divelab/Sys2Bench huggingface dataset:
Question: x is a set containing 8 different positive odd numbers. y is a set containing 7 different numbers, all of which are members of S. Which of the following statements CANNOT be true?
Options:
A)The range of y is even.
B)The mean of x is even.
C)The mean of y is even.
D)The range of x is equal to the range of y.
E)The median of y is equal to the mean of y.
Rationale: Given: x is a set containing 8 different positive odd numbers
Easily inferred: y is a set containing 7 differentoddnumbers
(A) The range of y is even. - Always even, since it is a difference of two odd numbers
(B) The mean of x is even. - can be. suppose 8 numbers consist of 4 pairs of (16n-1,16n+1) for different values of n. Adding these will give a multiple of 16, which when divided by 8 (to arrive at mean), will give an even number
(C) The mean of y is even. - Cannot be. Mean = (sum of 7 odd numbers)/7 = Odd number/odd number thus cannot be even.
(D) The range of x is equal to the range of y.
(E) The median of y is equal to the mean of y
Option C
Answer: C
Question: The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is
Options:
A)32
B)37
C)84
D)29
E)21
Rationale: Explanation:
Let the numbers be x and 4x. Then, \inline \fn_jvn x\times 4x =84\times 21\: \: \Leftrightarrow \: \: x^{2}=\frac{84\times 21}{4} \: \: \Leftrightarrow \: \: x=21
Hence Larger Number = 4x = 84
Answer: C) 84
Answer: C
Question: A horse is tethered to one corner of a rectangular grassy field 36 m by 20 m with a rope 18 m long. Over how much area of the field can it graze?
Options:
A)154 cm2
B)308 m2
C)254 m2
D)260 m2
E)None of these
Rationale: Area of the shaded portion
= 1⁄4 × π × (18)2 = 254 m2
Answer C
Answer: C
Question: Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to British stamps is 5 to 1, what is the ratio of US to British stamps?
Options:
A)10 : 5
B)15 : 2
C)20 : 2
D)25 : 2
E)None of these
Rationale: Explanation:
Indian stamps are common to both ratios. Multiply both ratios by factors such that the Indian stamps are represented by the same number.
US : Indian = 5 : 2, and Indian : British = 5 : 1. Multiply the first by 5, and the second by 2.
Hence the two ratios can be combined and US : British = 25 : 2
Now US : Indian = 25 : 10, and Indian : British = 10 : 2 Answer: D
Answer: D
Question: The average weight of A, B and C is 43 kg. If the average weight of A and B be 48 kg and that of B and C be 42 kg, then the weight of B is:
Options:
A)55 kg
B)53 kg
C)49 kg
D)52 kg
E)51 kg
Rationale: EXPLANATION
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (43 x 3) = 129 â€¦. (i)
A + B = (48 x 2) = 96 â€¦. (ii)
B + C = (42 x 2) = 84 â€¦.(iii)
Adding (ii) and (iii), we get: A + 2B + C = 180 â€¦. (iv)
Subtracting (i) from (iv), we get : B = 51.
Bâ€™s weight = 51 kg.
Answer E
Answer: E
Question: The average of six numbers is 3.95. The average of two of them is 3.3, while the average of the other two is 3.85. What is the average of the remaining two numbers?
Options:
A)4.5
B)4.6
C)4.7
D)4.8
E)4.9
Rationale: Sum of the remaining two numbers
= (3.95 × 6) – [(3.3 × 2) + (3.85 × 2)]
= 23.70 – (6.6 + 7.7)
= 23.70 – 14.3 = 9.40
∴ Required average = (9.4⁄2) = 4.7
Answer C
Answer: C
Question: For the positive numbers, n+3, n + 4, n + 6, n + 8, and n + 9, the mean is how much greater than the median?
Options:
A)0
B)1
C)n+l
D)n+2
E)n+3
Rationale: Let’s first calculate the mean (arithmetic average).
mean = sum/quantity
mean = (n + 3+n + 4 + n + 6 + n + 8 + n + 9)/5
mean = (5n + 30)/5
mean = n + 6
Next, we determine the median. The median is the middle value when the terms are ordered from least to greatest. The terms ordered from least to greatest are as follows:
n+3, n + 4, n + 6, n +8, n + 9
The median is n + 6
Finally we are asked how much greater the mean is than the median. To determine the difference we can subtract the smaller value (the median) from the larger value (the mean) and we get:
n + 6 – (n + 6) = n + 6 – n – 6 = 0
The answer is A
Answer: A
Question: The average of first six prime numbers which are odd is?
Options:
A)15.5
B)15.7
C)9.3
D)15.8
E)15.1
Rationale: Sum of first 6 prime no. which are odd = 56
Average = 56/6 = 9.3
Answer:C
Answer: C
Question: Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Options:
A)12
B)13
C)16
D)10
E)15
Rationale: Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Answer: D
Answer: D
Question: A can do a piece of work in 4 days. B can do it in 5 days. With the assistance of C they completed the work in 2 days. Find in how many days can C alone do it?
Options:
A)23 days
B)18 days
C)20 days
D)22 days
E)28 days
Rationale: C
20 days
C = 1/2 - 1/4 - 1/5 = 1/20 => 20 days
Answer: C
Question: A certain fruit stand sold apples for $0.70 each and bananas for $0.60 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
Options:
A)8
B)9
C)10
D)11
E)12
Rationale: Let's start with 1 apple for $0.70.
Let's subtract $0.70 from $6.30 until we get a multiple of $0.60.
$6.30, $5.60, $4.90, $4.20 = 7*$0.60
The customer purchased 7 bananas and 3 apples.
The answer is C.
Answer: C
Question: Sagar purchased 10 kg of rice at the rate of Rs. 15 per kg and 25 kg of rice at the rate Rs. 14 per kg. He mixed the two and sold the mixture. Approximately at what rate per kg should he sell the mixture to make 40 % profit in the transaction?
Options:
A)Rs. 20.00
B)Rs. 19.50
C)Rs. 15
D)Rs. 17.5
E)None of these
Rationale: By 40% profit on cost price,
Selling price of mixture = 140/100 * 500
= Rs. 700
Therefore, selling price per kg of mixture = 700/35 = Rs. 20
ANSWER A
Answer: A
Question: The cost of 3 pens and 5 pencils is Rs.200. Also the cost of one pen and one pencil is in the ratio of 5:1 respectively. What is the cost of one dozen pens?
Options:
A)Rs.200
B)Rs.250
C)Rs.300
D)Rs.150
E)Rs. 600
Rationale: Explanation:
Let the cost of one pen is ‘5x’ and pencil is ‘x’
3 X 5x + 5x = Rs.200
15x + 5x = Rs.200
X = 200/20 = 10
:. Cost of 1 pen = 5x = 5 x 10 = 50
:. Cost of 12 pens, i.e. (one dozen) = 50 x 12 = Rs.600
Answer: Option E
Answer: E
Question: The sum of two numbers is 16. The difference is 4. What are the two numbers?Let x be the first number. Ley y be the second number
x + y = 16
x - y = 4
Options:
A)10 and 6
B)10 and 5
C)10 and 4
D)10 and 3
E)10 and 2
Rationale: Solution
Let x be the first number. Ley y be the second number
x + y = 16
x - y = 4
Solve the system of equations by elimination
Adding the left sides and the right sides gives:
x + x + y + -y = 16 + 4
2x = 20
x = 10
Since x + y = 16, 10 + y = 16
10 + y = 16
10 - 10 + y = 16 - 10
y = 6
The numbers are 10 and 6
Answer A
Answer: A
Question: There are 28 balls which are red, blue or green. If 12 balls are green and the sum of red balls and green balls is less than 24, at most how many red balls are there?
Options:
A)10
B)12
C)11
D)13
E)14
Rationale: r+g+b=28
g=12
r + g < 24
=> r + 12 < 24
=> r < 12
=> at most 11 red balls
ANSWER:C
Answer: C
Question: At a certain restaurant, 30 percent of the servers reported their tips, while 40 percent of the servers who were tipped didn't report their tips. What percent of the servers were tipped?
Options:
A)30%
B)60%
C)50%
D)70%
E)80%
Rationale: Statement: 30% of Servers (S) Reported their Tips (RT). Of the servers who were tipped (T) 40% didn't report their tips (notRT). [Note: A Venn Diagram is useful to visualize the problem.]
Solution: Sample space is 100%.
S = 100%
RT/S = 30%
notRT/T = 40% => 100% - 40% => RT/T = 60%
T/S = (30)/(60)% = 50%
Answer: C
Answer: C
Question: The value of a machine depreciates at the rate of 10% per year. If its present is Rs.1,62,000 what will be its worth after 2 years ? What was the value of the machine 2 years ago ?
Options:
A)50000
B)100000
C)150000
D)200000
E)None of them
Rationale: Value of the machine after 2 years
=Rs.[162000*(1-(10/100))^2] = Rs.[162000*(9/10)*(9/10)]
=Rs. 131220
Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000
Answer is D.
Answer: D
Question: In the first 10 overs of a cricket game, the run rate was only 4.6. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
Options:
A)4.25
B)5.9
C)6.25
D)7
E)7.5
Rationale: Required run rate = 282 - (4.6x 10) =	236 236/40= 5.9
B)
Answer: B
Question: A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 98.01, the number of the member is the group is:
Options:
A)57
B)67
C)77
D)99
E)97
Rationale: Money collected = (98.01 x 100) paise = 9801 paise.
Number of members =Square root of 9801 = 99.
Answer: Option D
Answer: D
Question: A certain tire company can produce tires at a cost of $22,500 per batch plus $6 per tire. The company can sell tires to the wholesaler at a cost of $20 per tire. If a batch of 15,000
tires is produced and sold, what is the company’s profit per tire?
Options:
A)$9.00
B)$9.50
C)$11.00
D)$13.50
E)$12.50
Rationale: CP(15000 tires) = $22500 +($6×15000) =$112500
SP (15000 tires) = $20×15000 = $300000
Profit = SP-CP
= $300000-$112500= $187500
PROFIT/TIRE = $187500/15000 = $12.50
Answer E
Answer: E
Question: A rower can row 5 km/h in still water. When the river is running at 2 km/h, it takes the rower 1 hour to row to Big Rock and back. How many kilometers is it to Big Rock?
Options:
A)1.5
B)1.8
C)2.1
D)2.4
E)2.7
Rationale: Let x be the distance to Big Rock.
time = x/3 + x/7 = 1
x = 21/10 = 2.1 km
The answer is C.
Answer: C
Question: A garrison of 3300 men has provisions for 32 days when supplied at the rate of 850 g per head. At the end of 7 days, a reinforcement arrives, and it is found that the provisions can last for 17 days more when supplied at the rate of 825 g per head. What is the strength of the reinforcement?
Options:
A)1700
B)1000
C)3000
D)2700
E)2600
Rationale: 825*17*x=850*25*3300
=> 5000
strength of reinforcement = (5000-1300)= 1700
final answer 1700.
ANSWER:A
Answer: A
Question: 4, 8, 16, 28, 44, ?
Options:
A)64
B)56
C)74
D)57
E)32
Rationale: The pattern is + 4, + 8, + 12, + 16, +20
Answer : A.
Answer: A
Question: Priya and her father differ in their ages by 31 years. After 8 years. if the sum of their ages will be 69, their present ages are
Options:
A)11, 42
B)12, 24
C)15, 42
D)16, 19
E)33, 44
Rationale: A
11, 42
Let the age of Priya = x years
Then, the age of her father = (x+ 31) years
ATQ,
(x+ 8) + (x+ 31 + 8) = 69
2x+ 47 = 69
2x= 69 – 47 = 22
x= 11
The age of Priya = 11 years
The age of her father = 11 + 31 = 42 years
Answer: A
Question: A reduction of 10% in the price of tea enables a dealer to purchase 25 kg more tea for Rs. 22500. What is the reduced price per kg of tea?
Options:
A)Rs. 70
B)Rs. 80
C)Rs. 90
D)Rs. 100
E)None
Rationale: Solution: 1st Method:
Let the original price of tea be Rs. x/kg.
After reduction the price becomes = x -10% of x = 9x/10 per kg.
Now,
(22500/(9x/10))-22500/x = 25
Or, 22500[10/9x- 1/x] = 25
Or, 25*9x = 22500;
Or, x = (22500/2589) = Rs. 100.
Hence, new price = 90 per kg.
Thought process Method:
Let the original price be Rs. 100 per kg, he get tea = 22500/100 = 225 kg.
After reduction the price becomes = 90 per kg, he get tea = 22500/90 = 250 kg.
So, reduction price is Rs. 90 per kg as it enables him to buy 25 kg of more tea.
Answer: Option C
Answer: C
Question: Find (7x + 4y ) / (x-2y) if x/2y = 3/2 ?
Options:
A)24
B)25
C)26
D)27
E)29
Rationale: x/2y=3/2
=>x=6y/2=3y
=>(7x+4y)/(x-2y)=((7*(3y))+4y)/(3y-2y)
=>25y/y=25
ANSWER:B
Answer: B
Question: In how many ways can the letters of word IMAGINATION be arranged such that all the vowels as well as all the Consonants always appear together?
Options:
A)9!
B)5!*4!
C)5!*5!
D)5!*4!*2!
E)6!*5!*2!
Rationale: Since all the vowels and consonants have to be together there can only be there are 6! way to order the vowels and 5! to order the consonants. But they must be together either vowels come first or consonants so you multiply by 2.
So 6!5!2! (E)
Answer: E
Question: When positive integer N is divided by positive integer J, the remainder is 18. If N/J = 134.03, what is value of J?
Options:
A)22
B)56
C)78
D)112
E)600
Rationale: 1) We know that
decimal part of decimal quotient ={remainder/divisor}
So 0.03, the decimal part of the decimal quotient, must equal the remainder, 18, divided by the divisor J.
0.03 = 18/J
0.03*J = 18
J = 18/0.03 = 1800/3 = 600
So J = 600, answer = E.
Answer: E
Question: The rate of spin of a certain gyroscope doubled every 15 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 400 meters per second, what was the speed, in meters per second, when the stopwatch was started?
Options:
A)25/2
B)25/4
C)25/8
D)25/16
E)25/32
Rationale: Let x be the original speed when the stopwatch was started.
In 90 seconds, the speed doubled 6 times.
2^6 * x = 400
x = (2^4 * 25) / 2^6 = 25/4
The answer is B.
Answer: B
Question: A train covers a distance of 15 km in 10 min. If it takes 6 sec to pass a telegraph post, then the length of the train is?
Options:
A)m
B)m
C)m
D)m
E)m
Rationale: Speed = (15/10 * 60) km/hr
= (90 * 5/18) m/sec = 25 m/sec.
Length of the train = 25 * 6
= 150 m.
Answer:C
Answer: C
Question: A man purchases the milk at Rs. x/litre and sells it at Rs. 2x/liter stillhe mixes 2liters waterwith every 6liters of pure milk. What is the profit %?
Options:
A)123%
B)134%
C)145%
D)166.66%
E)186.66%
Rationale: Let the cost price of 1 litre pure milk be Re.1, then
\inline{6litres(milk)→CP=Rs.62litres(water)→CP=Rs.0}{6litres(milk)→CP=Rs.62litres(water)→CP=Rs.0}\rightarrow CP=Rs.6 only
and 8 litre mixture \rightarrow SP \rightarrow 8 x 2 = Rs.16
Profit %= 166.66%
D
Answer: D
Question: In the xy-plane, line k passes through point (3,-2) and intersect perpendicularly with line n that passes through points (0,3) and (3,5). What is the equation of line k?
Options:
A)3x+2y-5=0
B)3x+2y+5=0
C)2x-3y-12=0
D)2x+3y=0
E)4x+3y-6=0
Rationale: If two lines intersect perpendicularly, we get -1 when we multiply two slopes together. The slope of a line n is (5-3)/(3-0)=2/3. Hence, the slope of a line k should be -3/2. Since it crosses (3,-2), if we calculate we get 3x+2y=5. Hence, the correct answer is A.
Answer: A
Question: A can build up a wall in 8 days while B can break it in 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of wall?
Options:
A)13 1⁄3 days
B)7 1⁄3 days
C)6 1⁄3 days
D)7 days
E)None of these
Rationale: A’s one day’s work = 1⁄8th work.
B’s one day’s work = 1⁄3rd work.
∴ A’s 4 day’s work = 4 × 1⁄8 = 1⁄2nd work
∴ In next two days, total wall = 1⁄2 + 2(1⁄8) - 2(1⁄3) = 1⁄12th wall
Remaining wall = 1 - 1⁄12 = 11⁄12th
Now, 1⁄8th wall is built up by A in one day.
∴ 11⁄12th wall is built up by A in 8 × 11⁄12 = 71⁄3 days.
Answer B
Answer: B
Question: The probability that A speaks truth is 7/9 and that of B speaking truth is 9/13. What is the probability that they agree in stating the same fact?
Options:
A)18/35
B)111/117
C)157/195
D)192/235
E)None of these
Rationale: If both agree stating the same fact, either both of them speak truth of both speak false.
Probability = 7/9 * 9/13 + 6/9 * 8/13
= 63/117 + 48/117 = 111/117
ANSWER:B
Answer: B
Question: Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?
Options:
A)60
B)50
C)48
D)76
E)34
Rationale: Relative speed = 45 + 30 = 75 km/hr.
75 * 5/18 = 125/6 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 6/125 = 48 sec.
Answer: Option C
Answer: C
Question: James have 10 parrots, 4 fishes, 9 rabbits and 6 dogs.How many pets does he have in all?
Options:
A)17
B)11
C)9
D)15
E)29
Rationale: 10+4+9+6=29. Answer is E
Answer: E
Question: Scott's average (arithmetic mean) golf score on his first four rounds was 58.What score does he need on his fifth round to drop his average score by 2 points?
Options:
A)48
B)72
C)78
D)88
E)312
Rationale: Average score for first 4 rounds = 58
Average score after 5 rounds = 56
Score Scott needs on his 5th round to drop average score by 2 points = 58 - (2*5) = 48
Answer A
Answer: A
Question: Find the odd man out. 18, 7, 12, 24, 48, 34, 46
Options:
A)16
B)46
C)24
D)11
E)7
Rationale: Explanation :
7 is the only odd number in the given series
Answer : Option E
Answer: E
Question: In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth
Rs.8 per Kg?
Options:
A)5 : 4
B)2 : 1
C)3 : 2
D)4 : 3
E)None of these
Rationale: Explanation :
By the rule of alligation, we have
CP of 1 kg Rice of 1st kind CP of 1 kg Rice of 2nd kind
7.1 9.2
Mean Price
8
9.2 - 8 = 1.2 8 - 7.1 = .9
Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3
Answer : Option D
Answer: D
Question: A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:
Options:
A)12
B)13
C)14
D)15
E)18
Rationale: Dividend on Rs 20 = 9/100*20 = 9/5
is 12 income 100
9/5 is an income on Rs (100/12*9/5) = Rs 15
ANSWER D
Answer: D
Question: Taps A and B can fill a bucket in 10 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?
Options:
A)7 min. 30 sec.
B)8min. 20 sec.
C)7min. 25 sec
D)7 min.35 sec.
E)8min. 20 sec.
Rationale: Explanation:
Part filled in 3 minutes =
3∗(1/10+1/15)=3∗5/30=5/10
Remaining part =1−5/10=5/10
=>1/15:5/10=1:X
=>X=15/2
=>X=7.5mins.
So it will take further 7 mins 30 seconds to fill the bucket.
OptionA
Answer: A
Question: For what value of â€œkâ€ will the equation (2kx2 + 4kx +2)=0 have equal roots?
Options:
A)1
B)9/4
C)16/25
D)7/1
E)7/2
Rationale: for a 2nd degree equation ax2+bx_c=0 has equal roots the condition is b2-4ac =0
in the given equation
(4k)^2-4*2k*2=0
by solving this equation
we get k=0, k=1
ANSWER:A
Answer: A
Question: Which of the following numbers should not be included in the set?
331, 482, 551, 263, 386, 242, 111
Options:
A)263
B)386
C)242
D)111
E)None of these
Rationale: Solution
In each number except 386,the product of first and third digits is the middle one. Answer B
Answer: B
Question: If x is an integer such that 5 < x < 21, 7 < x < 18, 13 > x > 2, 12 > x > 9, and x + 1 < 13, then x is
Options:
A)3
B)5
C)7
D)9
E)11
Rationale: 5 < x < 21,
7 < x < 18,
2 < x < 13
9 < x < 12
x<13
From above: 10< x < 12 --> x = 11.
Answer: E.
Answer: E
Question: How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?
Options:
A)16.8
B)17.8
C)15.9
D)13.2
E)12.4
Rationale: D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec
ANSWER A
Answer: A
Question: The average of first 11 prime numbers is?
Options:
A)14.55
B)11.11
C)12.11
D)13.11
E)14.11
Rationale: Sum of 11 prime no. = 160
Average = 160/11 = 14.55
ANSWER: A
Answer: A
Question: On flipping 3 coins together what is the outcomes for at least three TAILl?
Options:
A)1/7
B)3/8
C)1/8
D)4/13
E)5/13
Rationale: Sample space, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Favorable outcomes for at least two heads
= favorable outcomes for exactly two heads and favorable outcomes for more than two heads
= {TTT}
1/8
C
Answer: C
Question: If 12!/3^x is an integer, what is the greatest possible value of x?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: This question is asking how many power of 3 exists in 12!
easiest way is to identify factor of three
12 -> 4*3 => 1 power of 3
9 -> 3*3=> 2 power of 3
6 -> 2*3=> 1 power of 3
3 -> 1*3=> 1 power of 3
Total equal 5
ANSWER:C
Answer: C
Question: Number of prime numbers between 70 and 100 is
Options:
A)4
B)5
C)6
D)7
E)9
Rationale: total 6=71, 73, 79, 83, 89, 97
ANSWER:C
Answer: C
Question: Carl is facing very difficult financial times and can only pay the interest on a $10,000 loan he has taken. The bank charges him a quarterly compound rate of 4%. What is the approximate interest he pays annually?
Options:
A)$1750
B)$2000
C)$2150
D)$2500
E)$12000
Rationale: An easy way too attempt this is by approximation:
For the first quarter, he pays 4% of 10,000 which is $400.
So for the four quarters in the year, he will approximately play 400*4=1600 over the entire year.
Because of compound interest it will be more than 1600. Approx. option A is correct.
Answer : A
Answer: A
Question: If x ≠ 0 and y/x=y+x, which of the following expressions is equal to its opposite?
Options:
A)x^2
B)x^2 – y^2
C)y^2
D)x^2 + xy - y^2
E)(x – y)^2
Rationale: y/x=(y+x)/y
=> y^2 = xy +x^2
=> x^2 + xy - y^2= 0
Answer D
Answer: D
Question: In a renowned city, the average birth rate is 5 people every two seconds and the death rate is 3 people every two seconds. Estimate the size of the population net increase that occurs in one day.
Options:
A) 85,400
B) 86,400
C) 87,400
D) 88,400
E) 89,400
Rationale: This question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds.
For this particular question : Increase in the population every 2 seconds = 5-3 = 2 people.
Total 2 second interval in a day = 24*60*60/2 = 43,200
Population increase = 43,200 * 2 = 86,400. Hence B.
Answer: B
Question: A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 3:4:5. If the total value of all the coins in the bag is Rs.350, find the number of 50 paise coins ?
Options:
A)498
B)488
C)375
D)528
E)418
Rationale: 50*3k + 20*4k + 10*5k = 35000
280k = 35000 => k = 125
50p coins = 3k = 3*125 = 375
ANSWER:C
Answer: C
Question: The cost price of three varieties of oranges namely A, B and C is Rs 20/kg, Rs 40/kg and Rs 50/kg. Find the selling price of one kg of orange in which these three varieties of oranges are mixed in the ratio of 2 : 3 : 5 such that there is a net profit of 20%?
Options:
A)Rs 48
B)Rs 48.6
C)Rs 49.2
D)Rs 49.8
E)Rs 48.8
Rationale: Cost price of one kg of orange in which the three varieties of oranges are mixed in the ratio 2 : 3 : 5 is equal to S where
S = 0.2 x 20 + 0.3 x 40 + 0.5 × 50
= 4 + 12 + 25
= Rs 41
Selling price per kg of oranges to ensure there is a net profit of 20%
= 1.2 x 41
= Rs 49.2
ANSWER:C
Answer: C
Question: A certain class of students is being divided into teams. The class can either be divided into 8 teams with an equal number of players on each team or 24 teams with an equal number of players on each team. What is the lowest possible number of students in the class?
Options:
A)6
B)36
C)48
D)24
E)72
Rationale: Let total no of students in the class be N
So, we are told that N is divisible by both 8 24
So, Lets find the least common multiple of 8 24 , ie 24
So our answer is (D) 24
Answer: D
Question: The compound ratio of 5:6, 3:4 and 4:5?
Options:
A)1:1
B)1:87
C)1:6
D)1:9
E)1:2
Rationale: 5/6 * 3/4 * 4/5 = 1/2
1:2
Answer: E
Answer: E
Question: A 8000 liter tank, half-full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes. At the same time, the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes. How many minutes does it take to fill the tank completely?
Options:
A)8
B)12
C)18
D)24
E)48
Rationale: In: we have: 1,000/2min = 500 litres per minute
Out: we have: 1,000/4 + 1,000/6
Then do: IN - OUT to figure out the net inflow per minute (you get 83.3). Then divide the total number of litres you need (4,000 by that net inflow to get the minutes) - 48 min. Answer E.
Answer: E
Question: If x = -3/7 and y = -2/3, what is the value of the expression -2x – y^2?
Options:
A)-33/22
B)23/85
C)26/63
D)58/41
E)-26/63
Rationale: x = -3/7 and y = -2/3
==> -2(-3/7) - (2/3)^2 = 6/7 - 4/9 = 26/63
Ans: C
Answer: C