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Applying alligation METHOD,

-25/3% +25/2%

0%

25/2 25/3

∴ Table : CHAIR = 1/2 : 1/3 = 3 : 2 

Let the CP of table and chair be 3x and 2x respectively.

⇒ 3x × 25/200 - 2x × 25/300 = 25

⇒ x = 120 

∴ Price of chair = 2x = 2 × 120 = Rs. 240 

Price of table = 3x = 3 × 120 = Rs. 360

LET the four digit number be 1000x + 100y + 10Z + n

Statement I

⇒ n = z/3

⇒ 3n = z

Statement II

⇒ n = y/5

⇒ 5n = y

&

⇒ n = x/2

⇒ 2n = x

Statement III

⇒ SUM of all the digits is 11.

Now, putting all values

⇒ 3n + 5n + 2n + n = 11

⇒ 11n = 11

⇒ n = 1

Hence, the number is 2531.

The pattern of the original number series is

⇒ 2

⇒ 2² = 4

⇒ 4² = 16

⇒ 16 → 1 + 6 = 7 → 7² = 49

⇒ 49 → 4 + 9 = 13 → 13² = 169

⇒ 169 → 1 + 6 + 9 = 16 → 16² = 256

According the pattern of the above series the number series starting from the GIVEN number 3 WOULD be as:

⇒ 3

⇒ A = 3² = 9

⇒ B = 9 → 9² = 81

⇒ C = 81 → 8 + 1 = 9 → 9² = 81

⇒ D = 81 → 8 + 1 = 9 → 9² = 81

⇒ E = 81 → 8 + 1 = 9 → 9² = 81

The pattern of the original number series is

⇒ 2² = 4,

⇒ 3² = 9,

⇒ 5² = 25,

⇒ 7² = 49,

⇒ 11² = 121,

⇒ 13² = 169,

Notice 2, 3, 5, 7, 11 & 13 are prime numbers.

According the pattern of the above series the number series starting from the given number 17 would be as:

⇒ 17² = 289,

⇒ A = 19² = 361

⇒ B = 23² = 529

⇒ C = 29² = 841

⇒ D = 31² = 961

⇒ E = 37² = 1369

Notice 17, 19, 23, 29, 31 & 37 are prime numbers.

∴ The REQUIRED number in place of A would be 361

The pattern of the original NUMBER series is

⇒ 7 + 2¹ = 9

⇒ 14 + 2² = 18

⇒ 21 + 2³ = 29

⇒ 28 + 2⁴ = 44

⇒ 35 + 2⁵ = 67

According the pattern of the above series the number series starting from the given number would be as:

⇒ A = 9 + 2¹ = 11

⇒ B = 18 + 2² = 22

⇒ C = 27 + 2³ = 35

⇒ D = 36 + 2⁴ = 52

⇒ E = 45 + 2⁵ = 77

WITHOUT any change in score,

Total score of PRAJVAL = 88 + 91 + 71 + 85 + 86 = 421

% score of Prajval = (421/500) × 100 = 84.2%

CHANGING in score,

Updated score of Math = Current score + 12.5% of current score = 88 + (1/8) × 88 = 88 + 11 = 99

Updated score of Science = current score – 14.29% of current score = 91 – (1/7) × 91 = 91 – 13 = 78

Total score of Prajval after changing = 99 + 78 + 71 + 85 + 86 = 419

⇒ % score of Prajval after change = (419/500) × 100 = 83.8%

∴ Difference in % = 84.2 – 83.8 = 0.4%

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.

Given expression is,

$(\frac{3}{4} + \frac{7}{8} + \frac{{11}}{{12}} + \frac{{13}}{{16}} = \frac{{? + 1}}{{48}})$

$(\Rightarrow \;\frac{{36 + 42 + 44 + 39}}{{48}} = \frac{{? + 1}}{{48}})$

⇒ 161/48 = (? + 1) /48

⇒ 161 = ? + 1

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.

Given expression is,

$(3\frac{2}{5} \times 12\frac{5}{8} - 2\frac{1}{5} \times 5\frac{1}{4} = ?)$

⇒ ? = 17/5 × 101/8 - 11/5 × 21/4

⇒ ? = (1717/40) – (231/20)

⇒ ? = (1717 – 462) /40

⇒ ? = 1255/40

Lets, PRODUCTION cost of 100 KG of goods = 100x

Packaging cost of 100 kg of goods = 15x

Actual cost price = 100x + 15x = 115x

New production cost = 100x × 90/100 = 90X

New Packaging cost = 15x × 140/100 = 21x 

New actual cost price of 100 kg goods = 90x + 21x = 111x 

∴ Required percentage decrease = (115x – 111x) /115x × 100 = 3.48%

The pattern of the GIVEN series is,

⇒ 16 – 14 = 2

⇒ 19 – 16 = 3

⇒ 24 – 19 = 5

⇒ 31 – 24 = 7

⇒ 42 – 31 = 11

⇒ 55 – 42 = 13

⇒ 55 + 17 = 72

Quantity 1:

Let the speed of the slower car be x, so the time taken by the two cars is-

T1 = 350/x and t2 = 350/x + 5 

And the difference between the time is 2 hrs 20 minutes = 7/3 

⇒ 350/x - 350/x + 5 = 7/3 

Solving this we get ⇒ x² + 5x - 750 

And solving this quadratic equation we get two values ⇒ - 30, 25 

∴ The speed of the slower car = 25 Km/h

Quantity 2:

2y² - 83y + 861 = 0 

⇒ 2y² - 41y - 42y + 861 = 0 

⇒ 2y(y- 21) - 41(y - 21) = 0 

⇒ (2y - 41)(y - 21) = 0 

⇒ y = 21, 20.5

The average age of 20 tourists = 14 YEARS 

Age of girl member = 14 – 2 = 12 year

New average (INCLUDING guide) = 16 years 

∴ Sum of their ages 

⇒ 20 × 14 + Age of guide = 21 × 16 

∴ Age of guide = 56 Years

∴ DIFFERENCE between the ages of guide and girl member = (56 - 12) = 44

According to the BODMAS RULE, the priority in which the operations should be done is:

⇒ 4.99² + 13.003² - 11.01² = (?)³ - 51.98

Approximating the value to the nearest INTEGER:

⇒ 5² + 13² - 11² = (?)³ - 52

⇒ 25 + 169 - 121 = (?)³ - 52

⇒ 25 + 48 + 52 = (?)³

(?)³ = 125

(?) = 5

By using the law of indices?

Laws of Indices?-

1-? a^m × aⁿ = a^{{m + n}}

2-? a^m ÷ aⁿ = a^{{m – n}}

3-? [(a^m)ⁿ] = a^mn

4-? (a)^(1/m) = ^m√a

5-? (a)^(-m) = 1/a^m

6-? (a)^(m/n) = ⁿ√a^m

7-? (a)⁰ = 1

The equation will be SOLVED as

25^4.1 ÷ 5^3.3 × 125^6.7 = 25^X

⇒ (5²)^4.1 ÷ 5^3.3 × (5³)^6.7 = (5²)^X

⇒ 5^8.2 ÷ 5^3.3 × 5^20.1 = 5^2X

⇒ 5^{(8.2 – 3.3)}× 5^20.1 = 5^2X

⇒ 5^4.9× 5^20.1 = 5^2X

⇒ 5^{(4.9 + 20.1)} = 5^2X

⇒5²⁵ = 5^2X

⇒ 25 = 2X

According to the BODMAS RULE, the PRIORITY in which the operations should be done is:

⇒ 4986.02 ÷ 216.004 - 3725.921 ÷ 207.004 = ??

APPROXIMATING the value to the nearest integer:

⇒ 4986 ÷ 216 - 3726 ÷ 207 = ?? 

⇒ 23 - 18 = ??

?? = 5 

(?) = 5³ 

(?) = 125

According to the BODMAS rule, the priority in which the operations should be DONE is:

⇒ (36.99 × 53) + 25% of 212.02 = (111.01 × 11.01) + (?)

APPROXIMATING the value to the nearest integer:

⇒ 37 × 53 + 25% of 212 = 111 × 11 + (?)

⇒ 1961 + 53 = 1221 + (?)

(?) = 2014 - 1221

(?) = 793

TOTAL production of SUGAR by the company in that year 

5000 + 5100 + 5200 + 5300 + … + 6100 = 66600 

∴ AVERAGE monthly production = 66600/12 = 5550 TONNES

Let us find out the number of female and MALE programmers for each language taking part in the competition.

Total no. of programmers = 1600

Number of female programmers = $(\frac{1}{4} \times 1600 = 400)$

Number of male programmers = $(\frac{3}{4} \times 1600 = 1200)$

Number of Python programmers = 25% of 1600 = 400

Number of C# programmers = 220

Number of C programmers = 10% of 1600 = 160

Number of Java programmers = 2 × 220 = 440

Number of PHP programmers = 1600 - (400 + 220 + 160 + 440) = 380

Number of female Python programmers = 30% of 400 = 120

∴ Number of male Python programmers = 400 - 120 = 280

Number of female C# programmers = HALF of female python programmers = $(\frac{1}{2} \times 120 = 60)$

∴ Number of male C# programmers = 220 - 60 = 160

Number of female C programmers = 10% of Java programmers = 0.1 × 440 = 44

∴ Number of male C programmers = 160 - 44 = 116

Number of female Java programmers = Number of female PHP programmers

$(= \frac{1}{2}\left[ {400 - \left( {120 + 60 + 44} \right)} \right] = \frac{1}{2}\left[ {400 - 224} \right] = 88)$

∴ Number of male Java Programmers = 440 - 88 = 352

∴ Number of male PHP programmers = 380 - 88 = 292

From above data,

According to the BODMAS rule, the PRIORITY in which the operations should be done is:

APPROXIMATING the values to the nearest integer:

(?) = 15% × 5840 + 9450 ÷ 150 + 453 

(?) = 15/100 × 5840 + 9450/150 + 453 

(?) = 876 + 63 + 453 

(?) = 1392

(√36)⁵ × [(1296)^(1/4)]⁷ ÷ 6¹⁰ = 6^? ÷ 216⁸

⇒ (6⁵) × (6)⁷ ÷ (6)¹⁰ = 6^? ÷ (6³)⁸

By using the LAW of indices,

a^m ÷ a^N = a^{{m – n}}

a^m × aⁿ = a^{{m + n}}

⇒ 6^{5 + 7 - 10} = 6^? ÷ 6²⁴

⇒ 6^{12 - 10} = 6^{? - 24}

⇒ 6² = 6^{? - 24}

⇒ 2 = ? - 24

Here, only three subjects CONSIDER i.e. Maths, SCIENCE and Social science;

For the % SCORE of Hitesh,

Total of Hitesh in 3 subject = 85 + 79 + 91 = 255

So, % of Hitesh in 3 subject = (255/300) × 100 = 85%

For the % of Foram,

Total of Foram in 3 subject = 92 + 91 + 63 = 246

So, % of Foram in 3 subject = (246/300) × 100 = 82%

⇒ Difference in % = 85 – 82 = 3%

4²² × 6⁴⁶ ÷ 6⁴⁰ = 4^? × 3⁶

By USING the LAW of INDICES,

(a^m )^N = a^mn

⇒ 4²² × 2⁴⁶ × 3⁴⁶ ÷ 6⁴⁰ = 4^? × 3⁶

⇒ 4²² × 2⁶ × 3⁶ × 6⁴⁰ ÷ 6⁴⁰ = 4^? × 3⁶

By using the law of indices,

a^m ÷ aⁿ = a^{{m – n}}

⇒ 4²² × (2²)³ × 3^{6 – 6} × 6^{40 – 40} = 4^?

⇒ 4²² × 4³ = 4^?

By using the law of indices,

a^m × aⁿ = a^{{m + n}}

⇒ 4^{22 + 3} = 4^?

⇒ 4²⁵ = 4^?

Here, two languages is Hindi and ENGLISH

Average score in Hindi = (79 + 80 + 89 + 90 + 85)/5 = 84.6

Average score in English = (83 + 82 + 68 + 72 + 71)/5 = 75.2

The SERIES is as FOLLOWS:$

8 × 1 = 8

8 × 3/2 = 12

12 × 2 = 24

24 × 5/2 = 60

60 × 3 = 180

180 × 7/2 = 630

Hence, there should be 24.

Rajesh’s TOTAL investment = 80000 × 6 + 70000 × 3 + 90000 × 3 = 960000

Mukesh’s total investment = 45000 × 4 + 70000 × 8 = 740000

RATIO of their profits = Ratio of their investments

⇒ 30000/Mukesh’s PROFIT = 960000/740000

Average of RISHI/Average of Kartik = 87.55 = 7/8

For the average of Rishi,

⇒ Total SCORE of Rishi = 87 + 82 + 72 + 90 + 89 = 420

⇒ Average marks of Rishi = 420/5 = 84%

From the ratio,

⇒ Average of Rishi/Average of Kartik = 7/8

⇒ (84 × 8)/7 = Average of Kartik

⇒ Average of Kartik = 96

For 96 average, Total score of kartik = 96 × 5 = 480

⇒ Current total score of Kartik = 90 + 89 + 82 + 80 + 76 = 417

45 + 2² = 49

49 - 3² = 40 

40 + 4² = 56

56 - 5² = 31

31 + 6² = (67)

Quantity 1:

Let TOTAL number of books be 100 

Urdu books = 100 - (20 + 20 + 18) = 42 

42 ⇒ 29400

1 ⇒ 29400/42

100 ⇒ 29400/42 × 100 = 70000

Quantity 2:

⇒ Value of 4 × 7⁵ = 4 × 16807 = 67228

Let amount of the mixture of salt and sugar is 1 KG

Mixture CONTAINS 4 PARTS of salt and 5 parts of sugar

⇒ Amount of salt in the mixture = 4/9 kg

⇒ Amount of sugar in the mixture = 5/9 kg

After replacing 1/4^th PART with salt,

⇒ Amount of salt in the new mixture = (4/9 – 4/9 × ¼ + 1/4) kg = (16 – 4 + 9) /36 kg = 7/12 kg

⇒ Amount of sugar in the new mixture = (5/9 – 5/9 × 1/4) kg = 5/12 kg

78% of 450 – 65% of 340 + 29% of 1200 = ?

⇒ [(78/100) × 450] – [(65/100) × 340] + [(29/100) × 1200] = ?

⇒ ? = 351 – 221 + 348

Suresh’s TOTAL investment = 60000 × 10 + 80000 × 2 = 760000

RATIO of their profits = Ratio of their investments = 960000? 740000? 760000 = 48 ? 37 ? 38

Sum of ratios = 48 + 37 + 38 = 123

Rajesh's profit = (48/123) × Total profit = Rs. 30000

It is given that principal = Rs. 30000

Rate = 7%

Let TIME PERIOD = n

Compound INTEREST = Rs. 4347

Amount = Principal + Compound interest = Rs. 34347

We know, Amount = Principal (1 + Rate/100)ⁿ

⇒ 34347 = 30000 (1 + 7/100)ⁿ

⇒ 11449/10000 = (107/100)ⁿ

⇒ (107/100)² = (107/100)^t

Let SAMSON’s salary last MONTH be 100

He spent 60% (60% of 100)

∴ His saving’s = 100 – 60 = 40 

Samson’s salary this month = 1.25 × 100 = 125 

Expenditure this month = 1.20 × 60 = 72 

SAVINGS this month = 125 – 72 = 53 

Percentage increase in savings = (53 – 40) /40 × 100 = 32.5%

28 + 256 = 284

284 + 128 = 412

412 + 64 = (476)

476 + 32 = 508

Invested amount of A and C are equal in 2013.

Hence the PROFIT ratio will be equal to their timing ratio.

∴ Profit ratio of A and C = timing ratio of A and C = 10 : 12 = 5 : 6

TOTAL profit = 5a + 6a = 11a

⇒ 11a = 11,000

⇒ a = 1,000

∴ Profit of A = 5a = 5,000

Profit of C = 6a = 6,000

Let amount invested by A for 10 months = z × 10 (in thousand)

∴ Amount invested by C for 12 months = z × 12 (in thousand)

∴ Total amount = 22z (in thousand)

B’s Amount for 8 months In 2013 = 20,000 × 8

(20000 × 8)/(22000 × z) = B’s Profit/11000

But we don’t have B’s Profit.