mathewhe/aqua_rat huggingface dataset:
Question: If 13 men do a work in 80 days, in how many days will 20 men do it?
(A) 18 days
(B) 38 days
(C) 42 days
(D) 48 days
(E) 52 days
Rationale: 13 * 80 = 20 * x
x = 52 days
ANSWER:E
Answer: (E) 52 days
Question: A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior?
(A) ²²C₁₀
(B) ²²C₁₀ + 1
(C) ²²C₉ + ¹⁰C₁
(D) ²²C₁₀ - 1
(E) None of these
Rationale: The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1
ANSWER:D
Answer: (D) ²²C₁₀ - 1
Question: A store sells nine different kinds of shirts. If we want to buy two different kinds of shirts, how many possible combinations of two shirts can we buy?
(A) 32
(B) 33
(C) 34
(D) 35
(E) 36
Rationale: 9C2=36.
The answer is E.
Answer: (E) 36
Question: Number N is randomly selected from a set of all primes between 10 and 40, inclusive. Number G is selected from a set of all multiples of 5 between 10 and 40 inclusive. What is the probability that N+G is odd?
(A)  1/2
(B)  2/3
(C)  3/4
(D)  4/7
(E)  5/8
Rationale: All primes except 2 are odd, thus N must be odd. For N + G = odd + G to be odd, G must be even.
There are 7 multiples of 5 between 10 and 40, inclusive: 10, 15, 20, 25, 30, 35, and 40. Out of them 3 (15, 25, and 35) are odd.
Therefore the probability that N + G is odd is 4/7.
Answer: D.
Answer: (D)  4/7
Question: If a person walks at 15 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:
(A) 50 km
(B) 56 km
(C) 60 km
(D) 40 km
(E) 80 km
Rationale: Let the actual distance travelled be x km.
x/10 = (x+20)/15
15x = 10x + 200
5x = 200
x = 40 km.
answer :D
Answer: (D) 40 km
Question: Every digit of a number written in binary is either 0 or 1. To translate a number from binary, multiply the nth digit (reading from right to left) by 2^(n-1)
What is the largest prime number (written in binary) that is a factor of both 10010000 and 100100000 ?
(A)  10
(B)  11
(C)  1001
(D)  1011
(E)  10001
Rationale: Binary Divison can provide a quick answer if you are comfortable with it.
as option E is the biggest binary number we try with it first :
100010000/ 1001 =10000
1000100000/ 1001 =100000
so answer is option is C
Answer: (C)  1001
Question: According to a recent student poll, 5/6 out of 30 members of the finance club are interested in a career in investment banking. If two students are chosen at random, what is the probability that at least one of them is interested in investment banking?
(A) 1/14
(B) 4/49
(C) 2/7
(D) 28/29
(E) 13/14
Rationale: 25 students are interested, 5 are not interested
Prob = 1 - 5C2/30C2 = 1 - (5*4/(30*29))=1 - 20/580
= 1- 1/29 = 28/29
Answer : D
Answer: (D) 28/29
Question: Look at this series: 53, 53, 39, 39, 25, 25, ... What number should come next?
(A) A)12
(B) B)11
(C) C)27
(D) D)53
(E) E)86
Rationale: In this series, each number is repeated, then 14 is subtracted to arrive at the next number.
Answer: B
Answer: (B) B)11
Question: B and C work together can complete work in 8 days A and B together can complete same work in 12 days and A& C together complete in 16 days in total how many days A,B and C complete the same work together?
(A) 3 9/13
(B) 7 5/13
(C) 7 5/12
(D) 3 5/12
(E) 4 5/12
Rationale: Explanation:
(A+B+C)’s 2 day work = (1/8+1/12+1/16 ) = 13/48
on day work =13/96
so they complete work in 96 /13 = 7 5/13 days
Answer: Option B
Answer: (B) 7 5/13
Question: One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
(A) 6/18
(B) 3/13
(C) 2/14
(D) 9/18
(E) 7/21
Rationale: Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = 12/52 = 3/13.
Answer B
Answer: (B) 3/13
Question: Two-third of a positive number and 16/216 of its reciprocal are equal. Find the positive number.
(A) 9/25
(B) 14/4
(C) 4 /12
(D) 144/25
(E) 144/29
Rationale: Explanation:
Let the positive number be x.
Then, 2/3 x = 16/216 * 1/x
x2 = 16/216 * 3/2
= 16/144
x = √16/144 =4 /12.
ANSWER:C
Answer: (C) 4 /12
Question: if a@b = a * b^(1/2) then 5@9 = ?
self made
(A) 12
(B) 16
(C) 13
(D) 15
(E) 14
Rationale: a@b = a * b^(1/2)
5@9 = 5*9^(1/2) = 5*3 = 15
Correct Option: D
Answer: (D) 15
Question: Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec. What is the length of the fast train?
(A) 35
(B) 24
(C) 27
(D) 27 7/9
(E) 67
Rationale: Relative speed = (40 - 20) = 20 km/hr.
= 20 * 5/ 18 = 50/9 m/sec.
Length of faster train = 50/9 * 5 = 250/9 = 27 7/9 m.
Answer: Option D
Answer: (D) 27 7/9
Question: Large, medium and small ships are used to bring water. 4 large ships carry as much water as 7 small ships; 3 medium ships carry as much water as 2 large ships and 1 small ship. If 15 large, 7 medium and 14 small ships, each made 36 journeys to bring a certain quantity of water, then in how many journeys will 12 large, 14 medium and 21 small ships bring the same quantity of water?
(A) 29 journeys
(B) 30
(C) 20
(D) 50
(E) 55
Rationale: Explanation :
Here, Large, Medium and Small ships are denoted by the symbol L, M and S.
Now, according to the question,
4L = 7S. ------(i)
3M = 2L + S. ------(ii)
Using above equations, the ratios of the capacity of the Large, Medium and Small ships are :-
7: 6 : 4.
Let, the number of journeys required be x. Since, quantity of water remains the same, so :-
=((15×7+7×6+ 14×4)36) = x (12×7+14×6+21×4).
=>((15×7+7×6+14×4)36) / (12×7+14×6+21×4)= x
=>x=7308/252
=>x=29.
Answer : A
Answer: (A) 29 journeys
Question: 30 square stone slabs of equal size were needed to cover a floor area of 50.7 sq.m. Find the length of each stone slab?
(A) 130cm
(B) 767cm
(C) 88cm
(D) 666cm
(E) 776cm
Rationale: Area of each slab
=50.7/30m2=1.69m2
Length of each slab √1.69=1.3m=130cm
Answer: (A) 130cm
Question: Six students wrote science exam. Their average marks are 70. Five Students got 65, 75, 55, 72 and 69 marks respectively. Therefore what is the marks of the sixth student?
(A) 84
(B) 68
(C) 85
(D) 75
(E) 42
Rationale: EXPLANATION
Total marks of 5 students = (65+75+55+72+69) = 336
Required marks = [ (70 x 6) – 336 ]
= (420 – 336)
= 84
Answer A
Answer: (A) 84
Question: Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
(A) 12
(B) 14
(C) 18
(D) 20
(E) 22
Rationale: a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.
B)
Answer: (B) 14
Question: Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is
(A) 50
(B) 55
(C) 60
(D) 65
(E) 70
Rationale: Explanation:
Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60
Option C
Answer: (C) 60
Question: If the probability of rain on any given day in Chicago during the summer is 25%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 7 inclusive?
(A) 1/16
(B) 3/32
(C) 5/32
(D) 3/64
(E) 7/64
Rationale: One possible case is: rainy-rainy-rainy-not rainy.
The probability of this case is 1/4*1/4*1/4*3/4 = 3/256
The number of possible cases is 4C3 = 4.
P(exactly 3 rainy days) = 4*3/256 = 3/64
The answer is D.
Answer: (D) 3/64
Question: The value of each of 5 numbers is at least 8. The average (arithmetic mean) of these 5 numbers is 24. If the average of two of the numbers is 18, what is the largest possible value that any of these 5 numbers can have?
(A) 28
(B) 50
(C) 68
(D) 76
(E) 84
Rationale: Average => (a1+a2+a3+a4+a5)/5 =24
Thus => a1+a2+a3+a4+a5 = 120
now least value is 8, so let a1 = 8
average of 2 numbers is 18
thus , let (a2+a3) /2 =18
a2+a3 = 36
This tells that sum of last 2 numbers should be :
120 -8-36 =76
Then by elimination , eliminate 84 and 76
highest possible value of a5 should be 68 from the answer choice
Answer : C
Answer: (C) 68
Question: 1, 6, 13, 22, 33, ?
(A) 35
(B) 46
(C) 48
(D) 49
(E) 38
Rationale: The pattern is + 5, + 7, + 9, + 11,....
Answer : B.
Answer: (B) 46
Question: QUANT 4-PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...
For every positive integer n that is greater than 1, the function f(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The f(n) could have any of the following units digits except…?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 9
Rationale: For every positive integer n that is greater than 1, the function f(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The f(n) could have any of the following units digits except…?
Used brute force on this one to calculate. Not sure if there is a quicker way.
1+3=4 C is out
1+3+....15= 81 A is out
1+3+5+7=16 D is out
1+3+5=9 E i out
Answer: B
Answer: (B) 2
Question: Jerry, an electrician, worked 9 months out of the year. What percent
of the year did he work? (round answer to the nearest hundredth)
What percent 12 is 9? 12 months = 1 year
(A) 58.33%
(B) 68.33%
(C) 78.33%
(D) 88.33%
(E) 75.00%
Rationale: 1. Multiply the opposites
9 x 100 = 900
100=9/12
Divide by the remaining number
75%
correct answer E
Answer: (E) 75.00%
Question: The twice the square of a natural number is increased by 8, the number is equal to 458more than the number.
(A) 3
(B) 6
(C) 8
(D) 12
(E) 15
Rationale: Explanation:
Let the number be x. Then,
2x2 +8 = 458
=> 2x2 = 450 => x2 - 225
=> x = 15
Answer: Option E
Answer: (E) 15
Question: Two dice are tossed simultaneously. Find the probability that the total is a prime number.
(A) 5/9
(B) 5/8
(C) 5/6
(D) 5/34
(E) 5/1
Rationale: ANSWER: A
Answer: (A) 5/9
Question: Statement: "You are appointed as a programmer with a probation period of 1 year and your performance will be reviewed at the end of the period for confirmation." Assumptions: The performance of an individual generally is not known at the time of appointment offer. Generally an individual tries to prove his worth in the probation period.
(A) Only assumption I is implicit
(B) Only assumption II is implicit
(C) Either I or II is implicit
(D) Neither I nor II is implicit
(E) Both I and II are implicit
Rationale: The performance of the individual has to be tested over a span of time as the statement mentions. So, I is implicit. The statement mentions that the individual's worth shall be reviewed (during probation period) before confirmation. So, II is also implicit.
E
Answer: (E) Both I and II are implicit
Question: In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 94% of the products of M3 III are not defective. What is the percentage of defective in the stockpile?
(A) 3.3%
(B) 5%
(C) 2.5%
(D) 4%
(E) 6%
Rationale: Explanation :
Let there be 100 products in the stockpile. 40% and 30% were manufactured by M1 and M2 respectively. So, 30% are manufactured by M3.
Products from M1 = 40, from M2 = 30 and from M3 = 30.
Number of defective products from M1 = 0.03 x 40 = 1.2, from M2 = 0.01 x 30 = 0.3 and from M3 = 0.06 x 30 = 1.8
Therefore, total number of defective products = 3.3
Answer : A
Answer: (A) 3.3%
Question: In a college the ratio of the numbers of boys to the girls is 8:5. If there are 190 girls, the total number of students in the college is?
(A) 562
(B) 356
(C) 452
(D) 494
(E) 512
Rationale: Let the number of boys and girls be 8x and 5x
Then , 5x = 190
x= 38
Total number of students = 13x = 13*38 = 494
Answer is D
Answer: (D) 494
Question: A sum was put at simple interest at a certain rate for 9 years Had it been put at 5% higher rate, it would have fetched 1350 more. Find the sum.
(A) 3000
(B) 3100
(C) 3200
(D) 3300
(E) 3400
Rationale: Difference in S.I. = P×T/100(R1−R2)
⇒ 1350 = P×9x5/100(∵ R1 - R2 = 2)
⇒ P = 1350×100/9x5=3000
Answer A
Answer: (A) 3000
Question: The remainder when W=1+3+3^2+3^3+..........+3^200 is divided 13.
(A) 12
(B) 7
(C) 0
(D) 5
(E) 3
Rationale: W=1+3+3^2+3^3+..................+3^200
Is a Geometric progression having common ratio as '3' and number of terms as '201'.
Since Sum to n terms in GP = a(r^n-1)/(r-1)
where a=First term and r =common ration
Hence,
1*(3^201 -1 )/(3-1)
Rem of (3^201-1)/2 divided by 13
3^201 -1 /26
WKT, 3^3 = 27 = 26+1
{(26+1)^67 - 1}/26
{1-1}/26
=>0.C
Answer: (C) 0
Question: Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is?
(A) 76 sec
(B) 88 sec
(C) 48 sec
(D) 44 sec
(E) 33 sec
Rationale: Relative speed = 60 + 90 = 150 km/hr.
= 150 * 5/18 = 125/3 m/sec.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 * 3/125 = 48 sec.
Answer: C
Answer: (C) 48 sec
Question: Find the least square number which is exactly divisible by 10,12,15 and 18.
(A) 900
(B) 120
(C) 103
(D) 140
(E) 217
Rationale: L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 22 * 32 * 5.
To make it a perfect square, it must be multiplied by 5.
Required number = (22 * 32 * 52) = 900.
ANSWER A
Answer: (A) 900
Question: Currently apples cost 75 cents/pound. Due to a disease affecting the apple trees it is expected that next month apples will cost 150% more than they do currently. How much are apples expected to cost next month?
(A) 195 cents /pound
(B) 192 cents /pound
(C) 187.50 cents /pound
(D) 187 cents /pound
(E) 190 cents /pound
Rationale: If a new cost is p percent greater than the old cost, then (new cost) = (old cost) + (p/100)(old cost).
In this case, (new cost) = 75 cents/pound + (150/100)(75 cents/pound)
= 75 cents/pound + 112.50 cents/pound
= 187.50 cents /pound
Answer : C
Answer: (C) 187.50 cents /pound
Question: Find the number of ways in which 10 different beads can be arranged to form a necklace.
(A) 181440
(B) 245890
(C) 457890
(D) 124560
(E) 231560
Rationale: Since in forming a necklace clockwise and anticlockwise arrangements are not different, therefore,10 beads can be arranged to form a necklace in(10-1)! /2 ways
= 9!/2 = 181440ways
Answer is A
Answer: (A) 181440
Question: A person takes 5hrs 45 min in walking to a place and ridingback. He would have gained 2hrs by riding both ways. The time he would take to walk both ways is?
(A) 5 hours 45 min
(B) 6 hours 30min
(C) 4 hours 50 min
(D) 7 hours 45 min
(E) 8 hours
Rationale: ----------------------------------------------------------------
Solution 1
----------------------------------------------------------------
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
In fact, you can do all these calculations mentally and save a lot of time
which will be a real benefit for you.
D
Answer: (D) 7 hours 45 min
Question: If ab+bc+ca=0,What is the value of a^2+b^2+c^2?
(A) (a^2+b^2+c^2)/2
(B) (a+b+c)^2
(C) (a^2+b^2+c^2)^2
(D) a+b+c
(E) c+a+b
Rationale: The option B is correct.
The formula (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
(a+b+c)^2=a^2+b^2+c^2+2(0).
a^2+b^2+c^2=(a+b+c)^2 is the answer
Answer: (B) (a+b+c)^2
Question: A bus starts from city X. The number of women in the bus is half of the number of men. In city Y, 12 men leave the bus and six women enter. Now, number of men and women is equal. In the beginning, how many passengers entered the bus ?
(A) 15
(B) 30
(C) 36
(D) 45
(E) 54
Rationale: Explanation:
Originally, let number of women = x. Then, number of men = 2x.
So, in city Y, we have : (2x - 12) = (x + 6) or x = 18.
Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 54.
Answer: E
Answer: (E) 54
Question: A and B can finish a work 30 days if they work together. They worked together for 20 days and then B left. A finished the remaining work in another 20 days. In how many days A alone can finish the work?
(A) 60
(B) 50
(C) 40
(D) 30
(E) 20
Rationale: Explanation :
Amount of work done by A and B in 1 day = 1/30
Amount of work done by A and B in 20 days = 20 × (1/30) = 20/30 = 2/3
Remaining work – 1 – 2/3 = 1/3
A completes 1/3 work in 20 days
Amount of work A can do in 1 day = (1/3)/20 = 1/60
=> A can complete the work in 60 days
Answer : Option A
Answer: (A) 60
Question: A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 46 years now, the son's age five years back was
(A) 16years
(B) 17years
(C) 18years
(D) 19years
(E) 20years
Rationale: Let the son's present age be x years. Then, (46 - x) = x
2x = 46.
x = 23.
Son's age 5 years back (23 - 5) = 18 years.
C)
Answer: (C) 18years
Question: A person was asked to state his age in years. His reply was, "Take my age three years hence, multiply it by 3 and then subtract three times my age three years ago and you will know how old I am." What was the age of the person?
(A) 18
(B) 87
(C) 26
(D) 27
(E) 29
Rationale: Let the present age of the person be x years.
Then, 3(x + 3) - 3(x - 3) = x
3x + 9 - 3x + 9 = x => x = 18'
Answer: A
Answer: (A) 18
Question: 8^100 is divisible by 17. Find the remainder?
(A) 8
(B) 12
(C) 16
(D) 18
(E) 20
Rationale: This is an extremely difficult problem to solve with out Fermat's little theorem. By applying Fermat's little theorem , We know that 816 when divided by 17, the remainder is 1.
So divide 100 by 16 and find the remainder. Remainder = 4
Therefore, 100 = (16 × 6) + 4
Now this problem can be written as 810017 = 816×6+417 = (816)6×8417
Now this problem simply boils down to (1)6×8417 = 8417
84 = 82×82, we need to find the remainder when 64 × 64 is divisible by 17. Or 13 × 13 = 169. When 169 is divided by 17, remainder is 16.
C
Answer: (C) 16
Question: Four of the five parts numbered (a), (b), (c), (d) and (e) in the following equation are exactly equal. Which of the parts is not equal to the other four? The number of that part is the answer.
(A) 371.587 + 46.32 – 217.907
(B) 4 × 125 – 75 × 4
(C) 58.25 × 4.5 – 65.875
(D) 25 × 12 – 2 × 5 × 10
(E) 121 × 3.5 – 2 × 111.75
Rationale: The other parts are equal to 200.
Answer C
Answer: (C) 58.25 × 4.5 – 65.875
Question: For any integer n greater than 1, n* denotes the product of all the integers from 1 to n, inclusive. How many multiples of 4 are there between 4* and 5*, inclusive?
(A) 5
(B) 6
(C) 20
(D) 24
(E) 25
Rationale: here, 4* = 4! and 5* = 5!
question is asking no. of terms in an AP whose first term is 24(4!), last term is 120(5!) and common difference is 4.
last term OR nth term = first term +(no. of tems -1)(common difference)
Tn = a + (n-1)d
120 = 24 + (n-1)4
96/4 = n-1
n = 25 .....E
Answer: (E) 25
Question: Two buses and a van were employed to transport a class of students on a field trip. 3/5 of the class boarded the first bus. 1/2 of the remaining students boarded the second bus, and the rest of the students boarded the van. When the second bus broke down, 1/2 of the students on the second bus boarded the first bus. What fraction of the class was on board the first bus?
(A) a) 1/2
(B) b) 7/10
(C) c) 11/15
(D) d) 23/30
(E) e) 4/5
Rationale: This is a pure ratio question (we aren't given any numbers for anything), so you can just choose any starting number you like, and use it to solve the problem. The best number to pick is almost always the product of all the denominators of your fractions, so here we could start with 30 students. Then we have:
• 3/5 of these, or 18 students, board the first bus
• there are 12 students left. 1/2 of these, or 6 students, board the second bus
• this bus breaks down, and 1/2 of the 6 students, or 3 students, board the first bus
• the first bus now contains 21 out of the original 30 students, so the answer is 21/30 = 7/10
Answer B
Answer: (B) b) 7/10
Question: The average of 11 numbers is 30. If the average of first six numbers is 17.5 and that of last six is 42.5, then what is the sixth number?
(A) 30
(B) 36
(C) 45
(D) 47
(E) 43
Rationale: Given: Average of 11 numbers = 30
Step 1: Calculate total of 11 numbers by multiplying it by average value 30 = 11 x 30 = 330
Step 2: Calculate total of first six members by multiplying it by average value 17.5 = 17.5 x 6 = 105
Step 3: Calculate total of last six members by multiplying it by average value 42.5 = 42.5 x 6 = 255
Therefore, we can find sixth number by adding value of first six and last six numbers and subtracting it from the total value of 11 numbers.
Sixth number =(105 + 255)- 330 = 30
ANSWER IS A
Answer: (A) 30
Question: Mrs. Evans gave a test to her freshmen economics class, which has 40 students enrolled and 30 of them answered question 1 correctly. If 29 answered question 2 correctly and 10 did not take the test then how many answered both questions correctly?
(A) 24
(B) 27
(C) 29
(D) 31
(E) 33
Rationale: Total number of enrolled students = 40
Number of students who did not take test = 10
Hence, number of students who took test = 40 - 10 = 30
Number of students who answered Q2 correctly =29, therefore, all students who took test answered Q2 correctly.
So, number of students who answered Q1 correctly, also answered Q2 correctly = 29.
Number of students who answered both Q1 & Q2 correctly = 29.
ANSWER:C
Answer: (C) 29
Question: For a finite sequence of non zero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -7, -6 ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: Given sequence: {1, -3, 2, 5, -4, -6}
The questions basically asks: how manypairs of consecutive termsare there in the sequence such that the product of these consecutive terms is negative.
1*(-3)=-3=negative;
-3*2=-6=negative;
2*5=10=positive;
5*(-4)=-20=negative;
(-4)*(-6)=24=positive.
So there are 5 pairs of consecutive terms of the sequence for which the product is negative.
Answer: E.
Answer: (E) 5
Question: Sonika deposited Rs.8000 which amounted to Rs.9200 after 3 years at simple interest. Had the interest been 2% more. She would get how much?
(A) 9680
(B) 2338
(C) 1278
(D) 1098
(E) 1279
Rationale: (8000*3*2)/100 = 480
9200
--------
9680
Answer: A
Answer: (A) 9680
Question: What is the units digit of the expression 14^7−19^4?
(A)  0
(B)  3
(C)  4
(D)  6
(E)  8
Rationale: I think answer on this one should be E too. Since we know that 14^7>19^4, as Will said one should always check if the number is positive.
Answer: (E)  8
Question: If x < y < 0, which of the following must be true?
(A) 2xy/3x is a negative integer
(B) x-y is a positive integer
(C) (-x)+ (-y) is positive integer
(D) x/y is a negative integer
(E) 2x+y is a negative integer
Rationale: Given x, y are less than 0. So both will be negative integers
negative + negative = negative
negative * negative = positive
negative / negative = positive
x-y is false since xSo, option C is true , because if we multiple negative number with negative it becomes positive(x, y are negative numbers)
Answer : C
Answer: (C) (-x)+ (-y) is positive integer
Question: A bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely. What is the value of (10^4 -10^2)(0.0014)?
(A) 13.86
(B) 14.86
(C) 15.86
(D) 12.86
(E) 11.86
Rationale: You get 10^2 (100-1) (.0014)
we know 14/10000 = .0014
10^2 (99) * (14/10000)
10^2 = 100 and 100^2 = 10000
Cancel out the 100 with the 10^2
Left with 99(14/100).
.99*14 = 13.86
Answer : A
Answer: (A) 13.86
Question: In a rectangular coordinate system, what is the area of a rhombus whose vertices have the coordinates (0, 3.5), (11, 0), (0, -3.5), (-11, 0)?
(A) 56
(B) 77
(C) 112
(D) 116
(E) 120
Rationale: Area of rhombus= 1/2* D1 *D2
Length of 1st diagonal = 11+11= 22
Length of 2nd diagonal= 3.5+3.5= 7
Area= 1/2*22*7= 77
B is the answer
Answer: (B) 77