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 ) 229 , b ) 787 , c ) 152 , d ) 128 , e ) 121 | c | multiply(add(const_1, const_4), 38) | one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill tank in 38 min , then the slower pipe alone will be able to fill the tank in ? | "let the slower pipe alone fill the tank in x min . then , faster pipe will fill it in x / 3 min . 1 / x + 3 / x = 1 / 38 4 / x = 1 / 38 = > x = 152 min . answer : c" | a = 1 + 4
b = a * 38
|
a ) 0.75 , b ) 1 , c ) 0.25 , d ) 0.5 , e ) 0.58 | c | divide(multiply(divide(1, const_2), divide(1, const_2)), multiply(const_1, const_1)) | a circular dartboard of radius 1 foot is at a distance of 20 feet from you . you throw a dart at it and it hits the dartboard at some point q in the circle . what is the probability that q is closer to the center of the circle than the periphery ? | radius of inner dartboard / overall radius of dartboard ( ( r / 2 ) ^ 2 ) / ( r ^ 2 ) = 1 / 4 = 0.25 answer : c | a = 1 / 2
b = 1 / 2
c = a * b
d = 1 * 1
e = c / d
|
a ) 2 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | b | subtract(25, 20) | a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 76 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | "we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 76 / 25 = 3 . . . . so the max is 3 76 / 20 = 3 . . . so the max is 3 76 - 75 = 1 answer : b" | a = 25 - 20
|
a ) 200 , b ) 250 , c ) 300 , d ) 350 , e ) 400 | a | subtract(add(300, 600), 700) | a , b and c have rs . 700 between them , a and c together have rs . 300 and b and c rs . 600 . how much does c have ? | "a + b + c = 700 a + c = 300 b + c = 600 - - - - - - - - - - - - - - a + b + 2 c = 900 a + b + c = 700 - - - - - - - - - - - - - - - - c = 200 answer : a" | a = 300 + 600
b = a - 700
|
a ) 14 , b ) 10 , c ) 21 , d ) 20 , e ) 25 | a | add(inverse(subtract(divide(const_1, 7), divide(const_1, 14))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 7 days . if a alone can do it in 14 days . in how many days can b alone do it ? | "a 14 1 / 7 Γ’ β¬ β 1 / 14 = 1 / 14 = > 14" | a = 1 / 7
b = 1 / 14
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 8 pm , b ) 6 am , c ) 7 am , d ) 10 am , e ) 8 am | e | add(7, divide(25, add(12, 13))) | a and b start walking towards each other at 7 am at speed of 12 kmph and 13 kmph . they were initially 25 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 25 / 13 + 12 = 25 / 25 = 1 hr after 7 pm = 8 am answer is e" | a = 12 + 13
b = 25 / a
c = 7 + b
|
a ) 360 , b ) 504 , c ) 510 , d ) 320 , e ) 280 | b | lcm(lcm(24, 36), 42) | find the lowest common multiple of 24 , 36 and 42 . | "lcm = 2 * 2 * 2 * 3 * 3 * 7 = 504 . answer is b" | a = math.lcm(24, 36)
b = math.lcm(a, 42)
|
a ) 72 / 21 , b ) 27 / 85 , c ) 34 / 47 , d ) 13 / 81 , e ) 16 / 21 | e | divide(multiply(7, 7), multiply(3, multiply(7, 2))) | if a / b = 1 / 7 , b / c = 3 , c / d = 2 / 3 , d / e = 2 and e / f = 1 / 2 , then what is the value of abc / def ? | "say a = 1 . then : a / b = 1 / 7 - - > b = 7 ; b / c = 3 - - > c = 7 / 3 ; c / d = 2 / 3 - - > d = 7 / 2 ; d / e = 2 - - > e = 7 / 4 ; e / f = 1 / 2 - - > f = 7 / 2 . abc / def = ( 1 * 7 * 7 / 3 ) / ( 7 / 2 * 7 / 4 * 7 / 2 ) = 16 / 21 . answer : e ." | a = 7 * 7
b = 7 * 2
c = 3 * b
d = a / c
|
a ) 4 , b ) 5 , c ) 2 , d ) 6 , e ) 7 | c | divide(divide(divide(lcm(43, 87), 87), const_4), const_4) | what is the least value of x . so that 43 x 87 is divisible by 3 ? | "the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 4 + 3 + x + 8 + 7 = 22 + x least value of x may be 2 therefore 22 + 2 = 24 is divisible by 3 . c" | a = math.lcm(43, 87)
b = a / 87
c = b / 4
d = c / 4
|
a ) a ) 443.173 , b ) b ) 752.804 , c ) c ) 714.642 , d ) d ) 629.906 , e ) of these | a | subtract(multiply(divide(675.987, const_100), 56.84), multiply(divide(const_1, const_3), multiply(divide(675.987, const_100), 56.84))) | 675.987 - ? + 56.84 = 289.654 | "explanation : 443.173 answer : option a" | a = 675 / 987
b = a * 56
c = 1 / 3
d = 675 / 987
e = d * 56
f = c * e
g = b - f
|
a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 5 | c | divide(800, 600) | if shares of two persons in profits are rs . 800 and rs . 600 then ratio of their capitals is | "total profit = 1000 ratio = 800 / 600 = 4 : 3 answer : c" | a = 800 / 600
|
a ) 5 , b ) 3 , c ) 1 , d ) 4 , e ) 2 | e | add(add(divide(7, 10), divide(4, 5)), divide(1, 2)) | if 7 / 10 of a pencil is green , 4 / 5 of the remaining is gold and the remaining 1 / 2 is white , what is the total length of the pencil ? | green is 7 / 10 gold is 4 / 5 which can also be written as 8 / 10 white is 1 / 2 which can also be written as 5 / 10 7 / 10 + 8 / 10 + 5 / 10 = 2 answer is e ) 2 | a = 7 / 10
b = 4 / 5
c = a + b
d = 1 / 2
e = c + d
|
a ) 30 days , b ) 31 days , c ) 33 days , d ) 41 days , e ) 52 days | c | divide(multiply(add(add(divide(lcm(const_2, const_3), const_1), divide(lcm(const_2, const_3), const_2)), divide(lcm(const_2, const_3), const_3)), 18), lcm(const_2, const_3)) | a takes twice as much time as b & c takes thrice as much time as b to finish a piece of work . working together they can finish the work in 18 days . the no . of days need for b to do the work alone is ? | let a takes x days to complete the work , then b takes x / 2 days and c takes 3 x / 2 days to complete the work . a , b and c together complete the work in 18 days in 1 day , all they together can do = > 1 / x + 2 / x + 2 / 3 x = 1 / 18 3 + 6 + 2 / 3 x = 1 / 18 3 x 23 x 2 = 18 x = 66 days the number of days b alone takes to finish the work = x / 2 = 33 days c | a = math.lcm(2, 3)
b = a / 1
c = math.lcm(2, 3)
d = c / 2
e = b + d
f = math.lcm(2, 3)
g = f / 3
h = e + g
i = h * 18
j = math.lcm(2, 3)
k = i / j
|
a ) 1 kmph , b ) 7 kmph , c ) 6 kmph , d ) 5 kmph , e ) 4 kmph | c | divide(subtract(20, 8), const_2) | a man can row his boat with the stream at 20 km / h and against the stream in 8 km / h . the man ' s rate is ? | "ds = 20 us = 8 s = ? s = ( 20 - 8 ) / 2 = 6 kmph answer : c" | a = 20 - 8
b = a / 2
|
a ) rs 10123.20 , b ) rs 10123.30 , c ) rs 10123.40 , d ) rs 10123.50 , e ) none of these | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on rs . 25000 after 3 years at the rate of 12 % per annum | "explanation : ( 25000 Γ ( 1 + 12 / 100 ) 3 ) = > 25000 Γ 28 / 25 Γ 28 / 25 Γ 28 / 25 = > 35123.20 so compound interest will be 35123.20 - 25000 = rs 10123.20 option a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | divide(70, add(3, 4)) | a certain airline ' s fleet consisted of 70 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ? | "let x be the number of years . 4 x > 70 - 3 x 7 x > 70 x > 10 the answer is e ." | a = 3 + 4
b = 70 / a
|
['a ) 2898 sq m', 'b ) 1788 sq m', 'c ) 4312 sq m', 'd ) 7656 sq m', 'e ) 1786 sq m'] | c | subtract(circle_area(divide(352, multiply(const_2, const_pi))), circle_area(divide(264, multiply(const_2, const_pi)))) | the circumferences of two circles are 264 meters and 352 meters . find the difference between the areas of the larger and the smaller circles ? | let the radii of the smaller and the larger circles be s m and l m respectively . 2 β s = 264 and 2 β l = 352 s = 264 / 2 β and l = 352 / 2 β difference between the areas = β l 2 - β s 2 = β { 1762 / β 2 - 1322 / β 2 } = 1762 / β - 1322 / β = ( 176 - 132 ) ( 176 + 132 ) / β = ( 44 ) ( 308 ) / ( 22 / 7 ) = ( 2 ) ( 308 ) ( 7 ) = 4312 sq m answer : c | a = 2 * math.pi
b = 352 / a
c = circle_area - (
|
a ) s . 12 , b ) s . 15 , c ) s . 18 , d ) s . 21 , e ) s . 45 | e | multiply(divide(const_100, 12), multiply(divide(9, const_100), 60)) | a man buys rs . 60 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | "dividend on rs . 60 = rs . 9 / 100 x 60 = rs . 5.4 . rs . 12 is an income on rs . 100 . rs . 5.4 is an income on rs . 100 / 12 x 5.4 = rs . 45 . answer : option e" | a = 100 / 12
b = 9 / 100
c = b * 60
d = a * c
|
a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 17 | e | divide(53, 3) | a number of 53 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 16 * 3 + 1 * 5 = 53 - - > 16 + 1 = 17 . answer : e ." | a = 53 / 3
|
a ) 80 yards , b ) 82 yards , c ) 84 yards , d ) 85 yards , e ) 90 yards | b | multiply(subtract(25, divide(subtract(power(25, const_2), power(20, const_2)), multiply(25, const_2))), const_4) | the rhombus ( afce ) is inscribed in rectangle ( abcd ) . the length of bf = de . if the rectangle has a width of 20 yards and a length of 25 yards , what would be the total length ( the perimeter ) of a fence along the sides defined by afce ? | the definition of a rhombus states that every side is of equal length . this definition is the key to solving this question . 1 ) we state that the hypotenuse of either side triangle is x . for this solution , i will be working with the left triangle . 2 ) we state that the bottom and top sides of the rhombus are also x , due to the definition of a rhombus . 3 ) we will define the bottom of the left triangle as y . that is , for the rectangle bottom itself , we can define it as 25 = y + x , where x is a side for the rhombus and y is the bottom part of the left triangle . 4 ) from ( 1 ) , we know that x = 202 + y 2 β β β β β β β β 202 + y 2 . 5 ) from ( 3 ) , we know x = 25 - y . 6 ) from ( 4 ) and ( 5 ) , we have the equation 202 + y 2 β β β β β β β β = 25 β y 202 + y 2 = 25 β y 7 ) solving for y , we get 4.5 . 8 ) 25 - 4.5 = 20.5 = x 9 ) since all sides of the rhombus are the same , 20.5 * 4 = 41 * 2 = 82 . the answer is b , 82 | a = 25 ** 2
b = 20 ** 2
c = a - b
d = 25 * 2
e = c / d
f = 25 - e
g = f * 4
|
a ) 2.04 % , b ) 6.12 % , c ) 8 % , d ) 8.25 % , e ) 10 % | e | multiply(2.4, const_4) | on the first of the year , james invested x dollars at proudstar bank in an account that yields 2.4 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ? | "if the interest were compounded annually instead of quarterly then in one year the interest would be 2.4 * 4 = 9.6 % . now , since the interest is compounded quarterly then there would be interest earned on interest ( very small amount ) thus the actual interest would be a little bit more than 9.6 % , answer : e" | a = 2 * 4
|
a ) 12 days , b ) 16 days , c ) 15 days , d ) 17 days , e ) 18 days | c | divide(const_1, add(divide(const_1, 60), divide(const_1, 20))) | a can do a piece of work 60 days . b can do work in 20 days . in how many days they will complete the work together ? | "lcm = 60 , ratio = 60 : 20 = 3 : 1 no of days = 60 / ( 3 + 1 ) = 90 / 4 = 15 days answer : c" | a = 1 / 60
b = 1 / 20
c = a + b
d = 1 / c
|
a ) 10 : 20 , b ) 2 : 4 , c ) 1 : 2 , d ) 4 : 8 , e ) it can not be determined from the information given | c | divide(10, 20) | if there are 10 apples and 20 oranges , what is the ratio of apples to oranges ? | a ratio should be reduced to the lowest common denominator . 10 : 20 means 10 x : 20 x which can be written as 1 ( 10 ) : 2 ( 10 ) . . . answer : c | a = 10 / 20
|
a ) rs . 500 , b ) rs . 600 , c ) rs . 700 , d ) rs . 1500 , e ) rs . 2500 | a | subtract(multiply(8000, const_4), subtract(multiply(9500, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 9500 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 9500 = 38000 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 6000 salary of may is rs . 6500 salary of january = rs . 500 answer : a" | a = 8000 * 4
b = 9500 * 4
c = b - 6500
d = a - c
|
a ) 94 , b ) 99 , c ) 26 , d ) 73 , e ) none of the above | a | add(multiply(const_10, add(subtract(10, 5), const_1)), 5) | what two - digit number is less than the sum of the square of its digits by 10 and exceeds their doubled product by 5 ? | "let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 10 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 10 x ^ 2 + y ^ 2 - 2 xy = 15 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . . thus the two numbers can be 15 or 94 thus the answer is a" | a = 10 - 5
b = a + 1
c = 10 * b
d = c + 5
|
a ) 75 % , b ) 72 % , c ) 70 % , d ) 73 % , e ) 74 % | c | multiply(divide(multiply(divide(20, const_100), subtract(const_1, divide(30, const_100))), divide(20, const_100)), const_100) | last week david spent 20 percent of his wages on recreation . this week , his wages are 30 percent less than last week ΚΌ s wages and he spent 20 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week | "say david ' s wages last week were $ 100 , so he spent 0.20 * 100 = $ 20 on recreation ; this week ' s wages is 0.70 * 100 = $ 70 , so he spends 0.2 * 70 = $ 14 on recreation ; 14 / 20 = 0.70 , hence the amount he spends on recreation this week is 70 % of the amount he spent on recreation last week : 20 * 0.7 = 14 answer : c" | a = 20 / 100
b = 30 / 100
c = 1 - b
d = a * c
e = 20 / 100
f = d / e
g = f * 100
|
a ) 450 , b ) 570 , c ) 480 , d ) 520 , e ) 550 | a | divide(180, divide(subtract(70, subtract(const_100, 70)), const_100)) | in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 180 votes . find the total number of valid votes ? | "let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 180 = > 4 x / 10 = 180 = > 4 x = 1800 = > x = 450 . answer : a" | a = 100 - 70
b = 70 - a
c = b / 100
d = 180 / c
|
a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | d | add(add(multiply(divide(multiply(2, 10), subtract(multiply(3, 5), multiply(2, 4))), 4), 10), multiply(divide(multiply(2, 10), subtract(multiply(3, 5), multiply(2, 4))), 3)) | in a can , there is a mixture of milk and water in the ratio 4 : 3 . if the can is filled with an additional 10 liters of milk , the can would be full and the ratio of milk and water would become 5 : 2 . find the capacity of the can ? | let c be the capacity of the can . ( 4 / 7 ) * ( c - 10 ) + 10 = ( 5 / 7 ) * c 4 c - 40 + 70 = 5 c c = 30 the answer is d . | a = 2 * 10
b = 3 * 5
c = 2 * 4
d = b - c
e = a / d
f = e * 4
g = f + 10
h = 2 * 10
i = 3 * 5
j = 2 * 4
k = i - j
l = h / k
m = l * 3
n = g + m
|
a ) 8 hours , b ) 6 hours , c ) 9 hours , d ) 7 hours 12 minutes , e ) none | b | multiply(90, divide(const_3, divide(90, const_2))) | a car runs at the speed of 55 km per hour when not serviced and runs at 90 kmph when serviced . after servicing the car covers a certain distance in 3 hours . how much time will the car take to cover the same distance when not serviced ? | explanation : time = 90 * 3 / 45 = 6 hours answer Γ’ β¬ β b | a = 90 / 2
b = 3 / a
c = 90 * b
|
a ) 18 kmph , b ) 6 kmph , c ) 8 kmph , d ) 10 kmph , e ) 12 kmph | c | subtract(24, divide(multiply(24, const_2), const_3)) | the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 24 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 24 / 3 = 8 kmph . answer : c" | a = 24 * 2
b = a / 3
c = 24 - b
|
a ) 1.5 , b ) 6.5 , c ) 7.5 , d ) 8 , e ) 2 | c | subtract(multiply(const_4, const_3), divide(divide(multiply(27000, multiply(const_4, const_3)), 36000), 2)) | a began business with rs . 27000 and was joined afterwards by b with rs . 36000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | "27 * 12 : 36 * x = 2 : 1 x = 4.5 12 - 4.5 = 7.5 answer : c" | a = 4 * 3
b = 4 * 3
c = 27000 * b
d = c / 36000
e = d / 2
f = a - e
|
a ) 1 / 5 , b ) 2 / 3 , c ) 3 / 7 , d ) 3 / 5 , e ) 4 / 5 | c | divide(3, add(const_3, const_4)) | a certain kind of fuel is 1515 high octane and 4545 regular octane . if high octane costs 3 times as much as regular octane , what fraction of the cost of the fuel is due to the high octane ? | let x be the amount of fuel high octane = x / 5 regular = 4 x / 5 price of regular = y price of high octante = 3 y cost of high octane in the fuel = 3 y * x / 5 total cost = 3 y * x / 5 + 4 x / 5 * y % = cost of high octante / total cost = 3 / 7 answer is c | a = 3 + 4
b = 3 / a
|
a ) 1 / 256 , b ) 1 / 32 , c ) 1 / 64 , d ) 1 / 128 , e ) 1 / 189 | b | divide(8, power(const_2, 8)) | 8 coins are tossed . whats is the probability that one and only one turns up head ? | explanation : possible outcomes ( sample space ) when 8 coins are tossed are : ( h , h , h , h , h , h , h , h ) , ( h , h , h , h , h , h , h , t ) , ( h , h , h , h , h , h , t , h ) , ( h , h , h , h , h , h , t , t ) , . . . . . . . ( t , t , t , t , t , t , t , t ) = 28 required outcomes : ( h , t , t , t , t , t , t , t ) , ( t , h , t , t , t , t , t , t ) , ( t , t , h , t , t , t , t , t ) , ( t , t , t , h , t , t , t , t ) , ( t , t , t , t , h , t , t , t ) , ( t , t , t , t , t , h , t , t ) , ( t , t , t , t , t , t , h , t ) , ( t , t , t , t , t , t , t , h ) = 8 so the required probability = 8 / 28 = 1 / 32 answer : b | a = 2 ** 8
b = 8 / a
|
a ) s . 3988 , b ) s . 4800 , c ) s . 3228 , d ) s . 3922 , e ) s . 3928 | b | multiply(multiply(subtract(add(110, 60), 10), 10), 3) | a rectangular lawn of dimensions 110 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "area = ( l + b β d ) d ( 110 + 60 β 10 ) 10 = > 1600 m 2 1600 * 3 = rs . 4800 answer : b" | a = 110 + 60
b = a - 10
c = b * 10
d = c * 3
|
a ) 2 , b ) 6 , c ) 1 , d ) 0 , e ) 7 | d | multiply(const_1, subtract(const_1, const_1)) | if you multiply all the numbers on your mobile phone , what is the answer ? | d 0 their is a zero in your phone . | a = 1 - 1
b = 1 * a
|
a ) 10.2 , b ) 12.1 , c ) 8.1 , d ) 15.6 , e ) 14.4 | e | add(12, multiply(divide(20, const_100), 12)) | if x is 20 percent greater than 12 , then x = | "x is 20 % greater than 12 means x is 1.2 times 12 ( in other words 12 + 20 / 100 * 12 = 1.2 * 12 ) therefore , x = 1.2 * 12 = 14.4 answer : e" | a = 20 / 100
b = a * 12
c = 12 + b
|
a ) 11 / 3 , b ) 11 / 2 , c ) 242 , d ) 3 ^ 10 , e ) 7 ^ 11 - 1 | e | subtract(power(7, add(9, const_2)), const_1) | if 7 ^ k + 7 ^ k = ( 7 ^ 9 ) ^ ( 7 ^ 9 ) - 7 ^ k , then k = ? | "7 ^ k + 7 ^ k = ( 7 ^ 9 ) ^ 7 ^ 9 - 7 ^ k 7 * ( 7 ^ k ) = 7 ^ ( 49 * 7 ^ 9 ) = 7 ^ ( 7 ^ 2 * 7 ^ 9 ) = 7 ^ ( 7 ^ 11 ) 7 ^ k + 1 = 7 ^ ( 7 ^ 11 ) so k + 1 = 7 ^ 11 so k = 7 ^ 11 - 1 answer is e" | a = 9 + 2
b = 7 ** a
c = b - 1
|
a ) 0.5 , b ) 0.4 , c ) 0.3 , d ) 0.2 , e ) 0.1 | d | divide(multiply(factorial(subtract(10, const_1)), factorial(const_2)), factorial(10)) | find the probability in a group of 10 students sitting in a row , two of them always sit next to each other | assume that these 2 students are 1 and then u can sit is 9 ! and these 2 students can sit by 2 ways then total ways are 9 ! * 2 and the probability is 9 ! * 2 / 10 ! = 1 / 5 = 0.2 answer : d | a = 10 - 1
b = math.factorial(a)
c = math.factorial(2)
d = b * c
e = math.factorial(10)
f = d / e
|
a ) 66 % , b ) 67 % , c ) 68 % , d ) 69 % , e ) 70 % | e | subtract(multiply(const_3, 75), add(74, 81)) | a student got 74 % in math and 81 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | "74 + 81 + x = 3 * 75 x = 70 the answer is e ." | a = 3 * 75
b = 74 + 81
c = a - b
|
a ) 7580 , b ) 7960 , c ) 8290 , d ) 1440 , e ) none | d | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(3, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 3 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 3 - 1 / 12 ) = 1 / 4 work done by the inlet in 1 min . = ( 1 / 4 Γ 1 / 60 ) = 0.004167 volume of 0.004167 part = 6 litres . therefore , volume of whole = 0.004167 x 6 βΉ = βΊ 1440 litres . answer d" | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 3 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 20 , b ) 24 , c ) 52 , d ) 96 , e ) 60 | d | sqrt(divide(8, add(power(6, 4), add(power(2, 4), power(4, 4))))) | the ratio of 2 numbers is 4 : 6 and their h . c . f . is 8 . their l . c . m . is ? | "let the numbers be 4 x and 6 x their h . c . f . = 8 so the numbers are 4 * 8 , 6 * 8 = 32,48 l . c . m . = 96 answer is d" | a = 6 ** 4
b = 2 ** 4
c = 4 ** 4
d = b + c
e = a + d
f = 8 / e
g = math.sqrt(f)
|
a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | b | subtract(35, reminder(5, 7)) | when positive integer n is divided by 4 , 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 35 ? | "when positive integer n is divided by 4 , the remainder is 1 i . e . , n = 4 x + 1 values of n can be one of { 1 , 5 , 9 , 13 , 17 , . . . . . . . . . . . 29 , 33 , 37 . . . . . . . . . . . . . . . . . . } 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 , . . . . . . . . } combining both the sets we get n = { 5,33 , . . . . . . . . . . . } what is the smallest positive integer p , such that ( n + p ) is a multiple of 35 or 35 x in case of n = 33 p = 2 so for min value of p , we take min value of n . b is the answer ." | a = 35 - reminder
|
a ) 13230.0 , b ) 12300.02 , c ) 16537.5 , d ) 15000.0 , e ) 16537.11 | a | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 12000) | sam invested rs . 12000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 12000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 12000 * ( 1 + 5 / 100 ) 2 ] = ( 12000 * 21 / 20 * 21 / 20 ) = rs . 13230.00 answer : a" | a = 10 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 12000
|
a ) $ 32 , b ) $ 34.31 , c ) $ 28.44 , d ) $ 67.54 , e ) $ 65.23 | a | divide(multiply(multiply(const_100, const_100), 18), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 18 , what was the price before the first discount ? | let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 18 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 18 solve for x x = $ 32 correct answer a | a = 100 * 100
b = a * 18
c = 100 - 25
d = c * 100
e = 100 - 25
f = e * 25
g = d - f
h = b / g
|
a ) 50 min , b ) 40 min , c ) 35 min , d ) 25 min , e ) 36 min | e | add(20, 16) | the jogging track in a sports complex is 1000 meters in circumference . deepak and his wife start from the same point and walk in opposite directions at 20 km / hr and 16 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 1000 m apart to be 20 + 16 = 36 km apart , they take 1 hour to be 1000 m apart , they take 36 * 1000 / 1000 = 36 min . answer is e" | a = 20 + 16
|
a ) 2.5 cm , b ) 3.6 cm , c ) 5 cm , d ) 6.43 cm , e ) 7.56 cm | a | divide(volume_cube(5), multiply(10, 5)) | a cube of edge 5 cm is immersed completely in a rectangular vessel containing water . if the dimensions of the base of vessel are 10 cm * 5 cm , find the rise in water level ? | "increase in volume = volume of the cube = 5 * 5 * 5 cm ^ 3 rise in water level = volume / area = 5 * 5 * 5 / 10 * 5 = 2.5 cm answer is a" | a = volume_cube / (
|
a ) 2 / 69 , b ) 2 / 15 , c ) 2 / 63 , d ) 2 / 29 , e ) 2 / 10 | b | multiply(divide(1, 3), divide(2, 5)) | two brother x and y appeared for an exam . the probability of selection of x is 1 / 3 and that of b is 2 / 5 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 3 , p ( b ) = 2 / 5 . let c be the event that both are selected . p ( c ) = p ( a ) Γ£ β p ( b ) as a and b are independent events : = ( 1 / 3 ) Γ£ β ( 2 / 5 ) = 2 / 15 answer : b ) 2 / 15" | a = 1 / 3
b = 2 / 5
c = a * b
|
a ) 18 : 19 , b ) 17 : 19 , c ) 15 : 19 , d ) 18 : 14 , e ) 18 : 11 | a | divide(subtract(multiply(12, 4), multiply(6, 5)), subtract(multiply(7, 5), multiply(4, 4))) | 4 men and 12 boys finish a piece of work in 4 days , 7 men and 6 boys do it in 5 days . the ratio between the efficiencies of a man and boy is ? | "4 m + 12 b - - - - - 4 days 7 m + 6 b - - - - - - - 5 days 16 m + 48 b = 35 m + 30 b 18 b = 19 m = > 5 m = 6 b m : b = 18 : 19 answer : a" | a = 12 * 4
b = 6 * 5
c = a - b
d = 7 * 5
e = 4 * 4
f = d - e
g = c / f
|
a ) 500 , b ) 400 , c ) 300 , d ) 200 , e ) 100 | c | sqrt(subtract(power(multiply(50, 10), const_2), power(400, const_2))) | a boatman trying to cross the river . the width of the river is 400 m . he is selling boat with 10 m / sec . he takes 50 sec to cross the river due to flow of water . calculate the drift . | the distance boatman travel to cross the river = 50 * 10 = 500 m he sell the boat diagonally . so height = 400 m , base be x and hypotenuse = 500 m so by pythagoras theorem , square of 400 + square of x = square of 500 so , square of x = 250000 - 160000 x = Γ’ Λ Ε‘ 250000 - 160000 , x = Γ’ Λ Ε‘ 90000 = 300 drift = 300 m answer c | a = 50 * 10
b = a ** 2
c = 400 ** 2
d = b - c
e = math.sqrt(d)
|
a ) 27 , b ) 36 , c ) 25 , d ) 64 , e ) 147 | c | multiply(divide(48, 112), 60) | the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 60 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 60 ounces of a substance has a volume of ( 48 / 112 ) * 60 = 25 cubic inches answer a alternatively , we can use estimation 112 ounces of a substance has a volume of 48 cubic inches 56 ounces of a substance has a volume of 24 cubic inches therefore , 60 will have a volume a little more than 24 , that is 25 c" | a = 48 / 112
b = a * 60
|
a ) 45 minutes , b ) 50 minutes , c ) 55 minutes , d ) 60 minutes , e ) none of these | a | subtract(add(multiply(subtract(add(1, multiply(add(const_4, const_2), const_2)), 9), const_60), 45), multiply(divide(348, 87), const_60)) | the average speed of a train is 87 kmph . the train was scheduled to start at 9 am in bangalore , and as per schedule it has to reach a destination of 348 km away from bangalore at 1 : 45 pm same day and a halt was scheduled onthe way . for how long was the halt scheduled ? | speed of train = 87 kmph and distance it covers is 348 km . time taken to cover the distance = distance / speed = 348 / 87 = 4 hours . but total time taken by train to reach destination is = 1 : 45 pm β 9 a . m = 4 hours 45 minutes . time of halt = total time β take taken without halt = 4 hours 45 minutes β 4 hours = 45 minutes a ) | a = 4 + 2
b = a * 2
c = 1 + b
d = c - 9
e = d * const_60
f = e + 45
g = 348 / 87
h = g * const_60
i = f - h
|
a ) 4 m , b ) 5 m , c ) 6 m , d ) 7 m , e ) 8 m | c | power(divide(10368, multiply(multiply(4, 3), 4)), divide(const_1, const_3)) | the height of the wall is 4 times its width and length of the wall is 3 times its height . if the volume of the wall be 10368 cu . m . its width is | "explanation : let width = x then , height = 4 x and length = 12 x 12 x Γ£ β 4 x Γ£ β x = 10368 x = 6 answer : c" | a = 4 * 3
b = a * 4
c = 10368 / b
d = 1 / 3
e = c ** d
|
a ) 6000 , b ) 8000 , c ) 8960 , d ) 2000 , e ) none of these | c | divide(multiply(multiply(24, const_100), multiply(14, const_100)), multiply(25, 15)) | a courtyard is 24 meter long and 14 meter board is to be paved with bricks of dimensions 25 cm by 15 cm . the total number of bricks required is : | "explanation : number of bricks = courtyard area / 1 brick area = ( 2400 Γ£ β 1400 / 25 Γ£ β 15 ) = 8960 option c" | a = 24 * 100
b = 14 * 100
c = a * b
d = 25 * 15
e = c / d
|
a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 4096 | e | power(3, add(multiply(3, 3), multiply(3, 3))) | when a = x + ( 3 / x ) and b = x - ( 3 / x ) , ( 2 ^ a ^ 2 ) / ( 2 ^ b ^ 2 ) = ? | "- - > ( 2 ^ a ^ 2 ) / ( 2 ^ b ^ 2 ) = { ( 2 ) ^ ( a ^ 2 - b ^ 2 ) } = 2 ^ ( a - b ) ( a + b ) . since a - b = 6 / x and a + b = 2 x , 2 ^ ( a - b ) ( a + b ) = 2 ^ ( 6 / x ) ( 2 x ) = 2 ^ 12 = 4096 therefore , the answer is e" | a = 3 * 3
b = 3 * 3
c = a + b
d = 3 ** c
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 70 % , e ) 90 % | a | multiply(20, const_1) | 20 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | to maximize the percentage of companies offering both , let ' s assume that all 20 % of companies which offer wireless internet also offer snacks . the answer is a . | a = 20 * 1
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | divide(180, multiply(const_0_2778, 54)) | in how many seconds will a train 180 meters long pass an oak tree , if the speed of the train is 54 km / hr ? | speed = 54 * 5 / 18 = 15 m / s time = 180 / 15 = 12 seconds the answer is b . | a = const_0_2778 * 54
b = 180 / a
|
a ) 63 , b ) 72 , c ) 144 , d ) 180 , e ) 400 | d | multiply(divide(multiply(40, const_3), 20), 30) | if 20 typists can type 40 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 40 letters , so 30 typists can type = 40 * 30 / 20 40 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 40 * 30 * 60 / 20 * 20 = 180 d is the answer" | a = 40 * 3
b = a / 20
c = b * 30
|
a ) 24 hrs , b ) 16 hrs , c ) 56 / 3 hrs , d ) 58 / 3 hrs , e ) 18 hrs | d | add(add(add(divide(300, 50), divide(subtract(300, multiply(divide(300, 50), 10)), add(50, 10))), divide(subtract(300, multiply(divide(300, 50), 10)), add(50, 10))), add(divide(subtract(300, subtract(subtract(300, multiply(divide(300, 50), 10)), multiply(10, 10))), add(50, 10)), divide(subtract(300, subtract(subtract(300, multiply(divide(300, 50), 10)), multiply(10, 10))), add(50, 10)))) | there are 4 people who has to cross a stretch of 300 km . they normally run at a speed of 10 kmph . one of them has a bike that travels at 50 kmph . the bike first takes one person alone and crosses the stretch while the other two keep running . then he comes back without wasting time and picks up another person from the way , drives him across the stretch , and does the same for the last person . how long does this whole process take ? | expl : time taken to carry 2 nd person = 300 / 50 = 6 hrs . time taken to meet 3 rd person = ( 300 - 6 * 10 ) / ( 50 + 10 ) = 4 hrs time taken to carry 3 rd person = 4 hours time taken to meet 4 th person = ( 300 β 140 ) / 60 = 8 / 3 total time = 6 + 4 + 4 + 8 / 3 + 8 / 3 = 58 / 3 hours answer : d | a = 300 / 50
b = 300 / 50
c = b * 10
d = 300 - c
e = 50 + 10
f = d / e
g = a + f
h = 300 / 50
i = h * 10
j = 300 - i
k = 50 + 10
l = j / k
m = g + l
n = 300 / 50
o = n * 10
p = 300 - o
q = 10 * 10
r = p - q
s = 300 - r
t = 50 + 10
u = s / t
v = 300 / 50
w = v * 10
x = 300 - w
y = 10 * 10
z = x - y
A = 300 - z
B = 50 + 10
C = A / B
D = u + C
E = m + D
|
a ) 16000 , b ) 18078 , c ) 11076 , d ) 11456 , e ) none of these | c | divide(multiply(multiply(18, const_100), multiply(12, const_100)), multiply(15, 13)) | a courtyard is 18 meter long and 12 meter board is to be paved with bricks of dimensions 15 cm by 13 cm . the total number of bricks required is : | "explanation : number of bricks = courtyard area / 1 brick area = ( 1800 Γ£ β 1200 / 15 Γ£ β 13 ) = 11076 option c" | a = 18 * 100
b = 12 * 100
c = a * b
d = 15 * 13
e = c / d
|
['a ) 12 y / 3', 'b ) 2 y / 2', 'c ) 6 y / 3', 'd ) 12 y / 4', 'e ) 4 y / 3'] | d | divide(square_area(multiply(4, const_3)), square_area(4)) | if one length of a square is tripled , and the original parameter is 4 y . what is the new length of one of the sides ? | if the original parameter is 4 y , each length of the square is 4 y / 4 . if the parameter is tripled , then each length becomes 3 ( 4 y ) / 4 . this is 12 y / 4 . answer option d . | a = 4 * 3
b = square_area / (
|
a ) 8 : 6 , b ) 8 : 7 , c ) 8 : 1 , d ) 8 : 2 , e ) 8 : 4 | b | divide(multiply(multiply(3, 2), const_4), multiply(7, 3)) | an order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the ? | "let the length and breadth of the carpet in the first case be 3 x units and 2 x units respectively . let the dimensions of the carpet in the second case be 7 y , 3 y units respectively . from the data , . 2 ( 3 x + 2 x ) = 2 ( 7 y + 3 y ) = > 5 x = 10 y = > x = 2 y required ratio of the areas of the carpet in both the cases = 3 x * 2 x : 7 y : 3 y = 6 x 2 : 21 y 2 = 6 * ( 2 y ) 2 : 21 y 2 = 6 * 4 y 2 : 21 y 2 = 8 : 7 answer : b" | a = 3 * 2
b = a * 4
c = 7 * 3
d = b / c
|
a ) 300 , b ) 288 , c ) 270 , d ) 100 , e ) 281 | d | divide(20, divide(20, const_100)) | 80 % of a number is added to 20 , the result is the same number . find the number ? | ( 80 / 100 ) * x + 20 = x 4 x + 100 = 5 x x = 100 answer : : d | a = 20 / 100
b = 20 / a
|
a ) 28 % , b ) 25 % , c ) 55 % , d ) 28 % , e ) 28.2 % | e | subtract(multiply(divide(const_100, 780), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | a dishonest dealer professes to sell goods at the cost price but uses a weight of 780 grams per kg , what is his percent ? | "780 - - - 220 100 - - - ? = > 28.2 % answer : e" | a = 100 / 780
b = 3 + 2
c = b * 2
d = 100 * c
e = a * d
f = e - 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 16 , e ) 30 | e | multiply(divide(subtract(840, divide(480, const_2)), 40), const_2) | angelina walked 840 meters from her home to the grocery at a constant speed . she then walked 480 meters to the gym at double the speed . she spent 40 seconds less on her way from the grocery to the gym than on her way from home to the grocery . what was angelina ' s speed , in meters per second , from the grocery to the gym ? | "let the speed be x . . . so time taken from home to grocery = 840 / x . . the speed to gym = 2 x . . so time taken = 480 / 2 x = 240 / x . . its given 840 / x - 240 / x = 40 . . 600 / x = 40 . . x = 15 m / secs . . so grocery to gym = 2 * 15 = 30 m / s . . . e" | a = 480 / 2
b = 840 - a
c = b / 40
d = c * 2
|
a ) a ) 87 , b ) b ) 99 , c ) c ) 68 , d ) d ) 82 , e ) e ) 76 | e | divide(add(add(add(add(74, 65), 82), 67), 90), add(const_2, const_3)) | david obtained 74 , 65 , 82 , 67 and 90 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 74 + 65 + 82 + 67 + 90 ) / 5 = 380 / 5 = 76 . answer : e" | a = 74 + 65
b = a + 82
c = b + 67
d = c + 90
e = 2 + 3
f = d / e
|
a ) 16 % , b ) 16.66 % , c ) 17.8 % , d ) 18 % , e ) 19.5 % | b | multiply(divide(subtract(70, 60), 60), const_100) | john makes $ 60 a week from his job . he earns a raise andnow makes $ 70 a week . what is the % increase ? | "increase = ( 10 / 60 ) * 100 = ( 1 / 6 ) * 100 = 16.66 % . b" | a = 70 - 60
b = a / 60
c = b * 100
|
a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many e acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn e = 4 * 94 = 376 correct option - c" | a = 1000 * 1
b = 10 * 3
c = b + 4
d = a + c
e = 5 + 2
f = e + 4
g = d / f
h = g * 4
|
a ) 5 β 7 , b ) 7 β 10 , c ) 1 β 3 , d ) 7 β 30 , e ) 5 β 70 | c | subtract(const_1, divide(2, 3)) | at an elementary school , 70 % of the faculty members are women and 60 % of the faculty members are married . if 2 β 3 of the men are single , what fraction of the men are married ? | "- - - - - - - - - - - - - - - - - - - - m - - - - - - w - - - - - - - - total marrried - - - - - - - - - - 10 - - - - - 50 - - - - - - - - - 60 not married - - - - - 20 - - - - - 20 - - - - - - - - - 40 total - - - - - - - - - - - - - 30 - - - - - 70 - - - - - - - - 100 need married man / total man , so 10 / 30 = 1 / 3 c" | a = 2 / 3
b = 1 - a
|
a ) 3 , b ) 5 , c ) 4 , d ) 6 , e ) 7 | c | subtract(5, reminder(4, 8)) | when n is divided by 24 , the remainder is 5 . what is the remainder when 4 n is divided by 8 ? | "let n = 5 ( leaves a remainder of 5 when divided by 24 ) 4 n = 4 ( 5 ) = 20 , which leaves a remainder of 4 when divided by 8 . answer c" | a = 5 - reminder
|
a ) 277 m , b ) 700 m , c ) 900 m , d ) 187 m , e ) 1678 m | c | divide(400, multiply(subtract(78, 1), const_0_2778)) | a train 400 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 400 + x ) / 60 = 65 / 3 x = 900 m . answer : c" | a = 78 - 1
b = a * const_0_2778
c = 400 / b
|
a ) 29.8 % , b ) 16 % , c ) 18 % , d ) 82 % , e ) 23 % | a | multiply(divide(subtract(64900, add(42000, 8000)), add(42000, 8000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 8000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 8000 = rs . 50000 and sp = rs . 64900 profit ( % ) = ( 64900 - 50000 ) / 50000 * 100 = 29.8 % answer : a" | a = 42000 + 8000
b = 64900 - a
c = 42000 + 8000
d = b / c
e = d * 100
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | e | divide(subtract(990, divide(990, add(divide(10, const_100), const_1))), 10) | machine p and machine q are each used to manufacture 990 sprockets . it takes machine p 10 hours longer to produce 990 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 990 / x = 990 / 1.1 x + 10 1.1 ( 990 ) = 990 + 11 x 11 x = 99 x = 9 the answer is e ." | a = 10 / 100
b = a + 1
c = 990 / b
d = 990 - c
e = d / 10
|
a ) 80 , b ) 90 , c ) 95 , d ) 105 , e ) 110 | a | subtract(100, subtract(power(2, divide(100, 40)), const_1)) | what is the greatest value of positive integer x such that 2 ^ x is a factor of 100 ^ 40 ? | "put in simple words , we need to find the highest power of 2 in 100 ^ 40 100 = 2 ^ 2 * 5 ^ 2 therefore 100 ^ 40 = ( 2 ^ 2 * 5 ^ 2 ) ^ 40 = 2 ^ 80 * 5 ^ 80 answer : a" | a = 100 / 40
b = 2 ** a
c = b - 1
d = 100 - c
|
a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(2, const_1) | if the product of 2 integers is negative , at most how many of the integers can be negative ? | "the product of 2 integers is negative thus an odd number of integers need to be negative to have a negative product we are asked at most how many are required . so , the highest odd integer before 6 , i . e . 1 correct option : a" | a = 2 - 1
|
a ) rs 62.10 , b ) rs 63.10 , c ) rs 64.10 , d ) rs 65.10 , e ) none of these | c | subtract(subtract(multiply(1000, power(add(divide(10, const_100), const_1), 4)), 1000), multiply(multiply(1000, divide(10, const_100)), 4)) | what will be the difference between simple and compound interest @ 10 % per annum on the sum of rs 1000 after 4 years | explanation : s . i . = 1000 β 10 β 4 / 100 = 400 c . i . = [ 1000 ( 1 + 10 / 100 ) 4 β 1000 ] = 464.10 so difference between simple interest and compound interest will be 464.10 - 400 = 64.10 option c | a = 10 / 100
b = a + 1
c = b ** 4
d = 1000 * c
e = d - 1000
f = 10 / 100
g = 1000 * f
h = g * 4
i = e - h
|
a ) 6 , b ) 7 , c ) 8 , d ) 4 , e ) 3 | d | divide(16, 4) | find k if 16 / k = 4 . | "since 16 / k = 4 and 16 / 4 = 4 , then k = 4 correct answer d" | a = 16 / 4
|
a ) rs . 962 , b ) rs . 672 , c ) rs . 546 , d ) rs . 876 , e ) none of these | d | multiply(divide(surface_cube(8), 16), 36.50) | the cost of the paint is rs . 36.50 per kg . if 1 kg of paint covers 16 sq . ft , how much will it cost to paint outside of a cube having 8 feet each side | "explanation : surface area of a cube = 6 x 82 = 384 sq . ft quantity of paint required = ( 384 / 16 ) = 24 kg cost of painting = 36.5 x 24 = rs . 876 answer : d" | a = surface_cube / (
b = a * 16
|
a ) 1960 , b ) 1965 , c ) 1970 , d ) 1975 , e ) 2040 | e | add(1950, add(1.3, add(4, 3))) | in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.3 times as old as robert ? | "in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 ro = 10 , ri = 40 now for each year we can calculate : 1960 : ri = 50 , ro = 20 1965 : ri = 55 , ro = 25 2040 : ri = 100 , ro = 130 thus ans : e" | a = 4 + 3
b = 1 + 3
c = 1950 + b
|
a ) 8981 , b ) 3799 , c ) 1200 , d ) 1600 , e ) 1732 | d | add(multiply(multiply(divide(800, 10), 5), const_3), multiply(divide(800, 10), 5)) | the simple interest on a sum of money will be rs . 800 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 800 p - - - 5 - - - - - 400 3 p - - - 5 - - - - - 1200 - - - - - - = > 1600 answer : d" | a = 800 / 10
b = a * 5
c = b * 3
d = 800 / 10
e = d * 5
f = c + e
|
a ) a . 138 , b ) b . 50 , c ) c . 48 , d ) d . 46 , e ) e . 44 | a | multiply(divide(66, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 66 seconds . at 12 : 00 , how many seconds elapse between the beginning of the first stroke and the end of the last stroke ? | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 66 seconds . thus x = 6 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 intervals so the total time is ( 12 + 11 ) x = 23 x = 138 seconds . answer a" | a = 1 + 4
b = 6 + a
c = 66 / b
d = 1 + 4
e = 6 + d
f = 12 + e
g = c * f
|
a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | c | add(divide(subtract(add(40, 2), 30), 1.5), 30) | each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , annie is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week annie worked a total of 50 hours . if harry and annie were paid the same amount last week , how many hours did harry work last week ? | "annie earned 40 x + 10 ( 2 x ) = 60 x let h be the number of hours that harry worked . harry earned 30 x + 1.5 x ( h - 30 ) = 60 x ( 1.5 x ) ( h ) = 75 x h = 50 hours the answer is c ." | a = 40 + 2
b = a - 30
c = b / 1
d = c + 30
|
a ) 80 , b ) 82 , c ) 84 , d ) 88 , e ) 90 | e | divide(multiply(multiply(25, 30), 15), volume_cube(divide(15, const_2))) | a box measuring 25 inches long by 30 inches wide by 15 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 5 is the biggest number that could divide all three , 25 , 30 and 15 . thus side of cube must be 5 , and total number of cubes = 25 / 5 * 30 / 5 * 15 / 5 = 90 ans e ." | a = 25 * 30
b = a * 15
c = 15 / 2
d = b / volume_cube
|
a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 1 | b | divide(10, add(10, 25)) | in lottery , there are 10 prizes and 25 blanks . a lottery is drawn at random . what is the probability of getting a prize ? | total number of outcomes possible , n ( s ) = 10 + 25 = 35 . total number of prize , n ( e ) = 10 p ( e ) = n ( e ) / n ( s ) = 10 / 35 = 2 / 7 answer is option b | a = 10 + 25
b = 10 / a
|
a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.11111 , e ) 1.1111 | d | multiply(divide(0.999991, 0.111119), const_100) | 0.999991 + 0.111119 = ? | "0.999991 + 0.111119 = 0.999991 + 0.11111 + 0.00009 = ( 0.999991 + 0.00009 ) + 0.11111 = 1 + 0.11111 = 1.11111 d" | a = 0 / 999991
b = a * 100
|
a ) 35 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | a | multiply(divide(const_100, add(const_100, 30)), 45) | from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 30 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "1.30 x = 45 - - > 13 / 10 * x = 45 - - > x = 45 * 10 / 13 = 450 / 13 = ~ 35 . answer : a ." | a = 100 + 30
b = 100 / a
c = b * 45
|
a ) $ 4 , b ) $ 8 , c ) $ 12 , d ) $ 16 , e ) $ 10 | e | multiply(10, const_1) | if $ 5,000 is invested in an account that earns 10 % interest compounded semi - annually , then the interest earned after one year would be how much greater than if the $ 5,000 had been invested at 8 % simple yearly interest ? | solution amount ( ci ) = p + ( 1 + r / n ) ^ nt = 5000 + ( 1 + 0.10 / 2 ) ^ 2 = 5410 amount ( si ) = p + ptr / 100 = 5000 + ( 5000 * 1 * 10 / 100 ) = 5400 difference = 5410 - 5400 = 10 $ e | a = 10 * 1
|
a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17 | b | add(subtract(divide(15, 5), const_2), add(divide(15, 3), divide(15, 5))) | line q has the equation 5 y β 3 x = 15 . if line s is perpendicular to q , has an integer for its y - intercept , and intersects q in the second quadrant , then how many possible line s β s exist ? ( note : intersections on one of the axes do not count . ) | "5 y - 3 x = 15 and so y = 3 x / 5 + 3 when x = 0 , then y = 3 . when y = 0 , then x = - 5 the slope is 3 / 5 , so the slope of line s is - 5 / 3 . through the point ( - 5 , 0 ) , 0 = - 5 ( - 5 ) / 3 + c the y - intercept is c = - 25 / 3 < - 8 . thus the perpendicular line s can have y - intercepts from - 8 up to 2 . the number of possible lines is 8 + 2 + 1 = 11 the answer is b ." | a = 15 / 5
b = a - 2
c = 15 / 3
d = 15 / 5
e = c + d
f = b + e
|
a ) 10 % , b ) 12 % , c ) 15 % , d ) 17 % , e ) 23 % | e | multiply(divide(subtract(divide(50, const_100), divide(35, const_100)), subtract(const_1, divide(35, const_100))), const_100) | mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 35 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "lets assume that candidate got 35 % votes and total votes is 100 . candidate won = 35 remaining = 65 to get 50 % , candidate requires 15 votes from 100 which is 15 % and 15 votes from 65 . 15 / 65 = 23.07 which is approx 23 % . hence the answer is e" | a = 50 / 100
b = 35 / 100
c = a - b
d = 35 / 100
e = 1 - d
f = c / e
g = f * 100
|
a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,300 , d ) $ 1,800 , e ) $ 2,200 | c | subtract(1,000, 900) | a family pays $ 900 per year for an insurance plan that pays 80 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total . | "upfront payment for insurance plan = 900 $ family needs to pay 20 % of first 1000 $ in expense = 200 $ total amount paid by family when medical expenses are equal to or greater than 1000 $ = 900 + 200 = 1100 $ total amount paid by insurance plan for first 1000 $ = 900 $ total amount paid by family will equal amount paid by plan when medical expense = 1300 $ ( since insurance plan will pay 100 % of amount that exceeds 1100 $ ) answer c" | a = 1 - 0
|
a ) 16.2 , b ) 16.4 , c ) 16.6 , d ) 16.8 , e ) 17.0 | e | multiply(sqrt(divide(28.9, 10.0)), 10.0) | at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 28.9 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | "let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 28.9 x ^ 2 = 2.89 x = 1.7 at 4 : 00 pm , the amount of bacteria was 10 ( 1.7 ) = 17 grams . the answer is e ." | a = 28 / 9
b = math.sqrt(a)
c = b * 10
|
a ) 4 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | a | divide(subtract(78, 58), subtract(58, 53)) | rahul played weel in this season . his current batting averagge is 53 . if he score 78 runs in today match . his batting average will become 58 . how many matches had he played in this season . | "53 x + 78 = 58 ( x + 1 ) = > 5 x = 20 = > x = 4 answer : a" | a = 78 - 58
b = 58 - 53
c = a / b
|
['a ) 140 sq . units', 'b ) 158 sq . units', 'c ) 187 sq . units', 'd ) 607 sq . units', 'e ) 815 sq . units'] | a | multiply(10, multiply(const_2, divide(sqrt(1225), divide(const_10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 1225 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | given that the area of the square = 1225 sq . units = > side of square = β 1225 = 35 units the radius of the circle = side of the square = 35 units length of the rectangle = 2 / 5 * 35 = 14 units given that breadth = 10 units area of the rectangle = lb = 14 * 10 = 140 sq . units answer : a | a = math.sqrt(1225)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | e | divide(multiply(add(33, divide(1, 3)), 390), const_100) | 33 1 / 3 % of 390 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 390 = 130 e )" | a = 1 / 3
b = 33 + a
c = b * 390
d = c / 100
|
a ) 5591 , b ) 2477 , c ) 2877 , d ) 2676 , e ) 1881 | a | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6600), 5) | sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6600 ? | "total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6600 x 6 ) - 34009 ] = rs . ( 39600 - 34009 ) = rs . 5591 answer : a" | a = 6835 + 9927
b = a + 6855
c = b + 7230
d = c + 6562
e = d / 5
f = e - 6600
g = f * 5
|
a ) 8 , b ) 5 , c ) 9 , d ) 6 , e ) 11 | e | divide(subtract(186, 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 $ 186 , 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 = $ 186 - $ 120 = $ 66 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 66 / $ 6 = 11 ans is e | a = 3 * 40
b = 186 - a
c = 3 * 2
d = b / c
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | 6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days c do they need to finish the same job ? | let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time c = 3 days to finish the task imo a | a = 6 * 8
b = 3 * 6
c = a - b
d = 6 + 4
e = c / d
|
a ) 5.98 , b ) 9.98 , c ) 9.06 , d ) 9.87 , e ) 9.82 | c | multiply(divide(multiply(add(9, 1.2), subtract(9, 1.2)), add(add(9, 1.2), subtract(9, 1.2))), const_2) | a man can row 9 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 9 s = 1.2 ds = 7.2 us = 10.8 x / 10.8 + x / 7.8 = 1 x = 4.53 d = 4.53 * 2 = 9.06 answer : c" | a = 9 + 1
b = 9 - 1
c = a * b
d = 9 + 1
e = 9 - 1
f = d + e
g = c / f
h = g * 2
|
a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 18 | e | multiply(12, 4) | jacob is 12 years old . he is 4 times as old as his brother . how old will jacob be when he is twice as old ? | "j = 12 ; j = 4 b ; b = 12 / 4 = 3 ; twice as old so b = 3 ( now ) + ( 3 ) = 6 ; jacob is 12 + 6 = 18 answer : e" | a = 12 * 4
|
a ) 1 / 5 , b ) 29 / 35 , c ) 5 / 6 , d ) 35 / 29 , e ) 35 / 12 | d | multiply(3, multiply(const_2, divide(const_1, 5))) | a type q machine can complete a job in 5 hours and a type b machine can complete the job in 7 hours . how many hours will it take 2 type q machines and 3 type b machines working together and independently to complete the job ? | now d should be the answer . q need 5 hours to complete and b needs 7 hours to compete so 2 q + 3 b will complete 2 / 5 + 3 / 7 or 29 / 35 portion of the job in 1 hour so the whole job will take 35 / 29 hours . . . . = d | a = 1 / 5
b = 2 * a
c = 3 * b
|
a ) 1050 , b ) 1220 , c ) 1250 , d ) 2000 , e ) 1110 | d | divide(1820, subtract(const_1, divide(multiply(3, 3), const_100))) | a fellow borrowed a certain sum of money at 3 % per annum at simple interest and in 3 years the interest amounted to rs . 1820 less than the sum lent . what was the sum lent ? | "p - 1820 = ( p * 3 * 3 ) / 100 p = 2000 answer : d" | a = 3 * 3
b = a / 100
c = 1 - b
d = 1820 / c
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.