RRB Clerk Prelims Memory Based Paper (Held On: 17 August 2024 Shift 4) (Previous Year Paper)

Given:

Six people – A, B, C, D, E, and F

Flat Alpha is to the West of Flat Beta.

1. D does not live on the same floor as A.

2. F lives below D but not in the same flat as D.

3. F does not live on the bottommost floor.

4. C lives below F in the same flat.

5. B does not live on an even-numbered floor.

6. E lives below F.

7. B lives above A, who lives in Flat Alpha.

(Case 1a will not be possible as it will contradict point 7)

Hence, F lives on the second floor of Flat Beta.

Given:

Six people – A, B, C, D, E, and F

Flat Alpha is to the West of Flat Beta.

1. D does not live on the same floor as A.

2. F lives below D but not in the same flat as D.

3. F does not live on the bottommost floor.

4. C lives below F in the same flat.

5. B does not live on an even-numbered floor.

6. E lives below F.

7. B lives above A, who lives in Flat Alpha.

(Case 1a will not be possible as it will contradict point 7)

The logic followed here is that all the given options are people living on odd numbered floors, except A.

Hence, A is the odd one out.

Given:

Six people – A, B, C, D, E, and F

Flat Alpha is to the West of Flat Beta.

1. D does not live on the same floor as A.

2. F lives below D but not in the same flat as D.

3. F does not live on the bottommost floor.

4. C lives below F in the same flat.

5. B does not live on an even-numbered floor.

6. E lives below F.

7. B lives above A, who lives in Flat Alpha.

(Case 1a will not be possible as it will contradict point 7)

Hence, the correct answer is option 3.

Given:

Six people – A, B, C, D, E, and F

Flat Alpha is to the West of Flat Beta.

1. D does not live on the same floor as A.

2. F lives below D but not in the same flat as D.

3. F does not live on the bottommost floor.

4. C lives below F in the same flat.

5. B does not live on an even numbered floor.

6. E lives below F.

7. B lives above A, who lives in Flat Alpha.

(Case 1a will not be possible as it will contradict point 7)

Hence, all the statements are correct.

Given:

Six people – A, B, C, D, E, and F

Flat Alpha is to the West of Flat Beta.

1. D does not live on the same floor as A.

2. F lives below D but not in the same flat as D.

3. F does not live on the bottommost floor.

4. C lives below F in the same flat.

5. B does not live on an even numbered floor.

6. E lives below F.

7. B lives above A, who lives in Flat Alpha.

(Case 1a will not be possible as it will contradict point 7)

After rearrangement:

Hence, E lives on the fourth floor after the rearrangement.

Given statements: U > V ≥ D; K ≥ T ≤ D

On combining: U > V ≥ D ≥ T ≤ K

Conclusions:

I. T ≤ V → True (as V ≥ D ≥ T → V ≥ T)

II. K > U → False (as U > V ≥ D ≥ T ≤ K thus clear relation between K and U cannot be determined)

Hence, only I is True.

Given statements: X < Q ≤ N = L; H > F > X

On combining: H > F > X < Q ≤ N = L

Conclusions:

I. H > N → False (as H > F > X < Q ≤ N there is opposite sign between H and N. So, It is false)

II. F ≤ L → False (as F > X < Q ≤ N = L there is opposite sign between F and L. So, It is false)

Hence, Neither I nor II is True.

Given Statements: A > B ≥ Q; G < C > D = Q

On combining: A > B ≥ Q = D < C > G

Conclusions:

I. A > C → False (as A > D < C → thus clear relation between A and C cannot be determined).

II. A > D → True ( as A > Q = D)

Therefore, only conclusion II is true

Given statements: A < B ≥ C > D ; C < F ; E ≤ D

**On combining:**A < B ≥ C > D ≥ E ; C < F

Conclusions:

 I. A < F → False ( As A < B ≥ C < F  →  thus clear relation between A and F cannot be determined)

II. E < F  →True (As F >  C > D  ≥  E  → E < F)

Therfore only conclusion II is True.

Given series is:

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

Arrangement required: Symbol → Vowel → Consonant

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

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

Given series is:

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

Arrangement required: Symbol → Consonant → Symbol

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

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

Given series is:

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

Arrangement required: Consonant → Number → Vowel

Clearly none of the number follows the given condition$.

Given series:

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

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

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

Given series:

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

Here the group is formed in which second element is second next to the third element, and the third element is to the just next of the first element.

Hence, 9 I O does not belong to the group$.

Hence, three pairs are there in the word ‘ENLIGHTEN’, each of which has as many letters between them in the word (both forward and backward direction) as they have between them in the English Alphabet.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, B2 sits third to the right of A4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Here A3 sits second to the right of B2.

Hence, “None of the above” is the correct answer.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, “B2 and A3” are the neighbors of B4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, A3 sits immediate right of B4.

According to the question,

1. Only one person sits between A4 and A1. 

2. B3 sits immediate left of A4.

3. A2 sits third to the right of B3.

4. B4 is not a neighbor of A1 and A2.

5. A3 is not immediate neighbor of A2.

6. B1 is not a neighbor of B4.

Hence, A4 sits immediate left with respect to B1.

The least possible diagram for the given statements is as follows

Conclusions:

I. Some winter is nights → False (It is possible but not definite)

II. No winter is night → False (It is possible but not definite)

Conclusion I and II is complementary pair.

Hence, Either conclusion I or conclusion II follows.

The least possible Venn diagram is shown below

Conclusions:

I. Some Purse are Bag → False (It is possible but not definite. As no direct relation between Bag and Purse is given).

So it is considered as Cannot be determined.

II. No Bag is Purse → False (It is possible but not definite. As no direct relation between Bag and Purse is given).

Part of Bag can go in Purse.

So, it is considered as Cannot be determined.

Thus conclusion I and II form a complementary pair.

Hence, Either conclusion I or II follow.

Additional Information:

Conditions for the Either Or:

1. Both the conclusion must be Cannot be determined.
2. Both the objects in both the conclusion must be same.
3. If one conclusion is positive, then other should be negative.

Complementary pairs:

1. All - Some not (Positions of the object should be same in the conclusion)
2. Some - No

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

Conclusions:

I. Some Daisy are not Camelia → False (It is possible but not definite)

II. All Pansy being Poppy is a possibility → False (As ‘Only a few Pansy are Poppy’, some parts of Pansy will never be Poppy.)

Additional Information:

‘Only a few Pansy are Poppy’ → here the first element is Pansy and the second element is Poppy. So, the statement means ‘some Pansy are Poppy and some parts of Pansy will never be Poppy but all Poppy can be Pansy.

So, only a few = some + some not.

Hence, neither conclusion I nor II follow.

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

Conclusions:

I. Some Dogs are Zebra ⇒ False ( As it is possible but not definite, as no direct relation between the two is given).

II. All Zebra can be dog ⇒ True ( As there is no relation between Zebra and Dog, so the possibility is true).

Hence, Only conclusion II follows.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

1. Replacing second letter of each word by immediate succeeding letter as per alphabetical series: BUS   WIN   CLU   PPQ   RUN   ZHN

2. Clearly, three English meaningful words BUS, WIN, RUN are formed.

Therefore, three English meaningful words are formed.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

1. Arranging given words as per English dictionary from left to right:

 Left side   BTS   CKU   POQ   RTN   WHN   ZGN   Right side

Hence, WHN will be 2^nd from the right side after rearrangement.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

Replacing the first letter to immediate preceding letter as per English alphabetical series:

ATS   VHN   BKU   OOQ   QTN   YGN

Clearly, three words VHN, QTN, YGN have no vowels after rearrangement.

Note: In English alphabetical series there are 5 vowels i.e. A, E, I, O, U else all are consonants.

Hence, three words have no vowels after rearrangement.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

Replacing each consonant of each word to its immediate preceding letter as per alphabetical series: ASR   VGM   BJU   OOP   QSM   YFM

Note: In English alphabetical series there are 5 vowels i.e. A, E, I, O, U else all are consonants.

Hence, one word has only one consonant after rearrangement.

Given words are: BTS   WHN   CKU   POQ   RTN   ZGN

Left side   BTS   WHN   CKU   POQ   RTN   ZGN   Right side

1. First letter of third word from left side: CKU → C

2. Last letter of third word from right side: POQ → Q

C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q

Hence, there are 13 letters between first letter of third word from left side and last letter of third word from right side in English alphabetical series.

i) Only three persons are sitting to the right of P.

ii) K is sitting to the right of P, but not to the immediate right.

Iii) Also K is not sitting at the extreme end.

iv) Two persons are sitting between K and N.

v) D is one of the neighbours of N.

vi) Only one person is sitting to the left of T and only one person is sitting between T and D.

vii) The number of persons sitting between K and N is the same as the number of persons sitting between P and J, who is sitting at one of the ends.

Hence, there are nine persons sitting in a row.

i) Only three persons are sitting to the right of P.

ii) K is sitting to the right of P, but not to the immediate right.

Iii) Also K is not sitting at the extreme end.

iv) Two persons are sitting between K and N.

v) D is one of the neighbours of N.

vi) Only one person is sitting to the left of T and only one person is sitting between T and D.

vii) The number of persons sitting between K and N is the same as the number of persons sitting between P and J, who is sitting at one of the ends.

Hence, the position of N is fifth from the right end.

i) Only three persons are sitting to the right of P.

ii) K is sitting to the right of P, but not to the immediate right.

Iii) Also K is not sitting at the extreme end.

iv) Two persons are sitting between K and N.

v) D is one of the neighbours of N.

vi) Only one person is sitting to the left of T and only one person is sitting between T and D.

vii) The number of persons sitting between K and N is the same as the number of persons sitting between P and J, who is sitting at one of the ends.

Hence, there are five persons sitting between K and T.

From the given instructions:

1. X is heavier than Q. Q is heavier than P. P is heavier than only three people. So, we have two cases;

Case 1:

X/ > X/ > Q > P > __ > __ > __

Case 2:

X > Q > __ > P > __ > __ > __

2. T is heavier than S but lighter than R. U is heavier than R but lighter than Q. So, Case 1 is invalid.

So, Final Arrangement,

X > Q > U > P > R > T > S

Hence, four boys are lighter than U.

From the given instructions:

1. X is heavier than Q. Q is heavier than P. P is heavier than only three people. So, we have two cases;

Case 1:

X/ > X/ > Q > P > __ > __ > __

Case 2:

X > Q > __ > P > __ > __ > __

2. T is heavier than S but lighter than R. U is heavier than R but lighter than Q. So, Case 1 is invalid.

So, Final Arrangement,

X > Q > U > P > R > T > S

Hence, three boys are there boys between Q and T.

Given word: ABNORMALITY

The first, third, fourth, and eighth letters are A, N. O, and L respectively.

There is only one word that can be formed using these letters i.e. LOAN.

Therefore, the second letter from the left end of the word 'LOAN' is O. 

Hence, the required letter is O.

German is taught in the last day of the week. English is teaches on Thursday.

Chemistry is taught five days before the day German is taught . Mathematics is taught three days before English. 

English is taught one of the day before French is taught.  

Sanskrit is taught one of the day after Physics. 

More than two subjects are taught in between Chemistry and Sanskrit. Hence case 2 gets eliminated. 

Final arrangement will be

Hence four subjects are taught after Physics.

German is taught in the last day of the week. English is teaches on Thursday.

Chemistry is taught five days before the day German is taught . Mathematics is taught three days before English. 

English is taught one of the day before French is taught.  

Sanskrit is taught one of the day after Physics. 

More than two subjects are taught in between Chemistry and Sanskrit. Hence case 2 gets eliminated. 

Final arrangement will be

Sanskrit is taught just after French class.

German is taught in the last day of the week. English is teaches on Thursday.

Chemistry is taught five days before the day German is taught . Mathematics is taught three days before English. 

English is taught one of the day before French is taught.  

Sanskrit is taught one of the day after Physics. 

More than two subjects are taught in between Chemistry and Sanskrit. Hence case 2 gets eliminated. 

Final arrangement will be

English is taught three days before German.

German is taught in the last day of the week. English is teaches on Thursday.

Chemistry is taught five days before the day German is taught . Mathematics is taught three days before English. 

English is taught one of the day before French is taught.  

Sanskrit is taught one of the day after Physics. 

More than two subjects are taught in between Chemistry and Sanskrit. Hence case 2 gets eliminated. 

Final arrangement will be

Sanskrit is taught on Saturday.

German is taught in the last day of the week. English is teaches on Thursday.

Chemistry is taught five days before the day German is taught . Mathematics is taught three days before English. 

English is taught one of the day before French is taught.  

Sanskrit is taught one of the day after Physics. 

More than two subjects are taught in between Chemistry and Sanskrit. Hence case 2 gets eliminated. 

Final arrangement will be

French was taught on Saturday is incorrect statement.

Calculation

A = 1.2B (20% more)

B = 0.75C (25% less)

Substitute:

A = 1.2 × 0.75C = 0.9C

Thus,

A : C=0.9

Or, A : C = 9  : 10

Given:

Average weight of 5 persons = 64 kg

Average weight of 3 girls = 58 kg

Calculations:

Total weight of 5 persons = 5 × 64 = 320 kg

Total weight of 3 girls = 3 × 58 = 174 kg

Total weight of the 2 boys = 320 - 174 = 146 kg

Average weight of the 2 boys = 146 / 2 = 73 kg

∴ The average weight of the two boys is 73 kg.

GIVEN :

Total Students = 1200

Boys : Girls = 5 : 3

Required ratio = 1 : 1

CONCEPT :

Ratio and its types

ASSUMPTION :

Let the number of original number of boys and girls in the school be 5x and 3x

CALCULATION :

According to the question

5x + 3x = 1200

8x = 1200

x = 150

5x = 5 × 150 = 750

3x = 3 × 150 = 450

Let the number of girls joined be y

Now,

\(\frac{{750}}{{450; + ;y}}; = ;\frac{1}{1}\)

750 = 450 + y

y = 300

∴, number of girls added was 300

Calculation

People on Tuesday + Wednesday = 320 + 280 = 600

People on Thursday + Friday = 360 + 400 = 760

Required ratio:

600 : 760 = 15 : 19

Answer: 15 : 19

Calculation

People attend on Friday = 400

Saturday = 20% more than Friday

Thus:

[ 400 + [(20/100) ×400] = [400 + 80] = 480

480 people on Saturday

Calculation

People on Monday + Wednesday =

480 + 280 = 760

People on Thursday + Tuesday =

360 + 320 = 680

Required difference =

760 − 680 = 80

Answer: 80

Calculation

On Tuesday, male : female = 5 : 3

Total people on Tuesday = 320

Number of females = 3 parts = [320 × 3 / 8] = 120

Given, number of females = 5t

Thus:

5t = 120

So, t = 24

Answer: 24

Calculation

On Monday, 75% of registered people attended the seminar.

Let total registered people = x

Given:

75% of x = 480

 X = [480 × 100] / 75 = 640

640 people registered

Given:

The ratio of milk to water in 630 liters of mixture is 4:3.

140 liters of the mixture is taken out.

Calculations:

The total parts of the mixture = 4 + 3 = 7 parts.

Quantity of milk in the mixture = (4/7) × 630 = 360 liters of milk.

Quantity of water in the mixture = (3/7) × 630 = 270 liters of water.

When 140 liters of the mixture is taken out, the ratio of milk and water in the 140 liters will also be 4:3.

Milk taken out = (4/7) × 140 = 80 liters.

So, the remaining milk = 360 - 80 = 280 liters.

∴ The quantity of milk now is 280 liters.

Given:

Rate of interest (r) = 8%

Time (t) = 10 years

Simple Interest (SI) = ₹17,600

Formula used:

SI = P × r × t / 100

Calculations:

17600 = P × 8 × 10 / 100

⇒ 17600 = P × 80 / 100

⇒ 17600 = P × 0.8

⇒ P = 17600 / 0.8

⇒ P = ₹22,000

∴ The principal is ₹22,000.

Given:

When the item is sold at Rs. 1400, profit percent is 40%.

Let the cost price be C.

Formula used:

Profit = Selling Price - Cost Price

Profit percent = (Profit / Cost Price) × 100

Calculations:

Profit percent = 40%, so:

(1400 - C) / C = 40 / 100

⇒ (1400 - C) = 0.4C

⇒ 1400 = 1.4C

⇒ C = 1400 / 1.4 = Rs. 1000

Now, when the item is sold at Rs. 1345, profit = 1345 - 1000 = Rs. 345

∴ The profit amount when the item is sold at Rs. 1345 is Rs. 345.

Given:

24 × 9 + 84 ÷ 14 = (?) + 6²

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

24 × 9 + 84 ÷ 14 = (?) + 6²

⇒ 216 + 6 = (?) + 36

⇒ ? = 222 - 36 = 186

∴ The value of ? is 186.

Given:

60% of 135 - (√729 + ?) = 4

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

60% of 135 - (√729 + ?) = 4

⇒ 135 × (60/100) - (27 + ?) = 4

⇒ 135 × (3/5) - 27 - ? = 4

⇒ 81 - 27 - 4 = ?

⇒ 50 = ?

∴ 50 will come in place of the question mark ('?').

Given

207 - 7√3 × 5√3 = 17 + ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

207 - 7√3 × 5√3 = 17 + ?

⇒ 207 - 35 × 3 = 17 + ?

⇒ 207 - 105 = 17 + ?

⇒ 102 - 17 = ?

⇒ ? = 85

∴ 85 will come in place of the question mark ('?').

Given:

(13)2 + √169 × 17 – 26 × √81 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(13)2 + √169 × 17 – 26 × √81 = ?

⇒ (13)² + √169 × 17 – 26 × √81 = ?

⇒ 169 + 221 – 234 = ?

⇒ ? = 156

∴ The value of ? is 156.

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ {16 × 2 + √ 2704}/(? + 4) = 12

⇒ {32 + 52}/(? + 4) = 12

⇒ 84/12 = ? + 4

⇒ 7 = ? + 4

⇒ ? = 3

∴ The value of ? is 3.

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ 64 × (?% of 250) + 432 = 3632

⇒ 64 × ?/100 × 250 = 3632 - 432

⇒ 16 × ? × 10 = 3200

⇒ ? = 20

∴ The value of ? is 20.

Given:

(12.25 + 8)1/2 × (1000)1/3 = ?

Concept Used:

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

Calculation:

(12.25 + 8)1/2 × (1000)1/3 = ?

⇒ (20.25)1/2 × (1000)1/3 = ?

⇒ 4.5 × (10) = ?

⇒ ? = 45

∴ The required value is 45

Concept Used:

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

Calculation:

⇒ 5/12 × 480 + 150 = ?% of 875

⇒ 200 + 150 = ?/100 × 875

⇒ 350 = ?/100 × 875

⇒ ? = 350 × 100/875

⇒ ? = 40

∴ The value of ? is 40

Given:

(1/10) of (12.5 + 7.5) - 5 = ? - 10

Concept:

Follow the BODMAS rule according to the table given below:

Calculation:

(1/10) of (12.5 + 7.5) - 5 = ? - 10

⇒ (1/10) × 20 - 5 = ? - 10

⇒ 2 - 5 = ? - 10

⇒ ? = 10 - 3

⇒ ? = 7

∴ 7 will come in place of the question mark ('?').

Given:  

20% of 400 + 40 = ? + 30

Concept:

Follow the BODMAS rule according to the table given below:

Calculation:

20% of 400 + 40 = ? + 30

⇒ (1/5) × 400 + 40 = ? + 30

⇒ 80 + 40 = ? + 30

⇒ 120 = ? + 30

⇒ ? = 120 - 30

⇒ ? = 90

∴ 90 will come in place of the question mark ('?').

Given:

48% of 250 + 3 × 7 - ? ÷ 9 = 137

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

48% of 250 + 3 × 7 - ? ÷ 9 = 137

⇒ (48/100) × 250 + 21 - ? ÷ 9 = 137

⇒ 120 + 21 - ? ÷ 9 = 137

⇒ 141 - ? ÷ 9 = 137

⇒ ? ÷ 9 = 4

⇒ ? = 36

∴ 36 should come in place of the question mark (?).

Given:

54% of 150 + 4 × ? = 9% of 2900

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

54% of 150 + 4 × ? = 9% of 2900

⇒ (54/100) × 150 + 4 × ? = (9/100) × 2900

⇒ 81 + 4 × ? = 261

⇒ 4 × ? = 180

⇒ ? = 45

∴ 45 should come in place of the question mark (?).

Concept used:

We have to follow the BODMAS rule

Calculation:

⇒ 20% of 225 + 62 = ?% of 180

⇒ 20/100 × 225 +  62 = ?/100 × 180

⇒ 45 + 36 = ?/10 × 18

⇒ 81 × 10/18 = ?

⇒ ? = 45

∴ The value of ? is 45.

Concept Used:

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

Calculation:

⇒  20 × 7.5 + √ 225 = ? × 5

⇒ 150 + 15 = ? × 5

⇒ 165 = ? × 5

⇒ ? = 33

∴ The value of ? is 33.

Concept Used:

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

Calculation:

⇒ (212 ÷ 27) × 5 + 180 = ?% of 680

⇒ (2^{12 - 7}) × 5 + 180 = (?/100) × 680

⇒ 2⁵ × 5 + 180 = (?/10) × 68

⇒ 32 × 5 + 180 = (?/10) × 68

⇒ 160 + 180 = (?/10) × 68

⇒ 340 × (10/68) = ?

⇒ ? = 50

∴ The value of ? is 50

Calculation

Let speed of boat = x km/h.

Then, speed of stream = x − 0.4x = 0.6x

Downstream speed = boat + stream 

⇒ x + 0.6x = 1.6x

Given 1.6x = 16.

Thus,

⇒ x = 16/ 1.6 ​= 10 km/h

Calculation

First, side of square:

Side = 64/4 = 16 m

Length of rectangle = 16 + 4 = 20 

Area of rectangle = Length × Breadth = 20 × 12 = 240

Calculation

Total work =  LCM of 20 and 30 = 60

Efficiency of B is 60 /30 = 2

Total Efficiency of A and B together is 60/20 = 3

So, efficiency of A is 3 -2 = 1

So, A alone complete the work in 60 / 1 = 60 days

Calculation

Total boys = 420 + 380 + 320 + 580 + 350 = 2050

Total girls = 400 + 200 + 400 + 500 + 450 = 1950

Difference = 2050 − 1950 = 100