divelab/Sys2Bench huggingface dataset:
Question: Spanish Language Broadcast records last 90 min on each of two sides. If it takes 3 hours to translate one hour of broadcast, how long will it take to translate 16 full records?
Options:
A)144 hours
B)124 hours
C)134 hours
D)154 hours
E)164 hours
Rationale: Records last 90 min on each of 2 sides,
==> record last 90*2 = 180 min = 3 hours
16 full records--> 16*3 = 48 hour broadcast
Given, 3 hours to translate 1 hour of broadcast
Let x be the time required to translate 48 hour broadcast(16 full records)
x = 48*3
=144 hours
ANSWER:A
Answer: A
Question: In a race where 12 cars are running, the chance that car X will win is 1/6, that Y will win is 1/10 and that Z will win is 1/8. Assuming that a dead heat is impossible, find the chance that one of them will win.
Options:
A)1/140
B)1/180
C)27/410
D)47/120
E)57/120
Rationale: Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive).
= 1/6 + 1/10 + 1/8
= 47/120
ANSWER:D
Answer: D
Question: I remember during the school days , the teacher asked the class "can you tell me the sum of the first 50 odd numbers ?".
I ran quickly to the teacher and told her "the answer is 2500". The teacher replied "lucky guess".
She then asked me "can you tell me the sum of first 71 odd numbers ?".
I wait for approx 10 seconds and replied with the correct answer.
How can i answer so quickly and whats the correct answer ?
Options:
A)8715
B)0152
C)3581
D)5041
E)9126
Rationale: D
5041
n^1
71 * 71 = 5625 (sum of first 71 odd numbers).
50 * 50 = 2500 (sum of first 50 odd numbers).
Answer: D
Question: A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for 5 days and the rest of the work is done by C in 2 days. If they get 450 for the whole work, how should they divide the money ?
Options:
A)225, 150, 75
B)250, 100, 100
C)200, 150, 100
D)175, 175, 100
E)None of these
Rationale: Work done by A and B in 5 days = (1⁄10 + 1⁄15) × 5 = 5⁄6
Work remaining = 1 - 5⁄6 = 1⁄6
∴ C alone can do the work in 6 × 2 = 12 days
Ratio of their share work = 5⁄10 : 5⁄15 : 2⁄12 = 3 : 2 : 1
Share of wages = 225, 150, 75
Answer A
Answer: A
Question: The average of 8 numbers is 25, that of the first 2 is 20 and of the next three is 26. The sixth number is less than the seventh by 4, and less than the eighth by 6. The last number is
Options:
A)30
B)32
C)40
D)36
E)38
Rationale: Explanation :
Sum of all the numbers = 25*8 = 200
Sum of first two numbers = 40
Sum of next three numbers(i.e., 3rd, 4th and 5th) = 78
So, sum of first five numbers = 118 and sum of 6th, 7th and 8th number = 82
Assuming sixth number to be x, we get 7th number = x+4 and 8th number = x+6.
So, x + (x+4) + (x+6) = 82
=> x = 24.
Hence, last number(8th number) = 30.
Answer : A
Answer: A
Question: In a hostel there were 100 students. To accommodate 20 more students the average is decreased by rupees 5. But total expenditure increased by Rs.400. Find the total expenditure of the hostel now?
Options:
A)54888
B)2777
C)5400
D)1278
E)2881
Rationale: 100x + 400 = 12(x – 5)
x = 50
100 * 50 + 400 = 5400
Answer: C
Answer: C
Question: Find number which is 80% less than 80.
Options:
A)18
B)22
C)28
D)26
E)16
Rationale: Explanation:
80% less is 20% of the given number
Therefore, 20% of 80 is 16.
ANSWER: E
Answer: E
Question: The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table?
Options:
A)2266
B)2987
C)2678
D)6725
E)2677
Rationale: CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.
Answer: D
Answer: D
Question: By using the numbers 1, 3, 4, 6 and 9 only once, how many five digit numbers can be made that are divisible by 8?
Options:
A)3
B)12
C)6
D)18
E)42
Rationale: A number to be divisible by 8 must end with 08, 16, 24,32 ,40 or 72.
So, there are the following cases possible:
349-16 --> the first three digits can be arranged in 3=6 ways.
Total = 6 = 6.
Answer: C.
Answer: C
Question: In certain year in Country C, x sets of twins and y sets of triplets were born. If there were z total babies born in Country C in this year, and x and y were both greater than 0, which of the following represents the fraction Q of all babies born who were NOT part of a set of twins or triplets?
Options:
A)z - 2x - 3y
B)(2x+3y)/z
C)(x+y)/z
D)1 - (2x+3y)/z
E)1 - (x+y)/z
Rationale: X set of twins implies 2x babies of twin category.
Y set of triplets means 3y babies of triplet category.
Let K be the babies other than twin or triplet category.
Total babies = z = 2x+3y+k.
or K=Z-(2x+3y)
=> Fraction of babies other than twin or triplet categories = K/Z = Z-(2x+3y)/Z.
Q= 1- (2x+3y)/Z. Hence answer is D.
Answer: D
Question: Diane find 1 can of paint are just enough to paint one third of her room. How many more cans of paint will she need to finish her room and paint a second room of the same size?
Options:
A)5
B)7 and a half
C)10
D)12 and a half
E)15
Rationale: She will need 2 cans to paint the rest of this room and 3 for the next room for a total of 5 cans.
A
Answer: A
Question: The sum of two numbers is 216 and their H.C.F. is 27. The number are:
Options:
A)27, 189
B)81,189,
C)108, 108
D)154, 162
E)none of these
Rationale: Let the required numbers be 27a and 27b.Then, 27a+27b=216
a+b=8
Now, co-primes with sum 8 are(1, 7)and(3,5).
required numbers are (27*1*, 27*7) and (27*3,27*5)
i.e., (27, 189)and (81,135)
out of these, the given one in the answer is the pair(27,189).
correct Option:A
Answer: A
Question: A committee has 6 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee?
Options:
A)127
B)300
C)882
D)272
E)323
Rationale: The number of ways to select two men and three women
= 6C₂ * ⁶C₃
= (6*5)/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 300
Answer: B
Answer: B
Question: The sum E of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is give by En = (n/2)*(2a + n-1)d. What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.
Options:
A)4345
B)4302
C)4258
D)4214
E)4170
Rationale: a method to do it under one minutes is to take advantage of the choices given...
lets work on the last digit as we have different units digit in each choice...
total sum of 1 to 100 inclusive will have 0 as the last digit..
this is so because the resultant will be 10*(sum of all single digits)... and since we are multiplying by 10,units digit will be 0...
now for single digit in sum E of even number from 25 to 63..
25 to 65 will have 4 times sum of single digit even int, 4*(2+4+6+8+0)=4*20..
here too the units digit is 0, but 64 has to be excluded from the total..
two ways from here on..
1) we are subtracting 0 from 0
so units digit should be 0, but we have to add 64..
so last/units digit =4..
2)we subtract 64 from sum of even int..
so units digit=80-4=76..
or units digit =6...
so our answer should have units digit as 10-6=4..
only D has 4 as units digit..
ans D
Answer: D
Question: In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?
Options:
A)24
B)48
C)56
D)72
E)96
Rationale: These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.
Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
ANSWER:D
Answer: D
Question: The original price of a car was $25,200. Because the car owner thought he could get more money for the car, he increased the price of the car to 110% of its original price. After a week, the car had not sold, so the owner then discount ed the price by 10%, and the car was finally sold. What price was the car sold for?
Options:
A)$25,200
B)$25,000
C)$24,948
D)$24,542
E)$23,658
Rationale: The car is sold for 1.1*0.9*25200
=0.99 * 25200
=(1 - 0.01) * 25200
=25200 - 252
=24948
C is the answer
Answer: C
Question: Each student at a certain university is given a four-character identification code, the rest two characters of which are digits between 0 and 9, inclusive, and the last two characters of which are selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?
Options:
A)135,200
B)67,600
C)64,000
D)60,840
E)58,500
Rationale: First two parts for the code, are digits between 0-9, therefore, 10 options for the first part of the code,
and as characters may be repeated, 10 options for the second part as well
Therefore, we have 10 X 10 possibilities for the first and second part of the code
Last two parts of the code, are characters selected from the 26 letters of the alphabet, therefore, 26 options for the third part of the code,
and as characters may be repeated, 26 options for the fourth part as well
Therefore, we have 26 X 26 possibilities for the third and fourth part of the code
so, in all total no. of different identification codes generated following these rules
= 10 X 10 X 26 X 26 = 67600
Answer B
Answer: B
Question: Pipe A can fill a tank in 2 hr, pipe B in 4hr and pipe C in 8hr. If all the pipes are open in how many hours will tank be filled?
Options:
A)1hr 27min
B)2hr 19min
C)3hr 14min
D)3hr
E)5hr 45min
Rationale: Part filled by (A+B+C) in 1hour = 1/2 + 1/4 + 1/8 = 7/8
All the 3 pipes together will fill the tank in 1hour 27min
Answer is A
Answer: A
Question: Find large number from below question The difference of two numbers is 1375. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder
Options:
A)1235
B)1345
C)1678
D)1767
E)1647
Rationale: Let the smaller number be x. Then larger number = (x + 1375).
x + 1375 = 6x + 15
5x = 1360
x = 272
Large number = 272+1375 = 1647
E
Answer: E
Question: The H.C.F of two numbers is 12 and their L.C.M is 6600. If one of the numbers is 288, then the other is?
Options:
A)255
B)260
C)265
D)270
E)275
Rationale: Other number = (12 * 6600)/288
= 275.
Answer: E
Answer: E
Question: Varma can read a book in k minutes. What part of the book can he read in 8 minutes?(k>8)
Options:
A)(k - 8)/k
B)k/8
C)8 + k
D)k-8
E)8/k
Rationale: Option E
Explanation:
Part of the book he can read in 1 minute = 1/k
Part of the book he can read in 8 minutes =8/k.
Answer: E
Question: If Rs. 527 be divided among A, B, C in such a way that A gets 2/3 of what B gets and B gets 1/4 of what C gets, then their shares are respectively?
Options:
A)s.300
B)s.360
C)s.389
D)s.368
E)s.372
Rationale: (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4
A:B = 2:3 and B:C = 1:4 = 3:12
A:B:C = 2:3:12
A;s share = 527 * 2/17 = Rs. 62
B's share = 527 * 3/17 = Rs. 93
C's share = 527 * 12/17 =Rs.372.
Answer:E
Answer: E
Question: If six numbers are randomly selected without replacement from set {1, 2, 3, 4, 5, 6}, what is the probability that the numbers are selected in ascending order?
Options:
A)1/280
B)1/640
C)1/720
D)1/240
E)1/120
Rationale: The total number of ways to arrange 6 numbers is 6! = 720.
There is only one way to arrange the numbers in ascending order.
P(ascending order) = 1/720
The answer is C.
Answer: C
Question: A man took loan from a bank at the rate of 12% p.a. S.I. After 20 years he had to pay Rs. 2100 interest only for the period. The principal amount borrowed by him was?
Options:
A)600
B)715
C)600
D)875
E)900
Rationale: Principal = (100 * 2100) / (12 * 20)
= Rs.875
Answer: D
Answer: D
Question: In a camp,there is a meal for 120 men or 200 children.If 150 children have taken the meal,how many men will be catered to with the remaining meal ?
Options:
A)20
B)30
C)40
D)50
E)60
Rationale: Solution
There is a metal for 200 children. 150 children have taken the meal.
Remaining meal is to be catered to 50 children.
Now, 200 children = 120 men
50 children =( 120/200x50)men= 30 men. Answer B
Answer: B
Question: The ratio of men to women in a class is 6 to 5. If 3 men drop out of the class and the ratio of men to women after the drop out is 1 to 1, then how many men are currently in the class?
Options:
A)13
B)14
C)15
D)18
E)22
Rationale: Equation on the basis of question
6x-3/5x=1/1
x=3
Original number of men = 6*3=18
Original Number of Women = 5*3=15
Current number of men = 18-3=15 (Answer)
Current number of women = 15
Ratio = 15/15= 1/1 ( Just to verify)
Ans C
Answer: C
Question: 40% of a number is added to 120, the result is the same number. Find the number?
Options:
A)300
B)277
C)266
D)200
E)121
Rationale: :
(40/100) * X + 120 = X
2X = 400
X = 200
Answer: D
Answer: D
Question: In a shop, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, find out approximately what percentage of the selling price is the profit?
Options:
A)180 %
B)120 %
C)90 %
D)80 %
E)70 %
Rationale: Let the CP = 100
Profit=(320 / 100) ×100=320
SP=CP+Profit=100+320=420
If the cost increases by 25%, New CP=(125/100) ×100=125
Selling Price is constant, hence New SP=420
Profit=SP–CP=420–125=295
Required Percentage =(295/420)×100
=2950 / 42
=1475 / 21 ≈ 70
ANSWER:E
Answer: E
Question: Two cars are driving toward each other. The first car is traveling at a speed of 60 km/h, which is 25% slower than the second car's speed. If the distance between the cars is 770 km, how many hours will it take until the two cars meet?
Options:
A)4
B)4.5
C)5
D)5.5
E)6
Rationale: The speed of the first car is 60 km/h.
The speed of the second car is 60 / 0.75 = 80 km/h.
The two cars complete a total of 140 km each hour.
The time it takes the cars to meet is 770 / 140 = 5.5 hours.
The answer is D.
Answer: D
Question: Aishwarya Rai walks into a bank to cash out her check.
By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars.
On the way home she spends 5 cent, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount ?
Options:
A)21.55
B)25.89
C)30.56
D)31.55
E)31.63
Rationale: E
The check was for dollars 31.63.
The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.
Let x be the dollars of the check, and y be the cent.
The check was for 100x + y cent
He was given 100y + x cent
Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
or 199x ≡ -5 (mod 98)
or 98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
this quickly leads to x = 31
Answer: E
Question: Insert the missing number. 2, 7, 10, 22, 18, 37, 26
Options:
A)42
B)52
C)46
D)62
E)68
Rationale: Explanation :
There are two series here
2, 10, 18, 26, ... (Increase by 8)
7, 22, 37, ... (Increase by 15)
Hence, next term is 37+15 = 52
Answer : Option B
Answer: B
Question: Working at a constant rate, P can finish a job in 4 hours. Q, also working at a constant rate, can finish the same job in 15 hours. If they work together for 3 hours, how many more minutes will it take P to finish the job, working alone at his constant rate?
Options:
A)5
B)8
C)10
D)12
E)15
Rationale: Each hour they complete 1/4 + 1/15 = 19/60 of the job.
In 3 hours, they complete 3(19/60) = 19/20 of the job.
The time for P to finish is (1/20) / (1/4) = (1/5) hour = 12 minutes
The answer is D.
Answer: D
Question: 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length if the wire in meters will be:
Options:
A)76 m
B)80 m
C)84 m
D)88 m
E)None of these
Rationale: Explanation:
Let the length of the wire be h
Radius=1/2mm=120cm
πr2h=66
22/7∗1/20∗1/20∗h=66
=>h=66∗20∗20∗7/22=8400cm=84m
Option C
Answer: C
Question: What is the sum of the first 23 natural numbers (starting from 1)?
Options:
A)256
B)266
C)276
D)286
E)296
Rationale: The sum of n natural numbers = n*(n+1) / 2 = 23*24/2 = 276
The answer is C.
Answer: C
Question: If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by
Options:
A)10%
B)20%
C)25%
D)30%
E)None
Rationale: Solution
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 6 = x × 7.50
x = 80 kg.
Reduction in consumption = 20%.
Answer B
Answer: B
Question: Of 40 applicants for a job, 32 had at least 5 years of prior work experience, 24 had advanced degrees, and 16 had at least 5 years of prior work experience and advanced degrees. How many of the applicants had neither 5 years of prior work experience nor advanced degrees?
Options:
A)0
B)2
C)4
D)8
E)16
Rationale: soln:
The 16 with both experience and degrees are the first group;
we can subtract the 16 from the 24 with advanced degrees to determine that 8 have advanced degrees but not 5 years of experience,
we can subtract the 16 from the 32 with experience to determine that there are 16 applicants with 5 years of experience but no advanced degrees
We can then subtract these three groups from the total pool of 40 applicants to determine how many lack both
experience and advanced degrees; 40-16-16-8=0,
so there were no such applicants among the 40
ANSWER:A
Answer: A
Question: What is the ratio whose term differ by 30 and the measure of which is 2/7?
Options:
A)A)12:42
B)B)16:56
C)C)71:85
D)D)32:39
E)E)41:52
Rationale: Let the ratio be x:(x+30)
Then, x/(x+30) = 2/7
x=12
required ratio = 12:42
Answer is A
Answer: A
Question: There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 530. The number of hares in the zoo is?
Options:
A)90
B)99
C)77
D)65
E)221
Rationale: 200*2 = 400
530
-----
130
1----2
?----130 = 65
Answer:D
Answer: D
Question: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
Options:
A)Rs. 20
B)Rs. 60.
C)Rs. 40
D)Rs. 30
E)Rs. 90
Rationale: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
C.P. of 1 ball = Rs. 720/12 = Rs. 60.
Answer:B
Answer: B
Question: What sum of money put at C.I amounts in 2 years to Rs.8840 and in 3 years to Rs.9261?
Options:
A)8054
B)8877
C)2877
D)2678
E)1011
Rationale: 8840 ---- 421
100 ---- ? => 4.76%
x *104.76/100 * 104.76/100 = 8840
x*1.0975=8840
x=8840/1.0975 => 8054.54
Answer: A
Answer: A
Question: The population of a bacteria colony doubles every day. If it was started 3 days ago with 3 bacteria and each bacteria lives for 12 days, how large is the colony today?
Options:
A)512
B)1024
C)768
D)2048
E)4096
Rationale: 9 days ago- 2
8 days ago- 4
7 days ago- 8
6 days ago- 16
5 days ago- 32
4 days ago- 64
3 days ago- 128
2 days ago- 256
Yesterday- 512
Today- 1024
Answer:B
Answer: B
Question: Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveler sets out from city X in the same direction as the first traveler and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveler be ahead of the first?
Options:
A)2 days
B)6 days
C)From the 2nd day after the 2nd traveler starts
D)From the 3rd day after the 2nd traveler starts
E)None
Rationale: Explanatory Answer
Travel Pattern of Traveller 2
Day 1: 1 km; Day 2: (1 + 2) = 3 km; Day 3: (1 + 2 + 2) = 5km and so on
i.e. on each of these days he covers the following distance Day 1: 1km, Day 2: 3km, Day 3: 5km and so on.
The distances covered each day by the first traveller is in an AP with the first term being 1 and the common difference being 2
After 3 days, the first traveler would have traveled 9 kms. He is ahead of the second traveler at the end of 3 days by 9 kms.
Travel Pattern of Traveller 2
The second traveler travels as follows: On his first day he travels 12 kms, his second day he travels 13 kms and so on.
The distances covered each day by the second traveller is in an AP with the first term being 12 and the common difference being 1
Step 2
Compute total distance covered by each of the travellers
Let the number of days travelled by the second traveller be n.
Therefore, the first traveller would have travelled for (n + 3) days. Traveler 1 has covered a total distance of 1+ 3 + 5 + 7 + ..... + (n + 3) km.
The total distance covered by traveler 1 is the summation of the AP.
Sum of an AP = (n/2)(2a1 + (n-1)d)
Total distance travelled by traveller 1 in (n + 3) days = n+3/2[2(1)+(n+3−1)(2)]
The above term simplifies as (n + 3)2
The total distance covered by the second traveler = 12 + 13 + 14 + ... + n terms
Total distance travelled by traveller 2 in n days = n2[2(12)+(n−1)(1)] = n(n+23)2
On the days that the second traveller is ahead of the first one, the total distance covered by the second traveller till that day will be more than the total distance covered by the first traveller.
Note: All values calculated are with respect to the day from which the second traveller started travelling
Therefore, let us solve the inequality n(n+23)/2 > (n + 3)2
n2 + 23n > 2(n2 + 6n + 9)
Or n2 - 11n + 18 < 0
The quadratic factorizes as (n - 2)(n - 9) < 0
Possibility 1: (n - 2) > 0 AND (n - 9) < 0
i.e., n > 2 AND n < 9
Therefore, 2 < n < 9
Possibility 2: (n - 2) < 0 AND (n - 9) > 0
i.e., n < 2 AND n > 9
Infeasible solution
Therefore, starting the 3rd day and up to the 8th day the second traveller was ahead of the first traveller
The correct answer is Choice (2). 6 days
Answer B
Answer: B
Question: After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
Options:
A)2 years
B)4 years
C)8 years
D)15 years
E)17 years
Rationale: Solution
Age decreased	= (5 x 3) years
15 years
So, required difference =15years.
Answer D
Answer: D
Question: At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. The 3 restaurants are the family buffet, the dining room, and the snack bar. Exactly 19 employees are trained to work in the family buffet, 16 are trained to work in the dining room, and 12 are trained to work in the snack bar. If 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants?
Options:
A)2
B)3
C)4
D)5
E)6
Rationale: 39 = 19 + 16 + 12 - 4 - 2x
2x = 19 + 16 + 12 - 4 - 39
= 43- 39
=4
x = 2
A
Answer: A
Question: If log1087.5 = 5.9421, then the number of digits in (875)10 is?
Options:
A)30
B)70
C)27
D)26
E)25
Rationale: X = (875)10 = (87.5 x 10)10
Therefore, log10X = 10(log1087.5 + 1)
= 10(5.9421 + 1)
= 10(6.9421) = 69.421
X = antilog(69.421)
Therefore, number of digits in X = 70.
ANSWER:B
Answer: B
Question: How many integers between 323,700 and 458,600 have tens digit 1 and units digit 3?
Options:
A)10,300
B)10,030
C)1,353
D)1,352
E)1,349
Rationale: There is one number in hundred with 1 in the tens digit and 3 in the units digit:13, 113, 213, 313, ...
The difference between 323,700 and 458,600 is 458,600-323,700=134,900 - one number per each hundred gives 134,900/100=1,349 numbers.
Answer: E.
Answer: E
Question: The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?
Options:
A)1:96
B)1:98
C)1:95
D)1:45
E)1:92
Rationale: Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 2
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x =1:96.
Answer: A
Answer: A
Question: A man invested Rs. 4940 in Rs. 10 shares quoted at Rs. 9.50. If the rate of dividend be 14%, his annual income is
Options:
A)708
B)718
C)738
D)748
E)728
Rationale: Market Value of a share = Rs.9.50
Investment = Rs.4940
Number of shares = 4940/9.50 = 520
Face Value of a share = Rs.10
dividend = 14%
dividend per share = (10×14)/100 = Rs. 1.4
His annual income = 520 × 1.4 = Rs.728
Answer is E.
Answer: E