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1.729 ml of combination has coffee and water in the ratio 7:2 how much more water is

to be included to get a new combination has coffee and water in the ratio 7:3?

Facts and ratios

a: b = a/b

Answer : 81 ml

Ratio of coffee and water in 729 ml is 7:2

->Coffee in 729 ml of combination = [(7/9)*(729)] ml = 567 ml

water in 729 ml of combination = 729 – 567 = 162

->If “a” be the some of water included to new combination, with the ratio 7:3

some of water in the new combination= (162 + a )

Then (7/3) = [ (567) / (162+a) ]

=> 7 (162 + a) = 3 x 567

=> 1134 + 7a = 1701

7a = 1701 – 1134

A = 567/7 = 81 ml

Some of water included to new combination = 81 ml.

2. Addition of Rs.312 was divided by100 guys and gals in such a way that the guy

gets Rs.3.60 and each gal Rs.2.40 the number of gals is

Solving simultaneous equations

Answer : 40

-> If “a” be the count of guys and “b” be the count of gals.

Total count of guys and gals = 100

a + b = 100 —Eqn(i)

-> A guy gets Rs.3.60 and a gal gets Rs. 2.40

The money given to 100 guys and gals = Rs. 312

3.60a +2.40b = 312 —Eqn(ii)

->Solving (i) and (ii)

a 3.60 => 3.60 a + 3.60b = 360 (-)

=> 3.60 a + 2.40b = 312

1.20b = 48

B = 48 / 1.20 = 40

Therefore, Count of gals = 40

3. Addition of Rs.36.90 is created by 180 coins which are possibly 10 paise coins or

25paise coins. The count of 10paise coins are

Conversion of Paise into Rupees

Answer : 54 coins

-> Total count of coins = 180

If “a” is count of 10paise coins and “b” be count of 25paise coins

A + b = 180 —Eqn (i)

-> Given 10paise coins and 25paise coins create the addition = Rs. 36.90

[ (10a/100) + (25b/100) ]= 36.90

=> 10a +25b = 3690 ———(ii)

-> Solving (i) and (ii)

a10 => 10 a + 10b = 1800 (-)

=> 10a +25b = 3690

– 15b = -1890

B = 1890 / 15 = 126

Substitute b value in eqn (i) , A = 180 – 126 = 54

Therefore Count of 10paise coins = 54

4. A fraudulent dairy guy declares to market his dairy at price range but he combined it

with water and therefore benefits 25%. The percentage of water in the combination is

Capability Price = { [ (100/(100+benefits%)]a S.P }

The rule of allegation

Needed ratio is less expensive amount: High expensive amount = (d-m): (m-c)

Answer : 20%

-> If Capability Price of 1 L of milk = Rs.1

And S.Price of 1 L of combination = Rs.1

Benefits = 20%

Capability Price of 1 L of combination =[ (100 / 100+25)*1 ]

= Rs. 100/125 = Rs. 4/5

-> From the rule of allegation

i. Capability Price of 1 L of water = 0

ii. Capability Price of 1 L of milk = 1

iii. (p) = 4/5

iv. d – m = 1 – 4/5 = 1/5

v. m – c = 4/5 – 0 = 4/5

Ratio of milk to water = 4/5 : 1/5 = 4:1

% of water in the combination = (1/5 x 100)% = 20%

5. A combination of 20 kg of milk and water contains 10% water. Then how much water

must be included to this combination to increase the % of water to 25%?

Facts of percentage

Answer : 4 kg

-> Some of water in 20 kg of combination is (10/100)*20= 2 kg

-> If “a” kg of water be included with the combination to raise 25% of water

[ (2+a) / (20+a) ] = (25 / 100 )

=> (2 + a ) 100 = 25 (20 + a)

=> 200 + 100a = 500 + 25a

=> 100a – 25a = 500 – 200

=> 75a = 300

A = 4 kg

4 Kg of water is included to this combination.

6. A drum has 40 kg of water, from this drum 4kg of water was taken out and

replaced by milk. This task was repeated nearly 2 times. How much water is now

contained by the drum?

If a drum contains “a” kg of water from which “b” kg is taken out and replaced by milk. After “n” operations, the amount of water = { x[ 1 – (b/a) ]n }kg

Answer : 29.16 kg

Some of water in the Drum a is 40 kg

Some of water taken out b is 4 kg

No.of times = 3

Some of water in the Drum is { a[ 1 – (b/a) ]n }kg

={ 40[ 1 – (4/40) ]3 }kg

= 40 (9/10)3 kg

= 40 ( 729/1000)

Some of water in the Drum = 29.16 kg

7. A drum has a combination of 2 liquids X and Y in the proprtion 7:5 when 9 L of

combination are taken out and the Drum is filled with Y, the proportion of X and Y

becomes 7:9. Then how many litres of liquids X was obtained by the Drum initially?

Facts of combinations and ratios

Answer : 21 L

=> If the Drum has 7a and 5a litres of combinations X and Y correspondingly. Amount of Combination taken out is 9 litres

Some of X in Combination left = [ 7a – (7/12)*9] litres

= [ 7a – (21/4) ] litres

Some of Y in combination left = [ 5a – (5/12)*9 ] litres

= [ 5a – (15/4) ] litres

=> Since the combination is in the ratio 7:9,

{ [7a – (21/4)] / [ 5a – (15/4) ]+9 } = (7/9) => [ (28a – 21) / (20a + 21) ] = (7/9)

=>252a – 189 = 140a + 147

=> 112a = 336

=> a = 3

Some of liquid X, has in the Drum initially is (7*3) = 21 litres

8. The proportion between the current ages of A and B is 6:7. If B is 4 years elder than A, what will be the proportion of the ages of A and B after 4 years?

a. 3 : 5 b. 3 : 4 c. 4 : 3 d. None of these

Facts of ratios

Answer : D) None of the above

If A’s age and B’s age = 6a and 7a years respectively

Then B is 4 years elder than A

=> 7a – 6a = 4

A = 4

After 4 years, A’s age = 6a + 4 years

And B’s age = 7a + 4

Needed ratio = 6a + 4 : 7a + 4

Replace the “a” value of 4

= 28 : 32

= 7 : 8

9. Right now, the proportion between the ages of Karthikh and Kumar is 4 : 3. After 6 years, Karthikh’s age will be 26 years. As of now what is the age of Kumar?

a. 15 years

b. 11 years

c. 16 years

d. 40 years

Answer : A) 15 Years

-> If the current age of Karthikh = 4a

and Kumar =3a correspondingly.

After 6 years , Karthikh’s age = 4a +6 years

-> After 6 years, Karthikh’s age will be 26 years.

4a + 6 = 26

A = (26 – 6) /4 = 5

Kumar’s current age is 3 x 5 = 15 years.

10. The length and breadth of a rectangle are raised in the proportion 3:4 and 4:5 correspondingly. What is the proportion of the previous place to the new?

Area of the rectangle = length * breadth

Answer: 3:5

-> If the real length = 3a

And the real breadth = 4a

Area = (3a)*(4a)

=12 a2

-> New Length is 4a and New breadth is 5a

New area is (4a)*(5a)

= 20 a2

Ratio of previous place to new place = 12 a2 : 20 a2

= 12 : 20

= 3 : 5