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 ) 1 hr , b ) 1 hr 15 min , c ) 50 min , d ) 1 hr 30 min , e ) 1 hr 45 min | b | divide(100, subtract(divide(100, 1), 20)) | a train covers a distance of 100 km in 1 hour . if its speed is decreased by 20 km / hr , the time taken by the car to cover the same distance will be ? | speed = 100 / 1 = 100 km / hr new speed = 100 - 20 = 80 km / hr time taken = 100 / 80 = 1 hr 15 min answer is b | a = 100 / 1
b = a - 20
c = 100 / b
|
a ) 2 / 27 , b ) 1 / 9 , c ) 1 / 3 , d ) 4 / 27 , e ) 2 / 9 | d | multiply(multiply(divide(3, add(add(3, 4), 2)), divide(2, add(add(3, 4), 2))), const_2) | a bag contains 3 red , 4 black and 2 white balls . what is the probability of drawing a red and a white ball in two successive draws , each ball being put back after it is drawn ? | p = 2 β 39 β 29 = 427 p = 2 β 39 β 29 = 427 we are multiplying by 2 as there are two possible wining scenarios rw and wr . answer : d . | a = 3 + 4
b = a + 2
c = 3 / b
d = 3 + 4
e = d + 2
f = 2 / e
g = c * f
h = g * 2
|
a ) 142 % , b ) 150 % , c ) 165 % , d ) 176 % , e ) 188 % | c | multiply(divide(multiply(divide(15, const_100), add(const_100, 10)), divide(multiply(10, const_100), const_100)), const_100) | last year sandy saved 10 % of her annual salary . this year , she made 10 % more money than last year , and she saved 15 % 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 0.1 x this year , sandy saved 0.15 * 1.1 x = 0.165 x 0.165 x / 0.1 x = 1.65 = 165 % the answer is c ." | a = 15 / 100
b = 100 + 10
c = a * b
d = 10 * 100
e = d / 100
f = c / e
g = f * 100
|
a ) 0.125 , b ) 0.25 , c ) 0.5 , d ) 0.75 , e ) not enough information to determine the rate | e | divide(840, 0.5) | the volume of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate t , in cubic meters per minutes , at which the water is flowing into the swimming pool ? | "the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e" | a = 840 / 0
|
a ) 4 , b ) 5 , c ) 45 , d ) 6970 , e ) none | d | subtract(multiply(const_10, 7), 7) | the difference between the place values of 7 and 3 in the number 527435 is | "sol . = ( place value of 7 ) β ( place value of 3 ) = ( 700 - 30 ) = 6970 answer d" | a = 10 * 7
b = a - 7
|
a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42 | c | divide(15, subtract(const_1, sqrt(divide(9, add(9, 16))))) | 15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container m ? | "if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 / 37.5 ; expressed as a fraction this becomes ( 45 / 2 ) / ( 75 / 2 ) , or 45 / 2 * 2 / 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 / 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 / 15 of a means there must be 6 / 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 / 16 ; 13.5 / 24 = ( 27 / 2 ) * ( 1 / 24 ) = 9 / 16 . choice c is correct ." | a = 9 + 16
b = 9 / a
c = math.sqrt(b)
d = 1 - c
e = 15 / d
|
a ) 2.4 days , b ) 8 days , c ) 1.2 days , d ) 1.0 days , e ) 5 days | a | add(inverse(subtract(divide(const_1, 2), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 2 days . a alone can do it in 12 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 2 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 2 ) - ( 1 / 12 ) = 2.4 days answer : option a" | a = 1 / 2
b = 1 / 12
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 5 , b ) 6 , c ) 7 , d ) 11 , e ) 12 | c | subtract(subtract(subtract(12, 3), const_1), const_1) | list k consists of 12 consecutive integers . if - 3 is the least integer in list k , what is the range of the positive integers in list k ? | "answer = c = 7 if least = - 3 , then largest = 8 range = 8 - 1 = 7" | a = 12 - 3
b = a - 1
c = b - 1
|
a ) 350 m , b ) 145 m , c ) 240 m , d ) 135 m , e ) 255 m | a | multiply(divide(multiply(60, const_1000), const_3600), 21) | a train running at the speed of 60 km / hr crosses a pole in 21 seconds . find the length of the train . | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 15 = 350 meter answer : a" | a = 60 * 1000
b = a / 3600
c = b * 21
|
a ) 97.959 % , b ) 98.959 % , c ) 99.959 % , d ) 96.959 % , e ) 95.959 % | a | subtract(const_100, multiply(divide(divide(7, 7), multiply(7, 7)), const_100)) | instead of multiplying a number by 7 , the number is divided by 7 . what is the percentage of error obtained ? | "let the number be x the right number is 7 x the wrong number is x / 7 error is ( 7 x - x / 7 ) = 48 x / 7 percentage of error is ( ( 48 x / 7 ) / 7 x ) * 100 = 97.959 % answer : a" | a = 7 / 7
b = 7 * 7
c = a / b
d = c * 100
e = 100 - d
|
a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 3 , d ) 1 / 2 , e ) 3 / 4 | a | inverse(power(2, const_3)) | if a coin is flipped , the probability that the coin will land tails is 1 / 2 . if the coin is flipped 3 times , what is the probability that it will land tails up on the first flip and not on the last 2 flips ? | ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 8 answer : a | a = 2 ** 3
b = 1/(a)
|
a ) 38 , b ) 43 , c ) 50 , d ) 53 , e ) 62 | b | add(divide(360, 9), 3) | a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 360 kilos of plankton , how many kilos did he consume on the sixth hour ? | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 360 / 9 = 40 , so the median is also 40 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in the next hour , so the answer is 43 . option b" | a = 360 / 9
b = a + 3
|
a ) 1375 toys , b ) 2170 toys , c ) 2375 toys , d ) 2175 toys , e ) 5375 toys | b | divide(4340, 2) | a factory produces 4340 toys per week . if the workers at this factory work 2 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 2 days ) by 2 . 4340 / 2 = 2170 toys correct answer b" | a = 4340 / 2
|
a ) 65.25 , b ) 62.6 , c ) 62.1 , d ) 62.7 , e ) 62.2 | a | divide(multiply(add(47.50, divide(multiply(47.50, 25), const_100)), const_100), subtract(const_100, 9)) | at what price must an article costing rs . 47.50 be marked in order that after deducting 9 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 91 / 100 ) = 59.375 mp = 65.25 answer : a" | a = 47 * 50
b = a / 100
c = 47 + 50
d = c * 100
e = 100 - 9
f = d / e
|
a ) 36.0 , b ) 36.5 , c ) 36.24 , d ) 36.12 , e ) 36.22 | e | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 34 was wrongly taken as 23 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 434 - 23 ) = 1825 . correct mean = = 1811 / 50 = 36.5 answer : e ) 36.22" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(subtract(subtract(3, 3), 1), 1) | in the coordinate plane , line a has a slope of - 1 and an x - intercept of 1 . line b has a slope of 3 and a y - intercept of - 3 . if the two lines intersect at the point ( a , b ) , what is the sum a + b ? | "the equation of line a is y = - x + 1 the equation of line b is y = 3 x - 3 3 x - 3 = - x + 1 x = 1 y = 0 the point of intersection is ( 1,0 ) and then a + b = 1 . the answer is b ." | a = 3 - 3
b = a - 1
c = b / 1
|
a ) - 14 , b ) 14 , c ) 33 , d ) 83 , e ) 93 | e | add(multiply(negate(11), negate(8)), 5) | if - 11 and - 8 are negative integers , then - 11 * - 8 + 5 is | answer : e | a = negate * (
b = a + negate
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(multiply(add(const_4, const_3), 3), 20) | if there are 20 apples and the apples have to be shared equally among 3 babies . what number of apples are to be added ? | given there are 20 apples . if 1 extra apple is added , then it becomes 21 which can be divided equally that is 7 apples to each baby . option a is correct . | a = 4 + 3
b = a * 3
c = b - 20
|
a ) 1 / 9 , b ) 1 / 8 , c ) 1 / 7 , d ) 7 / 8 , e ) 8 / 7 | b | divide(const_1, 8) | 8 people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire rest of the car , then the share of each of the remaining persons increased by ? | when there are eight people , the share of each person is 1 / 8 th the total cost answer : b | a = 1 / 8
|
a ) 7 / 4 , b ) 1 / 2 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3 | a | divide(add(3, 4), const_4) | a box contains 3 red balls and 4 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | my complicated version of your simple approach let the 5 black balls be bbbbb and 3 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 4 ! x 3 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 7 slots and 3 black ( bbb ) and 3 red ( rrr ) they can be arranged in ( 6 ! ) / ( 3 ! x 3 ! ) hence required probability = [ ( 7 ! ) / ( 4 ! x 3 ! ) ] / [ ( 6 ! ) / ( 3 ! x 3 ! ) ] ans = 7 / 4 a | a = 3 + 4
b = a / 4
|
a ) 11 , b ) 22 , c ) 77 , d ) 33 , e ) 88 | a | add(add(4, 2), 5) | in kaya ' s teacher ' s desk there are 4 pink highlighters , 2 yellow highlighters , and 5 blue highlighters . how many highlighters are there in all ? | "add the numbers of highlighters . 4 + 2 + 5 = 11 . answer is a ." | a = 4 + 2
b = a + 5
|
a ) 75 kg , b ) 50 kg , c ) 85 kg , d ) 90 kg , e ) 60 kg | d | add(multiply(10, 2.5), 65) | the average weight of 10 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg . what is the weight of the new person ? | "explanation : total increase in weight = 10 Γ£ β 2.5 = 25 if x is the weight of the new person , total increase in weight = x Γ’ Λ β 65 = > 25 = x - 65 = > x = 25 + 65 = 90 answer : option d" | a = 10 * 2
b = a + 65
|
a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | on a map , 1 inch represents 28 miles . how many z inches would be necessary to represent a distance of 383.6 miles ? | "z inches necessary to represent a distance of 383.6 miles = 383.6 / 28 = 13.7 answer c" | a = 383 / 6
|
a ) 5 , b ) 6 , c ) 7 , d ) 9 , e ) 8 | e | divide(150, multiply(add(60, 6), const_0_2778)) | a train 150 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 150 * 3 / 55 = 8 sec . answer : e" | a = 60 + 6
b = a * const_0_2778
c = 150 / b
|
a ) 276 , b ) 600 , c ) 699 , d ) 722 , e ) 745 | b | multiply(40, 15) | the h . c . f . of two numbers is 40 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is : | "the numbers are ( 40 x 11 ) and ( 40 x 15 ) . larger number = ( 40 x 15 ) = 600 . answer : b" | a = 40 * 15
|
a ) a ) 13.55 , b ) b ) 14.55 , c ) c ) 15.55 , d ) d ) 16.55 , e ) e ) 17.55 | c | subtract(const_60, multiply(const_60, divide(40, 54))) | excluding stoppages , the speed of a train is 54 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 14 / 54 * 60 = 15.55 answer : option c" | a = 40 / 54
b = const_60 * a
c = const_60 - b
|
a ) 5 , b ) 21 , c ) 7 , d ) 8 , e ) 9 | b | multiply(const_3600, divide(divide(385, const_1000), add(60, 6))) | a train 385 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 = [ 385 * 3 / 55 ] sec = 21 sec answer : option b" | a = 385 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 61 % , b ) 70 % , c ) 75 % , d ) 85 % , e ) 95 % | a | multiply(subtract(divide(multiply(multiply(const_100, add(const_1, divide(40, const_100))), add(const_1, divide(15, const_100))), const_100), const_1), const_100) | a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 15 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ? | "find the product of the two increases : ( 1.4 ) * ( . 15 ) , which is 1.61 and a 61 % increase . a" | a = 40 / 100
b = 1 + a
c = 100 * b
d = 15 / 100
e = 1 + d
f = c * e
g = f / 100
h = g - 1
i = h * 100
|
a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 22 litres , e ) 18 litres | e | multiply(9, const_1) | a mixture contains alcohol and water in the ratio 4 : 3 . if 9 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 9 ) = 4 / 5 20 x = 4 ( 3 x + 9 ) 8 x = 36 x = 4.5 quantity of alcohol = ( 4 x 4.5 ) litres = 18 litres . answer is e" | a = 9 * 1
|
a ) 10 days , b ) 30 days , c ) 20 days , d ) 80 days , e ) 40 days | c | inverse(add(divide(9, multiply(12, 54)), divide(39, multiply(20, 54)))) | if 12 men or 20 women can do a piece of work in 54 days , then in how many days can 9 men and 39 women together do the work ? | c 20 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 39 women = 15 women + 39 women = 54 women 20 women can do the work in 54 days . so , 54 women can do it in ( 20 * 54 ) / 54 = 20 days . | a = 12 * 54
b = 9 / a
c = 20 * 54
d = 39 / c
e = b + d
f = 1/(e)
|
a ) 357003 , b ) 357090 , c ) 37667 , d ) 379888 , e ) 287761 | b | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(79, const_100)) | in an election , candidate a got 79 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 Γ 476000 = 35700000 / 100 = 357090 answer : b" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 79 / 100
e = c * d
|
a ) 50 kmh , b ) 55 kmh , c ) 60 kmh , d ) 65 kmh , e ) 70 kmh | e | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(105, 35)), const_4), power(35, const_2))), 35), const_2) | if a car had traveled 35 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 105 km , at what speed did it travel ? | "time = distance / speed difference in time = 1 / 2 hrs 105 / x - 105 / ( x + 35 ) = 1 / 2 substitute the value of x from the options . - - > x = 70 - - > 105 / 70 - 105 / 105 = 3 / 2 - 1 = 1 / 2 answer : e" | a = 105 * 35
b = 2 * a
c = b * 4
d = 35 ** 2
e = c + d
f = math.sqrt(e)
g = f - 35
h = g / 2
|
['a ) 36', 'b ) 18', 'c ) 12', 'd ) 8', 'e ) 6'] | e | divide(multiply(divide(power(3, const_2), const_2), 4), const_3) | the base of pyramid p is an isosceles right triangle whose leg is 3 . if the height of p is 4 , what is the volume of p ? | volume of a pyramid is given by v = 1 / 3 ah , where a is the area of the triangular base and h is the height of the pyramid . area of triangular base = 1 / 2 * 3 * 3 = 9 / 2 so , v = 1 / 3 * 9 / 2 * 4 = 6 option e | a = 3 ** 2
b = a / 2
c = b * 4
d = c / 3
|
a ) 8545 , b ) 8770 , c ) 8515 , d ) 7999 , e ) 7989 | c | floor(divide(6514, multiply(divide(subtract(const_100, 10), const_100), divide(subtract(const_100, 15), const_100)))) | 10 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 6514 , how much was it in the beginning ? | "x * ( 90 / 100 ) * ( 85 / 100 ) = 6514 x = 8515 answer : c" | a = 100 - 10
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 6514 / e
g = math.floor(f)
|
a ) 3 , b ) 4 , c ) 5 , d ) 7 , e ) 8 | e | add(const_4, const_4) | if each year the population of the country grows by 10 % , how many years will elapse before the population of the country doubles ? | till year 2000 , population is 100 . year 2001 : population becomes 110 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 121 . . . . . . . . . . . . . 2 year elapsed year 2003 : population becomes > 133 . . . . . . . . . . . . . 3 year elapsed year 2004 : population becomes > 146 . . . . . . . . . . . . . . . . . . 4 year elapsed year 2005 : population becomes > 160 . . . . . . . . . . . . . . . . . . 5 year elapsed year 2006 : population becomes > 176 . . . . . . . . . . . . . . . . . . 6 year elapsed year 2007 : population becomes > 193 . . . . . . . . . . . . . . . . . . 7 year elapsed year 2008 : population > 200 . . . . . . . . . . . . . . . . . . 8 year elapsed answer : e | a = 4 + 4
|
a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 120 | b | divide(600, multiply(2, divide(const_100, 5))) | 5 percent of a certain grass seed is timothy . if the amount of the mixture needed to plant one acre contains 2 pounds of timothy , how many acres can be planted with 600 pounds of the seed mixture ? | if the mixture contains 2 pounds of timothy , the mixture must be 40 pounds . thus we need 40 pounds of mixture per acre . 600 / 40 = 15 acres the answer is b . | a = 100 / 5
b = 2 * a
c = 600 / b
|
a ) $ 400 , b ) $ 450 , c ) $ 500 , d ) $ 550 , e ) $ 600 | d | divide(subtract(multiply(multiply(const_100, multiply(add(const_2, const_3), const_2)), add(divide(6, const_100), const_1)), add(add(multiply(const_100, multiply(add(const_2, const_3), const_2)), multiply(multiply(add(const_2, const_3), const_2), 5)), multiply(const_2, const_3))), subtract(add(divide(6, const_100), const_1), add(divide(const_4.0, const_100), const_1))) | we invested a total of $ 1,000 . we invested one part of the money at 5 % and the rest of the money at 6 % . the total investment with interest at the end of the year was $ 1 , 054.50 . how much money did we invest at 5 % ? | "let x be the money invested at 5 % . 1.05 x + 1.06 ( 1000 - x ) = 1054.50 . 0.01 x = 1060 - 1054.50 . 0.01 x = 5.50 . x = 550 . the answer is d ." | a = 2 + 3
b = a * 2
c = 100 * b
d = 6 / 100
e = d + 1
f = c * e
g = 2 + 3
h = g * 2
i = 100 * h
j = 2 + 3
k = j * 2
l = k * 5
m = i + l
n = 2 * 3
o = m + n
p = f - o
q = 6 / 100
r = q + 1
s = 4 / 0
t = s + 1
u = r - t
v = p / u
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 7 | d | subtract(subtract(multiply(3, 16), add(subtract(19, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 19 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 19 a β d = 3 a + d = 19 2 d = 16 d = 8 answer d" | a = 3 * 16
b = 19 - 16
c = b + 3
d = a - c
e = d - 16
|
a ) 105 , b ) 120 , c ) 190 , d ) 210 , e ) 240 | b | add(add(add(add(add(add(add(5, 10), add(5, const_2)), add(5, const_1)), 5), 10), const_2), const_1) | if two integers x , y ( x > y ) are selected from - 10 to 5 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 5 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 16 c 2 = 16 * 15 / 2 = 120 . therefore , the answer is b ." | a = 5 + 10
b = 5 + 2
c = a + b
d = 5 + 1
e = c + d
f = e + 5
g = f + 10
h = g + 2
i = h + 1
|
a ) 80 m 2 , b ) 50 m 2 , c ) 83 m 2 , d ) 86 m 2 , e ) 55 m 2 | c | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 6), multiply(add(divide(25, const_100), 1), 8))), multiply(6, 8)) | a cistern 8 m long and 6 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 6 x 1.25 + 8 x 1.25 ) + 8 x 6 ] m 2 = 83 m 2 . answer : c | a = 25 / 100
b = a + 1
c = b * 6
d = 25 / 100
e = d + 1
f = e * 8
g = c + f
h = 2 * g
i = 6 * 8
j = h + i
|
a ) 63 % , b ) 64 % , c ) 65 % , d ) 66 % , e ) 67 % | a | divide(add(multiply(70, 60), multiply(70, subtract(90, 70))), 90) | in a certain accounting class of 90 students , 70 % of the students took the final exam on the assigned day while the rest of the students took the exam on a make - up date . if the students on the assigned day had an average score of 60 % , and the students on the make - up date had an average score of 70 % , what was the average score for the entire class ? | "70 % of the class scored 60 % and 30 % of the class scored 70 % . the difference between 60 % and 70 % is 10 % . the average will be 60 % + 0.3 ( 10 % ) = 63 % . the answer is a ." | a = 70 * 60
b = 90 - 70
c = 70 * b
d = a + c
e = d / 90
|
a ) 4462.5 , b ) 8032.5 , c ) 8925 , d ) none of these , e ) can not be determined | c | divide(divide(multiply(4016.25, const_100), 9), 5) | a sum fetched a total simple interest of 4016.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "explanation : principal = ( 100 x 4016.25 ) / ( 9 x 5 ) = 401625 / 45 = 8925 answer : option c" | a = 4016 * 25
b = a / 9
c = b / 5
|
a ) 3 / 4 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | divide(add(subtract(20, add(5, 9)), 9), 20) | a bag consists of 20 marbles , of which 5 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | "bag consists of 20 marbles , of which 5 are blue , 9 are red remainder are white . so , white = 20 - 5 - 9 = 6 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 6 / 20 = 15 / 20 = 3 / 4 hence , answer will be a ." | a = 5 + 9
b = 20 - a
c = b + 9
d = c / 20
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 18 | c | add(add(const_4, const_3), const_1) | how many different positive integers exist between 10 ^ 6 and 10 ^ 7 , the sum of whose digits is equal to 2 ? | "total cases = > 1000000 = > 6 cases for 1 being present at any of the 6 zero and last case 2000000 hence & cases answer : c" | a = 4 + 3
b = a + 1
|
a ) 45 , b ) 65 , c ) 98 , d ) 107 , e ) 198 | d | add(add(add(divide(lcm(lcm(lcm(1, 1), 1), 3), 1), divide(lcm(lcm(lcm(1, 1), 1), 3), 1)), divide(lcm(lcm(lcm(1, 1), 1), 3), 1)), divide(lcm(lcm(lcm(1, 1), 1), 3), 3)) | john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 1 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | "rakesh : rahul : ranjan : rohit = 1 / 1 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 1 x 60 ) = 60 . rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : total number of pencils are ( 60 x + 20 x + 15 x + 12 x ) = 107 x . for minimum number of pencils x = 1 . the person should have at least 107 pencils . d )" | a = math.lcm(1, 1)
b = math.lcm(a, 1)
c = math.lcm(b, 3)
d = c / 1
e = math.lcm(1, 1)
f = math.lcm(e, 1)
g = math.lcm(f, 3)
h = g / 1
i = d + h
j = math.lcm(1, 1)
k = math.lcm(j, 1)
l = math.lcm(k, 3)
m = l / 1
n = i + m
o = math.lcm(1, 1)
p = math.lcm(o, 1)
q = math.lcm(p, 3)
r = q / 3
s = n + r
|
a ) 16 : 15 , b ) 15 : 16 , c ) 15 : 8 , d ) 8 : 15 , e ) 8 : 17 | a | divide(add(8, 8), add(5, add(5, 5))) | a sum of rs . 1550 was lent partly at 5 % and partly at 8 % p . a . simple interest . the total interest received after 3 years was rs . 300 . the ratio of the money lent at 5 % to that lent at 8 % is : | explanation : let the partial amount at 5 % be x and the partial amount at 8 % be ( 1550 - x ) interest on x at 5 % for 3 years + interest on ( 1550 - x ) at 8 % for 3 years = 300 x Γ 5 Γ 3 / 100 + ( 1550 - x ) Γ 8 Γ 3 / 100 = 300 x Γ 5 / 100 + ( 1550 - x ) Γ 8 / 100 = 100 5 x + 8 ( 1550 β x ) = 10000 5 x + 12400 β 8 x = 10000 3 x = 2400 x = 800 required ratio = x : ( 1550 - x ) = 800 : ( 1550 - 800 ) = 800 : 750 = 16 : 15 answer : option a | a = 8 + 8
b = 5 + 5
c = 5 + b
d = a / c
|
a ) $ 0.60 , b ) $ 0.70 , c ) $ 0.80 , d ) $ 0.90 , e ) $ 1.00 | b | add(multiply(multiply(multiply(0.20, const_2), const_2), const_2), 0.30) | a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.20 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the third day it is overdue ? | "1 st day fine - 0.2 2 nd day fine - 0.2 * 2 = 0.4 ( as doubling gives lower value ) 3 rd day fine - 0.4 + 3 = 0.7 ( as doubling gives higher value ) answer : b ." | a = 0 * 20
b = a * 2
c = b * 2
d = c + 0
|
a ) 13 % , b ) 15 % , c ) 25 % , d ) 50 % , e ) 100 % | a | multiply(divide(divide(15, const_100), divide(add(15, const_100), const_100)), const_100) | a part - time employee β s hourly wage was increased by 15 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 15 % wage increase , the employee makes $ 1.15 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.15 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.15 to get : new # of hours = 100 / 1.15 β 87 hours so , the number of hours decreases from 100 hours to ( approximately ) 87 hours . this represents a 13 % decrease ( approximately ) . answer : a" | a = 15 / 100
b = 15 + 100
c = b / 100
d = a / c
e = d * 100
|
a ) 29.3 % , b ) 33.4 % , c ) 37.5 % , d ) 41.6 % , e ) 45.7 % | c | multiply(divide(60, add(60, const_100)), const_100) | if y is 60 % greater than x , than x is what % less than y ? | "y = 1.6 x x = y / 1.6 = 10 y / 16 = 5 y / 8 x is 3 / 8 less which is 37.5 % less than y . the answer is c ." | a = 60 + 100
b = 60 / a
c = b * 100
|
a ) 90 % , b ) 25 % , c ) 45 % , d ) 55 % , e ) 65 % | a | subtract(multiply(70, const_3), add(50, 70)) | a student gets 50 % in one subject , 70 % in the other . to get an overall of 70 % how much should get in third subject . | "let the 3 rd subject % = x 50 + 70 + x = 3 * 70 120 + x = 210 x = 210 - 120 = 90 answer : a" | a = 70 * 3
b = 50 + 70
c = a - b
|
a ) 51 , b ) 68 , c ) 57 , d ) 55 , e ) 37 | a | sqrt(2652) | the product of two successive numbers is 2652 . which is the smaller of the two numbers ? | "a 51 from the given alternatives , 51 Γ 52 = 2652 β΄ smaller number = 51" | a = math.sqrt(2652)
|
a ) 30 % , b ) 10 % , c ) 20 % , d ) 16 % , e ) 50 % | d | multiply(divide(subtract(2989, 2560), 2560), const_100) | a sum of money deposited at c . i . amounts to rs . 2560 in 3 years and to rs . 2989 in 4 years . find the rate percent ? | "2560 - - - 429 100 - - - ? = > 16 % answer : d" | a = 2989 - 2560
b = a / 2560
c = b * 100
|
a ) 0 , b ) 1 , c ) 45 , d ) 130 , e ) 520 | b | subtract(power(3, subtract(const_1, const_1)), power(subtract(const_1, const_1), 3)) | if k is a non - negative integer and 30 ^ k is a divisor of 929,260 then 3 ^ k - k ^ 3 = | 9 + 2 + 9 + 2 + 6 + 0 = 28 , so this number is not divisible by 3 and thus not divisible by 30 . therefore , k = 0 3 ^ k - k ^ 3 = 1 - 0 = 1 the answer is b . | a = 1 - 1
b = 3 ** a
c = 1 - 1
d = c ** 3
e = b - d
|
a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 23 % , e ) 40 % | d | subtract(multiply(divide(subtract(const_100, 18), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 18 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.82 * 150 x = 123 x hence total profit = 23 % option d" | a = 100 - 18
b = a / 100
c = 100 + 20
d = 100 + 25
e = d / 100
f = c * e
g = b * f
h = g - 100
|
a ) 2800 , b ) 1980 , c ) 2900 , d ) 2200 , e ) 2300 | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 5500) | in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 5500 , the number of valid votes that the other candidate got , was : | "b number of valid votes = 80 % of 5500 = 4400 . valid votes polled by other candidate = 45 % of 4400 = ( 45 / 100 x 4400 ) = 1980 ." | a = 20 / 100
b = 1 - a
c = 55 / 100
d = 1 - c
e = b * d
f = e * 5500
|
a ) 95 kg . , b ) 99 kg . , c ) 89 kg . , d ) 90 kg . , e ) 92 kg . | b | divide(add(multiply(7, 103), add(110, 60)), add(7, const_2)) | there are 7 players in a bowling team with an average weight of 103 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 103 * 7 + 110 + 60 ) / 9 = 99 kgs b is the answer" | a = 7 * 103
b = 110 + 60
c = a + b
d = 7 + 2
e = c / d
|
a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 45 | e | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | find the average of all numbers between 3 and 86 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 + 70 + 75 + 80 + 85 ) / 17 = 765 / 17 = 45 answer : option e" | a = 5 * 3
b = 5 * 2
c = 5 + b
d = a + c
e = 5 * 4
f = d + e
g = 4 + 1
h = g * 5
i = f + h
j = i / 5
|
a ) 12100 , b ) 15240 , c ) 12456 , d ) 16940 , e ) 10002 | d | multiply(multiply(divide(add(10, const_100), const_100), 14000), divide(add(10, const_100), const_100)) | if the annual increase in the population of a town is 10 % and the present number of people is 14000 , what will the population be in 2 years ? | "the required population is = 14000 ( 1 + 10 / 100 ) ^ 2 = 14000 * 11 / 10 * 11 / 10 = 16940 answer is d" | a = 10 + 100
b = a / 100
c = b * 14000
d = 10 + 100
e = d / 100
f = c * e
|
a ) 48 , b ) 51 , c ) 47 , d ) 49 , e ) 55 | e | subtract(multiply(11, 41), subtract(multiply(21, 44), multiply(48, 11))) | the average of 21 numbers is 44 . average of the first 11 of them is 48 and that of the last 11 is 41 . find the 11 th number ? | sum of all the 21 numbers = 21 * 44 = 924 sum of the first 11 of them = 11 * 48 = 528 sum of the last 11 of them = 11 * 41 = 451 so , the 11 th number = 528 + 451 - 924 = 55 . answer : e | a = 11 * 41
b = 21 * 44
c = 48 * 11
d = b - c
e = a - d
|
a ) 75 m , b ) 180 m , c ) 324 m , d ) can not be determined , e ) none of these | a | multiply(divide(multiply(30, const_1000), const_3600), 9) | a train running at the speed of 30 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 30 * 5 / 18 = 25 / 3 m / sec length of the train = speed * time = 25 / 3 * 9 = 75 m answer : a" | a = 30 * 1000
b = a / 3600
c = b * 9
|
a ) 13 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | d | subtract(divide(200, 4), divide(200, 6)) | in the next cricket world cup t - 20 , team t has decided to score 200 runs only through 4 s and 6 s . in how many ways can the team t score these 200 runs ? | team t can score a maximum of 50 fours and a minimum of 2 fours with an interval or spacing of 3 units to accommodate the 6 ' s . so the number of fours scored forms an ap 2 , 5 , 8 , . . . 50 with a common difference of 3 . number of ways of scoring 200 only through 4 ' s and 6 ' s = ( 50 - 2 ) / 3 + 1 = 17 answer : d alternate solution : 4 x + 6 y = 200 - - > 2 x + 3 y = 100 x = ( 100 - 3 y ) / 2 - - > 100 - 3 y should be even . this is possible when 3 y is even . there are 17 even multiples of 3 between 0 and 100 . answer : d | a = 200 / 4
b = 200 / 6
c = a - b
|
a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 2 , d ) 4 / 9 , e ) 2 / 3 | c | multiply(subtract(1, divide(1, 5)), subtract(1, divide(subtract(1, divide(const_2.0, 5)), 1))) | for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 1 / 5 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ? | "p ( of not burning out in a six mnth period ) = 1 / 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 1 / 5 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 1 / 5 = 4 / 5 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 4 / 5 = 4 / 10 = 2 / 5 - - - > p ( of burning out in a six mnth period ) = 1 - 2 / 5 = 3 / 5 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 4 / 5 * 3 / 5 = 1 / 2 ans c" | a = 1 / 5
b = 1 - a
c = 2 / 0
d = 1 - c
e = d / 1
f = 1 - e
g = b * f
|
a ) 4 , b ) 5.656 , c ) 5 , d ) 6 , e ) 7 | b | divide(multiply(8, sqrt(2)), 2) | the perimeter of an isosceles right triangle is 8 + 8 sq rt 2 . what is the length of the hypotenuse of the triangle ? | "side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 8 + 8 sqrt 2 or , a . ( 2 + sqrt 2 ) = 8 ( 1 + sqrt 2 ) , a = 8 * ( 1 + sqrt 2 ) / 2 + sqrt 2 = 8 * 2.414 / 3.414 = 0.707 * 8 then hypotenuse = 5.656 b" | a = math.sqrt(2)
b = 8 * a
c = b / 2
|
a ) - 4 , b ) - 2 , c ) 11 , d ) 13 , e ) 22 | c | divide(add(26, 18), add(3, const_1)) | when x is multiplied by 3 , the result is 18 more than the result of subtracting x from 26 . what is the value of x ? | "the equation that can be formed is : 3 x - 18 = 26 - x or , 4 x = 44 or , x = 11 . c answer ." | a = 26 + 18
b = 3 + 1
c = a / b
|
a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 39 , d ) 4 / 27 , e ) 3 / 23 | c | divide(choose(5, 2), choose(add(add(5, 6), 2), 2)) | a bag contains 5 red , 6 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , 5 c 213 c 2 = 6 c 213 c 2 = 10 / 78 = 5 / 39 c" | a = math.comb(5, 2)
b = 5 + 6
c = b + 2
d = math.comb(c, 2)
e = a / d
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | floor(add(reminder(add(subtract(multiply(divide(47, const_100), 1442), multiply(divide(36, const_100), 1412)), 61), const_10), const_1)) | find the value of x . ( 47 % of 1442 - 36 % of 1412 ) + 61 = x ? | b ) 2 | a = 47 / 100
b = a * 1442
c = 36 / 100
d = c * 1412
e = b - d
f = e + 61
g = reminder + (
h = math.floor(g, 1)
|
a ) 50 , b ) 75 , c ) 69 , d ) 79 , e ) 82 | a | gcd(100, 50) | the maximum number of students among them 100 pens and 50 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 100 number of pencils = 50 required number of students = h . c . f . of 100 and 50 = 50 answer is a" | a = math.gcd(100, 50)
|
a ) 14 years , b ) 1.7 years , c ) 1.9 years , d ) 2.8 years , e ) none | d | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 95 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . β΄ required average = 266 / 95 = 2.8 years . answer d" | a = 35 * 16
b = a + 21
c = b + 35
d = 35 * 16
e = c - d
|
a ) 52.2 mph , b ) 61.9 mph , c ) 44.4 mph , d ) 35.7 mph , e ) 65.6 mph | c | divide(add(divide(200, 5), divide(200, 4)), const_2) | rahul travels from a to b a distance of 200 miles in 5 hours . he returns to a in 4 hours . find his average speed ? | speed from a to b = 200 / 5 = 40 mph speed from b to a = 200 / 4 = 50 mph average speed = 2 * 40 * 50 / 90 = 44.4 mph answer is c | a = 200 / 5
b = 200 / 4
c = a + b
d = c / 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(14, subtract(14, 5)) | if 5 x * x + 14 * x + k = 0 having two reciprocal roots find the value of k . | product of reciprocal roots = 1 product of roots of quadratic eqn ax ^ 2 + bx + c = 0 is c / a . hence k / 5 = 1 k = 5 answer : e | a = 14 - 5
b = 14 - a
|
a ) 1600 , b ) 2998 , c ) 2778 , d ) 2788 , e ) 2991 | a | divide(multiply(400, const_100), subtract(add(const_100, 5), subtract(const_100, 15))) | a watch was sold at a loss of 15 % . if it was sold for rs . 400 more , there would have been a gain of 5 % . what is the cost price ? | "85 % 120 % - - - - - - - - 25 % - - - - 400 100 % - - - - ? = > rs . 1600 answer : a" | a = 400 * 100
b = 100 + 5
c = 100 - 15
d = b - c
e = a / d
|
a ) 154,889 , b ) 196,889 , c ) 238,889 , d ) 250,889 , e ) 264,889 | c | add(add(add(subtract(const_10, 1), multiply(subtract(const_100, const_10), const_2)), multiply(subtract(const_100, const_10), const_3)), multiply(subtract(const_100, const_10), const_4)) | meena wrote all the numbers from 1 to 49,999 inclusive . how many digits did she write in total ? | "1 - 9 = > 1 * 9 digits 10 - 99 = > 2 * 90 = 180 ( numbers between 10 - 99 is 90 where each has 2 digits ) 100 - 999 = > 3 * 900 = 2700 1000 - 9999 = > 4 * 9000 = 36,000 10000 - 49999 = > 5 * 40,000 = 200,000 the answer is 238,889 the answer is c ." | a = 10 - 1
b = 100 - 10
c = b * 2
d = a + c
e = 100 - 10
f = e * 3
g = d + f
h = 100 - 10
i = h * 4
j = g + i
|
a ) 19 , b ) 17 , c ) 13 , d ) 10 , e ) 12 | e | multiply(multiply(divide(7, 5), divide(12, 7)), 5) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 12 . if p invested the money for 5 months , find for how much time did q invest the money ? | "7 * 5 : 5 * x = 7 : 12 x = 12 answer : e" | a = 7 / 5
b = 12 / 7
c = a * b
d = c * 5
|
a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 400 | d | divide(multiply(770, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 770 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 770 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 770 Γ’ β‘ β y [ 12 / 10 + 1 ] = 770 Γ’ β‘ β 22 y / 10 = 770 Γ’ β‘ β 22 y = 7700 Γ’ β‘ β y = 700 / 22 = 700 / 2 = rs . 350 d )" | a = 1 + 4
b = a * 2
c = 770 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) 48 , b ) 52 , c ) 66 , d ) 68 , e ) 84 | b | divide(factorial(subtract(add(const_4, 1), const_1)), multiply(factorial(1), factorial(subtract(const_4, const_1)))) | how many positive integers less than 300 can be formed using the numbers 1 , 2 , 3 and 5 for the digits ? | "notice that we can find the number of 2 and 3 digit numbers by just assuming the first digit can also be zero : 0 1 1 1 2 2 2 3 3 2 5 5 number of possibilities = 3 * 4 * 4 = 48 . then , just add up the number of 1 digits numbers = 4 , so total is 48 + 4 = 52 . answer : b" | a = 4 + 1
b = a - 1
c = math.factorial(b)
d = math.factorial(1)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) - 7 , b ) - 6 , c ) - 3 , d ) 1 , e ) 6 | b | divide(subtract(10, add(power(3, 2), multiply(3, 3))), 3) | if 3 is one solution of the equation x ^ 2 + 3 x + k = 10 , where k is a constant , what is the other solution ? | "the phrase β 3 is one solution of the equation β means that one value of x is 3 . thus , we first must plug 3 for x into the given equation to determine the value of k . so we have 3 ^ 2 + ( 3 ) ( 3 ) + k = 10 9 + 9 + k = 10 18 + k = 10 k = - 8 next we plug - 8 into the given equation for k and then solve for x . x ^ 2 + 3 x β 8 = 10 x ^ 2 + 3 x β 18 = 0 ( x + 6 ) ( x - 3 ) = 0 x = - 6 or x = 3 thus , - 6 is the other solution . answer b ." | a = 3 ** 2
b = 3 * 3
c = a + b
d = 10 - c
e = d / 3
|
a ) 22378 , b ) 279889 , c ) 2788 , d ) 9000 , e ) 2882 | d | divide(multiply(multiply(35,000, const_12), 3), multiply(subtract(const_12, 5), 2)) | a starts business with rs . 35,000 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b β s contribution in the capital ? | "explanation : ratio in which profit is to be divided = 2 : 3 assume that b ' s contribution to the capital = b β 3500 Γ 12 : b Γ 7 = 2 : 3 β 3500 Γ 12 / 7 b = 2 / 3 β b = ( 3500 Γ 12 Γ 3 ) / ( 2 Γ 7 ) = 500 Γ 6 Γ 3 = 9000 answer : d" | a = 35 * 0
b = a * 3
c = 12 - 5
d = c * 2
e = b / d
|
a ) $ 200 , b ) $ 300 , c ) $ 150 , d ) $ 450 , e ) $ 500 | e | multiply(divide(3, add(add(1, 3), 3)), 600) | a person want to give his money of $ 600 to his 3 children a , b , c in the ratio 1 : 2 : 3 . what is the b + c share ? | "b ' s share = 600 * 2 / 6 = $ 200 c ' s share = 600 * 3 / 6 = $ 300 b + c = $ 500 answer is e" | a = 1 + 3
b = a + 3
c = 3 / b
d = c * 600
|
a ) 4 , b ) 8 , c ) 12 , d ) 13 , e ) 18 | d | divide(power(7, const_2), 4) | if the area of a square with sides of length 7 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 7 ^ 2 = l * 4 = > l = 49 / 4 = 13 answer d" | a = 7 ** 2
b = a / 4
|
a ) 20 % , b ) 18 % , c ) 15 % , d ) 12 % , e ) 14 % | a | subtract(const_100, divide(multiply(add(const_100, 15), subtract(const_100, 30)), const_100)) | the tax on a commodity is diminished by 30 % but its consumption is increased by 15 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 70 * 115 = 8050 10000 - - - - - - - 1950 100 - - - - - - - ? = 20 % answer : a" | a = 100 + 15
b = 100 - 30
c = a * b
d = c / 100
e = 100 - d
|
a ) 8 sec , b ) 10 sec , c ) 12 sec , d ) 14 sec , e ) 16 sec | c | divide(add(137, 163), divide(multiply(add(48, 42), const_1000), const_3600)) | two trains 137 meters and 163 meters in length are running towards each other on parallel lines , one at the rate of 42 kmph and another at 48 kmpb . in what time will they be clear of each other from the moment they meet | explanation : relative speed of the trains = ( 42 + 48 ) kmph = 90 kmph = ( 90 Γ 5 / 18 ) m / sec = 25 m / sec . time taken by the trains to β pass each other = time taken to cover ( 137 + 163 ) m at 25 m / sec = ( 300 / 25 ) sec = 12 sec answer : option c | a = 137 + 163
b = 48 + 42
c = b * 1000
d = c / 3600
e = a / d
|
a ) 2 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | e | divide(subtract(17, const_1), const_2) | arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 17 from the product of their ages a year from today . what is their age today ? | "ad = ( a + 1 ) ( d + 1 ) - 17 0 = a + d - 16 a + d = 16 a = d ( as they are twin brothers ) a = d = 8 e is the answer" | a = 17 - 1
b = a / 2
|
a ) 22.5 % , b ) 31.5 % , c ) 37.5 % , d ) 42.5 % , e ) 46.5 % | c | multiply(divide(divide(subtract(30, add(divide(divide(const_100, add(const_2, const_3)), const_2), divide(divide(divide(const_100, add(const_2, const_3)), const_2), const_2))), const_2), divide(const_100, add(const_2, const_3))), const_100) | a bowl contains equal numbers of red , orange , green , blue , and yellow candies . kaz eats all of the green candies and half of the orange ones . next , he eats half of the remaining pieces of each color . finally , he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 30 % of the original number . what percent of the red candies remain ? | "let x be the original number of each color so there are a total of 5 x candies . kaz eats all of the green candies and half of the orange ones . there are 0 green candies and 0.5 x orange candies remaining . he eats half of the remaining pieces of each color . there are 0.25 x orange candies , and 0.5 x each of red , yellow , and blue candies . he eats red and yellow candies in equal proportions . orange + blue + red + yellow = 0.75 x + red + yellow = 1.5 x red + yellow = 0.75 x red = 0.375 x , since red = yellow . the answer is c ." | a = 2 + 3
b = 100 / a
c = b / 2
d = 2 + 3
e = 100 / d
f = e / 2
g = f / 2
h = c + g
i = 30 - h
j = i / 2
k = 2 + 3
l = 100 / k
m = j / l
n = m * 100
|
a ) 1.6 , b ) 1.7 , c ) 1.8 , d ) 1.9 , e ) 2.0 | a | divide(divide(multiply(10, 40), 40), const_2) | mixture a is 10 percent alcohol , and mixture b is 40 percent alcohol . if the two are poured together to create a 4 - gallon mixture that contains 20 percent alcohol , approximately how many gallons of mixture a are in the mixture ? | "( 40 - 20 ) / ( 20 - 15 ) = qa / qb 20 / 10 = qa / qb 2 / 1 = qa / qb qa = ( 2 / 5 ) * 4 = 8 / 5 = 1.6 approx answer : a" | a = 10 * 40
b = a / 40
c = b / 2
|
a ) 246 , b ) 650 , c ) 110 , d ) 240 , e ) 360 | c | multiply(power(add(const_1, divide(3, const_100)), 4), 100) | rs . 100 amounts to rs . 320 in 4 years at simple interest . if the interest is increased by 3 % , it would amount to how much ? | "( 100 * 4 * 3 ) / 100 = 12 100 + 12 = 112 answer : c" | a = 3 / 100
b = 1 + a
c = b ** 4
d = c * 100
|
a ) 1 / 5 , b ) 1 / 3 , c ) 3 / 7 , d ) 2 / 7 , e ) 3 / 7 | b | divide(const_0_25, multiply(const_3, const_0_25)) | a family has two children . find the probability that both the children are girls given that at least one of them is a girl ? | let b stand for boy and g for girl . the sample space of the experiment is s = { ( g , g ) , ( g , b ) , ( b , g ) , ( b , b ) } let e and f denote the following events : e : β both the children are girls β f : β at least one of the child is a girl β then e = { ( g , g ) } and f = { ( g , g ) , ( g , b ) , ( b , g ) } now e n f = { ( g , g ) } thus p ( f ) = 3 / 4 and p ( e n f ) = 1 / 4 therefore p ( e | f ) = p ( e β© f ) / p ( f ) = ( 1 / 4 ) / ( 3 / 4 ) = 1 / 3 b ) | a = 3 * const_0_25
b = const_0_25 / a
|
a ) 12.4 sec , b ) 24.3 sec , c ) 62.4 sec , d ) 60.1 sec , e ) none | c | multiply(multiply(650, inverse(multiply(add(45, 30), const_0_2778))), const_2) | two good train each 650 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 . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 650 + 650 ) m = 1300 m . required time = ( 1300 x 6 / 125 ) sec = 62.4 sec . answer c" | a = 45 + 30
b = a * const_0_2778
c = 1/(b)
d = 650 * c
e = d * 2
|
a ) 1235 , b ) 1540 , c ) 1650 , d ) 2000 , e ) 2500 | e | subtract(negate(50), multiply(subtract(3,5, 7,9), divide(subtract(3,5, 7,9), subtract(1, 3,5)))) | 1 , 3,5 , 7,9 , . . 50 find term of sequnce for this ? | "this is an arithmetic progression , and we can write down a = 1 a = 1 , d = 2 d = 2 , n = 50 n = 50 . we now use the formula , so that sn = 12 n ( 2 a + ( n β 1 ) l ) sn = 12 n ( 2 a + ( n β 1 ) l ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 98 ) = 25 Γ ( 2 + 98 ) = 2500 = 2500 e" | a = negate - (
|
a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 95 | b | divide(130, const_2) | he total marks obtained by a student in physics , chemistry and mathematics is 130 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ? | "let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 130 + p c + m = 130 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 130 / 2 = 65 . answer : b" | a = 130 / 2
|
a ) 2 % decrease , b ) 8 % increase , c ) 9 % decrease , d ) 3 % decrease , e ) 2 % decrease | b | subtract(const_100, multiply(multiply(add(const_1, divide(20, 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 20 % . the effect on revenue is ? | "100 * 100 = 10000 90 * 120 = 10800 - - - - - - - - - - - 10000 - - - - - - - - - - - 800 100 - - - - - - - - - - - ? = > 8 % increase answer : b" | a = 20 / 100
b = 1 + a
c = 10 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 7620 , b ) 6200 , c ) 10020 , d ) 10080 , e ) 11000 | d | subtract(negate(240), multiply(subtract(12, 48), divide(subtract(12, 48), subtract(4, 12)))) | 4 , 12 , 48 , 240 , 1440 , ( . . . ) | "go on multiplying the given numbers by 3 , 4 , 5 , 6 , 7 4 x 3 = 12 ; 12 x 4 = 48 . . . . . 1440 x 7 = 10080 answer is d ." | a = negate - (
|
a ) 22 , b ) 120 , c ) 388 , d ) 377 , e ) 272 | b | multiply(subtract(1380, divide(multiply(9, 10000), const_100)), divide(const_1, const_4)) | a salesman gets commission on total sales at 9 % . if the sale is exceeded rs . 10000 he gets an additional commission as bonus of 3 % on the excess of sales over rs . 10000 . if he gets total commission of rs . 1380 , then the bonus he received is : | explanation : commission up to 10000 = \ inline 10000 \ times \ frac { 9 } { 100 } = 900 again after 10000 , commission : bonus 9 : 3 \ inline \ rightarrow 3 x : x \ therefore bonus = \ inline ( 1380 - 900 ) \ times \ frac { 1 } { 4 } = rs . 120 answer : b ) rs . 120 | a = 9 * 10000
b = a / 100
c = 1380 - b
d = 1 / 4
e = c * d
|
a ) 1 : 3 , b ) 4 : 3 , c ) 2 : 3 , d ) 2 : 1 , e ) 2 : 4 | c | divide(divide(const_1, multiply(add(20, const_2), const_10)), divide(const_1, multiply(20, const_10))) | a work can be finished in 20 days by 10 women . the same work can be finished in fifteen days by 20 men . the ratio between the capacity of a man and a woman is | "work done by 10 women in 1 day = 1 / 20 work done by 1 woman in 1 day = 1 / ( 10 Γ 20 ) work done by 20 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 Γ 20 ) ratio of the capacity of a man and woman = 1 / ( 15 Γ 20 ) : 1 / ( 10 Γ 20 ) = 1 / 15 : 1 / 10 = 1 / 3 : 1 / 2 = 2 : 3 option c" | a = 20 + 2
b = a * 10
c = 1 / b
d = 20 * 10
e = 1 / d
f = c / e
|
a ) 1752 , b ) 1799 , c ) 1719 , d ) 1750 , e ) 1800 | e | divide(divide(multiply(216, const_1000), divide(const_60, const_1)), const_2) | the length of a train and that of a platform are equal . if with a speed of 216 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 216 * 5 / 18 ] m / sec = 60 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 60 Γ£ Β¨ x = 60 * 60 / 2 = 1800 answer : e" | a = 216 * 1000
b = const_60 / 1
c = a / b
d = c / 2
|
a ) 1 β 3 , b ) 150 , c ) 80 , d ) 120 , e ) 180 | b | divide(multiply(2, const_60), divide(4, 5)) | if it takes a machine 4 β 5 minute to produce one item , how many items will it produce in 2 hours ? | 1 item takes 4 / 5 min so it takes 120 min to produce x 4 x / 5 = 120 the x = 150 answer : b | a = 2 * const_60
b = 4 / 5
c = a / b
|
a ) 34 , b ) 40 , c ) 68 , d ) 88 , e ) 32 | e | add(multiply(divide(120, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 120 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 120 sq . ft . so , b = 6 ft . length of fencing = ( l + 2 b ) = ( 20 + 12 ) ft = 32 ft . answer : e" | a = 120 / 20
b = a * 2
c = b + 20
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | what is the units digit of 31 ! + 50 ! + 2 ! + 3 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 2 + 6 = 8 the units digit is 8 . the answer is e ." | a = 4 + 3
b = a + 2
|
a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | c | add(const_3, const_4) | what is the smallest positive integer x such that 520 - x is the cube of a positive integer | "given 520 - x is a perfect cube so we will take 512 = 8 * 8 * 8 520 - x = 512 x = 520 - 512 = 8 correct option is c" | a = 3 + 4
|
a ) 20 , b ) 32 , c ) 100 , d ) 200 , e ) 400 | b | add(power(multiply(2, 2), const_2), power(multiply(2, 2), const_2)) | if equation | x / 2 | + | y / 2 | = 2 encloses a certain region on the coordinate plane , what is the area of this region ? | the equation can be reduced to intercept form as | x / 4 | + | y / 4 | = 1 , so these are lines in four quadrants with x and y intercept as 14 , so it is a rhombus with diagonals of 8 each and hence area = 1 / 2 * d 1 * d 2 = 1 / 2 * 8 * 8 = 32 . answer b . | a = 2 * 2
b = a ** 2
c = 2 * 2
d = c ** 2
e = b + d
|
a ) 700 cm 2 , b ) 600 cm 2 , c ) 500 cm 2 , d ) 400 cm 2 , e ) 100 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 25), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(25, 25)))), 25) | find the area of a rhombus one side of which measures 20 cm and one diagonal is 25 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 25 x 32 ) cm 2 = 400 cm 2 answer : option d" | a = 1 / 2
b = a * 25
c = 20 * 20
d = c * 4
e = 25 * 25
f = d - e
g = math.sqrt(f)
h = b * g
i = h + 25
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.