options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 1155 , b ) 2310 , c ) 4620 , d ) 30030 , e ) 60060 | b | divide(divide(multiply(multiply(multiply(10, add(10, const_1)), add(add(10, const_1), const_1)), add(add(add(add(10, const_1), const_1), const_1), const_1)), const_2), const_4) | if j is the least positive integer that is divisible by every integer from 10 to 15 inclusive , then j / 26 is = | the integer should be divisible by : 10 , 11 , 12 , 13 , 14 and 15 , that is 5 * 211 , 3 * 2 ^ 2 , 13 , 2 * 7 , and 3 * 5 . the least common multiple of these integers is the product of 13 , 11 , 7 , 5 , 3 and 2 ^ 2 . then , r / 26 is ( 13 * 11 * 7 * 5 * 3 * 2 ^ 2 ) / 2 * 13 = 11 * 7 * 5 * 3 * 2 = 2310 . answer : b . | a = 10 + 1
b = 10 * a
c = 10 + 1
d = c + 1
e = b * d
f = 10 + 1
g = f + 1
h = g + 1
i = h + 1
j = e * i
k = j / 2
l = k / 4
|
a ) s . 190 , b ) s . 1120 , c ) s . 1200 , d ) s . 1250 , e ) s . 1290 | c | subtract(multiply(20, divide(8000, const_100)), multiply(4, divide(15000, const_100))) | john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 20 % . overall how much he make aprofit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 20 / 100 ) = 8000 + 1600 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 1600 = rs . 1200 as this is positive , an overall profit of rs . 1200 was made . c" | a = 8000 / 100
b = 20 * a
c = 15000 / 100
d = 4 * c
e = b - d
|
a ) 52.6 , b ) 52.9 , c ) 52.8 , d ) 118.8 , e ) 52.2 | d | multiply(multiply(power(18, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | the area of sector of a circle whose radius is 18 metro and whose angle at the center is 42 â ° is ? | "42 / 360 * 22 / 7 * 18 * 18 = 118.8 m 2 answer : d" | a = 18 ** 2
b = 2 * 10
c = b + 2
d = 4 + 3
e = c / d
f = a * e
g = 3600 / 10
h = 42 / g
i = f * h
|
a ) 80 , b ) 70 , c ) 75 , d ) 16 , e ) 14 | c | multiply(divide(120, add(add(1, 2), 5)), 5) | the amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5 . if they worked a combined total of 120 hours , how many hours did the secretary who worked the longest spend on the project ? | "8 x = 120 = > x = 15 therefore the secretary who worked the longest spent 15 x 5 = 75 hours on the project option ( c )" | a = 1 + 2
b = a + 5
c = 120 / b
d = c * 5
|
a ) 10 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | divide(44, 3) | a number of 44 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 13 * 3 + 1 * 5 = 44 - - > 13 + 1 = 14 . answer b" | a = 44 / 3
|
a ) 13 , b ) 16 , c ) 18 , d ) 29 , e ) 32 | e | divide(power(multiply(const_3, power(192, const_2)), divide(const_1, const_3)), 1.5) | danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 192 , what is the area of the side face of the box ? | "lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 / 2 w * d ( l = 1 / 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 192 l * 1.5 l * 2 l = 192 l = 4 side face area = l * d = l * 2 l = 4 * 2 * 4 = 32 e is the answer" | a = 192 ** 2
b = 3 * a
c = 1 / 3
d = b ** c
e = d / 1
|
a ) 2 , b ) 20 , c ) 28 , d ) 14 , e ) 19 | a | divide(divide(1, 2), divide(1, 4)) | diana is painting statues . she has 1 / 2 of a gallon of paint remaining . each statue requires 1 / 4 gallon of paint . how many statues can she paint ? | "number of statues = all the paint ÷ amount used per statue = 1 / 2 ÷ 1 / 4 = 1 / 2 * 4 / 1 = 2 answer is a" | a = 1 / 2
b = 1 / 4
c = a / b
|
a ) 10 , b ) 120 , c ) 15 , d ) 8 , e ) 16 | d | divide(add(add(add(add(add(add(add(add(add(add(add(add(add(add(const_1, const_2), const_3), const_4), add(const_4, const_1)), add(add(const_1, const_2), const_3)), add(const_4, const_3)), add(const_4, const_4)), add(add(const_4, const_4), const_1)), const_10), add(const_10, const_1)), const_12), add(const_12, const_1)), add(add(const_12, const_1), const_1)), 15), 15) | find the average of first 15 natural numbers . | explanation : sum of first n natural numbers = n ( n + 1 ) / 2 hence , sum of first 15 natural numbers = ( 15 x 16 ) / 2 = 120 therefore , required average of = 120 / 15 = 8 answer : d | a = 1 + 2
b = a + 3
c = b + 4
d = 4 + 1
e = c + d
f = 1 + 2
g = f + 3
h = e + g
i = 4 + 3
j = h + i
k = 4 + 4
l = j + k
m = 4 + 4
n = m + 1
o = l + n
p = o + 10
q = 10 + 1
r = p + q
s = r + 12
t = 12 + 1
u = s + t
v = 12 + 1
w = v + 1
x = u + w
y = x + 15
z = y / 15
|
a ) 3 / 8 , b ) 13 / 36 , c ) 17 / 36 , d ) 19 / 36 , e ) 23 / 36 | b | divide(add(add(add(add(const_4, 2), 1), 2), const_4), multiply(6, 6)) | the “ a - number ” of a number x is defined as the ones digit of 2 ^ x . antony rolls a die with 6 sides labeled with the integers from 1 to 6 , each of which has an equal probability of landing face - up . he then takes 3 ^ a , where a is the a - number of the result of his die roll , and plots 3 ^ a on a number line as the point a . finally , he repeats this entire process , this time plotting the result as the point b . what is the probability that the distance between a and b is greater than the value of b ? | if you calculate 3 ^ a for 1 st roll , all 6 results will be 9 , 81 , 6561 , 729 , 9 , 81 . this result is the same for 2 nd roll . 9 , 81 , 6561 , 729 , 9 , 81 . about distance : if the first result is 9 and the second is also 9 , the distance is 9 - 9 = 0 which is smaller than 9 . if the first result is 9 and the second is 81 , the distance is 81 - 9 = 72 which is also smaller than b which has the value of 81 . if the first result is 81 and the second is 9 , the distance will be greater than b . distance 81 - 9 = 72 > 9 . on the first roll , the probability of getting result 9 is 2 / 6 . in this case no other alternative values for second roll which would make the distance greater than b . so probability is 0 . so next estimations are : probability of getting 81 on the first roll ( 2 / 6 ) * probability of getting 9 on the second roll ( 2 / 6 ) = 1 / 9 probability of getting 729 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 on the second roll ( 4 / 6 ) = 1 / 9 probability of getting 6561 on the first roll ( 1 / 6 ) * probability of getting 9 , 81 , 729 on the first roll ( 5 / 6 ) = 5 / 36 all together : 1 / 9 + 1 / 9 + 5 / 36 = 13 / 36 = b | a = 4 + 2
b = a + 1
c = b + 2
d = c + 4
e = 6 * 6
f = d / e
|
a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 10 , e ) $ 5 | d | add(divide(subtract(17, 3), const_2), 3) | you and your friend spent a total of $ 17 for lunch . your friend spent $ 3 more than you . how much did your friend spend on their lunch ? | "my lunch = l , my friends lunch = l + 3 ( l ) + ( l + 3 ) = 17 l + l + 3 - 3 = 17 - 3 2 l = 14 l = 7 my friends lunch l + 3 = 7 + 3 = $ 10 , the answer is d" | a = 17 - 3
b = a / 2
c = b + 3
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 25 % , e ) 28 % | b | multiply(divide(subtract(60, 50), 50), const_100) | a man buy a book in rs 50 & sale it rs 60 . what is the rate of profit ? ? ? | "cp = 50 sp = 60 profit = 60 - 50 = 10 % = 10 / 50 * 100 = 20 % answer : b" | a = 60 - 50
b = a / 50
c = b * 100
|
a ) 121 km , b ) 242 km , c ) 224 km , d ) 112 km , e ) 110 km | c | multiply(const_2, divide(multiply(multiply(2121, 2424), 1010), add(2121, 2424))) | a man complete a journey in 1010 hours . he travels first half of the journey at the rate of 2121 km / hr and second half at the rate of 2424 km / hr . find the total journey in km . | "solution 1 average speed = 2 ã — 21 ã — 2421 + 24 = 22.4 km / hr = 2 ã — 21 ã — 2421 + 24 = 22.4 km / hr total distance = 22.4 ã — 10 = 224 km answer is c" | a = 2121 * 2424
b = a * 1010
c = 2121 + 2424
d = b / c
e = 2 * d
|
a ) 11 , b ) 40 , c ) 18 , d ) 16 , e ) 12 | b | multiply(multiply(divide(7, 5), divide(10, 7)), 20) | the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 20 months , find for how much time did q invest the money ? | "7 * 5 : 20 * x = 7 : 10 x = 40 answer : b" | a = 7 / 5
b = 10 / 7
c = a * b
d = c * 20
|
a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | inverse(add(divide(const_1, 56), divide(7, 56))) | pipe a fills a tank in 56 minutes . pipe b can fill the same tank 7 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | a ' s rate is 1 / 56 and b ' s rate is 1 / 8 . the combined rate is 1 / 56 + 1 / 8 = 1 / 7 the pipes will fill the tank in 7 minutes . the answer is d . | a = 1 / 56
b = 7 / 56
c = a + b
d = 1/(c)
|
a ) 25 , b ) 30 , c ) 35 , d ) 30 , e ) none of these | d | divide(multiply(20, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | a contractor undertakes to built a walls in 50 days . he employs 20 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | 20 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 20 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 20 × 25 × 0.6 / 25 × 0.4 = 30 men answerd | a = 100 - 40
b = a / 100
c = 20 * b
d = 4 / 10
e = c / d
|
a ) 0 , b ) - 1 , c ) 1 , d ) 2 , e ) - 4 | e | add(135, 17) | find the value for m ? 19 ( m + n ) + 17 = 19 ( - m + n ) - 135 | 19 m + 19 n + 17 = - 19 m + 19 n - 135 38 m = - 152 = > m = - 4 e | a = 135 + 17
|
a ) 320 , b ) 345 , c ) 450 , d ) 380 , e ) 400 | c | multiply(divide(720, 5), 3) | there are 720 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 720 . hence 3 x + 5 x = 720 solve for x 8 x = 720 x = 90 number of boys 3 x = 3 × 90 = 270 number of girls 5 x = 5 × 90 = 450 c" | a = 720 / 5
b = a * 3
|
a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 200 | e | multiply(10, multiply(const_2, divide(sqrt(2500), divide(10, const_2)))) | the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 2500 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 2500 sq . units = > side of square = √ 2500 = 50 units the radius of the circle = side of the square = 50 units length of the rectangle = 2 / 5 * 50 = 20 units given that breadth = 10 units area of the rectangle = lb = 20 * 10 = 200 sq . units answer : option e" | a = math.sqrt(2500)
b = 10 / 2
c = a / b
d = 2 * c
e = 10 * d
|
a ) 130 ares . , b ) 160 ares . , c ) 180 ares . , d ) 230 ares . , e ) 250 ares . | b | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 1.6), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | convert 1.6 hectares in ares | "1.6 hectares in ares 1 hectare = 100 ares therefore , 1.6 hectares = 1.6 × 100 ares = 160 ares . answer - b" | a = 3 + 2
b = a * 2
c = 3 + 2
d = c * 2
e = b * d
f = e * 1
g = 3 + 2
h = g * 2
i = 3 + 2
j = i * 2
k = h * j
l = f / k
|
a ) a ) 110 , b ) b ) 122 , c ) c ) 90 , d ) d ) 85 , e ) e ) 75 | e | multiply(15, 5) | the average of 7 numbers is 15 . if each number be multiplied by 5 . find the average of new set of numbers ? | explanation : average of new numbers = 15 * 5 = 75 answer : option e | a = 15 * 5
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a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 11 kmph | e | divide(add(15, 7), const_2) | calculate the speed of a boat in still water ( in km / hr ) if in one hour , the boat goes 15 km / hr downstream and 7 km / hr upstream . | "speed in still water = ( 15 + 7 ) 1 / 2 kmph = 11 kmph . answer : e" | a = 15 + 7
b = a / 2
|
a ) 40 sec , b ) 198 sec , c ) 178 sec , d ) 665 sec , e ) 276 sec | a | divide(400, subtract(multiply(36, const_0_2778), multiply(36, const_0_2778))) | a and b go around a circular track of length 400 m on a cycle at speeds of 36 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 400 / ( 36 * 5 / 18 ) , 400 / ( 36 * 5 / 18 ) } = lcm ( 40 , 40 ) = 40 sec . answer : a" | a = 36 * const_0_2778
b = 36 * const_0_2778
c = a - b
d = 400 / c
|
a ) 108 ° , b ) 10 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | a | divide(multiply(subtract(const_100, add(add(add(add(9, 14), 10), 29), 8)), divide(const_3600, const_10)), const_100) | a circle graph shows how the megatech corporation allocates its research and development budget : 9 % microphotonics ; 14 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ? | "9 % microphotonics ; 14 % home electronics ; 10 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 9 + 14 + 10 + 29 + 8 ) = 30 % basic astrophysics . 30 % of 360 ° is 108 ° . answer : a ." | a = 9 + 14
b = a + 10
c = b + 29
d = c + 8
e = 100 - d
f = 3600 / 10
g = e * f
h = g / 100
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(subtract(multiply(40, divide(20, const_100)), multiply(40, divide(10, const_100))), divide(subtract(const_100, 20), const_100)) | a mixture of 40 liters of milk and water contains 10 % water . how much water should be added to this so that water may be 20 % in the new mixture ? | 4 ltr - water 36 ltr - milk let x water be added then the new mixture must have 20 % water then 4 + x = 20 / 100 ( 40 + x ) x = 5 answer : c | a = 20 / 100
b = 40 * a
c = 10 / 100
d = 40 * c
e = b - d
f = 100 - 20
g = f / 100
h = e / g
|
a ) 0 , b ) 5 , c ) 20 , d ) 15 , e ) 25 | c | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 50), const_1), 50), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 50), const_1), 50), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 50), const_1), 50), 11,0), const_10)), const_10)), const_10) | in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 50 , what is the greatest possible value of b × a ? | "you should notice that 50 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 4 = 20 ( next number wo n ' t have 110 as the first 3 digits so we have only two options 0 and 25 ) . answer : c . ! please post ps questions in the ps subforum : gmat - problem - solving - ps - 140 / please post ds questions in the ds subforum : gmat - data - sufficiency - ds - 141 / no posting of ps / ds questions is allowed in the mainmath forum . c" | a = 11 ** 0
b = a + 1000
c = b / 50
d = c + 1
e = d * 50
f = reminder * (
g = f - 10
|
a ) 16 , b ) 18 , c ) 24 , d ) 40 , e ) 12 | d | divide(multiply(40, 5), add(divide(40, 40), divide(multiply(2, 40), 20))) | a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 5 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 5 x / ( x / 8 ) = 40 kmph answer : d" | a = 40 * 5
b = 40 / 40
c = 2 * 40
d = c / 20
e = b + d
f = a / e
|
a ) 43 kms , b ) 45 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | a | multiply(speed(add(21, 22), const_60), 60) | riya and priya set on a journey . riya moves eastward at a speed of 21 kmph and priya moves westward at a speed of 22 kmph . how far will be priya from riya after 60 minutes | "total eastward distance = 21 kmph * 1 hr = 21 km total westward distance = 22 kmph * 1 hr = 22 km total distn betn them = 21 + 22 = 43 km ans 43 km answer : a" | a = 21 + 22
b = speed * (
|
a ) $ 35.83 , b ) $ 35.84 , c ) $ 35.85 , d ) $ 35.86 , e ) $ 35.87 | c | subtract(subtract(subtract(add(subtract(multiply(subtract(9.5, const_3), 9.5), const_0_25), const_0_25), const_12), const_12), const_2) | a family had dinner in a restaurant and paid $ 30 for food . they also had to pay 9.5 % sale tax and 10 % for the tip . how much did they pay for the dinner ? | solution they paid for food , sales tax and tip , hence total paid = $ 30 + 9.5 % * 30 + 10 % * 30 = $ 35.85 answer c | a = 9 - 5
b = a * 9
c = b - const_0_25
d = c + const_0_25
e = d - 12
f = e - 12
g = f - 2
|
a ) 78 years , b ) 22 years , c ) 88 years , d ) 66 years , e ) 24 years | e | divide(subtract(26, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 26 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "let ' s son age is x , then father age is x + 26 . = > 2 ( x + 2 ) = ( x + 26 + 2 ) = > 2 x + 4 = x + 28 = > x = 24 years answer : e" | a = 2 * 2
b = a - 2
c = 26 - b
d = 2 - 1
e = c / d
|
a ) 47.14 , b ) 45.15 , c ) 43.23 , d ) 41.44 , e ) 41.51 | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(40, 20)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 40 cm and breadth 20 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 40 + 20 ) = 120 cm parameter of the square = 120 cm i . e . 4 a = 120 a = 30 diameter of the semicircle = 30 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 30 ) = 1 / 2 ( 22 / 7 ) ( 30 ) = 660 / 14 = 47.14 cm to two decimal places answer : a" | a = square_edge_by_perimeter / (
b = circumface / (
|
a ) 18 , b ) 10.9 , c ) 10.4 , d ) 10.8 , e ) 10.1 | a | divide(add(200, 300), multiply(add(60, 40), const_0_2778)) | two trains 200 m and 300 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 200 + 300 = 500 m . required time = 500 * 9 / 250 = 18 secs answer : a" | a = 200 + 300
b = 60 + 40
c = b * const_0_2778
d = a / c
|
a ) 17 litres , b ) 23.3 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(35, 1399.45), multiply(35, 262.85)), subtract(3104.35, 1399.45)) | the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 35 litres of the mixed fruit juice . how many litres of the a ç ai berry juice does he need to add ? | "262.85 ( 35 ) + 3 , 104.35 x = 1 , 399.45 ( 35 + x ) solve the equation . 262.85 ( 35 ) + 3 , 104.35 x = 1 , 399.45 ( 35 + x ) 9 , 199.75 + 3 , 104.35 x = 48980.75 + 1 , 399.45 x 9 , 199.75 + 1 , 704.9 x = 48980.75 1 , 704.9 x = 39,781 x ≈ 23.3 answer is b ." | a = 35 * 1399
b = 35 * 262
c = a - b
d = 3104 - 35
e = c / d
|
a ) 2 , b ) 6 , c ) 3 , d ) 4 , e ) 1 | d | subtract(8, reminder(5, 6)) | when n is divided by 15 , the remainder is 8 . what is the remainder when 5 n is divided by 6 ? | "let n = 8 ( leaves a remainder of 8 when divided by 48 ) 5 n = 5 ( 8 ) = 40 , which leaves a remainder of 4 when divided by 6 . answer d" | a = 8 - reminder
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a ) 12.5 % , b ) 87.5 % , c ) 80 % , d ) 94.4 % , e ) 1 % | d | multiply(divide(subtract(18, const_1), 18), const_100) | a number x is 18 times another number y . the percentage that y is less than x is | "say y = 1 and x = 18 . then y = 1 is less than x = 18 by ( 18 - 1 ) / 18 * 100 = 17 / 18 * 100 = 94.4 % . answer : d ." | a = 18 - 1
b = a / 18
c = b * 100
|
a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 5 % | b | multiply(divide(multiply(subtract(add(add(10, multiply(divide(3, const_100), const_10)), add(10, multiply(divide(6, const_100), 10))), add(10, 10)), const_100), add(10, 10)), const_2) | a dozen eggs and 5 pounds of oranges are currently at the same price . if the price of a dozen eggs rises by 3 percent and the price of oranges rises by 6 % . how much more will it cost to buy a dozen eggs and 10 pounds of oranges . | say currently both a dozen eggs and 5 pounds of oranges cost $ 100 ( they are at the same price ) . so , to buy a dozen eggs and 5 pounds of oranges we need $ 100 . after the increase , the price of a dozen eggs will be $ 103 and the price of 5 pounds of oranges will be $ 106 . so after the increase , to buy a dozen eggs and 5 pounds of oranges we ' ll need $ 209 . increase = 9 % . answer : b . | a = 3 / 100
b = a * 10
c = 10 + b
d = 6 / 100
e = d * 10
f = 10 + e
g = c + f
h = 10 + 10
i = g - h
j = i * 100
k = 10 + 10
l = j / k
m = l * 2
|
a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | a number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a . what is the greatest two digit prime saturated integer ? | "clearly d a number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a . 96 has more number of smaller prime factor thats the clue ! ! = d" | a = 2 * 3
b = 100 - 10
c = a + b
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a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 6072 | e | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.60) | the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.60 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.60 ) = rs . 6072 . answer : e ) 6072" | a = 13 / 86
b = math.sqrt(a)
c = b * 100
d = circumface * (
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(multiply(add(6, const_1), 19), add(multiply(6, 19), multiply(6, 3))) | 3 years ago the average age of a class of 6 members was 19 years . a boy have been added , the average age of the class is the same today . what is the age of the boy ? | 6 * 22 = 132 7 * 19 = 133 ` 1 a | a = 6 + 1
b = a * 19
c = 6 * 19
d = 6 * 3
e = c + d
f = b - e
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a ) 237 , b ) 270 , c ) 177 , d ) 285.7 , e ) 111 | d | floor(divide(8200, add(28, divide(2.5, const_100)))) | find the number of shares that can be bought for rs . 8200 if the market value is rs . 28 each with brokerage being 2.5 % . | "explanation : cost of each share = ( 28 + 2.5 % of 28 ) = rs . 28.7 therefore , number of shares = 8200 / 28.7 = 285.7 answer : d" | a = 2 / 5
b = 28 + a
c = 8200 / b
d = math.floor(c)
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a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | d | multiply(sqrt(subtract(multiply(29, const_3.0), multiply(3, 3))), 2) | if ( x + 3 ) ^ 2 / ( 3 x + 29 ) = 2 , then the difference between the two possible values of x is : | "( x + 3 ) ^ 2 / ( 3 x + 29 ) = 2 ( x + 3 ) ^ 2 = 2 ( 3 x + 29 ) x ^ 2 + 6 x + 9 = 6 x + 58 x ^ 2 - 49 = 0 ( x - 7 ) ( x + 7 ) = 0 x = 7 or x = - 7 the answer is d ." | a = 29 * 3
b = 3 * 3
c = a - b
d = math.sqrt(c)
e = d * 2
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a ) 2898 , b ) 277 , c ) 500 , d ) 297 , e ) 590 | e | divide(710, multiply(subtract(78, 1), const_0_2778)) | a train 710 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 710 + x ) / 60 = 65 / 3 x = 590 m . answer : e" | a = 78 - 1
b = a * const_0_2778
c = 710 / b
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a ) 1 / 20 , b ) 1 / 10 , c ) 1 / 5 , d ) 2 / 5 , e ) 3 / 4 | d | add(divide(multiply(subtract(const_1, divide(40, multiply(20, 20))), 100), 400), divide(add(multiply(divide(20, multiply(20, 20)), 300), multiply(divide(40, multiply(20, 20)), 100)), 400)) | in a certain corporation , there are 300 male employees and 100 female employees . it is known that 20 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 400 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | "in this corporation , there are 400 total employees . there are 100 women . of the 300 men , 20 % have advanced degrees — - 10 % of 300 must be 30 , so 20 % of 300 must be 60 . add the women and the men with advanced degrees : 100 + 60 = 160 . this is the or region , full set of individuals that satisfy the condition “ has an advanced degree or is female . ” of the 400 employees , what ’ s the probability of picking one of the 160 in this particular group ? p = 160 / 400 = 16 / 40 = 4 / 10 = 2 / 5 answer = d" | a = 20 * 20
b = 40 / a
c = 1 - b
d = c * 100
e = d / 400
f = 20 * 20
g = 20 / f
h = g * 300
i = 20 * 20
j = 40 / i
k = j * 100
l = h + k
m = l / 400
n = e + m
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a ) 850 , b ) 900 , c ) 750 , d ) 950 , e ) 800 | b | add(multiply(divide(1350, add(45, multiply(60, divide(24, 28)))), 30), multiply(multiply(divide(1350, add(45, multiply(60, divide(24, 28)))), divide(24, 28)), 40)) | the price of 24 apples is equal to that of 28 oranges . the price of 45 apples and 60 oranges together is rs . 1350 . the total price of 30 apples and 40 oranges is | let the price of one apple = a and price of one orange = b price of 24 apples is equal to that of 28 oranges 24 a = 28 b = > 6 a = 7 b ⇒ b = 6 a / 7 - - - - - ( equation 1 ) price of 45 apples and 60 oranges together is rs . 1350 = > 45 a + 60 b = 1350 = > 3 a + 4 b = 90 ⇒ 3 a + 4 ( 6 a ) / 7 = 90 ( ∵ substituted the value of b from equation 1 ) ⇒ a + 4 ( 2 a ) / 7 = 30 = > 7 a + 8 a = 30 × 7 = 210 = > 15 a = 210 ⇒ a = 21015 = 423 = 14 b = 6 a 7 = ( 6 × 14 ) / 7 = 6 × 2 = 12 total price of 30 apples and 40 oranges = 30 a + 45 b = ( 30 × 14 ) + ( 40 × 12 ) = 420 + 480 = 900 answer is b . | a = 24 / 28
b = 60 * a
c = 45 + b
d = 1350 / c
e = d * 30
f = 24 / 28
g = 60 * f
h = 45 + g
i = 1350 / h
j = 24 / 28
k = i * j
l = k * 40
m = e + l
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a ) 5 / 32 , b ) 6 / 32 , c ) 5 / 16 , d ) 4 / 32 , e ) 8 / 32 | a | divide(choose(5, 4), power(const_2, 5)) | rhoda tosses a coin 5 times . find the probability of getting exactly 4 heads . | exactly 4 heads out of 5 would be : hhhht , hhthh , or many other combinations . therefore , no . of possible combinations are 5 ! / 4 ! * 1 ! = 5 probability = 5 / 2 ^ 5 = 5 / 32 answer : a | a = math.comb(5, 4)
b = 2 ** 5
c = a / b
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a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) r = 1 / 6 | e | multiply(divide(const_1, 2), divide(const_1, const_3)) | if two of the 4 expressions x + y , x + 5 y , x - y , and 5 x - y are chosen at random , what is the probability r that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | only ( x + y ) ( x - y ) pair will give the form x ^ 2 - ( by ) ^ 2 the probability of selecting these two pairs are 1 / 4 * 1 / 3 = 1 / 12 , assuming x + y is picked first , then x - y , but x - y can be picked first followed by x + y . so the probability r = 1 / 12 * 2 = 1 / 6 ans e | a = 1 / 2
b = 1 / 3
c = a * b
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a ) 87 , b ) 18 , c ) 17 , d ) 16 , e ) 30 | e | divide(divide(1600, 20), divide(1400, multiply(25, 21))) | 25 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 25 1400 21 x 1600 20 x / 25 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 30 answer : e" | a = 1600 / 20
b = 25 * 21
c = 1400 / b
d = a / c
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a ) rs . 9621 , b ) rs . 6921 , c ) rs . 9261 , d ) rs . 6261 , e ) rs . 6361 | c | multiply(8000, power(add(const_1, divide(5, const_100)), 3)) | the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 3 years is | "c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) ^ 3 = 8000 ( 21 / 20 * 21 / 20 * 21 / 20 ) = rs 9261 answer : c" | a = 5 / 100
b = 1 + a
c = b ** 3
d = 8000 * c
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a ) 45 , b ) 30 , c ) 24 , d ) 40 , e ) none of these | a | divide(multiply(divide(add(multiply(18, 3), multiply(18, 2)), 3), 3), 2) | the ratio of the capacity to do work of a and b is 3 : 2 . if they together can complete a work in 18 days , then how long does a take to complete the work alone ? | llet a and b take 3 x and 2 x days to complete the work 1 / 3 x + 1 / 2 x = 1 / 18 ⇒ x = 15 so a will take 45 days . answer : a | a = 18 * 3
b = 18 * 2
c = a + b
d = c / 3
e = d * 3
f = e / 2
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a ) 40 , b ) 77 , c ) 48 , d ) 52 , e ) none of these | b | add(add(multiply(5, divide(14, subtract(multiply(divide(6, 5), 5), 4))), divide(14, subtract(multiply(divide(6, 5), 5), 4))), 14) | in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 14 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 14 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 14 = 4 / 9 ( t - 14 ) 10 t = 1386 - 616 = > t = 77 . answer : b" | a = 6 / 5
b = a * 5
c = b - 4
d = 14 / c
e = 5 * d
f = 6 / 5
g = f * 5
h = g - 4
i = 14 / h
j = e + i
k = j + 14
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a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) 7 % | e | multiply(divide(divide(subtract(add(multiply(multiply(const_1000, const_100), 2), 97500), add(multiply(const_1000, const_100), 75000)), const_10), add(multiply(const_1000, const_100), 75000)), const_100) | the population of a town increased from 1 , 75000 to 2 , 97500 in a decade . the average percent increase of population per year is | solution increase in 10 years = ( 297500 - 175000 ) = 122500 increase % = ( 122500 / 175000 ã — 100 ) % = 70 % . required average = ( 70 / 10 ) % = 7 % . answer e | a = 1000 * 100
b = a * 2
c = b + 97500
d = 1000 * 100
e = d + 75000
f = c - e
g = f / 10
h = 1000 * 100
i = h + 75000
j = g / i
k = j * 100
|
a ) 480 , b ) 500 , c ) 520 , d ) 460 , e ) 440 | c | add(divide(multiply(divide(50, const_100), 20), subtract(divide(40, const_100), divide(50, const_100))), 20) | the workforce of company x is 40 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 50 % . how many employees did the company have after hiring the additional male workers ? | "let x be the total worker then 0.4 x = female worker and 0.6 x is male worker then 20 male worker added 04 x / ( 0.6 x + 20 ) = 50 / 100 or 40 x = 50 * ( 0.6 x + 100 ) = 30 x + 5000 or 10 x = 5000 , x = 500 total worker = 500 + 20 = 520 c" | a = 50 / 100
b = a * 20
c = 40 / 100
d = 50 / 100
e = c - d
f = b / e
g = f + 20
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a ) 5 : 6 , b ) 7 : 8 , c ) 8 : 7 , d ) 12 : 11 , e ) 14 : 13 | c | divide(8, add(6, const_1)) | the ratio of kamla and nisha is 6 : 5 and the sum of their ages is 44 . what will be the ratio of their ages after 8 years . | answer : let age of kamla is 6 x and age of nisha is 5 x . then , 6 x + 5 x = 4411 x = 44 , x = 4 after 8 year the ratio of their ages = 6 x + 8 / 5 x + 8 = 8 : 7 answer c | a = 6 + 1
b = 8 / a
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a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(divide(divide(1, 3), divide(4, const_6)), const_6) | two identical cubes if one of them is painted pink on its 4 side and blue on the remaining two side then how many faces painted pink to other cube so that probability of getting cube is 1 / 3 when we roll both the cube . | 0 faces should be painted pink i . e all faces blue answer : a | a = 1 / 3
b = 4 / 6
c = a / b
d = c * 6
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a ) a ) 11.79 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | a | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 25), divide(9, 55))), const_4) | due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 25 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 9 miles stretch = 9 * 60 / 55 = 9 * 12 / 11 = 9.81 new time in minutes to cross 9 miles stretch = 9 * 60 / 25 = 9 * 12 / 5 = 21.6 time difference = 11.79 ans : a" | a = 55 + 9
b = a - 1
c = 9 / 25
d = 9 / 55
e = c - d
f = b * e
g = max(f)
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a ) 17 % , b ) 12 % , c ) 22 % , d ) 24 % , e ) 21 | a | multiply(divide(subtract(2400, 1980), 2400), const_100) | the cost price of a radio is rs . 2400 and it was sold for rs . 1980 , find the loss % ? | "2400 - - - - 420 100 - - - - ? = > 17 % answer : a" | a = 2400 - 1980
b = a / 2400
c = b * 100
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a ) $ 100 , b ) $ 104 , c ) $ 108 , d ) $ 112 , e ) $ 116 | a | divide(multiply(divide(multiply(132, const_100), add(const_100, 20)), const_100), add(const_100, 10)) | a couple spent $ 132 in total while dining out and paid this amount using a credit card . the $ 132 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? | "let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 132 x = 100 the correct answer is a ." | a = 132 * 100
b = 100 + 20
c = a / b
d = c * 100
e = 100 + 10
f = d / e
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a ) 28 , b ) 19 , c ) 22 , d ) 20 , e ) 21 | a | add(7, divide(multiply(7, subtract(14000, 8000)), subtract(8000, 6000))) | the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 14000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 14000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 56000 = x = 28 . answer : a" | a = 14000 - 8000
b = 7 * a
c = 8000 - 6000
d = b / c
e = 7 + d
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a ) 10 , b ) 36 , c ) 38 , d ) 40 , e ) 42 | a | divide(subtract(52, subtract(power(multiply(3, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 3 inches wide on each side , the total area would have been ( m + 52 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 2 ) ( b + 2 ) = m - - - - ( 2 ) and ( l + 6 ) ( b + 6 ) = m + 52 - - - - - - - - - > ( l + 6 ) ( b + 6 ) - 52 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 2 ) ( b + 2 ) = ( l + 6 ) ( b + 6 ) - 52 lb + 2 l + 2 b + 4 = lb + 6 l + 6 b + 36 - 52 simplify 4 l + 4 b = 20 - - - - - - > 2 ( l + b ) = 10 ( check eq ( 1 ) ) answer is a" | a = 3 * 2
b = a ** 2
c = 1 * 2
d = c ** 2
e = b - d
f = 52 - e
g = f / 2
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a ) 22.86 , b ) 23.86 , c ) 25 , d ) 26 , e ) 27 | a | multiply(divide(30, const_100), 30) | uncle bruce is baking chocolate chip cookies . he has 50 ounces of dough ( with no chocolate ) and 40 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 30 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 70 % * total = 40 = > ( 7 / 10 ) total = 40 = > total = 400 / 7 = > total = 57.14 oz , from there , you must find 30 % of the total 57.14 oz of the mixture . 30 % * total = > ( 3 / 10 ) ( 57.14 ) = 17.14 oz choclate used , not forgetting that the question asks how much chocolate is left over we must subtract the chocolate used from the initial chocolate . 40 - 17.14 = 22.86 oz chocolate left over . answer : a" | a = 30 / 100
b = a * 30
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a ) 86.5 kg , b ) 62.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg | b | divide(add(64, add(61, const_1)), const_2) | in arun ' s opinion , his weight is greater than 61 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 64 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ? | let arun ' s weight be x kg . according to arun , 61 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 64 . the values satisfying all the above conditions are 62 and 63 . required average = ( 62 + 63 ) / 2 = 62.5 kg answer : b | a = 61 + 1
b = 64 + a
c = b / 2
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a ) $ 400 , b ) $ 489 , c ) $ 484 , d ) $ 365 , e ) % 342 | c | divide(multiply(multiply(1210, const_2), const_2), const_10) | a and b together have $ 1210 . if of a ' s amount is equal to of b ' s amount , how much amount does b have ? | ( 4 / 15 ) a = ( 2 / 5 ) b = > a = ( ( 2 / 5 ) * ( 15 / 4 ) ) b = > a = ( 3 / 2 ) b = > a / b = 3 / 2 a : b = 3 : 2 b ' s share = rs . ( 1210 * ( 2 / 5 ) ) = $ 484 option c | a = 1210 * 2
b = a * 2
c = b / 10
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a ) 20 % , b ) 28 % , c ) 60 % , d ) 26 % , e ) 30 % | a | subtract(divide(subtract(const_100, 20), divide(2, 3)), const_100) | what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 20 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 80 2 / 3 sp 1 = 80 sp 1 = 120 100 - - - 20 = > 20 % answer : a" | a = 100 - 20
b = 2 / 3
c = a / b
d = c - 100
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a ) - 3 , b ) 3 , c ) 5 , d ) 7 , e ) 11 | e | add(6, 5) | what is the minimum value of | x - 4 | + | x + 6 | + | x - 5 | ? | a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 6 | > = 0 - - > minimum occurs at x = - 6 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 6 - - > result = 10 + 0 + 11 = 21 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 10 + 1 = 11 x = 5 - - > result = 1 + 11 + 0 = 12 x = 6 - - > result = 2 + 12 + 1 = 15 so minimum value of the expression occurs at x = 4 and the resultant value = 11 answer : e | a = 6 + 5
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a ) 44648 , b ) 27844 , c ) 28642 , d ) 16864 , e ) 32458 | c | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 7) | how many five - digit numbers that do not contain the digits 4 or 7 are there ? | "we can have 7 digits ( 1 , 2,3 , 5,6 , 8,9 ) for the first place ( ten thousand ' s place ) . and similarly 8 digits for thousand ' s , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 5,6 , 8,9 ) so in total 7 * 8 * 8 * 8 * 8 = 28672 hence c" | a = 10 * 10
b = 10 - 1
c = a * b
d = c * 7
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a ) 90 ° , b ) 75 ° , c ) 60 ° , d ) 45 ° , e ) 30 ° | d | divide(90, const_2) | the larger interior angle of a parallelogram is 90 ° more than its smaller interior angle . what is the measure of the smaller interior angle of the parallelogram ? | let smaller be x , larger becomes x + 90 . . . total of all angles = 360 so 4 x + 180 = 360 and x = 45 answer : d | a = 90 / 2
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a ) 72 seconds , b ) 27 seconds , c ) 40 seconds , d ) 128 seconds , e ) 18 seconds | a | divide(add(400, 800), divide(multiply(60, const_1000), const_3600)) | a train is 400 meter long is running at a speed of 60 km / hour . in what time will it pass a bridge of 800 meter length ? | "speed = 60 km / hr = 60 * ( 5 / 18 ) m / sec = 150 / 9 m / sec total distance = 400 + 800 = 1200 meter time = distance / speed = 1200 * ( 9 / 150 ) = 72 seconds answer : a" | a = 400 + 800
b = 60 * 1000
c = b / 3600
d = a / c
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a ) rs . 5000 , b ) rs . 25000 , c ) rs . 7500 , d ) rs . 8500 , e ) rs . 9500 | a | divide(multiply(200, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100))))) | a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 200 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 – 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 200 = > 5000 answer : a" | a = 200 * 100
b = 100 - 20
c = 100 - 20
d = 20 / 100
e = c * d
f = b + e
g = 100 - f
h = a / g
|
a ) 40 cm , b ) 20 cm , c ) 30 cm , d ) 25 cm , e ) 35 cm | b | divide(add(multiply(20, 20), multiply(11, 20)), 31) | the average height of 20 students is 20 cm and the average height of 11 students is 20 cm . what is the average height of the 31 students ? | the total height of the 31 students = ( 20 * 20 ) + ( 11 * 20 ) = 400 + 220 = 620 cm the required average height = 620 / 31 = 20 cm answer : b | a = 20 * 20
b = 11 * 20
c = a + b
d = c / 31
|
a ) 1 / 3 , b ) 2 / 3 , c ) 5 / 7 , d ) 7 / 5 , e ) 3 / 2 | d | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_2, const_3)), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))), subtract(divide(const_1, const_2), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))) | a certain ball team has an equal number of right - and left - handed players . on a certain day , two - thirds of the players were absent from practice . of the players at practice that day , two - third were left handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of lefthanded players who were not at practice ? | say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 12 absent and 6 present . of the players at practice that day , one - third were left - handed - - > 6 * 2 / 3 = 4 were left - handed and 2 right - handed . the number of right - handed players who were not at practice that day is 9 - 2 = 7 . the number of left - handed players who were not at practice that days is 9 - 4 = 5 . the ratio = 7 / 5 . answer : d | a = 1 / 2
b = 2 / 3
c = 1 - b
d = 2 / 3
e = 1 - d
f = 2 / 3
g = e * f
h = c - g
i = a - h
j = 1 / 2
k = 2 / 3
l = 1 - k
m = 2 / 3
n = l * m
o = j - n
p = i / o
|
a ) 22 , b ) 15 , c ) 77 , d ) 28 , e ) 91 | b | subtract(const_60, multiply(divide(48, 64), const_60)) | excluding the stoppages , the speed of a bus is 64 km / hr and including the stoppages the speed o the bus is 48 km / hr . for how many minutes does the bus stop per hour ? | answer : b ) 15 minutes | a = 48 / 64
b = a * const_60
c = const_60 - b
|
a ) $ 3420 , b ) $ 3630 , c ) $ 3870 , d ) $ 4040 , e ) $ 4220 | b | multiply(3000, power(add(const_1, divide(10, const_100)), const_2)) | what amount does an investor receive if the investor invests $ 3000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 5000 = 1.21 * 5000 = 3630 the answer is b ." | a = 10 / 100
b = 1 + a
c = b ** 2
d = 3000 * c
|
a ) rs 8.82 , b ) rs 9.82 , c ) rs 10.82 , d ) rs 14.82 , e ) none of these | d | divide(multiply(12, add(const_100, 5)), subtract(const_100, 15)) | a fruit seller sells mangoes at the rate of rs . 12 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 85 : 12 = 105 : x x = ( 12 × 105 / 85 ) = rs 14.82 option d" | a = 100 + 5
b = 12 * a
c = 100 - 15
d = b / c
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a ) 237 , b ) 126 , c ) 971 , d ) 611 , e ) 115 | e | divide(divide(225, divide(add(const_100, 45), const_100)), divide(add(35, const_100), const_100)) | a sells a bicycle to b and makes a profit of 35 % . b sells the same bicycle to c at a profit of 45 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a . | "explanation : let cp be 100 a sells at 35 % profit so sp = 135 b sells at 45 % profit = 135 x ( 1 + 45 / 100 ) = 195.75 cp sp 100 - - - 195.75 x - - - 225 cp = 225 x 100 / 195.75 = 114.94 answer : e" | a = 100 + 45
b = a / 100
c = 225 / b
d = 35 + 100
e = d / 100
f = c / e
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a ) 10 , b ) 4 , c ) 4 , d ) 3 , e ) 2 | e | add(add(divide(10, add(const_4, const_1)), divide(subtract(10, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(10, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | how many zeros does 10 ! end with ? | "according to above 10 ! has 10 / 5 = 2 = 2 trailing zeros . answer : e" | a = 4 + 1
b = 10 / a
c = 4 + 1
d = 10 - c
e = 4 + 1
f = e ** 2
g = d / f
h = b + g
i = 4 + 1
j = 10 - i
k = 4 + 1
l = k ** 3
m = j / l
n = h + m
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a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 432 | e | divide(multiply(power(12, 2), power(6, 4)), 432) | find the value of y from ( 12 ) ^ 2 x 6 ^ 4 ÷ 432 = y ? | 432 e | a = 12 ** 2
b = 6 ** 4
c = a * b
d = c / 432
|
a ) rs . 40 , b ) rs . 50 , c ) rs . 60 , d ) rs . 70 , e ) none of these | c | multiply(multiply(multiply(add(multiply(multiply(multiply(5, 19), const_100), const_100), multiply(multiply(multiply(19, 19), const_100), multiply(add(19, 5), 5))), divide(add(multiply(6, 19), 5), 19)), divide(multiply(19, 19), multiply(5, multiply(5, 19)))), divide(const_1, const_100)) | find the simple interest on rs . 5000 at 6 % per annum for the period from 5 th feb to 19 th april , 2015 . | "explanation : given : 1 ) principal = rs . 5000 2 ) rate of interest = 6 % 3 ) time = 5 th feb to 19 th april , 2015 first find the time period 5 th feb to 19 th april , 2015 feb = 28 – 5 = 23 days march = 31 days april = 19 days total days = 23 + 31 + 19 = 73 days convert days into years , by dividing it by 365 time = 73 / 365 = 1 / 5 simple interest = ( p × r × t ) / 100 = [ 5000 × 6 × ( 1 / 5 ) ] / 100 = rs . 60 simple interest = rs . 60 answer is c" | a = 5 * 19
b = a * 100
c = b * 100
d = 19 * 19
e = d * 100
f = 19 + 5
g = f * 5
h = e * g
i = c + h
j = 6 * 19
k = j + 5
l = k / 19
m = i * l
n = 19 * 19
o = 5 * 19
p = 5 * o
q = n / p
r = m * q
s = 1 / 100
t = r * s
|
a ) 160 , b ) 240 , c ) 288 , d ) 277 , e ) 221 | a | multiply(20, multiply(36, const_0_2778)) | a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 36 km / hr . what is the length of the platform ? | "speed = 36 * 5 / 18 = 10 m / sec . length of the train = 10 * 20 = 200 m . let the length of the platform be x m . then , ( x + 200 ) / 36 = 10 = > x = 160 m . answer : a" | a = 36 * const_0_2778
b = 20 * a
|
a ) 11 : 6 , b ) 7 : 13 , c ) 13 : 7 , d ) 15 : 6 , e ) 13 : 6 | b | divide(multiply(42, 6), multiply(52, 9)) | car a runs at the speed of 42 km / hr & reaches its destination in 6 hr . car b runs at the speed of 52 km / h & reaches its destination in 9 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 42 × 6 = 252 km distance travelled by car b = 52 × 9 = 468 km ratio = 252 / 468 = 7 : 13 b" | a = 42 * 6
b = 52 * 9
c = a / b
|
a ) 72.5 kmph , b ) 75 kmph , c ) 87 kmph , d ) 56 kmph , e ) 86 kmph | a | divide(add(90, 55), const_2) | the speed of a car is 90 km in the first hour and 55 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 55 ) / 2 = 72.5 kmph answer : a" | a = 90 + 55
b = a / 2
|
a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | b | multiply(divide(const_1, multiply(add(const_100, 25), divide(const_1, subtract(const_100, 25)))), 15) | by selling 15 pencils for a rupee a man loses 25 % . how many for a rupee should he sell in order to gain 25 % ? | "75 % - - - 15 125 % - - - ? 75 / 125 * 15 = 9 answer : b" | a = 100 + 25
b = 100 - 25
c = 1 / b
d = a * c
e = 1 / d
f = e * 15
|
a ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 142,800 , e ) 168,000 | d | multiply(const_4, const_10) | a certain machine produces 850 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 7 days ? | "since 7 days consist of 24 * 7 hours the total hours of a week is 168 hours . since every hour the machine produces 850 units of product p the total product during 168 hours is 850 * 168 = 142,800 . correct option : d" | a = 4 * 10
|
a ) 4457 , b ) 4567 , c ) 4235 , d ) 4547 , e ) 7429 | e | multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 5) | the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 5 per metre approximately | "explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 5 ) = rs . 7429 . answer : option e" | a = 17 / 56
b = math.sqrt(a)
c = b * 100
d = circumface * (
|
a ) 105 , b ) 30 , c ) 35 , d ) 20 , e ) none of these | a | divide(multiply(70, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | a contractor undertakes to built a walls in 50 days . he employs 70 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | "70 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 70 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 70 × 25 × 0.6 / 25 × 0.4 = 105 men answera" | a = 100 - 40
b = a / 100
c = 70 * b
d = 4 / 10
e = c / d
|
a ) 99 , b ) 98 , c ) 97 , d ) 96 answer , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | a number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n . what is the greatest two digit prime saturated integer ? | "the square roots of all numbers from answer choices are between 9 and 10 , so the product of primes of the number we are looking should be less then 9 , so this number should have only 2 - s , only 3 - s , only 5 - s , or only 2 - s and 3 - s as its primes . only 96 satisfies this . answer : d ." | a = 2 * 3
b = 100 - 10
c = a + b
|
a ) 10 , b ) 15 , c ) 35 , d ) 1 , e ) 45 | d | divide(divide(40, 10), 4) | in how many no . between 10 and 40 exactly two of the digits is 4 ? | "it ' s simple can be solved by elimination of answer choices . option b and c are too large , not possible . even ae are large to have correct choice . ans : d" | a = 40 / 10
b = a / 4
|
a ) 1628.4 , b ) 110 , c ) 1492 , d ) 1496 , e ) none of these | b | multiply(divide(add(multiply(8, 10), multiply(subtract(10, 8), 4)), subtract(10, subtract(10, 8))), 10) | 10 people went to a hotel for combine dinner party 8 of them spent rs . 10 each on their dinner and rest spent 4 more than the average expenditure of all the 10 . what was the total money spent by them . | "solution : let average expenditure of 10 people be x . then , 10 x = 8 * 10 + 2 * ( x + 4 ) ; or , 10 x = 10 * 10 + 2 x + 8 ; or , x = 11 ; so , total money spent = 11 * 10 = rs . 110 . answer : option b" | a = 8 * 10
b = 10 - 8
c = b * 4
d = a + c
e = 10 - 8
f = 10 - e
g = d / f
h = g * 10
|
a ) 4.5 , b ) 7.5 , c ) 9.5 , d ) 17.5 , e ) 24.5 | d | divide(multiply(5, add(1, 6)), const_2) | a carpenter worked alone for 1 day on a job that would take him 6 more days to finish . he and another carpenter completed the job in 5 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | a carpenter worked only 1 day on something that takes him 6 more days . means ; carpenter finishes his work in 7 days . let his buddy finish the same task in x days . respective rates per day : 1 / 7 and 1 / x to complete 1 work : first guy worked for 5 days @ rate = 1 / 7 per day . second one worked for 5 days @ rate = 1 / x per day expression : days * rate = work 5 * 1 / 7 + 5 * 1 / x = 1 5 x + 35 = 7 x 2 x = 35 x = 17.5 days . ans : d | a = 1 + 6
b = 5 * a
c = b / 2
|
a ) a ) 1080 , b ) b ) 10080 , c ) c ) 10025 , d ) d ) 11080 , e ) e ) 12080 | b | add(divide(divide(multiply(floor(add(divide(subtract(multiply(const_10, const_1000), const_1), 18), const_1)), 18), const_4), const_100), const_2) | find the least number of five digits which is exactly divisible by 12 and 18 ? | "the smallest five digit numbers are 10025 , 10080,11080 10025 is not divisible by 12 10080 is divisible by both 12 and 18 answer : b" | a = 10 * 1000
b = a - 1
c = b / 18
d = c + 1
e = math.floor(d)
f = e * 18
g = f / 4
h = g / 100
i = h + 2
|
a ) 20 , b ) 1 , c ) 2 , d ) 16 , e ) 17 | a | add(subtract(75, const_100), const_4) | how many odd prime numbers are there less than 75 ? | "odd prime number less than 75 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 there is 20 the odd prime number answer is a" | a = 75 - 100
b = a + 4
|
a ) 23 days , b ) 25 days , c ) 21 days , d ) 26 days , e ) 29 days | c | multiply(divide(multiply(7, add(const_2, const_1)), const_2), const_2) | a is twice as good a workman as b and they took 7 days together to do the work b alone can do it in . | "c 21 days wc = 2 : 1 2 x + x = 1 / 7 x = 1 / 21 = > 21 days" | a = 2 + 1
b = 7 * a
c = b / 2
d = c * 2
|
a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | if x is a positive integer with fewer than 3 digits , what is the probability that c x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability c that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) ." | a = 100 / 4
b = a * 2
c = 100 / 5
d = c * 2
e = b + d
f = e - 4
g = 100 - 1
h = f / g
|
a ) 35 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 33 % | d | divide(const_100, divide(50, subtract(65, 50))) | if the cost price of 65 chocolates is equal to the selling price of 50 chocolates , the gain percent is : | "explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 50 chocolates = rs . 50 ; s . p . of 50 chocolates = rs . 65 . . ' . gain % = 15 * 100 / 50 = 30 % answer : d" | a = 65 - 50
b = 50 / a
c = 100 / b
|
a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | d | subtract(multiply(multiply(45, const_0_2778), 30), 155) | a train , 155 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 155 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 155 + x ) / 30 m / s = > 45 ã — ( 10 / 36 ) = ( 155 + x ) / 30 = > 45 ã — 10 ã — 30 / 36 = 155 + x = > 45 ã — 10 ã — 10 / 12 = 155 + x = > 15 ã — 10 ã — 10 / 4 = 155 + x = > 15 ã — 25 = 155 + x = 375 = > x = 375 - 155 = 220 answer : option d" | a = 45 * const_0_2778
b = a * 30
c = b - 155
|
a ) 24.58 , b ) 24.5 , c ) 24.3 , d ) 24.9 , e ) 24.1 | b | multiply(divide(7, subtract(9, 7)), 7) | sachin is younger than rahul by 7 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 7 , so ( x - 7 ) / x = 7 / 9 = > 9 x - 63 = 7 x = > 2 x = 63 = > x = 31.5 so sachin age is 31.5 - 7 = 24.5 answer : b" | a = 9 - 7
b = 7 / a
c = b * 7
|
a ) 512 , b ) 615 , c ) 810 , d ) 413 , e ) 123 | d | add(500, divide(multiply(500, 10), const_100)) | the present population of a town is 500 . population increase rate is 10 % p . a . find the population of town before 2 years ? | "p = 500 r = 10 % required population of town = p / ( 1 + r / 100 ) ^ t = 500 / ( 1 + 10 / 100 ) ^ 2 = 500 / ( 11 / 10 ) ^ 2 = 413 ( approximately ) answer is d" | a = 500 * 10
b = a / 100
c = 500 + b
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(add(const_4, const_2), const_1) | a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 5,000 and at the same time n is a perfect square and a perfect cube ? | "if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 . the answer is c ." | a = 4 + 2
b = a / 1
|
a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | c | add(2001, divide(add(divide(90, const_100), subtract(4.4, 4.2)), subtract(divide(30, const_100), subtract(4.4, 4.2)))) | the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 90 cents more than the commodity y ? | the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 90 = $ 1.10 cents on commodity y , which will take 11 years . the answer is c . | a = 90 / 100
b = 4 - 4
c = a + b
d = 30 / 100
e = 4 - 4
f = d - e
g = c / f
h = 2001 + g
|
a ) 1000 , b ) 1,250 , c ) 2,250 , d ) 3,250 , e ) 4,250 | d | multiply(divide(multiply(divide(650, 2), divide(650, 2)), subtract(725, 650)), 2) | shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 650 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 725 as interest . what was the value of his total savings before investing in these two bonds ? | so , we know that shawn received 20 % of the amount he invested in a year . we also know that in one year shawn received $ 325 , thus 0.2 x = $ 325 - - > x = $ 1625 since , he invested equal sums in his 2 bonds , then his total savings before investing was 2 * $ 1,625 = $ 3,250 answer d | a = 650 / 2
b = 650 / 2
c = a * b
d = 725 - 650
e = c / d
f = e * 2
|
a ) 4 / 25 , b ) 8 / 29 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 3 | b | divide(divide(multiply(40, 40), const_100), add(divide(multiply(40, 40), const_100), divide(multiply(70, subtract(const_100, 40)), const_100))) | at joel ’ s bookstore , the current inventory is 40 % historical fiction . of the historical fiction books , 40 % are new releases , while 70 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? | "let there be 100 books in all historic fiction books = 40 % of total = 40 other books = 60 new historic fiction = 40 % of 40 = 16 other new books = 70 % of 60 = 42 total new books = 58 fraction = 16 / 58 = 8 / 29 ans : b" | a = 40 * 40
b = a / 100
c = 40 * 40
d = c / 100
e = 100 - 40
f = 70 * e
g = f / 100
h = d + g
i = b / h
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(multiply(10, const_3), subtract(const_3, const_1)) | 10 years ago , the age of anand was one - third the age of bala at that time . the present age of bala is 10 years more than the present age of anand . find the present age of anand ? | let the present ages of anand and bala be ' a ' and ' b ' respectively . a - 10 = 1 / 3 ( b - 10 ) - - - ( 1 ) b = a + 10 substituting b = a + 12 in first equation , a - 10 = 1 / 3 ( a + 0 ) = > 3 a - 30 = a = > 2 a = 30 = > a = 15 . answer : d | a = 10 * 3
b = 3 - 1
c = a / b
|
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