mathewhe/aqua_rat huggingface dataset:
Question: In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
(A) 67 kg.
(B) 68 kg.
(C) 69 kg.
(D) Data inadequate
(E) None of these
Rationale: Explanation:
Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70.
According to Arun's mother, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
req average = (66 + 67 + 68) / 3 = 67 kg
Answer: A
Answer: (A) 67 kg.
Question: 7^1+7^2+7^3+...........+7^2015 find out how many numbers present which unit place contain 3 ?
(A) 504
(B) 505
(C) 506
(D) 507
(E) 508
Rationale: 7^3,7^7,7^11,7^15,7^19.........7^2015 contains last digit=3.
Now,we see
3,7,11,15....2015 are in AP
tn=a+(n-1)d
2015=3+(n-1)4
2015=3+4n-4
4n=2016
n=504
ANSWER:A
Answer: (A) 504
Question: Each of the cucumbers in 100 pounds of cucumbers is composed of 99% water, by weight. After some of the water evaporates, the cucumbers are now 93% water by weight. What is the new weight of the cucumbers, in pounds?
(A) 2
(B) 15
(C) 92
(D) 96
(E) 98
Rationale: Out of 100 pounds 99% or 99 pounds is water and 1 pound is non-water. After somewaterevaporates the cucumbers become 93% water and 7% of non-water, so now 1 pound of non-water composes 7% of cucucmbers, which means that the new weight of cucumbers is 1/0.07=15 pounds.
Answer: B.
Answer: (B) 15
Question: If each digit in the set of A = {1,2,3,4,5,6,7,8,9} is exactly used once, in how many ways can the digits be arranged such that the 2 and 5 are not adjacent?
(A) 9! - 8!
(B) 8!
(C) 9!
(D) 7*8!
(E) 9! / 2!
Rationale: The answer should be the (total arrangements) - (arrangements when 2 and 5 are adjacent).
The number of ways to arrange 9 units is 9!
If we think of 2 and 5 as one single unit, the number of arrangements is 8!
Note that 2 and 5 can have two arrangements, so we multiply by 2 to get 2*8!
The total number of valid arrangements is 9! - 2*8! = 9*8! - 2*8! = 7*8!
The answer is D.
Answer: (D) 7*8!
Question: The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector has a central angle of 76.5°, what fraction of the wheel’s area is represented by the smallest sector?
(A) 3/80
(B) 2/35
(C) 2/25
(D) 1/20
(E) 1/15
Rationale: The largest angle is 76.5.
Let d be the difference between any two angles in the progression.
The sum of all the angles will be:
76.5 + (76.5-d) + (76.5-2d) + ... + (76.5-7d) = 612 - 28d
The sum of all the central angles in a circle = 360
612 - 28d = 360
d = 252/28 = 9
The smallest sector is (76.5-7d) = 76.5-7*9 = 13.5
13.5/360 = 27/720 = 3/80
The answer is A.
Answer: (A) 3/80
Question: The volumes of three containers are in the ratio 4 : 5 : 6. If the sum of the volumes of the largest and the smallest container is is 20litres more than the weight of the third container, what is the volume of the smallest container?
(A) 11litres
(B) 19litres
(C) 15litres
(D) 16litres
(E) None of these
Rationale: Let the volumes of the three containers be 4k, 5k and 6k respectively.
4k + 6k = 5k + 20
=> 5k = 20 => k = 4
Therefore the weight of the lightest boy
= 4k = 4(4) = 16 litres.
ANSWER:D
Answer: (D) 16litres
Question: A merchant sells an item at a 20% discount, but still makes a gross profit of 20 percent of the cost. What percent of the cost would the gross profit on the item have been if it had been sold without the discount?
(A) 20%
(B) 40%
(C) 50%
(D) 60%
(E) 75%
Rationale: Soln:-
original SP = X
Cost =C
current selling price = .8x ( 20% discount)
.8x = 1.2C (20% profit)
x= 1.2/.8 *C
x= 3/2 C
Original selling price is 1.5C which is 50% profit
ANSWER:C
Answer: (C) 50%
Question: In the list 1, 2, 3, 4, 5, 5, 5, 5, 7, 11, 21, what fraction of the data is less than the mode?
(A) 2/9
(B) 1/3
(C) 2/5
(D) 4/11
(E) 7/9
Rationale: Mode: The mode of any set is the term which has the highest frequency (Occurrence)
Highest frequent term in the set is 5 (with frequency 4) hence
Mode = 5
four terms (1, 2, 3, 4) out of a total of 11 terms are less than mode of the set.
Fraction of set that are less than Mode of set = 4/11
Answer: Option D
Answer: (D) 4/11
Question: Find the odd man out.
36, 60, 6, 18, 72, 25, 48
(A) 60
(B) 72
(C) 6
(D) 36
(E) 25
Rationale: Explanation:
Each of the numbers except 25 is a divisible by 6.
Answer: E) 25
Answer: (E) 25
Question: If 4 less than x is 4 greater than y, then a number z that is 5 greater than x reduced by y is equal to
(A) 4
(B) 13
(C) 14
(D) y + 9
(E) y + 14
Rationale: 13 it is....nice one....
B
Answer: (B) 13
Question: A train of length 240 m crosses a bridge of length 150m in 20 seconds. What is the speed of train?
(A) 33
(B) 27
(C) 25
(D) 70.2
(E) 72
Rationale: Sol : (length of train+ length of bridge) = speed of train x Time
(240+150) = 20 x Speed
Speed = 390/20= 19.5 m/s =70.2 km/h
ANSWER=D
Answer: (D) 70.2
Question: At Annie's party, there were five times as many females as males. There were three times as many adults as children. Which of the following could NOT be the number of people at the party?
(A) 384
(B) 258
(C) 216
(D) 120
(E) 72
Rationale: we have to work on the info and understand what that info takes us to...
1) there were five times as many females as males At Annie's party..
this means F=5M.. so total number=6M..
2) There were three times as many adults as children..
A=3C...
or total = 4C..
so the total is 4C or 6M..
since C and M have to be positive integers, we can say for sure that Total is a multiple of 4 and 6..
or Total is multiple of LCM of 4,6..
LCM of 4,6= 12..
lets see which number is not multiple of 12..
A. 384...12*32
B. 258... no
C. 216....12*18
D. 120....12*10
E. 72......12*6..
so B is our answer..
Answer: (B) 258
Question: Income and expenditure of a person are in the ratio 8 : 7. If the income of the person is Rs. 40000, then find his savings?
(A) 5000
(B) 6000
(C) 7000
(D) 8000
(E) 9000
Rationale: Let the income and the expenditure of the person be Rs. 8x and Rs. 7x respectively.
Income, 8x = 40000
=> x = 5000
Savings = Income - expenditure
= 8x - 7x = x
So, savings =Rs.5000.
Answer: A
Answer: (A) 5000
Question: A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?
(A) 81/256
(B) 15/256
(C) 75/256
(D) 175/256
(E) 95/256
Rationale: The man will hit the target if he hits it once or twice or thrice or all four times in the four shots that he takes.
So, the only possibility when the man will not hit the target is when he fails to hit the target in even one of the four shots that he takes.
The event of not hitting the target even once is the complement of the event of hitting the target at least once.
The probability that he will not hit the target in any given shot = 1 – 1/4 = ¾
Therefore, the probability that he will not hit the target in any of the four shots =3/4*3/4*3/4*3/4=81/256
The probability that he will hit the target at least in one of the four shots=1-81/256
=175/256
ANSWER D 175/256
Answer: (D) 175/256
Question: A student has to obtain 33% of the total marks to pass. He got 59 marks and failed by 40 marks. The maximum marks are?
(A) 450
(B) 300
(C) 500
(D) 610
(E) 175
Rationale: Let the maximum marks be x
Then, 33% of x = 59+40
33x/100 = 99
x = 300
Answer is B
Answer: (B) 300
Question: A and B together can do a piece of work in 6 days and A alone can do it in 10 days. In how many days can B alone can do it?
(A) 12 days
(B) 18 days
(C) 15 days
(D) 21 days
(E) 22 days
Rationale: Explanation:
A and B can do work 1/6 in 1 day
A alone can do 1/10 work in 1 day
B alone can do (1/6 -1/10) =1/15 work in 1 day
=> complete work can be done in 15 days by B
Answer: Option C
Answer: (C) 15 days
Question: If (0.2)x = 2 and log 2 = 0.3010, then Find the value of x to the nearest tenth.
(A) -0.4
(B) -0.2
(C) -0.6
(D) 0.4
(E) 0.6
Rationale: (0.2)x = 2.
Taking log on both sides
log (0.2)x = log 2.
x log (0.2) = 0.3010, [since log 2 = 0.3010].
x log (2/10) = 0.3010.
x [log 2 - log 10] = 0.3010.
x [log 2 - 1] = 0.3010,[since log 10=1].
x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].
x[-0.699] = 0.3010.
x = 0.3010/-0.699.
x = -0.4306….
x = -0.4 (nearest tenth)
Answer is A.
Answer: (A) -0.4
Question: 51 + 52 + 53 + .........+100) is equal to
(A) 3775
(B) 3779
(C) 3629
(D) 3729
(E) 3719
Rationale: 51 + 52 + 53 + ...........+ 100
= (1 + 2 + 3 + .... + 100) - (! + 2 + 3 + ...... + 50)
=
= (5050 - 1275) = 3775
Answer;A
Answer: (A) 3775
Question: Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 7 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2011, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?
(A) 2013
(B) 2014
(C) 2015
(D) 2017
(E) 2016
Rationale: chart starting at 2013:
A+K=13
L=0.
Then I added 9 to a+k for each year and 14 to L for each year.
2014:AK=22 L=14
2015:AK=31 L=28
2016:AK=40 L=42
thus, 2016 is the correct answer. => E
Answer: (E) 2016
Question: A train 150 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 49
(B) 59
(C) 12
(D) 13
(E) 67
Rationale: Speed of the train relative to man = (150/10) m/sec = (15) m/sec. [(15) * (18/5)] km/hr = 54 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 54 ==> x = 59 km/hr.
Answer: Option B
Answer: (B) 59
Question: Cook can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Cook has in his herd?
(A) 336
(B) 180
(C) 120
(D) 456
(E) 155
Rationale: On first look A, D,E are not divisible either by 5 or 6 - ruled out.
C,B divisible by both 5,6, but 180 is divisible by 9.
Answer : C
Answer: (C) 120
Question: Divide Rs. 3250 among A, B and C so that A receives 1/5 as much as B and C together and B receives 2/3 as A and C together. B's share is?
(A) 6500
(B) 3250
(C) 1100
(D) 1300
(E) 1200
Rationale: A+B+C = 3250
A = 1/5(B+C); B = 2/3(A+C)
B/(A+C) = 2/3
B = 1/5 * 6500 => 1300
Answer: D
Answer: (D) 1300
Question: A9B, D100F, H361K,then find next ?
(A) M784P
(B) M900Q
(C) N841Q
(D) N756P
(E) L884P
Rationale: By giving the number to alphabets, we get
A + B = 1+2 = 3 --> 3^2 = 9
D + F = 4+6 = 10 --> 10^2 = 100
H + K = 8+11= 19 -->19^2 = 361
: :
: :
M + Q = 13+17=30 --->30^2 = 900
ANSWER:B
Answer: (B) M900Q
Question: In the manufacture of a certain product, 6 percent of the units produced are defective and 5 percent of the defective units are shipped for sale. What percent of the units produced are defective units that are shipped for sale?
(A) 0.125%
(B) 0.3%
(C) 0.8%
(D) 1.25%
(E) 2.0%
Rationale: 0.06*0.05 = 0.003 = 0.3%
The answer is B.
Answer: (B) 0.3%
Question: Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 3 ½ hours if they are walking in the same direction?
(A) 5 km
(B) 10 4/3 km
(C) 4 3/8 km
(D) 18 km
(E) 6 km
Rationale: Explanation:
RS = 5 ¾ - 4 ½ = 1 ¼
T = 3 ½ h.
D = 5/4 * 7/2 = 35/8 = 4 3/8 km
C)
Answer: (C) 4 3/8 km
Question: In a coconut grove, (x + 3) trees yield 60 nuts per year, x trees yield 120 nuts per year and (x – 3) trees yield 180 nuts per year. If the average yield per year per tree be 100, find x.
(A) 3
(B) 4
(C) 5
(D) 6
(E) None of the above
Rationale: (x+3)×60+x×120+(x−3)×180/(x+3)+x+(x−3)=100
⇒ 360x−360/3x = 100
⇒ 60x = 360 ⇒ x = 6
Answer D
Answer: (D) 6
Question: If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 10 inclusive?
(A) 7/32
(B) 11/64
(C) 35/128
(D) 49/128
(E) 65/128
Rationale: One possible case is: rainy-rainy-rainy-not rainy-not rainy-not rainy-not rainy.
The probability of this case is (1/2)^7 = 1/128
The number of possible cases is 7C3 = 35.
P(exactly 3 rainy days) = 35*1/128 = 35/128
The answer is C.
Answer: (C) 35/128
Question: A certain bacteria colony doubles in size every day for 16 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?
(A) 6.33
(B) 7.5
(C) 10
(D) 15
(E) 19
Rationale: If there is one bacteria colony, then it will reach the limit of its habitat in 16 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 15 times.
Answer: D.
Similar questions to practice:
Hope it helps.
Answer: (D) 15
Question: Steve is laying floor tiles in 2 rooms of his house. Each room requires 30 tiles, all the same size. How many of the following combinations could Steve buy if: A Small box (1 box@$8.00 w/tax) contains 10 tiles, and a Large box (1 box@$20.00 w/tax) contains 25 tiles. Steve only has $50 on him when he gets to the store.
(A) 6 Small Boxes
(B) 3 Large Boxes
(C) 3 Small Boxes and 1 Large Box
(D) A and C
(E) A, B and C
Rationale: The Correct Answer must total at least 60 tiles (2 rooms of 30) AND cost $50 or less
A=Total of 60 tiles (6*10) and costs $48 (6*$8.00) *Correct Answer*
B=Total of 75 tiles (3*25) (while sufficient amount of tiles, cost would be $60 which is $10 over budget)
C=Total of 55 tiles (3*10 +25) (while sufficient cash is on hand of $49 [3*8+25] he is 5 tiles short.
D=A is correct however C is not
E=A is correct however B and C are not
The Correct Answer is A
Answer: (A) 6 Small Boxes
Question: The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
(A) 10568
(B) 10569
(C) 10560
(D) 10561
(E) 10562
Rationale: Let the rate of interest be R% p.a.
4400{[1 + R/100]2 - 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years
= 44000(2)(12)/100
=Rs.10560
Answer:C
Answer: (C) 10560
Question: In a recent election, Ms. Robbins received 7,000 votes cast by independent voters, that is, voters not registered with a specific political party. She also received 10 percent of the votes cast by those voters registered with a political party. If N is the total number of votes cast in the election and 40 percent of the votes cast were cast by independent voters, which of the following represents the number of votes that Ms. Robbins received?
(A) 0.06N + 3,200
(B) 0.1N + 7,200
(C) 0.4N + 7,200
(D) 0.1N + 8,000
(E) 0.06N + 7,000
Rationale: total vote = N
vote cast by independent voters = 0.4 N
vote cast by registered voters = 0.6 N
vote received by Ms. Robbins = 7000 + 10% of vote be registered voters = 7000 + 0.06 N
Answer choice E
Answer: (E) 0.06N + 7,000
Question: If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is?
(A) Rs. 850
(B) Rs. 980
(C) Rs. 990
(D) Rs. 1000
(E) Rs. 1100
Rationale: Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A)
Answer: (A) Rs. 850
Question: A container has a capacity of 30 gallons and is full of spirit. 4 gallons of spirit is drawn out and the container is again filled with water. This process is repeated 5 times. Find how much spirit is left in the resulting mixture finally?
(A) 6(257/525) gallons
(B) 6(346/625) gallons
(C) 6.5 gallons
(D) 6.25 gallons
(E) None of these
Rationale: There is a direct formula to calculate such repetition:
Left QTY / Orig QTY = ( (Orig QTY - Rep QTY) / Orig QTY )^n
where n is number of times mixing done:
Left QTY / 30 = (30 -4)^5 / 20^5
Left QTY = 30 (26/20)^5
Left QTY = 30 ( 13/10 )^5
Left QTY = 74.2586
Answer is E
Answer: (E) None of these
Question: The G.C.D. of 1.08, 0.36 and 0.45 is?
(A) 0.03
(B) 0.09
(C) 0.18
(D) 0.108
(E) none of these
Rationale: Given numbers are 1.08,0.36 and 0.45.
H.C.F of 108, 36 and 45 is 9
H.C.F of given numbers=0.09.
Correct Options:B
Answer: (B) 0.09
Question: Rectangular tile each of size 80cm by 55cm must be laid horizontally on a rectangular floor of size 210cm by 330cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:
(A) 26
(B) 16
(C) 18
(D) 19
(E) 20
Rationale: Area of tile = 80*55 = 4400
Area of floor = 210*330 = 72600
No of tiles = 72600/4400 = 16.5
So, the no of tile = 16
ANSWER:B
Answer: (B) 16
Question: If 15,555 – n is divisible by 11, and 0 < n < 11, what is n?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
Rationale: 15,555 / 11 = 1414 with a remainder of 1.
We need to subtract the remainder to get a multiple of 11.
The answer is A.
Answer: (A) 1
Question: A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among seven colors produced enough color codes to identify all but 6 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)
(A) 25
(B) 26
(C) 27
(D) 28
(E) 29
Rationale: 7C2 = 21
The number of test subjects is 21 + 6 = 27
The answer is C.
Answer: (C) 27
Question: A bag holds 3 red marbles and 3 green marbles. If you removed two randomly selected marbles from the bag, without replacement, what is the probability that both would be red?
(A) 1/10
(B) 1/5
(C) 3/10
(D) 2/5
(E) 1/2
Rationale: Given: 3R and 3 G marbles
Required: Probability that 2 marbles removed without replacement are both red
Initially we have to pick one red from a total of 3 red and 3 green marbles
After one red has been picked, we need to pick 1 red from a total of 2 red and 3 green marbles.
P(Both red) = (3/6)*(2/5) = 1/5
Option B
Answer: (B) 1/5
Question: A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
(A) 42m
(B) 53m
(C) 37m
(D) 40m
(E) 29m
Rationale: Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x ————–1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2
divide 1 & 2 equations
we get 2(y+3)=3(y-2)
2y + 6 = 3y-6
y= 12
Put y=12 in eq.--2
12*10=3x
by solving we get x = 40
Answer is D
Answer: (D) 40m
Question: In country Z, 12% of the people do not have a university diploma but have the job of their choice, and 25% of the people who do not have the job of their choice have a university diploma. If 40% of the people have the job of their choice, what percent of the people have a university diploma?
(A) 35%
(B) 43%
(C) 55%
(D) 65%
(E) 75%
Rationale: Setting up a matrix is how I solve this one.
Diploma No Diploma Totals
Job of Choice w/Diploma
Job of Choice w/o Diploma = 12%
Job of Choice Total = 40%
Not Job of Choice with Diploma =.25X
Not Job of Choice w/o Diploma= .75X
Total Not Job of Choice = X
Total with Diploma
Total without Diploma
Total citizen = 100
If 40% of people have their job of choice, then 60% of people do NOT have their job of choice. 25% of 60% = 15%. We can also see that 28% of the people have their job of choice and a diploma (40%-12%=28%). 28% + 15% = 43%. Therefore 43% of the people in Country Z have a diploma.
Ans B
Answer: (B) 43%
Question: New age consultants have three consultants Gyani, Medha, and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with with Medha but without Gyani and 3 projects with Gyani but without Medha. The total number of projects for New age consultants is one less than twice the number of projects in which more than one consultant is involved.
What is the number of projects in which Medha alone is involved?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: gyani works with medha but with out buddhi=14-6=8.
buddhi with medha but without gyani=2
buddhi with gyani but without medha=3
all three involved projects=6
so number projects in which more than one consultant is involved=8+2+3+6=19
total no.projects for consultants is 2*19-1=37
among 37 projects in 19 projects more than one consultant is involved
so no. of projects done by onle one consultant=37-19=18
so b+g=16+x
b+g=17
x=1
ANSWER:A
Answer: (A) 1
Question: Compound interest earned on a sum for the second and the third years are Rs.2400 and Rs.2640 respectively. Find the rate of interest?
(A) 10% p.a.
(B) 20% p.a.
(C) 15% p.a.
(D) 12% p.a.
(E) None of these.
Rationale: Rs.2640 - 2400 = Rs.240 is the interest on Rs.2400 for one year.
Rate of interest = (240/2400) * 100 = 10% p.a
ANSWER:A
Answer: (A) 10% p.a.
Question: What is the possible value of 2*((1+1/100)^100)
(A) 5.4
(B) 6.4
(C) 7.4
(D) 8.4
(E) 4.4
Rationale: 1+1/100=1.01
1.o1^100==2.70
2.70*2
5.40 ans
ANSWER:A
Answer: (A) 5.4
Question: Running at the same constant rate, 6 identical machines can produce a total of 300 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
(A)  648
(B)  1,800
(C)  2,000
(D)  10,800
(E)  64,800
Rationale: 6 machines produce 300 bottles per minute;
1 machine produces 300/6=50 bottles per minute;
10 machines produce 50*10=500 bottles per minute;
In 4 minutes 10 machines produce 500*4=2,000 bottles.
Answer: C.
Answer: (C)  2,000
Question: n 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B the present age of B is
(A) 22
(B) 897
(C) 268
(D) 20
(E) 48
Rationale: A +10=2(B-10) ........(1)
A =B + 9 ......... (2)
from equations. 1 & 2
we get B = 39 A will be 39+9=48 years old.
Answer:E
Answer: (E) 48
Question: Each student is given a five-character identification code. The first four characters are selected from the numbers 0 to 9 inclusive, and the last character is selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?
(A) 240,000
(B) 245,000
(C) 250,000
(D) 255,000
(E) 260,000
Rationale: The number of possible codes is 10*10*10*10*26=260,000.
The answer is E.
Answer: (E) 260,000
Question: The average of first 12 natural numbers is?
(A) 7.6
(B) 6.5
(C) 6.1
(D) 6.8
(E) 7.2
Rationale: Sum of 12 natural no.
= 156/2 = 78
Average = 78/12
= 6.5
Answer:B
Answer: (B) 6.5
Question: Two pie pans of the same size remain on the counter. One pan has 1/2 of a pie, and the other
has 3/4 of a pie left in it. Mom wishes to divide all the pie into pieces that are each 1/8 of a pie. How
many pieces of pie will she have when she is finished?
(A) 20 pieces
(B) 9 pieces
(C) 10 pieces
(D) 19 pieces
(E) 15 pieces
Rationale: 1/2 + 3/4 = 2/4 + 3/4 = 5/4 pie
5/4 ÷ 1/8 = 40/32 ÷ 4/32 =
40 ÷ 4 = 10 pieces of pie
correct answer C
Answer: (C) 10 pieces