LIC ADO Prelims Memory Based Paper (Held On: 12 Mar 2023 Shift 1) (Previous Year Paper)

Facing South: P, Q, R, S, and T (Row – 1)

Facing North: A, B, C, D, and E (Row – 2)

1. S sits third to the right of Q where either of them is sitting on any of the extreme ends of the row

(Here, there are two possible cases, Case 1: Q sits at the extreme right end of the row, Case 2: S sits at the extreme left end of the row)

2. The one who faces Q sits second to the right of E.

3. Two persons are sitting between B and E.

(Here we can eliminate Case 1)

4. Neither A nor C sits at an extreme end of the row. 

(So D sits at the extreme end of row 2)

5. The immediate neighbour of A faces the person who sits immediately to the right of Q.

6)  R and T are immediate neighbours of each other.

7. T does not face the immediate neighbour of D.

The final arrangement will be as follows.

Hence, B is facing P.

Facing South: P, Q, R, S, and T (Row – 1)

Facing North: A, B, C, D, and E (Row – 2)

1. S sits third to the right of Q where either of them is sitting on any of the extreme ends of the row

(Here, there are two possible cases, Case 1: Q sits at the extreme right end of the row, Case 2: S sits at the extreme left end of the row)

2. The one who faces Q sits second to the right of E.

3. Two persons are sitting between B and E.

(Here we can eliminate Case 1)

 

4. Neither A nor C sits at an extreme end of the row. 

(So D sits at the extreme end of row 2)

5. The immediate neighbour of A faces the person who sits immediately to the right of Q.

6)  R and T are immediate neighbours of each other.

7. T does not face the immediate neighbour of D. 

The final arrangement will be as follows.

Hence, R faces one of the immediate neighbours of the D is a true statement.

Facing South: P, Q, R, S, and T (Row – 1)

Facing North: A, B, C, D, and E (Row – 2)

1. S sits third to the right of Q where either of them is sitting on any of the extreme ends of the row

(Here, there are two possible cases, Case 1: Q sits at the extreme right end of the row, Case 2: S sits at the extreme left end of the row)

2. The one who faces Q sits second to the right of E.

3. Two persons are sitting between B and E.

(Here we can eliminate Case 1)

 

4. Neither A nor C sits at an extreme end of the row. 

(So D sits at the extreme end of row 2)

5. The immediate neighbour of A faces the person who sits immediately to the right of Q.

6)  R and T are immediate neighbours of each other.

7. T does not face the immediate neighbour of D.

The final arrangement will be as follows.

Hence, C is facing T.

Facing South: P, Q, R, S, and T (Row – 1)

Facing North: A, B, C, D, and E (Row – 2)

1. S sits third to the right of Q where either of them is sitting on any of the extreme ends of the row

(Here, there are two possible cases, Case 1: Q sits at the extreme right end of the row, Case 2: S sits at the extreme left end of the row)

2. The one who faces Q sits second to the right of E.

3. Two persons are sitting between B and E.

(Here we can eliminate Case 1)

 

4. Neither A nor C sits at an extreme end of the row. 

(So D sits at the extreme end of row 2)

5. The immediate neighbour of A faces the person who sits immediately to the right of Q.

6) R and T are immediate neighbours of each other.

7. T does not face the immediate neighbour of D.

The final arrangement will be as follows.

Hence, C sits immediate left of A.

Facing South: P, Q, R, S, and T (Row – 1)

Facing North: A, B, C, D, and E (Row – 2)

1. S sits third to the right of Q where either of them is sitting on any of the extreme ends of the row

(Here, there are two possible cases, Case 1: Q sits at the extreme right end of the row, Case 2: S sits at the extreme left end of the row)

2. The one who faces Q sits second to the right of E.

3. Two persons are sitting between B and E.

(Here we can eliminate Case 1)

 

4. Neither A nor C sits at an extreme end of the row. 

(So D sits at the extreme end of row 2)

5. The immediate neighbour of A faces the person who sits immediately to the right of Q.

6) R and T are immediate neighbours of each other.

7. T does not face the immediate neighbour of D.

The final arrangement will be as follows.

S, P, D, and B are sitting at the extreme ends of the row. C sits in the middle of the row.

Hence, C does not belong to the group.

The pattern for the code is as follows,

Similarly,

Hence, PRINT will be written as RPKLV.

Given persons: S, P, L, Q, R, M, and I 

1) M lives on the floor immediately above S who does not live on an odd-numbered floor.

2. M does not live on the fifth floor.

(Here, there are two possible cases: Case 1: S lives on the 2^nd floor, and Case 2: S lives on the 6^th floor)

 

3) Three persons live between I and M.

4) I does not live on the first floor.

 

5) Two persons live between R and S.

(In case 2, there is no place for R, so we can eliminate case 2)

6) Q lives neither on the first floor nor on the fourth floor. 

7)  P neither lives on an odd-numbered floor nor on the topmost floor.

The final arrangement will be as follows:

 

Hence, P lives on the floor just above M.

Given persons: S, P, L, Q, R, M, and I 

1) M lives on the floor immediately above S who does not live on an odd-numbered floor.

2. M does not live on the fifth floor.

(Here, there are two possible cases: Case 1: S lives on the 2^nd floor, and Case 2: S lives on the 6^th floor)

 

3) Three persons live between I and M.

4) I does not live on the first floor.

 

5) Two persons live between R and S.

(In case 2, there is no place for R, so we can eliminate case 2)

6) Q lives neither on the first floor nor on the fourth floor. 

7)  P neither lives on an odd-numbered floor nor on the topmost floor.

The final arrangement will be as follows:

 

Hence, two persons live between L and P.

Given persons: S, P, L, Q, R, M, and I 

1) M lives on the floor immediately above S who does not live on an odd-numbered floor.

2. M does not live on the fifth floor.

(Here, there are two possible cases: Case 1: S lives on the 2^nd floor, and Case 2: S lives on the 6^th floor)

 

3) Three persons live between I and M.

4) I does not live on the first floor.

 

5) Two persons live between R and S.

(In case 2, there is no place for R, so we can eliminate case 2)

6) Q lives neither on the first floor nor on the fourth floor. 

7)  P neither lives on an odd-numbered floor nor on the topmost floor.

The final arrangement will be as follows:

 

Hence, L and I live on the first floor and the topmost floor respectively.

Given persons: S, P, L, Q, R, M, and I 

1) M lives on the floor immediately above S who does not live on an odd-numbered floor.

2. M does not live on the fifth floor.

(Here, there are two possible cases: Case 1: S lives on the 2^nd floor, and Case 2: S lives on the 6^th floor)

 

3) Three persons live between I and M.

4) I does not live on the first floor.

 

5) Two persons live between R and S.

(In case 2, there is no place for R, so we can eliminate case 2)

6) Q lives neither on the first floor nor on the fourth floor. 

7)  P neither lives on an odd-numbered floor nor on the topmost floor.

The final arrangement will be as follows:

 

Hence, I lives on the topmost floor.

Given persons: S, P, L, Q, R, M, and I 

1) M lives on the floor immediately above S who does not live on an odd-numbered floor.

2. M does not live on the fifth floor.

(Here, there are two possible cases: Case 1: S lives on the 2^nd floor, and Case 2: S lives on the 6^th floor)

 

3) Three persons live between I and M.

4) I does not live on the first floor.

 

5) Two persons live between R and S.

(In case 2, there is no place for R, so we can eliminate case 2)

6) Q lives neither on the first floor nor on the fourth floor. 

7)  P neither lives on an odd-numbered floor nor on the topmost floor.

The final arrangement will be as follows:

 

Hence, Third floor - M is true.

Given statement: E < F ≤ G > H ≥ K

Conclusions: 

I. F > H → False (as E < F ≤ G > H ≥ K → F ≤ G > H → no clear relation between F and H can be determined as there are opposite signs between F and H. So, It is false)

II. G < E → False (as E < F ≤ G > H ≥ K → E < F ≤ G → E < G. So, it is false)

Hence, Neither conclusion I nor II is true.

Given statement: A = B < F ≥ C > G > H

Conclusions:

I. A > G → False (Here A = B < F ≥ C > G there is no clear relation between A and G because of the opposite sign between A and G).

II. F > H → True (Here F ≥ C > G > H → F > H, so it is true).

Hence, Only II follow.

Given statement: I ≤ J ≥ K > L ≥ M = N < O

Conclusions:

I. O < J → False (as J ≥ K > L ≥ M = N < O → no clear relation between O and J can be determined)

II. K ≥ N → False (as K > L ≥ M = N → K > N)

Hence, neither conclusion I nor conclusion II is true.

Given statements:

A < B = C > E ≥ D ≥ F

Conclusions:

I. E > F → False (as E ≥ D ≥ F → E ≥ F)

II. E = F → False (as E ≥ D ≥ F → E ≥ F)

Note: Conclusion I and II are complementary pairs.

Hence, either conclusion I or II is true.

Given Series:

E W % I 5 V # D T Y N 9 M 6 4 3 Q @ * 0 B O ? 8 A 2 P ! U 7 1

Required order → Alphabet → Number → Alphabet

I 5 V,  N 9 M, and A 2 P.

Hence the correct answer is Three.

Given Series

E W % I 5 V # D T Y N 9 M 6 4 3 Q @ * 0 B O ? 8 A 2 P ! U 7 1

Symbols are there to the left of M in the given series is,

% and #

Hence, the answer is Two.

Given Series

E W % I 5 V # D T Y N 9 M 6 4 3 Q @ * 0 B O ? 8 A 2 P ! U 7 1

15^th to right of the element which is 10^th from the left end,

i.e 10 + 15

which is the 25^th from the left end.

25^th element is A

The element is 1^5th from the right of the element which is 10^th from the left end is A.

Given Series

E W % I 5 V # D T Y N 9 M 6 4 3 Q @ * 0 B O ? 8 A 2 P ! U 7 1

If all the symbols are deleted, then the series is

E W I 5 V  D T Y N 9 M 6 4 3 Q  0 B O 8 A 2 P U 7 1

element is 15^th from the right end in the new series is M.

Hence, M is the correct answer.

The least possible Venn diagram for the given statements is:

Conclusions:

I. All Envelopes are Pockets → False (It is possible but not definite. So, it is false)

II. All Boxes can never be Envelopes → True (no Envelopes is Gift, some Gift is Box. So, some Envelopes is definitely not Boxes  → all Boxes can never be Envelopes)

Hence, Only conclusion II follows.

The least possible Venn Diagram will be as follows:

Conclusions:

I. Some Bread is not Sauce → True (As Some Bread is Jam and No Jam is Sauce, the part of the bread that is common with Jam will not be Sauce. So it is true)

II. Some Sauce is Bread is a possibility → True (As there is no direct relation between Sauce and Bread, it is possible. So it is true)

The diagram of possibility will be as follows:

Hence, both I and II follow.

The least possible Venn diagram is:

Conclusions:

I. Some Low are High is a possibility → False (As, Only a few High are Low is given which means Some High are Low and Some High are not Low, which also states that Some Low are High, so this is a definite conclusion so the possibility is false).

II. All Boxes can be Low → True (The possibility is true, as no direct relationship between the two is given).

Hence, Only Conclusion II follows.

The least possible Venn diagram for the given statements is:

Conclusions:

I. All rectangle being circle is a possibility. → True ( As nothing specific about rectangle and circle is given and none of the given statements is contradicting this possibility, hence the given possibility is true. The possibility diagram is shown below:

II. Some square are triangle. → True ( As some square are circle and all the circle are triangle, hence it is definitely true.)

Both follow.

1. At least two people buy after Q.

2. Three people buy items between V and Q.

3. V neither buy at first nor at last

4. One person buys between T and Q. 

5. T buys an item before Q.

6) The number of persons buy items before T is same as the number of persons buy the item after W.

7)  Two people buy item between W and P.

8) Two people buy items between P and S (This eliminate case 1) 

9) R buys before U. 

Hence, T buys the item at first.

1. At least two people buy after Q.

2. Three people buy items between V and Q.

3. V neither buy at first nor at last

4. One person buys between T and Q. 

5. T buys an item before Q.

6) The number of persons buy items before T is same as the number of persons buy the item after W.

7)  Two people buy item between W and P.

8) Two people buy items between P and S (This eliminate case 1) 

9) R buys before U. 

Hence, Two people buys the item after U.

1. At least two people buy after Q.

2. Three people buy items between V and Q.

3. V neither buy at first nor at last

4. One person buys between T and Q. 

5. T buys an item before Q.

6) The number of persons buy items before T is same as the number of persons buy the item after W.

7)  Two people buy item between W and P.

8) Two people buy items between P and S (This eliminate case 1) 

9) R buys before U. 

Hence, Three person buy the item between R and W.

1. At least two people buy after Q.

2. Three people buy items between V and Q.

3. V neither buy at first nor at last

4. One person buys between T and Q. 

5. T buys an item before Q.

6) The number of persons buy items before T is same as the number of persons buy the item after W.

7)  Two people buy item between W and P.

8) Two people buy items between P and S (This eliminate case 1) 

9) R buys before U. 

Hence, The number of persons buy item before S is same as the number of persons buy item after V.

1. At least two people buy after Q.

2. Three people buy items between V and Q.

3. V neither buy at first nor at last

4. One person buys between T and Q. 

5. T buys an item before Q.

6) The number of persons buy items before T is same as the number of persons buy the item after W.

7)  Two people buy item between W and P.

8) Two people buy items between P and S (This eliminate case 1) 

9) R buys before U. 

Hence, Q buys immediately before R.

As per the given information,

1. Mr. Arjun starts walking from

2.  point T towards the south. After walking 6m he reached point X then turns to the left and walks for 5m to reach point L.

3. Then he turns to the right and walks for 9m to reach point V.

4. Then he turns to the right and walks for 12m to reach point P and then turns to the right and walks for 4m to reach point Q.

5. Then he turns to the left and walks for 9m to reach Point S.

6. then turn to the left and walks for 12m and stopped at point B.

The final diagram is, 

Hence, 15m towards the southwest is the shortest distance and direction of B with respect to Q.

As per the given information,

1. Mr. Arjun starts walking from

2)  point T towards the south. After walking 6m he reached point X then turns to the left and walks for 5m to reach point L.

3. Then he turns to the right and walks for 9m to reach point V.

4. Then he turns to the right and walks for 12m to reach point P and then turns to the right and walks for 4m to reach point Q.

5. Then he turns to the left and walks for 9m to reach Point S.

6. then turn to the left and walks for 12m and stopped at point B.

The final diagram is, 

Hence, 21m is the distance between S and A.

As per the given information,

1. Mr. Arjun starts walking from

 

2)  point T towards the south. After walking 6m he reached point X then turns to the left and walks for 5m to reach point L.

3. Then he turns to the right and walks for 9m to reach point V.

4. Then he turns to the right and walks for 12m to reach point P and then turns to the right and walks for 4m to reach point Q.

5. Then he turns to the left and walks for 9m to reach Point S.

6. then turn to the left and walks for 12m and stopped at point B.

The final diagram is, 

Hence, North-East is the direction of T with respect to point S

Things: M, N, O, P, Q, R and S

1. The M is kept second from the left end.

2. Nothing is kept in between M and N.

3. The N is not kept at any of the extreme ends.

4. The P is kept seven places away from the N.

5. Only three things are kept in between O and P.

6. Only two things are kept between O and N.

7. Only three things are kept in between P and Q.

8. The S is second to the right of the R, which is kept third to the right of the Q.

The final arrangement:

 

Hence, 19 places are there in a row.

Things: M, N, O, P, Q, R and S

1. The M is kept second from the left end.

2. Nothing is kept in between M and N.

3. The N is not kept at any of the extreme ends.

4. The P is kept seven places away from the N.

5. Only three things are kept in between O and P.

6. Only two things are kept between O and N.

7. Only three things are kept in between P and Q.

8. The S is second to the right of the R, which is kept third to the right of the Q.

The final arrangement:

 

Hence,

Things: M, N, O, P, Q, R and S

1. The M is kept second from the left end.

2. Nothing is kept in between M and N.

3. The N is not kept at any of the extreme ends.

4. The P is kept seven places away from the N.

5. Only three things are kept in between O and P.

6. Only two things are kept between O and N.

7. Only three things are kept in between P and Q.

8. The S is second to the right of the R, which is kept third to the right of the Q.

The final arrangement:

 

Hence, The O is 14^th from the right end.

Things: M, N, O, P, Q, R and S

1. The M is kept second from the left end.

2. Nothing is kept in between M and N.

3. The N is not kept at any of the extreme ends.

4. The P is kept seven places away from the N.

5. Only three things are kept in between O and P.

6. Only two things are kept between O and N.

7. Only three things are kept in between P and Q.

8. The S is second to the right of the R, which is kept third to the right of the Q.

The final arrangement:

 

Hence, Nothing is kept second to the left** of the R.**

Given 

215 + 369 – 60% of 900 + 5 = ?²

Concept Used 

Follow the BODMAS rule according to the table given below:

Calculation

215 + 369 – 60% of 900 + 5 = ?2

⇒ √(215 + 369 - 540 + 5) = ?

⇒ ? = √(589 - 540) = 7

∴ The required answer is 7

Given 

(252 – 106) ÷ 3 – 132 + 35 = ?

Concept used 

Follow the BODMAS rule according to the table given below:

Calculation 

(252 – 106) ÷ 3 – 132 + 35 = ?

⇒ (625 - 106) ÷ 3 - 169 + 35 = ?

⇒ 519 ÷ 3 - 169 + 35 = ? 

⇒ 173 - 169 + 35 = ?

⇒ ? = 39 

∴ the required answer is 39

Mistake Points ⇒ 173 - 169 + 35 = ?

there is no btacket after - sign before 169 so the calculation can't come -204.

Calculation 

(3.9 + 4.1) × 2.5 + 9 – 25 = ?²

⇒ 8 × 2.5 + 9 - 25 = ?²

⇒ 20 + 9 - 25 = ?²

⇒ ? = √4 = 2  

∴ The required answer is 2

The correct answer is: 4

Given:

377 ÷ 13 + √81 - 57 ÷ 19 = ?

Concept used:

The BODMAS rule should be followed as per the given table:

 

 

Calculation:

377 ÷ 13 + √81 - 57 ÷ 19 = ?

29 + √81 - 3 = ?

29 + 9 - 3 = ?

38 - 3 = ?

⇒ ? = 35

 

∴ The value of '?' is 35

The correct answer is : 3

Given:

12% of √5625 - ? % of √225 = 142 - 136

Concept used:

The BODMAS rule should be followed as per the given table: 

 

Calculation:

12% of √5625 - ? % of √225 = 142 - 136

⇒ 12/100 × 75  - ?/100  × 15 = 6

⇒ 3/25 × 75  - ?/100  × 15 = 6

⇒ 9 - ?/100  × 15 = 6 

⇒ 3 = ?/100  × 15

⇒ ? = (100×3)/15

⇒ ? = 20

 

∴ The value of '?' is 20

Given:

48% of 420 + 32% of 470 = 11% of ?

Concept Used:

a% of b = b% of a

Calculation:

Follow the BODMAS rule according to the table given below:

⇒ 48% of 420 + 32% of 470 = 11% of ?

⇒ 48% of 420 + 32% of (420 + 50) = 11% of ?

⇒ 48% of 420 + 32% of 420 + 32% of 50 = 11% of ?

⇒ (48% + 32%) of 420 + 50% of 32 = 11% of ?

⇒ 80% of 420 + 50% of 32 = 11% of ?

⇒ (80/100) × 420 + 1/2 × 32 = (11/100) of ?

⇒ 8 × 42 + 16 = (11/100) of ?

⇒ 336 + 16 = (11/100) of ?

⇒ 352 = (11/100) of ?

? = (352 × 100)/11

⇒ 32 × 100

⇒ 3200

∴ The value of '?' is 3200.

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

56 ÷ 14 × 4 – 316 + ?2  = 100

⇒ 4 × 4 – 316 + ?² = 100

⇒ 16 – 316 + ?² = 100

⇒ ?² = 100 + 316 – 16

⇒ ?² = 416 – 16

⇒ ?² = 400

⇒ ? = 20

∴ The value of ? is 20

Concept used:

Follow the BODMAS rule according to the table given below: 

Calculation:

√5625 ÷ 15 × 21 ÷ 3 + 5 = ?

⇒ 75 ÷ 15 × 21 ÷ 3 + 5 = ?

⇒ 5 × 21 ÷ 3 + 5 = ?

⇒ 5 × 7 + 5 = ?

⇒  35 + 5 = ?

⇒ ? = 40

∴ The value of ? is 40

Total number of bags from all the shops = 40 + 70 + 35 = 145

Total number of watches from all the shops = 60 + 50 + 42 = 152

So the difference between the total number of bags and watches from all the shops is 152 - 145 = 7.

The total number of watches in all the shops = 60 + 50 + 42 = 152

The ratio of male and female watches is 10:9,

Hence the total number of female watches = (9/(10 + 9)) x 152 = 72.

The total number of bags in all the shops = 40 + 70 + 35 = 145

The ratio of male and female bags is 10:19,

Hence the total number of female bags = (19/(10 + 19)) x 145 = 95.

Hence the total number of female items = 72 + 95 = 167.

Total amount generated by shop P =40 x 500 + 60 x 800 = Rs. 68000.

Total amount generated by shop Q =70 x 500 + 50 x 800 = Rs. 75000.

Total amount generated by shop R =25 x 500 + 42 x 800 = Rs. 46100.

Hence shop Q generate the most amount.

The total number of items in shop P = 40 + 60 = 100

The total number of items in shop Q = 70 + 50 = 120.

Hence by ((120 - 100)/100) x 100 = 20%, the total items in shop Q are more.

The total number bags in shops P and Q = 40 + 70 = 110

The total number of watches in shops Q and R = 50 + 42 = 92

So the ratio is 110:92 = 55:46.

Given:

18 , 22 , 24 , 28 , 34 , ?

Calculation:

Here the logic follows as the series of prime number from 13 added with 5

Prime numbers 13 , 17 , 19 , 23 , 29 , 31

= 13 + 5 , 17 + 5 , 19 + 5 , 23 + 5 , 29 + 5 , 31 + 5

= 18 , 22 , 24 , 28 , 34 , 36

\(\therefore\) The correct answer is 36.

Calculation:

The series follows the following pattern:

⇒ 7 × 0.5 = 3.5

⇒ 3.5 × 1 = 3.5

⇒ 3.5 × 2 = 7

⇒ 7 × 4 = 28

⇒ 28 × 8 = 224

∴ The value of ? is 224

Given:

12, 16, 25, 41, 66, ?

Calculation:

The series follows the following pattern

12 + 2² = 16

16 + 3² = 25

25 + 4² = 41

41 + 5² = 66

66 + 6² = 102

∴ The required value of ? is 102

Calculation:

Considering the above series,

1, 3, 11, ?, 239, 1439

The logic of the series can be explained as,

1 × 2 + 1 = 3

3 × 3 + 2 = 11

11 × 4 + 3 = ? = 47

47 × 5 + 4 = 239

239 × 6 + 5 = 1439

∴ The value of (?) is 47.

The pattern of the series is as follows: +14, +28, +42, +56, +70

8 + 14 = 22

22 + 28 = 50

50 + 42 = 92

92 + 56 = 148

148 + 70 = 218

∴ The required answer is 218.

Total quantity of mangoes is 300 kg.

Total cost incurred is 50 x 300 = 15000

Profit made is Rs. 9000,

So the total selling price of 300 kg mangoes = 15000 + 9000 = 24000.

So the selling price of 1 kg mango = 24000/300 = Rs. 80/kg

Hence the profit percentage = ((80 - 50)/50) x 100 = 60%.

The cost prices of mangoes in companies A and B be Rs. 60/kg and Rs. 40/kg.

The new cost prices are 1.2 x 60 = Rs. 72/kg and 1.25 x 40 = Rs. 50/kg respectively.

So the total cost price = 72 x 650 + 200 x 50 = 46800 + 10000 = Rs. 56800.

Total selling price = 80 x (650 + 200) = 68000.

So the profit = Rs. 11200.

Number of mangoes sold by company A in August 2024 = (120/100) × 650 = 780

Number of mangoes sold by company B in August 2024 = (140/100) × 200 = 280

So, new ratio will be = 780 : 280 = 39:14

Total number of mangoes sold from shop A in the present year (2023) = 500 + 650 + 750 + 800 + 500 = 3200

Total number of mangoes sold in next year will be = (220/100) ×  3200 = 7040

As the sales of mangoes except October and November remain same.

So,the number of mangoes sold in 2024 from July, August, and September = 500 + 650 + 750 = 1900

So, the number of mangoes sold in 2024 from October and November = 7040 – 1900 = 5140

Number of mangoes sold in 2024 in the month of November = 5140/2 = 2570

Hence, the total number of mangoes sold in the month of November for 2023 and 2024 = 500 + 2570 = 3070.

As per the data.

Total number of mangoes sold from company A = 500 + 650 + 750 + 800 + 500 = 3200

Total number of mangoes sold from company B = 150 + 200 + 300 + 400 + 500 = 1550

Hence, required ratio = 3200 : 1550 = 320 : 155 = 64 : 31

Given:

5 men and 3 women together or 9 men can cultivate a land in 60 days

Calculation:

According to the question,

(5m + 3w) = 9m

⇒ 3w = 4m

So, (8m + 9w) = 8m + 12m = 20m

9m can cultivate a land in 60 days

1m can cultivate a land in 60 × 9 days

20m can cultivate a land in (60 × 9)/20 days = 27 days.

Alternate Method(5m + 3w) × 60 = 9m × 60

⇒ m/w = 3/4

Total work = (5 × 3 + 3 × 4) × 60 = 27 × 60 =1620

Total efficiently of (8m + 9w) = 8 × 3 + 9 × 4 = 24 + 36 = 60

Required no of days = 1620/60 = 27 days

Shortcut Trick

Let the total salary be 100 units.

Remaining after 30% grocery = 70 units.

Expenses from remaining: 20% (Rent) + 50% (Loan) = 70%.

Savings from remaining = 30% of 70 = 21 units.

21 units = Rs. 14700 → 1 unit = Rs. 700 → 100 units = Rs. 70000.

∴ The correct answer is Rs. 70000.

Alternate Method

Given: Savings = Rs. 14700 | Grocery = 30% | Rent = 20% of remaining | Loan = 50% of remaining

Formula: Savings = Salary × (100 − %Spent₁)/100 × (100 − %Spent₂)/100

⇒ Let Salary be S.

⇒ Remaining after Grocery = S × (100 − 30)/100 = 0.70S

⇒ Total % spent from remaining = 20% + 50% = 70%

⇒ Savings from remaining = 100% − 70% = 30%

⇒ 30% of 0.70S = 14700

⇒ (30/100) × (70/100) × S = 14700

⇒ 0.21S = 14700

⇒ S = 14700 ÷ 0.21 = 70000

∴ The correct answer is Rs. 70000.

Additional Information

Remaining Balance Concept

When expenses are calculated on the remaining amount, use the multiplying factor method: Final = Initial × (1 − r₁/100) × (1 − r₂/100).

Successive Percentage Change

The overall percentage reduction from the original value is 100 − [100 × (1 − r₁/100) × (1 − r₂/100)].

Base Change Rule

Always identify if the percentage is of the total or of the remaining balance to avoid calculation errors.

Given:

Speed of train = 30 km/hr.

speed of man = 6 km/hr.

Time taken to cross the platform = 24 seconds

time taken to cross the man = 8 seconds.

 

Formula used:

Distance = speed x time

 

Calculation:

 

speed of the train relative to the man = 30 + 6 km/hr.

⇒ 36 × 5/18  m/s

⇒ 10 m/s

Distance travelled by the train in 8 seconds at this relative speed = 8 × 10 = 80m

Length of the train = 80m

 

Speed of the train = 30 km/hr. = 30 × 5/18 m/s

⇒ 25/3 m/s

 

distance travelled in 24 seconds = 24 × 25/3

⇒ 200m

This distance 200m covered is the sum of the length of the train and the length of the platform.

Length of train + length of platform = 200m

Length of platform = 200 - 80

⇒ 120m

 

∴The lengths of train and the platform are 80m and 120m respectively.

The correct answer is: 3

Given:

in 10 hours, John can go 30 km upstream and 44 km downstream.

in 13 hours, he can go 40 km upstream and 55 km downstream.

Formula used:

Time = distance/speed

Calculation:

let speed of john in still water i.e. boat = B

and the speed of stream = S

then, 30/B - S  +  44/B + S = 10     ----(1)

 &      40/B - S  +  55/B + S = 13     ----(2)

let 1/B - S = m     and    1/B + S = n

then, 30m + 44n = 10     ----(3)

and    40m + 55n = 13     ----(4)

by solving (3) and (4), we get,

m = 1/5 and n = 1/11

∴ B - S = 5   &  B + S = 11

by solving this, we get  2B = 16 

⇒ B = 8

∴ Speed of john in still water is 8 km/hr.

Given:

Average marks of a student in seven subjects = 75

Average marks in 6 subjects = 73

Calculations:

Total marks in seven subjects = 7 × 75 = 525

Now, excluding English

∴ Total marks in six subjects = 6 × 73 = 438

∴ Marks in English = 525 - 438

⇒ 87

Marks he scored in English is 87.

Given:

10 years ago, Ratio of age of Ram's father and Ram = 3 : 1

10 years later, Ratio of age of Ram's father and Ram = 2 : 1

Calculation:

Let 10 years ago Ram's age be x years.

then, 10 years ago the age of Ram's father = 3x years

The present age of Ram = (x + 10) years

And the present age of Ram's father = (3x + 10) years

According to the question,

3x + 10 + 10 = 2(x + 10 + 10)

⇒ 3x + 20 = 2(x + 20)

⇒ 3x + 20 = 2x + 40

⇒ 3x - 2x = 40 - 20

⇒ x = 20

Hence the present age of Ram = (20 + 10) years

⇒ 30 years.

∴ The present age of Ram is 30 years.

Given:

A basket contains 2 white, 3 red and 4 black balls, two balls are drawn at random

Concept used:

The classical definition of probability.

Formula used:

Probability = Favorable Outcome/Total Outcome.

P (E) = n(E)/n(S)

Calculation:

Total number of balls = 2 + 3 + 4

Let S be the sample space.

n(S) = Total number of ways of drawing 2 balls out of 9 

⇒ ⁹C₂

Let E = Event of drawing 2 balls, none of them is black

n(E) = Number of ways of drawing 2 balls, none of them is black

⇒ ⁵C₂

There are four black balls in the total nine balls. Total number of non-black balls = 9 - 4 = 5

P(E) = n(E)/n(S)

⁵C₂/⁹C₂

⇒ {(5 × 4)/(2 × 1)}/{(9 × 8)/(2 × 1)}

⇒ 10/36 = 5/18

∴ The probability of not any ball being drawn is black is 5/18.

Shortcut TrickOverall profit of A = Interest paid by B to A - Interest paid by A to Bank 

The overall profit of A = P × [R₂ - R₁] × T/100, 

The overall profit of A = 1500 × [15 - 10] × 3/100

The overall profit of A = Rs. 225

 

Alternate Method Given:

A takes a loan of Rs. 1500

Rate = 10% and Time = 3 years 

A lends Rs. 1500 to B 

Rate = 15% and Time = 3 years 

Formula Used:

S.I = (P × R × T)/100

Where, P = Principal, R = Rate, T = Time 

Calculation:

According to the question

Interest paid by A to the bank (S.I₁) = (1500 × 10 × 3)/100

⇒ S.I1 = Rs. 450    ----(i)

Now, 

Interest paid by B to A (S.I₂) = (1500 × 15 × 3)/100

⇒ S.I2 = Rs. 675    ----(ii)

Now, we have to find the total profit of A

S.I₂ - S.I1 

⇒ 675 - 450

⇒ 225

∴ The overall profit of A in the whole transaction is Rs.225.

Hint 100 : 121

121 = 24200

1 = 200

Principal= 20000 

Alternate MethodGiven:

Amount = Rs 24200

Rate = 10%

Time = 2 years

Formula used:

If the principal is P, Time period is T, and the rate of interest is r;

Then the amount, when compounded annually, becomes:

A = P (1 + r/100)^t

Calculation:

A = P [1 + (r/100)]^t

⇒ 24200 = P [1 + (10/100)]² = P (11/10)²

⇒ P = 24200 × (100/121) = Rs.20000 

∴ The sum is Rs.20000.

Shortcut Trick

Remaining Mixture = 50 − 10 = 40 litres.

Ratio 4 : 1 ⇒ Total 5 units = 40 litres ⇒ 1 unit = 8 litres.

Milk = 4 × 8 = 32 litres; Water = 1 × 8 = 8 litres.

Final Milk = 32 + 3 = 35 litres; Final Water = 8 + 5 = 13 litres.

Final Ratio = 35 : 13.

∴ The correct answer is 35 : 13.

Alternate Method

Given: Total mixture = 50 litres, Ratio M:W = 4 : 1.

Formula: Quantity = Total × (Ratio Part / Total Parts).

⇒ Quantity of Milk removed = 10 × (4 / 5) = 8 litres.

⇒ Quantity of Water removed = 10 × (1 / 5) = 2 litres.

⇒ Initial Milk = 50 × (4 / 5) = 40 litres.

⇒ Initial Water = 50 × (1 / 5) = 10 litres.

⇒ New Milk = (40 − 8) + 3 = 35 litres.

⇒ New Water = (10 − 2) + 5 = 13 litres.

⇒ Final Ratio = 35 : 13.

∴ The correct answer is 35 : 13.

Additional Information

Mixture Withdrawal Property

When a portion of a mixture is removed, the ratio of components in the removed part and the remaining part stays the same as the original ratio.

Alligation Rule

Used to find the proportion in which two or more ingredients at a given price/concentration are mixed to produce a mixture of a desired price/concentration.

Replacement Formula

If x units are removed from V volume and replaced with water n times: Final Pure Liquid = Initial Liquid × (1 − x/V)ⁿ.

Shortcut Trick

Profit = 40% ⇒ Profit/CP = 2/5. Let Cost Price (CP) = 5 units and Selling Price (SP) = 7 units.

Profit in units = 7 − 5 = 2 units.

Given SP − CP = Rs. 200 ⇒ 2 units = Rs. 200.

1 unit = Rs. 100.

Cost Price = 5 units × Rs. 100 = Rs. 500.

∴ The correct answer is Rs. 500.

Alternate Method

Given → MP : SP = 8 : 7, Profit% = 40%, SP − CP = Rs. 200. Formulas → SP = CP × (100 + P%)/100 | Profit = SP − CP.

⇒ Let CP be x. Profit = 40% of x = 0.4x.

⇒ Given: SP − CP = 200 ⇒ (CP + Profit) − CP = 200.

⇒ Profit = 200 ⇒ 0.4x = 200.

⇒ x = 200/0.4 = 2000/4 = 500.

∴ The correct answer is Rs. 500.

Additional Information

Marked Price & Discount

MP is the price printed on product. Discount = MP − SP. Discount% = (Discount/MP) × 100.

Relation between CP and MP

CP/MP = (100 − D%)/(100 + P%). This is a vital ratio for complex profit/loss problems.

Shortcut Trick

Perimeter of semicircle = r(π + 2) ⇒ 36 = r(3 + 2) ⇒ 36 = 5r

Radius (r) = 36 ÷ 5 = 7.2 cm

Breadth (B) = Radius (r) = 7.2 cm

Length (L) = Area ÷ Breadth = 216 ÷ 7.2 = 30 cm

∴ The correct answer is 30 cm.

Alternate Method

Given: Perimeter of semicircle = 36 cm, Area of rectangle = 216 cm², π = 3.

Formula: Perimeter of semicircle = πr + 2r; Area of rectangle = Length × Breadth.

⇒ 36 = r(3 + 2)

⇒ 36 = 5r

⇒ r = 7.2 cm

⇒ Breadth (B) = 7.2 cm

⇒ 216 = Length × 7.2

⇒ Length = 216 ÷ 7.2 = 30 cm

∴ The correct answer is 30 cm.

Additional Information

Area of Semicircle

The total region enclosed by a semicircle is given by half the area of a circle: Area = (πr²) ÷ 2.

Perimeter of Rectangle

The total boundary length of a rectangle is calculated as: Perimeter = 2 × (Length + Breadth).

Circle Properties

Circumference of a full circle = 2πr and Area of a full circle = πr².

The correct answer is option 2.

Key Points

To answer the question let's first look at the meanings of the given word and all the words given in the options - 

• Spotted - to have seen, noticed, or recognised (someone or something) that is difficult to detect or that one is searching for. (देखना)
• For Example -
• He spotted his blue shirt in the huge pile of clothes.
• Missed - failed to notice, hear, or understand (चुक होना)
• For Example  -
• We were standing right next to the pole, you missed us.
• Detected  - noticed or discovered something that is difficult to see, feel, etc. (देख पाना/का पता चला)
• For Example -
• He detected a faint smell of gas in the kitchen, so he immediately turned off the stove and opened the windows to let the air in.
• Collected - calm and in control of yourself, your feelings, thoughts, etc. (उत्तेजनाहीन)
• For Example -
• She is always calm and collected even in the toughest of situations.
• Terminated - to end or to make something end (समाप्त)
• For Example -
• Their contract was terminated last month because of some issues.
• Conducted - to organise and do something, especially research (संचालित)
• For Example -
• The training was conducted to improve our performance.
• After looking at the meanings of all the given words we can conclude that 'detected' is the only word that is similar in meaning to the given word 'spotted'.​

​

Therefore, the correct answer would be 'detected'.

The correct answer is option 5.

Key Points

To answer the question let's first look at the meanings of the given word and all the words given in the options - 

• Produced - to make something or bring something into existence (उत्पादित करना​)
• For Example -
• She produced enough food to feed all of us.
• Initiated - to start something (शुरू किया)
• For Example  -
• He initiated the conversation because she was shy.
• Proposed  - to formally suggest something as a possible plan or action (प्रस्तावित)
• For Example -
• This brilliant idea was proposed by my sister.
• Eliminated - to remove somebody/something that is not wanted or needed (सफाया)
• For Example -
• She eliminated all fatty foods from her diet.
• Nominated - to officially suggest someone for a position, an honour, or election (नामित)
• For Example -
• This movie has been nominated for the Oscars.
• Created - to cause something new to happen or exist (निर्माण करना)
• For Example -
• We are living in a world created by the God.
• ​After looking at the meanings of all the given words we can conclude that 'created' is the only word that is similar in meaning to the given word 'produced'.​

​

Therefore, the correct answer would be 'created'.

The correct answer is option 3.

Key Points

To answer the question, refer to the fourth paragraph of the given passage -

Overwhelmed with gratitude, the man offered a generous reward to his guide for helping him find the herb. He also vowed to use his wealth to help others who suffered from similar conditions. From that day on, the man became known for his acts of kindness and _____ towards those in need.

• Above lines talk about the wealthy man being overwhelmed with gratitude after his eye disease was cured by the potion made by the herb. He then offered to reward the guide for helping him find the herb and vowed to use his wealth to help others who suffered from similar conditions as him. And from that day onwards he became known for his kindness and goodness towards the needy.

 

Now, let's understand the meanings of the given options -

• Indulgence - the attitude of allowing yourself or someone else to have something enjoyable (अतिभोग)
• For Example -
• Considering you health conditions you should avoid excessive indulgence in sweets and canned drinks.
• Satisfaction - fulfilment of one's wishes, expectations, or needs, or the pleasure derived from this (संतुष्टि)
• For Example -
• I feel satisfaction in helping others.
• Generosity - the quality of being kind and generous (उदारता)
• For Example -
• I don't think I can ever thank you enough for your generosity and kindness.
• ​Displeasure - a feeling of annoyance or disapproval (अप्रसन्नता)
• For Example -
• He has been very vocal in his displeasure over the results of his exams.
• Wrath - extreme anger (क्रोध)
• For Example -
• A mother's wrath has no bounds when it comes to someone hurting her kids.​
• After understanding the above lines and the meanings of all the given options we can conclude that the most suitable filler for the given blank is 'generosity'.

 

Therefore, the correct answer would be 'generosity'.

The correct answer is option 1.

Key Points

• To answer this question refer to the third and fourth lines of the first paragraph of the given passage -
• 'One day, he learned about a herbal cure that could treat his ailment. However, the herb was only found deep in the forest, and it was not easy to locate.'
• After reading the above lines we can conclude that the herb that could cure the man's condition could only be found deep in the forest.
• Thus, the correct answer is option 1 i.e. It could only be found deep in the forest.

​

Additional Information

Let's understand some important vocab from the first paragraph of the given passage - 

• Severe - (of something bad or undesirable) very great; intense (गंभीर)
• For Example -
• Don't take it lightly, It is a severe disease.
• Seeking - attempt to find something (खोजना)
• For Example -
• I am seeking a better job opportunity.
• Ailment - an illness, typically a minor one (रोग)
• For Example -
• My friend got timely treatment for his ailment.

The correct answer is option 5.

Key Points

• To answer this question refer to the second and third lines of the second paragraph of the given passage -
• 'They embarked on a journey through the thick forest, facing many obstacles along the way. They had to navigate through dangerous terrain, cross treacherous rivers, and avoid wild animals.'
• After reading the above lines we can conclude that on their journey through the thick forest to find the herb they faced many challenges such as to navigate through dangerous terrain, cross treacherous rivers, and avoid wild animals.

Thus, the correct answer is option 5 i.e. All 1, 2, and 4.

 

Additional Information

​Let's understand some important vocab from the first paragraph of the given passage - 

• Gathered - bring together and take in from scattered places or sources (इकट्ठा)
• For Example -
• She gathered her books and left the class.
• Treacherous - (of ground, water, conditions, etc.) presenting hidden or unpredictable dangers 
• For Example -
• The snow made our hike all the more treacherous and difficult.
• Obstacles - things that block one's way or prevent or hinder progress (बाधाएं)
• For Example -
• She has faced many obstacles on her way to become the most successful woman in the country.

The correct answer is option 4.

Key Points

• To answer this question refer to the third and fourth lines of the third paragraph of the given passage -
• 'The man applied the potion to his eyes, and to his great relief, his vision cleared up, and his eye disease was cured.'
• After reading the above lines we can conclude that after the man applied the potion to his eyes, his vision cleared up.

Thus, the correct answer is option 4 i.e. his vision became clearer.​

 

Additional Information

Let's understand some important vocab given in the passage - 

• Overwhelmed - have a strong emotional effect on (अभिभूत)
• For Example -
• I was overwhelmed with emotions when I saw my little nephew for the first time.
• Gratitude - the quality of being thankful; readiness to show appreciation for and to return kindness (कृतज्ञता)
• For Example -
• He expressed his gratitude to everyone for their hard work.
• Generous - showing a readiness to give more of something, especially money, than is strictly necessary or expected (उदार)
• For Example -
• She has always been very generous toward everyone.