Ratio and Proportion Quiz-1

Solution:
(\frac{6}{10})R= (\frac{2}{5})J

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

R = (\frac{2}{3})J

\frac{R}{J} = \frac{2}{3}

R: J = 2:3

Therefore, Jitesh’s share = (\frac{3}{5})\times 1500= 900

Solution:

Let the third box be Z

Then, first box = 140% of z = \frac{140z}{100} = \frac{7z}{5}

Then, second box = 180% of z = \frac{180z}{100} = \frac{9z}{5}

Therefore, ratios of the two boxes will be \frac{7z}{5} : \frac{9z}{5} =  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

Solution:

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

Increased seats = (\frac{120}{100}\times 4y),(\frac{150}{100}\times 6y)

=\frac{48y}{10} , \frac{90y}{10}

= 48:90

The ratio will be 16: 30

= 8: 15

Explanation:
Let the original salaries of Rishabh Pant and Gautam Gambhir be Rs. 2x and Rs. 3x respectively.
Then,
\frac{2x+4000}{3x+4000}=\frac{40}{57}
57(2x+4000)=40(3x+4000)
6x=68,000
3x=34,000

Gautam Gambhir's salary=(3x+4000)
=(34,000+4,000)
=Rs.38,000

Explanation:
No.of staff is inversely proportional to the amount
N=\frac{k}{A}
49=\frac{k}{20}
k=49\times\frac{20}{3.5}
N=280

Solution:
Let the original salaries of Rohan and Salim be 4x and 6x respectively.

according to question,
\frac{4x+8000}{6x+8000}=\frac{40}{50}
=50\times(4x+8000)=40\times (6x+8000)
=40x=80000
x=2000

therefore salim's present salary=6x+8000
6\times2000+8000 =20000

Solution:

Y\times3=0.5\times 12

Y= \frac{6}{3}

Y=2

Solution:

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

Total amount given = Rs.320

=0.25\times 4x+ 0.50\times 6x = 320

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

X = 80

= 25 paisa coins = 320

= 50 paisa coins = 480

Solution:
Let the container contains 7x and 5x litres of mixtures in A and B respectively.

Quantity of mixture A left=7\times x-\frac{7}{12}\times 9=7\times x-\frac{21}{4} ltrs.
Quantity of mixture B left=5\times x-\frac{5}{12}\times 9=5\times x-\frac{15}{4} ltrs.

Now,
\frac{7\times x-\frac{21}{4}}{5\times x-\frac{15}{4}+9}=\frac{7}{9}
=>\frac{28x-21}{20x+21}=\frac{7}{9}
=>252x-189=140x+147
=>x=3

hence,A contains 7x=7\times 3=21 ltrs