Problem
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32
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stringlengths 1
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| options
stringlengths 37
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| correct
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848
| linear_formula
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| category
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|---|---|---|---|---|---|---|
a that box?
|
b 425 ( 68 ã · 16 ) ã — 100
|
a ) 336 , b ) 425 , c ) 275 , d ) 235 , e ) 689
|
b
|
subtract(multiply(multiply(68, const_4), const_2), const_100)
|
multiply(n0,const_4)|multiply(#0,const_2)|subtract(#1,const_100)
|
general
|
the from same?
|
drop = 12 inches / day 5 days ago = w , means now it ' s equal w - 60 and in 4 days = w - 60 - 48 = w - 108 answer a
|
a ) w − 108 , b ) w − 56 , c ) w − 14 , d ) w + 14 , e ) w + 126
|
a
|
multiply(12, divide(4, 5))
|
divide(n2,n1)|multiply(n0,#0)
|
gain
|
how many apples?
|
sol . apples 250 each carries 25 = 250 / 25 = 10 answer : d
|
a ) a ) 9 , b ) b ) 5 , c ) c ) 7 , d ) d ) 10 , e ) e ) none of the above
|
d
|
divide(250, 25)
|
divide(n0,n1)
|
general
|
how hr hr?
|
"15 km / hr = 15000 m / 3600 s = ( 150 / 36 ) m / s = ( 25 / 6 ) m / s time = 600 / ( 25 / 6 ) = 144 seconds the answer is c ."
|
a ) 128 , b ) 136 , c ) 144 , d ) 152 , e ) 160
|
c
|
divide(600, multiply(15, const_0_2778))
|
multiply(n1,const_0_2778)|divide(n0,#0)|
|
physics
|
if persons capitals is
|
"total profit = 1000 ratio = 600 / 300 = 2 : 1 answer : e"
|
a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 2 : 1
|
e
|
divide(600, 300)
|
divide(n0,n1)|
|
other
|
the observations is :
|
"correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer a"
|
a ) 36.5 , b ) 35 , c ) 34 , d ) 33 , e ) 32.5
|
a
|
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
|
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
|
general
|
a remaining exercise?
|
"let us suppose there are 100 people . 50 % of them donated $ 25000 ( 500 * 50 ) $ 25000 is 60 % of total amount . so total amount = 25000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 25000 * ( 100 / 60 ) * ( 40 / 100 ) = 50000 / 3 this amount has to be divided by 50 ( remaining people are 50 ) so per head amount is 50000 / 3 / 50 = 32000 / 180 = 333.33 ; answer : b"
|
a ) $ 200 , b ) $ 333.33 , c ) $ 100.25 , d ) $ 277.78 , e ) $ 377.00
|
b
|
divide(multiply(divide(multiply(divide(50, const_100), 500), divide(60, const_100)), divide(50, const_100)), divide(60, const_100))
|
divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)|
|
general
|
a bacteria limit?
|
"if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 18 times . answer : d . similar questions to practice : hope it helps ."
|
a ) 6.33 , b ) 7.5 , c ) 10 , d ) 18 , e ) 19
|
d
|
subtract(19, divide(19, 19))
|
divide(n0,n0)|subtract(n0,#0)|
|
physics
|
the of circle.
|
circumference = 2 * pi * r = 2 * pi * 4 / pi = > 8 a
|
['a ) 8', 'b ) 4 π', 'c ) 4', 'd ) 6', 'e ) 5']
|
a
|
circumface(divide(4, const_pi))
|
divide(n0,const_pi)|circumface(#0)
|
geometry
|
s.3
|
"given exp . = 0.3 * 0.3 + ( 0.3 * 0.3 ) = 0.09 + 0.09 = 0.18 answer is c ."
|
a ) 0.52 , b ) 0.42 , c ) 0.18 , d ) 0.64 , e ) 0.46
|
c
|
add(multiply(0.3, 0.3), multiply(0.3, 0.3))
|
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
|
general
|
how between280?
|
straight a way lets exclude all the even numbers between 260 and 280 . so now the number starts from 261 to 279 ( only odd ) 261 is a divisible of 3 and next odd divisible by 3 will be 261 + 6 = 267 + 6 = 273 + 6 = 279 . also we can eliminate numbers ending with ' 5 ' so in odd , the excluded numbers are 261 , 265,267 , 273,279 , which leave us with 263,269 , 271,277 . checked the above listed four numbers are divisible by any numbers till 20 . answer : e
|
a ) none , b ) one , c ) two , d ) three , e ) four
|
e
|
subtract(divide(subtract(280, 260), const_4), const_1)
|
subtract(n1,n0)|divide(#0,const_4)|subtract(#1,const_1)
|
general
|
one it time?
|
"first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 = 5 / 2 so they complete one robot in 2 / 5 th of an hour - ( 3 ) now the 5 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 3 so they complete one robot in 1 / 3 th of an hour - ( 4 ) now the 6 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 7 / 2 so they complete one robot in 2 / 7 th of an hour - ( 5 ) now the 7 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 9 / 2 so they complete one robot in 2 / 9 th of an hour - ( 6 ) and now we have 8 robots so total = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ( 5 ) + ( 6 ) = 146 1 / 7 minutes answer - e"
|
a ) 70 min , b ) 94 min , c ) 110 min , d ) 131 1 / 7 min , e ) 146 1 / 7 min
|
e
|
add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2))))
|
inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)|
|
physics
|
{ -0?
|
x = - 2.5 prob = 1 / 12 answer - a
|
a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2
|
a
|
divide(1, multiply(6, 2))
|
multiply(n1,n14)|divide(n5,#0)
|
general
|
a of hour?
|
"the distance between the nest and the ditch is 300 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 300 = 9000 . d = st 9000 / 1.5 = t , i think we can take 9000 meters as 9 km , then only we get t = 6 . ( 1000 meters = 1 km ) d )"
|
a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8
|
d
|
divide(divide(multiply(300, multiply(15, const_2)), const_1000), divide(15, const_10))
|
divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)|
|
physics
|
a and q?
|
"now given that q is set the consecutive integers between a and b . and q contains 9 multiples of 9 let take a as 36 . then 36 45 54 63 72 81 90 99 108 . . . so b will 108 . now let ' s check the multiples of 4 among this set 108 - 36 / 4 + 1 = > 18 + 1 = > 19 ans option b ."
|
a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22
|
b
|
subtract(multiply(9, const_2), const_1)
|
multiply(n1,const_2)|subtract(#0,const_1)|
|
physics
|
find (20 )
|
"20 c 20 = 1 ( 20 c 2 ) * ( 20 c 20 ) = 20 ! * 1 / 18 ! = 20 * 19 * 18 ! / 18 ! = 20 * 19 * 1 = 380 answer : b"
|
a ) 400 , b ) 380 , c ) 360 , d ) 350 , e ) 330
|
b
|
multiply(add(divide(18, 20), 20), 20)
|
divide(n1,n2)|add(n0,#0)|multiply(#1,n2)|
|
general
|
if test ).
|
"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 30 % - 50 % = 20 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.30 x 95 + 0.50 x 79 + 0.20 x 60 = 80 the class average ( rounded ) is 80 % final answer e ) 80 %"
|
a ) 76 % , b ) 77 % , c ) 78 % , d ) 79 % , e ) 80 %
|
e
|
divide(add(add(multiply(30, 95), multiply(50, 79)), multiply(subtract(const_100, add(30, 50)), 60)), const_100)
|
add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)|
|
gain
|
two and profit of
|
"the profit on the first kind of vodka = x % ; the profit on the second kind of vodka = y % . when they are mixed in the ratio 1 : 2 ( total of 3 parts ) the average profit is 10 % : ( x + 2 y ) / 3 = 10 . when they are mixed in the ratio 2 : 1 ( total of 3 parts ) the average profit is 20 % : ( 2 x + y ) / 3 = 20 . solving gives : x = 30 % and y = 0 % . after the individual profit percent on them areincreased by 4 / 3 and 5 / 3 times respectively the profit becomes 40 % and 0 % , on the first and te second kinds of vodka , respectively . if they are mixed in equal ratio ( 1 : 1 ) , then the mixture will fetch the profit of ( 40 + 0 ) / 2 = 20 % . answer : a"
|
a ) 20 % , b ) 40 % , c ) 18 % , d ) 23 % , e ) can not be determined
|
a
|
add(divide(multiply(10, 4), 6), add(10, 5))
|
add(n4,n8)|multiply(n4,n6)|divide(#1,const_2.0)|add(#0,#2)|
|
general
|
if6 there?
|
if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? a . 2 b . 3 c . 4 d . 5 e . 6 - > if m = y = 60 , 2 ( m + y ) = 240 is derived . if m = y = 69 , 2 ( m + y ) = 276 is derived , which makes 4,5 , 6,7 possible for the tens digit . therefore , the answer is c .
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
|
c
|
subtract(6, 2)
|
subtract(n0,n1)
|
physics
|
the way pencils?
|
"number of pens = 848 number of pencils = 630 required number of students = h . c . f . of 848 and 630 = 2 answer is c"
|
a ) 10 , b ) 4 , c ) 2 , d ) 14 , e ) 16
|
c
|
gcd(848, 630)
|
gcd(n0,n1)|
|
general
|
find of2?
|
1 : 2 answer : a
|
['a ) 1 : 2', 'b ) 2 : 3', 'c ) 2 : 9', 'd ) 2 : 1', 'e ) 2 : 2']
|
a
|
divide(1, 2)
|
divide(n0,n1)
|
geometry
|
n3 m?
|
"you have 6 digits : 1 , 2 , 3 , 7 , 8 , 9 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 1 and 9 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 2 * * and 3 * * ) . make the 2 * * number as large as possible and make the 3 * * number as small as possible . 298 and 317 ( difference is 19 ) or try and make a pair with ( 7 * * and 8 * * ) . make the 7 * * number as large as possible and make the 8 * * number as small as possible . we get 793 and 812 ( difference is 19 ) a"
|
a ) 19 , b ) 49 , c ) 58 , d ) 113 , e ) 131
|
a
|
subtract(subtract(const_100, multiply(subtract(8, 1), const_10)), const_1)
|
subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)|
|
general
|
the is team?
|
"explanation : let the average age of the whole team by x years . 11 x â € “ ( 29 + 32 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer e"
|
a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 26 years
|
e
|
divide(subtract(add(29, add(29, 3)), multiply(3, 3)), const_2)
|
add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)|
|
general
|
if is area?
|
"the triangle with sides 31 cm , 29 cm and 15 cm is right angled , where the hypotenuse is 31 cm . area of the triangle = 1 / 2 * 29 * 15 = 217.5 cm 2 answer : e"
|
a ) 220.75 cm 2 , b ) 258 cm 2 , c ) 225.50 cm 2 , d ) 222.25 cm 2 , e ) 217.5 cm 2
|
e
|
divide(multiply(29, 15), const_2)
|
multiply(n1,n2)|divide(#0,const_2)|
|
geometry
|
in obtain kg?
|
"let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 6.5 ) = 5 / 4 answer : c"
|
a ) 5 / 6 , b ) 5 / 9 , c ) 5 / 4 , d ) 5 / 3 , e ) 7 / 6
|
c
|
divide(divide(subtract(8.75, 7.50), subtract(8.75, 6.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6.5))))
|
subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|
|
other
|
a colour colour is
|
"drawing two balls of same color from seven green balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = ⁷ c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 7 / 15 answer : e"
|
a ) 7 / 16 , b ) 7 / 12 , c ) 7 / 19 , d ) 7 / 12 , e ) 7 / 15
|
e
|
add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 8), const_1))))
|
add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)|
|
other
|
the of salary?
|
"explanation : manager ' s monthly salary rs . ( 1500 * 21 - 1400 * 20 ) = rs . 3500 . answer : e"
|
a ) 3600 , b ) 3890 , c ) 88798 , d ) 2789 , e ) 3500
|
e
|
subtract(multiply(add(1400, 100), add(20, const_1)), multiply(1400, 20))
|
add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)|
|
general
|
4 many it?
|
"let 1 man ' s 1 day work = x and 1 woman ' s 1 day work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving these two equations , we get : x = 11 / 400 and y = 1 / 400 1 woman ' s 1 day work = ( 1 / 400 * 10 ) = 1 / 40 . hence , 10 women will complete the work in 40 days . answer : b"
|
a ) 21 days , b ) 40 days , c ) 27 days , d ) 18 days , e ) 17 days
|
b
|
inverse(multiply(divide(subtract(divide(const_1, 10), multiply(3, divide(subtract(divide(const_1, 8), multiply(divide(6, 7), divide(const_1, 10))), subtract(4, multiply(3, divide(6, 7)))))), 7), 8))
|
divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)|
|
physics
|
a the desk?
|
"purchase price = 180 selling price = x 180 + 0.4 * x = x 0.6 * x = 180 x = 300 profit = 300 - 180 = 120 answer : d"
|
a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 120 , e ) $ 100
|
d
|
divide(multiply(subtract(divide(180, subtract(const_1, divide(40, const_100))), 180), const_100), 180)
|
divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)|
|
gain
|
a in train?
|
"length = speed * time speed = l / t s = 400 / 10 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : b"
|
a ) 165 kmph , b ) 144 kmph , c ) 172 kmph , d ) 175 kmph , e ) 178 kmph
|
b
|
divide(divide(400, const_1000), divide(10, const_3600))
|
divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)|
|
physics
|
a at is?
|
"ds = 30 us = 14 s = ? s = ( 30 - 14 ) / 2 = 8 kmph answer : e"
|
a ) 1 kmph , b ) 4 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph
|
e
|
divide(subtract(30, 14), const_2)
|
subtract(n0,n1)|divide(#0,const_2)|
|
gain
|
3 work?
|
explanation : work done by 4 men and 6 women in 1 day = 1 / 8 work done by 3 men and 7 women in 1 day = 1 / 10 let 1 man does m work in 1 day and 1 woman does w work in 1 day . the above equations can be written as 4 m + 6 w = 1 / 8 - - - ( 1 ) 3 m + 7 w = 1 / 10 - - - ( 2 ) solving equation ( 1 ) and ( 2 ) , we get m = 11 / 400 and w = 1 / 400 amount of work 10 women can do in a day = 10 × ( 1 / 400 ) = 1 / 40 ie , 10 women can complete the work in 40 days answer : option b
|
a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10
|
b
|
inverse(multiply(divide(subtract(divide(const_1, 8), multiply(4, divide(subtract(divide(const_1, 10), multiply(divide(7, 6), divide(const_1, 8))), subtract(3, multiply(4, divide(7, 6)))))), 6), 10))
|
divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)
|
physics
|
what logarith4?
|
log ( 0.0000134 ) . since there are four zeros between the decimal point and the first significant digit , the characteristic is – 5 . answer : b
|
a ) 5 , b ) - 5 , c ) 6 , d ) - 6 , e ) 7
|
b
|
floor(divide(log(0.0000134), log(const_10)))
|
log(n0)|log(const_10)|divide(#0,#1)|floor(#2)
|
other
|
in. selection?
|
this is equivalent to : - 2 x * 4 y * 5 z = 16000 y / 2 = z ( given ) 2 x * 4 y * 5 y / 2 = 16000 2 x * y ^ 2 = 16000 / 10 2 x * y ^ 2 = 1600 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 2 as 2 * 2 * y ^ 2 = 1600 y ^ 2 = 400 y = 20 number of red chips = 2 hence b
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
|
b
|
divide(multiply(multiply(power(2, 4), power(2, const_3)), power(5, const_3)), multiply(power(const_2, multiply(2, const_3)), power(5, const_3)))
|
multiply(n0,const_3)|power(n0,n1)|power(n0,const_3)|power(n2,const_3)|multiply(#1,#2)|power(const_2,#0)|multiply(#4,#3)|multiply(#5,#3)|divide(#6,#7)
|
general
|
compound ).
|
"10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 2000 [ 1 + ( 5 / 100 ) ] ^ 3 - 2000 = 2000 [ ( 21 / 20 ) ^ 3 - 1 ] = 2000 ( 9261 - 8000 ) / 8000 = 2 * 1261 / 8 = 315 answer : d"
|
a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 315 , e ) none of the above
|
d
|
subtract(multiply(2000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 2000)
|
divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)|
|
gain
|
the height area?
|
"cone curved surface area = ï € rl 22 / 7 ã — 49 ã — 35 = 154 ã — 35 = 5390 m ( power 2 ) answer is b ."
|
a ) 5160 , b ) 5390 , c ) 6430 , d ) 6720 , e ) 7280
|
b
|
volume_cone(49, 35)
|
volume_cone(n0,n1)|
|
geometry
|
the of is :
|
"perimeter = distance covered in 10 min . = ( 12000 / 60 ) x 10 m = 2000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2000 or x = 200 . length = 600 m and breadth = 400 m . area = ( 600 x 400 ) m 2 = 240000 m 2 . answer : e"
|
a ) 153601 , b ) 153600 , c ) 153602 , d ) 153603 , e ) 240000
|
e
|
rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), 2))
|
add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|
|
physics
|
a the is?
|
"speed = ( 11 / 10 * 60 ) km / hr = ( 66 * 5 / 18 ) m / sec = 55 / 3 m / sec . length of the train = 55 / 3 * 6 = 110 m . answer : c"
|
a ) m , b ) m , c ) m , d ) m , e ) m
|
c
|
divide(11, subtract(divide(11, 10), 6))
|
divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)|
|
physics
|
on the class?
|
( 38 g + 20 b ) / ( g + b ) = 30 38 g + 20 b = 30 ( g + b ) 8 g = 10 b b / g = 4 / 5 the answer is d .
|
a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6
|
d
|
divide(30, 38)
|
divide(n2,n1)
|
general
|
in last last?
|
"i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 5 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 5 = 1000 ( m * n ) water available is 2000 ( p ) it will last 2 months ( p / m * n ) ans : d"
|
a ) 9 , b ) 5 , c ) 6 , d ) 2 , e ) 4
|
d
|
divide(2000, multiply(200, 5))
|
multiply(n0,n1)|divide(n2,#0)|
|
gain
|
sum is numbers?
|
"as two numbers are prime , only two options satisfy ie option c and d . but option c will not make the product of numbers i . e 99 answer : d"
|
a ) 8 and 12 , b ) 14 and 6 , c ) 19 and 1 , d ) 11 and 9 , e ) 12 and 9
|
d
|
add(99, 20)
|
add(n0,n1)|
|
physics
|
a train is?
|
"s = 180 / 20 * 18 / 5 = 32 kmph answer : c"
|
a ) 37 kmph , b ) 35 kmph , c ) 32 kmph , d ) 38 kmph , e ) 36 kmph
|
c
|
multiply(const_3_6, divide(180, 20))
|
divide(n0,n1)|multiply(#0,const_3_6)|
|
physics
|
rs of years?
|
"solution s . i . = rs . ( 956 - 925 ) = rs . 31 rate = ( 100 x 31 / 925 x 3 ) = 124 / 111 % new rate = ( 124 / 111 + 4 ) % = 568 / 111 % new s . i . = rs . ( 925 x 568 / 111 x 3 / 100 ) rs . 142 ∴ new amount = rs . ( 925 + 142 ) = rs . 1067 . answer c"
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1067 , d ) data inadequate , e ) none of these
|
c
|
add(925, divide(multiply(multiply(925, add(divide(multiply(subtract(956, 925), const_100), multiply(925, 3)), 4)), 3), const_100))
|
multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|
|
gain
|
ex 21?
|
"the total number of the people in the room must be a multiple of both 7 and 12 ( in order 3 / 7 and 5 / 12 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 12 , which is 84 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 84 people in the room . therefore there are 3 / 7 * 84 = 36 people in the room under the age of 21 . answer : c ."
|
a ) 21 , b ) 35 , c ) 36 , d ) 60 , e ) 65
|
c
|
divide(multiply(multiply(7, 12), 3), 7)
|
multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)|
|
general
|
s the is?
|
"3 x * 2 x = 2460 = > x = 20.24 2 ( 79.76 + 50 ) = 259.52 m 259.52 * 1 / 2 = rs . 129.76 answer : c"
|
a ) s . 122 , b ) s . 129 , c ) s . 129.76 , d ) s . 120 , e ) s . 121
|
c
|
divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(2460, 2), 3)), divide(2460, sqrt(divide(multiply(2460, 2), 3))))), const_100)
|
multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)|
|
physics
|
what between inclusive?
|
"to solve this problem , all you have to do is take every even number between 24 and 50 and add them together . so we have 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 , which is 481 . final answer : b"
|
a ) 592 , b ) 481 , c ) 330 , d ) 475 , e ) 483
|
b
|
add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 24), add(const_2, const_4))
|
add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)|
|
general
|
if, 29
|
r + b + c + 14 + 15 = 12 * 5 = 60 = > r + b + c = 60 - 29 = 31 r + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d
|
a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
|
d
|
divide(add(subtract(multiply(add(const_4, const_1), 12), add(14, 15)), 29), const_4)
|
add(const_1,const_4)|add(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|add(n3,#3)|divide(#4,const_4)
|
general
|
if b is :
|
expl : 15 % of a i = 30 % of b = 15 a / 100 = 30 b / 100 = 2 / 1 = 2 : 1 answer : e
|
a ) 1 : 4 , b ) 4 : 3 , c ) 6 : 7 , d ) 3 : 5 , e ) 2 : 1
|
e
|
divide(divide(30, const_100), divide(15, const_100))
|
divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1)
|
gain
|
let and )?
|
"p ( a and b ) = 1 / 2 * 1 / 5 = 1 / 10 the answer is b ."
|
a ) 1 / 18 , b ) 1 / 10 , c ) 1 / 5 , d ) 1 / 3 , e ) 1 / 2
|
b
|
multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2)))
|
subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)|
|
physics
|
what pm?
|
"required angle = 240 – 24 × ( 11 / 2 ) = 240 – 132 = 108 ° answer d"
|
a ) 100 ° , b ) 107 ° , c ) 106 ° , d ) 108 ° , e ) none of these
|
d
|
subtract(multiply(8, multiply(const_3, const_2)), 2)
|
multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)|
|
geometry
|
two immediately c?
|
let x per minute be the speed of c and y per minute be the speed of d . after meeting at a point , c travels for 32 mins and d travels for 50 mins . so distance covered by each of them post point of crossing c = 32 x and d = 50 y the distance covered by c and d before they cross each would be distance covered by d and c post crossing respectively . therefore distance covered by d before he meets c = 32 x time taken by d cover 32 x distance = 32 x / y mins therefore total time taken by d = 32 x / y + 50 mins . . . . . . . . . . . . . . . . . i we need to find value of x in terms of y to arrive at final answer . total distance = 32 x + 50 y combined speed of c and d = x + y therefore time taken before c and d meet en - route = ( 32 x + 50 y ) / ( x + y ) time taken by d reach destination after meeting c = 50 mins total travel time for d = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 mins . . . . . . . . . . . . . . . . . . . ii equate i and ii 32 x / y + 50 = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 ( 32 x + 50 y ) / y = ( 82 x + 100 y ) / ( x + y ) 32 x ^ 2 + 50 xy + 32 xy + 50 y ^ 2 = 82 xy + 100 y ^ 2 32 x ^ 2 + 82 xy - 82 xy + 50 y ^ 2 - 100 y ^ 2 = 0 32 x ^ 2 - 50 y ^ 2 = 0 32 x ^ 2 = 50 y ^ 2 16 x ^ 2 = 25 y ^ 2 taking square root . . ( since x and y denote speed , square root ca n ' t be negative ) 4 x = 5 y y = 4 x / 5 . . . . . . . . . . . . iii substitute in i = 32 x / ( 4 x / 5 ) + 50 = 32 x * 5 / 4 x + 50 = 40 + 50 = 90 mins a
|
a ) 90 , b ) 80 , c ) 75 , d ) 60 , e ) 65
|
a
|
add(sqrt(multiply(50, 32)), 50)
|
multiply(n0,n1)|sqrt(#0)|add(n1,#1)
|
physics
|
a to is :
|
"a 150 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 ."
|
a ) 150 , b ) 120 , c ) 130 , d ) 160 , e ) 210
|
a
|
divide(225, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))
|
divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)|
|
gain
|
the % workers?
|
"let x be the total worker then 0.5 x = female worker and 0.5 x is male worker then 20 male worker added 05 x / ( 0.5 x + 20 ) = 50 / 100 or 50 x = 50 * ( 0.5 x + 100 ) = 25 x + 5000 or 25 x = 5000 , x = 5000 / 25 = 200 total worker = 200 + 20 = 220 b"
|
a ) 225 , b ) 220 , c ) 230 , d ) 235 , e ) 240
|
b
|
add(divide(multiply(divide(50, const_100), 20), subtract(divide(const_60.0, const_100), divide(50, const_100))), 20)
|
divide(n2,const_100)|divide(const_60.0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)|
|
gain
|
x varies to :
|
"explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 5 ^ 2 = 9 / 25 answer : e"
|
a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 25
|
e
|
divide(multiply(1, power(3, const_2)), power(5, const_2))
|
power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)|
|
general
|
in play sports?
|
notice that 7 play both baseball and cricket does not mean that out of those 7 , some does not play football too . the same for cricket / football and baseball / football . [ color = # ffff 00 ] { total } = { baseball } + { cricket } + { football } - { hc + ch + hf } + { all three } + { neither } for more checkadvanced overlapping sets problems [ / color ] 50 = 20 + 15 + 11 - ( 7 + 4 + 5 ) + { all three } + 18 - - > { all three } = 2 ; those who play only baseball and cricket are 7 - 2 = 5 ; those who play only cricket and football are 4 - 2 = 2 ; those who play only baseball and football are 5 - 2 = 3 ; hence , 5 + 2 + 3 = 10 students play exactly two of these sports . answer : a .
|
a ) 10 , b ) 46 , c ) 67 , d ) 68 , e ) 446
|
a
|
add(subtract(5, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), add(subtract(7, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), subtract(4, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18)))))
|
add(n1,n2)|add(n4,n5)|add(n3,#0)|add(n6,#1)|subtract(#2,#3)|add(n7,#4)|subtract(n0,#5)|subtract(n4,#6)|subtract(n5,#6)|subtract(n6,#6)|add(#7,#8)|add(#10,#9)
|
other
|
the distance minutes?
|
"90 * 10 / 60 = 15 kmph answer : a"
|
a ) 15 , b ) 87 , c ) 99 , d ) 77 , e ) 55
|
a
|
multiply(divide(10, const_60), 90)
|
divide(n1,const_60)|multiply(n0,#0)|
|
physics
|
in the is :
|
"speed in still water = ( 11 + 5 ) / 2 = 8 kmph ans - c"
|
a ) 2 kmph , b ) 3 kmph , c ) 8 kmph , d ) 9 kmph , e ) 7 kmph
|
c
|
stream_speed(11, 5)
|
stream_speed(n0,n1)|
|
physics
|
a and x?
|
"sq rt ( 5 x / 3 ) = x = > 5 x / 3 = x ^ 2 = > x = 5 / 3 ans - d"
|
a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 5 / 3 , e ) 1 / 2
|
d
|
divide(5, 3)
|
divide(n0,n1)|
|
general
|
the y?
|
"suppose commodity x will cost 40 paise more than y after z years . then , ( 4.20 + 0.40 z ) - ( 6.30 + 0.15 z ) = 0.40 0.25 z = 0.40 + 2.10 z = 2.50 / 0.25 = 250 / 25 = 10 . therefore , x will cost 40 paise more than y 10 years after 2001 i . e . , 2011 . answer is d ."
|
a ) 2010 , b ) 2001 , c ) 2012 , d ) 2011 , e ) 2009
|
d
|
add(2001, divide(add(divide(40, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), divide(15, const_100))))
|
divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5)|
|
general
|
the find hence?
|
"a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 5 years , their ages will be 48 and 28 . sum of their ages = 48 + 28 = 76 . answer : d"
|
a ) 50 , b ) 60 , c ) 70 , d ) 76 , e ) 90
|
d
|
add(add(multiply(divide(60, 8), const_2), 8), add(divide(60, 8), 8))
|
divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)|
|
general
|
a his capital is
|
"explanation : capital = rs . x , then 5 / 7 x = 61.5 x = 87.86 answer : b ) rs . 87.86"
|
a ) 22.378 , b ) 87.86 , c ) 246.0 , d ) 78.88 , e ) 127.71
|
b
|
divide(61.50, divide(const_4, 7))
|
divide(const_4,n3)|divide(n4,#0)|
|
gain
|
the289 is?
|
"let the numbers be x and y . then , xy = 468 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 468 ) = 1225 x + y = 35 . option e"
|
a ) a ) 23 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35
|
e
|
sqrt(add(power(sqrt(subtract(289, multiply(const_2, 468))), const_2), multiply(const_4, 468)))
|
multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)|
|
general
|
the alcohol contain?
|
"soap : alcohol initial ratio soap : alcohol : water - - > 4 : 20 : 60 initial soap : alcohol = 4 / 20 = 4 : 20 after doubled soap : alcohol = 2 * 4 / 20 = 8 : 20 initial soap : water = 4 / 60 = 4 : 60 after halved soap : water : 1 / 2 * 4 / 60 = 2 / 60 = 2 : 60 after soap : alcohol : water - - > 8 : 20 : 240 - - > 2 : 5 : 60 given alcohol 100 cubic centimeter . ratio is 40 : 100 : 1200 ( 2 : 5 : 60 ) for 100 cubic centimeter of alcohol - - - 1200 cubic cm water is required ."
|
a ) 1200 , b ) 1250 , c ) 1300 , d ) 1400 , e ) 1450
|
a
|
divide(divide(divide(divide(divide(volume_rectangular_prism(100, 60, 20), const_3), const_2), 4), 4), 4)
|
volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,n0)|divide(#3,n0)|divide(#4,n0)|
|
geometry
|
if = )?
|
"since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 13 then f ( 5 ) = 13 - 5 = 8 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 8 - 6 = 2 . answer : d . questions on funtions to practice :"
|
a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 4
|
d
|
subtract(subtract(13, add(1, 4)), 6)
|
add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)|
|
general
|
what diameter diameter?
|
area of semicircle = ½ π r 2 = ½ × 22 ⁄ 7 × 7 × 7 = 77 m 2 answer b
|
['a ) 154 sq metres', 'b ) 77 sq metres', 'c ) 308 sq metres', 'd ) 22 sq metres', 'e ) none of these']
|
b
|
divide(circle_area(divide(14, const_2)), const_2)
|
divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)
|
geometry
|
the price table?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 2200 ( 100 / 110 ) = rs . 2000 answer : b"
|
a ) 2299 , b ) 2000 , c ) 2670 , d ) 6725 , e ) 2601
|
b
|
divide(2200, add(const_1, divide(10, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
one by weight?
|
"say the second solution ( which was 1 / 4 th of total ) was x % salt , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : b ."
|
a ) 24 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 %
|
b
|
multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(10, const_100), const_4), divide(10, const_100))), const_100)
|
divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)|
|
gain
|
if and a?
|
"the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 170 - - 2 subtracting 1 from 2 we have c - a = 80 ; answer : b"
|
a ) 25 , b ) 80 , c ) 90 , d ) 140 , e ) it can not be determined from the information given
|
b
|
subtract(multiply(85, const_2), multiply(45, const_2))
|
multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|
|
general
|
3 in day?
|
the required number of working hours per day x , more pumps , less working hours per day ( indirect ) less days , more working hours per day ( indirect ) pumps 4 : 3 , days 1 : 2 } : : 8 : x therefore 4 * 1 * x = 3 * 2 * 8 , x = ( 3 * 2 * 8 ) / 4 x = 12 correct answer ( d )
|
a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13
|
d
|
divide(multiply(multiply(3, 8), 2), 4)
|
multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)
|
physics
|
p was invested?
|
since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 810 = 44 interest earned for 3 years = 44 * 3 = 132 amount invested = 815 - 132 = 683 answer : b
|
a ) 670 , b ) 683 , c ) 698 , d ) 744 , e ) 700
|
b
|
subtract(810, multiply(divide(subtract(854, 810), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100)))
|
divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5)
|
gain
|
a.ght?
|
"800 * 9.2 7360.0 gm 7.36 kg answer : e"
|
a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.9 kg , e ) 7.36 kg
|
e
|
divide(multiply(9.2, 800), const_1000)
|
multiply(n0,n1)|divide(#0,const_1000)|
|
general
|
what sum25?
|
"1022 ã · 25 = 40 with remainder = 22 22 + 3 = 25 . hence 3 should be added to 1022 so that the sum will be divisible by 25 answer : option b"
|
a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 5
|
b
|
subtract(25, reminder(1022, 25))
|
reminder(n0,n1)|subtract(n1,#0)|
|
general
|
a at rate?
|
"a hostel had provisions for 250 men for 44 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 44 250 × 44 = 200 x 5 × 44 = 4 x x = 5 × 11 = 55 answer a"
|
a ) 55 , b ) 40 , c ) 50 , d ) 60 , e ) 65
|
a
|
divide(multiply(250, 44), subtract(250, 50))
|
multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)|
|
gain
|
the cost table?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 8339 ( 100 / 124 ) = rs . 6725 . answer : c"
|
a ) rs . 6825 , b ) rs . 6721 , c ) rs . 6725 . , d ) rs . 4298 , e ) rs . 6729
|
c
|
divide(8339, add(const_1, divide(24, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
63 ) =?
|
"63 + 5 * 12 / ( 180 / 3 ) = 63 + 5 * 12 / ( 60 ) = 63 + ( 5 * 12 ) / 60 = 63 + 1 = 64 . answer : d"
|
a ) 22 , b ) 77 , c ) 29 , d ) 64 , e ) 21
|
d
|
add(63, divide(multiply(5, 12), divide(180, 3)))
|
divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)|
|
general
|
in all cars?
|
12 * 1.75 + 0.45 * 12 * 55 = 318 hence - a
|
a ) 318 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $
|
a
|
multiply(multiply(0.45, 55), 12)
|
multiply(n1,n3)|multiply(n2,#0)|
|
general
|
if x x?
|
"since x is an integer , ( - 6 ) ^ 2 x is always positive . so , 6 ^ 2 x = 6 ^ ( 7 + x ) 2 x = 7 + x x = 7 answer : e"
|
a ) 5 , b ) 4 , c ) 3 , d ) 8 , e ) 7
|
e
|
divide(7, 6)
|
divide(n3,n0)|
|
general
|
a of duck?
|
"given ; 4 duck = 3 hen ; or , duck / hen = 3 / 4 ; let hen ' s 1 leap = 4 meter and ducks 1 leap = 3 meter . then , ratio of speed of hen and duck = 4 * 3 / 3 * 2 = 2 : 1 ' ' answer : 2 : 1 ;"
|
a ) 2 : 1 , b ) 3 : 4 , c ) 4 : 3 , d ) 1 : 4 , e ) 5 : 6
|
a
|
divide(divide(3, 2), divide(3, 4))
|
divide(n0,n1)|divide(n3,n2)|divide(#0,#1)|
|
other
|
pr0 capital?
|
"let hari ’ s capital be rs . x . then , 3640 * 12 / 7 x = 2 / 3 = > 14 x = 131040 = > x = 9360 . answer : e"
|
a ) s . 7500 , b ) s . 8000 , c ) s . 8500 , d ) s . 9000 , e ) s . 9360
|
e
|
divide(divide(3640, subtract(const_1, divide(5, const_12))), divide(2, 3))
|
divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)|
|
other
|
sal90 dressing?
|
"let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.20 0.2 x = 0.10 x = 0.5 = 50 % the answer is b ."
|
a ) 60 % , b ) 50 % , c ) 40 % , d ) 30 % , e ) 20 %
|
b
|
divide(subtract(30, 10), subtract(20, 10))
|
subtract(n0,n2)|subtract(n4,n2)|divide(#0,#1)|
|
gain
|
how q5?
|
1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e .
|
a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26
|
e
|
subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), floor(add(subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(add(const_10, 5), 2)), const_1))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5)))
|
add(n4,const_10)|divide(n1,n2)|divide(n0,n2)|divide(n1,n3)|divide(n0,n3)|divide(n1,n4)|divide(n0,n4)|subtract(n1,n0)|add(#7,const_1)|add(n2,#0)|subtract(#1,#2)|subtract(#3,#4)|subtract(#5,#6)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|subtract(#8,#13)|subtract(#14,#9)|subtract(#15,#0)|add(#17,const_1)|floor(#19)|subtract(#16,#20)|subtract(#21,#18)
|
other
|
rs certain years?
|
"solution s . i . = rs . ( 956 - 850 ) = rs . 106 rate = ( 100 x 106 / 850 x 3 ) = 212 / 51 % new rate = ( 212 / 51 + 4 ) % = 416 / 51 % new s . i . = rs . ( 850 x 416 / 51 x 3 / 100 ) rs . 208 . ∴ new amount = rs . ( 850 + 208 ) = rs . 1058 . answer c"
|
a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1058 , d ) data inadequate , e ) none of these
|
c
|
add(850, divide(multiply(multiply(850, add(divide(multiply(subtract(956, 850), const_100), multiply(850, 3)), 4)), 3), const_100))
|
multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)|
|
gain
|
a4imeters )
|
"let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 4 gms in 600,000 cm ^ 3 , y = 4 * 600,000 = 2 , 400,000 gms total gms in 1 m ^ 3 = 1 , 000,000 + 2 , 400,000 = 3 , 400,000 answer : a"
|
a ) 3 , 400,000 , b ) 2 , 800,000 , c ) 55,000 , d ) 28,000 , e ) 280
|
a
|
subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 4)), 4)), volume_cube(100))
|
multiply(const_2,n3)|volume_cube(n5)|divide(#1,const_10)|multiply(#2,n2)|multiply(#2,#0)|multiply(#3,n2)|multiply(#4,n3)|add(#5,#6)|subtract(#7,#1)|
|
geometry
|
it onth?
|
"the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 10 ) = 1 / 600 of the video per second . the combined rate is 1 / 15 + 1 / 600 = 41 / 600 of the video per second . the time required to load the video is 600 / 41 = 14.63 seconds . the answer is d ."
|
a ) 13.42 , b ) 13.86 , c ) 14.25 , d ) 14.63 , e ) 14.88
|
d
|
subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))
|
add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)|
|
physics
|
a awarded?
|
since the starting point is given as the $ 4000 scholarship , assume $ 4000 scholarships to be x by the given information , $ 2500 scholarships = 2 x and $ 1250 scholarships = 6 x gievn : total $ 1250 scholarships = $ 75000 6 x * 1250 = 75000 solve for x = 10 option d
|
a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 15
|
d
|
divide(divide(75000, 1250), multiply(const_2, 3))
|
divide(n8,n0)|multiply(n5,const_2)|divide(#0,#1)
|
general
|
in and. ]
|
"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 / 12 * 33 = 33 cents . answer : c ."
|
a ) 0.33 , b ) 0.44 , c ) 33 , d ) 44 , e ) 55
|
c
|
multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.33, 100))
|
divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|
|
general
|
a40 is?
|
"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 40 ) = > x / 2 = 33 = > x = 66 . answer : c"
|
a ) 18 , b ) 82 , c ) 66 , d ) 27 , e ) 77
|
c
|
multiply(subtract(73, 40), const_2)
|
subtract(n0,n1)|multiply(#0,const_2)|
|
general
|
a year years?
|
"let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 25 / 100 ) * 100 = 75 after second year , value of baseball card = ( 1 - 10 / 100 ) * 75 = 67.5 total percent decrease of the card ' s value over the two years = ( 100 - 67.5 ) / 100 * 100 % = 31.5 % answer c"
|
a ) 28 % , b ) 30 % , c ) 32.5 % , d ) 36 % , e ) 72 %
|
c
|
subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(25, const_100))), const_100))
|
divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)|
|
gain
|
if is8?
|
"total nos of ways in which we can choose n = 96 n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? case 1 : n = odd then n + 2 = odd & n + 1 will be even i . e this needs get divided by 8 , hence is a multiple of 8 so we have 8 . . 96 = 12 multiples to fill the n + 1 pos hence 12 ways case 2 : n is even then n + 2 will be even & the product will be divisible by 24 & thus 8 so nos of values that can be used for n = 2 . . . . 96 ( all even nos ) i . e 48 nos total = 48 + 12 = 60 ways so reqd p = 60 / 96 = 5 / 8 ; answer : d"
|
a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4
|
d
|
divide(add(divide(96, 2), divide(96, 8)), 96)
|
divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)|
|
general
|
a if.?
|
"let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 40 m per hour = 1.5 hours distance travelled by car a = 40 * 1.5 = 60 miles since car b catches up car a at 10 : 30 , time = 60 mins = 1 hour speed of car b = 60 / ( 1 ) = 60 miles per hour answer b"
|
a ) 45 , b ) 60 , c ) 53 , d ) 55 , e ) 50
|
b
|
divide(60, divide(add(multiply(subtract(10, 9), const_60), subtract(40, 10)), const_60))
|
subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)|
|
physics
|
1 =?
|
"1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ? 150 = 4 check the fourth eqn . answer : c"
|
a ) 1 , b ) 255 , c ) 4 , d ) 445 , e ) 235
|
c
|
divide(subtract(subtract(225, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2)
|
add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)|
|
general
|
a the listed?
|
"given cash discount - 16 % profit - 25 % items sold - 60 price sold at = list price of 50 assume list price = $ 10 total invoice = $ 500 - 16 % cash discount = $ 420 let cost price of 60 items be x so total cost = 60 * x given the shopkeeper had a profit of 25 % 60 * x * 125 / 100 = 420 or x = $ 7 * 4 / 5 = $ 28 / 5 which means his products were listed at $ 10 which is a 78 + ( 4 / 7 ) % markup over $ 28 / 5 answer e"
|
a ) 50 % , b ) 60 % , c ) 70 % , d ) 75 % , e ) 78 + ( 4 / 7 ) %
|
e
|
multiply(subtract(divide(divide(divide(add(const_100, 25), const_100), subtract(const_1, divide(subtract(60, 50), 60))), divide(subtract(const_100, 16), const_100)), const_1), const_100)
|
add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)|
|
gain
|
if is8?
|
as per question = > n = 8 p + 1 for some integer p hence 3 n = > 24 q + 3 = > remainder = > 3 for some integer q hence b
|
a ) 1 , b ) 3 , c ) 7 , d ) 5 , e ) 6
|
b
|
multiply(3, 1)
|
multiply(n1,n2)
|
general
|
the. surface.
|
a 3 = 2197 = > a = 13 6 a 2 = 6 * 13 * 13 = 1014 answer : b
|
a ) 864 , b ) 1014 , c ) 1299 , d ) 1268 , e ) 1191
|
b
|
surface_cube(cube_edge_by_volume(2197))
|
cube_edge_by_volume(n0)|surface_cube(#0)|
|
geometry
|
15 business place?
|
"there are 15 business exec and in each handshake 2 business execs are involved . hence 15 c 2 = 105 also , each of 15 exec will shake hand with every 3 other chairmen for total of 45 handshake . total = 45 + 105 = 150 ans : a"
|
a ) 150 , b ) 131 , c ) 115 , d ) 90 , e ) 45
|
a
|
add(divide(multiply(15, subtract(15, const_1)), const_2), multiply(15, 3))
|
multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)|
|
geometry
|
if pint language?
|
"1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . mumbo is 43426 . . . answer : a"
|
a ) 43426 , b ) 14236 , c ) 13436 , d ) 14263 , e ) 15263
|
a
|
divide(79523, add(const_3, const_3))
|
add(const_3,const_3)|divide(n0,#0)|
|
general
|
a and work?
|
"a can do the work in 18 / 2 i . e . , 9 days . a and b ' s one day ' s work = 1 / 9 + 1 / 18 = ( 2 + 1 ) / 18 = 1 / 6 so a and b together can do the work in 6 days . answer : d"
|
a ) 10 , b ) 16 , c ) 18 , d ) 6 , e ) 12
|
d
|
inverse(add(divide(const_1, 18), multiply(divide(const_1, 18), const_2)))
|
divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)|
|
physics
|
the. m?
|
"1 / 2 * 5.5 * 6 = 16.5 m 2 answer : b"
|
a ) 11 m 2 , b ) 16.5 m 2 , c ) 18.5 m 2 , d ) 19.5 m 2 , e ) 12 m 2
|
b
|
triangle_area(5.5, 6)
|
triangle_area(n0,n1)|
|
geometry
|
working b a?
|
"15 * a + 15 * b = x pages in 15 mins printer a will print = 15 / 45 * x pages = 1 / 3 * x pages thus in 15 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 3 more pages per min that printer a . in 15 mins printer b will print 45 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 45 = > x = 135 pages answer : b"
|
a ) 125 , b ) 135 , c ) 145 , d ) 155 , e ) 165
|
b
|
multiply(divide(3, subtract(divide(45, 15), const_1)), 45)
|
divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)|
|
physics
|
the the table?
|
"cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 125 ) = rs . 6720 . answer : d"
|
a ) rs . 5725 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6720 , e ) none of these
|
d
|
divide(8400, add(const_1, divide(25, const_100)))
|
divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)|
|
gain
|
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