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A collection of questions that typically appear in the Common Admission Test (CAT) from the topic Number Theory.

These questions will guide you through your CAT and other MBA entrance exam preparation.

1. When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

2. If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of ‘a’?

3. What is the remainder when 9^1 + 9^2 + 9^3 + …. + 9^8 is divided by 6?

2. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

4. ‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse?

5. Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

6. What is the reminder when 91 + 92 + 93 + …… + 99 is divided by 6?

7. For what value of ‘n’ will the remainder of 351^n and 352^n be the same when divided by 7?

8. How many keystrokes are needed to type numbers from 1 to 1000?

9. When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

10. Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.

11.Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

12. Find the value of 1.1! + 2.2! + 3.3! + ……+n.n!

13. Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

14. When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

15. A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

16. How many different factors are there for the number 48, excluding 1 and 48?

17. How many zeros contained in 100!?
18. Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

19. What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

20. What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?

Solve the following and check with the answers given at the end.

Q 1 . A, B, and C went to have a meal in the hotel, A has paid 50% more than the amount that B has paid, and C has paid 5/6 of that of A has paid. Also B has paid $2 more than that of C has paid. Then what is the total amount that all the three paid for their meal?

Q 2. If the cost of an article is x, first discount given is y% of cost, second discount given is z% of cost. The selling price of x is…?

Q 3.. In june a baseball team that played 60 games had won 30% of its game played. After a phenomenal winning streak this team raised its average to 50% .How many games must the team have won in a row to attain this average?

4. The price of a product is reduced by 30% . By what percentage should it be increased to make it 100% ?

5. If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain?

6. If on an item a company gives 25% discount, they earn 25% profit. If they now give 10% discount then what is the profit percentage?

7. If an item costs Rs.3 in ‘99 and Rs.203 in ‘00.What is the % increase in price?

8. A student gets 70% in one subject, 80% in the other. To get an overall of 75% how much should get in third subject?

9. The cost of an item is Rs 12.60. If the profit is 10% over selling price what is the selling price ?

10. A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs._______ for the horse and Rs.________ for the cart.

ANSWERS

Ans 1. $30
Ans 2. x (1-y/100)(1-z/100)
Ans 3. 24
Ans 4. 42.857%
Ans 5. 44%
Ans 6. 30%
Ans 7. 200/3 %
Ans 8. 75%
Ans 9. Rs 13.86/-
Ans 10. Cost price of horse = Rs. 400 & the cost price of cart = 200.

Questions along with illustrative explanations

Question:
When processing flower-nectar into honeybees’ extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?

(1) 1.5 kgs
(2) 1.7 kgs
(3) 3.33 kgs
(4) None of these

Correct Answer - (2)

Solution:
Flower-nectar contains 50% of non-water part.
In honey this non-water part constitutes 85% (100-15).
Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg
Therefore amount of flower-nectar needed = (0.85/0.5) * 1kg = 1.7 kg.
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Question:
A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?

(1) 17
(2) 23
(3) 77
(4) None of these

Correct Answer - (2)

Solution:
Let the number of apples be 100.
On the first day he sells 60% apples ie.,60 apples.Remaining apples =40.
He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has 40-6 = 34 apples
The next day he throws 50% of the remaining 34 apples i.e., 17.
Therefore in all he throws 6+17 =23 apples.

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Question:
If the cost price of 20 articles is equal to the selling price of 16 articles, What is the percentage of profit or loss that the merchant makes?

(1) 20% Profit
(2) 25% Loss
(3) 25% Profit
(4) 33.33% Loss

Correct Answer - (3)

Solution:
Let Cost price of 1 article be Re.1.
Therefore, Cost price of 20 articles = Rs. 20.

Selling price of 16 articles = Rs. 20
Therefore, Selling price of 20 articles = (20/16) * 20 = 25

Profit = Selling price - Cost price
= 25 - 20 = 5

Percentage of profit = Profit / Cost price * 100.
= 5 / 20 * 100 = 25% Profit
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Question:
Peter got 30% of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What was the passing marks in the examination?

(1) 35
(2) 250
(3) 75
(4) 85

Correct Answer - (4)

Solution:
Let ‘x’ be the maximum marks in the examination.

Therefore, Peter got 30% of x = 0.3x
And Paul got 40% of x = = 0.4x.
In terms of the maximum marks Paul got 0.4x - 0.3x = 0.1x more than Peter. —— (1)
The problem however, states that Paul got 15 marks more than the passing mark and Peter got 10 marks less than the passing mark.
Therefore, Paul has got 15 + 10 = 25 marks more than Peter. —— (2)

Equating (1) and (2), we get
0.1x = 25 => x = 250
‘x’ is the maximum mark and is equal to 250 marks.

We know that Peter got 30% of the maximum marks. Therefore, Peter got = 75 marks.
We also know that Peter got 10 marks less than the passing mark. Therefore, the passing mark will be 10 marks more than what Peter got = 75 + 10 = 85.

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Question:
If the price of petrol increases by 25% and Raj intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
1. 10%
2. 12%
3. 8%
4. 6.67%
Correct Answer - 8%. Choice (3)

Solution:
Let the price of 1 litre of petrol be Rs.x and let Raj initially buy ‘y’ litres of petrol.
Therefore, he would have spent Rs. xy on petrol.

When the price of petrol increases by 25%, the new price per litre of petrol is 1.25x.

Raj intends to increase the amount he spends on petrol by 15%.
i.e., he is willing to spend xy + 15% of xy = 1.15xy

Let the new quantity of petrol that he can get be ‘q’.
Then, 1.25x * q = 1.15xy
Or q = 0.92y.

As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what he used to get earlier.
Or a reduction of 8%.

Questions along with respective explanations.

Q.1. Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260 kms?

(1) 60 km/hr (2) 90 km/hr (3) 120 km/hr (4) 135 km/hr

Correct Answer - (3)

Solution:
All three trains meet at the same time between Delhi and Mumbai. Which means Train A and Train B are at the same point at that time. This will happen when Train B is overtaking Train A.

Train A starts 3 hours before Train B. Therefore, by the time Train B leaves Mumbai, Train A has covered 3 * 60 = 180 kms.

The relative speed between Train A and Train B = 90 - 60 = 30 kmph. Therefore, Train B will overtake Train A in = 6 hours from the time Train B leaves Mumbai. That is at 3 A.M, Train B will overtake Train A. The point between Mumbai and Delhi at which Train B overtakes Train A will be 6*90=540 kms from Mumbai.

Train C will also be at that point at 3 A.M while Train B is overtaking Train A. And Train C would have travelled 1260-540 = 720 kms in these 6 hours. Therefore, the speed of Train C = 120 km/hr.

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Q.2 A train travels at an average speed of 90 km/hr without any stoppages. However, its average speed decrease to 60km/hr on account of stoppages. On an average, how many minutes per hour does the train stop?

(1) 12 minutes (2) 18 minutes (3) 24 minutes (4) 20 minutes

Correct Answer - (4)

Solution:
If it travelled at 90 km / hr, it would have crossed 90 kms in an hour. However, it covered only 60 kms due to stoppages.

The distance it covered decreased by 1/3 or it covered only 2/3rd of the distance that it can cover for which the traveling time would have been 2/3rd of an hour. The remaining 1/3rd of an hour was spent in stoppages. Therefore, the train stops on an average for 20 minutes every hour.
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Q.3 A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?
(1) 3 hours (2) 3.5 hours (3) 2.25 hours (4) 1.5 hours

Correct Answer - (4)

Solution:
The average speed for the round trip = km/hr
Let the speed during the onward journey be ‘D’ km/hr. Let the speed of the boat in still water be ‘B’ km/hr.
Therefore, D = B + S => D = B + 10 (As the speed of the stream is 10 km/hr).
Let the speed during the return journey be ‘U’ km/hr.
Therefore, U = B - S = B - 10

As the distance between A and B is the same as the distance between B and A, the average speed is given by the formula =
=> => 3B2 - 300 = 80B.
=> 3B2 - 80B - 300 = 0 => 3B2 - 90B + 10B - 300 = 0
=> 3B(B - 30)+10(B - 30) = 0
=> (B-30)(3B+10) = 0
=> B = 30 or B = -10/3

As speed is a positive quantity, B = 30.
Therefore, D = 30 + 10 = 40 km/hr and U = 30 - 10 = 20 km/hr.

His onward journey was done at a speed of 40 km/hr. The distance covered was 60 kms.
Therefore, the time taken for the onward journey = 1.5 hours

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Q. 4 The speed of a motor boat itself is 20 km/h and the rate of flow of the river is 4 km/h. Moving with the stream the boat went 120 km. What distance will the boat cover during the same time going against the stream?
(1) 80 km (2) 180 km (3) 60 km (4) 100 km

Correct Answer - (1)

Solution:
Let the distance to be covered by the boat when it is travelling against the stream be x.
The boat goes down the river at a speed of 20 + 4 = 24 km/h and up the river at a speed of 20 – 4 = 16 km/h.
Since the time taken is same 120/24 = x/16
Therefore, x = 80 km.

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Q. 5 A passenger train covers the distance between stations X and Y, 50 minutes faster than a goods train. Find this distance if the average speed of the passenger train is 60 kmph and that of goods train is 20 kmph.
(1) 20 kms (2) 25 kms (3) 45 kms (4) 40 kms
Correct Answer - (2)

Solution:
Let ‘d’ be the distance between the stations X and Y.
Time taken by the passenger train to cover the distance ‘d’ = d/60 hour
Time taken by the goods train to cover the distance ‘d’ = d/20 hour
Time difference between these two trains is given by 50 minutes or 50/60 hour

i.e., d/20 –d/60 = 50/60

d = 25kms.

Solve this exercise & check your answers with those given at the end.

1. 2 hours after a freight train leaves Delhi a passenger train leaves the same station travelling in the same direction at an average speed of 16 km/hr. After travelling 4 hrs the passenger train overtakes the freight train. The average speed of the freight train was?
A. 30
B. 40
C.58
D. 60

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2. If a boat is moving in upstream with velocity of 14 km/hr and goes downstream with a velocity of 40 km/hr, then what is the speed of the stream ?

(a) 13 km/hr
(b) 26 km/hr
(c) 34 km/hr
(d) none of these

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3. Two trains move in the same direction at 50 kmph and 32 kmph respectively. A man in the slower train
observes the 15 seconds elapse before the faster train completely passes by him.
What is the length of faster train ?

(a) 100m
(b) 75m
(c) 120m
(d) 50m

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4. I drove 60 km at 30 kmph and then an additional 60 km at 50 kmph. Compute my average speed over my 120 km.

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5. If a car starts from A towards B with some velocity due to some problem in the engine after travelling 30km.If the car goes with 4/5 th of its actuval velocity the car reaches B 45min later to the actual time. If the car engine fails ofter travelling 45km, the car reaches the destination B 36min late to the actual time , what is the initial velocity of car and what is the distance between A and B in km

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6. Two trains are travelline at equilateral .Train A is travelling in the direction of earths spin.Other train B is travelling in opposite direction of earths spin.Which trains wheels will wear first?and why?

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7. A person, who decided to go to weekend trip should not exceed 8 hours driving in a day. Average speed of forward journey is 40 m/h. Due to traffic in Sundays, the return journey average speed is 30 m/h. How far he can select a picnic spot?

a) 120 miles
b) Between 120 and 140 miles
c) 160 miles

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8. A person was fined for exceeding the speed limit by 10mph. Another person was also fined for exceeding the same speed limit by twice the same. If the
second person was traveling at a speed of 35 mph, find the speed limit.

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9. A bus started from bus stand at 8.00am, and after 30 minutes staying at
destination, it returned back to the bus stand. The destination is 27 miles from
the bus stand. The speed of the bus is 18mph. In return journey bus travels
with 50% fast speed. At what time it returns to the bus stand?

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10. A train goes from a to b. If it travels with 50 km/s then it is late by 10 mins. When it travels with30km/s then it is late by 50 mins. Find the distance between a and b.

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ANSWERS

Ans 1. B
Ans 2. A
Ans 3: B
Ans 4: 37 1/2
Explanation : Time reqd for the first 60 km = 120 min.; Time reqd for the second 60 km = 72 min.; Total time reqd = 192 min
Avg speed = (60*120)/192 = 37 1/2
Ans 5. 20 & 130.
Ans 6:TRAIN B .Because of less centrifugal force.
Ans 7: 120 miles
Ans 8: Let ‘x’ be the speed limit

Person ‘A’ was fined for exceeding the speed limit by = 10mph
Person ‘B’ was fined for exceeding the speed limit by = twice of ‘A’
= 2*10mph=20mph
given that the second person was traveling at the speed of 35mph

=> 35mph – 20mph = 15mph
Therefore the speed limit is =15 mph
Ans 9: 11.00am

1. The time in a clock is 20 minute past 2. Find the angle between the hands of the clock.

Solution:
Time is 2:20. Position of the hands: Hour hand at 2 (nearly).
Minute hand at 4
Angle between 2 and 4 is 60 degrees [(360/12) * (4-2)]
Angle made by the hour hand in 20 minutes is 10 degrees, since it turns through ½ degrees in a minute.
Therefore, angle between the hands is 60 degrees - 10 degrees = 50 degrees.

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2. How often between 11 O’clock and 12 O’clock are the hands of the clock together at an integral number value?

Solution:
At 11 O’clock, the hour hand is 5 spaces apart from the minute hand. During the next 60 minutes, i.e. between 11′ O clock and 12′ O clock the hour hand will move five spaces [integral values as denoted by the 56 minute, 57 minute, 58 minute, 59 minute and 60 minute positions]. For each of these 5 positions, the minute hand will be at the 12th minute, 24th minute, 36th minute, 48th minute and 60th minute positions. Hence the difference between the positions of the hour hand and the minute hand will have an integral number of minutes between them.

i.e. 5 positions.

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3. At how many points between 10 O’clock and 11 O’clock are the minute hand and hour hand of a clock at an angle of 30 degrees to each other?

Solution:

Between 10 and 11, the minute hand and hour hand are at an angle of 30o to each at (12/11) * 45minutes past 10 = 49 1/11minutes past 10. The next time they will be at angle of 30o to each other will be at 11.

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4. A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time that the clock will show exactly 14 days from the time it was set right?

Solution:
The clock loses 1% time during the first week.
In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
If the clock loses 1% time during the first week, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.
Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.

As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
So the clock will show a time which is 1.68 hours more than 12 Noon two weeks from the time it was set right.
1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds.
i.e. 1 : 40 : 48 P.M.

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5. What is the angle between the minute hand and the hour hand when the time is 1540 hours?

Solution:
The total angle made by the minute hand during an hour is 360o. If it takes 360o for an hour, it will take (40/60) * 360= 240o. The angle between the hour hand the minute hand will therefore, be somewhere between 240 - 90 = 150o, as the hour hand is between 3 and 4.

The angle made by the hour hand when it moves from say 3 to 4 will be 30o. That is the hour hand makes 30o during the course of an hour.

The hour hand will therefore, move (40/60) * 30= 20o.
Therefore, the net angle between the hour hand and the minute hand will be 150 - 20 = 130o.

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6. Given that on 27th February 2003 is Thursday. What was the day on 27th February 1603?

Solution:
After every 400 years, the same day occurs.
Thus, if 27th February 2003 is Thursday, before 400 years i.e., on 27th February 1603 has to be Thursday.

1. Find the day of the week on 16th july, 1776.

Ans: Tuesday

Explanation:
16th july, 1776 means = 1775 years + period from 1st january to 16th july.
Now, 1600 years have 0 odd days.
100 years have 5 odd days.
75 years = 18 leap years + 57 ordinary years
= (36 + 57) odd days = 93 odd days
= 13 weeks + 2 odd days = 2 odd days
Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.
Now, days from 1st Jan to 16th july; 1776

Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days

= (28 weeks + 2 days) odd days
Therefore, total number of odd days = 2
Therefore, the day of the week was Tuesday
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2 .Find the angle between the minute hand and hour hand of a click when the time is 7.20?

Ans: 100deg

Sol: Angle traced by the hour hand in 12 hours = 360 degrees.
Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.
Angle traced by minute hand in 60 min = 360 deg.
Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.
Therefore, required angle = (220 - 120) = 100deg.
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3.The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?

Ans: the clock gains 10 10/43 minutes

Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes. To be together again, the minute hand must gain 60 minutes over the hour hand.
55 minutes are gained in 60 min.
60 min. are gained in [(60/55) * 60] min == 65 5/11 min.
But they are together after 65 min.
Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.
Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.
Therefore, the clock gains 10 10/43 minutes in 24 hours.
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4.A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?

Ans. 48 min. past 12.

Sol: Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145/6 hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock
= 28 hrs 48 min of the correct clock.

Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
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5. At what time between 2 and 3 o’ clock will the hands 0a a clock together?
Ans: 10 10/11 min. past 2.

Sol: At 2 o’ clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min space apart. To be together, the minute hand must gain 10 minutes over the other hand. Now, 55 minutes are gained by it in 60 min.
Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.

Therefore, the hands will coincide at 10 10/11 min. past 2.
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6. First day of 1999 is sunday what day is the last day
Ans.: Monday
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7. What was the day of the week on 12th January, 1979?
Ans: Friday

1. A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

Ans:27 days.

Sol: (A’s 1 day’s work): (B’s 1 day’s work) = 2:1.
(A + B)’s 1 day’s work = 1/18.
Divide 1/18 in the ratio 2:1.
Therefore A’s 1 day’s work = (1/18 * 2/3) = 1/27.
Hence, A alone can finish the work in 27 days.
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2. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?

Ans: 12 ½ days.
Sol: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work =y.
Then, 2x+3y=1/10 and 3x+2y=1/8.
Solving, we get: x=7/200 and y=1/100.
Therefore (2 men +1 boy)’s 1 day’s work = (2*7/200 + 1*1/100) = 16/200 = 2/25.
So, 2 men and 1 boy together can finish the work in 25/2 =12 ½ days.

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3. A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together?
Also, find the number of days taken by each to finish it working alone?

Ans:60 days

Sol: (A+B)’s one day’s work=1/12; (B+C)’s one day’s work=1/15 and (A+C)’s one day’s work=1/20.
Adding, we get: 2(A+B+C)’s one day’s work = (1/12+1/15+1/20)=1/5.
Therefore, (A+B+C)’s one day’s work=1/10.
Thus, A, B and C together can finish the work in 10 days.
Now, A’s one day’s work
= [(A+B+C)’s one day’s work] – [(B+C)’s one day’s work]
= 1/10-1/15)
= 1/30.

Therefore, A alone can finish the work in 30 days.
Similarly, B’s 1 day’s work = (1/10 -1/20) = 1/20.
Therefore, B alone can finish the work in 20 days.
And, C’s 1 day’s work= (1/10-1/12) = 1/60.
Therefore, C alone can finish the work in 60 days.

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4. Two men A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?

Ans:24hours.

Sol: (A+B)’s one hour’s work =1/8, A’s one hour’s work =1/12
Therefore, B’s one hour’s work = (1/8-1/12) =1/24.
Hence, B alone can dig the trench in 24 hours.

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5. If 20 men can build a wall 56m long in 6 days, what length of a similar wall can be built by 35 men in 3 days?

Ans. Length=49m.

Sol: Since the length is to be found out, we compare each item with the length as shown below.
More men, more length built (Direct).
Less days, less length built (Direct).
Men 20:35 :: 56: x

Similarly, days 6:3 :: 56: x.
Therefore, 20*6*x= 35*3*56 or x= 49.
Hence, the required length= 49m.

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6. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?

Ans.81

Sol: Remaining work = 1-4/7 =3/7.
Remaining period = (46-33) days =13 days.
Less work, less men (direct)
Less days, more men (indirect).
More hours per day, less men (indirect)
Therefore, work 4/7:3/7 ::117/x
Days 13:33 :: 117/x
Hrs/day 9:8:: 117/x
Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.
Therefore, additional men to be employed =(198-117) =81.

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7. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z?

Sol. In one day X can finish 1/15th of the work.
In one day Y can finish 1/10th of the work.
Let us say that in one day Z can finish 1/Zth of the work.
When all the three work together in one day they can finish 1/15 + 1/10 + 1/Z = 1/5th of the work.
Therefore, 1/Z = 1/30.
Ratio of their efficiencies = 1/15: 1/10: 1/30 = 2: 3: 1.Therefore Z receives 1/6th of the total money.
According to their efficiencies money is divided as 240: 360: 120.
Hence, the share of Z = Rs. 120.

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8. A group of workers can do a piece of work in 24 days. However as 7 of them were absent it took 30 days to complete the work. How many people actually worked on the job to complete it?

Solution:
Let the original number of workers in the group be ‘x’
Therefore, actual number of workers = x-7.
We know that the number of manhours required to do the job is the same in both the cases.
Therefore, x (24) = (x-7).30
24x = 30x - 210
6x = 210
x = 35.
Therfore, the actual number of workers who worked to complete the job = x - 7 = 35 -7 = 28.

1. In the first test of the semester, kiran scored 60. In the last test of the semester, kiran scored 75%. By what percent did kiran’s score improve?

Ans: 25%

Sol. In first test kiran got 60
In last test he got 75.
% increase in test ( 60(x+100))/100=75
0.6X+60=75
0.6X=15
X=15/0.6=25%
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2. Randy’s chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in 1999, by what percent did the number of cars sold decrease?

Ans: 4%

Sol. Let percentage of decrease is x , then 16400(100-x)/100=15744
16400-15744=164x
x=656/164=4%
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3. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?
A) 260%
B) 150%
C) 100%
D) 50%
E) None of these

Ans: 260%

Sol. Let x be original cost of the radio.
The solding price = (100+20)x=120x
If , it is sold for thrice the price ,then 3*120x=360x
So, gain percent is (360-100)=260%.
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4. The boys and girls in a college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is:??

Ans.78%

Sol: Suppose boys = 3x and girls = 2x
Not adults = (80*3x/100) + (75*2x/100) = 39x/10
Required percentage = (39x/10)*(1/5x)*100 = 78%
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5. 5/9 part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is:

Ans: (250/9)%

Sol: Let the population =x Males=(5/9)x
Married males = 30% of (5/9)x = x/6
Married females = x/6
Total females = (x-(5/9)x)=4x/9
Unmarried females = (4x/9 – x/6) = 5x/18
Required percentage = (5x/18 * 1/x * 100) = (250/9)%
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6. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?

Ans. Rs. 50.

Sol: Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.
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7. A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks.

Solution:
Let the maximum marks be x.
From the given statement pass percentage is 42% - 12% = 30%
By hypothesis, 30% of x – 20% of x = 10 (marks)
i.e., 10% of x = 10
Therefore, x = 100 marks.
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8. In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?

Solution:
Let the percentage of the total votes secured by Party D be x%
Then the percentage of total votes secured by Party R = (x – 12)%
As there are only two parties contesting in the election, the sum total of the votes secured by the two parties should total up to 100% i.e., x + x – 12 = 100
2x – 12 = 100 or 2x = 112 or x = 56%.
If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.
44% of the total votes = 132,000
i.e.T = 300,000 votes.

The margin by which Party R lost the election = 12% of the total votes = 12% of 300,000 = 36,000.
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9. If the price of petrol increases by 25%, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?

Solution:
Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = 100 * 1 = Rs.100
Now, the price of petrol increases by 25%. Therefore, the new price of petrol = Rs.125.
As he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.
Let ‘x’ be the number of litres of petrol he will use at the new price.

Therefore, 125*x = 100 => x = 0.8 litres.

He has cut down his petrol consumption by 0.2 litres = = 20% reduction.

**There is a short cut for solving this problem.

If the price of petrol has increased by 25%, it has gone up of its earlier price.
Therefore, the % of reduction in petrol that will maintain the amount of money spent on petrol constant = 20%

Questions along with proper explanations.

1. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

Ans. 250

Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.
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2. Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by train was ?

Ans.800km

Sol: Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.
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3. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr.In how much time will they cross each other if they are running in the same direction.

Ans: 252sec

Sol: Relative speed=45-40=5km/hr=25/18 mt/sec
Total distance covered =sum of lengths of trains =350mts.
So, time taken =350*18/25=252sec.
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4. From height of 8 mts a ball fell down and each time it bounces half the distance back. What will be the distance travelled
Ans.: 24

Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24
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5. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Ans:6km.

Sol: Let the required distance be x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.
Hence ,the required distance is 6 km
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6. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

Ans:50 min

Sol: New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) – usual time = 10min
Therefore Usual time = 50min
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7. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.

Ans. length of the train=160m
length of the platform=140 m.

Sol: Let the length of the train be x meters and length of the platform be y meters.
Speed of the train relative to man=(54-6) kmph =48 kmph.
=(48*5/18) m/sec =40/3 m/sec.

In passing a man, the train covers its own length with relative speed.
Therefore, length of the train=(Relative speed *Time)
=(40/3 * 12) m =160 m.

Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.
Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.
Therefore, Length of the platform=140 m.
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8. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.

Ans: length of train=120m
Speed of train=54kmph

Sol: Let the length of the train be x meters
Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.
Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120
Therefore Length of the train = 120m
Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph
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9. Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260 kms?

Solution:
All three trains meet at the same time between Delhi and Mumbai. Which means Train A and Train B are at the same point at that time. This will happen when Train B is overtaking Train A.

Train A starts 3 hours before Train B. Therefore, by the time Train B leaves Mumbai, Train A has covered 3 * 60 = 180 kms.

The relative speed between Train A and Train B = 90 - 60 = 30 kmph. Therefore, Train B will overtake Train A in = 6 hours from the time Train B leaves Mumbai. That is at 3 A.M, Train B will overtake Train A. The point between Mumbai and Delhi at which Train B overtakes Train A will be 6*90=540 kms from Mumbai.

Train C will also be at that point at 3 A.M while Train B is overtaking Train A. And Train C would have travelled 1260-540 = 720 kms in these 6 hours. Therefore, the speed of Train C = = 120 km/hr.
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10. Two trains A and B start simultaneously from stations X and Y towards each other respectively. After meeting at a point between X and Y, train A reaches station Y in 9 hours and train B reaches station X in 4 hours from the time they have met each other. If the speed of train A is 36 km/hr, what is the speed of train B?

Solution:
The ratio of the speed of the two trains A and B is given by a/b, where b is the time taken by train B to reach its destination after meeting train A and a is the time taken by train A to reach its destination after meeting train B.

In this case,
Speed of train B = 3/2 * Speed of train A = 3/2 * 36 = 54 km/hr
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