RRB Clerk Prelims Memory Based Paper (Held On: 12 August 2023 Shift 1) (Previous Year Paper)

Persons: S, T, U, V, W, X, Y and Z.

1. S and Z are immediate left of each other.

2. T sits opposite to Z. 

3. Y sits second to the left of U and both face in same direction.

4. U does not sit adjacnet to Z. 

5. V and W are adajcnet to each other.

6. Neither V nor X face away from the center. So, case 2 won’t be possible. 

Hence, W sits immediate right of Z.

Persons: S, T, U, V, W, X, Y and Z.

1. S and Z are immediate left of each other.

2. T sits opposite to Z. 

3. Y sits second to the left of U and both face in same direction.

4. U does not sit adjacnet to Z. 

5. V and W are adajcnet to each other.

6. Neither V nor X face away from the center. So, case 2 won’t be possible. 

Hence, T sits second to the left of X.

Persons: S, T, U, V, W, X, Y and Z.

1. S and Z are immediate left of each other.

2. T sits opposite to Z. 

3. Y sits second to the left of U and both face in same direction.

4. U does not sit adjacnet to Z. 

5. V and W are adajcnet to each other.

6. Neither V nor X face away from the center. So, case 2 won’t be possible. 

All face towards the center except S.

Hence, S does not belong to the group.

Persons: S, T, U, V, W, X, Y and Z.

1. S and Z are immediate left of each other.

2. T sits opposite to Z. 

3. Y sits second to the left of U and both face in same direction.

4. U does not sit adjacnet to Z. 

5. V and W are adajcnet to each other.

6. Neither V nor X face away from the center. So, case 2 won’t be possible. 

Hence, two persons sit between Y and V when counted from the right of Y.

Persons: S, T, U, V, W, X, Y and Z.

1. S and Z are immediate left of each other.

2. T sits opposite to Z. 

3. Y sits second to the left of U and both face in same direction.

4. U does not sit adjacnet to Z. 

5. V and W are adajcnet to each other.

6. Neither V nor X face away from the center. So, case 2 won’t be possible. 

If the directions are inverted:

Hence, T sits third to the left of S.

Hence, the correct answer is Y.

There are six pairs

Hence, there are more than five pairs.

Alternate Method:

We can use the place values of the letters and then we can find how many letters are there in between the letters. (we can use this method only when we remember all the letters place values)

When nothing specific about forward or backward is given in the question, we have to take both directions,

The least possible Venn diagram for the given statements is as follows:

Conclusions:

I. Some pens are digital → True ( as some pens are mobiles and all mobiles are digital)

II. All pens are digital → False ( it is possible but not definite)

Hence, only I follows.

Answer:- 4)

The possible Venn diagram according to the statement is as follows,

I. Some Eggs are Chicks → False( it is possible, but not definite because no relation between Eggs and chicks)

II. No eggs is Chicks → False ( it is possible, but not definite because no relation between Eggs and chicks)

Both conclusions are complementary to each other

Hence, Either I or II follows

Additional Information

What is complementary pair?

I. Both conclusions have to be False

II. Both Conclusions should have the same variables.

III. One conclusion should be positive and the other conclusions should be negative.

Complementary pairs:

(All and Some not)

(some and No)

The least possible diagram for the given statements is as follows

Conclusions:

I. Some Bright are Dark → True (It is definite)

II. Some Pictures are not Dark → True (It is definite as it is given that only a few Pictures are Dark)

Hence, both I and II follows.

From the given information,

Given Information –

1. A is the sister of B.

2. D is the wife of B.

3. E is the son of D.

4. G and E are brothers.

5. F is the father of B.

6. C is paternal grandmother of E.

We can draw following Family Tree from information given above

Hence, ‘Daughter’ is the correct answer

From the given information,

Given Information –

1. A is the sister of B.

2. D is the wife of B.

3. E is the son of D.

4. G and E are brothers.

5. F is the father of B.

6. C is paternal grandmother of E.

We can draw following Family Tree from information given above

Hence, ‘Grandson is the correct answer

From the given information,

Given Information –

1. A is the sister of B.

2. D is the wife of B.

3. E is the son of D.

4. G and E are brothers.

5. F is the father of B.

6. C is paternal grandmother of E.

We can draw following Family Tree from information given above

Hence, ‘Father- in-law’ is the correct answer.

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Consonant → Number → Vowel

Clearly none of the number follows the given condition$.

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Symbol → Consonant → Symbol

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Clearly there is one such pair i.e., < F ?

Given series is:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Arrangement required: Symbol → Vowel → Consonant

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

Clearly there are two such pairs they are**: @ E B** and £ O R

Given series:

Left Side N # S 3 @ E B 7 $ 9 I £ O R ! 6 M > 3 % D 5 X & Z 8 < F ? 4 G A Right Side

If all the symbols are dropped: N S 3 E B 7 9 I O R 6 M 3 D 5 X Z 8 F 4 G A.

Hence, 7^th letter from the left end is 9.$

BCAD → B + 1 = C, C - 2 = A, A + 3 = D.

WXVY → W + 1 = X, X - 2 = V, V + 3 = Y.

OPNQ → O + 1 = P, P - 2 = N, N + 3 = Q**.**

FHEG → F + 2 = H, H - 3 = E, E + 2 = G.

STRU → S + 1 = T, T - 2 = R, R + 3 = U.

Hence, FHEG is the odd one out.

Given words: ZEN WAR DIP LAW DEW

Interchange the first and third letter in each word.

New words: NEZ RAW PID WAL WED

Here we get two meaningful words RAW and WED.

Hence, the correct answer is 2.

Given words: ZEN WAR DIP LAW DEW

1. Arrange the words according to alphabetical series.

ZEN WAR LAW DIP DEW

2. Word immediate right of the third word from the right end.

ZEN WAR LAW DIP DEW

Here we get the word DIP which is immediate right of third word from the right end.

Hence, the correct answer is DIP.

Given words: ZEN WAR DIP LAW DEW

Interchange the second and third letter within the word.

ZNE WRA DPI LWA DWE

Here we get no words that end with a consonant.

Hence, the correct answer is 0.

Given words: ZEN WAR DIP LAW DEW

Consonant is changed to the previous letter in the alphabetical series and each vowel is changed to the next letter in the alphabetical series.

New words: YFM VBQ CJO KBV CFV

Here we get only one word which contains at least one vowel.

Hence, the correct answer is 1.

Boxes: A, B, C, G, H and K.

1. One box is placed in between C and K

2.  Two boxes are placed in between K and B, where the latter is below the former.

3. The number of boxes below B is one less than the number of boxes above A.

4. H is an immediate neighbour of A. So, CASE 1 and CASE 2B is eliminated.

Thus, the final arrangement is,

Thus, four boxes are placed in between box H and B.

Boxes: A, B, C, G, H and K.

1. One box is placed in between C and K

2.  Two boxes are placed in between K and B, where the latter is below the former.

3. The number of boxes below B is one less than the number of boxes above A.

4. H is an immediate neighbour of A. So, CASE 1 and CASE 2B is eliminated.

Thus, the final arrangement is,

Thus, box G and B are the immediate neighbour of box C.

Boxes: A, B, C, G, H and K.

1. One box is placed in between C and K

2.  Two boxes are placed in between K and B, where the latter is below the former.

3. The number of boxes below B is one less than the number of boxes above A.

4. H is an immediate neighbour of A. So, CASE 1 and CASE 2B is eliminated.

Thus, the final arrangement is,

The number of boxes placed below K is same as the number of boxes placed above G.

Boxes: A, B, C, G, H and K.

1. One box is placed in between C and K

2.  Two boxes are placed in between K and B, where the latter is below the former.

3. The number of boxes below B is one less than the number of boxes above A.

4. H is an immediate neighbour of A. So, CASE 1 and CASE 2B is eliminated.

Thus, the final arrangement is,

Thus, H and G is the odd one out.

Boxes: A, B, C, G, H and K.

1. One box is placed in between C and K

2.  Two boxes are placed in between K and B, where the latter is below the former.

3. The number of boxes below B is one less than the number of boxes above A.

4. H is an immediate neighbour of A. So, CASE 1 and CASE 2B is eliminated.

Thus, the final arrangement is,

All are true.

Thus, none of the following is false regarding box A.

Persons: A, B, C, D, E, F and G

Designations: Admiral, Vice Admiral, Rear Admiral, Commodore, Captain, Commander and Lieutenant

1. F works as Rear Admiral.

2. Two people rank between G and C, where C ranks lower than G.  3. G is not the Vice Admiral.

4. The number of persons who rank higher than D is equal to the number of persons who rank lower than C. So, CASE 1 is eliminated.

5. Three persons rank between B and D. 

6. A is neither the Captain nor Commodore. So, A works as Commander.
7. A ranks lower than E and B. So, E works as Vice Admiral.

Thus, the final arrangement is,

Thus, I and IV is true.

Persons: A, B, C, D, E, F and G

Designations: Admiral, Vice Admiral, Rear Admiral, Commodore, Captain, Commander and Lieutenant

1. F works as Rear Admiral.

2. Two people rank between G and C, where C ranks lower than G.  3. G is not the Vice Admiral.

4. The number of persons who rank higher than D is equal to the number of persons who rank lower than C. So, CASE 1 is eliminated.

5. Three persons rank between B and D. 

6. A is neither the Captain nor Commodore. So, A works as Commander.
7. A ranks lower than E and B. So, E works as Vice Admiral.

Thus, the final arrangement is,

Thus, E works as Vice Admiral.

Persons: A, B, C, D, E, F and G

Designations: Admiral, Vice Admiral, Rear Admiral, Commodore, Captain, Commander and Lieutenant

1. F works as Rear Admiral.

2. Two people rank between G and C, where C ranks lower than G.  3. G is not the Vice Admiral.

4. The number of persons who rank higher than D is equal to the number of persons who rank lower than C. So, CASE 1 is eliminated.

5. Three persons rank between B and D. 

6. A is neither the Captain nor Commodore. So, A works as Commander.
7. A ranks lower than E and B. So, E works as Vice Admiral.

Thus, the final arrangement is,

Thus, E and F is the odd one out.

Persons: A, B, C, D, E, F and G

Designations: Admiral, Vice Admiral, Rear Admiral, Commodore, Captain, Commander and Lieutenant

1. F works as Rear Admiral.

2. Two people rank between G and C, where C ranks lower than G.  3. G is not the Vice Admiral.

4. The number of persons who rank higher than D is equal to the number of persons who rank lower than C. So, CASE 1 is eliminated.

5. Three persons rank between B and D. 

6. A is neither the Captain nor Commodore. So, A works as Commander.
7. A ranks lower than E and B. So, E works as Vice Admiral.

Thus, the final arrangement is,

Thus, If D and G are interchanged in their designations, then three persons rank lower than D.

Persons: A, B, C, D, E, F and G

Designations: Admiral, Vice Admiral, Rear Admiral, Commodore, Captain, Commander and Lieutenant

1. F works as Rear Admiral.

2. Two people rank between G and C, where C ranks lower than G.  3. G is not the Vice Admiral.

4. The number of persons who rank higher than D is equal to the number of persons who rank lower than C. So, CASE 1 is eliminated.

5. Three persons rank between B and D. 

6. A is neither the Captain nor Commodore. So, A works as Commander.
7. A ranks lower than E and B. So, E works as Vice Admiral.

Thus, the final arrangement is,

Thus**, Only II is the answer.**

Given Statement: K = L > F ≥ G; B ≥ C > R = F

On combining: K = L > F = R < C ≤ B; K = L > F = R ≥ G; B ≥ C > R = F ≥ G     

Conclusion

I. B > G → True

II. L ≥ C → False (clear relation between L and C cannot be determined)

Hence, “Only I is true” is the correct answer.

Given Statements: G ≥ H = I ≤ M > N ≥ O ≥ P

Conclusions:

I. G > M → False (Relation cannot be determined between G and M)

II. H ≥ M → False (H = I ≤ M → H ≤ M)

Hence, Neither conclusion I nor II follows

Given Statements: W ≤ E ≤ R = T = U = V ≥ X

Conclusions:

I. R = V → True (as R = T = U = V → its clear R = V)

II.W ≥ T → False (W ≤ E ≤ R = T → W ≤ T)

Hence, only conclusion I follows.

Given Statements: X < P ≤ S ≥ Q ≥ Z >A

Conclusions:

I. S = Z → False (we can’t say because S ≥ Q ≥ Z → S ≥ Z)

II. S > Z → False (we can’t say because S ≥ Q ≥ Z → S ≥ Z)

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

Hence, either conclusion I or II follows.

Given statements: S ≥ P > Q = R > M

Conclusions:

I. S ≥ R → False (as S ≥ P > Q = R → S > R)

II. M < P → True (as P > Q = R > M → P > M)

Hence, only II is True.

1- C is sitting at one end. 

2- Only two people are sitting between C and A. 

3- Only Four people are sitting between A and R. 

Case-1

Case-2

4- Only one person sitting between A and B, and B is not an immediate neighbor of C. 

5- Only two people are sitting between R and D. 

6- Q is sitting second to the right of D. 

7- T is sitting second to the right of Q.  

Case-1 

Case-2

8- No one sits to the right of T. So case-2 gets eliminated. 

So, Final arrangement,

Hence, Sixteen  people are sitting the line.

1- C is sitting at one end. 

2- Only two people are sitting between C and A. 

3- Only Four people are sitting between A and R. 

Case-1

Case-2

4- Only one person sitting between A and B, and B is not an immediate neighbor of C. 

5- Only two people are sitting between R and D. 

6- Q is sitting second to the right of D. 

7- T is sitting second to the right of Q.  

Case-1 

Case-2

8- No one sits to the right of T. So case-2 gets eliminated. 

So, Final arrangement,

Hence, Seven people are sitting between A and D.

1- C is sitting at one end. 

2- Only two people are sitting between C and A. 

3- Only Four people are sitting between A and R. 

Case-1

Case-2

4- Only one person sitting between A and B, and B is not an immediate neighbor of C. 

5- Only two people are sitting between R and D. 

6- Q is sitting second to the right of D. 

7- T is sitting second to the right of Q.  

Case-1 

Case-2

8- No one sits to the right of T. So case-2 gets eliminated. 

So, Final arrangement,

Hence, R is fifth to the left of Q.

Calculation:

Total number of mobile purchased = 85 + 70 + 75 + 100 + 60

⇒ 390 

Total number of Laptop purchased= 60 + 55 + 50 + 35 + 80

⇒ 280 

Now, required difference = 390 - 280

⇒ 110 

∴  the difference between the total number of Mobile and the total number of Laptop purchased in all the given months is 110.

Calculation:

Number of  Laptop purchased in January and May month  = 60 + 80

⇒ 140 

Number Camera purchased in the month February, March and April months  = 50 + 85 + 65

⇒ 200 

Now, required ratio = 140 ∶ 200

⇒ 7 ∶ 10

∴ the ratio of Laptop purchased in January and May month together to Camera purchased in the month February, March and April months toge​ther is 2 ∶ 3.

Given:

the number of Mobile purchased in month June =  (3/5)th of the total number of items in month of April

Formula used:

Percentage = (Favourable value/Base value) × 100

Calculation:

the total number of items in month of April = 100 + 65 + 35

⇒ 200 

So, the number of Mobile purchased in month June = (3/5) × 200

⇒ 120 

the number of Mobile purchased in month March = 75

Now, required percentage = (75/120) × 100

⇒ 62.5%

the number of Mobile purchased in month March is 62.5% of the number of Mobile purchased in month June

Formula used:

Percentage = (Favourable value/Base value) × 100

Calculation:

 total number of items in month of Feb = 70 + 50 + 55

⇒ 175 

 total number of items in month of March = 75 + 85 + 50

⇒ 210 

Now, required difference = 210 - 175

⇒ 35 

Then, required percent = (35/175) × 100

⇒ 20%

 total number of items in month of March is 20% more than that in Month of February

Formula used:

Average = (Sum of values)/(Number of values)

Calculation:

Total number of items in month of January = 85 + 45 + 60 = 190 

Total number of items in month of February= 70 + 50 + 55 = 175 

Total number of items in month of March= 75 + 85 + 50 = 210 

Total number of items in month of April= 100 + 65 + 35 = 200 

Total number of items in month of May= 60 + 95 + 80 = 235 

So, total number of items in all the months = 190 + 175 + 210 + 200 + 235 = 1010 

Now, the required average = 1010/5

⇒ 202 items

∴ The average number of items in all the months together is 202.

Concept used:

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

Calculation:

263 – 12 × √25 + 462 ÷ 21 = ?

⇒ 263 – 12 × √25 + 22 = ?

⇒ 263 – 12 × 5 + 22 = ?

⇒ 263 – 60 + 22 = ?

⇒ 285 – 60 = ?

⇒ 225 = ?

∴ The value of ? 225

Concept used:

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

Calculations:

80% of 440 + 20% of 220 = ? × 10%

⇒ 352 + 44 = ? × (1/10)

⇒ 396 = ?/10

⇒ ? = 3960

∴ The value of ? is 3960

Concept used:

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

Calculations:

18 ÷ 7 ÷ 5 × 105 + 210 = ?

⇒ (18/35) × 105 + 210 = ?

⇒ 54 + 210 = ?

⇒ ? = 264

∴ The value of ? is 264

Concept used:

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

Calculations:

61 × 25 – 15 × 25 = ? + 30

⇒ 1525 – 375 = ? + 30

⇒ 1150 = ? + 30

⇒ ? = 1120

∴ The value of ? is 1120

Concept used:

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

Calculations:

1331 ÷ 11 + 90 × 104 = ?

⇒ 121 + 9360 = ?

⇒ 9481 = ?

∴ The value of ? is 9481

Solution:

We have to follow the BODMAS rule

Calculation:

25/100 × 160 + 30/100 × 200 = 20/100 × ?

40 + 60 = 1/5 × ?

100 = 1/5 × ?

⇒ ? = 500

Concept Used:

We have to follow the BODMAS rule,

Calculation:

(√1024 × 20) ÷ 16 + 14² = ?

⇒ (32 × 20) ÷ 16 + 196 = ?

⇒ 640 ÷ 16 + 196 = ?

⇒ 40 + 196 = ?

⇒ 236 = ?

∴ The value of ? is 236.

Given:

7 × 1.5 ÷ 4 – 2.8 × 0.5 

Concept used:

We have to follow the BODMAS rule

Calculation:

\(7 \times \frac{{15}}{{10}} \div 4 - \frac{{28}}{{10}} \times \frac{5}{{10}}\) = ?

⇒ \(7 \times \frac{3}{2} \div 4 - \frac{{14}}{5} \times \frac{1}{2}\) = ?

⇒ \(7 \times \frac{{3;}}{8} - \frac{7}{5}\) = ?

⇒ \(\frac{{21}}{8} - \frac{7}{5}\) = ?

⇒ \(\frac{{105 - 56}}{{40}}\) = ?

⇒ ? = \(\frac{{49}}{{40}}\)

∴ The value of ? is 49/40.

Given:

[(72)2 ÷ 24 × 8] ÷ 24 = 6 × ?

Concept used:

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

Calculation:

[(72)2 ÷ 24 × 8] ÷ 24 = 6 × ?

⇒ [(72 × 72) ÷ 24 × 8] ÷ 24 = 6 × ?

⇒ [(72 × 3) × 8] ÷ 24 = 6 × ?

⇒ [216 × 8] ÷ 24 = 6 × ?

⇒ 1728 ÷ 24 = 6 × ?

⇒ 72 = 6 × ?

⇒ 12 = ?

∴ The value of ? is 12

Given:

2160 ÷ 36 – 5 × 8.2 = ? 

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

Using the BODMAS rule

2160 ÷ 36 – 5 × 8.2 = ? 

⇒ 60 – 41 = ?

⇒ 19 = ?

∴ The value of ? is 19

Given:

90% of 100 – 20 + 60% of 30 – 16 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

90% of 100 – 20 + 60% of 30 – 16 = ?

⇒ (90/100) × 100 – 20 + (60/100) × 30 – 16 = ?

⇒ 90 – 20 + 18 -16

⇒ 108 – 36 = 72

∴ The value of ? is 72

Concept used: 

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

Calculation:

(∛1331 × √144) ÷ 2 + 3 = ?

⇒ (11 × 12) ÷ 2 + 3 = ?

⇒ 132 ÷ 2 + 3 = ?

⇒ 66 + 3 = ?

⇒ 69 = ?

∴ The value of ? is 69

Concept used:

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

Calculation:

1500 –  5% of 700 – 13 × 5 = ?

1500 – (5/100) × 700 – 13 × 5 = ?

1500 – 35 –  65 = ?

1500 –  100 = ?

? = 1400

∴ The value of the '?' is 1400.

Concept used:

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

Calculation:

15 × 12 ÷  6 ÷  30 + 22 = ?

15 × (12/6)/30 + 4 = ?

15 × 12/(6 × 30) + 4 = ?

15 × (1/15) + 4 = ?

1 + 4 = ?

? = 5

∴ The value of the '?' is 5.

Concept used:

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

Calculation:

∛729 ÷ 9 of 3 + √25 ÷ √9 = √?

9 ÷ 27 + 5 ÷ 3 = √?

9 ÷ 27 + 5 ÷ 3 = √?

9/27 + 5/3 = √?

(1 + 5)/3 = √?

2 = √?

? = 4

∴ The value of '?' is 4.

The series follows the following pattern:

23 + 1² = 24

23 + 2² = 27

23 + 3² = 32

23 + 4² = 39

23 + 5² = 48

23 + 6² = 59

Given:

4, 8, 17, 33, 60, ?

Calculation:

The series follows the following pattern:

∴ 100 will come in place of the question mark (?).

Given:

127, 138, 158, 187, 225, ?

Calculation:

The series follows the following pattern:

∴ 272 will come in place of the question mark (?).

Given:

12, ?, 6, 9, 18, 45, 135

Calculation:

12 × 0.5 = 6

6 × 1 = 6

6 × 1.5 = 9

9 × 2 = 18

18 × 2.5 = 45

45 × 3 = 135

∴ ? = 6

All numbers are consecutive prime numbers.

71, 73, 79, 83, 89, 97

Given:

Number of Jio users in Uttarakhand = 300

Number of Airtel users in Himachal = 150

Calculations:

Percentage difference = {(300 – 150)/150} × 100

⇒ (150/150) × 100

⇒ 100%

∴ Jio users in Uttarakhand are 100% more than Airtel users in Himachal.

Given:

Number of Jio users in Haryana = 250

Number of Airtel users in Delhi = 250

Calculations:

Difference of Jio users in Haryana and Airtel users in Delhi = 250 – 250

⇒ 0

∴ Difference of Jio users in Haryana and Airtel users in Delhi is 0

Given:

Number of Jio users in Delhi = 350

Number of Airtel users in UP = 250

Calculations:

Ratio of Jio users in Delhi to Airtel users in UP = 350 : 250

⇒ 7 : 5

∴ The ratio of Jio users in Delhi to Airtel users in UP is 7 : 5

Given:

Number of Jio users in Delhi = 350

Number of Jio users in Uttarakhand = 300

Number of Jio users in Haryana = 250

Formula used:

Average = Sum of values/Number of values

Calculations:

Average of Jio users in Delhi, Uttarakhand and Haryana = (350 + 300 + 250)/3

⇒ 900/3

⇒ 300

∴ The average of Jio users in Delhi, Uttarakhand and Haryana is 300

Given:

Number of Airtel users in UP = 250

Number of Airtel users in Uttarakhand = 200

Number of Airtel users in Himachal = 150

Number of Airtel users in Haryana = 350

Number of Airtel users in Delhi = 250

Calculations:

Total Airtel users in all five states together = 250 + 200 + 150 + 350 + 250

⇒ 1200

∴ Total Airtel users in all five states together are 1200

Area of a circle = πr²

1386 = 22/7*r²

1386*(7/22) = r²

r ² = 441

 r = 21 cm

Circumference of a circle = 2πr = 2*(22/7)*21 = 132 cm

Perimeter of the rectangle = 2*132 = 264 m

264 = 2*(l + 100)

264 = 2l + 200

2l = 264 - 200

l = 64/2

l = 32

Area of the rectangle = 32*100 = 3200m²

Given:

Renu = 12 months, Rs. 27,000

Shubham = Rs. 50,000, 9 months

Kavia = Rs. 63,000, 6 months

Calculation:

Total investment by renu = 12 × 27,000

⇒ Total investment by shubham = 50,000 × 9

⇒ Total investment by kavia = 63,000 × 6

Ratio = (12 × 27,000) : (50,000 × 9) : (63,000 × 6)

⇒ 54 : 75 : 63

⇒ 18 : 25 : 21

∴ Ratio of profit sharing of renu, shubham and kavia is 18 : 25 : 21

Hint

Instead of calculating investment by each and then finding the ratio, calculate the investment by putting them in ratio format to save the efforts

Given:

(A + B) can do work in = 12 days

(B + C) can do work in = 15 days

(A + C) can do work in = 20 days

Calculation:

(A + B) 1 day work = 1/12

(B + C) 1 day work = 1/15

(C + A)   day work = 1/20

⇒ 2(A + B + C)'s 1 day work = (1/12 + 1/15 + 1/20) = 12/60 = 1/5 

⇒ (A + B + C) `s 1 day work = ((1/2) × (1/5)) = 1/10

∴ working together they can complete the work in 10 days.

Given:

Time taken by A to complete the work = 42 days

Percentage more efficiency of B than A = 20%

Concept used:

Total time taken = Total work/(Efficiency of A + Efficiency of B)

Calculation:

We know that, if efficiency is more than the time taken is less.

Time taken by B to complete the work = 42 × {100/(100 + 20)}

⇒ 42 × {5/6}

⇒ 210/6

⇒ 35 days

Work is done by A in a day = 1/42

Work is done by B in a day = 1/35

Work is done by A and B both in a day = (1/42) + (1/35)

⇒ {(5 + 6)/210}

⇒ 11/210

Time taken by A and B to complete the work = 210/11

⇒ 19(1/11) days

∴ The time taken by A and B to complete the work is 19(1/11) days.

MP + SP = Rs 3700  ------ eq1

MP - SP = Rs 1300 ------eq2

Add eq1 and eq2, we get the value of MP

⇒2MP = 5000

⇒MP = 2500

Put the value of these in any eq we get SP of an article 

⇒SP = 3700 - 2500

⇒SP = 1200

SP of a bag is Rs 1200

MP is Rs 2500

Give two discounts of 25% and X%

⇒2500 × 25% = 2500 × 1/4 = 625

⇒2500 - 625 = 1875 

And again give discount of X% 

SP is 1875 - 1200 = 675 

X% discount = (675 / 1875 ) × 100

⇒36% 

Hence 36% is the correct answer.

Age of Diya in 5 years ago be 4x 

Present age of Diya is 4x +5

Present age of Shiva is 5x

After 7 years age of Diya is 4x+5+7 = 4x+12

and age of Shiva is 5x+7 years 

their ratio of age after 7 years be 4 : 3

⇒ (4x+12) / (5x+7) = 4 / 3

⇒ 12x+36 = 20x+28

⇒ 36- 28 = 20x - 12x 

⇒ 8x = 8

⇒ x= 1

Present age of Shiva is 5x = 5×1 → 5years 

Age after 10 years = 10+5 = 15 years 

Hence 15 years is the correct answer.