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a ) 28 , b ) 32 , c ) 36 , d ) 40 , e ) 44 | c | add(add(multiply(4, 4), multiply(const_2, multiply(4, add(1, divide(25, const_100))))), multiply(const_2, multiply(4, add(1, divide(25, const_100))))) | a cistern 4 meters long and 4 meters wide contains water up to a depth of 1 meter 25 cm . what is the total area of the wet surface ? | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 4 x 1.25 ) + 4 x 4 ] = 36 the answer is c ." | a = 4 * 4
b = 25 / 100
c = 1 + b
d = 4 * c
e = 2 * d
f = a + e
g = 25 / 100
h = 1 + g
i = 4 * h
j = 2 * i
k = f + j
|
a ) 2 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 32.5 meters . how many times did the ball bounce ? | "ans : 2 division of total diatance travelled will be 16 + 16 + 0.5 a" | a = 16 / 2
b = a / 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | d | subtract(25, 20) | a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 146 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ? | "we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 146 / 25 = 5 . . . . so the max is 5 146 / 20 = 7 . . . so the max is 7 146 - 145 = 1 ( 6 small box + 1 big box ) answer : d" | a = 25 - 20
|
a ) $ 2400 , b ) $ 3200 , c ) $ 6000 , d ) $ 6400 , e ) $ 3700 | e | multiply(multiply(const_100, add(const_3, const_2)), divide(111, subtract(const_100, add(add(30, 20), 35)))) | each month , after jill pays for rent , utilities , food , and other necessary expenses , she has one fifth of her net monthly salary left as discretionary income . of this discretionary income , she puts 30 % into a vacation fund , 20 % into savings , and spends 35 % on eating out and socializing . this leaves her with $ 111 dollar , which she typically uses for gifts and charitable causes . what is jill β s net monthly salary ? | "let x be the monthly salary 15 % of 1 / 5 * x = 111 x = 3700 answer e" | a = 3 + 2
b = 100 * a
c = 30 + 20
d = c + 35
e = 100 - d
f = 111 / e
g = b * f
|
a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(40, const_100), divide(subtract(const_100, 40), const_100))) | in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 40 percent of the male and 40 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | "percent of students who are 25 years old or older is 0.4 * 48 + 0.4 * 52 = 40 , so percent of people who are less than 25 years old is 100 - 40 = 60 . answer : b ." | a = 40 / 100
b = 100 - 40
c = b / 100
d = a * c
e = 1 - d
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 15)), const_4)), 15) | two pipes p and q can fill a cistern in 12 and 15 minutes respectively . both are opened together , but at the end of 4 minutes the first is turned off . how many more minutes will it take for the cistern to fill after the first pipe is turned off ? | let x be the total time it takes for the cistern to fill . 4 / 12 + x / 15 = 1 x / 15 = 2 / 3 x = 10 after the first pipe is turned off , it takes 6 more minutes to fill the cistern . the answer is a . | a = 1 / 12
b = 1 / 15
c = a + b
d = c * 4
e = 1 - d
f = e * 15
|
a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42 | c | divide(15, subtract(const_1, sqrt(divide(9, add(9, 16))))) | 15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container q ? | "if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 / 37.5 ; expressed as a fraction this becomes ( 45 / 2 ) / ( 75 / 2 ) , or 45 / 2 * 2 / 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 / 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 / 15 of a means there must be 6 / 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 / 16 ; 13.5 / 24 = ( 27 / 2 ) * ( 1 / 24 ) = 9 / 16 . choice c is correct ." | a = 9 + 16
b = 9 / a
c = math.sqrt(b)
d = 1 - c
e = 15 / d
|
a ) 10 % , b ) 50 % , c ) 25 % , d ) 52 % , e ) 18.5 % | e | subtract(multiply(divide(300, add(225, 28)), const_100), const_100) | a retailer buys a radio for rs 225 . his overhead expenses are rs 28 . he sellis the radio for rs 300 . the profit percent of the retailer is | "explanation : cost price = ( 225 + 28 ) = 253 sell price = 300 gain = ( 47 / 253 ) * 100 = 18.5 % . answer : e" | a = 225 + 28
b = 300 / a
c = b * 100
d = c - 100
|
a ) $ 900 , b ) $ 750 , c ) $ 600 , d ) $ 300 , e ) $ 60 | b | divide(multiply(multiply(divide(add(multiply(multiply(4, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), 4), 10), multiply(const_3, 4)) | mr . hernandez , who was a resident of state x for only 10 months last year , had a taxable income of $ 22,500 for the year . if the state tax rate were 4 percent of the year β s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez β s state x tax for last year ? | "total tax for the year = 22,500 x 4 % = 900 as stated annual tax is prorated as per the duration of stay . prorated tax = 900 ( 10 / 12 ) = 750 answer b" | a = 4 * 100
b = a * 100
c = 100 * const_0_25
d = c * 100
e = b + d
f = e / 100
g = f * 4
h = g * 10
i = 3 * 4
j = h / i
|
a ) 40 % , b ) 45 % , c ) 50 % , d ) 55 % , e ) 60 % | c | divide(multiply(12, const_4), add(const_4, const_1)) | the cost price of 12 articles is equal to the selling price of 8 articles . what is the profit percent ? | "8 * sp = 12 * cp sp = 1.5 * cp the profit percent is 50 % . the answer is c ." | a = 12 * 4
b = 4 + 1
c = a / b
|
['a ) 24 cm 2', 'b ) 25 cm 2', 'c ) 26 cm 2', 'd ) 27 cm 2', 'e ) 28 cm 2'] | d | multiply(divide(add(4, 5), const_2), 6) | two parallel sides of a trapezium are 4 cm and 5 cm respectively . the perpendicular distance between the parallel sides is 6 cm . find the area of the trapezium . | area of trapezium when height and two parallel sides are given = 1 / 2 Γ h Γ ( a + b ) = 1 / 2 x 6 x ( 4 + 5 ) = 27 cm 2 answer d | a = 4 + 5
b = a / 2
c = b * 6
|
a ) 8 / 15 , b ) 1 / 6 , c ) 1 / 3 , d ) 1 / 2 , e ) 5 / 6 | a | subtract(add(12, divide(12, 3)), divide(2, 5)) | in a certain lottery , the probability that a number between 12 and 20 , inclusive , is drawn is 1 / 5 . if the probability that a number 12 or larger is drawn is 2 / 3 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 12 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 12 or moreor 20 or lessor both ) 2 / 3 + p ( set 2 ) - 1 / 5 = 1 p ( set 2 ) = 8 / 15 answer ( a )" | a = 12 / 3
b = 12 + a
c = 2 / 5
d = b - c
|
['a ) 13 / 15', 'b ) 12 / 4', 'c ) 16 / 17', 'd ) 22 / 16', 'e ) 17 / 81'] | e | divide(subtract(circle_area(divide(18, const_2)), circle_area(divide(16, const_2))), circle_area(divide(18, const_2))) | a ring x of diameter 16 inches is placed inside another ring y of diameter of 18 inches . how much fraction of the ring x surface is not covered by the ring y ? | the questions asks us to find the surface which is not covered by the ring x i . e . , area of the surface not covered . where as circumference is the length along the edge of the circle , 2 * pi * r implies the length of the curve pi * r ^ 2 implies area enclosed by that curve . . hence area of the circle is considered for this problem area of the ring x = pi * r ^ 2 , where r = 8 = > pi * 64 area of the ring y = pi * 81 surface not covered by the ring x = pi * 81 - pi * 64 = pi * 17 fraction of ring y ' s surface is not covered by a ring x = pi * 17 / total area of the ring y = > pi * 17 / pi * 81 17 / 81 answer e | a = 18 / 2
b = circle_area - (
c = 16 / 2
d = b / circle_area
|
a ) 165 , b ) 195 , c ) 240 , d ) 270 , e ) 315 | d | divide(divide(152, subtract(multiply(divide(const_4.0, 9), divide(3, 5)), multiply(divide(const_2.0, 3), divide(7, 9)))), 5) | 7 / 9 of 3 / 5 of a number is greater than 2 / 9 of 5 / 6 of the same number by 152 . what is half of that number ? | "let no . be x 7 / 9 * 3 / 5 * x - 2 / 9 * 5 / 6 * x = 152 by further solving 21 x / 45 - 10 x / 54 = 152 114 x / 405 = 152 x = 540 we have to find x / 2 = 540 / 2 = 270 answer : d" | a = 4 / 0
b = 3 / 5
c = a * b
d = 2 / 0
e = 7 / 9
f = d * e
g = c - f
h = 152 / g
i = h / 5
|
a ) 829 , b ) 869 , c ) 859 , d ) 4320 , e ) 427 | e | subtract(divide(multiply(21, 45), const_2), subtract(45, const_10)) | the smallest number which when diminished by 3 is divisible by 21 , 2836 and 45 is . . . | 427 answer : e | a = 21 * 45
b = a / 2
c = 45 - 10
d = b - c
|
a ) 200 , b ) 160 , c ) 100 , d ) 65 , e ) 50 | b | multiply(divide(divide(75, const_100), divide(divide(45, const_100), divide(96, const_100))), const_100) | if 45 % of z is 96 % of y and y is 75 % of x , what percent of x is z ? | "( 45 / 100 ) z = ( 96 / 100 ) y and y = ( 75 / 100 ) x i . e . y = ( 3 / 4 ) x i . e . ( 45 / 100 ) z = ( 96 / 100 ) * ( 3 / 4 ) x i . e . z = ( 96 * 3 ) x / ( 45 * 4 ) i . e . z = ( 1.6 ) x = ( 160 / 100 ) x i . e . z is 160 % of x answer : option b" | a = 75 / 100
b = 45 / 100
c = 96 / 100
d = b / c
e = a / d
f = e * 100
|
a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | e | divide(log(divide(multiply(power(5, 15), power(9, 8)), 3)), log(15)) | if ( 5 ^ 15 ) ( 9 ^ 8 ) = 3 ( 15 ^ x ) , what is the value of x ? | "( 5 ^ 15 ) ( 9 ^ 8 ) = 3 ( 15 ^ x ) = > 5 ^ 15 * 3 ^ 16 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 15 * 3 ^ 16 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 15 answer e" | a = 5 ** 15
b = 9 ** 8
c = a * b
d = c / 3
e = math.log(d)
f = math.log(15)
g = e / f
|
a ) s . 738 , b ) s . 638 , c ) s . 650 , d ) s . 810 , e ) s . 735 | d | subtract(825, divide(multiply(subtract(830, 825), 3), 4)) | a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 830 in 4 years . the sum is ? | "s . i . for 1 year = ( 830 - 825 ) = rs . 5 s . i . for 3 years = 5 * 3 = rs . 15 principal = ( 825 - 15 ) = rs . 810 . answer : d" | a = 830 - 825
b = a * 3
c = b / 4
d = 825 - c
|
a ) 18 , b ) 30 , c ) 42 , d ) 68 , e ) 70 | b | divide(multiply(divide(multiply(subtract(subtract(300, divide(multiply(70, 300), const_100)), 45), const_100), subtract(const_100, 40)), 40), const_100) | at a particular graduation party with 300 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 45 males did not bring gifts to the party , how many females did bring gifts ? | "the correct method total = 300 . . 70 % of 300 = 210 got gifts . . 90 did not get gift , out of which 48 are males , so remaining 90 - 45 = 50 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 45 , 100 % = 45 * 100 / 60 = 75 . . ans 40 % of 75 = 30 b" | a = 70 * 300
b = a / 100
c = 300 - b
d = c - 45
e = d * 100
f = 100 - 40
g = e / f
h = g * 40
i = h / 100
|
a ) 2 , b ) 15 , c ) 7 , d ) 10 , e ) 9 | b | divide(subtract(50, 20), const_2) | after playing 40 games , the lesley basketball team had won 50 % of the games . the additional games played gave the team a 20 % total winning average . how many additional games must the team have won to obtain this average ? | explanation : identify the parts , set up equation , and solve . number of games initially won by the team : 50 % * 40 = 20 additional games won by the team leaving a 20 % average . ( let x represent the unknown number of additional games won . ) therefore , the total number of games played is 40 + x , and the total number of games won is 20 + x . this gives the equation as : 20 % * ( 40 + x ) = 20 + x solve : let ' s get rid of the percent sign 20 / 100 * ( 40 + x ) = 20 + x 1 / 5 * ( 40 + x ) = 20 + x get rid of the fraction : 5 [ 1 / 5 * ( 40 + x ) ] = 5 ( 20 + x ) 1 * ( 40 + x ) = 100 + 5 x 40 + x = 100 + 5 x ( 40 - 100 ) + x = ( 100 - 100 ) + 5 x - 60 + x = 5 x - 60 + ( x - x ) + 5 x - x - 60 = 4 x 15 = x answer : b | a = 50 - 20
b = a / 2
|
a ) 11123 , b ) 18153 , c ) 11282 , d ) 18825 , e ) 18823 | e | multiply(subtract(subtract(add(19845, 9156), 1090), 9088), divide(const_60, const_2)) | ( x ) + 9088 + 1090 - 9156 = 19845 . calculate the value of x | "( x ) + 9088 + 1090 - 9156 = 19845 = x + 9088 + 1090 = 19845 + 9156 = x + 10178 = 29001 = x = 29001 - 10178 = 18823 answer is e" | a = 19845 + 9156
b = a - 1090
c = b - 9088
d = const_60 / 2
e = c * d
|
a ) 32 : 75 , b ) 82 : 31 , c ) 32 : 45 , d ) 34 : 89 , e ) 35 : 21 | a | divide(add(multiply(8000, 4), multiply(divide(15000, const_3), multiply(2, 4))), add(multiply(15000, multiply(2, const_3)), multiply(subtract(15000, divide(15000, const_3)), multiply(2, const_3)))) | a and b invests rs . 8000 and rs . 15000 in a business . after 4 months , a withdraws half of his capital and 2 months later , b withdraws one - third of his capital . in what ratio should they share the profits at the end of the year ? | "a : b ( 8000 * 4 ) + ( 4000 * 8 ) : ( 15000 * 6 ) + ( 10000 * 6 ) 64000 : 150000 32 : 75 answer : a" | a = 8000 * 4
b = 15000 / 3
c = 2 * 4
d = b * c
e = a + d
f = 2 * 3
g = 15000 * f
h = 15000 / 3
i = 15000 - h
j = 2 * 3
k = i * j
l = g + k
m = e / l
|
a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 4,400 , e ) $ 800 | b | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 15), multiply(const_2, const_4)), multiply(divide(15, const_100), const_2))) | the market value of a certain machine decreased by 15 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | "b . market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 15 / 100 ) = 8000 - 1200 = $ 6800 market value in 1984 = market value in 1983 - ( 15 % of $ 8000 ) = 6800 - 1200 = $ 5600" | a = 100 * 10
b = 2 * 4
c = a * b
d = 100 * 15
e = 2 * 4
f = d * e
g = 15 / 100
h = g * 2
i = f * h
j = c - i
|
a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | d | divide(divide(const_4, add(const_2, const_3)), subtract(inverse(6), inverse(10))) | a water tank is four - fifths full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how many minutes will it take to empty or fill the tank completely ? | the combined rate of filling / emptying the tank = 1 / 10 - 1 / 6 = - 1 / 15 since the rate is negative , the tank will be emptied . a full tank would take 15 minutes to empty . since the tank is only four - fifths full , the time is ( 4 / 5 ) * 15 = 12 minutes the answer is d . | a = 2 + 3
b = 4 / a
c = 1/(6)
d = 1/(10)
e = c - d
f = b / e
|
a ) 8 % , b ) - 25 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | b | multiply(divide(subtract(multiply(divide(2, 15), 120), 12), multiply(divide(2, 15), 120)), const_100) | a doctor prescribed 12 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage lesser than the typical dosage ? | "typical dosage per 15 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 15 ) = 2 * 8 = 16 c . c dosage prescribed by doctor for 120 pound patient = 12 c . c % prescribed dosage greater than the typical dosage = ( 12 - 16 / 16 ) * 100 % = ( - 4 / 16 ) * 100 % = - 25 % answer b" | a = 2 / 15
b = a * 120
c = b - 12
d = 2 / 15
e = d * 120
f = c / e
g = f * 100
|
a ) 2 , b ) 6 , c ) 5 , d ) 3 , e ) 4 | a | divide(multiply(10, 19.8), subtract(const_100, const_1)) | mr . smith calculated the average of 10 3 digit numbers . but due to a mistake he reversed the digits of a number and thus his average increased by 19.8 the difference between the unit digit and hundreds digit of that number is : | since the average increased by 29.7 and there were a total of 10 numbers , it means the incorrect number was 198 greater than the correct number . say , the correct number was abc ( where a , b and c are the digits of the 3 digit number ) then the incorrect number was cba . 100 c + 10 b + a - ( 100 a + 10 b + c ) = 198 99 c - 99 a = 99 ( c - a ) = 198 198 = 99 * 2 = 99 ( c - a ) so c - a = 2 answer ( a ) | a = 10 * 19
b = 100 - 1
c = a / b
|
a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | b | subtract(add(6, 8), 4) | a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 4 years ago ? | "let ' s say age of richard is ` ` r ' ' age of david is ` ` d ' ' age of scott is ` ` s ' ' now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6 = ( s + 8 ) + 6 = s + 14 therefore , 2 s + 8 = s + 14 i . e . s = 6 i . e . r = 20 i . e . d = 14 now , how old was david 4 years ago ? i . e . d - 4 = 14 - 4 = 10 years answer : b" | a = 6 + 8
b = a - 4
|
a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 101 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 24) | cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 24 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 24 shows a cube of a number could be right . here 64 + 24 = 88 72 + 24 = 96 86 + 24 = 110 98 + 24 = 122 101 + 24 = 125 - - - ( 5 * 5 * 5 ) so we have 101 as the answer ! e" | a = 1 + 3
b = 1 + a
c = b ** 3
d = c - 24
|
a ) 750 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 10,000 | a | divide(30, divide(2, 50)) | in a certain pond , 30 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "this is a rather straight forward ratio problem . 1 . 30 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 30 / x thus , x = 750 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . a" | a = 2 / 50
b = 30 / a
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | floor(divide(subtract(multiply(subtract(inverse(divide(60, const_100)), const_1), subtract(100, 23)), multiply(2, 23)), add(3, subtract(inverse(divide(60, const_100)), const_1)))) | in an examination , there are 100 questions divided into 3 groups a , b and c such that each group contains at least one question . each question in group a carries 1 mark , each question in group b carries 2 marks and each question in group c carries 3 marks . it is known that the questions in group a together carry at least 60 % of the total marks . if group b contains 23 questions , then how many questions are there in group c ? | explanation : since group b contains 23 questions , the marks associated with this group are 46 . now , check for option ( 1 ) , if group ( c ) has one question , the marks associated with this group will be 3 . thus total marks will be 49 . total number of questions = 100 , group ( a ) will have 76 questions , the corresponding weightage being 76 marks . answer : a | a = 60 / 100
b = 1/(a)
c = b - 1
d = 100 - 23
e = c * d
f = 2 * 23
g = e - f
h = 60 / 100
i = 1/(h)
j = i - 1
k = 3 + j
l = g / k
m = math.floor(l)
|
a ) 229.623 , b ) 224.777 , c ) 233.523 , d ) 414.637 , e ) none of these | a | subtract(897.7, divide(573.07, 95.007)) | 897.7 β 573.07 β 95.007 = ? | "solution given expression = 897.7 - ( 573.07 + 95.007 ) = 897.7 - 668.077 = 229.623 . answer a" | a = 573 / 7
b = 897 - 7
|
a ) - 12 number , b ) - 14 , c ) 2 , d ) 8 , e ) 6 | a | subtract(negate(30), multiply(subtract(64, 48), divide(subtract(64, 48), subtract(78, 64)))) | 78 , 64 , 48 , 30 , 10 , ( . . . ) | "explanation : 78 - 14 = 64 64 - 16 = 48 48 - 18 = 30 30 - 20 = 10 10 - 22 = - 12 answer : option a" | a = negate - (
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a ) 4.50 days , b ) 3.50 days , c ) 6.0 days , d ) 7.5 days , e ) 5.50 days | d | divide(multiply(5, 15), subtract(15, 5)) | a and b finish the job in 15 days . while a , b and c can finish it in 5 days . c alone will finish the job in | "explanation : 5 = ( 15 * x ) / ( 15 + x ) 75 + 5 x = 15 x 10 x = 75 x = 7.5 answer : option d" | a = 5 * 15
b = 15 - 5
c = a / b
|
a ) 25 , b ) 36 , c ) 30 , d ) 67 , e ) 75 | c | multiply(divide(multiply(subtract(divide(multiply(multiply(const_4, const_4), const_100), 8), divide(multiply(multiply(const_4, const_4), const_100), add(8, divide(multiply(8, 50), const_100)))), 8), multiply(multiply(const_4, const_4), const_100)), const_100) | working together at their respective constant rates , machine a and machine b can produce 1,500 units in 8 hours . working alone , machine b would complete that same output in 50 % more time . if machine a were to work on its own for an 8 - hour shift , what percent of the 1,500 unit total would it produce ? | "1 / a + 1 / b = 1 / t 1 / a + 1 / 14 = 1 / 8 ( 50 % more of 8 is 14 ) 1 / a = 1 / 24 machine a can produce 1500 units in 24 hrs , so it can produce 1500 * 8 / 24 = 500 units is 8 hrs . 500 is 30 % of 1500 . c is the answer" | a = 4 * 4
b = a * 100
c = b / 8
d = 4 * 4
e = d * 100
f = 8 * 50
g = f / 100
h = 8 + g
i = e / h
j = c - i
k = j * 8
l = 4 * 4
m = l * 100
n = k / m
o = n * 100
|
a ) 83 % , b ) 80 % , c ) 23 % , d ) 17 % , e ) 12 % | c | subtract(30, const_1) | at a local appliance manufacturing facility , the workers received a 30 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | "let ' s say he works usually 10 hours and earns 100 per hour . 10 * 100 = 1000 10 * 130 = 1300 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 130 = 7.67 so he can reduce his work by 2.33 hours . which is > 23 % . answer c" | a = 30 - 1
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a ) 6 days , b ) 3 days , c ) 4 days , d ) 2 days , e ) 5 days | c | divide(multiply(multiply(5, 4), 35), multiply(25, 7)) | calculate how many days it will take for 7 boys to wash a 35 m long wall if 5 boys can wash a 25 m long wall in 4 days , | "the length of wall painted by one boy in one day = 25 / 5 * 1 / 4 = 1.25 m no . of days required to paint 50 m cloth by 8 boys = 35 / 7 * 1 / 1.25 = 4 days . c" | a = 5 * 4
b = a * 35
c = 25 * 7
d = b / c
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a ) 1182 , b ) 1769 , c ) 2456 , d ) 3538 , e ) 4209 | b | divide(factorial(61), multiply(factorial(subtract(61, const_2)), factorial(const_2))) | how many diagonals does a 61 - sided convex polygon have ? | "a 61 - sided convex polygon has 61 vertices . if we examine a single vertex , we can see that we can connect it with 58 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 61 vertices can be connected with 58 vertices to create a diagonal then the total number of diagonals would be ( 61 ) ( 58 ) = 3538 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3538 by 2 to get 1769 . the answer is b ." | a = math.factorial(61)
b = 61 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
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a ) 903 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(253, multiply(const_4, const_100)), add(multiply(subtract(253, multiply(const_4, const_100)), 2), const_1)) | balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 253 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the triangle did . how many balls made up the triangle ? | "as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 253 ) ( x - 8 ) ^ 2 = t + 253 - - - - - ( b ) now the hardest part of the question will be to solve these 2 equations and this looks like time consuming but the easy way will be plug and play . also , we ' ve to find a value of t ( from 5 optiosn given below ) which can make a square of a a number . one we know this , it will be a cake walk . we can see that option a fits this criteria in eq ( b ) . add - 903 + 253 = 1156 = 34 ^ 2 = ( x - 8 ) ^ 2 hence , x = 42 . cross check by putting in eq ( a ) = x ( x + 1 ) / 2 = t = > 42 * 43 / 2 = 903 hence , answer is a ." | a = 4 * 100
b = 253 - a
c = 4 * 100
d = 253 - c
e = d * 2
f = e + 1
g = b * f
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a ) $ 1,292 , b ) $ 1,733 , c ) $ 3,466 , d ) $ 12,917 , e ) $ 20,796 | a | subtract(9, multiply(4, const_2)) | a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 310,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | 0.04 x + 0.09 y = 310000 [ 1 ] 0.04 x = 0.80 * 310000 = 248000 [ 2 ] 248000 + 0.09 y = 310000 - - > 0.09 y = 62000 [ 3 ] 248000 / 12 = 258333.3333 [ 4 ] 62000 / 12 = 5166.67 [ 5 ] adding [ 4,5 ] we get : 25833 [ 6 ] dividing [ 6 ] / 2 to get an average we get 1.292 , ans a | a = 4 * 2
b = 9 - a
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a ) 15467118 , b ) 16777217 , c ) 17827343 , d ) 18047455 , e ) 19357579 | b | divide(add(power(16, 7), 16), 16) | what is ( 16 ^ 7 + 16 ) / 16 ? | "( 16 ^ 7 + 16 ) / 16 = 16 * ( 16 ^ 6 + 1 ) / 16 = 16 ^ 6 + 1 clearly this is a number which ends with a 7 in the units place . the answer is b ." | a = 16 ** 7
b = a + 16
c = b / 16
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a ) 15 , b ) 77 , c ) 25 , d ) 87 , e ) 171 | a | floor(sqrt(240)) | if the sum of a number and its square is 240 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 240 x 2 + x - 240 = 0 ( x + 16 ) ( x β 15 ) = 0 x = 15 answer : a" | a = math.sqrt(240)
b = math.floor(a)
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a ) 4 , b ) 7 , c ) 9 , d ) 15 , e ) 20 | c | divide(multiply(6, const_3), const_2) | youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 6 minutes more to walk to work than to ride his bike to work , then x equals ? | "please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 360 ( which is time in seconds . ) . thus the correct equation is : 60 * x - 20 * x = 360 - - - - > 40 x = 360 - - > x = 9 . option c" | a = 6 * 3
b = a / 2
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a ) 2 , b ) 5 , c ) 12 , d ) 16 , e ) 22 | c | subtract(62, 50) | two family reunions are happening at the leela hotel , the oates reunion and the hall reunion . all 100 guests at the hotel attend at least one of the reunions . if 50 people attend the oates reunion and 62 people attend the hall reunion , how many people attend both reunions ? | no of people in oates reunion = 50 no of people in hall reunion = 62 attending both = x all guests attend at least one . therefore , 100 = 50 + 62 - ( both ) both = 12 answer c | a = 62 - 50
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a ) 600 , b ) 700 , c ) 840 , d ) 940 , e ) 1440 | e | divide(multiply(9, 1600), add(1, 9)) | ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 1600 then loss of pyarelal ? | "let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 1600 , loss of pyarelal = 1600 * 9 x / 10 x = 1440 answer : e" | a = 9 * 1600
b = 1 + 9
c = a / b
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a ) 11 . , b ) 12 . , c ) 13 . , d ) 14 . , e ) 14.5 | d | add(divide(multiply(10, const_100), add(const_100, 25)), divide(multiply(9, const_100), add(const_100, 50))) | following an increase in prices , the price of a candy box was 10 pounds and the price of a can of soda was 9 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "price of candy before price increase = 10 / 1.25 = 8 price of soda before price increase = 9 / 1.5 = 6 total price = 8 + 6 = 14 d is the answer" | a = 10 * 100
b = 100 + 25
c = a / b
d = 9 * 100
e = 100 + 50
f = d / e
g = c + f
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a ) 20 , b ) 35 , c ) 48 , d ) 50 , e ) 59 | b | divide(subtract(multiply(add(25, 5), 60), multiply(25, 65)), 5) | in a factory , an average of 65 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 60 sets / day . the average production per day for day last 5 days is ? | "production during these 5 days = total production in a month - production in first 25 days . = 30 x 60 - 25 x 65 = 175 β΄ average for last 5 days = 175 / 5 = 35 aswer : b" | a = 25 + 5
b = a * 60
c = 25 * 65
d = b - c
e = d / 5
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | d | multiply(const_10, subtract(const_10, const_2)) | how many integers k greater than 100 and less than 1000 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ? | kk is 3 - digit integer . any 3 digit integer can be expressed as 100 a + 10 b + c 100 a + 10 b + c . now , we are told that ( 100 a + 10 b + c ) + 99 = 100 c + 10 b + a ( 100 a + 10 b + c ) + 99 = 100 c + 10 b + a - - > 99 c β 99 a = 9999 c β 99 a = 99 - - > c β a = 1 c β a = 1 - - > there are 8 pairs of cc and aa possible such that their difference to be 1 : ( 9,8 ) , ( 8,7 ) , . . . , and ( 2,1 ) ( note that aa can not be 0 as kk is 3 - digit integer and its hundreds digit , aa , is more than or equal to 1 ) . also , bb , tens digit of kk , could tale 10 values ( 0 , 1 , . . . , 9 ) so there are total of 8 * 10 = 80 such numbers possible . example : 102 - 201 ; 112 - 211 ; 122 - 221 ; . . . 899 - 998 . answer : d . | a = 10 - 2
b = 10 * a
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a ) 9 , b ) 8 , c ) 7 , d ) 6 , e ) 5 | e | add(const_2, const_3) | if 225 / 999 = 0.225 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ? | "0 . [ u ] 225 [ / u = 0.225225225 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 5 . since 81 is multiple of 3 , the 81 st digit is 5 . answer e" | a = 2 + 3
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a ) 1 / 12 , b ) 1 / 8 , c ) 1 / 6 , d ) 1 / 4 , e ) 1 / 2 | a | divide(divide(2, 15), divide(divide(add(20, const_100), const_100), divide(subtract(const_100, 25), const_100))) | if the numerator of a fraction is increased by 20 % and its denominator is diminished by 25 % value of the fraction is 2 / 15 . find the original fraction . | x * ( 120 / 100 ) - - - - - - - - - - - - - - - - = 2 / 15 y * ( 75 / 100 ) x / y = 1 / 12 answer : a | a = 2 / 15
b = 20 + 100
c = b / 100
d = 100 - 25
e = d / 100
f = c / e
g = a / f
|
a ) . 7 , b ) . 07 , c ) . 007 , d ) . 0007 , e ) none of them | a | divide(divide(007, const_1000), divide(01, const_100)) | . 007 / ? = . 01 | "let . 007 / x = . 01 ; then x = . 007 / . 01 = . 7 / 1 = . 7 answer is a" | a = 7 / 1000
b = 1 / 100
c = a / b
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a ) 19 , b ) 19.7 , c ) 21.3 , d ) 21.5 , e ) 24.7 | e | subtract(30, add(multiply(divide(30, const_100), divide(const_100, 6)), divide(30, const_100))) | jerry went to a shop and bought things worth rs . 30 , out of which 30 % went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 30 given that out of this rs . 30 , 30 % is given as tax = > total tax incurred = 30 % = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 6 % β΄ ( 30 β 30 / 100 β x ) 6 / 100 = 30 / 100 β 6 ( 30 β 0.3 β x ) = 30 β ( 30 β 0.3 β x ) = 5 β x = 30 β 0.3 β 5 = 24.7 e )" | a = 30 / 100
b = 100 / 6
c = a * b
d = 30 / 100
e = c + d
f = 30 - e
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a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | e | divide(250, multiply(subtract(27, 200), const_0_2778)) | a train 250 m long is running at a speed of 27 km / hr . in what time will it pass a bridge 200 m long ? | "speed = 27 * 5 / 18 = 15 / 2 m / sec total distance covered = 250 + 200 = 450 m required time = 450 * 2 / 15 = 60 sec answer : e" | a = 27 - 200
b = a * const_0_2778
c = 250 / b
|
a ) 10 % , b ) ( 8 ) 20 % , c ) 30 % , d ) 50 % , e ) 65 % | c | subtract(add(add(75, 35), 20), const_100) | if 75 percent of a class answered the first question on a certain test correctly , 35 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "i did n ' t understand how to connect 20 percent with the whole equation . why 20 is out of the percentile diagram ( circles ) but inside of 100 ? 20 % appeared in the exam ; however did not answer question aquestion b correctly so they are out of the two circles , however as they have appeared in the exam , they have to be taken into consideration for calculation purpose i am doing it this way : suppose the total class size is 100 , then 20 percent of the class answered incorrectly and rest 80 percent answered correctly in question 1 or 2 or both . now , 100 - [ ( 75 - x ) + x + ( 35 - x ) ] = 20 . . . . . . . . . . values are in percent x = 20 - 100 + 110 x = 30 ans : c" | a = 75 + 35
b = a + 20
c = b - 100
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a ) 3.75 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these | a | multiply(divide(divide(multiply(subtract(const_100, 90), 30), const_100), subtract(const_100, 20)), const_100) | fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 30 kg of fresh grapes ? | "the weight of non - water in 30 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 30 Γ’ Λ β 0.1 = x Γ’ Λ β 0.8 - - > x = 3.75 answer : a" | a = 100 - 90
b = a * 30
c = b / 100
d = 100 - 20
e = c / d
f = e * 100
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 23), 1056) | what least number must be added to 1056 , so that the sum is completely divisible by 23 ? | "23 ) 1056 ( 45 92 - - - 136 115 - - - 21 - - - required number = ( 23 - 21 ) = 2 . a )" | a = 4 * 10
b = a + 2
c = b * 23
d = c - 1056
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a ) 24 , b ) 34 , c ) 35 , d ) 36 , e ) 39 | a | divide(factorial(4), factorial(subtract(4, 3))) | 4 ! / ( 4 - 3 ) ! = ? | 4 ! / ( 4 - 3 ) ! = 4 ! / 1 ! = 4 * 3 * 2 = 24 . hence , the correct answer is a . | a = math.factorial(4)
b = 4 - 3
c = math.factorial(b)
d = a / c
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a ) 1.065 , b ) 1.125 , c ) 1.225 , d ) 1.375 , e ) 1.425 | b | divide(divide(multiply(add(4, 2), 1.5), 2), 4) | natasha climbs up a hill , and descends along the same way she went up . it takes her 4 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 1.5 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "let the distance to the top be x , so the total distance traveled by natasha is 2 x . the total time is 4 + 2 = 6 hours the average speed = total distance / total time taken = 2 x / 6 = x / 3 the average speed of the complete journey is 1.5 km / hour x / 3 = 1.5 x = 4.5 km the average speed while climbing = distance / time = 4.5 / 4 = 1.125 km / h the answer is b ." | a = 4 + 2
b = a * 1
c = b / 2
d = c / 4
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a ) 326838 , b ) 323758 , c ) 323298 , d ) 273258 , e ) 327382 | a | multiply(divide(5358, 61), const_100) | 5358 x 61 = ? | "a 5358 x 61 = 5358 x ( 60 + 1 ) = 5358 x 60 + 5358 x 1 = 321480 + 5358 = 326838 ." | a = 5358 / 61
b = a * 100
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a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(4), log(power(4, 11))) | if n = 4 ^ 11 β 4 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 4 ^ 3 - 4 . now 4 ^ 3 has last digit 4 . , we subtract 4 from 4 = 0 is the answer . option a | a = math.log(4)
b = 4 ** 11
c = math.log(b)
d = a / c
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a ) 11 / 21 , b ) 14 / 31 , c ) 7 / 15 , d ) 17 / 35 , e ) 1 / 2 | a | divide(add(3, 2), const_10) | company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 11 / 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 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 11 / 5 units . total time for basic would be 10 * 1 = 10 and total time for deluxe would be 5 * 11 / 5 = 11 - - > total time for both of them would be 10 + 11 = 21 - - > deluxe / total = 11 / 21 . answer : a ." | a = 3 + 2
b = a / 10
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a ) 128 , b ) 277 , c ) 800 , d ) 2688 , e ) 1991 | c | divide(608, subtract(const_1, divide(24, const_100))) | after decreasing 24 % in the price of an article costs rs . 608 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 608 cp = 8 * 100 = > cp = 800 answer : c" | a = 24 / 100
b = 1 - a
c = 608 / b
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a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 22 | a | subtract(multiply(30, const_2), subtract(add(add(multiply(30, const_2), multiply(22, const_4)), multiply(8, const_4)), 132)) | in a zoo there are 30 penguins , 22 zebras , and 8 tigers with some zookeepers . if the total number of heads are 132 less than the number of feet , the number of zookeepers is | explanation : let number of zookeepers be x . then , total number of feet = 2 x 30 + 4 x 22 + 4 x 8 + 2 x = 2 x + 180 . total number of heads = 30 + 22 + 8 + x = 60 + x . therefore ( 2 x + 180 ) - 132 = ( 60 + x ) + or x = 12 . answer : a | a = 30 * 2
b = 30 * 2
c = 22 * 4
d = b + c
e = 8 * 4
f = d + e
g = f - 132
h = a - g
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a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(21, const_3) | what is the greatest number of identical bouquets that can be made out of 21 white and 91 red tulips if no flowers are to be left out ? ( two bouquets are identical whenever the number of red tulips in the two bouquets is equal and the number of white tulips in the two bouquets is equal ) | since no flowers are to be left out , then the number of bouquets must be a factor of both 21 and 91 . for example , we can not have 2 bouquets since we can not divide 91 red tulips into 2 bouquets without one tulip left over . only answer choice which is a factor of 91 is e ( 7 ) . answer : e . | a = 21 / 3
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 1 | e | subtract(multiply(add(multiply(const_4, const_10), const_2), 25), 1049) | what least number must be added to 1049 , so that the sum is completely divisible by 25 ? | if we divide 1049 by 25 remainder is 24 25 - 24 = 1 answer : e | a = 4 * 10
b = a + 2
c = b * 25
d = c - 1049
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a ) 12 , b ) 22 , c ) 30 , d ) 88 , e ) 13 | c | divide(add(25, 35), const_2) | man can row upstream at 25 kmph and downstream at 35 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 35 m = ( 35 + 25 ) / 2 = 30 answer : c" | a = 25 + 35
b = a / 2
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a ) 25 , b ) 34 , c ) 30 , d ) 67 , e ) 100 | c | divide(multiply(6, 200), divide(200, const_10)) | according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 10 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 10 oz can = 300 oz / 10 oz = 30 answer c , 30 cans" | a = 6 * 200
b = 200 / 10
c = a / b
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a ) 30 , b ) 35 , c ) 40 , d ) 50 , e ) 60 | d | divide(14, subtract(213.28, add(const_100, add(multiply(const_4, const_10), const_2)))) | when positive integer n is divided by positive integer j , the remainder is 14 . if n / j = 213.28 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 213.28 here 213 is the quotient . given that remainder = 14 so , 213.28 = 213 + 14 / j so , j = 50" | a = 4 * 10
b = a + 2
c = 100 + b
d = 213 - 28
e = 14 / d
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a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 67 | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 30), 50)), divide(multiply(const_100, 30), 50))) | if the cost price of 50 articles is equal to the selling price of 30 articles , then the gain or loss percent is ? | "percentage of profit = 20 / 30 * 100 = 67 % answer : e" | a = 100 * 30
b = a / 50
c = 100 - b
d = 100 * 30
e = d / 50
f = c / e
g = 100 * f
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a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 percent of the distribution lies within one standard deviation d of the mean , what percent w of the distribution is less than m + d ? | "d the prompt says that 68 % of the population lies between m - d and m + d . thus , 32 % of the population is less than m - d or greater than m + d . since the population is symmetric , half of this 32 % is less than m - d and half is greater than m + d . thus , w = ( 68 + 16 ) % or ( 100 - 16 ) % of the population is less than m + d ." | a = 100 - 68
b = a / 2
c = 100 - b
|
a ) 3.1 kmph , b ) 5 kmph , c ) 6 kmph , d ) 4 kmph , e ) 7 kmph | a | divide(add(4.5, subtract(4.5, 2.5)), const_2) | a man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 2.5 kmph , find his average speed for the total journey ? | explanation : speed of man = 4.5 kmph speed of stream = 2.5 kmph speed in downstream = 7 kmph speed in upstream = 2 kmph average speed = ( 2 x 7 x 2 ) / 9 = 3.1 kmph . answer : a | a = 4 - 5
b = 4 + 5
c = b / 2
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a ) a . 4 , b ) b . 5 / 2 , c ) c . 2 , d ) d . 3 / 2 , e ) e . 20 / 11 | e | divide(add(multiply(3, const_4), 3), const_10) | nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 1.5 screws and 4.5 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 1.5 s and 4.5 n = 1.5 s + 4.5 * 1.5 s = 8.25 s . . ratio = 15 s / 8.25 s = 15 / 8.25 = 20 / 11 e" | a = 3 * 4
b = a + 3
c = b / 10
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a ) 3.76 % , b ) 4.76 % , c ) 5.76 % , d ) 6.76 % , e ) 7.76 % | b | divide(multiply(subtract(52.50, add(15.50, add(multiply(2, 6.45), multiply(2, 11.05)))), const_100), 52.50) | a camera lens filter kit containing 5 filters sells for $ 52.50 . if the filters are purchased individually , 2 of them are priced at $ 6.45 each , 2 at $ 11.05 each , 1 at $ 15.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ? | "cost of kit = $ 52.50 if filters are purchased individually - $ 6.45 * 2 + $ 11.05 * 2 + $ 15.50 = $ 50.50 amount saved = $ 52.50 - $ 50.50 = $ 2.50 required % age = ( $ 2.50 / $ 52.50 ) * 100 = 4.76 % so , the correct answer is b ." | a = 2 * 6
b = 2 * 11
c = a + b
d = 15 + 50
e = 52 - 50
f = e * 100
g = f / 52
|
a ) 133 years , b ) 177 years , c ) 119 years , d ) 212 years , e ) none | a | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 2 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . β΄ required average = 266 / 2 = 133 years . answer a" | a = 35 * 16
b = a + 21
c = b + 35
d = 35 * 16
e = c - d
|
a ) 4 : 3 , b ) 4 : 8 , c ) 4 : 1 , d ) 4 : 2 , e ) 4 : 7 | a | divide(4, 3) | if the sides of a cube are in the ratio 4 : 3 . what is the ratio of their diagonals ? | "a 1 : a 2 = 4 : 3 d 1 : d 2 = 4 : 3 answer : a" | a = 4 / 3
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a ) 8 , b ) 7 , c ) 5 , d ) 4 , e ) 3 | d | add(floor(5), const_1) | if 638521 is to be divisible by 5 , what is the least whole number that should be added to ? | "a number is divisible by 5 if the last digit is either 0 or 5 . here , 638521 = 1 ( last digit is neither 0 or 5 ) . the next multiple of 5 i . e to make the last digit 0 or 5 add 4 4 must be added to 638521 to make it divisible by 5 d" | a = math.floor(5)
b = a + 1
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a ) 17 litres , b ) 24.67 litres , c ) 11 litres , d ) 24.07 litres , e ) 38.67 litres | b | divide(subtract(multiply(37, 1399.45), multiply(37, 262.85)), subtract(3104.35, 1399.45)) | the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a Γ§ ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 37 litres of the mixed fruit juice . how many litres of the a Γ§ ai berry juice does he need to add ? | "262.85 ( 37 ) + 3 , 104.35 x = 1 , 399.45 ( 37 + x ) solve the equation . 262.85 ( 37 ) + 3 , 104.35 x = 1 , 399.45 ( 37 + x ) 9725.45 + 3 , 104.35 x = 51779.65 + 1 , 399.45 x 9725.45 + 1 , 704.9 x = 51779.65 1 , 704.9 x = 42054.2 x β 24.67 answer is b ." | a = 37 * 1399
b = 37 * 262
c = a - b
d = 3104 - 35
e = c / d
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['a ) 21', 'b ) 88937', 'c ) 269', 'd ) 279', 'e ) 2701'] | a | multiply(multiply(divide(sqrt(const_3), const_2), 12), const_2) | two equilateral triangle of side 12 cm are placed one on top another , such a 6 pionted star is formed if the six vertices lie on a circle what is the area of the circle not enclosed by the star ? | answer : a | a = math.sqrt(3)
b = a / 2
c = b * 12
d = c * 2
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a ) 1 : 2 , b ) 3 : 4 , c ) 3 : 5 , d ) 2 : 3 , e ) none of these | d | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | the average age of a class is 15.8 years . the average age of boys in the class is 16.4 years while that of girls is 15.4 years . what is the ratio of boys to girls in the class ? | let x = number of girls y = number of boys sum of girls ages / x = 15.4 sum of boys ages / y = 16.4 the sum of the girls ' ages is 15.4 x , and boys ' sum is 16.4 y ( sum of girls ages + sum of boys ages ) / ( x + y ) = 15.8 ( 15.4 x + 16.4 y ) / ( x + y ) = 15.8 15.4 x + 16.4 y = 15.8 ( x + y ) 15.4 x + 16.4 y = 15.8 x + 15.8 y 0.6 y = 0.4 x 0.6 y / x = 0.4 y / x = 0.4 / 0.6 y / x = 2 / 3 which is the ratio of boys to girls answer : d | a = 15 - 8
b = 16 - 4
c = a / b
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a ) 0.93 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 7), const_100) | subtracting 7 % of a from a is equivalent to multiplying a by how much ? | "answer let a - 7 % of a = ab . β ( 93 x a ) / 100 = ab β΄ b = 0.93 correct option : a" | a = 100 - 7
b = a / 100
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | divide(subtract(57, 12), subtract(57, 12)) | a number divided by 84 leaves remainder 57 what is the remainder when same number divided by 12 | "add 84 + 57 = 141 now 141 divided by 12 so we get 9 as reaminder answer : c" | a = 57 - 12
b = 57 - 12
c = a / b
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a ) 5 , b ) 7 , c ) 6 , d ) 31 , e ) 18 | c | floor(divide(divide(93, 3), 5)) | a children ' s gift store sells gift certificates in denominations of $ 3 and $ 5 . the store sold ' m ' $ 3 certificates and ' n ' $ 5 certificates worth $ 93 on a saturday afternoon . if ' m ' and ' n ' are natural numbers , how many different values can ' m ' take ? | explanatory answer step 1 : key data 1 . total value of all certificates sold = $ 93 . 2 . certificates sold were in denominations of $ 3 and $ 5 . 3 . both ' m ' and ' n ' are natural numbers . step 2 : approach the value of all certificates sold , 93 is divisible by 3 . so , a maximum of 31 $ 3 certificates and no $ 5 certificates could have been sold . however , the question states that both ' m ' and ' n ' are natural numbers . hence , at least 1 $ 5 certificate should have been sold . let us reduce the number of $ 3 certificates from theoretical maximum count of 31 by say ' x ' and correspondingly increase $ 5 certificates by ' y ' . evidently , 3 x = 5 y because the value of $ 3 certificates reduced should be the same as the value of $ 5 certificates increased . it means that x has to be a multiple of 5 and y has to be a multiple of 3 . or $ 3 certificates reduce in steps of 5 certificates . step 3 : list down possibilities the following combinations are possible 1 . m = 26 , n = 3 2 . m = 21 , n = 6 3 . m = 16 , n = 9 4 . m = 11 , n = 12 5 . m = 6 , n = 15 6 . m = 1 , n = 18 an alternative way to think of the same concept - replacing five $ 3 certificates with three $ 5 certificates leads to no change in the overall value of certficates sold and gives us a new combination each time . we need to see how many such combinations are possible . answer c | a = 93 / 3
b = a / 5
c = math.floor(b)
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a ) 72 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(45, const_4), multiply(9, divide(const_1000, const_3600))) | how long will a boy take to run round a square field of side 45 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 45 * 4 = 180 m time taken = 180 * 2 / 5 = 72 sec answer is a" | a = 45 * 4
b = 1000 / 3600
c = 9 * b
d = a / c
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a ) 5 , b ) w = 9 , c ) w = 10 , d ) w = 20 , e ) w = 30 | d | multiply(subtract(9, 10), 10) | what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? | "what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a = 9 - 10
b = a * 10
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a ) 20 , b ) 15 , c ) 10 , d ) 5 , e ) 25 | c | add(add(50, subtract(60, 50)), 50) | in a group of 100 people , 60 like volleyball , 50 like hockey . how many like both volleyball and hockey ? | "make a venn diagram , and enter your data . let the number of people who like both volleyball and hockey be x 60 - x + x + 50 - x = 100 x = 10 so number who like both volleyball and hockey = 10 answer c" | a = 60 - 50
b = 50 + a
c = b + 50
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a ) 1 / 7 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(2, 2), add(multiply(divide(2, 8), const_60), multiply(divide(2, 10), const_60))) | triathlete dan runs along a 2 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 10 miles per hour and swims at a rate of 8 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 4 miles round trip . running part : ( 2 / 10 = 1 / 5 * 60 = 12 minutes ) swimming part : ( 2 / 8 = 1 / 4 * 60 = 15 minutes ) 4 miles in ( 12 + 15 ) minutes 4 / 27 = 1 / 7 mile per minute answer : 1 / 7 mile per minute a" | a = 2 + 2
b = 2 / 8
c = b * const_60
d = 2 / 10
e = d * const_60
f = c + e
g = a / f
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a ) 40 sec , b ) 11 sec , c ) 37.5 sec , d ) 19 sec , e ) 10 sec | c | divide(360, multiply(subtract(48, 140), const_0_2778)) | a train 360 m long is running at a speed of 48 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 48 * 5 / 18 = 40 / 3 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 3 / 40 = 37.5 sec answer : c" | a = 48 - 140
b = a * const_0_2778
c = 360 / b
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a ) 1 hr , b ) 1 hr 25 min , c ) 50 min , d ) 1 hr 30 min , e ) 1 hr 45 min | b | divide(100, subtract(divide(100, 1), 30)) | a train covers a distance of 100 km in 1 hour . if its speed is decreased by 30 km / hr , the time taken by the car to cover the same distance will be ? | "speed = 100 / 1 = 100 km / hr new speed = 100 - 30 = 70 km / hr time taken = 100 / 70 = 1 hr 25 min answer is b" | a = 100 / 1
b = a - 30
c = 100 / b
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a ) 66 , b ) 70 , c ) 74 , d ) 78 , e ) 82 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 7), add(2, 2)), 6) | the ratio of the present age of sandy to that of molly is 7 : 2 . sixteen years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | let the present age of sandy be 7 x years and that of molly be 2 x years . ( 7 x + 16 ) / ( 2 x + 16 ) = 5 / 2 4 x = 48 x = 12 six years ago , sandy ' s age was 7 ( 12 ) - 6 = 78 the answer is d . | a = 10 + 6
b = 5 * a
c = 10 + 6
d = 2 * c
e = b - d
f = e * 7
g = 2 + 2
h = f / g
i = h - 6
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a ) 62 , b ) 52 , c ) 32 , d ) 12 , e ) 42 | e | subtract(50, multiply(multiply(12, 3), 2)) | evaluate : 50 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 50 - 12 * 3 * 2 = 50 - 8 = 42 correct answer e" | a = 12 * 3
b = a * 2
c = 50 - b
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a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 40 % | a | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(60, 100))), add(400, 100)), 100) | a grocer has 400 pounds of coffee in stock , 20 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated , what percent , by weight , of the grocer β s stock of coffee is decaffeinated ? | "1 . 20 % of 400 = 80 pounds of decaffeinated coffee 2 . 60 % of 100 = 60 pounds of decaffeinated coffee 3 . wt have 140 pounds of decaffeinated out of 500 pounds , that means 140 / 500 * 100 % = 28 % . the correct answer is a ." | a = 20 / 100
b = a * 400
c = 60 / 100
d = 100 * c
e = b + d
f = 400 + 100
g = e / f
h = g * 100
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a ) 21 , b ) 9 , c ) 11 , d ) 14 , e ) 7 | d | sqrt(divide(1764, const_2)) | the area of a parallelogram is 1764 sq m and its base is nine times size of the corresponding altitude . then the length of the base is ? | "9 x * x = 1764 = > x = 14 answer : d" | a = 1764 / 2
b = math.sqrt(a)
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a ) 12.5 % , b ) 12.9 % , c ) 18.5 % , d ) 11.5 % , e ) 12.3 % | a | divide(const_100, 8) | at what rate percent per annum will a sum of money double in 8 years . | "explanation : let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 8 ) ] % = 12.5 % per annum . answer : a ) 12.5 %" | a = 100 / 8
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['a ) rs . 2250', 'b ) rs . 3600', 'c ) rs . 2700', 'd ) rs . 1800', 'e ) none of these'] | d | multiply(divide(multiply(divide(60, 10), multiply(10, 10)), 20), 60) | paper charge is rs . 60 per kg . how much expenditure would be there to cover a cube of edge 10 m with a paper , if one kg of paper covers 20 sq . m . area ? | explanation : total surface area of the cube = 6 x ( 10 ) 2 = 600 sq . m 1 kg covers 20 sq . m area , for 600 we need = 600 / 20 = 30 kg paper . expenditure = rate x quantity = rs . 60 x 30 kg = rs . 1800 answer is d | a = 60 / 10
b = 10 * 10
c = a * b
d = c / 20
e = d * 60
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a ) 25 % , b ) 18 % , c ) 20 % , d ) 40 % , e ) 45 % | a | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 800), 800), const_100) | a shopkeeper sells his goods at cost price but uses a weight of 800 gm instead of kilogram weight . what is his profit percentage ? | "100 ( 1000 - 800 ) / 800 = 25 % profit percentage = 25 % answer a" | a = 4 + 1
b = 4 + 1
c = a + b
d = c * 100
e = d - 800
f = e / 800
g = f * 100
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(10, const_4) | how many odd numbers between 10 and 900 are the squares of integers ? | "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 13 numbers . answer : b ." | a = 10 + 4
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a ) 20 , b ) 24 , c ) 0 , d ) 32 , e ) 48 | c | multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1)) | what is the value of x ^ 2 yz β xyz ^ 2 , if x = 2 , y = 1 , and z = 2 ? | "4 * 1 * 2 - ( 2 * 1 * 4 ) = 8 + 8 = 0 ans : c" | a = negate * (
b = a * 1
c = b * 1
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a ) 11 , b ) 13 , c ) 15 , d ) 18 , e ) 12 | e | multiply(divide(const_1, multiply(add(const_100, 35), divide(const_1, subtract(const_100, 10)))), 18) | by selling 18 pencils for a rupee a man loses 10 % . how many for a rupee should he sell in order to gain 35 % ? | 90 % - - - 18 135 % - - - ? 90 / 135 * 18 = 12 answer : e | a = 100 + 35
b = 100 - 10
c = 1 / b
d = a * c
e = 1 / d
f = e * 18
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a ) 30 , b ) 40 , c ) 35 , d ) 60 , e ) 70 | c | divide(12.5, 0.35) | how many pieces of 0.35 meteres can be cut from a rod 12.5 meteres long | "explanation : we need so simple divide 42.5 / 0.85 , = ( 1250 / 35 ) = 35 option c" | a = 12 / 5
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a ) 12 , b ) 15 , c ) 16 , d ) 9 , e ) 22 | d | subtract(add(7, 5), 3) | in a class , 7 students like to play basketball and 5 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 7 + 5 - 3 = 9 d )" | a = 7 + 5
b = a - 3
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a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | e | divide(multiply(choose(6, 2), choose(3, 1)), choose(add(add(6, 4), 3), 3)) | a jar contains only red , yellow , and orange marbles . if there are 3 red , 6 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 3 / 22 . this helped me realise that for the 3 possible orderings the probabbility is the same . so , it should be ( 1 / 22 ) * ( 6 ) , which indeed is 6 / 22 . e" | a = math.comb(6, 2)
b = math.comb(3, 1)
c = a * b
d = 6 + 4
e = d + 3
f = math.comb(e, 3)
g = c / f
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Subsets and Splits