mathewhe/aqua_rat huggingface dataset:
Question: V is the volume of a cylinder; the radius of the cylinder is 2.4. The height of the cylinder is 550% more than the radius. Which of the following is true?
(A) 100 < V < 300
(B) 300 < V < 500
(C) 500 < V < 700
(D) 700 < V < 900
(E) 900 < V < 1100
Rationale: As we see the answers are in form of range we can use approximation
Volume of cylinder is πr^2h
Given π= 22/7 and r = 2.4 so r^2 ~ 6 and h = 6.5 * 2.4 ~ 16
So 22/7 * 6 * 16 ~ 302
so answer should be
B. 300 < V < 500
Answer : B
Answer: (B) 300 < V < 500
Question: An instrument manufactured by a company consists of two parts A and B. In manufacturing part A, 9 out of 100 are likely to be defective and in manufacturing part B, 5 out of 100 are likely to be defective. Calculate the probability that the instrument will not be defective.
(A) 0.86.
(B) 0.91
(C) 0.95
(D) 0.97
(E) None of these
Rationale: Explanation :
Let, A ≡ event that part A is non- defective.
B ≡ event that part B is non- defective.
A and B are independent events.
AB ≡ event that part A and part B both are non-defective so that instrument is not defective.
=> P(AB) = P(A) x P(B).
=> [ 1 - P(~A)] x [ 1 - P(~B)].
Here, P(~A)= 9/100 = probability of part A being defective
and, P(~B) = 5 / 100 = probability of part B being defective.
=> P(AB) = [1- (9 / 100 ) ] x [1- (5 / 100)].
=> 0.91 x 0.95.
=> 0.86.
Hence, the required probability is 0.86.
Answer : A
Answer: (A) 0.86.
Question: The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ration of the number of boys to the number of girls in the class is:
(A) 2:3
(B) 1:2
(C) 4:5
(D) 5:4
(E) 3:2
Rationale: Let the ratio be k : 1. Then,
k * 16.4 + 1 * 15.4 = (k + 1) * 15.8
= (16.4 - 15.8)k = (15.8 - 15.4)
= k = 0.4/0.6 = 2/3
Required ratio = 2/3 : 1 = 2:3.
ANSWER A
Answer: (A) 2:3
Question: What is the sum of odd integers from 35 to 55, inclusive?
(A) 495
(B) 1,500
(C) 1,240
(D) 1,120
(E) 1,100
Rationale: Number of odd integers = (55-35)/2 + 1
= 20/2 + 1
= 11
Sum of odd integers = (35+55)/2 * 11
= 495
Answer A
Answer: (A) 495
Question: In a sequence of 12 numbers, each term, except for the first one, is 12^15 less than the previous term. If the greatest term in the sequence is 12^16, what is the smallest term in the sequence?
Thus, A12 = A1+12-1d
Where d is the difference between each pairs of consecutive terms. Since each term is 1211 less than the previous term, d=-1211Click individual answer choices for specific explanations.
From what I undersltand the formula for an arithmetic sequence represents
some-number-in-the-sequence= first-number-in-sequence +the-place-of-that-number - 1x the-difference-of-each-#-in-the-sequence
(A) −12^11
(B) 0
(C) 12^11
(D) 11·12^11
(E) 12^15
Rationale: there are total 12 terms. andgreatestterm is 12^16. Each term is 12^15 less than previous one. What does this signify? This shows it is an AP - a decreasing one. (NoticeEach term is 12^11 less than previous one')
Therefore, 12^16 is the first term of the AP and -12^15 is the difference between successive terms, the smallest term should be the last one.
Last term = First term + (n-1)*d
therefore, Last term = 12^16+ 11*(-12^15)
=> Last term = 12^16 - 11*12^15
=> Last term = 12^15 (12-11) = 12^15
Which is the smallest term.
Hence Ans E it is.
Answer: (E) 12^15
Question: The average of 35 students in a class is 16 years. The average age of 21 students is 14. What is the average age of remaining 14 students?
(A) 18 years
(B) 21 years
(C) 12 years
(D) 19 years
(E) 12
Rationale: Sum of the ages of 14 students
= (16 * 35) - (14 * 21) = 560 - 294 = 266
Required average = (266/14)
= 19 years.
Answer: D
Answer: (D) 19 years
Question: A train crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train?
(A) 287m
(B) 180m
(C) 279m
(D) 267m
(E) 112m
Rationale: Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
Answer:B
Answer: (B) 180m
Question: The youngest of 4 children has siblings who are 3, 5, and 8 years older than she is. If the average (arithmetic mean) age of the 4 siblings is 20, what is the age of the youngest sibling?
(A) 14
(B) 15
(C) 16
(D) 17
(E) 18
Rationale: x + (x+3) + (x+5) + (x+8) = 80
4x + 16 = 80
4x = 64
x = 16
The answer is C.
Answer: (C) 16
Question: Find the fourth proportional to 1.4,3.2 and 6.5?
(A) 13.19
(B) 14.09
(C) 14.86
(D) 15.36
(E) 15.19
Rationale: Formula= Fourth proportional =(b x c)/a
a=1.4 , b=3.2 and c=6.5
(3.2 x 6.5)/1.4=14.86
C
Answer: (C) 14.86
Question: What would be the perimeter of the square whose side is equal to the diameter of the circle whose perimeter is 52.5 cm?
(A) 52.5
(B) 60.2
(C) 66.8
(D) 74.25
(E) 80.12
Rationale: Diameter of the circle= perimeter/Ï€ = 52.5x7/22 = 16.70
Perimeter of the square=16.70x4=66.8cm
Answer: C
Answer: (C) 66.8
Question: A jogger running at 9 kmph along side a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
(A) 3.6 sec
(B) 18 sec
(C) 36 sec
(D) 72 sec
(E) None
Rationale: CATEGORY
Formulae
Examples
DIRECTIONS : Problems based on Trains.
11.	A train 280 m long, running with a speed of 63 km/hr will pass a tree in
A. 15 sec.
B. 16 sec.
C. 18 sec.
D. 20 sec.
Solution
Speed	= (63 x 5/18) m/sec
= 35/2 msec
Time taken	= (280 x 2 /35)m/sec
= 16 sec.
12.	A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is ihe length of the train?
A. 120 metres
B. 180 metres
C. 324 metres
D. None of these
Solution
13.	A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
A. 3.6 sec
B. 18 sec
C. 36 sec
D. 72 sec
Solution
Speed of train relative to jogger	= (45 - 9) km/hr
= 36 km/hr
= (36 x 5/18) m/sec
= 10 m/sec
Distance covered	= (240+120) m
= 360 m
Time taken	= (360 / 10)sec
= 36 sec.
Answer C
Answer: (C) 36 sec
Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
(A) 100
(B) 110
(C) 179
(D) 209
(E) 219
Rationale: Explanation:
In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.
Hence we have 4 options as given below
We can select 4 boys ...(option 1)
Number of ways to this = 6C4
We can select 3 boys and 1 girl ...(option 2)
Number of ways to this = 6C3 × 4C1
We can select 2 boys and 2 girls ...(option 3)
Number of ways to this = 6C2 × 4C2
We can select 1 boy and 3 girls ...(option 4)
Number of ways to this = 6C1 × 4C3
Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]
=
6
×
5
2
×
1
+
6
×
5
×
4
3
×
2
×
1
×
4
=6×52×1+6×5×43×2×1×4
+
6
×
5
2
×
1
×
4
×
3
2
×
1
+
6
×
4
+6×52×1×4×32×1+6×4
=
15
+
80
+
90
+
24
=209
D)
Answer: (D) 209
Question: A is twice as good a workman as B and they took 9 days together to do the work B alone can do it in.
(A) 17 days
(B) 12 days
(C) 29 days
(D) 25 days
(E) 27 days
Rationale: WC = 2:1
2x + x = 1/9
x = 1/27 => 27 days
ANSWER:E
Answer: (E) 27 days
Question: The maximum numbers of students among them 671 Note books and 610 pencils can be distributed in such a way that each student gets the same number of note books and same number of pencils is
(A) 91
(B) 910
(C) 61
(D) 1911
(E) None
Rationale: olution
Required number of students.
= H.C.F of 671 and 610.
â€¹=â€º 61.
Answer C
Answer: (C) 61
Question: The area of a triangle will be when a = 1m, b = 2m, c = 3m, a, b, c being lengths of respective sides.
(A) 3
(B) 4
(C) 5
(D) 6
(E) 87
Rationale: S = (1 + 2 + 3)/2 = 3
=> No triangle exists.Answer: A
Answer: (A) 3
Question: 1,4,9,16,....
(A) 58
(B) 60
(C) 61
(D) 62
(E) 25
Rationale: 1*1=1
2*2=4
3*3=9
4*4=16
5*5=25
ANSWER:E
Answer: (E) 25
Question: What is the smallest integer E greater than 1 that leaves a remainder of 1 when divided by any of the integers 6, 8, and 10?
(A) 21
(B) 41
(C) E=121
(D) 241
(E) 481
Rationale: Or u can just use the answer choices here. Since the answers are already arranged in ascending order, the first number which gives remainder E as 1 for all three is the correct answer. In the given question, the first number which gives a remainder of 1 for 6,8 and 10 is 121.C
Answer: (C) E=121
Question: In jar A there are 3 white balls and 2 green ones, in jar B there is one white ball and three green ones. A jar is randomly picked, what is the probability of picking up a white ball out of jar A?
(A) 2/5
(B) 3/5.
(C) 3/10
(D) 3/4
(E) 2/3
Rationale: Prob of picking jar A = 1/2
Prob of picking a white ball from jar A = 3/5
Prob of picking up a white ball out of jar A = 1/2*3/5 = 3/10
Hence, answer will be C.
Answer: (C) 3/10
Question: A and B started a business jointly A's investment was thrice the investment of B and the period of his investment was two times the period of investment of B. If B received Rs. 4000 as profit, then their total profit is
(A) 23477
(B) 28000
(C) 28877
(D) 1987
(E) 1771
Rationale: Explanation:
Suppose B invested Rs. x for y months. Then, A invested Rs. 3x for 2y months.
So, A : B = (3x * 2y) : (x * y) = 6xy : xy = 6 : 1.
B's profit : Total profit = 1 : 7.
Let the total profit be Rs. x Then, 1/7 = 4000/x or x = 28000.
Answer: B) 28000
Answer: (B) 28000
Question: The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
(A) 20
(B) 21
(C) 22
(D) 23
(E) 25
Rationale: Number of members in the team = 11
Let the average age of of the team = x
=> (Sum of ages of all 11 members)/11=x
=> Sum of the ages of all 11 members = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper
=11x−26−29=11x−55
Average age of 9 members of the team excluding captain and wicket keeper
=(11x−55)/9
Given that (11x−55)/9=(x−1)
⇒11x−55=9(x−1)
⇒11x−55=9x−9
⇒2x=46
⇒x=46/2
=23 years
Answer is D.
Answer: (D) 23
Question: In a span of 24 hours, how many times would the two hands of a clock point exactly in opposite directions ?
(A) 10
(B) 11
(C) 23
(D) 22
(E) 24
Rationale: Explanation:
In every hour, the two hands (minute and hour) of the clock would be exactly opposite to each other once. But, from 5 o'clock to 6 o'clock and from 6 o'clock to 7 o'clock (a span of hours), they would be exactly opposite only once (at exact 6 o'clock).
Thus, during the span of 12 hours, the two hands would be exactly opposite 11 times. Hence, in the span of 24 hours they would be exactly opposite 22 times.
ANSWER D
Answer: (D) 22
Question: The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. What is the teacher's age in years ?
(A) 81 years
(B) 21 years
(C) 51 years
(D) 31 years
(E) 61 years
Rationale: Age of the teacher
= (37 * 15 - 36 * 14) years
= 51 years.
Answer: C
Answer: (C) 51 years
Question: If -1 < h < 1, which of the following has the greatest value?
(A) a) 1-h
(B) b) 1+h
(C) c) 1+h^2
(D) d) 1-1/h
(E) e) 1-1/(h^2)
Rationale: easiest way of solving this would be to substitute a value.
lets say h= - 1/2
a) 3/2
b) 1/2
c) 5/4
d) 3
e) - 3
the value you chose is wrong . h cannot be 0.9 as -1<h<0. if you chose -0.9 answer will still be D.
Answer is C.
Answer: (C) c) 1+h^2
Question: The H.C.F. and L.C.M. of two numbers are 11 and 5040 respectively If one of the numbers is 144, find the other number
(A) 400
(B) 256
(C) 385
(D) 420
(E) None of these
Rationale: Explanation:
Solve this question by using below formula.
Product of 2 numbers = product of their HCF and LCM
144 * x = 11 * 5040
x = (11*5040)/144 = 385
Option C
Answer: (C) 385
Question: A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle?
(A) 17 cm2
(B) 30 cm2
(C) 18 cm2
(D) 19 cm2
(E) 15 cm2
Rationale: The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle
= 6 * 5
= 30 cm2
Answer: B
Answer: (B) 30 cm2
Question: A candidate got 40% of the votes polled and he lost to his rival by 2000 votes. How many votes were cast?
(A) 10000
(B) 2028
(C) 2775
(D) 5496
(E) 6851
Rationale: 40%-----------L
60%-----------W
------------------
20%----------2000
100%---------? => 10000
Answer:A
Answer: (A) 10000
Question: Sam invested Rs. 15000 @ 10% per annum for one year. If the interest is compounded half-yearly, then the amount received by Sam at the end of the year will be?
(A) Rs.16537.76
(B) Rs.16537.98
(C) Rs.16537.50
(D) Rs.16537.56
(E) Rs.16537.51
Rationale: P = Rs. 15000; R = 10% p.a. = 5% per half-year; T = 1 year = 2 half-year
Amount = [15000 * (1 + 5/100)2]
= (15000 * 21/20 * 21/20)
= Rs.16537.50
Answer: C
Answer: (C) Rs.16537.50
Question: A bag contains 3 red and 2 black ball. Another bag contains 4 red and 5 black balls. A ball is drawn from the first bag and is placed in the second. A ball is then drawn from the second. What is the probability that this draw of red ball is due to the drawing of red ball from the first bag ?
(A) 3/5
(B) 2/5
(C) 4/25
(D) 3/10
(E) 15/23
Rationale: Probability of getting red ball from bag1 and then red from bag 2 = 3/5*5/10 = 15/50
Probability of getting black ball from bag1 and then red from bag 2 = 2/5*4/10 = 8/50
Now comes the condition, hence apply conditional probability:
P(Red)/ P(Red + Black)
(15/50)/(15/50+8/50) = 15/50 /23/50 = 15/23.
ANSWER:E
Answer: (E) 15/23
Question: How many seconds will a 900 meter long train moving with a speed of 63 km/hr take to cross a man walking with a speed of 3 km/hr in the direction of the train ?
(A) 48
(B) 36
(C) 54
(D) 11
(E) 18
Rationale: Explanation:
Here distance d = 900 mts
speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s
time t = = 54 sec.
Answer: C
Answer: (C) 54
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 400 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 240
(B) 246
(C) 252
(D) 292
(E) 298
Rationale: Number of newborns that can die in first month = 1/10 * 400 = 40
Survived = 360
Number of newborns that can die in second month = 1/10 * 360 = 36
Survived = 324
Number of newborns that can die in third month = 1/10 * 324 = 32
Survived = 292
Answer:D
Answer: (D) 292
Question: Find the total number of prime factors in the expression (4)^11 x (7)^5 x (11)^5
(A) 26
(B) 32
(C) 25
(D) 27
(E) 29
Rationale: (4)^11x (7)^5 x (11)^5
= (2 x 2)^11 x (7)^5 x (11)^5
= 2^11 x 2^11 x7^5x 11^5
= 2^22 x 7^5 x11^5
Total number of prime factors = (22 + 5 + 5) = 32.
Answer is B.
Answer: (B) 32
Question: Dawson completes a piece of work in 40 days, Rose completes the same work in 60 days. If both of them work together, then the number of days required to complete the work is?
(A) 18
(B) 20
(C) 21
(D) 22
(E) 24
Rationale: If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days.
That is, the required No. of days = 40 × 60/100 = 24 days
Answer is E
Answer: (E) 24
Question: Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1232?
(A) 1000
(B) 2217
(C) 1100
(D) 2777
(E) 2991
Rationale: 1232 = P [1 + (5*12/5)/100]
P= 1100
Answer: C
Answer: (C) 1100
Question: A mixture of 200 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?
(A) 11 liters
(B) 11.2 liters
(C) 12.5 liters
(D) 13.78 liters
(E) 13.33 liters
Rationale: Number of liters of water in 200 liters of the mixture = 20% of 120 = 20/100 * 200 = 40 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (40 + P) and total volume of mixture is (200 + P).
(40 + P) = 25/100 * (200 + P)
160+4P = 200+P
P = 13.33 liters.
Answer:E
Answer: (E) 13.33 liters
Question: The average of 11 numbers is 10.5. If the average of first six is 10.5 and that of the last six is 11.4 the sixth number is?
(A) 9
(B) 9.2
(C) 10
(D) 15.9
(E) 17
Rationale: Explanation:
1 to 11 = 11 * 10.5 = 115.5
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 115.5 = 15.9
6th number = 15.9
D)
Answer: (D) 15.9
Question: The length of a rectangle is twice the breadth. If the area is 128 cm2, determine the length and the breadth.
(A) b=2,l=9
(B) b=8,l=16
(C) b=9,l=9
(D) b=4,l=14
(E) b=3,l=9
Rationale: We are given length l=2b and l×b=128
∴2b×b=128
2b^2=128
b^2=64
b=±8
But breadth must be positive, therefore b=8 cm, and l=2b=16 cm.
Answer is B.
Answer: (B) b=8,l=16
Question: A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 49
(B) 50
(C) 12
(D) 13
(E) 67
Rationale: Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 ==> x = 50 km/hr.
Answer: Option B
Answer: (B) 50
Question: Find the odd man out
20, 40, 60, 81, 100, 120, 140, 160
(A) 120
(B) 140
(C) 180
(D) 160
(E) 81
Rationale: as 81 is only odd number.
ANSWER:E
Answer: (E) 81
Question: The length of the bridge, which a train 140 m long and traveling at 45 km/hr can cross in 30 sec is?
(A) 235
(B) 240
(C) 245
(D) 250
(E) 255
Rationale: Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (140 + x)/30 = 25/2
x = 235 m.
Answer: Option A
Answer: (A) 235
Question: A painting at an art gallery is framed such that the area of the square frame is 1/5 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 6, what is the area of the painting inside the frame?
(A) 10
(B) 15
(C) 30
(D) 40
(E) 50
Rationale: Suppose side of the painting(which is a square) is a and side of the outer square(painting+frame) is b
Area of painting = a^2 and we know that the area of the frame is (1/5) of that of the painting so
Area of frame = (1/5) * a^2
Area of Frame + Painting = a^2 + (1/5)*a^2 = (6/5)*a^2 which is equal to b^2
Line connecting the corners of the frame is the diagonal of the outer square which is equal to b*sqrt2
so, b * sqrt2 = 6
so, b = 3 * sqrt2
b^2 = 18
we know that b^2 = (6/5)*a^2
so, a^2 = (5/6)*b^2 = (5/6)*18 = 15
And area of painting = a^2 = 40
So, answer will be B.
Answer: (B) 15
Question: 0.41 + 21.4 + 6.3 + 280=?
(A) 708.71
(B) 408.71
(C) 508.71
(D) 308.71
(E) 208.71
Rationale: Add 280.00+
21.40+
6.30+
0.41=
308.71
correct answer D
Answer: (D) 308.71
Question: How many different ways can you select a pilot and a co pilot if you have 80 people to choose from?
(A)  199
(B)  200
(C)  398
(D)  6320
(E)  9900
Rationale: You are not just choosing two from 80. Each of the two people chosen fills a particular slot. The first filling the pilot slot and the second filling the copilot slot is different from the first filling the copilot slot and the second filling the pilot slot.
So the answer is in a sense (80C2) x 2 = 6320.
Ans D
Answer: (D)  6320
Question: Find the number of square tiles to cover the floor of a room measuring 4 m * 9 m leaving 0.25 m space around the room. A side of square tile is given to be 25 cms?
(A) 422
(B) 476
(C) 429
(D) 428
(E) 413
Rationale: 3 1/2 * 8 1/2 = 1/4 * 1/4 * x => x '
= 476
Answer:B
Answer: (B) 476
Question: There are 4 copies of 5 different scales. In how many ways can they be arranged on a shelf?
(A) 20!/4!
(B) 20!/5(4!)
(C) 20!/(4!)^5
(D) 20!
(E) 5!
Rationale: There are 4 copies of 5 different scales.20!/((4!)^5)
Answer - C
Answer: (C) 20!/(4!)^5
Question: What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 30%?
(A) 5%
(B) 28%
(C) 27%
(D) 40%
(E) 70%
Rationale: SP2 = 2/3 SP1
CP = 100
SP2 = 70
2/3 SP1 = 70
SP1 = 105
100 --- 5 => 5%
Answer:A
Answer: (A) 5%
Question: An analyst will recommend a combination of 3 industrial stocks, 2 transportation stocks, and 2 utility stocks. If the analyst can choose from 5 industrial stocks, 4 transportation stocks, and 3 utility stocks, how many different combinations of 7 stocks are possible?
(A) 12
(B) 19
(C) 60
(D) 180
(E) 720
Rationale: 5C3 * 4C2 * 3C2 = 10*6*3 = 180.
Answer: D
Answer: (D) 180
Question: Look at this series: 40, 44, 48, __, 52, 56, 40, 60, ... What number should fill the blank?
(A) 60
(B) 50
(C) 40
(D) 46
(E) 52
Rationale: C
40
In this series, 4 is added to the previous number; the number 40 is inserted as every third number.
Answer: (C) 40
Question: A shopkeeper sold an article for 360 after giving 20% discount on the labelled price and made 20% profit on the cost price. What would have been the percentage profit,had he not given the discount’?
(A) 25%
(B) 30%
(C) 23%
(D) 50%
(E) None of these
Rationale: Cost price = 360×100/120= 300
S.P. at no discount = 360×100/80 = 450
∴ % profit = 150×100/300 = 50 %
Answer D
Answer: (D) 50%
Question: A sum of Rs.4000 is invested at a compound interest for three years, the rate of interest being 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years.
(A) 2888
(B) 3120
(C) 2877
(D) 2600
(E) 1277
Rationale: Let A be the amount received at the end of the three years.
A = 4000[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4000 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.6600
So the interest = 6600 - 4000 = Rs.2600
Answer: D
Answer: (D) 2600
Question: The difference between two angles of a triangle is 24o. The average of the same two angles is 54o. Which one of the following is the value of the greatest angle of the triangle
(A) 45o
(B) 45o
(C) 66o
(D) 72o
(E) None of these
Rationale: Solution:
Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24o.
=> a – b = 24.
Since the average of the two angles is 54o, we have (a+b)/2 =54.
Solving for b in the first equation yields b=a–24, and substituting this into the second equation yields,
[{a+(a−24)}/2] =54
2a−24 = 54*2
2a−24 = 108
2a = 108 +24
2a =132
a =66
Also,
b=a−24=66−24=42.
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is
180o, a+b+c =180.
Putting the previous results into the equation yields 66+42+c =180.
Solving for c yields c =72
Hence, the greatest of the three angles a, b and c is c, which equ
Answer: Option D
Answer: (D) 72o
Question: The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square?
(A) 5:18
(B) 5:28
(C) 5:29
(D) 5:22
(E) 5:16
Rationale: Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 212
a = (212)1/2 = 26 = 64
L = 2a and b = a - 24
b : l = a - 24 : 2a = 40 : 128
= 5:16
Answer: E
Answer: (E) 5:16