RRB Clerk Prelims Memory Based Paper (Held On: 14 August 2022 Shift 1) (Previous Year Paper)

6 Friends: E, R, K, S, A and G

Birthday dates: Either on 11th or 18th

Months: March, May and November

1) Only two persons have a birthday before G.

2. R has a birthday on the 18th of the month.

3. Number of persons having birthdays before K is same as the number of persons having birthdays after R.

Case 1

 

Case 2

 

4. E has a birthday in a month having 30 days.

5. Not more than two persons have a birthday after S.

Case 1

 

Case 2

 

6. R and A have their birthdays in different months.

Thus, case 1 gets eliminated.

 

Hence, two persons have their birthdays between R and G.

6 Friends: E, R, K, S, A and G

Birthday dates: Either on 11th or 18th

Months: March, May and November

1) Only two persons have a birthday before G.

2. R has a birthday on the 18th of the month.

3. Number of persons having birthdays before K is same as the number of persons having birthdays after R.

Case 1

 

Case 2

 

4. E has a birthday in a month having 30 days.

5. Not more than two persons have a birthday after S.

Case 1

 

Case 2

 

6. R and A have their birthdays in different months.

Thus, case 1 gets eliminated.

 

Hence, both S and E birthday is between 11th May and 18th November.

6 Friends: E, R, K, S, A and G

Birthday dates: Either on 11th or 18th

Months: March, May and November

1) Only two persons have a birthday before G.

2. R has a birthday on the 18th of the month.

3. Number of persons having birthdays before K is same as the number of persons having birthdays after R.

Case 1

 

Case 2

 

4. E has a birthday in a month having 30 days.

5. Not more than two persons have a birthday after S.

Case 1

 

Case 2

 

6. R and A have their birthdays in different months.

Thus, case 1 gets eliminated.

 

Hence, R birthday is on 18th November.

6 Friends: E, R, K, S, A and G

Birthday dates: Either on 11th or 18th

Months: March, May and November

1) Only two persons have a birthday before G.

2. R has a birthday on the 18th of the month.

3. Number of persons having birthdays before K is same as the number of persons having birthdays after R.

Case 1

 

Case 2

 

4. E has a birthday in a month having 30 days.

5. Not more than two persons have a birthday after S.

Case 1

 

Case 2

 

6. R and A have their birthdays in different months.

Thus, case 1 gets eliminated.

 

Hence, A birthday is on 18th March.

6 Friends: E, R, K, S, A and G

Birthday dates: Either on 11th or 18th

Months: March, May and November

1) Only two persons have a birthday before G.

2. R has a birthday on the 18th of the month.

3. Number of persons having birthdays before K is same as the number of persons having birthdays after R.

Case 1

 

Case 2

 

4. E has a birthday in a month having 30 days.

5. Not more than two persons have a birthday after S.

Case 1

 

Case 2

 

6. R and A have their birthdays in different months.

Thus, case 1 gets eliminated.

 

Hence, five people have a birthday after K.

The least possible Venn diagram is as follows:

Conclusion:

I: All bags are cats: False (No direct relation between Bags and cats, it is possible but not definite)

II: Some bags are not cats: True (The part of Bags which is books can never be cats as no cats are books)

Hence, Only II follows.

The best possible Venn diagram is,

Conclusions:

I. All rat are tigers: False( As no direct relation is given between rat and tigers, it is possible but not definite)

II. Some rats are not tigers: False( As no direct relation is given between rat and tigers, it is possible but not definite)

Conclusion I and II form a complementary pair.

Hence, Either I or II follows

Key Points

Conditions for Either - Or 

1. Subject and predicate should be the same
2. Both the individual conclusions should be false
3. One conclusion should be positive and one conclusion should be negative
4. There should either one of the following pairs present in the conclusions

• All + Some not
• Some + No

Statements

Only a few animal are cat → Some part of Animal are overlapping with Cat and some part of animal are not cat

Some animal are dog → Some part of animal are overlapping with dog 

Venn diagram for the given statements are follows.

Conclusions

I. All Cat are animal → False ( some part of animal are not cat but remaining part of the animal can be cat so, this possible statement not definite hence false)

II. Some cat are Dog → False( some dog are cat or not is definite it can be a possible case but not definite)

Hence Neither I nor II follow is the correct answer.

The least possible Venn diagram is as follows:

Conclusions:

I. Some pink are black. → True (As it is given  all black are pink this imply that some pink are black.)

II. Some red are pink. → False (As there is no definite relation between red and pink. Therefore this is possible but not definite).

Hence, only 1 follows.

Given person: A, B, C, D, and E

1.  There is only one married couple.

2. B is the father of C who has one daughter.

3. C doesn’t have a sibling. 

4. E is the mother of A who is a female.

5. D has only one brother.

6. D is an unmarried female.

Therefore, the final arrangement is,

Hence, A is the granddaughter of B.

Given person: A, B, C, D, and E

1.  There is only one married couple.

2. B is the father of C who has one daughter.

3. C doesn’t have a sibling. 

4. E is the mother of A who is a female.

5. D has only one brother.

6. D is an unmarried female.

Therefore, the final arrangement is,

Hence, Three female members in the family.

Given person: A, B, C, D, and E

1.  There is only one married couple.

2. B is the father of C who has one daughter.

3. C doesn’t have a sibling. 

4. E is the mother of A who is a female.

5. D has only one brother.

6. D is an unmarried female.

Therefore, the final arrangement is,

Hence, C is the Nephew of D.

Persons: P, Q, R, S, T, U, and V

I. Q goes on the trip on Thursday. 

II. Only one person goes on the trip between Q and V. 

III. P goes on the trip immediately after V.

 

IV. The number of people who go on the trip before P is the same as the number of people who go on the trip after R. 

V. Only one person goes on the trip between R and S. 

VI. U go on the trip after S. Hence case 1 is eliminated.

The final arrangement is as:

                         

Hence, 3 people go on the trip between R and U.

Persons: P, Q, R, S, T, U, and V

I. Q goes on the trip on Thursday. 

II. Only one person goes on the trip between Q and V. 

III. P goes on the trip immediately after V.

IV. The number of people who go on the trip before P is the same as the number of people who go on the trip after R. 

V. Only one person goes on the trip between R and S. 

VI. U go on the trip after S. Hence case 1 is eliminated.

The final arrangement is as:

                         

There is only one person go on the trip between UP, QV, SU, and RS thus forms a group except for the TV.

Hence, TV is the odd one.

Persons: P, Q, R, S, T, U, and V

I. Q goes on the trip on Thursday. 

II. Only one person goes on the trip between Q and V. 

III. P goes on the trip immediately after V.

IV. The number of people who go on the trip before P is the same as the number of people who go on the trip after R. 

V. Only one person goes on the trip between R and S. 

VI. U go on the trip after S. Hence case 1 is eliminated.

The final arrangement is as:

Hence, P goes on the trip on Sunday.

Persons: P, Q, R, S, T, U, and V

I. Q goes on the trip on Thursday. 

II. Only one person goes on the trip between Q and V. 

III. P goes on the trip immediately after V.

 

IV. The number of people who go on the trip before P is the same as the number of people who go on the trip after R. 

V. Only one person goes on the trip between R and S. 

VI. U go on the trip after S. Hence case 1 is eliminated.

The final arrangement is as:

       

Hence, S goes on the trip on Wednesday.

Persons: P, Q, R, S, T, U, and V

I. Q goes on the trip on Thursday. 

II. Only one person goes on the trip between Q and V. 

III. P goes on the trip immediately after V.

 

IV. The number of people who go on the trip before P is the same as the number of people who go on the trip after R. 

V. Only one person goes on the trip between R and S. 

VI. U go on the trip after S. Hence case 1 is eliminated.

The final arrangement is as:

                  

Hence, V goes on the trip on Saturday.

According to the question,

Row 1: A3, A5, A7, A9 (facing south)

Row 2: A2, A4, A6, A8 (facing north)

1. No person sits to the right of A5.

2. The person who faces A5 is a neighbour of A6.

3. One person sits between A6 and A8.

4. The person who faces A4 is an immediate neighbour of A9.

5. A9 does not sit at any extreme end of a row.

6. A3 does not sit at any of the extreme end.

7)  A9 is sitting immediate right of A3.

From the above arrangement, Either A2 or A4 sits opposite to A5.

Hence, “Either (1) or (3)” is the correct answer.

According to the question,

Row 1: A3, A5, A7, A9 (facing south)

Row 2: A2, A4, A6, A8 (facing north)

1. No person sits to the right of A5.

2. The person who faces A5 is a neighbour of A6.

3. One person sits between A6 and A8.

4. The person who faces A4 is an immediate neighbour of A9.

5. A9 does not sit at any extreme end of a row.

6. A3 does not sit at any of the extreme end.

7)  A9 is sitting immediate right of A3.

Hence, “A3” faces the A4 who is the only neighbour of A8.

According to the question,

Row 1: A3, A5, A7, A9 (facing south)

Row 2: A2, A4, A6, A8 (facing north)

1. No person sits to the right of A5.

2. The person who faces A5 is a neighbour of A6.

3. One person sits between A6 and A8.

4. The person who faces A4 is an immediate neighbour of A9.

5. A9 does not sit at any extreme end of a row.

6. A3 does not sit at any of the extreme end

7)  A9 is sitting immediate right of A3.

From the above arrangement, all of them are placed at the end except A3.

Hence, “A3” is the odd one and doesn’t belong to the group.

According to the question,

Row 1: A3, A5, A7, A9 (facing south)

Row 2: A2, A4, A6, A8 (facing north)

1. No person sits to the right of A5.

2. The person who faces A5 is a neighbour of A6.

3. One person sits between A6 and A8.

4. The person who faces A4 is an immediate neighbour of A9.

5. A9 does not sit at any extreme end of a row.

6. A3 does not sit at any of the extreme end.

7)  A9 is sitting immediate right of A3.

Hence, A7 sits third to the left of A5.

According to the question,

Row 1: A3, A5, A7, A9 (facing south)

Row 2: A2, A4, A6, A8 (facing north)

1. No person sits to the right of A5.

2. The person who faces A5 is a neighbour of A6.

3. One person sits between A6 and A8.

4. The person who faces A4 is an immediate neighbour of A9.

5. A9 does not sit at any extreme end of a row.

6. A3 does not sit at any of the extreme end.

7) 

Hence, “A6” placed opposite to A9.

Given Statements: A ≥ S ≥ D ≤ F ≤ G ≤ I < K

Conclusions:

I. D < G → False (D ≤ F ≤ G → D ≤ G)

II. K > D → True (as D ≤ F ≤ G ≤ I < K → its clear K > D)

Hence, only conclusion II follows

Given Statement: Q > W > E ≥ R ≥ T = P < Y

Conclusions:

I. Q ≥ Y → False (Relation cannot be determined between Q and Y)

II. Y > E → False (Relation cannot be determined between Y and E)

Hence, Neither conclusion I nor II follows

Given Statement: D > T > S = C ≤ X ≤ E ≤ G

Conclusions:

I. S < E → False (we can’t say because S = C ≤ X ≤ E → S ≤ E)

II. E = S → False (we can’t say because S = C ≤ X ≤ E → S ≤ E)

Therefore, the conclusion I and II form a complementary pair.

Hence, either conclusion I or II follows.

According to the given question,

The forward pairs are - AC

Hence, there is only one pair.

Given Word: CALCULATIVE

Third, the fifth, the seventh, eighth and the tenth letter : L, U, A, T, V

The word that can be formed is: VAULT

The second last letter of the word: L

Hence, ‘L’ is the correct answrer.

Given sequence is:

684    145    529    574    169

As per the given information, all the digits are arranged in ascending order within the number.

The series is,

684 → 468

145 → 145

529 → 259

574 → 457

169 → 169

Therefore, the highest number is 468.

Hence, the correct answer is 684.

Given sequence is:

684    145    529    574    169

As per the given information, the first digit of each is dropped and the second and the third digit are interchanged then the series will be,

684 → 48

145 → 54

529 → 92

574 → 47

169 → 96

Therefore, the highest number is 96.

Hence, 169 is the correct answer.

Given sequence is:

684    145    529    574    169

As per the given information, one is subtracted to each number then the sequence is

683     144     528      573      168

Therefore, 144 is the number which is a perfect square of 12.

Hence, one is the correct answer.

Given sequence is:

684    145    529    574    169

As per the given information, a first digit and the last digit are interchanges within the number the sequence will be,

486      541        925       475        961

The number which is the second-highest is 925.

Hence, the correct answer is 529.

Given sequence is:

684    145    529    574    169

As per the given information, the first digit of the highest number is divided by the second digit of the second-lowest number.

Therefore, the increasing order of the given sequence is,

(Lowest number) 145   169     529   574   684 (Highest number)

The first digit of the highest number = 6

the second digit of the second-lowest number = 6

6/6 = 1

Hence, the resultant is 1.

Code for health: Either pox, asw, or ktr

Code for laptop: tar

Code for heated: nyc

Code for are: ctx

Code for sun: tux

Hence, Either Option 1 or 3 is the correct answer.

Code for well: fou

Code for but: ctd

Code for cool: sit

Hence, fou sit ctd is the correct answer.

Word for the code nor: Either try or pour

Word for the code tux: sun

Hence, try sun is the correct answer

Code for the word heated: nyc

Code for the word are: ctx

Code for the word home: pqr/ nxw

Code for the word well: fou

Code for the word cool: sit

Code for the word girls:  pqr/ nxw

Hence, nyc sit pqr ctx fou nxw is the correct answer.

Given number: 96235741

 

The 5^th digit from the left end is 4 and the 3^rd digit from the right end is 6.

So, Product = 4 × 6 = 24

Hence, 24 is the correct answer.

Given points: P, Q, R, S, T, U, V, and W.

1)Point P is 10m north of point Q. Point Q is 15m west of point R.

2. Point S is 10m north of point R. Point S is 25m north of point T.

3. Point W is 8m east of S. Point U is 8m east of point T.

4.  Point U is 20m south of point V.

Therefore, the diagram is,  

 

On applying the Pythagoras theorem, we get the shortest distance between point R and V.

$〖RV〗^2=64+25$

$=√89$

$=√89 m$

Hence**, $√89$ m is the shortest distance between points R and V.**

Given points: P, Q, R, S, T, U, V, and W.

1)Point P is 10m north of point Q. Point Q is 15m west of point R.

2. Point S is 10m north of point R. Point S is 25m north of point T.

3. Point W is 8m east of S. Point U is 8m east of point T.

4.  Point U is 20m south of point V.

Therefore, the diagram is,  

 

PQ: P is to the north of Q

SR: S is to the north of R

ST: S is to the north of T

VU: V is to the north U

RV: R is to the south west of V

Hence, RV is the odd one.

Given points: P, Q, R, S, T, U, V, and W.

1)Point P is 10m north of point Q. Point Q is 15m west of point R.

2. Point S is 10m north of point R. Point S is 25m north of point T.

3. Point W is 8m east of S. Point U is 8m east of point T.

4.  Point U is 20m south of point V.

Therefore, the diagram is,  

Hence, the shortest distance between point W and Point V is 5m.

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculation:

(3/8) of (4/5) of (5/6) of 7450 = ?

⇒ (3/8) × (4/5) × (5/6) × 7450 = ?

⇒ 7450/4 = ?

⇒ 1862.5 = ?

∴ The value of ? is 1862.5

Given:

75 + [(85 – 32) of 20] = ?

Concept used:

We have to follow the BODMAS rule

Calculation:

75 + [(85 – 32) of 20] = ?

⇒ 75 + (53 of 20) = ?

⇒ 75 + 1060 = ?

⇒ 1135 = ?

∴ The value of ? is 1135.

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculations:

55 × 35 – 25 × 35 = ? + 80

⇒ 35(55 - 25) = ? + 80

⇒ 35 × 30 = ? + 80

⇒ ? = 1050 - 80

⇒ ? = 970

∴ The value of ? is 970.

Follow BODMAS rule to solve this question, as per the order given below.

Calculation:

√1444 + √5184 = 22 × ?

⇒ 38 + 72 = 22 × ?

⇒ 110 = 22 × ?

⇒ ? = 5

∴ The value of ? is 5.

Given:

\(25{\rm{% ;of; }}2500 - 33\frac{1}{3}% ;{\rm{of; }}27\)

Concept used:

We have to follow the BODMAS rule

Calculation:

(25/100) of 2500 – (1/3) of 27 = ?

⇒ 625 – 9 = ?

⇒ ? = 616

∴ The value of ? is 616.

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculations:

16.67% of 366 – 12.5% of 96 = ? × 7 

⇒ (1/6) × 366 – (1/8) × 96 = ? × 7      [12.5% = 1/8, 16.67% = 1/6]

⇒ 61 – 12 = ? × 7 

⇒ 49 = ? × 7 

⇒ ? = 49/7 

⇒ ? = 7

∴ The value of ? is 7.

Given:

21.6 − 18.8 − 3 + 3.8 × 3 = ?

Concept Used:

Calculation:

21.6 – 18.8 – 3 + 3.8 × 3 = ?

⇒ 2.8 – 3 + 11.4 = ?

⇒ ? = 14.2 – 3

∴ ? = 11.2

Follow these BODMAS rules to solve the question

Given expression is,

60% of 80 ÷ 16 × 30% of 70 = ?

⇒ 48 ÷​ 16 ×​ 21 = ?

⇒ 3 ×​ 21 = ?​

∴ ? = 63

Calculation:

We use the BODMAS rule as follows:

⇒ (9/4) + (31/6) – (33/8) = ? + (13/12)

⇒ (27 + 62)/12 – (33/8) – (13/12) = ?

⇒ (89/12) – (13/12) – (33/8) = ?

⇒ (76/12) – (33/8) = ?

⇒ (19/3) – (33/8) = ?

⇒ (152/24) – (99/24) = ?

⇒ ? = 53/24

∴ The required answer is 53/24.

Given:  

29 × ∛27 + 113 = (?)2 + 79

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

29 × ∛27 + 113 = (?)2 + 79

⇒ 29 × 3 + 113 = (?)2 + 79

⇒ 87 + 113 - 79 = (?)²

⇒ 121 = (?)²

⇒ ? = √121

⇒ ? = 11

∴ 11 will come in place of the question mark ('?').

Given:  

(1/11) of 440 - (1/8) of 640 = 10 × ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(1/11) of 440 - (1/8) of 640 = 10 × ?

⇒ 40 - 80 = 10 × ?

⇒ -40 = 10 × ?

⇒ ? = -40/10

⇒ ? = -4

∴ -4 will come in place of the question mark ('?').

Given:

9 × 24 ÷ 6 + 10 × 6 = ?

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculation:

9 × 24 ÷ 6 + 10 × 6 = ?

⇒ 9 × 4 + 10 × 6 = ?

⇒ 36 + 60 = ?

⇒ ? = 96

Given

A can do a work in 18 Days

He started the work and worked for 9 days alone and after that completed the work with B in 4 days.

Formula Used 

Work = Time × Efficiency

Calculation

According to the question,

Let the efficiency of A be x and B be y

⇒ 18 × x = 9 × x + (x + y)× 4

⇒ 18x = 13x + 4y

⇒ 5x = 4y

⇒ x/y = 4/5

Per Day efficiency of A is 4 and B is 5

Total work = Man efficiency × Day

⇒ 4 × 18 = 72 unit

Complete the work together

⇒ Total work/Efficiency of A and B

⇒ 72/9

⇒ 8 Days

∴ The work completed together in 8 days.

Given:

Profit ratio of Pooja : seems : rutu = 5 : 7 : 8

Formula:

Profit = Investment × time

Calculations:

Let their investments be

Pooja Rs. x for 14 months,

Seema Rs. y for 8 months and

Rutu Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

⇒ 14x/8y ​= 5/7

⇒ 98x = 40y

⇒ Y = 49​x/20

⇒ And, 14x/7z ​= 5/8

⇒ 112x = 35z

⇒ z = 112​x/35 = 16​x/5

⇒ x : y : z = x : 49​x/20 : 16​x/5 = 20 : 49 : 64

∴ Ratio of their investments is 20 : 49 : 64

Given:

The diagonal of a square = √2 × (The diagonal of a rectangle)

The perimeter of the rectangle = 62 cm

The breadth of the rectangle = 7 cm

Formula Used:

(i) For Rectangle,

Diagonal of rectangle = √(L² + B²)

Area of rectangle = L × B

Perimeter of rectangle, P = 2(L + B)

Where, L = Length of rectangle 

B = Breadth of rectangle 

P = Perimeter of rectangle

(ii) For Square,

Diagonal of square = √2 × (Side of square)

Area of square = (side)²

Calculation:

P = 2(L + B)

⇒ 62 = 2(L + 7)

⇒ L = (62/2) - 7

⇒ L = 31 - 7

⇒ L = 24 cm

Length of the rectangle = 24 cm

Area of rectangle = L × B

⇒ Area of rectangle = 24 × 7

⇒ 168 cm²

Diagonal of rectangle = √(L2 + B2)

⇒ √{(24)² + (7)²}

⇒ √625

⇒ 25 cm

According to the question,

Diagonal of square = 25√2 cm

⇒ √2 × (Side of square) = 25√2

⇒ Side of square = 25 cm

Area of square = (25)²

⇒ 625 cm²

Required difference = (625 - 168) cm²

⇒ 457 cm²

40% of the required difference = 40% of 457

⇒ (40/100) × 457

⇒ 182.80 cm²

∴ 40% of the difference between the area of the square and the rectangle will be 182.80 cm².

Total sales of Lifestyles for four items = 40 + 35 + 25 + 18 = 118

Total sales of Trends for four items = 30 + 45 + 65 + 20 = 160

Required difference = 160 – 118 = 42

Total sales of Pantaloons for four items = 55 + 30 + 35 + 15 =135

Sales of Footwear of Pantaloons = 30

Required percentage =

⇒ 30/135 × 100

⇒22.22% ≈ 22%

Total sales of Cosmetics for four shops in hundreds = 65 + 20 + 35 + 25 = 145

Total sales of Cloths for four shops in hundreds = 30 + 35 + 55 + 40 = 160

Required ratio =

⇒ 145: 160

⇒ 29: 32

Total sales of Max for four items = 3500 + 5000 + 2000 + 1000 = 11500

Required average = 11500/4 = 2875

Total sales of Footwear for four shops(in hundreds) = 45 + 50 + 30 + 35 = 160

Total sales of Watches for four shops (in hundreds)= 20 + 10 + 15 + 18 = 63

Required percentage =

⇒(160 – 63)/160 × 100%

⇒ 60.625% ≈ 61%

Given:

Amount = Rs 1176

3/5 of first : 2/7 of second = 3 : 2

Calculation:

Let the parts be a and b

So, 3a/5 : 2b/7 = 3 : 2

⇒ 21a/10b = 3/2

⇒ a/b = 5/7

⇒ a = 1176/12 × 5 = 490

∴ The first part is Rs. 490.

Given:

Pipe P can fill a tank in 44 minutes

Pipe R can fill a tank in 33 minutes.

Calculation:

Total capacity of tank = 132 units

⇒ Tank filled in 1 minute by pipe P = 3 units (i.e. 132 / 44)

⇒ Tank filled in 1 minute by pipe R = 4 units (i.e. 132 / 33)

⇒ Tank filled in first minute by pipe P = 3 units

⇒ Tank filled in second minute by pipe P = 3 units

⇒ Tank filled in third minute by pipe R = 4 units

⇒ Tank filled in fourth minute by pipe R = 4 units

So, in 4 minutes tank is filled = 14 units

⇒ in first 36 minutes tank is filled = 126 units

remaining 6 units will be filled by pipe P in next 2 minutes.

∴ 38 minutes are required to fill that tank

Given:

Elected candidate = 85% of votes.

Formula used:

X% of Y = (X × Y)/100

Calculation:

⇒ 85% - 15% = 910

⇒ 70% = 910

⇒ 100% = (910/70) × 100 = 1300

∴ The total number of votes polled is 1300.

Given:

Speed of three cars are in the ratio 5 ∶ 6 ∶ 8

Concept used:

Speed = 1/time      ----(When distance is equal)

Calculation:

Let, the total distance be d

Speed of A car = 5x

Speed of B car = 6x

Speed of C car = 8x

Time taken to cover the distance by car A = d/5x

Time taken to cover the distance by car B = d/6x

Time taken to cover the distance by car C = d/8x

According to the question,

⇒ (d/5x) ∶ (d/6x) ∶ (d/8x)

⇒ (1/5) ∶ (1/6) ∶ (1/8)

⇒ 24 ∶ 20 ∶ 15

∴ Require time ratio = 24 ∶ 20 ∶ 15.

Given:

Average height of 5 friends = 162 cm, 2 friends join the group and their average height is 156 cm

Formula used:

Average height = total height/no of persons

Calculation:

After joining two friends

Total height = 5 × 162 + 2 × 156

⇒ 810 + 312

⇒ 1122

Total no of persons = 7

Average height of this group = 1122/7 = 160.3

∴ Average height of this group should be 160.3 cm.

The series follows the following pattern:

480 ÷ 5 = 96                                          

96 ÷ 4 = 24

24 ÷ 3 = 8

8 ÷ 2 = 4

4 ÷ 1 = 4

∴ The wrong term in the series is 1

The series follows the following pattern:

34 + 7 = 41

41 + 11 = 52

52 + 15 = 67

67 + 19 = 86

86 + 23 = 109

∴ The wrong term in the series is 54.

The series follows the following pattern:

17 + 3 = 20

20 + 5 = 25

25 + 7 = 32

32 + 11 = 43

43 + 13 = 56

∴ The wrong term in the series is 55.

The series follows the following pattern:

76 + 12 = 77

77 - 22 = 73

73 + 32 = 82

82 - 42 = 66

66 + 52 = 91

∴ The wrong term in the series is 78.

The series follows the following pattern:

6² – 1 = 35

7² + 1 = 50

82 – 1 = 63

92 + 1 = 82

102 – 1 = 99

112 + 1 = 122

∴ The wrong term in the series is 80.

Given:

The production of the Pulse by the state UK, UP, Delhi, JK, and HP in 2020 = 150, 200, 150, 450, 150

Formula used

Total production of the Pulse in 2020 = production by the state UK + Production by the state UP + Production by the state Delhi + Production by the state JK + Production by the state HP

Calculation

The total production of the Pulse in 2020 = 150 + 200 + 150 + 450 + 150

Then,

The total production of the Pulse in 2020 by the state UK, UP, Delhi, JK, and HP = 1100

Given:

The Production of the Pulse by the State UK, Delhi, and HP in 2019 = 100, 200, and 250

The production of the Pulse by the state UP, JK, and the UK in 2020 = 200, 450, and 150

Formula used

Ratio = Total production of the Pulse in 2019 by the state UK, Delhi, and HP ÷  total production of the Pulse in 2020 by the state UP, JK, and the UK

Calculation

Ratio = (100 + 200 + 250) ÷ (200 + 450 + 150)

Ratio = 550 ÷ 800

Ratio = 11 ÷ 16

Given:

The production of the Pulse by the state HP, Delhi, and JK in 2019 = 250, 200, and 350

Formula used

The Average tones production of the Pulse = total production of the Pulse in 2019 state HP, Delhi, and JK ÷ total number of the state

Calculation

The average tones production of the Pulse = (250 + 200 + 350) ÷ 3

The average tones production of the Pulse = 800 ÷ 3

The average tones production of the Pulse = 266.66 tones

Given:

The Production of the Pulse by the state UK in 2020 = 150 and increased 20%

The Production of the Pulse by the state Delhi in 2020 = 150 and increased 40%

Formula used

Total Production of the Pulse by state UK and Delhi in 2021 = after increased the production of the Pulse by State UK + after increased the production of the Pulse by state Delhi

The increased production of the Pulse = given Production of the Pulse × increased percentage ÷ 100

Calculation

 The total production increased by the state UK = 150 + 150 × 20 ÷ 100

The total production of the Pulse by the state UK = 150 + 30 = 180

The total production increased by the state Delhi = 150 + 150 × 40 ÷ 100

The total production of the Pulse by the state Delhi = 150 + 60 = 210

Then total production of the Pulse by the state UP and the sate Delhi in 2021 = 180 + 210 = 390 tones

Given:

The production of the Pulse by State UP = 300 in 2019

The production of the Pulse by State UP = 200 in 2020

Formula used

Percentage of the Pulse in UP decreased = (Greatest production of Pulse – Smallest production of the Pulse) ÷ Greater production of the Pulse × 100

Calculation

The Percentage of the Pulse in UP decreased = (300 - 200) ÷ 300 × 100

The percentage of Pulse in UP decreased = 100 ÷ 300 × 100

The percentage of Pulse in UP decreased = 33.33% 

∴ The Production of the Pulse by the State UP decreased by 33.33% from 2019 to 2020

if distance is same, then speed is inversely proportional to time taken

⇒ ratio between speed in upstream to downstream is 3 : 5

Let speed of boat in upstream and downstream be 3x and 5x respectively.

speed of stream = (5x - 3x)/2 = 4

⇒ x = 4

upstream speed = 3x = 3 × 4 = 12 km/hr

∴  time taken to cover 96km in upstream = 96/12 = 8 hours

Given: 

A sum of ₹ 2550 was lent partly at 7% and partly at 9% per annum simple interest.

The total interest received after 4 yr was ₹ 800

Calculation:

Let the Sum lent at 7% = P

Then, Sum lent at 9% = (2550 – P)

Then,

[(P x 7 x 4)/100] + [{(2550 - P) x 9 x 4}/100] = 800

⇒ 7P + 22950 – 9P = 80000/4

⇒ - 2P = 20000 – 22950

⇒ - 2P =  - 2950

⇒ P = Rs. 1475

Now,

Sum lent at 9% = 2550 – 1475

⇒ Rs. 1075

Required ratio = 1475 : 1075 = 59 : 43

Therefore the correct answer is 59 : 43

Given:

The ratio of milk and water in the first vessel = 3 : 5

The ratio of milk and water in the second vessel = 3 : 2

The ratio of milk and water in the final mixture = 2 : 3

Concept used:

(New quantity of milk/New quantity of water) = (Quantity of milk taken from first vessel + Quantity of milk taken from the second vessel)/(Quantity of water taken from first vessel + Quantity of water taken from the second vessel)

Calculation:

Let the mixture from the two-vessel be mixed in the ratio x : y.

Quantity of milk taken from the first vessel = 3x/8

Quantity of water taken from the first vessel = 5x/8

Quantity of milk taken from the second vessel = 3y/5

Quantity of water taken from the second vessel = 2y/5

Then, {(3x/8) + (3y/5}/{(5x/8) + (2y/5)} = 2/3

⇒ {(15x + 24y)/40}/{(25x + 16y)/40} = 2/3

⇒ 3(15x + 24y) = 2(25x + 16y)

⇒ 45x + 72y = 50x + 32y

⇒ 50x – 45x = 72y – 32y

⇒ 5x = 40y

⇒ x/y = 8/1

∴ The first and second ratio is mixed in the ratio 8 : 1.

GIVEN :

Rohit is as much elder than Virat as he is younger to Jadeja,

Sum of ages of Virat and Jadeja is 36 years.

CALCULATION :

Let the present age of Rohit be x years and he is younger to Jadeja by y years

Hence, Jadeja’s age = (x + y) years

Virat’s age = (x – y) years

According to the question,

Jadeja’s age + Virat’s age = 36

⇒ (x + y) + (x – y) = 36

⇒ 2x = 36

⇒ x = 18 years

∴ Rohit’s age is 18 years.

Shortcut Trick

Present age of Rohit = (sum of ages)/2 = 36/2 = 18 years

∴ Rohit’s age is 18 years.

Given:

S.P. of pen = 50

Loss % = 20% 

Formula used 

C.P. = (100/100 - loss%) × S.P.)

Calculations:  

⇒ C.P. Of Pen = (100/80 × 50)

⇒ C.P. Of Pen = 500/8

⇒ C.P. Of Pen =  Rs. 62.5

∴ The required answer is Rs. 62.5