mathewhe/aqua_rat huggingface dataset:
Question: If a randomly selected positive single digit multiple of 3 is multiplied by a randomly selected prime number less than 20, what is the probability W that this product will be a multiple of 45?
(A) 1/32
(B) 1/28
(C) 1/24
(D) 1/16
(E) 1/14
Rationale: There are 3 single digit multiple of 3, that is, 3,6,9.
There are 8 prime nos less than 20 - 2,3,5,7,11,13,17,19
Total outcome - 8*3 = 24
Favourable outcome = 1 (9*5)
Hence required probability W= 1/24. Answer C.
Answer: (C) 1/24
Question: If the money was put in for the same duration of time by A,B and C, three business partners and four times A’s capital is equal to 6 times B’s capital is equal to 10 times C’s capital. Determine that out of a total profit of Rs 4650 what is C’s share?
(A) Rs 9008
(B) Rs 906
(C) Rs 900
(D) Rs 905
(E) Rs 901
Rationale: Explanation:
It is given that business partnership is independent of time in this case as all the investors have invested their money for same amount of time.
Let A’s capital be x, B’s capital be y and C’s capital be z.
=>x+y+z = 4650
=> 4x = 6y = 10z
=>C’s share is z so let us solve the question w.r.t. z
=>x = 5z/2
=>y = 5z/3
=>(5z/2)+(5z/3)+(z) = 4650
=>31z = 4650*6
=>z = 900
ANSWER:C
Answer: (C) Rs 900
Question: Which of the following is equal to (x-1)!/(x-2)!?
(A) x-1
(B) 1+(1/x)
(C) 1
(D) x+1
(E) x-1
Rationale: (x - 1)! = (x - 2)! * (x-1)
(x - 1)!/(x-2)! = x - 1
Answer: E
Answer: (E) x-1
Question: A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.150 more. What was the Sum?
(A) s.120
(B) s.130
(C) s.140
(D) s.150
(E) s.300
Rationale: At 5% more rate, the increase in S.I for 10 years = Rs.150 (given)
So, at 5% more rate, the increase in SI for 1 year = 150/10 = Rs.15/-
i.e. Rs.15 is 5% of the invested sum
So, 1% of the invested sum = 15/5
Therefore, the invested sum = 15 × 100/5 = Rs.300 Answer: E
Answer: (E) s.300
Question: A cylindrical water tower with radius 14 m and height 20 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π m3. From noon, how many hours will it take to drain the entire tank?
(A) 420 2/7
(B) 420 6/7
(C) 421
(D) 421 3/7
(E) 420
Rationale: initial volume = (3/4)×∏×14²×20 = 2940∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr
Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“
So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain
What i did initially was formed an equation 2940∏ = 7∏×n (n is the number of hrs) so ended up with 420. E
Answer: (E) 420
Question: 50^51^52 /11
(A) 6
(B) 4
(C) 7
(D) 3
(E) 5
Rationale: We know that 6^1=6 or 6^2=36 so for all power of 6 unit digit is 6 now que is 50^51^52 now if we divide 50/11 then rem is 6 means 6^51=6 at unit place same for 6^52 salso give 6 at unit place now finally 50^51^52 gives 6 at unit place now 6/11=6
ANSWER:A
Answer: (A) 6
Question: If there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city, find the number of 4 wheelers parked there if the total number of wheels is 66?
(A) 11
(B) 12
(C) 13
(D) 16
(E) 17
Rationale: four wheeler=16*4=64(max)
2 wheel=1
so no of 4 wheeler=16
ANSWER:D
Answer: (D) 16
Question: There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
(A) 70
(B) 75
(C) 80
(D) 85
(E) 90
Rationale: Total ways of selecting 3 people out of 10 = 10C3 = 10*9*8/3*2 = 120
So final answer cannot be greater than 120.
In 3 people only 1 couple is possible. So way of selecting 1 couple out of 5 = 5C1 = 5
Way of selecting 1 person out of remaining 8 = 8C1 = 8
Total ways when there will be 1 couple and 1 different person = 8*5 = 40
Total number of ways in which no people will be from same couple = 120 - 40 = 80 ANS"C
Answer: (C) 80
Question: A bat is bought for Rs.400 and sold at a gain of 20% find its selling price
(A) Rs.460/-
(B) Rs.470/-
(C) Rs.480/-
(D) Rs.500/-
(E) Rs.520/-
Rationale: 100 % ------> 400 (100 * 4 = 400)
120 % ------> 480 (120 * 4 = 480)
Selling price = Rs.480/-
C
Answer: (C) Rs.480/-
Question: On dividing 181 by a number, the quotient is 9 and the remainder is 1. Find the divisor?
(A) A)12
(B) B)15
(C) C)16
(D) D)17
(E) E)20
Rationale: d = (D-R)/Q
= (181 - 1)/9
= 180/9 = 20
E)
Answer: (E) E)20
Question: Renuka bought an electric drill at 90% of the regular price. She paid $450 for the drill. What was the regular price?
(A) $400
(B) $500
(C) $300
(D) $200
(E) $100
Rationale: Regular price = 450/90*100 = 500
Answer : B
Answer: (B) $500
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train?
(A) 150 meter
(B) 876 meter
(C) 167 meter
(D) 719 meter
(E) 169 meter
Rationale: Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9
= 150 meter
Answer: A
Answer: (A) 150 meter
Question: X can finish a work in 25 days. Y can finish the same work in 20 days. Y worked for 12 days and left the job. how many days does X alone need to finish the remaining work?
(A) 2
(B) 5
(C) 6
(D) 7
(E) 8
Rationale: Work done by X in 1 day = 1/25
Work done by Y in 1 day = 1/20
Work done by Y in 12 days = 12/20 = 3/5
Remaining work = 1 – 3/5 = 2/5
Number of days in which X can finish the remaining work = (2/5) / (1/20) = 8
E
Answer: (E) 8
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?
(A) 3 hrs 45 min
(B) 3 hrs 45 min
(C) 3 hrs 42 min
(D) 3 hrs 65 min
(E) 3 hrs 35 min
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) 3 hrs 45 min
Question: What percent of the different arrangements of the letters of the word AFACUT are those in which the vowels appear together?
(A) 25%
(B) 20%
(C) 30%
(D) 40%
(E) 50%
Rationale: let's determine the number of total possibilities in arranging the letters. There are six spaces, so the total number of arrangements is 6!, or 360.
Next, we need to figure out how to determine the number of ways that we can arrange the 3 vowels together - simply place them together (as in AAU) and call that a single place.
Next, we must determine the number of ways to arrange the now 4 units (i.e., AAU, B, C, S). Like above, there are 4 units and 4 places so the number of arrangements is 4!, or 24.
Finally, we need to account for the number of ways we can arrange AAU. We can either write out each unique iteration (AAU, AUA and UAA) or calculate as 3!/2! and get 3.
Putting this all together, we get the number of ways to arrange the letters so that the vowels are together is 4! x 3 ==> 72
the number of total arrangements of all the letters is 6! ==> 360
72/360 = 1/5, or 20%
Correct answer is B
Answer: (B) 20%
Question: How many bricks, each measuring 100 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
(A) 1600
(B) 2400
(C) 5500
(D) 7400
(E) 3400
Rationale: Number of bricks = Volume Of the wall / Volume of 1 brick = (800 x 600 x 22.5) / (100 x 11.25 x 6) = 1600
ANSWER:A
Answer: (A) 1600
Question: The average (arithmetic mean) of the integers from 100 to 400, inclusive, is how much greater than the average of the integers from 50 to 250, inclusive?
(A)  100
(B)  175
(C)  200
(D)  125
(E)  300
Rationale: For an AP the mean or average of series is average of first and last term.
So, average of numbers between 100 to 400, inclusive = (100+400)/2 = 250
average of numbers between 50 to 250, inclusive = (50+250)/2 = 150
difference = 250 - 150 = 100
Answer is A
Answer: (A)  100
Question: If log10 2 = 0.3010, then log2 10 is equal to:
(A) 1000/301
(B) 1000/500
(C) 966/301
(D) 0.301
(E) 0.966
Rationale: log2 10 =	1/log10 2 = 1/0.3010 = 10000/3010 = 1000/301
ANSWER A
Answer: (A) 1000/301
Question: The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 98, what is the difference between the ages of Patrick and Monica?
(A) 32
(B) 48
(C) 45
(D) 72
(E) 18
Rationale: Ages of P and Mi=3x:5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=49x=98
x=2
Difference if ages of Pa and Mo=25x-9x=16x=16*2=32
Answer A
Answer: (A) 32
Question: At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?
(A) 4%
(B) 74%
(C) 64%
(D) 74%
(E) 94%
Rationale: Let sum = x. Then, S.I. = 2x/5, Time = 10 years.
Rate = (100 * 2x) / (x * 5 * 10) = 4%
Answer: A
Answer: (A) 4%
Question: The perimeter of one square is 48 cm and that of another is 20 cm. Find the perimeter and the diagonal of a square which is equal in area to these two combined?
(A) 12√2
(B) 13√2
(C) 13√3
(D) 13√1
(E) 13√1
Rationale: 4a = 48 4a = 20
a = 12 a = 5
a2 = 144 a2 = 25
Combined area = a2 = 169 => a = 13
d = 13√2
Answer:B
Answer: (B) 13√2
Question: Machine–A produces 40% of the total output and Machine-B produces 60% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 200 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective?
(A) 0.6
(B) 0.06
(C) 0.0066
(D) 0.0006
(E) 6.0e-05
Rationale: Let total Production be 10000 units....
A produces 4000 units and 36 units are defective
B produces 6000 units and 30 units are defective
So, Out of total 10,000 units 66 units are defective...
So the required probability = 66/10000 => 0.0066
Answer will be (C)
Answer: (C) 0.0066
Question: Anand and Deepak started a business investing Rs. 22,500 and Rs. 35,000 respectively. Out of a total profit of Rs. 13,800, Deepak's share is?
(A) 2877
(B) 1299
(C) 8400
(D) 2799
(E) 2698
Rationale: Ratio of their shares = 22500 : 35000 = 9:14
Deepak's share = 13800 * 14/23
= Rs.8400.
Answer: C
Answer: (C) 8400
Question: The area of a circle is four times its circumference. What is the diameter of the Circle?
(A) 4π
(B) 16
(C) 8
(D) 12
(E) 8π
Rationale: pi*r^2 = 4*2*pi*r
r*pi(8 - r) = 0
r = 8
Diameter = 2r = 16
B
Answer: (B) 16
Question: X starts a business with Rs.45000. Y joins in the business after 3 months with Rs.36000. What will be the ratio in which they should share the profit at the end of the year?
(A) 20:9
(B) 2:1
(C) 3:2
(D) 2:3
(E) 5:3
Rationale: Ratio in which they should share the profit
= Ratio of the investments multiplied by the time period
=45000×12:36000×9
=45×12:36×9
=15×12:9×9
=20:9
Answer is A.
Answer: (A) 20:9
Question: Sonika deposited Rs.8000 which amounted to Rs.9200 after 3 years at simple interest. Had the interest been 1% more. She would get how much?
(A) 9440
(B) 96288
(C) 26667
(D) 1662
(E) 2882
Rationale: (8000*3*1)/100 = 240
9200
--------
9440
Answer:A
Answer: (A) 9440
Question: The square root of 64009 is:
(A) 253
(B) 347
(C) 363
(D) 803
(E) 888
Rationale: 2|64009( 253
|4
|----------
45|240
|225
|----------
503| 1509
| 1509
|----------
| X
|----------
64009 = 253.
ANSWER :A
Answer: (A) 253
Question: QUANT 4-PACK SERIES Problem Solving Pack 4 Question 2 For every positive integer n...
For every positive integer n that is greater than 1, the function t(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The t(n) could have any of the following units digits except…?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 9
Rationale: For every positive integer n that is greater than 1, the function t(n) is defined to be the sum of all of the odd integers from 1 to n, inclusive. The t(n) could have any of the following units digits except…?
Used brute force on this one to calculate. Not sure if there is a quicker way.
1+3=4 C is out
1+3+....15= 81 A is out
1+3+5+7=16 D is out
1+3+5=9 E i out
Answer: B
Answer: (B) 2
Question: 10% of a number is added to 45, the result is the same number. Find the number?
(A) 50
(B) 288
(C) 270
(D) 129
(E) 281
Rationale: (10/100) * X + 45 = X
9X = 450
X =50
Answer:A
Answer: (A) 50
Question: Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word BOOK could be arranged to create a four-letter code?
(A) 4!
(B) 4! − ( 2!)
(C) 4! − (3! × 2!)
(D) 4!/( 2!)
(E) 4!/(3! × 2!)
Rationale: BOOK - four letters can be arranged in 4! ways
since 'O' repeats 2 times we need to divide the 4! ways by 2! to adjust repetition.
4!/(2!)
Ans. D) 4!/( 2!)
Answer: (D) 4!/( 2!)
Question: A certain sum is invested at simple interest at 15% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 720. Find the sum?
(A) s. 7000
(B) s. 9000
(C) s. 14000
(D) s. 12000
(E) s. 27000
Rationale: Let the sum be Rs. x.
(x * 15 * 2)/100 - (x * 12 * 2)/100 = 720
=> 30x/100 - 24x/100 =720
=> 6x/100 = 720
=> x = 12000.
ANSWER:D
Answer: (D) s. 12000
Question: In a group of 4 boys and 6 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be selected?
(A) 135
(B) 155
(C) 175
(D) 195
(E) 215
Rationale: Number of ways to select four children is 10C4 = 210
Number of ways to choose only girls is 6C4 = 15
Number of ways that at least one boy is selected is 210 - 15 = 195
The answer is D.
Answer: (D) 195
Question: A committee is reviewing a total of 20x black-and-white films and 8y color films for a festival. If the committee selects y/x% of the black-and-white films and all of the color films, what fraction of the selected films are in color?
(A) 40/41
(B) 20/41
(C) 30/41
(D) 60/41
(E) 80/41
Rationale: Say x=y=10. In this case we would have:
20x=200 black-and-white films;
8y=80 color films.
y/x%=10/10%=1% of the black-and-white films, so 2 black-and-white films and all 80 color films, thus total of 82 films were selected.
Color films thus compose 80/82=40/41 of the selected films.
Answer: A
Answer: (A) 40/41
Question: Complete the numerical series by replacing the question mark with the correct numbers.
5+3+2 = 151022
9+2+4 = 183652
8+6+3 = 482466
5+4+5 = 202541
9+5+5 = ??????
(A) 454585
(B) 454583
(C) 456685
(D) 454588
(E) 456585
Rationale: A
454585
Explanation:
Number is formed as (numA*numB)(numB*numC)(numC*numA)
Following the formular, 1st row will be like
5+3+2 => 5*3 3*2 2*5 => 151022
Simillarly
(9*5)(9*5)(9*5+9*5-5) => 454585
Answer: (A) 454585
Question: The current of a stream runs at the rate of 4 kmph. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of the boat in still water?
(A) 3
(B) 7
(C) 8
(D) 5
(E) 1
Rationale: S = 4
M = x
DS = x + 4
US = x - 4
6/(x + 4) + 6/(x - 4) = 2
x = 8
Answer:C
Answer: (C) 8
Question: The following list follows a peculiar pattern. Can you find it out and guess the next number?
2, 12, 36, 80, 150, 252, ?
(A) 383
(B) 321
(C) 392
(D) 651
(E) 823
Rationale: C
The next number is 392.
The first number = (Square of 1 + Cube of 1)
The second number = (Square of 2 + Cube of 2)
The third number = (Square of 3 + Cube of 3)
The fourth number = (Square of 4 + Cube of 4)
The fifth number = (Square of 5 + Cube of 5)
The sixth number = (Square of 6 + Cube of 6)
Therefore, the seventh number will be
Square of 7 + Cube of 7 = 392.
Answer: (C) 392
Question: If log1087.5 = 3.9421, then the number of digits in (875)10 is?
(A) 30
(B) 28
(C) 27
(D) 50
(E) 25
Rationale: X = (875)10 = (87.5 x 10)10
Therefore, log10X = 10(log1087.5 + 1)
= 10(3.9421 + 1)
= 10(4.9421) = 49.421
X = antilog(49.421)
Therefore, number of digits in X = 50.
ANSWER:D
Answer: (D) 50
Question: If a and b are positive integers, and a=5b+20, the greatest common divisor of a and b CANNOT be
(A) 2
(B) 4
(C) 5
(D) 6
(E) 10
Rationale: If b is 2, 4, 5, or 10, then GCD of a and b is 2, 4, 5, and 10 respectively. So, by POE the answer must be D.
Still: if b is a multiple of 6, then a is 20 greater than a multiple of 6, so not a multiple of 6, so both of them cannot be divisive by 6.
Answer: D.
Answer: (D) 6
Question: The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?
(A) 66 m
(B) 17 m
(C) 87 m
(D) 65 m
(E) 43 m
Rationale: Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b2 = 867
b2 = 289 = 172 (b > 0)
b = 17 m.
Answer:B
Answer: (B) 17 m
Question: Median C of price of the car is 52,000. which could be the median of the other two.
35000, 44000, x, y, 57000.
(A) 38000
(B) 47000
(C) 48000
(D) 51,000
(E) 56,000
Rationale: Median of the car is 52,000 therefore 52,000 is the middle number. Assuming Y >= X, X will have to be the median value i.e. 52,000. Therefore the remaining value would have to be between 52,000 and 57,000 inclusive. The only answer choice that fits is C 56,000.E
Answer: (E) 56,000
Question: A car travelling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
(A) 17 6/7 km/hr
(B) 25 km/hr
(C) 30 km/hr
(D) 35 km/hr
(E) 45 km/hr
Rationale: Time taken = 1 hr 40 min 48 sec = 1 hr 40 4/5 min. = 126/75 hrs.
Let the actual speed be x km/hr.
Then, (5/7)x X 126/75 = 42
--> x = 35 km/hr.
Answer : D.
Answer: (D) 35 km/hr
Question: In how many different number of ways 5 boys and 4 girls can sit on a bench?
(A) 700
(B) 710
(C) 720
(D) 362880
(E) 740
Rationale: npn = n!
9p9 = 9 X 8 X 7 X 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 = 362880
D
Answer: (D) 362880
Question: Three men start together to travel the same way around a circular track of 11 kilometers in circumference. Their speeds are 4, 5 and 8 kilometers per hour respectively. When will they meet at a starting point?
(A) 11 hours
(B) 12 hours
(C) 220 hours
(D) 22 hours
(E) 24 hours
Rationale: Time for one round = 11/4, 11/5, 11/8
LCM of 11/4, 11/5, 11/8 = 11/1 = 11
ANSWER:A
Answer: (A) 11 hours
Question: The batting average of a particular batsman is 58 runs in 46 innings. If the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs, find his highest score.
(A) 179
(B) 208
(C) 210
(D) 133
(E) 229
Rationale: Explanation:
Total runs scored by the batsman = 58*46 = 2668 runs
Now excluding the two innings the runs scored = 58*44 = 2552 runs
Hence the runs scored in the two innings = 2668 – 2552 = 116 runs.
Let the highest score be x, hence the lowest score = x – 150
x + (x - 150) = 116
2x = 266
x = 133runs
ANSWER D
Answer: (D) 133
Question: In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is 1/10. For a group of 600 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 438
(B) 446
(C) 452
(D) 462
(E) 470
Rationale: Number of newborns that can die in first month = 1/10 *600 = 60
Survived = 540
Number of newborns that can die in second month = 1/10 * 540 = 54
Survived = 486
Number of newborns that can die in third month = 1/10 * 486 = 48
Survived = 438
Answer:A
Answer: (A) 438
Question: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has five iron coins and five copper coins, how many different sums from 1¢ to 35¢ can he make with a combination of his coins?
(A) 20
(B) 21
(C) 35
(D) 33
(E) 31
Rationale: The total sum is 5*2 + 5*5 = 35¢. If you can make each sum from 1 to 35 (1¢, 2¢, 3¢, ..., 35¢), then the answer would be 35 (maximum possible).
Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 32¢ and 34¢ (since total sum is 35¢ we cannot remove 1¢ or 3¢ to get 34¢ or 32¢).
So, out of 35 sums 4 are for sure not possible, so the answer must be 35 - 4 = 31 sums or less. Only A fits.
Answer: E.
Answer: (E) 31
Question: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical Q present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical Q required to keep the reaction rate unchanged?
(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase
Rationale: Let Q and B be the concentratrations of the individual chemicals.
Rate of chemical reaction = Q^2/B
Since new B is increased by 100%
Rate = Q^2/(2B)
For the rate to be constant, new value of Q should be square root(2)
So new value of Q = 1.414Q
That means Q is increased by 41.4%
Nearest solution is D
Answer: (D) 40% increase
Question: The true discount on a bill of Rs. 2460 is Rs. 360. What is the banker's discount?
(A) Rs. 432
(B) Rs. 422
(C) Rs. 412
(D) Rs. 442
(E) None of these
Rationale: Explanation :
F = Rs. 2460
TD = Rs. 360
PW = F - TD = 2460 - 360 = Rs. 2100
True Discount is the Simple Interest on the present value for unexpired time
=>Simple Interest on Rs. 2100 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2460 for unexpired time
=360/2100 × 2460 = 0.17 × 2460 = Rs. 422 Answer : Option B
Answer: (B) Rs. 422
Question: According to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . How many 6 - ounce cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?
(A) 25
(B) 34
(C) 50
(D) 67
(E) 100
Rationale: Orange juice concentrate: water::1:3
total quantity of orange juice = 200*6 = 1200 oz
so Orange juice concentrate: water :: 300 oz:900 oz
no. of 6 oz can = 300 oz/6 oz = 50
Answer C, 50 cans
Answer: (C) 50
Question: A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 40 years now, the son's age five years back was
(A) 14years
(B) 15years
(C) 17years
(D) 18years
(E) 19years
Rationale: Let the son's present age be x years. Then, (40 - x) = x
2x = 40.
x = 20.
Son's age 5 years back (20 - 5) = 15 years.
B)
Answer: (B) 15years
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
(A) 75
(B) 161
(C) 224
(D) 180
(E) 150
Rationale: Speed=(60 x	5/18) m/sec	=(50/3) m/sec.
Length of the train = (Speed x Time).
Length of the train =(50/3x 9) m = 150 m.
Answer is E
Answer: (E) 150
Question: A train is 360 meter long is running at a speed of 72 km/hour. In what time will it pass a bridge of 140 meter length?
(A) 65 seconds
(B) 46 seconds
(C) 25 seconds
(D) 97 seconds
(E) 26 seconds
Rationale: Speed = 72 Km/hr = 72*(5/18) m/sec = 20 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (1/20)
= 25 seconds
Answer:C
Answer: (C) 25 seconds
Question: It costs $3 for the first 1/3 hour to use the laundry machine at the Laundromat. After the first ¼ hour it costs $12 per hour. If a certain customer uses the laundry machine for 2 hours and 35 minutes, how much will it cost him?
(A) $30.
(B) $31
(C) $32.
(D) $33.
(E) $34.
Rationale: 2 hrs 35 min = 155 min
First 20 min ------> $3
Time left is 135 min...
Now, 60 min costs $ 12
1 min costs $ 12/60
155 min costs $ 12/60*155 => $ 31
So, Total cost will be $31 + $3 => $ 34
Hence answer will be C
Answer: (C) $32.
Question: If the true discount on a certain sum due 6 months hence at 15% is Rs.120. What is the banker's discount on the same sum for the same time and at the same rate?
(A) Rs. 119
(B) Rs. 129
(C) Rs. 131
(D) Rs. 139
(E) None
Rationale: Solution
Sum Given, T.D =60, B.D = 72
= (B.D x T.D / B. D - T.D)
= Rs.(72x60/72-60)
= Rs.(72 x 60 / 12)
= Rs.(4320/ 12)
= Rs. 360.
Answer A
Answer: (A) Rs. 119
Question: The sum of two consecutive even integers is 26. What are the two numbers?
(A) 12-14
(B) 13-13
(C) 10-16
(D) 9-17
(E) 8-18
Rationale: Let 2n be the first even integer and let 2n + 2 be the second integer
2n + 2n + 2 = 26
4n + 2 = 26
4n + 2 - 2 = 26 - 2
4n = 24
n = 6
So the first even integer is 2n = 2 × 6 = 12 and the second is 12 + 2 = 14
correct answer is A)12-14
Answer: (A) 12-14