divelab/Sys2Bench huggingface dataset:
Question: A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction Y of time taken by the two wide pipes working together to fill the tank?
Options:
A) 1/2
B) 2/3
C) 3/4
D) 3/7
E) 4/7
Rationale: I always plug in for these questions.. I find them easier
Let 2 large pipes each work fill 10lts/hr
small ones = 1/2*10 = 5
total 5 = 10+10+5+5+5 = 35 lts/hrs
now assume the total capacity = 140 lts ( LCM of 20,35)
Qn :All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank
all working together will take 140/35 = 4 hrs
two large pipes working will take 140/20 = 7 hrs
hence ratio Y = 4/7 = E
Answer: E
Question: A football player just signed a new contract in which he will earn a 10% raise in base pay from the previous year. Last year, he did not earn a performance bonus; but this year, he will have the chance to earn a performance bonus of up to $20,000. If his new salary is effectively an 11% raise after earning the top performance bonus, what is his new salary?
Options:
A)2,120,000
B)2,170,000
C)2,210,000
D)2,220,000
E)2,360,000
Rationale: let x = the old salary
(1+.10)*x + 20,000 = (1+.11)*x
1.1x + 20,000 = 1.11x
20,000 = .01x
2,000,000 = x
the new salary is thus (1+.11)*2,000,000 = 1.11*2,000,000 = 2,220,000
D
Answer: D
Question: On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 50 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets?
Options:
A)30%
B)40%
C)50%
D)60%
E)65%
Rationale: Let T be the total number of passengers.
Let x be the number of people with round trip tickets.
0.2T had round trip tickets and took their cars.
0.5x had round trip tickets and took their cars.
0.5x = 0.2T
x = 0.4T
The answer is B.
Answer: B
Question: In how many ways can the letters of wordSOLUTIONbe arranged such that all the Consonants always appear together?
Options:
A)9!
B)5!*4!
C)5!*5!
D)5!*4!*2!
E)6!*4!
Rationale: Let the arrangment be: _ _ _ _ _ _ _ _ _ _ (8 letters, 4 vowels O, U, I, O and 4 consonant S, L, T, N)
For all consonants together, consider them as 1 group {S, L, T, N} O U I O --> this gives us 5! arrangements with {} group considered as 1 . Now {} group can in itself be arranged in 4! ways
Thus total arrangements with all consonants together = 5!*4! --> B is the correct answer.
Answer: B
Question: Out of first 50 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is -.
Options:
A)1/2
B)16/19
C)4/5
D)61/50
E)3/5
Rationale: n(S) = 50
n(Even no) = 25 = n(E)
n(Prime no) = 15 = n(P)
P(EᴜP) = 25/50 + 15/50 - 1/50 = 61/50
ANSWER:D
Answer: D
Question: A cistern 12 m long and 14 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
Options:
A)224 m2
B)250 m2
C)220 m2
D)230 m2
E)233 m2
Rationale: Area of the wet surface	= [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (14 x 1.25 + 12 x 1.25) + 12 x 14] m2
= 233 m2.
answer :E
Answer: E
Question: When 15 is divided by integer u, the remainder is 4. For how many values of u is this be true?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: When 15 is divided by u, the remainder is 4 i.e. 4 mangoes left over after grouping, so u must be greater than 4.
It also means that 11 is completely divisible by u.
Factors of 11 are 1 and 11.
Out of these, u could be 11.
Answer (A)
Answer: A
Question: in a rectangle 8 vertical and 4 horizontal lines are drawn find the number of rectangle formed ?
Options:
A)654
B)664
C)674
D)684
E)694
Rationale: 8 vertical lines plus the 2 lines of the existing rectangle(as its mentioned in the question "in a rectangle") accordingly 4 plus 2 horizontal lines.
therefore
10C2*6C2= 675 and if its asked for the new rectangle formed than
675-1=674 (excluding the existing rectangle)
ANSWER:C
Answer: C
Question: Two bullet trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each bullet train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?
Options:
A)13 sec.
B)15 sec.
C)12 sec.
D)17 sec.
E)19 sec.
Rationale: C
12 sec.
Speed of the first bullet train = 120/10 m/sec = 12 m/sec.
Speed of the second bullet train = 120/15 m/sec = 8 m/sec.
Relative speed = (12 + 8) = 20 m/sec.
Required time = (120 + 120)/20 sec = 12 sec.
Answer: C
Question: Which of the following is divisible by the first thousand positive even integer multiples of 101?
(i) 55650000
(ii) 234177623000
(iii)11111111111110000000000000000000000000
Options:
A)None
B)i only
C)ii only
D)iii only
E)i, ii and iii
Rationale: The digits of (iii) add to 13, so (iii) is not divisible by 3, so is not divisible by 3*101, and hence not divisible by the first thousand positive even integer multiples.
A positive integer x is divisible by the first thousand integer multiples of 101 if and only if it is divisible by their least common multiple. There are at least 12 primes greater than 10 but less than 1000 so to be divisible by the first thousand integer multiples of 101 x must be at least 10^12. Consquently, neither (i) not (ii) is divisible by the first thousand integer multiples of 101.
Answer A
Answer: A
Question: A can do a piece of work in 12 days and B can do a piece of work in 18 days how many days will the work be completed if both of them work together?
Options:
A)15
B)7.2
C)7.7
D)7.4
E)8
Rationale: A's 1 day work=1/12,B's 1 day work =1/18;
(A+B) 1 day work=1/12+1/18=5/36;
they can finish the work in=36/5=7.2
ANSWER:B
Answer: B
Question: Two trains of equal are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
Options:
A)50
B)99
C)88
D)71
E)65
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 - 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.
Answer:A
Answer: A
Question: Rs. 6000 is lent out in two parts. One part is lent at 7% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?
Options:
A)5:1
B)5:5
C)5:8
D)5:4
E)5:2
Rationale: Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 - x) = 600 - 3x/100
=> 600 - 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5:1
Answer:A
Answer: A
Question: If BELT is coded as 27 and JUMPED is coded as 54, then CHAIRS is coded as
Options:
A)10
B)21
C)40
D)92
E)None
Rationale: BELT= 2+5+12+20=39 i.e 3*9=27
JUMPED= 10+21+13+16+5+4=69 i.e 6*9=54
CHAIRS= 3+8+1+9+18+19 =58 5*8=40
ANSWER:C
Answer: C
Question: A girl has 7 jeans and 12 shirts. In how many different ways can she select a jeans and a shirts?
Options:
A)A)100
B)B)102
C)C)108
D)D)110
E)E)84
Rationale: The girl can select one jeans in 7 ways.
The girl can select one shirt in 12 ways.
The number of ways in which he can select one jeans and one shirt is 7 * 12 = 84 ways.
E
Answer: E
Question: The width of a rectangular hall is ½ of its length. If the area of the hall is 800 sq.m, what is the difference between its length and breadth?
Options:
A)8m
B)10m
C)12m
D)20m
E)17m
Rationale: Let the length of the hall be x m
Breadth of the hall = 1x/2 m
Area of the hall = Length * Breadth
800 = x * 1x/2
x² = 1600
x =40
Difference between the length and breadth of the hall = x - 1x/2 = x/2
=40/2 = 20m
ANSWER:D
Answer: D
Question: 60% of a number is added to 150, the result is the same number. Find the number?
Options:
A)375
B)277
C)266
D)99
E)121
Rationale: :
(60/100) * X + 150 = X
2X = 750
X = 375
Answer: A
Answer: A
Question: Tough and Tricky questions: Distance/Rate.
On a reconnaissance mission, a state-of-the-art nuclear powered submarine traveled 600 miles to reposition itself in the proximity of an aircraft carrier. This journey would have taken 2 hour less if the submarine had traveled 10 miles per hour faster. What was the average speed, in miles per hour, for the actual journey?
Options:
A)20
B)40
C)60
D)50
E)100
Rationale: Say, if speed is 50, 600/50 = 12 hrs and 600/60 = 10 hrs (a reduction of 2 hr -> Correct answer)
Answer (D)
Answer: D
Question: After 57 liters of petrol were poured into the tank, it was still 5% empty. How much petrol must be poured into the tank in order to fill it?
Options:
A)7 liters
B)4 liters
C)3 liters
D)2 liters
E)6 liters
Rationale: 95% ----------- 57
5% ------------ ?
=> 3 liters
Answer: C
Answer: C
Question: A sum of money becomes Rs. 2200 after three years and Rs. 4400 after six years on compound interest. The sum is
Options:
A)Rs. 1400
B)Rs. 1100
C)Rs. 1000
D)Rs. 1200
E)Rs. 1400
Rationale: Explanation:
Let the sum be P and rate of interest be R% per annum.
Amount after 3 years = 2200
P(1 + R/100)T = 2200
P(1 + R/100)3 = 2200 --- ( 1)
Amount after 6 years = 4400
P(1+R/100)T = 4400
P(1 + R/100)6 = 4400 --- (2)
(2) ÷ (1) => [P(1+R/100)6]/[P(1+R/100)3] = 4400/2200 = 2
=> (1+R/100)3 = 2 (Substituting this value in equation 1)
=> P × 2 = 2200
P = 22002 = 1100
Answer: Option B
Answer: B
Question: The greater of two numbers is twice the less, and the sum of the numbers is 96. What are the numbers?
Options:
A)64
B)68
C)70
D)79
E)80
Rationale: Let x = the less number
2x = the greater number
The sum of the numbers is 96
x+2x=96
3x=96
x=32
2x=64
Option A
Answer: A
Question: A man can row upstream at 15kmph and downstream at 17kmph. Find man's rate in still water ?
Options:
A)5km/hr
B)15km/hr
C)10km/hr
D)20km/hr
E)16km/hr
Rationale: Rate in still water = 1/2 (15+17) = 16km/hr
Answer is E
Answer: E
Question: The value of (2.75)3 - (2.00)3 - (0.75)3 is
Options:
A)12.375
B)6
C)9
D)8
E)2
Rationale: Explanation:
(2.75)3 - (2.00)3 - (0.75)3
a^3 + b^3 + c^3 = (a + b + c)(a^3 + b^3 + c^3 - ab - bc - ca)
If a + b + c = 0 then a^3 + b^3 + c^3 = 3abc
Hence 2.75 - 2.00 - 0.75 = 0
So, (2.75)3 - (2.00)3 - (0.75)3
= 3 * 2.75 * 2 * 0.75 = 3 * 1.5 * 2.75
= 4.5 * 2.75=12.375
A
Answer: A
Question: If a copier makes 10 copies every 5 seconds, then continues at this rate, how many minutes will it take to make 10,000 copies?
Options:
A)82.33
B)81.33
C)84.33
D)83.33
E)85.33
Rationale: 10 copies 5 seconds
10000 copies = (10000*5)/10 seconds
= 5000/60 minutes
= 83.33 minutes
Answer : D
Answer: D
Question: A cow leaps 8 leaps for every 3 leaps of a bull, but 6 leaps of the bull are equal to 4 leaps of the cow. What is the ratio of the speed of the cow to that of the bull?
Options:
A)4:1
B)1:1
C)3:2
D)4:5
E)7:8
Rationale: Given;
6 bull = 4 cow;
Or, bull/cow = 4/6;
Let cow's 1 leap = 6 meter and bulls 1 leap = 4 meter.
Then, ratio of speed of cow and bull = 6*8/4*3 = 4:1.''Answer: 4:1
Answer: A
Question: A certain automobile company’s best-selling model is the Speedster. The Speedster, like all of their other models, comes in coupe and convertible styles. 2/3 of the current inventory is Speedsters, of which 4/5 are convertibles. If there are 40 vehicles that are not Speedsters, how many Speedster convertibles are there?
Options:
A)30
B)64
C)60
D)70
E)90
Rationale: Total Vehicle = 2/3 of speedster + 1/3 of others.
Speedster convertibles = 2/3 total vehicle * 4 /5
Given : 1/3 constitutes 40 Vehicles. hence 2/3 constitutes 80
Speedster convertibls = 80 * 4/5 = 64
B
Answer: B
Question: A jar contains 8 marbles consisting of an equal number of red, green, and blue marbles. Four marbles are removed from the jar and discarded. What is the probability that only two colors will remain in the jar after the four marbles have been removed?
Options:
A)1/495
B)1/165
C)3/70
D)1/35
E)11/20
Rationale: Jar contains {4-red, 4-green, 4-blue} marbles. We are asked to find the probability that after removing 4 marbles only 2 colours remain in the jar, i.e., the probability of removing EITHER red, OR green, OR blue.
Sample space = 8C4 = 495.
Probability of choosing 4 red = 4C4/8C4 = 1/70.
Probability of choosing 4 green = 4C4/8C4 = 1/70.
Probability of choosing 4 blue = 4C4/8C4 = 1/70.
Required probability = 1+1+1/70 = 3/70 Ans (C).
Answer: C
Question: Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much E does she have in her account after two years?
Options:
A)$10079.44
B)E=$10815.83
C)$12652.61
D)$14232.14
E)$20598.11
Rationale: PS. I guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8%
Hence B the answer
Answer: B
Question: Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. In how much time will the trains cross each other?
Options:
A)89/7 sec
B)80/7 sec
C)80/8 sec
D)89/7 sec
E)80/7 sec
Rationale: Relative speed = (54 + 72)* 5/18 = 7 * 5 = 35 mps.
The time required = d/s = (100 + 100 + 200)/35
= 400/35 = 80/7 sec.
Answer:B
Answer: B
Question: If |7x+4|=|3x+5|, which of the following could be a value of x?
Options:
A)1
B)9/10
C)1/4
D)-2
E)10
Rationale: 7x + 4 = 3x + 5 or 7x + 4 = -3x - 5
4x = 1 or 10x = -9
x = 1/4 or x = -9/10
Answer: C
Answer: C
Question: The calendar for 1992 is the same as for
Options:
A)1997
B)1996
C)1995
D)1994
E)1992
Rationale: consider 1992 has 0 odd days
therefore sum of years which is equal to 0 odd days is
1992 + 1993 + 1994 + 1995 + 1996
( 2 + 1 + 1 + 1 + 2 ) = 7 or 0 odd days
therefore 1997 is the answer and its calender is similar to 1992
ANSWER:A
Answer: A
Question: Sandy is younger than Molly by 12 years. If their ages are in the respective ratio of 7:9, how old is Molly?
Options:
A)36
B)45
C)54
D)63
E)72
Rationale: S = M - 12
S / M = 7/9
9S = 7M
9(M-12) = 7M
M = 54
The answer is C.
Answer: C
Question: A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m?
Options:
A)1350
B)2999
C)2772
D)2672
E)2871
Rationale: Area = (l + b +2d) 2d
= (75 + 55 +2.5 * 2) 2*2.5 => 675
675 * 2 = Rs.1350
Answer:A
Answer: A
Question: How many of the following numbers are divisible by 132?
264, 396, 462, 792, 968, 2178, 5184, 6336
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: A number is divisible by 132, if it is divisible by each one of 11, 3 and 4.
Clearly, 968 is not divisible by 3. None of 462 and 2178 is divisible by 4.
Also, 5184 is not divisible by 11.
Each one of remaining 4 is divisible by each one of 11, 3 and 4 and therefore, by 132.
ANSWER:A
Answer: A
Question: The average price of 10 oranges is Rs. 12 while the average price of 8 of these oranges is Rs. 11.75. Of the remaining two oranges, if the price of one orange is 75% more than the price of the other, what is the price of each of these two oranges?
Options:
A)Rs 5, Rs 7.50
B)Rs 8, Rs 12
C)Rs 9.5, Rs 16.5
D)Rs 12, Rs 14
E)None of these
Rationale: Explanation :
Total price of the two oranges = Rs. [(12 x 10) (11.75 x 8)]
= Rs. (120 - 94) = Rs. 26.
Let the price of one orange be Rs. x.
Then, the price of other orange = Rs. (x + 75% of x) = Rs. x + 3/4x = Rs. 7x/4
So, x + 7x/4 = 26
11x = 104
x = 9.5
The prices of the two oranges are Rs. 9.5 and Rs. 16.5
Answer : C
Answer: C
Question: In how many K ways can a four-letter password be chosen, using the letters A, B, C, D, E, and/or F, such that at least one letter is repeated within the password?
Options:
A)720
B)864
C)900
D)936
E)1,296
Rationale: Total number of four letter passwords = 6*6*6*6= 1296 ------(1)
Total number of passwords in which no letter repeats = 6C4*4!= 15*24=360------(2)
therefore required value K= (1)-(2)= 1296-360=936.D
Answer: D
Question: If x is real, find the maximum value of the expression -2x^2 + 7x + 9.
Options:
A)6.125
B)9.125
C)12.125
D)15.125
E)18.125
Rationale: This is an equation of a downward facing parabola.
The maximum value is the top point of the parabola.
-2x^2+7x+9 = (-2x+9)(x+1)
The roots are 9/2 and -1.
The maximum value must be when x is halfway between these two points.
x = 1.75
The maximum value is -2(1.75)^2+7(1.75)+9 = 15.125
The answer is D.
Answer: D
Question: John went to the Hospital at the speed of 60 km per hour while returning for his home he covered the half of the distance at the speed of 10 per hour, but suddenly he realized that he was getting late so he increased the speed and reached the home by covering rest 1/2 of the distance at the speed of 30km per hour. The average speed of the John in the whole length of journey is?
Options:
A)19 km/hr
B)21 km/hr
C)24 km/hr
D)28 km/hr
E)34 km/hr
Rationale: Let distance between his home and hospital =
d
d km
Total distance travelled
=
d
+
d
=
2
d
=d+d=2d km
Total time taken
=
d
60
+
(
d
2
)
10
+
(
d
2
)
30
=d60+(d2)10+(d2)30
=
d
60
+
d
20
+
d
60
=
5
d
60
=
d
12
=d60+d20+d60=5d60=d12 hr
Average speed in the whole journey
=
2
d
(
d
12
)
=
2
d
×
12
d
=
24
=2d(d12)=2d×12d=24 km/hr
C
Answer: C
Question: Sakshi can do a piece of work in 20 days.Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work :
Options:
A)15
B)16
C)18
D)25
E)29
Rationale: Solution
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4 .
Suppose Tanya taken x days to do the work.
5 : 4 : : 20 : x ⇒ x = (20x4/5)=x 16 days.
Hence,Tanya takes 16 days is complete the work
Answer B
Answer: B
Question: David works at a resort from the beginning of March to the end of September. During the month of August this past year, he made 2 times the average (arithmetic mean) of his monthly totals in tips for the other months. His total tips for August were what fraction of his total tips for all of the months he worked?
Options:
A)1/6
B)1/7
C)1/8
D)1/4
E)1/3
Rationale: First notice the number of months for which he worked - March to Sept i.e. 7 months
Avg of monthly totals in tips for months other than August = x
Tips in August = 2x
Total tips for all months = 6*x + 2x = 8x
Tips for August/Total tips for all months = 2x/8x = 1/4
Answer : D
Answer: D
Question: P, Q, R enter into a partnership. P initially invests 25 lakh and adds another 10 lakh after one year. Q initially invests 35 lakh and withdraws 10 lakh after 2 years. R's investment is Rs 30 lakh. In what ratio should the profit be divided at the end of 3 years?
Options:
A)18:19:19
B)18:18:19
C)19:19:18
D)18:19:18
E)none
Rationale: Explanation:
P : Q : R
25×1+35×2):(35×2+25×1) :(30×3)
=95:95:90
=19:19:18
Answer: Option C
Answer: C
Question: A Man can row upstream at 98kmph and downward stream at 10 kmph. Find Man's rate still in water
Options:
A)6.8km/hr
B)54km/hr
C)90km/hr
D)3.8km/hr
E)7.9km/hr
Rationale: still in water=1/2(98+10)km/hr
=54km/hr
correct option is :B
Answer: B
Question: 7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
Options:
A)150
B)300
C)275
D)135
E)451
Rationale: 1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300
Ans: B
Answer: B
Question: Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:
Options:
A)9
B)9.6
C)10
D)10.2
E)10.8
Rationale: Relative speed of the trains=60+40=100km/hr=100*5/18 m/s=250/9
Time taken to cross each other=total distance/speed
300/250/9=54/5=10.8sec
ANSWER:E
Answer: E
Question: A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 390 m long?
Options:
A)40 sec
B)60 sec
C)45 sec
D)48 sec
E)50 sec
Rationale: Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 750 m
Required time = 750 * 2/25 = 40 sec
ANSWER:B
Answer: B
Question: 20.20/2000 is equal to:
Options:
A)1.01252677
B)0.013466667
C)0.125266677
D)0.126256677
E)0.12725776
Rationale: 20.20/2000 = 2525/200000
= 0.013466667
ANSWER:B
Answer: B
Question: The average of 1st 3 of 4 numbers is 6 and of the last 3 are 5. If the sum of the first and the last number is 13. What is the last numbers?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: A + B + C = 18
B + C + D = 15
A + D = 13
A – D = 3
A + D = 13
2D = 10
D = 5
ANSWER:B
Answer: B
Question: A banker paid Rs.5767.20 for a bill of Rs.5840, drawn of Apr 4 at 6 months. If the rate of interest was 7%, what was the day on which the bill was discounted?
Options:
A)3rd March
B)3rd September
C)3rd October
D)3rd August
E)None of these
Rationale: Explanation :
F = Rs.5840
R = 7%
BD = 5840 - 5767.20 = Rs.72.8
BD = FTR/100
⇒ 72.8 = (5840 × T × 7)/100
⇒ T = (72.8×100)/(7×5840) = (10.4×100)/5840 = 1040/5840 = 104/584 = 13/73 years
= (13 × 365)/73 Days = 65 Days
=> Unexpired Time = 65 Days
Given that Date of Draw of the bill = 4th April at 6 months
=> Nominally Due Date = 4th October
=> Legally Due Date = (4th October + 3 days) = 7th October
Hence, The date on which the bill was discounted
= (7th October - 65 days)
= (7th October - 7 days in October - 30 days in September - 28 days in August)
= 3rd August Answer : Option D
Answer: D
Question: Tea worth Rs.126 per kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs.163 per kg, the price of the third variety per kg will be
Options:
A)Rs.169.50
B)Rs.1700
C)Rs. 185.50
D)Rs. 180
E)None
Rationale: Solution
Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135/2) = Rs.130.50
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
x-163/22.50 = 1 =â€º x - 163 = 22.50 =â€º x=185.50.
Hence, price of the third variety = Rs.175.50 per kg.
Answer C
Answer: C
Question: FIND THE VALUE OF X ?
(2/7)^-8*(7/2)^-2=(2/7)^X
Options:
A)-5
B)-6
C)4
D)0
E)-4
Rationale: (2/7)^-8*(7/2)^-2=(2/7)^x
Now, (7/2)^8*(2/7)^2=(2/7)^x
on solving,
(7/2)^6=(2/7)^x
(2/7)^-6=(2/7)^x
Thus, x=-6.
ANSWER:B
Answer: B
Question: Which of the following numbers does NOT have a divisor greater than 1 that is the square of an integer?
Options:
A)82
B)27
C)64
D)48
E)50
Rationale: Prime Factorization
82=41*2 We got our candidate. No perfect squares here.
27= 3*3*3= 3^2*3. Includes a perfect square of 9.
64=8^2. Is a perfect square.
48=2*2*2*2*3=4^2*3. Includes a perfect square of 16.
50= 5*5*2=5^2*2. Includes a perfect square of 25.
All but 82 have perfect squares in them as a factor.
82=41*2 No perfect squares here.
27= 3*3*3= 3^2*3. Perfect square of 9.
64=8^2. Perfect square.
48=2*2*2*2*3=4^2*3. Perfect square of 16.
50= 5*5*2=5^2*2. Perfect square of 25.
Answer: A
Question: A committee that includes 9 members is about to be divided into 3 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is Annie also a member?
Options:
A)10%
B)20%
C)25%
D)40%
E)50%
Rationale: Michael is a member of one of the subcommittees.
P(Annie is on the same subcommittee) = 2/8 = 25%.
The answer is C.
Answer: C