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 ) 10 % , b ) 30 % , c ) 20 % , d ) 70 % , e ) 45 % | c | multiply(divide(multiply(const_2, divide(70, const_10)), 70), const_100) | what percentage of numbers from 1 to 70 have 1 or 9 in the unit ' s digit ? | d 20 % clearly , the numbers which have 1 or 9 in the unit ' s digit , have squares that end in the digit 1 . such numbers from 1 to 70 are 1 , 9 , 11 , 19 , 21 , 29 , 31 , 39 , 41 , 49 , 51 , 59 , 61 , 69 . number of such number = 14 required percentage = ( 14 x 100 / 70 ) % = 20 % . | a = 70 / 10
b = 2 * a
c = b / 70
d = c * 100
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a ) 2.8 units , b ) 6 units , c ) 3 units , d ) 5 units , e ) 12 units | a | divide(triangle_area_three_edges(8, 15, 19), divide(triangle_perimeter(8, 15, 19), const_2)) | what is the measure of the radius of the circle inscribed in a triangle whose sides measure 8 , 15 and 19 units ? | "sides are 8 , 15 and 19 . . . thus it is right angle triangle since 19 ^ 2 = 8 ^ 2 + 15 ^ 2 therefore , area = 1 / 2 * 15 * 8 = 60 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 19 + 15 + 8 / 2 = 21 thus , 60 = 21 * r and hence r = in - radius = 2.8 option a" | a = triangle_area_three_edges / (
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a ) 6 , b ) 6.67 , c ) 12 , d ) 13 , e ) 15 | e | multiply(divide(12, 80), const_100) | 12 is what % of 80 ? | "we assume that 80 is 100 % assume ' x ' is value we looking for here , 80 = 100 % and x % = 12 therefore , 100 / x = 80 / 12 100 / x = 6.6667 x = 15 e" | a = 12 / 80
b = a * 100
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a ) $ 255 , b ) $ 275 , c ) $ 510 , d ) $ 612 , e ) $ 2,550 | d | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 1.2) | when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 1.20 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | total no . of boxes = 3060000 / ( 20 × 20 × 15 ) = 510 total cost = 510 × $ 1.20 = $ 612 answer d | a = 1000 * 1000
b = 3 * 6
c = 20 * 20
d = c * 15
e = b / d
f = e * 1
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a ) 500 , b ) 277 , c ) 222 , d ) 297 , e ) 400 | e | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 40)) | the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 40 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 140 = rs . 400 answer : e" | a = 616 * 100
b = 100 + 10
c = a / b
d = 100 + 10
e = c * d
f = 100 + 40
g = e / f
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a ) 5964 , b ) 5664 , c ) 4964 , d ) 4664 , e ) 5555 | a | multiply(multiply(multiply(const_3, power(const_2, const_2)), divide(divide(divide(852, const_2), const_2), const_3)), add(const_4, const_3)) | find l . c . m of 852 and 1491 | 852 ) 1491 ( 1 852 639 ) 852 ( 1 639 213 ) 639 ( 3 639 0 h . c . f of 852 and 1491 is 213 : . l . c . m = 852 * 1491 / 213 = 5964 answer is a . | a = 2 ** 2
b = 3 * a
c = 852 / 2
d = c / 2
e = d / 3
f = b * e
g = 4 + 3
h = f * g
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a ) 31 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | d | divide(factorial(subtract(add(const_4, 17), const_1)), multiply(factorial(17), factorial(subtract(const_4, const_1)))) | how many positive integers less than 300 are there such that they are multiples of 17 or multiples of 16 ? | "300 / 17 = 17 ( plus remainder ) so there are 17 multiples of 17 300 / 16 = 18 ( plus remainder ) so there are 18 multiples of 16 we need to subtract 1 because 17 * 16 is a multiple of both so it was counted twice . the total is 17 + 18 - 1 = 34 the answer is d ." | a = 4 + 17
b = a - 1
c = math.factorial(b)
d = math.factorial(17)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
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a ) 2 : 3 , b ) 5 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | b | divide(sqrt(25), sqrt(9)) | two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 25 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 25 : â ˆ š 9 = 5 : 3 . answer b" | a = math.sqrt(25)
b = math.sqrt(9)
c = a / b
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a ) 784596 , b ) 1126538 , c ) 804670 , d ) 784596 , e ) 864520 | b | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 700), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 700 miles into meters ? | "1 mile = 1609.34 meters 700 mile = 700 * 1609.34 = 1126538 meters answer is b" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 700
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | e | add(reminder(19, 19), const_1) | when the positive integer a is divided by 11 , the quotient is b and the remainder 3 . when a is divided by 19 , the remainder is also 3 . what is the remainder when b is divided by 19 ? | any number which when divided by divisor d 1 , d 2 , etc . leaving same remainderrtakes the form ofk + r where k = lcm ( d 1 , d 2 ) in this case the divisors are 1119 and remainder is 3 . so lcm ( 11,19 ) = 209 so n = 209 + 3 = 212 also a = d 1 q + 3 ; which means d 1 q = 209 d 1 = 11 therefore q = 19 and ( b divided by 19 ) 19 / 19 leaves remainder 0 . answer is e | a = reminder + (
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a ) 654 , b ) 655 , c ) 611 , d ) 657 , e ) 658 | c | multiply(divide(subtract(const_100, 35), const_100), 940.00) | yearly subscription to professional magazines cost a company $ 940.00 . to make a 35 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 35 / 100 = 329 so the cut in amount is 329 the less amount to be spend is 940 - 329 = 611 answer : c" | a = 100 - 35
b = a / 100
c = b * 940
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a ) 50 % , b ) 22.22 % , c ) 25 % , d ) 16.66 % , e ) 12.5 % | b | multiply(subtract(divide(2, 3), multiply(divide(2, 3), divide(1, 2))), const_100) | a certain article of clothing was discounted during a special sale to 2 / 3 of its original retail price . when the clothing did n ' t sell , it was discounted even further to 1 / 2 of its original retail price during a second sale . by what percent did the price of this article of clothing decrease from the first sale to the second sale ? | "say the original retail price of the item was $ 200 . the price after the first sale = 2 / 3 * $ 200 = $ 400 / 3 . the price after the second sale = 1 / 2 * $ 200 = $ 100 . the percent change from the first sale to the second = ( 400 / 3 - 100 ) / 150 = 1 / 3 = 22.22 % . answer : b ." | a = 2 / 3
b = 2 / 3
c = 1 / 2
d = b * c
e = a - d
f = e * 100
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a ) $ 1200 , b ) $ 1500 , c ) $ 1400 , d ) $ 1250 , e ) $ 1350 | c | add(add(multiply(100, 10), multiply(20, 5)), multiply(multiply(30, 5), const_2)) | rates for having a manuscript typed at a certain typing service are $ 10 per page for the first time a page is typed and $ 5 per page each time a page is revised . if a certain manuscript has 100 pages , of which 20 were revised only once , 30 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "50 pages typed 1 x 20 pages typed 2 x ( original + one revision ) 30 pages typed 3 x ( original + two revisions ) 50 ( 10 ) + 20 ( 10 + 5 ) + 30 ( 10 + 5 + 5 ) = 500 + 300 + 600 = 1400 answer - c" | a = 100 * 10
b = 20 * 5
c = a + b
d = 30 * 5
e = d * 2
f = c + e
|
a ) 19 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | subtract(add(add(20, 40), 60), add(add(multiply(7, const_3), 10), 70)) | the average ( arithmetic mean ) of 20 , 40 , and 60 is 7 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 7 = 33 sum of second list = 33 * 3 = 99 therefore the number = 99 - 80 = 19 answer : a" | a = 20 + 40
b = a + 60
c = 7 * 3
d = c + 10
e = d + 70
f = b - e
|
a ) 50.78 kg , b ) 49.32 kg , c ) 61.67 kg , d ) 58.88 kg , e ) 70.89 kg | d | divide(add(multiply(50, 50), multiply(40, 70)), add(50, 40)) | there are 2 sections a and b in a class , consisting of 50 and 40 students respectively . if the average weight of section a is 50 kg and that of section b is 70 kg , find the average of the whole class ? | "total weight of 50 + 40 students = 50 * 50 + 40 * 70 = 2500 + 2800 average weight of the class is = 5300 / 90 = 58.88 kg answer is d" | a = 50 * 50
b = 40 * 70
c = a + b
d = 50 + 40
e = c / d
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | divide(add(20, multiply(4, 7)), subtract(7, const_1)) | dan ' s age after 20 years will be 7 times his age 4 years ago . what is the present age of dan ? | "let dan ' s present age be x . x + 20 = 7 ( x - 4 ) 6 x = 48 x = 8 the answer is b ." | a = 4 * 7
b = 20 + a
c = 7 - 1
d = b / c
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a ) 2 days , b ) 3 days , c ) 4 days , d ) 5 days , e ) 6 days | c | inverse(add(inverse(12), multiply(const_2, inverse(12)))) | a work as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work in ? | ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 6 b ' s 1 day work = 1 / 12 a + b 1 day work = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 a and b can finish the work in 4 days answer is c | a = 1/(12)
b = 1/(12)
c = 2 * b
d = a + c
e = 1/(d)
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a ) 187 , b ) 190 , c ) 192 , d ) 195 , e ) 213 | d | subtract(divide(factorial(add(6, 4)), multiply(factorial(4), factorial(6))), divide(factorial(6), multiply(factorial(const_2), factorial(4)))) | in a group of 6 boys & 4 girls a committee of 4 persons is to be formed . in how many different ways can it be done so that the committee has at least 1 girl ? | the committee of 4 persons is to be so formed that it has at least 1 woman the different ways that we can choose to form such a committee are : ( i ) lw . 3 m in t 6 x 5 x 4 ' x 6 c 3 = 4 x — — 80 3 x 2 x 1 x 6 c 2 = 42 : : x 26 : : = 90 ( ii ) 2 w . 2 m in ° c 2 ( iii ) 3 w . 1 m in 4 c 3 x 6 c 1 = 4 x 6 = 24 ( iv ) 4 w in 6 c 4 = 1 total no . of different ways in which a committee of 4 persons can be formed so that it has at least one woman . = 80 + 90 + 24 + 1 = 195 d | a = 6 + 4
b = math.factorial(a)
c = math.factorial(4)
d = math.factorial(6)
e = c * d
f = b / e
g = math.factorial(6)
h = math.factorial(2)
i = math.factorial(4)
j = h * i
k = g / j
l = f - k
|
['a ) 1.967', 'b ) 1.963', 'c ) 2.388', 'd ) 3.388', 'e ) 2.382'] | a | subtract(triangle_area_three_edges(multiply(3.5, const_2), multiply(3.5, const_2), multiply(3.5, const_2)), divide(circle_area(3.5), const_2)) | three circles of radius 3.5 cm are placed in such a way that each circle touches the other two . the area of the portion enclosed by the circles is | explanation : required area = ( area of an equilateral triangle of side 7 cm ) - ( 3 * area of sector with ã = 60 degrees and r = 3.5 cm ) \ inline { \ color { black } ( \ frac { \ sqrt { 3 } } { 4 } \ times 7 \ times 7 ) - ( 3 \ times \ frac { 22 } { 7 } \ times 3.5 \ times 3.5 \ times \ frac { 60 } { 360 } ) } sq cm = \ inline { \ color { black } ( \ frac { \ sqrt { 3 } } { 4 } \ times 49 ) - ( 11 \ times 0.5 \ times 3.5 ) } sq cm = 1.967 sq cm answer : a ) 1.967 | a = 3 * 5
b = 3 * 5
c = 3 * 5
d = triangle_area_three_edges - (
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a ) 2 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | d | add(9, 2) | 2 a 33 + b 9 ____ 99 if a and b represent positive single digits in the correctly worked computation above , what is the value of a + 3 b ? | adding the digits in unit ' s place , a + 3 + 9 = 19 ( since a and b are positive single digits ) = > a = 7 now adding the digits in ten ' s place , 2 + 3 + b + 1 = 9 ( 1 has been carried over from nit ' s place addition ) = > b = 3 a + 3 b = 2 + 2 * 3 = 11 answer d | a = 9 + 2
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a ) - 4 , b ) - 3 , c ) - 2 , d ) - 1 , e ) 0 | e | subtract(multiply(5, 2), 5) | if ( 5 - x ) / ( 7 + x ) = x , what is the value of x ^ 2 + 8 x - 5 ? | "( 5 - x ) = x * ( 7 + x ) ( 5 - x ) = 7 x + x ^ 2 0 = x ^ 2 + 8 x - 5 the answer is e ." | a = 5 * 2
b = a - 5
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a ) 25 % , b ) 40 % , c ) 30 % , d ) 8 % , e ) 12 % | c | multiply(divide(300, 1), const_100) | what percent is 300 gm of 1 kg ? | "1 kg = 1000 gm 300 / 1000 ã — 100 = 30000 / 1000 = 30 % answer is c" | a = 300 / 1
b = a * 100
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a ) 1.21 % , b ) 2.56 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | a | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 11)), 675958), multiply(divide(const_100, subtract(const_100, 11)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 11)), 675958), multiply(divide(const_100, subtract(const_100, 11)), 675958))), const_100) | a man two flats for $ 675958 each . on one he gains 11 % while on the other he loses 11 % . how much does he gain or lose in the whole transaction ? | "in such a case there is always a loss loss % = ( 11 / 10 ) ^ 2 = 23 / 19 = 1.21 % answer is a" | a = 100 + 11
b = 100 / a
c = b * 675958
d = 100 - 11
e = 100 / d
f = e * 675958
g = c + f
h = 675958 + 675958
i = g - h
j = 100 + 11
k = 100 / j
l = k * 675958
m = 100 - 11
n = 100 / m
o = n * 675958
p = l + o
q = i / p
r = q * 100
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a ) 3 : 8 , b ) 3 : 6 , c ) 3 : 7 , d ) 2 : 1 , e ) 3 : 3 | d | divide(2, divide(2, 2)) | what is the ratio between perimeters of two squares one having 2 times the diagonal then the other ? | "d = 2 d d = d a √ 2 = 2 d a √ 2 = d a = 2 d / √ 2 a = d / √ 2 = > 2 : 1 answer : d" | a = 2 / 2
b = 2 / a
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a ) 20 mph , b ) 24 mph , c ) 30 mph , d ) 32 mph , e ) 55 mph | e | divide(60, subtract(divide(subtract(60, 2), 48), divide(2, 24))) | richard traveled the entire 60 miles trip . if he did the first 2 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 48 miles per hour , what is the his average speed , in miles per hour ? | average speed = sum of distance / sum of time . if he travelled the first 2 miles at 24 miles / hr , it would take 0.083 hr . for the remaining trip , if he went at 48 miles / 1 hr , it would take 1 hour . then , the average speed is 60 miles / ( 0.083 + 1 ) hrs = 55 miles / 1 hr . therefore , the answer is e . | a = 60 - 2
b = a / 48
c = 2 / 24
d = b - c
e = 60 / d
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a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 39 | a | divide(subtract(multiply(8, divide(2, 1)), 8), subtract(const_1, multiply(divide(2, 5), divide(2, 1)))) | a man ’ s current age is ( 2 / 5 ) of the age of his father . after 8 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | a 40 let , father ’ s current age is a years . then , man ’ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 8 ] = ( 1 / 2 ) ( a + 8 ) 2 ( 2 a + 40 ) = 5 ( a + 8 ) a = 40 | a = 2 / 1
b = 8 * a
c = b - 8
d = 2 / 5
e = 2 / 1
f = d * e
g = 1 - f
h = c / g
|
a ) 40 % , b ) 60 % , c ) 50 % , d ) 45 % , e ) 55 % | b | multiply(divide(const_3, add(const_3, const_2)), const_100) | a feed store sells two varieties of birdseed : brand a , which is 60 % millet and 40 % sunflower , and brand b , which is 65 % millet and 35 % safflower . if a customer purchases a mix of the two types of birdseed that is 50 % millet , what percent of the mix is brand a ? | yes there is a simple method : consider the following method brand a : 60 % millet and 40 % sunflower brand b : 65 % millet and 35 % safflower mix : 50 % millet here the weighted average is 50 % , now brand a has 60 % millet , which is 10 % more than the weighted average of mix = + 0.10 a - - - - - - - - - - - - - - - i similarly , brand b has 65 % millet , which is 15 % more than the weighted average of mix = + 0.15 b - - - - - - - - - - - - ii now , both brand a and brand b are combined to give a 50 % mix containing millet , so equate i and ii implies , 0.10 a = 0.15 b therefore a / b = 0.15 / 0.10 = 3 / 2 a : b : ( a + b ) = 3 : 2 : ( 3 + 2 ) = 3 : 2 : 5 we have to find , percent of the mix is brand a i . e . a : ( a + b ) = 3 : 5 = ( 3 / 5 ) * 100 = 60 % here is a pictorial representation : brand a = 60 % - - - - - - - - - - - - - - - - - - - - - - - - 10 % or 0.10 above average , a times - - - - - - - - - - - - - - - - - total above = + 0.10 a - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - average = 50 % or 0.50 brand b = 65 % - - - - - - - - - - - - - - - - - - - - - - - - - - 15 % or 0.15 above average , b times - - - - - - - - - - - - - - - - - total above = + 0.15 b since the amount below the average has to equal the average above the average ; therefore , 0.10 a = 0.15 b a / b = 3 / 2 a : b : total = 3 : 2 : 5 therefore a / total = 3 : 5 = 60 % answer : b | a = 3 + 2
b = 3 / a
c = b * 100
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a ) 33.33 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 5 % profit | a | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 45), 30)), divide(multiply(const_100, 45), 30))) | if the cost price of 30 articles is equal to the selling price of 45 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 30 articles = 30 * 1 = $ 30 the selling price of 45 articles = cost price of 30 articles = $ 30 . now , we know the selling price of 45 articles . let us find the cost price of 45 articles . cost price of 45 articles = 45 * 1 = $ 45 . therefore , profit made on sale of 45 articles = selling price of 45 articles - cost price of 45 articles = 30 - 45 = - $ 15 . as the profit is in the negative , the merchant has made a loss of $ 15 . therefore , % loss = loss / cp * 100 % loss = - 15 / 45 * 100 = 33.33 % loss . a" | a = 100 * 45
b = a / 30
c = 100 - b
d = 100 * 45
e = d / 30
f = c / e
g = 100 * f
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a ) 164 , b ) 152 , c ) 100 , d ) 182 , e ) 195 | a | subtract(multiply(21, 21), add(multiply(5, 14), multiply(9, 16))) | the average age of 21 students of a class is 21 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years , the age of the 21 th student is | "explanation : age of the 21 th student = [ 21 * 21 - ( 14 * 5 + 16 * 9 ) ] = ( 378 - 214 ) = 164 years . answer : a" | a = 21 * 21
b = 5 * 14
c = 9 * 16
d = b + c
e = a - d
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a ) 18 , b ) 56 , c ) 12 , d ) 17 , e ) 10 | e | subtract(40, multiply(10, 3)) | the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 40 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 40 + q = > q = 10 answer : e" | a = 10 * 3
b = 40 - a
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a ) 158 , b ) 200 , c ) 255 , d ) 400 , e ) 280 | c | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | the difference between a number and its two - fifth is 510 . what is 30 % of that number ? | "let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 10 % of 850 = 255 . answer : c" | a = 4 + 1
b = 510 * a
c = 1 + 2
d = b / c
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a ) 3 , b ) 3 1 / 3 , c ) 3 1 / 2 , d ) 4 , e ) 4 1 / 3 | e | divide(add(subtract(18, 15), subtract(15, subtract(multiply(divide(40, const_60), 18), multiply(divide(40, const_60), 15)))), subtract(multiply(divide(40, const_60), 18), multiply(divide(40, const_60), 15))) | john and jacob set out together on bicycle traveling at 18 and 15 miles per hour , respectively . after 40 minutes , john stops to fix a flat tire . if it takes john one hour to fix the flat tire and jacob continues to ride during this time , how many hours will it take john to catch up to jacob assuming he resumes his ride at 18 miles per hour ? ( consider john ' s deceleration / acceleration before / after the flat to be negligible ) | "john ' s speed - 18 miles / hr jacob ' s speed - 15 miles / hr after 40 min ( i . e 2 / 3 hr ) , distance covered by john = 18 x 2 / 3 = 12 miles . jacob continues to ride for a total of 1 hour and 40 min ( until john ' s bike is repaired ) . distance covered in 1 hour 40 min ( i . e 5 / 3 hr ) = 15 x 5 / 3 = 25 miles . now , when john starts riding back , the distance between them is 13 miles . jacob and john are moving in the same direction . for john to catch jacob , the effective relative speed will be 18 - 15 = 3 miles / hr . thus , to cover 13 miles at 3 miles / hr , john will take 13 / 3 = 4 1 / 3 hours answer e" | a = 18 - 15
b = 40 / const_60
c = b * 18
d = 40 / const_60
e = d * 15
f = c - e
g = 15 - f
h = a + g
i = 40 / const_60
j = i * 18
k = 40 / const_60
l = k * 15
m = j - l
n = h / m
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a ) 27 , b ) 36 , c ) 42 , d ) 24 , e ) 147 | d | multiply(divide(48, 112), 56) | the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 56 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 56 ounces of a substance has a volume of ( 48 / 112 ) * 56 = 24 cubic inches answer d" | a = 48 / 112
b = a * 56
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a ) 1200 , b ) 1210 , c ) 1180 , d ) 1190 , e ) 1680 | e | add(800, multiply(800, divide(110, const_100))) | 800 is increased by 110 % . find the final number . | explanation final number = initial number + 110 % ( original number ) = 800 + 110 % ( 800 ) = 800 + 880 = 1680 . answer e | a = 110 / 100
b = 800 * a
c = 800 + b
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a ) 66 % , b ) 75 % , c ) 82 % , d ) 116 % , e ) 150 % | c | multiply(divide(add(100, 40), add(100, divide(add(100, 40), const_2))), 100) | company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 40 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw would be approximately what percent of these combined assets ? | "let the price of company a ' s assets be 100 price of assets of kw is 40 % more than company a ' s assets which is 140 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 70 a + b = 170 kw = 140 kw / ( a + b ) * 100 = 140 / 170 * 100 = 82.35 % or 82 % c" | a = 100 + 40
b = 100 + 40
c = b / 2
d = 100 + c
e = a / d
f = e * 100
|
a ) 11 : 17 , b ) 8 : 27 , c ) 5 : 9 , d ) 2 : 9 , e ) none | b | divide(multiply(multiply(2, 4), 5), multiply(multiply(3, 5), 9)) | if a : b = 2 : 3 , b : c = 4 : 5 and c : d = 5 : 9 then a : d is equal to : | "solution : a / d = ( a / b ) * ( b * c ) * ( c / d ) = ( 2 / 3 ) * ( 4 / 5 ) * ( 5 / 9 ) = ( 2 * 4 * 5 ) / ( 3 * 5 * 9 ) = 8 / 27 . answer : option b" | a = 2 * 4
b = a * 5
c = 3 * 5
d = c * 9
e = b / d
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a ) 1 / 7 , b ) 3 / 14 , c ) 1 / 2 , d ) 2 / 7 , e ) 3 / 4 | d | divide(const_2, 7) | if an integer n is to be selected at random from 1 to 105 , inclusive , what is probability n ( n + 1 ) will be divisible by 7 ? | "for n ( n + 1 ) to be a multiple of 7 , either n or n + 1 has to be a multiple of 7 . thus n must be of the form 7 k or 7 k - 1 . the probability is 2 / 7 . the answer is d ." | a = 2 / 7
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a ) 2 : 5 , b ) 3 : 5 , c ) 4 : 5 , d ) 6 : 5 , e ) 7 : 5 | c | subtract(const_100, multiply(divide(add(20, const_100), add(50, const_100)), const_100)) | two numbers are respectively 20 % and 50 % more than a third number . the ratio of the two numbers is | "explanation : let the third number be x . first number ( 120 / 100 ) * x = 6 x / 5 second number ( 150 / 100 ) * x = 3 x / 2 ratio = 6 x / 5 : 3 x / 2 = > 4 : 5 option c" | a = 20 + 100
b = 50 + 100
c = a / b
d = c * 100
e = 100 - d
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a ) 5,050 , b ) 7,500 , c ) 8,050 , d ) 15,000 , e ) 19,600 | c | multiply(divide(add(200, 101), const_2), add(divide(subtract(200, 101), const_2), const_1)) | the sum of the first 50 positive even integers is 2,000 . what is the sum of the odd integers from 101 to 200 , inclusive ? | "101 + 103 + . . . . . . . 199 if we remove 100 from each of these it will be sum of 1 st 100 odd numbers . so 101 + 103 + . . . . . . . 199 = 50 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) sum of 1 st 100 natural numbers = ( 100 * 101 ) / 2 = 5050 sum of 1 st 50 positive even integers = 2000 sum of 1 st 100 odd numbers = 5050 - 2000 = 3050 so 101 + 103 + . . . . . . . 199 = 50 * 100 + ( 1 + 3 + 5 + 7 + . . . . . . ) = 5000 + 3050 = 8050 c is the answer ." | a = 200 + 101
b = a / 2
c = 200 - 101
d = c / 2
e = d + 1
f = b * e
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['a ) 3 hours and 12 minutes .', 'b ) four hours and 30 minutes', 'c ) four hours and 45 minutes', 'd ) five hours and 10 minutes', 'e ) five hours and 30 minutes'] | b | divide(13.5, 3) | rain is falling at a rate of 3 centimeters per hour all over springfield . somewhere downtown in springfield a group of pigeons is waiting for the rain to stop . if the rain filled a round puddle the with a base area of 350 square centimeters and a depth of 13.5 centimeters , how long did the pigeons wait for the rain to stop ? | the volume of the puddle is irrelevant and only height matters since rain fell all over the city . thus , it takes only . 13.5 / 3 = 4.5 hours of rain to fill the puddle answer is : b , 4 hours and 30 mins | a = 13 / 5
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a ) - 1 , b ) 6 , c ) 7 , d ) 12 , e ) 14 | c | add(add(const_4, 3), subtract(3, const_4)) | what is the sum of all possible solutions to | x + 3 | ^ 2 + | x + 3 | = 20 ? | "denote | x + 3 | as y : y ^ 2 + y = 20 - - > y = - 5 or y = 4 . discard the first solution since y = | x + 3 | , so it ' s an absolute value and thus can not be negative . y = | x + 3 | = 4 - - > x = 7 . the sum = 7 . answer : c ." | a = 4 + 3
b = 3 - 4
c = a + b
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a ) 1000 , b ) 1250 , c ) 1300 , d ) 2000 , e ) 2250 | e | divide(add(divide(multiply(800, const_100), 30), 800), divide(80, const_100)) | a small pool filled only with water will require an additional 800 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 800 gallons of water will increase the amount of water in the pool by 30 % , what is the total capacity of the pool in gallons ? | "since pumping in additional 800 gallons of water will increase the amount of water in the pool by 30 % , then initially the pool is filled with 1,000 gallons of water . so , we have that 1,000 + 800 = 0.8 * { total } - - > { total } = 2,250 . answer : e ." | a = 800 * 100
b = a / 30
c = b + 800
d = 80 / 100
e = c / d
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a ) a ) 8239 , b ) b ) 2900 , c ) c ) 4500 , d ) d ) 2393 , e ) e ) 5000 | e | multiply(multiply(subtract(4, 3), 2500), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 2500 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 2500 = > x = 2500 . b ' s share = rs . 2 x = 2 * 2500 = rs . 5000 . answer : e" | a = 4 - 3
b = a * 2500
c = b * 3
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a ) 178.27 cm , b ) 182.29 cm , c ) 978.29 cm , d ) 178.89 cm , e ) 176.29 cm | b | floor(divide(add(subtract(multiply(35, 184), 166), 106), 35)) | the average height of 35 boys in a class was calculated as 184 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? | "calculated average height of 35 boys = 184 cm . wrong total height of 35 boys = 184 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 184 cm - ( 166 cm - 106 cm ) / 35 = 184 cm - 60 / 35 cm = 184 cm - 1.71 cm = 182.29 cm . answer : b" | a = 35 * 184
b = a - 166
c = b + 106
d = c / 35
e = math.floor(d)
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a ) 25 , b ) 40 , c ) 35 , d ) 30 , e ) 28 | b | multiply(subtract(27, 7), const_2) | all the students of class are told to sit in circle shape . here the boy at the 7 th position is exactly opposite to 27 th boy . total number of boys in the class ? | "as half the circle shape consist of 27 - 7 = 20 boys , so total number of boys in full circle = 2 * 20 = 40 answer : b" | a = 27 - 7
b = a * 2
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a ) 130 cm , b ) 126 cm , c ) 120 cm , d ) 116 cm , e ) 112 cm | c | subtract(132, divide(multiply(132, 10), const_100)) | on my sister ' s birthday , she was 132 cm in height , having grown 10 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.1 x = 132 x = 120 the answer is c ." | a = 132 * 10
b = a / 100
c = 132 - b
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a ) 18 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 80 % | a | multiply(divide(subtract(72.95, 59.95), 72.95), const_100) | a $ 72.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | listed selling price of chair = 72.95 $ discounted selling price of chair = 59.95 $ discount = 72.95 - 59.95 = 13 $ % decrease in price of chair = ( 13 / 72.95 ) * 100 % = 18 % approx answer a | a = 72 - 95
b = a / 72
c = b * 100
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a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 1 : 2 | e | divide(400, 800) | if shares of two persons in profits are rs . 400 and rs . 800 then ratio of their capitals is | "profit = time * capital so 400 : 800 = 1 : 2 answer : e" | a = 400 / 800
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a ) $ 1100.25 , b ) $ 1100.75 , c ) $ 1101.25 , d ) $ 1101.75 , e ) $ 1102.50 | e | multiply(1000, power(add(const_1, divide(divide(10, const_2), const_100)), const_2)) | if an investor puts $ 1000 in a savings account that earns 10 percent annual interest compounded semiannually , how much money will be in the account after one year ? | "1.05 * 1.05 * 1000 = $ 1102.50 the answer is e ." | a = 10 / 2
b = a / 100
c = 1 + b
d = c ** 2
e = 1000 * d
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a ) 45 / 21 , b ) 46 / 2 , c ) 13 / 30 , d ) 23 / 6 , e ) 4 / 6 | c | divide(add(multiply(7, const_0_33), multiply(3, divide(const_2, const_3))), add(7, 3)) | at an international conference , “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is 7 : 3 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed , then what is the fraction of the participants who are left - handed ? | "red : blue = 7 : 3 let red = 7 x and blue = 3 x 1 / 3 of red are left handed = > 1 / 3 * 7 x = 7 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 3 x = 6 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( 7 x / 3 + 6 x / 3 ) / ( 7 x + 3 x ) = 13 / 30 answer : c" | a = 7 * const_0_33
b = 2 / 3
c = 3 * b
d = a + c
e = 7 + 3
f = d / e
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a ) s . 2.99 , b ) s . 2.4 , c ) s . 2.5 , d ) s . 2.2 , e ) s . 2.1 | a | multiply(divide(divide(multiply(divide(23, const_100), 100), 10), multiply(divide(23, const_100), 100)), const_100) | a reduction of 23 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 23 / 100 ) = 23 - - - 10 ? - - - 1 = > rs . 2.3 100 - - - 77 ? - - - 2.3 = > rs . 2.99 answer : a" | a = 23 / 100
b = a * 100
c = b / 10
d = 23 / 100
e = d * 100
f = c / e
g = f * 100
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a ) 50 , b ) 100 , c ) 30 , d ) 75 , e ) 5 | a | multiply(const_100, divide(add(27, multiply(divide(10, const_100), 30)), 60)) | if 60 % of a number is greater than 30 % of 10 by 27 , what is the number ? | explanation : 60 / 100 * x - 30 / 100 * 10 = 27 60 / 100 * x - 3 = 27 60 / 100 * x = 30 x = 30 * 100 / 60 x = 50 answer : option a | a = 10 / 100
b = a * 30
c = 27 + b
d = c / 60
e = 100 * d
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a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | divide(53, const_10) | how many integers from 0 to 53 , inclusive , have a remainder of 1 when divided by 3 ? | "my ans is also c . 17 . explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 52 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term of an ap , a is the first term and d is the common difference . so , 52 = 1 + ( n - 1 ) 3 or , ( n - 1 ) 3 = 51 or , n - 1 = 17 or , n = 18 d" | a = 53 / 10
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a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | b | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(3, add(add(3, 2), 2))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(2, add(add(5, 2), 2)))) | a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 5 : 2 : 2 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 4 : 3 , respectively . how many more acres of land were planted with tobacco under the new system ? | "originally ( 2 / 9 ) * 1350 = 300 acres were planted with tobacco . in the new system ( 3 / 9 ) * 1350 = 450 acres were planted with tobacco . the answer is b ." | a = 100 * 10
b = 3 * 100
c = a + b
d = 5 * 10
e = c + d
f = 3 + 2
g = f + 2
h = 3 / g
i = e * h
j = 100 * 10
k = 3 * 100
l = j + k
m = 5 * 10
n = l + m
o = 5 + 2
p = o + 2
q = 2 / p
r = n * q
s = i - r
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a ) 0.68 , b ) 0.086 , c ) 0.86 , d ) 0.068 , e ) none of them | c | divide(subtract(power(0.96, 3), power(0.1, 3)), add(add(power(0.96, 2), 0.096), power(0.1, 2))) | ( 0.96 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.96 ) ( power 2 ) + 0.096 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.96 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.96 ) ( power 2 ) + ( 0.96 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.96 - 0.1 ) = 0.86 answer is c ." | a = 0 ** 96
b = 0 ** 1
c = a - b
d = 0 ** 96
e = d + 0
f = 0 ** 1
g = e + f
h = c / g
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a ) 36 , b ) 72 , c ) 120 , d ) 144 , e ) 108 | e | divide(450, add(divide(100, const_60), divide(150, const_60))) | a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 450 bolts ? | "old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 450 bolts 450 * 6 / 25 = 108 mins ans e" | a = 100 / const_60
b = 150 / const_60
c = a + b
d = 450 / c
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a ) 10 , b ) 20 , c ) 16 , d ) 30 , e ) 40 | c | subtract(subtract(21, 2.5), 2.5) | a man ' s speed with the current is 21 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 21 km / hr = > speed of the man + speed of the current = 21 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 21 - 2.5 = 18.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 18.5 - 2.5 = 16 km / hr answer is c ." | a = 21 - 2
b = a - 2
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a ) 11 sec , b ) 15 sec , c ) 16 sec , d ) 20 sec , e ) none | a | multiply(divide(275, multiply(90, const_1000)), const_3600) | a train 275 m long , running with a speed of 90 km / hr will pass a tree in | "sol . speed = ( 90 x 5 / 18 ) m / sec . = 25 m / sec . time taken = ( 275 x 1 / 25 ) sec = 11 sec answer a" | a = 90 * 1000
b = 275 / a
c = b * 3600
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a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54 | b | divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2) | the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if steve and danny have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ? | "if steve and danny submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combined no discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) ." | a = 80 * 0
b = 2 * a
c = 100 - 25
d = c / 100
e = 160 * d
f = e * 0
g = b - f
h = g / 2
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a ) 25 , b ) 30 , c ) 50 , d ) 55 , e ) 70 | e | multiply(divide(28, 40), const_100) | 28 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ? | "i came up with ( e ) 70 think of 100 people total : from the first fact , 28 of these are women with fair hair . from the second fact , these 20 women make up 40 % of the total fair haired population . we can then make a ratio of 60 : 40 fair haired men to fair haired women . this means that ( 60 / 40 ) * 28 equals the number of fair haired men , which is 42 men with fair hair . add this 42 to the 28 women and get 70 fair haired men and women out of 100 total men and women . 70 % e" | a = 28 / 40
b = a * 100
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a ) 88 , b ) 27 , c ) 36 , d ) 80 , e ) 25 | e | divide(add(150, 100), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | a jogger running at 9 km / hr along side a railway track is 150 m ahead of the engine of a 100 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 150 + 100 = 250 m . time taken = 250 / 10 = 25 sec . answer : e" | a = 150 + 100
b = 45 - 9
c = 10 / 2
d = 45 - 9
e = d / 2
f = c / e
g = b * f
h = a / g
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a ) 200 m , b ) 178 m , c ) 186 m , d ) 168 m , e ) 150 m | a | multiply(divide(multiply(60, const_1000), const_3600), 12) | a train running at the speed of 60 km / hr crosses a pole in 12 sec . what is the length of the train ? | "speed = 60 * 5 / 18 = 50 / 3 m / sec length of the train = speed * time = 50 / 3 * 12 = 200 m answer : a" | a = 60 * 1000
b = a / 3600
c = b * 12
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a ) 2515 , b ) 2525 , c ) 2535 , d ) 2545 , e ) 2555 | b | divide(5, 1) | if x + 1 / x = 5 then find out the value of x ^ 5 + 1 / x ^ 5 = ? | "given , ( x + 1 / x ) = 5 x ^ 2 + 1 / x ^ 2 = ( x + 1 / x ) ^ 2 - 2 * x * 1 / x = 5 ^ 2 - 2 = 23 - - - - - - - ( 1 ) x ^ 3 + 1 / x ^ 3 = ( x + 1 / x ) ^ 3 - 3 * x * 1 / x * ( x + 1 / x ) = 5 ^ 3 - 3 * 5 = 110 - - - - - - - ( 2 ) multiplying ( 1 ) & ( 2 ) , we get = > ( x ^ 2 + 1 / x ^ 2 ) * ( x ^ 3 + 1 / x ^ 3 ) = 23 * 110 = > x ^ 5 + 1 / x + x + 1 / x ^ 5 = 2530 = > x ^ 5 + 1 / x ^ 5 + ( x + 1 / x ) = 2530 = > x ^ 5 + 1 / x ^ 5 + 5 = 2530 = > ( x ^ 5 + 1 / x ^ 5 ) = 2525 answer : b" | a = 5 / 1
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a ) 120 , b ) 140 , c ) 160 , d ) 180 , e ) 200 | e | divide(40, subtract(subtract(const_1, inverse(30)), divide(const_1, const_2))) | a student traveled 30 percent of the distance of the trip alone , continued another 40 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.3 x + 40 miles + 0.5 x = x 40 miles = 0.2 x x = 200 miles the answer is e ." | a = 1/(30)
b = 1 - a
c = 1 / 2
d = b - c
e = 40 / d
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a ) 22 : 00 , b ) 20 : 00 , c ) 21 : 00 , d ) 18 : 00 , e ) 17 : 00 | b | multiply(divide(const_3, 4), divide(const_1, subtract(divide(const_1, 4), divide(const_1, 6)))) | pipe a fills a swimming pool in 4 hours . pipe b empties the pool in 6 hours . if pipe a was opened at 10 : 00 am and pipe b at 11 : 00 am , at what time will the pool be full ? | "pipe a fills the pool in 4 hrs . 1 hour ' s work : 1 / 4 pipe b empties the pool in 6 hrs . 1 hour ' s work : 1 / 6 together if they work , 1 hour ' s work = 1 / 4 - 1 / 6 = 1 / 12 given : pipe a started at 10 : 00 a . m and pipe b at 11 : 00 a . m pool filled after 1 hour by pipe a : 1 / 4 or 3 / 12 after 11 : 00 a . m pool filled after 1 hour with both the pipes on : 1 / 12 pool filled after 9 hours with both pipes on : 9 / 12 pool filled in 1 hour + pool filled in 9 hours = 3 / 12 + 9 / 12 = 1 therefore , it takes 10 hrs to fill the pool as pipe a started at 10 : 00 a . m , pool is full at 20 : 00 hrs answer : b" | a = 3 / 4
b = 1 / 4
c = 1 / 6
d = b - c
e = 1 / d
f = a * e
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a ) 1 / 5 , b ) 2 / 9 , c ) 3 / 19 , d ) 4 / 19 , e ) 7 / 9 | d | divide(subtract(20, 18), 18) | a number , x is chosen at random from the set of positive integers less than 20 . what is the probability that ( 18 / x ) > x ? | "number x has to be chosen from numbers 1 - 19 ( 18 / x ) > x = > 18 > x ^ 2 = > x ^ 2 - 18 < 0 x can have 2 values only 1 , 2 , 3 , 4 therefore , probability = 4 / 19 answer d" | a = 20 - 18
b = a / 18
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a ) 127 , b ) 160 , c ) 287 , d ) 237 , e ) 111 | b | multiply(520, divide(8, add(add(7, 8), 11))) | a , b , c hired a car for rs . 520 and used it for 7 , 8 and 11 hours respectively . hire charges paid by b were ? | a : b : c = 7 : 8 : 11 . hire charges paid by b = rs . ( 520 * 8 / 26 ) = rs . 160 . answer : b | a = 7 + 8
b = a + 11
c = 8 / b
d = 520 * c
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a ) 12 , b ) 14 , c ) 16 , d ) 15 , e ) 11 | a | subtract(multiply(log(divide(power(4, 4), const_2)), const_2), 4) | the population of locusts in a certain swarm doubles every two hours . if 4 hours ago there were 1,000 locusts in the swarm , in approximately how many hours will the swarm population exceed 256,000 locusts ? | "- 4 hours : 1,000 - 2 hours : 2,000 now : 4,000 + 2 hours : 8,000 + 4 hours : 16,000 + 6 hours : 32,000 + 8 hours : 64,000 + 10 hours : 128,000 + 12 hours : 256,000 answer : a" | a = 4 ** 4
b = a / 2
c = math.log(b)
d = c * 2
e = d - 4
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a ) 2 , b ) 4 , c ) 12 , d ) 7 , e ) 6 | e | add(2, 4) | there are 2 white plates , 6 green plates , 8 red plates , 4 pink plates , and 10 purple plates in the cabinet . what is the least number of plates that you have to pull out to make sure that you will have a matching pair ? ( assume that all the plates are identical except for the color ) | since there are 5 colors , if you take 5 plates , you could still have 1 plate per color and not have a match . therefore , upon taking the 6 th plate , you will definitely have a match since the 6 th plate will form at least a pair with the 1 st 5 so 5 + 1 = 6 answer is e | a = 2 + 4
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a ) 1008 , b ) 1015 , c ) 1022 , d ) 1032 , e ) none | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 7) | the smallest number which when diminished by 7 , is divisible by 12 , 16 , 18 , 21 and 28 is | "solution required numbers = ( l . c . m of 12 , 16,18 , 21,28 ) + 7 ‹ = › 1008 + 7 = 1015 . answer b" | a = 3 ** 2
b = 2 ** 0
c = a * b
d = 3 + 4
e = c * d
f = e + 7
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a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | b | multiply(divide(const_100, add(const_100, 15)), 45) | from january 1 , 1991 , to january 1 , 1993 , the number of people enrolled in health maintenance organizations increased by 15 percent . the enrollment on january 1 , 1993 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 1991 ? | "soln : - 15 x = 45 - - > 23 / 20 * x = 45 - - > x = 45 * 20 / 23 = 900 / 23 = ~ 39 . answer : b ." | a = 100 + 15
b = 100 / a
c = b * 45
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a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 7,750 , e ) $ 4,000 | d | divide(add(divide(subtract(360, multiply(divide(8, const_100), 1,000)), subtract(divide(8, const_100), divide(8, const_100))), divide(subtract(360, multiply(divide(8, const_100), 1,000)), subtract(divide(8, const_100), divide(8, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 360 plus 8 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of a ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ? | "sometime , setting up an equation is an easy way to go with : 350 + 0.04 ( x - 1000 ) = 0.08 x x = 7,750 ans : d" | a = 8 / 100
b = a * 1
c = 360 - b
d = 8 / 100
e = 8 / 100
f = d - e
g = c / f
h = 8 / 100
i = h * 1
j = 360 - i
k = 8 / 100
l = 8 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
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a ) 5 , b ) 12 , c ) 7 , d ) 9 , e ) 5 | b | divide(220, multiply(add(60, 4), const_0_2778)) | a train 220 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 4 = 64 km / hr . = 64 * 5 / 18 = 160 / 9 m / sec . time taken to pass the men = 220 * 9 / 160 = 12 sec . answer : b" | a = 60 + 4
b = a * const_0_2778
c = 220 / b
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a ) 175 , b ) 176 , c ) 177 , d ) 178 , e ) 179 | b | add(multiply(14, 12), 8) | what is the dividend . divisor 14 , the quotient is 12 and the remainder is 8 | b = d * q + r b = 14 * 12 + 8 b = 168 + 8 b = 176 | a = 14 * 12
b = a + 8
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a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | a | divide(add(multiply(3, 60), 170), add(7, 3)) | jhon works for 60 days . for the day he is present he is paid 7.00 / - and for the day he is absent he is paid 3.00 / - per day . if he is paid 170 / - . how many days he is present . | let no . days be = x and no . days he is absent = 60 - x , so 7 x - 3 ( 60 - x ) = 170 x = 35 answer : a | a = 3 * 60
b = a + 170
c = 7 + 3
d = b / c
|
a ) 24 kg , b ) 23 kg , c ) 28 kg , d ) 25 kg , e ) 27 kg | a | multiply(subtract(const_1, divide(30, const_100)), multiply(subtract(const_1, divide(40, const_100)), multiply(140, subtract(const_1, divide(60, const_100))))) | a statue is being carved by a sculptor . the original piece of marble weighed 140 kg . in the first week 60 percent is cut away . in the second week 40 percent of the remainder is cut away . in the third week the statue is completed when 30 percent of the remainder is cut away . what is the weight of the final statue ? | a 24 kg 140 ã — 0.4 ã — 0.6 ã — 0.7 = 24 kg . | a = 30 / 100
b = 1 - a
c = 40 / 100
d = 1 - c
e = 60 / 100
f = 1 - e
g = 140 * f
h = d * g
i = b * h
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | e | divide(add(sqrt(add(multiply(multiply(171, const_2), const_4), const_1)), const_1), const_2) | if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 171 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 171 n * ( n - 1 ) = 342 = 19 * 18 ( trial and error ) the answer is e ." | a = 171 * 2
b = a * 4
c = b + 1
d = math.sqrt(c)
e = d + 1
f = e / 2
|
['a ) 16 sqrt 2', 'b ) 8 sqrt 2', 'c ) 32 sqrt 2', 'd ) sqrt 72', 'e ) sqrt 82'] | d | sqrt(divide(multiply(multiply(circle_area(1), 12), 6), const_pi)) | a circular metal plate of even thickness has 12 holes of radius 1 cm drilled into it . as a result the plate lost 1 / 6 th its original weight . the radius of the circular plate is | area of 12 holes = 12 * pi * 1 ^ 2 = 12 * pi as 12 * pi = ( 1 / 6 ) th weight , so total area of plate = 12 * pi / ( 1 / 6 ) = 72 * pi if r = radius of the plate , then its area = pi * r ^ 2 = 72 * pi , so r = sqrt 72 answer : d | a = circle_area * (
b = a * 12
c = b / 6
d = math.sqrt(c)
|
['a ) 1 / 4', 'b ) 3 / 8', 'c ) 1 / 2', 'd ) 3 / 5', 'e ) 2'] | d | divide(multiply(const_2, const_3), multiply(add(multiply(multiply(const_1, const_2), const_2), multiply(const_1, const_1)), const_2)) | what is the ratio g of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled ? | one side surface area of a cube = x * x = x ^ 2 total 6 sides = 6 x ^ 2 as for the rectangular , height ( h ) and width ( w ) are same as cube , x . only length = 2 x . l x h = 2 x * x = 2 x ^ 2 - - - - > 4 sides = 2 x ^ 2 * 4 = 8 x ^ 2 w * h = x * x = x ^ 2 - - - - - - > 2 sides = x ^ 2 * 2 = 2 x ^ 2 total 6 sides = 8 x ^ 2 + 2 x ^ 2 = 10 x ^ 2 ratio of cube area to rectangular area g = 6 x ^ 2 / 10 x ^ 2 - - - - > 6 / 10 - - - - > 3 / 5 ( d ) | a = 2 * 3
b = 1 * 2
c = b * 2
d = 1 * 1
e = c + d
f = e * 2
g = a / f
|
a ) 76.5 , b ) 58 , c ) 39.8 , d ) 90.3 , e ) 66.67 | a | divide(add(add(add(add(subtract(multiply(62, add(const_4, const_1)), add(add(add(28, 50), 78), 104)), 48), 62), 98), 124), add(const_4, const_1)) | if the mean of numbers 28 , x , 50 , 78 and 104 is 62 , what is the mean of 48 , 62 , 98 , 124 and x ? | "mean = ( sum of all no . in series ) / ( no . in series ) m 1 = 62 = ( x + 28 + 50 + 78 + 104 ) / ( 5 ) 62 x 5 = 260 + x x = 50 therefore mean of 2 nd series m 2 = ( x + 48 + 62 + 98 + 124 ) / 5 m 2 = 76.5 answer = a" | a = 4 + 1
b = 62 * a
c = 28 + 50
d = c + 78
e = d + 104
f = b - e
g = f + 48
h = g + 62
i = h + 98
j = i + 124
k = 4 + 1
l = j / k
|
a ) 16.12 % , b ) 16.66 % , c ) 16.56 % , d ) 17.66 % , e ) 13.33 % | e | multiply(divide(subtract(68, 60), 60), const_100) | john makes $ 60 a week from his job . he earns a raise andnow makes $ 68 a week . what is the % increase ? | "increase = ( 8 / 60 ) * 100 = ( 2 / 15 ) * 100 = 13.33 % . e" | a = 68 - 60
b = a / 60
c = b * 100
|
a ) rs . 9718.27 , b ) rs . 9750 , c ) rs . 10123.20 , d ) rs . 10483.20 , e ) none | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on a sum of rs . 24,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 24000 * ( 1 + 12 / 100 ) 3 ] = 24000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 33718.27 c . i . = ( 33718.27 - 24000 ) = rs . 9718.27 answer : a" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
|
a ) 14 kmph , b ) 12 kmph , c ) 18 kmph , d ) 1 kmph , e ) none of these | c | divide(multiply(36, const_2), add(const_3, const_1)) | a man can row 36 kmph in still water . it takes him thrice as long to row up as to row down the river . find the rate of the stream ? | explanation : let man ' s rate upsteam be x kmph then his rate of downstream = 3 x kmph rate still water = 1 / 2 ( 3 x + x ) = 2 x 2 x = 36 x = 18 rate of upstream = 18 rate of downstream = 54 rate of stream 1 / 2 ( 54 - 18 ) = 18 kmph answer : option c | a = 36 * 2
b = 3 + 1
c = a / b
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | a | add(reminder(multiply(reminder(51, const_4), 93), const_10), reminder(35, const_10)) | the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 51 ) is : | "the units digit of powers of 3 , cycles in a group of 4 : { 3 , 9 , 7 , 1 } 51 has the form 4 k + 3 , so the units digit of 93 ^ 51 is 7 . the units digit of powers of 5 is always 5 . 7 + 5 = 12 , so the units digit is 2 . the answer is a ." | a = reminder * (
b = reminder + (
|
a ) 9 , b ) 18 , c ) 17 , d ) 13 , e ) can not be determined | d | add(divide(subtract(75, 5), 5), const_1) | how many multiples of 5 are there between 5 to 75 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 . answer : d" | a = 75 - 5
b = a / 5
c = b + 1
|
a ) 191 , b ) 192 , c ) 193 , d ) 212 , e ) 213 | b | divide(765, 4) | to be considered for “ movie of the year , ” a film must appear in at least 1 / 4 of the top - 10 - movies lists submitted by the cinematic academy ’ s 765 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for “ movie of the year ” ? | "total movies submitted are 765 . as per question we need to take 1 / 4 of 765 to be considered for top 10 movies = 191.25 approximate the value we 192 . answer : option b is the correct answer . ." | a = 765 / 4
|
a ) 28 , b ) 30 , c ) 32 , d ) 35 , e ) 37 | b | divide(multiply(multiply(24, 40), 16), volume_cube(divide(16, const_2))) | a box measuring 24 inches long by 40 inches wide by 16 inches deep is to be filled entirely with identical cubes . no space is to be left unfilled . what is the smallest number of cubes that can accomplish this objective ? | "least number of cubes will be required when the cubes that could fit in are biggest . 8 is the biggest number that could divide all three , 24 , 40 and 16 . thus side of cube must be 8 , and total number of cubes = 24 / 8 * 40 / 8 * 16 / 8 = 30 ans b" | a = 24 * 40
b = a * 16
c = 16 / 2
d = b / volume_cube
|
a ) 180 , b ) 220 , c ) 260 , d ) 300 , e ) 340 | e | subtract(1000, multiply(add(add(multiply(add(1, 11), divide(10, const_2)), divide(10, const_2)), const_1), const_10)) | there are 11 baskets numbered from 1 to 11 and filled with apples . 10 children are asked to pick apples one by one from each basket such that the number of apples picked by each child from each basket is equal to the number marked on the basket . if there were 1000 apples in total and the baskets were filled in such a way that none of the basket became empty in the process , how many apples were left in the end ? | each child takes a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 11 * 12 / 2 = 66 apples . the total number of apples taken by 10 children is 10 * 66 = 660 the number of apples left in the end is 1000 - 660 = 340 apples . the answer is e . | a = 1 + 11
b = 10 / 2
c = a * b
d = 10 / 2
e = c + d
f = e + 1
g = f * 10
h = 1000 - g
|
a ) 26 : 11 , b ) 23 : 11 , c ) 21 : 11 , d ) 24 : 11 , e ) 25 : 11 | c | power(3, 3) | two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 3 : 1 and 6 : 4 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | "vessel a = 300 gallons - - > milk = 225 , water = 75 ; vessel b = 500 gallons - - > milk = 300 , water = 200 ; vessel a + b = 800 gallons - - > milk = 525 , water 275 the ratio = 525 / 275 - - > 21 : 11 answer : c" | a = 3 ** 3
|
a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 8 | e | divide(subtract(508, multiply(22, divide(subtract(multiply(508, const_2), 656), subtract(multiply(22, const_2), 8)))), 16) | suzie ’ s discount footwear sells all pairs of shoes for one price and all pairs of boots for another price . on monday the store sold 22 pairs of shoes and 16 pairs of boots for $ 508 . on tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $ 656 . how much more do pairs of boots cost than pairs of shoes at suzie ’ s discount footwear ? | "let x be pair of shoes and y be pair of boots . 22 x + 16 y = 508 . . . eq 1 8 x + 32 y = 656 . . . . eq 2 . now multiply eq 1 by 2 and sub eq 2 . 44 x = 1016 8 x = 656 . 36 x = 360 = > x = 10 . sub x in eq 2 . . . . we get 80 + 32 y = 656 . . . then we get 32 y = 576 then y = 18 differenece between x and y is 8 answer : e" | a = 508 * 2
b = a - 656
c = 22 * 2
d = c - 8
e = b / d
f = 22 * e
g = 508 - f
h = g / 16
|
a ) 6 , b ) 56 , c ) 15 , d ) 40 , e ) 22 | b | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(35, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(35, const_10))), const_10)) | a rectangular block 10 cm by 20 cm by 35 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 10 * 20 * 35 = 7000 cm ^ 3 side of the largest cube = h . c . f of 10 , 20,35 = 5 cm volume of the cube = 5 * 5 * 5 = 125 cm ^ 3 number of cubes = 7000 / 125 = 56 answer is b" | a = rectangle_area / (
b = a - 10
c = 3 * 2
d = 1000 * c
e = b + d
f = 3 * 1000
g = 35 * 10
h = f + g
i = e / h
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | e | add(floor(divide(subtract(subtract(40, 5), subtract(20, 5)), const_2)), 1) | the scoring system in a certain football competition goes as follows : 3 points for victory , 1 point for a draw , and 0 points for defeat . each team plays 20 matches . if a team scored 5 points after 5 games , what is the least number of the remaining matches it has to win to reach the 40 - point mark by the end of the tournament ? | "to get 40 points as end of season we need another 35 points or more from remaining 15 matches : option a = 6 * 3 + 9 * 1 = 27 option b = 7 * 3 + 8 * 1 = 29 option c = 8 * 3 + 7 * 1 = 31 option d = 9 * 3 + 6 * 1 = 33 option e = 10 * 3 + 5 * 1 = 35 hence option e - 10" | a = 40 - 5
b = 20 - 5
c = a - b
d = c / 2
e = math.floor(d)
f = e + 1
|
a ) 90 , b ) 122.5 , c ) 132.5 , d ) 114.5 , e ) 212.5 | a | subtract(divide(multiply(multiply(2000, 11.5), 3), const_100), divide(multiply(multiply(2000, 10), 3), const_100)) | if a lends rs . 2000 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 2000 * 1.5 * 3 ) / 100 = > 90 answer : a" | a = 2000 * 11
b = a * 3
c = b / 100
d = 2000 * 10
e = d * 3
f = e / 100
g = c - f
|
['a ) 260', 'b ) 262', 'c ) 270', 'd ) 272', 'e ) none of these'] | c | multiply(quadrilateral_area(12, 1, 4), 9) | a swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side . its volume is : | explanation : volume will be length * breadth * height , but in this case two heights are given so we will take average , volume = ( 12 ∗ 9 ∗ ( 1 + 4 / 2 ) ) m 312 ∗ 9 ∗ 2.5 m 3 = 270 m 3 option c | a = quadrilateral_area * (
|
a ) 24 kmph , b ) 25 kmph , c ) 26 kmph , d ) 30 kmph , e ) 28 kmph | a | divide(divide(multiply(72, const_2), const_3), const_2) | the time taken by mr . dhoni to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 72 kmph , find the speed of the stream ? | the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 72 / 3 = 24 kmph . answer : a | a = 72 * 2
b = a / 3
c = b / 2
|
a ) 7 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | subtract(subtract(divide(divide(multiply(24, 8), const_10), const_2), const_0_25), const_0_25) | a math teacher has 24 cards , each of which is in the shape of a geometric figure . half of the cards are rectangles , and a third of the cards are rhombuses . if 8 cards are squares , what is the maximum possible number of cards that re circles . | "a square is a special kind of rhombus ( sides are perpendicular ) a square is a special kind of rectangles ( sides with same length ) among the 24 cards with have : 15 rectangles 10 rhombus 8 squares among the 15 rectangles , there could be 8 special ones ( with sides of same length ) that are squares . that lets at least 7 rectangles that are not square . among the 10 rectangles , there could be 8 special ones ( with sides perpendicular ) that are squares . that lets at least 2 rhombus that are not square . we have 8 squares . so the minimum different cards that represent a square , a rhombus or a rectangle is 2 + 7 + 8 = 17 which means that the maximum number of circles that you could have is 24 - 17 = 7 answer ( a )" | a = 24 * 8
b = a / 10
c = b / 2
d = c - const_0_25
e = d - const_0_25
|
a ) 31.25 , b ) 37.5 , c ) 30.0 , d ) 32.75 , e ) 32.25 | e | multiply(subtract(multiply(add(divide(15, const_100), const_1), add(divide(15, const_100), const_1)), const_1), const_100) | increasing the original price of a certain item by 15 percent and then increasing the new price by 15 percent is equivalent to increasing the original price by what percent ? | "we ' re told that the original price of an item is increased by 15 % and then that price is increased by 15 % . . . . if . . . . starting value = $ 100 + 15 % = 100 + . 15 ( 100 ) = 115 + 15 % = 115 + . 15 ( 115 ) = 115 + 17.25 = 132.25 the question asks how the final price relates to the original price . this is essentially about percentage change , which means we should use the percentage change formula : percentage change = ( new - old ) / old = difference / original doing either calculation will yield the same result : 32.25 / 100 = 32.25 % final answer : e" | a = 15 / 100
b = a + 1
c = 15 / 100
d = c + 1
e = b * d
f = e - 1
g = f * 100
|
a ) 1 / 3 , b ) ¼ , c ) 9 / 25 , d ) 5 / 16 , e ) 0 | d | divide(add(3, const_2), multiply(const_4, const_4)) | if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability t that n is divisible by 3 ? | "prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 / 16 . ans d . took me 3 : 20 mins ." | a = 3 + 2
b = 4 * 4
c = a / b
|
a ) 3000 , b ) 4000 , c ) 5000 , d ) 5500 , e ) 6000 | b | divide(subtract(multiply(divide(15, 125), 12000), 1360), subtract(divide(15, 125), divide(12, 120))) | raj invests a part of rs . 12000 in 12 % stock at rs . 120 and the remainder in 15 % stock at rs . 125 . if his total dividend per annum is rs . 1360 , how much does he invest in 12 % stock at rs . 120 ? | investment in 12 % stock be x then investment 15 % stock = ( 12000 - x ) 12 / 120 * x + 15 / 125 * ( 12000 - x ) = 1360 x = 4000 answer b | a = 15 / 125
b = a * 12000
c = b - 1360
d = 15 / 125
e = 12 / 120
f = d - e
g = c / f
|
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.24, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.24, subtract(const_0_25, divide(const_1, const_100))), 14)) | if 42.24 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ? | 42.24 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.24 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.24 = km / 50 . . . . . . 24 / 100 = 3 m / 50 . . . . . . m = 4 k + m = 3 + 4 = 7 answer : b | a = 1 / 100
b = const_0_25 - a
c = b * 50
d = 1 / 100
e = const_0_25 - d
f = 42 - 24
g = f / 14
h = c / g
i = 1 / 100
j = const_0_25 - i
k = 42 - 24
l = k / 14
m = h + l
|
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