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 ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | c | multiply(sqrt(divide(divide(400, 3), const_3)), const_3) | the length of a rectangular floor is more than its breadth by 200 % . if rs . 400 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 400 / 3 = 133.33 sq m l b = 133.33 i . e . , l * l / 3 = 133.33 l ^ 2 = 400 > l = 20 . c" | a = 400 / 3
b = a / 3
c = math.sqrt(b)
d = c * 3
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(320, divide(subtract(320, multiply(320, divide(subtract(const_100, 20), const_100))), multiply(divide(subtract(const_100, 20), const_100), const_2))) | a baker charges n dollars for a birthday cake . next week , she intends to earn 320 dollars selling cakes . if she were to reduce the price of the cake by 20 percent , she would have to sell two more cakes to earn the same amount of revenue . how many cakes does she intend to sell next week ? | we ' re told that by selling cakes at n dollars / cake , a baker will make $ 320 . reducing the price by 20 % and selling 2 more cakes will also make her $ 320 . we ' re asked for the original number of cakes that the baker intends to sell . $ 320 is a rather interesting number . based on the answer choices , we ' re a... | a = 100 - 20
b = a / 100
c = 320 * b
d = 320 - c
e = 100 - 20
f = e / 100
g = f * 2
h = d / g
i = 320 / h
|
a ) 30 % , b ) 31 % , c ) 34 % , d ) 19 % , e ) 50 % | b | multiply(divide(subtract(2402, 1820), 1820), const_100) | a sum of money deposited at c . i . amounts to rs . 1820 in 4 years and to rs . 2402 in 5 years . find the rate percent ? | "1820 - - - 582 100 - - - ? = > 31 % answer : b" | a = 2402 - 1820
b = a / 1820
c = b * 100
|
a ) 40 , b ) 44 , c ) 48 , d ) 52 , e ) 56 | a | divide(multiply(60, const_2), add(divide(60, 30), const_1)) | an assembly line produces 30 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this ... | "the time to produce the first 60 cogs is 60 / 30 = 2 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / 3 hours = 40 cogs per hour . the answer is a ." | a = 60 * 2
b = 60 / 30
c = b + 1
d = a / c
|
a ) 83.55 % , b ) 83.33 % , c ) 80 % , d ) 83.39 % , e ) 84.33 % | b | divide(const_100, divide(24, subtract(44, 24))) | if the cost price of 44 chocolates is equal to the selling price of 24 chocolates , the gain percent is : | explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 24 chocolates = rs . 24 ; s . p . of 24 chocolates = rs . 44 . . ' . gain % = 20 * 100 / 24 = 83.33 % answer : b | a = 44 - 24
b = 24 / a
c = 100 / b
|
a ) 5 , b ) 15 , c ) 85 , d ) 90 , e ) 95 | c | subtract(90, 5) | 90 percent of your marbles are a solid color . 5 percent of your marbles are solid yellow . what percentage of your marbles are a solid color other than yellow ? | 90 percent are solid colors including solid yellow . 5 percent are solid yellow . 90 % - 5 % = 85 % , so 85 percent are a solid color other than yellow = > ( c ) | a = 90 - 5
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a ) 432 sq m , b ) 356 sq m , c ) 452 sq m , d ) 588 sq m , e ) 525 sq m | d | multiply(multiply(divide(112, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(112, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | the length of rectangle is thrice its breadth and its perimeter is 112 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 112 l = 42 b = 14 lb = 42 * 14 = 588 answer : d" | a = 3 * 2
b = 1 * 2
c = a + b
d = 112 / c
e = d * 3
f = 3 * 2
g = 1 * 2
h = f + g
i = 112 / h
j = e * i
|
a ) 0.9 % , b ) 0.8 % , c ) 0.3 % , d ) 0.2 % , e ) 0.5 % | b | subtract(subtract(5, 4), divide(multiply(5, 4), const_100)) | in measuring the sides of a rectangle , one side is taken 5 % in excess and the other 4 % in deficit . find the error percent in the area , calculate from the those measurements . | "explanation : let x and y be the sides of the rectangle then correct area = = error % = answer : b ) 0.8 %" | a = 5 - 4
b = 5 * 4
c = b / 100
d = a - c
|
a ) a 216 , b ) b 196 , c ) c 170 , d ) d 140 , e ) e 121 | a | add(cube_edge_by_volume(144), 144) | we have a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 144 inches and the width is 72 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | "l = 144 / 12 = 12 ft w = 72 / 12 = 6 ft area of paper = 72 area of cube = 12 * side ^ 2 side of cube = 6 v of cube = 216" | a = cube_edge_by_volume + (
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a ) a . 10 , b ) b . 12 , c ) c . 34 , d ) d . 18 , e ) e . 24 | c | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 7)), const_1), const_1), const_1), const_1), 7), const_1), const_1), const_1), const_1), const_1), const_1) | working at constant rate , pump x pumped out half of the water in a flooded basement in 7 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constan... | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 34 hours c" | a = 3 + 7
b = 1 + a
c = b + 1
d = c + 1
e = d + 1
f = e + 1
g = f + 7
h = g + 1
i = h + 1
j = i + 1
k = j + 1
l = k + 1
m = l + 1
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(log(divide(1000, 0.00010101)), log(10)) | if k is an integer and 0.00010101 x 10 ^ k is greater than 1000 , what is the least possible value of k ? | "0.00010101 * 10 ^ k > 1000 we need to move the decimal point to the right 7 places to get 1010.1 this is equivalent to multiplying by 10 ^ 7 . the answer is b ." | a = 1000 / 0
b = math.log(a)
c = math.log(10)
d = b / c
|
a ) 30 kg , b ) 31 kg , c ) 37 kg , d ) 39 kg , e ) 41 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 31 . b ' s wei... | a = 40 * 2
b = 43 * 2
c = a + b
d = 45 * 3
e = c - d
|
a ) $ 503 , b ) $ 504 , c ) $ 515 , d ) $ 528 , e ) $ 537 | d | add(divide(multiply(multiply(22, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(22, const_1000), subtract(const_1, divide(10, const_100))))) | a car is purchased on hire - purchase . the cash price is $ 22 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ? | "explanation : cash price = $ 22 000 deposit = 10 % ã — $ 22 000 = $ 2200 loan amount = $ 22000 â ˆ ’ $ 2200 number of payments = 60 = $ 19800 i = p * r * t / 100 i = 11880 total amount = 19800 + 11880 = $ 31680 regular payment = total amount / number of payments = 528 answer : d" | a = 22 * 1000
b = 10 / 100
c = 1 - b
d = a * c
e = d / 60
f = 12 / 100
g = f / 12
h = 22 * 1000
i = 10 / 100
j = 1 - i
k = h * j
l = g * k
m = e + l
|
a ) 21 , b ) 20 , c ) 63 , d ) 30 , e ) 45 | c | subtract(divide(multiply(add(divide(44, const_2), 5), 5), 2), 5) | 5 is added to a certain number , the sum is multiplied by 2 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 44 . what is the number ? | "let number is x . when 5 added to it , = ( x + 5 ) 2 multiplied to sum , = 2 * ( x + 5 ) now , = [ { 2 * ( x + 5 ) } / 5 ] and , = [ { 2 * ( x + 5 ) } / 5 ] - 5 according to question , [ { 2 * ( x + 5 ) } / 5 ] - 5 = half of 44 [ ( 2 x + 10 ) / 5 ) = 22 + 5 2 x + 10 = 27 * 5 2 x - 135 - 10 x = 125 / 2 = 62.5 = 63 so ,... | a = 44 / 2
b = a + 5
c = b * 5
d = c / 2
e = d - 5
|
a ) 2 : 9 , b ) 2 : 7 , c ) 3 : 6 , d ) 2 : 0 , e ) 2 : 1 | b | divide(subtract(9, 7), 7) | the ratio between the sale price and the cost price of an article is 9 : 7 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 7 x and s . p . = rs . 9 x . then , gain = rs . 2 x required ratio = 2 x : 7 x = 2 : 7 answer : b" | a = 9 - 7
b = a / 7
|
a ) 14 , b ) 20 , c ) 26 , d ) 28 , e ) 30 | a | divide(196, subtract(15, const_1)) | find the number which when multiplied by 15 is increased by 196 . | let the number be x . then , 15 x - x = 196 < = > 14 x = 196 < = > x = 14 . answer : a | a = 15 - 1
b = 196 / a
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a ) 20 , b ) 33 1 / 3 , c ) 35 , d ) 40 , e ) 80 | e | multiply(divide(divide(divide(multiply(const_100, 3), 5), 3), subtract(const_100, divide(multiply(const_100, 3), 4))), const_100) | one night a certain hotel rented 3 / 4 of its rooms . including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | consider total # of rooms to be 100 ; as 3 / 5 of the rooms are air conditioned then # of rooms that are air conditioned is 3 / 5 * 100 = 60 ; 3 / 4 rooms were rented - - > 1 / 4 * 100 = 25 were not rented ; 2 / 3 of air conditioned rooms were rented - - > 1 / 3 * 60 = 20 air conditioned room were not rented ; 20 / 25 ... | a = 100 * 3
b = a / 5
c = b / 3
d = 100 * 3
e = d / 4
f = 100 - e
g = c / f
h = g * 100
|
a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 72), const_2) | two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 72 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 72 seconds , the relative difference in distance traveled is 72 * 25 / 9 = 200 meters this distance is twice the length of each train . the length of each train is 200 / 2 = 100 meters the answer is d ." | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 72
e = d / 2
|
a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 81 | e | add(80, divide(multiply(5, 12), divide(180, 3))) | 80 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 80 + 5 * 12 / ( 180 / 3 ) = 80 + 5 * 12 / ( 60 ) = 80 + ( 5 * 12 ) / 60 = 80 + 1 = 81 . answer : e" | a = 5 * 12
b = 180 / 3
c = a / b
d = 80 + c
|
a ) 8 : 3 , b ) 3 : 2 , c ) 4 : 3 , d ) 2 : 3 , e ) 3 : 8 | b | divide(const_2, add(const_1, divide(const_1, const_3))) | ann and bob drive separately to a meeting . ann ' s average driving speed is greater than bob ' s avergae driving speed by one - third of bob ' s average driving speed , and ann drives twice as many miles as bob . what is the ratio t of the number of hours ann spends driving to the meeting to the number of hours bob sp... | say the rate of bob is 3 mph and he covers 6 miles then he needs 6 / 3 = 2 hours to do that . now , in this case the rate of ann would be 3 + 3 * 1 / 3 = 4 mph and the distance she covers would be 6 * 2 = 12 miles , so she needs 12 / 4 = 3 hours for that . the ratio t of ann ' s time to bob ' s time is 3 : 2 . answer :... | a = 1 / 3
b = 1 + a
c = 2 / b
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a ) 213 , b ) 58 , c ) 84 , d ) 122 , e ) 168 | a | divide(multiply(multiply(10, 14), 6), const_4) | if 6 x = 10 y = 14 z , then what is a possible sum of positive integers x , y , and z ? | answer = d = 122 6 x = 10 y = 14 z 3 x = 5 y = 7 z 3 ( 5 * 7 ) = 5 ( 3 * 7 ) = 7 ( 3 * 5 ) addition = 35 + 21 + 15 = 71 answer would be multiple of 71 which is 213 answer : a | a = 10 * 14
b = a * 6
c = b / 4
|
a ) 171.4 , b ) 171.5 , c ) 171.6 , d ) 171.7 , e ) none of the above | a | multiply(divide(add(const_100, 20), subtract(const_100, 30)), const_100) | tough and tricky questions : percents . over the course of a year , a certain microbrewery increased its beer output by 20 percent . at the same time , it decreased its total working hours by 30 percent . by what percent did this factory increase its output per hour ? | lets assume the initial production was 100 litres of beer for 100 hr . with the 20 % increase the total amount of beer production will be 120 litres and with 30 % decrease in total hours will be reduced to 70 hr . 100 hr - - - - > 100 lts 1 hr - - - - - > 1 lts 70 hr - - - - - > 120 lts 1 hr - - - - - > 1.714 lts total... | a = 100 + 20
b = 100 - 30
c = a / b
d = c * 100
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | divide(8, subtract(divide(5, 4), const_1)) | walking with 4 / 5 of my usual speed , i arrive at the bus stop 8 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 8 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 32 the answer is d ." | a = 5 / 4
b = a - 1
c = 8 / b
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a ) 2000 , b ) 2300 , c ) 2700 , d ) 2500 , e ) 250 | d | divide(multiply(multiply(const_2, const_100), const_100), multiply(0.08, const_100)) | an inspector rejects 0.08 % of the meters as defective . how many will be examine to project ? | let the number of meters to be examined be x . then , 0.08 % of x = 2 [ ( 8 / 100 ) * ( 1 / 100 ) * x ] = 2 x = [ ( 2 * 100 * 100 ) / 8 ] = 2500 answer is d . | a = 2 * 100
b = a * 100
c = 0 * 8
d = b / c
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a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 8 days | c | divide(const_1, add(divide(const_1, 15), divide(const_1, 10))) | a can do a piece of work in 15 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 * 5 = 1 / 2 remaining work = 1 â ˆ ’ 1 / 2 = 1 / 2 a can finish work = 15 â ˆ — 1 / 2 = 7.5 days answer is c" | a = 1 / 15
b = 1 / 10
c = a + b
d = 1 / c
|
a ) 3 / 4 , b ) 1 / 2 , c ) 5 / 8 , d ) 1 , e ) 1 / 4 | c | subtract(1, multiply(divide(factorial(3), factorial(2)), power(divide(1, 2), 3))) | a couple decides to have 3 children . if they succeed in having 3 children and each child is equally likely to be a boy or a girl , what is the probability that they will have exactly 1 girl and 2 boys ? | sample space = 2 ^ 3 = 8 favourable events = { bgg } , { bgb } , { bbb } , { ggg } , { gbg } , probability = 5 / 8 = 5 / 8 . ans ( c ) . | a = math.factorial(3)
b = math.factorial(2)
c = a / b
d = 1 / 2
e = d ** 3
f = c * e
g = 1 - f
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a ) $ 20000 , b ) $ 14000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | b | divide(multiply(multiply(add(const_2, const_3), const_1000), 6), const_2) | if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 60 / r years . if pat ' s parents invested $ 7,000 in a long - term bond that pays 6 percent interest , compounded annually , what will be the approximate total amount of the investment 20 years ... | "since investment doubles in 60 / r years , then for r = 6 it ' ll double in 60 / 6 = ~ 10 years ( we are not asked about the exact amount so such an approximation will do ) . thus after 20 years investment will become $ 7,000 * 2 = $ 14,000 . answer : b" | a = 2 + 3
b = a * 1000
c = b * 6
d = c / 2
|
a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 12 , 6,16 , d ) 9 , 6,12 , e ) 9 , 2,23 | c | divide(multiply(4, 3), 3) | find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 28 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 28 = > 3 x + 4 x = 28 = > x = 4 a , b , c are 3 x , 2 x , 4 x a , b , c are 12 , 8 , 16 . answer : c" | a = 4 * 3
b = a / 3
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a ) 8 / 3 , b ) 3 / 8 , c ) 38 / 15 , d ) 5 / 8 , e ) 5 / 3 | c | divide(divide(570, 3), divide(300, 4)) | eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 570 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy :... | "distance traveled by eddy = 570 km time taken by eddy = 3 hours average speed of eddy = 570 / 3 = 190 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 190 / 75 = 38 / 15 answer c" | a = 570 / 3
b = 300 / 4
c = a / b
|
a ) 21.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 15 sec | a | divide(add(180, 360), multiply(add(60, 30), const_0_2778)) | two trains 180 m and 360 m long run at the speed of 60 kmph and 30 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 30 = 90 kmph * 5 / 18 = 25 m / s distance covered in crossing each other = 180 + 360 = 540 m required time = 540 * 1 / 25 = 21.6 sec answer is a" | a = 180 + 360
b = 60 + 30
c = b * const_0_2778
d = a / c
|
a ) 16 , b ) 14 , c ) 12 , d ) 10 , e ) 18 | d | divide(log(multiply(power(2, 10), power(5, 10))), log(10)) | if 5 ^ 10 x 2 ^ 10 = 10 ^ n what is the value of n ? | "5 ^ 10 * 2 ^ 10 = 10 ^ n or 10 ^ 10 = 10 ^ n n = 10 d" | a = 2 ** 10
b = 5 ** 10
c = a * b
d = math.log(c)
e = math.log(10)
f = d / e
|
a ) 37 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(55, 28), multiply(28, 55)), add(55, 28)) | the average of 55 results is 28 and the average of other 28 results is 55 . what is the average of all the results ? | "answer sum of 83 result = sum of 55 result + sum of 28 result . = 30 x 20 + 20 x 30 = 3080 correct option : a" | a = 55 * 28
b = 28 * 55
c = a + b
d = 55 + 28
e = c / d
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a ) 87 , b ) 89 , c ) 90 , d ) 98 , e ) 95 | d | subtract(multiply(4, add(90, 2)), multiply(90, 3)) | jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 90 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 90 * 3 = 270 jerry wants the average to be = 92 hence total score on 4 tests should be = 92 * 4 = 368 score required on the fourth test = 368 - 270 = 98 option d" | a = 90 + 2
b = 4 * a
c = 90 * 3
d = b - c
|
a ) 24 , b ) 77 , c ) 88 , d ) 72 , e ) 21 | d | divide(multiply(18, 36), 9) | 36 men can complete a piece of work in 18 days . in how many days will 9 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 9 : 36 : : 18 : x x = 72 answer : d ) 72 days" | a = 18 * 36
b = a / 9
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a ) 6.85 % , b ) 7.25 % , c ) 7.76 % , d ) 8.33 % , e ) 8.75 % | d | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 4,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 6000 = 1 / 12 which is about 8.33 % the answer is d ." | a = 100 * 100
b = 5 / 100
c = a * b
d = 100 * 100
e = 2 + 3
f = 2 + 3
g = f * 2
h = g * 100
i = e * h
j = 2 + 3
k = j * 100
l = i + k
m = d - l
n = c / m
o = n * 100
|
a ) 80 m 2 , b ) 84 m 2 , c ) 88 m 2 , d ) 86 m 2 , e ) none of these | c | multiply(5, multiply(multiply(multiply(4, divide(22, 7)), divide(1.4, 4)), 4)) | the diameter of a garden roller is 1.4 m and it is 4 m long . how much area will it cover in 5 revolutions ? ( use ï € = 22 â „ 7 ) | "required area covered in 5 revolutions = 5 ã — 2 ï € rh = 5 ã — 2 ã — 22 â „ 7 ã — 0.7 ã — 4 = 88 m 2 answer c" | a = 22 / 7
b = 4 * a
c = 1 / 4
d = b * c
e = d * 4
f = 5 * e
|
a ) 3 / 4 , b ) 1 / 2 , c ) 1 / 4 , d ) 3 / 8 , e ) 5 / 16 | a | divide(2, add(8, 2)) | at a certain high school , the senior class is twice the size of the junior class . if 7 / 8 of the seniors and 1 / 2 of the juniors study japanese , what fraction of the students in both classes study japanese ? | "start by deciding on a number of students to represent the number of students in the senior class . for this example i will choose 200 students . that would make the number of students in the junior class 100 . then we can find out how many students are taking japanese in each grade and add them together . ( 7 / 8 ) *... | a = 8 + 2
b = 2 / a
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(divide(multiply(400, multiply(15, const_2)), const_1000), divide(15, const_10)) | a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 400 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | "the distance between the nest and the ditch is 400 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 400 = 12000 . d = st 12000 / 1.5 = t , i think we can take 12000 meters as 12 km , then only we get t = 8 . ( 10... | a = 15 * 2
b = 400 * a
c = b / 1000
d = 15 / 10
e = c / d
|
a ) $ 3.36 , b ) $ 6.85 , c ) $ 8.36 , d ) $ 10.08 , e ) $ 11.85 | c | add(multiply(2, 2.44), multiply(4, 0.87)) | what is the total cost of 2 sandwiches at $ 2.44 each and 4 sodas at $ 0.87 each ? | "answer = c 2 * 2.44 + 4 * 0.87 = 2 ( 2.50 - 0.06 ) + 4 ( 1.00 - 0.13 ) = 5 + 4 - 0.12 - 0.52 = 9 - 0.64 = 8.36" | a = 2 * 2
b = 4 * 0
c = a + b
|
a ) 35 , b ) 20 , c ) 40 , d ) 30 , e ) 67 | b | subtract(multiply(5, 12), multiply(10, const_4)) | average of 5 numbers is 12 . if one number is excluded the average becomes 10 . the excluded number is | explanation : number is ( 5 * 12 ) - ( 4 * 10 ) = 60 - 40 = 20 answer : option b | a = 5 * 12
b = 10 * 4
c = a - b
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a ) 1000 , b ) 1429 , c ) 1977 , d ) 2778 , e ) 2711 | b | divide(multiply(200, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | a watch was sold at a loss of 10 % . if it was sold for rs . 200 more , there would have been a gain of 4 % . what is the cost price ? | "90 % 104 % - - - - - - - - 14 % - - - - 200 100 % - - - - ? = > rs . 1429 answer : b" | a = 200 * 100
b = 100 + 4
c = 100 - 10
d = b - c
e = a / d
|
a ) 850 , b ) 1250 , c ) 1650 , d ) 2050 , e ) 2450 | b | multiply(subtract(62, 12), add(divide(subtract(60, 12), const_2), const_1)) | set a contains all the even numbers between 12 and 60 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 50 more than the corresponding term in set a . the difference of the sums = 25 * 50 = 1250 . the answer is b ." | a = 62 - 12
b = 60 - 12
c = b / 2
d = c + 1
e = a * d
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a ) 28 , b ) 29 , c ) 31 , d ) 34 , e ) 36 | b | add(multiply(divide(20, subtract(20, 15)), const_2), const_1) | a worker earns $ 20 on the first day and spends $ 15 on the second day . the worker earns $ 20 on the third day and spends $ 15 on the fourth day . if this pattern continues , on which day will the worker first reach a net total of $ 90 ? | "every two days , the net total is $ 5 . after 28 days , the worker will have $ 70 . on day 29 , the worker will receive $ 20 for a net total of $ 90 . the answer is b ." | a = 20 - 15
b = 20 / a
c = b * 2
d = c + 1
|
a ) 629 , b ) 729 , c ) 829 , d ) 929 , e ) 727 | b | power(9, 3) | log 3 n + log 9 n what is 3 digit number n that will be whole number | "no of values n can take is 1 9 ^ 3 = 729 answer : b" | a = 9 ** 3
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a ) 15 % , b ) 16 2 ⁄ 3 % , c ) 20 % , d ) 22 % , e ) 24 % | c | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | an employee ’ s annual salary was increased $ 15,000 . if her new annual salary now equals $ 90,000 , what was the percent increase ? | "new annual salary = $ 90,000 salary increase = $ 15,000 . original salary = $ 90,000 - $ 15,000 . = $ 75,000 % increase = ( $ 15,000 / $ 75,000 ) * 100 = 20 % hence c ." | a = 100 - 10
b = a * 1000
c = 100 - 10
d = c * 1000
e = const_0_25 * 100
f = e * 1000
g = d - f
h = b / g
i = h - 1
j = i * 100
|
a ) 20 , b ) 24 , c ) 25 , d ) 12 , e ) 32 | d | add(add(divide(50, add(const_4, const_1)), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(50, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros does 50 ! end with ? | "according to above 50 ! has 50 / 5 + 50 / 25 = 10 + 2 = 12 trailing zeros . answer : d ." | a = 4 + 1
b = 50 / a
c = 4 + 1
d = 50 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 50 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
|
a ) 209 , b ) 250 , c ) 265 , d ) 280 , e ) 225 | a | add(multiply(multiply(4, 5), const_10), 1) | when 1 + 2 = 23 , 2 + 3 = 65 , 3 + 4 = 127 , then 4 + 5 = ? | "1 + 2 = > 1 x 2 = 2 & 1 + 2 = 3 = > 2 & 3 = > 23 2 + 3 = > 2 ã — 3 = 6 & 2 + 3 = 5 = > 6 & 6 = > 65 3 + 4 = > 3 ã — 4 = 12 & 3 + 4 = 7 = > 12 & 7 = > 127 then 4 + 5 = > 4 ã — 5 = 20 & 4 + 5 = 9 = > 20 & 9 = > 209 answer : a" | a = 4 * 5
b = a * 10
c = b + 1
|
a ) 13,400 , b ) 14,800 , c ) 16,400 , d ) 15,800 , e ) 19,600 | a | add(subtract(add(10, const_4), const_1), divide(const_4, const_10)) | paul sells encyclopedias door - to - door . he earns $ 150 on every paycheck , regardless of how many sets he sells . in addition , he earns commission as follows : commission sales 10 % $ 0.00 - $ 1000.00 5 % $ 1000.01 - - - > he does not earn double commission . that is , if his sales are $ 12000 , he earns 10 % on t... | his pay check was $ 1320 . out of this , $ 150 was his fixed salary so the total commission he earned was $ 1320 - $ 150 = $ 1170 he earns 10 % on the sales of first $ 10,000 which gives a commission of $ 1000 . he earns 5 % on every subsequent dollar . since he earns another $ 170 , he must have had sales of another 1... | a = 10 + 4
b = a - 1
c = 4 / 10
d = b + c
|
a ) $ 900 , b ) $ 1200 , c ) $ 1500 , d ) $ 1800 , e ) $ 2100 | b | multiply(divide(84, 12), const_100) | if an article is sold at 19 % profit instead of 12 % profit , then the profit would be $ 84 more . what is the cost price ? | "7 % * cost price = $ 84 1 % * cost price = $ 84 / 7 = $ 12 the cost price is $ 1200 . the answer is b ." | a = 84 / 12
b = a * 100
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a ) 0.3408 , b ) 3.408 , c ) 34.08 , d ) 340.8 , e ) none of these | d | multiply(16, 21.3) | if 213 × 16 = 3408 , then 16 × 21.3 is equal to : | "solution 16 × 21.3 = ( 16 x 213 / 10 ) = ( 16 x 213 / 10 ) = 3408 / 10 = 340.8 . answer d" | a = 16 * 21
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a ) 1000 , b ) 2000 , c ) 3000 , d ) 4000 , e ) 5000 | d | divide(multiply(21600, const_10), subtract(subtract(add(36, 24), 5), const_1)) | a , b and c invest in a partnership in the ratio : 7 / 24 / 36 / 5 . after 4 months , a increases his share 50 % . if the total profit at the end of one year is rs . 21600 , then what is b ' s share in the profit ? | ratio of the initial investment = 7 / 2 : 4 / 3 : 6 / 5 = 105 : 40 : 36 therefore , let the initial investments of a , b and c be 105 x , 40 x and 36 x respectively a increases his share 50 % after 4 months . hence the ratio of their investments = ( 105 x × 4 ) + ( 105 x × 150 / 100 × 8 ) : 40 x × 12 : 36 x × 12 = 105 ... | a = 21600 * 10
b = 36 + 24
c = b - 5
d = c - 1
e = a / d
|
a ) 0 , b ) 3 , c ) 6 , d ) 9 , e ) 12 | d | power(3, 2) | if x and y are numbers such that ( x + 3 ) ( y - 3 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2 | "from ( x + 3 ) ( y - 3 ) = 0 it follows that either x = - 3 or y = 3 . thus either x ^ 2 = 9 or y ^ 2 = 9 . now , if x ^ 2 = 9 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 9 + 0 = 9 . similarly if y ^ 2 = 9 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 0 +... | a = 3 ** 2
|
a ) 3 / 4 , b ) 1 / 3 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(3), factorial(subtract(3, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 3)), factorial(subtract(add(6, 3), const_2))), factorial(const_2))) | a bag contains 6 black and 3 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 3 ) = 9 . number of white balls = 3 . p ( drawing a white ball ) = 3 / 9 = 1 / 3 . option b ." | a = math.factorial(3)
b = 3 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(6)
h = 6 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 6 + 3
o = math.factorial(n)
p = 6 + 3
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t... |
a ) 1437 , b ) 1250 , c ) 1540 , d ) 1600 , e ) 1635 | a | multiply(divide(subtract(1200, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1200 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1200 = 6 x + 15 5 x = 1185 x = 237 large number = 237 + 1365 = 1437 a" | a = 1200 - 15
b = 6 - 1
c = a / b
d = c * 6
|
a ) a ) 4.99 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | a | max(multiply(subtract(add(55, 8), const_1), subtract(divide(8, 35), divide(8, 55))), const_4) | due to construction , the speed limit along an 8 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 8 miles stretch = 8 * 60 / 55 = 8 * 12 / 11 = 8.72 new time in minutes to cross 8 miles stretch = 8 * 60 / 35 = 8 * 12 / 7 = 13.71 time difference = 4.99 ans : a" | a = 55 + 8
b = a - 1
c = 8 / 35
d = 8 / 55
e = c - d
f = b * e
g = max(f)
|
a ) 12 , b ) 13 , c ) 16.6 , d ) 17 , e ) 8 | e | subtract(subtract(13, const_2), add(add(2, const_4), const_1)) | set a consists of all the prime numbers between 2 and 13 . what is the range of set a ? | "the range of a set of data is the difference between the highest and lowest values in the set in this set , highest number = 11 lowest number = 3 range = highest - lowest = 11 - 3 = 8 option e" | a = 13 - 2
b = 2 + 4
c = b + 1
d = a - c
|
a ) 16 , b ) 14 , c ) 25 , d ) 87 , e ) 171 | b | floor(sqrt(210)) | if the sum of a number and its square is 210 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 210 x 2 + x - 210 = 0 ( x + 15 ) ( x – 14 ) = 0 x = 14 answer : b" | a = math.sqrt(210)
b = math.floor(a)
|
a ) 33 , b ) 27 , c ) 25 , d ) 70.2 , e ) 72 | d | multiply(const_3_6, divide(add(240, 150), 20)) | a train of length 240 m crosses a bridge of length 150 m in 20 seconds . what is the speed of train ? | "sol : ( length of train + length of bridge ) = speed of train x time ( 240 + 150 ) = 20 x speed speed = 390 / 20 = 19.5 m / s = 70.2 km / h answer = d" | a = 240 + 150
b = a / 20
c = const_3_6 * b
|
a ) 4 , b ) 0 , c ) 6 , d ) 8 , e ) 12 | b | subtract(add(add(20, 22), 20), 50) | three interviewers , a , b , and c are interviewing 50 applicants . only with three interviewers ' admission can an applicant be admitted . if interviewer a admitted 20 applicants , b admitted 22 applicants , and c admitted 20 applicants , at least how many applicants get the admission ? | "if a admitted 20 are overlapping with b admission of 22 but c does not overlap with anybody . then no student will get nod from all the 3 . hence 0 student will get admission . answer : b" | a = 20 + 22
b = a + 20
c = b - 50
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a ) 5500 , b ) 2005 , c ) 10000 , d ) 5800 , e ) 4960 | c | add(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2)))) | find the sum of first 100 odd numbers | "explanation : n 2 = 1002 = 10000 answer : option c" | a = 100 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 2
f = 100 / e
g = f / 2
h = 100 / 2
i = h / 2
j = i / 2
k = j / 2
l = k / 2
m = 100 / l
n = m / 2
o = math.sqrt(n)
p = 1 + o
q = g + p
|
a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 30 | e | divide(subtract(6, multiply(6, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 6 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 6 ] = ( 1 / 2 ) ( a + 6 ) 2 ( 2 a + 30 ) = 5 ( a + 6 ) a = 30 e" | a = 1 / 2
b = 6 * a
c = 6 - b
d = 1 / 2
e = 2 / 5
f = d - e
g = c / f
|
a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 6.00 , e ) $ 5.00 | d | add(1.50, multiply(1.50, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 1.50 + 1.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 1.50 + 1.50 ( x − 2 ) = 1.50 + 1.5 * 3 = 1.5 + 4.5 = 6 $ answer d" | a = 18 - 2
b = a / 4
c = b + 1
d = c - 2
e = 1 * 50
f = 1 + 50
|
a ) 8 , b ) 13 , c ) 17 , d ) 15 , e ) none of these | a | add(3, divide(const_100, 20)) | a fraction in reduced form is such that when it is squared and then its numerator is reduced by 33 ( 1 / 3 ) % and denominator is reduced to 20 % , its result is twice the original fraction . the sum of numerator and denominator is : | solution : let the fraction be x / y . when fraction is squared its numerator is reduced by 33 ( 1 / 3 ) and denominator is reduced by 20 % . according to question , ( x / y ) 2 * 33 ( 1 / 3 ) % / 20 % = 2 ( x / y ) . or , ( x / y ) 2 * ( 2 / 3 ) / ( 1 / 5 ) = 2 ( x / y ) . or , x / y = 3 / 5 . sum of numerator and den... | a = 100 / 20
b = 3 + a
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a ) 1.5 , b ) 2.0 , c ) 2.5 , d ) 3.0 , e ) 4.0 | e | divide(add(24, 8), subtract(58, 50)) | car a is 24 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 24 miles and drive 8 miles ahead so to drive 32 miles it ' ll need 32 / 8 = 4 hours . answer : e ." | a = 24 + 8
b = 58 - 50
c = a / b
|
a ) 90 , b ) 114 , c ) 125 , d ) 130 , e ) 144 | e | divide(divide(multiply(200, const_100), 60), const_2) | a rainstorm increased the amount of water stored in state j reservoirs from 180 billion gallons to 200 billion gallons . if the storm increased the amount of water in the reservoirs to 60 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the... | after the reservoir is filled to 200 gallons the amount of water is at 60 % - which means that 40 % of the reservoir is empty . to figure out what that 40 % is approximate : 200 gallons / 60 percent = x gallons / 40 percent , therefore , x = 133.33 gallons , answer choices a , b , c , d are below 133.33 . we know that ... | a = 200 * 100
b = a / 60
c = b / 2
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a ) rs . 300 , b ) 3 rs . 20 , c ) rs . 420 , d ) rs . 520 , e ) rs . 460 | b | multiply(2000, divide(8, const_100)) | find the simple interest for rs . 2000 at 8 % p / a for 2 years . | "s . i = ( 2000 * 8 * 2 ) / 100 p = 320 answer : b" | a = 8 / 100
b = 2000 * a
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a ) 144 , b ) 131 , c ) 232 , d ) 90 , e ) 45 | c | add(divide(multiply(16, subtract(16, const_1)), const_2), multiply(16, 7)) | 16 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 16 business exec and in each handshake 2 business execs are involved . hence 16 c 2 = 120 also , each of 16 exec will shake hand with every 7 other chairmen for total of 112 handshake . total = 120 + 112 = 232 ans : c" | a = 16 - 1
b = 16 * a
c = b / 2
d = 16 * 7
e = c + d
|
a ) 10 , b ) 30 , c ) 40 , d ) 50 , e ) 70 | e | subtract(120, add(multiply(30, subtract(divide(120, add(40, 30)), 1)), multiply(40, subtract(divide(120, add(40, 30)), 1)))) | if two trains are 120 miles apart and are traveling toward each other at constant rate of 30 miles per hour and 40 miles per hour , respectively , how far apart will they be 1 hour before they meet ? | they would meet in 120 / 70 = 12 / 7 hours 1 hour earlier 5 / 7 hours distance between trains = total distance - ( distance covered by train a + distance covered by train b ) = 120 - ( 5 / 7 ( 40 + 30 ) ) [ distance of a = 5 / 7 * 40 ; like wise distance by b ] = 70 km answer : e | a = 40 + 30
b = 120 / a
c = b - 1
d = 30 * c
e = 40 + 30
f = 120 / e
g = f - 1
h = 40 * g
i = d + h
j = 120 - i
|
a ) 22 , b ) 134 , c ) 62 , d ) 122 , e ) 672 | b | multiply(2, divide(divide(2211, add(const_10, const_1)), const_3)) | a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 2211 , what is the value of n ? | "factorise 2211 . . 3 * 11 * 67 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 61 = 134 . . ans : b" | a = 10 + 1
b = 2211 / a
c = b / 3
d = 2 * c
|
a ) 1 , b ) 3.5 , c ) 30 , d ) 49 , e ) 30 | c | divide(1, divide(1, 30)) | if 30 honey bees make 30 grams of honey in 30 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 30 : : 30 : x honey 30 : 1 = > 1 x 30 x x = 30 x 1 x 30 = > x = 30 . answer : c" | a = 1 / 30
b = 1 / a
|
a ) 20 . , b ) 21 . , c ) 22 . , d ) 23 . , e ) 24 . | c | add(divide(1430, add(divide(1430, add(add(const_10, const_10), const_2)), 45)), 9) | a basket of 1430 apples is divided equally among a group of apple lovers . if 45 people join the group , each apple lover would receive 9 apples less . how many q apples did each person get before 45 people joined the feast ? | "before solving it algebraically , let us prime factorize 1430 = 2 * 5 * 11 * 13 . since number of apples per person * total persons q = 1430 , the answer should be a factor of 1430 . only c is . and that ' s your answer . c" | a = 10 + 10
b = a + 2
c = 1430 / b
d = c + 45
e = 1430 / d
f = e + 9
|
a ) 69 , b ) 72 , c ) 75 , d ) 89 , e ) 93 | b | divide(multiply(16, 19), const_2) | if the sides of a triangle are 9 cm , 16 cm and 19 cm , what is its area ? | "the triangle with sides 9 cm , 16 cm and 19 cm is right angled , where the hypotenuse is 19 cm . area of the triangle = 1 / 2 * 9 * 16 = 72 cm 2 answer : option b" | a = 16 * 19
b = a / 2
|
a ) 59 , b ) 57 , c ) 55 , d ) 58 , e ) 60 | a | divide(divide(add(floor(divide(subtract(multiply(4, 1000), 1), 17)), 1), const_2), const_2) | the number of positive integers valued pairs ( x , y ) satisfying 4 x - 17 y = 1 and x < = 1000 ( x is less than or equal to 1000 ) is | "we are asked to find integer solutions for x , y . there is no need to apply euclidian algorithm here , we can get particular solutions simply by plugging in some values . as y reaches 3 we get : y ’ = 3 and x ’ = 13 . we need only to solve for x . general solution for x will be : x = x ’ + bn = 13 + 17 n hence lower ... | a = 4 * 1000
b = a - 1
c = b / 17
d = math.floor(c)
e = d + 1
f = e / 2
g = f / 2
|
a ) 30 th minute , b ) 31 st minute , c ) 37 th minute , d ) 32 nd minute , e ) 33 th minute | c | add(multiply(multiply(const_4, 2), 2), 1) | a monkey ascends a greased pole 20 meters high . he ascends 2 meters in the first minute and then slips down 1 meter in the alternate minute . if this pattern continues until he climbs the pole , in how many minutes would he reach at the top of the pole ? | "the money is climbing 1 meter in 2 min . this pattern will go on till he reaches 10 meters . i mean this will continue for first 18 * 2 = 36 mins . he would have reached 18 meters . after that he will climb 2 meters and he will reach the pole . so total time taken = 36 + 1 = 37 mins . so , asnwer will be c" | a = 4 * 2
b = a * 2
c = b + 1
|
a ) $ 14.00 , b ) $ 16.00 , c ) $ 17.00 , d ) $ 19.00 , e ) $ 23.00 | d | divide(subtract(multiply(10, const_100.0), add(add(20, 22), 25)), 7) | a retailer sells 10 shirts . the first 3 he sells for $ 20 , $ 22 and $ 25 . if the retailer wishes to sell the 10 shirts for an overall average price of over $ 20 , what must be the minimum average price of the remaining 7 shirts ? | "first 3 shirts are sold for $ 20 , $ 22 and $ 25 = $ 67 . to get average price of $ 20 , total sale should be 10 * $ 20 = $ 200 so remaining 7 shirts to be sold for $ 200 - $ 67 = $ 133 answer should be 133 / 7 = $ 19.00 that is d" | a = 10 * 100
b = 20 + 22
c = b + 25
d = a - c
e = d / 7
|
a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum z of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum z = average * total count = 200 * 101 = 20,200 answer is b" | a = 301 - 1
b = 99 + 1
c = a - b
d = c / 2
e = d + 1
|
a ) 200 , b ) 175 , c ) 150 , d ) 125 , e ) 100 | c | divide(subtract(multiply(300, divide(99, const_100)), multiply(300, divide(98, const_100))), subtract(const_1, divide(98, const_100))) | there are 300 employees in a room . 99 % are managers . how many managers must leave the room to bring down the percentage of managers to 98 % ? | there are 297 managers and 3 others . the 3 others would compose 2 % of the total number of people if there were 150 people in the room . thus 150 managers must leave . the answer is c . | a = 99 / 100
b = 300 * a
c = 98 / 100
d = 300 * c
e = b - d
f = 98 / 100
g = 1 - f
h = e / g
|
a ) 11 , b ) 28 , c ) 24 , d ) 29 , e ) 32 | b | floor(multiply(divide(3, 8), 75)) | the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 75 black cars in the lot , how many red cars are there in the lot ? | "b is correct r / b = 3 / 8 and b = 75 r = 75 * 3 / 8 = 28" | a = 3 / 8
b = a * 75
c = math.floor(b)
|
a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 8 , e ) 8 / 7 | a | divide(divide(divide(20, const_100), divide(50, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(50, const_100))) | a total of 50 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 20 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a c... | let ' take the number of geese to be 100 . male = 50 . female = 50 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 20 % are male i . e 20 / 100 * 20 = 4 geese ( males ) and now we know out of the total 20 geese , 4 are male , then 16 have to... | a = 20 / 100
b = 50 / 100
c = a / b
d = 2 * 4
e = d * 10
f = e / 100
g = 50 / 100
h = f / g
i = c / h
|
a ) 47.6 , b ) 48.6 , c ) 49.6 , d ) 50.6 , e ) 51.6 | b | subtract(56.7, divide(56.7, 6)) | mukesh weight 56.7 kg . if he reduces his weight in the ratio 7 : 6 . new weight of mukesh is ? | given ratio is 7 : 6 ; let weight before and after reduction be 7 x and 6 x respectively . a / q 7 x = 56.7 = > x = 56.7 / 7 = > x = 8.1 reduced weight = 6 x = 6 * 8.1 = 48.6 ; hence reduced weight = 48.6 answer : b | a = 56 / 7
b = 56 - 7
|
a ) 15.92 % , b ) 16.92 % , c ) 17.92 % , d ) 18.92 % , e ) none of these | b | subtract(const_100, multiply(divide(add(8, const_100), add(30, const_100)), const_100)) | a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 8 % is | "explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 8 % , so selling price = rs 108 discount = 130 - 108 = 22 discount % = ( 22 / 130 ) * 100 = 16.92 % option b" | a = 8 + 100
b = 30 + 100
c = a / b
d = c * 100
e = 100 - d
|
a ) 12 , b ) 14 , c ) 15 , d ) 13 , e ) 11 | a | multiply(3, divide(36, add(3, 6))) | maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 36 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 6 km / h . what is the distance traveled by maxwell when the... | consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 3 xt = 36 - 6 xt ( i . e distance = speed x time ) = > 9 t =... | a = 3 + 6
b = 36 / a
c = 3 * b
|
a ) 40 - 42 , b ) 39 - 41 , c ) 38 - 40 , d ) 37 - 39 , e ) 36 - 37 | e | add(multiply(32.50, divide(15, const_100)), 32.50) | a meal cost $ 32.50 and there was no tax . if the tip was more than 10 pc but less than 15 pc of the price , then the total amount paid should be : | "10 % ( 32.5 ) = 3.25 15 % ( 32.5 ) = 4.875 total amount could have been 32.5 + 3.25 and 32.5 + 4.875 = > could have been between 35.75 and 37.375 = > approximately between 36 and 37 answer is e ." | a = 15 / 100
b = 32 * 50
c = b + 32
|
a ) 6.5 mile , b ) 5.7 mile , c ) 7.0 mile , d ) 8.0 mile , e ) 12 mile | b | multiply(divide(subtract(15, multiply(24, divide(add(1, 5), const_60))), add(5, add(1, 5))), 5) | stacy and heather are 15 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 1 mile per hour fast than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the tw... | "ss - stacy ' s speed = 6 m / hr sh - heather ' s speed = 5 m / hr in 24 minutes stacy will cover = ( 24 / 60 ) * 6 = 2.4 miles now since both are walking in opposite directions , add their speeds - 6 + 5 = 11 m / hr and distance to cover is 15 - 2.4 = 12.6 time taken = distance / speed = 12.6 / 11 = 1.145 hrs heather ... | a = 1 + 5
b = a / const_60
c = 24 * b
d = 15 - c
e = 1 + 5
f = 5 + e
g = d / f
h = g * 5
|
a ) 57 , b ) 67 , c ) 37 , d ) 87 , e ) 97 | c | sqrt(multiply(13.69, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 13.69 , the number of the member is the group is : | "money collected = ( 13.69 x 100 ) paise = 1369 paise numbers of members = 1369 squareroot = 37 answer c" | a = 13 * 69
b = math.sqrt(a)
|
a ) 99 , b ) 132 , c ) 16 , d ) 10 , e ) 15 | b | divide(multiply(40, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 40 m long and 16 1 / 2 m board ? | "40 * 33 / 2 = 5 / 2 * 2 * x = > x = 132 answer : b" | a = 1 / 2
b = 16 + a
c = 40 * b
d = 1 / 2
e = 2 + d
f = e * 2
g = c / f
|
a ) 389 , b ) 356 , c ) 380 , d ) 350 , e ) 349 | e | add(multiply(13, add(add(const_10, const_12), const_4)), 11) | when a number is divided by 13 , the remainder is 11 . when the same number is divided by 17 , the remainder is 9 . what is the number ? | "x = 13 p + 11 and x = 17 q + 9 13 q + 11 = 17 q + 9 17 q - 13 q = 2 q = 2 + 13 q / 17 the least value of p for which q = 2 + 13 q / 17 is a whole number is p = 26 x = ( 13 * 26 * 11 ) = ( 338 + 11 ) = 349 . answer is e" | a = 10 + 12
b = a + 4
c = 13 * b
d = c + 11
|
a ) - 1 , b ) 3 , c ) 1 , d ) 2 , e ) 5 | e | subtract(add(5, 1), 1) | if 1 / ( x + 5 ) + 1 / ( x - 5 ) = 1 / ( x - 5 ) , what is the value of x ? | "if we solve the question , we get x = 5 . option : e" | a = 5 + 1
b = a - 1
|
a ) 12 , b ) 15 , c ) 18 , d ) 22 , e ) 25 | e | multiply(divide(20, 10), divide(60, 10)) | how many liters of water must be added to 20 liters of milk and water containing 10 % water to make it 60 % water ? | "by rule of alligation : 60 % - 10 % = 50 % 100 % - 60 % = 40 % quantity of pure water : quantity of the mixture = 5 : 4 there are 20 liters of mixture , so we need to add 25 liters of pure water . the answer is e ." | a = 20 / 10
b = 60 / 10
c = a * b
|
a ) $ 250000 , b ) $ 430000 , c ) $ 120000 , d ) $ 170000 , e ) $ 150000 | c | add(multiply(multiply(subtract(1, divide(20, const_100)), subtract(1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | the value of a machine depreciates at 20 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "c $ 1 , 20,000 the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = $ 96,000 sp such that a profit of $ 24,000 is made = 96,000 + 24,000 = $ 1 , 20,000" | a = 20 / 100
b = 1 - a
c = 20 / 100
d = 1 - c
e = b * d
f = 100 * 100
g = math.sqrt(100)
h = f * g
i = math.sqrt(100)
j = i / 2
k = j * 100
l = k * 100
m = h + l
n = e * m
o = 20 + 2
p = o * 100
q = math.sqrt(100)
r = p * q
s = n + r
|
a ) 42 , b ) 70 , c ) 140 , d ) 165 , e ) 462 | e | multiply(multiply(12, 3), 7) | a certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 12 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidat... | "ans : 462 7 c 1 * 12 c 2 answer e )" | a = 12 * 3
b = a * 7
|
a ) a ) 100 , b ) b ) 120 , c ) c ) 250 , d ) d ) 200 , e ) e ) 160 | d | divide(multiply(20, const_100), 10) | an inspector rejects 10 % of the meters as defective . how many will he examine to reject 20 ? | "then , 10 % of x = 20 ( 10 / 100 ) x = 20 x = ( 20 * 100 * ) / 10 = 200 answer is d" | a = 20 * 100
b = a / 10
|
a ) 67 , b ) 88 , c ) 4.5 , d ) 26 , e ) 12 | c | divide(multiply(6, 3), subtract(6, 3)) | pipe a can fill a tank in 3 hours . due to a leak at the bottom , it takes 6 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 3 - 1 / x = 1 / 6 = > 1 / x = 1 / 2 - 1 / 6 = 1 / 4.5 = > x = 4.5 answer : c" | a = 6 * 3
b = 6 - 3
c = a / b
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | b | divide(subtract(divide(60, const_2), sqrt(subtract(multiply(divide(60, const_2), divide(60, const_2)), multiply(const_4, 200)))), const_2) | the area of a rectangular field is equal to 200 square meters . its perimeter is equal to 60 meters . find the width of this rectangle . | "l * w = 200 : area , l is the length and w is the width . 2 l + 2 w = 60 : perimeter l = 30 - w : solve for l ( 30 - w ) * w = 200 : substitute in the area equation w = 10 and l = 20 : correct answer b" | a = 60 / 2
b = 60 / 2
c = 60 / 2
d = b * c
e = 4 * 200
f = d - e
g = math.sqrt(f)
h = a - g
i = h / 2
|
a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.0 feet , e ) 3.5 feet | d | divide(add(add(multiply(add(2, 10), 2), 10), 2), add(2, 10)) | carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 2 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 2 = 36 36 / 12 = 3.0 feet answer : d" | a = 2 + 10
b = a * 2
c = b + 10
d = c + 2
e = 2 + 10
f = d / e
|
a ) s . 57.40 , b ) s . 57.22 , c ) s . 51.219 , d ) s . 56.18 , e ) s . 53.11 | a | subtract(add(add(divide(multiply(divide(56, multiply(divide(5, const_100), 2)), 5), const_100), divide(56, multiply(divide(5, const_100), 2))), divide(multiply(add(divide(multiply(divide(56, multiply(divide(5, const_100), 2)), 5), const_100), divide(56, multiply(divide(5, const_100), 2))), 5), const_100)), divide(56, m... | if the simple interest on a sum of money for 2 years at 5 % per annum is rs . 56 , what is the compound interest on the same sum at the rate and for the same time ? | "explanation : sum = ( 56 * 100 ) / ( 2 * 5 ) = rs . 560 amount = [ 560 * ( 1 + 5 / 100 ) 2 ] = rs . 617.4 c . i . = ( 617.4 - 560 ) = rs . 57.4 . answer : a" | a = 5 / 100
b = a * 2
c = 56 / b
d = c * 5
e = d / 100
f = 5 / 100
g = f * 2
h = 56 / g
i = e + h
j = 5 / 100
k = j * 2
l = 56 / k
m = l * 5
n = m / 100
o = 5 / 100
p = o * 2
q = 56 / p
r = n + q
s = r * 5
t = s / 100
u = i + t
v = 5 / 100
w = v * 2
x = 56 / w
y = u - x
|
a ) 22 , b ) 66 , c ) 88 , d ) 100 , e ) 281 | a | subtract(divide(multiply(divide(multiply(18, 8), 30), 50), 6), 18) | 18 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 18 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 40 40 – 18 = 22 answer : a" | a = 18 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 18
|
['a ) 400 square meters', 'b ) 4000 square meters', 'c ) 20 square meters', 'd ) 2000 square meters', 'e ) 1000 square meters'] | b | multiply(multiply(power(divide(280, multiply(add(5, 2), const_2)), const_2), 5), 2) | the perimeter of a rectangle is equal to 280 meters . the ratio of its length to its width is 5 : 2 . find the area of the rectangle . | solution if the ratio of the length to the width is 5 : 2 , then the measure l of the length and and the measure w of the with can be written as l = 5 x and w = 2 x we now use the perimeter to write 280 = 2 ( 2 l + 2 w ) = 2 ( 5 x + 2 x ) = 14 x solve for x 280 = 14 x x = 280 / 14 = 20 the area a of the rectangle is gi... | a = 5 + 2
b = a * 2
c = 280 / b
d = c ** 2
e = d * 5
f = e * 2
|
a ) 291.67 , b ) 105 , c ) 301.05 , d ) 288 , e ) 249.55 | a | divide(multiply(35, 3), divide(36, const_100)) | at the end of year x , automobile installment credit accounted for 36 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 35 billion of credit , or 1 / 3 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at ... | "system of equations a = ( 36 / 100 ) c ( 1 / 3 ) a = 35 - - > a = 105 substitution 105 = ( 36 / 100 ) c c = ( 100 / 36 ) 105 - you can do the ugly calculation 105 / 36 * 100 the correct answer is a . the correct answer is 291.67" | a = 35 * 3
b = 36 / 100
c = a / b
|
a ) 200 m , b ) 600 m , c ) 400 m , d ) 800 m , e ) 900 m | c | multiply(divide(400, divide(add(15, 5), const_2)), const_10) | the cross - section of a tunnel is a trapezium in shape . if the tunnel is 15 m wide at the top and 5 m wide at the bottom and the area of cross - section is 400 sq m , the depth of tunnel is ? | 1 / 2 * d ( 15 + 5 ) = 400 d = 400 answer : c | a = 15 + 5
b = a / 2
c = 400 / b
d = c * 10
|
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