mathewhe/aqua_rat huggingface dataset:
Question: A girl multiplies 987 by a certain number and obtains 559981 as her answer. If in the answer, both 9’s are wrong but the other digits are correct, then the correct answer will be
(A) 553681
(B) 555181
(C) 555681
(D) 556581
(E) None of these
Rationale: 987 = 3 x 7 x 47.
So, required number must be divisible by each one of 3, 7, 47.
None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.
ANSWER:C
Answer: (C) 555681
Question: In a village there are 150 men and 90 women in present time.if in next year population will be P=(a^2 +b^2)^1/2 , and in every year men are reduces 10%.what is population of after 2 year.
(A) 140
(B) 141
(C) 142
(D) 143
(E) 144
Rationale: next year total population=[150^2+90^2]^.5=174.92=175
man decreased by 10% so total man =150*.9=135
women will be= 175-135=40
so population after two years= [135^2+40^2]^.5=140.80=141
so population after two year= 141
ANSWER:B
Answer: (B) 141
Question: How many factors does 25^2 have?
(A) 2
(B) 8
(C) 24
(D) 25
(E) 26
Rationale: 36^2 = 6*6*6*6 = 2^4*3^4
Total factors = (4+1)*(4+1) = 4*2 = 8
Answer B.
Answer: (B) 8
Question: P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
(A) 20
(B) 22
(C) 25
(D) 27
(E) 30
Rationale: Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days
Answer is C.
Answer: (C) 25
Question: sum of the digits of 5^2010*8^673
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Rationale: 5^2010 * 8^673 (2^3=8) therefore 673*3=2019
=5^2010 * 2^2019
=10^2010 * 2^9 (2^9=512)
=10^2010 * 512= 512000..... upto 2010 zeroes.So the sum is 8.
ANSWER:B
Answer: (B) 8
Question: A solution of 130 litres contains milk and water in the ratio 7:x. If thirty litres of water is added to the solution, the ratio becomes 3:2, find the value of x?
(A) 2.6
(B) 2.4
(C) 2.9
(D) 2.8
(E) 3
Rationale: Exp. Total new quantity = original sol + water =130+30 =160
New ratio = 3:2, New quantity of milk =3/5*160 = 96 Lit,
New quantity of water = 2/5*160= 64 Lit
Water present initially = (64-30) = 34 Lit
Ratio = 96/34 =7/x There for x = 2.4
Answer: B
Answer: (B) 2.4
Question: In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 5 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers:
(A) 36
(B) 37
(C) 38
(D) 39
(E) 40
Rationale: M/NM > 5x/24x
M = 8
8/NM > 5x/24x
8/NM - 5x/24x > 0
(192x - 5x * NM)/(24x * NM) > 0
192x - 5x*NM > 0
5x*NM < 192x
NM < 192x/5x
NM < 38.4
Maximum number of non managers = 38
Answer: C
Answer: (C) 38
Question: In how many ways Q can the crew of a ten oared boat be arranged ,when of the ten persons available, 2 of whom can row only on the bow side and 3 of whom can row only on the stroke side?
(A) 1. 10!/(2! * 3!)
(B) 2. 10!/(8! * 7!)
(C) 3. 5!/(3! * 2!)
(D) 4. (5!)^3/(3! * 2!)
(E) 5. 5!/(8! * 7!)
Rationale: Ans is D.
Going by the assumtion that both the Bow side and Stroke side will have 5 person each.
Five(3+2) of them had allready been selected(according to their preference), so from the remaining 5 we have to select 2 for the stroke side or 3 for the bow side.
and the 5 crew of each side can be arranged in Q= 5! ways.
Hence answer is 5C3 * 5! * 5!= (5!)^3/ 3!2!=D
Answer: (D) 4. (5!)^3/(3! * 2!)
Question: The present ages of three persons are in proportions 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages.
(A) 16,28,36
(B) 8,10,20
(C) 5,10,12
(D) 30,32,45
(E) 6,8,10
Rationale: Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
20x = 80 => x = 4
Their present ages are 16, 28 and 36 years respectively.
ANSWER A
Answer: (A) 16,28,36
Question: What is the probability of flipping a fair coin four times and the coin landing on heads on exactly one flip?
(A) 1/3
(B) 1/4
(C) 1/6
(D) 1/12
(E) 1/16
Rationale: Probability = (Favorable combination)/(Total combination)
Favorable combination = (HTTT, THTT, TTHT,TTTH) ==> 4
Total combination = 16 (2^4)
Probability = (4)/(16)=1/4
(B)
Answer: (B) 1/4
Question: Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?
(A) 11 sec
(B) 15 sec
(C) 19 sec
(D) 12 sec
(E) 20 sec
Rationale: D
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec
Answer: (D) 12 sec
Question: There are three vessels of equal capacity .Vessel A contains Milk and water in the ratio 12:10 ;Vessel B contains Milk and water in the ratio 2:1 and Vessel C contains Milk and water in the ratio 3:2 .If the mixture in all the three vessels is mixed up.What will be the ratio of water to milk ?
(A) 547/398
(B) 392/598
(C) 598/392
(D) 210/279
(E) 398/547
Rationale: 12:10=>12x+10x=22x
2:1=>2y+1y=3y
3:2=>3z+2z=5z
22x=3y=5z
Take LCM of 22,3,5=330
Or simply;
x=15
y=110
z=66
So,
Ratio of Water:Milk=(10x+y+2z)/(12x+2y+3z)=392/598
Ans:B
Answer: (B) 392/598
Question: If v and y are integers, then vy(v + y) is
(A) Always positive
(B) Always negative
(C) Positive when v is positive
(D) Negative when v is negative
(E) It cannot be determined by the information given
Rationale: Given v and y are integers, then vy(v + y)
Let's take even as 2 / -2 and odd as 3 / -3..we get the following.. and IMO option D is correct answer..
I presume its Option E if am not missing anything.
v=-3 and y=1
vy(v + y) is positive = 6
Answer: (E) It cannot be determined by the information given
Question: A person jogged 10 times along the perimeter of a rectangular field at the rate of 12 kilometers per hour for 30 minutes. If field has a length that is twice its width, find the area of the field in square meters.
(A) 56,000 square meters
(B) 32,000 square meters
(C) 65,000 square meters
(D) 20,000 square meters
(E) 43,000 square meters
Rationale: Let us first find the distance d jogged
distance = rate * time = (12 km / hr)* 30 minutes
= (12 km/hr) * 0.5 hr = 6 km
The distance of 6 km corresponds to 10 perimeters and therefore 1 perimeter is equal to
6 km / 10 = 0.6 km = 0.6 * 1000 meters = 600 meters
Let L and W be the length and width of the field. The length is twice the width. Hence
L = 2 W
The perimeter is 600 meters and is given by
2 (L + W) = 600
Substitute L by 2 W
2 (2 W + W) = 600
Simplify and solve for W
4 W + 2 W = 600
6 W = 600
W = 100
Find L
L = 2 W = 200
Find the area A of the rectangle
A = L * W = 200 * 100 = 20,000 square meters correct answer D
Answer: (D) 20,000 square meters
Question: In expressing a length of 55.471 km as nearly as possible with the three significant digits, find the percentage error
(A) 0.52%
(B) 0.24%
(C) 0.052%
(D) 0.024%
(E) None of these
Rationale: Explanation:
Error = (55.5 - 55.471) = 0.029
Required percentage =
0.029/ 55.471Ã—100=0.052
Option C
Answer: (C) 0.052%
Question: Two trains start from A & B and travel towards each other at
speed of 40kmph and 50kmph resp. At the time of the meeting the
second train has traveled 130km more than the first.the distance
between them.
(A) 1170 km
(B) 1180 km
(C) 1270 km
(D) 1280 km
(E) 1350 km
Rationale: Let the distance traveled by the first train be x km
then distance covered by the second train is x + 130km
x/40 = x+130 / 50
x= 520
so the distance between A & B is x + x + 130 = 1170 km
Answer is A.
Answer: (A) 1170 km
Question: The total of the ages of Jayant, Prem and Saransh is 93 years. Ten years ago, the ratio of their ages was 2 : 3 : 4. What is the present age of Saransh?
(A) 24 years
(B) 31 years
(C) 34 years
(D) 38 years
(E) None
Rationale: Solution
Let the ages of Jayant, Prem and Saransh 10 years ago be 2x, 3x and 4x years respectively. Then,
(2x + 10)+(3x + 10)+(4x + 10)	= 93.
‹=›9x = 63
x = 7.
∴ Prem's present age = 3x + 10 = 31 years.
Answer B
Answer: (B) 31 years
Question: A man sells an article at 15% loss. Had he sold it for Rs. 420/- more he could have gained 20%. What is the cost price of the article?
(A) Rs.420/-
(B) Rs.400/-
(C) Rs.900/-
(D) Rs.1200/-
(E) None of these
Rationale: Explanation:
Short cut method:
Loss =====> 15 %
Profit ====> 20 %
~
----------------
35 % ---------> Rs. 420/-
100 % --------> Rs. 1200/-
Answer: Option D
Answer: (D) Rs.1200/-
Question: Average of first six multiples of 15 is
(A) 9.5
(B) 44.5
(C) 52.5
(D) 15.75
(E) 17.7
Rationale: Explanation:
Average=15(1+2+3+4+5+6) / 6=315 / 6=52.5
Option C
Answer: (C) 52.5
Question: A bag contains 6 black and 6 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
(A) 4
(B) 4/3
(C) 4/5
(D) 4/9
(E) 1/2
Rationale: Let number of balls = (6 + 6) = 12.
Number of white balls = 6.
P (drawing a white ball) = 6/12 = 1/2
Hence Answer is E
Answer: (E) 1/2
Question: For a race a distance of 224 meters can be covered by P in 8 seconds and Q in 32 seconds. By what distance does P defeat Q eventually?
(A) 672m
(B) 6738m
(C) 634m
(D) 671m
(E) 636m
Rationale: Explanation:
This is a simple speed time problem. Given conditions:
=>Speed of P= 224/8 = 28m/s
=>Speed of Q=224/32 = 7m/s
=>Difference in time taken = 24 seconds
Therefore, Distance covered by P in that time= 28m/s x 24 seconds = 672 metres
ANSWER: A
Answer: (A) 672m
Question: In a game of billiards, A can give B 40 points in 80 and he can give C 60 points in 80. How many points can B give C in a game of 100?
(A) 50
(B) 55
(C) 60
(D) 65
(E) 70
Rationale: A scores 80 while B score 40 and C scores 20.
The number of points that C scores when B scores 100 = (100 * 20)/40 = 50.
In a game of 100 points, B gives (100 - 50) = 50 points to C
Answer:A
Answer: (A) 50
Question: A certain quantity of 45% concentration solution is replaced with 20% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?
(A) 1:3
(B) 1:2
(C) 2:3
(D) 2:1
(E) 3:1
Rationale: 20% - - - - - - - - - 35% - - - - - - 45%
From 20 to 35 =15
From 35 to 45 = 10
So the ratio is 15/10 = 3 to 2
C it is
Answer: (C) 2:3
Question: If P and Q are positive integers, P - 1 is divisible by 2 and Q - 3 is divisible by 4, all of the following could be the value of P - Q Except?
(A) 9
(B) 2
(C) 4
(D) 6
(E) 12
Rationale: P-1 = 2Q
=> P = 2Q + 1 = 1,3,5,7,9,11,13,15,17,19,21,23,25
Q-3 = 4P
=> Q = 4P + 3 = 3,7,11,15,19,23,27,31,35,39,43,47
2=5-3
4=11-7
6=17-11
12=23-11
So , we can see that all the answer choices are possible here except A
Answer: (A) 9
Question: Three boys picked up 30 mangoes .In how many ways can they divide them if all mangoes be identical?
(A) A)4^30
(B) B)2^30
(C) C)1^30
(D) D)3^30
(E) E)6^30
Rationale: Each Mango can be given to any one of the four people or in other words..1 mango can be divided into 3 ways...so all 30 can be divided in 3^30 ways
ANS:D
Answer: (D) D)3^30
Question: Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
(A) Rs. 150
(B) Rs. 300
(C) Rs. 250
(D) Rs. 200
(E) None of these
Rationale: Explanation :
Let the amount paid to X per week = x
and the amount paid to Y per week = y
Then x + y = 550
But x = 120% of y = 120y/100 = 12y/10
∴ 12y/10 + y = 550
⇒ y[(12/10) + 1] = 550
⇒ 22y/10 = 550
⇒ 22y = 5500
⇒ y = 5500/22 = 500/2 = Rs.250
Answer : Option C
Answer: (C) Rs. 250
Question: In expressing a length of 61.472 km as nearly as possible with the three significant digits, find the percentage error ?
(A) 0.35%
(B) 0.4%
(C) 0.045%
(D) 0.6%
(E) None of these
Rationale: Explanation :
Error = (61.5 - 61.472) = 0.028
Required percentage = 0.028/61.472100=0.045
Answer : C
Answer: (C) 0.045%
Question: The cost per unit of electricity during a 4 year period is as follows:
Year 1, $1 per unit;
Year 2, $1.5 per unit;
Year 3, $2 per unit;
Year 4, $2.5 per unit.
If a factory spent equal dollars on electricity each year, what was the average cost of a unit of electricity during the 4 year period?
(A) $1.66
(B) $1.46
(C) $1.36
(D) $1.56
(E) $1.26
Rationale: Let money spent each year = $15
Units bought first year = 15
Units bought second year = 10
Units bought third year = 7.5
Units bought fourth year = 6
Total number of units = 38.5
Total money spent = 15*4 = 60
Average = 60/38.5 = 1.56
Answer : D
Answer: (D) $1.56
Question: By selling 10 pencils for a rupee a man loses 25%. How many for a rupee should he sell in order to gain 25%?
(A) 6
(B) 9
(C) 10
(D) 89
(E) 81
Rationale: 75% --- 10
125% --- ?
75/125 * 10 = 6
Answer: A
Answer: (A) 6
Question: A does a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work?
(A) 2 days
(B) 6 days
(C) 5 days
(D) 7 days
(E) 1 days
Rationale: B
6 days
A's 1 day work = 1/10; B's 1 day work = 1/15
(A + B)'s 1 day work = (1/10 + 1/15) = 1/6
So, both together will finish the work in 6 days.
Answer: (B) 6 days
Question: 1,257
1,275
1,527
........
........
+7,521
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum E of the 24 integers ?
(A)  26,996
(B)  44,406
(C)  60,444
(D)  66,660
(E)  99,990
Rationale: This is the way i solved it :
Total no. of different combos = 24, total digits 4 therefore each digit must repeat 24/4 = 6 times in each row .. So the right most row would add up to 1x6 + 2x6 + 5x6 + 7x6 = 6+12+30+42 = 90 .. Each row would add up to 90, so 90 in the first means we have 9 that carries over and we get 0 , the second time its 90+9 and 9 stays and one 9 goes to the row to the left, so the last two digits of the SUM E should be 90 (E) .. We could go on and solve the exact number but since only one answer choice has the last digits as 90 we needn't go any further..
Answer: (E)  99,990
Question: A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 3 more days. How many days would it have taken the second carpenter to do the complete job working alone?
(A) 4 2/3
(B) 7
(C) 9
(D) 14
(E) 24
Rationale: A carpenter worked only 1 day on something that takes him 6 MORE days.
Means;
Carpenter finishes his work in 7 days.
Let his buddy finish the same task in x days.
Respective rates per day:
1/7 AND 1/x
To complete 1 work:
First guy worked for 5 days @ rate=1/7 per day.
Second one worked for 3 days @ rate=1/x per day
Expression:
Days*Rate=Work
4*1/7+3*1/x=1
4x+21=7x
3x=21
x=7 days.
Ans:B
Answer: (B) 7
Question: A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- hours.
(A) 12 hours
(B) 6 hours
(C) 8 hours
(D) 10 hours
(E) 15 hours
Rationale: Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12
Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hours
Option A
Answer: (A) 12 hours
Question: Beth received 2/10 of the votes cast in a certian election. What franction of the other votes cast would she have needed in order to have received 1/2 of the votes cast?
(A) 1/5
(B) 3/7
(C) 3/10
(D) 7/20
(E) 1/2
Rationale: Let the total votes be 10. Beth received 2 votes, other votes 10-2=8. In order to have received 1/2 of the votes she needed 5 votes so 3 more, which is 3/7 of the other votes.
Answer: B.
Answer: (B) 3/7
Question: A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability L that the first ball picked up is odd numbered?
(A) 	0
(B) 	1/3
(C) 	1/2
(D) 	2/3
(E) 	1
Rationale: Answer-D
Selecting the balls either Even or Odd is having probability 50/100=1/2
We have already selected 3 balls with 2 odd numbers and 1 even number.
So we have 3 combinations OOE, OEO, EOO. We have 3 outcomes and 2 are favourable as in 2 cases 1st number is odd. So probability L is 2/3.D
Answer: (D) 	2/3
Question: If x and y are different integers and x^2 = xy, which of the following must be true ?
I. x = 2
II. x = 0
III. x = -y
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Rationale: Answer choice A is correct.
The question gives you x^2=xy
you can simplify that as x^2-xy=0
x(x-y)=0 the solution will be either x=0 or x=y, since x and y are different integers x cannot equal y, that leaves you x=0 to be the only answer.
Answer B
Answer: (B) II only
Question: In how many different number of ways a combination of 3 men and 2 women can be selected from 6 men and 5 women.
(A) 10
(B) 20
(C) 30
(D) 35
(E) 40
Rationale: 6C3
= 6!/(3! . 3!)
= (6 × 5 × 4)/(3 × 2)
= 5 × 4 = 20.
B)
Answer: (B) 20
Question: If anangletis defined as 1 percent of 1 degree, then how many anglets are there in a sixth of a circle?
(A) 0.36
(B) 3.6
(C) 360
(D) 6000
(E) 36,000
Rationale: 1 degree * 1/100 = 1 anglet
So 1 degree = 100 anglets
=> 60 degrees = 6000 anglets
Answer - D
Answer: (D) 6000
Question: Mrs. Napier has 86 stickers to give to 10 students for a reward. How many
stickers will each student get? Will there be any stickers left over?
(A) 8-6
(B) 2-4
(C) 2-6
(D) 2-3
(E) 2-8
Rationale: 86/10 = 8 R 6 Mrs. Napier will give each student 8 stickers and there
will be 6 left over.
correct answer A
Answer: (A) 8-6
Question: Two persons A and B can complete a piece of work in 18 days and 24 days respectively. If they work together, what part of the work will be completed in 2 days?
(A) 7/36
(B) 9/35
(C) 4/37
(D) 8/35
(E) 5/3
Rationale: A's one day's work = 1/18
B's one day's work = 1/24
(A + B)'s one day's work = 1/18 + 1/24 = 7/72
The part of the work completed in 3 days = 2 (7/72) = 7/36.
Answer:A
Answer: (A) 7/36
Question: The length of the longest tape in cm which can be used to measure exactly, the length 2m; 400cm; and 5m is:
(A) 130
(B) 500
(C) 100
(D) 400
(E) 200
Rationale: The three lengths in cm are 200, 400& 500.
HCF of 200, 400& 500 is 100
Hence, the answer is 100 cm.
ANSWER:C
Answer: (C) 100
Question: The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 1500 resolutions?
(A) 2112 m
(B) 2704 m
(C) 2278 m
(D) 2288 m
(E) 2378 m
Rationale: In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 1500 resolutions.
= 1500 * 2 * 22/7 * 22.4 = 211200 cm
= 2112 m
Answer: A
Answer: (A) 2112 m
Question: What is the value of 3x^2 − 1.8x + 0.5 for x = 0.6?
(A) −0.3
(B) 0
(C) 0.3
(D) 1.08
(E) 0.5
Rationale: 3x^2-1.8x+0.5 for x=0.6
=3(0.6*0.6)-3*0.6*(0.6)+0.5
=0+0.5
=0.5
ANSWER:E
Answer: (E) 0.5
Question: Tom and Marry each have a bag of marbles, each of which contains 4 Red marbles, 10 blue marbles, and 6 yellow marbles. If Tom and Marry each select one marble from their respective bags, what is the probability that either Tom or Marry select a blue marble?
(A) 1/7
(B) 2/4
(C) 3/4
(D) 1/8
(E) 1/16
Rationale: Probability that either selects a red marble,P(Either) = 1- probability that neither selects a blue marble, P(Neither)
Probability that Tom selects non blue = 10c1 / 20c1 = 1/2
Probability that Marry selects non blue = 10c1 / 20c1 = 1/2
=> P(Either)=1- (1/2*1/2) = 1-1/4= 3/4.
C)
Answer: (C) 3/4
Question: In a class of 50 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German?
(A) 30
(B) 10
(C) 18
(D) 28
(E) 32
Rationale: Total=English+German-Both+Neither --> 50=English+22-12+0 --> English=40 --> Only English=English-Both=40-12=28.
Answer: D.
Answer: (D) 28
Question: A, B and C can do a piece of work in 14, 30 and 40 days respectively. They start the work together but C leaves 4 days before the completion of the work. In how many days is the work done?
(A) 7.6 days
(B) 7.7 days
(C) 6.7 days
(D) 5.7 days
(E) 8.7 days
Rationale: x/14 + x/30 + x/40 = 1
x = 7.7 days
ANSWER:B
Answer: (B) 7.7 days
Question: Of all the homes on Gotham Street, 1/3 are termite-ridden, and 4/5 of these are collapsing. What fraction of the homes are termite-ridden, but NOT collapsing?
(A) a) 1/15
(B) b) 3/15
(C) c) 4/5
(D) d) 2/5
(E) e)1/5
Rationale: let total homes be 15
termite ridden= 1/3(15)= 5
termite ridden and collapsing = 4/5(5)=4
thus homes that are termite ridden, but not collapsing = 5-4=1
thus required ratio = 1/15
Answer A
Answer: (A) a) 1/15
Question: If x, y, and z are consecutive positive integers greater than 1, not necessarily in that order, then which of the following is (are) true?
I. x > z
II. 3(x+ y) > 3z
III. yz > xz
IV. xy > y + z
(A)  I only
(B)  II only
(C)  II and III only
(D)  II and IV only
(E)  III and IV only
Rationale: we do not know the values or the relative values but that the three are consecutive..
so the max difference in any two number is 2 and min 1..
lets see the choices..
I. x > z...
we do not know the values or the relative values... need not be true..
II. 3(x+ y) > 3z or x+ y > z
we take the worst case that z is teh largest and x and y are lowest possible..... so 2+3>4.. 5>4.. always true
III. yz > xz
we do not know the values or the relative values between x and y..need not be true
IV. xy > y + z
xy-y>z or y(x-1)>z...
again lets take the worst scenario..
we take the worst case that z is the largest and x and y are lowest possible..... so (2-1)3>4.. 3>4.. not true
only ii is true
ans B
Answer: (B)  II only
Question: if in one hour, a canoe rows at 16 km/hr downstream and 9 km/hr upstream, Calculate the speed of a boat in still water (in km/hr).
(A) 18 kmph
(B) 15.5 kmph
(C) 12.5 kmph
(D) 13.5 kmph
(E) 22.5 kmph
Rationale: Speed in still water = (16 + 9)1/2 kmph = 12.5 kmph.
answer :C
Answer: (C) 12.5 kmph
Question: 8+88+888+.....+8888........8888. There are 21 “8” digits in the last term of the series.
Find the last three digits of the sum.
(A) 958
(B) 968
(C) 978
(D) 988
(E) 999
Rationale: As their is 21 no of 8 will add in first summation then 20 and then 19.
so 21*8=168 from that 8 will come 16 as a carry
and 20*8=160 , 160 + 16 (carry from first term) = 176 and from that 6 will come 17 as a carry
at last 19*8=152 , 152+17=169 , so from that 9 will come
and answer become 968
ANSWER:B
Answer: (B) 968
Question: A gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same. If there are 2 rows or 4 rows or 4 rows, then no tree will be left. Find the least number of trees required
(A) 22
(B) 60
(C) 4
(D) 8
(E) 10
Rationale: Explanation:
The least number of trees that are required = LCM(2,4,4) = 4.
Answer: C
Answer: (C) 4
Question: After getting two successive discounts, a shirt with a list price of Rs. 300 is available at Rs. 210. if the second discount is 12.5%, find the first discount.
(A) 15%
(B) 20%
(C) 25%
(D) 12.5%
(E) None of these
Rationale: Explanation :
Solution: let the first discount be x%.
Then, 87.5% of(100-x)% of 300 =210.
87.5/100 * (100-x)/100 of 300 =210 ==> 100-x = 210*100*100/(300*87.5) = 80 ==> x = (100-80) = 20.
.'. First discount = 20%.
Answer : B
Answer: (B) 20%
Question: In how many ways can 5 boysand 2 girls be arranged at a table if 2 girls are never together?
(A) 346
(B) 356
(C) 378
(D) 390
(E) 480
Rationale: Total ways of arranging 7 person in a circle = 6! = 720 ways
If we consider two women as one, then we have 5+1=6 persons. Arranging these six persons in circle will take = 5! = 120 ways
And two women can arrange themselves in 2! ways
Total ways in which two women sit together in circle = 5!*2! = 240 ways
Total ways in which two women doesn’t sit together in circle = 6! - 5!*2! = 720 - 240 = 480 ways
E
Answer: (E) 480