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Quantity A:

Let present age of Son be x

⇒ Present age of FATHER = 3x

Now, 5 years ago

⇒ Father’s age = 3x – 5

⇒ Son’s age = x – 5

According to QUESTION,

⇒ x – 5 = (1/5) × (3x – 5)

⇒ 5X – 25 = 3x – 5

⇒ 2x = 20

⇒ x = 10

⇒ Father’s present age = 3x = 3 × 10 = 30 years

∴ Present age of mother = 30 – 5 = 25 years

Quantity B: 30

Quantity A < Quantity B

Let the numbers be X and y.

2436 = 3 × 4 × 7 × 29

$(\THEREFORE \FRAC{x}{y} = \frac{{4k\; \times 7 \times 29}}{{3k \times 7 \times 29}} = \frac{4}{3})$

As LCM is the least common factor, we would be using 1 as the value of k.

∴ x = 812 and y = 609

HCF (812, 609) = $(\frac{{812 \times 609}}{{2436}} = 203)$

Alternative METHOD,

Let the two number be 3x and 4x.

Then, the LCM of 3x and 4x = 12x

⇒ 12x = 2436

⇒ x = 203

The HCF(highest common factor) of 3x and 4x is x.

Hence, HCF (x) = 203 will be the correct answer.

Let MONTHLY salary of R be Rs. x. 

Total profit = Rs. 240000 

R has invested his money for 8 months. 

∴ Profit to be shared = Rs. (240000 – 8X) 

Ratio of investments by P and V = 2 : 3 = 4 : 6 

Ratio of investments by V and R = 6 : 5 

Ratio of investments by P, V and R = 4 : 6 : 5

Profit will be shared in the ratio of (4 × 12) : (6 × 12) : (5 × 8) = 6 : 9 : 5 

V’s share = Rs. 90000 = 9/20 × (240000 – 8x) 

⇒ 240000 – 8x = 200000

⇒ 8x = 40000

⇒ x = Rs. 5000

Let x be the fourth proportional to 3,4 and 15

⇒ 3 ? 4 ?? 15 ? x

Since, PRODUCT of the EXTREMES = Product of the means

∴ 3x = 15 x 4

∴ x = 20

Quantity 1

⇒ Sum of RATIO = 2x + 3x + 5x = 10X

⇒ 10x = 1200

⇒ x = 120

Required sum = 2x + 5x = 7 × 120 = 840

Quantity 2

⇒ Sum of the ratio = 5x + 6X + 11x = 22x

⇒ 22x = 1100

⇒ x = 50

⇒ 2 × middle TERM = 2 × 6x = 600

∴ Quantity 1 > Quantity 2

Number of GIRLS selected for interview were one-third of TOTAL girls

So, total number of girls who WENT for interview = 600/(1/3) = 1800

Number of BOYS who got selected after interview = 1560 – 600 = 960

So, total number of boys who went for interview = 960/(3/5) = 1600

$(\FRAC{A}{3} = \frac{{3B}}{4} = \frac{{5C}}{6} = \frac{{7D}}{8})$

∴ A ? D = 21 ? 8

Ratio of amounts invested by MAN to his brother = 100000/125000 = 4 ? 5

Ratio of TIME PERIODS for the man and his brother = 12/8 = 3 ? 2

Let the distance will be covered by TEMPO be a km.

Given,

An auto rickshaw travels a distance of 36km.

RATIO of distance covered by auto rickshaw and tempo = 36 : a

Given,

Ratio of TIME taken by auto rickshaw and tempo = 2 : 3

Ratio of SPEEDS of auto rickshaw and tempo = 9 : 8

Use compounded ratio = Speed × Time

⇒ (9 : 8) and (2 : 3)

⇒ (9 × 2) : (8 × 3)

⇒ 3 : 4

Distance = speed × time

⇒ (36 : a) = (3 : 4)

⇒ 36/a = 3/4

⇒ a = 36 × 4/3 = 48

∴ The distance covered by Tempo in 3 hours is 48 km.

Let incomes of wife and husband be Rs. 3x and Rs. 7X respectively.

SAVINGS of husband = Rs. (3x - 15000)

Savings of wife = Rs. (7x - 45000)

Ratio of savings = $(\frac{{3x - 15000}}{{7x - 45000}} = \frac{3}{5})$

⇒ 15x - 75000 = 21x - 135000

⇒ 6x = 60000

∴ 3x = 30000

⇒ 7x = 70000

∴ Their RESPECTIVE incomes are Rs. 30000 and Rs. 70000 respectively.

We know that, M₁ × T₁ = M₂ × T₂

⇒ 40 × 15 = (40 + 10) × T₂

⇒ 50 × T₂ = 600

⇒ T₂ = 12 DAYS

If TOGETHER the three had 30 grapes more, each of them would have had 120 grapes on an average.

$(\Rightarrow \frac{{60 + p + q + 30}}{3} = 120)$

⇒ p + q = 270------- (1)

If Shama gives her 15 grapes to Radha, she and Inaya have grapes in the ratio 7 ? 10.

$(\Rightarrow \frac{{p - 15}}{q} = \frac{7}{{10}})$

⇒ 10p - 150 = 7Q

∴ 10p - 7q = 150------- (2)

Multiplying the first EQUATION with 7 and adding it with equation second :

7P + 7q + 10p - 7q = 2040

⇒ 17p = 2040

∴ p = 120

Putting the value of p in equation (1):

⇒ q = 150

∴ p ? q = 4 ? 5

As per given data-

50% of a certain number is EQUAL to one-third of second number.

Let the 1^st number be X & 2^nd number be Y.

X/2 = Y/3

3X/2 = Y

To find ratio in between 2 numbers-

X ? Y = 2 : 3

Let’s assume that the marks OBTAINED by Tina in Arts = x

? The marks SCORED in History are equal to 1/3^rd of the marks scored in Arts

⇒ Marks received by her in history = x/3

∴ TOTAL marks obtained by her in Arts and History = x + (x/3) = (4x/3)

⇒ 160 = (4x/3)

⇒ x = 160 × (3/4)

Ratio of salaries of DANNY, Bravo and Casillas = 3 : 5 : 9

Part of salary of Danny = 3x

Part of salary of Bravo = 5x

Part of salary of Casillas = 9x

Increased part of salary of Danny = 55% of 3x = $(\frac{{55}}{{100}} \times 3x = 1.65x)$

Increased part of salary of Bravo = 50% of 5x = (50/100) × 5x = 2.5x

Increased part of salary of Casillas = 75% of 9x = (75/100) × 9x = 6.75x

Ratio of profit made by 3 of them = 10 × 12x : 8 × 12x : 12 × 6X = 10X : 8x : 6x

Difference between profit of Kishan and SAURABH = Rs 2,40,000 i.e. 8x – 6x

Profit of Saurabh = 8x = Rs 9,60,000

Profit of Kishan = 6x = Rs 7,20,000

Profit of Deepak = 10x = Rs 12,00,000

TOTAL MONEY which is divided = Rs. 430

As per the given information:

5 × A’s share = 6 × B’s share = 9 × C’s share = k

∴ A’s share = k/5

∴ B’s share = k/6

∴ C’s share = k/9

∴ total amount $(= \FRAC{k}{5} + \frac{k}{6} + \frac{k}{9})$

$(\begin{array}{l} \RIGHTARROW \;\frac{k}{5} + \frac{k}{6} + \frac{k}{9} = 430\\ \Rightarrow \frac{{k\;\left( {54 + 45 + 30} \right)}}{{270}} = 430 \end{array})$

⇒ 129 k = 430 × 270

$(\Rightarrow k = \frac{{430 \times 270}}{{129}} = 900)$

The employer REDUCES the NUMBER of employees in the ratio 8 : 7

Let the COMMON multiple be ‘x’

∴ the number of employees before and after will be 8X and 7x respectively

The employer increases the wages in the ratio 17 : 24

Let the common multiple be y

∴ The wages before and after the hike will be 17y and 24y respectively

Ratio of expenses = $(\FRAC{{old\;number\;of\;employees \times old\;wages}}{{new\;number\;of\;employees\; \times new\;wages}})$

⇒ Ratio of expenses = $(\frac{{8x \times 17y}}{{7x \times 24y}})$

<P>The ratio of their ages is 4:5 so let their respective ages be p and Q so we have,

p/q = 4/5

⇒ 5p = 4q

⇒ 5p/4 = q

Four years LATER

(p + 4) + 5 = q + 4

⇒ p + 5 = 5p/4

⇒ p/4 = 5

As per GIVEN data-

Initial fruit SYRUP to water ratio is 3 ? 5.

Let X is multiplication CONSTANT in ratio.

∴ Barrel contains 3X quantity of fruit syrup & 5X quantity of water.

When 4 liters of water is added in barrel NEW ratio of syrup to water becomes 1 ? 2

3X/(5X + 4) = 1/2

∴ 3X × 2 = 5X + 4

∴ 6X = 5X + 4

∴ X = 4 liters

∴ Quantity of fruit syrup = 3X = 3 × 4 = 12 liters

Let the broken pieces out of 12 be x

⇒ The Glass not broken = (12 - x)

Ratio of Unbroken : Broken = x : (12 - x)

⇒ Now For x = 1 ⇒ Ratio = 1 : 11

⇒ Now For x = 2 ⇒ Ratio = 2 : 10 = 1 : 5

⇒ Now For x = 3 ⇒ Ratio = 3 : 9 = 1 : 3

⇒ Now For x = 4 ⇒ Ratio = 4 : 8 = 1 : 2

⇒ Now For x = 5 ⇒ Ratio = 5 : 7

⇒ Now For x = 6 ⇒ Ratio = 6 : 6 = 1 : 1

⇒ Now For x = 7 ⇒ Ratio = 7 : 5

⇒ Now For x = 8 ⇒ Ratio = 8 : 4 = 2 : 1

⇒ Now For x = 9 ⇒ Ratio = 9 : 3 = 3 : 1

⇒ Now For x = 10 ⇒ Ratio = 10 : 2 = 5 : 1

⇒ Now For x = 11 ⇒ Ratio = 11 : 1

It is understood that SCHOOL can be open for only FIXED number of hours

No. of school hours = 6 × 60 = 360 minutes

Let duration of each period in new schedule be X minutes

⇒ 360 = 9 × X

⇒ X = 360/9

∴ X = 40 minutes

LET the FIRST NUMBER be x and second number be y

 If 25% of x is subtracted from y i.e. $(y - \frac{{25}}{{100}}x,)$ then y becomes $(\frac{5}{6}y)$

i.e. $(y - \frac{{25}}{{100}}x = \;\frac{5}{6}y)$

$(\Rightarrow y - \frac{5}{6}y = \frac{x}{4})$

$(\Rightarrow \frac{y}{6} = \frac{x}{4})$

Share of Suhas = (8/21) × 172452 = Rs 65696

Share of Kiran = (13/21) × 172452 = Rs 106756

∴ $TWICE of Kiran’s share = 106756 × 2 = Rs 213512

∴ $THRICE of Suhas’s share = 65696 × 3 = Rs 197088

∴ REQUIRED difference = Rs (213512 - 197088) = Rs 16424

Suppose the ages of Aman and his grandfather are ‘A’ and ‘B’

⇒ [B – 16] = 9 × [A – 16]

⇒ 9A – B = 128

And

⇒ [B + 8] = 3 × [A + 8]

⇒ B – 3A = 16

Solving both equations,

⇒ A = 24 and B = 88

∴ Ratio of the PRESENT age of Aman’s and his grandfather’s = 24 ? 88 = 3 ? 11

⇒ A : B = a : b or it can be ALSO INFERRED that A/B = a/b ... (i)

⇒ B : C = p : q or it can be also inferred that B/C = p/q ... (II)

Multiplying Eq. (i) & (ii)

We get

GIVEN,

Let present age of AMBUJ and Avinash be A years and B years respectively.

⇒ (A – 3) = 3(B – 3)

⇒ A – 3 = 3B – 9

⇒ 3B – A = 6

Given,

A : B = 8 : 3

⇒ A = 8B/3

Solving EQUATIONS,

⇒ 3B – 8B/3 = 6

⇒ 9B – 8B = 18

⇒ B = 18

⇒ A = (8 × 18)/3

⇒ A = 48

Difference between the present ages of A and B =

= 48 – 18

= 30

According to the question,

A = 2B/3 and B = C/4

⇒ A/B = 2/3 and B/C = 1/4

To make the RATIO common. MULTIPLY B : C by 3 i.e. B : C = 3 : 12

⇒ A : B : C = 2 : 3 : 12

∴ A's share $(= 510 \times \;\frac{2}{{2 + 3 + 12}}\; = \;510\; \times \frac{2}{{17}})$ = Rs. 60

∴ B's share $(= 510 \times \;\frac{3}{{2 + 3 + 12}}\; = \;510\; \times \frac{3}{{17}})$ = Rs. 90

Anu’s present age = x years

Naman’s present age = x - 6 years

Bittu’s present age = x + 10 years

According to the question,

After 6 years

$(\BEGIN{array}{L} \frac{{x + 16}}{{x - 6 + 6}} = \frac{3}{2}\\ \Rightarrow \frac{{x + 16}}{x} = \frac{3}{2} \end{array})$

⇒ 2X + 32 = 3x

⇒ 3x - 2x = 32

⇒ x = 32years

∴ Age of Anu is 32 years

Now, speed and time are inversely proportional

∴ (s1)(t1) = d1

And, (s2)(T2) = D2

Also, d2 = 3(d1)

∴ $(\frac{{\LEFT( {s1} \RIGHT)\left( {t1} \right)}}{{\left( {s2} \right)\left( {t2} \right)}} = \frac{1}{3})$

∴ $(t2 = \frac{{3\left( {50} \right)\left( {96} \right)}}{{72}} = 200)$ MINUTES

As PER given data-

Ratio of male to female employees in a company is 7 ? 8

Let’s assume common multiplication constant be k.

∴ Number of male employees = 7k

When 20% male employees were REDUCED, NEW number will be 0.8 × 7k = 5.6k

& Number of female employees = 8k

When 30% female employees were reduced, new number will be 0.7 × 8k

∴ New number of female employees = 5.6k

∴ New ratios of male to female employees-

∴ 5.6k ? 5.6k = 1 ? 1

Ganguly’s present age = 24 - 10 = 14 years

∴ Tendulkar’s age = 14 + x

And Sehwag’s age = 14 – x

According to the QUESTION,

(14 + x)/(14 - x) = 9/5

⇒ 70 + 5X = 126 – 9x

⇒ 14x = 56

⇒ x = 4

Sehwag’ age = 14 - x = 10 years

LET Rs. 1270 be divided into two parts x and y

Given,

We can WRITE now,

(12% of x) + (7% of y) = 8% of (x + y)

⇒ 0.12x + 0.07y = 0.08x + 0.08y

⇒ 0.04x = 0.01y

⇒ x/y = ¼

So, x = ¹/₅ × 1270 = 254, y = ⁴/₅ × 1270 = 1016

Given, A puts 126 oxen in the park for 3 months, B puts in 162 oxen for 5 months and C puts in 216 oxen in 4 months.

∴ For A, usage = 126 × 3 = 378

For B, usage = 162 × 5 = 810

For C, usage = 216 × 4 = 864

Ratio of A : B :C = 384 : 810 : 864

If the TOTAL rent is 570, rent paid by B $(= \;\frac{{810}}{{378\; + \;810\; + 864\;}} \times 570\;)$

⇒ Rent paid by B = $(\frac{{810}}{{\;2052\;}} \times 570)$

Let the two digit number be 10x + y where x and y are 10^th place digit and unit place digit respectively.

Given,

Unit place digit and tenth place digit of a two digit number are in the RATIO 2 : 3

⇒ y : x = 2 : 3

⇒ y = 2X/3

Given,

sum of tenth place digit and unit place digit is 10

⇒ x + y = 10

⇒ x + 2x/3 = 10

⇒ 3x + 2x = 30

⇒ 5x = 30

⇒ x = 6

⇒ y = 2x/3 = 12/3 = 4

Two digit number = 10x + y

= 10(6) + 4

= 64

3/4^th of SQUARE root of two digit number = (3/4) × (64)^1/2

= (¾) × 8 = 6

∴ 3/4^th of square root of two digit number is 6.

Let ‘x’ be the NUMBER of months for which Mahesh INVESTED his Rs. 111.

$(\begin{array}{l} \THEREFORE \frac{{555 \times 12}}{{111 \times x}} = \frac{{15}}{1}\\ \therefore \frac{{60}}{x} = 15 \end{array})$

⇒ x = 4

⇒ Mahesh invested his money for 4 months, i.e 12 - 4 = 8 months after MANOJ invested his money