RRB Clerk Prelims Memory Based Paper (Held On: 19 Sept, 2020 Shift 1) (Previous Year Paper)

Given names: A, B, C, D, E, F, and G.

Face direction: North

1. E has equal number of persons to his left and right side.

2. F sits on the extreme left end of the row.

3. There are three persons between F and B.

4. G sits second from one of the ends.

(Here, there will be two cases, Case 1: when G is second from the right end and Case 2: when G is second from the left end).

5. A sits immediate right of C.

6. A does not sit at the corner.

(In case 2 A will be at the corner to adjust the statement 5, so we can eliminate case 2 here)

Hence the final arrangement will be as follows:

Hence, C is sitting second to the left of E.

Given names: A, B, C, D, E, F, and G.

Face direction: North

1. E has equal number of persons to his left and right side.

2. F sits on the extreme left end of the row.

3. There are three persons between F and B.

4. G sits second from one of the ends.

(Here, there will be two cases, Case 1: when G is second from the right end and Case 2: when G is second from the left ed).

5. A sits immediate right of C.

6. A does not sit at the corner.

(In case 2 A will be at the corner to adjust the statement 5, so we can eliminate case 2 here.)

Hence the final arrangement will be as follows:

Hence, two persons are sitting between A and G.

Given names: A, B, C, D, E, F, and G.

Face direction: North

1. E has equal number of persons to his left and right side.

2. F sits on the extreme left end of the row.

3. There are three persons between F and B.

4. G sits second from one of the ends.

(Here, there will be two cases, Case 1: when G is second from the right end and Case 2: when G is second from the left ed).

5. A sits immediate right of C.

6. A does not sit at the corner.

(In case 2 A will be at the corner to adjust the statement 5, so we can eliminate case 2 here.)

Hence the final arrangement will be as follows:

Hence, B is sitting third from the right end.

Given names: A, B, C, D, E, F, and G.

Face direction: North

1. E has equal number of persons to his left and right side.

2. F sits on the extreme left end of the row.

3. There are three persons between F and B.

4. G sits second from one of the ends.

(Here, there will be two cases, Case 1: when G is second from the right end and Case 2: when G is second from the left ed).

5. A sits immediate right of C.

6. A does not sit at the corner.

(In case 2 A will be at the corner to adjust the statement 5, so we can eliminate case 2 here.)

Hence the final arrangement will be as follows:

Hence, five persons are sitting to the left of G.

Given names: A, B, C, D, E, F, and G.

Face direction: North

1. E has equal number of persons to his left and right side.

2. F sits on the extreme left end of the row.

3. There are three persons between F and B.

4. G sits second from one of the ends.

(Here, there will be two cases, Case 1: when G is second from the right end and Case 2: when G is second from the left ed).

5. A sits immediate right of C.

6. A does not sit at the corner.

(In case 2 A will be at the corner to adjust the statement 5, so we can eliminate case 2 here.)

Hence the final arrangement will be as follows:

Hence, A is sitting exactly between F and B.

Given number: 45392276

Even digits: 4, 2, 2, 6

Odd digits: 5, 3, 9, 7

1 is added to each even digit: 5, 3, 3, 7

2 is subtracted from each odd digit: 3, 1, 7, 5

New number: 53173357

Left side 53173357 Right side

2^nd number from left: 3

6^th number from right: 1

Their sum is: 3 + 1 = 4.

Hence, 4 is the correct answer.

Let us draw the diagram according to the information given in the question,

Hence, person J is in North-East of Person L.

Let us draw the diagram according to the information given in the question,

LO = 18 m and NO = 15 m

By Pythagoras Theorem

LN² = LO² + ON²

LN² = 18² + 15²

LN² = 324 + 225

LN² = 549 (by taking square roots on both the sides)

LN² = √549

Hence, the shortest distance between L and N is √549.

Let us draw the diagram according to the information given in the question,

Hence, K is in East direction with respect to L.

Given names: A, B, C, D, E, and F

Dates given: 7^th and 12^th

Months given: January, February and March

1. A was born on the even number date of a month having 31 days.

(Here, we have two cases, Case 1: when A is born on 12 January and Case 2: when A is born on 12 March).

2. Only one person was born between A and B.

Case 1: when A is born on 12 January

Case 2: when A is born on 12 March

3. C is the youngest person.

(Here, we can eliminate case 2 as in this case A is the youngest person).

4. E was born just before C.

5. At least two persons are born between E and D.

Hence, the final arrangement will be as follows:

Hence, B was born on 12^th February.

Given names: A, B, C, D, E, and F

Dates given: 7^th and 12^th

Months given: January, February and March

1. A was born on the even number date of a month having 31 days.

(Here, we have two cases, Case 1: when A is born on 12 January and Case 2: when A is born on 12 March).

2. Only one person was born between A and B.

Case 1: when A is born on 12 January

Case 2: when A is born on 12 March

3. C is the youngest person.

(Here, we can eliminate case 2 as in this case A is the youngest person).

4. E was born just before C.

5. At least two persons are born between E and D.

Hence, the final arrangement will be as follows:

Hence, A was born just before F.

Given names: A, B, C, D, E, and F

Dates given: 7^th and 12^th

Months given: January, February and March

1. A was born on the even number date of a month having 31 days.

(Here, we have two cases, Case 1: when A is born on 12 January and Case 2: when A is born on 12 March).

2. Only one person was born between A and B.

Case 1: when A is born on 12 January

Case 2: when A is born on 12 March

3. C is the youngest person.

(Here, we can eliminate case 2 as in this case A is the youngest person).

4. E was born just before C.

5. At least two persons are born between E and D.

Hence, the final arrangement will be as follows:

Hence, two persons were born between A and E.

Given names: A, B, C, D, E, and F

Dates given: 7^th and 12^th

Months given: January, February and March

1. A was born on the even number date of a month having 31 days.

(Here, we have two cases, Case 1: when A is born on 12 January and Case 2: when A is born on 12 March).

2. Only one person was born between A and B.

Case 1: when A is born on 12 January

Case 2: when A is born on 12 March

3. C is the youngest person.

(Here, we can eliminate case 2 as in this case A is the youngest person).

4. E was born just before C.

5. At least two persons are born between E and D.

Hence, the final arrangement will be as follows:

Hence, E – 7 – March is the correct name – date – month pair.

Given names: A, B, C, D, E, and F

Dates given: 7^th and 12^th

Months given: January, February and March

1. A was born on the even number date of a month having 31 days.

(Here, we have two cases, Case 1: when A is born on 12 January and Case 2: when A is born on 12 March).

2. Only one person was born between A and B.

Case 1: when A is born on 12 January

Case 2: when A is born on 12 March

3. C is the youngest person.

(Here, we can eliminate case 2 as in this case A is the youngest person).

4. E was born just before C.

5. At least two persons are born between E and D.

Hence, the final arrangement will be as follows:

Hence, five persons were born after D.

Consider the series given:

S 4 5 4 S S 4 5 4 S 7 S W 7 5 7 5 W 4 4 4 W C C 5 5 7 C C 7

We need to find out combinations having

Number Alphabet

Thus, six numbers are immediately succeeded by an alphabet in the series.

Hence, six is the correct answer.

Consider the series:

S 4 5 4 S S 4 5 4 S 7 S W 7 5 7 5 W 4 4 4 W C C 5 5 7 C C 7

Total number of C → 4

Total number of S → 5

Total number of 4 → 7

Total number of 5 → 6

Total number of W → 3

Hence, W has the lowest frequency.

Consider the following series: _____

S 4 5 4 S S 4 5 4 S 7 S W 7 5 7 5 W 4 4 4 W C C 5 5 7 C C 7

Numbers are 454, 454, 757, 575, 444

Hence, five is the correct answer.

Consider the following series:

S 4 5 4 S S 4 5 4 S 7 S W 7 5 7 5 W 4 4 4 W C C 5 5 7 C C 7

Hence, according to the above series, three consecutive repeated pairs are there: SS, CC, CC 

Hence three is the correct answer.

Consider the given series:

S 4 5 4 S S 4 5 4 S 7 S W 7 5 7 5 W 4 4 4 W C C 5 5 7 C C 7

Hence, five numbers are there which are immediately preceded by 5:- 54, 54, 57, 55, 57

Hence, five is the correct answer.

The given word MACHINE and its positional value given in the table and its pairs in backward and forward directions:

Forward pairs (Left to Right): HI

Backward pairs (Right to Left): IM, EH

There are three pairs in both Backward and Forward direction.

Hence, three is the correct answer.

Note:

I. Only few feasts are vultures.

This implies that

All feasts are not vultures.

Some feasts are vultures.

Some feasts are not vultures.

II. Only feasts are cards.

The above statement implies that

All cards are feasts.

No card is vulture.

No card is veg.

The least possible Venn diagram is given below

Conclusions:

I. Some veg are feasts.→ False (it is possible but not definite)

II. Some feasts are cards. → True (as only feasts are cards)

Hence, only conclusion II follows.

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

Conclusions:

I. Some roses are smell → False (It is possible but not definite)

II. No roses are smell → False (It is possible but not definite)

Conclusion I and II forms complementary pair.

Thus, either conclusion I or conclusion II follow.

Note:

I. None of the fumes are volcano.

The above statement implies that

No fume is volcano.

II. A few volcanoes are active.

The above statement implies that

Some volcanoes are active.

The least possible Venn diagram is given below

Conclusions:

I. Some volcanoes are violent. → True (those volcanoes that are active are also violent)

II. Some active are not fumes.→ True (those active that are volcanoes are not fumes)

Hence, both the conclusion I and II follow.

Note:

I. Only Ds are Fi.

The above statement implies that

All Fi are Ds.

No Fi is Qw.

No Fi is Cv.

III. Few Cv are Qw.

The above statement implies that

Some Cv are Qw.

The least possible Venn diagram is given below

Conclusions:

I. No Ds is Qw. → False (as Some Qw are Cv and All Cv are Ds)

II. All Fi are Qw.→ False (as only Ds are Fi.)

Hence, none of the conclusion follows.

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

I. No Mango is Grape → True (As all mango is banana and no banana is grape, so no mango is grape.)

II. Some Kiwi is Mango → True (This is definitely true)

Hence, ‘Both I and II follows’ is the correct answer.

1. F sits third from one of the extreme ends.

(Here, there will be two cases, Case 1: when F is third from the right end and Case 2: when F is third from the left end).

2. D sits fifth from the left end.

3. There are more than two persons between F and D.

(As, in case 2 there is only one person between F and D, we can eliminate case 2 here).

4. Only four persons are there between A and F.

5. There are 3 persons sitting between A and B. A does not sit second to the right of D.

6. Five persons are sitting between B and D.

7. E sits exactly between A and B.

8. One person sits between B and C.

9. G sits third to the left of D.

Hence, the final arrangement will be as follows:

Hence, 22 persons are sitting in the row.

1. F sits third from one of the extreme ends.

(Here, there will be two cases, Case 1: when F is third from the right end and Case 2: when F is third from the left end).

2. D sits fifth from the left end.

3. There are more than two persons between F and D.

(As, in case 2 there is only one person between F and D, we can eliminate case 2 here).

4. Only four persons are there between A and F.

5. There are 3 persons sitting between A and B. A does not sit second to the right of D.

6. Five persons are sitting between B and D.

7. E sits exactly between A and B.

8. One person sits between B and C.

9. G sits third to the left of D.

Hence, the final arrangement will be as follows:

Hence, five persons are sitting between A and C.

1. F sits third from one of the extreme ends.

(Here, there will be two cases, Case 1: when F is third from the right end and Case 2: when F is third from the left end).

2. D sits fifth from the left end.

3. There are more than two persons between F and D.

(As, in case 2 there is only one person between F and D, we can eliminate case 2 here).

4. Only four persons are there between A and F.

5. There are 3 persons sitting between A and B. A does not sit second to the right of D.

6. Five persons are sitting between B and D.

7. E sits exactly between A and B.

8. One person sits between B and C.

9. G sits third to the left of D.

Hence, the final arrangement will be as follows:

Hence, the position of G from the left end is second.

​​

1. F sits third from one of the extreme ends.

(Here, there will be two cases, Case 1: when F is third from the right end and Case 2: when F is third from the left end).

2. D sits fifth from the left end.

3. There are more than two persons between F and D.

(As, in case 2 there is only one person between F and D, we can eliminate case 2 here).

4. Only four persons are there between A and F.

5. There are 3 persons sitting between A and B. A does not sit second to the right of D.

6. Five persons are sitting between B and D.

7. E sits exactly between A and B.

8. One person sits between B and C.

9. G sits third to the left of D.

Hence, the final arrangement will be as follows:

Hence, eight persons are sitting between G and B.

Given numbers: 378 256 431 875 762

On adding 2 to the second digit: 398 276 451 895 782

On arranging the numbers in descending order: 895 782 451 398 276

451 is the third-lowest number and the corresponding number of that in the actual series is 431

Hence, 451 is the third lowest number which is obtained from 431.

Given numbers: 378 256 431 875 762

Sum of first and second digits,

762 = 7 + 6 = 13

378 = 3 + 7 = 10

256 = 2 + 5 = 7

431 = 4 + 3 = 7

875 = 8 + 7 = 15

Hence, 15 is more than 13 which is obtained from 875.

Given numbers: 378 256 431 875 762

On subtracting 1 from each number: 377 255 430 874 761

Hence, two numbers i.e., 255 and 430 are divisible by 5.

Given numbers: 378 256 431 875 762

On subtracting first and third digits,

378 = 8 – 3 = 5

256 = 6 – 2 = 4

431 = 4 – 1 = 3

875 = 8 – 5 = 3

762 = 7 – 2 = 5

Hence, 4 is the second highest resultant which is obtained from 256.

Eight people A, B, C, D, J, K, L and M

1) C sits second to the right of L.

2) Only one person sits between C and M when counted from the right of C.

3) A is not an immediate neighbor of M. B faces D who sits immediately to the left of K.

4) K sits at one of the corners.

Therefore, case 1 gets eliminated.

Thus, the final arrangement is:

Hence, K sits third to the left of J.

Eight people A, B, C, D, J, K, L and M

1) C sits second to the right of L.

2) Only one person sits between C and M when counted from the right of C.

3) A is not an immediate neighbor of M. B faces D who sits immediately to the left of K.

4) K sits at one of the corners.

Therefore, case 1 gets eliminated.

Thus, the final arrangement is:

​

There are 4 people between C and D when counted clockwise from D.

Hence, "More than three" is the correct answer.

Eight people A, B, C, D, J, K, L and M

1) C sits second to the right of L.

2) Only one person sits between C and M when counted from the right of C.

3) A is not an immediate neighbor of M. B faces D who sits immediately to the left of K.

4) K sits at one of the corners.

Therefore, case 1 gets eliminated.

Thus, the final arrangement is:

​

C, K, L, M are sitting at the corners of the table whereas B is sitting at the middle of one of the sides.

Hence, B does not belong to the group.

Eight people A, B, C, D, J, K, L and M

1) C sits second to the right of L.

2) Only one person sits between C and M when counted from the right of C.

3) A is not an immediate neighbor of M. B faces D who sits immediately to the left of K.

4) K sits at one of the corners.

Therefore, case 1 gets eliminated.

Thus, the final arrangement is:

​

1) A sits immediate right of K. - True

2) A sits second to the right of B - True

3) A is an immediate neighbor of C - False (B and J are immediate neighbours of C)

Hence, "Both 1 and 2" is the correct answer.

Eight people A, B, C, D, J, K, L and M

1) C sits second to the right of L.

2) Only one person sits between C and M when counted from the right of C.

3) A is not an immediate neighbor of M. B faces D who sits immediately to the left of K.

4) K sits at one of the corners.

Therefore, case 1 gets eliminated.

Thus, the final arrangement is:

​

M sits immediate right of D.

K sits second to the right of M.

Hence, K is the correct answer.

Given names: A, B, C, D, E, F, and G.

1. B is the second lightest person.

2. The one who is the lightest person has 65 kg of weight.

3. Three persons have their weight between B and G.

4. F is the heaviest of all.

5. D has an equal number of persons who are heavier and lighter than him.

(Hence, D will be in the middle of the arrangement).

6. D has a weight of 85 kgs.

7. E is just heavier than B.

8. A is heavier than C.

Hence the final arrangement will be as follows:

Weight of D = 85 kg

Weight of C = 65 kg

Average weight = (85 + 65) / 2 = 150 / 2 = 75

Hence, the average weight of D and C is 75 kg.

Given names: A, B, C, D, E, F, and G.

1. B is the second lightest person.

2. The one who is the lightest person has 65 kg of weight.

3. Three persons have their weight between B and G.

4. F is the heaviest of all.

5. D has an equal number of persons who are heavier and lighter than him.

(Hence, D will be in the middle of the arrangement).

6. D has a weight of 85 kgs.

7. E is just heavier than B.

8. A is heavier than C.

Hence the final arrangement will be as follows:

The weight of E should be between the weight of D (85) and C (65).

Hence, 76 kg is the possible weight of E.

Given

Discount = Rs. 50.

Profit percentage = 25%

Calculation

Let cost price be 100x.

Hence, marked price = 150% of 100x = 150x

Now, selling price = MP - Discount = 150x - 50      ---- (1)

Also, if profit is 25%, then selling price = 125% of CP = 125x

From (1), we get

150x - 50 = 125x

⇒ 25x = 50

⇒ x = 2

∴ Selling price = 125x = 125 × 2 = Rs. 250.

Given

Rate of interests = 10% and 20%

Interest received = 1000 and 4000.

Formula

SI = P × R × T/100

Calculation

(P × 10 × T)/100 = 1000

⇒ P = 10000/T      ---- (1)

Also, [P × 20 × (T + 4)/100 = 4000

Putting P from (1), we get

(10000/T) × 20 × (T + 4) = 400000

⇒ 2T + 8 = 4T

⇒ 2T = 8

⇒ T = 4 years.

Given

Man rows 24 km upstream in 6 hours and 35 km downstream in 7 hours.

Formula

Downstream speed = M + S

Upstream speed = M - S

where M = Speed of man in still water and S = Speed of stream.

Calculation

Downstream speed = M + S = 35/7 = 5 km/hr      ---- (1)

Upstream speed = M - S = 24/6 = 4km/h      ---- (2)

Adding (1) and (2), we get

2M = 9

⇒ M = 4.5 km/hr

Given

Share of A in profit = 3200

Total profit = 6800

Concept

Ratio of share of profit = Ratio of product of investment and time for which investment has been done.

Here, as time is same for both hence ratio of share in profit will depend only on ratio of investment.

Calculation

Investment by A/Total investment = 3200/6800

⇒ x/(2x + 800) = 8/17

⇒ 17x = 16x + 6400

⇒ x = Rs. 6400.

Alternate Method

Given:

Total profit is Rs. 6,800

A's share in this is 3,200

A & B has invested Rs. x and Rs. (x + 800) respectively

Concept Used:

Amount invested is directly proportional to Profit earned

Calculation:

B's share in Profit,

⇒ Total Profit – A's share

⇒ 6,800 – 3,200 = 3,600

Now as per the concept,

⇒ x/(x + 800) = 3,200/3,600

⇒ x/(x + 800) = 8/9

⇒ 9x = 8x + 6,400

⇒ x = 6,400

∴ The value of x is Rs. 6,400.

Given

The Ratio of milk and water in the mixture is 3 : 1.

Calculation

Let the initial quantity of milk and water be 3x and x.

Quantity of milk taken out in 20 lt = 20 × (3/4) = 15 lt

Quantity of water taken out in 20 lt = 20 × (1/4) = 5 lt

Final quantity of milk = 3x - 15 and final quantity of water = x - 5

According to question

(3x - 15) - (x - 5) = 70

⇒ 2x - 10 = 70

⇒ 2x = 80

⇒ x = 40

∴ Initial quantity of mixture = 3x + x = 4x = 4 × 40 = 160 lt.

Number of clothes received by shop O on all days = 54 + 40 + 23 + 53 = 170

Number of clothes received by all shops on Monday = 39 + 42 + 35 + 54 = 170

∴ Required percentage = (170/170) × 100 = 100%.

Number of clothes received by all shops on Tuesday = 35 + 28 + 37 + 40 = 140

 Number of clothes received by all shops on Wednesday = 53 + 69 + 45 + 23 = 190

∴ Required difference = 190 - 140 = 50.

Number of clothes received by shop M on all days = 42 + 28 + 69 + 41 = 180

Number of clothes received by shop N on all days = 35 + 37 + 45 + 33 = 150

∴ Required percentage = [(180 - 150)/150] × 100 = 20%.

Number of clothes received by shops L and N together on Tuesday = 35 + 37 = 72

Number of clothes received by shops M and O together on Monday = 42 + 54 = 96

∴ Required ratio = 72 : 96 = 3 : 4.

Total number of clothes received on Monday, Tuesday, and Thursday by shop L = 39 + 35 + 43 = 117

∴ Required average = 117/3 = 39.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

5376 ÷ 24 + 100 = x²

⇒ 224 + 100 = x²

⇒ 324 = x²

⇒ x = √324 = 18.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

√576 ÷ √144 – x = 200

⇒ 24 ÷ 12 - x = 200

⇒ 2 - x = 200

⇒ x = 2 - 200 = -198.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

8 + (112/4) + (72/3) = x

⇒ 8 + 28 + 24 = x

⇒ 60 = x.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

80% of x ÷ 25 = 12

⇒ 80/100 × x ÷ 25 = 12

⇒ 4x/5 ÷ 25 = 12

⇒ 4x/5 × 1/25 = 12

⇒ x = (12 × 5 × 25)/4 = 3 × 5 ×  25 = 375 

⇒ x = 375.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

x² = 3/8 of 1440 + 85

⇒ x² = 3 × 180 + 85

⇒ x² = 540 + 85

⇒ x² = 625

⇒ x = √625 = 25.

Given:

Speed of train = 72 km/hr

Time is taken by train to cross a pole = 45 sec

Concept:

The distance covered by a train to cross a pole or person is equal to the length of the train.

When we convert speed into m/sec from km/hr we multiply by 5/18.

Formula used:

Distance = speed × time

Calculation:

The speed of the train in m/sec = 72 × (5/18) = 20 m/sec

Now the distance covered by train in 45 sec = 20 × 45 = 900 m.

Thus, the length of train = 900 m.

Now, the speed of train = 54 × (5/18) = 15 m/sec.

So, the time is taken by train to cross pole = 900/15 = 60 sec.

∴ Now the train crosses a pole in 60 sec.

Given:

The present age of Q is 2 years more than P.

The ratio of the age of Q after 2 years and the present age of P = 7: 5

Calculation:

Let the present age of P be x, then the present age of Q will be (x + 2).

After two years the age of Q = x + 4

The ratio of the age of Q after 2 years and the present age of P = (x + 4) ∶ x

Now we compare this ratio with the given ratio of their age after two years

⇒ (x + 4)/x = 7/5

⇒ 5x + 20 = 7x 

⇒ x = 10 years

Now the age of P before 6 years = 10 - 6 = 4 year

And the present age of Q = 10 + 2 = 12 years

Now to find the ratio between the present age of Q and the age of P before 6 years = 12 ∶ 4 = 3 ∶ 1

∴ The ratio between the present age of Q and the age of P before 6 years = 3 ∶ 1

Given:

The circumference of the circle = 2 cm + (The perimeter of the rectangle)

The length of the rectangle = The diameter of the circle

The breadth of the rectangle = The radius of the circle

Formula used:

The circumference of the circle = 2πr

where, r = radius of circule 

The diameter of the circle,d = 2r

The perimeter of the rectangle = 2 × (length + breadth)

Calculation:

The length of the rectangle = The diameter of the circle = 2r

The breadth of the rectangle = The radius of the circle = r

The circumference of the circle = 2 cm + (The perimeter of the rectangle)

⇒ 2πr = 2 + [2 × (length + breadth)]

⇒ (44/7) × r = 2 + [2 × (2r + r)]

⇒ 44r/7 = 2 + 6r

⇒ 44r = 14 + 42r

⇒ 2r = 14

⇒ r = 7 cm.

∴ The radius of the circle is 7 cm.

Given

Pipes A and B together take 20 hours, B and C together take 15 hours, and C and A together take 12 hours to fill a tank.

Formula

Efficiency = Work done/Time taken

Calculation

Let capacity of tank be 60 units. (LCM of 20, 15 and 12)

Now, efficiency of A and B = 60/20 = 3 units/hr      ---- (1)

Efficiency of B and C = 60/15 = 4 units/hr      ---- (2)

Efficiency of A and C = 60/12 = 5 units/hr      ---- (3)

Adding (1), (2) and (3), we get

2(A + B + C) = 12

⇒ A + B + C = 6

∴ Time taken all three pipes to fill tank = 60/6 = 10 hours.

Given

8 women can complete the certain amount of work in 72 days.

Concept

Work done = Efficiency × Time taken

Calculation

Let efficiency of 1 woman = 1 unit/day

Hence, total work = 72 × 8 = 576

Now, 2/3^rd of total work = 576 × (2/3) = 384

∴ Time taken by 12 women to complete 2/3^rd of work = 384/12 = 32 days.

12 × 0.5 = 6

6 × 1 = 6

6 × 1.5 = 9

9 × 2 = 18

∴ Missing number is 6.

12 × 0.5 + 1 = 7

7 × 1 + 1 = 8

8 × 1.5 + 1 = 13

13 × 2 + 1 = 27

∴ Missing number is 27.

113 - 49 = 64 = 8²

194 - 113 = 81 = 9²

294 - 194 = 100 = 10²

415 - 294 = 121 = 11²

∴ Missing number is 294.

19 - 34 = -15

27 - 19 = 8

12 - 27 = -15

20 - 12 = 8

Here, difference of two consecutive numbers are repeating in -15 and 8.

∴ Missing number is 20.

61.5 - 59.7 = 1.8 = 1.8 × 1

65.1 - 61.5 = 3.6 = 1.8 × 2

72.3 - 65.1 = 7.2 = 1.8 × 4

86.7 - 72.3 = 14.4 =  = 1.8 × 8

∴ Missing number is 86.7.

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

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket.

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is:

1123 + 1029 – 2831 + 702 = x

⇒ 2854 – 2831 = x

⇒ 23 = x

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

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket.

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is:

15% of 60 + 50% of 84 = x

⇒ (3/20) × 60 + (1/2) × 84 = x

⇒ 9 + 42 = x

⇒ 51 = x

∴ The value of ? is 51

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

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket.

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is:

√(6² + 7² + 36) = x – 50

⇒ √(36 + 49 + 36) = x – 50

⇒ √ 121 = x – 50

⇒ 11 = x – 50

⇒ x = 61

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

25% of 48 + 20% of x = 60

⇒ (25/100) × 48 + (20/100) × x = 60

⇒ 12 + x/5 = 60

⇒ x/5 = 48

⇒ x = 240.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

(1/8 of 2320) + (5² + 7²) = x² - 36

⇒ 290 + 25 + 49 = x² - 36

⇒ x² = 364 + 36

⇒ x² = 400

⇒ x = √400 = 20.

Total number of children = 60 + 45 + 55 + 40 + 50 = 250

∴ Required average = 250/5 = 50.

Number of children in coach B and C together = 45 + 55 = 100

Number of children in coach D and E together = 40 + 50 = 90

∴ Required ratio = 100 : 90 = 10 : 9.

Number of boys in coach B = (3/11) × Total number of children in coach C

⇒ Number of boys in coach B = (3/11) × 55

⇒ Number of boys in coach B = 15

∴ Number of girls in coach B = 45 - 15 = 30.

Number of children in coach C and E together = 55 + 50 = 105

Number of children in coach A and D together = 60 + 40 = 100

∴ Required percentage = [(105 - 100)/100] × 100 = 5%.

Number of children in A and B  = 60 + 45 = 105

Number of children in C, D, and E = 55 + 40 + 50 = 145

∴ Required difference = 145 - 105 = 40.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now,

Calculations:

1/6 + 1/12 + 1/18 = 22/x

⇒ [(6 + 3 + 2)/36] = 22/x

⇒ 11/36 = 22/x

⇒ x = 72.

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

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculation:

180% of 2048 + 79%1840 – 440% of 970 = x

⇒ (180 × 2048)/100 + (79 × 1840)/100 - (440 × 970)100

⇒ 1.8 × 2048 + 7.9 × 184 - 44 × 97

⇒ 3686.4 + 1453.6 - 4268

⇒ 5140 - 4268

⇒ 872

∴ The value of x is 872.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

(319.111 + 256.889)^1/2 ÷ 6 = x% of 40

⇒ 576^1/2 ÷ 6 = x × (40/100)

⇒ 24 ÷ 6 = x × (40/100)

⇒ 4 = 2x/5

⇒ x = 10.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now,

Calculations:

16² + 12² = x² + 39

⇒ 256 + 144 = x² + 39

⇒ 400 = x² + 39

⇒ x² = 361

⇒ x = √361 = 19.

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now,

Calculations:

12.5% of 6400 – 200 × 3 = x 

⇒ 6400 × (1/8) - 600 = x

⇒ 800 - 600 = x

⇒ x = 200.