### Find solutions for all the questions at the end of each set:

### Set A

1. Out of 19 persons, 18 persons spent Rs.25 each for their meals. The 19th one spent Rs. 15 more than the average expenditure of all the other. The total money spent by all of them was (in Rs.)

(a) 465

(b) 196

(c) 300

(d) 490

2. In a school with 600 students, the average age of boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years and 9 months, then the number of the girls in the school is

(a) 450

(b) 250

(c) 150

(d) 350

3. The contents of two vessels containing water & milk are in the ratio 3 : 2 and 5 : 3 are mixed in the ratio of 1 : 2. The resulting mixture will have water and milk in the ratio of

(a) 37 : 23

(b) 21 : 13

(c) 4 : 5

(d) 5 : 4

4. The students in three classes are in the ratio 4 : 6 : 9. If 12 students are increased in each class, the ratio changes to 7 : 9 : 12. Then the total number of students in the three classes before the increase is

(a) 114

(b) 76

(c) 95

(d) 100

5. There is a ratio of 4 : 3 between two numbers. If 10% of the first is 6 then 20% of the second is how much times of 5% of the first number?

(a) 9

(b) 3

(c) 6

(d) 4.5

6. The sum of the squares of three numbers is 532 and the ratio of the first to the second and the ratio of second to the third is 3 : 2. The second number is –

(a) 4

(b) 2

(c) 6

(d) 12

7. A man purchased a table and a chair for Rs. 1300. He sold the table at a profit of 20% and the chair at a profit of 25%. In this way, his total profit was 23 1/3%. The cost price of the table is (in Rs.)

(a) 750

(b) 800

(c) 500

(d) 433.3

8. A and B invested in a business in the ratio 3 : 5. If 20% of the total profit goes to charity and then A’s share is Rs. 12000, then the total profit is (in Rs.)

(a) 32000

(b) 8000

(c) 40000

(d) 15000

9. Two numbers A and B are 20% and 28% less then a third number C. Find by what percentage is the number B less than the number A?

(a) 10%

(b) 9%

(c) 12%

(d) 8%

10. The population of a town is 3,11,250. The ratio between women and men is 43 : 40. If there are 24% literate among men and 8% literate among women, the total number of illiterate persons in the town is

(a) 269450

(b) 262350

(c) 254450

(d) 211650

Solutions

1. (d); Total money spent by all of them = (18 × 25) + {1 × (25 + 15)}

= 490

2. (c); Let x = no. of girls

So,

600 × 11 yr 9 months

= x × 11 yrs + (600 – x) × 12 yrs

⇒ 7050 yr. = (11x + 7200 – 12x) years

⇒ x = 7200 – 7050 = 150

3. (a); Vessel ‘1’ Vessel ‘2

Water Milk Water Milk

3 : 2 5 : 3

Ratio of contents of vessel ‘1’ to vessel ‘2’ to mix with each other = 1 : 2

⇒ Ratio of water and milk in the resulting Mixture

= (3/5 + (5/8 × 2))/(2/5 + (3/8 × 2)) = ((24 + 50)/40)/((16 + 30)/40) = 74/46 =37/23

4. (b); Initial Ratio = 4 : 6 : 9

Initial numbers = 4x, 6x, 9x

Final Ratio = 7 : 9 : 12

(on increasing 12 students in each class)

⇒ (4x + 12)/(6x + 12) = 7/9

X=4

⇒ Total no. of students in the three classes before the increase = 4x + 6x + 9x = 19x = 76

5. (b); According to question

10% of 1st no. = 6

Or, 1/10 of 1st no. 6 ⇒ 1st no. = 60

And, ∵ Ratio of 1st number to 2nd numbers = 4 : 3

⇒ Nos. are → 60 & 45

Now, 20% of 2nd no. = 20% of 45 = 9

& 5% of 1st no. = 5% of 60 = 3

And, ∵ 9 is 3 times of 3.

⇒ 20% of 2nd no. is 3 times of 5% of the first no.

6. (d); 1st no. : 2nd no. & 2nd no. : 3rd no.

= 3 : 2 = 3 : 2

⇒ 1st no. : 2nd no. : 3rd no.

= 9 : 6 : 4

So, Let the no. are 9x, 6x & 4x

ATQ, (9x)2 + (6x)2 + (4x)2 + 532

⇒ 133×2 = 532

⇒ x = 2

So, the 2nd no.

i.e. 6x = 6 × 2 = 12

7. (d); Let, x = C.P. of table

⇒ (1300 – x) = C.P. of chair

Now, ATQ,

Profit on table + Profit on chair = Total profit

or, 20% of x = 25% of (1300 – x) = 23 1/3% of 1300

⇒ x = (65 × 20)/3 = Rs. 433.33

8. (c); A’s share = Rs. 12000

= 3/8 of (100% – 20%) of total profit

= 3/8 of 80% of total profit

⇒80% of total profit = Rs. 12000 × 8/3

So, Total profit (i.e. 100%)

= Rs. 12000 × 8/3 × 100/8 = Rs. 40000

9. (a); Let the third no. be 100.

So, A and B will be 80 and 72 respectively.

So, required % = (difference of A &B )/ (value of A) × 100%

= 8/80 × 100% = 10%

10.(b) Total no. of illiterate persons in the town = no. of literate men + no. of illiterate women

= {(100% – 24%) of 40/83 × 311250} + {(100% – 8%) of 43/83 × 311250}

= 262350

### Find solutions for all the questions at the end of each set:

### Set B

Q1. Supposing that telegraph poles on a railway track are 50 metres apart. How many pole will be passed by a train in 2 hours. If the speed of the train is 45 km/hr

(a) 1801

(b) 1800

(c) 1601

(d) 1600

S1. Ans.(a)

Sol. Distance travelled by train in 2 hours = 45 × 2 = 90 kms = 90000 m

No. of pole passed = 90000/50+1

= 1800 + 1

= 1801 poles

Q2. A train 150 metre long, passes a pole in 15 seconds and another train of the same length travelling in the opposite direction in 12 seconds. The speed of the second train is

(a) 45 km/hr

(b) 48 km/hr

(c) 52 km/hr

(d) 54 km/hr

S2. Ans.(a)

Sol. When distance is same, speed is indirectly proportional to the time

Speed of the first train = 150/15 = 10 m/s

Speed of the second train = 10/12×15

= 150/12 m/s

=150/12×18/5

= 45 km/hr

Q3. A carriage driving in a fog passed a man who was walking at the rate of 3 kms an hour in the same direction. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 m. What was the speed of the carriage?

S3. Ans.(a)

Sol. The distance travelled by the man in 4 minutes = S × T

= 3 × 5/18 × 4 × 60 = 200 m.

(3 is multiplied by 5/18 to change it into m./sec. and 4 multiplied by 60 to change it into seconds)

∴ distance travelled by the carriage in 4 minutes = (200 + 100) = 300 metres.

∴ Speed of carriage = D/T=(300 m.)/(4 min.)

= (300/1000)/(4/60) km./hr. = 9/2 km per hour

Q4. A train 100 metres long moving at a speed of 50 kmph crossed a train 120 metres long coming from opposite direction in 6 seconds. The speed of the second train is

(a) 82 km/hr

(b) 84 km/hr

(c) 86 km/hr

(d) Can’t be determined

S4. Ans.(a)

Sol. Total speed = (100 + 120)/6

= 220/6 m/sec.

= 220/6×18/5 km/hr.

= 132 km/hr.

Speed of second train = 132 – 50 = 82 km/hr.

Q5. A boat covers 12 km. upstream and 18 km downstream in 3 hours while it covers 36 km upstream and 24 km downstream in 6 hours, what is the velocity of the stream?

(a) 1.5 km/hr

(b) 1 km/hr

(c) 2 km/hr

(d) 2.5 km/hr

S5. Ans.(c)

Sol. 12/y+18/x = 3 …(i)

36/y+24/x=13/2 …(ii)

To equate (i) & (ii)

Multiply (i) by 3 first

36/y+54/x = 9 …(iii)

36/y+24/x=13/2 …(iv)

Subtracting (iv) from (iii)

30/x=9-13/2=5/2

⇒ x = 12km/hr.

Now, put value of x in equation (i), we get y = 8 km./hr.

∴ Velocity of stream(Sc)=(12 – 8)/2 = 2 km./hr.

Q6. In a flight of 600 kms, an aircraft was slowed down due to bad weather. Its average speed for the trip is reduced by 200 kms/hr. and the time of flight increased by 30 min. The duration of the flight is –

(a) 1 hr.

(b) 2 hrs.

(c) 3 hrs.

(d) 4 hrs.

S6. Ans.(a)

Sol. Let the duration of flight be t hours.

S = D/T

S1-S2 = 200 kms/hr.

600/t-600/(t+1/2) = 200

600/t-(2×600)/(2t+1)=200

(2t + 1) 600 – t × 1200 = 200t (2t + 1)

3(2t + 1) – 6t = t (2t + 1)

6t + 3 – 6t = 2t^2 + t

2t^2 + t – 3 = 0

2t^2 + 3t – 2t – 3 = 0

t (2t + 3) – 1(2t + 3)

(2t + 3) (t – 1) = 0

t = 1 hour.

Q7. Without any stoppage a person travels a certain distance at an average speed of 42 km/hr. and with stoppage he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop?

(a) 20 min.

(b) 30 min.

(c) 21 min.

(d) 23 min.

S7. Ans.(a)

Sol. Here x = 42 and y = 28

∴ Stoppage time/hr. = (x – y)/x

⇒(42 – 28)/42⇒1/3 hr. ⇒ 20 min.

Q8. The speed of a car increases by 2 km/hr after every hour. If the distance travelled in the first one hour was 35 km, the total distance travelled in 12 hours was-

(a) 456 km

(b) 482 km

(c) 552 km

(d) 556 km

S8. Ans.(c)

Sol. Distance travelled in first hour = 35 km

Distance travelled in second hour = 37 km

common difference = 2 km

So, distance travelled in 12 hour = 12/2 [2 × 35 + (12 – 1) × 2]

= 12/2 × (70 + 22)

= 12 × 46

= 552 km

Q9. Walking at 6/7th of his usual speed a man is 12 minutes late. The usual time taken by him to cover that distance is-

(a) 1 hr.

(b) 1 hr. 12 min.

(c) 1 hr. 15 min.

(d) 1 hr. 20 min.

S9. Ans.(b)

Sol. Reduced in speed = (1-6/7)/1=1/7

So, increase in time 1/(7 – 1)=1/6 time=12

total time = 6 × 12

= 1 hr. 12 min.

Q10. A starts from a place P to go to a place Q. At the same time B starts from Q for P. If after meeting each other A and B took 4 and 9 hours more respectively to reach their destinations, the ratio of their speeds is

(a) 3:2

(b) 5:2

(c) 9:4

(d) 9:13

S10. Ans.(a)

Sol. (Speed of A)/(Speed of B)=√(9/4 )

(Speed of A)/(Speed of B)=3/2

Ratio=3:2

Q11. I have to be at a certain place at a certain time and I find that I shall be 15 minutes too late, if I walks at 4 km and hour; and 10 minutes too soon, if I walks at 6 km an hour. How far have I to walk?

(a) 3 km

(b) 5 km

(c) 6 km

(d) 8 km

S11. Ans.(b)

Sol. The difference = 15 – (–10) = 25 m

Increase in speed = 4 km to 6 km

= 2/4=1/2

So,

1/(2 + 1) × time = 25 minute

Time = 75 minute

Total walk = 4 × 75/60

= 5 km.

Q12. The ratio of copper and zinc in a 63 kg alloy is 4:3. Some amount of copper is extracted from the alloy and the ratio becomes 10:9. How much copper is extracted.

(a) 6 kg

(b) 8 kg

(c) 12 kg

(d) 10 kg

Q13. A bag contains Rs. 410 in the form of Rs. 5, Rs. 2 and Rs. 1 coins. The number of coins are in ratio 4 : 6 : 9. So, find the number of 2 Rs. coins.

(a) 40

(b) 50

(c) 60

(d) 70

Q14. Two numbers are in ratio 4:5 respectively. If each number is reduced by 25, then the ratio becomes 3:4. Find the 2nd numbers.

(a) 120

(b) 125

(c) 130

(d) 135

Q15. The ratio between the male and female existing employees in an organization is 8:3. The ratio between the males and female new recruits of 240 is 5:7. What will be new ratio between male and female employees after the recruits join the organization?

(a) 6:5

(b) 5:4

(c) 3:2

(d) Can’t be determined

S15. Ans.(d)

Sol. In given question no. of existing employee not given.

So, answer will be can’t be determine.

### Find solutions for all the questions at the end of each set:

### Set C

Q1. A spends 90% of his salary and B spends 85% of his salary. But saving of both are equal. Find the income of B. If the sum of their income is Rs. 5000

(a) Rs. 2000

(b) Rs. 2500

(c) Rs. 3000

(d) Rs. 3500

S1. Ans.(a)

Sol. Saving of A = 10% of his salary

Saving of B = 15% of his salary

According to question,

10% of A salary = 15% of B salary

A’s salary : B’s salary = 15 : 10 = 3 : 2

B salary =5000×2/5=2000

Q2. Ratio of milk and water in a mixture is 4 : 1. If 5 litre of milk is added in the mixture, new ratio become 5 : 1. The quantity of milk in the mixture originally was.

(a) 10 litres

(b) 15 litres

(c) 20 litres

(d) 25 litres

S2. Ans.(c)

Sol. On adding 5 litre of milk its ratio in the mixture is increased by 1unit

So, original quantity in the milk = 4 × 5 = 20 litres

Q3. Ratio between the annual income of A and B is 5 : 4 and between their expenditure is 4 : 3. If they save Rs. 1000 each. Find A annual income

(a) 6000

(b) 3000

(c) 4000

(d) 5000

Q4. Two vessels A and B contain milk and water in the ratio 5 : 3 and 8 : 6. Find the ratio in which the quantities be taken from the two vessels so that the ratio of milk and water in the new mixture is 13 : 9

(a) 8 : 15

(b) 4 : 7

(c) 9 : 13

(d) 13 : 9

Q5. The population of a town is 20,000. In a particular year, male population is incased by 5% and female population by 8%. If the population at the end of the year is 21,180. Find was the size male population in the town in the beginning of the year.

(a) 16,000

(b) 15,000

(c) 14,000

(d) 13,000

Q6. A clock buzzes 1 time at 1 o’clock, 2 times at 2 o’clock, 3 times at 3’o clock and so on. What will be the total number of buzzes in a day (12 hours)?

(a) 128.

(b) 98.

(c) 148.

(d) 78.

S6. Ans.(d)

Sol. Total number of buzzes = 2(1 + 2 + 3 + 4… + 12)

This is an A.P. in which a = 1, d = 1, n = 12 and l = 12

(1 + 2 + 3 + … + 12) =12/2.(1+12)=78.

Q7. A bouncing tennis ball rebounds each time to a height equal to one half the height of the previous bounce. If it is dropped from a height of 16 m, find the total distance it has travelled when it hits the grounds for the 10th time:

S7. Ans.(b)

Sol. Distance travelled before the first hit = 16 m

Distance travelled before the second hit =16+1/2×16×2=(16+16)m

Distance travelled before the third hit

=16+16+1/2×8×2=(16+16+8)m

Distance travelled before the fourth hit

= 16 + 16 + 8 + 1/2×4×2 = (16 + 16 + 8 + 4)m

and so on.

Hence the required distance

=16+16+8+4+2+1+1/2+1/4+1/8+1/16

=16+(16+8+4+⋯+1/16)

=16+16(1-(1/2)^9 )/(1-1/2)=16+16×(511/512)×2/1

=16+511/16=47 15/16

Q8. A bucket contains a mixture of two liquids A and B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B, then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket?

(a) 18L

(b) 20L

(c) 15L

(d) 10L

S8. Ans.(c)

Sol. 16 litres of mixture is taken out it means liquid A and liquid B also taken out as proportion 5 : 3 so, A will be taken out 16 × (5/8) = 10 L

B will be taken out 16 × (3/8) = 6L

Then, ((5x-10))/((3x-6+16) )=3∶5

((5x-10))/((3x+10) )=3/5 ⇒ x = 5

Liquid B = 3x = 3 × 5 = 15L

Q9. The average of 11 results is 50. If the average of first six results is 49 and that of last six is 52, find the sixth result.

(a) 56

(b) 112

(c) 48

(d)64

S9. Ans.(a)

Sol. The total of 11 results = 11 × 50 = 550

The total of first 6 results = 6 × 49 = 294

The total of last 6 results = 6 × 52 = 312

The 6th result is common to both;

Sixth result = 294 + 312 – 550 = 56

Q10. A person divides his total route of journey into three equal parts and decides to travel the three parts with speeds of 20, 15 and 10 km/hr respectively. Find the average speed during the whole journey.

(a) 160/13 km/hr

(b) 120/13 km/hr

(c) 280/13 km/hr

(d) 180/13 km/hr

Q11. In an examination, a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination?

(a) 8

(b) 9

(c) 10

(d) 11

S11. Ans.(d)

Sol. Let three be n papers

Total marks of all papers = 63n

65n – 63n = (20 + 2)

⇒ 2n = 22

⇒ n = 11

So, there were 11 papers in the examination.

Q12. An article is sold at 20% profit. If its cost price is increased by Rs. 50 and at the same time if its selling price is also increased by Rs. 30, the percentage of profit decreases by 3 1/3%. Find the cost price.

(a) Rs. 850

(b) Rs. 450

(c) Rs. 550

(d) Rs. 750

S12. Ans.(a)

Sol. Suppose the cost price = Rs. x

Then SP =120/100 x=6/5 x

Now, new CP = Rs. (x + 50)

New SP =[120/100 x+30]=(6/5 x+30)

Now, the new% profit =20-3 1/3=16 2/3=50/3%

Thus, (100+50/3)% of (x+50)=6/5 x+30

Or, 350/300 (x+50)=6/5 x+30

Or, 7/6 x+175/3=6/5 x+30

Or, (6/5-7/6)x=175/3-30

Or, 1/30 x=85/3

x = Rs. 850

Q13. A man sold a cow at a loss of 7%. Had he been able to sell it at a gain of 9%, it would have fetched Rs 80 more than it did. What was the cost price?

(a) Rs. 450

(b) Rs. 500

(c) Rs. 750

(d) Rs. 600

S13. Ans.(b)

Sol. Here 109% of cost – 93% of cost = Rs. 80

16% of cost = Rs. 80

Cost =(80×100)/16 = Rs. 500

Q14. A dishonest fruit vendor professes to sell his goods at a loss of 40% but he uses a weight of 12gm for 15gm. Find his gain per cent.

(a) 25% loss

(b) 25% Gain

(c) 40% loss

(d) 40% Gain

S14. Ans.(a)

Sol. gain% due to faulty weight =(15/12-1)×100=25%

Total gain or loss percentage after selling at loss of 40% = (25%-40%-(25 × 40)/10%)

= -25%

= 25% loss

Q15. A shopkeeper sells his goods at 25% profit on original price. Due to increase in demand, he further increases by price by 10%. How much % profit will he get?

(a) 33.3%

(b) 40%

(c) 25%

(d) 37.5%

S15. Ans.(d)

Sol. percentage profit

={25+10+((25×10)/100)}=37.5%

Percentage profit = 37.5%