LIC AAO Prelims Memory Based Paper Shift 1 (Held on: 20 Feb 2023) (Previous Year Paper)

Eight people A, B, C, D, E, F, G and H.

Bidirectional.

1- B is sitting at the end and A is second to the left of B.

2- Only two people are sitting between F and B.

3- D is an immediate neighbor of F.

 

 

 

4- Number of people left to the D is one less than people sitting to the right of F. 

5- A is sitting immediately to the right of C.

6- Number of people sitting between A and H is two more than the number of people sitting to the left of C. 

 

 

  

 

7- H is facing North and is opposite to E.

8- E is sitting second to the right of G and both of them are facing the same directions. 

9- A is facing the opposite direction to G.

 

  

 

 

10- More than two people are sitting between F and E. So, case-1 gets eliminated.

 

So, Final arrangement is:

 

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A is sitting fourth to the right of H.

Eight people A, B, C, D, E, F, G and H.

Bidirectional.

1- B is sitting at the end and A is second to the left of B.

2- Only two people are sitting between F and B.

3- D is an immediate neighbor of F.

 

 

 

 

4- Number of people left to the D is one less than people sitting to the right of F. 

5- A is sitting immediately to the right of C.

6- Number of people sitting between A and H is two more than the number of people sitting to the left of C. 

 

 

7- H is facing North and is opposite to E.

8- E is sitting second to the right of G and both of them are facing the same directions. 

9- A is facing the opposite direction to G.

 

 

10- More than two people are sitting between F and E. So, case-1 gets eliminated.

 

So, Final arrangement is:

 

 

Four person are facing south.

Eight people A, B, C, D, E, F, G and H.

Bidirectional.

1- B is sitting at the end and A is second to the left of B.

2- Only two people are sitting between F and B.

3- D is an immediate neighbor of F.

 

 

4- Number of people left to the D is one less than people sitting to the right of F. 

5- A is sitting immediately to the right of C.

6- Number of people sitting between A and H is two more than the number of people sitting to the left of C. 

 

 

 

7- H is facing North and is opposite to E.

8- E is sitting second to the right of G and both of them are facing the same directions. 

9- A is facing the opposite direction to G.

 

 

10- More than two people are sitting between F and E. So, case-1 gets eliminated.

 

So, Final arrangement is:

 

 

D is sitting immediate left of G.

Eight people A, B, C, D, E, F, G and H.

Bidirectional.

1- B is sitting at the end and A is second to the left of B.

2- Only two people are sitting between F and B.

3- D is an immediate neighbor of F.

 

 

 

4- Number of people left to the D is one less than people sitting to the right of F. 

5- A is sitting immediately to the right of C.

6- Number of people sitting between A and H is two more than the number of people sitting to the left of C. 

 

 

 

 

7- H is facing North and is opposite to E.

8- E is sitting second to the right of G and both of them are facing the same directions. 

9- A is facing the opposite direction to G.

 

 

10- More than two people are sitting between F and E. So, case-1 gets eliminated.

 

So, Final arrangement is:

 

 

Four person are sitting right of G.

Eight people A, B, C, D, E, F, G and H.

Bidirectional.

1- B is sitting at the end and A is second to the left of B.

2- Only two people are sitting between F and B.

3- D is an immediate neighbor of F.

 

 

4- Number of people left to the D is one less than people sitting to the right of F. 

5- A is sitting immediately to the right of C.

6- Number of people sitting between A and H is two more than the number of people sitting to the left of C. 

 

 

7- H is facing North and is opposite to E.

8- E is sitting second to the right of G and both of them are facing the same directions. 

9- A is facing the opposite direction to G.

 

 

 

10- More than two people are sitting between F and E. So, case-1 gets eliminated.

 

So, Final arrangement is:

 

 

The number of people who are sitting to the right of A are same as the number of people are sitting between F and H.

Given:

Ten persons: A, B, C, D, E, F, G, H, I and J

Date: 7th or 28th

Months: January, February, March, April and May.

1. Only two persons attend the meeting before E. 

2. B attends the meeting on the 28th of the month which has the least number of days.

3. Three persons attend the meeting between B and H.

 

4) D attends on the month which has 30 days.

5. A and D attend on odd dates. 

6) Two persons attend the meeting between J and D.

Case 1:

 

Case 2:

 

Case 3:

 

7) F attends the meeting in March. 

8. J attends the meeting immediately after I (this eliminates case 3)

9. Two persons attend the meeting between G and I (this eliminates case 2).

The final table is as follows:

 

C attends the meeting on 28^th January.

Given:

Ten persons: A, B, C, D, E, F, G, H, I and J

Date: 7th or 28th

Months: January, February, March, April and May.

1. Only two persons attend the meeting before E. 

2. B attends the meeting on the 28th of the month which has the least number of days.

3. Three persons attend the meeting between B and H.

 

4) D attends on the month which has 30 days.

5. A and D attend on odd dates. 

6) Two persons attend the meeting between J and D.

Case 1:

 

Case 2:

 

Case 3:

 

7) F attends the meeting in March. 

8. J attends the meeting immediately after I (this eliminates case 3)

9. Two persons attend the meeting between G and I (this eliminates case 2).

The final table is as follows:

 

 Four persons attend the meeting between B and I.

Given:

Ten persons: A, B, C, D, E, F, G, H, I and J

Date: 7th or 28th

Months: January, February, March, April and May.

1. Only two persons attend the meeting before E. 

2. B attends the meeting on the 28th of the month which has the least number of days.

3. Three persons attend the meeting between B and H.

 

4) D attends on the month which has 30 days.

5. A and D attend on odd dates. 

6) Two persons attend the meeting between J and D.

Case 1:

 

Case 2:

 

Case 3:

 

7) F attends the meeting in March. 

8. J attends the meeting immediately after I (this eliminates case 3)

9. Two persons attend the meeting between G and I (this eliminates case 2).

The final table is as follows:

 

 G attends the meeting immediately after F.

Given:

Ten persons: A, B, C, D, E, F, G, H, I and J

Date: 13th or 24th

Months: January, February, March, April and May.

1. Only two persons attend the meeting before E. 

2. B attends the meeting on the 24th of the month which has the least number of days.

3. Three persons attend the meeting between B and H.

 

4) D attends on the month which has 30 days.

5. A and D attend on odd dates. 

6) Two persons attend the meeting between J and D.

Case 1:

 

Case 2:

 

Case 3:

 

7) F attends the meeting in March 

8. J attends the meeting immediately after I (this eliminates case 3)

9. Two persons attend the meeting between G and I (this eliminates case 2).

The final table is as follows:

 

 J attends the meeting in the month of May.

Given:

Ten persons: A, B, C, D, E, F, G, H, I and J

Date: 7th or 28th

Months: January, February, March, April and May.

1. Only two persons attend the meeting before E. 

2. B attends the meeting on the 28th of the month which has the least number of days.

3. Three persons attend the meeting between B and H.

 

4) D attends on the month which has 30 days.

5. A and D attend on odd dates. 

6) Two persons attend the meeting between J and D.

Case 1:

 

Case 2:

 

Case 3:

 

7) F attends the meeting in March. 

8. J attends the meeting immediately after I (this eliminates case 3)

9. Two persons attend the meeting between G and I (this eliminates case 2).

The final table is as follows:

 

A attends the meeting immediately before C.

Given word:- TECHNIQUE

Forward Pair - EI

Backward Pair - No such pair.

Thus, there is only one such pair of letters in the word “TECHNIQUE” which has as many letters between them as in the English alphabetical series (both in forward and backward directions).

The correct answer is One.

The least possible Venn Diagram for the given statements will be as follows,

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I. All paint can be colour → False (as only a few paint is colour which means there is some paint which is not colour so, all paint cannot be colour)

II. Some colour is print → False (as no code is colour and some colour is code. Therefore, some colour which is code is not print)

 Neither I nor II follow.

The least possible Venn Diagram for the given statements will be as follows,

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I. Some question can be gift → True (as there is no direct relation given between question and gift so, the possibility holds true)

II. No life is question → False (as there is no direct relation given between life and question so nothing can be said as specific. That is why it will be considered as false)

Only I follow.

The least possible Venn Diagram for the given statements will be as follows,

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I. All cat can be dog → False (as only a few cat is dog means there is some cat which is not dog. So, all cat can never be dog)

II. No mouse is dog → False (as there is no direct relation given between mouse and dog so we cannot say anything specifically. That’s why the answer is not known and it will be considered as false)

Neither I nor II follow.

The least possible Venn Diagram for the given statements will be as follows,

 

 

I.  Some pen is paper →False (as no direct relation given between pen and paper so we cannot say anything specifically. That’s why the answer is not known and it will be considered as false)

II.  All paper can be pencil →True (as in statement it is mentioned that some pencil is not paper but we do not know that any paper is pencil or not so, the possibility holds true)

Only II follows.

The pattern used here is:-

Thus, the correct answer is “XJQ”.

Eight Persons: E, F, G, H, I, J, K, and L

Designations: MD, GM, AGM, DGM, Manager, AM, PO, and Clerk

1) Only four persons are junior to H who is senior to K.

2. Three designations are there between I and K.

3. As many persons are senior to I as junior to J.

4) F is three posts senior to L who is junior to J.

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5) G is a senior to E who is not the GM in the bank.

As the above condition is not satisfied in case 2, it will get canceled.

Final Arrangement:

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 I is the GM in the bank.

Eight Persons: E, F, G, H, I, J, K, and L

Designations: MD, GM, AGM, DGM, Manager, AM, PO, and Clerk

1) Only four persons are junior to H who is senior to K.

2. Three designations are there between I and K.

3. As many persons are senior to I as junior to J.

4) F is three posts senior to L who is junior to J.

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5) G is a senior to E who is not the GM in the bank.

As the above condition is not satisfied in case 2, it will get canceled.

Final Arrangement:

 

 F is the Manager in the bank.

Eight Persons: E, F, G, H, I, J, K, and L

Designations: MD, GM, AGM, DGM, Manager, AM, PO, and Clerk

1) Only four persons are junior to H who is senior to K.

2. Three designations are there between I and K.

3. As many persons are senior to I as junior to J.

4) F is three posts senior to L who is junior to J.

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5) G is a senior to E who is not the GM in the bank.

As the above condition is not satisfied in case 2, it will get canceled.

Final Arrangement:

 

E is the AGM in the bank.

Eight Persons: E, F, G, H, I, J, K, and L

Designations: MD, GM, AGM, DGM, Manager, AM, PO, and Clerk

1) Only four persons are junior to H who is senior to K.

2. Three designations are there between I and K.

3. As many persons are senior to I as junior to J.

4) F is three posts senior to L who is junior to J.

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5) G is a senior to E who is not the GM in the bank.

As the above condition is not satisfied in case 2, it will get canceled.

Final Arrangement:

 

J is the PO in the bank.

Eight Persons: E, F, G, H, I, J, K, and L

Designations: MD, GM, AGM, DGM, Manager, AM, PO, and Clerk

1) Only four persons are junior to H who is senior to K.

2. Three designations are there between I and K.

3. As many persons are senior to I as junior to J.

4) F is three posts senior to L who is junior to J.

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5) G is a senior to E who is not the GM in the bank.

As the above condition is not satisfied in case 2, it will get canceled.

Final Arrangement:

Two designations are there between G and H.

Given statements: A > C < D < B, D > E < F

On combining: A > C < D > E < F, B > D > E < F

Conclusions:

I. A > E → False (as A > C < D > E, so no definite relationship can be determined between A and E)

II. E < B → True (as B > D > E, so E is less than B)

Only II follow.

Given statements: P ≥ M > V ≤ U, T ≤ U > L

On combining: P ≥ M > V ≤ U > L, P ≥ M > V ≤ U ≥ T

Conclusions:

I. L < M → False (as M > V ≤ U > L, so no definite relationship can be determined between L and M)

II. V ≤ T → False (as V ≤ U ≥ T, so no definite relationship can be determined between V and T)

Neither I nor II follow.

Given statements: M > N = O < P ≤ Q, R > S ≤ Q

On combining: M > N = O < P ≤ Q ≥ S < R

Conclusions:

I. N ≤ S → False (as N = O < P ≤ Q ≥ S, so no definite relationship can be determined between N and S)

II. N > S → False (as N = O < P ≤ Q ≥ S, so no definite relationship can be determined between O and S)

There are only three relations can be established between N and S. S can be greater than N, S can be less than N and S can be equal to N. These two conclusions are satisfying all three possible relations. That is why the answer will be either or between S and N.

Either I or II follow.

Given statements: C ≤ D < E ≥ F < G, D > O ≥ K

On combining: C ≤ D > O ≥ K, K ≤ O < D < E ≥ F < G

Conclusions:

I. D > K → True (as D > O ≥ K, so D is greater than K)

II. E > K → True (as K ≤ O < D < E, so clearly E is greater than K)

Both I and II follows.

Given Word:- CHEMISTRY

The fourth letter is 'M', the fifth letter is 'I', the seventh letter is 'T' and the eighth letter is 'R' from the left end.

Meaningful four-letter English words can be formed is -

1) TRIM which means to cut something off.

Thus, only one such word can be formed.

The correct answer is One.

Boxes – 8; A, B, C, D, E, F, G and H;

1. At least two and at most four boxes are placed above box D.

From this we get the possible arrangements as follows.

 

2. Three boxes are placed between boxes D and F.
3. Number of boxes placed above box F is same as the number of boxes placed below box H.

4. Only two boxes are placed between boxes H and A.

From this case 2 and case 3 are eliminated, as we cannot place box A.

5. Box A is placed three boxes away from box C.

6. Box E is not placed at the top but above box B.

 

If all the boxes are arranged in alphabetical order from bottom to top, then we get 

 

Here, only box E remains in the same position.

Therefore, one is the correct option.

Boxes – 8; A, B, C, D, E, F, G and H;

1. At least two and at most four boxes are placed above box D.

    From this we get the possible arrangements as follows.

 

2. Three boxes are placed between boxes D and F.
3. Number of boxes placed above box F is same as the number of boxes placed below box H.

4. Only two boxes are placed between boxes H and A.

      From this case 2 and case 3 are eliminated, as we cannot place box A.

5. Box A is placed three boxes away from box C.

6. Box E is not placed at the top but above box B.

 

Box G is placed at the top.

Boxes – 8; A, B, C, D, E, F, G and H;

1. At least two and at most four boxes are placed above box D.

    From this we get the possible arrangements as follows.

 

2. Three boxes are placed between boxes D and F.
3. Number of boxes placed above box F is same as the number of boxes placed below box H.

4. Only two boxes are placed between boxes H and A.

      From this case 2 and case 3 are eliminated, as we cannot place box A.

5. Box A is placed three boxes away from box C.

6. Box E is not placed at the top but above box B.

 

Five boxes are placed between boxes H and C.

Boxes – 8; A, B, C, D, E, F, G and H;

1. At least two and at most four boxes are placed above box D.

    From this we get the possible arrangements as follows.

 

2. Three boxes are placed between boxes D and F.
3. Number of boxes placed above box F is same as the number of boxes placed below box H.

4. Only two boxes are placed between boxes H and A.

      From this case 2 and case 3 are eliminated, as we cannot place box A.

5. Box A is placed three boxes away from box C.

6. Box E is not placed at the top but above box B.

 

Box E is fourth from top.

Boxes – 8; A, B, C, D, E, F, G and H;

1. At least two and at most four boxes are placed above box D.

    From this we get the possible arrangements as follows.

 

2. Three boxes are placed between boxes D and F.
3. Number of boxes placed above box F is same as the number of boxes placed below box H.

4. Only two boxes are placed between boxes H and A.

      From this case 2 and case 3 are eliminated, as we cannot place box A.

5. Box A is placed three boxes away from box C.

6. Box E is not placed at the top but above box B.

 

Number of boxes above box C is same as number of boxes below box G, and number of boxes above box D is same as number of boxes below box B.

The number of boxes above box A is the same as the number of boxes below box E.

The followed pattern is,

Z - 4 = V, V - 4 = R, R - 4 = N

U - 4 = Q, Q - 4 = M, M - 4 = I

S - 4 = O, O - 4 = K, K - 4 = G

T - 4 = P, P - 4 = L, L - 4 = H

A - 4 = W, W - 4 = S, S - 4 = O

As logic implemented above “UQMJ” is odd one Out.

1. The person first goes 5 km towards the east.

2. Then he turns left and travels in the north direction.

3. Then he turns right and travels in the east direction.

4. Then again he turns right and travels in the south direction.

5. He then turns left and travels in the east direction and after taking one more left he travels in the north direction.

6. He finally takes a left turn and travels in the west direction.

The least possible graphical representation of the given data is:

Point N is to the north-west of point Q.

1. The person first goes 5 km towards the east.

2. Then he turns left and travels in the north direction.

3. Then he turns right and travels in the east direction.

4. Then again he turns right and travels in the south direction.

5. He then turns left and travels in the east direction and after taking one more left he travels in the north direction.

6. He finally takes a left turn and travels in the west direction.

The least possible graphical representation of the given data is:

The starting point is L and the finishing point is S which is 8 km to the east of L.

The shortest distance between the starting and the finishing point is 8 km.

1. The person first goes 5 km towards the east.

2. Then he turns left and travels in the north direction.

3. Then he turns right and travels in the east direction.

4. Then again he turns right and travels in the south direction.

5. He then turns left and travels in the east direction and after taking one more left he travels in the north direction.

6. He finally takes a left turn and travels in the west direction.

The least possible graphical representation of the given data is:

Point S is 2 km to the south of point O.

Given:

The investment ratio of Shivani and Payal = 7 : 9

The ratio of their time period = 3 : 4         

Formula used:

Profit = Money invested × Time 

The investment ratio is equal to the profit ratio

Calculation:

Ratio of their profit will be = (7 × 3) : (9 × 4)

⇒ 7 : 12

According to question,

Profit = Rs. 39,140

So, the profit share of Shivani = 39140 × {7/(7 + 12)}

⇒ 2060 × 7

⇒ Rs. 14,420 

∴ Shivani profit will be Rs 14,420 .

Given:

Selling Price = Rs. 3360

Discount = 8%

Loss = 16%

Profit = 15%

Formula Used:

C.P = S.P/(100 – loss%) × 100

M.P = C.P × (100 + profit%)/(100 – discount%)

Calculation:

∴ according to the formula:

C.P = S.P/(100 – loss%) × 100

⇒ C.P = 3360/(100 - 16) × 100

⇒ C.P = (3360/84) × 100

⇒ C.P = 40 × 100

⇒ C.P = Rs. 4000

∴ Marked Price = C.P × (100 + profit%)/(100 – discount%)

⇒ 4000 × (100 + 15)/(100 – 8)

⇒ 4000 × (115/92)

⇒ 4000 × 1.25

⇒ Rs. 5000

∴ The marked price of the saree should be Rs. 5,000.

Given:

Akash is twice effcient of Bikram

Chetan takes twice as much time as Bikram takes

Time taken by Bikram to completes the work = 8 days

Formula used:

Efficiency = Work/(Time Taken)

Calculation:

Efficiency of Akash : Efficiency of Bikram = 2 : 1

Then, Time taken by Akash : Time taken by Bikram = 1 : 2

Time taken by Akash = (1/2) × Time taken by Bikram

⇒ (1/2) × 8 days

⇒ 4 days

And, Time taken by Bikram : Time taken by Chetan = 1 : 2

Then, Time taken by Chetan = 2 × Time taken by Bikram

⇒ 2 × 8 days

⇒ 16 days

Total work = LCM of 4, 8 and 16 = 16 units

Efficiency of Akash = (16/4) = 4 units/day

And Efficiency of Chetan = (16/16) = 1 unit/day

In 1 day, Akash and Chetan complete = (4 + 1) units = 5 units

Now, Required time = 16/5 days

∴ Akash and Chetan together complete the work in (16/5) days.

Let P = principal, R = rate of interest and N = time period

⇒ Simple interest = PNR/100

Given,

⇒ R = 12%, N = 3

⇒ Simple interest = Rs.5400

⇒ 5400 = (P × 12 × 3)/100

⇒ 5400 × 100 = 36P

⇒ P = 15000

∴Principal amount borrowed by him is Rs. 15000

Let the amount invested for 3 yrs = x

And amount invested for 5 yrs = y

We can write,

⇒ x (1 + 5/100)³ = y (1 + 5/100)⁵ [given that amount is equal after 3 yrs and 5 yrs]

⇒ x/y = (1 + 5/100)²

⇒ x/y = (21/20)²

⇒ x/y = 441/400      ----(1)

⇒ x + y = 5046      ----(2)

Let x = 441k

⇒ y = 400k

Putting the value of x and y in equation 2 we have

⇒ 841k = 5046

⇒ K = 6

∴ Amount invested for 3 years = 441k = 441 × 6 = Rs. 2646

Let the speed of train B x km/h

Speed of train A = 120 km/h

Time taken by the train A to cross a pole = 30 seconds = 30/(60 × 60) h = 1/120 h

Length of train A = 120 × 1/120 km = 1 km

Length of platform = 800 m = 0.8 km

Time taken by train A to cross the platform = (1 + 0.8)/120 h = 0.015 h

Time taken by train B to cross the pole = 45 seconds = 45/(60 × 60) h = 1/80 h

Length of train B = x × 1/80 km = x/80 km

Length of platform = 280 m = 0.28 km

Time taken by train B to cross the platform = (x/80 + 0.28)/x h = 0.015 h (given)

⇒ x/80 + 0.28 = 0.015x

⇒ x + 22.4 = 1.2x

⇒ 0.2x = 22.4

⇒ x = 112

∴ The speed of train B is 112 km/h

On Monday:

The total number of type X and Y cars sold is 120.

The ratio of type X and Y cars is 5:7,

So the total number of type X and Y cars is 50 and 70 respectively.

On Tuesday:

The total number of type X and Y cars sold is 150.

The ratio of type X and Y cars is 7:8,

So the total number of type X and Y cars is 70 and 80 respectively.

On Wednesday:

The total number of type X and Y cars sold is 140.

The ratio of type X and Y cars is 5:9,

So the total number of type X and Y cars is 50 and 90 respectively.

On Thursday:

The total number of type X and Y cars sold is 170.

The ratio of type X and Y cars is 9:8,

So the total number of type X and Y cars is 90 and 80 respectively.

On Friday:

The total number of type X and Y cars sold is 110.

The ratio of type X and Y cars is 5:6,

So the total number of type X and Y cars is 50 and 60 respectively.

The total number of type X cars sold on Tuesday and Wednesday = 70 + 50 = 120

The total number of type Y cars sold on Friday and Wednesday = 60 + 90 = 150

So,

By ((150 - 120)/150) x 100 = 20% the type X cars sold in two days is less than the total number of type Y cars in two days.

Hence by 20%, the type X cars sold in two days is less than the total number of type Y cars in two days.

On Monday:

The total number of type X and Y cars sold is 120.

The ratio of type X and Y cars is 5:7,

So the total number of type X and Y cars is 50 and 70 respectively.

On Tuesday:

The total number of type X and Y cars sold is 150.

The ratio of type X and Y cars is 7:8,

So the total number of type X and Y cars is 70 and 80 respectively.

On Wednesday:

The total number of type X and Y cars sold is 140.

The ratio of type X and Y cars is 5:9,

So the total number of type X and Y cars is 50 and 90 respectively.

On Thursday:

The total number of type X and Y cars sold is 170.

The ratio of type X and Y cars is 9:8,

So the total number of type X and Y cars is 90 and 80 respectively.

On Friday:

The total number of type X and Y cars sold is 110.

The ratio of type X and Y cars is 5:6,

So the total number of type X and Y cars is 50 and 60 respectively.

On Monday,

The total percentage of sedan cars of type X and type Y is 30% and 40%, 

So the total number of Sedan cars of type X is 30% of 50 = 15

So the total number of hatchback cars is 50 - 15 = 35.

The total number of Sedan cars of type Y is 40% of 70 = 28

So the total number of hatchback cars is 70 - 28 = 42.

So the difference between the hatchback cars is 42 - 35 = 7 cars.

Hence the difference between the hatchback cars is 7 cars.

On Monday:

The total number of type X and Y cars sold is 120.

The ratio of type X and Y cars is 5:7,

So the total number of type X and Y cars is 50 and 70 respectively.

On Tuesday:

The total number of type X and Y cars sold is 150.

The ratio of type X and Y cars is 7:8,

So the total number of type X and Y cars is 70 and 80 respectively.

On Wednesday:

The total number of type X and Y cars sold is 140.

The ratio of type X and Y cars is 5:9,

So the total number of type X and Y cars is 50 and 90 respectively.

On Thursday:

The total number of type X and Y cars sold is 170.

The ratio of type X and Y cars is 9:8,

So the total number of type X and Y cars is 90 and 80 respectively.

On Friday:

The total number of type X and Y cars sold is 110.

The ratio of type X and Y cars is 5:6,

So the total number of type X and Y cars is 50 and 60 respectively.

The total number of type X cars on Monday and Wednesday = 50 + 50 = 100

The total number of type Y cars on Monday and Wednesday = 70 + 90 = 160

Hence the ratio of type X and type Y cars is 100:160 = 5:8.

On Monday:

The total number of type X and Y cars sold is 120.

The ratio of type X and Y cars is 5:7,

So the total number of type X and Y cars is 50 and 70 respectively.

On Tuesday:

The total number of type X and Y cars sold is 150.

The ratio of type X and Y cars is 7:8,

So the total number of type X and Y cars is 70 and 80 respectively.

On Wednesday:

The total number of type X and Y cars sold is 140.

The ratio of type X and Y cars is 5:9,

So the total number of type X and Y cars is 50 and 90 respectively.

On Thursday:

The total number of type X and Y cars sold is 170.

The ratio of type X and Y cars is 9:8,

So the total number of type X and Y cars is 90 and 80 respectively.

On Friday:

The total number of type X and Y cars sold is 110.

The ratio of type X and Y cars is 5:6,

So the total number of type X and Y cars is 50 and 60 respectively.

On Thursday

<br>

The total number of X type of cars bought by females = 90

30% of the type Y cars are bought by females,

So the total number of type Y cars bought by females = 30% of 80 = 24 cars.

So the total number of cars that are bought by females = 90 + 24 = 114.

Hence the total number of cars that are bought by females is 114.

On Monday:

The total number of type X and Y cars sold is 120.

The ratio of type X and Y cars is 5:7,

So the total number of type X and Y cars is 50 and 70 respectively.

On Tuesday:

The total number of type X and Y cars sold is 150.

The ratio of type X and Y cars is 7:8,

So the total number of type X and Y cars is 70 and 80 respectively.

On Wednesday:

The total number of type X and Y cars sold is 140.

The ratio of type X and Y cars is 5:9,

So the total number of type X and Y cars is 50 and 90 respectively.

On Thursday:

The total number of type X and Y cars sold is 170.

The ratio of type X and Y cars is 9:8,

So the total number of type X and Y cars is 90 and 80 respectively.

On Friday:

The total number of type X and Y cars sold is 110.

The ratio of type X and Y cars is 5:6,

So the total number of type X and Y cars is 50 and 60 respectively.

The weight of type X car is 1000 kg and the weight of type Y car is 1200 kg,

So the total weight of both types of car on Friday = 50 x 1000 + 60 x 1200 = 50000 + 72000 = 122000.

Hence the total weight of both cars on Friday is 122000 kg.

Given:

5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

⇒ 6% of 800 + ?% of 10 = 11 × 3 + √100 × 3

⇒ (6/100) × 800 + (?/100) × 10 = 33 + 10 × 3

⇒ 48 + (?/10) = 33 + 30

⇒ 48 + (?/10) = 63

⇒ (?/10) = 63 - 48 

⇒ (?/10) = 15 

⇒ ? = 15 × 10 = 150

∴ The value of ? is 150.

Given:

​9.99% of 29.906 + 299.84% of 54.908 – 59.86% of 149.59 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

​9.99% of 29.906 + 299.84% of 54.908 – 49.86% of 149.59 = ?

⇒ ​10% of 30 + 300% of 55 – 50% of 150 = ?

⇒ ​(10/100) × 30 + (300/100) × 55 – (50/100) × 150 = ?

⇒ 3 + 165 – 75 = ?

⇒ ? = 93

∴ The value of ? is 93.

Given:  

63.76 ÷ √(255.79) × {24.99% of (2.89)3} = (?)3

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

63.92 ÷ √(255.89) × {24.91% of (2.99)3} = (?)³

⇒ 64 ÷ √(256) × {25% of (3)3} = (?)³

⇒ 64 ÷ 16 × {(25/100) × 27} = (?)³

⇒ 4 × (27/4) = (?)³

⇒ 27 = (?)³

⇒ ? = 3

∴ 3 will come in place of the question mark ('?').

Given:

728.921/3 + √1155.78 + 6.042 ÷ 2.962 + 1.990 = √?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

Since we need to find out the approximate value, we can write these values to their nearest integers.

728.821/3 + √1155.98 + 6.142 ÷ 2.992 + 1.970 = √?

⇒ ∛729 + √1156 + 62 ÷ 32 + 2 = √?

⇒ 9 + 34 + 36/9 + 2 = √?

⇒ 9 + 34 + 4 + 2 = √?

⇒ √? = 49

⇒ ? = 49²

⇒ ? = 2401

∴ The value of ‘?’ Is 2401.

Given:

11.11% of 99.11 + 22.22% of 98.97 – 9.99% of 100 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

Since we need to find out the approximate value, we can write these values to their nearest integers.

11.11% of 99.17 + 22.22% of 98.87 - 9.89% of 100.12 = ?

⇒ 11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ 11 + 22  - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

<br>

Alternate Method

11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ {(1/9) + (2/9)} × 99 - 10 = ?

⇒ (1/3) × 99 - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

The total number of mobile phones in each of the shops is greater than the total number of laptops.

 

For shop A:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (1)

Also, it is given that,

Mobile phones - Laptops = 20...................................... (2)

Adding both equations,

2 x Mobile Phones = 130

So,

Mobile phones = 65

Put this in equation (2),

65 - Laptops = 20

So,

the number of laptops = 65 - 20 = 45.

For shop B:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (3)

Also, it is given that,

Mobile phones - Laptops = 10...................................... (4)

Adding both equations,

2 x Mobile Phones = 120

So,

Mobile phones = 60

Put this in equation (4),

60 - Laptops = 10

So,

the number of laptops = 60 - 10 = 50.

For shop C:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

80 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 160................................... (5)

Also, it is given that,

Mobile phones - Laptops = 40...................................... (6)

Adding both equations,

2 x Mobile Phones = 200

So,

Mobile phones = 100

Put this in equation (6),

100 - Laptops = 40

So,

the number of laptops = 100 - 40 = 60.

For shop D:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

70 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 140................................... (7)

Also, it is given that,

Mobile phones - Laptops = 30...................................... (8)

Adding both equations,

2 x Mobile Phones = 170

So,

Mobile phones = 85

Put this in equation (8),

85 - Laptops = 30

So,

the number of laptops = 85 - 30 = 55.

From the above data:

<br>

The ratio of Lenovo and Samsung mobile phones in shop C is 9:11,

So the total number of Samsung mobile phones = (11/(9 + 11)) x 100 = 55

The ratio of Lenovo and Samsung Laptops in shop C is 11:4,

So the total number of Lenovo Laptops = 11/(11 + 4)x 60 = 44

Hence the difference between the total number of Samsung mobile phones and Lenovo laptops is 55 - 44 = 11.

The total number of mobile phones in each of the shops is greater than the total number of laptops.

 

For shop A:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (1)

Also, it is given that,

Mobile phones - Laptops = 20...................................... (2)

Adding both equations,

2 x Mobile Phones = 130

So,

Mobile phones = 65

Put this in equation (2),

65 - Laptops = 20

So,

the number of laptops = 65 - 20 = 45.

For shop B:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (3)

Also, it is given that,

Mobile phones - Laptops = 10...................................... (4)

Adding both equations,

2 x Mobile Phones = 120

So,

Mobile phones = 60

Put this in equation (4),

60 - Laptops = 10

So,

the number of laptops = 60 - 10 = 50.

For shop C:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

80 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 160................................... (5)

Also, it is given that,

Mobile phones - Laptops = 40...................................... (6)

Adding both equations,

2 x Mobile Phones = 200

So,

Mobile phones = 100

Put this in equation (6),

100 - Laptops = 40

So,

the number of laptops = 100 - 40 = 60.

For shop D:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

70 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 140................................... (7)

Also, it is given that,

Mobile phones - Laptops = 30...................................... (8)

Adding both equations,

2 x Mobile Phones = 170

So,

Mobile phones = 85

Put this in equation (8),

85 - Laptops = 30

So,

the number of laptops = 85 - 30 = 55.

From the above data:

<br>

The total number of laptops in shop D = 55

The total number of laptops and mobile phones in shop C = 100 + 60 = 160

So the required percentage is 55/160 x 100 = 34.375%.

Hence the total number of laptops in shop D is 34.375% of the total number of laptops and mobile phones in shop C.

The total number of mobile phones in each of the shops is greater than the total number of laptops.

 

For shop A:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (1)

Also, it is given that,

Mobile phones - Laptops = 20...................................... (2)

Adding both equations,

2 x Mobile Phones = 130

So,

Mobile phones = 65

Put this in equation (2),

65 - Laptops = 20

So,

the number of laptops = 65 - 20 = 45.

For shop B:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (3)

Also, it is given that,

Mobile phones - Laptops = 10...................................... (4)

Adding both equations,

2 x Mobile Phones = 120

So,

Mobile phones = 60

Put this in equation (4),

60 - Laptops = 10

So,

the number of laptops = 60 - 10 = 50.

For shop C:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

80 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 160................................... (5)

Also, it is given that,

Mobile phones - Laptops = 40...................................... (6)

Adding both equations,

2 x Mobile Phones = 200

So,

Mobile phones = 100

Put this in equation (6),

100 - Laptops = 40

So,

the number of laptops = 100 - 40 = 60.

For shop D:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

70 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 140................................... (7)

Also, it is given that,

Mobile phones - Laptops = 30...................................... (8)

Adding both equations,

2 x Mobile Phones = 170

So,

Mobile phones = 85

Put this in equation (8),

85 - Laptops = 30

So,

the number of laptops = 85 - 30 = 55.

From the above data:

<br>

The total number of mobile phones in shop E is 20% more than the total number of mobile phones in shop D,

So the total number of mobile phones in shop E = 1.2 x 85 = 102.

The total number of laptops in shop E is 10% less than the total number of laptops in shop C,

So the total number of laptops in shop E = 0.9 x 60 = 54.

Hence the difference between the total number of laptops and mobile phones sold in shop E is 102 - 54 = 48.

The total number of mobile phones in each of the shops is greater than the total number of laptops.

 

For shop A:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (1)

Also, it is given that,

Mobile phones - Laptops = 20...................................... (2)

Adding both equations,

2 x Mobile Phones = 130

So,

Mobile phones = 65

Put this in equation (2),

65 - Laptops = 20

So,

the number of laptops = 65 - 20 = 45.

For shop B:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (3)

Also, it is given that,

Mobile phones - Laptops = 10...................................... (4)

Adding both equations,

2 x Mobile Phones = 120

So,

Mobile phones = 60

Put this in equation (4),

60 - Laptops = 10

So,

the number of laptops = 60 - 10 = 50.

For shop C:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

80 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 160................................... (5)

Also, it is given that,

Mobile phones - Laptops = 40...................................... (6)

Adding both equations,

2 x Mobile Phones = 200

So,

Mobile phones = 100

Put this in equation (6),

100 - Laptops = 40

So,

the number of laptops = 100 - 40 = 60.

For shop D:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

70 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 140................................... (7)

Also, it is given that,

Mobile phones - Laptops = 30...................................... (8)

Adding both equations,

2 x Mobile Phones = 170

So,

Mobile phones = 85

Put this in equation (8),

85 - Laptops = 30

So,

the number of laptops = 85 - 30 = 55.

From the above data:

<br>

The total number of mobile phones in all the shops = 65 + 60 + 100 + 85 = 310

The total number of laptops in all the shops = 45 + 50 + 60 + 55 = 210

Hence the required difference is 310 - 210 = 100.

The total number of mobile phones in each of the shops is greater than the total number of laptops.

 

For shop A:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (1)

Also, it is given that,

Mobile phones - Laptops = 20...................................... (2)

Adding both equations,

2 x Mobile Phones = 130

So,

Mobile phones = 65

Put this in equation (2),

65 - Laptops = 20

So,

the number of laptops = 65 - 20 = 45.

For shop B:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

55 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 110................................... (3)

Also, it is given that,

Mobile phones - Laptops = 10...................................... (4)

Adding both equations,

2 x Mobile Phones = 120

So,

Mobile phones = 60

Put this in equation (4),

60 - Laptops = 10

So,

the number of laptops = 60 - 10 = 50.

For shop C:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

80 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 160................................... (5)

Also, it is given that,

Mobile phones - Laptops = 40...................................... (6)

Adding both equations,

2 x Mobile Phones = 200

So,

Mobile phones = 100

Put this in equation (6),

100 - Laptops = 40

So,

the number of laptops = 100 - 40 = 60.

For shop D:

Average of laptops and phone = (Laptops + Mobile Phones)/2 

So,

70 x 2 = Laptops + Mobile Phones

So,

Laptops + Mobile phones = 140................................... (7)

Also, it is given that,

Mobile phones - Laptops = 30...................................... (8)

Adding both equations,

2 x Mobile Phones = 170

So,

Mobile phones = 85

Put this in equation (8),

85 - Laptops = 30

So,

the number of laptops = 85 - 30 = 55.

From the above data:

The total number of mobile phones in shop B = 60

The total number of laptops in shop A = 45

So,

By ((60 - 45)/45) x 100 = 33.33%, the total number of mobile phones in shop B is more. 

Hence by 33.33%, the total number of mobile phones in shop B is more than the total number of laptops in shop A.

Let's denote the age of the younger son as Y

and the age of the elder son as E.

We know that E = Y + 6,

and both Y and E are prime numbers.

Also, 30 < Y + E < 50. Considering the information above,

we can try different prime numbers for Y and E:

• If Y = 11, then E = 17 (Y + 6). Their ages sum to 28, which is not within the given range.
• If Y = 13, then E = 19 (Y + 6). Their ages sum to 32, which is within the given range.
• If Y = 17, then E = 23 (Y + 6). Their ages sum to 40, which is within the given range.
• If Y = 19, then E = 25 (Y + 6). However, 25 is not a prime number.

 

So, we have two possible pairs of ages for the sons:

(13, 19) and (17, 23).

We're given that the father's age will be the sum of their ages after 6 years.

For the first pair (13, 19),

the sons' ages after 6 years will be 19 and 25, which sums to 44.

For the second pair (17, 23),

the sons' ages after 6 years will be 23 and 29, which sum to 52.

Since the father's age should be the sum of their ages after 6 years,

the father is either 44 or 52 years old now.

Shortcut Trick

Ratio of Difference to Sum = 3 : 10

Given, Difference = 24 years ⇒ 3 units = 24 years ⇒ 1 unit = 8 years.

Sum = 10 units = 10 × 8 = 80 years.

Father's age = (Sum + Difference) ÷ 2 = (80 + 24) ÷ 2 = 52 years.

∴ The correct answer is 52 years.

Alternate Method

Given:

Ratio of (F − S) : (F + S) = 3 : 10

Difference (F − S) = 24 years

Calculations:

⇒ 3 units = 24 years

⇒ 1 unit = 24 ÷ 3 = 8 years

⇒ Sum (F + S) = 10 units = 10 × 8 = 80 years

⇒ F + S = 80

⇒ F − S = 24

Adding both equations:

⇒ 2F = 104

⇒ F = 104 ÷ 2 = 52

∴ The correct answer is 52 years.

Additional Information

Sum and Difference Rule

If the sum (a+b) and difference (a−b) of two numbers are known, then the larger number is (Sum + Diff) ÷ 2 and the smaller number is (Sum − Diff) ÷ 2.

Age Difference Property

The difference between the ages of two persons remains constant throughout their lives, regardless of the number of years passed or preceding.

Componendo and Dividendo

If (a+b)/(a−b) = x/y, then a/b = (x+y)/(x−y). In this case, F/S = (10+3)/(10−3) = 13/7.

Given:

length + breadth = 34 cm

Area = 280 sq.cm

Formula used:

Area of a rectangle = length × breadth

Calculation:

Let the length be L cm, and

the breadth be B cm.

L + B = 34 

⇒ B = 34 - L

So, the area becomes:

280 = L × B

⇒ 280 = L × (34 - L)

⇒ 280 = 34L - L²  

⇒ L² - 34L + 280 = 0

⇒ (L - 20)(L - 14) = 0

⇒ L = 20, 14

∴ The length of the longer side is 20 cm

Alternate Method

Given:

Height = 7 m

Volume = 7128 cu.m

Formula used:

Volume of cylinder = πr²h

C.S.A = 2πrh

Calculation:

πr2h = 7128

⇒ r² = (7128 × 7)/(22 × 7 )

⇒ r = √324 = 18 m

C.S.A = 2 × 22/7 × 18 × 7 = 792 m²

∴ the area is 792 m²

 

Shortcut Trick

If we calculate the radius of the cylinder, then 

The area of the sheet required = CSA of Cylinder = 2πrh =  2/r × (πr2h)

So, Required area = 2/18 × (7128) = 1/9 × (7128) = 792 m2

GIVEN:

Speed of boat = 9 km/h

Speed of stream = 3 km/h

FORMULA USED:

Speed = distance/speed

CALCULATION:

Downstream speed = (9 + 3) = 12 km/h

Upstream speed = (9 - 3) = 6 km/h

Let the distance travelled is x km

(x/12) + (x/6) = 1

⇒ (x + 2x)/12 =1

⇒ 3x = 12

⇒ x = 4 km

∴ Distance travelled is 4 km.

Shortcut Trick

Equate the speed of the stream in both ratios to compare the boats.

Ratio 1 (Stream : Boat₁) = 4 : 7

Ratio 2 (Stream : Boat₂) = 2 : 5 × 2 → 4 : 10

Comparing the Boat speeds directly when Stream speed is 4 units.

Boat₁ : Boat₂ = 7 : 10

∴ The correct answer is 7 : 10.

Alternate Method

Given: Ratio of stream (s) to boat 1 (b₁) = 4 : 7. Ratio of stream (s) to boat 2 (b₂) = 2 : 5.

Calculations:

⇒ Let Speed of Stream = s, Speed of Boat 1 = b₁, and Speed of Boat 2 = b₂.

⇒ From first ratio: s/b₁ = 4/7 ⇒ b₁ = 7s/4

⇒ From second ratio: s/b₂ = 2/5 ⇒ b₂ = 5s/2

⇒ Ratio of Boat 1 to Boat 2 = b₁ : b₂ = (7s/4) : (5s/2)

⇒ Multiply by 4 to clear denominators: 7s : (5s/2 × 4) = 7s : 10s

⇒ b₁ : b₂ = 7 : 10

∴ The correct answer is 7 : 10.

Additional Information

Relative Speed in Water

Downstream Speed = Speed of Boat + Speed of Stream. Upstream Speed = Speed of Boat − Speed of Stream.

Common Variable Method

When comparing two ratios with a common element, multiply the terms of the ratios so that the common element has the same numerical value.

Given:

Average age of 14 people = 50

Calculations:

Total age of all the people in the family = 50 × 14 = 700

Newborn joins the family and 4 years later the total age of all the members = 700 + (15 × 4)

⇒ 700 + 60

⇒ 760

∴ Required average = 760/15

⇒ 50.6

The average age after 4 years is 50.6.

Shortcut Trick

Target total marks for 6 subjects at an average of 85 = 6 × 85 = 510.

Total marks obtained in 5 subjects = 435.

Required marks in sixth subject = 510 − 435 = 75.

∴ The correct answer is 75 marks.

Alternate Method

Given:

Total marks in 5 subjects = 435

Required Average of 6 subjects = 85

Formula Used:

Total Marks = Average × Number of Subjects

Calculations:

⇒ Total marks for 6 subjects = 85 × 6

⇒ Total marks for 6 subjects = 510

⇒ Marks in 6th subject = Total marks for 6 subjects − Total marks for 5 subjects

⇒ Marks in 6th subject = 510 − 435

⇒ Marks in 6th subject = 75

∴ The correct answer is 75 marks.

Additional Information

Average Concept

Average is the ratio of the Sum of observations to the Number of observations. Formula: Average = (Sum) ÷ (n).

Deviation Method

If the average of 5 subjects is 435/5 = 87, and the target average is 85, the 6th subject must compensate for the +2 deviation in 5 subjects: 85 − (5 × 2) = 75.

Given:

I. x² - 19x + 90 = 0

II. y² - 5y + 6 = 0

Calculation:

From I

x² - 19x + 90 = 0

⇒ x² - 9x - 10x + 90 = 0

⇒ x(x - 9) - 10(x - 9) = 0

⇒ (x - 9)(x - 10) = 0

⇒ x = 9, 10

From II

y² - 5y + 6 = 0

⇒ y² - 3y - 2y + 6 = 0

⇒ y(y - 3) - 2(y - 3) = 0

⇒ (y - 3)(y - 2) = 0

⇒ y = 3, 2

Comparison between x and y (via Tabulation)

<br>

∴ x > y

Important Points

Given:

I. x² - 16x + 63 = 0

II. y² + 5y - 84 = 0

Calculation:

From I

x² - 16x + 63 = 0

⇒ x² - 9x - 7x + 63 = 0

⇒ x(x - 9) - 7(x - 9) = 0

⇒ (x - 9)(x - 7) = 0

⇒ x = 9, 7

From II

y² + 5y - 84 = 0

⇒ y² + 12y - 7y -84 = 0

⇒ y(y + 12) - 7(y + 12) = 0

⇒ (y + 12)(y - 7) = 0

⇒ y = -12, 7

Comparison between x and y (via Tabulation)

<br>

∴ x ≥ y

Important Points

Given:

I. x2 - 11x + 24 = 0

II. y2 - 13y + 40 = 0

Calculation:

From I,

I. x2 - 11x + 24 = 0

⇒ x² - 8x - 3x + 24 = 0

⇒ x(x - 8) - 3(x - 8) = 0

⇒ (x - 8)(x - 3) = 0

Taking,

⇒ (x - 8) = 0 or (x - 3) = 0

⇒ x = 8 or 3

From II,

y2 - 13y + 40 = 0

⇒ y2 - 8y - 5y + 40 = 0

⇒ y(y - 8) - 5(y - 8) = 0

⇒ (y - 8)(y - 5) = 0

Taking,

⇒ (y - 8) = 0 or (y - 5) = 0

⇒ y = 8 or 5

Comparison between x and y (via Tabulation):

 

∴ x = y or relation between x and y can not be established.

Calculation:

I. 2x2 + 15x + 7 = 0

⇒ 2x² + 14x + x + 7 = 0

⇒ 2x(x + 7) + 1(x + 7) = 0

⇒ (x + 7)(2x + 1) = 0

⇒ x = (-7), (-1/2)

II. y² + 17y + 72 = 0

⇒ y² + 9y + 8y + 72 = 0

⇒ y(y + 9) + 8(y + 9) = 0

⇒ (y + 9)(y + 8) = 0

⇒ y = (-9), (-8)

 

When we compared the values of ‘x’ and ‘y’ in the table above, we found that there is only ONE relation between X and Y i.e. >. So, a relation between x and y is “x > y”.

∴ After comparison, all the values of x and y the relation is “x > y”.

Calculation:

I. x2 + 7x + 6 = 0

⇒ x² + 6x + x + 6 = 0

⇒ x(x + 6) + 1(x + 6)= 0

⇒ (x + 6)(x + 1) = 0

⇒ x = (-6), (-1)

II. y2 - 6y + 8 = 0

⇒ y² - 4y - 2y + 8 = 0

⇒ y(y - 4) - 2(y - 4) =0

⇒ (y - 4)(y - 2) = 0

⇒ y = 4, 2

Comparison between x and y (via Tabulation):

<br>

When we compared the values of ‘x’ and ‘y’ in the table above, we found that there is only ONE relation between X and Y i.e. <. So, a relation between x and y is “x < y”.

∴ After comparison, all the values of x and y the relation is “x < y”.

Shortcut Trick

I. x2 + 7x + 6 = 0

There are three terms = a, b, and c

Where 'a' = value of x², 'b' = Value of x, 'c' = 6

If the value of 'c' is a positive value and the value of 'b' is also a positive value, then both values will be NEGATIVE.

II. y2 - 6y + 8 = 0

There are also three terms = a, b, and c

Where 'a' = value of y², 'b' = Value of y, 'c' = 8

If the value of 'c' is a positive value and the value of 'b' is a negative value, then both values will be POSITIVE.

∴ After comparison, all the values of x and y, the relation between x and y is “x < y”.