mathewhe/aqua_rat huggingface dataset:
Question: When tossed, a certain coin has an equal probability of landing on either side. If the coin is tossed 5 times, what is the probability that it will land on the same side each time?
(A) 1/6
(B) 1/8
(C) 1/12
(D) 1/16
(E) 1/32
Rationale: On the first toss, the coin will land on one side or the other.
On the second toss, the probability of landing on the same side is 1/2.
On the third toss, the probability of landing on the same side is 1/2.
On the fourth toss, the probability of landing on the same side is 1/2.
On the fifth toss, the probability of landing on the same side is 1/2.
P(same side all five times)=1/2*1/2*1/2*1/2=1/16.
The answer is D.
Answer: (D) 1/16
Question: The volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2. What is the length of the wire?
(A) 2 : 5
(B) 2 : 6
(C) 2 : 0
(D) 2 : 5
(E) 2 : 1
Rationale: The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r12h1/r22h2 => 1/10 = (1)2h1/(2)2h2
=> h1/h2 = 2/5
i.e. h1 : h2 = 2 : 5
Answer: A
Answer: (A) 2 : 5
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) 140
(B) 130
(C) 150
(D) 170
(E) 160
Rationale: Speed=(60x5/8)m/sec = (50/3) m/sec.
Length of the train = (Speed x Time).
Length of the train = (50/3 x 9)m = 150 m.
Answer is C.
Answer: (C) 150
Question: Consider the set S = {1, 2, 3, …., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and with 1000 and have at least 3 elements?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Explanation :
We know that it should be an AP, 1st element is 1, and last element 1000.
a = 1, a + (n-1) d = 1000 ; (n-1) d = 999.
So 999 is the product of (n-1) and d. So, we need to see the factors of 999 to see what all values /how many of those values (n-1) and d can assume.The prime factors are 3, 3, 3, and 37.
(999 = 3 * 3 * 3 * 37). (Other than 1 and 999)
So, we can have
(n - 1) = 3, d = (3 * 3 * 37)
(n - 1) = 3 * 3, d = (3 * 37)
(n - 1) = 3 * 3 * 3, d = (37)
(n - 1) = 37, d = (3 * 3 * 3)
(n - 1) = 37 * 3, d = (3 * 3)
(n - 1) = 37 * 3 * 3, d = (3)
Other than this, we can have 1 and 999 also as factors of 999. If we have n-1 as 1, n is 2. But that means only 2 terms in AP. (Not possible according to the question). If we have n-1 = 999, n = 1000. d = 1. This is possible. So, we have totally, 7 possible combinations.
Answer : D
Answer: (D) 6
Question: A, B and C can complete a piece of work in 24,6 and 12 days respectively.Working together, they will complete the same work in:
(A) 22/9 days
(B) 23/8 days
(C) 24/7 days
(D) 23/8 days
(E) 29/8 days
Rationale: Explanation:
(A+B+C)'s 1 day's work =\inline {\color{Black}\left ( \frac{1}{24}+\frac{1}{6}+\frac{1}{12} \right )=\frac{7}{24}}
so, A,B and C together will complete the work in 24/7 days.
Answer: C) 24/7 days
Answer: (C) 24/7 days
Question: The average weight of 25 persons sitting in a boat had some value. A new person added to them whose weight was 46 kg only. Due to his arrival, the average weight of all the persons decreased by 5 kg. Find the average weight of first 25 persons?
(A) 172
(B) 267
(C) 265
(D) 176
(E) 282
Rationale: 25x + 46 = 26(x – 5)
x = 176
Answer:D
Answer: (D) 176
Question: An article is sold at a certain price. By selling it at 2/3 of that price one loses 10%. Find the gain percent at original price.
(A) 15%
(B) 35%
(C) 25%
(D) 50%
(E) 30%
Rationale: Let the original S.P be Rs. X. Then new S.P = Rs. 2/3 X, Loss =10%
So C.P = Rs. [100/90*2/3 X ] = 20 X/27.
Now C.P = Rs. 20X/27, S.P =Rs. X, Gain = Rs. [ X -20X/27 ] =Rs.7X/27.
Gain % = [ 7X/27 *27/20X *100 ]% =35%
ANSWER:B
Answer: (B) 35%
Question: If log 27 = 1.431, then the value of log 9 is:
(A) 0.934
(B) 0.945
(C) 0.954
(D) 0.958
(E) 0.987
Rationale: log 27 = 1.431
log (33 ) = 1.431
3 log 3 = 1.431
log 3 = 0.477
log 9 = log(32 ) = 2 log 3 = (2 x 0.477) = 0.954.
Answer: Option C
Answer: (C) 0.954
Question: How many litres of water must be added to 16 liters of milk and water containing 10% water to make it 20% water in it?
(A) 2 litre
(B) 4 litre
(C) 6 litre
(D) 7 litre
(E) 9 litre
Rationale: By rule of alligation,
% Concentration
of water in pure water (100)	% Concentration
of water in the given mixture (10)
Mean % concentration
(20)
20 - 10 = 10	100 - 20 = 80
=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8
Here quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
=> Quantity of water
=
16
×
1
8
=
2
=16×18=2 litre
A)
Answer: (A) 2 litre
Question: The H.C.F. of two numbers is 12 and their L.C.M. is 600. If one of the numbers is 720, then the other is:
(A) 8
(B) 10
(C) 12
(D) 4
(E) 6
Rationale: Other number =	(12 x 600)/720 = 10.
ANSWER :C
Answer: (C) 12
Question: The Expenditure of an woman increase consistently by 30% per year. If her present expenditure is Rs. 1000 then what will her expenditure be after 3 years ?
(A) 2107
(B) 2197
(C) 2187
(D) 2177
(E) 2167
Rationale: Explanation :
Expenditure = 1000 x 1.3 x 1.3 x 1.3 = 2197
Answer : Option B
Answer: (B) 2197
Question: find the least number that when divided by 16,18 and 20 leaves the same remainder 4 in each case but it is completely divisble by 7
(A) 2884
(B) 2994
(C) 2784
(D) 3254
(E) 3897
Rationale: LCM of 16,18,20 is 720...
(720+4)is not divisible by 7...
(720*2+4)not divisible by7..
.
.
(720*4+4)divisible by 7....
so the ans is... 2880+4=2884
ANSWER:A
Answer: (A) 2884
Question: The total car population in a Chennai is 900 out of which 500 are made by Honda. Out of every 100 Honda cars, 90 are red in color, but only 60% of the total car population is red. Find the % of non-Honda cars that are red
(A) 20.5%
(B) 22.5%
(C) 21.5%
(D) 10.5%
(E) 32.5%
Rationale: Total Honda brand red cars = 90/100 *500 = 450
Total red cars = 60/100*900 = 540
No. of Non-Honda cars = 900-500 = 400
No. of Non-Honda red cars = 540 - 450 = 90
% of Non-Honda red cars = 90/400*100 = 22.5%
Answer : B
Answer: (B) 22.5%
Question: If 2kg of mangoes and 3kg of apples cost Rs.102 and 3 kg of mangoes and 1kg of apple cost Rs. 76. What is the cost of per kg of mango?
(A) Rs.20
(B) Rs.19
(C) Rs.16
(D) Rs.22
(E) Rs.18
Rationale: Let the cost of each pen and pencil be 'p' and 'q' respectively.
2x + 3y= 102 --- (1)
3x + y = 76
9x+ 3y = 228 --- (2)
(1) - (2) => 7x = 126
=> x = 18
ANSWER:E
Answer: (E) Rs.18
Question: 517 x 517 + 483 x 483=?
(A) 79698
(B) 80578
(C) 500578
(D) 81268
(E) None of them
Rationale: =(517)^2 + (483)^2
=(500+17)^2 + (500-17)^2
=2[(500)^2+(17)^2]
=2[250000+289]
=2 x 250289 = 500578
Answer is C
Answer: (C) 500578
Question: A certain quantity is measured on two different scales, the L-scale and the S-scale, that are related linearly. Measurements on the L-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the L-scale corresponds to a measurement of 100 on the S-scale?
(A)  20
(B)  36
(C)  48
(D)  60
(E)  84
Rationale: First, we have to understand what linearly means. It's not a straight ratio (since 6:30 does NOT equal 24:60). We need to look at the increases in each measurement to see what the scalar actually is.
From 6 to 24 we have an increase of 18. From 30 to 60 we have an increase of 30. Therefore, the increase ratio is 18:30 or 3:5. In other words, for every 3 that L increases, S increases by 5.
We know that S is 100. To get from 60 to 100, we went up by 40, or 8 jumps of 5; therefore, L will go up by 8 jumps of 3.
24 + 8(3) = 24 + 24 = 48=C
Answer: (C)  48
Question: 4,25,49,121,169,289,361,529,
(A) 149
(B) 169
(C) 841
(D) 920
(E) 219
Rationale: 29^2= 841 because follow sequence of square of the prime numbers
ANSWER:C
Answer: (C) 841
Question: If the set S is composed of the following numbers {99, 100, 100, 105, 106, 111, 128}, which of the following is largest?
(A) The average (arithmetic mean) of set S
(B) The median of set S
(C) The mode of set S
(D) The range of set S
(E) The standard deviation of set S
Rationale: A) 99, 100, 100, 105, 106, 111, 128 = 749 and its mean is 749/7 = 107.
B) The list is of 7 numbers and median is 105.
C) Mode is number of repetition of a number in the list , here it is 100.
D) Range = Max value - min value = 128 -99 = 29.
E) SD = square root of (sum of ( mean of series - number)^2 ) / n =>< 100
IMO option A is correct answer..
Answer: (A) The average (arithmetic mean) of set S
Question: J and M were each paid x dollars in advance to do a certain job together. J worked on the job for 8 hours and M worked for 4 hours less than J. If M gave J y dollars of her payment so that they would have received the same hourly wage, what was the dollar amount, in terms of y, that J was paid in advance?
(A) a)4y
(B) b)3y
(C) c)6y
(D) d)8y
(E) e)9y
Rationale: J and M were both paid x dollars
If M gave Y dollars from her X to J then at the end
J will have X+Y dollars
M will have X- Y dollars
Now since the hourly wages have to be same then
Hourly wage of J = (X+Y) / 8 (Since J worked for 8 hrs)
Hourly wage of M = (X-y) /4 (Since M worked 4 hrs less than J)
Equating both
(X+Y)/8=(X-Y)/4
on simplifying X=3y
Answer B
Answer: (B) b)3y
Question: Working at their respective constant rates, machine A makes 100 copies in 10 minutes and machine B makes 150 copies in 15 minutes. If these machines work simultaneously at their respective rates for 30 minutes, what is the total number of copies that they will produce?
(A)  250
(B)  600
(C)  675
(D)  700
(E)  750
Rationale: Machine A can produce 100 *30/10= 300 copies
And, Machine B can produce 150* 30/15= 300 Copies
Total producing 600 Copies.
B is the answer
Answer: (B)  600
Question: A work can be finished in 20 days by 10 women. The same work can be finished in fifteen days by 20 men. The ratio between the capacity of a man and a woman is
(A) 1:3
(B) 4:3
(C) 2:3
(D) 2:1
(E) 2:4
Rationale: Work done by 10 women in 1 day = 1/20
Work done by 1 woman in 1 day = 1/(10×20)
Work done by 20 men in 1 day = 1/15
Work done by 1 man in 1 day = 1/(15×20)
Ratio of the capacity of a man and woman =1/(15×20) : 1/(10×20) = 1/15 : 1/10
= 1/3 :1/2 = 2:3
Option C
Answer: (C) 2:3
Question: A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
(A) 8925
(B) 8960
(C) 8970
(D) 8900
(E) 8990
Rationale: Principal= Rs. ( (100x4016.25)/(9x5)
=Rs. 401625/45
= Rs. 8925.
answer :A
Answer: (A) 8925
Question: What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
(A) 42%
(B) 43%
(C) 45%
(D) 44%
(E) 46%
Rationale: Let original length =l
original breadth =b
Then original area =lb
Length is increased by 20%
⇒ New length =l×120100=1.2l
Breadth is increased by 20%
⇒ New breadth =b×120100=1.2b
New area =1.2l×1.2b=1.44lb
Increase in area = new area - original area
=1.44lb−lb=0.44lb
Percentage increase in area
=increase in areaoriginal area×100=0.44lblb×100=44%
Answer is D.
Answer: (D) 44%
Question: In a class,7 students like to play Basketball and 8 like to play Cricket. 3 students like to play on both Basketball and Cricket. How many students like to play Basketball or Cricket or both?
(A) 12
(B) 15
(C) 16
(D) 18
(E) 22
Rationale: Draw a Venn Diagram yourself !
B + C - BC = Number of students that play either Basketball or Cricket
7 +8 - 3 = 12
A)
Answer: (A) 12
Question: The second of two numbers is two less than three times the first. Find the numbers if there sum is
66.
(A) 7-19
(B) 8-20
(C) 10-16
(D) 17-49
(E) 19-55
Rationale: We are looking for two numbers.
#1 - x
#2 - 3x â€“ 2
The sum is 66. #1 + #2 = 66
Substituting x + 3x â€“ 2 = 66
4x â€“ 2 = 66
4x = 68
x = 17
The first number is 17, the second number is two less than three times 17 or 49.
correct answer D
Answer: (D) 17-49
Question: If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 5 strokes ?
(A) 31/32
(B) 7/8
(C) 1/4
(D) 1/8
(E) 1/16
Rationale: Left After 1st stroke = 1/2
Left After 2nd stroke = 1/2 * 1/2 = 1/4
Left After 3rd stroke = 1/2 * 1/4 = 1/8
Left After 4th stroke = 1/2 * 1/8 = 1/16
Left After 5th stroke = 1/2 * 1/16 = 1/32
So removed = 1- 1/32 = 31/32
Answer: (A) 31/32
Question: John left home and drove at the rate of 55 mph for 2 hours. He stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination. How many miles did John drive?
(A) 235 miles.
(B) 245 miles.
(C) 255 miles.
(D) 265 miles.
(E) 275 miles.
Rationale: The total distance D traveled by John is given by
D = 55 * 2 + 3 * 55 = 275 miles.
Answer E
Answer: (E) 275 miles.
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 72 sec. The length of each train is?
(A) 50
(B) 26
(C) 27
(D) 28
(E) 100
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/72 = 25/9 => x = 100.
Answer: E
Answer: (E) 100
Question: The original price of a certain TV set is discounted by g percent, and the reduced price is then discounted by 2g percent. If P is the original price of the TV Set, which of the following represents the price of the television set after the two successive discounts?
(A) P(1 - 0.03x + 0.02x^2)
(B) P(1 - 0.03g + 0.0002g^2)
(C) P(1 - 0.03x + 0.002x^2)
(D) P(1 - 2x^2)
(E) P(1 - 3x + 2x^2)
Rationale: Youcansolve this problem with number plugging.
Say the original price was $10 and g=50. Then after the first reduction the price would become $5 and after the second reduction of 2*50=100% the rprice would become $0.
Now, since P is not zero, then the expression in the brackets must be zero for g=50. Only answer choice B works.
Answer: B.
On this problem - number pluggin is not giving me the answer.. I initially used g = 10, then g = 20 and P = 100. Answer should after both consecutive discounts = 72. I plug in the respective values and I keep getting 68. Can you double check my math.
100 (1-0.03(10) + 0.0002 (10)^2)
100 (1-0.3 + 0.0002 (100))
100 (0.7 + 0.02)
100 (0.68) = 68????
double check my math. Am I missing something? I also plugged in your numbers and still did not get zero as final answer with choice B..
Answer: (B) P(1 - 0.03g + 0.0002g^2)
Question: The sum of ages of 5 children born at the intervals of 3 years each is 55 years. What is the age of the youngest child?
(A) 5 years
(B) 4 years
(C) 6 years
(D) 7 years
(E) 8 years
Rationale: Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 55
5x = 25
x = 5.
Age of the youngest child = x = 5 years.
A)
Answer: (A) 5 years
Question: An alloy of copper and zinc contains copper and zinc in the ratio 5 : 4. Another alloy of copper and zinc contains copper and zinc in the ratio 3 : 6. In what ratio should the two alloys be mixed so that the resultant alloy contains equal proportions of copper and zinc?
(A) 4 : 1
(B) 6 : 1
(C) 1 : 3
(D) 2 : 1
(E) 4 : 1
Rationale: Let Alloy_1 be x units, and let Alloy_2 be y units.
So, fraction of copper in Alloy_1 = 5x/9, and fraction of zinc in Alloy_1 = 4x/9.
Similarly, fraction of copper in Alloy_2 = 3y/9, and fraction of zinc in Alloy_2 = 6y/9.
Mixing them, we get
copper = 5x/9 +3y/9; zinc = 4x/9 + 6y/9
So, 5x+3y = 4x+6y
-> x = 3y
-> x/y =1/3
So, they must be mixed in the ratio 1 : 3
Answer : C
Answer: (C) 1 : 3
Question: A salesperson receives a base salary of $1000 per month and a commission of 7.5% of the selling price of each item sold above the first 50 item. If this month she sold 290 items for $150 each, what will her monthly salary be?
(A) $1,500
(B) $1,800
(C) $2,100
(D) $3,700
(E) $2,800
Rationale: Ans : 1000+ 7.5 % * 150 * (290-50) = $ 3700
ANSWER:D
Answer: (D) $3,700
Question: In how many years will a sum of Rs 800 and 10% per annum completed semiannually becomes Rs.926.10?
(A) 1 years
(B) 2 years
(C) 1 years 6 months
(D) 2 years 6 months
(E) 3 years
Rationale: 800 ------------->926.10926.10-800=126.10Ist six month =40IInd 6 month=40+2=42IIIrd 6 month=40+4.1=44.140+42+44.1=126.1Answer C
Answer: (C) 1 years 6 months
Question: The sum of the even numbers between 1 and n is 95*96, where n is an odd number, then n=?
(A) 167
(B) 175
(C) 183
(D) 191
(E) 199
Rationale: Let n-1 = 2a.
2+4+...+2a = 2*(1+2+...+a) = 2*(a)(a+1)/2 = (a)(a+1) = 95*96
Then a = 95 and n = 191.
The answer is D.
Answer: (D) 191
Question: A train 540 meters long is running with a speed of 54 kmph. The time taken by it to cross a tunnel 180 meters long is?
(A) 41 sec
(B) 42 sec
(C) 43 sec
(D) 48 sec
(E) 47 sec
Rationale: D
48 sec
D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec
Answer: (D) 48 sec
Question: If the average (arithmetic mean) of a list of numbers is 12 and the standard deviation of that list of numbers is 1.3, then which of the following numbers is more than two standard deviations from the mean?
I. 14.3
II. 12.3
III. 9.2
(A) I only
(B) I and II only
(C) II only
(D) III only
(E) I and III only
Rationale: Mean = 12
SD = 1.3
2 SD above mean = 12+2*1.3 = 14.6
2 SD below mean = 12-2*1.3 = 9.4
12.3 and 14.3 lies in the range of 2SD from mean
Answer: Option D
Answer: (D) III only
Question: For 2consecutive yrs, my incomes are in the ratio of 4:7 and expenses in the ratio of 3:5. If my income in the 2nd yr is Rs. 42000 & my expenses in the first yr in Rs.21000, my total savings for the two-year is
(A) Rs. 8000
(B) Rs. 9000
(C) Rs. 9900
(D) Rs. 98900
(E) Rs. 10000
Rationale: Sol. Income in first year = * x 42000 = Rs. 24000
Expenses in second year = \ x 21000 = Rs. 35000 Total savings = Total income - Total expenses = (42000 + 24000)- (21000 + 35000)
= 66000 - S6000 = Rs. 10000
E
Answer: (E) Rs. 10000
Question: What is the ratio R of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled?
(A) 1/4
(B) 3/8
(C) 1/2
(D) 3/5
(E) 2
Rationale: Let X be the side of the cube. Therefore X^2*6= surface area.
the rectangle is the same other than length is 2X. The width and height are the same as the cube. 2*W*H+2*W*L+2*H*L= 2X^2+4X^2+4X^2= 10X^2.
6X^2/10X^2 = 3/5= R.
Answer = D
Answer: (D) 3/5
Question: The length of the rectangular field is double its width. Inside the field there is square shaped pond 8m long. If the area of the pond is 1/8 of the area of the field. What is the length of the field?
(A) 22
(B) 32
(C) 44
(D) 66
(E) 88
Rationale: A/8 = 8 * 8 => A = 8 * 8 * 8
x * 2x = 8 * 8 * 8
x = 16 => 2x = 32 .Answer: B
Answer: (B) 32
Question: Plane leaves from Chelsea every 30 minutes from West Ham airport. Air hostess told Frank Lampard that the plane had left 25 minutes ago and the next plane for Chelsea will leave at 7:00 p.m.
At what time air hostess convey this information to Frank Lampard?
(A) 5:35 p.m
(B) 6:55 p.m
(C) 7:35 p.m
(D) 8:55 p.m
(E) 9:35 p.m
Rationale: Explanation:
Time of Frank Lampard last missed plane = 7:00 p.m - 30 min = 6:30 p.m
As this information has been given to Frank Lampard 25 minutes after the plane has already left, the air hostess gave the information at 6:30 p.m + 25 min = 6:55 p.m
Answer B
Answer: (B) 6:55 p.m
Question: The H.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77,find the other.
(A) 99
(B) 88
(C) 77
(D) 66
(E) 55
Rationale: Other number = 11 X 693/77 = 99
Answer is A.
Answer: (A) 99
Question: Three models (F, G, and H) of cars are distributed among three showrooms. The number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom. There are 19 cars of Model F, 17 cars of Model G, and 15 cars of Model H. What is the maximum number of cars of model F in any showroom ?
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
Rationale: The total number of cars is 51. So each showroom has 17 cars(since the number of cars in each showrooms should be equal 51/3=17).
Moreover that the number of model F is maximum means that the numbers of model G and H should be minimum. Since each model must be represented by at least one car in every showroom that minimum number should be 1.
So Maximum number of model F is 17-2= 15.
The answer is (C)
Answer: (C) 15
Question: In a question on division with zero remainder, a candidate took 20 as divisor instead of 10. The quotient obtained by him was 35. The correct quotient is?
(A) A)70
(B) B)24
(C) C)28
(D) D)29
(E) E)30
Rationale: Number = (35 * 20) = 700
Correct quotient = 700/10 = 70
A)
Answer: (A) A)70
Question: The 46th digit after the decimal in the decimal expansion of 11996 is
(A) 0
(B) 1
(C) 4
(D) 6
(E) 8
Rationale: Let dj denote the jth digit of 1/1996 after the decimal. In other words, we write
1/1996=0.d1d2d3....The problem is to determine d46. Observe that
(*)
10in45*1/1996=d1d2d3...d45d46d47...
We compute 1045 modulo 1996. Since 1996 5 = (2000 - 4) 5 = 104 -20, we get
that 104* 20 (mod 1996). Since 1996 4 = (2000- 4) 4 = 8000 - 16, we obtain
1012*(104)3 *203 * 8000*6 (mod 1996). Continuing in this manner, we get 1036 *
(1012)3* 16*212 * 4096 * 104 (mod 1996), 1044* 1036 104 104 *104 20 20 *
2080 20*84 20 * 1680 (mod 1996), and 1045 * 1044 10* 16800* 8 (1996 +
4) + 800 * 832 (mod 1996). This means that the remainder when we divide 1045 by 1996
is 832. If q denotes the quotient, we deduce that
1045/1996= q +832/1996= q.4168...
where the digits after the decimal are computed by long division. We are only interested
in the first of these, the digit 4 since it follows from () that this digit must be d46.
correct answer C
Answer: (C) 4
Question: The probability of Sam passing the exam is 1/4. The probability of Sam passing the exam and Michael passing the driving test is 1/5. What is the probability of Michael passing his driving test?
(A) 1/24.
(B) 1/2.
(C) 1/3.
(D) 2/3.
(E) 4/5
Rationale: 1/4 *1/m= 1/5 , 1/m = 4/5
Answer E
Answer: (E) 4/5
Question: A type W machine can complete a job in 5 hours and a type B machine can complete the job in 7 hours. How many hours will it take 2 type W machines and 3 type B machines working together and independently to complete the job?
(A) 1/5
(B) 29/35
(C) 5/6
(D) 35/29
(E) 35/12
Rationale: now D should be the answer.
W need 5 hours to complete
and B needs 7 hours to compete
so 2W + 3B will complete 2/5 + 3/7 or 29/35 portion of the job in 1 hour
so the whole job will take 35/29 hours....=D
Answer: (D) 35/29
Question: A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work?
(A) 12 days
(B) 9 days
(C) 6 days
(D) 5 days
(E) 8 days
Rationale: Sol. (A + B)'s 1 day's work = (1/4). A's 1 day's work = (1/12)
B's 1 day's work =((1/4)-(1/12))=(1/6)
Hence, B alone can complete the work in 6 days.
Ans: C
Answer: (C) 6 days
Question: One year ago, the ratio of Gaurav' and Sachin's age was 6 : 7 respectively Four years hence , this raio would become 7 : 8. How old is Sachin ?
(A) 34 years
(B) 36 years
(C) 37 years
(D) 38 years
(E) 39 years
Rationale: Explanation :
Sol Let Gaurav's and Sachins ages one year ago be 6xand7x6xand7x 7ears respectively . Then
Gaurav's age 4 years hence = (6x+1) +4=(6x+5)(6x+1) +4=(6x+5) years
Sachin's age 4 years hence = (7x+1)+4=(7x+5)(7x+1)+4=(7x+5) years.
6X+5 / 7X+5 =7/8
â‡” 8(6x+5)= 7(7x+5)
â‡” 48x+40=49x+35
â‡” x=5
Hence , Schin's present age = (7x+1)(7x+1) = 36 years.
ANSWER IS B
Answer: (B) 36 years
Question: Marge has E candies, where E is an integer such that 20 <E< 50. If Marge divides the candies equally among 5 children, she will have 2 candies remaining. If she divides the candies among 6 children, she will have 1 candy remaining. How many candies will remain if she divides the candies among 7 children?
(A)  0
(B)  1
(C)  2
(D)  3
(E)  4
Rationale: If Marge divides the candies equally among 5 children, she will have 2 candies remaining.
This tells us that the candies are of 5x + 2 type and so 22, 27, 32, 37, 42, or 47
If she divides the candies among 6 children, she will have 1 candy remaining.
this tells us it is if 6x+1 type and so 25, 31, 37, 43, or 49
ONLY 37 is common, so candies are 37 in number..
if 37 is divided in 7 kids, remainder E= 2
C
Answer: (C)  2