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | there were totally 100 men . 81 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 81 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 19 - 25 - 15 - 30 = 100 - 89 = 11 answer : a" | a = 100 - 81
b = 100 - 75
c = a + b
d = 100 - 85
e = c + d
f = 100 - 70
g = e + f
h = 100 - g
|
a ) 40 , b ) 25 , c ) 35 , d ) 30 , e ) 50 | d | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | nicky and cristina are running a 400 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . d" | a = 3 * 12
b = a / 3
c = 3 * 12
d = 5 - 3
e = c / d
f = b + e
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.