mathewhe/aqua_rat huggingface dataset:
Question: In digging a pond 20 m * 10 m * 8 m the volumes of the soil extracted will be?
(A) 3387
(B) 1600
(C) 2866
(D) 2787
(E) 1121
Rationale: 20 * 10 * 8 = 1600
Answer: B
Answer: (B) 1600
Question: A man buys an article and sells it at a profit of 20%. If he had bought it at 20% less and sold it for Rs.55 less, he could have gained 25%. What is the cost price?
(A) 197
(B) 375
(C) 275
(D) 278
(E) 268
Rationale: CP1 = 100 SP1 = 120
CP2 = 80 SP2 = 80 * (125/100) = 100
20 ----- 100
55 ----- ? => 275
Answer:C
Answer: (C) 275
Question: In how many ways can 21 books on English and 19 books on French be placed in a row on a shelf so that two books on French May not be Together ?
(A) 3990
(B) 1540
(C) 1995
(D) 3672
(E) None of these
Rationale: In order that two books on French are never together, we must place all books as under
X E X E X E X E X E .....X E X
where E denotes the position of an English book and X that of a French book.
Since there are 21 books on English, the number of places marked X are therefore, 22.
Now, 19 places out of 22 can be chosen in 22C19 = 22C3 = (22x21x20)/(3x2x1) = 1540 ways
Answer B.
Answer: (B) 1540
Question: total 100 members are writing exam. In the 48 members are writing first exam. 45 members are writing second exam. 38 members are writing third exam. 5 members are writing all the three exams.How many members are writing 2 exams?
(A) 20
(B) 21
(C) 22
(D) 23
(E) 24
Rationale: It's basic set theory problem using venn diagram.
Draw three intersecting circles 1,2,3. and name all the parts a,b,c,d,e,f,g. 'e' being common in all circles.
these will be the equation.
a+b+c+d+e+f+g= 100 ........(1)
a+b+d+e=48.......................(2) (members are writing first exam)
b+e+f+c=45........................(3) (members are writing second exam)
d+e+f+g=38........................(4) (members are writing third exam)
e=5......................................(5) (members are writing all the three exams)
we have to find the value of d+b+f=?
add eq 2,3,4 and put the value of a+c+g from 1.
you will get d+b+f=21.
ANSWER:B
Answer: (B) 21
Question: A single discount equivalent to the discount series of 20%, 10% and 5% is?
(A) 31.5
(B) 31.9
(C) 31.6
(D) 31.3
(E) 31.1
Rationale: 100*(80/100)*(90/100)*(95/100) = 68.4
100 - 68.4 = 31.6
Answer: C
Answer: (C) 31.6
Question: Dacid obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks?
(A) 75
(B) 45
(C) 23
(D) 28
(E) 96
Rationale: Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.
Answer: A
Answer: (A) 75
Question: On dividing 16698 by a certain number, we get 89 as quotient and 14 as remainder. What is the divisor?
(A) 743
(B) 154
(C) 187
(D) 741
(E) 785
Rationale: Divisor *quotient+remainder=dividend
divisor=(dividend)-(remainder)/quotient
(16698-14)/89=187
Answer(C)
Answer: (C) 187
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 148, what is the difference between the ages of Patrick and Monica?
(A) 27
(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=47x=148
x=3
Difference if ages of Pa and Mo=25x-9x=16x=16*3=48
Answer B
Answer: (B) 48
Question: In a class of 10 students there are 3 girls. In how many ways can they be arranged in a row such that no three girls are consecutive?
(A) 32*6!
(B) 42*8!
(C) 25*5!
(D) 45*5!
(E) 52*9!
Rationale: Number of girls = 3
Number of boys = 7
7 boys can be arranged in a row in 7! ways
Now there are 8 places for 3 girls
3 girls can be arranged in 8P3 ways
Required number of ways = 8P3 * 7! = 8!*7!/5! = 42*8!
Answer is B
Answer: (B) 42*8!
Question: A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?
(A) 378
(B) 288
(C) 287
(D) 646
(E) 217
Rationale: 840 = P [1 + (10*3)/100]
P = 646
Answer: D
Answer: (D) 646
Question: A number whose fifth part increased by 7 is equal to its fourth part diminished by 7 is ?
(A) 160
(B) 180
(C) 200
(D) 280
(E) None
Rationale: Answer
Let the number be N.
Then, (N/5) + 7 = (N/4) - 7
â‡’ (N/4) - (N/5) = 14
â‡’ (5N - 4N)/20 = 14
âˆ´ N = 280
Option: D
Answer: (D) 280
Question: If the product of the integers from 1 to n is divisible by 630, what is the least possible value of n?
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
Rationale: 630 = 2 x 3 x 3 x 5 x 7
N must include at least up to the number 7.
The answer is A.
Answer: (A) 7
Question: At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
(A) 3
(B) 7
(C) 4
(D) 2
(E) 1
Rationale: 150 = (750*5*R)/100
R = 4%
Answer: C
Answer: (C) 4
Question: By using the numbers 0, 2, 3, 4 and 5 only once, how many five digit numbers can be made that are divisible by 4?
(A) 20
(B) 40
(C) 1120
(D) 60
(E) 30
Rationale: A number to be divisible by 4 must end with 04, 08, 12,16 ,20 ,24.....
So, there are the following cases possible:
235-04 --> the first three digits can be arranged in 3=6 ways.
035-24 --> the first three digits can be arranged in 3=6 ways.
045-32 --> the first three digits can be arranged in 3=6 ways.
235-40 --> the first three digits can be arranged in 3=6 ways.
034-52 --> the first three digits can be arranged in 3=6 ways
Total = 6+6+6+6+6 = 30.
Answer: E.
Answer: (E) 30
Question: Roberto has three children: two girls and a boy. All were born on the same date in different years. The sum of the ages of the two girls today is smaller than the age of the boy today, but a year from now the sum of the ages of the girls will equal the age of the boy. four years from today, the difference between the age of the boy and the combined ages of the girls will be
(A) 1
(B) 2
(C) -3
(D) –2
(E) –1
Rationale: Approach I (Plugin's)
Girl I ............... Girl II ................ Boy
1 ....................... 1 ........................ 3 (Assume the current ages)
1 + 1 < 3 .......... (Satisfies the given condition)
1 Year later there ages are
2 ....................... 2 ......................... 4
2 + 2 = 4 ............ (Satisfies the given condition)
After 4 years there ages are
5 ....................... 5 ............................ 7
Difference = 7 - (5+5) = 7 - 10 = -3
Answer = C
Answer: (C) -3
Question: How many minutes does it take John to type u words if he types at the rate of x words per minute?
(A) x/y
(B) u/x
(C) xy
(D) 60x/y
(E) y/(60x)
Rationale: time*rate=job --> time*x=u --> time=u/x.
Answer: B.
Answer: (B) u/x
Question: Let q represent the integer length of a side of a triangle. If r represents the number of distinct values for q such that we can create a triangle with lengths q, 9, and 12, what is the value of r?
(A) 5
(B) 17
(C) 18
(D) 22
(E) 29
Rationale: From the triangle inequality rule, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So, the three inequalities are:
1. q + 9 > 12
2. 9 + 12 > q
3. q + 12 > 9
Equation 3 is true for any positive value of q.
From equation 1, q > 3
From equation 2, 21 > q
Then 21 > q > 3 so q can be any value from 4 to 20 inclusive.
The number of possible values is 17.
The answer is B.
Answer: (B) 17
Question: John makes $60 a week from his job. He earns a raise andnow makes $78 a week. What is the % increase?
(A) 16.12%
(B) 16.66%
(C) 30%
(D) 17.66%
(E) 18.1%
Rationale: Increase = (18/60)*100 = (3/10)*100 = 30%.
C
Answer: (C) 30%
Question: Tom bought 2 shares and sold them for $60 each. If he had a profit of 25% on the sale of one of the shares but a loss of 40% on the sale of the other share, then on the sale of both shares Tom had
(A)  a profit of $50
(B)  a loss of $50
(C)  a profit of $60
(D)  a loss of $60
(E)  neither a profit nor a loss
Rationale: total CP = 6000/125 + 6000/60 = 80 + 100=180
SP=60*2=120
loss= Rs 60/-
Answer: D
Answer: (D)  a loss of $60
Question: The cost price of a radio is Rs.1890 and it was sold for Rs.1500, find the loss %?
(A) 18
(B) 20
(C) 77
(D) 66
(E) 41
Rationale: 1890 ---- 390
100 ---- ? => 20%
Answer: B
Answer: (B) 20
Question: John has 1,210 Dogs in his house. If exactly 40 percent of all dogs in home are tagged, what percent of the untagged dogs must be tagged so that half of all dogs in the home are tagged?
(A) 25.5
(B) 30.33
(C) 33
(D) 37
(E) 16.6
Rationale: All dogs =1210
Currently Tagged = 1210*40/100 = 484
So Untagged dogs = 1210 - 484 =726
Half of all dogs =605
The number of dogs to be tagged to make half of all dogs tagged = 605-484 = 121
So now the question remains - 121 is how much percentage of untagged dogs (726)
= 121 * 100 /726
= 16.6%
Answer (E)
Answer: (E) 16.6
Question: The unit's digit in the product 274 × 318 × 577 × 313 is
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: Solution
Required digit	= Unit digit in(4×8×7×3).
= 2.
Answer A
Answer: (A) 2
Question: Pipes A and B can fill a tank in 7 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in
(A) 1 13/17 hours
(B) 2 8/11 hours
(C) 3 9/17 hours
(D) 4 8/19 hours
(E) None
Rationale: Solution
Net part filled in 1 hour	= (1/7 + 1/6 -1/12)
= 19/84.
Therefore, the tank will be full in 84/19 hrs.	â€¹=â€º 4 8/19 hrs.
Answer D
Answer: (D) 4 8/19 hours
Question: Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
(A) 1/18
(B) 64/4032
(C) 63/64
(D) 1/9
(E) 1/4
Rationale: Explanation :
No of ways of selecting two squares from 64 squares on the chess board is 64C2.
Three cases arise:
case 1: when first square is any of the four corner ones
In this case second square can be chosen in 2 ways
no of ways of selecting two squares: 4*2 = 8
case2: when first square is any of the 24 squares on the side of the chess board other than the corner ones
the second square can be chosen in 3 ways
no of ways of selecting two squares: 24*3 = 72
case 3: the first square is any of the 36 remaining squares
the second square can be chosen in 4 ways
no of ways of selecting two squares: 36*4 = 144
Total no of ways of selecting two adjacent squares = 8 + 72 + 144 = 224
So required probability = 224/64C2 = 224/4032 = 1/18
Answer : A
Answer: (A) 1/18
Question: The length of a rectangular plot is 10mtr more than its width. The cost of fencing the plot along its perimeter at the rate of Rs. 6.5mtr is Rs.2210. The perimeter of the plot is?
(A) 126
(B) 156
(C) 340
(D) 321
(E) 260
Rationale: Sol. Let width = x, Length = (10+x)
Perimeter = 2(x+(10+x))
= 2(2x=10)
& 2(2x+10)*6.5 = 2210
X = 80
Required perimeter = 2(80+90) = 340
C
Answer: (C) 340
Question: The radius of the two circular fields is in the ratio 4: 10 the area of the second field is what percent greater than the area of the first?
(A) 72%
(B) 96%
(C) 94%
(D) 84%
(E) 82%
Rationale: r = 4 Ï€r2 = 16
r = 10 Ï€r2 = 100
100 Ï€ â€“ 84 Ï€
100 ---- ? =>84%.Answer:D
Answer: (D) 84%
Question: If y = 35p, and p is prime, what is the greatest common factor of y and 42p, in terms of p?
(A) p
(B) 2p
(C) 5p
(D) 7p
(E) p^2
Rationale: y = 35p = 5*7*p
42p = 2*3*7*p
The greatest common factor of 30p and 20p is the product of all the common prime factors, using the lower power of repeated factors. The greatest common factor is 7*p = 7p
The answer is D.
Answer: (D) 7p
Question: A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?
(A) 11
(B) 28
(C) 27
(D) 29
(E) 271
Rationale: Explanation:
One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
Answer: A) 11 days
Answer: (A) 11
Question: The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
(A) 1008
(B) 1015
(C) 1022
(D) 1032
(E) 1122
Rationale: Explanation :
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
ANSWER IS B
Answer: (B) 1015
Question: If every 2 out of 3 readymade shirts need alterations in the sleeves, and every 4 out of 5 need it in the body, how many alterations will be required for 60 shirts ?
(A) 88
(B) 123
(C) 133
(D) 143
(E) 153
Rationale: no.of Alteration required in 1 shirt is (2/3)+(4/5)+(3/4)=133/60
for 60 shirts it will 60*(133/60)=133
ANSWER:C
Answer: (C) 133
Question: I bought two books; for Rs.480. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss?
(A) 280
(B) 670
(C) 230
(D) 67
(E) 120
Rationale: A
x*(85/100) = (480 - x)119/100
x = 280
Answer: (A) 280
Question: A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
(A) 198 ways
(B) 128 ways
(C) 100 ways
(D) 108 ways
(E) 208 ways
Rationale: The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 =108 ways
Answer:D
Answer: (D) 108 ways
Question: A and B invest in a business in the ratio 3: 2. Assume that 5% of the total profit goes to charity. If A's share is Rs. 855, what is the total profit?
(A) 1200
(B) 1100
(C) 1500
(D) 1700
(E) 1600
Rationale: Assume that the total profit is x
Since 5% goes for charity, 95% of x will be divided between A and B in the ratio 3:2
Therefore, A's profit =(95x/100)×(3/5)
Given that A's share is Rs. 855. Therefore,
(95x/100)×(3/5)=855
⇒(95x/100)=855×5/3=285×5=1425
⇒x=(1425×100)/95=(285×100)/19=1500
Hence the total profit = 1500
Answer is C.
Answer: (C) 1500
Question: In a dairy farm, 50 cows eat 50 bags of husk in 50 days. In how many days one cow will eat one bag of husk ?
(A) 34 days
(B) 40 days
(C) 46 days
(D) 50 days
(E) None of these
Rationale: Explanation:
Less Cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)
[Cows1 50
Bags 50 1]::50:x
=>x∗50∗1=50∗1∗50
=>x=50
Option D
Answer: (D) 50 days
Question: Given that 2x + 13 > 5 and 5x - 13 < 7, all values of x must be between which of the following pairs of integers?
(A) -4 and -1
(B) -4 and 4
(C) -4 and 1
(D) -2 and 5
(E) 2 and 5
Rationale: 2x + 13 > 5
i.e 2x >5-13
i.e. 2x > -8
i.e. x > -4
Also, 5x - 13 < 7
i.e. 5x <7+13
i.e 5x < 20
i.e x < 4
i.e. -4 < x < 4
Answer: Option B
Answer: (B) -4 and 4
Question: If (2/3)^n=(3/2)^3, what is the value of n?
(A) -3
(B) -2
(C) -1
(D) 0
(E) 1
Rationale: (2/3)^n = (3/2)^3
= (2/3)^-3
n = -3 (Answer = A)
Answer: (A) -3
Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
(A) Rs. 2000
(B) Rs. 10,000
(C) Rs. 15,000
(D) Rs. 20,000
(E) Rs. 25,000
Rationale: Principal =(100 x 5400)/(12 x 3) = Rs. 15000.
Answer: Option C
Answer: (C) Rs. 15,000
Question: Q. How many 8-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6, 6?
(A) 180
(B) 240
(C) 210
(D) 270
(E) 300
Rationale: the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2
i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _
case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6
when the units digit is 2 i.e 1 _ _ _ _ _ 2
number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60
when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30
case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)
number of ways this can be done would be 5! / (2!) for 2 =60
and 5! / 2! for 6 ... = 60
adding all of these ... gives the answer 270 ...
the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat
D
Answer: (D) 270
Question: If a and b are odd integers, which of the following is an even integer?
(A) 3(2a+b)
(B) 3(2a+b)+2a
(C) 3(2a+b)+a+b
(D) 3(2a+b)+7a-b
(E) 3(2a+b)+ab
Rationale: 3(2a+b)+ab=odd(odd)+odd=odd+odd=even
The answer is E.
Answer: (E) 3(2a+b)+ab
Question: If the probability of Sita getting selected to a school is 20% and she applied to 3 schools. What is the probability that she will get selected in at least one school?
(A) 24/125
(B) 21/125
(C) 61/125
(D) 12/125
(E) 18/125
Rationale: The probability that she will be selected by at least one school equals to the sum of the probabilities of the following three events:
1. She is selected by only one school: P(SRR)=3!/2!∗1/5∗4/5∗4/5=48/125 (S stands for selected and R stands for rejected). We are multiplying by 3!/2! since SRR scenario can occur in several ways: SRR, RSR, RRS, (so 3!/2! is # of permutations of 3 letters SRR out of which 2 R's are identical);
2. She is selected by only two school: P(SSR)=3!/2!∗1/5∗1/5∗4/5=12/125, the same reason of multiplying by 3!/2!;
3. She is selected by all three school: P(SSS)=(15)^3=1125, we are not multiplying this time since SSS can occur only in one way.
So P=48/125+12/125+1/125=61/125
Answer: C
Answer: (C) 61/125
Question: Find the largest number of four digits which is exactly divisible by 27,18,12,15
(A) 9700
(B) 9710
(C) 9720
(D) 9520
(E) 7500
Rationale: Explanation:
LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720
Option C
Answer: (C) 9720
Question: Julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest. After two years she earned $120 and $124 from the simple interest account and the compound interest account respectively. If the interest rates for both accounts were the same, what was the amount of Julie's initial savings?
(A) 600
(B) 720
(C) 1080
(D) 1200
(E) 1800
Rationale: $120 for 2 years = $60 per year.
Extra $4 yearned with the compound interest is the percent yearned on percent. So, $4 is yearned on $60, which means that the interest = 15%.
This on the other hand means that half of the savings = 60*15 = $900. Twice of that = $1,800.
Answer: E.
Answer: (E) 1800
Question: A can do a piece of work in 12 days. He worked for 15 days and then B completed the remaining work in 10 days. Both of them together will finish it in.
(A) 12 1/2 days
(B) 5 days
(C) 8.5 days
(D) 2 days
(E) 11.5 days
Rationale: Answer: Option A
15/25 + 10/x = 1 => x = 25
1/25 + 1/25 = 2/25
25/2 = 12 1/2 days
Answer: (A) 12 1/2 days
Question: A can do a piece of work in 30 days. He works at it for 5 days and then B finishes it in 20 days. In what time can A and B together it?
(A) 13 1/0
(B) 13 1/3
(C) 13 1/2
(D) 13 1/1
(E) 13 1/7
Rationale: 5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days
Answer:B
Answer: (B) 13 1/3
Question: Students at a school were on average 180 cm tall. The average female height was 170 cm, and the average male height was 181 cms. What was the ratio of men to women?
(A)  5:2
(B)  5:1
(C)  4:3
(D)  4:1
(E)  10:1
Rationale: We're given a few facts to work with:
1) The average height of the females is 170cm
2) The average height of the males is 181cm
3) The average of the GROUP is 180cm
We're asked for the RATIO of men to women.
W = number of Women
M = number of Men
(170W + 181M)/(W+M) = 180
170W + 181M = 180W + 180M
1M = 10W
M/W = 10/1
The ratio of men to women is 10 to 1.
E
Answer: (E)  10:1
Question: The greatest common factor of two numbers is 4!. Which of the following can be the second number, if one of the numbers is 8!?
(A) 3(5!)
(B) 4(5!)
(C) 5(5!)
(D) 6(5!)
(E) 4(4!)
Rationale: GCF is the product of common factors of the numbers involved.
GCF = 4!
a = 8! = 8*7*6*5*4!
b will certainly have 5! and cannot have any more common factors with a (as this will increase the GCF)
Looking at the answers only 4 (4!) and 8! will have GCF as 4!
Ans E
Answer: (E) 4(4!)
Question: Thomas bought a pack of eleven computers for Rs. 500000 and sold it for Rs.550000. What is the % of profit he is making per computer?
(A) 10
(B) 14
(C) 19
(D) 21
(E) 25
Rationale: The cost of buying a single computer= Rs(500000/11)=Rs. 45454.5
The cost of selling a selling computer=Rs(550000/11)=Rs. 50000.0
Profit per computer=Rs.(50000-45454.5)=Rs. 4545.5
The % of profit in selling a single computer=(4545.5/45454.5)*100=10%
Answer:A
Answer: (A) 10
Question: Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for 1 1/2 years.
(A) 306.08
(B) 306.03
(C) 306.04
(D) 306.01
(E) 306.09
Rationale: Explanation:
A = 5000(51/50)3
= 5306.04
5000
-----------
306.04
Answer:C
Answer: (C) 306.04
Question: Convert 600 miles into meters?
(A) 784596
(B) 845796
(C) 965604
(D) 784596
(E) 864520
Rationale: 1 mile = 1609.34 meters
600mile = 600*1609.34 = 965604 meters
Answer is C
Answer: (C) 965604
Question: Insert the missing number.
3, 7, 6, 5, 9, 3, 12, 1, 15, (....)
(A) 3
(B) 7
(C) -1
(D) 8
(E) -45
Rationale: Explanation:
There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.
The next number is 1 - 2 = -1.
Answer: C) -1
Answer: (C) -1
Question: If BOMBAY is written as MYMYMY, how will TAMIL NADU be written in that code?
(A) TIATIATIA
(B) IATIATIAT
(C) MNUMNUMNU
(D) ALDALDALD
(E) None of these
Rationale: Letters at multiples of 3 is repeated in the code.
So,
In the code BOMBAY, 3rd and 6th letter repeated for the code.
Like that, in the code TAMIL NADU, 3rd 6th and 9th letter must be repeated.
So, TAMILNADU = MNUMNUMNU
ANSWER:C
Answer: (C) MNUMNUMNU
Question: An bus covers a certain distance at aspeed of 240 kmph in 5 hours. To cover the same distance in 1hr, it must travel at a speed of?
(A) 700 km/hr
(B) 650 km/hr
(C) 690 km/hr
(D) 710 km/hr
(E) 720 km/hr
Rationale: Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed = 1200 x 3 km/hr = 720 km/hr.
5
E
Answer: (E) 720 km/hr
Question: What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
(A) 128
(B) 277
(C) 120
(D) 888
(E) 171
Rationale: Let the present worth be Rs. x. Then,
S.I. = (132 - x)
(x * 5 * 2) / 100 = (132 - x)
10x = 13200 - 100x
110x = 13200 => x = 120.
Answer: C
Answer: (C) 120
Question: A man can row downstream at the rate of 26 Kmph and upstream at 12 Kmph. Find the man’s rate in still water and rate of current?
(A) 8.5
(B) 8.0
(C) 9.5
(D) 7.0
(E) 8.25
Rationale: Rate of still water = 1/2 (down stream + upstream)
= 1/2 (26 + 12) = 19 Kmph
rate of current = 1/2 (down stream - upstream)
= 1/2 (26 - 12)
= 1/2 (14)
= 7 kmph
Answer is D.
Answer: (D) 7.0
Question: A car is running at a speed of 120kmph. What distance will it cover in 12sec?
(A) 396m
(B) 267m
(C) 180m
(D) 200m
(E) 250m
Rationale: Speed = 120kmph = 120*5/18 = 33 m/s
Distance covered in 12sec = 33*12 = 396m
Answer is A
Answer: (A) 396m
Question: The average of 11 results is 48, if the average of first six results is 49 and that of the last six is 52. Find the sixth result?
(A) 21
(B) 56
(C) 78
(D) 25
(E) 23
Rationale: 1 to 11 = 11 * 48 = 528
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 â€“ 528 = 78
Answer:C
Answer: (C) 78
Question: Tom opened a shop investing Rs.30000. Jose joined him 2 months later, investing Rs.45000. They earned a profit of Rs.54000 after completion of one year. What will be Jose's share of profit?
(A) 12000
(B) 23000
(C) 24000
(D) 25000
(E) 30000
Rationale: SoL = ~s- So Anju’s share = [5/9] x 54000 = 30000
E
Answer: (E) 30000
Question: A trader mixes 25 kg of rice at Rs. 20 per kg with 25 kg of rice of other variety at Rs. 30 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
(A) No profit, no loss
(B) 5%
(C) 8%
(D) 20%
(E) None of these
Rationale: C.P. of 50 kg rice = Rs. (25 x 20 + 25 x 30) = Rs. (500 + 750) = Rs. 1250
S.P. of 50 kg rice = Rs. (50 x 30) = Rs. 1500
Gain =250/1250x100% = 20%.
Answer: Option D
Answer: (D) 20%