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RATIO of selected to unselected candidates = 9 ? 2

Let the number of selected candidates be 9x and unselected candidates be 2X

⇒ TOTAL number of applied candidates = 9x + 2x = 11x

New ratio = 5 ? 1

Selected = 9x – 20

⇒ Unselected = (11x – 80) – (9x – 20) = 2x – 60

⇒ (9x – 20)/(2x – 60) = 5/1

9x – 20 = 10X – 300

x = 280

Total number of candidates who appeared for army = 11x = 11 × 280 = 3080

As we know that, FOURTH proportional = (b × C)/a 

Where a,b and c are the 1^st, 2^nd and 3^rd elements of PROPORTION respectively a = 0.5, b = 1.12 and c = 1.3

(1.12 × 1.3)/0.5

1.456/0.5 = 2.912

The RATIO of boys and girls is 5 : 3

NUMBER of girls = (3/8) × number of students

The ratio of girls and teachers is 7 : 1

Number of girls/ Number of teachers = 7/1

[(3/8) × Number of students]/Number of teachers = 7/1

Number of students/Number of teachers = 56/3

∴ The ratio of students and teachers 56 ? 3

THIRD PROPORTION can be GIVEN as 25²/10 = 625/10

∴ The third proportion of 10 and 25 is 62.5

Let the AGE of children’ s be x, x + 2, x + 4, x + 6, x + 8

⇒ Sum of age = 60

⇒ x + x + 2 + x + 4 + x + 6 + x + 8 = 60

⇒ 5X + 20 = 60

⇒ 5x = 40

⇒ x = 8

As given-

∴ Simplified ratio of 1/3 : 2/5 : 1/6 = 10 ? 12 ? 5.

One can see that 1^st employee got 10 shares amongst 27 shares of RS. 729.

Money received by 1^st worker = (10/27) × 729 = 270

∴ 1^st worker received = 270 Rs.

Given that,

Ratio of the WAGES of SUNNY and Mickey = 2 : 3

Ratio of the wages of Mickey & VICKY = 7 : 8

⇒ Ratio of the wages of Sunny, Mickey and Vicky are 14 : 21 : 24

Let the wages of Mickey be 21X and that of Vicky be 24x

According to the question,

21x × 24x = 50400

⇒ 504x² = 50400

⇒ x = 10

So, the wages of Mickey = RS. 210 and the wage of the Vicky is Rs. 240

So, the wage of Sunny = 210 × (2/3) = Rs. 140

GIVEN,

Ratio of Populations of villages X and Y = X ? Y = 34 ? 43

Ratio of Populations of villages X and Y if the population of village Y INCREASE by 125000 = X ? (Y + 125000) = 17 ? 24

X ? Y = 34 ? 43

⇒ X/Y = 34/43

⇒ Y = 43X/34

Now,

X ? (Y + 125000) = 17 ? 24

⇒ X/ (Y + 125000) = 17/24

⇒ X/ (43X/34 + 125000) = 17/24

⇒ 24X = 43X/2 + 2125000

⇒ X = 850000

The population of village Y = 43 × 850000/34 = 1075000

∴ The population of village Y = 1075000

After 10 days, the PROVISION is left for 3000 men and is for 20 days. when 1000 more men join, the total number in the garrison are 4000.

“If 3000 men can CONSUME FOOD in 20 days, then 4000 will consume it in how many days”

3000 men consume food in 20 days

1 man consume food in 20 × 3000 days (more days)

4000 men consume food in $(= \;\frac{{20 \times 3000}}{{4000}} = 15\;{\rm{days}})$

FIRST Share is X

2nd share is y

3rd share is z

As per the Question

$4X = 3y = 3z = k $ 

x = k/4

y = k/3

z = k/3

x + y + z = 1703

k/4 + k/3 + k/3 = 1703

11k/12 = 1703

k = 20436/11

value of 2nd share is: k / 3 = 20436/33 = 619.27

Given that,

u : v = v : w

v ² = UW

v ⁴ = (uw)²

then,

INCOME = EXPENDITURE + savings

Income = 6 + 1 = 7 units = 42000

1 UNIT = 6000

Let the seven numbers be x, 3X, 7X, 12x, 13x, 21x and 23X respectively.

⇒ x + 3x + 7x + 12x + 13x + 21x + 23x = 3280

⇒ x = 3280/80 = 41

Sum of highest THREE numbers = 41 × (13 + 21 + 23) = 41 × 57 = 2337

⇒ The average of the highest three numbers = 2337/3 = 779

∴ The average of the highest three numbers = 779

According to the given information,

Number of total workers in the farm = 320

Ratio of MEN and women = 9 : 11

Number of men workers = total number of workers $(\times \;\frac{{9}}{{9 + 11}})$

Number of men workers $(= \;320\; \times \frac{{9}}{{9 + 11}}{\rm{\;}} = {\rm{\;}}144)$

Number of female workers = 320 – 144 = 176

According to the given information,

8 female workers leave the work,

the remaining number of female workers = 176 – 8 = 168

New ratio of male to female workers = 144 : 168 = 6 : 7

CONVERT the number in decimal form-

1/3rd$ of 24 = 1/3 × 24 = 8

4/5th $of 15 = 4/5 × 15 = 12

3/7th $of 21 = 3/7 × 21 = 9

Fourth PROPORTION = 8 ? 12 ?? 9 ? x

8/12 = 9/x

x = 27/2 = 13.5

We know that,

a : b :: c : X

Then fourth proportional x = (bc)/a

From the GIVEN data,

As we KNOW that, mean proportional = √a × B

Where a and b are outer elements of proportion

a = 5 and b = 45 

√5 × 45 = 15

Let TWO number be 5x and 4X

According to the QUESTION,

⇒ (5x)³ - (4x)³ = 61

⇒ 125x³ - 64x³ = 61

⇒ 61x³ = 61

⇒x³ = 1

⇒ x = 1

Given ratio of INCOME of P : Q = 5 : 3, Q : R = 5 : 2

⇒ In both the ratios, the common term is ‘Q’. Make the value with respect to ‘Q’ same in both the ratio

⇒ Multiply both numerator and denominator on other side of the ratio (P: Q) with 5 and of ratio (Q: R) with 3, we get

⇒ P: Q = (5 × 5): (3 × 5), Q: R = (5 × 3): (2 × 3)

⇒ P: Q = 25 : 15, Q : R = 15 : 6

⇒ Combining both the ratios we get, P : Q : R = 25 : 15 : 6

⇒ LET the incomes be 25x, 15x, 6x

⇒ From the condition, ¼ (25x) = 500 + 6x

⇒ (25x/4) - 6x = 500

⇒ (25x - 24X)/4 = 500

⇒ (x/4) = 500

⇒ x = 500 × 4 = Rs. 2000

We have

$(\FRAC{a}{3} = \frac{b}{5} = \frac{c}{7})$ 

⇒ a = 3b/5, and c = 7b/5

Hence,

$(\frac{{a + b + c}}{b}\frac{{a + b + c}}{b} = \frac{{\frac{3}{5}b + b + \frac{7}{5}b}}{b} = 3)$ 

Let SHARE of B = X/3

Total share = (1/2 + 1/3 + 1/4)x = 13x/12

Share of B out of 7800 = (x/3) / (13x/12) × 7800 = 4/13 × 7800 = 2400

Share of B (in Rs) = Rs. 2400

Let the ratio of selected(x) to UNSELECTED candidates (y) = x/y = 3m/m

Then the TOTAL NUMBER of candidates = 3m + m = 4M

Also given that 60 less had APPLIED and 30 less selected, the ratio of selected to unselected would have been 5 : 1

⇒ 4m - 60 have been applied, 3m - 30 has selected,

Total number of unselected candidates = Total number of applied candidates – total selected candidates

Total number of unselected candidates = (4m – 60) – (3m – 30) = m - 30

From the given data,

Selected candidates/Unselected candidates = 5/1

(3m - 30)/(m - 30) = 5/1

3m - 30 = 5(m - 30)

2m = 120

m = 60

∴ total no of candidates = 4m = 4(60) = 240

20% of A = 30% of B = 1/6 of C

⇒ 20/100 × A = 30/100 × B = 1/6 × C

⇒ A/5 = 3B/10 = C/6

Let salary of Ananya be ‘4x’ and salary of Krish be ‘5x’

⇒ According to the condition GIVEN in the problem,

$(\frac{{4x + 5000}}{{5x + 5000}} = \frac{{67}}{{71}})$

⇒ 71(4x + 5000) = 67(5x + 5000)

⇒ 284x + 355000 = 335x + 335000

⇒ 51x = 20000

⇒ x = RS. 392.15

Ratio of Chocolates A, B and C = 1/63 ? 1/77 ? 1/99 = 11 ? 9 ? 7

Quantity of the chocolates A after selling = 11 × 10/11 = 10

Quantity of the chocolates B after selling = 9 × 8/9 = 8

Quantity of the chocolates C after selling = 7 × 6/7 = 6

Milk contents in the mixture $(= \frac{7}{{10}} \TIMES 30 = 21\;{\rm{liters}})$

Water = 9 liters

In the new mixture, the ratio of milk to water is 3 : 7

In 21 liters of milk water is $(\frac{7}{3} \times 21 = 49\;{\rm{liters}})$

Let the number of 5 rupees COINS and 2 rupees coins be 4T and 7T, respectively.

We know, amount of money in form of 5 rupees is 42 rupees more than amount of money in form of 2 rupees.

⇒ (4T × 5) – (7T × 2) = 42

⇒ T = 42/6 = 7

Angle S is FOURTH proportional of angles P, Q and R.

⇒ P/Q = R/S

Ratio of angles Q and P is 5 : 4, and angle S is 125°.

So, P/Q = 4/5

⇒ 4/5 = R/125

⇒ R = 4 × 25 = 100

∴ MEASURE of angle R is 100°.

Let the C.P. of each book be Rs. 1.

Total C.P. of 25 BOOKS = 25

S.P. of books = 20

Loss percent $(= \left[ {\frac{{25 - 20}}{{25}}} \right] \times {\RM{\;}}100)$

$(= \frac{5}{{25}} \times {\rm{\;}}100{\rm{\;}} = {\rm{\;}}20{\rm{\% }})$

Let their SHARES be A, B and C respectively.

Given,

7A = 9B = 11C

⇒ A : B = 9 : 7

⇒ B : C = 11 : 9

The LCM of 7 and 11 is 77. Hence, we MULTIPLY the FIRST ratio by 11 and the second by 7.We have

A : B = 99 : 77

B : C = 77 : 63

Let the shares of M, N and O be REPRESENTED by m, n and o respectively.

According to given information,

M’s SHARE = m = 3n/9 = n/3

O’s share = o = 4m/7 = (4/7) × (n/3) = 4n/21

Ratio of their shares = m ? n ? o = (1/3) ? 1 ? (4/21) = 7 ? 21 ? 4

Share of O = 960 × [4/(7 + 21 + 4)] = Rs. 120

As we know that, Fourth proportional = (B × c)/a 

Where a,b and c are the 1^st, 2^nd and 3^rd ELEMENTS of proportion respectively

a = 250, b = 458 and c = 125

(458 × 125)/250

57250/250 = 229

LET the required NUMBER be X.

3 ? X = X ? 27

⇒ X² = 3 × 27

⇒ X = 9

So, the mean PROPORTIONAL between 3 and 27 = 9

Let the COST PRICE be RS. 1000 PER kg.

Selling price of 930 gm sugar instead of 1 kg = Rs. 1000

Selling price of 15 kg sugar same way = Rs. 15000

Cost price of 930 gm sugar = Rs. 930

So, cost of 930 × 15 = Rs. 13950

ACTUAL money with the person = 0.5((6/9) × 45) + 1((2/9) × 45) + 2((1/9) × 45) = 15 + 10 + 10 = Rs. 35

Total money in FIRST case:

Total VALUE of coins = 0.25((6/9) × 45) + 0.50((2/9) × 45) + 1((2/9) × 45) = 7.5 + 5 + 10 = Rs. 22.50

Total money in Second case:

Total value of coins = 0.25((6/9) × 90) + 0.50((2/9) × 90) + 1((2/9) × 90) = Rs. 45

Total money in Third case:

Total value of coins = 0.25((6/14) × 70) + 0.50((5/14) × 70) + 1((3/14) × 70) = Rs. 35

In third case, the amount is same

Let the monthly income of A and B be 4x and 3X.

Let the monthly expenses of A and B be 3y and 2y.

Income = Expenditure + Saving

⇒ 4x = 3y + 6000----(1)

⇒ 3x = 2y + 6000----(2)

By solving eq. (1) and (2), we get

⇒ X = 6000 and y = 6000

As we know, if a? B = b? c, then c is the THIRD proportional of a and b

Let a = 2(X² – y²), b = (xy² + x²y), c = [8(x + y)/(x – y)]

⇒ 2(x² – y²) : (xy² + x²y) = (xy² + x²y) : [8(x + y)/(x – y)]

⇒ 2(x² – y²) × [8(x + y)/(x – y)] = (xy² + x²y) × (xy² + x²y)

? (x² – y²) = (x + y)(x – y)

⇒ 2(x + y)(x – y) × [8(x + y)/(x – y)] = xy(x + y) × xy(x + y)

⇒ 16 × (x + y)² = (xy)² × (x + y)²

⇒ 16 = (xy)²

Let the number of COINS be 3x, 2x, 3x respectively

Then total amount will be given by

3x × 1 + 2x × 0.5 + 3x × 0.1 = 12.9

⇒ 3x + x + 0.3x = 12.9

⇒ 4.3x = 12.9

⇒ x = 3

Number of 50 paise coins = 2x = 2 × 3 = 6

⇒ LET the income and expenditures be Rs. 10X and Rs. 7X respectively

⇒ Given Expense is Rs. 11,200

⇒ 7x = 11,200

⇒ x = 11,200/7 = 1600

⇒ SAVING = income - expenditure

⇒ Saving = 10x - 7x

⇒ Saving = 3x

⇒ PUTTING the value of x in above we get

⇒ Saving = 3 × 1600

Let the number to be added be ‘m’, then according to the question,

$(\frac{{45 + m}}{{13 + m}} = \frac{{33 + m}}{{9 + m}})$

⇒ 405 + 9m + 45m + m² = 429 + 33m + 13m + m²

⇒ 54m – 46m = 429 – 405

⇒ 8m = 24

⇒ m = 3

LET the number of male and female WORKERS be 2x and 5x

⇒ 2x + 5x = 1400

⇒ 7X = 1400, x = 200

Thus, number of female workers = 5x = 1000

⇒ Number of male workers = 2x = 400

⇒ Now the ratio of female to male changes to 5 : 4 when 100 more FEMALES and a few more male workers joined the company.

Let the number of male workers who joined be x.

⇒ Then, (1000 + 100)/ (400 + x) = 5 : 4

⇒ 1100/ (400 + x) = 5 : 4

⇒ 4400 = 2000 + 5x

⇒ 5x = 2400

∴ x = 480

Let the third proportional be x

⇒ 16 ? 20 ? 20 ? x

⇒ 4/5 = 20/x

∴ x = 100/4 = 25

To find the third proportion we use a formula b² = AC where a and b are two numbers and C is the third proportion of them.

⇒ b² = ac

⇒ b² = 27 × 12

For cylinder A height is h1 and Volume is V1

Similarly,

For cylinder B height is h2 and Volume is V2

$(\frac{{{\rm{V1}}}}{{{\rm{V2}}}}{\rm{ = }}\frac{{{\rm{\PI (r1}}{{\rm{)}}^{\rm{2}}}{\rm{h1}}}}{{{\rm{\pi (r2}}{{\rm{)}}^{\rm{2}}}{\rm{h2}}}})$

$(\Rightarrow \frac{{\rm{5}}}{{{\rm{36}}}}{\rm{ = }}\frac{{{{{\rm{(r1)}}}^{\rm{2}}}}}{{{{({\rm{r2)}}}^{\rm{2}}}}}{\rm{ \TIMES }}\frac{{\rm{5}}}{{\rm{4}}})$

$(\Rightarrow \frac{{\rm{5}}}{{{\rm{36}}}}{\rm{ = }}\frac{{{{{\rm{(r1)}}}^{\rm{2}}}}}{{{{{\rm{(r2)}}}^{\rm{2}}}}}{\rm{ \times }}\frac{{\rm{5}}}{{\rm{4}}})$

$( \Rightarrow \frac{{{{{\rm{(r1)}}}^{\rm{2}}}}}{{{{{\rm{(r2)}}}^{\rm{2}}}}}{\rm{ = }}\frac{{\rm{5}}}{{{\rm{36}}}}{\rm{ \times }}\frac{{\rm{4}}}{{\rm{5}}})$

$(\Rightarrow \frac{{{{{\rm{(r1)}}}^{\rm{2}}}}}{{{{{\rm{(r2)}}}^{\rm{2}}}}}{\rm{ = }}\frac{1}{9})$

⇒ r1/r2 = 1/3

∴ Ratio of r1 and r2 is 1 ? 3

Let TOTAL coins in the purse are x.

⇒ 1-rupee, 50-paise and 10-paise coins would be 6X, 9x and 10x respectively.

Total amount is Rs. 34.5

⇒ 6x × 1 + 9x × 0.5 + 10x × 0.1 = 34.5

⇒ 6x + 4.5x + x = 34.5

⇒ 11.5x = 34.5

⇒ x = 3

⇒ Let the age of son be x

⇒ then age of FATHER = 3X

⇒ after 10 years father will be twice as old as son

⇒ Father age = 3x + 10

⇒ Son age = x + 10

⇒ 3x + 10 = 2(x + 10)

⇒ 3x + 10 = 2X + 20

⇒ x = 10

⇒ SUM of their ages = x + 3x

⇒ Sum of their ages = 10 + 30

Volume of each BOX = 2 × 5 × 10 = 100 cm³

Volume of WATER transferred to the containers = 5 × 100 = 500 cm³

Let the VOLUMES of the containers be 5x, 9x and 11x. Then,

5x + 9x + 11x = 500

⇒ 25x = 500

⇒ x = 20

∴ Volume of third container = 11x = 11 × 20 = 220 cm³ 

Given,

⇒ (7x – 3Y) ? (4x + y) = 5 ? 4

⇒ [7(x/y) – 3] / [4(x/y) + 1] = 5/4

⇒ 28(x/y) – 12 = 20(x/y) + 5

⇒ x/y = 17/8