Here, persons are K, L, M, N, O, P and Q.

1. Three persons give interview between M and O who is scheduled for interview after M.

2. Q gives interview two days before P.

3. Q gives interview before M.

4. L and N give interview on adjacent days.

5. Not more than one person gives interview between N and K. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Hence, two persons give interview between P and O.

Here, persons are K, L, M, N, O, P and Q.

1. Three persons give interview between M and O who is scheduled for interview after M.

2. Q gives interview two days before P.

3. Q gives interview before M.

4. L and N give interview on adjacent days.

5. Not more than one person gives interview between N and K. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, Q gives interview just before M. Therefore, statement I is correct.

P and L gives interview on adjacent days. Therefore, statement II is correct.

N gives interview just after L not before. Therefore, statement III is incorrect.

Hence, the correct statement is only I and II.

Here, persons are K, L, M, N, O, P and Q.

1. Three persons give interview between M and O who is scheduled for interview after M.

2. Q gives interview two days before P.

3. Q gives interview before M.

4. L and N give interview on adjacent days.

5. Not more than one person gives interview between N and K. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, each combination is incorrectly matched except K who gives interview on Sunday.

Hence, the odd one is K - Sunday.

Here, persons are K, L, M, N, O, P and Q.

1. Three persons give interview between M and O who is scheduled for interview after M.

2. Q gives interview two days before P.

3. Q gives interview before M.

4. L and N give interview on adjacent days.

5. Not more than one person gives interview between N and K. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, P gives interview on Wednesday. Therefore, statement I is correct.

K is the last person who gives interview. Therefore, statement II is correct.

No one gives interview between N and O. Therefore, statement III is correct.

Hence, the incorrect statement is None of these.

Here, persons are K, L, M, N, O, P and Q.

1. Three persons give interview between M and O who is scheduled for interview after M.

2. Q gives interview two days before P.

3. Q gives interview before M.

4. L and N give interview on adjacent days.

5. Not more than one person gives interview between N and K. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Hence, O give interview on Saturday.

1. Point A is 5 Km to the south of Point B.

2) Point C is 7 Km to the west of point B.

3. Point C is 3 Km to the north of point D.

4. Point D is 2 Km to the east of point E.

5. Point E is 4 Km to the north of point F.

6. Point G is 9 Km to the east of point F.

Hence, the shortest distance between A and G is 2 Km.

1. Point A is 5 Km to the south of Point B.

2) Point C is 7 Km to the west of point B.

3. Point C is 3 Km to the north of point D.

4. Point D is 2 Km to the east of point E.

5. Point E is 4 Km to the north of point F.

6. Point G is 9 Km to the east of point F.

Since, Point A, Point G and Point B makes a straight line. Therefore, statement I is correct.

Point E is in northwest direction of point A not southwest. Therefore, statement II is incorrect.

Point G is in Southeast direction of point D. Therefore, statement III is correct.

Hence, the correct answer is only I and III.

1. Point A is 5 Km to the south of Point B.

2) Point C is 7 Km to the west of point B.

3. Point C is 3 Km to the north of point D.

4. Point D is 2 Km to the east of point E.

5. Point E is 4 Km to the north of point F.

6. Point G is 9 Km to the east of point F.

Since, point E is in Southwest direction of point B.

Hence, the correct answer is Southwest.

Statement: X = C, C ≤ G, G < S, S > K

On arrangement: X = C ≤ G < S > K

I. X = G → not definitely true (X = C ≤ G < S → X ≤ G)

II. X < G → not definitely true (X = C ≤ G < S → X ≤ G)

Hence, either I or II is true.

Given statement: C ≥ L, L > U, U = G, I < G

After combining: C ≥ L > U = G > I

I. C > U → True (C ≥ L > U = G > I → C > U)

II. C < U → False (C ≥ L > U = G > I → U < C)

Hence, only conclusion I is true.

I. G < L → False (N < E = L ≤ G = S → L ≤ G)

II. E = G → False (N < E = L ≤ G = S → E ≤ G)

Hence, Neither I nor II is true

Here, friends are G, H, I, J, K, L, M and N.

1. M sits second to the right of the one who sits immediate right to K.

2. N sits second to the right of K. 

3. N is the neighbor of H and G.

4. H does not sit immediate right to K.

5. Two persons sit between G and J.

6. L and I sit immediate left to each other.

7. J sits to the left of I and both of them facing same direction. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

Hence, N sits opposite to I.

Here, friends are G, H, I, J, K, L, M and N.

1. M sits second to the right of the one who sits immediate right to K.

2. N sits second to the right of K. 

3. N is the neighbor of H and G.

4. H does not sit immediate right to K.

5. Two persons sit between G and J.

6. L and I sit immediate left to each other.

7. J sits to the left of I and both of them facing same direction. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

Since, H sits immediate left to J. Therefore, statement I is correct.

N sits immediate right not left to H. Therefore, statement II is incorrect.

H and K faces in opposite direction not same direction. Therefore, statement III is incorrect.

Hence, the correct statement is Only I.

Here, friends are G, H, I, J, K, L, M and N.

1. M sits second to the right of the one who sits immediate right to K.

2. N sits second to the right of K. 

3. N is the neighbor of H and G.

4. H does not sit immediate right to K.

5. Two persons sit between G and J.

6. L and I sit immediate left to each other.

7. J sits to the left of I and both of them facing same direction. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

Since, each pair represents persons sitting opposite to each other except whereas N and L do not sit opposite to each other.

Hence, the odd one NL.

Here, friends are G, H, I, J, K, L, M and N.

1. M sits second to the right of the one who sits immediate right to K.

2. N sits second to the right of K. 

3. N is the neighbor of H and G.

4. H does not sit immediate right to K.

5. Two persons sit between G and J.

6. L and I sit immediate left to each other.

7. J sits to the left of I and both of them facing same direction. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

Since, K faces outward direction. Therefore, statement I is correct.

H sits second to the right of L. Therefore, statement II is correct.

G and N sits immediate left to each other. Therefore, statement III is correct.

Hence, the incorrect statement is none of these.

Here, friends are G, H, I, J, K, L, M and N.

1. M sits second to the right of the one who sits immediate right to K.

2. N sits second to the right of K. 

3. N is the neighbor of H and G.

4. H does not sit immediate right to K.

5. Two persons sit between G and J.

6. L and I sit immediate left to each other.

7. J sits to the left of I and both of them facing same direction. Therefore, case 1 is eliminated. Hence, the final arrangement is as follows

Since, M sits third to the right of J and G sits second to the left of M.

Hence, the correct answer is G.

Given word:- INTRODUCE

Since, 'O' comes sixth from the left end.

Hence, the correct answer is 'O'.

Given series is: (Left) X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H (Right)

11^th from the left → G

15^th from right → #

Terms between G and # are $, M, 4, J, D and 4 is exactly in middle of them.

Hence, 4 is the correct answer.

Given series is: (Left)  X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H            (Right)

20^th element from the left = K

9^th element to the left of K = G

(Left)   X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H          (Right)

Hence, G is the correct answer.

Given series is: X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H

Arrangement required: Consonant → Symbols → Number

(Left)   X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H           (Right)

Hence, only one such pair is there, i.e. K @ 1.

Given series is: X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H

Arrangement required: Consonant → Number → Consonant

(Left)   X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H           (Right)

Hence, three such pairs are there i.e. M 4 J, P 8 K and X 9 H

Arrangement required: Number → Consonant → Vowel

Given series is: X L 5 I € V O 7 % R G $ M 4 J D # P 8 K @ 1 0 E U 2 α 6 X 9 H

Clearly no such pair is there.

Hence, none is the correct answer$.

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

Conclusions:

 I. Some pagers are not laptop → it is definitely true.

II. All laptop being pagers is a possibility → its possibility is true as shown in the diagram below.

Thus, both conclusion I and II follows.

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

Conclusions:

I. All chairs are tables → it’s possible but not definite, hence false.

II. All beds being table is a possibility → its possibility is true when all the three coincide as shown in the diagram below.

Thus, only conclusion II follows.

The least possible Venn diagram is as follows:

Conclusions:

I. Some folders are not paper → True (because no paper is a file and all files are folder. Therefore Some folder are not paper)

II. Some files are data → False (It is possible but not definite)

Hence, only conclusions I follows.

According to the given information:

Since, from the above diagram it is clear the 'Rony' is coded as 'qm' and 'brings' is coded as 'wt'.

Hence, the words for the code 'wt qm' is 'Rony brings'.

According to the given information:

Since, the code for the word 'snowfalls' can be 'fl' and the word for the code 'za' can be either 'good' or 'luck'. Therefore, the possible words for the code 'fl za' can be either 'snowfalls good' or 'luck snowfalls'.

Hence, the correct answer is either 2 or 4.

According to the given information:

Since, from the given figure it is clear the code for the word 'winter' is 'oc'.

Hence, the correct answer is 'oc'.

According to the given information:

Since, the possible code for 'snowfalls' can be 'fl' and the word 'sleet' is not given in the coded information so its code will be different from the given codes.

Hence, the possible code for the words 'Snowfalls sleet' is 'fl ut'.

According to the given information:

Since, 'likes a snow' is coded as 'sk al js' then the possible code for 'makes' can be 'vs' and code for 'season' is 'bp'

Hence, the possible code for the word 'makes season' is 'vs bp'.

Since, the first letters are added by 2 with the previous first letter of the given clusters, the second letters of clusters are added by 4 with the previous second letter of clusters and number are arranged in prime number series.

Hence, the correct answer is XS13.

Here, persons are A, B, C, D, E and F; months June, September and December; dates 15 and 28.

1. Two persons join between E and D who joins after E.

2. B and C join company on same month.

3. B and D join company immediate next to each other.

4. A joins the company before F.

5. C joins company before F. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Hence, C joins company on 15^th September.

Here, persons are A, B, C, D, E and F; months June, September and December; dates 15 and 28.

1. Two persons join between E and D who joins after E.

2. B and C join company on same month.

3. B and D join company immediate next to each other.

4. A joins the company before F.

5. C joins company before F. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, in each pair one person joins company between them except AE where both of them joins company immediate next to each other.

Hence, the odd one is AE.

Here, persons are A, B, C, D, E and F; months June, September and December; dates 15 and 28.

1. Two persons join between E and D who joins after E.

2. B and C join company on same month.

3. B and D join company immediate next to each other.

4. A joins the company before F.

5. C joins company before F. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, A joins the company in the on 15^th June. Therefore, statement I is correct.

D and F joins company in the December month. Therefore, statement II is correct.

E joins company on 28^th June not 15^th June. Therefore, statement III is incorrect.

Hence, the correct answer is only I and II.

Here, persons are A, B, C, D, E and F; months June, September and December; dates 15 and 28.

1. Two persons join between E and D who joins after E.

2. B and C join company on same month.

3. B and D join company immediate next to each other.

4. A joins the company before F.

5. C joins company before F. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, C joins company just before not after B. Therefore, statement I is incorrect.

B joins company in the month of September not December. Therefore, statement II is incorrect.

D joins company after B not before. Therefore, statement III is incorrect.

Hence, the incorrect statement is all of these.

Here, persons are A, B, C, D, E and F; months June, September and December; dates 15 and 28.

1. Two persons join between E and D who joins after E.

2. B and C join company on same month.

3. B and D join company immediate next to each other.

4. A joins the company before F.

5. C joins company before F. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Hence, E joins the company just before C.

Here, family members are K, M, O, P, W, X and Y.

1. O is the father – in – law of X.

2. P is the brother – in – law of X.

3. P is unmarried.

4. Y is the uncle of W.

5. K is the brother – in – law of Y.

6. M has two sons. Therefore, case 2 is eliminated.

7. K has one son. Hence, the final arrangement is as follows

Since, each person except X is a male member whereas X is a female member.

Hence, the odd one X.

Here, family members are K, M, O, P, W, X and Y.

1. O is the father – in – law of X.

2. P is the brother – in – law of X.

3. P is unmarried.

4. Y is the uncle of W.

5. K is the brother – in – law of Y.

6. M has two sons. Therefore, case 2 is eliminated.

7. K has one son. Hence, the final arrangement is as follows

Since, Y is the brother of X who is the sister - in - law of P. Therefore, statement I is incorrect.

K is the husband of X. Therefore, statement II is correct.

O is the grandfather of W. Therefore, statement III is correct.

Hence, the correct answer is only II and III.

Here, family members are K, M, O, P, W, X and Y.

1. O is the father – in – law of X.

2. P is the brother – in – law of X.

3. P is unmarried.

4. Y is the uncle of W.

5. K is the brother – in – law of Y.

6. M has two sons. Therefore, case 2 is eliminated.

7. K has one son. Hence, the final arrangement is as follows

Hence, K is the son of M.

Forward pair: CG

Backward pair: AC

Thus, there are two such pairs formed which have as many letters between them as in the English alphabetical series in the given word "CHANGE".

Hence, the correct answer is two such pairs.

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ Projects_F = (68 + 56) ÷ 2 = 62

⇒ Avg cost_F = (52 + 35) ÷ 2 = 87 ÷ 2 = 43.5 (₹ Lakhs)

⇒ Total Cost_F = 62 × 43.5 = 2697 (₹ Lakhs)

Hence, the correct answer is ₹2,697 Lakhs

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ B = 3796 ; E = 2976

⇒ Difference = 3796 − 2976 = 820

⇒ Percent more = (820 ÷ 2976) × 100 ≈ 0.2755 × 100 = 27.55% ≈ 27.5% more

Hence, the correct answer is 27.5% more.

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

⇒ C Total Cost = 3640

⇒ C Revenue = 4368 ⇒ Profit_C = 4368 − 3640 = 728 (₹ Lakhs)

Hence, the correct answer is ₹728 Lakhs

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

For E — 50% High-Priority, remaining 50% Standard

Number of Standard Projects = 62 × 50 / 100 = 31

Hence, the correct answer is 31.

General Solution:

Given:

Managers A, B, C, D, E with Number of Projects and Average Cost (₹ Lakhs):

A: 68 projects, ₹45

B: 73 projects, ₹52

C: 56 projects, ₹65

D: 84 projects, ₹35

E: 62 projects, ₹48

Total Costs:

⇒ A: 68 × 45 = 3060 (₹ Lakhs)

⇒ B: 73 × 52 = 3796 (₹ Lakhs)

⇒ C: 56 × 65 = 3640 (₹ Lakhs)

⇒ D: 84 × 35 = 2940 (₹ Lakhs)

⇒ E: 62 × 48 = 2976 (₹ Lakhs)

Calculations:

Total Cost A : Total Cost D

⇒ 3060 : 2940 = divide by 60 ⇒ 51 : 49

Hence, the correct answer is 51 : 49.

Given:

Total capital invested by A and B = Rs. 75,000

A invests for 8 months

B invests for 12 months

Both receive equal profit ⇒ Their effective capital (Capital × Time) must be equal.

Let A's capital = x

Then B's capital = 75,000 − x

Condition for equal profit:

A_capital × A_time = B_capital × B_time

x × 8 = (75,000 − x) × 12

Solve:

8x = 12(75,000 − x)

8x = 900,000 − 12x

8x + 12x = 900,000

20x = 900,000

x = 900,000 ÷ 20 = 45,000

A's investment = Rs. 45,000

Hence, the correct answer is Rs. 45,000

Given:

Total milk = 28 litres

Initial ratio A : B : C = 4 : 1 : 2

Find actual quantities:

Total ratio parts = 4 + 1 + 2 = 7

⇒ 1 part = 28 ÷ 7 = 4 litres

⇒ A = 4 × 4 = 16 L

⇒ B = 1 × 4 = 4 L

⇒ C = 2 × 4 = 8 L

4 litres added to container B:

New B = 4 + 4 = 8 L

New ratio:

A : B : C = 16 : 8 : 8

Divide by common factor 8:

= 2 : 1 : 1

Hence, the correct answer is 2 : 1 : 1

Given:

C.P of 12 article = Rs. 2424

Profit = 25%

Concept:

Cost price × (100 + Profit %)/100 = Selling price

Where C.P = Cost price, S.P = Selling price

Calculation:

C.P of 12 article = Rs. 2424

C.P of each article:

⇒ 2424/12

⇒ 202

S.P of an article = 202 × (125/100)

⇒ 202 × 5/4

⇒ 252.5

Profit = S.P – C.P

⇒ 252.5 – 202

⇒ 50.5

∴ Rs. 50.5 gained by the shopkeeper on each article.

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

37.5 × 2 + 15 ÷ 3 – 14/3 of (?) = 10 

⇒ 75 + 5 – 14/3 of (?) = 10 

⇒ 70 = 14/3 of (?) 

⇒ ? = 15

Concept:

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

CALCULATION :

Given expression :

3 × 5/9 of ( 36 ÷ 4 ) × 6 - ? + 5 = 4

⇒ 3 × 5/9 of (9) × 6 - ? + 5 = 4

⇒ 3 × 5 × 6 - ? + 5 = 4

⇒ 90 - ? + 5 = 4 

⇒ 90 - 4 + 5 = ?

⇒ ? = 91 

∴ ? = 91

Given expression : 

4 × 22.5 ÷ 1.5 + ? –  6 = 30

⇒ 4 × 15 + ? – 6 = 30

⇒ 60 + ? - 6 = 30

 ∴ ? = 30 - 60 + 6 = -24

Concept used:

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

Calculation:

Now,

[(275 - 124) + 42 - 92 - 36)] + 14 = ?

⇒ [151 + 16 - 81 - 36] + 14 = ? 

⇒ 50 + 14

⇒ 64 = ?

⇒ ? = 64

∴ The required value  is 64

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

75

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,

75 - 3 × {(62 ÷ 6) ÷ 3} ÷ \(\frac{1}{3}\) + 13 =?

⇒ 75 - 3 × {(62 × 1/6) ÷ 3} ÷ \(\frac{1}{3}\) + 13 = ?

⇒ 75 - 3 × {6 ÷ 3} ÷ \(\frac{1}{3}\) + 13 = ?

⇒ 75 - 3 × 2 × 3/1 + 13 = ?

⇒ 75 - 18 + 13 = ?

⇒ 88 - 18 = ?

⇒ ? = 70

∴ The required value in the place of the question mark (?) is 70

Solution:

We have to follow the BODMAS rule

Calculation:

220 ÷ 5 ÷ 0.4 + 20 × 3 = ?

⇒ 44 ÷ 0.4 + 20 × 3 = ?

⇒110 + 20 × 3 = ?

⇒ 110 + 60 = ?

⇒ ? = 170

∴ 170 will come in place of question mark

Given:

(42 × 6 × 6) ÷ 36 × 9 = 20% of ?

Solution:

Follow the BODMAS rule according to the table given below:

Calculation:

(42 × 6 × 6) ÷ 36 × 9 = 20% of ?

⇒ (96 × 6) ÷ 36 × 9 = 20% of ?

⇒ 576 ÷ 36 × 9 = 20% of ?

⇒ 16 × 9 = 20% of ?

⇒ (144 × 100)/20 = ?

⇒ ? = 720

∴ 720 will come in place of the question mark

Concept used:

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

Calculation:

Considering the given equation

\(13\frac{6}{7} + 8\frac{2}{5} - 1\frac{5}{7} × 2\frac{1}{3} = ;?\)

We can transform each term within the given equation in the p/q form as follows: -

⇒ 97/7 + 42/5 - 12/7 × 7/3 = ?

⇒ 97/7 + 42/5 - 12/3 = ?

⇒ 97/7 + 42/5 - 4 = ?

⇒ 779/35 - 4 = ?

⇒ (779 - 140)/35 = ?

⇒ 639/35 = ? 

⇒ ? = 639/35

⇒ ? = \(18\frac{9}{{35}}\)

∴ The required value of "?" in the given equation is ​\(18\frac{9}{{35}}\)​

Concept used:

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

Calculation:

Considering the given equation

[(121)² ÷ 11 × 2] ÷ 44 × 2 – 57 = ?²

⇒ [14641 ÷ 11 × 2] ÷ 44 × 2 – 57 = ?²

⇒ [1331 × 2] ÷ 44 × 2 – 57 = ?²

⇒ [2662] ÷ 44 × 2 – 57 = ?²

⇒ 2662 ÷ 44 × 2 – 57 = ?²

⇒ 60.5 × 2 – 57 = ?

⇒ 121 – 57 = ?

⇒ 64 = ?²

⇒ 8 = ?

∴ ? = 8

Concept used:

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

Calculation:

Considering the given equation

12167 ÷ 23 × 2 + 386 + ?² = 2888

⇒ 529 × 2 + 386 + ?² = 2888

⇒ 1058 + 386 + ?² = 2888

⇒ 1444 + ?² = 2888

⇒ ?² = 2888 – 1444

⇒ ?² = 1444

⇒ ? = 38

∴ ? = 38

Concept used:

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

Calculations:

∛6859 + √441 – ∛4096 – √576 = ?

⇒ 19 + 21 – 16 – 24 = ?

⇒ 19 + 21 – 16 – 24 = ?

⇒ 40 – 40 = ?

⇒ ? = 0

∴ The value of ? is 0

Concept used:

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

Step 1: Part 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 part of the equation that contain ‘Addition’ and ‘Subtraction’ should be calculated.

Now,

Calculations:

Considering the given equation,

[12²+ (14² – 40)] × 2.5 = ? × 25

⇒ [12² + (196 – 40)] × 2.5 = ? × 25

⇒ [12² + 156] × 2.5 = ? × 25

⇒ [144 + 156] × 2.5 = ? × 25

⇒ 300 × 2.5 = ? × 25

⇒ 750 = ? × 25

∴ ? = 30

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:

Considering the following question

25% of 88 × ? ÷ 2 = 66

 ⇒ ¼ × 88 × ? ÷ 2 = 66

⇒ 22 × ? ÷ 2 = 66

⇒ 11 × ? = 66

∴ ? = 6

Concept used:

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

Calculation:

90 × 1/3 + 49 ÷ 7 – 12 ÷ 3 = ?

⇒ 90 × 1/3 + 7 – 4 = ?

⇒ 30 + 7 – 4 = ?

⇒ 37 – 4 = ?

⇒ ? = 33

∴ The value of ? is 33

\({8^{\frac{2}{3}}} × \frac{7}{2} + ;\sqrt {49} ; - ;\sqrt[3]{{729}}\) = ?

⇒\({({2^3})^{\frac{2}{3}}} × \frac{7}{2} + 7 - 9\) = ?

 ⇒ \({2^{3 × }}^{\frac{2}{3}} × \frac{7}{2} + 7 - 9\) = ?

⇒ 2² × (7/2) + 7 - 9 = ?

⇒ 14 + 7 - 9 = ?

⇒ 21 - 9 = ?

⇒ ? =  12

∴ The required value of "?" is 12.

Given:

Speed₁ = 10 km/hr

Speed₂ = 20 km/hr

Formula Used:

Speed = Distance/Time

Average speed = Total distance/Total Time

Calculation:      

Let's assume that the distance is d km

Time₁ = d/10 hour

Time₂ = d/20 hour

Total time = d/10 + d/20

Total distance = d + d = 2d km

Average speed = \(\frac{2d}{\frac{d}{10}+\frac{d}{20}}\)

Average speed = \(2d \times \frac{20}{3d}\)

Average speed = 40/3 km/hr

Average speed = \(13\frac{1}{3}\) km/hr

Average speed is \(13\frac{1}{3}\) km/hr

The correct option is 1 i.e. \(13\frac{1}{3}\) km/hr

Given:

X and Y can complete a piece of work in 30 days and 12 days respectively

They started the work while working on alternate days, starting from X

Concept Used:

Work done = efficiency × time

Calculation:

Total work = LCM of (30,12) = 60 unit

Efficiency of X = 60/30 = 2 unit/day

Efficiency of Y = 60/12 = 5 unit/day

While working on alternate days,

In 2 days total work done = 2 + 5 = 7 units

Now multiplying with 8

So, in (8 × 2) = 16 days total work done = (7 × 8) = 56 units

On 17^th day X will work, so work done = 56 + 2 = 58 units

Remaining work = 60 - 58 = 2 units, which will be done by Y

Y will do the 2 unit work in = 2/5 days

So, the total work will be completed in 17 + 2/5 = 17\(\frac{2}{5}\) days

∴ The correct answer is option 3

General Solution:

Given:

Hotel A: Rooms = 39, Guests = 78

Hotel B: Rooms = 50, Guests = 105

Hotel C: Rooms = 45, Guests = 100

Hotel D: Rooms = 45, Guests = 117

Hotel E: Rooms = 60, Guests = 150

Calculations:

⇒ Total rooms = 39

⇒ Group P rooms = 2/3 × 39 = 26

⇒ Group Q rooms = 39 − 26 = 13

⇒ Group P guests = 52 ⇒ avg in P = 52 ÷ 26 = 2

⇒ Total guests = 78 ⇒ Group Q guests = 78 − 52 = 26

⇒ Avg in Q = 26 ÷ 13 = 2

Hence, the correct answer is 2.

General Solution:

Given:

Hotel A: Rooms = 39, Guests = 78

Hotel B: Rooms = 50, Guests = 105

Hotel C: Rooms = 45, Guests = 100

Hotel D: Rooms = 45, Guests = 117

Hotel E: Rooms = 60, Guests = 150​

Calculations:

⇒ Total rooms = 45

⇒ Deluxe rooms = 5/9 × 45 = 25

⇒ Standard rooms = 45 − 25 = 20

⇒ Total guests = 100 ⇒ Deluxe guests = 60% × 100 = 60

⇒ Standard guests = 100 − 60 = 40

⇒ Avg per Standard room = 40 ÷ 20 = 2

Hence, the correct answer is 2

General Solution:

Given:

Hotel A: Rooms = 39, Guests = 78

Hotel B: Rooms = 50, Guests = 105

Hotel C: Rooms = 45, Guests = 100

Hotel D: Rooms = 45, Guests = 117

Hotel E: Rooms = 60, Guests = 150

Calculations:

⇒ B avg = 105 ÷ 50 = 2.1 = 21/10

⇒ E avg = 150 ÷ 60 = 2.5 = 5/2

⇒ Ratio = (21/10) : (5/2) = (21/10) × (2/5) = 42/50 = 21/25

Hence, the correct answer is 21 : 25

General Solution:

Given:

Hotel A: Rooms = 39, Guests = 78

Hotel B: Rooms = 50, Guests = 105

Hotel C: Rooms = 45, Guests = 100

Hotel D: Rooms = 45, Guests = 117

Hotel E: Rooms = 60, Guests = 150

Calculations:

Hotel B revenue:

= (Room charge × Rooms) + (Guest fee × Guests)

= (1200 × 50) + (300 × 105)

= 60,000 + 31,500

= ₹91,500

Hotel C revenue:

= (1500 × 45) + (250 × 100)

= 67,500 + 25,000

= ₹92,500

Difference = 92,500 – 91,500 = ₹1,000

Hence, the correct answer is ₹1,000.

General Solution:

Given:

Hotel A: Rooms = 39, Guests = 78

Hotel B: Rooms = 50, Guests = 105

Hotel C: Rooms = 45, Guests = 100

Hotel D: Rooms = 45, Guests = 117

Hotel E: Rooms = 60, Guests = 150

Calculations:

⇒ Rooms in F = average of A & C = (39 + 45) ÷ 2 = 42

⇒ Guests in F = 20% more than B = 1.2 × 105 = 126

⇒ Avg in F = 126 ÷ 42 = 3

⇒ Avg in A = 78 ÷ 39 = 2

⇒ Difference = 3 − 2 = 1

Hence, the correct answer is 1

Given:

A and B have equal principal = P each.

A invests at 8% SI for 2 years.

B invests at 14% SI for 2 years.

Total interest earned together = Rs. 3,520

Simple Interest formula: SI = (P × R × T) / 100

Calculate interest of each:

Interest by A:

SI_A = (P × 8 × 2) / 100 = 16P / 100 = 0.16P

Interest by B:

SI_B = (P × 14 × 2) / 100 = 28P / 100 = 0.28P

Total interest:

0.16P + 0.28P = 0.44P

Given:

0.44P = 3,520

Solve for P:

P = 3,520 / 0.44

P = 3,520 ÷ 0.44

P = 8,000

So, B’s principal = Rs. 8,000

Hence, the correct answer is Rs. 8,000

Given:

Pens = 40% more than Pencils

Erasers = 25% less than Pens

Total (Pens + Pencils + Erasers) = 690

Formula used:

Pens = Pencils × 1.4

Erasers = Pens × 0.75

Total = Pencils + Pens + Erasers

Calculations:

Let Pencils = x

⇒ Pens = 7/5 × x

⇒ Erasers = 0.75 × Pens = 3/4 × (7/5 x) = 21/20 × x

⇒ Total = x + 7/5 x + 21/20 x

⇒ Total = (20/20 + 28/20 + 21/20) x = 69/20 × x

⇒ 69/20 × x = 690

⇒ x = 690 × 20 ÷ 69 = 200 (Pencils)

⇒ Erasers = 21/20 × x = 21/20 × 200 = 21 × 10 = 210

Hence, the correct answer is 210.

Given,

The ratio of the present ages of Vishal and Himanshu is 5 : 7.

The difference between their ages is 6 years

Calculations: 

Let the present age of Vishal and Himanshu be 5X and 7X respectively.

Given that

7X - 5X = 6

⇒ 2X = 6

⇒ X = 3

Then  Vishal present age is

⇒ X = 5(3) = 15

∴ The required answer is 15 years

Given:

Length of rectangle = 8 cm

Breadth of rectangle =7 cm

Formula:

Perimeter of square  = 4 × a

Perimeter of rectangle = 2 × (L + B)

Calculations:

Perimeter of a square = 2 × perimeter of a rectangle

Perimeter of a square = 4 × side

Perimeter of a rectangle = 2(length + breadth)

Given Rectangle of length = 8 cm and breadth = 7 cm

Perimeter of a rectangle = 2(8 + 7) = 2 × 15 = 30 cm

So, Perimeter of a square = 4 × side = 2 × perimeter of a rectangle

⇒ 4 × side = 2 × 30

⇒ Side = 15 cm

∴ Side of the square is 15 cm

The series follows the following pattern:

3 × 2 = 6

6 × 3 = 18

18 × 5 = 90

90 × 7 = 630

630 × 11 = 6930

∴ The wrong term in the series is 590.

The series follows the following pattern:

7 + 2² = 11

11 + 4² = 27

27 + 6² = 63

63 + 8² = 127

127 + 10² = 227

∴ The wrong term in the series is 61.

The series follows the following pattern:

11 + 2 = 13

13 + 3 = 16

16 + 5 = 21

21 + 7 = 28

28 + 11 = 39

∴ The wrong term in the series is 37.

The series follows the following pattern 

36 × 0.5 = 18

18 × 1 = 18 

18 × 1.5 = 27

27 × 2 = 54 

54 × 2.5 = 135 

∴ The wrong number in the series is 19.

The series follows the following pattern:

2880 ÷ 2 = 1440

1440 ÷ 3 = 480

480 ÷ 4 = 120

120 ÷ 5 = 24

24 ÷ 6 = 4

∴ The wrong term in the series is 1400.

Given:

The average score of X and Y is 41.5

The average score of Y and  Z is 62.5

The average score of Z and X is 54.95

Formula used:

Total Sum = Average × No of terms

Calculation:

Total score of X and Y = 41.5 × 2

X + Y = 83  -------(1)

Total score of Y and Z = 62.5 × 2

Y + Z = 125  -------(2)

Total score of Z and X = 54.95 × 2

Z + X = 109.9  --------(3)

By adding equations (1), (2) and (3)

2 (X + Y + Z) = 83 + 125 + 109.9

X + Y + Z = 317.9/2

X + Y + Z = 158.95   ---------(4)

Subtracting equation (3) from (4),

X + Y + Z - (Z + X) = 158.95 - 109.9

Y = 49.05

Score of W = Y + 19.45 marks

= 68.5

\(\therefore\) The correct answer is 68.5.