mathewhe/aqua_rat huggingface dataset:
Question: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is?
(A) 3/4
(B) 3/2
(C) 3/6
(D) 3/3
(E) 3/1
Rationale: Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train = 27 x meters, and length of the second train = 17 y meters. (27 x + 17 y) / (x + y) = 23 ==> 27 x + 17 y = 23 x + 23 y ==> 4 x
= 6 y ==> x/y = 3/2.
Answer: B
Answer: (B) 3/2
Question: Three partners shared the profit in a business in the ratio 4:6:8. They had partnered for 12 months, 8 months and 6 months respectively. What was the ratio of their investments?
(A) 4 : 9 : 19
(B) 4 : 9 : 12
(C) 4 : 9 : 11
(D) 4 : 9 : 16
(E) 4 : 9 : 77
Rationale: Let their investments be Rs. x for 12 months, Rs. y for 8 months and Rs. z for 6 months respectively.
Then, 12x : 8y : 6z = 4 : 6 : 8
Now, 12x/8y = 4/6 <=> 9x=4y <=> y=9x/4
And, 12x/6z = 4/8 <=> 4x=z <=> z=4x
Therefore, x : y: z = x : 9x/4: 4x = 4 : 9 : 16
Answer: D
Answer: (D) 4 : 9 : 16
Question: After decreasing 50% in the price of an article costs Rs.620. Find the actual cost of an article?
(A) 2777
(B) 2987
(C) 1200
(D) 1240
(E) 1671
Rationale: CP* (50/100) = 620
CP= 12.4 * 100 => CP = 1240
Answer: D
Answer: (D) 1240
Question: How many pounds of salt at 50 cents/lb must be mixed with 40 lbs of salt that costs 25 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?
(A) 20
(B) 15
(C) 40
(D) 60
(E) 25
Rationale: Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)
Basically, you need to mix 25 cents/lb (Salt 1) with 50 cents/lb (Salt 2) to get a mixture costing 40 cents/lb (Salt Avg)
weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (50 - 40)/(40 - 25) = 2/3
We know that weight of salt 1 is 40 lbs. Weight of salt 2 must be 60 lbs.
Answer (D)
Answer: (D) 60
Question: During a behavioral experiment in a psychology class, each student is asked to compute his or her lucky number by raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the class will be a non-integer?
(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%
Rationale: Student's age is integral value. In question, we are multiplying student's age by 2 and adding some 3* 7^x something over there.
It means Even (2* age of students) + Odd (3) * 7 ^x (where x is day of week).
We know that 7^x is always odd. (e.g. 7, 49, 243, 1701, and repeat).
=> 3* odd is again odd.
2 * age of students is always even.
So final equation is even +odd = odd.
Avg of any number of odd digits or any number of even digits can never be non integral value. So 0% is the answer A.
Answer: (A) 0%
Question: Three numbers are in the ratio 3:5:7. The largest number value is 42. Find difference between Smallest & largest number is?
(A) 21
(B) 22
(C) 24
(D) 25
(E) 26
Rationale: == 3:5:7
Total parts = 15
= The largest number value is 42
= The largest number is = 7
= Then 7 parts -----> 42 ( 7 * 6 = 42 )
= smallest number = 3 & Largest number = 7
= Difference between smallest number & largest number is = 7 - 3 = 4
= Then 4 parts -----> 24 (4 * 6 = 24)
C
Answer: (C) 24
Question: A can do a piece of work in 6 days. B can do it in 6 days. With the assistance of C they completed the work in 2 days. Find in how many days can C alone do it?
(A) 8 days
(B) 2 days
(C) 6 days
(D) 7 days
(E) 4 days
Rationale: C = 1/2 - 1/6 - 1/6= 1/6
=> 6 days
Answer: C
Answer: (C) 6 days
Question: In a candy dish the ratio of red to yellow candies is 2:5, the ratio of red to green candies is 3:8, and the ratio of yellow ot blue candies is 9:2.what is the maximum total number E of yellow and green candies combined if the maximum combined number of red and blue candies is fewer than 85?
(A) 144
(B) 189
(C) 234
(D) 279
(E) 309
Rationale: Given that Red is to yellow is 2:5, Red is to green is 3:8 and Yellow is to blue is 9:2.
Therefore, the total number of red, yellow, blue and green balls will be 18x, 45x, 10x and 48x respectively, where x is a constant.
If the combined number of red and blue balls is fewer than 85, i.e. max 84 balls, then the maximum number of yellow and green balls will be 279.
(10x+18x) < 85.
28x < 85, i.e. 28x <= 84 (Since number of balls cannot be in fraction). Thus, x<=3.
(45x+48x) = 93x.
Max (93x) E= 279.
Answer is D.
Answer: (D) 279
Question: A man buys a cycle for Rs. 1400 and sells it at a loss of 5%. What is the selling price of the cycle?
(A) s. 1090
(B) s. 1160
(C) s. 1190
(D) s. 1202
(E) s. 1330
Rationale: since,c.p=1400
loss%=(c.p-s.p)/c.p*100
5=(1400-s.p)/1400*100
so,after solving answer=1330.
ANSWER:E
Answer: (E) s. 1330
Question: If ab represents a positive two-digit number, where a and b are one-digit integers, which of the following CANNOT be true?
(A) ab = 0
(B) a + b = 0
(C) a – b = 0
(D) b – a = 0
(E) B, C and D
Rationale: The (tens place digit, a) cannot be zero, but the (ones place digit, b) can be zero.
A is possible
B is NOT possible
C is NOT possible
D is NOT possible
Answer: E
Answer: (E) B, C and D
Question: A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in?
(A) 7
(B) 6
(C) 8
(D) 9
(E) 5
Rationale: (A + B + C)'s 1 day work = 1/6;
(A + B)'s 1 day work = 1/8
(B + C)'s 1 day work = 1/12
(A + C)'s 1 day work = (2 * 1/6) - (1/8 + 1/12)
= (1/3 - 5/24) = 1/8
So, A and C together will do the work in 8 days.
Answer:C
Answer: (C) 8
Question: Sandy invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Sandy got back 3 times the original investment. What is the value of n?
(A) 20
(B) 25
(C) 30
(D) 35
(E) 40
Rationale: n* 0.08 * principal + principal = 3 * principal
n * 0.08 = 2
n = 25 years
The answer is B.
Answer: (B) 25
Question: A shopkeeper sells 20% of his stock at 10% profit ans sells the remaining at a loss of 5%. He incurred an overall loss of Rs. 450. Find the total worth of the stock?
(A) 20029
(B) 22500
(C) 20289
(D) 20027
(E) 20026
Rationale: Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50
The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50
Total SP = 11x/50 + 38x/50 = 49x/50
Overall loss = x - 49x/50 = x/50
x/50 = 450 => x = 22500
Answer: B
Answer: (B) 22500
Question: 17! is equal to which of the following?
(A) 355,687,428,096,622
(B) 355,687,428,096,245
(C) 355,687,428,096,000
(D) 355,687,428,096,924
(E) 355,687,428,096,736
Rationale: After 4!, the units digit of every factorial is 0.
5!=120
6!=720
etc...
The answer is C.
Answer: (C) 355,687,428,096,000
Question: What will come in place of the x in the following Number series? 279936, 46656, 7776, 1296, 216, x
(A) 24
(B) 26
(C) 36
(D) 42
(E) 45
Rationale: Go on dividing by 6 to the next number
C)
Answer: (C) 36
Question: If c and t are positive integers, ct +c + t cannot be
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Rationale: Let ct + t + c = x
Add 1 on both sides:
ct + t + c + 1 = x + 1
t(c + 1) + c + 1 = x + 1
(c + 1)(t + 1) = x + 1
Minimum value of (c + 1) = 2
Minimum value of (t + 1) = 2
Hence x + 1 cannot be prime
Substitute x from the given options: 6 + 1 = 7 --> prime --> ct + t + s cannot be 6
Answer: B
Answer: (B) 6
Question: The HCF and LCM of two numbers are 21 and 4641 respectively. If one of number lies between 200 and 300, the two numbers are
(A) 273, 357
(B) 273, 361
(C) 273, 359
(D) 273, 363
(E) None
Rationale: Note that 273 is common in all the options
HCF × LCM Product of 2 numbers
21 × 4641 =273x, x = 357
Answer: A.
Answer: (A) 273, 357
Question: Find the invalid no.from the following series 15, 17, 34, 105, 424, 2124, 12756
(A) 17
(B) 34
(C) 105
(D) 424
(E) 2124
Rationale: 2nd term = (1st term) * 1 + 1 = 15 *1 + 1 = 16.
3rd term = (2nd term) * 2 + 2 = 16 * 2 + 2 = 34
4th term = (3th term) *3 + 3 = 34 * 3 + 3 = 105
5th term = (4th term) * 4 + 4 = 105 * 4 + 4 = 424
6th term = (5th term) * 5 + 5 = 424 * 5 + 5 = 2125
2nd term should 16 instead of 17.
A
Answer: (A) 17
Question: If the average of 12x and 8y is greater than 200, and x is twice y, what is the least integer value of x ?
(A) 20
(B) 30
(C) 40
(D) 25
(E) 35
Rationale: Substitution can be used in the following way:
Always start with the equation: x = 2y. It is more straight forward to manage as compared to the inequality.
Substitute y = x/2, not the other way because you need to find the minimum value of x. So you can get rid of y.
Now go on to the inequality. So 8y = 8x/2 = 4x
Now average of 12x and 4x is greater than 200. Average of 12x and 4x is 8x. So,
8x > 200
x > 25
Answer : D
Answer: (D) 25
Question: Look at this series: 2, 1, (1/2), (1/4), ... What number should come next?
(A) 1/8
(B) 1/16
(C) 1/6
(D) 1/4
(E) 190
Rationale: 4/2 = 2
2/2 = 1
1/2 = 1/2
(1/2)/2 = 1/4
(1/4)/2 = 1/8
Answer: (E) 190
Question: The smallest number when increased by " 3 " is exactly divisible by 5,10,45,90,and 105 is:
(A) 630
(B) 633
(C) 150
(D) 140
(E) 627
Rationale: LCM = 630
630-3= 627
ANSWER:E
Answer: (E) 627
Question: Alex and Brian start a business with Rs.7000 each, and after 8 months, Brian withdraws half of his capital . How should they share the profits at the end of the 18 months?
(A) 5:4
(B) 4:3
(C) 18:11
(D) 3:2
(E) 18:13
Rationale: Alex invests Rs.7000 for 18 months, but Brian invests Rs.7000 for the first 8 months and then withdraws Rs.3500. So, the investment of Brian for remaining 10 months is Rs.3500 only.
Alex:Brian
7000*18 : (7000*8) + (3500*10)
126000 : 91000
Alex:Brian = 18:13
Answer: E
Answer: (E) 18:13
Question: if a cow eat 3kg fodder which is mean of 7 days.if he eat 2 kg,1kg,3kg,4kg,2kg,3kg
on mon,tue,wed,thurs,fri and Saturday.how much cow eat on Sunday?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: solution
let cow eat x kg on Sunday
(2+1+3+4+2+3+x)/7=3
15+x=21
x=6
answer D
Answer: (D) 6
Question: In May, Xiang sold 15 used cars. For these 15 cars, the range of the selling price was $15,000 and the lowest selling price was $4,500. In June, Xiang sold 10 used cars. For these 10 cars, the range of the selling prices was $16,500 and the lowest selling price was $6,700. What was the range of the selling prices of the 25 used cars sold by Xiang in May and June?
(A) 1. $15,600
(B) 2. $15,750
(C) 3. $16,820
(D) 4. $18,100
(E) 5. $18,700
Rationale: Edited the question as there was a typo.
Range = Highest Value - Lowest Value.
Range in May was 15,000 and the lowest price in May was 4,500, thus the highest price in May was 15,000+4,500=19,500.
Range in June was 16,500 and the lowest price in June was 6,700, thus the highest price in June was 16,500+6,700=23,200.
Lowest price of a car in May-June was 4,500 (May) and the highest price of a car in May-June was 23,200, thus the range for May-June is 23,200-4,500=18,700.
Answer: E.
Answer: (E) 5. $18,700
Question: Average of first five multiples of 6 is
(A) 9
(B) 11
(C) 13
(D) 15
(E) 18
Rationale: Explanation:
Average=6(1+2+3+4+5)/5=18
Answer: Option E
Answer: (E) 18
Question: If a rectangular billboard has an area of 120 square feet and a perimeter of 46 feet, what is the length of each of the shorter sides?
(A) 4
(B) 7
(C) 8
(D) 13
(E) 26
Rationale: This question can be solved algebraically or by TESTing THE ANSWERS.
We're told that a rectangle has an area of 120 and a perimeter of 46. We're asked for the length of one of the SHORTER sides of the rectangle.
Since the answers are all INTEGERS, and the area is 120, the shorter side will almost certainly be less than 10 (since 10x10 = 100, but we're NOT dealing with a square). Answer B (7) does NOT divide evenly into 120,
Let's TEST Answer C: 8
IF...
The shorter side = 8...
The area = 120....120/8 = 15 = the longer side
Perimeter = 8+8+15+15 = 46
C
Answer: (C) 8
Question: If 0.75 : x :: 5 : 8, then x is equal to:
(A) 1.5
(B) 1.9
(C) 1.2
(D) 1.4
(E) None of them
Rationale: ((x) x 5) = (0.75 x 8)
= x = 6/5
= 1.20
Answer is C.
Answer: (C) 1.2
Question: 12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days.
(A) 15
(B) 18
(C) 22
(D) 25
(E) 26
Rationale: One man’s one day’s work = 1/48
One woman’s one day’s work = 1/60
6 men’s 2 day’s work = ((6/48) x 2) = 1/4
Remaining work = 3/4
Now, 1/60 work s done in 1 day by 1 woman.
Thus, 3/4 work will be done in 3 days by (60 x (3/4) x (1/3)) = 15 women.
ANSWER:A
Answer: (A) 15
Question: When positive integer A is divided by positive integer B, the remainder is 6. If A/B = 58.3, what is the value of B?
(A) 15
(B) 20
(C) 18
(D) 19
(E) 45
Rationale: When positive integer A is divided by positive integer B, the remainder is 6 --> A=qB+6;
A/B=58.3 --> A=58B+0.3B (so q above equals to 58);
0.3B=6 --> B=20.
Answer: B.
Answer: (B) 20
Question: If f(x)= 5x^2- 2x+6 and g(y)= 3y-8, then g(f(x))
(A) 82x^2-9x+38
(B) 15x^2-6x+10
(C) 58x^2-4x+58
(D) 87x^2-5x+96
(E) 98x^2-7x+94
Rationale: g(f(x))=3(5x^2-2x+6)-8
=15x^2-6x+18-8
=15x^2-6x+10
the answer is B
Answer: (B) 15x^2-6x+10
Question: Alex deposited x dollars into a new account that earned 8 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?
(A) w/(1+1.08)
(B) w/(1.08+1.16)
(C) w/(1.16+1.24)
(D) w/(1.08+1.08^2)
(E) w/(1.08^2+1.08^2)
Rationale: Account at the end of the first year would be 1.08x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.08x+x) dollars. Account at the end of the second year would be (1.08x+x)*1.08=w --> x(1.08^2+1.08)=w --> x=w/(1.08+1.08^2).
Answer: D.
Answer: (D) w/(1.08+1.08^2)
Question: 734619
110907
524232
995001
678297
1548??
now what could be the missing area in the question
(A) 80
(B) 81
(C) 82
(D) 83
(E) 84
Rationale: there is a specific addition/subtraction in each two digits
73-27=46 46-27=19
11-2=9 9-2=7
same
67+15=82 82+15=97
15+33=48 48+33=81
ANSWER:B
Answer: (B) 81
Question: A set has exactly five consecutive positive integers starting with 1. What is the percentage decrease in the average of the numbers when the greatest one of the numbers is removed from the set?
(A) 8.54
(B) 12.56
(C) 15.25
(D) 16.66
(E) None of these
Rationale: Explanation :
The average of the five consecutive positive integers 1,2,3,4 and 5 is:
(1+2+3+4+5)/5=15/5=3.
After removing the the greatest number i.e 5, the new average becomes:
(1+2+3+4)/4=104=2.5.
Now, % drop in the average =(Old average – New average)×100/Old average.
=>(3−2.5)×100/3.
=>100/6=16.66.
Answer : D
Answer: (D) 16.66
Question: In Cricket game, one player throws two fair, six-sided die at the same time. If the player receives at least a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
(A) 1/3
(B) 4/9
(C) 5/9
(D) 2/3
(E) 3/4
Rationale: In Cricket game Probably the easiest approach would be to find the probability of the opposite event and subtract it from 1:
P(win) = 1- P(not win) = 1 - 4/6*4/6 = 5/9.
Answer: C.
Answer: (C) 5/9
Question: Reena took a loan of $. 1200 with simple interest for as many years as the rate of interest. If she paid $ 108 as interest at the end of the loan period, what was the rate of interest?
(A) 3.6
(B) 6
(C) 18
(D) 3
(E) None of these
Rationale: Let rate = R% and time = R years.
Then,1200 x R x R/100= 108
12R2 = 108
R2 = 9
R = 3.
Answer:D
Answer: (D) 3
Question: There are 2 positive integers x and y. What is the probability that x+y is odd?
(A) 1/4
(B) 1/2
(C) 1/3
(D) 1/5
(E) 1/9
Rationale: Explanation :
The addition of two integer results in an odd number only if any one of the integers is an odd number.
The probability of an odd among the two integer is 1/2.
Hence, the required probability is 1/2.
Answer : B
Answer: (B) 1/2
Question: In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 52, the dividend is:
(A) 5336
(B) 5340
(C) 5342
(D) 6345
(E) 6812
Rationale: Divisor = (5 * 52) = 260
= 10 * Quotient = Divisor
=> Quotient = 260/10 = 26
Dividend = (Divisor * Quotient) + Remainder
Dividend = (260 * 26) + 52 = 6812.
E
Answer: (E) 6812
Question: The product of two numbers is 120 and the sum of their squares is 289. The sum is:
(A) 23
(B) 21
(C) 43
(D) 56
(E) 30
Rationale: xy = 120 and x2 + y2 = 289.
(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529
x + y = 529 = 23.
ANSWER A
Answer: (A) 23
Question: 2,7,36,4,14,225,6,21,?
(A) 475
(B) 576
(C) 676
(D) 776
(E) 876
Rationale: 2,7,(x)2, x+3=2+7=9 x=6
4,14,(y)2 y+3=14+4=18 y=15
6,21,(z)2 z+3=21+6=27 z=24 answer=24*24=576
ANSWER:B
Answer: (B) 576
Question: How many digits are in (12×10^14)(10×10^15)?
(A) 24
(B) 25
(C) 26
(D) 27
(E) 32
Rationale: The question simplfies to (12×10^14)(10^16)
=> 12*10^30
=> Will contain 30 zeros +2 digit 12
=>32
Ans E
Answer: (E) 32
Question: An army’s recruitment process included N rounds of selection tasks. For the first A rounds, the rejection percentage was 60 percent per round. For the next B rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 100000 people who applied for the army and 175 were finally selected, what was the value of N?
(A) 4
(B) 9
(C) 6
(D) 8
(E) 10
Rationale: fastly i reduce 60% till it gets closer to our required 2000 candidates
step(1) 40000 accepted.
step(2) another 40%of 40000=16000 accepted.
here it is quiet observable that if we further deduct candidate by 60% it would change our probablity of easy going 2000 candidate.
so i would get to second stage of recruitment where 50 % is accepted
step(3) 50% of 16000=8000
step(4) 50% of 8000=4000
Step(5) 50% of 4000=2000
Step(6) 50% of 2000=1000
Step(7) 50% of 1000=500
Step(8) 50% of 500=250
Step(9) 70% of 350=175(our target)
total 9 steps required.
Ans B
Answer: (B) 9
Question: If x and y are non-zero integers, and 36x^4 – 4y^4 = 6x^2 + 2y^2, which of the following could be the value of x2 in terms of y?
(A) 2y^2+1/6
(B) 2y^2+1/8
(C) 2y^2+1
(D) 2y^2
(E) 2y
Rationale: 36x^4 – 4y^4 = 6x^2 + 2y^2
(6x^2 + 2y^2) (6x^2 - 2y^2) = 6x^2 + 2y^2
6x^2 - 2y^2 = 1
6x^2 = 2y^2+1
x^2 = 2y^2+1/6
Answer : A
Answer: (A) 2y^2+1/6
Question: A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 100 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?
(A) 1 hour
(B) 1 hour 10 minutes
(C) 2 hours 30 minutes
(D) 1 hour 20 minutes
(E) 2 hours 10 minutes
Rationale: v1 and V2 are speeds.
v1.t /100 = v2
v2.t/40 = v1
v1/v2 = 5/2
which train A would 100. 2/5 mins to cover the same distance
40 + 40 = 80 mins (ANS D)
Answer: (D) 1 hour 20 minutes
Question: A spirit and water solution is sold in a market. The cost per liter of the solution is directly proportional to the part (fraction) of spirit (by volume) the solution has. A solution of 1 liter of spirit and 1 liter of water costs 50 cents. How many cents does a solution of 1 liter of spirit and 2 liters of water cost?
(A) 13
(B) 33
(C) 50
(D) 51
(E) 52
Rationale: The cost per liter of the solution is directly proportional to the part (fraction) of spirit (by volume) the solution has.
It means there is no effect of change in volume of water to the costing of the solution
So a solution of 1 liter of spirit and 2 liters of water will cost 50 cents
Answer = 50
ANSWER:C
Answer: (C) 50
Question: 3 friends A,B,C went for week end party to McDonald's restaurant and there they measure there weights in some order in 7 rounds. A, B, C, AB, BC, AC, ABC. Final round measure is 155 kg, then find the average weight of all the 7 rounds?
(A) 56
(B) 66
(C) 77
(D) 88
(E) 99
Rationale: Its finally a easy one..haha
(A+B+C+AB+BC+CA+ABC)/7 ...as given ABC=155
so A+B+C also equal to 155 and AB+BC+CA=2ABC=(2*155)
finally..(155+2*155+155)/7=88.57
ANSWER:D
Answer: (D) 88
Question: In a recent election, James received 15 percent of the 10000 votes cast. To win the election, a candidate needed to receive more than 51 percent of the vote. How many additional votes would James have needed to win the election?
(A) 1004
(B) 1000
(C) 3600
(D) 1201
(E) 3601
Rationale: James = (15/100)*10000 = 1500
Votes to win = (51/100)*Total Votes +1 = (51/100)*10000 +1 = 5101
Remaining Voted needed to win election = 5101 - 1500 =3601
Answer: option E
Answer: (E) 3601
Question: If cost of sugar increases by 44%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed?
(A) 32.55
(B) 35.55
(C) 28.55
(D) 29.55
(E) 30.55
Rationale: 100
144
-----
144 ----- 44
100 ------ ? => 30.55%
Answer:E
Answer: (E) 30.55
Question: A 230 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
(A) 230
(B) 270
(C) 260
(D) 256
(E) 298
Rationale: Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 2340)/9 = 500/9 => x = 270.
Answer: B
Answer: (B) 270
Question: In a forest 190 deer were caught, tagged with electronic markers, then released. A week later, 50 deer were captured in the same forest. Of these 50 deer, it was found that 5 had been tagged with the electronic markers. If the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest, and if no deer had either left or entered the forest over the preceding week, what is the approximate number of deer in the forest?
(A) 150
(B) 750
(C) 1,250
(D) 1,900
(E) 2,500
Rationale: The percentage of tagged deer in the second sample = 5/50*100=10%.
So, 190 tagged deers comprise 10% of total # of deers --> total # of deers = 190*10 = 1,900.
Answer: D.
Answer: (D) 1,900
Question: Harold needs to buy a ticket to attend a conference for work. His own department contributes $4 less than half the price of the ticket. The HR department will contributes $1 more than a third of the price of the ticket. With these two contributions, Harold has to pay only $10 out of his own pocket to cover the cost of the ticket. What was the price of the ticket?
(A) $36
(B) $42
(C) $48
(D) $54
(E) $60
Rationale: Say price of ticket = x
Own department contribution =x/2−4
HR contribution =x/3+1
Self contribution = 10
Setting up the equation
x=x/2−4+x3+1+10
x=42
Answer = B
Answer: (B) $42
Question: 33 1/3% OF 270?
(A) 80
(B) 90
(C) 110
(D) 120
(E) 130
Rationale: 33 1/3 % = 1/3
1/3 × 270 = 90
B)
Answer: (B) 90
Question: The largest 5 digit number exactly divisible by 91 is:
(A) 25698
(B) 581396
(C) 369741
(D) 99918
(E) 1258
Rationale: Explanation:
Largest 5-digit number = 99999
91) 99999 (1098
91
---
899
819
----
809
728
---
81
---
Required number = (99999 - 81)
= 99918.
Answer:D
Answer: (D) 99918
Question: John had a stock of 620 books in his bookshop. He sold 50 on Monday, 82 on Tuesday, 60 on Wednesday, 48 on Thursday and 40 on Friday. What percentage of the books were not sold?
(A) 54.84%
(B) 36.5%
(C) 46.5%
(D) 56.5%
(E) 66.5%
Rationale: Let N be the total number of books sold. Hence
N = 50 + 82 + 64 + 78 + 135 = 280
Let M be the books NOT sold
M = 620 - N = 620 - 280 = 340
Percentage
Books not sold / total number of books = 340/620= 0.55 = 54.84%
correct answer A
Answer: (A) 54.84%