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DI NEW PATTERN SET-1 I. Study the following graph carefully to answer the given questions

Down Stream Distance = 200 km

Up Stream Distance = 300

km

Days Monday Tuesday Wednesday Thursday Friday

Still Water speed 5 _ 4 _ 3

Stream Speed _ 6 _ 4 _

1. A boatman rows down stream at 6kmph on Wednesday. Find the speed of upstream and Speed of current? 1.7 kmph 2.2 kmph 3.3 kmph 4.5 kmph 5.None of these 2|Page

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Answer & Explanation Answer – 2.2 kmph Explanation : Downstream Speed = 6kmph Upstream Speed = 8kmph – 6kmph = 2kmph Speed of Current = (6-2)/2 = 2kmph 2. A boatman rows up stream at 4kmph, 2 kmph and 8 kmph on Monday, Wednesday and Thursday respectively. Then find the ratio between the down stream speed of boat on Monday, Wednesday, and Thursday together and the rate of current on the same days together? 1.4:7 2.3:1 3.4:1 4.4:3 5.None of these Answer & Explanation Answer – 3.4:1 Explanation : Downstream Speed of boat on Monday = 10 – 4 = 6kmph. similarly Downstream Speed of boat on Wednesday and Thursday = 6 kmph, 16kmph Rate of current = 1kmph, 2kmph and 4kmph Ratio = 28:7 = 4 : 1 3. A boatman rows up stream at 16 kmph and 2 kmph on Tuesday and Friday. Then find the difference between the down stream speed of boat on Tuesday and Friday together and the speed of boat in still water on the same days together? 1.7 kmph 2.11 kmph 3|Page

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3.5 kmph 4.3 kmph 5.None of these Answer & Explanation Answer – 1.7 kmph Explanation : down stream speed of boat on Tuesday and Friday = 32 kmph Still water speed = 25 kmph Difference = 32 – 25 = 7 kmph 4. What time will be taken by a boat to cover a distance on Tuesday along the stream, if the up stream of boat is 3 kmph? 1.4 hours 2.5 hours 3.2 hours 4.3 hours 5.None of these Answer & Explanation Answer – 3.2 hours Explanation : upstream of boat = 3 kmph down stream of boat =15 kmph Down Stream Distance = 15% of 200 = 30; Time = 30/15 = 2 hr 5. What is the sum of total distance travelled by the boat on Monday, Wednesday and Friday together? 1.316 2.328 3.356 4.296 5.336 4|Page

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Answer & Explanation Answer – 2.328 Explanation : Total distance travelled by the boat on Monday = 48 + 90 = 138 Similarly Total distance travelled by the boat on Wednesday and Friday = 102, 88 Total distance travelled by the boat on Monday, Wednesday and Friday = 138 + 102 + 88 = 328 km 6. Find the ratio between the down stream distance travelled by boat on Monday, Wednesday, Thursday and Friday together and the Up stream distance travelled by the boat on the same days together? 1.15:27 2.14:21 3.20:27 4.17:24 5.None of these Answer & Explanation Answer – 4.17:24 Explanation : Down stream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 48 + 30 + 40 + 52 = 170 Up stream distance travelled by boat on Monday, Wednesday, Thursday and Friday together = 90 + 72 + 42 + 36 = 240 17 : 24. 7. Find the difference between the up stream distance travelled by boat on Tuesday and Friday together and the down stream distance travelled by boat on the same days together? 1.12 km 2.18 km 5|Page

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3.16 km 4.14 km 5.None of these Answer & Explanation Answer – 4.14 km Explanation : Difference = 96 – 82 = 14 8. Total downstream distance travelled by boat on Monday, Tuesday and Wednesday together is what percentage of total upstream distance travelled by boat on the same days together? 1.48.6% 2.46.7% 3.48.5% 4.47.5% 5.None of these Answer & Explanation Answer –1.48.6% Explanation : Total downstream distance travelled by boat on Monday, Tuesday and Wednesday = 108 Total up stream distance travelled by boat on Monday, Tuesday and Wednesday = 222 % = 108/222 * 100 = 48.6% 9. A boatman rows down stream at 6kmph on Wednesday. Find the time taken by the boat to cover upsream distance on the same day? 1.50 2.40 3.20 6|Page

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4.36 5.None of these Answer & Explanation Answer – 4.36 Explanation : Downstream Speed = 6kmph Upstream Speed = 8kmph – 6kmph = 2kmph Time taken by the boat to cover upstream distance = 72/2 = 36 hours 10.A boatman rows up stream at 16 kmph on Tuesday. Find the approximate time taken by the boat to cover downstream distance on the same day? 1.1 2.5 3.8 4.6 5.None of these Answer & Explanation Answer – 1.1 Explanation : Downstream speed = 28 kmph; Time taken by the boat to cover downstream distance = 30/28 = 1 hr SET-2 (Q.no: 1-5). Refer to the table and answer the given questions. NA refers the data which is not needed for a particular question. The number of working days of A, B, C, D is given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on.

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A _ _ 16 NA NA

B 25 _ 64/5 NA 9

C NA _ 32 50 8

D NA NA NA _ 3

Ratio Efficiency (A : B) = 5 : 4 NA NA _ NA

Total Number of Working Days NA _ NA _ _

1. A started a work alone and then B joined her 5 days before actual completion of the work. For how many days A worked alone? A. 9 B. 11 C. 10 D. 25 E. 12 Answer & Explanation Answer – B. 11 Explanation : Efficiency (A : B) = 5 : 4 Number of days(A : B) = 4x : 5x = 4x : 25 ∴ Number of days required by A to finish the work alone = 4x = 4 x 5 = 20.

A and B work together for last 5 days = 5 x 9 = 45% Efficiency of A = 5% and B’s efficiency = 4% ∴ No. of days taken by A to complete 55% work = 55/5 = 11days

2. Working together B and C take 50% more number of days than A, B and C together take and A and B working together, take 8/3 more number of days than A, B and C take together. If A, B and C all have worked together till the completion of the work and B has received Rs.120 out of total earnings of Rs. 480 then in how many days did A, B and C together complete the whole work? A. 2 days B. 4 days 8|Page

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C. 6 days D. 8 days E. 5 days Answer & Explanation Answer – E. 5 days Explanation : The days ratio of (A + B + C) : (B + C) = X:3X/2 = 2X:3x; Efficiency ratio = 3X:2X Efficiency of A = x. (480/3X) = Rs.160 Amount received by B = Rs.120 & C = 200 160:120:200 =4:3:5 1/4:1/3:1/5= 15:20:12; (1/15+1/12+1/20)*Y = 1 Y = 5 days 3. All of them started to work together but A leaves after 4 days. B leaves the job 3 days before the completion of the work. How long would the work last? A. 6 days B. 9 days C. 18 days D. 5 days E. None of these Answer & Explanation Answer – B. 9 days Explanation : Let the work lasted for x days, A’s 4 day’s work + B (x – 3) day’s work + C’s x day’s work = 1 ⇒ (4/16) + (x – 3) / (64/5) + x/32 = 1 9|Page

⇒ 5(x – 3)/64 + x/32 = 1 – 1/4

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⇒ [5(x – 3) + 2x] / 64 = 3/4 ⇒ 7x – 15 = 48

∴ x = (48 + 15)/7 = 63/7 = 9 days

4. C started the work and left after some days, when 25% work was done. After it D joined and completed it working for 25 days. In how many days C and D can do the complete work, working together? A. 6 B. 8 C. 10 D. 12 E. 20 Answer & Explanation Answer – E. 20 Explanation : Efficiency of C = (100/50) = 2% Rest work = 75% ∴ Efficiency of D = 75/25 = 3%

∴ Combined efficiency of C and D = 5%

∴ Number of days required by C and D to work together = 100/5 = 20 days.

5. B and C started the work. After 3 days D joined them. What is the total number of days in which they had completed the work? A. 12 B. 8 C. 4 D. 6 E. None of these Answer & Explanation 10 | P a g e

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Answer – C.4 days Explanation : Efficiency of B and C = 11.11 + 5.55 = 16.66% Work done in 3 days = 3 x 16.66 = 50% Rest work done by B, C and D = 50/50 = 1 day Work can be completed in 4 days. (Q.no: 6 -10). Refer to the table and answer the given questions.

NA refers the data which is not needed for a particular question. The price details of 5 products are given below and some of them are missing. Take the data of the first row for the first question and the second row for the second question and so on. Product Mobile Investment 2 Bags TV Machine

MP NA NA NA 40% NA

CP 8400 NA __ 54000 NA

SP __ NA __ __ NA

Profit(%) 5% NA 14% __ 30%

Loss(%) 5% 40% 14% NA 20%

Profit/Loss _ NA NA NA NA

6. Rahul purchased a mobile and sold it for a loss(loss % given in the table). From that money, he purchased another article and sold it for a gain of (Profit % given in the table). What is the overall gain or loss? A. Profit of Rs.21 B. Profit of Rs.24 C. Loss of Rs.21 D. Loss of Rs.24 E. None of these Answer & Explanation Answer – C. Loss of Rs.21 Explanation : 11 | P a g e

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CP = 8400 SP = 8400 * 95/100 = 7980 CP = 7980 SP = 7980 * 105/100 = 8379 Difference = 8400 – 8379 = 21 7. A, B and C invests rupees 8000, 12000 and 10000 respectively in a business. At the end of the year the balance sheet shows a loss of initial investment. Find the share of loss of B. A.4000 B.4500 C.4800 D.5000 E.None of these Answer & Explanation Answer – C.4800 Explanation : Total loss after one year = 30000*40/100 = 12000 share of B = (40/10)*12000 = 4800 8. A Shop Keeper sells two bags for Rs. 500 each. On one, he gets % profit(as per the table) and on the other he gets % loss(as per the table). His profit or loss in the entire transaction was? A. 64/25 % Gain B. 49/25 % Gain C. 64/25 % Loss D. 49/25 % Loss E. None of these Answer & Explanation Answer – D. 49/25 % Loss Explanation : 12 | P a g e

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%=x Loss % = x²/100 = 196/100 = 49/25% 9. The price of TV was marked up by %(as per the table).It was sold at a discount of 20% on the marked price. What was the profit percent of the cost price? A. 10% B. 11% C. 15% D. 12% E. None of these Answer & Explanation Answer – D. 12% Explanation : Explanation: 40 – 20 + [40 * (-20)/100] = 20 – 8 = 12% 10.Preethi sold a machine to Shalini at a profit of %(as per table). Shalini sold this machine to Arun at a loss of %(as per table). If Preethi paid Rs.5000 for this machine, then find the cost price of machine for Arun? A. 6200 B. 5200 C. 4800 D. 4750 E. None of these Answer & Explanation Answer – B. 5200 Explanation : R1 = 30% R2 = 20% 5000 * 130/100 * 80/100 = Rs. 5200 13 | P a g e

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DI NEW PATTERN SET-3 Directions : A train runs through cities A, B, C, D, E, F, G and H. The line graph indicates the time schedule of the train including times of arrival and departure.

Question 1: The overall average speed of the entire trip excluding stoppage time is nearly (a) 46 km/h (b) 75 km/h (c) 81 km/h (d) 65 km/h Question 2: What percentage of time of the entire trip was actually spent travelling between the cities? (a) 76% (b) 42.4% (c) 7.6% (d) 92% Question 3: The total stoppage time at the cities in the first half and second half of the total distance is in the ratio (a) 2 : 1 (b) 1 : 3 (c) 3 : 1 (d) 1 : 2 Question 4: Between how many pairs of consecutive stations does the train run below the overall average speed of the entire trip? (a) 3 14 | P a g e

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DI NEW PATTERN (b) 2 (c) 1 (d) 4

Answers and Explanations

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SET-4 Directions for questions 1 to 3 : A team of 5 players Arpit, Bimal, Chatur, Dinu and Elan participated in a ‘Freaket’ tournament and played four matches (1 to 4). The following table gives partial information about their individual scores and the total runs scored by the team in each match. Each column has two values missing. These are the runs scored by the two lowest scorers in that match.None of the two missing values is more than 10% of the total runs scored in that match.

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Match-1 Runs scored by player

Arpit Bimal

88

Match-2

Match-4

100

53

65

52

Chatur

110

Dinu

72

Elan

60

Total

Match-3

270

75

20

56

78 300

240

200

1) What is the maximum possible percentage contribution of Arpit in the total runs scored in the four matches? A.

19.7%

B.

19.9%

C.

20.1%

D.

20.2%

Answer: Option A Explanation: Maximum possible runs scored by Arpit in Match-1 = 27 Maximum possible runs scored by Arpit in Match-3 = 19 Maximum possible percentage contribution: (27+100+19+53)/(270+300+240+200)x100% = 199/1010x100% = 19.7% 2) If the absolute difference between the total runs scored by Arpit and Chatur in the four matches is minimum possible then what is the absolute difference between total runs scored by Bimal and Elan in the four matches? 17 | P a g e

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A.

32

B.

37

C.

27

D.

Cannot be determined

Answer: Option B Explanation: Maximum possible total runs scored by Chatur in the four matches = 27 + 30 + 110 + 20 = 187. In such a case minimum possible total runs scored by Arpit in the four matches = 23 + 100 + 13 + 53 = 189. Difference = 189 – 187 = 2 (minimum possible) Subsequently total runs scored by Bimal in the four matches = 88 + 65 + 19 + 52 = 224. Also, total runs scored by Elan in the four matches = 60 + 30 + 78 + 19 = 187 Absolute difference = 224 – 187 = 37 3) The players are ranked 1 to 5 on the basis of the total runs scored by them in the four matches, with the highest scorer getting Rank 1. If it is known that no two players scored the same number of total runs, how many players are there whose rank can be exactly determined? A.

0

B.

1

C.

3

D.

5

Answer: Option C Explanation:

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Individual ranges for total score: Arpit-> 189-199 Bimal-> 218-224 Chatur-> 182-187 Dinu-> 223 Elan-> 187-188 Least total will be of Chatur (Rank 5) 2nd least will be Elan (Rank 4) Rank 3 must be of Arpit. It is not possible to determine the exact ranks of Bimal and Dinu. SET-5 DIRECTIONS FOR QUESTIONS 1 to 5 : RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS

Station Name

Arrival Time

Departure Time

Distance(In KM)

New Delhi

--

17:05

0

Kanpur Central

21:48

21:53

440

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Mughal Sarai JN

01:55

02:05

786

Gaya JN

04:18

04:21

989

Koderma

05:22

05:24

1066

Bokaro Steel City

07:35

07:40

1191

Tatanagar JN

10:35

10:40

1342

Kharagpur JN

12:25

12:40

1476

Balasore

14:02

14:04

1595

Bhadrak

15:12

15:14

1657

Cuttack

16:33

16:35

1772

Bhubaneswar

17:30

--

1800

Answer : Station Name

New Delhi - Kanpur Central

Distance(In

Time Taken(In

Speed(In

Km)

hours)

kmph)

440

4.72

93.22

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Kanpur Central - Mughal Sarai

346

4.03

85.86

Mughal Sarai JN - Gaya JN

203

2.22

91.44

Gaya JN - Koderma

77

1.98

38.89

Koderma - Bokaro Steel City

125

2.18

57.34

Bokaro Steel City - Tatanagar

151

2.92

51.71

Tatanagar JN - Kharagpur JN

134

1.75

76.57

Kharagpur JN- Balasore

119

1.37

86.86

Balasore - Bhadrak

62

1.13

54.87

Bhadrak - Cuttack

115

1.32

87.12

Cuttack - Bhubaneswar

28

0.92

30.43

JN

JN

1.The longest run for the train between the two successive halts is (a) Mughal Sarai JN - Gaya JN

(b) New Delhi - Kanpur Central

(c) Kanpur Central - Mughal Sarai JN

(d) Balasore - Balasore

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Answer : Option b Explanation : With the run is of 440 km, the longest run is between New Delhi - Kanpur Central. 2. The average speed that the train maintained between two successive stations was the highest between (a) Kanpur Central - Mughal Sarai JN (c) New Delhi - Kanpur Central

(b) Mughal Sarai JN - Gaya JN (d) Bokaro Steel City - Tatanagar JN

Answer : Option c Explanation : The average speed of 93.22 kmph is the highest between New Delhi Kanpur Central. 3. The average speed that the train maintained between New Delhi and Bhubaneswar was nearly equal to (a) 72 kmph

(b) 74 kmph

(c) 75 kmph

(d) 82 kmph

Answer : Explanation : The average speed that the train maintained between New Delhi and Bhubaneswar = 1800 km/25 hrs and 25 min = 70.82 kmph. 4. If we consider a journey that begins in New Delhi and ends in Bhubaneswar, the train has the longest halt at (a) Kanpur Central

(b) Mughal Sarai JN

(c) Tatanagar JN

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(d) Kharagpur JN

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Explanation : The train has the longest halt of 15 minutes at Kharagpur JN. 5. The train begins its return journey from Bhubaneswar to New Delhi Seventeen hours after it has arrived at Bhubaneswar. If the train left New Delhi on Tuesday on what day will it have returned to New Delhi? (Assume that on the return journey that train maintains the same average speed as on the onward journey). (a) Thursday

(b) Friday

(c) SaturdayAnchor

(d) SundayAnchor

Answer : Option b Explanation :Friday SET-6 Direction (1-5): Study the following charts and answer the questions that follow: PERSON NO.OF DAYS THEY WORKED A

4

B

3

C

6

D

7.5

E

6

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PERCENTAGE OF WORK DONE BY 5 PEOPLE TO COMLETE THE PROJECT A-20% B-10% C-25% D-30% E-15%

1. A and B started doing the work. After 2 days they both left, and C joined the work. He completed his part of work. Now the remaining work was completed by F in 7 days. In how many days can F complete whole work? A) 18 days B) 16 days C) 12 days D) 20 days E) 10 days View Answer Option C Solution: A does 20% work in 4 days. so 100% work in 100*4/20 = 20 days Similarly B can complete 100% work in 100*3/10 = 30 days They worked for 2 days, so did [1/20 + 1/30]*2 = 1/6 work Now C completed 25% = 1/4 work So now remaining work = 1 – (1/6 + 1/4) = 7/12 F complete 7/12 work in 7 days, so complete work in 12 days 2. G who can complete whole work in 30 days replaced A and did A’s part of work. He left and then B also worked for same number of days as G. If remaining work was completed by M who can do complete work in one-fourth the number of days in which E can complete the work, then in how many days was the whole work completed? A) 22 days B) 14 days C) 21 days D) 18 days E) 12 days

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View Answer Option D Solution: A’s part of work = 20% = 1/5 So G did 1/5 of work, Whole work in 30 days, s 1/5 work in 1/5 * 30 = 6 days Now B also worked for 6 days. B can complete whole work in 30 days, so In 6 days, completed 6/30 = 1/5 of work Now remaining work = 1 – (1/5 + 1/5) = 3/5 [After G and B did work] Now E can complete whole work in 40 days [100*6/15] So M can complete work in 10 days. So completed 3/5 work in 3/5 * 10 = 6 days So total number of days = 6+6+6 = 18 days 3. All people decided to complete work in less number of days. So they divided the work equally among themselves. In how many days will the work be completed this way? (They all worked individually) A) 25 4/5 days B) 27 4/5 days C) 23 5/6 days D) 24 3/5 days E) None of these View Answer Option B Solution: 5 people equally divided the work so each did 1/5 work now So A 1/5 work in 4 days as earlier B == 1/10 work in 3 days , so 1/5 work in 6 days C == 1/4 work in 6 days , so 1/5 work in 24/5 days D == 3/10 work in 7.5 days , so 1/5 work in 5 days E == 3/20 work in 6 days , so 1/5 work in 8 days So complete work in = 4 + 6 + 24/5 + 5 + 8 = 27.8 days 4. P is 20% more efficient than B and Q is 60% more efficient than C. They worked together for 5 days and left the work, after which the remaining work was completed by C and E together. Had all worked together, in how many less days they could have completed the work? A) 6 3/13 days B) 5 days C) 4 3/5 days D) 7 days E) 8 4/11 days

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View Answer Option A Solution: P and B:: P == 120 : 100 = 6 : 5. So days = 5 : 6 6 = 30, so 1 = 5, so 5 = 25. P can complete work in 25 days Q and C:: Q == 160 : 100 = 8 : 5. So days = 5 : 8 8 = 24, so 1 = 3, so 5 = 15. Q can complete work in 15 days They worked for 5 days. So [1/25 + 1/15]*5 = 8/15 work Reaming 7/15 by C and E So [1/24 + 1/40]*x = 7/15 x = 7 days Total = 5+7 = 12 days Now had they did work together:: [1/25 + 1/15 + 1/24 + 1/40] = 13/75 So 75/13 days So less days = 12 – 75/13 = 81/13 = 6 3/13 days 5. J can complete the whole work in number of days equal to the average of number of days in which A and B can complete the work. J, C, D, and E all started the work and after 5 days they were replaced by A and B. A and B completed the remaining work in how many days? A) 3 1/5 days B) 2 2/3 days C) 1 2/3 days D) 2 3/5 days E) 4 2/5 days View Answer Option A Solution: A in 20 days, B in 30 days. So J in (20+30)/2 = 25 days J, C, D, and E started work:: [1/25 + 1/24 + 1/25 + 1/40]= 11/75. Worked for 5 days so did 11/75 * 5 = 11/15 of work Remaining work = 4/15 [1/20 + 1/130]*x = 4/15 Solve. x = 48/15 days = 3 1/5 days SET-7 Directions (1 – 5): Study the following table carefully and answer the questions that follow:

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DI NEW PATTERN The table shows the discount % given by stores on different items. The Market Price of any article on all stores is same. Some values are missing. Answer the questions on the basis of given table and information in question.

1. If the average SP of article II by in all the stores is Rs 2568, Find the MP of article II. A) Rs 3200 B) Rs 4500 C) Rs 3600 D) Rs 4300 E) Rs 4000 View Answer Option C Solution: SP by store A = (100-28)/100 * MP = 72% of MP, by B = 82% of MP, by C = 60% of MP So (72+82+60)/3 * MP/100 = 2568 Solve, MP = Rs 3600 2. Difference between SP of article I by stores A and B is Rs 486, Find the SP of same article by store C. A) Rs 3506 B) Rs 4005 C) Rs 4808 D) Rs 4104 E) Rs 3205 View Answer

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DI NEW PATTERN Option D Solution: Difference in SP = 93% of MP – 84% of MP So 9% of MP = 486 Solve, MP = 5400 So SP by store C = 76% of 5400 = Rs 4104 3. Average SP of article III by stores A and B is Rs 3608, by stores B and C is Rs 3300. Find the SP of article III by store C. A) Rs 2984 B) Rs 3122 C) Rs 3080 D) Rs 2764 E) Rs 3452 View Answer Option C Solution: Let x% discount by store B So [84 + (100-x)]/2*100 * MP = 3608 And [70 + (100-x)]/2*100 * MP = 3300 Put value of (100-x) from 1 equation to another and solve for MP MP = Rs 4400 So SP by store C = 70/100 * 4400 = Rs 3080 4. Store A earned 10% profit by selling product V. If CP of articles at all articles is same, find the ratio of profits by B and C in selling V. A) 6 : 15 B) 9 : 16 C) 8 : 15 D) 6 : 13 E) 9 : 14 View Answer Option E Solution: MP = x SP by A = 88% of x So CP by A = 100/110 * 88/100 *x = 4x/5 Now CP is same SP by B = 89x/100, so profit of B = 89x/100 – 4x/5 = 9x/100 SP by C = 94x/100, so profit of B = 94x/100 – 4x/5 = 14x/100 So required ratio = 9 : 14 28 | P a g e

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DI NEW PATTERN 5. Ratio of discounts on article IV by stores A and B is 2/3. Difference in SP of article IV by stores A and C is Rs 432. If SP of article IV by store A is Rs 216 more than that by store B, find the SP of article IV by store C? A) Rs 2138 B) Rs 2687 C) Rs 2736 D) Rs 2522 E) Rs 2544 View Answer Option C Solution: x/18 = 2/3 So discount by A = 12% Now 88% of MP = 82% of MP + 216 Solve, MP = Rs 3600 Now: let y% discount by store C [88 – (100-y)]/100 * MP = 432 MP = Rs 3600 Solve, y = 24% So SP by C = 76% of 3600 = Rs 2736 Directions (6 – 10): Study the following pie-chart and table carefully and answer the questions that follow: Some values are missing. Answer the questions on the basis of given table and information in question. Speed of stream is same for both upstream and downstream distance on respective days

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6. Time taken to cover the upstream distance on Friday is same as time taken to cover the downstream distance on Thursday. Total speed of still water on Thursday and Friday is 10 km/hr. Find the ratio of speed of still water on Thursday and Friday. A) 6 : 13 B) 4 : 11 C) 7 : 13 D) 4 : 15 E) 7 : 12 View Answer Option C Solution: Distances upstream: Monday = 18/100 * 150 = 27 km, Tuesday = 12/100 * 150 = 18 km. 30 | P a g e

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DI NEW PATTERN Wednesday = 39 km, Thursday = 45 km, Friday = 21 km Similarly Distances downstream: Monday = 27 km, Tuesday = 45 km. Wednesday = 18 km, Thursday = 36 km, Friday = 54 km Let speed of still water upstream on Friday = x, then downstream on Thursday = (10-x) Now 21/(x-3) = 36/[(10-x)+2.5)] Solve, x = 6.5 So required ratio is 3.5 : 6.5 = 7 : 13 7. On Monday, the boat takes a total of 4 hrs 30 minutes to cover both upstream and downstream distance. Ratio of speed of boat in still water in going upstream to downstream is 4 : 5. Find the speed of boat in still water while going downstream. A) 13 km/hr B) 15 km/hr C) 9 km/hr D) 10 km/hr E) 12 km/hr View Answer Option B Solution: speeds – 4x and 5x So on Monday 27/(4x-3) + 27/(5x+3) = 9/2 Solve, x = 3 So downstream speed = 5x = 15 km/hr 8. On Tuesday, Ratio of speed of boat in still water in going upstream to downstream is 3 : 8. Also difference in speed of boat in still water in going upstream and downstream is 5 km/hr. If the total time taken by boat to cover upstream and downstream distance is 14 hours on Tuesday, find the speed of stream. A) 2 km/hr B) 1 km/hr C) 3 km/hr D) 2.5 km/hr E) 1.5 km/hr View Answer Option B Solution: 8x and 3x, Also 8x – 3x = 5 31 | P a g e

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DI NEW PATTERN So x = 1, speeds are 8 and 3 km/hr So for Tuesday 18/(3-b) + 45/(8+b) = 14 Solve, b = 1 km/hr 9. On Wednesday, Ratio of speed of boat in still water in going upstream to downstream is 4 : 5. The difference between time to cover upstream distance and downstream distance is 5 hours, find the total time taken to cover upstream distance and downstream distance. A) 5 hours B) 4 hours C) 7 hours D) 8 hours E) 6 hours View Answer Option D Solution: On Wednesday 39/(4x-2) – 18/(5x+2) = 10 Solve Solve, x = 2 km/hr So required distance : 39/(8-2) + 18/(10+2) = 8 hours 10.Time taken to cover the upstream distance on Thursday is 12 hours more than time taken to cover the downstream distance on Friday. Total speed of still water on Thursday and Friday is 11 km/hr. Find the ratio of speed of still water on Thursday and Friday. A) 6 : 7 B) 4 : 5 C) 7 : 9 D) 5 : 6 E) 7 : 8 View Answer Option D Solution: Let speed of still water upstream on Thursday = x, then downstream on Friday = (11-x) Now 45/(x-2.5) – 54/[(11-x)+3)] = 12 15/(x-2.5) – 18/(14-x) = 4 The time is integral value, so by putting values higher than 2.5, x can be found 32 | P a g e

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DI NEW PATTERN x=5 So required ratio is 5 : 6

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