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a ) 18 , b ) 21 , c ) 22 , d ) 20 , e ) 30 | a | add(10, multiply(const_4, const_2)) | a owner of a mart earns an income of re 1 on the first day of his business . on every subsequent day , he earns an income which is just double of that made on the previous day . on the 10 th day of business , he earns an income of : | 2 nd day he earns = 2 ( 2 β 1 ) 3 rd day he earns = 2 ( 3 β 1 ) on 10 th day he earns 2 ( 10 - 1 ) = 18 rupees answer : a | a = 4 * 2
b = 10 + a
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(multiply(multiply(divide(243, divide(405, add(const_4, const_1))), 16), divide(125, add(const_4, const_1))), multiply(const_100, const_3)) | if p = 125 Γ 243 Γ 16 / 405 , how many digits are in p ? | p = 125 * 243 * 16 / 405 p = 5 ^ 3 * 3 * 9 ^ 2 * 4 ^ 2 / ( 5 * 9 ^ 2 ) p = 5 ^ 2 * 3 * 4 ^ 2 p = 20 ^ 2 * 3 = 1200 answer d | a = 4 + 1
b = 405 / a
c = 243 / b
d = c * 16
e = 4 + 1
f = 125 / e
g = d * f
h = 100 * 3
i = g / h
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 25 | e | add(divide(18, const_2), subtract(34, 18)) | jane started baby - sitting when she was 18 years old . whenever she baby - sat for a child , that child was no more than half her age at the time . jane is currently 34 years old , and she stopped baby - sitting 10 years ago . what is the current age of the oldest person for whom jane could have baby - sat ? | "check two extreme cases : jane = 18 , child = 9 , years ago = 34 - 18 = 16 - - > child ' s age now = 9 + 16 = 25 ; jane = 24 , child = 12 , years ago = 34 - 24 = 10 - - > child ' s age now = 12 + 10 = 22 . answer : e ." | a = 18 / 2
b = 34 - 18
c = a + b
|
a ) 3.0 , b ) 3.36 , c ) 23.93 , d ) 25.0 , e ) 31.36 | c | divide(multiply(28, const_100), add(const_100, 17)) | from the sale of sleeping bags , a retailer made a gross profit of 17 % of the wholesale cost . if each sleeping bag was sold for $ 28 , what was the wholesale cost per bag ? | "cost price * 1.17 = selling price - - > cost price * 1.17 = $ 28 - - > cost price = $ 23.93 . answer : c ." | a = 28 * 100
b = 100 + 17
c = a / b
|
a ) 10 , b ) 99 , c ) 27 , d ) 22 , e ) 5.6 | e | multiply(divide(multiply(4, 5), subtract(multiply(6.5, 4), multiply(3, 4))), 4) | mixture contains alcohol and water in the ratio 4 : 3 . if 6.5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 26 x = 4 ( 3 x + 5 ) 14 x = 20 x = 1.4 quantity of alcohol = ( 4 x 1.4 ) litres = 5.6 litres . answer : e | a = 4 * 5
b = 6 * 5
c = 3 * 4
d = b - c
e = a / d
f = e * 4
|
a ) 7 m , b ) 7.5 m , c ) 8 m , d ) 8.5 m , e ) none of these | b | divide(multiply(divide(810, 4.5), divide(75, const_100)), 18) | the cost of carpeting a room 18 m long with a carpet 75 cm wide at 4.50 per metre is 810 . the breadth of the room is : | length of the carpet = totalcost / rate / m = ( 8100 β 45 ) m = 180 m . area of the room = area of the carpet = ( 180 Γ 75 β 100 ) m 2 = 135 m 2 β΄ breadth of the room = arealength = 135 / 18 m = 7.5 m answer b | a = 810 / 4
b = 75 / 100
c = a * b
d = c / 18
|
a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | e | divide(multiply(1350, const_3), add(add(5, 2), 2)) | a farmer with 1350 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 : 2 : 5 , 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 ( 5 / 9 ) * 1350 = 750 acres were planted with tobacco . thus 750 - 300 = 450 more acres were planted with tobacco . the answer is e . | a = 1350 * 3
b = 5 + 2
c = b + 2
d = a / c
|
a ) s . 1014 , b ) s . 1140 , c ) s . 1998 , d ) s . 1085 , e ) s . 1020 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 3.00), 3) | the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 3.00 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 β 6 = 666 * 3.0 = 1998 answer : c" | a = math.sqrt(3136)
b = a * 4
c = 2 * 1
d = b - c
e = d * 3
f = e * 3
|
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | multiply(2, divide(negate(multiply(add(add(5, 4), 2), 9)), subtract(multiply(add(add(5, 4), 2), const_2), multiply(add(add(5, 4), 2), 4)))) | the ratio by weight , measured in pounds , of books to clothes to electronics in a suitcase initially stands at 5 : 4 : 2 . someone removes 9 pounds of clothing from the suitcase , thereby doubling the ratio of books to clothes . how many pounds do the electronics in the suitcase weigh ? | "the weights of the items in the suitcase are 5 k , 4 k , and 2 k . if removing 9 pounds of clothes doubles the ratio of books to clothes , then 9 pounds represents half the weight of the clothes . 2 k = 9 pounds and then k = 4.5 pounds . the electronics weigh 2 ( 4.5 ) = 9 pounds . the answer is c ." | a = 5 + 4
b = a + 2
c = b * 9
d = negate / (
e = 5 + 4
f = e + 2
g = f * 2
h = 5 + 4
i = h + 2
j = i * 4
k = g - j
l = 2 * d
|
a ) 800 m , b ) 838 m , c ) 834 m , d ) 831 m , e ) 836 m | a | subtract(224, multiply(7, speed(224, 32))) | for a race a distance of 224 meters can be covered by p in 7 seconds and q in 32 seconds . by what distance does p defeat q eventually ? | "explanation : this is a simple speed time problem . given conditions : = > speed of p = 224 / 7 = 32 m / s = > speed of q = 224 / 32 = 7 m / s = > difference in time taken = 25 seconds therefore , distance covered by p in that time = 32 m / s x 25 seconds = 800 metres answer : a" | a = 7 * speed
b = 224 - a
|
a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | d | subtract(add(70, 55), subtract(const_100, 35)) | in a particular state , 70 % of the counties received some rain on monday , and 55 % of the counties received some rain on tuesday . no rain fell either day in 35 % of the counties in the state . what percent of the counties received some rain on monday and tuesday ? | 70 + 55 + 35 = 160 % the number is 60 % above 100 % because 60 % of the counties were counted twice . the answer is d . | a = 70 + 55
b = 100 - 35
c = a - b
|
a ) rs . 19 , b ) rs . 22 , c ) rs . 20 , d ) rs . 21 , e ) none of these | d | add(20, divide(power(20, const_2), 400)) | the present worth of a certain bill due sometime hence is rs . 400 and the true discount is rs . 20 . what is the banker ' s discount ? | "explanation : bg = ( td ) 2 / pw = 202 / 400 = rs . 1 bg = bd β td = > 1 = bd - 20 = > bd = 1 + 20 = rs . 21 answer : option d" | a = 20 ** 2
b = a / 400
c = 20 + b
|
a ) rs . 432 , b ) rs . 422 , c ) rs . 416 , d ) rs . 442 , e ) none of these | c | multiply(divide(360, subtract(2660, 360)), 2660) | the true discount on a bill of rs . 2660 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2660 td = rs . 360 pw = f - td = 2660 - 360 = rs . 2300 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2300 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2160 for unexpired time = 360 / 2300 Γ 2660 = 0.16 Γ 2660 = rs . 416 answer : option c" | a = 2660 - 360
b = 360 / a
c = b * 2660
|
a ) 206 , b ) 216 , c ) 226 , d ) 256 , e ) 246 | b | multiply(multiply(divide(multiply(add(48, multiply(const_3, const_2)), divide(15, const_60)), multiply(const_3, const_2)), 48), const_2) | a train covered half of the distance between stations a and b at the speed of 48 km / hr , but then it had to stop for 15 min . to make up for the delay , it increased its speed by 53 m / sec and it arrived to station b on time . find the distance between the two stations and the speed of the train after the stop . | first let us determine the speed of the train after the stop . the speed was increased by 53 m / sec = 5 β
60 β
6031000 km / hr = 6 km / hr . therefore , the new speed is 48 + 6 = 54 km / hr . if it took x hours to cover the first half of the distance , then it took x β 1560 = x β 0.25 hr to cover the second part . so the equation is : 48 β
x = 54 β
( x β 0.25 ) 48 β
x = 54 β
x β 54 β
0.25 48 β
x β 54 β
x = β 13.5 β 6 x = β 13.5 x = 2.25 h . the whole distance is 2 Γ 48 Γ 2.25 = 216 km . answer is b . | a = 3 * 2
b = 48 + a
c = 15 / const_60
d = b * c
e = 3 * 2
f = d / e
g = f * 48
h = g * 2
|
a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 7,000 , d ) d ) 11,000 , e ) e ) 11,100 | c | multiply(multiply(const_4, const_2), const_100) | a certain city with a population of 84,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | "let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 84 k x = 7,000 answer : c" | a = 4 * 2
b = a * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | if the number 892 , 132,27 x is divisible by 11 , what must be the value of x ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 132,27 x sum of digits at odd places = 8 + 2 + 3 + 2 + x = 15 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 7 = 19 ( ii ) ( i ) - ( ii ) = 15 + x - 19 = x - 4 hence x should be = 4 to make this a multiple of 11 ( 0 ) option d" | a = 3 * 10
b = a + 1
c = b * 2
d = c * 4
e = d * 11
|
a ) 68.8 , b ) 73.6 , c ) 75.2 , d ) 76.8 , e ) 77.76 | e | multiply(add(add(7.5, 8.0), 8.8), 3.2) | in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.5 , 8.0 , 9.0 , 6.0 , and 8.8 , what was the point value of the dive ? | "degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.0 , 8.8 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 8 + 8.8 ) = 24.3 point of value of the dive = 24 * 3.2 = 77.76 answer e" | a = 7 + 5
b = a + 8
c = b * 3
|
['a ) 11 β 2 Ο', 'b ) 10 β 2 Ο', 'c ) 9 β 2 Ο', 'd ) 8 β 2 Ο', 'e ) 7 β 2 Ο'] | b | multiply(multiply(const_2, const_pi), divide(sqrt(add(power(divide(40, const_4), const_2), power(divide(40, const_4), const_2))), const_2)) | square p is inscribed in circle q . if the perimeter of p is 40 , what is the circumference of q ? | square forms two right angled triangles . any time we have a right angle triangle inside a circle , the hypotenuse is the diameter . hypotenuse here = diagonal of the square = 10 sqrt ( 2 ) = diameter = > radius = 5 sqrt ( 2 ) circumference of the circle = 2 pi r = 10 pi sqrt ( 2 ) answer is b . | a = 2 * math.pi
b = 40 / 4
c = b ** 2
d = 40 / 4
e = d ** 2
f = c + e
g = math.sqrt(f)
h = g / 2
i = a * h
|
a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 2,470 | e | divide(divide(multiply(multiply(19, add(19, const_1)), add(multiply(19, 2), const_1)), 6), divide(multiply(multiply(19, add(19, const_1)), add(multiply(19, 2), const_1)), 6)) | the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n ^ 3 / 3 + c * n ^ 2 + n / 6 , where c is a constant . what is the sum of the first 19 positive perfect squares ? | "first we need to find the constant ' c ' . the easiest way to find this is for the sum of the first two perfect squares for 1 and 2 = 1 and 4 respectively . hence lhs = 1 + 4 and plug n = 2 for rhs and simplify to get c = 1 / 2 . plug values of n = 19 and c = 1 / 2 into the equation and simplify to get the answer 2470 . option e ." | a = 19 + 1
b = 19 * a
c = 19 * 2
d = c + 1
e = b * d
f = e / 6
g = 19 + 1
h = 19 * g
i = 19 * 2
j = i + 1
k = h * j
l = k / 6
m = f / l
|
a ) 20 , b ) 15 , c ) 25 , d ) 18 , e ) 19 | c | 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 faulty meter that weighs 800 grams . find the profit percent . | "explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 800 100 + g / 100 + 0 = 1000 / 800 100 + g = 5 / 4 * 100 g = 25 answer : c" | a = 4 + 1
b = 4 + 1
c = a + b
d = c * 100
e = d - 800
f = e / 800
g = f * 100
|
a ) 2 , b ) 1 , c ) 6 , d ) 8 , e ) 10 | b | subtract(multiply(multiply(multiply(293, 567), 917), 343), subtract(multiply(multiply(multiply(293, 567), 917), 343), add(const_4, const_4))) | the unit digit in the product ( 293 * 567 * 917 * 343 ) is : | "explanation : unit digit in the given product = unit digit in ( 3 * 7 * 7 * 4 ) = 1 answer : b" | a = 293 * 567
b = a * 917
c = b * 343
d = 293 * 567
e = d * 917
f = e * 343
g = 4 + 4
h = f - g
i = c - h
|
a ) 150 % , b ) 134 1 / 3 % , c ) 135 1 / 3 % , d ) 140 1 / 3 % , e ) 143 1 / 3 % | a | multiply(divide(120, 80), const_100) | what percent is 120 of 80 ? | "120 / 80 = 3 / 2 3 / 2 Γ 100 = 300 / 2 = 150 % a" | a = 120 / 80
b = a * 100
|
a ) 4 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | divide(subtract(100, multiply(20, const_3)), const_4) | a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 20 more blue balls than white balls , and thrice as many red balls as blue balls , how many white balls does the box has ? | "x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 20 more blue balls than white balls . . . equation # 2 : y = 20 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 3 y solve equation # 2 for z : z = y - 20 now , we can replace both x and z with y in equation # 1 3 y + y + ( y - 20 ) = 100 5 y - 20 = 100 5 y = 120 y = 24 there are 24 blue balls . this is 20 more than the number of white balls , so z = 4 . that ' s the answer . just as a check , x = 72 , and 72 + 24 + 4 = 100 . answer = 4 , ( a )" | a = 20 * 3
b = 100 - a
c = b / 4
|
a ) 228 , b ) 287 , c ) 277 , d ) 188 , e ) 400 | e | subtract(divide(subtract(multiply(const_3, 1600), multiply(const_2, 1200)), subtract(multiply(const_3, 3), multiply(const_2, 2))), divide(subtract(1200, multiply(2, divide(subtract(multiply(const_3, 1600), multiply(const_2, 1200)), subtract(multiply(const_3, 3), multiply(const_2, 2))))), 3)) | the cost of 2 chairs and 3 tables is rs . 1600 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | explanation : 2 c + 3 t = 1600 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 400 = > t - c = 400 answer : e | a = 3 * 1600
b = 2 * 1200
c = a - b
d = 3 * 3
e = 2 * 2
f = d - e
g = c / f
h = 3 * 1600
i = 2 * 1200
j = h - i
k = 3 * 3
l = 2 * 2
m = k - l
n = j / m
o = 2 * n
p = 1200 - o
q = p / 3
r = g - q
|
a ) 9 % , b ) 10 % , c ) 11 % , d ) 35 % , e ) 15 % | d | multiply(subtract(divide(18, const_100), divide(subtract(6, multiply(divide(18, const_100), multiply(18, 2))), subtract(multiply(15, 3), multiply(18, 2)))), const_100) | fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 6 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ? | "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 6 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 6 or 45 * 18 - 45 * x + 36 * x = 6 * 100 or 9 x = - 6 * 100 + 45 * 18 x = 210 / 6 = 35 % d" | a = 18 / 100
b = 18 / 100
c = 18 * 2
d = b * c
e = 6 - d
f = 15 * 3
g = 18 * 2
h = f - g
i = e / h
j = a - i
k = j * 100
|
a ) 125 , b ) 150 , c ) 212 , d ) 250 , e ) 500 | c | sqrt(divide(multiply(90, const_100), divide(20, const_100))) | 90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 20 % of the total school population of x students , what is x ? | 90 = x / 100 * 20 / 100 * x = > x ^ 2 = 9 * 10000 / 2 = > x = 212 c | a = 90 * 100
b = 20 / 100
c = a / b
d = math.sqrt(c)
|
a ) 237 , b ) 126 , c ) 971 , d ) 611 , e ) 150 | e | divide(divide(225, divide(add(const_100, 25), const_100)), divide(add(20, const_100), const_100)) | a sells a bicycle to b and makes a profit of 20 % . b sells the same bicycle to c at a profit of 25 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a . | "explanation : answer : e" | a = 100 + 25
b = a / 100
c = 225 / b
d = 20 + 100
e = d / 100
f = c / e
|
a ) 32 , b ) 87 , c ) 30 , d ) 99 , e ) 77 | a | add(floor(divide(multiply(multiply(21, 8), multiply(16, 3)), multiply(multiply(21, 2), 6))), const_1) | 16 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | "3 w = 2 m 16 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 16 * 21 * 8 = 14 * x * 6 x = 32 answer : a" | a = 21 * 8
b = 16 * 3
c = a * b
d = 21 * 2
e = d * 6
f = c / e
g = math.floor(f)
h = g + 1
|
a ) 3.75 days , b ) 3.7 days , c ) 3.6 days , d ) 2.4 days , e ) 5.75 days | d | inverse(add(inverse(4), inverse(6))) | a and b complete a work in 4 days . a alone can do it in 6 days . if both together can do the work in how many days ? | "1 / 4 + 1 / 6 = 5 / 12 12 / 5 = 2.4 days answer : d" | a = 1/(4)
b = 1/(6)
c = a + b
d = 1/(c)
|
a ) 122 , b ) 210 , c ) 216 , d ) 217 , e ) 220 | c | add(factorial(5), multiply(factorial(subtract(5, const_1)), subtract(5, const_1))) | a 5 digit number divisible by 3 is to be formed using the digits 0 , 1 , 2 , 3 , 4 and 5 without repetitions . that total no of ways it can be done is ? | first step : we should determine which 5 digits from given 6 , would form the 5 digit number divisible by 3 . we have six digits : 0,1 , 2,3 , 4,5 . their sum = 15 . for a number to be divisible by 3 the sum of the digits must be divisible by 3 . as the sum of the six given numbers is 15 ( divisible by 3 ) only 5 digits good to form our 5 digit number would be 15 - 0 = ( 1 , 2,3 , 4,5 ) and 15 - 3 = ( 0 , 1,2 , 4,5 ) . meaning that no other 5 from given six will total the number divisible by 3 . second step : we have two set of numbers : 1 , 2,3 , 4,5 and 0 , 1,2 , 4,5 . how many 5 digit numbers can be formed using this two sets : 1 , 2,3 , 4,5 - - > 5 ! as any combination of these digits would give us 5 digit number divisible by 3 . 5 ! = 120 . 0 , 1,2 , 4,5 - - > here we can not use 0 as the first digit , otherwise number wo n ' t be any more 5 digit and become 4 digit . so , total combinations 5 ! , minus combinations with 0 as the first digit ( combination of 4 ) 4 ! - - > 5 ! - 4 ! = 96 120 + 96 = 216 answer : c . | a = math.factorial(5)
b = 5 - 1
c = math.factorial(b)
d = 5 - 1
e = c * d
f = a + e
|
a ) 30 % , b ) 33 1 / 3 % , c ) 37 1 / 2 % , d ) 40 % , e ) 50 % | c | multiply(const_100, divide(add(multiply(8, divide(30, const_100)), multiply(2, divide(30, const_100))), add(add(multiply(8, divide(30, const_100)), multiply(2, divide(30, const_100))), add(subtract(multiply(8, divide(70, const_100)), 2), multiply(2, divide(70, const_100)))))) | solution y is 30 percent liquid x and 70 percent water . if 2 kilograms of water evaporate from 8 kilograms of solutions y and 2 kilograms of solution y are added to the remaining 6 kilograms of liquid , what percent of this new liquid solution is liquid x ? | "at the beginning , you have 8 kg of solution y , i . e . 70 % * 8 kg = 5,6 kg of water 30 % * 8 kg = 2,4 kg of x 2 kg of water evaporate : 5,6 kg - 2 kg = 3,6 kg of water 2,4 kg of x 2 kg of liquid are added : 3,6 kg + 2 * 0,70 = 5 kg of water 2,4 kg + 2 * 0,30 = 3 kg of x so you have 3 kg of x in 8 kg ( 3 + 5 ) solution . therefore x concentration is 3 / 8 = 37,5 % answer : c" | a = 30 / 100
b = 8 * a
c = 30 / 100
d = 2 * c
e = b + d
f = 30 / 100
g = 8 * f
h = 30 / 100
i = 2 * h
j = g + i
k = 70 / 100
l = 8 * k
m = l - 2
n = 70 / 100
o = 2 * n
p = m + o
q = j + p
r = e / q
s = 100 * r
|
a ) 8 , b ) 9 , c ) 16 , d ) 17 , e ) 18 | b | divide(add(add(add(add(1, const_4), add(1, const_4)), add(const_4, const_4)), 50), 5) | the sum of ages of 5 children born 1 year different each is 50 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) and ( x + 4 ) years . then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) = 50 5 x = 40 x = 8 x + 4 = 8 + 1 = 9 answer : b" | a = 1 + 4
b = 1 + 4
c = a + b
d = 4 + 4
e = c + d
f = e + 50
g = f / 5
|
a ) $ 146.6 , b ) $ 120 , c ) $ 180 , d ) $ 220 , e ) $ 260 | a | divide(multiply(22, const_100), subtract(multiply(add(12, 1), const_10), add(const_100, 15))) | right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al β s account were to increase by 10 % and eliot β s account were to increase by 15 % , then al would have exactly $ 22 more than eliot in his account . how much money does eliot have in his account right now ? | lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a = 22 + 1.15 e = = > 11 a = 220 + 11.5 e - - - - - - - ( 2 ) substituting ( 1 ) in ( 2 ) : 13 e = 220 + 11.5 e = = > 1.5 e = 220 or e = 440 / 3 = 146.6 a | a = 22 * 100
b = 12 + 1
c = b * 10
d = 100 + 15
e = c - d
f = a / e
|
a ) 5 , b ) 6 , c ) 7 , d ) 3 , e ) 2 | d | divide(divide(20, 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 20 - meter platform and caught it after it had traveled 52.5 meters . how many times did the ball bounce ? | "going down = 20 m going up = 10 - - > total = 30 going down = 10 - - > total = 40 going up = 5 - - > total = 45 going down = 5 - - > total = 50 going up = 2.5 - - > total = 52.5 ( caught ) no of bouncing = 3 answer : d" | a = 20 / 2
b = a / 2
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a ) 2 , b ) 4 , c ) 16 , d ) 38 , e ) 76 | e | divide(subtract(power(21, const_2), power(17, const_2)), const_2) | the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 21 - inch screen would be how many square inches greater than the area of a square 17 - inch screen ? | "pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 21 ^ 2 / 2 - 17 ^ 2 / 2 = 152 / 2 = 76 ans e ." | a = 21 ** 2
b = 17 ** 2
c = a - b
d = c / 2
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a ) 3 % , b ) 8 % , c ) 7 % , d ) 10 % , e ) 15 % | c | subtract(const_100, add(add(add(subtract(const_100, 80), subtract(const_100, 82)), subtract(const_100, 70)), subtract(const_100, 75))) | in a urban village of india named ` ` owlna ' ' , 80 % people have refrigerator , 82 % people have television , 70 % people got computers and 75 % got air - conditionor . how many people ( minimum ) got all these luxury . | "c 7 % 100 - [ ( 100 - 80 ) + ( 100 - 82 ) + ( 100 - 70 ) + ( 100 - 75 ) ] = 100 - ( 20 + 18 + 30 + 25 ) = 100 - 93" | a = 100 - 80
b = 100 - 82
c = a + b
d = 100 - 70
e = c + d
f = 100 - 75
g = e + f
h = 100 - g
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a ) 4 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | b | subtract(40, add(8, 24)) | in a certain alphabet , 8 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | "we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 8 + 24 = 32 ; next , we find the total # of letters without line : 40 - 32 = 8 ; finally , we find the # of letters that contain a dot but do not contain a straight line : 8 - 0 = 8 . b" | a = 8 + 24
b = 40 - a
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a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | c | subtract(multiply(const_4, const_2), 7) | ( 18 ) 7 x ( 5832 ) - 2 Γ· ( 324 ) - 1 = ( 18 ) 7 | "explanation : ( 18 ) 7 x ( 183 ) - 2 Γ· ( 182 ) - 1 ( 18 ) 7 x ( 18 ) - 6 Γ· ( 18 ) - 2 ( 18 ) 7 - 6 + 2 = ( 18 ) 3 answer : option c" | a = 4 * 2
b = a - 7
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | add(divide(const_10, const_2), 0) | if a , b , c , d , e and f are integers and ( ab + cdef ) < 0 , then what is the maximum number w of integers that can be negative ? | "minimuum should be 1 maximum should be 4 : 1 out of a or b to make the multiplication negative 3 out of c , d , e or f to make the multiplication negative . negative + negative < 0 answer : c maximum will be 5 . . you dont require both the multiplicatin to be negative for entire equation to be negative . . . any one a or b can be negative to make ab negative and it can still be more ( away from 0 ) than the multiplication of 4 other - ve numbers . . . actually by writing minimum required as 1 out of 6 , you are actually meaning 5 out of 6 also possible as you will see w = 5 or 1 will give you same equation . . ans d" | a = 10 / 2
b = a + 0
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a ) 24 , b ) 66 , c ) 88 , d ) 27 , e ) 91 | a | add(add(multiply(const_3, const_2), 8), const_10) | rajan is sixth from the left end and vinay is tenth from the right end in a row of boys . if there are 8 boys between rajan and vinay , how many boys are there in the row ? | explanation : number of boys in the row = ( 6 + 10 + 8 ) = 24 answer : a ) 24 | a = 3 * 2
b = a + 8
c = b + 10
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a ) 2.98 , b ) 2.88 , c ) 2.82 , d ) 2.86 , e ) 2.81 | b | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 6 s = 1.2 ds = 6 + 1.2 = 7.2 us = 6 - 1.2 = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 answer : b" | a = 6 + 1
b = 6 - 1
c = a * b
d = 6 + 1
e = 6 - 1
f = d + e
g = c / f
h = g * 2
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a ) 6 , b ) 7 , c ) 8 , d ) 4 , e ) 3 | c | divide(64, 8) | find k if 64 / k = 8 . | "since 64 / k = 8 and 64 / 8 = 8 , then k = 8 correct answer c" | a = 64 / 8
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a ) 2 , b ) 0 , c ) 6 , d ) 5 , e ) 1 | b | subtract(multiply(multiply(multiply(445, 534), 999), 234), subtract(multiply(multiply(multiply(445, 534), 999), 234), add(const_4, const_4))) | the unit digit in the product ( 445 * 534 * 999 * 234 ) is : | "explanation : unit digit in the given product = unit digit in ( 5 * 4 * 9 * 4 ) = 0 answer : b" | a = 445 * 534
b = a * 999
c = b * 234
d = 445 * 534
e = d * 999
f = e * 234
g = 4 + 4
h = f - g
i = c - h
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a ) 230 , b ) 120 , c ) 95 , d ) 310 , e ) 320 | b | divide(add(1097, 17), subtract(10, const_1)) | the difference of two numbers is 1097 . on dividing the larger number by the smaller , we get 10 as quotient and the 17 as remainder . what is the smaller number ? | "solution : let the smaller number be x . then larger number = ( x + 1097 ) x + 1097 = 10 x + 17 9 x = 1080 x = 120 answer b" | a = 1097 + 17
b = 10 - 1
c = a / b
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a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 1 | b | divide(add(divide(20, 2), divide(20, 5)), const_2) | a boat running downstream covers a distance of 20 km in 2 hours while for covering the same distance upstream , it takes 5 hours . what is the speed of the boat in still water ? | explanation : rate downstream = ( 20 / 2 ) kmph = 10 kmph ; rate upstream = ( 20 / 5 ) kmph = 4 kmph speed in still water = 1 / 2 ( 10 + 4 ) kmph = 7 kmph answer : b | a = 20 / 2
b = 20 / 5
c = a + b
d = c / 2
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a ) 30 , b ) 60 , c ) 120 , d ) 240 , e ) 480 | b | divide(multiply(multiply(6, 5), 4), const_2) | a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 4 rows or 5 rows or 6 rows , then no tree will be left . find the least number of trees required . | "the least number of trees that are required = lcm ( 4 , 5 , 6 ) = 60 . answer : b" | a = 6 * 5
b = a * 4
c = b / 2
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a ) 10 , b ) 25 , c ) 60 , d ) 8 , e ) 20 | d | divide(multiply(6, 40), subtract(40, 10)) | a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 6 days , a finished the remaining work . in how many days a alone can finish the job ? | "a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 6 days whole work will be done by a in 6 * 4 / 3 = 8 days answer is d" | a = 6 * 40
b = 40 - 10
c = a / b
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a ) 877 m , b ) 600 m , c ) 167 m , d ) 176 m , e ) 546 m | b | multiply(multiply(72, const_0_2778), 30) | what distance will be covered by a bus moving at 72 kmph in 30 seconds ? | "explanation : 72 kmph = 72 * 5 / 18 = 20 mps d = speed * time = 20 * 30 = 600 m . answer : b" | a = 72 * const_0_2778
b = a * 30
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a ) 53.33 mph , b ) 56.67 mph , c ) 60 mph , d ) 64 mph , e ) 66.67 mph | c | add(divide(add(multiply(80, 2), multiply(40, 2)), add(2, 2)), subtract(divide(const_100, 2), const_0_33)) | steve traveled the first 2 hours of his journey at 40 mph and the last 2 hours of his journey at 80 mph . what is his average speed of travel for the entire journey ? | "average speed = total distance / total time = ( 40 * 2 + 80 * 2 ) / ( 2 + 2 ) = 240 / 4 = 60 answer : c" | a = 80 * 2
b = 40 * 2
c = a + b
d = 2 + 2
e = c / d
f = 100 / 2
g = f - const_0_33
h = e + g
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a ) 130 feet , b ) 22 feet , c ) 20 feet , d ) 15 feet , e ) 10 feet | a | add(multiply(divide(600, 10), const_2), 10) | a rectangular field is to be fenced on three sides leaving a side of 10 feet uncovered . if the area of the field is 600 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 10 * breadth = 600 = > breadth = 60 feet area to be fenced = 2 b + l = 2 * 60 + 10 = 130 feet answer : option a" | a = 600 / 10
b = a * 2
c = b + 10
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a ) 18 , b ) 8 , c ) 12 , d ) 10 , e ) none of these | d | divide(subtract(40, 20), const_2) | find the greatest number which leaves the same remainder when it divides 20 , 40 and 90 . | "90 - 40 = 50 40 - 20 = 20 90 - 20 = 70 the h . c . f of 20 , 50 and 70 is 10 . answer : d" | a = 40 - 20
b = a / 2
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a ) rs . 10111.00 , b ) rs . 10123.20 , c ) rs . 10123.00 , d ) rs . 10100.00 , e ) rs . 10110.00 | b | 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 . 25,000 after 3 years at the rate of 12 p . c . p . a ? | "= rs . ( 25000 x ( 1 + 12 / 100 ) Β³ = rs . ( 25000 x 28 / 25 x 28 / 25 x 28 / 25 ) = rs . 35123.20 . c . i = rs ( 35123.20 - 25000 ) = rs . 10123.20 answer : b" | 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
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a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 5 | c | divide(const_2, choose(add(2, 2), 2)) | a bag contains 2 white marbles and 2 black marbles . if each of 2 girls and 2 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | "first , total ways to select for all boys and girls , i . e 4 ! / ( 2 ! * 2 ! ) = 4 * 3 * 2 * 1 / 2 * 1 * 2 * 1 = 6 then there are one two way girls can have all same colors , either white or black . the number of ways in which 2 girls can select 2 white balls = 2 c 2 = 1 the number of ways in which 2 girls can select 2 black balls = 2 c 2 = 1 therefore , total favorable outcomes / total outcomes = 2 / 6 = 1 / 3 c" | a = 2 + 2
b = math.comb(a, 2)
c = 2 / b
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a ) 40000 , b ) 50000 , c ) 70000 , d ) 60000 , e ) 80000 | e | subtract(108000, multiply(const_60, const_100)) | a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 100000 . if the profit at the end of a year is rs . 108000 , then the share of b is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 100000 * 6 ) = 7 : 20 total profit = rs . 108000 share of b = 20 / 27 ( 108000 ) = rs . 80000 answer : e" | a = const_60 * 100
b = 108000 - a
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a ) 21 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | b | divide(10, subtract(const_1, multiply(10, divide(const_1, 20)))) | matt and peter can do together a piece of work in 20 days . after they have worked together for 10 days matt stops and peter completes the remaining work in 10 days . in how many days peter complete the work separately . | "together they complete the job in 20 days means they complete 10 / 20 of the job after 10 days . peter completes the remaining ( 10 / 20 ) of the job in 10 days which means that the whole job ( 1 ) can be completed in x days . < = > 8 / 20 - > 10 < = > x = 10 / ( 10 / 20 ) = 20 b" | a = 1 / 20
b = 10 * a
c = 1 - b
d = 10 / c
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a ) rs . 1991 , b ) rs . 2991 , c ) rs . 3991 , d ) rs . 4991 , e ) rs . 5991 | e | subtract(multiply(add(5, const_1), 7500), add(add(add(add(7435, 7927), 7855), 8230), 7562)) | a grocer has a sale of rs . 7435 , rs . 7927 , rs . 7855 , rs . 8230 and rs . 7562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 7500 ? | "explanation : total sale for 5 months = rs . ( 7435 + 7927 + 7855 + 8230 + 7562 ) = rs . 39009 . required sale = rs . [ ( 7500 x 6 ) Γ’ β¬ β 39009 ] = rs . ( 45000 Γ’ β¬ β 39009 ) = rs . 5991 . answer e" | a = 5 + 1
b = a * 7500
c = 7435 + 7927
d = c + 7855
e = d + 8230
f = e + 7562
g = b - f
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a ) 12 sec , b ) 30 sec , c ) 86 sec , d ) 15 sec , e ) 18 sec | d | divide(150, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 150 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 150 * 3 / 50 = 15 sec . answer : d" | a = 63 - 3
b = a * const_0_2778
c = 150 / b
|
a ) 10 , b ) 11 , c ) 13 , d ) 14 , e ) 16 | c | add(multiply(add(3, 3), 3), floor(divide(subtract(150, multiply(divide(multiply(3, subtract(10, 1)), const_2), 10)), 10))) | in a certain supermarket , a triangular display of cans is arranged in 10 rows , numbered 1 through 10 from top to bottom . each successively numbered row contains 3 more cans than the row immediately above it . if there are fewer than 150 cans in the entire display , how many cans are in the fifth row ? | "let x be the number of cans in row 1 . the total number of cans is x + ( x + 3 ) + . . . + ( x + 27 ) = 10 x + 3 ( 1 + 2 + . . . + 9 ) = 10 x + 3 ( 9 ) ( 10 ) / 2 = 10 x + 135 since the total is less than 150 , x must equal 1 . the number of cans in the 5 th row is 1 + 3 ( 4 ) = 13 the answer is c ." | a = 3 + 3
b = a * 3
c = 10 - 1
d = 3 * c
e = d / 2
f = e * 10
g = 150 - f
h = g / 10
i = math.floor(h)
j = b + i
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a ) 7 : 5 , b ) 12 : 11 , c ) 17 : 15 , d ) 22 : 17 , e ) 27 : 25 | e | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(2, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(2, 1)), 2))) | an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 2 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | "let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 2 y and y . 2 ( 3 x + 2 x ) = 2 ( 2 y + y ) 5 x = 3 y ( 5 / 3 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 2 y * y = 6 x ^ 2 : 2 y ^ 2 = 6 x ^ 2 : 2 * ( 25 / 9 ) * x ^ 2 = 54 : 50 = 27 : 25 the answer is e ." | a = rectangle_area / (
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a ) 10 , b ) 6.8 , c ) 27 , d ) 22 , e ) 29 | b | multiply(divide(multiply(4, 5), subtract(multiply(6, 4), multiply(3, 4))), 4) | mixture contains alcohol and water in the ratio 4 : 3 . if 6 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 24 x = 4 ( 3 x + 5 ) 12 x = 20 x = 1.7 quantity of alcohol = ( 4 x 1.7 ) litres = 6.8 litres . answer : b" | a = 4 * 5
b = 6 * 4
c = 3 * 4
d = b - c
e = a / d
f = e * 4
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a ) 24 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | a | divide(const_1, subtract(divide(const_1, 8), divide(const_1, 12))) | a cistern is filled by pipe a in 8 hours and the full cistern can be leaked out by an exhaust pipe b in 12 hours . if both the pipes are opened , in what time the cistern is full ? | time taken to full the cistern = ( 1 / 8 - 1 / 12 ) hrs = 1 / 24 = 24 hrs answer : a | a = 1 / 8
b = 1 / 12
c = a - b
d = 1 / c
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a ) 220 : 113 , b ) 201 : 200 , c ) 210 : 201 , d ) 100 : 99 , e ) 113 : 77 | b | divide(add(300, multiply(divide(300, add(subtract(400, const_100), 300)), add(238, multiply(const_2, add(const_3, const_2))))), subtract(400, const_100)) | a and b put in rs . 300 and rs . 400 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 238 where as b does not . in what ratio should they divide the second year ' s profit ? | "explanation : 3 : 4 a = 3 / 7 * 238 = 102 402 : 400 201 : 200 answer : b" | a = 400 - 100
b = a + 300
c = 300 / b
d = 3 + 2
e = 2 * d
f = 238 + e
g = c * f
h = 300 + g
i = 400 - 100
j = h / i
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a ) $ 42.25 . , b ) $ 40.25 . , c ) $ 38.25 . , d ) $ 36.25 . , e ) $ 34.25 | c | add(add(multiply(15, 2), 2), multiply(divide(15, const_60), divide(25, 6))) | it costs $ 2 for the first 1 / 6 hour to use the laundry machine at the laundromat . after the first ΒΌ hour it costs $ 15 per hour . if a certain customer uses the laundry machine for 2 hours and 25 minutes , how much will it cost him ? | "2 hrs 25 min = 145 min first 10 min - - - - - - > $ 2 time left is 135 min . . . now , 60 min costs $ 15 1 min costs $ 15 / 60 145 min costs $ 15 / 60 * 145 = > $ 36.25 so , total cost will be $ 36.25 + $ 2 = > $ 38.25 hence answer will be c" | a = 15 * 2
b = a + 2
c = 15 / const_60
d = 25 / 6
e = c * d
f = b + e
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a ) 96 , b ) 87 , c ) 75 , d ) 25 , e ) 12 | c | divide(9, subtract(86.12, floor(86.12))) | when positive integer x is divided by positive integer y , the remainder is 9 . if x / y = 86.12 , what is the value of y ? | "by the definition of a remainder , the remainder here is equal to 9 / y . the remainder in decimal form is given as . 12 therefore , 9 / y = . 12 solve for y and get 75 . c" | a = math.floor(86, 12)
b = 86 - 12
c = 9 / b
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a ) 480 minutes , b ) 492 minutes , c ) 504 minutes , d ) 528 minutes , e ) 540 minutes | d | power(divide(70, divide(divide(10, const_2), const_2)), const_2) | a truck driver starts with the speed of 70 kmph with the truck driver decreasing speed every two hours by 10 kmph . the truck driver ' s speed does not drop below 50 kmph . in how many hours will it take the truck driver to travel 500 kms ? | distance covered in first two hours = 70 Γ£ β 2 = 140 km distance covered in next two hours = 60 Γ£ β 2 = 120 km distance covered in next two hours = 50 Γ£ β 2 = 100 km distance covered in next two hours = 50 Γ£ β 2 = 100 km distance covered in first eight hours 140 + 120 + 100 + 100 = 460 km remaining distance = 500 Γ’ β¬ β 460 = 40 km . now , this distance will be covered at the speed of 50 km / hr . Γ’ Λ Β΄ time taken = 40 Γ’ Β β 50 = 4 Γ’ Β β 5 hour . total time = 8 + 4 Γ’ Β β 5 = 8 4 Γ’ Β β 5 hour answer d | a = 10 / 2
b = a / 2
c = 70 / b
d = c ** 2
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a ) 57 , b ) 76 , c ) 77 , d ) 87 , e ) 97 | b | sqrt(multiply(57.76, const_100)) | a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 57.76 , the number of the member is the group is : | "money collected = ( 57.76 x 100 ) paise = 5776 paise numbers of members = 5776 squareroot = 76 answer b" | a = 57 * 76
b = math.sqrt(a)
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a ) 5 , b ) 6 , c ) 7 , d ) 3 , e ) 9 | d | 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 44 meters . how many times did the ball bounce ? | "going down = 16 m going up = 8 - - > total = 24 going down = 8 - - > total = 32 going up = 4 - - > total = 36 going down = 4 - - > total = 40 going up = 2 - - > total = 42 going down = 2 - - > total = 44 ( caught ) no of bounce = 3 . . answer : d" | a = 16 / 2
b = a / 2
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['a ) 296', 'b ) 289', 'c ) 270', 'd ) 125', 'e ) 278'] | d | divide(volume_cube(const_10), volume_cube(2)) | how many cubes of edge 2 dm can be cut out of a meter cube ? | 1 * 1 * 1 = 2 / 10 * 2 / 10 * 2 / 10 * x x = 125 answer : d | a = volume_cube / (
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a ) 59 m , b ) 60 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(add(divide(5300, 26.50), multiply(const_2, 20)), const_4) | length of a rectangular plot is 20 mtr more than its breadth . if the cost of fencing the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 18 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 18 ) + x ] = 200 2 x + 18 = 100 2 x = 82 x = 41 . hence , length = x + 18 = 59 m a" | a = 5300 / 26
b = 2 * 20
c = a + b
d = c / 4
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a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | d | divide(const_1, subtract(divide(const_1, 16), divide(const_1, 20))) | a cistern is filled by pipe a in 16 hours and the full cistern can be leaked out by an exhaust pipe b in 20 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 16 - 1 / 20 ) hrs = 1 / 80 = 80 hrs answer : d" | a = 1 / 16
b = 1 / 20
c = a - b
d = 1 / c
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a ) 15 , b ) 17 , c ) 11 , d ) 18 , e ) 13 | b | divide(subtract(76000, 42000), add(800, 1200)) | village x has a population of 76000 , which is decreasing at the rate of 1200 per year . village y has a population of 42000 , which is increasing at the rate of 800 per year . in how many years will the population of the two villages be equal ? | "let the population of two villages be equal after p years then , 76000 - 1200 p = 42000 + 800 p 2000 p = 34000 p = 17 answer is b ." | a = 76000 - 42000
b = 800 + 1200
c = a / b
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a ) a ) 70 , b ) b ) 76 , c ) c ) 78 , d ) d ) 89 , e ) e ) 88 | d | subtract(multiply(add(34, 5), add(10, const_1)), multiply(10, 34)) | average of 10 matches is 34 , how many runs one should should score to increase his average by 5 runs . | "explanation : average after 11 innings should be 39 so , required score = ( 11 * 39 ) - ( 10 * 34 ) = 429 - 340 = 89 answer : option d" | a = 34 + 5
b = 10 + 1
c = a * b
d = 10 * 34
e = c - d
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a ) 110 hours , b ) 112 hours , c ) 115 hours , d ) 100 hours , e ) none of these | b | divide(multiply(divide(multiply(14, 16), const_2), subtract(multiply(divide(add(14, 16), const_2), divide(16, const_2)), divide(multiply(14, 16), const_2))), subtract(multiply(divide(add(14, 16), const_2), divide(16, const_2)), divide(multiply(14, 16), const_2))) | two pipes can fill a cistern in 14 and 16 hours respectively . the pipes are opened simultaneously and it is found that due to leakage in the bottom , 32 minutes extra are taken for the cistern to be filled up . if the cistern is full , in what time would the leak empty it ? | cistern filled by both pipes in one hour = 1 β 14 + 1 β 16 = 15 β 112 th β΄ both pipes filled the cistern in 112 β 15 hrs . now , due to leakage both pipes filled the cistern in 112 β 15 + 32 β 60 = 8 hrs . β΄ due to leakage , filled part in one hour 1 β 8 β΄ part of cistern emptied , due to leakage in one hour = 15 β 112 - 1 β 8 = 1 β 112 th β΄ in 112 hr , the leakage would empty the cistern . answer b | a = 14 * 16
b = a / 2
c = 14 + 16
d = c / 2
e = 16 / 2
f = d * e
g = 14 * 16
h = g / 2
i = f - h
j = b * i
k = 14 + 16
l = k / 2
m = 16 / 2
n = l * m
o = 14 * 16
p = o / 2
q = n - p
r = j / q
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a ) 18 , b ) 16 , c ) 26 , d ) 17 , e ) 11 | c | multiply(divide(subtract(1500, 1110), 1500), const_100) | the cost price of a radio is rs . 1500 and it was sold for rs . 1110 , find the loss % ? | explanation : 1500 - - - - 390 100 - - - - ? = > 26 % answer : c | a = 1500 - 1110
b = a / 1500
c = b * 100
|
a ) 23.5 % , b ) 32.5 % , c ) 35 % , d ) 18.75 % , e ) 20 % | d | multiply(multiply(subtract(const_1, divide(const_1, const_4)), divide(const_1, const_4)), const_100) | a restaurant spends one quarter of its monthly budget for rent and quarter of the rest for food and beverages . what percentage of the budget does the restaurant spend for food and beverages ? | spend on rent = 1 / 4 spend on food and beverage = 1 / 4 of remaining = 1 / 4 * 3 / 4 = 3 / 16 so 3 / 16 = 18.75 % d is the answer | a = 1 / 4
b = 1 - a
c = 1 / 4
d = b * c
e = d * 100
|
a ) 10 % , b ) 12 % , c ) 14 % , d ) 16 % , e ) 18 % | b | multiply(subtract(const_1, multiply(add(divide(10, const_100), const_1), divide(80, const_100))), const_100) | a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 10 % so that the customer could not buy the required amount of the product . the customer managed to buy only 80 % of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ? | "let x be the amount of money paid for the first purchase . the second time , the customer paid 0.8 ( 1.1 x ) = 0.88 x . the difference is 12 % . the answer is b ." | a = 10 / 100
b = a + 1
c = 80 / 100
d = b * c
e = 1 - d
f = e * 100
|
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | a | divide(multiply(78, 3), add(add(multiply(4, const_2), 4), const_1)) | of 3 numbers , the third is twice the second and the second is 4 times the first . if their average is 78 , the smallest of the 3 numbers is : | explanation : let first number be x . so , 2 nd no . = 4 x & 3 rd no . = 8 x . so , x + 4 x + 8 x = 78 Γ 3 = 234 . 13 x = 234 x = 234 / 13 hence , smallest number x = 18 . answer : a | a = 78 * 3
b = 4 * 2
c = b + 4
d = c + 1
e = a / d
|
a ) 5 : 12 , b ) 5 : 13 , c ) 2 : 9 , d ) 1 : 15 , e ) 1 : 5 | d | divide(subtract(sqrt(625), 20), multiply(sqrt(625), const_2)) | the area of a square is 625 sq cm . find the ratio of the breadth and the length of a rectangle whose length is thrice the side of the square and breadth is 20 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 625 a = 25 l = 3 a and b = a - 20 b : l = a - 20 : 3 a = 5 : 75 = 1 : 15 answer : d" | a = math.sqrt(625)
b = a - 20
c = math.sqrt(625)
d = c * 2
e = b / d
|
a ) 590 liters , b ) 540 liters , c ) 820 liters , d ) 900 liters , e ) 580 liters | d | multiply(multiply(inverse(subtract(add(add(divide(const_1, 20), divide(const_1, 15)), divide(const_1, 45)), divide(const_1, 15))), const_3), 15) | two pipes a and b can separately fill a tank in 20 and 15 minutes respectively . a third pipe c can drain off 45 liters of water per minute . if all the pipes are opened , the tank can be filled in 15 minutes . what is the capacity of the tank ? | "1 / 20 + 1 / 15 - 1 / x = 1 / 15 x = 12 20 * 45 = 900 answer : d" | a = 1 / 20
b = 1 / 15
c = a + b
d = 1 / 45
e = c + d
f = 1 / 15
g = e - f
h = 1/(g)
i = h * 3
j = i * 15
|
a ) 32 % , b ) 36 % , c ) 40 % , d ) 44 % , e ) 48 % | c | multiply(divide(subtract(divide(multiply(2, subtract(const_100, 50)), const_100), multiply(0.5, 2)), subtract(2, multiply(0.5, 2))), const_100) | a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 50 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 2 hours ? | "the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.5 ( 6 ) = 3 gallons at the start , the amount of water is 0.5 ( 6 ) = 3 gallons after 2 hours , the amount of water is 3 - 0.5 ( 2 ) = 2 gallons the concentration of water is 2 / ( 3 + 2 ) = 2 / 5 = 40 % the answer is c ." | a = 100 - 50
b = 2 * a
c = b / 100
d = 0 * 5
e = c - d
f = 0 * 5
g = 2 - f
h = e / g
i = h * 100
|
a ) 5.5 , b ) 4.5 , c ) 3.5 , d ) 2.5 , e ) 1.5 | a | divide(55, multiply(36, const_0_2778)) | in what time will a train 55 m long cross an electric pole , it its speed be 36 km / hr ? | "speed = 36 * 5 / 18 = 10 m / sec time taken = 55 / 10 = 5.5 sec . answer : a" | a = 36 * const_0_2778
b = 55 / a
|
a ) 2345 , b ) 2350 , c ) 2457 , d ) 4657 , e ) 5760 | e | multiply(multiply(multiply(6, 4), const_60), 4) | a leak in the bottom of a tank can empty the tank in 6 hrs . an pipe fills water at the rate of 4 ltrs / minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hrs . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 e" | a = 6 * 4
b = a * const_60
c = b * 4
|
a ) 72 , b ) 224 , c ) 320 , d ) 512 , e ) 1,600 | a | gcd(20, const_4) | if m and n are positive integers and m ^ 2 + n ^ 2 = 20 , what is the value of m ^ 3 + n ^ 3 ? | "you need to integers which squared are equal 20 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 4 ^ 2 = 16 stop . the integers ca n ' t be greater than 4 or we will score above 20 . the second integer need to be picked up the same way . 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 4 ^ 2 = 16 the only pair that matches is 4 ^ 2 + 2 ^ 2 = 20 . so 4 ^ 3 + 2 ^ 3 = 72 . answer a . )" | a = math.gcd(20, 4)
|
a ) 0.5 , b ) 1 , c ) 1.5 , d ) 2 , e ) 2.5 | b | divide(multiply(multiply(10, divide(6, 7)), inverse(subtract(const_1, divide(6, 7)))), const_60) | a train is moving at 6 / 7 of its usual speed . the train is 10 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | "new time = d / ( 6 v / 7 ) = 7 / 6 * usual time 10 minutes represents 1 / 6 of the usual time . the usual time is 1 hour . the answer is b ." | a = 6 / 7
b = 10 * a
c = 6 / 7
d = 1 - c
e = 1/(d)
f = b * e
g = f / const_60
|
a ) 15500 , b ) 16000 , c ) 15000 , d ) 17000 , e ) 20000 | c | divide(400, subtract(multiply(divide(10, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(20, const_100), divide(20, const_100)))) | a shopkeeper sells 20 % of his stock at 20 % profit ans sells the remaining at a loss of 10 % . he incurred an overall loss of rs . 400 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 6 / 5 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.90 = 19 x / 25 = 36 x / 50 total sp = 12 x / 50 + 36 x / 50 = 48 x / 50 overall loss = x - 48 x / 50 = 2 x / 50 2 x / 50 = 600 = > x = 15000 answer : c" | a = 10 / 100
b = 100 - 20
c = b / 100
d = a * c
e = 20 / 100
f = 20 / 100
g = e * f
h = d - g
i = 400 / h
|
a ) 5 / 14 , b ) 8 / 21 , c ) 12 / 21 , d ) 43 / 91 , e ) 47 / 91 | d | add(multiply(divide(8, add(6, 8)), divide(subtract(8, const_1), subtract(add(6, 8), const_1))), multiply(divide(subtract(6, const_1), subtract(add(6, 8), const_1)), divide(6, add(6, 8)))) | a bag contains 6 green balls and 8 white balls . if two balls are drawn simultaneously , what is the probability that both balls are the same colour ? | the total number of ways to draw two balls is 14 c 2 = 91 the number of ways to draw two green balls is 6 c 2 = 15 the number of ways to draw two white balls is 8 c 2 = 28 p ( two balls of the same colour ) = 43 / 91 the answer is d . | a = 6 + 8
b = 8 / a
c = 8 - 1
d = 6 + 8
e = d - 1
f = c / e
g = b * f
h = 6 - 1
i = 6 + 8
j = i - 1
k = h / j
l = 6 + 8
m = 6 / l
n = k * m
o = g + n
|
a ) 1200 km , b ) 1500 km , c ) 2000 km , d ) 2700 km , e ) 3600 km | d | multiply(30, 45) | a walks at 30 kmph and 30 hours after his start , b cycles after him at 45 kmph . how far from the start does b catch up with a ? | "suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 30 hours . x / 30 - x / 45 = 30 x = 2700 km answer is d" | a = 30 * 45
|
['a ) 190', 'b ) 180', 'c ) 170', 'd ) 160', 'e ) 148'] | e | subtract(320, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(320, 10.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | a snooker tournament charges $ 40.00 for vip seats and $ 10.00 for general admission ( β regular β seats ) . on a certain night , a total of 320 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | let no of sits in vip enclosure is x then x * 40 + 10 ( 320 - x ) = 7500 or 25 x = 7500 - 3200 , x = 4300 / 25 = 172 vip = 172 general = 320 - 172 = 148 e | a = 3 + 4
b = a * 1000
c = 2 + 3
d = c * 100
e = b + d
f = 320 * 10
g = e - f
h = 2 + 3
i = 2 + 3
j = h * i
k = g / j
l = 320 - k
|
a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 12 | e | add(divide(6, 2), const_2) | if a and b are positive integers , and a = 2 b + 6 , the greatest common divisor of a and b can not be | "since a = 2 b + 6 , so we can say b = ( a / 2 - 3 ) . so we need to find not possible gcd values for a , ( a / 2 - 3 ) . a . 1 , we can easily get this value by making a = 8 . b . 2 . we can again get this value as gcd by keeping a = 10 c . 3 we will get this as gcd by keeping a = 12 d . 6 we can get gcd 6 by keeping ( a / 2 - 3 ) = 6 and a as 18 . e . 12 this is not possible as for 12 ( 2 ^ 2 * 3 ) to be gcd = 2 ^ 2 * 3 both a and a / 2 - 3 should be divisible by 4 and 3 . so a has to be a multiple of 4 . this means a / 2 has to be even and even - odd will be odd and odd number wont be divisible by 4 . this 12 cant be the gcd . answer e ." | a = 6 / 2
b = a + 2
|
a ) 3 hours . , b ) 4 hours . , c ) 4 hours and 20 minutes . , d ) 5 hours and 50 minutes , e ) 6 hours . | e | divide(60, const_10) | naomi drives to the beauty parlor in 60 minutes . on the way back , her average speed is half the average speed as it was to the way to the parlor . how much time will it take naomi to travel two round trips to the beauty parlor ? | s 1 = 2 s 2 since , speed is inversely proportional to time we have , s 1 / s 2 = 2 / 1 = t 2 / t 1 therefore , the time taken for one trip = 60 + 120 = 180 minutes total = 180 * 2 = 360 minutes = 6 hours answer : e | a = 60 / 10
|
a ) 40 m 2 , b ) 45 m 2 , c ) 49 m 2 , d ) 51 m 2 , e ) 55 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(39, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 12 metro and whose angle at the center is 39 Β° is ? | "39 / 360 * 22 / 7 * 12 * 12 = 49 m 2 answer : c" | a = 12 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 39 / g
i = f * h
|
a ) 10,000 , b ) 11,600 , c ) 12,000 , d ) 14,000 , e ) 16,400 | c | add(divide(10000, 2000), 6) | jerome anticipated that the webweb . com stock price would fall and sold all his webweb . com stocks for $ 5 per stock . he paid $ 10000 tax on the revenue . a week later , jerome became convinced that the webweb . com stock price would rise , and he used the money that he had gotten from selling the webweb . com stocks to purchase them again , this time at $ 6 per stock . if jerome ended up having 2000 webweb . com stocks fewer than the number he had before selling them , how many webweb . com stocks did he have originally ? | let the number of shares be x . 5 * x - 10000 ( money paid in taxes ) = 6 ( x - 2000 ) solving for x , we get the shares as 12000 . ans : ( option c ) | a = 10000 / 2000
b = a + 6
|
a ) 100 , b ) 250 , c ) 300 , d ) 330 , e ) none of these | a | divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 175), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2))))) | the radius of the wheel of a bus is 175 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is | "radius of the wheel of bus = 175 cm . then , circumference of wheel = 2 Γ― β¬ r = 350 Γ― β¬ = 1100 cm distance covered by bus in 1 minute = 66 Γ’ Β β 60 Γ£ β 1000 Γ£ β 100 cms distance covered by one revolution of wheel = circumference of wheel = 1100 cm Γ’ Λ Β΄ revolutions per minute = 6600000 / 60 Γ£ β 1100 = 100 answer a" | a = 1000 * 66
b = 100 * a
c = const_60 * 1
d = b / c
e = 2 * 175
f = 3 * 4
g = 2 + f
h = 4 * 2
i = 2 + h
j = i ** 2
k = g / j
l = 3 + k
m = e * l
n = d / m
|
a ) 200 , b ) 278 , c ) 282 , d ) 202 , e ) 600 | e | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 60)), 1000), multiply(divide(const_100, add(const_100, 60)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 60)), 1000), multiply(divide(const_100, add(const_100, 60)), 1000))), const_100) | a shopkeeper buys two articles for rs . 1000 each and then sells them , making 60 % profit on the first article and 60 % loss on second article . find the net profit or loss percent ? | "profit on first article = 60 % of 1000 = 600 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : e" | a = 100 - 60
b = 100 / a
c = b * 1000
d = 100 + 60
e = 100 / d
f = e * 1000
g = c + f
h = 1000 + 1000
i = g - h
j = i * 100
k = 100 - 60
l = 100 / k
m = l * 1000
n = 100 + 60
o = 100 / n
p = o * 1000
q = m + p
r = j / q
s = r * 100
|
a ) 100 , b ) 120 , c ) 140 , d ) 160 , e ) 180 | b | divide(500, multiply(15, const_0_2778)) | how many seconds does sandy take to cover a distance of 500 meters , if sandy runs at a speed of 15 km / hr ? | "15 km / hr = 15000 m / 3600 s = ( 150 / 36 ) m / s = ( 25 / 6 ) m / s time = 500 / ( 25 / 6 ) = 120 seconds the answer is b ." | a = 15 * const_0_2778
b = 500 / a
|
a ) 1 : 7 , b ) 6 : 5 , c ) 6 : 7 , d ) 4 : 7 , e ) 6 : 9 | c | divide(divide(multiply(5, 3), multiply(7, 2)), divide(multiply(3, 4), multiply(2, 5))) | the compound ratio of 5 : 7 , 3 : 2 and 4 : 5 ? | "5 / 7 * 3 / 2 * 4 / 5 = 6 / 7 6 : 7 answer : c" | a = 5 * 3
b = 7 * 2
c = a / b
d = 3 * 4
e = 2 * 5
f = d / e
g = c / f
|
a ) 320 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $ | d | multiply(multiply(0.65, 55), 12) | in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.65 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | "total cost = ( 1.75 * 12 ) + ( 0.65 * 12 * 55 ) = 21 + 429 = > 450 hence answer will be ( d ) 450" | a = 0 * 65
b = a * 12
|
a ) $ 7600 , b ) $ 7000 , c ) $ 8000 , d ) $ 9000 , e ) $ 5000 | b | divide(multiply(add(divide(subtract(1000, multiply(divide(7, 6), 1000)), subtract(divide(7, 6), divide(6, 5))), 1000), 7), 6) | the ratio of the incomes of rajan and balan is 7 : 6 and the ratio of their expenditure is 6 : 5 . if at the end of the year , each saves $ 1000 then the income of rajan is ? | let the income of rajan and balan be $ 7 x and $ 6 x let their expenditures be $ 6 y and $ 5 y 7 x - 6 y = 1000 - - - - - - - 1 ) 6 x - 5 y = 1000 - - - - - - - 2 ) from 1 ) and 2 ) x = 1000 rajan ' s income = 7 x = 7 * 1000 = $ 7000 answer is b | a = 7 / 6
b = a * 1000
c = 1000 - b
d = 7 / 6
e = 6 / 5
f = d - e
g = c / f
h = g + 1000
i = h * 7
j = i / 6
|
a ) 54 , b ) 66 , c ) 80 , d ) 36 , e ) 96 | d | multiply(12, divide(divide(36, const_2), 6)) | two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 12 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | d 36 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 12 mph = 36 miles | a = 36 / 2
b = a / 6
c = 12 * b
|
a ) 550 , b ) 289 , c ) 350 , d ) 882 , e ) 281 | a | subtract(multiply(divide(300, 18), 51), 300) | a 300 meter long train crosses a platform in 51 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "speed = [ 300 / 18 ] m / sec = 50 / 3 m / sec . let the length of the platform be x meters . then , x + 300 / 51 = 50 / 3 3 ( x + 300 ) = 2550 Γ¨ x = 550 m . answer : a" | a = 300 / 18
b = a * 51
c = b - 300
|
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