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 ) 51 : 52 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 55 , e ) 51 : 52 | e | divide(add(const_100, 2), add(const_100, 4)) | the cash difference between the selling prices of an article at a profit of 2 % and 4 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 102 % of x ) / ( 104 % of x ) = 102 / 104 = 51 / 52 = 51 : 52 . answer : e" | a = 100 + 2
b = 100 + 4
c = a / b
|
a ) 5 days , b ) 6 days , c ) 8 days , d ) 10 days , e ) 12 days | c | inverse(add(inverse(40), inverse(10))) | jose completes a piece of work in 10 days , raju completes the same work in 40 days . if both of them work together , then the number of days required to complete the work is | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 10 × 40 / 50 = 8 days . c ) | a = 1/(40)
b = 1/(10)
c = a + b
d = 1/(c)
|
a ) 6 : 8 , b ) 6 : 4 , c ) 6 : 2 , d ) 6 : 5 , e ) 6 : 2 | d | divide(subtract(multiply(12, 4), multiply(6, 5)), subtract(multiply(7, 5), multiply(5, 4))) | 5 men and 12 boys finish a piece of work in 4 days , 7 men and 6 boys do it in 5 days . the ratio between the efficiencies of a man and boy is ? | 5 m + 12 b - - - - - 4 days 7 m + 6 b - - - - - - - 5 days 20 m + 48 b = 35 m + 30 b 18 b = 15 m = > 5 m = 6 b m : b = 6 : 5 answer : d | a = 12 * 4
b = 6 * 5
c = a - b
d = 7 * 5
e = 5 * 4
f = d - e
g = c / f
|
['a ) 89.8 cm ³', 'b ) 92.5 cm ³', 'c ) 132.8 cm ³', 'd ) 144.5 cm ³', 'e ) none'] | a | divide(multiply(multiply(multiply(multiply(const_1, add(multiply(7, const_3), const_1)), divide(7, const_2)), divide(7, const_2)), 7), multiply(const_3, 7)) | the volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is : | sol . volume of the block = ( 10 * 5 * 2 ) cm ³ . volume of the cone carved out = [ 1 / 3 * 22 / 7 * 3 * 3 * 7 ] cm ³ = 66 cm ³ ∴ wood wasted = ( 100 - 66 ) % = 34 % . answer a | a = 7 * 3
b = a + 1
c = 1 * b
d = 7 / 2
e = c * d
f = 7 / 2
g = e * f
h = g * 7
i = 3 * 7
j = h / i
|
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | divide(140, multiply(add(54, 72), const_0_2778)) | two trains of length 100 meters and 200 meters are 140 meters apart . they are moving towards each other on parallel tracks , at speeds of 54 km / h and 72 km / h . after how many seconds will the trains meet ? | "the speeds are 54000 / 3600 = 15 m / s and 72000 / 3600 = 20 m / s the relative speed is 35 m / s . time = 140 / 35 = 4 seconds the answer is b ." | a = 54 + 72
b = a * const_0_2778
c = 140 / b
|
a ) 60 , b ) 64 , c ) 68 , d ) 72 , e ) 76 | a | multiply(divide(140, 4), divide(140, add(divide(140, 4), divide(140, 3)))) | two trains start simultaneously from opposite ends of a 140 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 140 - km trip in 4 hours . train y , travelling at a constant rate , completes the 140 - km trip in 3 hours . how many kilometers had train x traveled when it met train y ? | "if the two trains cover a total distance d , then train x travels ( 3 / 7 ) * d while train y travels ( 4 / 7 ) * d . if the trains travel 140 km to the meeting point , then train x travels ( 3 / 7 ) * 140 = 60 km . the answer is a ." | a = 140 / 4
b = 140 / 4
c = 140 / 3
d = b + c
e = 140 / d
f = a * e
|
a ) 1998 , b ) 2001 , c ) 2002 , d ) 2003 , e ) none of these | c | add(1991, subtract(1991, 1990)) | the year next to 1991 having the same calendar as that of 1990 is – | we go on counting the odd days from 1991 onward still the sum is divisible by 7 . the number of such days are 14 upto the year 2001 . so , the calender for 1991 will be repeated in the year 2002 . answer c | a = 1991 - 1990
b = 1991 + a
|
a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | d | divide(add(multiply(multiply(divide(3, 2), 2), 7), 6), subtract(multiply(divide(3, 2), 2), 2)) | find ( 7 x + 6 y ) / ( x - 2 y ) if x / 2 y = 3 / 2 ? | "x / 2 y = 3 / 2 = > x = 6 y / 2 = 3 y = > ( 7 x + 6 y ) / ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 6 y ) / ( 3 y - 2 y ) = > 27 y / y = 27 answer : d" | a = 3 / 2
b = a * 2
c = b * 7
d = c + 6
e = 3 / 2
f = e * 2
g = f - 2
h = d / g
|
a ) 76 . , b ) 77 . , c ) 78 . , d ) 79 . , e ) 80 . | a | divide(add(multiply(divide(20, add(20, 40)), 72), multiply(divide(40, add(20, 40)), 81)), divide(add(20, 40), const_60)) | a car was driving at 72 km / h for 20 minutes , and then at 81 km / h for another 40 minutes . what was its average speed ? | driving at 72 km / h for 20 minutes , distance covered = 72 * 1 / 3 = 24 km driving at 90 km / h for 40 minutes , distance covered = 81 * 2 / 3 = 54 km average speed = total distance / total time = 78 / 1 = 78 km / h answer : a | a = 20 + 40
b = 20 / a
c = b * 72
d = 20 + 40
e = 40 / d
f = e * 81
g = c + f
h = 20 + 40
i = h / const_60
j = g / i
|
a ) 33 , b ) 77 , c ) 66 , d ) 11 , e ) 99 | d | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 4)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 4))), add(divide(const_1, 24), divide(const_1, 30)))), 4) | a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 4 days before the completion of the work . in how many days was the work completed ? | "one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 40 = 1 / 10 work done by a and b together in the last 4 days = 4 * ( 1 / 24 + 1 / 30 ) = 3 / 10 remaining work = 7 / 10 the number of days required for this initial work = 7 days . the total number of days required = 4 + 7 = 11 days . answer : d" | a = 1 / 24
b = 1 / 30
c = a + b
d = c * 4
e = 1 - d
f = 2 + 3
g = 2 * 4
h = f * g
i = 1 / h
j = 1 / 24
k = 1 / 30
l = j + k
m = i + l
n = e / m
o = n + 4
|
a ) 15 seconds , b ) 18 seconds , c ) 25 seconds , d ) 36 seconds , e ) 45 seconds | d | divide(36, subtract(const_4, const_3)) | nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 36 meter head start . if cristina runs at a pace of 4 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 36 for cristina = c = 5 * t @ t = 36 , n = 144 c = 144 right answer ans : d | a = 4 - 3
b = 36 / a
|
a ) 9 , b ) 7 , c ) 3 , d ) 8 , e ) 12 | d | subtract(divide(power(6, const_2), power(2, const_2)), const_1) | what is the largest integral value of ' k ' for which the quadratic equation x 2 - 6 x + k = 0 will have two real and distinct roots ? | explanatory answer any quadratic equation will have real and distinct roots if the discriminant d > 0 the discriminant ' d ' of a quadratic equation ax 2 + bx + c = 0 is given by b 2 - 4 ac in this question , the value of d = 62 - 4 * 1 * k if d > 0 , then 36 > 4 k or k < 9 . therefore , the highest integral value that k can take is 8 . correct choice is ( d ) | a = 6 ** 2
b = 2 ** 2
c = a / b
d = c - 1
|
a ) $ 320 , b ) $ 420 , c ) $ 720 , d ) $ 220 , e ) $ 200 | c | add(add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), 20)), subtract(multiply(add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), 20), const_3), add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)))) | a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 20 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 20 a = 20 + b ( 1 ) = = = > c = 60 + 3 b - b = 2 b + 60 = = > 2 b - c = - 60 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 20 = 2.5 b - 20 = = > 2.5 b - c = 20 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 80 b = $ 160 a = $ 180 c = 540 - 160 = $ 380 total amount = 180 + 160 + 380 = $ 720 answer : c" | a = 3 * 20
b = a * 2
c = 2 * 20
d = b + c
e = 3 * 20
f = e * 2
g = 2 * 20
h = f + g
i = h + 20
j = d + i
k = 3 * 20
l = k * 2
m = 2 * 20
n = l + m
o = n + 20
p = o * 3
q = 3 * 20
r = q * 2
s = 2 * 20
t = r + s
u = p - t
v = j + u
|
a ) 45 , b ) 79 , c ) 50 , d ) 55 , e ) 60 | b | add(subtract(divide(multiply(multiply(3, 5), 8), 3), divide(multiply(multiply(3, 5), 8), 5)), divide(multiply(multiply(3, 5), 8), 8)) | if x , y , and z are positive integers and 3 x = 5 y = 8 z , then the least possible value of x + y + z is | "given 3 x = 5 y = 8 z x + y + z in terms of x = x + ( 3 x / 5 ) + ( 3 x / 8 ) = 79 x / 40 now checking with each of the answers and see which value gives a minimum integer value . a x = 40 / 79 * 45 , not an integer c , d , e can be ruled out similarly . b is minimum value as x = 79 * 40 / 79 = 79 answer is b" | a = 3 * 5
b = a * 8
c = b / 3
d = 3 * 5
e = d * 8
f = e / 5
g = c - f
h = 3 * 5
i = h * 8
j = i / 8
k = g + j
|
a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 13 | a | divide(subtract(add(multiply(2, 5), 4), add(multiply(3, 3), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] – 4 = f ( 3 x – 6 ) | "answer = a = 3 f ( x ) = 3 x – 5 2 * [ f ( x ) ] – 4 = f ( 3 x – 6 ) 2 ( 3 x - 5 ) - 4 = 3 ( 3 x - 6 ) - 5 6 x - 14 = 9 x - 23 x = 3" | a = 2 * 5
b = a + 4
c = 3 * 3
d = c + 5
e = b - d
f = 2 * 3
g = 3 * 1
h = f - g
i = e / h
|
a ) 21 ½ days , b ) 11 3 / 7 days , c ) 23 ½ days , d ) 12 ½ days , e ) none of these | b | add(divide(const_1, 8), divide(const_1, 20)) | a can do a job in 8 days and b can do it in 20 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "explanation : 1 / 8 + 1 / 20 = 7 / 40 40 / 7 = 40 / 7 * 2 = 11 3 / 7 days answer : b" | a = 1 / 8
b = 1 / 20
c = a + b
|
['a ) 17', 'b ) 45 / 78', 'c ) 72 / 17', 'd ) 17 / 72', 'e ) 72'] | c | divide(multiply(8, 9), add(8, 9)) | in an electric circuit , two resistors with resistances 8 ohms and 9 ohms are connected in parallel . in this case , if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of two resistors . what is r value ? | the wording is a bit confusing , though basically we are told that 1 / r = 1 / 8 + 1 / 9 , from which it follows that r = 72 / 17 . answer : c . | a = 8 * 9
b = 8 + 9
c = a / b
|
a ) 76 , b ) 79 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 35)), multiply(10, 35)) | the average of runs of a cricket player of 10 innings was 35 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 35 so , total = average x no . of innings = 35 x 10 = 350 . now increase in avg = 4 runs . so , new avg = 35 + 4 = 39 runs total runs = new avg x new no . of innings = 39 x 11 = 429 runs made in the 11 th inning = 429 - 350 = 79 answer : b" | a = 10 + 1
b = 4 + 35
c = a * b
d = 10 * 35
e = c - d
|
a ) a ) 12 , b ) b ) 14 , c ) c ) 16 , d ) d ) 18 , e ) e ) 22 | c | floor(divide(161, 10)) | on dividing 161 by a number , the quotient is 10 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 161 - 1 ) / 10 = 160 / 10 = 16 c )" | a = 161 / 10
b = math.floor(a)
|
a ) 7 days , b ) 14 days , c ) 6 days , d ) 8 days , e ) 9 days | d | divide(const_1, add(multiply(6, divide(divide(const_1, 6), 24)), multiply(8, divide(divide(const_1, 6), 16)))) | 16 boys or 24 girls can construct the wall in 6 days . the number of days that 8 boys and 6 girls will take to construct ? | "explanation : 16 boys = 24 girls , 1 boy = 24 / 16 girls 1 boy = 6 / 4 girls 8 boys + 6 girls = 8 ã — 6 / 4 + 12 = 12 + 6 = 18 girls 8 days to complete the work answer : option d" | a = 1 / 6
b = a / 24
c = 6 * b
d = 1 / 6
e = d / 16
f = 8 * e
g = c + f
h = 1 / g
|
a ) 9 , b ) 9.2 , c ) 10 , d ) 13.7 , e ) 12 | d | subtract(add(multiply(11.4, 6), multiply(10.5, 6)), multiply(11, 10.7)) | the average of 11 numbers is 10.7 . if the average of first 6 is 10.5 and that of the last 6.00001 is 11.4 the sixth number is ? | explanation : 1 to 11 = 11 * 10.7 = 117.7 1 to 6 = 6 * 10.5 = 63 6 to 11 = 6 * 11.4 = 68.4 63 + 68.4 = 131.4 – 117.7 = 13.7 6 th number = 13.7 d ) | a = 11 * 4
b = 10 * 5
c = a + b
d = 11 * 10
e = c - d
|
a ) 25 / 33 , b ) 31 / 39 , c ) 37 / 45 , d ) 43 / 51 , e ) 53 / 61 | d | divide(divide(subtract(multiply(6, add(5, 8)), 7), subtract(7, 6)), add(divide(subtract(multiply(6, add(5, 8)), 7), subtract(7, 6)), 8)) | the denominator of a fraction is 8 greater than the numerator . if the numerator and the denominator are increased by 5 , the resulting fraction is equal to 6 â „ 7 . what is the value of the original fraction ? | "let the numerator be x . then the denominator is x + 8 . x + 5 / x + 13 = 6 / 7 . 7 x + 35 = 6 x + 78 . x = 43 . the original fraction is 43 / 51 . the answer is d ." | a = 5 + 8
b = 6 * a
c = b - 7
d = 7 - 6
e = c / d
f = 5 + 8
g = 6 * f
h = g - 7
i = 7 - 6
j = h / i
k = j + 8
l = e / k
|
a ) 102500 , b ) 112584 , c ) 117600 , d ) 118450 , e ) 128450 | c | divide(50, divide(47, 50)) | evaluate 50 ! / 47 ! | "evaluate 50 ! / 47 ! = 50 * 49 * 48 * ( 47 ! ) / 47 ! = 50 * 49 * 48 = 117600 answer : c" | a = 47 / 50
b = 50 / a
|
a ) 155.68 , b ) 160 , c ) 159.38 , d ) 180 , e ) 175 | c | subtract(multiply(divide(425, 40), 55), 425) | a 425 meter long train crosses a platform in 55 seconds while it crosses a signal pole in 40 seconds . what is the length of the platform ? | "speed = [ 425 / 40 m / sec = 42.5 / 4 m / sec . let the length of the platform be x meters . then , x + 425 / 55 = 42.5 / 4 4 ( x + 425 ) = 2337.5 è x = 159.38 m . answer : c" | a = 425 / 40
b = a * 55
c = b - 425
|
a ) 133.33 , b ) 882 , c ) 772 , d ) 252 , e ) 121 | a | multiply(divide(multiply(60, const_1000), const_3600), 8) | a train running at the speed of 60 km / hr crosses a pole in 8 seconds . find the length of the train . | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 8 = 133.33 meter . answer : a" | a = 60 * 1000
b = a / 3600
c = b * 8
|
a ) 3.625 , b ) 3.5 , c ) 3 , d ) 3.04 , e ) 4 | d | add(divide(subtract(divide(40, 12), 2.75), const_2), 2.75) | annika hikes at a constant rate of 12 minutes per kilometer . she has hiked 2.75 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 40 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exactly 40 minutes , for how many kilometers total did she hike east ? | "set up two r x t = d cases . 1 . 1 / 12 km / min x t = 2.75 from which t = 33 mins . we know total journey time now is 40 + 33 = 73 . the rate is the same ie 1 / 12 km / min . set up second r x t = d case . 1 / 12 km / min x 73 = 6.08 km now the total journey would be halved as distance would be same in each direction . 6.08 / 2 = 3.04 d ." | a = 40 / 12
b = a - 2
c = b / 2
d = c + 2
|
a ) 3000 , b ) 4000 , c ) 5000 , d ) 6900 , e ) 6677 | a | divide(800, subtract(add(multiply(700, divide(subtract(const_1, divide(const_1, const_3)), add(700, 300))), divide(divide(const_1, const_3), const_2)), add(multiply(300, divide(subtract(const_1, divide(const_1, const_3)), add(700, 300))), divide(divide(const_1, const_3), const_2)))) | tom and jerry enter into a partnership by investing $ 700 and $ 300 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if tom received $ 800 more than jerry did , what was the profit made by their business in that year ? | say the profit was $ x . tom share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.7 jerry share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.3 thus ( x - x / 3 ) * 0.7 - ( x - x / 3 ) * 0.3 = 800 - - > x = 3000 . answer is a | a = 1 / 3
b = 1 - a
c = 700 + 300
d = b / c
e = 700 * d
f = 1 / 3
g = f / 2
h = e + g
i = 1 / 3
j = 1 - i
k = 700 + 300
l = j / k
m = 300 * l
n = 1 / 3
o = n / 2
p = m + o
q = h - p
r = 800 / q
|
a ) 0.3504 , b ) 0.4209 , c ) 0.4889 , d ) 0.6412 , e ) 0.4612 | a | add(multiply(0.72, 0.43), multiply(0.12, 0.34)) | simplify : 0.72 * 0.43 + 0.12 * 0.34 | given exp . = 0.72 * 0.43 + ( 0.12 * 0.34 ) = 0.3096 + 0.0408 = 0.3504 answer is a | a = 0 * 72
b = 0 * 12
c = a + b
|
a ) 28 kmph , b ) 52 kmph , c ) 20 kmph , d ) 23 kmph , e ) 65 kmph | d | divide(280, multiply(divide(3, 2), 8)) | a jeep takes 8 hours to cover a distance of 280 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 8 distance = 280 3 / 2 of 8 hours = 8 * 3 / 2 = 12 hours required speed = 280 / 12 = 23 kmph d )" | a = 3 / 2
b = a * 8
c = 280 / b
|
a ) 16.32 , b ) 16.9 , c ) 12.25 , d ) 13.21 , e ) none | a | divide(multiply(multiply(34, 8), 6), const_100) | the simple interest on rs . 34 for 8 months at the rate of 6 paise per rupeeper month is | "sol . s . i . = rs . [ 34 * 6 / 100 * 8 ] = rs . 16.32 answer a" | a = 34 * 8
b = a * 6
c = b / 100
|
a ) 1.5 , b ) 1.75 , c ) 2.0 , d ) 2.25 , e ) 2.5 | a | divide(3, subtract(4, 2)) | a swimmer can swim in still water at 4 km / h . if the speed of the water current is 2 km / h , how many hours will the swimmer take to swim against the current for 3 km ? | "the swimmer can swim against the current at a speed of 4 - 2 = 2 km / h . the time it will take is 3 / 2 = 1.5 hours . the answer is a ." | a = 4 - 2
b = 3 / a
|
a ) 8164 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(8880, power(add(subtract(divide(9261, 8880), const_1), const_1), 2)) | what sum of money put at c . i amounts in 2 years to rs . 8880 and in 3 years to rs . 9261 ? | "8880 - - - - 381 100 - - - - ? = > 4.29 % x * 104.29 / 100 * 104.29 / 100 = 8880 x * 1.0876 = 8880 x = 8880 / 1.0876 = > 8164.38 answer : a" | a = 9261 / 8880
b = a - 1
c = b + 1
d = c ** 2
e = 8880 / d
|
a ) 25 , b ) 30 , c ) 45 , d ) 64 , e ) 60 | e | divide(400, multiply(const_0_2778, subtract(30, 6))) | how many seconds will a 400 metre long train take to cross a man running with a speed of 6 km / hr in the direction of the moving train if the speed of the train is 30 km / hr ? | "explanation : speed of train relatively to man = ( 30 - 6 ) km / hr = 24 km / hr = ( 24 x 5 / 18 ) m / sec = 6.66 m / sec time taken to pass the man = ( 400 / 6.66 ) sec = 60 sec . answer : e" | a = 30 - 6
b = const_0_2778 * a
c = 400 / b
|
a ) 5 , b ) 4 , c ) 16 , d ) 20 , e ) 25 | c | divide(multiply(16, 16), multiply(4, 4)) | what is the maximum number of pieces of birthday cake of size 4 ” by 4 ” that can be cut from a cake 16 ” by 16 ” ? | "the prompt is essentially asking for the maximum number of 4 x 4 squares that can be cut from a larger 16 by 16 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 4 ' pieces ' each , we have ( 4 ) ( 4 ) = 16 total smaller squares ( at maximum ) . c" | a = 16 * 16
b = 4 * 4
c = a / b
|
a ) 240 , b ) 300 , c ) 360 , d ) 420 , e ) 480 | b | divide(add(80, 10), divide(30, const_100)) | a student needs 30 % of the marks on a test to pass the test . if the student gets 80 marks and fails the test by 10 marks , find the maximum marks set for the test . | "30 % = 90 marks 1 % = 3 marks 100 % = 300 marks the answer is b ." | a = 80 + 10
b = 30 / 100
c = a / b
|
a ) 2 , b ) 3 , c ) 3 1 / 3 , d ) 4 , e ) 4 4 / 5 | b | add(9, divide(multiply(6, 9), add(6, 12))) | a work crew of 6 men takes 9 days to complete one - half of a job . if 12 men are then added to the crew and the men continue to work at the same rate , how many days will it take the enlarged crew to do the rest of the job ? | "suppose 1 man can do work in x days . . so 6 men will do in . . 6 / x = 1 / 9 * 1 / 2 as half job is done x = 108 now 12 more are added then 18 / 108 = 1 / 2 * 1 / d for remaining half job d = 3 number of days b" | a = 6 * 9
b = 6 + 12
c = a / b
d = 9 + c
|
a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | b | add(add(15, 6), const_1) | in a class , 6 students can speak gujarati , 15 can speak hindi and 6 can speak marathi . if two students can speak two languages and one student can speak all the 3 languages , then how many students are there in the class | n ( aubuc ) = 6 + 15 + 6 - 2 - 2 - 2 + 1 = 22 answer : b | a = 15 + 6
b = a + 1
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a ) 41 / 45 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | a | divide(divide(41, 4.5), divide(41, 4.1)) | jack and jill are marathon runners . jack can finish a marathon ( 41 km ) in 4.5 hours and jill can run a marathon in 4.1 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 41 / ( 9 / 2 ) = 82 / 9 average speed of jill = 41 / ( 4.1 ) = 10 ratio of average speed of jack to jill = ( 82 / 9 ) / 10 = 82 / 90 = 41 / 45 answer a" | a = 41 / 4
b = 41 / 4
c = a / b
|
a ) 25 % , b ) 40 % , c ) 60 % , d ) 65 % , e ) 50 % | c | subtract(100, 40) | if the selling price of 100 articles is equal to the cost price of 40 articles , then the loss or gain percent is : | let c . p . of each article be re . 1 . then , c . p . of 100 articles = rs . 100 ; s . p . of 100 articles = rs . 40 . loss % = 60 / 100 * 100 = 60 % answer : c | a = 100 - 40
|
a ) 511 , b ) 546 , c ) 552 , d ) 562 , e ) 570 | a | multiply(multiply(multiply(700, subtract(1, divide(1, 10))), subtract(1, divide(1, 10))), subtract(1, divide(1, 10))) | in a certain animal population , for each of the first 3 months of life , the probability that an animal will die during that month is 1 / 10 . for a group of 700 newborn members of the population , approximately how many would be expected to survive the first 3 months of life ? | "number of newborns that can die in first month = 1 / 10 * 700 = 70 survived = 630 number of newborns that can die in second month = 1 / 10 * 630 = 63 survived = 567 number of newborns that can die in third month = 1 / 10 * 567 = 56 survived = 511 answer : a" | a = 1 / 10
b = 1 - a
c = 700 * b
d = 1 / 10
e = 1 - d
f = c * e
g = 1 / 10
h = 1 - g
i = f * h
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a ) 40 , b ) 60 , c ) 50 , d ) 30 , e ) 70 | b | multiply(divide(300, 100), divide(40, const_2)) | the ratio , by volume , of bleach to detergent to water in a certain solution is 2 : 40 : 100 . the solution will be altered so that the ratio of bleach ( b ) to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 300 liters of water , how many liters of detergent will it contain ? | "b : d : w = 2 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 40 ) = ( 1 / 60 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 300 dnew = wnew / 5 = 300 / 5 = 60 so , answer will be b b" | a = 300 / 100
b = 40 / 2
c = a * b
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a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 75 | e | subtract(multiply(subtract(60, 20), const_3), multiply(subtract(60, 50), const_3)) | in x game of billiards , x can give y 20 points in 60 and he can give z 50 points in 60 . how many points can y give z in x game of 100 ? | "x scores 60 while y score 40 and z scores 10 . the number of points that z scores when y scores 100 = ( 100 * 50 ) / 40 = 25 . in x game of 100 points , y gives ( 100 - 25 ) = 75 points to c . e" | a = 60 - 20
b = a * 3
c = 60 - 50
d = c * 3
e = b - d
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a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 21 , d ) 3 / 8 , e ) 2 / 3 | c | inverse(divide(factorial(7), multiply(factorial(2), factorial(5)))) | jack and jill work at a hospital with 5 other workers . for an internal review , 2 of the 7 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "1 / 7 c 2 = 1 / 21 . answer : c ." | a = math.factorial(7)
b = math.factorial(2)
c = math.factorial(5)
d = b * c
e = a / d
f = 1/(e)
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a ) 3 cm , b ) 5 cm , c ) 6 cm , d ) 8 cm , e ) none | b | sqrt(divide(multiply(multiply(const_pi, multiply(10, divide(10, const_2))), const_2), multiply(const_pi, const_4))) | the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 10 cm each . the radius of the sphere is | "solution 4 î r 2 = 2 î 5 x 10 â ‡ ’ r 2 = ( 5 x 10 / 2 ) â ‡ ’ 25 â ‡ ’ r = 5 cm . answer b" | a = 10 / 2
b = 10 * a
c = math.pi * b
d = c * 2
e = math.pi * 4
f = d / e
g = math.sqrt(f)
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a ) 67 , b ) 68 , c ) 69 , d ) 70 , e ) 71 | c | divide(add(18, multiply(3, 86)), const_4) | find a number such that it exceeds 18 by 3 times the number by which it is less than 86 ? | x - 18 = 3 [ 86 - x ] x = 69 answer : c | a = 3 * 86
b = 18 + a
c = b / 4
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a ) 40 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(subtract(divide(multiply(3600, 12), 25), divide(divide(multiply(3600, 12), 25), 4)), subtract(3600, divide(divide(multiply(3600, 12), 25), 4))), const_100) | of the 3600 employees of company x , 12 / 25 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | let ' s see , the way i did it was 12 / 25 are clerical out of 3600 so 1728 are clerical 1728 reduced by 1 / 4 is 1728 * 1 / 4 so it reduced 432 people , so there is 1296 clerical people left but since 432 people left , it also reduced from the total of 3600 so there are 3168 people total since 1296 clerical left / 3168 people total you get ( a ) 40 % | a = 3600 * 12
b = a / 25
c = 3600 * 12
d = c / 25
e = d / 4
f = b - e
g = 3600 * 12
h = g / 25
i = h / 4
j = 3600 - i
k = f / j
l = k * 100
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a ) 12 % , b ) 26 % , c ) 29 % , d ) 31 % , e ) 60 % | d | multiply(divide(add(multiply(divide(25, const_100), 300), multiply(divide(40, const_100), 200)), add(300, 200)), const_100) | for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 25 percent of the seeds in the first plot germinated and exactly 40 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "in the first plot 25 % of 300 seeds germinated , so 0.25 x 300 = 75 seeds germinated . in the second plot , 40 % of 200 seeds germinated , so 0.4 x 200 = 80 seeds germinated . since 75 + 80 = 155 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 155 / 500 ) x 100 % , or 31 % . answer : d ." | a = 25 / 100
b = a * 300
c = 40 / 100
d = c * 200
e = b + d
f = 300 + 200
g = e / f
h = g * 100
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a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 360 cm | e | add(triangle_perimeter(60, 60, 60), triangle_perimeter(60, 60, 60)) | an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 60 cm , find the sum of the perimeters of all the triangles . | "we have 60 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 30 and continues . so we have 60 , 30,15 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 60 / 1 - ( 1 / 2 ) = 120 equilateral triangle perimeter is 3 a = 3 * 120 = 360 . so option e ." | a = triangle_perimeter + (
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a ) 25 % , b ) 30 % , c ) 50 % , d ) 75 % , e ) 80 % | c | multiply(divide(subtract(divide(66, const_100), multiply(divide(74, const_100), divide(2, const_3))), divide(1, const_3)), const_100) | in an election , candidate douglas won 66 percent of the total vote in counties x and y . he won 74 percent of the vote in county x . if the ratio of people who voted in county x to county y is 2 : 1 , what percent of the vote did candidate douglas win in county y ? | given voters in ratio 2 : 1 let x has 200 votersy has 100 voters for x 66 % voted means 74 * 200 = 148 votes combined for xy has 300 voters and voted 66 % so total votes = 198 balance votes = 198 - 148 = 50 as y has 100 voters so 50 votes means 50 % of votes required ans c | a = 66 / 100
b = 74 / 100
c = 2 / 3
d = b * c
e = a - d
f = 1 / 3
g = e / f
h = g * 100
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a ) rs . 4250 , b ) rs . 4228 , c ) rs . 4128 , d ) rs . 4988 , e ) rs . 4192 | a | add(4200, divide(subtract(1500, 1200), multiply(const_2, const_3))) | amithab ' s average expenditure for the january to june is rs . 4200 and he spends rs . 1200 in january and rs . 1500 in july . the average expenditure for the months of febraury to july is | amithab ' s total expenditure for jan - june = 4200 x 6 = 25200 expenditure for feb - june = 25200 - 1200 = 24000 expenditure for the months of feb - july = 24000 + 1500 = 25500 the average expenditure = { 25500 } / { 6 } = 4250 answer : a | a = 1500 - 1200
b = 2 * 3
c = a / b
d = 4200 + c
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a ) 18 , b ) 16 , c ) 12 , d ) 14 , e ) 8 | e | add(16, multiply(20, divide(50, const_100))) | one week , a certain truck rental lot had a total of 20 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 16 trucks on the lot that saturday morning , what is the greatest number of different trucks that could have been rented out during the week ? | "n - not rented trucks ; r - rented trucks n + r = 20 n + r / 2 = 16 r = 8 e" | a = 50 / 100
b = 20 * a
c = 16 + b
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a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72 | b | divide(80, divide(add(divide(80, 40), divide(80, 120)), const_2)) | a person walks from one end to the other of a 80 - meter long moving walkway at a constant rate in 40 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 120 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | "let v be the speed of the person and let x be the speed of the walkway . 40 ( v + x ) = 80 then 120 ( v + x ) = 240 120 ( v - x ) = 80 when we add the two equations : 240 v = 320 v = 4 / 3 time = 80 / ( 4 / 3 ) = 60 seconds the answer is b ." | a = 80 / 40
b = 80 / 120
c = a + b
d = c / 2
e = 80 / d
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a ) 60 years , b ) 33 , c ) 77 , d ) 66 , e ) 101 | a | add(multiply(4, divide(subtract(multiply(const_2, add(12, 12)), add(12, 12)), subtract(4, const_2))), 12) | neha ' s mother was 4 times her age 12 years ago . she will be twice as old as neha 12 years from now . what is the present age of neha ' s mother ? | explanation : let neha ' s present age be ' x ' years . her age 12 years ago = ( x – 12 ) years therefore , her mother ' s age 12 years ago = 4 ( x – 12 ) her mother ' s present age = 4 x – 48 + 12 = 4 x – 36 neha ' s age after 12 years = x + 12 her mother ' s age after 12 years = 4 x – 36 + 12 = 4 x – 24 4 x – 24 = 2 ( x + 12 ) 4 x – 2 x = 24 + 24 x = 24 therefore , neha ' s present age = 24 her mother ' s present age = 4 x – 36 = 4 ( 24 ) – 36 = 60 years answer : a | a = 12 + 12
b = 2 * a
c = 12 + 12
d = b - c
e = 4 - 2
f = d / e
g = 4 * f
h = g + 12
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a ) 3.5 , b ) 3 , c ) 5 , d ) 4 , e ) 2.5 | e | divide(triangle_area_three_edges(3, 4, 5), divide(triangle_perimeter(3, 4, 5), const_2)) | what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 3 , 4 , and 5 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circum raduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 5 / 2 = 2.5 ans . e" | a = triangle_area_three_edges / (
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a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | c | divide(multiply(1, 5), const_4) | what is the sum of the numbers between 1 and 5 , inclusive ? | "sol . add numbers between 1 and 5 . answer = c , 15 ." | a = 1 * 5
b = a / 4
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a ) 73.41 , b ) 74.31 , c ) 73.42 , d ) 73.43 , e ) can not be determined | c | divide(subtract(multiply(35, 72), subtract(96, 46)), 35) | a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 46 instead of 96 then find the correct average marks up to two decimal places . | "total marks = 35 x 72 = 2520 corrected total marks = 2520 - 46 + 96 = 2570 correct average = 2570 / 35 = 73.42 answer : c" | a = 35 * 72
b = 96 - 46
c = a - b
d = c / 35
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a ) 320 cm 2 , b ) 360 cm 2 , c ) 340 cm 2 , d ) 310 cm 2 , e ) 300 cm 2 | c | divide(multiply(34, 20), const_2) | if the sides of a triangle are 36 cm , 34 cm and 20 cm , what is its area ? | "the triangle with sides 36 cm , 34 cm and 20 cm is right angled , where the hypotenuse is 36 cm . area of the triangle = 1 / 2 * 34 * 20 = 340 cm 2 answer : c" | a = 34 * 20
b = a / 2
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a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | multiply(multiply(3, const_2), divide(const_3, const_2)) | a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 3 miles per day and it will take them two days , how many miles long is the route down the mountain ? | "on the way down , the rate is 1.5 * 3 = 4.5 miles per day . the distance of the route down the mountain is 2 * 4.5 = 9 miles . the answer is c ." | a = 3 * 2
b = 3 / 2
c = a * b
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a ) 16.67 , b ) 30 , c ) 50 , d ) 60.33 , e ) 70 | e | divide(subtract(multiply(divide(40, const_100), 70), multiply(divide(20, const_100), 70)), subtract(divide(40, const_100), divide(20, const_100))) | how many ounces of a 60 % salt solution must be added to 70 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ? | let x = ounces of 60 % salt solution to be added . 2 * 70 + . 6 x = . 4 ( 70 + x ) x = 70 answer e | a = 40 / 100
b = a * 70
c = 20 / 100
d = c * 70
e = b - d
f = 40 / 100
g = 20 / 100
h = f - g
i = e / h
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a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | c | subtract(subtract(multiply(3, 16), add(subtract(15, 16), 3)), 16) | the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 15 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 13 a – d = 3 a + d = 15 2 d = 12 d = 6 answer c" | a = 3 * 16
b = 15 - 16
c = b + 3
d = a - c
e = d - 16
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a ) 18 , b ) 750 , c ) 21 , d ) 120 , e ) none of these | c | divide(circle_area(divide(33, multiply(2, const_pi))), 4) | how many plants will be there in a circular bed whose outer edge measure 33 cms , allowing 4 cm 2 for each plant ? | "circumference of circular bed = 33 cm area of circular bed = ( 33 ) 2 â „ 4 ï € space for each plant = 4 cm 2 â ˆ ´ required number of plants = ( 33 ) 2 â „ 4 ï € ã · 4 = 21.65 = 21 ( approx ) answer c" | a = 2 * math.pi
b = 33 / a
c = circle_area / (
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a ) 5 / 4 , b ) 25 / 16 , c ) 4 / 5 , d ) 16 / 25 , e ) 16 / 5 | d | divide(power(4, const_2), power(5, const_2)) | rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 4 / 5 , what is the ratio of rectangle a ’ s area to rectangle b ’ s area ? | "the area of rectangle a is ab . c = 5 a / 4 and d = 5 b / 4 . the area of rectangle b is cd = 25 ab / 16 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 25 ab / 16 ) = 16 / 25 . the answer is d ." | a = 4 ** 2
b = 5 ** 2
c = a / b
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a ) 13 : 00 , b ) 14 : 00 , c ) 15 : 00 , d ) 17 : 00 , e ) 16 : 00 | c | multiply(divide(const_3, 4), divide(const_1, subtract(divide(const_1, 4), divide(const_1, 6)))) | pipe a fills a swimming pool in 4 hours . pipe b empties the pool in 6 hours . if pipe a was opened at 5 : 00 am and pipe b at 6 : 00 am , at what time will the pool be full ? | "pipe a fills the pool in 4 hrs . 1 hour ' s work : 1 / 4 pipe b empties the pool in 6 hrs . 1 hour ' s work : 1 / 6 together if they work , 1 hour ' s work = 1 / 4 - 1 / 6 = 1 / 12 given : pipe a started at 5 : 00 a . m and pipe b at 6 : 00 a . m pool filled after 1 hour by pipe a : 1 / 4 or 3 / 12 after 6 : 00 a . m pool filled after 1 hour with both the pipes on : 1 / 12 pool filled after 9 hours with both pipes on : 9 / 12 pool filled in 1 hour + pool filled in 9 hours = 3 / 12 + 9 / 12 = 1 therefore , it takes 10 hrs to fill the pool as pipe a started at 5 : 00 a . m , pool is full at 15 : 00 hrs answer : c" | a = 3 / 4
b = 1 / 4
c = 1 / 6
d = b - c
e = 1 / d
f = a * e
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a ) 22 , b ) 64 , c ) 365 , d ) 29 , e ) 36 | b | multiply(divide(subtract(divide(5, 3), divide(3, 5)), divide(5, 3)), const_100) | a student multiplied a number by 3 / 5 instead of 5 / 3 , what is the percentage error in the calculation ? | "let the number be x . then , ideally he should have multiplied by x by 5 / 3 . hence correct result was . by mistake he multiplied x by 3 / 5 . hence the result with error = then , error = error % = = 64 % answer : b" | a = 5 / 3
b = 3 / 5
c = a - b
d = 5 / 3
e = c / d
f = e * 100
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a ) 288 , b ) 497 , c ) 168 , d ) 127 , e ) 664 | b | subtract(multiply(84, 6), subtract(92, 85)) | a team of 6 entered for a shooting competition . the best marks man scored 85 points . if he had scored 92 points , the average scores for . the team would have been 84 . how many points altogether did the team score ? | explanation : 6 * 84 = 504 - 7 = 497 answer : b | a = 84 * 6
b = 92 - 85
c = a - b
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a ) 600 , b ) 620 , c ) 650 , d ) 700 , e ) 720 | a | divide(multiply(60, 500), 50) | if 60 percent of 500 is 50 percent of x , then x = ? | "0.6 * 500 = 0.5 * x x = 6 / 5 * 500 = 600" | a = 60 * 500
b = a / 50
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a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 7.5 , e ) 8.5 | b | divide(220, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 220 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 220 kilometers ? 220 ÷ 40 = 5.5 × 1 gallon = 5.5 gallons correct answer b" | a = 220 / 40
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a ) 30 , b ) 20 , c ) 58 , d ) 48 , e ) 68 | a | divide(98, add(add(divide(2, 3), const_1), divide(8, 5))) | the sum of 3 numbers is 98 . if the ratio between first and second be 2 : 3 and that between second and third be 5 : 8 , then the second number is ? | let the numbers be x , y and z . then , x + y + z = 98 , x / y = 2 / 3 and y / z = 5 / 8 therefore , x = 2 y / 3 and z = 8 y / 5 . so , 2 y / 3 + y + 8 y / 5 = 98 . 49 y / 15 = 98 y = 30 . answer : a | a = 2 / 3
b = a + 1
c = 8 / 5
d = b + c
e = 98 / d
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a ) 22877 , b ) 27778 , c ) 20000 , d ) 22500 , e ) 17799 | d | divide(multiply(multiply(25, const_100), multiply(18, const_100)), multiply(20, 10)) | a courtyard is 25 meter long and 18 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ? | "number of bricks = courtyard area / 1 brick area = ( 2500 × 1800 / 20 × 10 ) = 22500 answer : d" | a = 25 * 100
b = 18 * 100
c = a * b
d = 20 * 10
e = c / d
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['a ) 20 m', 'b ) 24 m', 'c ) 28 m', 'd ) 40 m', 'e ) 43 m'] | d | multiply(const_4, sqrt(add(square_area(divide(24, const_4)), square_area(divide(32, const_4))))) | the perimeter of one square is 24 m & that of another is 32 m . perimeter of a square whose area is equal to the sum of the areas of the 2 squares will be | perimeter of square = 4 * side perimeter of first square = 24 m . side of first square = ∴ area of first square = perimeter of second square = 32 m . side of second square = ∴ area of second square = sum of the areas of two squares = 36 + 64 = 100 m 2 ∴ side of square = ∴ perimeter of square = 4 * 10 = 40 m . d | a = 24 / 4
b = square_area + (
c = 32 / 4
d = math.sqrt(b)
e = 4 * d
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a ) 500 , b ) 625 , c ) 1,000 , d ) 1,500 , e ) 2,500 | b | multiply(divide(2500, 2), divide(const_1, 2)) | the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 2500 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ? | "x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 2500 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 2500 ) * t = = = > 3 d = ts + 2500 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 2500 t - 2 d = - 2 st + 2500 t 2 d = 2 st - 2500 t d = st - 1250 t x 13 : d = s * t st - 1250 t = s * t s - 1250 = s - 625 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 2500 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 2500 2 = 1250 d = s * t d = 1250 * ( 1 / 2 ) d = 625 answer : b" | a = 2500 / 2
b = 1 / 2
c = a * b
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a ) − 0.3 , b ) 0 , c ) 0.06 , d ) 1.08 , e ) 2.46 | c | subtract(multiply(divide(divide(subtract(power(3, 2), power(2.4, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(2.4, 0.3)), const_1000), const_1000)) | what is the value of 3 x ^ 2 − 2.4 x + 0.3 for x = 0.6 ? | "3 x ^ 2 - 2.4 x + 0.3 for x = 0.6 = 3 ( 0.6 * 0.6 ) - 4 * 0.6 * ( 0.6 ) + 0.3 = - 0.6 * 0.6 + 0.3 = 0.06 correct option : c" | a = 3 ** 2
b = 2 ** 4
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 2 ** 4
i = g - h
j = i / 1000
k = j / 1000
l = f - k
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a ) 5 / 2 , b ) 3 / 2 , c ) 4 / 5 , d ) 2 / 5 , e ) 0 | c | divide(add(1, const_2.0), add(2, 3)) | if x = 1 - 3 t and y = 2 t - 3 , then for what value of t does x = y ? | "we are given x = 1 – 3 t and y = 2 t – 3 , and we need to determine the value for t when x = y . we should notice that both x and y are already in terms of t . thus , we can substitute 1 – 3 t for x and 2 t – 3 for y in the equation x = y . this gives us : 1 – 3 t = 2 t – 3 4 = 5 t 4 / 5 = t the answer is c ." | a = 1 + 2
b = 2 + 3
c = a / b
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a ) rs 25 , b ) rs 30 , c ) rs 35 , d ) rs 40 , e ) none of these | a | divide(divide(multiply(multiply(divide(add(multiply(add(const_4, const_3), multiply(add(const_4, const_1), const_2)), const_3), add(divide(multiply(const_60, const_60), multiply(add(const_4, const_1), const_2)), add(const_4, const_1))), 2000), 25), const_4), const_100) | find the simple interest on the rs . 2000 at 25 / 4 % per annum for the period from 4 th feb 2005 to 18 th april 2005 | explanation : one thing which is tricky in this question is to calculate the number of days . always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted . so lets calculate the number of days now , time = ( 24 + 31 + 18 ) days = 73 / 365 years = 1 / 5 years p = 2000 r = 25 / 4 % s . i . = = 2000 × 254 × 5 × 100 = 25 answer : a | a = 4 + 3
b = 4 + 1
c = b * 2
d = a * c
e = d + 3
f = const_60 * const_60
g = 4 + 1
h = g * 2
i = f / h
j = 4 + 1
k = i + j
l = e / k
m = l * 2000
n = m * 25
o = n / 4
p = o / 100
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['a ) 120 l', 'b ) 180 l', 'c ) 240 l', 'd ) 300 l', 'e ) 360 l'] | b | subtract(multiply(multiply(6, const_10), const_4), multiply(6, const_10)) | volume of a right circular cylinder is 6 o l . if radius of cylinder is doubled , then what will be the increase in volume ? | pi ∗ r ^ 2 ∗ h = 60 double the radius . . . . . pi ∗ ( 2 r ) ^ 2 ∗ h = pi ∗ 4 ∗ r ^ 2 ∗ h = 4 ∗ 60 = 240 important . . we are looking at the increase so 240 - 60 = 180 answer : b | a = 6 * 10
b = a * 4
c = 6 * 10
d = b - c
|
a ) 13 % , b ) 8.2 % , c ) 9 % , d ) 14 % , e ) 12 % | b | multiply(subtract(const_1, multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(2, const_100)))), const_100) | you enter a weight loss challenge game and manage to lose 10 % of your body weight . for the final weigh in you are forced to wear clothes that add 2 % to your weight . what percentage of weight loss is measured at the final weigh in ? | "( 100 % - 10 % ) * ( 100 % + 2 % ) = 0.90 * 1.02 = 8.2 % the weigh in records your weight loss at 8.2 % ! the answer is b" | a = 10 / 100
b = 1 - a
c = 2 / 100
d = 1 + c
e = b * d
f = 1 - e
g = f * 100
|
a ) 18 , b ) 92 , c ) 27 , d ) 26 , e ) 30 | e | add(multiply(5, 3), multiply(5, 3)) | a person was asked to state his age in years . his reply was , ` ` take my age 5 years hence , multiply it by 3 and subtract 3 times my age 5 years ago and you will know how old i am . ' ' what was the age of the person ? | "explanation : let the present age of person be x years . then , 3 ( x + 5 ) - 3 ( x - 5 ) = x < = > ( 3 x + 15 ) - ( 3 x - 15 ) = x < = > x = 30 . . answer : e" | a = 5 * 3
b = 5 * 3
c = a + b
|
a ) 32.9 , b ) 32.4 , c ) 32.2 , d ) 32.1 , e ) 32.7 | b | add(divide(circumface(6.3), const_2), multiply(6.3, const_2)) | the radius of a semi circle is 6.3 cm then its perimeter is ? | "36 / 7 r = 6.3 = 32.4 answer : b" | a = circumface / (
b = a + 2
|
a ) 90 seconds , b ) 85 seconds , c ) 40 seconds , d ) 100 seconds , e ) 120 seconds | d | divide(add(300, 1200), divide(multiply(54, const_1000), const_3600)) | a train is 300 meter long is running at a speed of 54 km / hour . in what time will it pass a tunnel of 1200 meter length ? | speed = 54 km / hr = 54 * ( 5 / 18 ) m / sec = 15 m / sec total distance = 300 + 1200 = 1500 meter time = distance / speed = 1500 * ( 15 ) = 100 seconds answer : d | a = 300 + 1200
b = 54 * 1000
c = b / 3600
d = a / c
|
a ) 2 / 3 , b ) 3 / 5 , c ) 4 / 7 , d ) 8 / 9 , e ) 1 / 3 | e | divide(const_2, add(4, const_2)) | in a single throw of a die , what is probability of getting a number greater than 4 ? | "e = { 5,6 } n ( e ) = 2 p ( s ) = n ( e ) / n ( s ) = 2 / 6 = 1 / 3 ans : e" | a = 4 + 2
b = 2 / a
|
a ) 0.09 , b ) 0.15 , c ) 0.54 , d ) 0.8 , e ) 0.91 | d | divide(add(add(22, 18), 8), 60) | a certain bag contains 60 balls — 22 white , 18 green , 8 yellow , 5 red , and 7 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 18 + 8 ) / 60 = 48 / 60 = 0.8 . answer : d ." | a = 22 + 18
b = a + 8
c = b / 60
|
['a ) 1 / 6', 'b ) 1 / 5', 'c ) 3 / 10', 'd ) 1 / 3', 'e ) 4 / 10'] | e | multiply(const_2, divide(const_2, 10)) | a 10 meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | a square with an area of 4 has a perimeter of 8 . for the area to be > 4 , the longer piece must be > 8 . the wire must be cut within 2 meters from either end . the probability of this is 4 / 10 = 2 / 5 . the answer is e . | a = 2 / 10
b = 2 * a
|
a ) 40 , b ) 36.5 , c ) 46.5 , d ) 20.5 , e ) 40.6 | e | divide(add(multiply(40, 50), subtract(subtract(50, const_2), 15)), 50) | the mean of 50 observations was 40 . it was found later that an observation 45 was wrongly taken as 15 the corrected new mean is | "explanation : correct sum = ( 40 * 50 + 45 - 15 ) = 2030 correct mean = = 2030 / 50 = 40.6 answer : e" | a = 40 * 50
b = 50 - 2
c = b - 15
d = a + c
e = d / 50
|
a ) a ) 7 , b ) b ) 2 , c ) c ) 8.73 , d ) d ) 5.77 , e ) e ) 3 | d | divide(power(4, 4), 4) | the current of a stream runs at the rate of 4 kmph . a boat goes 6 km and back to the starting point in 4 hours , then find the speed of the boat in still water ? | "s = 4 m = x ds = x + 4 us = x - 4 6 / ( x + 4 ) + 6 / ( x - 4 ) = 4 x = 5.77 answer : d" | a = 4 ** 4
b = a / 4
|
a ) 28 , b ) 26 , c ) 25 , d ) 23 , e ) 21 | e | multiply(divide(const_1, multiply(add(const_100, 20), divide(const_1, subtract(const_100, 20)))), 14) | by selling 14 pencils for a rupee a man loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "80 % - - - 14 120 % - - - ? 80 / 120 * 14 = 21 answer : e" | a = 100 + 20
b = 100 - 20
c = 1 / b
d = a * c
e = 1 / d
f = e * 14
|
a ) 3.5 , b ) 3.75 , c ) 4 , d ) 4.25 , e ) 4.5 | a | divide(350, add(divide(multiply(3, 1000), const_100), divide(multiply(1400, 5), const_100))) | a money lender lent rs . 1000 at 3 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years . | ( 1000 xtx 3 / 100 ) + ( 1400 xtx 5 / 100 ) = 350 → t = 3.5 answer a | a = 3 * 1000
b = a / 100
c = 1400 * 5
d = c / 100
e = b + d
f = 350 / e
|
a ) 5 , b ) 10 , c ) 20 , d ) 41 , e ) 45 | d | divide(subtract(70.50, 50), subtract(1.50, 1.00)) | caleb spends $ 70.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | "solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 70.50 . by solving the equations y = 41 . ans d ." | a = 70 - 50
b = 1 - 50
c = a / b
|
a ) 28.36 mph , b ) 26.55 mph , c ) 28.57 mph , d ) 25.56 mph , e ) 28.45 mph | c | divide(100, add(divide(50, subtract(100, 50)), divide(50, 20))) | tom traveled the entire 100 miles trip . if he did the first 50 miles of at a constant rate 20 miles per hour and the remaining trip of at a constant rate 50 miles per hour , what is the his average speed , in miles per hour ? | "avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 50 + 50 ) / ( ( 50 / 20 ) + ( 50 / 50 ) ) = 60 * 2 / 3 = 28.57 mph c" | a = 100 - 50
b = 50 / a
c = 50 / 20
d = b + c
e = 100 / d
|
a ) 2 , b ) 1 , c ) 0 , d ) - 1 , e ) - 2 | d | subtract(multiply(7, divide(subtract(multiply(2, 4), 5), subtract(multiply(6, 3), multiply(2, 11)))), add(multiply(11, divide(subtract(multiply(2, 4), 5), subtract(multiply(6, 3), multiply(2, 11)))), 4)) | when positive integer x is divided by 11 , the quotient is y and the remainder is 4 . when 2 x is divided by 6 , the quotient is 3 y and the remainder is 5 . what is the value of 7 y – x ? | "( 1 ) x = 11 y + 4 ( 2 ) 2 x = 18 y + 5 let ' s subtract equation ( 1 ) from equation ( 2 ) . 7 y + 1 = x 7 y - x = - 1 the answer is d ." | a = 2 * 4
b = a - 5
c = 6 * 3
d = 2 * 11
e = c - d
f = b / e
g = 7 * f
h = 2 * 4
i = h - 5
j = 6 * 3
k = 2 * 11
l = j - k
m = i / l
n = 11 * m
o = n + 4
p = g - o
|
a ) 1 / 4 , b ) 7 / 12 , c ) 2 / 3 , d ) 1 , e ) 8 / 7 | d | divide(divide(divide(30, const_100), divide(30, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(30, const_100))) | a total of 30 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 30 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a certain sex = ( number of geese of that sex migrating ) / ( total number of geese of that sex ) ] | "let ' take the number of geese to be 100 . male = 30 . female = 70 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 20 % are male i . e 20 / 100 * 30 = 6 geese ( males ) and now we know out of the total 20 geese , 6 are male , then 14 have to be female . now the ratio part , male geese ratios = 6 / 30 = 1 / 5 . - a female geese ratios = 14 / 70 = 1 / 5 - b cross multiply equations a and b and you get = 1 . ans d" | a = 30 / 100
b = 30 / 100
c = a / b
d = 2 * 4
e = d * 10
f = e / 100
g = 30 / 100
h = f / g
i = c / h
|
['a ) 7 : 15', 'b ) 7 : 5', 'c ) 7 : 25', 'd ) 7 : 45', 'e ) 7 : 55'] | b | divide(add(10, divide(4, multiply(const_3, const_4))), add(71, divide(9, multiply(const_3, const_4)))) | a clock hangs on the wall of a railway station , 71 ft 9 in long and 10 ft 4 in high . those are the dimension of the wall , not the clock ! while waiting for a train we noticed that the hands of the clock were pointing in opposite directions , and were parallel to one of the diagonals of the wall , what was the exact time ? | it can have four possiblities time could be 7 : 5 1 : 35 11 : 25 5 : 55 answer : b | a = 3 * 4
b = 4 / a
c = 10 + b
d = 3 * 4
e = 9 / d
f = 71 + e
g = c / f
|
a ) 18 , b ) 28 , c ) 26 , d ) 27 , e ) 23 | a | subtract(divide(multiply(60, 50), const_100), divide(multiply(40, 30), const_100)) | how much 60 % of 50 is greater than 40 % of 30 ? | "( 60 / 100 ) * 50 – ( 40 / 100 ) * 30 30 - 12 = 18 answer : a" | a = 60 * 50
b = a / 100
c = 40 * 30
d = c / 100
e = b - d
|
a ) $ 9 , b ) $ 3 , c ) $ 4 , d ) $ 6 , e ) $ 11 | e | add(divide(subtract(19, 3), const_2), 3) | you and your friend spent a total of $ 19 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 ) = 19 l + l + 3 - 3 = 19 - 3 2 l = 16 l = 8 my friends lunch l + 3 = 8 + 3 = $ 11 , the answer is e | a = 19 - 3
b = a / 2
c = b + 3
|
a ) 33.3 % , b ) 32.3 % , c ) 31.3 % , d ) 30.3 % , e ) 29.3 % | a | multiply(divide(30, 90), const_100) | by selling 90 pens , a trader gains the cost of 30 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 90 pens = rs . 90 profit = cost of 30 pens = rs . 30 profit % = 30 / 90 * 100 = 33.3 % answer : a" | a = 30 / 90
b = a * 100
|
a ) 4 hours , b ) 20 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours | a | inverse(subtract(divide(const_1, 2), divide(const_1, add(2, 2)))) | a cistern is normally filled in 2 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 2 - 1 / x = 1 / 4 , solving 1 / x = 1 / 4 so in 4 hours full cistern will be empty . answer : a" | a = 1 / 2
b = 2 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) 70 / 7 sec , b ) 80 / 7 sec , c ) 40 / 7 sec , d ) 60 / 7 sec , e ) 90 / 7 sec | b | divide(100, multiply(add(54, 72), const_0_2778)) | two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 72 kmph . in how much time will the trains cross each other ? | "b relative speed = ( 54 + 72 ) * 5 / 18 = 7 * 5 = 35 mps . the time required = d / s = ( 100 + 100 + 200 ) / 35 = 400 / 35 = 80 / 7 sec ." | a = 54 + 72
b = a * const_0_2778
c = 100 / b
|
a ) 6.09 % , b ) 6.10 % , c ) 6.12 % , d ) 6.14 % , e ) none of these | a | subtract(multiply(power(add(divide(divide(6, const_2), const_100), const_1), const_2), const_100), const_100) | effective annual rate of interest corresponding to nominal rate of 6 % per annum compounded half yearly will be | explanation : let the amount rs 100 for 1 year when compounded half yearly , n = 2 , rate = 6 / 2 = 3 % amount = 100 ( 1 + 3 / 100 ) 2 = 106.09 effective rate = ( 106.09 - 100 ) % = 6.09 % option a | a = 6 / 2
b = a / 100
c = b + 1
d = c ** 2
e = d * 100
f = e - 100
|
a ) 3 , b ) 5 , c ) 8 , d ) 11 , e ) 17 | b | divide(1365, multiply(multiply(add(const_2, const_3), add(add(const_2, const_3), const_2)), add(const_10, const_3))) | in a certain business school class , p students are accounting majors , q students are finance majors , r students are marketing majors , and s students are strategy majors . if pqrs = 1365 , and if 1 < p < q < r < s , how many students in the class are finance majors ? | "pqrs = 1365 = 3 * 5 * 7 * 13 since 1 < p < q < r < s , the number of students who are finance majors is q = 5 . the answer is b ." | a = 2 + 3
b = 2 + 3
c = b + 2
d = a * c
e = 10 + 3
f = d * e
g = 1365 / f
|
a ) 298 m , b ) 300 m , c ) 800 m , d ) 967 m , e ) 1181 m | c | divide(500, multiply(subtract(78, 1), const_0_2778)) | a train 500 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 , ( 500 + x ) / 60 = 65 / 3 x = 800 m . answer : c" | a = 78 - 1
b = a * const_0_2778
c = 500 / b
|
a ) 22 , b ) 30 , c ) 25 , d ) 38 , e ) 27 | c | add(divide(add(21, 23), const_2), multiply(const_1, 3)) | the average age of 3 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 3 * 2 ) years = 6 years . sum of ages of two new men = ( 21 + 23 + 6 ) years = 50 years average age of two new men = ( 50 / 2 ) years = 25 years . answer : c" | a = 21 + 23
b = a / 2
c = 1 * 3
d = b + c
|
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