InfyTQ Ratio and Proportion Quiz 4

Correct answer: Option II

\frac{6}{10}R=\frac{2}{5}J
R=\frac{2}{5}\times \frac{10}{6}J
R=\frac{2}{3}J

R/J = 2/3

R: J = 2:3

Therefore, Jitesh’s share = 2/5 *1500 = 600

Correct answer: Option I

Solution:

Let the third box be Z

Then, first box = 140% of z = 140z/100 = 7/5

Then, second box = 180% of z = 180z/100 = 9/5

Therefore, ratios of the two boxes will be 7z/5 : 9z/5 = 35z : 45z =  7:9

Solution:

Let the shares be = 6x : 3x : 3x : 2x respectively

Then 3x- 2x = 1000

X = 1000

Therefore, Bharat’s share = 3x

= 3 * 1000 = 3000

Correct answer: Option II

Solution:

Let the number of seats left in the classes be = 4y and 6y

Increased seats = (120/100* 4y) (150/100*6y)

= 48y/10 , 9y

= 48:90

The ratio will be 16: 30,

= 8: 15

Correct answer: Option I

Solution:

Quantity of juice = (30 *4/6) liters = 20 litres

Quantity of water = (30- 20) = 10 litres

Another ratio = 2:4

Let the quantity of pulp to be added further be x times

Then, juice: pulp (20/10+x)

Now, (20/10+x) = 2/4

80 = 20 + 2x

x = 60/2 = 30

Correct answer: Option IV

Explanation:

Let the number of blue and red balls in the box be 4x and 8x respectively.

Their increased number is (140% of 4x) and (120% of 8x).

{140/100*4x} and {120/100*8x}

The new ratio will be = 35x/5: 15x/5

= 7: 3

Let the original salaries of Ronit and Sahil be Rs. 4x and Rs. 6x respectively

Check formula used on this page here

(Quick Method)

S.P = 27.50

P = 25 %

S.P = C.P x 125/100

27.50 = C.P x 125/100

C.P = 11 Rs.

24                    0

\              /

\       /

22

/      \

/              \

22                    2

 

W : S = 11:1 Ans

(Long Solution)



 

Solution:

Let the number of coins be 4x and 6x.

Total amount given = Rs.320

=> (.25)(4x) + (.50)(6x) = 320

Hence we get x = 1x + 3x = 320

X = 80

= 25 paisa coins = 320

= 50 paisa coins = 480

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =7x - 7/12 x 9 litres = 7x - 21/4 litres.

Quantity of B in mixture left = 5x - 5/12 x 9 litres = 5x - 15/4 litres.

=> 7x - 21/4 / 5x - 15/4 = 7/9
=> 252x - 189 = 140x + 147
=> x = 3.
i.e 7x = 7 x 3

So, the can contained 21 litres of A.