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 |
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a ) 45 , b ) 36 , c ) 41 , d ) 51 , e ) 48 | a | add(28, const_1) | the average age of 28 students in a group is 16 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ? | "sol . age of the teacher = ( 29 ã — 17 â € “ 28 ã — 16 ) years = 45 years . answer a" | a = 28 + 1
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a ) 2.15 pm , b ) 1.15 pm , c ) 4.15 pm , d ) 3.15 pm , e ) 12.15 pm | a | subtract(add(divide(add(390, 35), add(65, 35)), 10), add(10, const_2)) | a and b are two stations 390 km apart . a train starts from a at 10 a . m . and travels towards b at 65 kmph . another train starts from b at 11 a . m . and travels towards a at 35 kmph . at what time do they meet ? | suppose they meet x hours after 10 a . m . then , ( distance moved by first in x hrs ) + [ distance moved by second in ( x - 1 ) hrs ] = 390 . 65 x + 35 ( x - 1 ) = 390 = > 100 x = 425 = > x = 17 / 4 so , they meet 4 hrs . 15 min . after 10 a . m i . e . , at 2.15 p . m . answer a | a = 390 + 35
b = 65 + 35
c = a / b
d = c + 10
e = 10 + 2
f = d - e
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a ) 8 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | a | divide(16, subtract(const_4, const_2)) | in a group of cows and chickens , the number of legs was 16 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 16 more than twice the number of heads . therefore , 2 × ( x + y ) + 16 = 4 x + 2 y . or , 2 x + 2 y + 16 = 4 x + 2 y . or , 2 x + 16 = 4 x [ subtracting 2 y from both sides ] . or , 16 = 4 x – 2 x [ subtracting 2 x from both sides ] . or , 16 = 2 x . or , x = 8 [ dividing by 2 on both sides ] . therefore , the number of cows = 7 . correct answer : a ) 8" | a = 4 - 2
b = 16 / a
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a ) 1978 , b ) 2707 , c ) 7728 , d ) 8160 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(1000, add(5000, 1000)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 5000 and b rs . 1000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "5000 : 1000 = > 5 : 1 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 5 / 6 = 7200 + 960 = 8160 answer : d" | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 5000 + 1000
e = 1000 / d
f = c * e
g = 9600 - f
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a ) 12 , b ) 14 , c ) 16 , d ) 17 , e ) 18 | a | subtract(divide(multiply(90, 40), const_100), divide(multiply(80, 30), const_100)) | how much 90 % of 40 is greater than 80 % of 30 ? | "( 90 / 100 ) * 40 – ( 80 / 100 ) * 30 36 - 24 = 12 answer : a" | a = 90 * 40
b = a / 100
c = 80 * 30
d = c / 100
e = b - d
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a ) $ 115,000 , b ) $ 190,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | b | add(multiply(20, 10), 10) | an auction house charges a commission of 20 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "say the price of the house was $ x , then 0.2 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 190,000 ( 20 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : b ." | a = 20 * 10
b = a + 10
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a ) 2 / 1 , b ) 4 / 1 , c ) 2 / 3 , d ) 3 / 4 , e ) 3 / 1 | e | divide(subtract(45, divide(45, add(4, 1))), add(divide(45, add(4, 1)), 3)) | in a mixture of 45 litres the ratio of milk to water is 4 : 1 . additional 3 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 4 x / x and 4 x + x = 45 - - > x = 9 . thus milk = 4 x = 36 liters and water = x = 9 liters . new ratio = 36 / ( 9 + 3 ) = 36 / 12 = 3 / 1 . answer : e ." | a = 4 + 1
b = 45 / a
c = 45 - b
d = 4 + 1
e = 45 / d
f = e + 3
g = c / f
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a ) 19828.29 , b ) 19828.2 , c ) 19828.8 , d ) 19828.88 , e ) 19828.21 | c | add(divide(2828.80, subtract(power(add(const_1, divide(8, const_100)), const_2), const_1)), 2828.80) | the compound interest earned by sunil on a certain amount at the end of two years at the rate of 8 % p . a . was rs . 2828.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned ? | "let the sum be rs . p p { [ 1 + 8 / 100 ] 2 - 1 } = 2828.80 p ( 8 / 100 ) ( 2 + 8 / 100 ) = 2828.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 2828.80 / ( 0.08 ) ( 2.08 ) = 1360 / 0.08 = 17000 principal + interest = rs . 19828.80 answer : c" | a = 8 / 100
b = 1 + a
c = b ** 2
d = c - 1
e = 2828 / 80
f = e + 2828
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a ) rs . 1846 , b ) rs . 18846 , c ) rs . 13846 , d ) rs . 2000 , e ) rs . 14846 | c | divide(divide(multiply(5400, const_100), 3), 13) | a man took loan from a bank at the rate of 13 % p . a . s . i . after 3 years he had to pay rs . 5400 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 5400 ) / ( 13 * 3 ) = rs . 13846 answer : c" | a = 5400 * 100
b = a / 3
c = b / 13
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a ) rs . 4000 , b ) rs . 5000 , c ) rs . 4500 , d ) rs . 4800 , e ) rs . 5800 | a | divide(4410, power(add(divide(5, const_100), const_1), const_2)) | a sum amounts to rs . 4410 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | ci = 4410 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 4410 = p [ 21 / 20 ] ^ 2 4410 [ 20 / 21 ] ^ 2 4000 answer : a | a = 5 / 100
b = a + 1
c = b ** 2
d = 4410 / c
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a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9 | b | divide(divide(add(18, 8), const_2), const_2) | a man can row downstream at 18 kmph and upstream at 8 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 8 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 26 = > x = 13 , y = 5 . answer : b" | a = 18 + 8
b = a / 2
c = b / 2
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a ) 7 / 98 , b ) 1 / 48 , c ) 1 / 98 , d ) 3 / 350 , e ) 1 / 68 | d | inverse(multiply(2, 7)) | the compound ratio of 2 / 3 , 6 / 7 , 1 / 5 and 1 / 8 is given by ? | "2 / 3 * 6 / 7 * 1 / 5 * 1 / 8 = 12 / 1400 = 3 / 350 answer : d" | a = 2 * 7
b = 1/(a)
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a ) 4 , b ) 3 , c ) 1 , d ) 2 , e ) 0 | c | divide(21, 14) | how many of the positive factors of 14 are not factors of 21 ? | "factors of 14 - 1 , 2 , 7 , 14 , factors of 21 - 1 , 3 , 7 , 21 , comparing both , we have three factors of 14 which are not factors of 21 - 2,14 the answer is c" | a = 21 / 14
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a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) 21 litres | c | divide(60, divide(divide(32, 3), subtract(divide(20, 1.5), divide(32, 3)))) | how much water must be added to 60 litres of milk at 1.5 litres for rs . 20 so as to have a mixture worth rs . 32 / 3 a litre ? | c . p . of 1 litre of milk = rs . 20 × 2 / 3 = rs . 40 / 3 . c . p . of 1 litre of water = rs 0 . mean price = rs . 32 / 3 . by the rule of alligation , we have : c . p of 1 litre c . p of 1 litre of water of milk ( 0 ) ( rs . 40 / 3 ) \ / mean price ( rs . 32 / 3 ) / \ 40 / 3 − 32 / 3 32 / ( 3 − 0 ) 8 / 3 32 / 3 the ratio of water and milk = 8 / 3 : 32 / 3 . = 8 : 32 = 1 : 4 . thus , quantity of water to be added to 60 litres of milk : = ( 1 / 4 ) × 60 litres . = 15 litres . answer : c | a = 32 / 3
b = 20 / 1
c = 32 / 3
d = b - c
e = a / d
f = 60 / e
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a ) 24 % , b ) 25 % , c ) 30 % , d ) 36 % , e ) 43 % | e | multiply(divide(255, subtract(850, 255)), const_100) | a cricket bat is sold for $ 850 , making a profit of $ 255 . the profit percentage would be | "255 / ( 850 - 255 ) = 255 / 595 = 51 / 119 = 43 % . answer : e ." | a = 850 - 255
b = 255 / a
c = b * 100
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a ) 104 kmph , b ) 194 kmph , c ) 109 kmph , d ) 174 kmph , e ) 240 kmph | e | divide(624, add(2, divide(3, 5))) | a car covers a distance of 624 km in 2 3 / 5 hours . find its speed ? | 624 / 2 3 / 5 = 240 kmph answer : e | a = 3 / 5
b = 2 + a
c = 624 / b
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a ) 5 square meters , b ) 36 square meters , c ) 42 square meters , d ) 45 square meters , e ) 120 square meters | d | divide(subtract(subtract(300, 180), 30), const_2) | three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 30 square meters . what is the area that is covered with three layers of wallpaper ? | "300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 30 = 90 sq m of the wallpaper overlaps in three layers . 90 sq m makes two extra layers hence the area over which it makes two extra layers is 45 sq m . answer ( d ) ." | a = 300 - 180
b = a - 30
c = b / 2
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a ) 10 , b ) 20 , c ) 30 , d ) 27.7 , e ) 50 | d | add(20, const_1) | the average of first three prime numbers greater than 20 is ? | "23 + 29 + 31 = 83 / 3 = 27.7 answer : d" | a = 20 + 1
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a ) $ 50 , b ) $ 64 , c ) $ 75 , d ) $ 96 , e ) can not be determined | a | divide(add(60, 40), 2) | if greg buys 3 shirts , 4 trousers and 2 ties , the total cost is $ 40 . if greg buys 7 shirts , 2 trousers and 2 ties , the total cost is $ 60 . how much will it cost him to buy 3 trousers , 5 shirts and 2 ties ? | "solution : 3 x + 4 y + 2 z = 40 7 x + 2 y + 2 z = 60 adding both the equations = 10 x + 6 y + 4 z = 100 5 x + 3 y + 2 z = 50 ans a" | a = 60 + 40
b = a / 2
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a ) 120 cm , b ) 150 cm , c ) 160 cm , d ) 190 cm , e ) of these | a | multiply(divide(150, subtract(add(4, 6), 5)), 4) | the heights of 3 individuals are in the ratio 4 : 5 : 6 . if the sum of the heights of the heaviest and the lightest boy is 150 cm more than the height of the third boy , what is the weight of the lightest boy ? | let the heights of the three boys be 4 k , 5 k and 6 k respectively . 4 k + 6 k = 5 k + 150 = > 5 k = 150 = > k = 30 therefore the height of the lightest boy = 4 k = 4 ( 30 ) = 120 cm . answer : a | a = 4 + 6
b = a - 5
c = 150 / b
d = c * 4
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a ) 9 , b ) 15 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(29, 5), 9) | according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 29 , by how much does the temperature in degrees celsius ( c ) increase ? | "you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 29 = 61 - - > c = 5 / 9 * 29 = 16.11 . increase = 16.11 degrees . answer : b ." | a = 29 * 5
b = a / 9
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a ) rs . 80 , b ) rs . 100 , c ) rs . 120 , d ) rs . 145 , e ) rs . 156 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | a man engaged a servant on the condition that he would pay him rs . 900 and auniform after 1 year service . he served only for 9 months and received uniform and rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a = 1 * 900
b = 3 * 4
c = b * 9
d = a - c
e = 3 * 4
f = e * 1
g = d / f
h = g * 4
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a ) rs 800 , b ) rs 820 , c ) rs 860 , d ) rs 850 , e ) none of these | d | divide(subtract(1190, multiply(1190, divide(12, const_100))), add(divide(25, const_100), const_1)) | the sale price of a trolley bag including the sale tax is rs . 1190 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is : | "explanation : 112 % of s . p . = 1190 s . p . = rs . ( 1190 x 100 / 112 ) = rs . 1062.50 . c . p . = rs ( 100 / 125 x 1062.50 ) = rs 850 answer : d" | a = 12 / 100
b = 1190 * a
c = 1190 - b
d = 25 / 100
e = d + 1
f = c / e
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a ) 22 , b ) 77 , c ) 70 , d ) 98 , e ) 72 | e | add(add(multiply(divide(60, 6), const_2), 6), add(divide(60, 6), 6)) | the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 6 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 6 years , their ages will be 46 and 26 . sum of their ages = 46 + 26 = 72 . answer : e" | a = 60 / 6
b = a * 2
c = b + 6
d = 60 / 6
e = d + 6
f = c + e
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a ) 920 , b ) 930 , c ) 940 , d ) 950 , e ) 960 | d | divide(divide(divide(3800, divide(add(5, 3), const_2)), 3), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 5 m wide at the top and 3 m wide at the bottom and the area of cross - section is 3800 sq m , the depth of cannel is ? | "1 / 2 * d ( 5 + 3 ) = 3800 d = 950 answer : d" | a = 5 + 3
b = a / 2
c = 3800 / b
d = c / 3
e = d / 2
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | divide(add(74, 6), add(4, 6)) | maxwell leaves his home and walks toward brad ' s house . one hour later , brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 74 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h . what is the total time it takes maxwell before he meets up with brad ? | "total distance = 74 kms maxwell speed = 4 kms / hr maxwell travelled for 1 hour before brad started , therefore maxwell traveled for 4 kms in 1 hour . time taken = total distance / relative speed total distance after brad started = 70 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken to meet brad after brad started = 70 / 10 = 7 hrs distance traveled by maxwell = maxwell ' s speed * time taken = 4 * 7 = 28 + 4 = 32 kms . . . therefore total time taken by maxwell to meet brad = distance travelled by maxwell / maxwell ' s speed = 32 / 4 = 8 hrs . . . answer e" | a = 74 + 6
b = 4 + 6
c = a / b
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['a ) 45', 'b ) 50', 'c ) 60', 'd ) 70', 'e ) 80'] | c | subtract(multiply(divide(add(sqrt(subtract(power(add(68, 32), const_2), multiply(divide(add(sqrt(subtract(power(multiply(68, 32), const_2), multiply(power(960, const_2), const_4))), multiply(68, 32)), const_2), const_4))), add(68, 32)), const_2), const_2), add(68, 32)) | the two sides of a triangle are 32 and 68 . the area is 960 sq . cm . find the third side of triangle ? | we see 68 ^ 2 - 32 ^ 2 = ( 68 + 32 ) * ( 68 - 32 ) = 100 * 36 = 60 ^ 2 now ( 1 / 2 ) * 60 * 32 = 960 ( match with given options ) ( i . e area of a right angled triangle whose sides are 32 , 60,68 ) third side = 60 answer : c | a = 68 + 32
b = a ** 2
c = 68 * 32
d = c ** 2
e = 960 ** 2
f = e * 4
g = d - f
h = math.sqrt(g)
i = 68 * 32
j = h + i
k = j / 2
l = k * 4
m = b - l
n = math.sqrt(m)
o = 68 + 32
p = n + o
q = p / 2
r = q * 2
s = 68 + 32
t = r - s
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a ) 80 , b ) 126 , c ) 95 , d ) 100 , e ) 108 | b | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 40), const_100)) | from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 40 percent . what percent of the stock â € ™ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 40 % = 90 * 1.2 = 126 thus ratio = 126 / 100 = 1.26 ( in % terms = 126 % ) . thus b is the correct answer ." | a = 100 + 20
b = a / 100
c = 100 * b
d = 100 - 25
e = d / 100
f = c * e
g = 100 + 40
h = g / 100
i = f * h
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a ) - 4 , b ) - 2 , c ) 0 , d ) 2 , e ) 80 | e | multiply(negate(multiply(divide(80, 2), 2)), 9) | if 9 a - b = 10 b + 80 = - 12 b - 2 a , what is the value of 9 a - 11 b ? | "this implies 9 a - b = 10 b + 80 , 9 a - b = - 12 b - 2 a , 10 b + 80 = - 12 b - 2 a manipulating the second equation gives us 9 a - b = 10 b + 80 = = > 9 a - 11 b = 80 answer is e" | a = 80 / 2
b = a * 2
c = negate * (
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a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 40 % | e | multiply(divide(subtract(70, 50), 50), const_100) | john makes $ 50 a week from his job . he earns a raise and now makes $ 70 a week . what is the % increase ? | "increase = ( 20 / 50 ) * 100 = 40 % . e" | a = 70 - 50
b = a / 50
c = b * 100
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a ) 1 / 2 , b ) 2 , c ) 1 / 6 , d ) 3 , e ) 1 / 4 | c | divide(1, divide(add(subtract(5, 1), 1), 1)) | the vertex of a rectangle are ( 1 , 0 ) , ( 5 , 0 ) , ( 1 , 1 ) and ( 5 , 1 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | "if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 3 , 0.5 ) . the slope of a line passing through ( 0,0 ) and ( 3 , 0.5 ) is 0.5 / 3 = 1 / 6 . the answer is c ." | a = 5 - 1
b = a + 1
c = b / 1
d = 1 / c
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a ) 1 / 5 , b ) 1 / 3 , c ) 8 / 27 , d ) 1 / 27 , e ) 7 / 8 | c | divide(power(const_2, 3), power(3, const_3)) | on a game show , there are 3 tables . each table has 3 boxes ( one zonk ! , one cash prize , and one grand prize ) . the contestant can choose one box from each table for a total of 3 boxes . if any of the boxes are a zonk ! , the contestant loses everything . what is the probability of getting no zonk ! from any of the 3 boxes chosen ? | no zonk ! : 1 st box = no zonk ! = 2 / 3 2 nd box = no zonk ! = 2 / 3 3 rd box = no zonk ! = 2 / 3 ( 2 / 3 ) ( 2 / 3 ) ( 2 / 3 ) = 8 / 27 8 / 27 is the probability of no zonk ! , so . . . answer : c | a = 2 ** 3
b = 3 ** 3
c = a / b
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a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 40 kg , e ) 37 kg | d | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(42, const_3)) | the average weight of a , b and c is 42 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 = ( 42 x 3 ) = 126 . . . . ( 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 = 40 . b ' s weight = 40 kg . d" | a = 40 * 2
b = 43 * 2
c = a + b
d = 42 * 3
e = c - d
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a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 40 / 27 , e ) 5 ⁄ 16 | d | multiply(divide(8, 3), divide(5, 9)) | dividing by 3 ⁄ 8 and then multiplying by 5 ⁄ 9 is the same as dividing by what number ? | say x / 3 / 8 * 5 / 9 = x * 8 / 3 * 5 / 9 = x * 40 / 27 d | a = 8 / 3
b = 5 / 9
c = a * b
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | subtract(divide(100, const_10), const_2) | how many integers n greater than and less than 100 are there such that , if the digits of n are reversed , the resulting integer is n + 9 ? | ( 10 x + y ) = ( 10 y + x ) + 9 = > 9 x - 9 y = 9 = > x - y = 1 and only 8 numbers satisfy this condition and the numbers are 21,32 , 43,54 , 65,76 , 87,98 answer : d | a = 100 / 10
b = a - 2
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a ) 75 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | d | multiply(50, const_2) | a runs twice as fast as b and gives b a start of 50 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 50 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 50 m is 50 / 1 = 50 m so the total time taken by a and b to reach = 2 * 50 = 100 m answer : d" | a = 50 * 2
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a ) 20 % , b ) 25 % , c ) 13 1 / 30 % , d ) 32 % , e ) 13 % | d | subtract(divide(subtract(const_100, 12), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 12 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 88 2 / 3 sp 1 = 88 sp 1 = 132 100 - - - 32 = > 32 % answer : d" | a = 100 - 12
b = 2 / 3
c = a / b
d = c - 100
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a ) 76.6 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | divide(add(add(add(add(65, 67), 76), 80), 95), add(const_4, const_1)) | reeya obtained 65 , 67 , 76 , 80 and 95 out of 100 in different subjects , what will be the average | explanation : ( 65 + 67 + 76 + 80 + 95 / 5 ) = 76.6 option a | a = 65 + 67
b = a + 76
c = b + 80
d = c + 95
e = 4 + 1
f = d / e
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a ) 2.215 . , b ) 2.55 , c ) 2.775 . , d ) 3.2 . , e ) 3.5 . | b | multiply(divide(divide(600, const_1000), divide(12, const_60)), subtract(const_1, divide(15, const_100))) | an ant walks an average of 600 meters in 12 minutes . a beetle walks 15 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 600 meters in 12 minutes 600 meters in 1 / 5 hours the beetle walks 15 % less distance = 600 - 90 = 510 meters in 12 minutes 0.510 km in 12 / 60 = 1 / 5 hours speed = 0.510 * 5 = 2.55 km / h correct answer b = 2.55" | a = 600 / 1000
b = 12 / const_60
c = a / b
d = 15 / 100
e = 1 - d
f = c * e
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a ) 440 , b ) 490 , c ) 640 , d ) 740 , e ) 250 | b | divide(multiply(450, 4), divide(add(multiply(3, 3), 2), 3)) | an aeroplane covers a certain distance of 450 kmph in 4 hours . to cover the same distance in 3 2 / 3 hours , it must travel at a speed of | "speed of aeroplane = 450 kmph distance travelled in 4 hours = 450 * 4 = 1800 km speed of aeroplane to acver 1800 km in 11 / 3 = 1800 * 3 / 11 = 490 km answer b ." | a = 450 * 4
b = 3 * 3
c = b + 2
d = c / 3
e = a / d
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a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 9 | d | subtract(72, 72) | when the number 72 y 6139 is exactly divisible by 11 , then the smallest whole number that can replace y is ? | "the given number = 72 y 6139 sum of the odd places = 9 + 1 + y + 7 sum of the even places = 3 + 6 + 2 = 11 ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) ( 16 + y ) - ( 11 ) = divisible by 11 y � 5 = divisible by 11 . y must be 6 , to make given number divisible by 11 . d" | a = 72 - 72
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a ) $ 4100 , b ) $ 4300 , c ) $ 4500 , d ) $ 4700 , e ) $ 4900 | b | multiply(divide(subtract(42000, multiply(divide(23700, 8), 12)), 12), 8) | a 12 month project had a total budget of $ 42000 . after 8 months , the project had spent $ 23700 . at this point , how much was the project under budget ? | each month , the project should spend $ 42,000 / 12 = $ 3500 . in 8 months , the project should spend 8 * $ 3500 = $ 28,000 . the project is under budget by $ 28,000 - $ 23,700 = $ 4300 . the answer is b . | a = 23700 / 8
b = a * 12
c = 42000 - b
d = c / 12
e = d * 8
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(multiply(4, const_4), 2) | a is the average ( arithmetic mean ) of the first 7 positive multiples of 4 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 – b ^ 2 is zero , what is the value of n ? | "if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 16 , which also equals b . b is the second positive multiple of n , thus n = 16 / 2 = 8 . the answer is d ." | a = 4 * 4
b = a / 2
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a ) 102 m sqaure , b ) 100 m sqaure , c ) 152 m sqaure , d ) 164 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(85, const_100), 1), 6), multiply(add(divide(85, const_100), 1), 8))), multiply(6, 8)) | a cistern 8 m long and 6 m wide contains water up to a breadth of 1 m 85 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 1.85 + 8 * 1.85 ) ] + 8 * 6 = 100 m square option b" | a = 85 / 100
b = a + 1
c = b * 6
d = 85 / 100
e = d + 1
f = e * 8
g = c + f
h = 2 * g
i = 6 * 8
j = h + i
|
a ) 65 , b ) 60 , c ) 55 , d ) 40 , e ) 45 | a | multiply(divide(60, subtract(13, 1)), 13) | the ratio of buses to cars on river road is 1 to 13 . if there are 60 fewer buses than cars on river road , how many cars are on river road ? | "b / c = 1 / 13 c - b = 60 . . . . . . . . . > b = c - 60 ( c - 60 ) / c = 1 / 13 testing answers . clearly eliminate bcde put c = 65 . . . . . . . . . > ( 65 - 60 ) / 65 = 5 / 65 = 1 / 13 answer : a" | a = 13 - 1
b = 60 / a
c = b * 13
|
a ) 27 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | a | add(subtract(multiply(multiply(const_3, const_3), multiply(2, const_2)), add(multiply(const_3, const_3), multiply(2, const_2))), multiply(2, const_2)) | if a takes x days to do a work then b takes 2 x days to do the same work then with in how many day ' s a will alone complete this work ? | 1 / x + 1 / 2 x = 1 / 18 = > 3 / 2 x = 1 / 18 = > x = 27 days . hence , a alone can finish the work in 27 days . answer : a | a = 3 * 3
b = 2 * 2
c = a * b
d = 3 * 3
e = 2 * 2
f = d + e
g = c - f
h = 2 * 2
i = g + h
|
a ) 64 , b ) 128 , c ) 152 , d ) 154 , e ) 256 | d | subtract(volume_cube(add(cube_edge_by_volume(62), const_2)), 62) | 62 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "62 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 62 = 154 hence , d is the answer ." | a = cube_edge_by_volume + (
b = volume_cube - (
|
a ) 1 / 16 , b ) 2 / 14 , c ) 1 / 12 , d ) 1 / 19 , e ) 1 / 17 | c | add(add(divide(const_1, power(const_6, const_2)), divide(const_1, power(const_6, const_2))), divide(const_1, power(const_6, const_2))) | if you throw two dice at a same time . can you find the probability of getting sum as 10 of the two numbers shown ? | c 1 / 12 all possible cases can be 36 ( 6 * 6 ) case we need : [ ( 4,6 ) , ( 5,5 ) , ( 6,4 ) ] = 3 probability = > 3 / 36 = 1 / 12 | a = 6 ** 2
b = 1 / a
c = 6 ** 2
d = 1 / c
e = b + d
f = 6 ** 2
g = 1 / f
h = e + g
|
a ) 480 , b ) 48 , c ) 50 , d ) 144 , e ) 160 | b | divide(subtract(divide(multiply(divide(16, divide(10, const_100)), 90), const_100), multiply(divide(16, divide(10, const_100)), divide(const_1, const_2))), multiply(const_2, const_4)) | in an office , 30 percent of the workers have at least 5 years of service , and a total of 16 workers have at least 10 years of service . if 90 percent of the workers have fewer than 10 years of service , how many of the workers have at least 5 but fewer than 10 years of service ? | "( 10 / 100 ) workers = 16 = > number of workers = 160 ( 30 / 100 ) * workers = x + 16 = > x = 48 answer b" | a = 10 / 100
b = 16 / a
c = b * 90
d = c / 100
e = 10 / 100
f = 16 / e
g = 1 / 2
h = f * g
i = d - h
j = 2 * 4
k = i / j
|
a ) a ) 2 , b ) b ) 3 , c ) c ) 1 , d ) d ) 5 , e ) e ) 6 | c | subtract(820, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | the least number which must be subtracted from 820 to make it exactly divisible by 9 is : | "on dividing 820 by 9 , we get remainder = 1 therefore , required number to be subtracted = 1 answer : c" | a = 4 + 1
b = a * 10
c = 4 + 2
d = b + c
e = d * 9
f = 820 - e
|
['a ) $ 1', 'b ) $ 2', 'c ) $ 4', 'd ) $ 8', 'e ) $ 16'] | e | multiply(multiply(4, const_2), divide(power(const_2, const_2), const_2)) | can n and can в are both right circular cylinders . the radius of can n is twice the radius of can b , while the height of can n is half the height of can b . if it costs $ 4.00 to fill half of can b with a certain brand of gasoline , how much would it cost to completely fill can n with the same brand of gasoline ? | let x be the radius of b and 2 h be the height of b . therefore , radius of n = 2 x and height = h vol of b = 3.14 * x ^ 2 * 2 h vol of a = 3.14 * 4 x ^ 2 * h cost to fill half of b = $ 4 - - > cost to fill full b = $ 8 - - > 3.14 * x ^ 2 * 2 h = 8 - - > 3.14 * x ^ 2 * h = 4 - - > 4 * ( 3.14 * x ^ 2 * h ) = $ 16 ans e | a = 4 * 2
b = 2 ** 2
c = b / 2
d = a * c
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 27 % , e ) 40 % | d | divide(divide(multiply(40, 60), multiply(120, 60)), add(divide(multiply(40, 60), multiply(120, 60)), 8)) | a certain car can travel 40 minutes on a gallon of gasoline at 60 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 120 miles at 60 mph ? | "total time for travelling 120 miles @ 60 mph = 120 / 60 = 2 hour = 120 minutes . given , the car uses 1 gallon for every 40 minutes of driving @ 60 mph . thus in 120 minutes it will use = 3 gallons . thus , full tank = 3 + 8 = 11 gallons - - - > 3 / 11 = 27 % of the fuel used . d is the correct answer ." | a = 40 * 60
b = 120 * 60
c = a / b
d = 40 * 60
e = 120 * 60
f = d / e
g = f + 8
h = c / g
|
a ) 120 cm 2 , b ) 765 cm 2 , c ) 216 cm 2 , d ) 270 cm 2 , e ) 275 cm 2 | d | divide(multiply(36, 15), const_2) | if the sides of a triangle are 39 cm , 36 cm and 15 cm , what is its area ? | "the triangle with sides 39 cm , 36 cm and 15 cm is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 36 * 15 = 270 cm 2 answer : d" | a = 36 * 15
b = a / 2
|
a ) 136 , b ) 137 , c ) 138 , d ) 139 , e ) 140 | d | add(multiply(19, 7), 6) | what is the dividend . divisor 19 , the quotient is 7 and the remainder is 6 | "d = d * q + r d = 19 * 7 + 6 d = 133 + 6 d = 139" | a = 19 * 7
b = a + 6
|
a ) 6.67 % , b ) 12.5 % , c ) 18.75 % , d ) 25 % , e ) 33.3 % | d | subtract(subtract(multiply(multiply(divide(circumface(120), 120), const_3), const_2), const_4), const_2) | obra drove 120 π meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 π square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 π square meters so , π ( r ^ 2 ) = 57,600 π - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 π r = 480 π therefore , part of circumference covered = 120 π / 480 π = 25 % hence , answer is d ." | a = circumface / (
b = a * 120
c = b * 3
d = c - 2
e = d - 4
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | c | multiply(multiply(20, divide(const_1, const_2)), divide(const_1, const_2)) | some persons can do a piece of work in 20 days . two times the number of these people will do half of that work in ? | "20 / ( 2 * 2 ) = 5 days answer : c" | a = 1 / 2
b = 20 * a
c = 1 / 2
d = b * c
|
['a ) $ 1.60', 'b ) $ 16.00', 'c ) $ 96.00', 'd ) $ 108.00', 'e ) $ 196.00'] | a | multiply(divide(3.2, 1200), surface_cube(10)) | if paint costs $ 3.20 per quart , and a quart covers 1200 square feet , how much will it cost to paint the outside of a cube 10 feet on each edge ? | total surface area = 6 a ^ 2 = 6 * 10 * 10 = 600 each quart covers 20 sqr ft thus total number of quarts = 600 / 1200 = 0.5 cost will be 0.5 * 3.2 = $ 1.6 ans : a | a = 3 / 2
b = a * surface_cube
|
['a ) 20 ft', 'b ) 25 ft', 'c ) 750 ft', 'd ) 900 ft', 'e ) 1650 ft'] | e | multiply(30, add(divide(multiply(30, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 33 ft long . what is the distance traveled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 33 . ( formula for calculating the distance traveled by each wheel is : # of revolutions * circumference . ) solving this eqn . gives x = 50 . sub x = 50 either in ( x + 5 ) * 30 or in x * 33 to get the distance , which is 1650 . so the correct choice is e . | a = 10 / 2
b = 30 * a
c = b / 3
d = 10 / 2
e = c + d
f = 30 * e
|
a ) w = 16 , b ) w = 8 √ 2 , c ) w = 8 , d ) w = 2 √ 2 , e ) ( √ 2 ) / 3 | d | sqrt(divide(multiply(4, 4), const_2)) | the two lines y = x and x = - 4 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length w of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 4 , - 4 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 4 × 4 ) = 8 ; so z = 8 . the next step requires us to find the length of a side of a cube that has a face area equal to 8 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 8 . since the area of a square is equal to s ² , where s is the length of one of its side , we can write and solve the equation s ² = 8 . clearly s = √ 8 = 2 √ 2 , oranswer choice ( d ) ." | a = 4 * 4
b = a / 2
c = math.sqrt(b)
|
a ) 50.5 % , b ) 42.8 % , c ) 22.2 % , d ) 33.3 % , e ) 25 % | b | multiply(divide(subtract(50, 20), add(50, 20)), const_100) | if 50 % of ( x - y ) = 20 % of ( x + y ) , then what percent of x is y ? | "50 % of ( x - y ) = 20 % of ( x + y ) 50 / 100 ( x - y ) = 20 / 100 ( x + y ) 3 x = 7 y required percentage = y / x * 100 = 3 y / 7 y * 100 = 42.85 % answer is b" | a = 50 - 20
b = 50 + 20
c = a / b
d = c * 100
|
a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | d | divide(subtract(34, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 34 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "34 % is 9 % - points above 25 % and 6 % - points below 40 % . thus the ratio of mixture y to mixture x is 2 : 3 . the percent of mixture x is 3 / 5 = 60 % . the answer is d ." | a = 25 + 1
b = 34 - a
c = 40 / 100
d = 25 + 1
e = d / 100
f = c - e
g = b / f
|
['a ) 1 : 1', 'b ) 3 : 2', 'c ) 4 : 1', 'd ) 3 : 8', 'e ) 2 : 5'] | d | divide(subtract(6, divide(multiply(8, 6), 10)), subtract(8, divide(multiply(8, 6), 10))) | in a triangle abc , ab = 6 , bc = 8 and ac = 10 . a perpendicular dropped from b , meets the side ac at d . a circle of radius bd ( with centre b ) is drawn . if the circle cuts ab and bc at p and q respectively , then ap : qc is equal to | explanation : triangle abc and triangle adb are similar , ac / ab = bc / bd 10 / 6 = 8 / r r = 24 / 5 bp = bq = bd = r = 24 / 5 ap = ab - r = 6 - 24 / 5 = 6 / 5 cq = bc - r = 8 - 24 / 5 = 16 / 5 ap : cq = 6 / 16 = 3 : 8 answer : d | a = 8 * 6
b = a / 10
c = 6 - b
d = 8 * 6
e = d / 10
f = 8 - e
g = c / f
|
a ) 59 , b ) 61 , c ) 63 , d ) 65 , e ) 67 | c | divide(add(49, 81), const_2) | find the mean proportional between 49 & 81 ? | formula = √ a × b a = 49 and b = 81 √ 49 × 81 = 7 × 9 = 63 c | a = 49 + 81
b = a / 2
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a ) 41100 , b ) 42000 , c ) 21100 , d ) 42200 , e ) 21000 | c | subtract(add(6050, 8000), 7050) | the average monthly income of p and q is rs . 6050 . the average monthly income of q and r is rs . 7050 and the average monthly income of p and r is rs . 8000 . the monthly income of p + q + r is : | "explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 6050 x 2 ) = 12100 . . . . ( i ) q + r = ( 7050 x 2 ) = 14100 . . . . ( ii ) p + r = ( 8000 x 2 ) = 16000 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 42200 or p + q + r = 21100 . . . . ( iv ) ( p + q + r ) ' s monthly income = rs . 21100 . answer : c" | a = 6050 + 8000
b = a - 7050
|
a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 320 | d | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 2 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with two 2 - s , 2 x 2 x 1 * 5 * 1 * 4 = 20 . thus there are 320 - 20 = 300 such numbers . answer : d ." | a = 4 + 1
b = 4 + 1
c = a * b
d = 4 * 4
e = c * d
f = 4 + 1
g = 4 + 1
h = f * g
i = h * 4
j = e - i
|
a ) 2 , b ) 4 , c ) 7 , d ) 12 , e ) 15 | c | divide(add(9, 5), const_2) | in one hour , a boat goes 9 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 9 + 5 ) kmph = 7 kmph . answer c" | a = 9 + 5
b = a / 2
|
a ) 35 , b ) 20 , c ) 28 , d ) 36 , e ) 30 | c | subtract(70, divide(70, add(divide(2, 3), const_1))) | a 70 cm long wire is to be cut into two pieces so that one piece will be 2 / 3 th of the other , how many centimeters will the shorter piece be ? | "1 : 2 / 3 = 3 : 2 2 / 5 * 70 = 28 answer : c" | a = 2 / 3
b = a + 1
c = 70 / b
d = 70 - c
|
a ) 66 liters , b ) 32 liters , c ) 41 liters , d ) 50 liters , e ) 34 liters | a | multiply(divide(165, add(3, 2)), 2) | 165 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 165 = 99 liters water = 66 liters 99 : ( 66 + p ) = 3 : 4 198 + 3 p = 396 = > p = 66 66 liters of water are to be added for the ratio become 3 : 4 . answer : a" | a = 3 + 2
b = 165 / a
c = b * 2
|
a ) 129 , b ) 287 , c ) 196 , d ) 188 , e ) 112 | c | add(divide(add(multiply(54, 150), multiply(36, 125)), add(36, 54)), multiply(divide(add(multiply(54, 150), multiply(36, 125)), add(36, 54)), divide(40, const_100))) | raman mixed 54 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 54 ( 150 ) + 36 ( 125 ) ] / ( 54 + 36 ) = rs . 140 sp = cp [ ( 100 + profit % ) / 100 ] = 140 * [ ( 100 + 40 ) / 100 ] = rs . 196 . answer : c" | a = 54 * 150
b = 36 * 125
c = a + b
d = 36 + 54
e = c / d
f = 54 * 150
g = 36 * 125
h = f + g
i = 36 + 54
j = h / i
k = 40 / 100
l = j * k
m = e + l
|
a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1200 | a | divide(multiply(subtract(850, 800), 4), divide(25, const_100)) | average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 25 % the new mean is $ 850 . what was original rent of friend whose rent is increased ? | "0.25 x = 4 ( 850 - 800 ) 0.25 x = 200 x = 800 answer a" | a = 850 - 800
b = a * 4
c = 25 / 100
d = b / c
|
a ) 50 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | a | divide(subtract(divide(multiply(multiply(5000, 5), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 5 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 4 × 2 / 100 = 400 he also lends it at 5 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 5 × 2 / 100 = 500 his overall gain in 2 years = rs . 500 - rs . 400 = rs . 100 his overall gain in 1 year = 100 / 2 = rs . 50 answer : option a" | a = 5000 * 5
b = a * 2
c = b / 100
d = 5000 * 4
e = d * 2
f = e / 100
g = c - f
h = g / 2
|
a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | d | divide(15, subtract(30, 15)) | a chemist mixes one liter of pure water with x liters of a 30 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 30 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 30 - 15 ) / ( 15 - 0 ) = 1 / 1 1 / x = 1 / 1 x = 1 answer : d" | a = 30 - 15
b = 15 / a
|
a ) 110 degree , b ) 115 degree , c ) 112 1 / 2 degree , d ) 117 degree , e ) 157 1 / 2 degree | e | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between two hands at 3.45 is | "theta degree = 11 / 2 m - 30 h = 11 / 2 ( 45 ) - 30 ( 3 ) = 247.5 - 90 = 157.5 or 157 1 / 2 degree answer : e" | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 14 min , b ) 18 min , c ) 25 min , d ) 32 min , e ) 40 min | e | multiply(add(add(1, 2), 2), divide(2000, subtract(add(200, multiply(50, 2)), multiply(25, 2)))) | pipe a and pipe b fill water into a tank of capacity 2000 litres , at a rate of 200 l / min and 50 l / min . pipe c drains at a rate of 25 l / min . pipe a is open for 1 min and closed , then pipe b is open for 2 min and closed . further the pipe c is opened and drained for another 2 min . this process is repeated until the tank is filled . how long will it take to fill the tank ? | "tank capacity : 2000 l , 1 st - 200 l / min for 1 min , volume filled : 200 l 2 nd - 100 l / min for 2 min , volume filled : 100 l 3 rd ( water draining ) : 25 l / min * 2 : 50 l total : ( 200 + 100 ) - 50 = 250 l filled for 1 cycle number of 250 in 2000 l tank : 2000 / 250 = 8 time taken to fill : 8 * total time = 8 * 5 = 40 ( option e )" | a = 1 + 2
b = a + 2
c = 50 * 2
d = 200 + c
e = 25 * 2
f = d - e
g = 2000 / f
h = b * g
|
a ) 52.9 % , b ) 55.4 % , c ) 58.3 % , d ) 61.7 % , e ) 64.5 % | a | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(9)))) | a wooden cube whose edge length is 9 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least one red face ? | "there are a total of 9 * 9 * 9 = 729 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 7 * 7 * 7 = 343 cubes . the number of cubes with at least one side painted red is 729 - 343 = 386 cubes the probability that a cube has at least one side painted red is 386 / 729 which is about 52.9 % the answer is a ." | a = 1 * 4
b = volume_cube / (
c = 1 - b
d = 100 * c
|
a ) 91.5 , b ) 91.4 , c ) 91.7 , d ) 82.3 , e ) 91.1 | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | the sector of a circle has radius of 21 cm and central angle 110 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 110 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 40.3 + 42 = 82.3 cm answer : d" | a = 21 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 21
|
a ) 27 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | e | subtract(subtract(subtract(25, 2), const_1), const_1) | how many positive integers less than 25 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 2 odd multiples of 5 : { 5 , 15 } 10 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 } notice , that 5 is in two sets , thus total # of integers satisfying the given conditions is 9 + 2 + 10 - 1 = 20 . answer : e ." | a = 25 - 2
b = a - 1
c = b - 1
|
a ) 6.25 , b ) 6.27 , c ) 6.23 , d ) 6.29 , e ) 6.39 | a | divide(subtract(282, multiply(10, 3.2)), 40) | in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = [ 282 - ( 3.2 * 10 ) ] / 40 = 250 / 40 = 6.25 ' answer : a" | a = 10 * 3
b = 282 - a
c = b / 40
|
a ) 5 / 22 , b ) 4 / 22 , c ) 3 / 22 , d ) 9 / 22 , e ) 7 / 22 | e | divide(7, add(add(divide(multiply(multiply(3, 5), 2), 3), 7), 5)) | company m produces two kinds of stereos : basic and deluxe . of the stereos produced by company m last month , 2 / 3 were basic and the rest were deluxe . if it takes 7 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos ? | # of basic stereos was 3 / 4 of total and # of deluxe stereos was 1 / 4 of total , let ' s assume total = 16 , then basic = 12 and deluxe = 4 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 7 / 5 units . total time for basic would be 12 * 1 = 12 and total time for deluxe would be 4 * 7 / 5 = 28 / 5 - - > total time for both of them would be 12 + 28 / 5 = 88 / 5 - - > deluxe / total = 28 / 5 / 88 / 5 = 28 / 88 = 7 / 22 e | a = 3 * 5
b = a * 2
c = b / 3
d = c + 7
e = d + 5
f = 7 / e
|
a ) 6 , b ) 2 , c ) 8 , d ) 9 , e ) 7 | d | divide(multiply(130, 230), multiply(80, 40)) | rectangular tile each of size 80 cm by 40 cm must be laid horizontally on a rectangular floor of size 130 cm by 230 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 80 * 40 = 3200 area of floor = 130 * 230 = 29900 no of tiles = 29900 / 3200 = 9.34 so , the no of tile = 9 answer : d" | a = 130 * 230
b = 80 * 40
c = a / b
|
a ) 1235 , b ) 1345 , c ) 1678 , d ) 1767 , e ) 1671 | e | multiply(divide(subtract(1395, 15), subtract(6, const_1)), 6) | find large number from below question the difference of two numbers is 1395 . 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 + 1395 ) . x + 1395 = 6 x + 15 5 x = 1380 x = 276 large number = 276 + 1395 = 1671 e" | a = 1395 - 15
b = 6 - 1
c = a / b
d = c * 6
|
a ) rs . 1000 , b ) rs . 1550 , c ) rs . 1510 , d ) rs . 1500 , e ) none of these | d | divide(1740, add(divide(multiply(divide(add(multiply(3, 5), 3), 5), 5), const_100), const_1)) | find the principle on a certain sum of money at 5 % per annum for 3 1 / 5 years if the amount being rs . 1740 ? | "explanation : 1740 = p [ 1 + ( 5 * 16 / 5 ) / 100 ] p = 1500 answer : option d" | a = 3 * 5
b = a + 3
c = b / 5
d = c * 5
e = d / 100
f = e + 1
g = 1740 / f
|
a ) 20 , b ) 25 , c ) 40 , d ) 45 , e ) 75 | b | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(45, const_4)), 45) | the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 45 , what is the possible maximum range of the temperatures ? | "average = 50 , sum of temperatures = 50 * 5 = 250 as the min temperature is 45 , max would be 250 - 4 * 45 = 70 - - > the range = 70 ( max ) - 45 ( min ) = 25 answer : b ." | a = 2 + 3
b = 50 * a
c = 45 * 4
d = b - c
e = d - 45
|
a ) s . 440 , b ) s . 850 , c ) s . 450 , d ) s . 900 , e ) s . 950 | d | subtract(multiply(add(5, const_1), 850), add(add(add(add(800, 900), 1000), 700), 800)) | a grocer has a sale of rs . 800 , rs . 900 , rs . 1000 , rs . 700 and rs . 800 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 850 ? | "total sale for 5 months = rs . ( 800 + 900 + 1000 + 700 + 800 ) = rs . 4200 required sale = rs . [ ( 850 x 6 ) - 4200 ] = rs . ( 5100 - 4200 ) = rs . 900 . option d" | a = 5 + 1
b = a * 850
c = 800 + 900
d = c + 1000
e = d + 700
f = e + 800
g = b - f
|
a ) 565 , b ) 444 , c ) 676 , d ) 767 , e ) 408 | e | add(multiply(divide(60, subtract(18, 16)), 16), multiply(divide(60, subtract(18, 16)), 18)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 18 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 2 ? - - - - - - 60 12 h rs = 16 + 18 = 34 t = 12 d = 34 * 12 = 408 answer : e" | a = 18 - 16
b = 60 / a
c = b * 16
d = 18 - 16
e = 60 / d
f = e * 18
g = c + f
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | divide(subtract(const_1, add(multiply(divide(const_1, const_4.0), const_2), multiply(divide(const_1, 16), const_2))), divide(const_1, 16)) | a can finish a piece of work in 5 days . b can do it in 16 days . they work together for two days and then a goes away . in how many days will b finish the work ? | "2 / 4 + ( 2 + x ) / 16 = 1 = > x = 6 days answer : c" | a = 1 / 4
b = a * 2
c = 1 / 16
d = c * 2
e = b + d
f = 1 - e
g = 1 / 16
h = f / g
|
a ) 100 , b ) 200 , c ) 300 , d ) 600 , e ) 900 | c | multiply(subtract(54, 100), 100) | what is the greatest positive integer n such that 3 ^ n is a factor of 54 ^ 100 ? | "54 = 3 ^ 3 * 2 . 54 ^ 100 = 3 ^ 300 * 2 ^ 100 the answer is c ." | a = 54 - 100
b = a * 100
|
a ) 36 / 625 , b ) 48 / 625 , c ) 64 / 625 , d ) 98 / 625 , e ) 128 / 625 | e | subtract(1, divide(const_2, 5)) | when a random experiment is conducted , the probability that event a occurs is 1 / 5 . if the random experiment is conducted 5 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 5 * 1 / 5 * 4 / 5 * 4 / 5 * 4 / 5 = 64 / 3125 the total number of possible cases is 5 c 2 = 10 p ( event a occurs exactly twice ) = 10 * ( 64 / 3125 ) = 128 / 625 the answer is e ." | a = 2 / 5
b = 1 - a
|
a ) 45 , b ) 56 , c ) 47 , d ) 57 , e ) 49 | c | subtract(48, const_1) | in a lake , there is a patch of lily pads . every day , the patch doubles in size . if it takes 48 days for the patch to cover the entire lake , how long would it take for the patch to cover half of the lake ? | "explanation : with all the talk of doubling and halves , your brain jumps to the conclusion that to solve the problem of when the lily patch covers half the lake , all you have to do is divide the number of days it took to fill the lake ( 48 ) in half . it ' s understandable but wrong . the problem says that the patch doubles in size every day , which means that on any day , the lily patch was half the size the day before . so if the patch reaches the entire size of the lake on the 48 th day , it means the lily pad was half the size of the lake on day 47 . correct answer is c ) 47" | a = 48 - 1
|
a ) 115 , b ) 230 , c ) 460 , d ) 575 , e ) 690 | b | multiply(multiply(2, add(const_3, const_2)), divide(divide(divide(690, const_2), const_3), add(const_3, const_2))) | the least common multiple of positive integer d and 3 - digit integer n is 690 . if n is not divisible by 3 and d is not divisible by 2 , what is the value of n ? | the lcm of n and d is 690 = 2 * 3 * 5 * 23 . d is not divisible by 2 , thus 2 goes to n n is not divisible by 3 , thus 3 goes to d . from above : n must be divisible by 2 and not divisible by 3 : n = 2 * . . . in order n to be a 3 - digit number it must take all other primes too : n = 2 * 5 * 23 = 230 . answer : b . | a = 3 + 2
b = 2 * a
c = 690 / 2
d = c / 3
e = 3 + 2
f = d / e
g = b * f
|
a ) 172,700 , b ) 172,800 , c ) 172,900 , d ) 173,000 , e ) 173,100 | b | multiply(multiply(subtract(7, 3), const_3600), const_12) | in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day . | "this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 7 - 3 = 4 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 4 = 172,800 . hence b ." | a = 7 - 3
b = a * 3600
c = b * 12
|
a ) 50 , b ) 26 , c ) 75 , d ) 28 , e ) 21 | c | divide(multiply(54, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 54 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 54 = 25 / 9 = > x = 75 . answer : c" | a = 46 - 36
b = a * 1000
c = b / 3600
d = 54 * c
e = d / 2
|
a ) 10 , b ) 6 , c ) 8 , d ) none of these , e ) can not be determined | a | divide(subtract(84, multiply(const_3, 8)), multiply(const_3, const_2)) | a number is doubled and 8 is added . if the resultant is trebled , it becomes 84 . what is that number ? | "solution let the number be x . then , 3 ( 2 x + 8 ) ‹ = › 84 ‹ = › 2 x + 8 = 28 ‹ = › 2 x = 20 x = 10 . answer a" | a = 3 * 8
b = 84 - a
c = 3 * 2
d = b / c
|
a ) 16 kmph , b ) 30 kmph , c ) 54 kmph , d ) 18 kmph , e ) 19 kmph | b | multiply(const_3_6, divide(100, 12)) | a train 100 m in length crosses a telegraph post in 12 seconds . the speed of the train is ? | "s = 100 / 12 * 18 / 5 = 30 kmph answer : b" | a = 100 / 12
b = const_3_6 * a
|
a ) 6 : 5 , b ) 6 : 2 , c ) 6 : 4 , d ) 7 : 4 , e ) 7 : 2 | d | divide(multiply(subtract(5, const_1), 2), multiply(3, 2)) | 50 liters of a mixture contains milk and water in the ratio 3 : 2 . if 5 liters of this mixture be replaced by 5 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 50 liters if mix = 50 * 3 / 5 = 30 liters quantity of milk in 55 liters of new mix = 30 + 5 = 35 liters quantity of water in it = 55 - 35 = 20 liters ratio of milk and water in new mix = 35 : 20 = 7 : 4 answer is d" | a = 5 - 1
b = a * 2
c = 3 * 2
d = b / c
|
a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | d | multiply(2, 48) | the area of a square garden is q square feet and the perimeter is p feet . if q = 2 p + 48 , what is the perimeter of the garden in feet ? | "let x be the length of one side of the square garden . x ^ 2 = 8 x + 48 x ^ 2 - 8 x - 48 = 0 ( x - 12 ) ( x + 4 ) = 0 x = 12 , - 4 p = 4 ( 12 ) = 48 the answer is d ." | a = 2 * 48
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(5, divide(subtract(36, 26), 10)) | the average of 10 numbers is calculated as 5 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "10 * 5 + 36 – 26 = 60 / 10 = 6 answer : e" | a = 36 - 26
b = a / 10
c = 5 + b
|
a ) 9 , b ) 14 , c ) 21 , d ) 28 , e ) 42 | d | multiply(14, const_2) | if ( 10 ^ 4 * 3.456789 ) ^ 14 is written as a single term , how many digits would be to the right of the decimal place ? | 3.456789 ^ 14 has 6 * 14 = 84 decimal places . 10 ^ 56 moves the decimal place to the right 56 places . ( 10 ^ 4 * 3.456789 ) ^ 14 has 84 - 56 = 28 digits after the decimal point . the answer is d . | a = 14 * 2
|
a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 4 : 5 , e ) 2 : 3 | d | multiply(divide(2, 3), multiply(divide(2, 3), divide(6, 3))) | find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 5 ) is | "required ratio = 2 / 3 * 6 / 11 * 11 / 5 = 2 / 1 = 4 : 5 answer is d" | a = 2 / 3
b = 2 / 3
c = 6 / 3
d = b * c
e = a * d
|
a ) 196 , b ) 158 , c ) 192 , d ) 200 , e ) none | c | add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(1, const_100), const_1), const_3)) | how many digits are required to write numbers between 1 to 100 . | "explanation : single digits are from 1 to 9 = 9 digits doubt digits are from 10 to 99 = 90 x 2 = 180 digits 100 needs 3 digits . total 192 digits correct option : c" | a = 10 - 1
b = 10 - 1
c = b * 10
d = c * 2
e = a + d
f = 1 - 100
g = f + 1
h = g * 3
i = e + h
|
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