divelab/Sys2Bench huggingface dataset:
Question: A tap can fill a tank in 6 hours. After half the tank is filled three more similar taps are opened. What is the total time taken to fill the tank completely?
Options:
A)3 hrs 15 min
B)3 hrs 45 min
C)4 hrs
D)4 hrs 15 min
E)5 hrs
Rationale: Time taken by one tap to fill the tank = 3 hrs.
Part filled by the taps in 1 hour = 4 * 1/6 = 2/3
Remaining part = 1 - 1/2 = 1/2
2/3 : 1/2 :: 1 : x
x = 1/2 * 1 * 3/2 = 3/4 hrs. i.e., 45 min
So, total time taken = 3 hrs 45 min.
ANSWER:B
Answer: B
Question: If p is an even integer and q is an odd integer, which of the following must be an odd integer?
Options:
A)p/q
B)pq
C)2p+q
D)2(p+q)
E)3p/q
Rationale: A. p/q
p = 6, q = 3 => p/q is odd integer
p = 4, q=3 =>p/q is not integer
B. pq
Must be even
C. 2p+q
2p must be even, q is odd=> 2p+q must be ODD: CORRECT
D. 2(p+q)
must be even
E. 3p/q
Same A
Answer: C
Answer: C
Question: There are 6 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?
Options:
A)10
B)20
C)40
D)60
E)12
Rationale: Method 2:
Each person is one participant of 4 games. So there are in all 4*6 = 24 instances of one participant games. But each game has 2 participants so total number of games = 24/2 = 12
E
Answer: E
Question: Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.18 of money?
Options:
A)1800
B)1992
C)9921
D)2798
E)2789
Rationale: P = 18(100/10)2
=> P = 1800
Answer: A
Answer: A
Question: A can give B 80 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
Options:
A)130.12
B)130.43
C)130.22
D)111.0
E)111.12
Rationale: A runs 1000 m while B runs 920 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/920
= 869.56 m
B can give C = 1000 - 869.56
=130.43m.
Answer:B
Answer: B
Question: A man is trying to fill up his pool and he has three different water sources, if he uses faucet A it takes 5 hours to fill up the pool, if he used faucet B it takes 20 hours to fill it up. He first runs the water from source A for 2 hours, then turning that off, he continues with faucet B for 10 hours. Then turning that off as well he uses Faucet C and in 3 hours the pool is filled up. How long does it take to fill up the pool only using the faucet C?
Options:
A)10 hours
B)20 hours
C)30 hours
D)40 hours
E)5 hours
Rationale: 2/5 +10/20 + 3/x = 1 x = 30ANSWER =C
Answer: C
Question: The age of Somu is one-third his father's. 5 years back he was one-fifth of his father's age. What is his persent age ?
Options:
A)11
B)13
C)14
D)12
E)10
Rationale: Explanation :
Let Somu's age be x and that of his father be 3x.
So, x - 5 = 3x - 5/5 = x = 10
Answer : Option E
Answer: E
Question: It was calculated that 75 men could complete a piece of work in 10 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work?
Options:
A)10 days.
B)20 days.
C)25 days.
D)12.5 days.
E)15 days.
Rationale: One day work = 1 / 10
One man’s one day work = 1 / ( 10 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 10 * 75)
The total no. of days required to complete the work = (75 * 10) / 50 = 15
Answer:E
Answer: E
Question: A and B invests Rs.6000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
Options:
A)32:99
B)82:31
C)24:45
D)34:89
E)35:21
Rationale: A : B
(6000*4)+(3000*8) : (9000*6)+(6000*6)
48000 : 90000
24 : 45
Answer:C
Answer: C
Question: If y exceeds x by 20%, then x is less than y by?
Options:
A)16 2/38%
B)16 3/3%
C)16 2/3%
D)18 2/3%
E)126 2/3%
Rationale: X=100	y=120
120------20
100-------? => 16 2/3%
Answer: C
Answer: C
Question: Calculate the fourth proportional to 7.4, 8.1 and 9.8?
Options:
A)10.73
B)13.73
C)15.73
D)11.73
E)12.73
Rationale: Explanation:
Formula = Fourth propotional = (b × c)/a
A = 7.4 , B = 8.1 and C = 9.8
(8.1 × 9.8)/7.4 = 10.73
Answer: Option A
Answer: A
Question: The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:
Options:
A)20
B)30
C)60
D)70
E)10
Rationale: B
30
Let the three numbers be x, y and z.
Sum of the numbers is 98.
x + y + z = 98………………(i)
The ratio of the first to the second is 2/3.
x/y = 2/3.
x = 2/3 × y.
x = 2y/3.
The ratio of the second to the third is 5/8.
y/z = 5/8.
z/y = 8/5.
z = 8/5 × y.
z = 8y/5.
Put the value of x = 2y/3 and z = 8y/5 in (i).
2y/3 + y + 8y/5 = 98
49y/15 = 98.
49y = 98 × 15.
49y = 1470.
y = 1470/49.
y = 30 .
Therefore, the second number is 30.
Answer: B
Question: Dominos charges $6.25 for a veg pizza of std size that costs a total of $2.85 to make. Pizza Hut charges $5.80 for a similar pizza that costs $1.80 to make. If Dominos sells 2,000 pizzas, how many must Pizza Hut sell to make at least as much in total gross profit as its competitor does?
Options:
A)858
B)1875
C)14,667
D)63,840
E)146,667
Rationale: Dominos : Profit/pizza = 6.25-2.85 = 3.75
: No of pizza = 2,000
: Gross Profit = 2,000*3.75 = 7,500
Pizza Hut : Profit/pizza = 4
: Gross Profit = 7,500
: No of Cups = 7,500/4 = 1875
Answer B
Answer: B
Question: What should come in the place of the question mark(?) in the following equation?
45(2)×27(2)/135(2)=?
Options:
A)81
B)1
C)243
D)9
E)None of these
Rationale: ? = 45×45×27×27 /135×135=81
Answer A
Answer: A
Question: What is the least number of squares tiles required to pave the floor of a room 4 m 5 cm long and 2 m 25 cm broad?
Options:
A)14
B)20
C)40
D)44
E)45
Rationale: Length of largest tile = H.C.F. of 405 cm and 225 cm = 45 cm.
Area of each tile = (45 x 45) cm2.
Required number of tiles =405 x 225/(45^2)= 45.
ANSWER:E
Answer: E
Question: If n=4p where p is a prime number greater than 3, how many different positive even divisors does n have, including n?
Options:
A)2
B)6
C)4
D)8
E)10
Rationale: We're told that N = 4P and that P is a prime number greater than 3. Let's TEST P = 4; so N = 16
The question now asks how many DIFFERENT positive EVEN divisors does 16 have, including 16?
16:
1,16
2,8
4,4
How many of these divisors are EVEN? 2,8,4,16…..4 even divisors.
ANSWER:C
Answer: C
Question: If Henry were to add 6 gallons of water to a tank that is already 3/4 full of water, the tank would be 7/8 full. How many gallons of water would the tank hold if it were full?
Options:
A)25
B)48
C)64
D)80
E)96
Rationale: 7/8x-3/4x=6galls
1/8*x=6gallons
x=48 gallons
Answer B
Answer: B
Question: Three machines operating independently, simultaneously, and at the same constant rate can fill a certain production order in 52 hours. If one additional machine were used under the same operating conditions, in how manyfewerhours of simultaneous operation could the production order be fulfilled?
Options:
A)6
B)9
C)12
D)27
E)13
Rationale: If 3 Machines can do the work in 36 hr then 4 Machines can do the work in 3/4*52 = 39 Hrs. hence time saved will be 13hr
option E is the correct answer
Answer: E
Question: The average of four consecutive even numbers is 27. find the largest of these numbers?
Options:
A)10
B)20
C)30
D)40
E)50
Rationale: let the numbers be x,x+2,x+4 andx+6. then,
(x+(x+2)+(x+4)+(x+6))/4) = 27
(4x+12)/4 = 27
x+3=27
x=24.
Therefore the largest number=(x+6)=24+6=30.
Option C
Answer: C
Question: There are 2 sections with 6 questions each. If two questions are selected from each section, the chance of getting different questions is________
Options:
A)10
B)100
C)240
D)225
E)6000
Rationale: we have to select two questions from each section so the chances are 6C2.6C2=225
ANSWER:D
Answer: D
Question: Find the least number which when divided by 35 and 11 leaves a remainder of 1 in each case.
Options:
A)384
B)391
C)388
D)397
E)386
Rationale: The least number which when divided by different divisors leaving the same remainder in each case
= LCM(different divisors) + remainder left in each case.
Hence the required least number
= LCM(35, 11) + 1 = 386.
ANSWER:E
Answer: E
Question: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 28 white marbles. How many red marbles could be in bag A?
Options:
A)1
B)3
C)4
D)6
E)7
Rationale: # of Red marbles in Bag A can be either 2 or 6. No 2 in the choices, so 6. D.
Bag A:
R:W:B = 2:6:9
Bag B
R:W = 1:4
6X + 4Y = 28 i.e 3X + 2Y = 14
X has to be odd to make an odd sum from the eq.
X = 2 , Y = 4 OR X = 4, Y = 1
So R can be 2X i.e 4 or 8.
Answer C
Answer: C
Question: From a corner of a ground of 20m*30m, a boy hits a ball that travels maximum 14m. What is the area of the football ground through which the ball can pass by the hit of that boy from that corner ?
Options:
A)132
B)140
C)150
D)154
E)168
Rationale: Area covered by ball = Pi*r^2/4 ( here we divide by 4 because ball is hit from the corner of the ground and only 1/4 part the ball can go)
where r= 14 m = length of pass
So area = (22/7)*14*14/4 =154 sq m
answer :D
Answer: D
Question: If n is a multiple of 5, n=p^2q, where p and q are prime numbers. Which of the following must be a multiple of 25?
Options:
A)p^2
B)q^2
C)pq
D)p^2q^2
E)p^3q
Rationale: You are given that n = p^2 * q, which is a multiple of 5. Therefore p^2 * q is a multiple of 5. Since 5 is a prime number, p OR q must be a multiple of 5. Therefore, in order to be a multiple of 25, you must have a MINIMUM of p^2 * q^2 in the answer, since you don't know whether p or q is the multiple of 5. The only answer which satisfies this condition is D. The answer is D: p^2*q^2.
Answer: D
Question: (x —1)(x + 3) > 0
(x +5)(x—4) < 0
Which of the following values of x satisfy both inequalities shown?
I. 3
II. -6
III. 2
IV. 5
Options:
A)I and II only
B)I and III only
C)II and III only
D)II and IV only
E)I, II, III, and IV
Rationale: (x - 1)(x + 3) > 0 --> x<-3 or x>1.
(x + 5)(x - 4) < 0 --> -5<x<4.
Among choices only 3 and 2 satisfy both inequalities.
Answer: B.
Answer: B
Question: 6persons enter a lift on the ground floor of a 9floor apartment. Assuming that each of them independently & with equal probability can leave the lift at any floor beginning with the first, what is the probability that all the 6persons are leaving the lift at different floors?
Options:
A)7P6/8^6
B)5P6/8^6
C)8P6/8^6
D)9P6/8^2
E)8P6/8^3
Rationale: Apart from the ground floor, there are 8 floors
Let's find out the total number of ways in which all the six persons can leave the
lift at eight different floors
The 1st person can leave the lift in any of the 8 floors (8 ways)
The 2nd person can leave the lift in any of the remaining 7 floors (7 ways)
The 3nd person can leave the lift in any of the remaining 6 floors (6 ways)
...
The 6th person can leave the lift in any of the remaining 3 floors (3 ways)
Total number of ways = 8 × 7 × 6 × 5 × 4 × 3 = 8P6
(In fact, from the definition of permutations itself,
we will be able to directly say that the number of ways in which all the six persons
can leave the lift at 8 different floors = 8P6)
n(E)= Total Number of ways in which all the six persons can leave the lift at eight
different floors = 8P6
Now we will find out the total number of ways in which each of the six persons
can leave the lift at any of the eight floors
The 1st person can leave the lift in any of the 8 floors (8 ways)
The 2nd person can leave the lift in any of the 8 floors (8 ways)
The 3nd person can leave the lift in any of the 8 floors (8 ways)
...
The 6th person can leave the lift in any of the 8 floors (8 ways)
Total number of ways = 8 × 8 × 8 × 8 × 8 × 8 = 86
i.e., The total number of ways in which each the six persons can leave the lift
at any of the eight floors = n(S) = 86
P(E) = n(E)n(S)=8P6/8^6
C
Answer: C
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 20000 people who applied for the army and 196 were finally selected, what was the value of N?
Options:
A)4
B)5
C)6
D)8
E)10
Rationale: step(1) 8000 accepted.
step(2) another 40%of 8000=3200 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 3200=1600
step(4) 50% of 1600=800
step(5) 50% of 800=400
Step(6) 50% of 400=200
70% of 400=280
and last step of accepting 70% of 280=196(our target)
total 8 steps required.
Ans D
Answer: D
Question: 1254,1254,627,209,52.25,?
Options:
A)1.29
B)10.45
C)3.58
D)4.96
E)3.87
Rationale: Explanation :
Terms are continuously divided by 1,2,3 ,4, and so on
1254
1254 Ã· 1 = 1254
1254 Ã· 2 = 627
627 Ã· 3 = 209
209 Ã· 4= 52.25
52.25Ã·5=10.45
Answer : Option B
Answer: B
Question: How many integer values R are there for x such that 1 < 3x + 5 < 17?
Options:
A)Two
B)Three
C)Four
D)Five
E)Six
Rationale: 1 < 3x + 5 < 17
=> -4 < 3x < 12
=> -4/3 < x < 4
x can take integer values R -1,0 , 1 , 2 , 3
Answer D
Answer: D
Question: Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is
Options:
A)6/19%
B)6/17%
C)5 5/11%
D)3 5/11%
E)None of these
Rationale: Explanation:
Whenever we get this type of question, we should have formula in mind that,
Gain% = (gain/cost)*100
Cost = 4700 + 800 = Rs. 5500
S.P. = Rs. 5800
Profit = 5800 - 5500 = 300
Gain%=300/5500∗100=5 5/11%
Option C
Answer: C
Question: How many options are there for license plate numbers if each license plate can include 2 digits and 3 letters (in that order), or 3 digits and 2 letters (in that order)? (Note: there are 26 letters in the alphabet)
Options:
A)37×260^2
B)36×260^2
C)36×270^2
D)10^2×26^3
E)10^3×26^2
Rationale: We are given that License plate could be either of the form DDLLL or DDDLL
Now, every D could be filled in 10 ways and every L could be filled in 26 ways.
or Number of ways for DDLLL = 10 * 10 * 26 * 26 * 26 --(1)
Number of ways for DDDLL = 10 *10*10*26*26 --(2)
Total ways = (1) +
(2)
= \(36×260^2\)
Answer s B.
Answer: B
Question: The average of seven numbers is 18. The average of first three numbers is 14 and the average of last three numbers is 19. What is the middle number?
Options:
A)26
B)27
C)29
D)32
E)36
Rationale: The total of seven numbers = 7X18 = 126
The total of first 3 and last 3 numbers is = 3 X 14+3 X 19 = 99
So, the middle number is (126 - 99 ) = 27
B
Answer: B
Question: If -2x+3y=8 and 4x-3y=20, what is the sum of x and y?
Options:
A)24
B)28
C)18
D)26
E)32
Rationale: Given -2x+3y=8 ---Eq 1
4x -3y =20 -- Eq 2
Sum both eqns we get 2x = 28 => x = 14
Sub 2x in eq 1 => -28 + 3y = 8.
=> y = 12
Now X+Y = 12 + 14 = 26
Option D is correct answer.
Answer: D
Question: If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum W of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
Options:
A) 6x
B) 4x
C) 5x/3
D) 3x/2
E) 3x/5
Rationale: Note that we have 5positivetemperatures.
Next, as the average is x then the sum of the temperatures is 5x and as all the temperatures are positive then the sum of the 3 greatest must be more than (or equal to) 3x (as the average of the 3 greatest must be at least x) and less than 5x: 3x<W<5x --> only option B fits.
Answer: B.
Answer: B
Question: In how many different number of ways 4 men and 2 women can sit on a bench?
Options:
A)690
B)700
C)720
D)740
E)760
Rationale: npn = n!
6p6 = 6 × 5 × 4 × 3 × 2 × 1 = 720
C
Answer: C
Question: If 5 men can dig a field in 20 days, how much time it will take 4 men to dig the same field?
Options:
A)21
B)22
C)23
D)24
E)25
Rationale: E
Less the number of men, more time it will take to dig the field (inversely proportion)
5 : 4 :: N : 20
N = 20 X 5 / 4 = 25
Answer: E
Question: If |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?
Options:
A)x-y> 0
B)x-y< 0
C)x+y> 0
D)xy>0
E)xy<0
Rationale: |x| - |y| = |x+y|
=> (|x| - |y|)^2 = |x+y|^2 = (x+y)^2
=> x^2 - 2|x||y| + y^2 = x^2 + 2xy + y^2
=> |xy| = -xy
because xy != 0 so xy <0
Ans: E
Answer: E
Question: The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 40 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel?
Options:
A)20 ft
B)25 ft
C)600ft
D)900 ft
E)1000 ft
Rationale: Point to note: Both the wheels would have traveled the same distance.
Now consider, no. of revolutions made by back wheel as x, which implies that the number of revolutions made by the front wheel is (x+5).
Equating the distance traveled by front wheel to back wheel: (x+5)*30 = x * 40. (Formula for calculating the distance traveled by each wheel is: # of revolutions * circumference.)
Solving this eqn. gives x=15.
Sub x=15 either in (x+5)*30 or in x * 40 to get the distance, which is 600. So the correct choice is C.
Answer: C
Question: A train speeds past a pole in 15 seconds and a platform 130 meters long in 25 seconds. What is the length of the train (in meters)?
Options:
A)175
B)185
C)195
D)205
E)215
Rationale: Let the length of the train be x meters.
The speed of the train is x / 15.
Then, x + 130 = 25*(x / 15)
10x = 1950
x = 195 meters
The answer is C.
Answer: C
Question: How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 390 m in length?
Options:
A)18.9 sec
B)88.9 sec
C)22.9 sec
D)30.00 sec
E)72.0 sec
Rationale: D = 110 + 390 = 500 m
S = 60 * 5/18 = 50/3
T = 500 * 3/50 = 30 sec
Answer: D
Answer: D
Question: When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 16 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?
Options:
A)40
B)34
C)42
D)46
E)50
Rationale: Total # of students 16+28+18-(9+7+10)+4=40.
Answer: A.
Answer: A
Question: If 5x + y = 15, 5y + z = 25 and 2z + x = 2, what is the value of x ?
Options:
A)2
B)5
C)6
D)7
E)1
Rationale: Explanation:
5x + y = 15 –-------I * 5
5y + z = 25 –------ II
2z + x = 2 -------- III
Subtract (I) and (II)
25x + 5y + 0 = 75
- 0 + 5y + z = 25
-------------------------------
25x + 0 – z = 50 ….......... IV
Multiply (IV)*2, we get 50x – 2z = 100
Adding (III) and (IV) we get,
50x – 2z = 100
+ x + 2z = 2
------------------------
51 x = 102
x = 2
The value of x = 2
ANSWER: A
Answer: A
Question: In what ratio should a variety of rice costing Rs. 4.5 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg?
Options:
A)5/6
B)5/9
C)5/1
D)5/3
E)5/12
Rationale: Let us say the ratio of the quantities of cheaper and dearer varieties
= x : y
By the rule of allegation, x/y
= (8.75 - 7.50) / (7.50 - 4.5) =5/12
Answer: E
Answer: E
Question: If x is the product of three consecutive positive integers, which of the following must be true?
I. x is an integer multiple of 3.
II. x is an integer multiple of 6
III. x is an integer multiple of 5
Options:
A)a) I only
B)b) II only
C)c) I and II only
D)d) I and III only
E)e) I,II and III
Rationale: The answer should be D i.e. X will be an integer and multiple of 3 and 6.
Let us take example n, n+1, n+2 as 3 three consecutive positive integers.
In a sequence of consecutive integers a number is multiple of 3 after every interval of 2 numbers i.e 3,4,5,6 Or 8,9,10,11,12
Hence in a product of 3 consecutive integers, the product is always divisible by 3.
Now, in a consecutive sequence every alternate is an even number, and when an even number is multiplied by 3 we will have 6 as one of the multiple also.
Now for a number to be a multiple of 4 we need at least 2 2's. this is only possible if the first number of three consecutive positive integers is an even number so that 3 is also even and we have 2 2's. But incase the sequence starts with odd we will have one 2 hence, the divisibility by 4 depends on the first number to be even
Answer C
Answer: C
Question: In his wardrobe, Dexter has three trousers. One of them is black the second is blue, and the third brown. In his wardrobe, he also has four shirts. One of them is black and the other 3 are white. He opens his wardrobe in the dark and picks out one shirt and one trouser pair without examining the colour. What is the likelihood that neither the shirt nor the trouser is black?
Options:
A)1/12
B)1/6
C)1/4
D)1/2
E)None of these
Rationale: Solution:
Probability that trouser is not black =2/3=2/3
Probability that shirt is not black =3/4=3/4
Required probability =2/3Ã—3/4=1/2
Answer is D
Answer: D
Question: The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years
Options:
A)of these
B)B)3:5
C)C)4:3
D)inadequate
E)A)3:4
Rationale: Let P's age and Q's age be 6x and 7x years respectively.
Then, 7x - 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8
ANSWER:A
Answer: A
Question: 15 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would 21 women take to do the same. If 3 women do as much work as 2 men?
Options:
A)20
B)10
C)30
D)40
E)50
Rationale: C
30
3W = 2M
15M ------ 21 * 8 hours
21 W ------ x * 6 hours
14 M ------ x * 6
15 * 21 * 8 = 14 * x * 6
x = 30
Answer: C
Question: If the ratio of the roots of the equation 4x2-px+8=0 is equal to that of the roots 2x2-rx+16=0, then:
Options:
A)4r=p
B)r=p
C)2r=p
D)2p=r
E)None
Rationale: 4x2-px+8=0 for this eqn... roots are R1=p/4 R2=-2
2x2-rx+16=0 for this eqn... roots are R3=r/2 R4=-8
As per Question, p/-8=r/-16
=> 2p=r
ANSWER:D
Answer: D
Question: The present worth of $3600 due in 2 years at 20% per annum compound interest is?
Options:
A)$1200
B)$1850
C)$2010
D)$3200
E)$2500
Rationale: Present worth = 3600/ (1 + 20/100)^2 = 3600 * 5/6 * 5/6 = $2500
Answer is E
Answer: E
Question: The total marks obtained by a student in Physics, Chemistry and Mathematics is 150 more than the marks obtained by him in Physics. What is the average mark obtained by him in Chemistry and Mathematics?
Options:
A)75
B)87
C)68
D)99
E)32
Rationale: Let the marks obtained by the student in Physics, Chemistry and Mathematics be P, C and M respectively.
P + C + M = 150 + P
C + M = 150
Average mark obtained by the student in Chemistry and Mathematics = (C + M)/2 = 150/2
= 75.
Answer:A
Answer: A
Question: During 2003, a company produced an average of 3,000 products per month. How many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 300% over its 2003 average?
Options:
A)520,000
B)540,000
C)525,000
D)530,000
E)550,000
Rationale: Company produced 12*3000 = 36,000 products in 2003. If company produces X products from 2004 to 2007, then total amount of product produced in 4 years (2003 through 2007) is X+36,000. The gives the average of (X+36,000)/4.
This average needs to be 300% higher than that in 2003. In math terms, 36,000+300%(36,000) = 144,000. So:
(X+36,000)/4 = 144,000
X+36,000 = 576,000
X = 540,000
The answer is B.
Answer: B
Question: Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.441 in 2 years?
Options:
A)399
B)400
C)2887
D)277
E)1213
Rationale: 441 = P(21/20)2
P = 400.Answer: B
Answer: B
Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 18. The larger of the two numbers is:
Options:
A)276
B)414
C)322
D)345
E)355
Rationale: Clearly, the numbers are (23 x 13) and (23 x 18).
Larger number = (23 x 18) = 414. Answer: Option B
Answer: B