Explore topic-wise InterviewSolutions in .

Let the incomes of P and Q be RS. ‘4X’ and Rs. ‘7x’ respectively

Let the EXPENDITURES of P and Q be Rs. ‘3y’ and Rs. ‘7y’ respectively

Now, SAVINGS = Income – Expenditure

⇒ Savings of P = 4x –3y = 10000 ----(1)

⇒ Savings of Q = 7x – 7y = 7000

⇒ x – y = 1000

⇒ y = x – 1000

Substituting in (1),

⇒ 4x – 3(x – 1000) = 10000

⇒ 4x – 3x + 3000 = 10000

⇒ x = 10000 – 3000 = 7000

Let a = 3x and b = 5x, then according to the question,

(3a + 4b) : (5A + 3b) = (3a + 4b)/(5a + 3b)

⇒ (3a + 4b) : (5a + 3b) = (9x + 20x)/(15x + 15x) = 29x/30x = 29/30

⇒ (3a + 4b) : (5a + 3b) = 29 : 30

Hence, the required ratio is 29 : 30.

Alternate Method:

(3a + 4b) : (5a + 3b)

$(= \frac{{3\frac{a}{b} + 4}}{{5\frac{a}{b} + 3}} = \frac{{3 \times \frac{3}{5} + 4}}{{5 \times \frac{3}{5} + 3}} = \frac{{\LEFT( {9 + 20} \right) \times \frac{1}{5}}}{{3 + 3}})$

= 29/(6× 5) = 29/30 = 29 : 30

Hence, the required ratio is 29 : 30.

Third PROPORTION of 6 and 8 is $(\FRAC{{8 \times 8}}{6} = \frac{{32}}{3})$

Fourth proportion of 3,4 and 10 is $(\frac{{4 \times 10}}{3} = \frac{{40}}{3})$

Required ratio $(= \frac{{32}}{3}\;:\;\frac{{40}}{3} = 32\;:\;40\;{\rm{or}}\;4\;:\;5)$

SUPPOSE the number of people that were absent = a

∴ 36 × 5X = (36 – a) × 6x

⇒ 180 = 216 – 6A

⇒ 6a = 36

⇒ a = 6

Let, the RATIO = x ? y

According to problem,

⇒ 50X + 40y = 43(x + y)

⇒ 7x = 3y

⇒ x ? y = 3 ? 7

∴ Two types of sugar should be MIXED in the ratio = 3 ? 7

Sum of money to be divided = 1087

TOTAL money withdrawn from A, B and C = (10 + 12 + 15) = 37

Now remainders = 1087 - 37 = 1050 which will be divided in the proportion 5, 7, 9.

SHARE of B out of 1050 $(= \frac{7}{{21}} \times 1050 = 7 \times 50 = 350)$

From GIVEN data-

BLUE, red, GREEN and black in the ratio of 3 ? 8 ? 5 ? 4

Let the common multiplication constant be X.

∴ We can say there are 3X blue pens, 8X red, 5X green and 4X black pens in the collection.

As given- green pens are 40 more than blue pens.

∴ 5X - 3X = 40

∴ X = 20

∴ Black pens are 4X

∴ Number of black pens are = 4 × 20 = 80

Assuming the total no of MEMBERS to be 100

? 28% members are married

⇒ Married = 28% of 100 = (28/100) × 100 = 28

⇒ UNMARRIED = Total Members – Unmarried = 100 – 28 = 72

⇒ Married/Unmarried = 28/72 = 7/18

Let the salary of HAAMID be Rs.X.

GADADHAR's salary is 3/2 times of Haamid's and SARVESH's is 4/3 times of Haamid's.

So, salary of Gadadhar = Rs. x × (3/2) = Rs. 3x/2

And, the salary of Sarvesh = Rs. x × (4/3) = Rs. 4x/3

Given,

Speed of person from point N to L = 65km/hr

Let Speed of person from point L to N be a km/hr.

AVERAGE speed for the whole journey = 100 km/hr.

If distance of whole journey is same then average speed of journey is =

⇒ 100 = (2 × a × 65)/(65 + a)

⇒ 6500 + 100A = 130a

⇒ 6500 = 30a

⇒ a = 6500/30

⇒ a = 216.67

∴ Speed of person from point L to N is 216.67 km/hr.

$(\because\frac{{{\RM{x\;}} + {\rm{\;y}}}}{{{\rm{x\;}}-{\rm{\;y\;}}}}{\rm{\;}} = \frac{{15}}{1})$

Using the property of compenendo and dividendo:

$(\THEREFORE \frac{{\rm{x}}}{{\rm{y}}} = \frac{8}{7})$

∴ x² – y² = 8² – 7² = 64 – 49 = 15

FIRST Case:

PRINCIPAL AMOUNT = P

Amount = 2P

Time = 7 years

Rate = r%

⇒ $(2P = P \times {\left( {1 + \frac{r}{{100}}} \right)^7})$

⇒ 2 = (1 + 0.01r)⁷

⇒ $({2^{\frac{1}{7}}} = 1 + 0.01r)$----(1)

Second Case:

Principal = P

Amount = 16P

Time = t

Rate = r%

⇒ $(16P = P \times {\left( {1 + 0.01r} \right)^t})$

⇒ $(16 = {\left( {1 + 0.01r} \right)^t})$

From EQ. (1), we get

⇒ $({2^4} = {\left( {{2^{\frac{1}{7}}}} \right)^t})$

⇒ 4 = t/7 ⇒ t = 4 × 7 = 28

∴ Sum will be 16 times after 28 years.

⇒ Let the ratio of money with the PRIVATE banks be 4p, 7p and 5p

⇒ Now, ICICI bank gives 1/7^th i.e. 7p/7 = p to HDFC bank and 2/7^th i.e. 7p × 2/7 = 2P to AXIS bank

⇒ So, the new ratio of money will be 5p, 4p and 7p.

⇒ Now, HDFC bank gives 1/5^th i.e. 5p/5 = p to Axis bank.

∴ The final ratio will be 4p : 4p : 8p or 1 : 1 : 2

INCOMES of A and B are 4x and 3x

Expenses of A and B are 3y and 2y

SAVINGS of A is 4x - 3y and B is 3x - 2y.

Since the savings of each is the same

4x - 3y = 3x - 2y or x = y

4x - 3y = 600

4x - 3x = 600

X = 600

a : B = 2 : 3

⇒ a/b = 2/3

⇒ b = 3a/2

b : c = 4 : 5

⇒ b/c = 4/5

$(\RIGHTARROW \frac{{3a}}{2}\;:c = 4\;:5\;)$

⇒ c = 15A/8

c : d = 6 : 7

⇒c/d = 6/7

$(\Rightarrow \frac{{\frac{{15a}}{8}}}{d} = \frac{6}{7})$

$(\Rightarrow \frac{a}{d} = \frac{8}{{15}} \TIMES \frac{6}{7})$

⇒ a/d = 16/35

<P>⇒ Given, P = RS. 90,000, R = 16.5% and T = 3 years

⇒ S.I = (P ? R ? T)/100

⇒ S.I = (90,000 ? 16.5 ? 3)/100

⇒ S.I = 90 ? 165 ? 3

Let the monthly salaries of A and B be 11x and 21X.

According to the condition,

⇒ [(11x + 4000)/(21x + 4000)] = 3/5

⇒ 55x + 20000 = 63x + 12000

⇒ 63x - 55x = 20000 - 12000

⇒ 8x = 8000

⇒ X = 8000/8 = 1000

Let the share of Sunil, Anil and jamil be x, y and z respectively.

Hence, x + y + z = 15525

22, 35 and 48 are DIMINISHED from the shares of x, y and z respectively

∴ Their share now is (x – 22),(y – 35) and (z – 48)

Total share = x + y + z – 22 – 35 – 48 = 15525 – 105 = 15420

It is given that from this total diminished share ratio of shares of Sunil, anil and Jamil is 7 ? 10 ? 13.

∴ Share of Sunil = 7/(7 + 10 + 13) × 15420 = 3598

∴ Share of Anil = 5140

And share of Jamil = 6682

∴ The original share for

Sunil is? x = 3598 + 22 = 3620

Anil is? y = 5140 + 35 = 5175

Jamil is? z = 6682 + 48 = 6730

Now 16, 77 and 37 is added respectively to x, y and z

Hence the new ratio = (3620 + 16) ? (5175 + 77) ? (6730 + 37) = 3636 ? 5252 ? 6767

Let Vaibhav’s age = a and AMIT’s age = 5 + a

(Amit's age after 4 years)/(Vaibhav's age after 4 years) = 5/4

⇒ (5 + a + 4)/(a + 4) = 5/4

⇒ 20 + 4a + 16 = 5A + 20

⇒ 5a – 4a = 16

⇒ a = 16

∴ Amit’s age = 5 + a = 5 + 16 = 21

$(\frac{1}{2}\;:\;\frac{1}{3}\;:\;\frac{1}{4} = 6\;:\;4\;:\;3)$

(Multiply each RATIO by 12 i.e. LCM of 2, 3 and 4)

INDIRECT PROPORTION: Less cows, More DAYS

Direct proportion: Less bags, Less days

Let the number of days be X

1 × 30 × x = 30 × 1 × 30

Let the age of FATHER, MOTHER and son be 4x, 3x, x.

Then, Age of daughter = 6x/5

Age of Grandmother = 2 × 3x = 6x

Sum of ages of family = 4x + 3x + x + 6x/5 + 6x = 76x/5

⇒ 76x/5 = 45.6 × 5

⇒ x = 15

⇒ Age of grandmother = 15 × 6 = 90 YEARS

One year AGO the ratio between A’s SALARY and B’s salary was 5 ? 6

Ratio of last year to present year salary of A = 2 ? 3

⇒ Present year salary = 3/2 TIMES the last year

Ratio of last year to present year salary of B = 3 ? 4

⇒ Present year salary = 4/3 times the last year

One year ago the ratio between A’s salary and B’s salary was 5 ? 6

⇒ Present year salary ratio = 5 × 3/2 ? 6 × 4/3 = 15 ? 16

Sum of current salary of A and B = Rs. 6200

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 = 1/3, b = 2/5 and c = 8.4

$(\frac{{\frac{2}{5} \times 8.4}}{{\frac{1}{3}}})$ 

3.36 / $(\frac{1}{3})$ = 10.08

As per GIVEN data-

Ratio of male to FEMALE employees in COMPANY is 1 ? 2.

Let’s assume common multiplication constant be k.

∴ Total number of male employees in company is k & total number of female employees in company is 2k.

When 2 males and 2 females LEFT from job new ratio becomes 1 ? 3.

(k - 2)/ (2k - 2) = 1/3

∴ 3k - 6 = 2k - 2

∴ k = 4

∴ Number of male employees is 4 and female employees are 8.

LET the income of Amit be 9x and that of Bunny be 8x.

Let the expenditure of Amit be 7Y and that of Bunny be 6y.

Hence, savings of Amit = 8000 = 9x – 7y

Savings of Bunny = 8000 = 8x – 6y

Hence,

9x – 7y = 8x – 6y

⇒x = y

Substituting this in ONE of the above equations, we have

8000 = 2x

⇒x = 4000

Hence, 9x = 36000 and 8x = 32000

Since Rs. 20 is less with all the three and it is less from the original amount so we have

Total money left = 960 - (20 + 20 + 20) = Rs.900

According to the question

5x + 7x + 4X = 900

⇒ 16x = 900

⇒ X = 56.25

SHARE of A = 5 × 56.25 = 281.25

Share of B = 7 × 56.25 = 393.75

Share of C = 4 × 56.25 = 225

Let present age of sheetal and Divya be 6x and 5x respectively

After 8 years,

⇒ Sheetal’s age = 6x + 8

⇒ Divya’s age = 5x + 8

According to QUESTION,

⇒ (6x + 8) / (5x + 8) = 22/19

⇒ 114x + 152 = 110x + 176

⇒ 4x = 24

⇒ x = 6

Now as a²b, bc, x and C² are in proportion

We can write

a²b: bc = x : c²

∴ a² : c = x : c²

∴ a² = x : c

Since the RATIO of selected to unselected was 2 : 1;

Selected CANDIDATES = 2x, Unselected candidates = x and total applied candidates = 3x

If 50 LESS had applied and 25 less selected, the ratio of selected to unselected would have been 9 : 4

∴ Total applied candidates = (3x – 50)

And Selected candidates = (2x – 25) and Unselected candidates = x – 50 + 25

∴ (2x – 25) / (x – 25) = 9/4

⇒ x = 125

Let the sum received by A, B and C be x/10, x/6 and x/15 respectively

According to the QUESTION,

x/10 + x/6 + x/15 = 25000

⇒ 10x/30 = 25000

⇒ x = 75000

LET the three numbers be 4x, 5x and 6x.

The sum of their squares = 1925

⇒ (4x)² + (5x)² + (6x)² = 1925

⇒ 16x² + 25x² + 36x² = 1925

⇒ 77x² = 1925

⇒ x² = 1925/77 = 25

⇒ x = 5

The three numbers are 4x = 20, 5x = 25 and 6x = 30

Hence, the three numbers are 20, 25 and 30.

$(\BEGIN{array}{l} \frac{{\sqrt {3{\rm{x\;}} + {\rm{\;}}4} {\rm{\;}} + {\rm{\;}}\sqrt {3{\rm{x}}\; - \;5} }}{{\sqrt {3{\rm{x\;}}+ {\rm{\;}}4{\rm{\;}}}\; - \;\sqrt {3{\rm{x}}\; - \;5} }} = 9\\ \RIGHTARROW {\rm{\;}}\frac{{\sqrt {3{\rm{x\;}} + {\rm{\;}}4} {\rm{\;}} + {\rm{\;}}\sqrt {3{\rm{x}} \;-\; 5} }}{{\sqrt {3{\rm{x\;}} + {\rm{\;}}4{\rm{\;}}} \;-\; \sqrt {3{\rm{x}}\; - \;5} }}\; = \;\frac{9}{1} \end{array})$

using componendo and dividendo, we get

$(\frac{{(\sqrt {3{\rm{x\;}} + {\rm{\;}}4} {\rm{\;}} + {\rm{\;}}\sqrt {3{\rm{x}} - 5)} {\rm{\;}} + {\rm{\;}}(\sqrt {3{\rm{x\;}} + {\rm{\;}}4} - \sqrt {3{\rm{x}} - 5)} }}{{(\sqrt {3{\rm{x\;}} + {\rm{\;}}4{\rm{\;}}} {\rm{\;}} + {\rm{\;}}\sqrt {3{\rm{x}} - 5)} - (\sqrt {3{\rm{x\;}} + {\rm{\;}}4} - \sqrt {3{\rm{x}} - 5)} }} = \frac{{9{\rm{\;}} + {\rm{\;}}1}}{{9 - 1}}{\rm{\;}})$

$(\Rightarrow {\rm{\;}}\frac{{2\sqrt {3{\rm{x\;}} + {\rm{\;}}4} }}{{2\sqrt {3{\rm{x}} - 5} }} = \frac{{10}}{8} \Rightarrow \;\frac{{\sqrt {3{\rm{x\;}} + {\rm{\;}}4} }}{{\sqrt {3{\rm{x}} - 5} }} = \frac{5}{4})$

$(\Rightarrow {\rm{\;}}\frac{{3{\rm{x\;}} + {\rm{\;}}4}}{{3{\rm{x}} - 5}} = \frac{{25}}{{16}})$ (On squaring both sides)

⇒ 75x – 125 = 48x + 64

⇒ 75x – 48x = 64 + 125

⇒ 27x = 189 ⇒ x = 7.

⇒ (X + y)/(x – y) = 5/2,

⇒ 2(x + y) = 5(x – y)

⇒ 2x + 2y = 5x – 5y

⇒ 3X = 7y

⇒ x = (7/3)y

Initially number of apples = 7a and ORANGES = 5a

When 25 apples were rotten, NEW ratio becomes (7a – 25) / (5a + 25) = 1/5

⇒ 35a – 125 = 5a + 25

⇒ 30a – 125 = 25

⇒ a = 5

ATQ, Angles of the quadrilateral are 3x, 5X, 12x and 10X respectively

SUM of the angle of the quadrilateral is 360°

⇒ 3x + 5x + 12x + 10x = 360

⇒ 30x = 360

⇒ x = 12

⇒ Largest angle = 12x = 12 × 12 = 144°

⇒ SMALLEST angle = 3x = 12 × 3 = 36°

⇒ Thrice the smallest angle = 36 × 3 = 108°

⇒ One third of the largest angle = 144/3 = 48°

∴ Required sum = 108 + 48 = 156°

Let the two numbers be X and Y. According to question:

$(\Rightarrow \FRAC{X}{Y} = \frac{5}{7})$

⇒ 7X = 5Y ….(1)

On diminishing each of them by 40 we get the new numbers as (X - 40) and (Y - 40). Hence, according to the question the new ratio is:

 $(\Rightarrow \frac{{X - 40}}{{Y - 40}} = \frac{{17}}{{27}})$ 

⇒ 27X – 17Y – 400 = 0 …(2)

SOLVING the above two EQUATIONS we get,

⇒ X = 125 and Y = 175

Hence, the difference of the numbers = 175 – 125

= 50

Total Number of employees = 100

Ratio of number of directors, SUBJECT e x perts and INTERNS = 1 : 2 : 7

∴ Number of directors = 10

Number of subject e x perts = 20

Number of interns = 70

Ratio of entrance tickets = 2 : 3 : 5

⇒ Total ticket costs = (10 x 2) + (20 x 3) + (70 x 5) = 430

When Total amount = Rs 430, interns pay (70 x 5) = Rs 350

Trick: One thing is clear that A is elder than B and B is elder than C

Now going through OPTIONS LET's assume A's age is 9 and according to the question STATEMENT B's age is 1.5 and C's would be 0.5

This is not possible as average would be less than 22 which contradicts the other statement

Same will be valid for options 2 and 3 i.e. 3 and 6.

Hence eliminating options 2, 3 and 4

Therefore correct option is 1

Alternative solution:

Total of all the three ages = 22 ? 3 = 66

⇒ A = 6B

⇒ A/B = 6....(1)

⇒ B = 3C

⇒ B/C = 3

Multiply both numerator and denominator of equation (1) by 3

⇒ A/B = 18/3

⇒ A : B : C = 18 : 3 : 1

Let the PRICE be RS. x /kg

Consumption be y kg

Expenditure = price × consumption

⇒ Expenditure = x × y

Price of SUGAR is increased by 10%

New price of sugar = 1.1 x /kg

Let new consumption be z kg

New Expenditure = (1.1x) × z

New Expenditure = old Expenditure

⇒ (1.1x) × z = x × y

⇒ z = y/1.1

Reduction in consumption = (y - z) = y - y/1.1 = y/11

Percentage reduction in consumption=(reduction in consumption/original consumption)× 100

Percentage reduction in consumption = [(y/11)/y] × 100 = 100/11 = 9.09%

Shortcut method :

If the price of a commodity INCREASES by r%, then the reduction in consumption so as not to increase the expenditure = [r/(100 + r)] × 100%

Percentage reduction in consumption = [10/(100 + 10)] × 100 = 1000/110 = 9.09%

RATIO of cost PRICE and selling price = 32 : 37

∴ PROFIT % = (37 - 32) /32 × 100 = 5/32 × 100 = 15.625 = 15.63%

GIVEN that AMOUNT paid by A: amount paid by B = 1 : 2

And amount paid by B: amount paid by C = 3 : 2

⇒ RATIO of amount paid by A, B and C is 3 : 6 : 4

⇒ Amount paid by A = 3k

⇒ Amount paid by B = 6k

⇒ Amount paid by C = 4K

⇒ 3k + 6k + 4k = 260

⇒ 13k = 260

⇒ k = 20

Given,

P = 5000

R = 12%

T = 2

$(\Rightarrow CI = P \times {\LEFT( {1 + \FRAC{R}{{100}}} \right)^T})$

$(\Rightarrow CI = 5000 \times {\left( {1 + \frac{{12}}{{100}}} \right)^2} = 6272)$

Given, x³ α y

⇒ x³ = k × y

∴ 2³ = k × 28

∴ k = 8/28 = 2/7